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4400 | https://www.geeksforgeeks.org/maths/exhaustive-events/ | Exhaustive Events
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Exhaustive Events are a set of events where at least one of the events must occur while performing an experiment. Exhaustive events are a set of events whose union makes up the complete sample space of the experiment.
In this article, we will understand the meaning of exhaustive events, its definition, Venn diagram of exhaustive events, collective exhaustive events, and examples of exhaustive events.
Table of Content
What are Exhaustive Events?
Exhaustive Event Venn Diagram
Collectively Exhaustive Events
Examples of Exhaustive Events
Calculation of Probability for Exhaustive Events
What are Exhaustive Events?
Exhaustive events, in the context of probability, refer to a set of events that collectively cover all possible outcomes of an experiment or situation. In other words, when we say that a set of events is exhaustive, it means that one of those events must occur. There are no other possible outcomes left.
For example, consider flipping a fair coin. The possible outcomes are heads (H) or tails (T). In this case, "getting heads" and "getting tails" are exhaustive events because they cover all possible outcomes when you flip the coin. The sample space, in this case, is {H, T}, and the events "getting heads" and "getting tails" are exhaustive since there are no other possible outcomes.
Mathematically, if E1, E2, ......., En are exhaustive events, their union (E1 ∪ E2 ∪ ...... ∪En) equals the entire sample space (S).
Definition of Exhaustive Events
Exhaustive events in probability refer to a collection of events that together cover all possible outcomes of an experiment or situation.
In simpler terms, exhaustive events ensure that one of the events in the collection must occur, leaving no room for other outcomes. This concept is fundamental in probability theory as it ensures a comprehensive understanding and analysis of possible outcomes.
What are Events in Probability?
In probability theory, events refer to outcomes or occurrences that can happen as a result of an experiment or observation. A simple example of rolling a six-sided die. The possible outcomes when rolling the die are the numbers 1, 2, 3, 4, 5, and 6.
The two events:
Event A: Rolling an even number.
Event B: Rolling a number greater than 4.
For Event A, the possible outcomes are 2, 4, and 6. So, if you roll the die and get any of these numbers, you have experienced Event A.
For Event B, the possible outcomes are 5 and 6. If you roll the die and get either 5 or 6, you have experienced Event B.
Events in probability are essentially sets of outcomes, and their occurrence depends on the specific conditions or criteria set for each event. Analyzing events helps in calculating probabilities and understanding the likelihood of different outcomes in a given experiment or situation.
Collectively Exhaustive Events
Collectively exhaustive events, in probability theory, refer to a set of events that cover all possible outcomes and no outcome is counted more than once across the events. Here, the events do not overlap with each other.
Consider flipping a fair coin. The possible outcomes are either heads (H) or tails (T). In this case, the events "getting heads" and "getting tails" are collectively exhaustive because one of these outcomes must happen when the coin is flipped. There are no other possible outcomes besides heads or tails.
Mutually Exclusive and Exhaustive Events
Mutually exclusive events and exhaustive events are two important concepts in probability theory.
| Aspect | Mutually Exclusive Events | Exhaustive Events |
---
| Definition | Events that cannot occur simultaneously. | Events that cover all possible outcomes of an experiment. |
| Occurrence | Cannot occur together in the same trial. | At least one of the events must occur. |
| Probability Relationship | The intersection of mutually exclusive events has zero probability i.e., P(A ⋂ B) = 0. | The union of exhaustive events covers the entire sample space i.e., A1 ⋃ A2 ⋃ ... ⋃ An = S. |
| Example | Rolling an even number and rolling an odd number on a fair six-sided die. | Rolling a number less than 3 and rolling a number greater than or equal to 3 on a fair six-sided die. |
Example: Consider rolling a fair six-sided die. The possible outcomes are numbers 1 through 6. Now, let's define two events:
Event A: Rolling an odd number {1, 3, 5}
Event B: Rolling an even number {2, 4, 6}
Solution:
Events A and B are mutually exclusive because if you roll an odd number (Event A), you cannot roll an even number (Event B), and vice versa.
Events A and B are also collectively exhaustive because together, they cover all possible outcomes of rolling the die. You will always roll either an odd number (Event A) or an even number (Event B).
Exhaustive Event Venn Diagram
Exhaustive events in probability cover all possible outcomes of an experiment, leaving no room for other outcomes. In a Venn diagram, representing exhaustive events involves using circles to encompass all possible outcomes.
Consider the exhaustive events when tossing a fair coin. The possible outcomes are either getting heads (H) or tails (T). In the Venn diagram:
Event A represents the event of getting heads (H).
Event B represents the event of getting tails (T).
Both events together cover all possible outcomes (H and T), making them exhaustive events. The entire rectangle enclosing them represents the sample space, indicating all potential outcomes of the coin toss.
Examples of Exhaustive Events
Examples of Exhaustive Events can be coin tossing, rolling a dice and drawing cards from a deck of cards. Explanation of each of them is given below:
Coin Tossing
When tossing a fair coin, the possible outcomes are either heads (H) or tails (T).
The events "getting heads" and "getting tails" are collectively exhaustive because one of these outcomes must occur.
For example, if you toss a coin, you will either get heads or tails. There are no other possible outcomes.
Rolling a Dice
When rolling a fair six-sided die, the possible outcomes are numbers 1 through 6.
The events "rolling a 1," "rolling a 2," and so on up to "rolling a 6" are collectively exhaustive because one of these outcomes must occur.
For instance, if you roll a die, you will get a number between 1 and 6. There are no other possible outcomes besides these six numbers.
Drawing Cards from a Deck
When drawing cards from a standard deck of 52 playing cards, the possible outcomes include the various ranks (2 through 10, Jack, Queen, King, Ace) and suits (hearts, diamonds, clubs, spades).
The events "drawing a heart," "drawing a diamond," "drawing a club," and "drawing a spade" are collectively exhaustive because one of these outcomes must occur.
Additionally, the events "drawing a 2," "drawing a 3," and so on up to "drawing an Ace" for each suit are also collectively exhaustive.
For example, if you draw a card from a deck, it will be either a heart, diamond, club, or spade, and it will also be one of the ranks from 2 to Ace. There are no other possible outcomes besides these.
Calculation of Probability for Exhaustive Events
Calculating the probability for exhaustive events involves determining the likelihood of each individual event occurring and then using the concept of exhaustiveness to find the probability of at least one of the events happening. The steps to calculate probability for exhaustive events are given below:
Step 1: Determine Individual Probabilities
Calculate the probability of each individual event occurring. This can be done by dividing the number of favorable outcomes (for each event) by the total number of possible outcomes in the sample space.
Step 2: Verify Exhaustiveness
Ensure that the events are collectively exhaustive, meaning that together they cover all possible outcomes.
Step 3: Use the Principle of Exhaustiveness
Since the events are collectively exhaustive, the probability of at least one of them occurring is equal to 1 (or 100%). This is because if one event doesn't happen, then another event in the set must happen. Therefore, the sum of the probabilities of all exhaustive events equals 1.
Example: Consider rolling a fair six-sided die. The possible outcomes are numbers 1 through 6.
Events A: Rolling an odd number {1, 3, 5}
Events B: Rolling an even number {2, 4, 6}
Solution:
Step 1: Calculate Individual Probabilities
Probability of Event A:
P(A) = 3/6 = 1/2
Probability of Event B:
P(B) = 3/6 =1/2
Step 2: Verify Exhaustiveness
Together, Events A and B cover all possible outcomes (1 through 6), making them collectively exhaustive.
Step 3: Use the Principle of Exhaustiveness
Since the events are collectively exhaustive, the probability of at least one of them occurring is 1 (or 100%).
So, in this example, the probability of rolling either an odd number or an even number (or both) on a fair six-sided die is 1. This means that you are guaranteed to roll an odd or an even number when you roll the die.
Read More,
Probability Distribution
Types of Events in Probability
Dependent and Independent Events
Solved Examples on Exhaustive Events
Example 1: You roll two fair six-sided dice. Define events for the sum of the numbers rolled (2 through 12). Are these events collectively exhaustive? What is the probability of getting any sum from 6 to 12?
Solution:
Sum of the numbers rolled can range from 2 (when both dice show a 1) to 12 (when both dice show a 6).
Here are the events for each possible sum:
Event for the sum of 2: {(1, 1)}
Event for the sum of 3: {(1, 2), (2, 1)}
Event for the sum of 4: {(1, 3), (2, 2), (3, 1)}
Event for the sum of 5: {(1, 4), (2, 3), (3, 2), (4, 1)}
Event for the sum of 6: {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
Event for the sum of 7: {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
Event for the sum of 8: {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
Event for the sum of 9: {(3, 6), (4, 5), (5, 4), (6, 3)}
Event for the sum of 10: {(4, 6), (5, 5), (6, 4)}
Event for the sum of 11: {(5, 6), (6, 5)}
Event for the sum of 12: {(6, 6)}
These events cover all possible sums from 2 to 12. Therefore, they are collectively exhaustive.
To find the probability of getting any sum from 6 to 12, we need to calculate the probability of each event and then add them up.
Since each die is fair and has 6 sides, there are a total of (6 × 6 = 36) equally likely outcomes when rolling two dice.
calculate the probabilities for each event:
Probability of sum = 6: P(6) = 5/36
Probability of sum = 7: P(7) = 6/36
Probability of sum = 8: P(8) = 5/36
Probability of sum = 9: P(9) = 4/36
Probability of sum = 10: P(10) = 3/36
Probability of sum = 11: P(11) = 2/36
Probability of sum = 12: P(12) = 1/36
Now, to find the probability of getting any sum from 6 to 12, we add up the probabilities of all these events:
P(sum from 6 to 12)=P(6)+P(7)+P(8)+…+P(12)
⇒ P(sum from 6 to 12)= 365+366+365+364+363+362+361
⇒ P(sum from 6 to 12)= 365+6+5+4+3+2+1
⇒ P(sum from 6 to 12)=26/36
⇒ P(sum from 6 to 12)=13/18
So, the probability of getting any sum from 6 to 12 when rolling two fair six-sided dice is 13/18 .
Example 2: You flip three fair coins. Define events for the number of heads obtained (0, 1, 2, or 3). Are these events collectively exhaustive? What is the probability of getting any number of heads from 0 to 3?
Solution:
To define events for the number of heads obtained when flipping three fair coins, we need to consider all possible combinations of outcomes. Each coin can either show heads (H) or tails (T).
Here are the events for each possible number of heads:
Event for 0 heads (all tails): {TTT}
Event for 1 head: {HTT, THT, TTH}
Event for 2 heads: {HHT, HTH, THH}
Event for 3 heads (all heads): {HHH}
These events cover all possible outcomes when flipping three fair coins and represent the number of heads obtained (0, 1, 2, or 3). Therefore, they are collectively exhaustive.
Calculate the probability of getting any number of heads from 0 to 3.
Since each coin is fair and has 2 equally likely outcomes (heads or tails), there are a total of 23=8 equally likely outcomes when flipping three coins.
Probability of getting 0 heads: There is only 1 outcome (TTT) with 0 heads.P(0 heads)= 1/8
Probability of getting 1 head: There are 3 outcomes (HTT, THT, TTH) with 1 head. P(1 head)= 3/8
Probability of getting 2 heads: There are 3 outcomes (HHT, HTH, THH) with 2 heads. P(2 heads)=3/8
Probability of getting 3 heads: There is only 1 outcome (HHH) with 3 heads. P(3 heads)= 1/8
To find the probability of getting any number of heads from 0 to 3, we add up the probabilities of these events:
(0 to 3 heads)=P(0 heads)+P(1 head)+P(2 heads)+P(3 heads)
⇒ P(0 to 3 heads)=81+83+83+81
⇒ P(0 to 3 heads)=88=1
So, the probability of getting any number of heads from 0 to 3 when flipping three fair coins is 1 or 100%. This is because one of these outcomes is guaranteed to occur when you flip the coins.
Practice Questions on Exhaustive Events
Question 1: You flip a fair coin. Define two events:
Event A: Getting heads (H)
Event B: Getting tails (T)
Are these events collectively exhaustive? What is the probability of getting either heads or tails?
Question 2: You roll a fair six-sided die. Define six events, one for each possible outcome (numbers 1 through 6). Are these events collectively exhaustive? What is the probability of rolling any number from 1 to 6?
Question 3: You draw a card from a standard deck of 52 playing cards. Define four events:
Event A: Drawing a heart
Event B: Drawing a diamond
Event C: Drawing a club
Event D: Drawing a spade
Are these events collectively exhaustive? What is the probability of drawing a card of any suit?
Question 4: You flip two fair coins. Define three events:
Event A: Getting two heads (HH)
Event B: Getting two tails (TT)
Event C: Getting one head and one tail (HT or TH)
Are these events collectively exhaustive? What is the probability of getting either two heads, two tails, or one head and one tail?
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why this sequence is a n=2 n−1?a n=2 n−1?
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Let postive integers sequence {a n}{a n},if for any postive integers m m we have
∑i=1 a m a i=(2 m−1)2∑i=1 a m a i=(2 m−1)2
show that
a n=2 n−1 a n=2 n−1
It seem can use mathematical induction? Now only following
∑i=1 a m a i−∑i=1 a m−1 a i=a a m−1+1+a a m−1+2+⋯+a a m∑i=1 a m a i−∑i=1 a m−1 a i=a a m−1+1+a a m−1+2+⋯+a a m
Thanks in advance!
sequences-and-series
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edited Sep 6, 2015 at 13:31
Michael Hardy
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asked Sep 6, 2015 at 13:11
user253631 user253631
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Don't you mean ∑m i=1∑i=1 m?Hagen von Eitzen –Hagen von Eitzen 2015-09-06 13:15:01 +00:00 Commented Sep 6, 2015 at 13:15
i think there is a typo in your formula Dr. Sonnhard Graubner –Dr. Sonnhard Graubner 2015-09-06 13:15:57 +00:00 Commented Sep 6, 2015 at 13:15
@HagenvonEitzen,it is ∑a m i=1∑i=1 a m not ∑m i=1∑i=1 m user253631 –user253631 2015-09-06 13:17:14 +00:00 Commented Sep 6, 2015 at 13:17
A dumb but workable way: try the ansatz a n=c 2 n 2+c 1 n+c 0 a n=c 2 n 2+c 1 n+c 0, sum the LHS and equate coefficients.krvolok –krvolok 2015-09-06 13:27:23 +00:00 Commented Sep 6, 2015 at 13:27
Would this answer your question ?Lucian –Lucian 2015-09-06 21:47:00 +00:00 Commented Sep 6, 2015 at 21:47
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First choose m=1 m=1 so :
a 1+…+a a 1=1 a 1+…+a a 1=1
but the a i a i's are positive integers so it must be that a 1=1 a 1=1 .
The sequence in obviously one-to-one because if for some i,j i,j we have a i=a j a i=a j then the sums on the LHS are the same so (2 i−1)2=(2 j−1)2(2 i−1)2=(2 j−1)2 which leads to i=j i=j .
The sequence is also strictly increasing because if i>j i>j then the sum on the i i-LHS is bigger than the sum on the j j-LHS so a i>a j a i>a j .As a corollary we get :
a n≥n a n≥n
for every n n .
If a 2=2 a 2=2 then choose m=2 m=2 which leads to a 1+a 2=9 a 1+a 2=9 which is false .
Also if a 2≥4 a 2≥4 then putting m=2 m=2 we get :
9≥a 1+a 2+a 3+a 4≥1+2+3+4=10 9≥a 1+a 2+a 3+a 4≥1+2+3+4=10 ,false so a 2=3 a 2=3 .
Now assume that we established a i=2 i−1 a i=2 i−1 for every i i between 1 1 and n n .Denote a n+1=x a n+1=x .
Subtract the equations for m=n m=n and m=n+1 m=n+1 to get :
a 2 n+…+a x=(2 n+1)2−(2 n−1)2=8 n a 2 n+…+a x=(2 n+1)2−(2 n−1)2=8 n
with x≥2 n x≥2 n .
Assume that x≥2 n+2 x≥2 n+2 .Using the strictly increasing condition we get :
a 2 n≥a n+n=3 n−1 a 2 n≥a n+n=3 n−1 , a 2 n+1≥3 n a 2 n+1≥3 n , a 2 n+2≥3 n+1 a 2 n+2≥3 n+1 so plugging back :
8 n≥a 2 n+a 2 n+1+a 2 n+2≥9 n 8 n≥a 2 n+a 2 n+1+a 2 n+2≥9 n
clearly false .
So it must be that x≤2 n+1 x≤2 n+1 .
Now assume that x=2 n x=2 n so a 2 n=8 n a 2 n=8 n .
Now use m=2 n m=2 n and subtract it with m=n m=n :
(a 1+…+a 8 n)−(a 1+…+a 2 n−1)=(4 n−1)2−(2 n−1)2=4 n(3 n−1)(a 1+…+a 8 n)−(a 1+…+a 2 n−1)=(4 n−1)2−(2 n−1)2=4 n(3 n−1)
a 2 n+…+a 8 n=4 n(3 n−1)a 2 n+…+a 8 n=4 n(3 n−1)
Use again the strictly increasing property to get :
a 2 n+d≥a 2 n+d=8 n+d a 2 n+d≥a 2 n+d=8 n+d
for every d d so plugging back we get :
4 n(3 n−1)≥8 n+(8 n+1)+…+14 n=11 n(6 n+1)4 n(3 n−1)≥8 n+(8 n+1)+…+14 n=11 n(6 n+1)
which is clearly false .
It follows that a n+1=x=2 n+1 a n+1=x=2 n+1 and from the inductive hypothesis that
a n=2 n−1 a n=2 n−1
for every n n .
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edited Sep 6, 2015 at 13:54
answered Sep 6, 2015 at 13:44
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We have {a n}{a n} with a n>0 a n>0 and
∑i=1 a m a i=(2 m−1)2∑i=1 a m a i=(2 m−1)2
Hence
∑i=1 a m a i=a 1+a 2+⋅⋅⋅+a a m=(2 m−1)2∑i=1 a m a i=a 1+a 2+⋅⋅⋅+a a m=(2 m−1)2
∑i=1 a m+1 a i=a 1+a 2+⋅⋅⋅+a a m+1=(2 m+1)2∑i=1 a m+1 a i=a 1+a 2+⋅⋅⋅+a a m+1=(2 m+1)2
Suppose a m=r,a m+1=s a m=r,a m+1=s with r≥s;m≥1 r≥s;m≥1 so that a 1+a 2+⋅⋅⋅+a r=(2 m−1)2≥a 1+a 2+⋅⋅⋅+a s=(2 m+1)2⇒−8 m≥0 a 1+a 2+⋅⋅⋅+a r=(2 m−1)2≥a 1+a 2+⋅⋅⋅+a s=(2 m+1)2⇒−8 m≥0 This can't be true because m>0 m>0. Hence the sequence {a n}{a n} is strictly increasing. Besides
∑i=1 a m+1 a i−∑i=1 a m−1 a i=8 m∑i=1 a m+1 a i−∑i=1 a m−1 a i=8 m
a 1=k 1=?a 1=k 1=?
∑i=1 a 1 a i=a 1+a 2+⋅⋅⋅+a k 1=(2⋅1−1)2=1⇒k 1=1⇒a 1=1∑i=1 a 1 a i=a 1+a 2+⋅⋅⋅+a k 1=(2⋅1−1)2=1⇒k 1=1⇒a 1=1
because a n>0 a n>0.
∑i=1 a 2 a i=a 1+a 2+a 3+⋅⋅⋅+a a 2=(2⋅2−1)2=9∑i=1 a 2 a i=a 1+a 2+a 3+⋅⋅⋅+a a 2=(2⋅2−1)2=9
where 1+2+3+4+5+⋅⋅⋅+a a 2≤a 1+a 2+a 3+⋅⋅⋅+a a 2 1+2+3+4+5+⋅⋅⋅+a a 2≤a 1+a 2+a 3+⋅⋅⋅+a a 2. Therefore 2≤a 2≤3 2≤a 2≤3 because 1+2+3<9<1+2+3+4 1+2+3<9<1+2+3+4. In order to have a n=2 n−1 a n=2 n−1 only a 2=3 a 2=3 agrees in which case a 1+a 2+a 3=9 a 1+a 2+a 3=9 hence (a 1,a 2,a 3)=(1,3,5)(a 1,a 2,a 3)=(1,3,5) Now
∑i=1 a 3 a i=a 1+a 2+a 3+a 4+a 5=(2⋅3−1)2=25∑i=1 a 3 a i=a 1+a 2+a 3+a 4+a 5=(2⋅3−1)2=25
i.e.
a 4+a 5=25−9=16 a 4+a 5=25−9=16
where a 4≥6 a 4≥6 so (a 4,a 5)=(6,10),(7,9)(a 4,a 5)=(6,10),(7,9) and only (7,9)(7,9) agrees. Similarly
∑i=1 a 4 a i=a 1+a 2+a 3+a 4+a 5+a 6+a 7=49∑i=1 a 4 a i=a 1+a 2+a 3+a 4+a 5+a 6+a 7=49
i.e.
a 6+a 7=49−25=24 a 6+a 7=49−25=24
where a 6≥10⇒(a 6,a 7)=(10,14),(11,13)a 6≥10⇒(a 6,a 7)=(10,14),(11,13) and (a 6,a 7)=(11,13)(a 6,a 7)=(11,13).
NOTATION: For easy, ∑i=1 a n a i=S(a n)∑i=1 a n a i=S(a n)
I use these first seven calculated values of a n a n to infer an expression of S (a_n) as a function of n.
S(a 1)=a 1=1 S(a 1)=a 1=1
S(a 2)=S(a 1)+a 2+a 3=9⇒a 2+a 3=8⋅1 S(a 2)=S(a 1)+a 2+a 3=9⇒a 2+a 3=8⋅1
S(a 3)=S(a 2)+a 4+a 5=25⇒a 4+a 5=8⋅2 S(a 3)=S(a 2)+a 4+a 5=25⇒a 4+a 5=8⋅2
S(a 4)=S(a 3)+a 6+a 7=49⇒a 6+a 7=8⋅3 S(a 4)=S(a 3)+a 6+a 7=49⇒a 6+a 7=8⋅3
S(a 5)=S(a 4)+a 8+a 9=81⇒a 8+a 9=8⋅4 S(a 5)=S(a 4)+a 8+a 9=81⇒a 8+a 9=8⋅4
S(a 6)=S(a 5)+a 10+a 11=121⇒a 10+a 11=8⋅5 S(a 6)=S(a 5)+a 10+a 11=121⇒a 10+a 11=8⋅5
S(a 7)=S(a 6)+a 12+a 13=169⇒a 12+a 13=8⋅6 S(a 7)=S(a 6)+a 12+a 13=169⇒a 12+a 13=8⋅6
These seven equalities successively generated show at first sight that a 1=1,a 2=3,…..,a 7=13 a 1=1,a 2=3,…..,a 7=13 because, according to the statement, the last index of the ∑a n i=1 a i∑i=1 a n a i must be a 1,a 2,a 3,….a 7 a 1,a 2,a 3,….a 7 respectively. Each value of an a i a i determines the value of a i+1=a i+2 a i+1=a i+2 which then verify that a n=2 n−1 a n=2 n−1. On the other hand, we have by easy induction on S(a n)−S(a n−1)=8(n−1)S(a n)−S(a n−1)=8(n−1),
S(a n)=1+8[1+2+3+⋅⋅⋅+(n−1)]S(a n)=1+8[1+2+3+⋅⋅⋅+(n−1)]
i.e.
S(a n)=1+4 n(n−1)=(2 n−1)2⟺∑i=1 a m a i=(2 m−1)2 S(a n)=1+4 n(n−1)=(2 n−1)2⟺∑i=1 a m a i=(2 m−1)2
(BIS proof).-Optionally, using induction hypothesis ∑a m i=1 a i=(2 m−1)2∑i=1 a m a i=(2 m−1)2, having ∑a 1 i=1 a i=(2⋅1−1)2=1∑i=1 a 1 a i=(2⋅1−1)2=1, it follows S(a n+1)−S(a n)=8 n⇒S(a n+1)=S(a n)+8 n=(2 n−1)2+8 n=(2 n+1)2 S(a n+1)−S(a n)=8 n⇒S(a n+1)=S(a n)+8 n=(2 n−1)2+8 n=(2 n+1)2, thus a n+1=2 n+1⟺a n+1=2(n+1)−1 a n+1=2 n+1⟺a n+1=2(n+1)−1.
NOTE.- Finally, it was very difficult to interpret the statement at the beginning; I also thought, as well as distinguished followers of Stack Exchange there must be some typo. What I find remarkable is the equivalence for natural integers ∑a n i=1 a i=(2 n−1)2⟺a n=2 n−1∑i=1 a n a i=(2 n−1)2⟺a n=2 n−1 which everyone knows only one implication.
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answered Sep 13, 2015 at 17:26
AtaulfoAtaulfo
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4403 | https://math.stackexchange.com/questions/777493/do-the-kolmogorovs-axioms-permit-speaking-of-frequencies-of-occurence-in-any-me | probability - Do the Kolmogorov's axioms permit speaking of frequencies of occurence in any meaningful sense? - Mathematics Stack Exchange
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Do the Kolmogorov's axioms permit speaking of frequencies of occurence in any meaningful sense?
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It is frequently stated (in textbooks, on Wikipedia) that the "Law of large numbers" in mathematical probability theory is a statement about relative frequencies of occurrence of an event in a finite number of trials or that it "relates the axiomatic concept of probability to the statistical concept of frequency". Isn't this is a methodological mistake of ascribing an interpretation to a mathematical term, perhaps relying too much on the colorful language, that does not at all follow from how this term is mathematically defined? Recall the typical derivation of the WLLN:
Let X 1,X 2,...,X n X 1,X 2,...,X n be a sequence of n independent and identically distributed random variables with the same finite mean μ μ, and with variance σ 2 σ 2 and let:
X¯¯¯¯=1 n(X 1+⋯+X n)X¯=1 n(X 1+⋯+X n)
We have:
E[X¯¯¯¯]=E[X 1+...+X n]n=E[X 1]+...+E[X n]n=n μ n=μ E[X¯]=E[X 1+...+X n]n=E[X 1]+...+E[X n]n=n μ n=μ V a r[X¯¯¯¯]=V a r[X 1+...+X n]n 2=V a r[X 1]+...+V a r[X n]n 2=n σ 2 n 2=σ 2 n V a r[X¯]=V a r[X 1+...+X n]n 2=V a r[X 1]+...+V a r[X n]n 2=n σ 2 n 2=σ 2 n
And from Chebyshev's inequality:
P(|X¯¯¯¯−μ|>ϵ)≤σ 2 n ϵ 2 P(|X¯−μ|>ϵ)≤σ 2 n ϵ 2
And so X is said to converge in probability to μ μ.
Now consider what is strictly speaking the meaning of this expression in the axiomatic framework it is derived in:
P(|X¯¯¯¯−μ|>ϵ)≤σ 2 n ϵ 2 P(|X¯−μ|>ϵ)≤σ 2 n ϵ 2
P()P(), everywhere it occurs in the derivation, is known only to be a number satisfying Kolmogorov's axioms, so a number between 0 and 1, and so forth, but none of the axioms introduce any theoretical equivalent of the intuitive notion of frequency. If additional assumptions about P()P() are not made, the sentence can obviously not be interpreted at all, but what is also important the theoretical mean μ μ is not necessarily the mean value in an infinite number of trials, X¯¯¯¯X¯ is not necessarily the mean value from n trials, and so forth. Consider an experiment of tossing a fair coin repeatedly - quite obviously, nothing in Kolmogorov's axioms enforces using 1/2 for the probability of heads, you could just as well use 1/π−−√1/π, yet the derivation continues to "work", except the meaning of the various variables is not in agreement with their intuitive interpretations. The P()P() might still mean something, it might be a quantification of an absurd belief of mine, the mathematical derivation continues be true regardless, in the sense that as long as the initial P()′s P()′s satisfy axioms, theorems about other P()′s P()′s follow, and with Kolmogorov's axioms providing only weak constraints on and not a definition of P()P(), it's basically only symbol manipulation.
This "relative frequency" interpretation frequently given seems to rest on an additional assumption, and this assumption seems to be a form of the law of large numbers itself. Consider this fragment from Kolmogorov's Grundbegriffe on applying the results of probability theory to the real world:
We apply the theory of probability to the actual world of experiment in the following manner:
...
4) Under certain conditions, which we shall not discuss here, we may assume that the event A which may or may not occur under conditions S, is assigned a real number P(A) which has the following characteristics:
a) One can be practically certain that if the complex of conditions S is repeated a large number of times, n, then if m be the number of occurrences of event A, the ratio m/n will differ very slightly from P(A).
Which seems equivalent to introducing the weak law of large numbers in a particular, slightly different form, as an additional axiom.
Meanwhile, many reputable sources contain statements that seem completely in opposition to the above reasoning, for example Wikipedia:
It follows from the law of large numbers that the empirical probability of success in a series of Bernoulli trials will converge to the theoretical probability. For a Bernoulli random variable, the expected value is the theoretical probability of success, and the average of n such variables (assuming they are independent and identically distributed (i.i.d.)) is precisely the relative frequency.
This seem to be mistaken already in claiming that from a mathematical theorem anything can follow about empirical probability (the page on which defines it as the relative frequency in actual experiment), but there are many more subtle claims that technically also seem erroneous from the above considerations:
The LLN is important because it "guarantees" stable long-term results for the averages of random events.
Note that the Wikipedia article about LLN claims to be about the mathematical theorem, not about the empirical observation, which was also historically sometimes been called the LLN. It seems to me that LLN does nothing to "guarantee stable long-term results", for as stated above those stable long-term results have to be assumed in the first place for the terms occuring in the derivation to have the intuitive meaning we typically ascribe to them, not to mention something has to be done to at all interpret P()P() in the first place. Another instance from Wikipedia:
According to the law of large numbers, if a large number of six-sided die are rolled, the average of their values (sometimes called the sample mean) is likely to be close to 3.5, with the precision increasing as more dice are rolled.
Does this really follow from the mathematical theorem? In my opinion, the interpretation of the theorem that is used here, rests on assuming this fact. There is a particularly vivid example in the "Treatise on probability" by Keynes of what happens when one follows the WLLN with even a slight deviation from this initial assumptions of p's being the relative frequencies in the limit of an infinite number of trials:
The following example from Czuber will be sufficient for the purpose of illustration. Czuber’s argument is as follows: In the period 1866–1877 there were registered in Austria
m = 4,311,076 male births
n = 4,052,193 female births
s = 8,363,269
for the succeeding period, 1877–1899, we are given only
m' = 6,533,961 male births;
what conclusion can we draw as to the number n of female births? We can conclude, according to Czuber, that the most probable value
n' = nm'/m = 6,141,587
and that there is a probability P = .9999779 that n will lie between the limits 6,118,361 and 6,164,813. It seems in plain opposition to good sense that on such evidence we should be able with practical certainty P = .9999779 = 1 − 1/45250 to estimate the number of female births within such narrow limits. And we see that the conditions laid down in § 11 have been flagrantly neglected. The number of cases, over which the prediction based on Bernoulli’s Theorem is to extend, actually exceeds the number of cases upon which the à priori probability has been based. It may be added that for the period, 1877–1894, the actual value of n did lie between the estimated limits, but that for the period, 1895–1905, it lay outside limits to which the same method had attributed practical certainty.
Am I mistaken in my reasoning above, or are all those really mistakes in the Wikipedia? I have seen similar statements all over the place in textbooks, and I am honestly wondering what I am missing.
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edited May 1, 2014 at 20:52
Jarosław RzeszótkoJarosław Rzeszótko
asked May 1, 2014 at 20:04
Jarosław RzeszótkoJarosław Rzeszótko
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This is a much more concrete version of the question I asked earlier math.stackexchange.com/questions/775788/…, and that I would ask the dear moderators to delete, as it was too vague to be useful. Please forgive me partially reposting something, I hope you will understand making a complicated reasoning clear does not always come easy or quickly. I will not post anything similar again.Jarosław Rzeszótko –Jarosław Rzeszótko 2014-05-01 20:07:01 +00:00 Commented May 1, 2014 at 20:07
1 The law of large numbers is a red herring, I think: you're stuck on the idea of expressing "physical" quantities (such as the result of a frequency-measuring experiment) as random variables.user14972 –user14972 2014-05-01 21:18:57 +00:00 Commented May 1, 2014 at 21:18
1 You can express a frequency-measuring experiment as X-dash as defined above regardless of what P() is, but the moment you take expectations, and multiply the P()'s of particular values of the random variable by the actual values, you end up with a statement about what we intuitively think as the mean value from repetitions of the experiment only with additional assumptions about P()'s that are not in the axioms of Kolmogorov. That is indeed where my disagreement with Wikipedia and its interpretation of LLN has its roots, but you seem to claim I am simply misunderstanding something here, right?Jarosław Rzeszótko –Jarosław Rzeszótko 2014-05-01 21:53:10 +00:00 Commented May 1, 2014 at 21:53
better work the example out what is really the variance.Willemien –Willemien 2014-05-02 06:43:48 +00:00 Commented May 2, 2014 at 6:43
@Willemien I am not sure what you mean?Jarosław Rzeszótko –Jarosław Rzeszótko 2014-05-02 06:58:31 +00:00 Commented May 2, 2014 at 6:58
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I. I agree with you that no version of the Law of Large Numbers tells us something about real life frequencies, already for the reason that no purely mathematical statement tells us anything about real life at all, without first giving the mathematical objects in it a "real life interpretation" (which never can be stated, let alone "proven", within mathematics itself).
Rather, I think of the LLN as something which, within any useful mathematical model of probabilities and statistical experiments, should hold true! In the sense that: If you show me a new set of axioms for probability theory, which you claim have some use as a model for real life dice rolling etc.; and those axioms do not imply some version of the Law of Large Numbers -- then I would dismiss your axiom system, and I think so should you.
II. Most people would agree there is a real life experiment which we can call "tossing a fair coin" (or "rolling a fair die", "spinning a fair roulette wheel" ...), where we have a clearly defined finite set of outcomes, none of the outcomes is more likely than any other, we can repeat the experiment as many times as we want, and the outcome of the next experiment has nothing to do with any outcome we have so far.
And we could be interested in questions like: Should I play this game where I win/lose this much money in case ... happens? Is it more likely that after a hundred rolls, the added number on the dice is between 370 and 380, or between 345 and 350? Etc.
To gather quantitative insight into answering these questions, we need to model the real life experiment with a mathematical theory. One can debate (but again, such a debate happens outside of mathematics) what such a model could tell us, whether it could tell us something with certainty, whatever that might mean; but most people would agree that it seems we can get some insight here by doing some kind of math.
Indeed, we are looking for two things which only together have any chance to be of use for real life: namely, a "purely" mathematical theory, together with a real life interpretation (like a translation table) thereof, which allows us to perform the routine we (should) always do:
Step 1: Translate our real life question into a question in the mathematical model.
Step 2: Use our math skills to answer the question within the model.
Step 3: Translate that answer back into the real life interpretation.
The axioms of probability, as for example Kolmogorov's, do that: They provide us with a mathematical model which will give out very concrete answers. As with every mathematical model, those concrete answers -- say, P(X¯100∈[3.45,3.5])>P(X¯100∈[3.7,3.8])P(X¯100∈[3.45,3.5])>P(X¯100∈[3.7,3.8]) -- are absolutely true within the mathematical theory (foundational issues a la Gödel aside for now). They also come with a standard interpretation (or maybe, a standard set of interpretations, one for each philosophical school). None of these interpretations are justifiable by mathematics itself; and what any result of the theory (like P(X¯100∈[3.45,3.5])>P(X¯100∈[3.7,3.8])P(X¯100∈[3.45,3.5])>P(X¯100∈[3.7,3.8])) tells us about our real life experiment is not a mathematical question. It is philosophical, and very much up to debate. Maybe a frequentist would say, this means that if you roll 100 dice again and again (i.e. performing kind of a meta-experiment, where each individual experiment is already 100 "atomic experiments" averaged), then the relative frequency of ... is greater than the relative frequency of ... . Maybe a Bayesian would say, well it means that if you have some money to spare, and somebody gives you the alternative to bet on this or that outcome, you should bet on this, and not that. Etc.
III. Now consider the following statement, which I claim would be accepted by almost everyone:
( ∗∗ ) "If you repeat a real life experiment of the above kind many times, then the sample means should converge to (become a better and better approximation of) the ideal mean".
A frequentist might smirkingly accept (∗∗), but quip that it's is true by definition, because he might claim that any definition of such an "ideal mean" beyond "what the sample means converge to" is meaningless. A Bayesian might explain the "ideal mean" as, well you know, the average -- like if you put it in a histogram, see, here is the centre of weight -- the outcome you would bet on -- you know! And she might be content with that. And she would say, yes, of course that is related to relative frequencies exactly in the sense of (∗∗).
I want to strees that (∗∗) is not a mathematical statement. It is a statement about real life experiments, which we claim to be true, although we might not agree on why we do so: depending on your philosophical background, you can see it as a tautology or not, but even if you do it is not a mathematical tautology (it's not a mathematical statement at all), just maybe a philosophical one.
And now let's say we do want a model-plus-translation-table for our experiments from paragraph II. Such a model should contain an object which models [i.e. whose "real life translation" is] one "atomic" experiment: that is the random variable X X, or to be precise, an infinite collection of i.i.d. random variables X 1,X 2,...X 1,X 2,....
It contains something which models "the actual sample mean after 100,1000,...,n 100,1000,...,n trials": that is X¯n:=1 n∑n 1 X i X¯n:=1 n∑1 n X i.
And it contains something which models "an ideal mean": that is μ=E X μ=E X.
So with that model-plus-translation, we can now formulate, within such model, a statement (or set of related statements) which, under the standard translation, appear to say something akin to (∗∗).
And that is the (or are the various forms of the) Law of Large Numbers. And they are true within the model, and they can be derived from the axioms of that model.
So I would say: The fact that they hold true e.g. in Kolmogorov's Axioms means that these axioms pass one of the most basic tests they should pass: We have a philosophical statement about the real world, (∗∗), which we believe to be true, and of the various ways we can translate it into the mathematical model, those translations are true in the model. The LLN is not a surprising statement on a meta-mathematical level for the following reason: Any kind of model for probability which, when used as model for the above real life experiment, would not give out a result which is the mathematical analogy of statement (∗∗), should be thrown out!
In other words: Of course good probability axioms give out the Law of Large Numbers. They are made so that they give them out. If somebody proposed a set of mathematical axioms, and a real-life-translation-guideline for the objects in there, and any model-internal version of (∗∗) would be wrong -- then that model should be deemed useless (both by frequentists and Bayesians, just for different reasons) to model the above real life experiments.
IV. I want to finish by pointing out one instance where your argument seems contradictory, which, when exposed, might make what I write above more plausible to you.
Let me simplify an argument of yours like this:
(A) A mathematical statement like the LLN in itself can never make any statement about real life frequencies.
(B) Many sources claim that LLN does make statements about real life frequencies. So they must be implicitly assuming more.
(C) As an example, you exhibit a Kolmogorov quote about applying probability theory to the real world, and say that it "seems equivalent to introducing the weak law of large numbers in a particular, slightly different form, as an additional axiom."
I agree with (A) and (B). But (C) is where I want you to pause and think: Were we not in agreement, cf. (A), that no mathematical statement can ever tell us something about real life frequencies? Then what kind of "additional axiom" would say that? Whatever the otherwise mistaken sources in (B) are implicitly assuming, and Kolmogorov himself talks about in (C), it cannot just be an "additional axiom", at least not a mathematical one: Because one can throw in as many mathematical axioms as one wants, they will never bridge the fundamental gap in (A).
I claim the thing that all the sources in (B) are implicitly assuming, and what Kolmogorov talks about in (C), is not an additional axiom within the mathematical theory. It is the meta-mathematical translation / interpretation that I talk about above, which in itself is not mathematical, and in particular cannot be introduced as an additional axiom within the theory.
I claim, indeed, most sources are very careless, in that they totally forget the translation / interpretation part between real life and mathematical model, i.e. the bridge we need to cross the gap in (A); i.e. steps 1 and 3 in the routine explained in paragraph II. Of course it is taught in any beginner's class that any model in itself (i.e. without a translation, without steps 1 and 3) is useless, but it is commonly forgotten already in the non-statistical sciences, and more so in statistics, which leads to all kind of confusions. We spend so much time and effort on step 2 that we often forget steps 1 and 3; also, step 2 can be taught and learned and put on exams, but steps 1 and 3 not so well: they go beyond mathematics, seem to fit better into a science or philosophy class (although I doubt they get a good enough treatment there either). However, if we forget them, we are left with a bunch of axioms linking together almost meaningless symbols; and the remnants of meaning which we, as humans, cannot help applying to these symbols, quickly seem to be nothing but circular arguments.
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edited Apr 3, 2023 at 16:13
answered Aug 19, 2021 at 20:29
Torsten SchoenebergTorsten Schoeneberg
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1 your distinction between "mathematical model" and "physical object /process" is the key point ! (+1)G Cab –G Cab 2022-03-28 18:58:31 +00:00 Commented Mar 28, 2022 at 18:58
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I understand OP's concern and I want to illustrate it by an example from geometry.
Pythagorean theorem. Let V V be a two-dimensional real vector space with inner product ⟨⋅,⋅⟩⟨⋅,⋅⟩ and induced norm ∥⋅∥‖⋅‖. In linear algebra we learn that ⟨a,b⟩=0⟨a,b⟩=0 implies ∥a∥2+∥b∥2=∥a+b∥2‖a‖2+‖b‖2=‖a+b‖2, and this result is called Pythagorean theorem. This is Step 2 of the routine in Torsten's answer . Clearly, we would like to know what the connection of this Pythagorean theorem is with right-angled triangles drawn on a real sheet of paper. So we need to think about Steps 1 and 3 of the routine. Generations of students have drawn right-angled triangles and measured the lengths of the legs a,b a,b and the hypothenuse c c, arriving at the identity a 2+b 2=c 2 a 2+b 2=c 2 to some degree of precision. Based on the amount of data, it is plausible to assume that there exists an empirical Pythagorean theorem in the real world. Now we can use the empirical Pythagorean theorem and the standard interpretation (using a rectangular coordinate system) to 'identify' ∥a∥‖a‖ with the length of a a. In this way we obtain an interpretation of the Pythagorean theorem (in V V) in terms of lengths. Doesn't it then feel wrong to say that the empirical Pythagorean theorem is a consequence of the Pythagorean theorem (in V V) under the above interpretation?
There is an empirical result often called the empirical law of large numbers or the stability of frequencies, which states, for example, that the relative frequencies of heads in a long sequence of coin tosses 'converge' to some value p p. In my opinion it is this empirical law which Kolmogorov refers to in the excerpt cited by OP. Afterwards OP argues that since we are using stability of frequencies to interpret probabilities and thereby the law of large numbers (LLN), it feels wrong to say that LLN 'guarantees' stability of frequencies.
I agree that it seems unnecessary to say that LLN is responsible for stability of frequencies whenever stability of frequencies is used to provide an interpretation.
However, stability of frequencies only looks good on paper. Since we cannot make infinitely many observations, this empirical law isn't of much help for determining probabilities in practice. On top of that many problems of practical interest are not as reproducible as a coin toss. I am not a probabilist or statistician, so from here on I have to rely on the opinion of other people.
First of all, let me quote Mark Kac. OP has provided a short excerpt in . Here is how Kac continues.
The applicability of such a theory [probability theory] to natural sciences must ultimately be tested by an experiment. But this is true of all mathematical theories when applied outside the realm of mathematics, and the vague feeling of discomfort one encounters (mostly among philosophers!) when first subjected to statistical reasoning must be attributed to the relative novelty of the ideas.
To me there is no methodological distinction between the applicability of differential equations to astronomy and of probability theory to thermodynamics or quantum mechanics.
It works! And brutally pragmatic as this point of view is, no better substitute has been found. (, p.5)
What Kac suggests is a more pragmatic point of view, that of a physicist. Let me quote Krzysztof Burdzy.
[Compared with mathematicians,] physicists have a different idea of a 'proof' – you start with a large number of unrelated assumptions, you combine them into a single prediction, and you check if the prediction agrees with the observed data. If the agreement is within 20%, you call the assumptions proved. (, p.41)
I think that Burdzy has made up the figure of 20%, but I am not a physicist. More importantly, we can apply probability theory (including LLN) with any assumption that we deem fit, which makes stability of frequencies in some sense obsolete. As long as we can produce predictions that can be tested, we don't have to worry about the 'vague' link between the model and the real world. Over time and by doing a lot of experiments, we acquire a certain confidence in our claims (if they agree with the observations) and then they become accepted by the math/science community.
All of this is rather difficult to comprehend for a beginner in probability/statistics. In a first probability course, the statistical tools that are needed for the predictions only enter very late or not at all, which may be the reason why students don't see this 'pragmatic' approach to applied probability/statistics. On the other hand, stability of frequencies may still be useful for gaining intuition.
A layman (like myself) gets lost easily in the big philosophical frequentist vs. Bayesian debate(s). As a mathematician, I can accept that the only definition of probability appears in the Kolmogorov axioms and I don't need to know its 'true' meaning in order to learn and apply the theory. My goal in writing this was to provide some consolation for a specific group of people (including myself), i.e. those who have gone through a similar thought process as OP.
Torsten Schoeneberg (Aug 19, 2021)
Mark Kac, "Probability and related topics in physical sciences"
Krzysztof Burdzy, "The search for certainty: on the clash of science and philosophy of probability" (suggested by Bjørn Kjos-Hanssen in )
Logical issues with the weak law of large numbers and its interpretation
Is probability and the Law of Large Numbers a huge circular argument?
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edited Mar 29, 2022 at 11:12
answered Mar 28, 2022 at 16:41
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Kolmogorov's axioms, if one were to make an assumption about the distribution of the random variable X i X i, could be used to derive the distribution of the random variable X¯X¯. Notice in the last statement that since X i X i is a random variable, X¯X¯ is also a random variable. The fact that X¯X¯ is a random variable means that there is a probability measure for the random variable X¯X¯. The beauty of the WLLN is that so long as both μ μ and σ 2 σ 2 are finite, no assumptions about the measure P()P() must be made in order to derive that X n¯X n¯ converges in probability to μ μ. I agree with Hurkyl. Perhaps this post will help with the concept of a random variable
You do make a good point, however, about whether or not the assumptions that the X X's are independent and identically distributed random variables may not be true in practice, which is the problem alluded to in the Keynes example.
The example regarding dice appears to rely on the assumption that the die is fair, which may or may not be reasonable depending on how the die is constructed and rolled. However, it seems reasonable to assume that there exists appropriate setups of a dice rolling experiments for which the rolls are i.i.d i.i.d random variables with a probability measure P P. In such a case, it does follow from the WLLN that X¯X¯ would indeed converge to μ μ.
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edited Apr 13, 2017 at 12:44
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1 I have no doubts that X-dash converges to mu in the framework of the Kolmogorov's axioms, but the question is whether this allows to draw any interpretable conclusions. Based only on the axioms, mu and X-dash are not interpretable as the average value from a large result of trials, they are simply weighted averages of some set of values using the, arbitrary to some extent, measure P(). Similarly when "relative frequency" is mentioned in the context of the theory, I think it does not really translate into real world relative frequency, unless the WLLN is assumed as true a priori.Jarosław Rzeszótko –Jarosław Rzeszótko 2014-05-03 13:04:11 +00:00 Commented May 3, 2014 at 13:04
In other words, it seems to me people widely fail to notice that when "relative frequency" is spoken of in the context of probability theory, it only corresponds to our intuitive notion of "relative frequency", if one makes assumptions additional to the Kolmogorov's axioms, and the assumption needed is the WLLN itself. Hence no conclusions about real world situations follow purely from the WLLN as derived from the axioms.Jarosław Rzeszótko –Jarosław Rzeszótko 2014-05-03 13:05:24 +00:00 Commented May 3, 2014 at 13:05
By the way, the trials in the Keynes examples are independent and identically distributed, the problem is that the probabilities are slightly off from the ideal theoretical relative frequency in an infinite limit of trials. While such P()'s satisfy the axioms, and the formal mathematics stays "true", you see that the result does not seem to be true anymore, and that is because the intuitive interpretation of the various terms in the derivation does not hold anymore. This example shows the WLLN has to be assumed a priori for the usual real world interpretation to hold.Jarosław Rzeszótko –Jarosław Rzeszótko 2014-05-03 13:19:23 +00:00 Commented May 3, 2014 at 13:19
2 I do now have a reference for this too: "It is also odd, if we begin with frequency as the definition of probability, that we should then expend great effort to prove the law of large numbers the theorem that the probability of an event will almost certainly be approximated by the event's relative frequency. This was seen as a real problem by the frequentists of the nineteenth century (Porter 1986)." (The following paragraphs are also very interesting, too little space here) glennshafer.com/assets/downloads/articles/article46.pdfJarosław Rzeszótko –Jarosław Rzeszótko 2014-05-03 20:08:43 +00:00 Commented May 3, 2014 at 20:08
1 Hey, I am not criticizing the foundations of probability theory here, I just think due to the suggestive terminology people fail to see those subtle distinction and make logically erroneous conclusions about what follows from what. This is important, for the axioms / assumptions have to be checked with the real world, while the theorems follow by logic alone from those axioms.Jarosław Rzeszótko –Jarosław Rzeszótko 2014-05-03 20:23:09 +00:00 Commented May 3, 2014 at 20:23
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You are correct. The Law of Large Numbers does not actually say as much as we would like to believe. Confusion arises because we try to ascribe too much philosophical importance to it. There is a reason that the Wikipedia article puts quotes around 'guarantees' because nobody actually believes that some formal theory (on its own) guarantees anything about the real world. All LLN says is that some notion of probability, without interpretation, approaches 1 -- nothing more, nothing less. It certainly doesn't prove for a fact that relative frequency approaches some probability (what probability?). The key to understanding this is to note that the LLN, as you pointed out, actually uses the term P() in its own statement. I will use this version of the LLN:
"The probability of a particular sampling's frequency distribution resembling the actual probability distribution (to a degree) as it gets large approaches 1."
Interpreting "probability" in the frequentist sense, it becomes this:
Interpret "actual probability distribution": "Suppose that as we take larger samples, they converge to a particular relative frequency distribution..."
Interpret the statement: "... Now if we were given enough instances of n-numbered samplings, the ratio of those that closely resemble (within ϵ ϵ) the original frequency distribution vs. those that don't approaches 1 to 0. That is, the relative frequency of the 'correct' instances converges to 1 as you raise both n and the number of instances."
You can imagine it like a table. Suppose for example that our coin has T-H with 50-50 relative frequency. Each row is a sequence of coin tosses (a sampling), and there are several rows -- you're kind of doing several samples in parallel. Now add more columns, i.e. add more tosses to each sequence, and add more rows, increasing the amount of sequences themselves. As we do so, count the number of rows which have a near 50-50 frequency distribution (within some ϵ ϵ) , and divide by the total number of rows. This number should certainly approach 1, according to the theorem.
Now some might find this fact very surprising or insightful, and that's pretty much what's causing the whole confusion in the first place. It shouldn't be surprising, because if you look closely at our frequentist interpretation example, we assumed "Suppose for now that our coin has T-H with 50-50 relative frequency." In other words, we have already assumed that any particular sequence of tossings will, with logical certainty, approach a 50-50 frequency split. So is should not be surprising when we say with logical certainty that a progressively larger proportion of these tossing-sequences will resemble 50-50 splits if we toss more in each, and recruit more tossers? It's almost a rephrasing or the original assumption but at a meta-level (we're talking about samples of samples).
So this certainty about the real world (interpreted LLN) only comes from another, assumed certainty about the real world (interpretation of probability).
First of all, with a frequentist interpretation, it is not the LLN that states that a sample will approach the relative frequency distribution -- it's the frequentist interpretation/definition of P()P() that says this. It sure is easy to think that, though, if we interpret the whole thing inconsistently -- i.e. if we lazily interpret the outer "probability that ... approaches 1" to mean "... approaches certainty" in LLN but leave the inner statement "relative frequency dist. resembles probability dist." up to (different) interpretation. Then of course you get "relative frequency dist. resembles probability dist. in the limit". It's kind of like if you have a limit of an integral of an integral, but you delete the outer integral and apply the limit to the inner integral.
Interestingly, if you interpret probability as a measure of belief, you might get something that sounds less trivial than the frequentist's version: "The degree of belief in 'any sample reflects actual belief measures in its relative frequencies within ϵ ϵ error' approaches certainty as we choose bigger samples." However this is still different from "Samples, as they get larger, approach actual belief measures in their relative frequencies." As an illustration, imagine if you have two sequences f n f n and p n p n. I am sure you can appreciate the difference between l i m n→∞P(|f n−p n|<ϵ)=1 l i m n→∞P(|f n−p n|<ϵ)=1 and l i m n→∞|f n−p n|=0 l i m n→∞|f n−p n|=0. The latter implies l i m n→∞f n l i m n→∞f n = l i m n→∞p n l i m n→∞p n (or =p=p taking p n p n to be a constant for simplicity), whereas this is not true for the former. The latter is a very powerful statement, and probability theory cannot prove it, as you suspected.
In fact, you were on the right track with the "absurd belief" argument. Suppose that probability theory were indeed capable of proving this amazing theorem, that "a sample's relative frequency approaches the probability distribution". However, as you've found, there are several interpretations for probability which conflict with each other. To borrow terminology from mathematical logic: you've essentially found two models of probability theory; one satisfies the statement "the rel. frequency distribution approaches 1/2:1/2 1/2:1/2", and another satisfies the statement "the rel. frequency distribution approaches 1/π:(1−1/π)1/π:(1−1/π)". So the statement "frequency approaches probability" is neither true nor false: it is independent as either one is consistent with the theory. Thus, Kolmogorov's probability theory is not powerful enough to prove a statement in the form "frequency approaches probability". (Now, if you were to force the issue by saying "probability should equal relative frequency" you've essentially trivialized the issue by baking frequentism into the theory. The only possible model for this probability theory would be frequentism or something isomorphic to it, and the statement becomes obvious.)
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answered Aug 7, 2017 at 7:37
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What you're missing is that the derivation of the WLLN is allowed to use, not only the Kolmogorov axioms, but also the assumption stated in the theorem: "The X 1,X 2,…,X n X 1,X 2,…,X n are a sequence of n n independent and identically distributed random variables with the same finite mean μ, and with variance σ 2 σ 2". So, for example, if we are tossing a fair coin, we know that μ=1/2 (this is what "fair coin" means in probability theory), not 1/π−−√1/π. And likewise, in a Bernoulli trial, we are given the actual mean to which the observed probabilities are supposed to converge. And Keynes/Czuber's example isn't a valid application of the LLN because we are not given the actual mean and standard deviation.
So the first two claims in the Wikipedia article are basically correct (except that "will converge to the theoretical probability" should read "will converge in probability to the theoretical probability"; the probability that the observed values do not converge to the theoretical value is 0; but it might happen anyway).
However, the third claim, "According to the law of large numbers, if a large number of six-sided die are rolled, the average of their values (sometimes called the sample mean) is likely to be close to 3.5, with the precision increasing as more dice are rolled." doesn't follow, since we don't know a priori that rolling a six-sided die constitutes a Bernoulli trial. Looking at the context, it seems that the fairness of the die is meant as an ambient assumption, since one of the preceding sentences is "For example, a single roll of a six-sided die produces one of the numbers 1, 2, 3, 4, 5, or 6, each with equal probability."
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answered Aug 29, 2014 at 16:41
Ben StandevenBen Standeven
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4404 | http://emlab.illinois.edu/ece451/appnotes/r-waveguide.pdf | WAVEGUIDES z y x Maxwell's Equation 2 2 E E 0 (A) 2 2 H H 0 (B) For a waveguide with arbitrary cross section as shown in the above figure, we assume a plane wave solution and as a first trial, we set Ez = 0. This defines the TE modes. From t H E , we have y x z z y x E H E j E j H y z t (1) y x z z x y H E E j E j H z x t (2) y y x x z z E E E E H j H x y t x y (3) From j H E, we get ˆ ˆ ˆ x y z j x y z H H H x y z E - 2 - y z z x z y x H H H j E j H j E y z y (4) x z z y z x y H H H j E j H j E z x x (5) 0 y x H H x y (6) We want to express all quantities in terms of Hz. From (2), we have z x y E H (7) in (4) 2 x z z x E H j j E y (8) Solving for Ex 2 2 z x z H j E y (9) From (1) z y x E H (10) in (5) 2 z y z y E H j j E x (11) so that 2 2 z y z H j E x (12) - 3 - 2 2 z z x z j H H x (13) 2 2 z z y z j H H y (14) Ez = 0 (15) Combining solutions for Ex and Ey into (3) gives 2 2 2 2 2 2 z z z z H H H x y (16) RECTANGULAR WAVEGUIDES a b z y x If the cross section of the waveguide is a rectangle, we have a rectangular waveguide and the boundary conditions are such that the tangential electric field is zero on all the PEC walls. TE Modes The general solution for TE modes with Ez=0 is obtained from (16) y y x x z j y j y j x j x j z z H e Ae Be Ce De (17) 2 2 y y x x z j y j y j x j x j z x y z E e Ae Be Ce De (18) 2 2 y y x x z j y j y y j x j x j z x z E e Ae Be Ce De (19) At y=0, Ex=0 which leads to C=D At x=0, Ey=0 which leads to A=B - 4 - cos cos z j z z o x y H H e x y (20) 2 2 sin cos z j z x y o x y z j E H e x y (21) 2 2 cos sin z y j z x o x y z j E H e x y (22) At x=a, Ey=0; this leads to x m a At y=b, Ex=0; this leads to y n b The dispersion relation is obtained by placing (20) in (16) 2 2 2 2 x y z (23) 2 2 2 2 z m n a b (24) and 2 2 2 z m n a b (25) The guidance condition is 2 2 2 m n a b (26) or f > fc where fc is the cutoff frequency of the TEmn mode given by the relation 2 2 1 2 c m n f a b (27) The TEmn mode will not propagate unless f is greater than fc. Obviously, different modes will have different cutoff frequencies. - 5 - TM Modes The transverse magnetic modes for a general waveguide are obtained by assuming Hz =0. By duality with the TE modes, we have 2 2 2 2 2 2 z z z z E E E x y (28) with general solution y y x x z j y j y j x j x j z z E e Ae Be Ce De (29) The boundary conditions are At x=0, Ez=0 which leads to A=-B At y=0, Ez=0 which leads to C=-D At x=a, Ez=0 which leads to x m a At y=b, Ez=0 which leads to y n b so that the generating equation for the TMmn modes is sin sin z j z z o x y E E e x y (30) NOTE: THE DISPERSION RELATION, GUIDANCE CONDITION AND CUTOFF EQUA-TIONS FOR A RECTANGULAR WAVEGUIDE ARE THE SAME FOR TE AND TM MODES. For additional information on the field equations see Rao (6th Edition), page 607, Table 9.1. There is no TE00 mode There are no TMm0 or TM0n modes The first TE mode is the TE10 mode The first TM mode is the TM11 mode - 6 - Impedance of a Waveguide For a TE mode, we define the transverse impedance as y x gTE x y z E E H H From the relationship for z and using 2 2 2 1 4 c m n f a b (31) we get 2 2 1 gTE c f f (32) where is the intrinsic impedance Analogously, for TM modes, it can be shown that 2 2 1 c gTM f f (33) Power Flow in a Rectangular Waveguide (TE10) The time-average Poynting vector for the TE10 mode in a rectangular waveguide is given by 2 2 1 ˆ Re sin 2 2 o z E x a P E×H z (34) 2 2 0 0 sin 2 a b o z E x Power dxdy a (35) 10 2 2 4 4 o o z gTE E E ab ab Power (36) Therefore, the time-average power flow in a waveguide is proportional to its cross-section area. |
4405 | https://pubchem.ncbi.nlm.nih.gov/compound/6374 | Iodoform | CHI3 | CID 6374 - PubChem
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Search PubChem
compound Summary
Iodoform
PubChem CID
6374
Structure
Primary Hazards
Laboratory Chemical Safety Summary (LCSS) Datasheet
Molecular Formula
CHI 3
Synonyms
IODOFORM
75-47-8
Triiodomethane
Methane, triiodo-
Carbon triiodide
View More...
Molecular Weight
393.732 g/mol
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Dates
Create: 2005-03-26
Modify: 2025-09-27
Description
Iodoform appears as bright yellow or yellow powder or crystals. Penetrating odor. Unctuous feel. Odor threshold 0.4 ppb. (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
Iodoform is a member of iodomethanes.
ChEBI
Iodoform is an organoiodine compound with the formula CHI3 and a tetrahedral molecular geometry. It is a relatively water-insoluble yellow solid that is chemically reactive in free-radical reactions. Due to its antimicrobial properties following topical administration, minimal levels of iodoform may be found in disinfectants and it is more primarily used for veterinary purposes. Iodoform has also been found in dental paste and root canal filling materials in combination with other intracanal medications due to its radiopacity. Since the beginning of the 20th century, iodoform has been commonly used as a healing and antiseptic dressing or powder for wounds and sores, however such clinical use to this date is limited. Iodoform is soluble in fatty acids and decomposes releasing iodine in nascent state (96,7% of iodine) when in contact with secretions or endodontic infections.
DrugBank
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1 Structures
1.1 2D Structure
Structure Search
Get Image
Download Coordinates
Chemical Structure Depiction
Full screen Zoom in Zoom out
PubChem
1.2 3D Conformer
PubChem
1.3 Crystal Structures
1 of 4 items
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CCDC Number
889546
Associated Article
DOI:10.1016/j.molstruc.2012.11.016
Crystal Structure Data
DOI:10.5517/ccyvn1g
Crystal Structure Depiction
The Cambridge Structural Database
2 Names and Identifiers
2.1 Computed Descriptors
2.1.1 IUPAC Name
iodoform
Computed by Lexichem TK 2.7.0 (PubChem release 2025.04.14)
PubChem
2.1.2 InChI
InChI=1S/CHI3/c2-1(3)4/h1H
Computed by InChI 1.07.2 (PubChem release 2025.04.14)
PubChem
2.1.3 InChIKey
OKJPEAGHQZHRQV-UHFFFAOYSA-N
Computed by InChI 1.07.2 (PubChem release 2025.04.14)
PubChem
2.1.4 SMILES
C(I)(I)I
Computed by OEChem 2.3.0 (PubChem release 2025.04.14)
PubChem
2.2 Molecular Formula
CHI 3
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Australian Industrial Chemicals Introduction Scheme (AICIS); CAMEO Chemicals; PubChem
2.3 Other Identifiers
2.3.1 CAS
75-47-8
Australian Industrial Chemicals Introduction Scheme (AICIS); CAMEO Chemicals; CAS Common Chemistry; ChemIDplus; DrugBank; DTP/NCI; EPA Chemicals under the TSCA; EPA DSSTox; European Chemicals Agency (ECHA); FDA Global Substance Registration System (GSRS); Hazardous Substances Data Bank (HSDB); International Fragrance Association (IFRA); New Zealand Environmental Protection Authority (EPA); NIAID ChemDB; Occupational Safety and Health Administration (OSHA); Risk Assessment Information System (RAIS); The National Institute for Occupational Safety and Health (NIOSH)
2.3.2 European Community (EC) Number
200-874-5
European Chemicals Agency (ECHA)
685-642-5
European Chemicals Agency (ECHA)
2.3.3 UNII
KXI2J76489
FDA Global Substance Registration System (GSRS)
2.3.4 AIDS Number
017640
NIAID ChemDB
2.3.5 ChEBI ID
CHEBI:37758
ChEBI
2.3.6 ChEMBL ID
CHEMBL1451116
ChEMBL
2.3.7 DrugBank ID
DB13813
DrugBank
2.3.8 DSSTox Substance ID
DTXSID4020743
EPA DSSTox
2.3.9 HMDB ID
HMDB0253528
Human Metabolome Database (HMDB)
2.3.10 KEGG ID
D01910
KEGG
C13383
KEGG
2.3.11 Metabolomics Workbench ID
52204
Metabolomics Workbench
2.3.12 NCI Thesaurus Code
C77050
NCI Thesaurus (NCIt)
2.3.13 Nikkaji Number
J1.456G
Japan Chemical Substance Dictionary (Nikkaji)
2.3.14 NSC Number
26251
DTP/NCI
2.3.15 RTECS Number
PB7000000
The National Institute for Occupational Safety and Health (NIOSH)
2.3.16 Wikidata
Q412393
Wikidata
2.3.17 Wikipedia
Iodoform
Wikipedia
2.4 Synonyms
2.4.1 MeSH Entry Terms
iodoform
tri-iodomethane
Medical Subject Headings (MeSH)
2.4.2 Depositor-Supplied Synonyms
IODOFORM
75-47-8
Triiodomethane
Methane, triiodo-
Carbon triiodide
Jodoform
Trijodmethane
Dezinfekt V
Iodoformum
NCI-C04568
DTXSID4020743
NSC-26251
CHEBI:37758
KXI2J76489
DTXCID40743
tri-iodomethane
RefChem:5907
200-874-5
685-642-5
CHI3
MFCD00001069
NCGC00091389-01
Jodoform [Czech]
Iodoform [JAN]
Trijodmethane [Czech]
CAS-75-47-8
CCRIS 346
HSDB 4099
Iodoform [USP:JAN]
EINECS 200-874-5
NSC 26251
Carbontriiodide
UNII-KXI2J76489
methyl triiodide
AI3-52396
Iodoform (TN)
tris(iodanyl)methane
Iodoform, 99%
IODOFORM [HSDB]
IODOFORM [MI]
WLN: IYII
IODOFORMUM [HPUS]
Iodoform (JP17/USP)
IODOFORM [MART.]
IODOFORM [WHO-DD]
Iodoform, SAJ first grade
SCHEMBL59824
SCHEMBL59825
CHEMBL1451116
IODOFORM [USP MONOGRAPH]
MSK14870
NSC26251
Tox21_111124
Tox21_202389
Tox21_302774
AKOS009031506
Iodoform, purum, >=99.0% (AT)
DB13813
Iodoform, puriss., 99.0-100.5%
s12111
NCGC00091389-02
NCGC00091389-03
NCGC00256394-01
NCGC00259938-01
AS-14199
NS00006693
D01910
A838427
Q412393
Carbon triiodide; Dezinfekt V; Iodoform; NSC 26251
PubChem
3 Chemical and Physical Properties
3.1 Computed Properties
Property Name
Property Value
Reference
Property Name
Molecular Weight
Property Value
393.732 g/mol
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
XLogP3-AA
Property Value
2.7
Reference
Computed by XLogP3 3.0 (PubChem release 2025.04.14)
Property Name
Hydrogen Bond Donor Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Hydrogen Bond Acceptor Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Rotatable Bond Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Exact Mass
Property Value
393.7212 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
Monoisotopic Mass
Property Value
393.7212 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
Topological Polar Surface Area
Property Value
0 Ų
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Heavy Atom Count
Property Value
4
Reference
Computed by PubChem
Property Name
Formal Charge
Property Value
0
Reference
Computed by PubChem
Property Name
Complexity
Property Value
8
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Isotope Atom Count
Property Value
0
Reference
Computed by PubChem
Property Name
Defined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Defined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Covalently-Bonded Unit Count
Property Value
1
Reference
Computed by PubChem
Property Name
Compound Is Canonicalized
Property Value
Yes
Reference
Computed by PubChem (release 2025.04.14)
PubChem
3.2 Experimental Properties
3.2.1 Physical Description
Iodoform appears as bright yellow or yellow powder or crystals. Penetrating odor. Unctuous feel. Odor threshold 0.4 ppb. (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
Yellow to greenish-yellow powder or crystalline solid with a pungent, disagreeable odor. [antiseptic for external use]; [NIOSH]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Yellow to greenish-yellow powder or crystalline solid with a pungent, disagreeable odor.
Occupational Safety and Health Administration (OSHA)
Yellow to greenish-yellow powder or crystalline solid with a pungent, disagreeable odor. [antiseptic for external use]
The National Institute for Occupational Safety and Health (NIOSH)
3.2.2 Color / Form
Yellow powder or crystals
Budavari, S. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 1996., p. 863
Hazardous Substances Data Bank (HSDB)
YELLOW HEXAGONAL PRISMS OR NEEDLES FROM ACETONE
Weast, R.C. (ed.) Handbook of Chemistry and Physics. 69th ed. Boca Raton, FL: CRC Press Inc., 1988-1989., p. C-351
Hazardous Substances Data Bank (HSDB)
3.2.3 Odor
Characteristic, disagreeable odor
Budavari, S. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 1996., p. 863
Hazardous Substances Data Bank (HSDB)
3.2.4 Boiling Point
424 °F at 760 mmHg (Sublimes) (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
218
MSDS
DrugBank
218 °C
Lide, D.R. (ed.). CRC Handbook of Chemistry and Physics. 79th ed. Boca Raton, FL: CRC Press Inc., 1998-1999., p. 3-208
Hazardous Substances Data Bank (HSDB)
424 °F (sublimes)
Occupational Safety and Health Administration (OSHA)
410 °F (Decomposes)
The National Institute for Occupational Safety and Health (NIOSH)
3.2.5 Melting Point
248 °F (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
119-122
MSDS
DrugBank
119 °C
Lide, D.R. (ed.). CRC Handbook of Chemistry and Physics. 79th ed. Boca Raton, FL: CRC Press Inc., 1998-1999., p. 3-208
Hazardous Substances Data Bank (HSDB)
Enthalpy of melting @ mp: 3.9 kcal/mole; enthalpy of sublimation @ 298 K: 16.7 kcal/mole; specific heat @ 400 K: 19.60 cal/K.mole, @ 600 K: 21.52 cal/K.mole, @ 800 K: 22.64 cal/K.mole, @ 1000 K: 23.38 cal/K.mole
Dean, J.A. Handbook of Organic Chemistry. New York, NY: McGraw-Hill Book Co., 1987., p. 5-74
Hazardous Substances Data Bank (HSDB)
246 °F
Occupational Safety and Health Administration (OSHA); The National Institute for Occupational Safety and Health (NIOSH)
3.2.6 Solubility
less than 1 mg/mL at 75 °F (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
Slightly soluble
MSDS
DrugBank
SOL IN ACETIC ACID
Weast, R.C. (ed.) Handbook of Chemistry and Physics. 69th ed. Boca Raton, FL: CRC Press Inc., 1988-1989., p. C-351
Hazardous Substances Data Bank (HSDB)
13.6 g/100 ml ether @ 25 °C
Kirk-Othmer Encyclopedia of Chemical Technology. 3rd ed., Volumes 1-26. New York, NY: John Wiley and Sons, 1978-1984., p. 13(81) 668
Hazardous Substances Data Bank (HSDB)
7.8 g/100 ml ethanol @ 25 °C
Kirk-Othmer Encyclopedia of Chemical Technology. 3rd ed., Volumes 1-26. New York, NY: John Wiley and Sons, 1978-1984., p. 13(81) 668
Hazardous Substances Data Bank (HSDB)
One gram dissolves in 60 ml cold alcohol, 16 ml boiling alcohol, 10 ml chloroform, 7.5 ml ether, 80 ml glycerol, 3 ml carbon disulfide, 34 ml olive oil; freely sol in benzene, acetone, slightly sol in petr ether.
Budavari, S. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 1996., p. 863
Hazardous Substances Data Bank (HSDB)
In water, 100 mg/l @ room temperature.
Yalkowsky SH, Dannenfelser RM; Aquasol Data Base of Water Solubility Ver 5, Tuscon, AZ: Univ Arizona, College of Pharmacy (1992)
Hazardous Substances Data Bank (HSDB)
0.01%
The National Institute for Occupational Safety and Health (NIOSH)
3.2.7 Density
4.008 at 68 °F (NTP, 1992) - Denser than water; will sink
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
Specific Gravity: 4.008 g/cu cm @ 25 °C
Lide, D.R. (ed.). CRC Handbook of Chemistry and Physics. 76th ed. Boca Raton, FL: CRC Press Inc., 1995-1996., p. 3-208
Hazardous Substances Data Bank (HSDB)
4.008
Occupational Safety and Health Administration (OSHA)
4.01
The National Institute for Occupational Safety and Health (NIOSH)
3.2.8 Vapor Density
4.1 (NTP, 1992) - Heavier than air; will sink (Relative to Air)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
13.6 (Air= 1)
Patty, F. (ed.). Industrial Hygiene and Toxicology: Volume II: Toxicology. 2nd ed. New York: Interscience Publishers, 1963., p. 1263
Hazardous Substances Data Bank (HSDB)
4.1
Occupational Safety and Health Administration (OSHA)
3.2.9 Vapor Pressure
0.04 [mmHg]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
3.2.10 Decomposition
DECOMPOSES VIOLENTLY @ 400 °F (204 °C)
Lewis, R.J., Sr (Ed.). Hawley's Condensed Chemical Dictionary. 12th ed. New York, NY: Van Nostrand Rheinhold Co., 1993, p. 642
Hazardous Substances Data Bank (HSDB)
DECOMPOSES @ HIGH TEMP WITH EVOLUTION OF IODINE.
Budavari, S. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 1996., p. 863
Hazardous Substances Data Bank (HSDB)
WHEN HEATED TO DECOMPOSITION ... EMITS TOXIC FUMES OF /HYDROGEN IODIDE/.
Lewis, R.J. Sax's Dangerous Properties of Industrial Materials. 9th ed. Volumes 1-3. New York, NY: Van Nostrand Reinhold, 1996., p. 1932
Hazardous Substances Data Bank (HSDB)
3.2.11 Heat of Combustion
161.9 kg cal/g mol wt at 20 °C (solid)
Weast, R.C. (ed.) Handbook of Chemistry and Physics. 69th ed. Boca Raton, FL: CRC Press Inc., 1988-1989., p. D-277
Hazardous Substances Data Bank (HSDB)
3.2.12 Odor Threshold
Odor Threshold Low: 0.000019 [ppm]
Odor Threshold High: 1.1 [ppm]
Odor threshold from AIHA
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
5.0 ppb (detection in air, purity not specified)
Fazzalari, F.A. (ed.). Compilation of Odor and Taste Threshold Values Data. ASTM Data Series DS 48A (Committee E-18). Philadelphia, PA: American Society for Testing and Materials, 1978., p. 90
Hazardous Substances Data Bank (HSDB)
Low: 0.0062 mg/cu m; High: 0.0833 mg/cu m
Ruth JH; Am Ind Hyg Assoc J 47: A-147 (1986)
Hazardous Substances Data Bank (HSDB)
3.2.13 Refractive Index
INDEX OF REFRACTION: 1.8 @ 20 °C
Clayton, G.D., F.E. Clayton (eds.) Patty's Industrial Hygiene and Toxicology. Volumes 2A, 2B, 2C, 2D, 2E, 2F: Toxicology. 4th ed. New York, NY: John Wiley & Sons Inc., 1993-1994., p. 4068
Hazardous Substances Data Bank (HSDB)
3.2.14 Kovats Retention Index
Standard non-polar
1173 , 1196 , 1221 , 1243.1
NIST Mass Spectrometry Data Center
3.2.15 Other Experimental Properties
Iodoform oxidizes arsenites to arsenates, antimonites to antimonates, and stannites to stannates.
Kirk-Othmer Encyclopedia of Chemical Technology. 3rd ed., Volumes 1-26. New York, NY: John Wiley and Sons, 1978-1984., p. 13(81) 670
Hazardous Substances Data Bank (HSDB)
Enthalpy of formation (gas) @ 25 °C: 50.40 kcal/mole; Gibbs (free) energy of formation (gas) @ 25 °C: 42.54 kcal/mole; entropy (gas) @ 25 °C: 84.97 cal/deg.mole; heat capacity (gas) @ 25 °C: 17.94 cal/deg.mole
Dean, J.A. Handbook of Organic Chemistry. New York, NY: McGraw-Hill Book Co., 1987., p. 5-41
Hazardous Substances Data Bank (HSDB)
Unctuous touch; characteristic, disagreeable odor. Volatile with steam.
Budavari, S. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 1996., p. 863
Hazardous Substances Data Bank (HSDB)
Decomposes at high temperatures with evolution of iodine.
Kirk-Othmer Encyclopedia of Chemical Technology. 4th ed. Volumes 1: New York, NY. John Wiley and Sons, 1991-Present., p. V14 729
Hazardous Substances Data Bank (HSDB)
3.3 SpringerMaterials Properties
Gibbs energy
Band gap energy
Band structure
Boiling point
Charge carrier mobility
Chemical bond
Corrosion
Crystal structure
Crystallographic point group
Defect energy
Density
Density functional theory
Diamagnetic susceptibility
Dielectric constant
Effective mass
Electron diffraction
Enthalpy
Entropy
Formation energy
Formation enthalpy
Formation entropy
Formula unit
Fusion temperature
Gross formula
Heat capacity
Heat of sublimation
Internuclear distance
Lattice stiffness
Magnetic anisotropy
Magnetic susceptibility
Melting temperature
Mobility
Molecular structure
Nuclear quadrupole resonance spectroscopy
Optical coefficient
Phase transition
Quadrupole coupling
Reaction coordinate
Refractive index
Space group
Symmetry axis
Thermal expansion coefficient
Transition enthalpy
Transition entropy
Unit cell
Unit cell parameter
Vapor pressure
Viscosity
SpringerMaterials
3.4 Chemical Classes
Other Uses -> Pharmaceuticals
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
3.4.1 Drugs
Pharmaceuticals -> Listed in ZINC15
S55 | ZINC15PHARMA | Pharmaceuticals from ZINC15 | DOI:10.5281/zenodo.3247749
NORMAN Suspect List Exchange
3.4.2 Food Contact Substances
FCS -> FDA Inventory of Food Contact Substances Listed in 21 CFR
FDA Packaging & Food Contact Substances (FCS)
3.4.3 Fragrances
Fragrance Ingredient (Methane, triiodo-) -> IFRA transparency List
International Fragrance Association (IFRA)
4 Spectral Information
4.1 1D NMR Spectra
1D NMR Spectra
NMR: 1:64D (Aldrich Library of Mass Spectra, Aldrich Chemical Co, Milwaukee, WI)
1D NMR Spectra
NMR: 6753 (Sadtler Research Laboratories Spectral Collection)
Hazardous Substances Data Bank (HSDB)
4.1.1 1H NMR Spectra
Instrument Name
Varian A-60
Source of Sample
Fluka Chemie AG, Buchs, Switzerland
Copyright
Copyright © 2009-2025 John Wiley & Sons, Inc. All Rights Reserved.
Thumbnail
SpectraBase
4.1.2 13C NMR Spectra
Source of Sample
Aldrich Chemical Company, Inc., Milwaukee, Wisconsin
Copyright
Copyright © 1980, 1981-2025 John Wiley & Sons, Inc. All Rights Reserved.
Thumbnail
SpectraBase
4.2 Mass Spectrometry
4.2.1 GC-MS
1 of 5 items
View All
NIST Number
228071
Library
Main library
Total Peaks
16
m/z Top Peak
267
m/z 2nd Highest
394
m/z 3rd Highest
140
Thumbnail
NIST Mass Spectrometry Data Center
2 of 5 items
View All
NIST Number
290756
Library
Replicate library
Total Peaks
16
m/z Top Peak
267
m/z 2nd Highest
127
m/z 3rd Highest
394
Thumbnail
NIST Mass Spectrometry Data Center
4.3 UV Spectra
MAX ABSORPTION (IN CHLORFORM): 244 NM (LOG E= 3.0), 307 NM (LOG E= 3.2); 347 NM (LOG E= 3.2)
Weast, R.C. (ed.). Handbook of Chemistry and Physics. 60th ed. Boca Raton, Florida: CRC Press Inc., 1979., p. C-375
Hazardous Substances Data Bank (HSDB)
UV: 5-1 (Organic Electronic Spectral Data, Phillips et al, John Wiley & Sons, New York)
Weast, R.C. and M.J. Astle. CRC Handbook of Data on Organic Compounds. Volumes I and II. Boca Raton, FL: CRC Press Inc. 1985., p. V1 843
Hazardous Substances Data Bank (HSDB)
UV absorption maximum at 349 nm; molar extinction coefficient 2170.
Calvert JG, Pitts JN Jr; Photochemistry NY, NY: Wiley pp. 523 (1966)
Hazardous Substances Data Bank (HSDB)
4.4 IR Spectra
IR Spectra
IR: 2:49H (Aldrich Library of Infrared Spectra, Aldrich Chemical Co, Milwaukee, WI)
IR Spectra
IR: 4553 (Sadtler Research Laboratories Prism Collection)
Hazardous Substances Data Bank (HSDB)
4.4.1 FTIR Spectra
Instrument Name
Bruker IFS 85
Technique
KBr-Pellet
Source of Sample
E. Merck AG, Darmstadt
Copyright
Copyright © 1989, 1990-2025 Wiley-VCH GmbH. All Rights Reserved.
Thumbnail
SpectraBase
4.4.2 ATR-IR Spectra
Source of Sample
Aldrich
Catalog Number
109452
Copyright
Copyright © 2018-2025 Sigma-Aldrich Co. LLC. - Database Compilation Copyright © 2018-2025 John Wiley & Sons, Inc. All Rights Reserved.
Thumbnail
SpectraBase
4.4.3 Vapor Phase IR Spectra
Source of Spectrum
Sigma-Aldrich Co. LLC.
Source of Sample
Sigma-Aldrich Co. LLC.
Catalog Number
109452
Copyright
Copyright © 2021-2025 Sigma-Aldrich Co. LLC. - Database Compilation Copyright © 2021-2025 John Wiley & Sons, Inc. All Rights Reserved.
Thumbnail
SpectraBase
4.5 Raman Spectra
Catalog Number
109452
Copyright
Copyright © 2017-2025 Sigma-Aldrich Co. LLC. - Database Compilation Copyright © 2017-2025 John Wiley & Sons, Inc. All Rights Reserved.
Thumbnail
SpectraBase
5 Related Records
5.1 Related Compounds with Annotation
Follow these links to do a live 2D search or do a live 3D search for this compound, sorted by annotation score. This section is deprecated (see the neighbor discontinuation help page for details), but these live search links provide equivalent functionality to the table that was previously shown here.
PubChem
5.2 Related Compounds
Same Connectivity Count
3
Same Parent, Connectivity Count
21
Same Parent, Exact Count
19
Mixtures, Components, and Neutralized Forms Count
1054
Similar Compounds (2D)
View in PubChem Search
Similar Conformers (3D)
View in PubChem Search
PubChem
5.3 Substances
5.3.1 PubChem Reference Collection SID
481107247
PubChem
5.3.2 Related Substances
All Count
1236
Same Count
153
Mixture Count
1083
PubChem
5.3.3 Substances by Category
PubChem
5.4 Entrez Crosslinks
PubMed Count
183
Taxonomy Count
2
Gene Count
3
PubChem
6 Chemical Vendors
PubChem
7 Drug and Medication Information
7.1 Drug Indication
No approved therapeutic indications.
DrugBank
7.2 Drug Labels
Active ingredient and drug
DailyMed
7.3 Clinical Trials
7.3.1 ClinicalTrials.gov
ClinicalTrials.gov
7.4 Therapeutic Uses
Iodoform /was used/ as an antiseptic wound powder for a long time: today its use is limited.
Kirk-Othmer Encyclopedia of Chemical Technology. 3rd ed., Volumes 1-26. New York, NY: John Wiley and Sons, 1978-1984., p. 7(79) 803
Hazardous Substances Data Bank (HSDB)
FORMERLY WAS EMPLOYED AS A TOPICAL AND INTRAVITREAL ANTISEPTIC.
Grant, W.M. Toxicology of the Eye. 3rd ed. Springfield, IL: Charles C. Thomas Publisher, 1986., p. 524
Hazardous Substances Data Bank (HSDB)
ANTI-INFECTIVE (TOPICAL)
Budavari, S. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 1996., p. 863
Hazardous Substances Data Bank (HSDB)
MEDICATION (VET): ANTISEPTIC, DISINFECTANT FOR SUPERFICIAL LESIONS & IN THE FEMALE REPRODUCTIVE TRACT
Budavari, S. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 1996., p. 863
Hazardous Substances Data Bank (HSDB)
THERAP CAT: Anti-infective (topical)
Budavari, S. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 1996., p. 863
Hazardous Substances Data Bank (HSDB)
8 Food Additives and Ingredients
8.1 FDA Food Contact Substances (FCS)
Substance
IODOFORM
Document Number (21 eCFR)
175.105
177.2600
FDA Packaging & Food Contact Substances (FCS)
9 Pharmacology and Biochemistry
9.1 Pharmacodynamics
Iodoform exhibits antibacterial activities after topical application. In a comparative study of wound dressing agents, iodoform gauze exerted an antibacterial effect 3 hours after the start of bacterial growth of E. coli and subsequently maintained the strong antibacterial effectiveness. A study demonstrated that direct and indirect exposure to high concentrations of iodoform induces a cytotoxic effect on cultures of macrophages and epithelial cells in vitro, while cell proliferation was enhanced at low concentrations of iodoform. This cytotoxic effect of iodoform in root canals may further lead to long-term local irritation to follicles of permanent successors and formation of cyst-like radiolucent defects.
DrugBank
9.2 ATC Code
D - Dermatologicals
D09 - Medicated dressings
D09A - Medicated dressings
D09AA - Medicated dressings with antiinfectives
D09AA13 - Iodoform
WHO Anatomical Therapeutic Chemical (ATC) Classification
ATCvet Code
QD - Dermatologicals
QD09 - Medicated dressings
QD09A - Medicated dressings
QD09AA - Medicated dressings with antiinfectives
QD09AA13 - Iodoform
WHO ATCvet - Classification of Veterinary Medicines
9.3 Absorption, Distribution and Excretion
Absorption
Iodoform is reported to be absorbed through denuded skin, wounds or mucous membranes.
DrugBank
Route of Elimination
No pharmacokinetic data available.
DrugBank
9.4 Metabolism / Metabolites
It is expected to be oxidized to iodine.
DrugBank
HALOFORMS ARE METABOLIZED TO CARBON MONOXIDE BY HEPATIC MICROSOMAL MIXED FUNCTION OXIDASES & THIS REACTION IS MARKEDLY STIMULATED BY SULFHYDRYL COMPOUNDS. /HALOFORMS/
PMID:526325
STEVENS JL, ANDERS MW; BIOCHEM PHARM 28: 3189-94 (1979)
Hazardous Substances Data Bank (HSDB)
TRIHALOMETHANES (HALOFORMS) WERE METABOLIZED TO CARBON MONOXIDE BY A RAT LIVER MICROSOMAL FRACTION REQUIRING BOTH NADPH AND MOLECULAR OXYGEN FOR MAXIMAL ACTIVITY. THE METABOLISM OF HALOFORMS TO CARBON MONOXIDE FOLLOWED THE HALIDE ORDER; IODOFORM YIELDED THE GREATEST AMOUNT OF CARBON MONOXIDE, WHEREAS CHLOROFORM YIELDED THE SMALLEST AMOUNT.
PMID:15814
AHMED AE ET AL; DRUG METAB DISPOS 5 (2): 198-204 (1977)
Hazardous Substances Data Bank (HSDB)
ADMINISTRATION OF HALOFORMS (TRIHALOMETHANES) TO RATS LED TO SUBSTANTIAL ELEVATIONS IN BLOOD CARBON MONOXIDE LEVELS. /TRIHALOMETHANES/
PMID:30605
ANDERS MW ET AL; DRUG METAB DISPOS 6 (5): 556-60 (1978)
Hazardous Substances Data Bank (HSDB)
9.5 Biological Half-Life
No pharmacokinetic data available.
DrugBank
9.6 Mechanism of Action
While the mechanism of action of iodoform remains unclear, it is proposed that iodoform releases iodine, which denatures bacterial proteins by oxidation of the free iodine. Iodoform may also play a role in chemical debridement for effective necrotic wound healing and tissue damage repair via collagen fibrinolysis; upon treatment in necrotic tissue, iodoform reduced the size of the macromolecules containing collagen I in wound surface proteins. In human gingival fibroblasts in vitro, high concentrations of iodoform was shown to decrease the viability of macrophages and epithelial cells and reduced the secretion of P. gingivalis-induced TNFα. P. gingivalis is an anaerobic bacteria present in anaerobic oral niches including periapical sites and periodontal pockets.
DrugBank
10 Use and Manufacturing
10.1 Uses
EPA CPDat Chemical and Product Categories
The Chemical and Products Database, a resource for exposure-relevant data on chemicals in consumer products, Scientific Data, volume 5, Article number: 180125 (2018), DOI:10.1038/sdata.2018.125
EPA Chemical and Products Database (CPDat)
Sources/Uses
Used as a disinfectant and veterinary antiseptic; [Merck Index, #5033]
Merck Index - O'Neil MJ, Heckelman PE, Dobbelaar PH, Roman KJ (eds). The Merck Index, An Encyclopedia of Chemicals, Drugs, and Biologicals, 15th Ed. Cambridge, UK: The Royal Society of Chemistry, 2013., #5033
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Industrial Processes with risk of exposure
Using Disinfectants or Biocides [Category: Clean]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Iodoform is used as a component of adhesives and as a polymerization control agent
21 CFR 175.105 (4/1/86)
Hazardous Substances Data Bank (HSDB)
Wound dressing and sensitizing agent in certain printing processes.
Kirk-Othmer Encyclopedia of Chemical Technology. 4th ed. Volumes 1: New York, NY. John Wiley and Sons, 1991-Present., p. V13 729
Hazardous Substances Data Bank (HSDB)
MEDICATION
Budavari, S. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 1996., p. 863
Hazardous Substances Data Bank (HSDB)
MEDICATION: VET
Budavari, S. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 1996., p. 863
Hazardous Substances Data Bank (HSDB)
10.1.1 Use Classification
Fragrance Ingredients
International Fragrance Association (IFRA)
10.2 Methods of Manufacturing
Prepn from acetone, sodium hypochlorite, potassium iodide, and sodium hydroxide: Glass, Quart. J. Pharm. Pharmacol. 8, 351 (1935); from chloroform + methyl iodide: Soroos, Hinkamp, J. Am. Chem. Soc. 67, 1642 (1945); by electrolysis: Glasstone, Ind. Chem. 7, 315 (1931)
Budavari, S. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 1996., p. 863
Hazardous Substances Data Bank (HSDB)
Electrolysis of an iodide solution in dilute alcohol or acetone.
Kirk-Othmer Encyclopedia of Chemical Technology. 4th ed. Volumes 1: New York, NY. John Wiley and Sons, 1991-Present., p. V14 729
Hazardous Substances Data Bank (HSDB)
10.3 Formulations / Preparations
GRADE: TECHNICAL, NF
Lewis, R.J., Sr (Ed.). Hawley's Condensed Chemical Dictionary. 12th ed. New York, NY: Van Nostrand Rheinhold Co., 1993, p. 642
Hazardous Substances Data Bank (HSDB)
Purified grades
Kuney, J.H. (ed.). CHEMCYCLOPEDIA 90. Washington, DC: American Chemical Society, 1990., p. 305
Hazardous Substances Data Bank (HSDB)
BIPP gauze: sterile ribbon gauze impregnated with a sterile paste consisting of iodoform 40%, bismuth subnitrate 20%, and liquid paraffin 40%.
Reynolds, J.E.F., Prasad, A.B. (eds.) Martindale-The Extra Pharmacopoeia. 28th ed. London: The Pharmaceutical Press, 1982., p. 865
Hazardous Substances Data Bank (HSDB)
Bismuth subnitrate& iodoform paste: bismuth subnitrate 1, iodoform 2, sterilized liquid paraffin 1, by wt, prepared aseptically.
Reynolds, J.E.F., Prasad, A.B. (eds.) Martindale-The Extra Pharmacopoeia. 28th ed. London: The Pharmaceutical Press, 1982., p. 865
Hazardous Substances Data Bank (HSDB)
For more Formulations/Preparations (Complete) data for IODOFORM (8 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
10.4 U.S. Production
(1977) PROBABLY GREATER THAN 3.18X10+6 G
SRI
Hazardous Substances Data Bank (HSDB)
(1979) PROBABLY GREATER THAN 3.18X10+6 G
SRI
Hazardous Substances Data Bank (HSDB)
10.5 General Manufacturing Information
EPA TSCA Commercial Activity Status
Methane, triiodo-: ACTIVE
EPA Chemicals under the TSCA
11 Identification
11.1 Analytic Laboratory Methods
A gravimetric determination of iodoform drug substance, iodoform in ointments, and iodoform on gauze requires the addition of silver nitrate and nitric acid. The silver iodide formed is filtered, collected and dried to constant weight. One g AgI= 0.5590 g iodoform.
Association of Official Analytic Chemists. Official Methods of Analysis of the AOAC. 14th ed. Arlington, VA: Association of Official Analytic Chemists, Inc., 1984., p. 671 36.032
Hazardous Substances Data Bank (HSDB)
Gas chromtographic separation of halogenated compounds on non-polar and polar wide bore capillary columns.
Castello G et al; J Chromatogr 454: 129-43 (1988)
Hazardous Substances Data Bank (HSDB)
The selection of optimal and std methods for detn of org compds contg I depends on the I content (5-97%). A combustion method in an oxygen flask is recommended. However, for drugs with a low I content (0.1-5%), micro methods, such as spectrophotometric or titrimetric, are preferable.
Listov SA, Arzamastsev AP; Khim Farm Zh 22 (6): 753-59 (1988)
Hazardous Substances Data Bank (HSDB)
12 Safety and Hazards
12.1 Hazards Identification
12.1.1 GHS Classification
1 of 5 items
View All
Note
This chemical does not meet GHS hazard criteria for < 0.1% (1 of 1363) of reports.
Pictogram(s)
Signal
Warning
GHS Hazard Statements
H302 (> 99.9%): Harmful if swallowed [Warning Acute toxicity, oral]
H312 (99.1%): Harmful in contact with skin [Warning Acute toxicity, dermal]
H315 (99.9%): Causes skin irritation [Warning Skin corrosion/irritation]
H319 (99.9%): Causes serious eye irritation [Warning Serious eye damage/eye irritation]
H332 (99%): Harmful if inhaled [Warning Acute toxicity, inhalation]
H335 (99.8%): May cause respiratory irritation [Warning Specific target organ toxicity, single exposure; Respiratory tract irritation]
Precautionary Statement Codes
P261, P264, P264+P265, P270, P271, P280, P301+P317, P302+P352, P304+P340, P305+P351+P338, P317, P319, P321, P330, P332+P317, P337+P317, P362+P364, P403+P233, P405, and P501
ECHA C&L Notifications Summary
Aggregated GHS information provided per 1363 reports by companies from 15 notifications to the ECHA C&L Inventory. Each notification may be associated with multiple companies.
Reported as not meeting GHS hazard criteria per 1 of 1363 reports by companies.
There are 14 notifications provided by 1362 of 1363 reports by companies with hazard statement code(s).
Information may vary between notifications depending on impurities, additives, and other factors. The percentage value in parenthesis indicates the notified classification ratio from companies that provide hazard codes. Only hazard codes with percentage values above 10% are shown. For more detailed information, please visit ECHA C&L website.
European Chemicals Agency (ECHA)
12.1.2 Hazard Classes and Categories
Acute Tox. 4 (> 99.9%)
Acute Tox. 4 (99.1%)
Skin Irrit. 2 (99.9%)
Eye Irrit. 2 (99.9%)
Acute Tox. 4 (99%)
STOT SE 3 (99.8%)
European Chemicals Agency (ECHA)
View More...
12.1.3 Health Hazards
Excerpt from NIOSH Pocket Guide for Iodoform:
Exposure Routes: Inhalation, skin absorption, ingestion, skin and/or eye contact
Symptoms: Irritation eyes, skin; lassitude (weakness, exhaustion), dizziness, nausea, incoordination, central nervous system depression; dyspnea (breathing difficulty); liver, kidney, heart damage; visual disturbance
Target Organs: Eyes, skin, respiratory system, liver, kidneys, heart (NIOSH, 2024)
CAMEO Chemicals
12.1.4 Fire Hazards
Literature sources indicate that this chemical is nonflammable. (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
12.1.5 Hazards Summary
Cases of CNS depression, delirium, and hallucinations were reported when iodoform was used as a topical antiseptic and absorbed through the wound. It causes severe liver injury in animal experiments. 2021 TLV Basis: thyroid effects and fetal/neonatal damage; [ACGIH] See Iodine for the TLV of Iodides.
ACGIH - Documentation of the TLVs and BEIs, 7th Ed. Cincinnati: ACGIH Worldwide, 2020.
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
12.2 Safety and Hazard Properties
12.2.1 Flammable Limits
Flammability
Noncombustible Solid
The National Institute for Occupational Safety and Health (NIOSH)
12.2.2 OSHA Standards
Vacated 1989 OSHA PEL TWA 0.6 ppm (10 mg/cu m) is still enforced in some states.
NIOSH. NIOSH Pocket Guide to Chemical Hazards. DHHS (NIOSH) Publication No. 97-140. Washington, D.C. U.S. Government Printing Office, 1997., p. 366
Hazardous Substances Data Bank (HSDB)
12.2.3 NIOSH Recommendations
Recommended Exposure Limit: 10 Hr Time-Weighted avg: 0.6 ppm (10 mg/cu m).
NIOSH. NIOSH Pocket Guide to Chemical Hazards. DHHS (NIOSH) Publication No. 97-140. Washington, D.C. U.S. Government Printing Office, 1997., p. 172
Hazardous Substances Data Bank (HSDB)
12.3 First Aid Measures
12.3.1 First Aid
EYES: First check the victim for contact lenses and remove if present. Flush victim's eyes with water or normal saline solution for 20 to 30 minutes while simultaneously calling a hospital or poison control center. Do not put any ointments, oils, or medication in the victim's eyes without specific instructions from a physician. IMMEDIATELY transport the victim after flushing eyes to a hospital even if no symptoms (such as redness or irritation) develop.
SKIN: IMMEDIATELY flood affected skin with water while removing and isolating all contaminated clothing. Gently wash all affected skin areas thoroughly with soap and water. If symptoms such as redness or irritation develop, IMMEDIATELY call a physician and be prepared to transport the victim to a hospital for treatment.
INHALATION: IMMEDIATELY leave the contaminated area; take deep breaths of fresh air. If symptoms (such as wheezing, coughing, shortness of breath, or burning in the mouth, throat, or chest) develop, call a physician and be prepared to transport the victim to a hospital. Provide proper respiratory protection to rescuers entering an unknown atmosphere. Whenever possible, Self-Contained Breathing Apparatus (SCBA) should be used; if not available, use a level of protection greater than or equal to that advised under Protective Clothing.
INGESTION: DO NOT INDUCE VOMITING. If the victim is conscious and not convulsing, give 1 or 2 glasses of water to dilute the chemical and IMMEDIATELY call a hospital or poison control center. Be prepared to transport the victim to a hospital if advised by a physician. If the victim is convulsing or unconscious, do not give anything by mouth, ensure that the victim's airway is open and lay the victim on his/her side with the head lower than the body. DO NOT INDUCE VOMITING. IMMEDIATELY transport the victim to a hospital. (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
(General first aid procedures)
Eye: Irrigate immediately - If this chemical contacts the eyes, immediately wash (irrigate) the eyes with large amounts of water, occasionally lifting the lower and upper lids. Get medical attention immediately.
Skin: Soap wash immediately - If this chemical contacts the skin, immediately wash the contaminated skin with soap and water. If this chemical penetrates the clothing, immediately remove the clothing, wash the skin with soap and water, and get medical attention promptly.
Breathing: Respiratory support
Swallow: Medical attention immediately - If this chemical has been swallowed, get medical attention immediately.
The National Institute for Occupational Safety and Health (NIOSH)
12.4 Fire Fighting
Fires associated with this material can be controlled with a dry chemical, carbon dioxide or Halon extinguisher. (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
12.5 Accidental Release Measures
12.5.1 Disposal Methods
SRP: At the time of review, criteria for land treatment or burial (sanitary landfill) disposal practices are subject to significant revision. Prior to implementing land disposal of waste residue (including waste sludge), consult with environmental regulatory agencies for guidance on acceptable disposal practices.
Hazardous Substances Data Bank (HSDB)
12.5.2 Preventive Measures
SRP: The scientific literature for the use of contact lenses in industry is conflicting. The benefit or detrimental effects of wearing contact lenses depend not only upon the substance, but also on factors including the form of the substance, characteristics and duration of the exposure, the uses of other eye protection equipment, and the hygiene of the lenses. However, there may be individual substances whose irritating or corrosive properties are such that the wearing of contact lenses would be harmful to the eye. In those specific cases, contact lenses should not be worn. In any event, the usual eye protection equipment should be worn even when contact lenses are in place.
Hazardous Substances Data Bank (HSDB)
The worker should immediately wash the skin when it becomes contaminated.
NIOSH. NIOSH Pocket Guide to Chemical Hazards. DHHS (NIOSH) Publication No. 97-140. Washington, D.C. U.S. Government Printing Office, 1997., p. 172
Hazardous Substances Data Bank (HSDB)
Work clothing that becomes wet or significantly contaminated should be removed or replaced.
NIOSH. NIOSH Pocket Guide to Chemical Hazards. DHHS (NIOSH) Publication No. 97-140. Washington, D.C. U.S. Government Printing Office, 1997., p. 172
Hazardous Substances Data Bank (HSDB)
Workers whose clothing may have become contaminated should change into uncontaminated clothing before leaving the work premises.
NIOSH. NIOSH Pocket Guide to Chemical Hazards. DHHS (NIOSH) Publication No. 97-140. Washington, D.C. U.S. Government Printing Office, 1997., p. 172
Hazardous Substances Data Bank (HSDB)
12.6 Handling and Storage
12.6.1 Nonfire Spill Response
SMALL SPILLS AND LEAKAGE: Should a spill occur while you are handling this chemical, FIRST REMOVE ALL SOURCES OF IGNITION, then you should dampen the solid spill material with 60-70% ethanol and transfer the dampened material to a suitable container. Use absorbent paper dampened with 60-70% ethanol to pick up any remaining material. Seal the absorbent paper, and any of your clothes, which may be contaminated, in a vapor-tight plastic bag for eventual disposal. Solvent wash all contaminated surfaces with 60-70% ethanol followed by washing with a soap and water solution. Do not reenter the contaminated area until the Safety Officer (or other responsible person) has verified that the area has been properly cleaned.
STORAGE PRECAUTIONS: You should protect this material from exposure to light, and store it in a refrigerator. (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
12.6.2 Storage Conditions
Store in a cool place in airtight containers. Protect from light.
Reynolds, J.E.F., Prasad, A.B. (eds.) Martindale-The Extra Pharmacopoeia. 28th ed. London: The Pharmaceutical Press, 1982., p. 865
Hazardous Substances Data Bank (HSDB)
12.7 Exposure Control and Personal Protection
12.7.1 Recommended Exposure Limit (REL)
REL-TWA (Time Weighted Average)
0.6 ppm (10 mg/m³)
Occupational Safety and Health Administration (OSHA)
TWA 0.6 ppm (10 mg/m 3)
The National Institute for Occupational Safety and Health (NIOSH)
12.7.2 Permissible Exposure Limit (PEL)
none See Appendix G
The National Institute for Occupational Safety and Health (NIOSH)
12.7.3 Immediately Dangerous to Life or Health (IDLH)
N.D.
See: IDLH INDEX
The National Institute for Occupational Safety and Health (NIOSH)
12.7.4 Threshold Limit Values (TLV)
0.001 [ppm], as elemental iodine, inhalable fraction and vapor
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
8 hr Time Weighted Avg (TWA): 0.6 ppm.
American Conference of Governmental Industrial Hygienists TLVs and BEIs. Threshold Limit Values for Chemical Substances and Physical Agents and Biological Exposure Indices. Cincinnati, OH, 2008, p. 35
Hazardous Substances Data Bank (HSDB)
Excursion Limit Recommendation: Excursions in worker exposure levels may exceed 3 times the TLV-TWA for no more than a total of 30 minutes during a work day, and under no circumstances should they exceed 5 times the TLV-TWA, provided that the TLV-TWA is not exceeded.
American Conference of Governmental Industrial Hygienists TLVs and BEIs. Threshold Limit Values for Chemical Substances and Physical Agents and Biological Exposure Indices. Cincinnati, OH, 2008, p. 5
Hazardous Substances Data Bank (HSDB)
TLV-TWA (Time Weighted Average)
0.001 ppm (inhalable fraction and vapor)
Occupational Safety and Health Administration (OSHA)
12.7.5 Other Standards Regulations and Guidelines
Australia: 0.6 ppm (1990); United Kingdom: 0.6 ppm, 10-min STEL 1.0 ppm (1991).
American Conference of Governmental Industrial Hygienists, Inc. Documentation of the Threshold Limit Values and Biological Exposure Indices. 6th ed. Volumes I, II, III. Cincinnati, OH: ACGIH, 1991., p. 802
Hazardous Substances Data Bank (HSDB)
12.7.6 Personal Protective Equipment (PPE)
Excerpt from NIOSH Pocket Guide for Iodoform:
Skin: PREVENT SKIN CONTACT - Wear appropriate personal protective clothing to prevent skin contact.
Eyes: PREVENT EYE CONTACT - Wear appropriate eye protection to prevent eye contact.
Wash skin: WHEN CONTAMINATED - The worker should immediately wash the skin when it becomes contaminated.
Remove: WHEN WET OR CONTAMINATED - Work clothing that becomes wet or significantly contaminated should be removed and replaced.
Change: DAILY - Workers whose clothing may have become contaminated should change into uncontaminated clothing before leaving the work premises. (NIOSH, 2024)
CAMEO Chemicals
Wear appropriate personal protective clothing to prevent skin contact.
NIOSH. NIOSH Pocket Guide to Chemical Hazards. DHHS (NIOSH) Publication No. 97-140. Washington, D.C. U.S. Government Printing Office, 1997., p. 172
Hazardous Substances Data Bank (HSDB)
Wear appropriate eye protection to prevent eye contact.
NIOSH. NIOSH Pocket Guide to Chemical Hazards. DHHS (NIOSH) Publication No. 97-140. Washington, D.C. U.S. Government Printing Office, 1997., p. 172
Hazardous Substances Data Bank (HSDB)
12.7.7 Respirator Recommendations
Important additional information about respirator selection
The National Institute for Occupational Safety and Health (NIOSH)
12.8 Stability and Reactivity
12.8.1 Air and Water Reactions
Insoluble in water.
CAMEO Chemicals
12.8.2 Reactive Group
Halogenated Organic Compounds
CAMEO Chemicals
12.8.3 Reactivity Profile
IODOFORM decomposes at high temperatures. Decomposes slowly in light at room temperature. Reacts violently with lithium. Is incompatible with mercuric oxide, calomel, silver nitrate, tannin, and balsam Peru. Is also incompatible with strong bases, strong oxidizing agents and magnesium. Vigorous reactions occur with acetone in the presence of solid potassium hydroxide or calcium hydroxide, hexamethylenetetramine at 352 °F, mercury(I) fluoride and finely divided reduced silver. (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
12.8.4 Hazardous Reactivities and Incompatibilities
INCOMPATIBILITIES: MERCURIC OXIDE, CALOMEL, SILVER NITRATE, TANNIN, BALSAM PERU DIRECTLY MIXED.
Budavari, S. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 1996., p. 863
Hazardous Substances Data Bank (HSDB)
Incompatibile with ... lithium, acetone.
Lewis, R.J. Sax's Dangerous Properties of Industrial Materials. 9th ed. Volumes 1-3. New York, NY: Van Nostrand Reinhold, 1996., p. 1932
Hazardous Substances Data Bank (HSDB)
Incompatibilites: alkalis, oxidizing agents, lead salts
Reynolds, J.E.F., Prasad, A.B. (eds.) Martindale-The Extra Pharmacopoeia. 28th ed. London: The Pharmaceutical Press, 1982., p. 865
Hazardous Substances Data Bank (HSDB)
Strong oxidizers, lithium, metallic salts (e.g., mercuric oxide, silver nitrate), strong bases, calomel, tannin.
NIOSH. NIOSH Pocket Guide to Chemical Hazards. DHHS (NIOSH) Publication No. 97-140. Washington, D.C. U.S. Government Printing Office, 1997., p. 172
Hazardous Substances Data Bank (HSDB)
For more Hazardous Reactivities and Incompatibilities (Complete) data for IODOFORM (6 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
12.9 Regulatory Information
The Australian Inventory of Industrial Chemicals
Chemical: Methane, triiodo-
Australian Industrial Chemicals Introduction Scheme (AICIS)
New Zealand EPA Inventory of Chemical Status
Iodoform: Does not have an individual approval but may be used under an appropriate group standard
New Zealand Environmental Protection Authority (EPA)
12.9.1 Clean Water Act Requirements
Toxic pollutant designated pursuant to section 307(a)(1) of the Federal Water Pollution Control Act and is subject to effluent limitations. /Halomethanes/
40 CFR 401.15 (7/1/99)
Hazardous Substances Data Bank (HSDB)
12.9.2 FDA Requirements
Iodoform is an indirect food additive for use only as a component of adhesives. Limitation: for use only as polymerization-control agent.
21 CFR 175.105 (4/1/99)
Hazardous Substances Data Bank (HSDB)
12.10 Other Safety Information
12.10.1 Special Reports
USEPA; Ambient Water Quality Criteria Doc: Halomethanes (1980) EPA 440/5-80-051
Hazardous Substances Data Bank (HSDB)
Bioassay of Iodoform for Possible Carcinogenicity (1978) Technical Rpt Series No. 110 DHEW Pub No. (NIH) 78-1365, U.S. Department of Health Education and Welfare, National Cancer Institute, Bethesda, MD 20014
Hazardous Substances Data Bank (HSDB)
13 Toxicity
13.1 Toxicological Information
13.1.1 NIOSH Toxicity Data
The National Institute for Occupational Safety and Health (NIOSH)
13.1.2 Carcinogen Classification
Substance
Iodoform
NTP Technical Report
TR-110: Bioassay of Iodoform for Possible Carcinogenicity (CASRN 75-47-8) (1978 )
Peer Review Date
04/26/78
Conclusion for Male Rat
No Evidence
Conclusion for Female Rat
No Evidence
Conclusion for Male Mice
No Evidence
Conclusion for Female Mice
No Evidence
Summary
Under the conditions of this bioassay, no convincing evidence was provided for the carcinogenicity of iodoform in Osborne-Mendel rats or B6C3F1 mice.
NTP Technical Reports
13.1.3 Exposure Routes
inhalation, skin absorption, ingestion, skin and/or eye contact
The National Institute for Occupational Safety and Health (NIOSH)
13.1.4 Signs and Symptoms
irritation eyes, skin; lassitude (weakness, exhaustion), dizziness, nausea, incoordination, central nervous system depression; dyspnea (breathing difficulty); liver, kidney, heart damage; visual disturbance
The National Institute for Occupational Safety and Health (NIOSH)
13.1.5 Target Organs
Eyes, skin, respiratory system, liver, kidneys, heart
The National Institute for Occupational Safety and Health (NIOSH)
13.1.6 Adverse Effects
Occupational hepatotoxin - Secondary hepatotoxins: the potential for toxic effect in the occupational setting is based on cases of poisoning by human ingestion or animal experimentation.
ACGIH Carcinogen - Not Classifiable.
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
13.1.7 Acute Effects
ChemIDplus
13.1.8 Toxicity Data
LC50 (rat) = 165 ppm/7h
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
13.1.9 Human Toxicity Excerpts
SYSTEMIC INTOXICATION AND VISUAL DISTURBANCES RESULTED FROM ABSORPTION OF EXCESSIVE AMT /OF IODOFORM/ APPLIED TO WOUNDS OR ABSCESSES, OR FROM INGESTION OF LARGE QUANTITIES, BUT NOT FROM APPLICATION TO THE EYE. ... MOST CHARACTERISTICALLY VISION WAS IMPAIRED BY RETROBULBAR NEURITIS WITH ACCOMPANYING CENTRAL SCOTOMA. IN RARE INSTANCES TRANSITORY COMPLETE BLINDNESS OCCURRED. IN SOME CASES THE BULBAR PORTION OF THE OPTIC NERVE WAS INVOLVED, WITH NEURORETINITIS & OCCASIONAL RETINAL HEMORRHAGES. AS A RULE, RECOVERY WAS SLOW ... IN MOST CASES VISION WAS COMPLETELY OR PARTIALLY RECOVERED, BUT RESIDUAL PALLOR OF THE TEMPORAL PART OF THE OPTIC NERVEHEAD WAS COMMON. EXCEPTIONALLY THE WHOLE NERVEHEAD BECAME ATROPHIC AND WHITE AND LITTLE VISION WAS RECOVERED.
Grant, W.M. Toxicology of the Eye. 3rd ed. Springfield, IL: Charles C. Thomas Publisher, 1986., p. 524
Hazardous Substances Data Bank (HSDB)
Both di- and tri-halogenated methane derivatives have been found to produce increased blood levels of methemoglobin; the greatest increase caused by iodo-, followed by bromo- and chloro- compounds. CNS functional disturbances are reported, including depression of rapid eyemovement sleep, as seen in carbon monoxide exposures. /Di- and tri-halogenated methane derivatives/
Fodor GG, Roscovanu A; Zentrialbl Bakteriol (Orig B) 162: 34 (1976) as cited in USEPA; Ambient Water Quality Criteria Doc: Halomethanes p.C-40 (1980) EPA 440/5-80-051
Hazardous Substances Data Bank (HSDB)
POISONING IS OFTEN DUE TO ABSORPTION THROUGH WOUND WHEN IODOFORM DRESSINGS ARE USED (NO MORE THAN 2 G IODOFORM SHOULD BE SO USED). MAY CAUSE DERMATITIS. SYSTEMIC EFFECTS INCL VOMITING & ALL DEGREES OF CEREBRAL DEPRESSION OR EXCITATION, INCL DELIRIUM, HALLUCINATIONS, COMA, & DEATH. VERY RAPID PULSE IS CHARACTERISTIC, WITH OR WITHOUT SLIGHT FEVER.
Gosselin, R.E., R.P. Smith, H.C. Hodge. Clinical Toxicology of Commercial Products. 5th ed. Baltimore: Williams and Wilkins, 1984., p. II-387
Hazardous Substances Data Bank (HSDB)
... Severe poisoning, which may be fatal, is characterized by headache, somnolence, delirium, and rapid feeble pulse.
Reynolds, J.E.F., Prasad, A.B. (eds.) Martindale-The Extra Pharmacopoeia. 28th ed. London: The Pharmaceutical Press, 1982., p. 865
Hazardous Substances Data Bank (HSDB)
13.1.10 Non-Human Toxicity Excerpts
... LESION PRODUCED BY IODOFORM /WAS COMPARED/ TO THAT PRODUCED BY CARBON TETRACHLORIDE. ... MORPHOLOGICALLY, LESIONS WERE QUITE COMPARABLE. IN ADDITION, LIPID PEROXIDATION OCCURRED /IN LIVER/ ... BEING ASSOC WITH DEPRESSION IN GLUCOSE-6-PHOSPHATASE ACTIVITY & CACLIUM FLUX. ... ALSO INCR IN CELL SAP RNA. THESE FINDINGS ARE ESSENTIALLY IDENTICAL TO THOSE OBSERVED AFTER CARBON TETRACHLORIDE INTOXICATION.
Doull, J., C.D.Klassen, and M.D. Amdur (eds.). Casarett and Doull's Toxicology. 3rd ed., New York: Macmillan Co., Inc., 1986., p. 296
Hazardous Substances Data Bank (HSDB)
/Iodoform toxicity/ ... is characterized by the production of both fatty liver and necrosis ...
Doull, J., C.D.Klassen, and M.D. Amdur (eds.). Casarett and Doull's Toxicology. 3rd ed., New York: Macmillan Co., Inc., 1986., p. 646
Hazardous Substances Data Bank (HSDB)
ORAL DOSES OF IODOFORM EQUIV TO 2600 MICROMOLES OF CARBON TETRACHLORIDE/100 G OF RAT CAUSED EARLY CENTRILOBULAR SUPPRESSION OF GLUCOSE-6-PHOSPHATASE & TRANSITORY INFLUX OF CALCIUM INTO MIDZONAL LIVER PARENCHYMAL CELLS. CARBON TETRAIODIDE AFFECTED ENZYME WITHIN 1ST 8 HR.
REYNOLDS ES, YEE AG; LAB INVEST 19 (3): 273-81 (1968)
Hazardous Substances Data Bank (HSDB)
LESIONS OF LIVER MEMBRANOUS CELLULAR COMPONENTS FOLLOWING IODOFORM.
PMID:5768872
Full text:
SELL DA, REYNOLDS ES; J CELL BIOL 41 (3): 736-52 (1969)
Hazardous Substances Data Bank (HSDB)
For more Non-Human Toxicity Excerpts (Complete) data for IODOFORM (11 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
13.1.11 Non-Human Toxicity Values
LD50 Mouse subcutaneous 1.6 mmoles/kg
Budavari, S. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 1996., p. 863
Hazardous Substances Data Bank (HSDB)
LDLo Rabbit subcutaneous 50 mg/kg
American Conference of Governmental Industrial Hygienists. Documentation of the Threshold Limit Values and Biological Exposure Indices. 5th ed. Cincinnati, OH: American Conference of Governmental Industrial Hygienists, 1986., p. 324
Hazardous Substances Data Bank (HSDB)
LDLo Canis familiaris (dog) oral 1000 mg/kg
American Conference of Governmental Industrial Hygienists. Documentation of the Threshold Limit Values and Biological Exposure Indices. 5th ed. Cincinnati, OH: American Conference of Governmental Industrial Hygienists, 1986., p. 324
Hazardous Substances Data Bank (HSDB)
View More...
13.1.12 Ongoing Test Status
The following link will take the user to the National Toxicology Program (NTP) Test Agent Search Results page, which tabulates all of the "Standard Toxicology & Carcinogenesis Studies", "Developmental Studies", and "Genetic Toxicity Studies" performed with this chemical. Clicking on the "Testing Status" link will take the user to the status (i.e., in review, in progress, in preparation, on test, completed, etc.) and results of all the studies that the NTP has done on this chemical. [
Available from:
Hazardous Substances Data Bank (HSDB)
13.1.13 National Toxicology Program Studies
A bioassay for possible carcinogenicity of technical grade iodoform was conducted using Osborne-Mendel rats and B6C3F1 mice. Iodoform in corn oil was administered by gavage, at either of two dosages, to groups of 50 male and 50 female animals of each species. Administration of the chemical occurred 5 days/wk, for a period of 78 wk, followed by an observation period of 34 wk for rats and 13 or 14 wk for mice. For each species, 20 animals of each sex were placed on test as vehicle controls. These animals were gavaged with pure corn oil at the same rate as the high dose group of the same sex. Twenty animals of each sex were placed on test as untreated controls for each species. ... Under the conditions of this bioassay, no convincing evidence was provided for the carcinogenicity of iodoform in Osborne-Mendel rats or B6C3F1 mice. Levels of Evidence of Carcinogenicity: Male Rats: Negative; Female Rats: Negative; Male Mice: Negative; Female Mice: Negative.
Bioassay of Iodoform for Possible Carcinogenicity (1978) Technical Rpt Series No. 110 DHEW Pub No. (NIH) 78-1365, U.S. Department of Health Education and Welfare, National Cancer Institute, Bethesda, MD 20014
Hazardous Substances Data Bank (HSDB)
13.1.14 Protein Binding
No pharmacokinetic data available.
DrugBank
13.2 Ecological Information
13.2.1 Ecotoxicity Values
LC50 Pimephales promelas (fathead minnows) 2.92 mg/l/96 hr, flow-through bioassay @ 25.1 °C, pH 7.7, water hardness 44.2 mg/l CaCO3 /Purity 98%/
Geiger D.L., Call D.J., Brooke L.T. (eds). Acute Toxicities of Organic Chemicals to Fathead Minnows (Pimephales Promelas). Vol. IV. Superior Wisconsin: University of Wisconsin-Superior, 1988., p. 40
Hazardous Substances Data Bank (HSDB)
EC50 Pimephales promelas (fathead minnows) 2.01 mg/l/96 hr, flow-through bioassay @ 25.1 °C, pH 7.7, water hardness 44.2 ml/l CaCO3 /Purity 98%/. /EC50 was based upon loss of equlibrium as manifested by the fish's inability to maintain an upright position when swimming./
Geiger D.L., Call D.J., Brooke L.T. (eds). Acute Toxicities of Organic Chemicals to Fathead Minnows (Pimephales Promelas). Vol. IV. Superior Wisconsin: University of Wisconsin-Superior, 1988., p. 40
Hazardous Substances Data Bank (HSDB)
13.2.2 Environmental Fate / Exposure Summary
Iodoform's production and use as an antiseptic and sensitizing agent in certain printing processes may result in its release to the environment through various waste streams. Its present use is rather limited. It is sometimes formed in drinking water during the chlorination process. If released to air, an estimated vapor pressure of 0.040 mm Hg at 25 °C indicates iodoform will exist solely as a vapor in the ambient atmosphere. Vapor-phase iodoform will be degraded in the atmosphere by reaction with photochemically-produced hydroxyl radicals; the half-life for this reaction in air is estimated to be 56 days. Iodoform absorbs light >290 nm and therefore there is a potential for direct photolysis. If released to soil, iodoform is expected to have very high mobility based upon an estimated Koc of 35. Volatilization from moist soil surfaces is expected to be an important fate process based upon an estimated Henry's Law constant of 3.1X10-5 atm-cu m/mole. Iodoform's biodegradation potential in soil or water is unknown. If released into water, iodoform is not expected to adsorb to suspended solids and sediment based upon the estimated Koc. Volatilization from water surfaces is expected to be an important fate process based upon this compound's estimated Henry's Law constant. Estimated volatilization half-lives for a model river and model lake are 40 hours and 25 days, respectively. An estimated BCF of 43 suggests the potential for bioconcentration in aquatic organisms is low. Hydrolysis is not expected to be an important fate process because the estimated hydrolysis rate is low. Occupational exposure to iodoform may occur through inhalation and dermal contact with this compound at workplaces where iodoform is produced or used. The general population may be exposed to iodoform in some drinking water. Direct human exposure occurs when iodoform is applied to a wound. (SRC)
Hazardous Substances Data Bank (HSDB)
13.2.3 Artificial Pollution Sources
Iodoform's production and use as an antiseptic and sensitizing agent in certain printing processes may result in its release to the environment through various waste streams(1,2). However its present use is rather limited. Iodoform may be formed in finished drinking water during the chlorination process(3). When iodinated trihalomethanes occur, a medicinal taste or musty odor is imparted to the water. While there are only a few reports of iodoform in treated drinking water in the literature, strong medicinal odors have long been known to occur in such water.
(1) Budavari S; The Merck Index 12th ed; Merck and Co, Inc Rahway, NJ p. 863 (1996)
(2) Lauterbach A; Kirk-Othmer Encycl Chem Tech 4th Ed NY, NY: John-Wiley 14: 729 (1995)
(3) Bruchet A et al; Micropollut Aquat Emviron., Proc. Eur Supp., 6th. pp. 371-83 (1990)
Hazardous Substances Data Bank (HSDB)
13.2.4 Environmental Fate
TERRESTRIAL FATE: Based on a classification scheme(1), an estimated Koc value of 35(SRC), determined from a structure estimation method(2), indicates that iodoform is expected to have very high mobility in soil(SRC). Volatilization of iodoform from moist soil surfaces is expected to be an important fate process(SRC) given an estimated Henry's Law constant of 3.1X10-5 atm-cu m/mole(SRC), using a fragment constant estimation method(3). Iodoform is not expected to volatilize from dry soil surfaces(SRC) based upon an estimated vapor pressure of 0.040 mm Hg(SRC), determined from a fragment constant method(4). Data on the biodegradability of iodoform in soil is not available(SRC).
(1) Swann RL et al; Res Rev 85: 17-28 (1983)
(2) Meylan WM et al; Environ Sci Technol 26: 1560-67 (1992)
(3) Meylan WM, Howard PH; Environ Toxicol Chem 10: 1283-93 (1991)
(4) Lyman WJ; p. 31 in Environmental Exposure From Chemicals Vol I, Neely WB, Blau GE, eds, Boca Raton, FL: CRC Press (1985)
Hazardous Substances Data Bank (HSDB)
AQUATIC FATE: Based on a classification scheme(1), an estimated Koc value of 35(SRC), determined from an estimation method(2), indicates that iodoform is not expected to adsorb to suspended solids and sediment(SRC). Volatilization from water surfaces is expected(3) based upon an estimated Henry's Law constant of 3.1X10-5 atm-cu m/mole(SRC), developed using a fragment constant estimation method(4). Using this Henry's Law constant and an estimation method(3), volatilization half-lives for a model river and model lake are 40 hours and 25 days, respectively(SRC). According to a classification scheme(5), an estimated BCF of 43(SRC), from an estimated log Kow of 3.03(6) and a regression-derived equation(7), suggests the potential for bioconcentration in aquatic organisms is low. Data on the biodegradability of iodoform in water is not available(SRC).
(1) Swann RL et al; Res Rev 85: 17-28 (1983)
(2) Meylan WM et al; Environ Sci Technol 26: 1560-67 (1992)
(3) Lyman WJ et al; Handbook of Chemical Property Estimation Methods. Washington, DC: Amer Chem Soc pp. 4-9, 15-1 to 15-29 (1990)
(4) Meylan WM, Howard PH; Environ Toxicol Chem 10: 1283-93 (1991)
(5) Franke C et al; Chemosphere 29: 1501-14 (1994)
(6) Meylan WM, Howard PH; J Pharm Sci 84: 83-92 (1995)
(7) Meylan WM et al; Environ Toxicol Chem 18: 664-72 (1999)
Hazardous Substances Data Bank (HSDB)
ATMOSPHERIC FATE: According to a model of gas/particle partitioning of semivolatile organic compounds in the atmosphere(1), iodoform, which has an estimated vapor pressure of 0.396 mm Hg at 25 °C(2), is expected to exist solely as a vapor in the ambient atmosphere. Vapor-phase iodoform is degraded in the atmosphere by reaction with photochemically-produced hydroxyl radicals(SRC); the half-life for this reaction in air is estimated to be 56 days(SRC), from its rate constant of 0.289X10-12 cu cm/molecule-sec at 25 °C(3) determined using a structure estimation method(3). Iodoform absorbs radiation >290 nm(4) and therefore may be susceptible to direct photolysis(SRC). Photolysis rates for iodoform were not located.
(1) Bidleman TF; Environ Sci Technol 22: 361-367 (1988)
(2) Lyman WJ; p. 31 in Environmental Exposure From Chemicals Vol I, Neely WB, Blau GE, eds, Boca Raton, FL: CRC Press (1985)
(3) Meylan WM, Howard PH; Chemosphere 26: 2293-99 (1993)
Hazardous Substances Data Bank (HSDB)
13.2.5 Environmental Abiotic Degradation
The rate constant for the vapor-phase reaction of iodoform with photochemically-produced hydroxyl radicals has been estimated as 0.288X10-12 cu cm/molecule-sec at 25 °C(SRC) using a structure estimation method(1). This corresponds to an atmospheric half-life of about 55 days at an atmospheric concentration of 5X10+5 hydroxyl radicals per cu cm(1). A base-catalyzed second-order hydrolysis rate constant of 1.72X10-8 L/mole-sec(SRC) was estimated using a structure estimation method(2); this corresponds to half-lives of more than a million years at pH values of 7 to 8(2). Iodoform has a broad UV absorption band with a maximum at 349 nm(3) and therefore may be susceptible to direct photolysis. However photolysis reaction rates for iodoform were not located. The rate constant for the vapor-phase reaction of iodoform with photochemically-produced NO3 radicals is 1.76X10+17 cu cm/molecule-sec under simulated atmospheric conditions(4).
(1) Meylan WM, Howard PH; Chemosphere 26: 2293-99 (1993)
(2) Mill T et al; Environmental Fate and Exposure Studies Development of a PC-SAR for Hydrolysis: Esters, Alkyl Halides and Epoxides. EPA Contract No. 68-02-4254. Menlo Park, CA: SRI International (1987)
(3) Calvert JG, Pitts JN Jr; Photochemistry NY, NY: Wiley pp. 523 (1966)
(4) Carlier P et al; pp. 133-41 in Tropospheric NOx Chem -- Gasphase Multiphase Aspects. Risol Natl Lab ISSRISOE-M-2630 (1988)
Hazardous Substances Data Bank (HSDB)
13.2.6 Environmental Bioconcentration
An estimated BCF of 43 was calculated for iodoform(SRC), using an estimated log Kow of 3.03(1,SRC) and a regression-derived equation(2). According to a classification scheme(3), this BCF suggests the potential for bioconcentration in aquatic organisms is low.
(1) Meylan WM, Howard PH; J Pharm Sci 84: 83-92 (1995)
(2) Meylan WM et al; Environ Toxicol Chem 18: 664-72 (1999)
(3) Franke C et al; Chemosphere 29: 1501-14 (1994)
Hazardous Substances Data Bank (HSDB)
13.2.7 Soil Adsorption / Mobility
Using a structure estimation method based on molecular connectivity indices(1), the Koc for iodoform can be estimated to be 35(SRC). According to a classification scheme(2), this estimated Koc value suggests that iodoform is expected to have very high mobility in soil.
(1) Meylan WM et al; Environ Sci Technol 26: 1560-67 (1992)
(2) Swann RL et al; Res Rev 85: 17-28 (1983)
Hazardous Substances Data Bank (HSDB)
13.2.8 Volatilization from Water / Soil
The Henry's Law constant for iodoform is estimated as 3.1 atm-cu m/mole(SRC) using a fragment constant estimation method(1). This Henry's Law constant indicates that iodoform is expected to volatilize from water surfaces(2). Based on this Henry's Law constant, the volatilization half-life from a model river (1 m deep, flowing 1 m/sec, wind velocity of 3 m/sec)(2) is estimated as 40 hours(SRC). The volatilization half-life from a model lake (1 m deep, flowing 0.05 m/sec, wind velocity of 0.5 m/sec)(2) is estimated as 25 days(SRC). Iodoform's Henry's Law constant(1) indicates that volatilization from moist soil surfaces may occur(SRC). Iodoform is not expected to volatilize from dry soil surfaces(SRC) based upon an estimated vapor pressure of 0.040 mm Hg(SRC), determined from a fragment constant method(3).
(1) Meylan WM, Howard PH; Environ Toxicol Chem 10: 1283-93 (1991)
(2) Lyman WJ et al; Handbook of Chemical Property Estimation Methods. Washington, DC: Amer Chem Soc pp. 15-1 to 15-29 (1990)
(3) Lyman WJ; p 31 in Environmental Exposure From Chemicals Vol I, Neely WB, Blau GE(eds), Boca Raton, FL: CRC Press (1985)
Hazardous Substances Data Bank (HSDB)
13.2.9 Environmental Water Concentrations
Iodoform was found in two French water treatment plants at 80 ng/l and 10-20 ng/l, respectively. This is associated with the presence of chloramines in the water being treated. In the absence of chloramines, chlorine tends to quickly react with organic precursors to preferentially form chlorinated, brominated, and iodated trihalomethanes. One of the French treatment plants used ammonia-rich ground water(1).
(1) Bruchet A et al; Micropollut Aquat Emviron., Proc. Eur Supp., 6th. pp. 371-83 (1990)
Hazardous Substances Data Bank (HSDB)
13.2.10 Probable Routes of Human Exposure
Iodoform is used as an antiseptic wound powder; however, today its use is rather limited(1,2). Therefore, direct human exposure occurs through the application of the drug when applied to a wound(SRC). Workers involved in formulating and dispensing the drug may be exposed through dermal contact or inhalation of iodoform-containing dust(SRC). People may also be exposed to iodoform in some chlorinated drinking water(3). Iodoform-containing drinking water would generally have a medicinal taste or musty odor.
(1) Budavari S; The Merck Index 12th ed; Merck and Co, Inc Rahway, NJ p. 863 (1996)
(2) Lauterbach A; Kirk-Othmer Encycl Chem Tech 4th Ed NY: John-Wiley 14: 729 (1995)
(3) Bruchet A et al; pp. 371-83 in Micropollut Aquat Emviron., Proc. Eur Supp., 6th. (1990)
Hazardous Substances Data Bank (HSDB)
14 Associated Disorders and Diseases
Comparative Toxicogenomics Database (CTD)
15 Literature
15.1 Consolidated References
PubChem
15.2 NLM Curated PubMed Citations
Medical Subject Headings (MeSH)
15.3 Springer Nature References
Springer Nature
15.4 Thieme References
Thieme Chemistry
15.5 Chemical Co-Occurrences in Literature
PubChem
15.6 Chemical-Gene Co-Occurrences in Literature
PubChem
15.7 Chemical-Disease Co-Occurrences in Literature
PubChem
15.8 Chemical-Organism Co-Occurrences in Literature
PubChem
16 Patents
16.1 Depositor-Supplied Patent Identifiers
PubChem
Link to all deposited patent identifiers
PubChem
16.2 WIPO PATENTSCOPE
Patents are available for this chemical structure:
PATENTSCOPE (WIPO)
16.3 Chemical Co-Occurrences in Patents
PubChem
16.4 Chemical-Disease Co-Occurrences in Patents
PubChem
16.5 Chemical-Gene Co-Occurrences in Patents
PubChem
16.6 Chemical-Organism Co-Occurrences in Patents
PubChem
17 Interactions and Pathways
17.1 Chemical-Target Interactions
Comparative Toxicogenomics Database (CTD)
18 Biological Test Results
18.1 BioAssay Results
PubChem
19 Taxonomy
LOTUS - the natural products occurrence database; Natural Product Activity and Species Source (NPASS)
20 Classification
20.1 MeSH Tree
Medical Subject Headings (MeSH)
20.2 NCI Thesaurus Tree
NCI Thesaurus (NCIt)
20.3 ChEBI Ontology
ChEBI
20.4 KEGG: Drug
KEGG
20.5 KEGG: ATC
KEGG
20.6 KEGG: JP15
KEGG
20.7 KEGG: Animal Drugs
KEGG
20.8 WHO ATC Classification System
WHO Anatomical Therapeutic Chemical (ATC) Classification
20.9 ChemIDplus
ChemIDplus
20.10 CAMEO Chemicals
CAMEO Chemicals
20.11 UN GHS Classification
GHS Classification (UNECE)
20.12 EPA CPDat Classification
EPA Chemical and Products Database (CPDat)
20.13 NORMAN Suspect List Exchange Classification
NORMAN Suspect List Exchange
20.14 EPA DSSTox Classification
EPA DSSTox
20.15 EPA TSCA and CDR Classification
EPA Chemicals under the TSCA
20.16 LOTUS Tree
LOTUS - the natural products occurrence database
20.17 EPA Substance Registry Services Tree
EPA Substance Registry Services
20.18 MolGenie Organic Chemistry Ontology
MolGenie
20.19 Chemicals in PubChem from Regulatory Sources
PubChem
20.20 ATCvet Classification
WHO ATCvet - Classification of Veterinary Medicines
21 Information Sources
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4406 | https://www.lenovo.com/us/en/glossary/what-is-binary/?srsltid=AfmBOorzSksAApyaWvF5fOwotKUFdW1d6qdMKsHRebAgoOBHQeidrZS9 | Introduction to Binary: Basics and Importance | Lenovo US
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What is binary?
Binary is a numerical system that uses two digits, usually represented as 0 and 1. It is the basis of all digital computers and is used to represent data or instructions in a machine-readable form. Binary is known as a “base 2” system because it uses two numbers to represent any quantity; in contrast, decimal systems use 10 digits (0–9). Binary data is stored in computer memory as binary numbers, which are then converted into other forms such as text or images for display onscreen. In programming languages, binary code consists of sequences of 0s and 1s that tell the computer what to do – making it an essential component of modern computing.
Are binary files executable?
Binary files are a type of executable file. They are special because they are not written in a human-readable language – instead, they consist of binary data that is written in a machine-readable form. Binary files often contain instructions for specific programs and can be opened by any application capable of executing them, such as Windows operating systems. When a user executes a binary file, the instructions within it are converted from their binary form into machine code which the computer's processor can handle. This allows the computer to perform tasks according to those instructions, making binary files an essential part of modern computing.
What do binary sequences represent?
Binary sequences can represent a wide variety of information, from simple values like numbers and strings to complex data such as images, audio files, and executable code. Each bit (0 or 1) in a binary sequence represents either one or zero, which translates into the traditional data types found in most programming languages. By using algorithms such as Huffman coding, binary sequences can also be used to represent text, image files, audio files, and even executable code. Binary sequences offer great flexibility when it comes to representing different kinds of data, making them ideal for use in computing applications.
Can binary search be used for strings?
Binary search can be used to quickly and efficiently search for strings in large amounts of data. It works by dividing the data into smaller pieces and searching those pieces individually. By doing this, it reduces the amount of time needed to find the desired string. This is especially useful when searching for a specific string within a large database or text file. Binary search is also an effective way to compare strings against a predefined list, as it allows for quick comparisons between different elements of the list. In addition, binary search can be used to store data in such a way that ensures fast retrieval of information when needed.
How does binary work?
Binary works by using combinations of ones and zeros to represent different values. Each one or zero is referred to as a "bit", and each combination of bits creates a "binary code". This binary code can represent any number, character, data type, or instruction that a computer can use. When combined, these binary sequences form the building blocks for information processing in computers. Binary works by performing operations on the combinations of ones and zeros and then providing a result - either true or false. By combining and manipulating many of these sequences together, complex calculations can be made to solve problems or execute instructions.
What is binary all about?
Binary is a way of representing data and instructions by using ones and zeros. It is the backbone of all information processing within computers. Each one or zero is referred to as a bit, and every combination of bits can represent a different number, character, data type or instruction used by computers. By combining bits together in various combinations, complex calculations can be made to solve many problems or process instructions. Binary is often used in cryptography and security systems to securely communicate data between two parties without them having to share the same physical space. It has also been used extensively in software development for allowing programs to store vast amounts of data efficiently.
Does binary start at 0 or 1?
Binary usually starts at 0. This is because the binary number system works differently from the decimal system we are used to in everyday life. In the decimal system, counting starts from 1 and each number has a value of 10 times the previous number. In binary, counting each number has a value of double the previous one - for example two is two ones, three is two ones with a zero, four would be two zeros with a one and so on. The first number in binary is always zero = 00 and this continues on until infinity. As such, when talking about binary numbers it's always assumed to start with 0 as this will represent to most accurate translation from binary into decimal numbers.
Where is the binary number system used?
The binary number system is used extensively in computers and other digital devices. Binary code is the language of computers, with all data and instructions being represented by combinations of ones and zeros. This allows them to store large amounts of data efficiently while providing a simple way to process and analyse it. Binary is also used in cryptography as a way to securely communicate between two parties without having to share the same physical space. Additionally, binary is often used in embedded systems, as they commonly rely on binary signals. It can also be used in data transmission over fibre optics or other digital forms of communication. Regardless of the application, binary plays an important role in the workings of computers and digital systems today.
Why is binary important?
Binary is important because it allows for the efficient storage and processing of data. It also provides a universal language for communication between computers and other digital devices, allowing them to interact with one another. Binary code enables us to quickly identify errors in data processing or transmission, as well as identifying potential areas of improvement. Additionally, binary code makes it easier to design complex systems that are compatible with different hardware and software components. Furthermore, binary is often used in cryptography as a way to securely communicate between two parties without having the need to share the same physical space. In short, binary plays an essential role in modern computing and data analysis, and its importance will continue to grow as technology advances.
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4407 | https://www.youtube.com/watch?v=AUBX_aHStQY | Factoring higher degree polynomials | Algebra 2 | Khan Academy
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Factoring a partially factored polynomial and factoring a third degree polynomial by grouping.
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Transcript:
there are many videos on khan academy where we talk about factoring polynomials but what we're going to do in this video is do a few more examples of factoring higher degree polynomials so let's start with a little bit of a warm up let's say that we wanted to factor six x squared plus nine x times x squared minus four x plus four pause this video and see if you can factor this into the product of even more expressions all right now let's do this together and the way that this might be a little bit different than what you've seen before is this is already partially factored this polynomial this higher degree polynomial is already expressed as the product of two quadratic expressions but as you might be able to tell we can factor this further for example 6x squared plus 9x both 6x squared and 9x are divisible by 3x so let's factor out a 3x here so this is the same thing as 3x times 3x times what is 6x squared well 3 times 2 is 6 and x times x is x squared and then 3x times what is 9x well 3x times 3 is 9x and you can verify that if we were to distribute this 3x you would get 6x squared plus 9x and then what about this second expression right over here can we factor this well you might recognize this as a perfect square some of you might have said hey i need to come up with two numbers whose product is 4 and whose sum is negative 4 and you might say hey that's negative 2 and negative 2. and so this would be x minus 2 we could write x minus 2 squared or we could write it as x minus 2 times x minus 2. if what i just did is unfamiliar i encourage you to go back and watch videos on factoring perfect square quadratics and things like that but there you have it i think we have factored this as far as we could go so now let's do a slightly trickier higher degree polynomial so let's say we wanted to factor x to the third minus 4 x squared plus 6 x minus 24 and i just like always pause this video and see if you can have a go at it and i'll give you a little bit of a hint you can factor in this case by grouping and in some ways it's a little bit easier than what we've done in the past historically when we've learned factoring by grouping we've looked at a quadratic and then we looked at the middle term the x term of the quadratic and we broke it up so that we had four terms here we already have four terms and see if you could have a go at that all right now let's do it together so you can't always factor a third degree polynomial by grouping but sometimes you can so it's good to look for it so when we see it written like this we say okay x to the third minus 4x squared is there a common factor here well yeah both x to the third and negative four x squared are divisible by x squared so what happens if we factor out an x squared so that's x squared times x minus four and what about these second two terms is there a common factor between 6x and negative 24 yeah they're both divisible by 6. so let's factor out a 6 here so plus 6 times x minus 4. and now you are probably seeing the home stretch where you have something times x minus 4 and then something else times x minus 4. and so you can sometimes i like to say undistribute the x minus 4 or factor out the x minus 4. and so this is going to be x minus 4 times x squared x squared plus 6. and we are done |
4408 | https://tasks.illustrativemathematics.org/content-standards/tasks/1475 | Engage your students with effective distance learning resources. ACCESS RESOURCES>>
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Alignments to Content Standards: 7.NS.A.1 7.EE.B.4
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Task
At the beginning of the month, Evan had \$24 in his account at the school bookstore. Use a variable to represent the unknown quantity in each transaction below and write an equation to represent it. Then represent each transaction on a number line. What is the unknown quantity in each case?
First he bought some notebooks and pens that cost \$16.
Then he deposited some more money and his account balance was \$28.
Then he bought a book for English class that cost \$34.
Then he deposited exactly enough money so that he paid off his debt to the bookstore.
Explain why it makes sense to use a negative number to represent Evan's account balance when he owes money.
IM Commentary
Many people like to use debt to help students understand negative numbers. When doing so, it is important to keep in mind that representing debt with negative numbers is a convention, and that it would be possible to talk about debt without ever using negative numbers. For example, if I borrow \$30 from my cousin, I owe her +30 dollars. If I give her \$40, then she owes me +10 dollars. The reason it is so convenient to represent debt with negative numbers is that it allows us to use arithmetic and algebra to keep track of who owes what to whom. Despite the fact that it is convenient to do so, it is not required simply by setting up a problem involving debt.
The purpose of this task is for students to use algebra and the number line to understand why it makes sense that we sometimes represent debt using negative numbers. If we agree that depositing money in an account adds a positive number to the balance, and buying somethings subtracts a positive number from the balance, then it is natural to represent debt with negative numbers.
The number line is an important mathematical model which helps students visualize numbers and make sense of addition and subtraction of numbers. In the middle school grades, it is important for them to develop the ability to perform arithmetic in a more abstract setting, paving the way for the development of algebraic reasoning. Thus in 6th grade they ''Apply and extend previous understandings of arithmetic to algebraic expressions'' (6.EE.A) and in the 7th grade students ''Solve real-life and mathematical problems using numerical and algebraic expressions and equations'' (7.EE.4). This task provides an opportunity for students to use the number line to explain arithmetic with the integers while at the same time representing these transactions with algebraic expressions. In other words, students are both increasing their understanding of arithmetic with numbers while working toward arithmetic with expressions which will be more fully developed in the 8th grade and in high school algebra.
This task assumes that students have already used the number line to represent signed numbers and that they know how to represent addition and subtraction of positive numbers on the number line. In other words, students should know that:
For every number $a$ on the number line there is a number $-a$ on the other side of zero that is equally distant from 0. These numbers are called opposites or additive inverses.
To add a positive number $b$ to a number $a$, you move $b$ units to the right of $a$ on the number line.
To subtract a positive number $b$ from a number $a$, you move $b$ units to the left of $a$ on the number line.
Putting this together with the fact that depositing money in the bookstore account means adding an amount to the account balance, and buying an item from the bookstore means subtracting an amount from the account balance leads us to see why it makes sense to represent debt with negative numbers.
Solution
He started with \$24 and spent \$16. Let $a$ represent his account balance after he bought the notebook and pens. Then $$24 €“ 16 = a$$
His new account balance is \$8.
He has \$8 in his account and then he deposited $b$ dollars. His account balance is now \$28. Then $$8 + b = 28$$
His deposited \$20 in his account.
He has \$28 and spent \$34. Let $c$ represent his account balance after he bought the book. Then $$28 €“ 34 = c$$
His new account balance is -\$6.
He started with an account balance of -\$6 and paid the debt off so his account balance is 0. If $d$ is the amount money he deposited to pay off his debt, then $$-6 + d = 0$$
He deposited \$6.
2. It makes sense to use a negative number to represent Evan's account balance when he owes money. Here are two ways to look at it:
(1) We always take the current account balance and add a positive number to that if he makes a deposit or subtract a positive number from that if he buys items at the bookstore. If he spends more than he has in his account, then we are subtracting a bigger number from a smaller number, and the result is negative.
(2) We can also see in this context that if he deposits just enough money to pay off his debt, that means we are adding a positive number to his account balance and the result is 0. If you add two numbers together to get 0, then the numbers must be opposites. The opposite of a positive number is a negative number, so it makes sense to represent his account balance with a negative number when he owes money to the bookstore.
Bookstore Account
At the beginning of the month, Evan had \$24 in his account at the school bookstore. Use a variable to represent the unknown quantity in each transaction below and write an equation to represent it. Then represent each transaction on a number line. What is the unknown quantity in each case?
First he bought some notebooks and pens that cost \$16.
Then he deposited some more money and his account balance was \$28.
Then he bought a book for English class that cost \$34.
Then he deposited exactly enough money so that he paid off his debt to the bookstore.
Explain why it makes sense to use a negative number to represent Evan's account balance when he owes money.
Print Task |
4409 | https://www.caee.utexas.edu/prof/bhat/ABSTRACTS/new_CML_3July2014.pdf | The Composite Marginal Likelihood (CML) Inference Approach with Applications to Discrete and Mixed Dependent Variable Models Chandra R. Bhat The University of Texas at Austin Department of Civil, Architectural and Environmental Engineering 301 E. Dean Keeton St. Stop C1761, Austin TX 78712, USA Phone: 512-471-4535; Fax: 512-475-8744; Email: bhat@mail.utexas.edu and King Abdulaziz University, Jeddah 21589, Saudi Arabia ABSTRACT This monograph presents the basics of the composite marginal likelihood (CML) inference approach, discussing the asymptotic properties of the CML estimator and the advantages and limitations of the approach. The composite marginal likelihood (CML) inference approach is a relatively simple approach that can be used when the full likelihood function is practically infeasible to evaluate due to underlying complex dependencies. The history of the approach may be traced back to the pseudo-likelihood approach of Besag (1974) for modeling spatial data, and has found traction in a variety of fields since, including genetics, spatial statistics, longitudinal analyses, and multivariate modeling. However, the CML method has found little coverage in the econometrics field, especially in discrete choice modeling. This monograph fills this gap by identifying the value and potential applications of the method in discrete dependent variable modeling as well as mixed discrete and continuous dependent variable model systems. In particular, it develops a blueprint (complete with matrix notation) to apply the CML estimation technique to a wide variety of discrete and mixed dependent variable models. 1 1. INTRODUCTION 1.1. Background The need to accommodate underlying complex interdependencies in decision-making for more accurate policy analysis as well as for good forecasting, combined with the explosion in the quantity of data available for the multidimensional modeling of inter-related choices of a single observational unit and/or inter-related decision-making across multiple observational units, has resulted in a situation where the traditional frequentist full likelihood function becomes near impossible or plain infeasible to evaluate. As a consequence, another approach that has seen some (though very limited) use recently is the composite likelihood (CL) approach. While the method has been suggested in the past under various pseudonyms such as quasi-likelihood (Hjort and Omre, 1994; Hjort and Varin, 2008), split likelihood (Vandekerkhove, 2005), and pseudolikelihood or marginal pseudo-likelihood (Molenberghs and Verbeke, 2005), Varin (2008) discusses reasons why the term composite likelihood is less subject to literary confusion. At a basic level, a composite likelihood (CL) refers to the product of a set of lower-dimensional component likelihoods, each of which is a marginal or conditional density function. The maximization of the logarithm of this CL function is achieved by setting the composite score equations to zero, which are themselves linear combinations of valid lower-dimensional likelihood score functions. Then, from the theory of estimating equations, it can be shown that the CL score function (and, therefore, the CL estimator) is unbiased (see Varin et al., 2011). In this monograph, we discuss these theoretical aspects of CL methods, with an emphasis on an overview of developments and applications of the CL inference approach in the context of discrete dependent variable models. The history of the CL method may be traced back to the pseudo-likelihood approach of Besag (1974) for modeling spatial data, and has found traction in a variety of fields since, including genetics, spatial statistics, longitudinal analyses, and multivariate modeling (see Varin et al., 2011 and Larribe and Fearnhead, 2011 for reviews). However, the CL method has found little coverage in the econometrics field, and it is the hope that this monograph will fill this gap by identifying the value and potential applications of the method in econometrics. 1.2. Types of CL Methods To present the types of CL methods, assume that the data originate from a parametric underlying model based on a random ( H ~ × 1) vector Y with density function ) , ( θ y f , where θ is an unknown K ~ -dimensional parameter vector (technically speaking, the density function ) , ( θ y f refers to the conditional density function ) , ( θ y X | Y f of the random variable Y given a set of explanatory variables X, though we will use the simpler notation ) , ( θ y f for the conditional density function). Each element of the random variable vector Y may be observed directly, or may be observed in a truncated or censored form. Assume that the actual observation vector corresponding to Y is given by the vector ) , , , , ( ~ 3 2 1 H m m m m m , some of which may take a continuous form and some of which may take a limited-dependent form. Let the likelihood 2 corresponding to this observed vector be ).
; ( m θ L Now consider the situation where computing ) ; ( m θ L is very difficult. However, suppose evaluating the likelihood functions of a set of E ~ observed marginal or conditional events determined by marginal or conditional distributions of the sub-vectors of Y is easy and/or computationally expedient. Let these observed marginal events be characterized by ( ) ( ..., , ) ( , ) ( ~ 2 1 m m m E A A A ). Let each event ) (m e A be associated with a likelihood object ) ( ; ) ; ( m θ m θ e e A L L , which is based on a lower-dimensional marginal or conditional joint density function corresponding to the original high-dimensional joint density of Y. Then, the general form of the composite likelihood function is as follows: E e e E e e CL e e A L L L ~ 1 ~ 1 ) ( ; ( ) ; ( ) , ( m θ m θ m θ , (1.1) where e is a power weight to be chosen based on efficiency considerations. If these power weights are the same across events, they may be dropped. The CL estimator is the one that maximizes the above function (or equivalently, its logarithmic transformation). The events ) (m e A can represent a combination of marginal and conditional events, though composite likelihoods are typically distinguished in one of two classes: the composite conditional likelihood (CCL) or the composite marginal likelihood (CML). In this monograph, we will focus on the CML method because it has many immediate applications in the econometrics field, and is generally easier to specify and estimate. However, the CCL method may also be of value in specific econometric contexts (see Mardia et al., 2009 and Varin et al., 2011 for additional details). 1.3. The Composite Marginal Likelihood (CML) Inference Approach In the CML method, the events ) (m e A represent marginal events. The CML class of estimators subsumes the usual ordinary full-information likelihood estimator as a special case. For instance, consider the case of repeated unordered discrete choices from a specific individual. Let the individual’s discrete choice at time t be denoted by the index t d , and let this individual be observed to choose alternative t m at choice occasion t ). ..., , 3 , 2 , 1 ( T t Then, one may define the observed event for this individual as the sequence of observed choices across all the T choice occasions of the individual. Defined this way, the CML function contribution of this individual becomes equivalent to the full-information maximum likelihood function contribution of the individual:1 ) ,..., , , ( Prob ) , ( ) , ( 3 3 2 2 1 1 1 T T CML m d m d m d m d L L m θ m θ . (1.2) 1 In the discussion below, for presentation ease, we will ignore the power weight term ωe. In some cases, such as in a panel case with varying number of observational occasions on each observation unit, the choice of ωe can influence estimator asymptotic efficiency considerations. But it does not affect other asymptotic properties of the estimator. 3 However, one may also define the events as the observed choices at each choice occasion for the individual. Defined this way, the CML function is: ) ( Prob ...
) ( Prob ) Prob( ) ( Prob ) , ( 3 3 2 2 1 1 2 T T CML m d m d m d m d L m θ . (1.3) This CML, of course, corresponds to the case of independence between each pair of observations from the same individual. As we will indicate later, the above CML estimator is consistent even when there is dependence among the observations of the individual. However, this approach, in general, does not estimate the parameters representing the dependence effects across choices of the same individual (i.e., only a subset of the vector θ is estimable). A third approach to estimating the parameter vector θ in the repeated unordered choice case is to define the events in the CML as the pairwise observations across all or a subset of the choice occasions of the individual. For presentation ease, assume that all pairs of observations are considered. This leads to a pairwise CML function contribution of individual q as follows: ) , ( Prob ) , ( 1 1 1 3 t t t t T t T t t CML m d m d L m θ . (1.4) Almost all earlier research efforts employing the CML technique have used the pairwise approach, including Apanasovich et al. (2008), Varin and Vidoni (2009), Bhat and Sener (2009), Bhat et al. (2010a), Bhat and Sidharthan (2011), Vasdekis et al. (2012), Ferdous and Bhat (2013), and Feddag (2013). Alternatively, the analyst can also consider larger subsets of observations, such as triplets or quadruplets or even higher dimensional subsets (see Engler et al., 2006 and Caragea and Smith, 2007). However, the pairwise approach is a good balance between statistical and computational efficiency (besides, in almost all applications, the parameters characterizing error dependency are completely identified based on the pairwise approach). Importantly, the pairwise approach is able to explicitly recognize dependencies across choice occasions in the repeated choice case through the inter-temporal pairwise probabilities. 1.4. Asymptotic Properties of the CML Estimator with many independent replicates The asymptotic properties of the CML estimator for the case with many independent replicates may be derived from the theory of unbiased estimating functions. For ease, we will first consider the case when we have Q independent observational units (also referred to as individuals) in a sample , ,..., , , 3 2 1 Q Y Y Y Y each q Y (q=1,2,…,Q) being a H ~ × 1 vector. That is, ).
,..., , ( ~ 2 1 H q q q Y Y Y q Y H ~ in this context may refer to multiple observations of the same variable on the same observation unit (as in the previous section) or a single observation of multiple variables for the observation unit (for example, expenditures on groceries, transportation, and leisure activities for an individual). In either case, Q is large relative to H ~ (the case when Q is small is considered in the next section). We consider the case when observation is made directly on each of the continuous variables , qh Y though the discussion in this section is easily modified 4 to incorporate the case when observation is made on some truncated or censored form of qh Y (such as in the case of a discrete choice variable). Let the observation on the random variable q Y be ).
,..., , ( ~ 2 1 H q q q y y y q y Define ).
,..., , ( 2 1 Q y y y y Also, we will consider a pairwise likelihood function as the CML estimator, though again the proof is generalizable in a straightforward manner to other types of CML estimators (such as using triplets or quadruplets rather than couplets in the CML). For the pairwise case, the estimator is obtained by maximizing (with respect to the unknown parameter vector θ , which is of dimension K ~ ) the logarithm of the following function: Q q Q q h q qh H h h qh H h h h qh H h H h h h q qh Q q H h h q h q qh qh H h h CML y y f L L y y f y Y y Y L 1 1 1 ~ 1 ~ 1 1 ~ 1 ~ 1 1 1 ~ 1 ~ 1 ) , ( where , ) , ( ) , ( Prob ) , ( y θ (1.5) Under usual regularity conditions (these are the usual conditions needed for likelihood objects to ensure that the logarithm of the CML function can be maximized by solving the corresponding score equations; the conditions are too numerous to mention here, but are listed in Molenberghs and Verbeke, 2005, page 191), the maximization of the logarithm of the CML function in the equation above is achieved by solving the composite score equations given by: , ) , , ( ) , ( log ) , ( 1 1 ~ 1 ~ 1 0 Q q h q qh h qh H h H h h CML y y L θ s y θ y θ sCML (1.6) where .
log ) , , ( θ θ s h qh h q qh h qh L y y Since the equations ) , ( y θ sCML are linear combinations of valid likelihood score functions ) , , ( h q qh h qh y y θ s associated with the event probabilities forming the composite log-likelihood function, they immediately satisfy the requirement of being unbiased. While this is stated in many papers and should be rather obvious, we provide a formal proof of the unbiasedness of the CML score equations (see also Yi et al., 2011). In particular, we need to prove the following: 0 Q q h q qh h qh H h H h h Q q h q qh h qh H h H h h y y E y y E E 1 1 ~ 1 ~ 1 1 1 ~ 1 ~ 1 ) , , ( ) , , ( )] , ( [ θ s θ s y θ sCML , (1.7) where the expectation above is taken with respect to the full distribution of ).
,..., , ( ~ 2 1 H Y Y Y Y The above equality will hold if )] , , ( [ h q qh h qh y y E θ s =0 for all pairwise combinations h h and for each q. To see that this is the case, we write: 5 qd d q y h q qh h q qh h qh y h qh h q qh h qh dy dy y y f L f L y y E h qh -h qh -y y q q dy y , θ )dy y θ θ s d qd -q ) , ( log ( log )] , , ( [ , (1.8) where h qh -y represents the subvector of q y with the elements qh y and h q y excluded. Continuing, 0 ) 1 ( 1 log ) , ( log ) , ( log )] , , ( [ θ θ θ θ θ θ dy y , θ θ s h qh -h qh -y h qh -qh h q qh h q qh h q qh h q qh h q qh h q y y h q qh h qh y y h q qh h qh y y h q qh h qh h qh h qh y y h q qh h qh h qh y y h q qh h q qh h qh y h q qh h q qh y h qh h q qh h qh dy dy L dy dy L dy dy L L L dy dy L L dy dy y y f L dy dy y y f L y y E (1.9) Next, consider the asymptotic properties of the CML estimator. To derive these, define the mean composite score function across observation units in the sample as follows: ), , ( 1 ) , ( 1 q q y θ s y θ s Q q Q where ) , ( q q y θ s ) , , ( 1 ~ 1 ~ 1 h q qh h qh H h H h h y y θ s . Then, 0 ) , , ( ) , ( 1 ~ 1 ~ 1 h q qh h qh H h H h h y y E E θ s y θ s q q for all values of θ . Let 0 θ be the true unknown parameter vector value, and consider the score function at this vector value and label it as ).
, ( q 0 q y θ s Then, when drawing a sample from the population, the analyst is essentially drawing values of ) , ( q 0 q y θ s from its distribution in the population with zero mean and variance given by ) , ( q 0 q y θ s J Var , and taking the mean across the sampled values of ) , ( q 0 q y θ s to obtain ).
, ( y θ s 0 Invoking the Central Limit Theorem (CLT), we have ) 0,J y θ s 0 ( ) , ( ~ K d MVN Q (1.10) where (.,.) ~ K MVN stands for the multivariate normal distribution of K ~ dimensions. Next, let CML θ ˆ be the CML estimator, so that, by design of the CML estimator, .
) , ˆ ( 0 y θ s CML Expanding ) , ˆ ( y θ s CML around ) , ( y θ s 0 in a first-order Taylor series, we obtain 0 CML 0 0 CML θ θ y θ s y θ s y θ s ˆ ) , ( ) , ( ) , ˆ ( 0 , or equivalently, 1 ) , ( ˆ y θ s θ θ 0 0 CML Q Q ) , ( y θ s 0 . (1.11) 6 From the law of large numbers (LLN), we also have that ) , ( y θ s 0 , which is the sample mean of ), , ( q 0 q y θ s converges to the population mean for the quantity. That is, ) , ( y θ s 0 ) , ( y θ s H 0 E d (1.12) Using Equations (1.10) and (1.12) in Equation (1.11), applying Slutsky’s theorem, and assuming non-singularity of J and H , we finally arrive at the following limiting distribution: H HJ G G θ θ -1 -1 0 CML where ), ( ˆ ~ 0, K d MVN Q (1.13) where G is the Godambe (1960) information matrix. Thus, the asymptotic distribution of CML θ ˆ is centered on the true parameter vector 0 θ . Further, the variance of CML θ ˆ reduces as the number of sample points Q increases. The net result is that CML θ ˆ converges in probability to 0 θ as Q (with H ~ fixed), leading to the consistency of the estimator. In addition, CML θ ˆ is normally distributed, with its covariance matrix being .
/Q -1 G However, both J and H , and therefore G, are functions of the unknown parameter vector 0 θ . But J and H may be estimated in a straightforward manner at the CML estimate CML θ ˆ as follows: CML θ θ θ J ˆ , , 1 log log 1 ˆ q CML q CML Q q L L Q , where h qh H h H h h q CML L L log log 1 ~ 1 ~ 1 , , (1.14) and CML CML CML θ θ θ q q θ θ θ s y θ s H ˆ 1 1 ~ 1 ~ 1 2 1 1 ~ 1 ~ 1 ˆ 1 ˆ log 1 ) , , ( 1 ) , ( 1 ˆ Q q H h H h h qh Q q H h H h h q qh d qd Q q L Q y y Q Q (1.15) If computation of the second derivative is time consuming, one can exploit the second Bartlett identity (Ferguson, 1996, page 120), which is valid for each observation unit’s likelihood term in the composite likelihood. That is, using the condition that , , , ( , , ( , , ( h q qh h qh h q qh h qh h q qh h qh y y E y y E y y Var 0 0 q 0 q θ s θ s H θ s J (1.16) an alternative estimate for H ˆ is as below: 7 Q q H h H h h qh h qh Q q H h H h h q qh h qh h q qh h qh Q q H h H h h q qh h qh L L Q y y y y Q y y Var Q 1 1 ~ 1 ~ 1 ˆ 1 1 ~ 1 ~ 1 ˆ 1 1 ~ 1 ~ 1 ˆ log log 1 , , ( , , ( 1 , , ( 1 ˆ CML CML CML θ θ 0 0 θ 0 θ θ θ s θ s θ s H (1.17) Finally, the covariance matrix of the CML estimator is given by .
ˆ ˆ ˆ ˆ Q Q 1 -1 -1 -H J H G The empirical estimates above can be imprecise when Q is not large enough. An alternative procedure to obtain the covariance matrix of the CML estimator is to use a jackknife approach as follows (see Zhao and Joe, 2005): , θ θ θ θ θ CML CML CML CML CML ˆ ˆ ˆ ˆ 1 ) ˆ ( Cov ) ( ) ( 1 q q Q q Q Q (1.18) where ) ( ˆ q CML θ is the CML estimator with the qth observational unit dropped from the data. However, this can get time-consuming, and so an alternative would be to use a first-order approximation for ) ( ˆ q CML θ with a single step of the Newton-Raphson algorithm with CML θ ˆ as the starting point. 1.5. Asymptotic Properties of the CML Estimator for the Case of Very Few or No Independent Replicates Even in the case when the data include very few or no independent replicates (as would be the case with global social or spatial interactions across all observational units in a cross-sectional data in which the dimension of H ~ is equal to the number of observational units and Q=1), the CML estimator will retain the good properties of being consistent and asymptotically normal as long as the data is formed by pseudo-independent and overlapping subsets of observations (such as would be the case when the social interactions taper off relatively quickly with the social separation distance between observational units, or when spatial interactions rapidly fade with geographic distance based on an autocorrelation function decaying toward zero; see Cox and Reid, 2004 for a technical discussion).2 The same situation holds in cases with temporal processes; the CML estimator will retain good properties as long as we are dealing with a stationary time series with short-range dependence (the reader is referred to Davis and Yau, 2011 and Wang et al., 2013 for additional discussions of the asymptotic properties of the CML estimator for the case of time-series and spatial models, respectively). 2 Otherwise, there may be no real solution to the CML function maximization and the usual asymptotic results will not hold. 8 The covariance matrix of the CML estimator needs estimates of J and H. The “bread” matrix H can be estimated in a straightforward manner using the Hessian of the negative of ) ( log θ CML L , evaluated at the CML estimate θ ˆ. This is because the information identity remains valid for each pairwise term forming the composite marginal likelihood. But the estimation of the “vegetable” matrix J is more involved. Further details of the estimation of the CML estimator’s covariance matrix for the case with spatial data are discussed in Section 2.3. 1.6. Relative Efficiency of the CML Estimator The CML estimator loses some asymptotic efficiency from a theoretical perspective relative to a full likelihood estimator, because information embedded in the higher dimension components of the full information estimator are ignored by the CML estimator. This can also be formally shown by starting from the CML unbiased estimating functions 0 )] , ( [ y θ s 0 CML E , which can be written as follows (we will continue to assume continuous observation on the variable vector of interest, so that Y is a continuous variable, though the presentation is equally valid for censored and truncated observations on Y ): 0 0 θ θ y θ θ y 0 CML dy θ y)dy θ y θ s ML CML CML L L f L E log ( log )] , ( [ 0 (1.19) Take the derivative of the above function with respect to θ to obtain the following: 0 θ θ y dy θ θ ML CML L L log 2 0 θ y CML L log 0 θ θ dy θ ML ML L L log (1.20) ) , ( ) , ( )] , ( [ y θ s y θ s y θ s 0 ML 0 CML 0 CML E E , where y) θ s 0 ML , ( is the score function of the full likelihood. From above, we get the following: )] , ( ), , ( [ Cov ) , ( y θ s y θ s y θ s H 0 CML 0 ML 0 CML E , and ] ), , ( ), , ( [ Cov )) , ( ( Var )] , ( ), , ( [ Cov 1 y θ s y θ s y θ s y θ s y θ s G 0 ML 0 CML 0 CML 0 CML 0 ML (1.21) Then, using the multivariate version of the Cauchy-Schwartz inequality (Lindsay, 1988), we obtain the following: .
)] , ( [ Var G y θ s IFISHER 0 ML (1.22) Thus, from a theoretical standpoint, the difference between the regular ML information matrix (i.e., IFISHER) and the Godambe information matrix (i.e., G ) is positive definite, which implies that the difference between the asymptotic variances of the CML estimator and the ML estimator is positive semi-definite (see also Cox and Reid, 2004). However, many studies have found that the efficiency loss of the CML estimator (relative to the maximum likelihood (ML) 9 estimator) is negligible to small in applications. These studies are either based on precise analytic computations of the information matrix IFISHER and the Godambe matrix G to compare the asymptotic efficiencies from the ML and the CML methods, or based on empirical efficiency comparisons between the ML and CML methods for specific contexts by employing a simulation design with finite sample sizes. A brief overview of these studies is presented in the next section. 1.6.1. Comparison of ML and CML Estimator Efficiencies Examples of studies that have used precise analytic computations to compare the asymptotic efficiency of the ML and CML estimators include Cox and Reid (2004), Hjort and Varin (2008), and Mardia et al. (2009). Cox and Reid (2004) derive IFISHER and G for some specific situations, including the case of a sample of independent and identically distributed vectors, each of which is multivariate normally distributed with an equi-correlated structure between elements. In the simple cases they examine, they show that the loss of efficiency between IFISHER and G is of the order of 15%. They also indicate that in the specific case of Cox’s (1972) quadratic exponential distribution-based multivariate binary data model, the full likelihood function and a pairwise likelihood function for binary data generated using a probit link are equivalent, showing that the composite likelihood estimator can achieve the same efficiency as that of a full maximum likelihood estimator. Hjort and Varin (2008) also study the relationship between the IFISHER and G matrices, but for Markov chain models, while Mardia et al. (2007) and Mardia et al. (2009) examine efficiency considerations in the context of multivariate vectors with a distribution drawn from closed exponential families. These studies note special cases when the composite likelihood estimator is fully efficient, though all of these are rather simplified model settings. Several papers have also analytically studied efficiency considerations in clustered data, especially the case when each cluster is of a different size (such as in the case of spatially clustered data from different spatial regions with different numbers of observational units within each spatial cluster, or longitudinal data on observational units with each observational unit contributing a different number of sample observations). In such situations, the unweighted CML function will give more weight to clusters that contribute more sample observations than those with fewer observations. To address this situation, a weighted CML function may be used. Thus, Le Cessie and Van Houwelingen (1994) suggest, in their binary data model context, that each cluster should contribute about equally to the CML function. This may be achieved by power-weighting each cluster’s CML contribution by a factor that is the inverse of the number of choice occasions minus one. The net result is that the composite likelihood contribution of each cluster collapses to the likelihood contribution of the cluster under the case of independence within a cluster. In a general correlated panel binary data context, Kuk and Nott (2000) confirmed the above result for efficiently estimating parameters not associated with dependence within clusters for the case when the correlation is close to zero. However, their analysis suggested that the unweighted CML function remains superior for estimating the correlation (within cluster) parameter. In a relatively more recent paper, Joe and Lee (2009) theoretically 10 studied the issue of efficiency in the context of a simple random effect binary choice model. They indicate that the weights suggested by Le Cessie and Van Houwelingen (1994) and Kuk and Nott (2000) can provide poor efficiency even for non-dependence parameters when the correlation between pairs of the underlying latent variables for the “repeated binary choices over time” case they studied is moderate to high. Based on analytic and numeric analyses using a longitudinal binary choice model with an autoregressive correlation structure, they suggest that using a weight of 1 1 )] 1 ( 5 .
0 1 [ ) 1 ( q q T T for a cluster appears to do well in terms of efficiency for all parameters and across varying dependency levels ( q T is the number of observations contributed by unit or individual q). Further, the studies by Joe and Lee (2009) and Varin and Vidoni (2006), also in the context of clustered data, suggest that the inclusion of too distant pairings in the CML function can lead to a loss of efficiency. A precise analytic computation of the asymptotic efficiencies of the CML and full maximum likelihood approaches, as just discussed, is possible only for relatively simple models with or without clustering. This, in turn, has led to the examination of the empirical efficiency of the CML approach using simulated data sets for more realistic model contexts. Examples include Renard et al. (2004), Fieuws and Verbeke (2006), and Eidsvik et al. (2014). These studies indicate that the CML estimator performs well relative to the ML estimator. For instance, Renard et al. (2004) examined the performance of CML and ML estimators in the context of a random coefficients binary choice model, and found an average loss of efficiency of about 20% in the CML parameter estimates relative to the ML parameter estimates. Fiews and Verbeke (2006) examined the performance of the CML and ML estimators in the context of a multivariate linear model based on mixing, where the mixing along each dimension involves a random coefficient vector followed by a specification of a general covariance structure across the random coefficients of different dimensions. They found that the average efficiency loss across all parameters was less than 1%, and the highest efficiency loss for any single parameter was of the order of only 5%. Similarly, in simulated experiments with a spatial Gaussian process model, Eidsvik et al. (2014) used a spatial blocking strategy to partition a large spatially correlated space of a Gaussian response variable to estimate the model using a CML technique. They too found rather small efficiency losses because of the use of the CML as opposed to the ML estimator. However, this is an area that needs much more attention both empirically and theoretically. Are there situations when the CML estimator’s loss is less or high relative to the ML estimator, and are we able to come up with some generalizable results from a theoretical standpoint that apply not just to simple models but also more realistic models used in the field? In this regard, is there a “file drawer” problem where results are not being reported when the CML estimator in fact loses a lot of efficiency? Or is the current state of reporting among scholars in the field a true reflection of the CML estimator’s loss in efficiency relative to the ML? So far, the CML appears to be remarkable in its ability to pin down parameters, but there needs to be much more exploration in this important area. This opens up an exciting new direction of research and experimentation. 11 1.6.2. Comparison of Maximum Simulated Likelihood (MSL) and CML Estimator Efficiencies The use of the maximum likelihood estimator is feasible for many types of models. But the estimation of many other models that incorporate analytically intractable expressions in the likelihood function in the form of integrals, such as in mixed multinomial logit models or multinomial probit models or count models with certain forms of heterogeneity or large-dimensional multivariate dependency patterns (just to list a few), require an approach to empirically approximate the intractable expression. This is usually done using simulation techniques, leading to the MSL inference approach (see Train, 2009), though quadrature techniques are also sometimes used for cases with 1-3 dimensions of integrals in the likelihood function expression. When simulation methods have to be used to evaluate the likelihood function, there is also a loss in asymptotic efficiency in the maximum simulated likelihood (MSL) estimator relative to a full likelihood estimator. Specifically, McFadden and Train (2000) indicate, in their use of independent number of random draws across observations, that the difference between the asymptotic covariance matrix of the MSL estimator obtained as the inverse of the sandwich information matrix and the asymptotic covariance matrix of the ML estimator obtained as the inverse of the cross-product of first derivatives is theoretically positive semi-definite for finite number of draws per observation. Consequently, given that we also know that the difference between the asymptotic covariance matrices of the CML and ML estimators is theoretically positive semi-definite, it is difficult to state from a theoretical standpoint whether the CML estimator efficiency will be higher or lower than the MSL estimator efficiency. However, in a simulation comparison of the CML and MSL methods for multivariate ordered response systems, Bhat et al. (2010b) found that the CML estimator’s efficiency was almost as good as that of the MSL estimator, but with the benefits of a very substantial reduction in computational time and much superior convergence properties. As they state “….any reduction in the efficiency of the CML approach relative to the MSL approach is in the range of non-existent to small”. Paleti and Bhat (2013) examined the case of panel ordered-response structures, including the pure random coefficients (RC) model with no autoregressive error component, as well as the more general case of random coefficients combined with an autoregressive error component. The ability of the MSL and CML approaches to recover the true parameters is examined using simulated datasets. The results indicated that the performances of the MSL approach (with 150 scrambled and randomized Halton draws) and the simulation-free CML approach were of about the same order in all panel structures in terms of the absolute percentage bias (APB) of the parameters and empirical efficiency. However, the simulation-free CML approach exhibited no convergence problems of the type that affected the MSL approach. At the same time, the CML approach was about 5-12 times faster than the MSL approach for the simple random coefficients panel structure, and about 100 times faster than the MSL approach when an autoregressive error component was added. Thus, the CML appears to lose relatively little by way of efficiency, while also offering a more stable and much faster estimation approach in the panel ordered-ordered-response context. Similar results of substantial computational 12 efficiency and little to no finite sample efficiency loss (and sometimes even efficiency gains) have been reported by Bhat and Sidharthan (2011) for cross-sectional and panel unordered-response multinomial probit models with random coefficients (though the Bhat and Sidharthan paper actually combines the CML method with a specific analytic approximation method to evaluate the multivariate normal cumulative distribution function). Finally, the reader will note that there is always some simulation bias in the MSL method for finite number of simulation draws, and the consistency of the MSL method is guaranteed only when the number of simulation draws rises faster than the square root of the sample size (Bhat, 2001, McFadden and Train, 2000). The CML estimator, on the other hand, is unbiased and consistent under the usual regularity conditions, as discussed earlier in Section 1.4. 1.7. Robustness of Consistency of the CML Estimator As indicated by Varin and Vidoni (2009), it is possible that the “maximum CML estimator can be consistent when the ordinary full likelihood estimator is not”. This is because the CML procedures are typically more robust and can represent the underlying low-dimensional process of interest more accurately than the low dimensional process implied by an assumed (and imperfect) high-dimensional multivariate model. Another way to look at this is that the consistency of the CML approach is predicated only on the correctness of the assumed lower dimensional distribution, and not on the correctness of the entire multivariate distribution. On the other hand, the consistency of the full likelihood estimator is predicated on the correctness of the assumed full multivariate distribution. Thus, for example, Yi et al. (2011) examined the performance of the CML (pairwise) approach in the case of clustered longitudinal binary data with non-randomly missing data, and found that the approach appears quite robust to various alternative specifications for the missing data mechanism. Xu and Reid (2011) provided several specific examples of cases where the CML is consistent, while the full likelihood inference approach is not. 1.8. Model Selection in the CML Inference Approach Procedures similar to those available with the maximum likelihood approach are also available for model selection with the CML approach. The statistical test for a single parameter may be pursued using the usual t-statistic based on the inverse of the Godambe information matrix. When the statistical test involves multiple parameters between two nested models, an appealing statistic, which is also similar to the likelihood ratio test in ordinary maximum likelihood estimation, is the composite likelihood ratio test (CLRT) statistic. Consider the null hypothesis 0 τ τ : 0 H against 0 τ τ : 1 H , where is a subvector of θ of dimension d ~ ; i.e., ) , ( α τ θ . The statistic takes the familiar form shown below: )], ˆ ( log ) ˆ ( [log 2 R θ θ CML CML L L CLRT (1.23) 13 where R θ ˆ is the composite marginal likelihood estimator under the null hypothesis )) ( ˆ , ( 0 0 τ α τ CML . More informally speaking, θ ˆ is the CML estimator of the unrestricted model, and R θ ˆ is the CML estimator for the restricted model. The CLRT statistic does not have a standard chi-squared asymptotic distribution. This is because the CML function that is maximized does not correspond to the parametric model from which the data originates; rather, the CML may be viewed in this regard as a “mis-specification” of the true likelihood function because of the independence assumption among the likelihood objects forming the CML function (see Kent, 1982, Section 3). To write the asymptotic distribution of the CLRT statistic, first define 1 )] ( [ θ G and 1 )] ( [ θ H as the d d ~ ~ submatrices of 1 )] ( [ θ G and 1 )] ( [ θ H , respectively, which correspond to the vector τ . Then, the CLRT has the following asymptotic distribution: 2 ~ 1 ~ ~ i i d i W CLRT , (1.24) where 2 ~ i W for i = 1, 2, …,d ~ are independent 2 1 variates and d ...
2 1 are the eigenvalues of the matrix 1 )] ( )][ ( [ θ G θ H evaluated under the null hypothesis (this result may be obtained based on the (profile) likelihood ratio test for a mis-specified model; see Kent, 1982, Theorem 3.1 and the proof therein). Unfortunately, the departure from the familiar asymptotic chi-squared distribution with d ~ degrees of freedom for the traditional maximum likelihood procedure is annoying. Pace et al. (2011) have recently proposed a way out, indicating that the following adjusted CLRT statistic, ADCLRT, may be considered to be asymptotically chi-squared distributed with d ~ degrees of freedom: CLRT ADCLRT ) ( )] ( [ ] ) ( [ ) ( )] ( )][ ( [ )] ( [ ] ) ( [ 1 1 1 θ S θ H θ S θ S θ H θ G θ H θ S τ τ τ τ τ τ τ τ (1.25) where ) (θ Sτ is the 1 ~ d submatrix of ) (θ S θ θ) ( log CML L corresponding to the vector τ , and all the matrices above are computed at R θ ˆ . The denominator of the above expression is a quadratic approximation to CLRT, while the numerator is a score-type statistic with an asymptotic 2 ~ d null distribution. Thus, ADCLRT is also very close to being an asymptotic 2 ~ d distribution under the null. Alternatively, one can resort to parametric bootstrapping to obtain the precise distribution of the CLRT statistic for any null hypothesis situation. Such a bootstrapping procedure is rendered simple in the CML approach, and can be used to compute the p-value of the null hypothesis test. The procedure is as follows: 14 1. Compute the observed CLRT value as in Equation (1.23) from the estimation sample. Let the estimation sample be denoted as obs y ~ , and the observed CLRT value as ).
~ ( obs y CLRT 2. Generate C sample data sets C y y y y ~ ,..., ~ , ~ , ~ 3 2 1 using the CML convergent values under the null hypothesis 3. Compute the CLRT statistic of Equation (1.23) for each generated data set, and label it as ).
~ ( c y CLRT 4. Calculate the p-value of the test using the following expression: , 1 ) ~ ( ) ~ ( 1 1 C CLRT CLRT I p C c obs c y y where 1 } { A I if A is true. (1.26) The above bootstrapping approach has been used for model testing between nested models in Varin and Czado (2010), Bhat et al. (2010b), and Ferdous et al. (2010). When the null hypothesis entails model selection between two competing non-nested models, the composite likelihood information criterion (CLIC) introduced by Varin and Vidoni (2005) may be used. The CLIC takes the following form3: 1 ) ˆ ( ˆ ) ˆ ( ˆ ) ˆ ( log ) ˆ ( log θ H θ J θ θ tr L L CML CML (1.27) The model that provides a higher value of CLIC is preferred. 1.9. Positive-Definiteness of the Implied Multivariate Covariance Matrix In cases where the CML approach is used as a vehicle to estimate the parameters in a higher dimensional multivariate covariance matrix, one has to ensure that the implied multivariate covariance matrix in the higher dimensional context is positive definite. For example, consider a multivariate ordered-response model context, and let the latent variables underlying the multivariate ordered-response model be multivariate normally distributed. This symmetric covariance (correlation) matrix Σ has to be positive definite (that is, all the eigenvalues of the matrix should be positive, or, equivalently, the determinant of the entire matrix and every principal submatrix of Σ should be positive). But the CML approach does not estimate the entire correlation matrix as one single entity. However, there are three ways that one can ensure the positive-definiteness of the Σ matrix. The first technique is to use Bhat and Srinivasan’s (2005) strategy of reparameterizing the correlation matrix Σ through the Cholesky matrix, and then using these Cholesky-decomposed parameters as the ones to be estimated. That is, the Cholesky of an initial positive-definite specification of the correlation matrix is taken before starting the optimization routine to maximize the CML function. Then, within the optimization procedure, one can reconstruct the Σ matrix, and then pick off the appropriate elements of this matrix to 3 This penalized log-composite likelihood is nothing but the generalization of the usual Akaike’s Information Criterion (AIC). In fact, when the candidate model includes the true model in the usual maximum likelihood inference procedure, the information identity holds (i.e., H(θ) = J(θ)) and the CLIC in this case is exactly the AIC [ ) ˆ ( log θ ML L (# of model parameters)]. 15 construct the CML function at each iteration. This is probably the most straightforward and clean technique. The second technique is to undertake the estimation with a constrained optimization routine by requiring that the implied multivariate correlation matrix for any set of pairwise correlation estimates be positive definite. However, such a constrained routine can be extremely cumbersome. The third technique is to use an unconstrained optimization routine, but check for positive-definiteness of the implied multivariate correlation matrix. The easiest method within this third technique is to allow the estimation to proceed without checking for positive-definiteness at intermediate iterations, but check that the implied multivariate correlation matrix at the final converged pairwise marginal likelihood estimates is positive-definite. This will typically work for the case of a multivariate ordered-response model if one specifies exclusion restrictions (i.e., zero correlations between some error terms) or correlation patterns that involve a lower dimension of effective parameters. However, if the above simple method of allowing the pairwise marginal estimation approach to proceed without checking for positive definiteness at intermediate iterations does not work, then one can check the implied multivariate correlation matrix for positive definiteness at each and every iteration. If the matrix is not positive-definite during a direction search at a given iteration, one can construct a “nearest” valid correlation matrix (for example, by replacing the negative eigenvalue components in the matrix with a small positive value, or by adding a sufficiently high positive value to the diagonals of a matrix and normalizing to obtain a correlation matrix; see Rebonato and Jaeckel, 1999, Higham, 2002, and Schoettle and Werner, 2004 for detailed discussions of these and other adjusting schemes; a review of these techniques is beyond the scope of this monograph). The values of this “nearest” valid correlation matrix can be translated to the pairwise correlation estimates, and the analyst can allow the iterations to proceed and hope that the final implied convergent correlation matrix is positive-definite. 1.10. The Maximum Approximate Composite Marginal Likelihood Approach In many application cases, the probability of observing the lower dimensional event itself in a CML approach may entail multiple dimensions of integration. For instance, in the case of a multinomial probit model with I choice alternatives per individual (assume for ease in presentation that all individuals have all I choice alternatives), and a spatial dependence structure (across individuals) in the utilities of each alternative, the CML approach involves compounding the likelihood of the joint probability of the observed outcomes of pairs of individuals. However, this joint probability itself entails the integration of a multivariate normal cumulative distribution (MVNCD) function of dimension equal to ) 1 ( 2 I . The evaluation of such an integration function cannot be pursued using quadrature techniques due to the curse of dimensionality when the dimension of integration exceeds two (see Bhat, 2003). In this case, the MVNCD function evaluation for each agent has to be evaluated using simulation or other analytic approximation techniques. Typically, the MVNCD function is approximated using simulation techniques through the use of the Geweke-Hajivassiliou-Keane (GHK) simulator or the Genz-Bretz (GB) simulator, which are among the most effective simulators for evaluating the MVNCD function 16 (see Bhat et al., 2010b for a detailed description of these simulators). Some other sparse grid-based techniques for simulating the multivariate normal probabilities have also been proposed by Heiss and Winschel (2008), Huguenin et al. (2009), and Heiss (2010). In addition, Bayesian simulation using Markov Chain Monte Carlo (MCMC) techniques (instead of MSL techniques) have been used in the literature (see Albert and Chib, 1993, McCulloch and Rossi, 2000, and Train, 2009). However, all these MSL and Bayesian techniques require extensive simulation, are time-consuming, are not very straightforward to implement, and create convergence assessment problems as the number of dimensions of integration increases. Besides, they do not possess the simulation-free appeal of the CML function in the first place. To accommodate the situation when the CML function itself may involve the evaluation of MVNCD functions, Bhat (2011) proposed a combination of an analytic approximation method to evaluate the MVNCD function with the CML function, and labeled this as the Maximum Approximate Composite Marginal Likelihood (MACML) approach. While several analytic approximations have been reported in the literature for MVNCD functions (see, for example, Solow, 1990, Joe, 1995, Gassmann et al., 2002, and Joe, 2008), the one Bhat proposes for his MACML approach is based on decomposition into a product of conditional probabilities. Similar to the CML approach that decomposes a large multidimensional problem into lower level dimensional components, the analytic approximation method also decomposes the MVNCD function to involve only the evaluation of lower dimensional univariate and bivariate normal cumulative distribution functions. Thus, there is a type of conceptual consistency in Bhat’s proposal of combining the CML method with the MVNCD analytic approximation. The net result is that the approximation approach is fast and lends itself nicely to combination with the CML approach. Further, unlike Monte-Carlo simulation approaches, even two to three decimal places of accuracy in the analytic approximation is generally adequate to accurately and precisely recover the parameters and their covariance matrix estimates because of the smooth nature of the first and second derivatives of the approximated analytic log-likelihood function. The MVNCD approximation used by Bhat for discrete choice mode estimation itself appears to have been first proposed by Solow (1990) based on Switzer (1977), and then refined by Joe (1995). However, the focus of the earlier studies was on computing a single MVNCD function accurately rather than Bhat’s use of the approximation for choice model estimation where multiple MVNCD function evaluations are needed. To describe the MVNCD approximation, let ) ,..., , , ( 3 2 1 I W W W W be a multivariate normally distributed random vector with zero means, variances of 1, and a correlation matrix Σ . Then, interest centers on approximating the following orthant probability: ) ..., , , , ( Pr ) ( Pr 3 3 2 2 1 1 I I w W w W w W w W w W . (1.28) The above joint probability may be written as the product of a bivariate marginal probability and univariate conditional probabilities as follows (I ≥ 3): 17 . ) ..., , , , | ( Pr ) , ( Pr ) ( Pr 1 1 3 3 2 2 1 1 I 3 i 2 2 1 1 i i i i w W w W w W w W w W w W w W w W (1.29) Next, define the binary indicator i I ~ that takes the value 1 if i i w W and zero otherwise. Then ) ( ) ~ ( i i w I E , where (.) is the univariate normal standard cumulative distribution function. Also, we may write the following: , )] ( 1 )[ ( ) ( ) ( ) ~ ( ) ~ , ~ ( Cov ), ( ) ( ) , , ( ) ~ ( ) ~ ( ) ~ ~ ( ) ~ , ~ ( Cov 2 2 i i i i i i i j i ij j i j i j i j i w w w w I Var I I j i w w w w I E I E I I E I I (1.30) where ij is the ijth element of the correlation matrix Σ . With the above preliminaries, consider the following conditional probability: . ) 1 ~ ..., , 1 ~ , 1 ~ , 1 ~ | ~ ( ) ..., , , , | ( Pr 1 3 2 1 1 1 3 3 2 2 1 1 i i i i i i I I I I I E w W w W w W w W w W (1.31) The right side of the expression may be approximated by a linear regression model, with i I ~ being the “dependent” random variable and ) ~ ,...
~ , ~ ( ~ 1 2 1 i I I I i I being the independent random variable vector.4 In deviation form, the linear regression for approximating Equation (1.31) may be written as: ~ )] ~ ( ~ [ ) ~ ( ~ i i I I α E I E I i i , (1.32) where α is the least squares coefficient vector and ~ is a mean zero random term. In this form, the usual least squares estimate of α is given by: i i i α , 1 Ω Ω ˆ , where (1.33) 4 This first-order approximation can be continually improved by increasing the order of the approximation. For instance, a second-order approximation would approximate the right side of Equation (1.31) by the expectation from a linear regression model that has i I ~ as the “dependent” random variable and ) , ~ , ~ , ~ , , ~ , ~ , ~ , ~ , ~ ( 1 , 2 1 , 2 24 23 1 , 1 13 12 1 2 1 i i i i i I I I I I I I I I I i I as the independent random variable vector, where .
~ ~ ~ j i j i I I I Essentially this adds second-order interactions in the independent random variable vector (see Joe, 1995). However, doing so entails trivariate and four-variate normal cumulative distribution function (CDF) evaluations (when I >4) as opposed to univariate and bivariate normal CDF evaluations in the first-order approximation, thus increasing computational burden. As discussed in Bhat (2011) and shown in Bhat and Sidharthan (2011), the first-order approximation is more than adequate (when combined with the CML approach) for estimation of MNP models. Thus, in the rest of this monograph, we will use the term approximation to refer to the first-order approximation evaluation of the MVNCD function. 18 ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) , ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) , ( Cov 1 1 3 1 2 1 1 1 1 3 3 3 2 3 1 3 1 2 3 2 2 2 1 2 1 1 3 1 2 1 1 1 i i i i i i i i I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I i i i I I Ω , and (1.34) ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) ~ , ~ ( Cov ) , ( Cov 1 3 2 1 i i i i i I I I I I I I I i i i i I I , Ω . Finally, putting the estimate of α ˆ back in Equation (1.32), and predicting the expected value of i I ~ conditional on 1 i I ~ (i.e., ) 1 ~ , 1 ~ , 1 ~ 1 2 1 i I I I , we get the following approximation for Equation (1.31): ) ) ( 1 )...
( 1 ), ( 1 ( ) ( ) ( ) ..., , , | ( Pr 1 2 1 1 1 2 2 1 1 i i i i i i w w w w w W w W w W w W i i i , 1 Ω Ω (1.35) This conditional probability approximation can be plugged into Equation (1.29) to approximate the multivariate orthant probability in Equation (1.28). The resulting expression for the multivariate orthant probability comprises only univariate and bivariate standard normal cumulative distribution functions. One remaining issue is that the decomposition of Equation (1.28) into conditional probabilities in Equation (1.29) is not unique. Further, different permutations (i.e., orderings of the elements of the random vector ) ,..., , , ( 3 2 1 I W W W W W ) for the decomposition into the conditional probability expression of Equation (1.29) will lead, in general, to different approximations. One approach to resolve this is to average across the 2 / !
I permutation approximations. However, as indicated by Joe (1995), the average over a few randomly selected permutations is typically adequate for the accurate computation of the multivariate orthant probability. In the case when the approximation is used for model estimation (where the integrand in each individual’s log-likelihood contribution is a parameterized function of the β and Σ parameters), even a single permutation of the W vector per choice occasion may suffice, as several papers in the literature have now shown (see later chapters). 19 2. APPLICATION TO TRADITIONAL DISCRETE CHOICE MODELS In this section, we will develop a blueprint (complete with matrix notation) for the use of the CML inference method to estimate traditional discrete choice models. The focus will be on two specific kinds of discrete choice models: Ordered-response models and unordered-response models. In the case when there are only two alternatives to choose from (the binary choice case), the ordered-response and the unordered-response formulations collapse to the same structure. But these formulations differ when extended to the multinomial (more than two alternatives) choice case. The next section provides a brief overview of ordered-response and unordered-response model systems. Section 2.2 then focuses on aspatial specifications within each type of discrete choice model, while Section 2.3 focuses on spatial specifications. Section 2.4 discusses applications of the CML method to count models. In each of Sections 2.2, 2.3, and 2.4, we provide a list of references of applications after presenting the formulation and CML estimation approach. Doing so allows us to present the model structure and estimation without unnecessary interspersing with references. The contents of the individual sections do inevitably draw quite substantially from the corresponding references of applications. Also, codes to estimate most of the models presented are available at (these codes are in the GAUSS matrix programming language). 2.1. Ordered and Unordered-Response Model Systems Ordered-response models are used when analyzing discrete outcome data with a finite number of mutually exclusive categories that may be considered as manifestations of an underlying scale that is endowed with a natural ordering. Examples include ratings data (of consumer products, bonds, credit evaluation, movies, etc.), or likert-scale type attitudinal/opinion data (of air pollution levels, traffic congestion levels, school academic curriculum satisfaction levels, teacher evaluations, etc.), or grouped data (such as bracketed income data in surveys or discretized rainfall data). In all of these situations, the observed outcome data may be considered as censored (or coarse) measurements of an underlying latent continuous random variable. The censoring mechanism is usually characterized as a partitioning or thresholding of the latent continuous variable into mutually exclusive (non-overlapping) intervals. The reader is referred to McKelvey and Zavoina (1975) and Winship and Mare (1984) for some early expositions of the ordered-response model formulation. The reader is also referred to Greene and Hensher (2010) for a comprehensive history and treatment of the ordered-response model structure. These reviews indicate the abundance of applications of the ordered-response model in the sociological, biological, marketing, and transportation sciences, and the list of applications only continues to grow rapidly. Unordered-response models are used when analyzing discrete outcome data with a finite number of mutually exclusive categories that do not represent any kind of ordinality. Examples include mode choice data or brand choice data or college choice data. In general, unordered-response models will include valuations (by decision-makers) of attributes that are alternative-specific. Most unordered-response models in economics and other fields are based on the 20 concept of utility-maximizing. That is, the attributes and individual characteristics are assumed to be translated into a latent utility index for each alternative, and the individual chooses the alternative that maximizes utility. The reader is referred to Train (2009) for a good exposition of the unordered-response model formulation. In general, the ordered-response formulation may be viewed as originating from a decision-rule that is based on the horizontal partitioning of a single latent variable, while the unordered-response formulation may be viewed as originating from a decision-rule that is based on the vertical comparison of multiple latent variables (one each for each alternative, that represents the composite utility of each alternative) to determine the maximum. A detailed theoretical comparison of the two alternatives is provided in Bhat and Pulugurta (1998). 2.2. Aspatial Formulations 2.2.1. Ordered-Response Models The applications of the ordered response model structure are quite widespread. The aspatial formulations of this structure may take the form of a cross-sectional univariate ordered-response probit (CUOP), a cross-sectional multivariate ordered-response probit (CMOP), or a panel multivariate ordered-response probit (PMOP). Within each of these formulations, many different versions are possible. In the discussion below, we present each formulation in turn in a relatively general form. 2.2.1.1 The CUOP Model Most applications of the ordered-response model structure are confined to the analysis of a single outcome at one point in time (that is, a cross-sectional analysis). Let q be an index for observation units or individuals (q = 1, 2,…, Q, where Q denotes the total number of individuals in the data set), and let k be the index for ordinal outcome category (k =1, 2,…, K). Let the actual observed discrete (ordinal) level for individual q be q m ( q m may take one of the K values; i.e., q m {1, 2,…, K}). In the usual ordered response framework notation, we may write the latent propensity ( q y ) for the ordered-response variable as a function of relevant covariates and relate this latent propensity to the ordinal outcome categories through threshold bounds: k y y q q q , q q x β if k q q k q y , 1 , , (2.1) where q x is an (L×1) vector of exogenous variables (not including a constant), q β is a corresponding (L×1) vector of individual-specific coefficients to be estimated, q is an idiosyncratic random error term that we will assume in the presentation below is independent of the elements of the vectors q β and q x , and k q ψ , is the individual-specific upper bound threshold for discrete level k ( 0 , q and q K q q q K q 1 , 2 , 1 , , ...
; in the usual ordered response fashion). The q terms are assumed independent and identical across 21 individuals. The typical assumption for q is that it is either normally or logistically distributed, though non-parametric or mixtures-of-normal distributions may also be considered. In this monograph, we will consider a normal distribution for q , because this has substantial benefits in estimation when q β is also considered to be multivariate normally distributed (or skew normally distributed, or mixtures of normal distributed). For identification reasons, the variance of q is normalized to one.5 Next, consider that the individual-specific thresholds are parameterized as a non-linear function of a set of variables q z (which does not include a constant), ) ( , q k k q f z . The non-linear nature of the functional form should ensure that (1) the thresholds satisfy the ordering condition (i.e., < ), 1 , 2 1 K q q q and (2) allows identification for any variables that are common in q x and q z . There are several plausible reasons provided in the ordered-response literature to motivate such varying thresholds across observation units, all of which originate in the realization that the set of thresholds represents a dimension to introduce additional heterogeneity over and beyond the heterogeneity already embedded in the latent variable q y . For instance, the threshold heterogeneity may be due to a different triggering mechanism (across individuals) for the translation (mapping) of the latent underlying q y propensity variable to observed ordinal data or different perceptions (across respondents) of response categories in a survey. Such generalized threshold models are referred to by different names based on their motivating origins, but we will refer to them in the current monograph as generalized ordered-response probit (GORP) models. Following Eluru et al. (2008), we parameterize the thresholds as: ) exp( 1 , , q k k q k q α z γk (2.2) In the above equation, k is a scalar, and k γ is a vector of coefficients associated with ordinal level 1 ,..., 2 , 1 K k . The above parameterization immediately guarantees the ordering condition on the thresholds for each and every individual, while also enabling the identification of parameters on variables that are common to the q x and q z vectors. For identification reasons, we adopt the normalization that ,1 q = 1 exp( ) for all q (equivalently, all elements of the vector 1 γ are normalized to zero, which is innocuous as long as the vector q x is included in the risk propensity equation). 5 The exclusion of a constant in the vector xq of Equation (2.1) is an innocuous normalization as long as all the intermediate thresholds (ψ1 through ψK–1) are left free for estimation. Similarly, the use of the standard normal distribution rather than a non-standard normal distribution for the error term is also an innocuous normalization (see Zavoina and McKelvey, 1975; Greene and Hensher, 2010). 22 Finally, to allow for unobserved response heterogeneity among observations, the parameter q β is defined as a realization from a multivariate normal distribution with mean vector b and covariance matrix , L L Ω where L is the lower-triangular Cholesky factor of Ω.6 Then, we can write , ~ q q β b β where ) , ( ~ ~ Ω 0 L q MVN β ( L MVN represents the multivariate normal distribution of dimension L). If this multivariate distribution becomes degenerate, then q q b β , and the Random Coefficients-Generalized Ordered Response Probit (RC-GORP) model collapses to the Generalized Ordered Response Probit (GORP) model. Further, in the GORP model, if all elements of k γ are zero for all k, the result is the standard ordered-response probit (SORP) model. The CUOP model of Equation (2.1) may be written as: k y y q q q q , ~ q q x β x b if k q q k q y , 1 , . (2.3) Then, the latent variable is univariate normally distributed as ), , ( ~ 2 q q q B N y where q x b q B and .
1 2 q q x x Ω q (2.4) Estimation is straightforward in this case using the maximum likelihood method. The parameter vector to be estimated in the model is , ) , , , , ( α γ b θ Ω where Ω is a column vector obtained by vertically stacking the upper triangle elements of the matrix Ω, , ) ,..., , ( 1 3 2 -I γ γ γ γ and ) ,..., , ( 1 2 1 K α α α α . The likelihood function ) (θ L for the CUOP model takes the following form: , ) ( ) ( 1 , , 1 1 q q m q q q m q Q q q q Q q B B m y P L q q θ (2.5) where (.) is the univariate cumulative standard normal distribution function. To ensure the positive definiteness of the covariance matrixΩ, the likelihood function is rewritten in terms of the Cholesky-decomposed matrix L of Ω. The maximum simulated likelihood approach then proceeds by optimizing with respect to the elements of L rather than Ω. Once convergence is achieved, the implied covariance matrix Ω may be reconstructed from the estimated matrix L. The estimation of the CUOP model presented above is very straightforward, and there have been many applications of the model or its more restrictive variants. In addition, there is a sprinkling of applications associated with two and three correlated ordered-response outcomes. Studies of two correlated ordered-response outcomes include Scotti (2006), Mitchell and Weale 6 For ease of presentation, we will treat all elements of βq as random, but this is not necessary; the researcher can fix some elements of βq and let the remaining elements be random. Also, it should be noted that, while random coefficients on exogenous variables can be estimated with cross-sectional data, it is generally easier to estimate random coefficients with panel or repeated-choice data where the random coefficients on the exogenous variables are specified to be individual-specific and the overall residual error term is specified to be choice-occasion specific. 23 (2007), Scott and Axhausen (2006), and LaMondia and Bhat (2011). The study by Scott and Kanaroglou (2002) represents an example of three correlated ordered-response outcomes. But the examination of more than two to three correlated outcomes is rare, mainly because the extension to an arbitrary number of correlated ordered-response outcomes entails, in the usual likelihood function approach, integration of dimensionality equal to the number of outcomes. On the other hand, there are many instances when interest may be centered around analyzing more than three ordered-response outcomes simultaneously, such as in the case of the number of episodes of each of several activity purposes, or satisfaction levels associated with a related set of products/services, or multiple ratings measures regarding the state of health of an individual/organization (we will refer to such outcomes as cross-sectional multivariate ordered-response outcomes). There are also instances when the analyst may want to analyze time-series or panel data of ordered-response outcomes over time, and allow flexible forms of error correlations over these outcomes. For example, the focus of analysis may be to examine rainfall levels (measured in grouped categories) over time in each of several spatial regions, or individual stop-making behavior over multiple days in a week, or individual headache severity levels at different points in time (we will refer to such outcomes as panel multivariate ordered-response outcomes). In the analysis of cross-sectional and panel ordered-response systems with more than three outcomes, the norm has been to apply numerical simulation techniques based on a maximum simulated likelihood (MSL) approach (for example, see Bhat and Zhao, 2002, Greene, 2009, and Greene and Hensher, 2010) or a Bayesian inference approach (for example, see Müller and Czado, 2005 and Girard and Parent, 2001). However, such simulation-based approaches become impractical in terms of computational time, or even infeasible, as the number of ordered-response outcomes increases. Even if feasible, the numerical simulation methods do get imprecise as the number of outcomes increase, leading to convergence problems during estimation (see Bhat et al. 2010a and Müller and Czado, 2005). As a consequence, another approach that has seen some (though very limited) use recently is the composite marginal likelihood (CML) approach, as discussed next. References for the CUOP Model There have been many applications of the cross-sectional generalized ordered-response model. The reader is referred to Greene and Hensher (2010) and Eluru et al. (2008). 2.2.1.2. The CMOP Model In many cases, a whole set of ordinal variables may be inter-related due to unobserved factors. For instance, the injury severity levels sustained by the occupants of a vehicle in a specific crash may be inter-related due to unobserved crash factors (in addition to being related due to observed crash factors), as may be the injury severity level of all occupants across all vehicles involved in a crash. Similarly, the evaluation ratings of a student of a professor on multiple dimensions (such as “interest in student learning”, “course well communicated”, and “tests returned promptly) may 24 also be correlated. The estimation of such multivariate ordered outcome models are discussed in this section. As earlier, let q be an index for individuals (q = 1, 2,…, Q, where Q denotes the total number of individuals in the data set), and let i be an index for the ordered-response variable (i = 1, 2,…, I, where I denotes the total number of ordered-response variables for each individual). Let i k be the index for ordinal outcome category ).
,..., 2 , 1 ( i i K k . Let the actual observed discrete (ordinal) level for individual q and variable i be mqi (mqi may take one of Ki values; i.e., mqi {1, 2,…, Ki} for variable i). In the usual ordered response framework notation, we write: i qi qi qi k y y , q qi x β if i k q qi i k q i i y , 1 , , (2.6) where all notations are as earlier except for the addition of the index i. Define , ) ,..., , ( 2 1 qI q q y y y q y q q x I IDEN x ~ (I×IL matrix; I IDEN is an identity matrix of size I), , ~ qi qi β b β i ) ,..., , ( vector), 1 × ( ) ~ ..., , ~ , ~ ( ~ 2 1 I 2 1 b b b b β β β β IL qI q q q (IL×1 vector), 1 ( ) ,..., , ( , 2 , 1 , 2 I I m q m q m q qI q qi up ψq vector), ) 1 ( ) ,..., , ( 1 , 2 1 , 1 1 , 2 I I m q m q m q qI q qi low ψq vector, and let ) , ( ~ ~ Ω 0 L I q MVN β . Also, let ) exp( 1 , , qk kiz γ ki i k q i k q α , and define , ) ,..., , ( , ) ,..., , ( I i , -K i i i γ γ γ γ γ γ γ γ i 2 1 1 3 2 .
) ,..., , ( and , ) ,..., , ( 1 2 1 I i α α α α α 2 1 i K i i α α The qi terms are assumed independent and identical across individuals (for each and all i). For identification reasons, the variance of each qi term is normalized to 1. However, we allow correlation in the qi terms across variables i for each individual q. Specifically, we define , )' , , , , ( 3 2 1 qI q q q q ε and assume that q ε is multivariate normal distributed with a mean vector of zeros and a correlation matrix as follows: 12 13 1 21 23 2 1 2 3 1 0 1 0 ~ , , or 1 0 I I q I I I N (2.7) q ~ , N 0 Σ The off-diagonal terms of Σ, along with the covariance matrix Ω, capture the error covariance across the underlying latent continuous variables; that is, they capture the effects of common unobserved factors influencing the underlying latent propensities. These are the so-called polychoric covariances between pairs of observed ordered-response variables. Then, we can write: ) , ( ~ q q q B y Ξ I MVN , where b Bq q x ~ and Σ x Ω x Ξ q q ~ ~ q . Let the vector of actual observed ordinal outcomes for individual q be stacked into an (I×1) vector ) ..., . , , ( 2 1 qI q q q m m m m . Also let ) ..., . , , ( 2 1 qI q q q y y y y . The parameter vector to be estimated 25 in the CMOP model is . ) , , , , ( α γ b θ Σ Ω The likelihood function for individual q takes the following form: , ) , | ( ) ( ) ( q y D I d f P L q q q q y B y m y θ Ξ (2.8) where q y D is the integration domain defined as } : { y low low ψ ψ q q q q y y q D , and (.) I f is the multivariate normal density function of dimension I . The likelihood function above involves I-dimensional rectangular integrals for each individual q. As indicated earlier, models that require integration of more than three dimensions (I >3) in a multivariate ordered-response model are typically estimated using maximum simulation likelihood (MSL) approaches. Balia and Jones (2008) adopt such a formulation in their eight-dimensional multivariate probit model of lifestyles, morbidity, and mortality. They estimate their model using a Geweke-Hajivassiliou-Keane (GHK) simulator. Yet another MSL method to approximate the MVNCD function in the likelihood functions of Equation (2.8) is based on the Genz-Bretz (GB) algorithm (see Bhat et al., 2010b for a discussion). Alternatively, Chen and Dey (2000), Herriges et al. (2008), Jeliazkov et al. (2008), and Hasegawa (2010) have considered a Bayesian estimation approach for the multivariate ordered response system through the use of standard Markov Chain Monte Carlo (MCMC) techniques. In particular, the Bayesian approach is based on assuming prior distributions on the non-threshold parameters, reparameterizing the threshold parameters, imposing a standard conjugate prior on the reparameterized version of the error covariance matrix and a flat prior on the transformed threshold, obtaining an augmented posterior density using Baye’s Theorem for the reparameterized model, and fitting the model using a Markov Chain Monte Carlo (MCMC) method. Unfortunately, the method remains cumbersome, requires extensive simulation, and is time-consuming. Further, convergence assessment becomes difficult as the number of dimensions increase (see Müller and Czado, 2005). In this regard, both the MSL and the Bayesian approaches are “brute force” simulation techniques that are not very straightforward to implement and can create numerical stability, convergence, and precision problems as the number of dimensions increase. The CML estimation of the CMOP model, on the other hand, can be very effective and fast. In particular, the pairwise likelihood function for individual q is formed by the product of likelihood contributions of pairs of ordinal variables as follows: ) , Pr( ) ( 1 1 1 , qg qg qi qi I i g I i CMOP q CML m y m y L θ 1 1 1 1 , 1 , 2 , 1 , 2 1 , , 2 , , 2 ) , , ( ) , , ( ) , , ( ) , , ( I i I j g qig g m q i m q qig g m q i m q qig g m q i m q qig g m q i m q qg qi qg qi qg qi qg qi (2.9) 26 where ) (.,., 2 qig is the standard bivariate normal cumulative distribution function with correlation qig , , Var Var , Cov , Var , , qg qi qg qi qjg qi i m q i m q y y y y y qi qi qi ' x b and the Var qi y , , Var qg y and , Cov qg qi y y terms are obtained by picking off the appropriate 2 2sub-matrix of the larger covariance matrix q Ξ of ) ..., , , ( 2 1 qI q q y y y . The pairwise marginal likelihood function is ) ( ) ( , θ θ CMOP q CML q CMOP CML L L . The asymptotic covariance matrix estimator is , ˆ ˆ ˆ ˆ Q Q 1 -1 -1 -H J H G with CML CML θ θ θ θ θ θ θ H ˆ 2 1 1 1 1 ˆ 1 1 1 1 , 2 ) , Pr( log 1 ) ( log 1 ˆ qg qg qi qi I i g I i Q q Q q I i I i g CMOP q CML m y m y Q L Q CML θ θ θ J ˆ 1 1 1 1 1 1 1 ) , Pr( log ) , Pr( log 1 ˆ I i I i g qg qg qi qi I i I i g qg qg qi qi Q q m y m y m y m y Q (2.10) An alternative estimator for H ˆ is as below: Q q I i I i g qg qg qi qi qg qg qi qi m y m y m y m y Q 1 1 1 1 ˆ ) , Pr( log ) , Pr( log 1 ˆ CML θ θ θ H (2.11) One final issue. The covariance matrix Ξ has to be positive definite, which will be the case if the matrices Ωand Σ are positive definite. The simplest way to ensure the positive-definiteness of these matrices is to use a Cholesky-decomposition and parameterize the CML function in terms of the Cholesky parameters (rather than the original covariance matrices). Also, the matrix Σ is a correlation matrix, which can be maintained by writing each diagonal element (say the aath element) of the lower triangular Cholesky matrix of Σ as 1 1 2 1 a j aj l , where the aj l elements are the Cholesky factors that are estimated. References for the CML Estimation of the CMOP Model Archer, M., Paleti, R., Konduri, K.C., Pendyala, R.M., Bhat, C.R., 2013. Modeling the connection between activity-travel patterns and subjective well-being. Transportation Research Record 2382, 102-111. 27 Bhat, C.R., Varin, C., Ferdous, N., 2010. A comparison of the maximum simulated likelihood and composite marginal likelihood estimation approaches in the context of the multivariate ordered response model. In Advances in Econometrics: Maximum Simulated Likelihood Methods and Applications, Vol. 26, Greene, W.H., Hill, R.C. (eds.), Emerald Group Publishing Limited, 65-106. Feddag, M.-L., 2013. Composite likelihood estimation for multivariate probit latent traits models. Communications in Statistics - Theory and Methods 42(14), 2551-2566. Katsikatsou, M., Moustaki, I, Yang-Wallentin, F., and Jöreskog, K.G., 2012. Pairwise likelihood estimation for factor analysis models with ordinal data. Computational Statistics and Data Analysis 56(12), 4243-4258. LaMondia, J.J., Bhat, C.R., 2011. A study of visitors’ leisure travel behavior in the northwest territories of Canada, Transportation Letters: The International Journal of Transportation Research 3(1), 1-19. Seraj, S., Sidharthan, R., Bhat, C.R., Pendyala, R.M., Goulias, K.G., 2012. Parental attitudes towards children walking and bicycling to school. Transportation Research Record 2323, 46-55. 2.2.1.3. The PMOP Model As earlier, let q be an index for individuals (q = 1, 2, …, Q), and let t be an index for the tth observation on individual q (t = 1, 2, …, T, where T denotes the total number of observations on individual q).7 Let the observed discrete (ordinal) level for individual q at the tth observation be mqt (mqt may take one of K values; i.e., mqt {1, 2,…, K}). In the usual random-coefficients ordered response framework notation, we write the latent variable ( qt y ) as a function of relevant covariates as: k y y qt qt qt , qt ' qx β if k t q qj k t q y , , 1 , , , (2.12) where qt x is a (L×1)-vector of exogenous variables (including a constant now), q β is an individual-specific (L×1)-vector of coefficients to be estimated that is a function of unobserved individual attributes, qt is a standard normal error term uncorrelated across individuals q and across observations of the same individual, and k t q , , is the upper bound threshold for ordinal discrete level k (k=1,2,…,K) for individual q at choice occasion t. The thresholds are written as ) exp( 1 , , , , qt k k t q k t q α z γk for k=2,3,…,K-1, with . 0, , ; ...
, , 1 , , 0 , , , , 1 , , 2 , , 1 , , 0 , , K t q t q t q K t q K t q t q t q t q Assume that the q β vector in Equation (2.12) is a time-invariant realization from a multivariate normal distribution with a mean vector b and covariance matrix , L L Ω where L is the lower-triangular Cholesky 7 We assume here that the number of panel observations is the same across individuals. Extension to the case of different numbers of panel observations across individuals does not pose any substantial challenges, and will be discussed later. 28 factor of Ω.8 Also, assume that the qt term, which captures the idiosyncratic effect of all omitted variables for individual q at the tth choice occasion, is independent of the elements of the q β and qt x vectors. Define ) ..., . , , ( 2 1 qT q q q y y y y matrix) 1 ( T , ) ..., . , , ( 2 1 qT q q q ε matrix) 1 ( T ) matrix 1 ( ) ..., , , ( 2 1 T y y y qT q q q y , matrix), ( )' ,..., , ( L T qT q q q x x x x 2 1 vector), 1 ( ) ,..., , ( , , , 2 , , 1 , 2 T qT q qi m T q m q m q up ψq ) 1 ( ) ,..., , ( 1 , , 1 , 2 , 1 , 1 , 2 T qJ q qi m T q m q m q low ψq vector. Also, let the vector of actual observed ordinal outcomes for individual q be stacked into a (T×1) vector ) ..., . , , ( 2 1 qT q q q m m m m . Then, we may write ) ( and where ), ( ~ T T MVN IDEN Ω Ξ Ξ q q q q q q q q x x b x B , B y , and the parameter vector to be estimated in the PMOP model is ) , , , ( α γ b θ Ω , where ) ,..., , ( 1 -K γ γ γ γ 3 2 and .
) ,..., , ( 1 3 2 K α α α α The likelihood function for individual q takes the following form: , ) | ( ) ( ) ( q y D T d f P L q q q q q q y , B y m y θ Ξ (2.13) where y D is the integration domain defined as } : { y low low ψ ψ q q q q y y q D , and (.) T f is the multivariate normal density function of dimension T . The likelihood function above involves T-dimensional rectangular integrals for each individual q. The above model is labeled as a mixed autoregressive ordinal probit model by Varin and Czado (2010), who examined the headache pain intensity of patients over several consecutive days. In this study, a full information likelihood estimator would have entailed as many as 815 dimensions of rectangular integration to obtain individual-specific likelihood contributions, an infeasible proposition using the computer-intensive simulation techniques. As importantly, the accuracy of simulation techniques is known to degrade rapidly at medium-to-high dimensions, and the simulation noise increases substantially. On the other hand, the CML approach is easy to apply in such situations, through a pairwise marginal likelihood approach that takes the following form: 1 1 1 , , Pr ) ( T t T t g qg qg qt qt PMOP q CML m y m y L θ 1 1 1 1 , , 1 , , 2 , , 1 , , 2 1 , , , , 2 , , , , 2 ) , , ( ) , , ( ) , , ( ) , , ( T t T t g qtg m g q m t q qtg m g q m t q qtg m g q m t q qtg m g q m t q qg qt qg qt qg qt qg qt (2.14) 8 More general autoregressive structures can also be considered for qt and q β to accommodate fading and time-varying covariance effects in the latent variables qt y (see Bhat, 2011 and Paleti and Bhat, 2013). This does not complicate the econometrics of the CML estimation method, but can lead to substantial number of additional parameters and may be asking too much from typical estimation data sets. In this paper, we present the case of independent qt across choice occasions and time-invariant random coefficients. 29 where , , , , Var Var , Cov and Var qg qt qg qt qtg qt m t q m t q y y y y y qt qt qt ' x b In the above expression, the Var qt y , Var qg y , and , Cov qg qt y y terms are obtained by picking off the appropriate ( 2 2)-sub-matrix of the larger covariance matrix q Ξ of ) ..., , , ( 2 1 qT q q y y y . The pairwise marginal likelihood function is ) ( ) ( , θ θ PMOP q CML q PMOP CML L L . The covariance matrix of the estimator can be obtained exactly as in the CMOP case. The analysis above assumes the presence of a balanced panel; that is, it assumes the same number of choice instances per individual. In the case when the number of choice instances varies across individuals, Joe and Lee (2009) proposed placing a power weight for individual q as 1 1 )] 1 ( 5 .
0 1 [ ) 1 ( q q q T T w (where the number of observations from individual q is q T ) and constructing the marginal likelihood contribution of individual q as: q qg qj qg qj qg qj qg qj w T t J t g qtg m g q m t q qtg m g q m t q qtg m g q m t q qtg m g q m t q PMOP q CML L 1 1 1 1 , , 1 , , 2 , , 1 , , 2 1 , , , , 2 , , , , 2 , ) , , ( ) , , ( ) , , ( ) , , ( ) ( θ (2.15) References for the CML Estimation of the PMOP Model Paleti, R., Bhat, C.R., 2013. The composite marginal likelihood (CML) estimation of panel ordered-response models. Journal of Choice Modelling 7, 24-43. Varin, C., Czado, C., 2010. A mixed autoregressive probit model for ordinal longitudinal data. Biostatistics 11(1), 127-138. Varin, C. Vidoni, P., 2006. Pairwise likelihood inference for ordinal categorical time series. Computational Statistics and Data Analysis 51(4), 2365-2373. Vasdekis, V.G.S., Cagnone, S., Moustaki, I., 2012. A composite likelihood inference in latent variable models for ordinal longitudinal responses. Psychometrika 77(3), 425-441. 2.2.2. Unordered-Response Models In the class of unordered-response models, the “workhorse” multinomial logit model introduced by Luce and Suppes (1965) and McFadden (1974) has been used extensively in practice for econometric discrete choice analysis, and has a very simple and elegant structure. However, it is also saddled with the familiar independence from irrelevant alternatives (IIA) property – that is, the ratio of the choice probabilities of two alternatives is independent of the characteristics of other alternatives in the choice set. This has led to several extensions of the MNL model through the relaxation of the independent and identically distributed (IID) error distribution (across alternatives) assumption. Two common model forms of non-IID error distribution include the generalized extreme-value (GEV) class of models proposed by McFadden (1978) and the multinomial probit (MNP) model that allows relatively flexible error covariance structures (up to certain limits of identifiability; see Train, 2009, Chapter 5). Both of these non-IID kernel 30 structures (or even the IID versions of the GEV and the MNP models, which lead to the MNL and the independent MNP models, respectively) can further be combined with continuous mixing error structures. While many different continuous distributions can be used to accommodate these additional structures, it is most common to adopt a normal distribution. For instance, when introducing random coefficients, it is typical to use the multivariate normal distribution for the mixing coefficients, almost to the point that the terms mixed logit or mixed GEV or mixed probit are oftentimes used synonymously with normal mixing (see Fiebig et al., 2010, Dube et al., 2002).9 In the context of the normal error distributions just discussed, the use of a GEV kernel structure leads to a mixing of the normal distribution with a GEV kernel, while the use of an MNP kernel leads once again to an MNP model. Both structures have been widely used in the past, with the choice between a GEV kernel or an MNP kernel really being a matter of “which is easier to use in a given situation” (Ruud, 2007). In recent years, the mixing of the normal with the GEV kernel has been the model form of choice in the economics and transportation fields, mainly due to the relative ease with which the probability expressions in this structure can be simulated (see Bhat et al., 2008 and Train, 2009 for detailed discussions). On the other hand, the use of an MNP kernel has not seen as much use in recent years, because the simulation estimation is generally more difficult. In any case, while there have been several approaches proposed to simulate these models with a GEV or an MNP kernel, most of these involve pseudo-Monte Carlo or quasi-Monte Carlo simulations in combination with a quasi-Newton optimization routine in a maximum simulated likelihood (MSL) inference approach (see Bhat, 2001, 2003). As has been discussed earlier, in such an inference approach, consistency, efficiency, and asymptotic normality of the estimator is critically predicated on the condition that the number of simulation draws rises faster than the square root of the number of individuals in the estimation sample. Unfortunately, for many practical situations, the computational cost to ensure good asymptotic estimator properties can be prohibitive and literally infeasible (in the context of the computation resources available and the time available for estimation) as the number of dimensions of integration increases. The Maximum Approximate Composite Marginal Likelihood (MACML) inference approach proposed by Bhat (2011), on the other hand, allows the estimation of models with both GEV and MNP kernels using simple, computationally very efficient, and simulation-free estimation methods. In the MACML inference approach, models with the MNP kernel, when combined with additional normal random components, are much easier to estimate because of the conjugate addition property of the normal distribution (which puts the structure resulting from the addition of normal components to the MNP kernel back into an MNP form). On the other hand, the MACML estimation of models obtained by superimposing normal error 9 It has been well known that using non-normal distributions can lead to convergence/computational problems, and it is not uncommon to see researchers consider non-normal distributions only to eventually revert to the use of a normal distribution (see, for example, Bartels et al., 2006 and Small et al., 2005). However, one appealing approach is to use a multivariate skew-normal (MSN) distribution for the response surface, as proposed by Bhat and Sidharthan (2012). 31 components over a GEV kernel requires a normal scale mixture representation for the extreme value error terms, and adds an additional layer of computational effort (see Bhat, 2011). Given that the use of a GEV kernel or an MNP kernel is simply a matter of convenience, we will henceforth focus in this monograph on the MNP kernel within the unordered-response model structure. The aspatial formulations of the unordered-response structure may take the form of a cross-sectional multinomial probit (CMNP), or a cross-sectional multivariate multinomial probit (CMMNP), or a panel multinomial probit (PMNP). 2.2.2.1. The CMNP Model In the discussion below, we will assume that the number of choice alternatives in the choice set is the same across all individuals. The case of different numbers of choice alternatives per individual poses no complication, since the only change in such a case is that the dimensionality of the multivariate normal cumulative distribution (MVNCD) function changes from one individual to the next. Consider the following specification of utility for individual q and alternative i: ) , ( ~ ~ , ~ ; Ω 0 L qi qi MVN U q q q qi q β β b β x β , (2.16) where qi x is an ) 1 ( L -column vector of exogenous attributes (including a constant for each alternative, except one of the alternatives), and q β is an individual-specific ) 1 ( L -column vector of corresponding coefficients that varies across individuals based on unobserved individual attributes. Assume that the q β vector is a realization from a multivariate normal distribution with a mean vector b and covariance matrix L L Ω . We also assume that qi is independent and identically normally distributed across q, but allow a general covariance structure across alternatives for individual q. Specifically, let ) ,..., , ( 2 1 qI q q q ξ ( 1 I vector). Then, we assume Λ) , 0 ( ~ I MVN q ξ . As usual, appropriate scale and level normalization must be imposed on Λ or identifiability. Specifically, only utility differentials matter in discrete choice models. Taking the utility differentials with respect to the first alternative, only the elements of the covariance matrix 1 Λ of ) 1 ( ~ 1 1 i q qi qi are estimable. However, the MACML inference approach proposed here, like the traditional GHK simulator, takes the difference in utilities against the chosen alternative during estimation. Thus, if individual q is observed to choose alternative q m , the covariance matrix q m Λ is desired for the individual. However, even though different differenced covariance matrices are used for different individuals, they must originate in the same matrix Λ . To achieve this consistency, Λ is constructed from 1 Λ by adding an additional row on top and an additional column to the left. All elements of this additional row and additional column are filled with values of zeros. An additional normalization needs to be imposed on Λ because the scale is also not identified. For this, we normalize the 32 element of Λ in the second row and second column to the value of one. Note that these normalizations are innocuous and are needed for identification. The Λ matrix so constructed is fully general. Also, in MNP models, identification is tenuous when only individual-specific covariates are used (see Keane, 1992 and Munkin and Trivedi, 2008). In particular, exclusion restrictions are needed in the form of at least one individual characteristic being excluded from each alternative’s utility in addition to being excluded from a base alternative (but appearing in some other utilities). But these exclusion restrictions are not needed when there are alternative-specific variables. The model above may be written in a more compact form by defining the following vectors and matrices: ) ,..., , ( 2 1 qI q q q U U U U 1 ( I vector), ) ,..., , , ( qI q q q q x x x x x 3 2 1 L I ( matrix), b xq q V 1 ( I vector), q q x x Ω Ωq ~ ) matrix ( I I , and I I q ( ~ ~ Λ Ω Ξq matrix). Then, we may write, in matrix notation, q q q ξ V U and ).
~ , ( ~ q q q Ξ V U I MVN Also, let ) ( ) , , ( 2 1 q qI q q q m i u u u u be an (I–1)×1 vector, where q m is the actual observed choice of individual q, and ).
( q qm qi qi m i U U u q Then, , 1 I q 0 u because alternative q m is the chosen alternative by individual q. To develop the likelihood function, define q M as an identity matrix of size I-1 with an extra column of ‘-1’ values added at the th q m column (thus, q M is a matrix of dimension )).
( ) ( I 1 -I Then, q u is distributed as follows: ) Ξ u q q B , ( ~ 1 I q MVN , where q qV M q B and q q M Ξ M Ξ q q ~ . The parameter vector to be estimated is . ) , , ( Λ Ω b θ Let q Ξ ω be the diagonal matrix of standard deviations of q Ξ . Using the usual notations as described earlier, the likelihood contribution of individual q is as below: ), ), ( ( ) ( 1 1 Ξ Ξ ω q q B θ q I q L (2.17) where .
1 1 q q q q Ξ Ξ ω Ξ ω Ξ The MVNCD approximation discussed earlier is computationally efficient and straightforward to implement when maximizing the likelihood function of Equation (2.17).10 As such, the MVNCD approximation can be used for any value of K and any value of I, as long as there is data support for the estimation of parameters. The positive-definiteness of Σ can be ensured by using a Cholesky-decomposition of the matrices Ω and Λ , and estimating these Cholesky-decomposed parameters. Note that, to obtain the Cholesky factor for Λ , we first obtain the Cholesky factor for 1 Λ , and then add a column of zeros as the first column and a row 10As indicated earlier, the CML class of estimators subsumes the usual ordinary full-information likelihood estimator as a special case. It is this characteristic of the CML approach that leads us to the label MACML for the estimation approach proposed here. Specifically, even in cross-sectional MNP contexts, when our approach involves only the approximation of the MVNCD function in the maximum likelihood function, the MACML label is appropriate since the maximum likelihood function is a special case of the CML function. 33 of zeros as the first row to the Cholesky factor of 1 Λ . The covariance matrix in this CMOP case is obtained using the usual Fisher information matrix, since the full (approximate) likelihood is being maximized. Bhat and Sidharthan (2011) apply the MACML estimation approach for estimating the CMNP model with five random coefficients and five alternatives, and compare the performance of the MSL and MACML approaches (though, in their simulations, they constrain Λ to be an identity matrix multiplied by 0.5). They conclude that the MACML approach recovers parameters much more accurately than the MSL approach, while also being about 50 times faster than the MSL approach. They also note that as the number of random coefficients and/or alternatives in the unordered-response model increases, one can expect even higher computational efficiency factors for the MACML over the MSL approach. References for the CML Estimation of the CMNP Model Bhat, C.R., 2011. The maximum approximate composite marginal likelihood (MACML) estimation of multinomial probit-based unordered response choice models. Transportation Research Part B 45(7), 923-939. Bhat, C.R., Sidharthan, R., 2011. A simulation evaluation of the maximum approximate composite marginal likelihood (MACML) estimator for mixed multinomial probit models. Transportation Research Part B 45(7), 940-953. Bhat, C.R., Sidharthan, R., 2012. A new approach to specify and estimate non-normally mixed multinomial probit models. Transportation Research Part B 46(7), 817-833. 2.2.2.2. The CMMNP Model Let there be G nominal (unordered multinomial response) variables for an individual, and let g be the index for variables (g = 1, 2, 3,…, G). Also, let Ig be the number of alternatives corresponding to the gth nominal variable (Ig3) and let g i be the corresponding index ( g i = 1, 2, 3,…, g I ). Note that g I may vary across individuals. Also, it is possible that some nominal variables do not apply for some individuals, in which case G itself is a function of the individual q. However, for presentation ease, we assume that all the G nominal variables are relevant for each individual, and that all the alternatives g I are available for each variable g. Consider the gth variable and assume that the individual q chooses the alternative qg m . Also, assume the usual random utility structure for each alternative g i . , g g g qgi qgi qg qgi U x β (2.18) where g qgi x is a (Lg×1)-column vector of exogenous attributes, qg β is a column vector of corresponding coefficients, and g qgi is a normal error term. Assume that the qg β vector is a realization from a multivariate normal distribution with a mean vector g b and covariance matrix 34 g gL L g Ω , where g L is the lower-triangular Cholesky factor of g Ω. While one can allow covariance among the qg β vectors across the coefficients of the different unordered-response variables for each individual, this specification will be profligate in the parameters to be estimated. So, we will assume that the qg β vectors are independent across the unordered-response dimensions for each individual. We also assume that g qgi is independent and identically normally distributed across individuals q, but allow a general covariance structure across alternatives for individual q. Specifically, let ) ,..., , ( 2 1 g qgI qg qg qg ξ ( 1 g I vector). Then, we assume ) Λg , 0 ( ~ I MVN qg ξ . Let ) ( qg g qgm qgi m qgi m i U U u qg g qg g , where qg m is the chosen alternative for the gth unordered-response variable by individual q, and stack the latent utility differentials into a vector qg g m qgI m qg m qg m i u u u qg g qg qg ; ,..., , 2 1 qg u ] vector 1 ) 1 [( g I . Let ) ,..., , , ( 3 2 1 q qgI qg qg qg qg x x x x x L I g ( matrix), g V b xqg qg 1 ( g I vector) and qg qg x x g qg Ω Ω ~ ) matrix ( g g I I . Define qg M as an identity matrix of size 1 g I , with an extra column of ‘-1’ values added at the th qg m column. Also, construct the matrices qg qg qg V M B , qg qg qg qg M Ω M Ω ~ , and .
qg g qg qg M Λ M Λ When there are G unordered-response variables, consider the stacked vector 1 ) 1 ( 1 G g g I ' ,..., , ' qG ' q ' q q u u u u 2 1 , each of whose element vectors is formed by differencing utilities of alternatives from the chosen alternative qg m for the gth variable. Also, form a block diagonal covariance matrix q Ω of size , ) 1 ( ) 1 ( 1 1 G g g G g g I I each block diagonal holding the matrix qg Ω , and the following matrix of the same size as q Ω : qG qG2 qG1 q2G q2 q21 q1G q12 q1 q Λ .
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Λ Λ Λ (2.19) The off-diagonal elements in q Λ capture the dependencies across the utility differentials of different variables, the differential being taken with respect to the chosen alternative for each variable. It must be ensured that q Λ across individuals is derived from a common covariance 35 matrix Λ for the original G g g I 1 -error term vector ) ,..., , ( qG q q q ξ ξ ξ ξ 2 1 . Appropriate identification considerations will have to be placed on the elements of Λ . The parameter vector to be estimated is . ) , ,..., , , ,...
, ( 2 1 Λ Ω Ω Ω 2 1 G G b b b θ Using the notations as described earlier, and defining ) ,..., , ( 2 1 qG q q q B B B B and , Λ Ω Ξ q q q the likelihood contribution of individual q is as below: ), ), (-( ) ( 1 ~ Ξ Ξ ω q B θ q q I q L (2.20) where 1 1 q q q q Ξ Ξ ω Ξ ω Ξ and G g g I I 1 ) 1 ( ~ The above likelihood function involves the evaluation of a G g g I 1 ) 1 ( -dimensional integral for each individual, which can be very expensive if there are several variables and/or if each variable can take a large number of values. But, once again the Maximum Approximated Composite Marginal Likelihood (MACML) approach of Bhat (2011) can be used gainfully in this context, in which the MACML function only involves the computation of univariate and bivariate cumulative distributive functions. Specifically, consider the following (pairwise) composite marginal likelihood function formed by taking the products (across the G nominal variables) of the joint pairwise probability of the chosen alternatives qg m for the gth variable and ql m for the lth variable for individual q. 1 1 1 , ) , Pr( ) ( G g G g l ql ql qg qg CMMNP q CML m d m d L θ , (2.21) where qg d is an index for the individual’s choice for the gth variable. One can also write: ), ~ ), ~ (-( ) , Pr( 1 ~ Ξ Ξ ω qgl B qgl qgl I ql ql qg qg m d m d (2.22) where 2 l g I I I ( g I is the number of alternatives for the gth variable), , ~ , ~ ' Δ Ξ Δ Ξ Δ qgl qgl qgl qgl q qgl qgl B B , ~ ~ 1 ~ 1 ~ qgl qgl Ξ Ξ ω Ξ ω Ξ qgl qgl and qgl Δ is a I I ~ -selection matrix with an identity matrix of size ( 1 g I ) occupying the first ( 1 g I ) rows and the th g j j I 1 ) 1 ( 1 1 through th g j j I 1 ) 1 ( columns (with the convention that 0 ) 1 ( 0 1 j j I ), and another identity matrix of size ( 1 l I ) occupying the last ( 1 l I ) rows and the th l j j I 1 ) 1 ( 1 1 36 through th l j j I 1 ) 1 ( columns. The net result is that the pairwise likelihood function now only needs the evaluation of a I -dimensional cumulative normal distribution function (rather than the I ~ -dimensional cumulative distribution function in the maximum likelihood function). This can lead to substantial computation efficiency, and can be evaluated using the MVNCD approximation of the MACML procedure. The MACML estimator MACML θ ˆ , obtained by maximizing the logarithm of the function Q q G g G g l qgl I CMMNP q MACML CMMNP q MACML CMMNP MACML L where L L 1 1 1 1 1 ~ , , ) ~ ), ~ (-( ) ( ), ( ) ( Ξ Ξ ω qgl B θ θ θ q (with the MVNCD approximation), is asymptotically normal distributed with mean θ and covariance matrix that can be estimated as: , ˆ ˆ ˆ ˆ Q Q 1 -1 -1 -H J H G (2.23) with MACML q θ θ θ B H ˆ 1 1 1 1 1 ~ 2 ) ~ ), ~ (-( log 1 ˆ Q q G g G g l qgl qgl I Q Ξ Ξ ω MACML q q θ θ B θ B J ˆ 1 1 1 1 ~ 1 1 1 1 ~ 1 ) ~ ), ~ (-( log ) ~ ), ~ (-( log 1 ˆ G g G g l qgl qgl I G g G g l qgl qgl I Q q Q Ξ Ξ Ξ ω Ξ ω (2.24) An alternative estimator for H ˆ is as below: Q q G g G g l qgl qgl I qgl qgl I Q 1 1 1 1 ˆ 1 ~ 1 ~ ) ~ ), ~ (-( log ) ~ ), ~ (-( log 1 ˆ MACML q q θ θ B θ B H Ξ Ξ Ξ ω Ξ ω (2.25) There are two important issues that need to be dealt with during estimation, each of which is discussed in turn below. Identification The estimated model needs to be theoretically identified. Suppose one considers utility differences with respect to the first alternative for each of the G variables. Then, the analyst can restrict the variance term of the top left diagonal of the covariance matrix (say ) g Λ of error differences ,...
, 1 1 3 1 2 qg qgI qg qg qg qg g to 1 to account for scale invariance. However, note that the matrix g Λ is different from the matrix g Λ , which corresponds to the covariance of utility differences taken with respect to the chosen alternative for the individual. 37 Next, create a matrix of dimension G g g G g g I I 1 1 ) 1 ( ) 1 ( similar to that of g Λ in Equation (2.19), except that the matrix is expressed in terms of utility differences with respect to the first alternative for each nominal variable: G G2 G1 2G 2 21 1G 12 1 Λ .
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Λ Λ Λ (2.26) In the general case, this allows the estimation of G g g g I I 1 1 2 ) 1 ( variance terms across all the G variables (originating from 1 2 ) 1 ( g g I I terms embedded in each g Λ matrix; g=1,2,…G), and 1 1 1 ) 1 ( ) 1 ( G g G g l l g I I covariance terms in the off-diagonal matrices of the Λ matrix characterizing the dependence between the latent utility differentials (with respect to the first alternative) across the variables (originating from ) 1 ( ) 1 ( l g I I estimable covariance terms within each off-diagonal matrix gl Λ in Λ ). To construct the general covariance matrix Λ for the original G g g I 1 -error term vector q ξ , while also ensuring all parameters are identifiable, zero row and column vectors are inserted for the first alternatives of each unordered dependent variable in Λ . To do so, define a matrix D of size G g g G g g I I 1 1 ) 1 ( . The first 1 I rows and ) 1 ( 1 I columns correspond to the first variable. Insert an identity matrix of size ) 1 ( 1 I after supplementing with a first row of zeros into this first 1 I rows and ) 1 ( 1 I columns of D. The rest of the columns for the first 1 I rows and the rest of the rows for the first ) 1 ( 1 I columns take a value of zero. Next, rows ) 1 ( 1 I through ) ( 2 1 I I and columns ) ( 1 I through ) 2 ( 2 1 I I correspond to the second variable. Again position an identity matrix of size ) 1 ( 2 I after supplementing with a first row of zeros into this position. Continue this for all G nominal variables. Thus, for the case with two nominal variables, one nominal variable with 3 alternatives and the second with four alternatives, the matrix D takes the form shown below: 38 5 7 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 (2.27) Then, the general covariance matrix may be developed as .
D Λ D Λ All parameters in this matrix are identifiable by virtue of the way this matrix is constructed based on utility differences and, at the same time, it provides a consistent means to obtain the covariance matrix q Λ that is needed for estimation (and is with respect to each individual’s chosen alternative for each variable). Specifically, define a matrix q M ~ of size G g g G g g I I 1 1 ) 1 ( . The first ) 1 ( 1 I rows and 1 I columns correspond to the first nominal variable. Insert an identity matrix of size ) 1 ( 1 I after supplementing with a column of ‘-1’ values in the column corresponding to the chosen alternative. The rest of the columns for the first ) 1 ( 1 I rows and the rest of the rows for the first 1 I columns take a value of zero. Next, rows ) ( 1 I through ) 2 ( 2 1 I I and columns ) 1 ( 1 I through ) ( 2 1 I I correspond to the second nominal variable. Again position an identity matrix of size ) 1 ( 2 I after supplementing with a column of ‘-1’ values in the column corresponding to the chosen alternative. Continue this procedure for all G nominal variables. With the matrix q M ~ as defined, the covariance matrix q Λ for any individual is given by .
~ ~ q q q M Λ M Λ Positive Definiteness The matrices q Λ and q Ω have to be positive definite. The simplest way to guarantee the positive definiteness of q Λ is to ensure that Λ is positive definite. To do so, the Cholesky matrix of Λ may be used as the matrix of parameters to be estimated. However, note that the top diagonal element of each Λ g is normalized to one for identification, and this restriction should be recognized when using the Cholesky factor of Λ . This can be achieved by appropriately parameterizing the diagonal elements of the Cholesky decomposition matrix. Thus, consider the lower triangular Cholesky matrix L of the same size as Λ . Whenever a diagonal element (say the kkth element) of Λ is to be normalized to one, the first element in the corresponding row of 39 L is written as k j kj l 2 2 1 , where the kj l elements are the Cholesky factors that are to be estimated. With this parameterization, Λ obtained as L L is positive definite and adheres to the scaling conditions. Using this, one constructs Λ , and subsequently obtains q Λ as discussed earlier. The resulting q Λ is positive definite. The positive definiteness of q Ω is ensured by writing g gL L g Ω . References for the CML Estimation of the CMMNP Model Bhat, C.R., Paleti, R., Pendyala, R.M., Lorenzini, K., Konduri, K.C., 2013. Accommodating immigration status and self selection effects in a joint model of household auto ownership and residential location choice. Transportation Research Record 2382, 142-150. Feddag, M.-L., 2013. Composite likelihood estimation for multivariate probit latent traits models. Communications in Statistics - Theory and Methods 42(14), 2551-2566. Kortum, K., Paleti, R., Bhat, C.R., Pendyala, R.M., 2012. Joint model of residential relocation choice and underlying causal factors, Transportation Research Record, 2303, 28-37. 2.2.2.3. The Panel MNP (PMNP) Model Consider the following model with ‘t’ now being an index for choice occasion: . ..., , 2 , 1 , ..., , 2 , 1 , ..., , 2 , 1 , ) , ( ~ , I i T t Q q MVN U qti qit Ω b β x β q qti q (2.28) For ease, we assume that all alternatives are available at each choice instance of each individual, and that we have a balanced panel (that is, we have the same number of choice instances from each individual). The first assumption is innocuous and helps in presentation. The relaxation of the second assumption only requires a different weight per individual, exactly as discussed earlier for the ordered-response case. qti x is a ) 1 ( L -column vector of exogenous attributes whose first (I-1) elements correspond to alternative specific constants for (I-1) alternatives (with one of the alternatives being the base alternative) and the remaining variables being the non-constant variables. q β is an individual-specific ) 1 ( L -column vector of corresponding coefficients that varies across individuals based on unobserved individual attributes. Assume that the q β vector is a realization from a multivariate normal distribution with a mean vector b and covariance matrix L L Ω , where L is the lower-triangular Cholesky factor of Ω. Thus, as in the case of the panel ordered-response model, the coefficients q β are considered constant over choice situations of a given decision maker. We also assume that qti is independent and identically normally distributed across individuals and choice occasions, but allow a general covariance structure across alternatives for each choice instance of each individual. Specifically, 40 let ) ,...
, ( 2 1 qtI qt qt qt ξ ( 1 I vector). Then, we assume Λ) , 0 ( ~ I MVN qt ξ . As usual, appropriate scale and level normalization must be imposed on Λ for identifiability. To do so, we follow the same exact procedure as in the CMNP model. Specifically, only utility differentials matter at each choice occasion. Taking the utility differentials with respect to the first alternative, only the elements of the covariance matrix 1 Λ of ) 1 ( ~ 1 1 i qt qti qti are estimable, and Λ is constructed from 1 Λ by adding an additional row on top and an additional column to the left. All elements of this additional row and additional column are filled with values of zeros. We also normalize the element of Λ in the second row and second column to the value of one. Define the following vectors and matrices: ) ,..., , ( 2 1 qtI qt qt qt U U U U 1 ( I vector), ) ,..., , ( 2 1 qI q q q U U U U 1 ( TI vector), ) ,...
, ( qT q2 q1 q ξ ξ ξ ξ 1 ( TI vector), ) ,..., , , ( 3 2 1 qtI qt qt qt qt x x x x x ( L I matrix), ) ,..., , ( 2 1 qT q q q x x x x ( L TI matrix), b xq q V 1 ( TI vector), q q q x x Ω Ω ~ ) matrix ( TI TI , and TI TI T q ( ) ( ~ ~ Λ IDEN Ω Ξq matrix). Then, we may write, in matrix notation, q q q ξ V U and ).
~ , ( ~ q q q Ξ V U TI MVN Let the individual q choose alternative qt m at the tth choice occasion. To develop the likelihood function, define q M as an ] [ )] 1 ( [ TI I T block-diagonal matrix, each block diagonal being of size )) ( ) ( I 1 -I and containing the matrix qt M . qt M itself is an identity matrix of size (I-1) with an extra column of ‘-1’ values added at the th qt m column. Let q q q V M B and q q M Ξ M Ξ q q ~ . The parameter vector to be estimated is . ) , , ( Λ Ω b θ The likelihood contribution of individual q is as below: ), ), ( ( ) ( 1 ~ Ξ Ξ ω q q B θ q J q L (2.29) where ), 1 ( ~ I T J and .
1 1 q q q q Ξ Ξ ω Ξ ω Ξ The simulation approaches for evaluating the panel likelihood function involve integration of dimension )] 1 ( [ I T . Consider the following (pairwise) composite marginal likelihood function formed by taking the products (across the T choice occasions) of the joint pairwise probability of the chosen alternatives qt m for the tth choice occasion and qg m for the gth choice occasion for individual q. 1 1 1 , ) , Pr( ) ( T t T t g qg qg qt qt PMNP q CML m d m d L θ , (2.30) where qt d is an index for the individual’s choice on the tth choice occasion. One can also write: ), ~ ), ~ (-( ) , Pr( 1 ~ qtg qtg B qgl Ξ ωΞ J qg qg qt qt m d m d (2.31) 41 where ) 1 ( 2 I J , , ~ , ~ ' Δ Ξ Δ Ξ Δ qtg qtg qtg qtg qtg q qtg B B , ~ ~ 1 ~ 1 ~ qtg qtg Ξ Ξ ω Ξ ω Ξ qtg qtg and qtg Δ is a J J ~ -selection matrix with an identity matrix of size ( 1 I ) occupying the first ( 1 I ) rows and the th I t 1 ) 1 ( ) 1 ( through th I t ) 1 ( columns, and another identity matrix of size ( 1 I ) occupying the last ( 1 I ) rows and the th I g 1 ) 1 ( ) 1 ( through th I g ) 1 ( columns. The pairwise likelihood function now only needs the evaluation of a J -dimensional cumulative normal distribution function (rather than the I ~ -dimensional cumulative distribution function in the maximum likelihood function). The MACML estimator MACML θ ˆ is obtained by maximizing the logarithm of the function Q q T t T t g J PMNP q MACML PMNP q MACML PMNP MACML L L L 1 1 1 1 1 ~ , , ) ~ ), ~ (-( ) ( where ), ( ) ( qtg qtg B θ θ θ qtg Ξ ωΞ (with the MVNCD approximation). The covariance matrix is estimated as: , ˆ ˆ ˆ ˆ Q Q 1 -1 -1 -H J H G with MACML qtg θ qtg qtg θ θ B H ˆ 1 1 1 1 1 ~ 2 ) ~ ), ~ (-( log 1 ˆ Q q T t T t g J Q Ξ ωΞ MACML qtg qtg θ qtg qtg qtg qtg θ B θ B J ˆ 1 1 1 1 ~ 1 1 1 1 ~ 1 ) ~ ), ~ (-( log ) ~ ), ~ (-( log 1 ˆ T t T t g J T t T t g J Q q Q Ξ ω Ξ ω Ξ Ξ (2.32) An alternative estimator for H ˆ is as below: Q q T t T g J J Q 1 1 1 1 ˆ 1 ~ 1 ~ ) ~ ), ~ (-( log ) ~ ), ~ (-( log 1 ˆ MACML qtg qtg θ qtg qtg qtg qtg θ B θ B H Ξ ω Ξ ω Ξ Ξ (2.33) References for the CML Estimation of the PMNP Model Bhat, C.R., 2011. The maximum approximate composite marginal likelihood (MACML) estimation of multinomial probit-based unordered response choice models. Transportation Research Part B 45(7), 923-939. Bhat, C.R., Sidharthan, R., 2011. A simulation evaluation of the maximum approximate composite marginal likelihood (MACML) estimator for mixed multinomial probit models. Transportation Research Part B 45(7), 940-953. Bhat, C.R., Sidharthan, R., 2012. A new approach to specify and estimate non-normally mixed multinomial probit models. Transportation Research Part B 46(7), 817-833. 42 2.3. Spatial Formulations In the past decade, there has been increasing interest and attention on recognizing and explicitly accommodating spatial (and social) dependence among decision-makers (or other observation units) in urban and regional modeling, agricultural and natural resource economics, public economics, geography, marketing, sociology, political science, and epidemiology. The reader is referred to a special issue of Regional Science and Urban Economics entitled “Advances in spatial econometrics” (edited by Arbia and Kelejian, 2010) and another special issue of the Journal of Regional Science entitled “Introduction: Whither spatial econometrics?” (edited by Patridge et al., 2012) for a collection of recent papers on spatial dependence, and to Elhorst (2010), Anselin (2010), Ferdous and Bhat (2013), and Bhat et al. (2014a) for overviews of recent developments in the spatial econometrics field. Within the past few years, there has particularly been an explosion in studies that recognize and accommodate spatial dependency in discrete choice models. The typical way this is achieved is by applying spatial structures developed in the context of continuous dependent variables to the linear (latent) propensity variables underlying discrete choice dependent variables (see reviews of this literature in Fleming, 2004, Franzese and Hays, 2008, LeSage and Pace, 2009, Hays et al. 2010, Brady and Irwin, 2011, and Sidharthan and Bhat, 2012). The two dominant techniques, both based on simulation methods, for the estimation of such spatial discrete models are the frequentist recursive importance sampling (RIS) estimator (which is a generalization of the more familiar Geweke-Hajivassiliou-Keane or GHK simulator; see Beron and Vijverberg, 2004) and the Bayesian Markov Chain Monte Carlo (MCMC)-based estimator (see LeSage and Pace, 2009). However, both of these methods are confronted with multi-dimensional normal integration of the order of the number of observational units in ordered-response models, and are cumbersome to implement in typical empirical contexts with even moderate estimation sample sizes (see Bhat, 2011 and Franzese et al., 2010). The RIS and MCMC methods become even more difficult (to almost infeasible) to implement in a spatial unordered multinomial choice context because the likelihood function entails a multidimensional integral of the order of the number of observational units factored up by the number of alternatives minus one (in the case of multi-period data, the integral dimension gets factored up further by the number of time periods of observation). Recently, Bhat and colleagues have suggested a composite marginal likelihood (CML) inference approach for estimating spatial binary/ordered-response probit models, and the maximum approximate composite marginal likelihood (MACML) inference approach for estimating spatial unordered-response multinomial probit (MNP) models. These methods are easy to implement, require no simulation, and involve only univariate and bivariate cumulative normal distribution function evaluations, regardless of the number of alternatives, or the number of choice occasions per observation unit, or the number of observation units, or the nature of social/spatial dependence structures. In the spatial analysis literature, the two workhorse specifications to capture spatial dependencies are the spatial lag and the spatial error specifications (Anselin, 1988). The spatial lag specification, in reduced form, allows spatial dependence through both spatial spillover 43 effects (observed exogenous variables at one location having an influence on the dependent variable at that location and neighboring locations) as well as spatial error correlation effects (unobserved exogenous variables at one location having an influence on the dependent variable at that location and neighboring locations). The spatial error specification, on the other hand, assumes that spatial dependence is only due to spatial error correlation effects and not due to spatial spillover effects. The spatial error specification is somewhat simpler in formulation and estimation than the spatial lag model. But, as emphasized by McMillen (2010), the use of a parametric spatial error structure is “troublesome because it requires the researcher to specify the actual structure of the errors”, while it is much easier to justify a parametric spatial lag structure when accommodating spatial dependence. Beck et al. (2006) also find theoretical and conceptual issues with the spatial error model and refer to it as being “odd”, because the formulation rests on the “hard to defend” position that “space matters in the error process but not in the substantive portion of the model”. As they point out, the implication is that if a new independent variable is added to a spatial error model “so that we move it from the error to the substantive portion of the model”, the variable magically ceases to have a spatial impact on neighboring observations. Of course, the spatial lag and spatial error specifications can be combined together in a Kelejian-Prucha specification (see Elhorst, 2010), or the spatial lag could be combined with spatially lagged exogenous variable effects in a Spatial Durbin specification (see Bhat et al., 2014a). In all of these cases, the spatial dependence leads also to spatial heteroscedasticity in the random error terms. In this monograph, we will assume the spatial lag structure as the specification of spatial dependency. However, it is very straightforward to extend our approach to other dependency specifications. Indeed, there is no conceptual difficulty in doing so, nor is there much impact on coding or computational burden. The focus on the spatial lag structure is simply for uniformity and notational ease. In addition to the spatial lag-based and resulting heteroscedasticity effect, it is also likely that there is heterogeneity (i.e., differences in relationships between the dependent variable of interest and the independent variables across decision-makers or spatial units (see, Fotheringham and Brunsdon, 1999, Bhat and Zhao, 2002, Bhat and Guo, 2004). When combined with the spatial lag effect, the unobserved heterogeneity effects get correlated over decision agents based on the spatial (or social) proximity of the agents’ locations, which is then referred to as spatial drift (see Bradlow et al., 2005 for a discussion). But such spatial drift effects have been largely ignored thus far in the literature (but see Bhat et al., 2014a). We explicitly incorporate such drift effects in the models discussed below. All notations from previous sections carry over to the sections below. 2.3.1 Spatial Ordered Response Models 2.3.1.1 The Spatial CUOP Model The spatial CUOP (SCUOP) is an extension of the aspatial CUOP model from Section 2.2.1.1, and may be written as follows: 44 q q Q q q qq q y w y x βq 1 ' ' ' , k yq if k q q k q y , 1 , , (2.34) where the ' qq w terms are the elements of an exogenously defined distance-based spatial (or social) weight matrix W corresponding to individuals q and q (with 0 qq w and 1 q q q w ), and ) 1 0 ( is the spatial autoregressive parameter. The weights ' qq w can take the form of a discrete function such as a contiguity specification ( q q w =1 if the individuals q and q are adjacent and 0 otherwise) or a specification based on a distance threshold ( ' ' ' , / q qq qq q q c c w where ' qq c is a dummy variable taking the value 1 if the individual q is within the distance threshold and 0 otherwise). It can also take a continuous form such as those based on the inverse of distance q q d and its power functions ), 0 ( 1 ) 1 ( 1 n /d /d w q n q q n q q q q the inverse of exponential distance, and the shared edge length q q d ~ between individuals (or observation units) ' ' ' ' ' , ~ ~ / ~ ~ q qq qq qq qq q q d c d c w (where ' ~ qq c is a dummy variable taking the value 1 if q and q are adjoining based on some pre-specified spatial criteria, and 0 otherwise). All of these functional forms for the weight matrix may be tested empirically. The latent propensity representation of Equation (2.34) can be written equivalently in vector notation as: ε β ~ ~ b y y x x W , (2.35) where ) ..., , , ( 2 1 y Q y y y and ) ,..., , ( 2 1 Q ε are (Q×1) vectors, ) ..., , , ( 2 1 Q x x x x is a (Q×L) matrix of exogenous variables for all Q individuals, x ~ is a (Q×QL) block-diagonal matrix with each block-diagonal of size (1×L) being occupied by the vector q x ( Q q ,..., 2 , 1 ), and ) ..., , , ( 2 1 Q β ~ β ~ β ~ β ~ is a (QL×1) vector. Through simple matrix algebra manipulation, Equation (2.35) may be re-written as: ε β ~ ~ b y x x S , (2.36) where 1 -Q W IDEN S is a (Q×Q) matrix. The vector y is multivariate normally distributed as ) , ( ~ Ξ B y Q MVN , where b B Sx and S IDEN x Ω IDEN x S Ξ Q Q ~ ~ . (2.37) The likelihood function ) (θ L for the SCUOP model takes the following form: 45 , ) , | ( ) ( ) ( y D Q d f P L y B y m y θ Ξ (2.38) where ) ,..., , ( 2 1 Q y y y y , ) ,..., , ( 2 1 Q m m m m is the corresponding (Q×1) vector of the actual observed ordinal levels, y D is the integration domain defined as } ..., , 2 , 1 , : { , 1 , y Q q y D q q m q q m q y , and (.) Q f is the multivariate normal density function of dimension Q . The rectangular integral in the likelihood function is of dimension Q, which can become problematic from a computational standpoint. Further, the use of traditional numerical simulation techniques can lead to convergence problems during estimation even for moderately sized Q (Bhat et al., 2010a; Müller and Czado, 2005). The alternative is to use the composite marginal likelihood (CML) approach. Using a pairwise CML method, the function to be maximized is: ) , , ( ) , , ( ) , , ( ) , , ( , is That .
] [ ] [ , ] [ ] [ where , ) ( ' ' 2 ' ' 2 ' ' 2 ' ' 2 ' ' 1 1 1 ' SCUOP qq q q qq q q qq q q qq q q q q q q q q q q Q q Q q q q q CML ν μ μ ν μ ν μ ν L P L L L m y m y θ (2.39) where ' ' ' 1 , , ] [ ] [ ] [ , ] [ ] [ , ] [ ] [ q q qq qq qq qq q m q q qq q m q q ν μ q q Σ Σ Σ Σ Σ B B . In the above expression, q ] [B represents the th q element of the column vector B, while ' qq ] [Σ represents the th q q element of the matrix Σ . The pairwise marginal likelihood function of Equation (2.39) comprises 2 / ) 1 ( Q Q pairs of bivariate probability computations, which can itself become quite time consuming. However, previous studies (Varin and Vidoni, 2009, Bhat et al., 2010a, Varin and Czado, 2010) have shown that spatial dependency drops quickly with inter-observation distance. Therefore, there is no need to retain all observation pairs because the pairs formed from the closest observations provide much more information than pairs far from one another. The “optimal” distance for including pairings can be based on minimizing the trace of the asymptotic covariance matrix. Thus, the analyst can start with a low value of the distance threshold (leading to a low number of pairwise terms in the CML function) and then continually increase the distance threshold up to a point where the gains from increasing the distance threshold is very small or even drops. To be specific, for a given threshold, construct a Q×Q matrix R ~ with its th q column filled with a Q×1 vector of zeros and ones as follows: if the observational unit q is not within the specified threshold distance of unit q, the th q row has a value of zero; otherwise, the th q row has a value of one. By construction, the th q row of the th q column has a value of one. Let q q R ~ be the th qq element of the matrix R ~ , and let 1 1 1 .
~ ~ Q q Q q q q q W R Define a set q C ~ of 46 all individuals (observation units) that have a value of ‘1’ in the vector , ~ q R where q R ~ is the qth column of the vector R ~ . Then, the CML function is as follows: 1 1 ~ 1 ' .
) ( Q q Q C q q q q q CML q L L θ (2.40) The covariance matrix of the CML estimator is W W ~ ˆ ˆ ˆ ~ ˆ 1 1 -1 -H J H G , where CML q q q Q C q q q Q q L W θ θ θ H ˆ 2 ~ 1 1 1 log ~ 1 ˆ , or alternatively, (2.41) 1 1 ~ 1 ˆ log log ~ 1 ˆ Q q Q C q q q q q q q q L L W CML θ θ θ H (2.42) However, the estimation of the “vegetable” matrix J is more difficult in this case. One cannot empirically estimate J as the sampling variance of the individual contributions to the composite score function (as was possible when there were Q independent contributions) because if the underlying spatial dependence in observation units. But a windows resampling procedure (see Heagerty and Lumley, 2000) may be used to estimate J . This procedure entails the construction of suitable overlapping subgroups of the sample that may be viewed as independent replicated observations. Then, J may be estimated empirically. While there are several ways to implement this, Bhat (2011) suggests overlaying the spatial region under consideration with a square grid providing a total of Q ~ internal and external nodes. Then, select the observational unit closest to each of the Q ~ grid nodes to obtain Q ~ observational units from the original Q observational units ( ).
~ , , 3 , 2 , 1 ~ Q q Let q R~ ~ be the 1 Q matrix representing the th q ~ column vector of the matrix R ~ , let q ~ ~ C be the set of all individuals (observation units) that have a value of ‘1’ in the vector q R~ ~ , and let q y~ be the sub-vector of y with values of ‘1’ in the rows of q R~ ~ . Let q N ~ be the sum (across rows) of the vector q R~ ~ (that is, q N ~ is the cardinality of q ~ ~ C ), so that the dimension of q y~ is .
1 ~ q N Let q l~ be the index of all elements in the vector q y~ , so that q l~ =1,2,… q N~ . Next, define .
2 / ) 1 ( ~ ~ ~ q q q N N C Then, the J matrix maybe empirically estimated as: .
log log 1 ~ 1 ˆ ˆ ~ 1 ~ 1 1 1 1 1 1 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ CML θ θ θ Q q N l N l l l l N l N l l l l q q q q q q q q q q q L L C Q J (2.43) 47 To ensure the constraints on the autoregressive term , the analyst can parameterize )] ~ exp( 1 /[ 1 . Once estimated, the ~ estimate can be translated back to estimates of an estimate of . References for the CML Estimation of the Spatial CUOP (or SCUOP) Model Ferdous, N., Pendyala, R.M., Bhat, C.R., Konduri, K.C., 2011. Modeling the influence of family, social context, and spatial proximity on use of nonmotorized transport mode, Transportation Research Record 2230, 111-120. Spissu, E., Eluru, N., Sener, I.N., Bhat, C.R., Meloni, I., 2010. Cross-clustered model of frequency of home-based work participation in traditionally off-work hours. Transportation Research Record 2157, 138-146. Whalen, K.E., Paez, A., Bhat, C., Moniruzzaman, M., Paleti, R., 2012. T-communities and sense of community in a university town: evidence from a student sample using a spatial ordered-response model. Urban Studies 49(6), 1357-1376. 2.3.1.2 The Spatial CMOP Model We start with Equation (2.6) of the aspatial CMOP model in Section 2.2.1.2, and now add a spatial lag formulation: i qi qi Q q i q qq i qi k y y w y , 1 ' ' ' q qi x β if i k q qi i k q i i y , 1 , . (2.44) Define , vector), 1 × ( ) ,..., , ( 2 1 I y y y qI q q q y ] ) ( ,..., ) ( , ) ( , ) [( 3 2 1 Q y y y y y (QI×1 vector), vector), 1 × ( ) ,..., , ( 2 1 I y y y qI q q q y ] ) ( ,..., ) ( , ) ( , ) [( 3 2 1 Q y y y y y (QI×1 vector), ) ..., . , , ( 2 1 qI q q q m m m m vector) 1 × (I , ) ,..., , ( 2 1 Q m m m m (QI×1 vector) , q q x I IDEN x ~ (I×IL matrix; I IDEN is an identity matrix of size I), ) ~ ,...
~ , ~ , ~ ( ~ 3 2 1 Q x x x x x (QI×IL matrix), , ~ qi qi β b β i vector), 1 × ( ) ~ ..., ~ , ~ ( ~ 2 1 IL qI q q q β β β β ) , 0 ( ~ ~ Ω L I q MVN β (the q β ~ random coefficients are independent across individuals), vector), 1 × (Q ) ~ ..., ~ , ~ ( ~ 2 1 IL Q β β β β ) ,..., , ( 2 1 I b b b b (IL×1 vector), , )' , , , , ( 3 2 1 qI q q q q ε )' , , , , ( 3 2 1 Q ε ε ε ε ε (QI×1 vector), ) ..., , , , ( 3 3 1 I δ (I×1 vector), and δ δ Q 1 ~ (QI×1 vector; Q 1 is a vector of size Q with all elements equal to 1). Also, define the following matrix: matrix). ( ~ ~ ~ ~ QIL QI Q 3 2 1 x 0 0 0 0 0 0 x 0 0 0 0 0 x 0 0 0 0 0 x x (2.45) 48 Collect all the weights ' qq w into a row-normalized spatial weight matrix W. All other notations from Section 2.2.1.2 are carried over to this section, including the multivariate standard normal distribution specification for q ε with mean zero and correlation matrix Σ (see Equation 2.7). With these definitions, Equation (2.44) may be re-written in matrix form as: ε x x IDEN W β b y δ y ~ ~ )] ( .
~ [ I , (2.46) where the operation ' '. in the equation above is used to refer to the element by element multiplication. After further matrix manipulation, we obtain: ), ~ ( ~ ε β b y x S x S where .
.
~ 1 I QI IDEN W IDEN S δ (2.47) The expected value and variance of y may be obtained from the above equation after developing the covariance matrix for the error vector ) ~ ( ε β x S . This may be written as S Σ IDEN x Ω IDEN x S Ξ Q Q . Then, we obtain ) , ( ~ Ξ B y QI MVN , where b B x S~ . The parameter vector to be estimated in the SCMOP model is . ) , , , , , ( δ α γ b θ Σ Ω Let 1 ( ) ,..., , ( , 2 , 1 , 2 I I m q m q m q qI q qi up q ψ vector), ) 1 ( ) ,..., , ( 1 , 2 1 , 1 1 , 2 I I m q m q m q qI q qi low q ψ vector, 1 ( ) ,..., , ( QI up Q up 2 up 1 up ψ ψ ψ ψ vector), and 1 ( ) ,..., , ( QI low Q low 2 low 1 low ψ ψ ψ ψ vector). The likelihood function for the SCMOP model is: , ) , | ( ) ( ) ( y D QI d f P L y B y m y θ Ξ (2.48) where : { up low ψ ψ y y y D , and (.) QI f is the multivariate normal density function of dimension QI. The dimensionality of the rectangular integral in the likelihood function is QI, which is very difficult to evaluate using existing estimation methods. The alternative is to use the pairwise composite marginal likelihood (CML) approach: Q q Q q q I i I i i i i q q CML L L 1 1 θ with , when i i q q where ) , ~ , ~ ( ) , ~ , ~ ( ) , ~ , ~ ( ) , ~ , ~ ( 2 2 2 2 i i q q i q i q i i q q i q i q i i q q i q i q i i q q i q i q i i q q L , (2.49) , ~ , ~ ) 1 ( , ) 1 ( ) 1 ( 1 , ) 1 ( , ) 1 ( ) 1 ( , i I q i I q i I q i m q i q i I q i I q i I q i m q i q qi qi Ξ Ξ B B and i I q i I q i I q i I q i I q i I q i i q q ) 1 ( , ) 1 ( ) 1 ( , ) 1 ( ) 1 ( , ) 1 ( Ξ Ξ Ξ . 49 The CML estimator is obtained by maximizing the logarithm of the function in Equation (2.49). The number of pairings in the CML function above is .
2 / ) 1 ( QI QI But again the number of pairings can be reduced by determining the “optimal” distance for including pairings across individuals based on minimizing the trace of the asymptotic covariance matrix (as discussed in the previous section).11 To do so, define a set q C ~ as in the previous section that includes the set of individuals q’ (including q) that are within a specified threshold distance of individual q. Then, the CML function reduces to the following expression: Q q Q C q q q I i I i i i i q q CML q L L 1 ~ 1 θ with .
when i i q q (2.50) Let W ~ be the total number of pairings used in the above CML function (after considering the distance threshold). The covariance matrix of the CML estimator is W W ~ ˆ ˆ ˆ ~ ˆ 1 1 -1 -H J H G , where , when log ~ 1 ˆ ˆ 1 ~ 1 2 i i q q L W CML q Q q Q C q q q I i I i i i i q q θ θ θ H (2.51) or alternatively, CML θ θ θ H ˆ 1 ~ 1 log log ~ 1 ˆ Q q Q C q q q i i q q i i q q I i I i i q L L W .
when i i q q (2.52) The sandwich matrix, J ˆ , may be computed by selecting Q ~ ) ~ ,..., 2 , 1 ~ ( Q q observational units from the original Q observational units as discussed in the earlier section. Let q C~ ~ be the set of individuals (observation units) within the specified threshold distance, and let q N ~ be the cardinality of q C~ ~ . Let q l~ be an index so that q l~ =1,2,… q N~ . Next, define .
2 / 1 ~ ~ ~ I N I N C q q q Then, the J matrix maybe empirically estimated as: 11 Technically, one can consider a threshold distance separately for each ordinal variable, so that the individual pairings within each variable are based on this variable-specific threshold distance and the individual-variable pairings across variables are based on different thresholds across variables. But this gets cumbersome, and so we will retain a single threshold distance across all ordinal variables. 50 CML θ θ θ ˆ ~ 1 ~ 1 1 1 1 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ log log 1 ~ 1 ˆ Q q i i l l N l N l l I i I i i N l N l l i i l l I i I i i q q q q q q q q q q q L L C Q J (2.53) There is another way that the analyst can consider cutting down the number of pairings even after using a threshold distance as a cut-off. That is by ignoring the pairings among different individuals (observation units) across the I ordinal variables. This will reduce the number of pairings quite substantially, while also retaining the pairings across individuals for each ordinal variable (that enables the estimation of the parameters of the vector δ ) and the pairings across ordinal variables within the same individual (that enables the estimation of the parameters in the correlation matrix Σ of q ε ). The CML is: Q q I i I i i i qi Q q Q C q q q I i i q q CML L L L q 1 1 1 1 1 1 ~ 1 1 θ (2.54) The number of pairings W ~ in the CML function above is much smaller than the CML function in Equation (2.50). The elements of the covariance matrix may be estimated as follows: , log log ~ 1 ˆ ˆ 1 1 1 1 2 1 1 ~ 1 1 2 CML q Q q I i I i i i qi Q q Q C q q q I i i q q L L W θ θ θ θ θ H (2.55) or alternatively, .
log log log log ~ 1 ˆ ˆ 1 1 1 1 1 1 ~ 1 1 CML q Q q I i I i i i qi i qi Q q Q C q q q I i i q q i q q L L L L W θ θ θ θ θ H (2.56) For estimating the J ˆ matrix define q C~ ~ and q N ~ be defined as earlier and let Q I I I N N C q q q 2 / ) 1 ( 2 / ) 1 ( ~ ~ ~ and .
log log ˆ 1 1 1 1 1 1 ~ ~ ~ ~ ~ ~ ~ ~ ~ CML θ q θ θ S q q q q q q q q N l I i I i i i i l l N l N l l i l l I i L L Then, .
1 ~ 1 ˆ ~ 1 ~ ~ ~ ~ Q q q C Q q qS S J (2.57) The positive-definiteness of the matrices Σ Ωand are ensured as discussed in Sections 2.3.1.1 and 2.2.1.2. 51 References for the CML Estimation of the Spatial CMOP (or SCMOP) Model No known applications. But the spatial cross-sectional multivariate count model of Narayanamoorthy et al. (2013) is very similar to the SCMOP model. 2.3.1.3. The Spatial PMOP (SPMOP) Model All notations from Section 2.2.1.3 are carried over. To include spatial dependency in the PMOP model, rewrite Equation (2.12) as follows: if , 1 ' k y y w y qt qt Q q t q q q qt qt ' qx β k t q qj k t q y , , 1 , , , (2.58) Define ) ..., . , , ( 2 1 qT q q q y y y y matrix) 1 ( T , ) ..., . , , ( 2 1 qT q q q ε matrix) 1 ( T , ) matrix 1 ( ) ..., , , ( 2 1 T y y y qT q q q y , matrix), ( )' ,..., , ( L T qT q q q x x x x 2 1 1 ( ) ,..., , ( , , , 2 , , 1 , 2 T qJ q qi m T q m q m q up q ψ vector), ) 1 ( ) ,..., , ( 1 , , 1 , 2 , 1 , 1 , 2 T qJ q qi m T q m q m q low q ψ vector. Also, let the vector of actual observed ordinal outcomes for individual q be stacked into a (T×1) vector ) ..., . , , ( 2 1 qT q q q m m m m . To write the equation system in (2.58) compactly, we next define several additional vectors and matrices. Let ] ) ( ,..., ) ( , ) ( , ) [( 3 2 1 Q y y y y y (QT×1 vector), ] ) ( ,..., ) ( , ) ( , ) [( 3 2 1 Q y y y y y (QT×1 vector), ) ,..., , ( 2 1 Q m m m m (QT×1 vector), ) ,...
, , ( Q x x x x x 3 2 1 (QT×L matrix), , ~ q q β b β ) , 0 ( ~ ~ Ω L q MVN β (the q β ~ random coefficients are independent across individuals), vector), 1 × (Q ) ~ ..., , ~ , ~ ( ~ 2 1 L Q β β β β )' , , , ( 3 2 1 Q ε ε ε ε ε , (QT×1 vector), Q 3 2 1 x x x x x 0 0 0 0 0 0 0 0 0 0 0 0 QL QT ( block diagonal matrix), (2.59) Also, collect all the weights q q w into a spatial weight matrix W. The vector β ~ above has a mean vector of zero and a covariance matrix Ω DEN I Q (of size QT×QT), while the vector ε has a mean vector of zero and a covariance matrix .
QT DEN I Using the vector and the matrix notations defined above, Equation (2.58) may be re-written compactly as: ε x x IDEN W β b y y ~ )] ( [ T , (2.60) After further matrix manipulation, we obtain: 52 ), ~ ε β b y x S( Sx where .
1 T QT IDEN W IDEN S (2.61) Next, we obtain ) , ( ~ Ξ B y QI MVN , where b B Sx and S IDEN x Ω IDEN x S Ξ QT Q (2.62) The parameter vector to be estimated in the SPMOP model is . ) , , , , ( α γ b θ Ω Let 1 ( ) ,..., , ( QT up Q up 2 up 1 up ψ ψ ψ ψ vector), and 1 ( ) ,..., , ( QT low Q low 2 low 1 low ψ ψ ψ ψ vector). The likelihood function for the SPMOP model is: , ) , | ( ) ( ) ( y D QT d f P L y B y m y θ Ξ (2.63) where : { up low ψ ψ y y y D , and (.) QT f is the multivariate normal density function of dimension QT. The much simpler pairwise composite marginal likelihood (CML) function is: Q q Q q q T t T t t t t q q CML L L 1 1 θ with , when t t q q where ) , ~ , ~ ( ) , ~ , ~ ( ) , ~ , ~ ( ) , ~ , ~ ( 2 2 2 2 t t q q t q qt t t q q t q qt t t q q t q qt t t q q t q qt t t q q L , , ~ , ~ ) 1 ( , ) 1 ( ) 1 ( 1 , , ) 1 ( , ) 1 ( ) 1 ( , , t I q t T q t T q m t q qt t T q t T q t T q m t q qt qt qt Ξ Ξ B B and (2.64) t T q t T q t T q t T q t T q t T q t t q q ) 1 ( , ) 1 ( ) 1 ( , ) 1 ( ) 1 ( , ) 1 ( Ξ Ξ Ξ . To reduce the number of pairings, define a set q C ~ as in the previous section that includes the set of individuals q’ (including q) that are within a specified threshold distance of individual q. Then, the CML function reduces to the following expression: Q q Q C q q q T t T i t t t q q CML q L L 1 ~ 1 θ with .
when t t q q (2.65) Let W ~ be the total number of pairings used in the above CML function (after considering the distance threshold). The covariance matrix of the CML estimator is W W ~ ˆ ˆ ˆ ~ ˆ 1 1 -1 -H J H G , where 53 , when log ~ 1 ˆ ˆ 1 ~ 1 2 t t q q L W CML q Q q Q C q q q T t T t t t t q q θ θ θ H (2.66) or alternatively, CML θ θ θ H ˆ 1 ~ 1 log log ~ 1 ˆ Q q Q C q q q t t q q t t q q T t T t t q L L W .
when t t q q (2.67) Defining q C~ ~ , q N ~ , and 2 / 1 ~ ~ ~ I N I N C q q q as in the previous section, the J matrix maybe empirically estimated as: CML θ θ θ ˆ ~ 1 ~ 1 1 1 1 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ log log 1 ~ 1 ˆ Q q t t l l N l N l l T t T t t N l N l l t t l l T t T t t q q q q q q q q q q q L L C Q J (2.68) One can also ignore the pairings among different individuals (observation units) across the T time periods. The CML then is: Q q T t T t t t qt Q q Q C q q q T t t q q CML L L L q 1 1 1 1 1 1 ~ 1 1 θ (2.69) The elements of the covariance matrix in this case may be estimated as follows: , log log ~ 1 ˆ ˆ 1 1 1 1 2 1 1 ~ 1 1 2 CML q Q q T t T t t t qt Q q Q C q q q T t t q q L L W θ θ θ θ θ H (2.70) or alternatively, , log log log log ~ 1 ˆ ˆ 1 1 1 1 1 1 ~ 1 1 CML q Q q T t T t t t qt t qt Q q Q C q q q T t t q q t q q L L L L W θ θ θ θ θ H (2.71) For estimating the J ˆ matrix, define q C~ ~ and q N ~ as earlier and let Q I I I N N C q q q 2 / ) 1 ( 2 / ) 1 ( ~ ~ ~ and .
log log ˆ 1 1 1 1 1 1 ~ ~ ~ ~ ~ ~ ~ ~ ~ CML θ q θ θ S q q q q q q q q N l T i T t t t t l l N l N l l t l l T t L L Then, 54 .
1 ~ 1 ˆ ~ 1 ~ ~ ~ ~ Q q q C Q q qS S J (2.72) References for the CML Estimation of the Spatial PMOP (SPMOP) Model Castro, M., Paleti, R., Bhat, C.R., 2013. A spatial generalized ordered response model to examine highway crash injury severity. Accident Analysis and Prevention 52, 188-203. Ferdous, N., Bhat, C.R., 2013. A spatial panel ordered-response model with application to the analysis of urban land-use development intensity patterns. Journal of Geographical Systems 15(1), 1-29. Paleti, R., Bhat, C.R., Pendyala, R.M., Goulias, K.G., 2013. Modeling of household vehicle type choice accommodating spatial dependence effects. Transportation Research Record 2343, 86-94. 2.3.2. Unordered-Response Models 2.3.2.1. The Spatial CMNP (SCMNP) Model The formulation in this case is similar to the aspatial case in Section 2.2.2.1, with the exception that a spatial lag term is included. Of course, this also completely changes the model structure from the aspatial case. , 1 | | ); , ( ~ ~ , ~ ; Ω 0 L qi i q q q q qi MVN U w U q q q qi q β β b β x β (2.73) where all notations are the same as in Section 2.2.2.1.12 Let ) ,..., , ( 2 1 qI q q q ξ ( 1 I vector). Then, we assume ) , 0 ( ~ Λ I MVN q ξ . As usual, appropriate scale and level normalization must be imposed on Λ for identifiability, as discussed in Section 2.2.2.1. The model above may be written in a more compact form by defining the following vectors and matrices: ) ,..., , ( 2 1 qI q q q U U U U 1 ( I vector), ) ,..., , ( Q U U U U 2 1 ( 1 QI vector), ) ( 2 1 Q ξ ξ ξ ξ ,..., , ( 1 QI vector), ) ,..., , , ( qI q q q q x x x x x 3 2 1 L I ( matrix), ) ( x 2 1 Q x x x ,..., , ( L QI matrix), and Q β β β β ~ ,..., ~ , ~ ~ 2 1 ( 1 QL vector). Also, define the following matrix: 12 One can allow the spatial lag dependence parameter δ to vary across alternatives i. However, due to identification considerations, one of the alternatives should be used as the base (with a zero dependence parameter). But doing so while also allowing the dependence parameters to vary across the remaining alternatives creates exchangeability problems, since the model estimation results will not be independent of the decision of which alternative is considered as the base. Hence, we prefer the specification that restricts the dependence parameter to be the same across alternatives i. 55 matrix), ( QL QI Q x x x x 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x 3 2 1 (2.74) Then, we can write Equation (2.73) in matrix form as: , ~ ξ β b x x S U (2.75) where matrix), ( ) ( QI QI I QI 1 IDEN W IDEN S and Wis the ) ( Q Q weight matrix with the weights q q w as its elements. Also, ), ~ , ( ~ Ξ V U QI MVN where b Sx V and .
S Λ) IDEN x Ω IDEN x S Ξ Q Q ( ) ( ~ Let ) ( ) , , ( 2 1 q qI q q q m i u u u u be an (I–1)×1 vector for individual q, where q m is the actual observed choice of individual q and ).
( q qm qi qi m i U U u q Stack the q u vectors across individuals (observation units): ] Vector 1 ) 1 ( [ ) ,..., , ( 2 1 I Q Q u u u u . The distribution of u may be derived from the distribution of U by defining a ] [ )] 1 ( [ QI I Q block diagonal matrix M, with each block diagonal having ) 1 ( I rows and I columns corresponding to each individual q. This I I ) 1 ( matrix for individual q corresponds to an ) 1 ( I identity matrix with an extra column of ‘ 1 ’ values added as the q m th column. For instance, consider the case of I = 4 and Q = 2. Let individual 1 be observed to choose alternative 2 and individual 2 be observed to choose alternative 1. Then M takes the form below. 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 M . (2.76) With the above construction for matrix M, it is easy to see that ), , ( ~ ) 1 ( Ξ u B I Q MVN where .
~ and M Ξ M Ξ V M B The likelihood of the observed sample (i.e., individual 1 choosing alternative 1 m , individual 2 choosing alternative 2 m , …, individual Q choosing alternative Q m ) may then be written succinctly as ] [ Prob 1) ( 0 u I Q . The parameter vector to be estimated is . ) , , , ( Λ Ω θ b Using the usual notations, the likelihood function is: 56 ), ), ( ( ) ( 1 ) 1 ( Ξ Ξ B ω θ I Q L (2.77) where .
1 1 Ξ Ξ ω Ξ ω Ξ This is cumbersome and impractical (if not infeasible) for most realistically-sized sample sizes. However, one can use the MACML technique. To do so, write the pairwise CML function corresponding to the full likelihood of Equation (2.77) as: , ) , Pr( where , ) ( 1 1 1 Q q Q q q q q q q q q q q SCMNP CML m d m d L L L θ (2.78) q d is an index for the individual’s choice, and ), ), (-( ) , Pr( 1 ~ q q q q B q q Ξ ωΞ J q q q q m d m d (2.79) where ) 1 ( 2 I J , , , ' q q Δ Ξ Δ Ξ Δ q q q q q q q q B B , 1 ~ 1 ~ q q q q Ξ q q Ξ q q ω Ξ ω Ξ and q q Δ is a ) 1 ( I Q J -selection matrix with an identity matrix of size ( 1 I ) occupying the first ( 1 I ) rows and the th I q 1 ) 1 ( ) 1 ( through th I q ) 1 ( columns, and another identity matrix of size ( 1 I ) occupying the last ( 1 I ) rows and the th I q 1 ) 1 ( ) 1 ( through th I q ) 1 ( columns. The number of pairings in the CML expression of Equation (2.78) can be reduced as explained in Section 2.3.1.1. Specifically, define a set q C ~ as in the previous section that includes the set of individuals q’ (including q) that are within a specified threshold distance of individual q. Then, the CML function reduces to the following expression: 1 1 ~ 1 ' .
) ( Q q Q C q q q q q SCMNP CML q L L θ (2.80) The expressions to obtain the covariance matrix are exactly the same as in Section 2.3.1.1, with ).
), (-( 1 ~ q q q q B q q Ξ ωΞ J q q L References for the CML Estimation of the Spatial CMNP (SCMNP) Model Bhat, C.R., 2011. The maximum approximate composite marginal likelihood (MACML) estimation of multinomial probit-based unordered response choice models. Transportation Research Part B 45(7), 923-939. Bhat, C.R., Sidharthan, R., 2011. A simulation evaluation of the maximum approximate composite marginal likelihood (MACML) estimator for mixed multinomial probit models. Transportation Research Part B 45(7), 940-953. Sener, I.N., Bhat, C.R., 2012. Flexible spatial dependence structures for unordered multinomial choice models: formulation and application to teenagers’ activity participation. Transportation 39(3), 657-683. 57 Sidharthan, R., Bhat, C.R, Pendyala, R.M., Goulias, K.G., 2011. Model for children's school travel mode choice: accounting for effects of spatial and social interaction. Transportation Research Record 2213, 78-86. 2.3.2.2. The Spatial CMMNP Model Rewrite Equation (2.18) from Section 2.2.2.2 to include spatial dependency in the utility that individual q attributes to alternative g i ( g i =1,2,..., ) g I for the gth variable. , 1 ' ' g g g qgi qgi Q q qq g qgi U w U g qgi qg x β (2.81) with all notations as earlier. g qgi x is an 1 g L -column vector of exogenous attributes, ), , ( ~ g g qg b β Ω q L MVN and ) , 0 ( ~ g qg ξ Λ I MVN ( ) ,...
, ( 2 1 g qgI qg qg qg ξ ( 1 g I vector)). As in Section 2.2.2.2, we will assume that the ) ~ ( qg g qg β b β vectors are independent across the unordered-response dimensions for each individual. We also assume that g qgi is independent and identically normally distributed across individuals q. Let qg m be the actual chosen alternative for the gth unordered-response variable by individual q. Define the following: ) ,..., , ( 2 1 g qgI qg qg qg U U U U ( 1 g I vector), ) ,...
, ( 2 1 qG q q q U U U U 1 ~ G vector , ~ 1 G g g I G ) ,...
, ( qG q2 q1 q ξ ξ ξ ξ ( 1 ~ G vector), ) ,...
, ( 2 1 Q U U U U ( 1 ~ G Q vector), ) ,..., , ( Q ξ ξ ξ ξ 2 1 ( 1 ~ G Q vector) , ) ,..., , ( G b b b b 2 1 ( 1 ~ L vector) , ~ 1 G g g L L ) ,..., , ( 2 1 g qgI qg qg qg x x x x ( g g L I matrix), ), matrix ~ ~ ( 0 0 0 0 0 0 0 0 0 0 0 0 0 .
0 0 0 3 2 1 L G qG q q q q x x x x x matrix), ~ ~ ( L Q G Q Q 3 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x x x x x ) ,..., , ( 1 Q 2 x x x x ( L G Q ~ ~ matrix), 1 ~ ( ~ ,..., ~ , ~ ~ L qG q2 q1 q β β β β vector), and Q β β β β ~ ,..., ~ , ~ ~ 2 1 ( 1 ~ L Q vector) . Let ), , ( ~ ~ ~ Λ 0G G MVN q ξ where the covariance matrix Λ is 58 to be constructed as discussed in Section 2.2.2.2. Then, ) , ( ~ ~ Λ IDEN 0 Q G Q G Q MVN ~ ξ . Also, define g g I I ( ~ qg g qg qg x Ω x Ω matrix), and the following matrices: ), matrix ~ ~ ( , ~ 0 0 0 0 0 0 ~ 0 0 0 0 0 ~ 0 0 .
0 0 0 ~ ~ ), matrix ~ ~ ( ~ 0 0 0 0 0 0 ~ 0 0 0 0 0 ~ 0 0 .
0 0 0 ~ ~ G Q G Q G G Q 3 2 1 qG q3 q2 q1 q Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω and ), matrix ~ ~ ( 0 0 0 0 0 0 0 0 0 0 0 0 0 .
0 0 0 ~ 3 2 1 G G G δ Equation (2.81) may then be written in matrix form as: , ~ ξ β b x x S U (2.82) where , ) ( ).
~ ( ~ ~ G QQ G Q IDEN W δ 1 IDEN S W is the ) ( Q Q weight matrix with the weights q q w , and “ . ” refers to the element-by-element multiplication of the two matrices involved. Also, ), ~ , ( ~ ~ Ξ V U G Q MVN where b Sx V and .
( ~ ~ S Λ) IDEN Ω S Ξ Q 13 To develop the likelihood function, construct a matrix M as follows. First, for each unordered variable g and individual q, construct a matrix qg M with ) 1 ( g I rows and g I columns. This matrix corresponds to an ) 1 ( g I identity matrix with an extra column of ‘ 1 ’ values added as the th qg m column. Then, define the following: 13 One can also obtain Ω ~ as x Ω Ω Ω Ω IDEN x Ω G 3 2 1 Q ~ 0 0 0 0 0 0 ~ 0 0 0 0 0 ~ 0 0 .
0 0 0 ~ ~ 59 G g g G I G G G 1 3 2 1 ), 1 ( where , ) matrix ~ ( 0 0 0 0 0 0 0 0 0 0 0 0 0 .
0 0 0 q q q q q M M M M M and (2.83) ).
matrix ~ ( 0 0 0 0 0 0 0 0 0 0 0 0 0 .
0 0 0 3 2 1 G Q G Q Q M M M M M (2.84) The parameter vector to be estimated is . ) ~ ~ ~ , ~ , ( 2 1 δ , Λ Ω ,..., Ω Ω θ G b Using the usual notations, the likelihood function is: 1 1 1 and , ~ , where ), ), ( ( ) ( Ξ Ξ Ξ ω Ξ ω Ξ M Ξ M Ξ MV Ξ ω θ B B G Q L (2.85) The likelihood function is of a very high dimensionality. Instead, consider the (pairwise) composite marginal likelihood function. Further, as in Section 2.1.2.2, we can reduce the pairings by testing different distance bands and determining the “optimal” distance for including pairings across individuals based on minimizing the trace of the asymptotic covariance matrix. Define a set q C ~ that includes the set of individuals q’ (including q) that are within a specified threshold distance of individual q. Then, the CML function reduces to the following expression: Q q Q C q q q G g G g g g g q q CML q L L 1 ~ 1 θ with , when g g q q where (2.86) ), ), (-( ) , Pr( 1 ~ g g q q g g q q B g g q q Ξ ωΞ g g J g q g q qg qg g g q q m d m d L and , 2 g g g g I I J , , ' g g q q Δ Ξ Δ Ξ Δ g g q q g g q q g g q q g g q q g g q q B B , 1 ~ 1 ~ g g q q g g q q g g q q g g q q Ξ Ξ ω Ξ ω Ξ and g g q q Δ is a G Q J -selection matrix with an identity matrix of size ( 1 g I ) occupying the first ( 1 g I ) rows and the th g l l I G q 1 ) 1 ( 1 1 through th g l l I G q 1 ) 1 ( columns, and another identity matrix of size ( 1 g I ) occupying the last ( 1 g I ) rows and the 60 th g l l I G q 1 ) 1 ( 1 1 through th g l l I G q 1 ) 1 ( columns (with the convention that ).
0 0 1 l l I The model can now be estimated using the MACML method. The computation of the covariance matrix is identical to the case in Section 2.2.2.2, with the use of g g q q L as in Equation (2.86) above. Once again, the analyst can consider further cutting down the number of pairings by ignoring the pairings among different individuals (observation units) across the G variables. References for the CML Estimation of the Spatial Cross-Sectional Multivariate MNP (SCMMNP) Model No known applications thus far. 2.2.2.3 The Spatial Panel MNP Model Consider the following model with ‘t’ now being an index for choice occasion: . ..., , 2 , 1 , ..., , 2 , 1 , ..., , 2 , 1 , ) , ( ~ , I i T t Q q MVN U w δ U L qti q ti q q q qit Ω b β x β q qti q (2.87) We assume that qit is independent and identically normally distributed across individuals and choice occasions, but allow a general covariance structure across alternatives for each choice instance of each individual. Specifically, let ) ,...
, ( 2 1 qtI qt qt qt ξ ( 1 I vector). Then, we assume ) , 0 ( ~ Λ I MVN qt ξ . As usual, appropriate scale and level normalization must be imposed on Λ for identifiability. Next, define the following vectors and matrices: ) ,..., , ( 2 1 qtI qt qt qt U U U U 1 ( I vector), ) ,..., , ( 2 1 qI q q q U U U U 1 ( TI vector), ) ,...
, ( qT q q q ξ ξ ξ ξ 2 1 1 ( TI vector), ) ,..., , ( 2 1 qtI qt qt qt x x x x ( L I matrix), ) ..., , ( 2 1 qT q q q x x x x ( L TI matrix), ) ,...
, ( 2 1 Q U U U U , ) ,...
, ( Q ξ ξ ξ ξ 2 1 ( 1 QTI vectors), and ) ,..., , ( 2 1 Q x x x x ( L QTI matrix). Let ) , ( ~ ~ , ~ Ω 0 L MVN q q q β β b β , Q β β β β ~ ,..., ~ , ~ ~ 2 1 , and ), matrix ( 0 0 0 0 0 0 0 0 0 0 0 0 0 .
0 0 0 ~ 3 2 1 QK QTI Q x x x x x (2.88) Then, we can write Equation (2.87) in matrix notation as: 61 ξ β b ~ ~ x x S U , (2.89) with ).
matrix ( ) ( 1 QTI QTI I T QTI IDEN IDEN W IDEN S Then, ), ~ , ( ~ Ξ V U QTI MVN where b Sx V and .
( ~ S Λ) Ω IDEN S Ξ Q To develop the likelihood function, define M as an ] [ )] 1 ( [ QTI I QT block diagonal matrix, with each block diagonal having ) 1 ( I rows and I columns corresponding to the tth observation time period on individual q. This I I ) 1 ( matrix for parcel q and observation time period t corresponds to an ) 1 ( I identity matrix with an extra column of “ 1 ” values added as the qt m th column. For instance, consider the case of Q = 2, T = 2, and I = 4. Let individual 1 be observed to choose alternative 2 in time period 1 and alternative 1 in time period 2, and let individual choose alternative 3 in time period 1 and in alternative 4 in time period 2. Then M takes the form below. 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 M (2.90) Let .
~ and M Ξ M Ξ V M B The parameter vector to be estimated is , ) , , , ( Λ Ω θ b and the likelihood function is: ), ), ( ( ) ( 1 ) 1 ( Ξ Ξ B ω θ I QT L (2.91) where .
1 1 Ξ Ξ ω Ξ ω Ξ Now, consider the following (pairwise) composite marginal likelihood function formed by taking the products (across the T choice occasions) of the joint pairwise probability of the chosen alternatives qt m for the tth choice occasion and qg m for the gth choice occasion for individual q. To reduce the number of pairings, define a set q C ~ as in the previous section that includes the set of individuals q’ (including q) that are within a specified threshold distance of individual q. Then, the CML function reduces to the following expression: 62 Q q Q C q q q T t T i t t t q q CML q L L 1 ~ 1 θ with t t q q when , where (2.92) ), ), (-( ) , Pr( 1 ~ ) 1 ( 2 t t q q t t q q B t t q q Ξ ωΞ I t q t q qt qt t t q q m d m d L where , 1 ~ 1 ~ t t q q t t q q Ξ t t q q Ξ t t q q ω Ξ ω Ξ , , ' t t q q t t q q Δ Ξ Δ Ξ Δ t t q q t t q q t t q q t t q q B B and t t q q Δ is a ) 1 ( ) 1 ( 2 I QT I -selection matrix with an identity matrix of size ( 1 I ) occupying the first ) 1 ( I rows and the th I t T I q 1 ) 1 ( ) 1 ( ) 1 ( ) 1 ( through th I t T I q ) 1 ( ) 1 ( ) 1 ( columns, and another identity matrix of size ( 1 I ) occupying the last ( 1 I ) rows and the th I t T I q 1 ) 1 ( ) 1 ( ) 1 ( ) 1 ( through th I t T I q ) 1 ( ) 1 ( ) 1 ( columns. The model can now be estimated using the MACML method. The computation of the covariance matrix is identical to the case in Section 2.2.2.2 with the use of t t q q L as in Equation (2.92) above. The analyst can consider further cutting down the number of pairings by ignoring the pairings among different individuals (observation units) across the T time periods. References for the CML Estimation of the Spatial Panel MNP (SPMNP) Model Bhat, C.R., 2011. The maximum approximate composite marginal likelihood (MACML) estimation of multinomial probit-based unordered response choice models. Transportation Research Part B 45(7), 923-939. Bhat, C.R., Sidharthan, R., 2011. A simulation evaluation of the maximum approximate composite marginal likelihood (MACML) estimator for mixed multinomial probit models. Transportation Research Part B 45(7), 940-953. Sidharthan, R., Bhat, C.R., 2012. Incorporating spatial dynamics and temporal dependency in land use change models. Geographical Analysis 44(4), 321-349. 2.4. Application to Count Models Count data models are used in several disciplines to analyze discrete and non-negative outcomes without an explicit upper limit. Applications of such count data models abound in the scholarly literature, both in number (a count in and of itself!) as well as diversity of topics. Applications include the analysis of (a) the number of doctor visits, the number of homes affected by cholera, the number of cancer incidents, and the number of milk formula bottles supplied to infants by breastfeeding mothers in the medicine field, (b) the number of crimes and the number of drug possession convictions in the criminology field, (c) the number of mergers and acquisitions of foreign direct investments, the number of faults in a bolt, the frequency of contract change orders, and the number of jobs by space unit in the economics field, (d) the number of harbor seals hauled out on glacial ice and the count of birds at sanctuaries in the ecology field, and (e) 63 roadway crash frequency, counts of flights from airports, and the number of drinking under intoxication (DUI) infractions in the transportation field. Count data models assume a discrete probability distribution for the count variables, followed by the parameterization of the mean of the discrete distribution as a function of explanatory variables. The two most commonly used discrete probability distributions are the Poisson and the negative binomial (NB) distributions, though other distributions such as the binomial and logarithmic distributions have also been occasionally considered. Several modifications and generalizations of the Poisson and negative binomial distributions have also been used. For example, in many count data contexts, there are a large number of zero count values. The most commonly used approach to accommodate this issue is the zero-inflated approach. The approach identifies two separate states for the count generating process – one that corresponds to a “zero” state in which the expected value of counts is so close to zero as being indistinguishable from zero, and another “normal” state in which a typical count model (with either a Poisson or NB distribution) operates. Effectively, the zero-inflated approach is a discrete-mixture model involving a discrete error distribution that modifies the probability of the zero outcome. Another similar approach to account for excess zeros is the hurdle-count approach (in which a binary outcome process of the count being below or above a hurdle (zero) is combined with a truncated discrete distribution for the count process being above the hurdle (zero) point. While the modifications and generalizations such as those just described have been effective for use with univariate count models, they are difficult to infeasible to implement in the case when there are inter-related multivariate counts at play (see Castro, Paleti and Bhat, 2012 (or CPB hereafter) and Herriges et al., 2008 for discussions). Also, including spatial dependence within the framework of traditional count formulations is very cumbersome. To address these situations, we can re-formulate the traditional count models as a special case of a generalized ordered-response probit (GORP) formulation (see CPB). Indeed, in this re-formulation, any count model can be formulated as a special case of a GORP formulation. Once this is achieved, all the GORP-related formulations in the earlier sections immediately carry over to count models. In this section, we will consider a single count variable based on a negative binomial distribution and show its aspatial GORP formulation, because extension to include multivariate and spatial contexts exactly mirror the previous GORP discussions. Consider the recasting of the count model using a specific functional form for the random-coefficients generalized ordered-response probit (GORP) structure of Section 2.2.1.1 as follows: k y y q q q , q q x β if k q q k q y , 1 , , (2.93) where q x is an (L×1) vector of exogenous variables (not including a constant), q β is a corresponding (L×1) vector of individual-specific coefficients to be estimated, q is an idiosyncratic random error term that we will assume in the presentation below is independent of the elements of the vectors q β and q x , and k q ψ , is the individual-specific upper bound threshold for discrete level k . The q terms are assumed independent and identically standard normally 64 distributed across individuals. The typical assumption for q is that it is either normally or logistically distributed, though non-parametric or mixtures-of-normal distributions may also be considered. Also, , ~ q q β b β where ) , 0 ( ~ ~ Ω L q MVN β . q y is an underlying latent continuous variable that maps into the observed count variable q y q l through the q ψ vector (which is a vertically stacked column vector of thresholds .
) ,..., , , , ( 2 1 0 1 , q q q q The k q ψ , thresholds are parameterized as a function of a vector of observable covariates q z (including a constant) as follows (see Bhat et al., 2014b): k k r r q q k q c r r c 0 1 , !
) ( ) ( 1 , q q q c , and q z γ e q . (2.94) In the above equation, [.] 1 is the inverse function of the univariate cumulative standard normal. is a parameter that provides flexibility to the count formulation, and, as we will see later, serves the same purpose as the dispersion parameter in a traditional negative binomial model (>0). ) ( is the traditional gamma function; 0 1 ) ( h h dh e h . The threshold terms in the q ψ vector satisfy the ordering condition (i.e., ) ....
2 1 0 , q q, q, q, q 1 as long as .
....
2 1 0 1 The presence of these terms provides substantial flexibility to accommodate high or low probability masses for specific count outcomes, beyond what can be offered by traditional treatments using zero-inflated or related mechanisms. For identification, we set , , 1 1 q q, and .
0 0 In addition, we identify a count value ......}) , 2 , 1 , 0 { ( e e above which ......}) , 2 , 1 , 0 { ( e e is held fixed at e ; that is, e e if , e e where the value of e can be based on empirical testing. For later use, let ) , , ( 2 1 e φ ( 1 e vector). The specification of the GORP model in the equation above provides a very flexible mechanism to model count data. It subsumes the traditional count models as very specific and restrictive cases. In particular, if the vector q β is degenerate with all its elements taking the fixed value of zero, and all elements of the φ vector are zero, the model in Equation (2.93) collapses to a traditional negative binomial model with dispersion parameter θ. To see this, note that the probability expression in the GORP model of Equation (2.93) with the restrictions may be written as: 65 , !
) ( ) ( 1 !
) ( ) ( 1 !
) ( ) ( 1 !
) ( ) ( 1 !
) ( ) ( 1 ] [ 0 1 0 1 0 1 1 0 1 k q q k r r q q k r r q q k r r q q q k r r q q q c k k c c r r c c r r c c r r c y c r r c P k y P (2.95) which is the probability expression of the negative binomial count model. If, in addition, , the result can be shown to be the Poisson count model. In an empirical context of crash counts at intersections, CPB interpret the GORP recasting of the count model as follows. There is a latent “long-term” crash propensity q y associated with intersection q that is a linear function of a set of intersection-related attributes q x On the other hand, there may be some specific intersection characteristics (embedded in q z within the threshold terms) that may dictate the likelihood of a crash occurring at any given instant of time for a given long-term crash propensity q y . Thus, two intersections may have the same latent long-term crash propensity q y , but may show quite different observed number of crashes over a certain time period because of different q y - to -q y mappings through the cut points ( q y is the observed count variable). CPB postulated that factors such as intersection traffic volumes, traffic control type and signal coordination, driveways between intersections, and roadway alignment are likely to affect “long-term” latent crash propensity at intersections and perhaps also the thresholds. On the other hand, they postulate that there may be some specific intersection characteristics such as approach roadway types and curb radii at the intersection that will load more on the thresholds that affect the translation of the crash propensity to crash outcomes. Of course, one can develop similar interpretations of the latent propensity and thresholds in other count contexts (see, for example, the interpretation provided by Bhat et al., 2014a, in a count context characterized by the birth of new firms in Texas counties). To summarize, the GORP framework represents a generalization of the traditional count data model, has the ability to retain all the desirable traits of count models and relax constraints imposed by count models, leads to a much simpler modeling structure when flexible spatial and temporal dependencies are to be accommodated, and may also be justified from an intuitive/conceptual standpoint. Indeed, all the spatial, multivariate, and panel-based extensions discussed under ordered-response models immediately apply to count models based on the count reformulation as a GORP model. 66 References for the CML Estimation of Count Models Castro, M., Paleti, R., Bhat, C.R., 2012. A latent variable representation of count data models to accommodate spatial and temporal dependence: application to predicting crash frequency at intersections. Transportation Research Part B 46(1), 253-272. Bhat, C.R., Paleti, R., Singh, P., 2014a. A spatial multivariate count model for firm location decisions. Journal of Regional Science 54(3):462-502. Bhat, C.R., Born, K., Sidharthan, R., Bhat, P.C., 2014b. A count data model with endogenous covariates: formulation and application to roadway crash frequency at intersections. Analytic Methods in Accident Research 1, 53-71. Narayanamoorthy, S., Paleti, R., Bhat, C.R., 2013. On accommodating spatial dependence in bicycle and pedestrian injury counts by severity level. Transportation Research Part B 55, 245-264. 67 3. APPLICATION TO JOINT MIXED MODEL SYSTEMS The joint modeling of data of mixed types of dependent variables (including ordered-response or ordinal variables, unordered-response or nominal variables, count variables, and continuous variables) is of interest in several fields, including biology, economics, epidemiology, social science, and transportation (see a good synthesis of applications in de Leon and Chough, 2013). For instance, in the transportation field, it is likely that households that are not auto-oriented choose to locate in transit and pedestrian friendly neighborhoods that are characterized by mixed and high land use density, and then the good transit service may also further structurally influence mode choice behaviors. If that is the case, then it is likely that the choices of residential location, vehicle ownership, and commute mode choice are being made jointly as a bundle. That is, residential location may structurally affect vehicle ownership and commute mode choice, but underlying propensities for vehicle ownership and commute mode may themselves affect residential location in the first place to create a bundled choice. This is distinct from a sequential decision process in which residential location choice is chosen first (with no effects whatsoever of underlying propensities for vehicle ownership and commute mode on residential choice), then residential location affects vehicle ownership (which is chosen second, and in which the underlying propensity for commute mode does not matter), and finally vehicle ownership affects commute mode choice (which is chosen third). The sequential model is likely to over-estimate the impacts of residential location (land use) attributes on activity-travel behavior because it ignores self-selection effects wherein people who locate themselves in mixed and high land use density neighborhoods were auto-disoriented to begin with. These lifestyle preferences and attitudes constitute unobserved factors that simultaneously impact long term location choices, medium term vehicle ownership choices, and short term activity-travel choices; the way to accurately reflect their impacts and capture the “bundling” of choices is to model the choice dimensions together in a joint equations modeling framework that accounts for correlated unobserved lifestyle (and other) effects as well as possible structural effects. There are many approaches to model joint mixed systems (see Wu et al., 2013 for a review), but the one we will focus on here is based on accommodating jointness through the specification of a distribution for the unobserved components of the latent continuous variables underlying the discrete (ordinal, nominal, or count) variables and the unobserved components of observed continuous variables. Very generally speaking, one can consider a specific marginal distribution for each of the unobserved components of the latent continuous variables (underlying the discrete variables) and the observed continuous variable, and then generate a joint system through a copula-based correlation on these continuous variables. However, here we will assume that the marginal distributions of the latent and observed continuous variables are all normally distributed, and assume a Gaussian Copula to stitch the error components together. This is equivalent to assuming a multivariate normal distribution on the error components. But the procedures can be extended to non-normal marginal distributions and non-Gaussian copulas in a relatively straightforward fashion. 68 From a methodological perspective, the simulation-based likelihood estimation of joint mixed models can become quite cumbersome and time-consuming. However, the use of the MACML estimation technique has once again opened up possibilities because of the dramatic breakthrough in the ease and computational feasibility of estimating joint mixed systems. 3.1. Joint Mixed Dependent Variable Model Formulation In the following presentation, for ease in exposition, we assume fixed coefficients on variables, though extension to the case of random coefficients in conceptually very straightforward (as in earlier sections). We will also suppress the notation for individuals, and assume that all error terms are independent and identically distributed across individuals. Finally, we will develop the formulation in the context of ordinal, nominal, and continuous variables, though the formulation is immediately applicable to count variables too because count variables may be modeled as a specific case of the GORP-based formulation for ordinal variables. Let there be N ordinal variables for an individual, and let n be the index for the ordinal variables ) ..., , 2 , 1 ( N n . Also, let n J be the number of outcome categories for the nth ordinal variable ) 2 ( n J and let the corresponding index be n j ) ..., , 2 , 1 ( n n J j . Let n y be the latent underlying variable whose horizontal partitioning leads to the observed choices for the nth ordinal variable. Assume that the individual under consideration chooses the th n a ordinal category. Then, in the usual ordered response formulation: , if , 1 n k ql n k n n n y k j y w δn (3.1) where w is a fixed and constant vector of exogenous variables (not including a constant), n δ is a corresponding vector of coefficients to be estimated, the ψ terms represent thresholds, and n is the standard normal random error for the nth ordinal variable. We parameterize the thresholds as: ) exp( 1 z γkn kn n k n k α (3.2) In the above equation, kn is a scalar, and kn γ is a vector of coefficients associated with ordinal level 1 ,..., 2 , 1 K k for the nth ordinal variable. The above parameterization immediately guarantees the ordering condition on the thresholds for each and every crash, while also enabling the identification of parameters on variables that are common to the w and z vectors. For identification reasons, we adopt the normalization that .
) exp( 1 1 n α n n Stack the N latent variables n y into an ) 1 ( N vector y , and let y Ξ f y , ~ N , where w δ w δ w δ N ,..., , ( 2 1 f and y Σ is the covariance matrix of ) ..., , , ( 2 1 N ε . Also, stack the lower thresholds corresponding to the actual observed outcomes for the n ordinal variables into an ) 1 ( N vector low ψ and the upper thresholds into another vector .
up ψ For later use, define 69 , ) ,..., , ( , ) ,..., , ( N n , -J n n n γ γ γ γ γ γ γ γ n 2 1 1 3 2 , ) ,..., , ( , ) ,..., , ( , 1 2 1 N n α α α α α 2 1 n J n n n α α and .
) ,..., , ( 2 1 N δ δ δ δ Let there be G nominal (unordered-response) variables for an individual, and let g be the index for the nominal variables (g = 1, 2, 3,…, G). Also, let Ig be the number of alternatives corresponding to the gth nominal variable (Ig3) and let ig be the corresponding index (ig = 1, 2, 3,…, Ig). Consider the gth nominal variable and assume that the individual under consideration chooses the alternative mg. Also, assume the usual random utility structure for each alternative ig. , g g g gi gi gi U x bg (3.3) where g gi x is a 1 L column vector of exogenous attributes, g b is a column vector of corresponding coefficients, and g gi is a normal error term. Let ) ,...
, ( 2 1 g gI g g g ξ ( 1 g I vector), ) , 0 ( ~ g Λ g I MVN g ξ . Let ) ,..., , ( 2 1 g gI g g g U U U U 1 ( g I vector), ) ,..., , , ( 3 2 1 q gI g g g x x x x xg L I g ( matrix), g V b xg g 1 ( g I vector). Then ).
, ( ~ g Λ V U g I g g MVN Under the utility maximization paradigm, g g gm gi U U must be less than zero for all g g m i , since the individual chose alternative g m . Let ) ( g g gm gi m gi m i U U u g g g g , and stack the latent utility differentials into a vector g g m gI m g m g m i u u u g g g g ; ,..., , 2 1 g u . As usual, only the covariance matrix of the error differences is estimable. Taking the difference with respect to the first alternative, only the elements of the covariance matrix g Λ of , ) ,..., , ( 3 2 g gI g g g ς where 1 g gi gi ( 1 i ), are estimable. However, the condition that 1 g I 0 u g takes the difference against the alternative g m that is chosen for the nominal variable g. Thus, during estimation, the covariance matrix g Λ (of the error differences taken with respect to alternative g m is desired). Since g m will vary across households, g Λ will also vary across households. But all the g Λ matrices must originate in the same covariance matrix g Λ for the original error term vector g ξ . To achieve this consistency, g Λ is constructed from g Λ by adding an additional row on top and an additional column to the left. All elements of this additional row and column are filled with values of zeros. Also, an additional scale normalization needs to be imposed on g Λ . For this, we normalize the first element of g Λ to the value of one. The discussion above focuses on a single nominal variable g. When there are G nominal variables, define G g g I G 1 and G g g I G 1 ) 1 ( ~ . Further, let , ,..., , 1 1 3 1 2 g gI g g g g U U U U U U g g u G u u u u ,..., , 2 1 , and G u u u u ,..., , 2 1 (so u is the vector of utility 70 differences taken with respect to the first alternative for each nominal variable, while u is the vector of utility differences taken with respect to the chosen alternative for each nominal variable). Now, construct a matrix of dimension G G ~ ~ that represents the covariance matrix of u : G 2G 1G 2G 2 12 1G 12 1 Λ .
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Λ Λ Σ u (3.4) In the general case, this allows the estimation of G g g g I I 1 1 2 ) 1 ( terms across all the G nominal variables (originating from 1 2 ) 1 ( g g I I terms embedded in each g Λ matrix; g=1,2,…G) and the 1 1 1 ) 1 ( ) 1 ( G g G g l l g I I covariance terms in the off-diagonal matrices of the u Σ matrix characterizing the dependence between the latent utility differentials (with respect to the first alternative) across the nominal variables (originating from ) 1 ( ) 1 ( l g I I estimable covariance terms within each off-diagonal matrix in u Σ ). For later use, define the stacked 1 G vectors G U U U U , ... , , 2 1 , and G V V V V 2 , ... , , 1 . Finally, let there be H continuous variables ) ..., , , ( 2 1 H y y y with an associated index h ) ..., , 2 , 1 ( H h . Let h h h y s h λ in the usual linear regression fashion, and .
) ,..., , ( 2 1 H λ λ λ λ Stacking the H continuous variables into a ) 1 ( H -vector y, one may write ), , ( y h MVN Σ c y where ' 2 1 ,..., , H s s s c H λ λ λ 2 1 , and y Σ is the covariance matrix of H ,.....
, 2 1 η . 3.2. The Joint Mixed Model System and the Likelihood Formation The jointness across the different types of dependent variables may be specified by writing the covariance matrix of y y u y , , as: Var ) ( y y y y u y y y y u y u y u u Σ Σ Σ Σ Σ Σ Σ Σ Σ Ω y , (3.5) 71 where y Σ u is a N G ~ matrix capturing covariance effects between the u vector and the y vector, y Σu is a H G ~ matrix capturing covariance effects between the u vector and the y vector, and y y Σ is an H N matrix capturing covariance effects between the y vector and the y vector. All elements of the matrix above are identifiable. However, the matrix represents the covariance of latent utility differentials taken with respect to the first alternative for each of the nominal variables. For estimation, the corresponding matrix with respect to the latent utility differentials with respect to the chosen alternative for each nominal variable, say Ω ~ , is needed. For this purpose, first construct the general covariance matrix Ω for the original 1 H N G vector y y , , U UY , while also ensuring all parameters are identifiable (note that Ω is equivalently the covariance matrix of ) , , ( η ξ ε τ . To do so, define a matrix D of size H N G H N G ~ . The first 1 I rows and ) 1 ( 1 I columns correspond to the first nominal variable. Insert an identity matrix of size ) 1 ( 1 I after supplementing with a first row of zeros in the first through ) 1 ( 1 I th columns of the matrix. The rest of the elements in the first 1 I rows and the first ) 1 ( 1 I columns take a value of zero. Next, rows ) 1 ( 1 I through ) ( 2 1 I I and columns ) ( 1 I through ) 2 ( 2 1 I I correspond to the second nominal variable. Again position an identity matrix of size ) 1 ( 2 I after supplementing with a first row of zeros into this position. Continue this for all G nominal variables. Put zero values in all cells without any value up to this point. Finally, insert an identity matrix of size N+H into the last N+H rows and N+H columns of the matrix D. Thus, for the case with two nominal variables, one nominal variable with 3 alternatives and the second with four alternatives, one ordinal variable, and one continuous variable, the matrix D takes the form shown below: (3.6) Then, the general covariance matrix of UY may be developed as .
D Ω D Ω All parameters in this matrix are identifiable by virtue of the way this matrix is constructed based on utility differences and, at the same time, it provides a consistent means to obtain the covariance matrix 7 9 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 72 Ω ~ that is needed for estimation (and is with respect to each individual’s chosen alternative for each nominal variable). Specifically, to develop the distribution for the vector y y u y , , ~ , define a matrix M of size H N G H N G ~ . The first ) 1 ( 1 I rows and 1 I columns correspond to the first nominal variable. Insert an identity matrix of size ) 1 ( 1 I after supplementing with a column of ‘-1’ values in the column corresponding to the chosen alternative. The rest of the columns for the first ) 1 ( 1 I rows and the rest of the rows for the first 1 I columns take a value of zero. Next, rows ) ( 1 I through ) 2 ( 2 1 I I and columns ) 1 ( 1 I through ) ( 2 1 I I correspond to the second nominal variable. Again position an identity matrix of size ) 1 ( 2 I after supplementing with a column of ‘-1’ values in the column corresponding to the chosen alternative. Continue this procedure for all G nominal variables. Finally, insert an identity matrix of size N +H into the last N +H rows and N +H columns of the matrix M. With the matrix M as defined, the covariance matrix Ω ~ is given by .
M MΩ Ω ~ Next, define ' ' y , u u ~ and .
, ~ f V g ) (M Also, partition Ω ~ so that ~ ~ ~ ~ ~ ~ y y y y u y y y y u y u y u u Σ Σ Σ Σ Σ Σ Σ Σ Σ Ω (3.7) Let y y Σ Σ Σ Σ Σ u y u u u ~ ~ ~ ~ ~ and ~ ~ ~ ~ ~ ~ ~ ) ~ Var( y y u y u u Σ Σ Σ Σ Ω y , where H N G y y y u y u ) ~ ( ~ ~ ~ Σ Σ Σ matrix. Also, supplement the threshold vectors defined earlier as follows: low low ψ ψ , ~ ~ G , and up up ψ 0 ψ , ~ ~ G , where G ~ is a ) 1 ~ ( G -column vector of negative infinities, and G ~ 0 is another ) 1 ~ ( G -column vector of zeros. The conditional distribution of u ~ given y, is multivariate normal with mean d y g g 1 ~ ~ ~ ~ ~ y y u Σ Σ and variance y u y y u u u ~ 1 ~ ~ ~ ~ ~ ~ ~ ~ Σ Σ Σ Σ Σ . Next, let θ be the collection of parameters to be estimated: . )] ( Vech ); ( Vech ; ); ( Vech ; , ; ..., , , [ ~ ~ 2 1 y Σ Σ λ Σ α γ, δ u y u G b b b θ Then the likelihood function for the household may be written as: , ~ ~ ~ Pr ) | ( ) ( up low ψ ψ Σ u c y θ y H L (3.8) , ~ ) ~ ~ , ~ ~ | ~ ( ) | ( ~ ~ ~ u d f u N G D y H u Σ Σ g u d y 73 where the integration domain } ~ ~ ~ : ~ { ~ up low ψ u ψ u u D is simply the multivariate region of the elements of the u ~ vector determined by the range ) 0 , ( for the nominal variables and by the observed outcomes of the ordinal variables, and (.) ~ N G f is the multivariate normal density function of dimension .
~ N G The likelihood function for a sample of Q observations is obtained as the product of the observation-level likelihood functions. The above likelihood function involves the evaluation of a N G ~ -dimensional rectangular integral for each household, which can be computationally expensive. So, the Maximum Approximate Composite Marginal Likelihood (MACML) approach of Bhat (2011) may be used. 3.3. The Joint Mixed Model System and the MACML Estimation Approach Consider the following (pairwise) composite marginal likelihood function formed by taking the products (across the N ordinal variables and G nominal variables) of the joint pairwise probability of the chosen alternatives for an individual, and computed using the analytic approximation of the multivariate normal cumulative distribution (MVNCD) function. .
) , Pr( ) , Pr( ) , Pr( ) | ( ) ( 1 1 1 1 1 1 1 1 ' G g N n n n g i G g G g g g i g i N n N n n n n n n H MACML a j m d m d m d a j a j L g g g y Σ c y θ (3.9) where g i d is an index for the individual’s choice for the gth nominal variable. The net result is that the pairwise likelihood function now only needs the evaluation of ~ and , ~ , ~ gn n n g g G G G dimensional cumulative normal distribution functions (rather than the N G ~ -dimensional cumulative distribution function in the maximum likelihood function), where ~ and 2, ~ , 2 ~ ' g gn n n g g g g I G G I I G . This leads to substantial computational efficiency. However, in cases where there are several alternatives for one or more nominal variables, the dimension gn g g G G ~ and ~ can still be quite high. This is where the use of an analytic approximation of the MVNCD function comes in handy. The resulting maximum approximated composite marginal likelihood (MACML) is solely based on bivariate and univariate cumulative normal computations. Also note that the probabilities in the MACML function in Equation (3.9) can be computed by selecting out the appropriate sub-matrices of the mean vector g ~ ~ and the covariance matrix u ~ ~ ~ Σ of the vector u ~ , and the appropriate sub-vectors of the threshold vectors .
~ and ~ up low ψ ψ The covariance matrix of the parameters θ may be estimated as: , ˆ ˆ ˆ ˆ Q Q 1 -1 -H J H G 1 - (3.10) 74 with MACML θ θ θ θ H ˆ , 2 1 ) ( log 1 ˆ q MACML Q q L Q MACML θ θ θ θ θ J ˆ , , 1 ) ( log ) ( log 1 ˆ q MACML q MACML Q q L L Q (3.11) An alternative estimator for H ˆ is as below: θ θ θ θ θ θ θ c y θ c y H ) , Pr( log ) , Pr( log ) , Pr( log ) , Pr( log ) , Pr( log ) , Pr( log ) | ( log ) | ( log 1 ˆ 1 1 1 1 1 1 1 1 1 ' ' n n g i n n g i G g N n g i g i g i g i G g G g g n n n n n n n n N n N n n H H Q q a j m d a j m d m d m d m d m d a j a j a j a j Q g g g g g g y y Σ Σ 3.4. Positive Definiteness The matrix Ω ~ for each household has to be positive definite. The simplest way to guarantee this is to ensure that the matrix Ω is positive definite. To do so, the Cholesky matrix of Ω may be used as the matrix of parameters to be estimated. However, note that the top diagonal element of each g Λ in Ω is normalized to one for identification, and this restriction should be recognized when using the Cholesky factor of Ω . Further, the diagonal elements of y Σ in Ω are also normalized to one. These restrictions can be maintained by appropriately parameterizing the diagonal elements of the Cholesky decomposition matrix. Thus, consider the lower triangular Cholesky matrix L of the same size as Ω . Whenever a diagonal element (say the kkth element) of Ω is to be normalized to one, the corresponding diagonal element of L is written as 1 1 2 1 a j kj d , where the kj d elements are the Cholesky factors that are to be estimated. With this parameterization, Ω obtained as L L is positive definite and adheres to the scaling conditions. References for the CML Estimation of the Mixed Variable Model Bhat, C.R., Born, K., Sidharthan, R., Bhat, P.C., 2014b. A count data model with endogenous covariates: formulation and application to roadway crash frequency at intersections. Analytic Methods in Accident Research 1, 53-71. 75 Khan, M., Paleti, R., Bhat, C.R., Pendyala, R.M., 2012. Joint household-level analysis of individuals' work arrangement choices. Transportation Research Record 2323, 56-66. Paleti, R., Bhat, C.R., Pendyala, R.M., 2013. Integrated model of residential location, work location, vehicle ownership, and commute tour characteristics. Transportation Research Record 2382, 162-172. Paleti, R., Pendyala, R.M., Bhat, C.R., Konduri, K.C., 2011. A joint tour-based model of tour complexity, passenger accompaniment, vehicle type choice, and tour length. Technical paper, School of Sustainable Engineering and the Built Environment, Arizona State University. Singh, P., Paleti, R., Jenkins, S., Bhat, C.R., 2013. On modeling telecommuting behavior: option, choice, and frequency. Transportation 40(2), 373-396. 76 4. CONCLUSIONS This monograph presents the basics of the composite marginal likelihood (CML) inference approach, discussing the asymptotic properties of the CML estimator and possible applications of the approach for a suite of different types of discrete and mixed dependent variable models. The approach can be applied using simple optimization software for likelihood estimation. In the case of models with complex and analytically intractable full likelihoods, the CML also represents a conceptually and pedagogically simpler simulation-free procedure relative to simulation techniques, and has the advantage of reproducibility of the results. For instance, in a panel application, Varin and Czado (2010) examine the headache pain intensity of patients over several consecutive days. In this study, a full information likelihood estimator would have entailed as many as 815 dimensions of integration to obtain individual-specific likelihood contributions, an infeasible proposition using computer-intensive simulation techniques. In another panel spatial application, Sidharthan and Bhat (2012) examine the case of spatial dependence in land-use of spatial grids, and the full information likelihood estimator would have entailed integration of the order of 4800 dimensions. Despite advances in simulation techniques and computational power, the evaluation of such high dimensional integrals is literally infeasible using traditional frequentist and Bayesian simulation techniques. For instance, in frequentist methods, where estimation is typically undertaken using pseudo-Monte Carlo or quasi-Monte Carlo simulation approaches (combined with a quasi-Newton optimization routine in a maximum simulated likelihood (MSL) inference), the computational cost to ensure good asymptotic estimator properties becomes prohibitive for the number of dimensions just discussed. Similar problems arise in Bayesian Markov Chain Monte Carlo (MCMC) simulation approaches, which remain cumbersome, require extensive simulation, are time consuming, and pose convergence assessment problems as the number of dimensions increases (see Ver Hoef and Jansen, 2007, and Franzese et al., 2010 for discussions). Even when the full likelihood involves a lower and more practically feasible dimensionality of integration, the accuracy of simulation techniques is known to degrade rapidly as the dimensionality increases, and the simulation noise increases substantially. This leads to convergence problems during estimation, unless a very high number of simulation draws is used. Several studies have demonstrated so in a variety of econometric modeling contexts (see, for example, Bhat and Sidharthan, 2011 and Paleti and Bhat, 2013). Besides, an issue generally ignored in simulation-based approaches is the accuracy (or lack thereof) of the covariance matrix of the estimator, which is critical for good inference even if the asymptotic properties of the estimator are well established. Thus, the CML can present a very attractive alternative to the traditional MSL method in many situations. Of course, there are some special cases where the MSL approach may be preferable to the CML approach. For example, consider a panel binary discrete choice case with J choice occasions per individual and K random coefficients on variables. Let the kernel error term be normally distributed and assume that the random coefficients are multivariate normally distributed, so that the overall error is also normally distributed. Here, when K < J, and K ≤ 3, the 77 MSL estimation with the full likelihood function is likely to be preferable to the CML. This is because integrating up to three dimensions is quite fast and accurate using quasi-Monte Carlo simulation techniques. This is particularly so when J is also large, because the number of pairings in the CML is high. For the case when K < J and K > 3, or K ≥ J > 3, the CML is likely to become attractive, because of the MSL-related problems mentioned earlier for moderate dimensions of integration. For example, when K = J =5, the CML is fast since it entails the evaluation of only 10 probability pairings for each individual (each pairing involving bivariate normal cumulative distribution function evaluations) rather than a five-dimensional integration for each individual in the MSL estimation. Note that one may be tempted to think that the CML loses this edge when J becomes large. For instance, when J = 10, there would be 45 probability pairings for each individual in a pairwise likelihood approach. But the surrogate likelihood function in the CML estimation can be formulated in many different ways rather than the full pairings approach presented here. Thus, one could consider only the pairing combinations of the first five (or five randomly selected) choice occasions for each individual, and assume independence between the remaining five choice occasions and between each of these remaining choice occasions and the choice occasions chosen for the pairings. Basically, the CML approach is flexible, and allows customization based on the problem at hand. The issue then becomes one of balancing between speed gain/convergence improvement and efficiency loss. Besides, the CML can also use triplets or quadruplets rather than the couplets considered here. If the probabilities of the lower dimensional events in the CML approach themselves have a multivariate normal cumulative distribution (MVNCD) form, then one can use the MACML approach proposed by Bhat to evaluate the MVNCD function using an analytic approximation. One potential limitation of the CML approach is the need to compute the Godambe information matrix to compute the asymptotic standard errors of parameters. However, even when an MSL method is used, the Godambe matrix is recommended to accommodate the simulation error that accrues because of the use of a finite number of draws. Another limitation of the CML approach is the need to compute the ADCLRT statistic, which is somewhat more complicated than the traditional likelihood ration test (LRT) statistic. It is hoped that such practical issues will be resolved once standard econometric software packages start accommodating the CML inference approach as an option for high dimensional model systems. In summary, the CML inference approach (and the associated MACML approach) can be very effective for the estimation and analysis of high-dimensional heterogeneous data. This has been shown in many recent studies, and there are many more empirical contexts that can gainfully use the CML approach using the formulations discussed in this monograph. In terms of future research on the CML approach itself, one wide open area pertains to how best to form a CML function in a given modeling and empirical context (especially because a precise theoretical analysis of the properties of the CML estimator is not possible except for the simplest of models). 78 ACKNOWLEDGEMENTS This research was partially supported by the U.S. Department of Transportation through the Data-Supported Transportation Operations and Planning (D-STOP) Tier 1 University Transportation Center. The author would also like to acknowledge support from a Humboldt Research Award from the Alexander von Humboldt Foundation, Germany. Finally, the author is grateful to Lisa Macias for her help in formatting this document, and individuals who provided useful comments on an earlier version of the monograph. REFERENCES Albert, J.H., Chib, S., 1993. Bayesian analysis of binary and polychotomous response data. Journal of the American Statistical Association 88(422), 669-679. 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In Frontiers in Econometrics, 105-142, Zarembka, P., (ed.), Academic Press, New York. McFadden, D., 1978. Modeling the choice of residential location. Transportation Research Record 672, 72-77. McFadden, D., Train, K., 2000. Mixed MNL models for discrete response. Journal of Applied Econometrics 15(5), 447-470. McKelvey, R.D., Zavoina, W., 1975. A statistical model for the analysis of ordinal level dependent variables. Journal of Mathematical Sociology 4(summer), 103-120. McMillen, D.P. 2010. Issues in spatial analysis. Journal of Regional Science 50(1), 119-141. Mitchell, J., Weale, M., 2007. The reliability of expectations reported by British households: Micro evidence from the BHPS. National Institute of Economic and Social Research discussion paper. Molenberghs, G., Verbeke, G., 2005. Models for Discrete Longitudinal Data. Springer Series in Statistics, Springer Science + Business Media, Inc., New York. Müller, G., Czado, C., 2005. An autoregressive ordered probit model with application to high frequency financial data. Journal of Computational and Graphical Statistics, 14(2), 320-338. Munkin, M.K., Trivedi, P.K., 2008. Bayesian analysis of the ordered probit model with endogenous selection. Journal of Econometrics, 143(2), 334-348. Pace, L., Salvan A., Sartori, N., 2011. Adjusting composite likelihood ratio statistics. Statistica Sinica 21(1), 129-148. Paleti, R., Bhat, C.R., 2013. The composite marginal likelihood (CML) estimation of panel ordered-response models. Journal of Choice Modelling 7, 24-43. Partridge, M.D., Boarnet, M., Brakman, S., Ottaviano, G., 2012. Introduction: Whither Spatial Econometrics? Journal Of Regional Science 57(2), 167-171. Rebonato, R., Jaeckel, P., 1999. The most general methodology for creating a valid correlation matrix for risk management and option pricing purposes. The Journal of Risk 2(2), 17-28. Ruud, P.A., 2007. 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International Finance Discussion Papers 875, Board of Governors of the Federal Reserve System (U.S.). Sidharthan, R., Bhat, C.R., 2012. Incorporating spatial dynamics and temporal dependency in land use change models. Geographical Analysis 44(4), 321-349. Small, K.A., Winston, C., Yan, J., 2005. Uncovering the distribution of motorists’ preferences for travel time and reliability. Econometrica 73(4), 1367-1382. Solow, A.R., 1990. A method for approximating multivariate normal orthant probabilities. Journal of Statistical Computation and Simulation 37(3-4), 225-229. Switzer, P., 1977. Estimation of spatial distribution from point sources with application to air pollution measurement. Bulletin of the International Statistical Institute 47(2), 123-137. Train, K. 2009. Discrete Choice Methods with Simulation, 2nd ed., Cambridge University Press, Cambridge. Vandekerkhove, P., 2005. Consistent and asymptotically normal parameter estimates for hidden markov mixtures of markov models. 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4410 | https://www.project-management-prepcast.com/pmp-formulas | Category: Project Management Professional (PMP)® Exam using A Guide to the Project Management Body of Knowledge (PMBOK® Guide)
PMP® Formulas and Calculations - The Complete Guide
When I speak to Project Management Professional (PMP)® students, there's often one thing on their mind: How are they going to learn all the PMP® formulas they need for their Project Management Institute (PMI)® exam?
There's no getting away from it. There are a lot of these formulas and calculations that you have to learn for the exam.
PMP® Formulas - The Complete Guide.
In this (almost) complete guide we'll go through the PMP formulas with examples. Stick with me and by the end, you'll see that learning the PMI® formulas isn't that bad. Read on and you'll learn a lot about how to crack formula-based PMP questions to help pass your PMP exam.
In This Article...
Chapter 1What Formula Question Types are on the PMP Exam?
Chapter 2Formula Question Example
Chapter 3The Earned Value Formulas
Chapter 4The Earned Value Calculation
Chapter 5Is There A Formula for Planned Value?
Chapter 6How to Calculate Schedule Variance
Chapter 7What is the Cost Variance (CV) Formula?
Chapter 8How Do You Calculate the Cost Performance Index (CPI)?
Chapter 9Challenge Yourself With a Really Hard CPI Sample PMP Exam Question
Chapter 10Using Estimate at Completion (EAC)
Chapter 11What is Schedule Performance Index (SPI)?
Chapter 12How to Use the Variance at Completion (VAC) Formula
Chapter 13What is the Estimate to Complete Formula (ETC)?
Chapter 14Understanding the To Complete Performance Index (TCPI)
Chapter 15PMP Formulas PDF Free Download
Chapter 16Five Formulas Explained
Chapter 17The Formula of Standard Deviation
Chapter 18The Advantage of PERT Formula
Chapter 19How To Use The Communication Channels Formula
Chapter 20How to Practice PMP Formulas
Chapter 21Learn More about PMP Exam Formulas
Chapter 22Conclusion and Recommendation: Practice. Practice. Practice!
What Formula Question Types are on the PMP Exam?
For the PMP exam, you must know how to correctly answer questions with formulas about earned value, communications, procurement, probability, network diagrams, project selection, depreciation, and some mathematical basics. You also have to know a lot of acronyms.
Here is a list with the types of questions you have to expect:
Apply a formula: These are straightforward questions where you are given values and are expected to apply the correct formula.
Apply two formulas: In these questions you get a set of values and asked to calculate a result. At first these look as if you can simply apply one formula. But as you are applying this first formula you suddenly realize that one value is missing. This missing value must then first be calculated via a second formula.
Invert a formula: These questions test your ability to take a basic formula and invert it. For instance instead of asking “4+6=?” the question would be “4+?=10” and it is your job to invert the formula and calculate “10-4=6”.
Result interpretation: In these types of questions you are given a result and asked, “What does this result mean for the project?”.
Find the correct formula: For these types of questions you are given a scenario and various formulas as options. Your task is to select the formula which best applies to the given scenario.
Use a formula based on keywords: There is more than one way to calculate earned value results. Which formula to use depends on the progression (or health) of your project. These are scenario-based questions that contain certain keywords. You must recognize these keywords and apply the correct formula
So you can easily see that being "mathematically ready" for your PMP exam means more than simply knowing your earned value (EV) formulas. Yes, you'll definitely see at least one EV question on your exam but that's simply not all there is.
Formula Question Example
Here is a formula-based sample PMP exam question for you:
Sample Question
Answer and Explanation
Tab
Sample Question
There are 20 stakeholders on a project. What is the total number of potential communication channels in this project?
A. 10
B. 45
C. 190
D. 100
Correct Answer: C. 190
Explanation: The total number of communication channels can be calculated using the formula n(n – 1)/2, where n = number of stakeholders. So in this case the total number of communication channels is (2019)/2 = 190.
More questions like this one...
The Earned Value Formulas
Many PMP aspirants find the concepts behind earned value management (EVM) hard to understand and the formula even harder, so that's where we are going to start.
First, let's define the term:
Earned Value (EV)1. The measure of work completed expressed in terms of the budget authorized for that work.
A Guide to the Project Management Body of Knowledge (PMBOK® Guide)
The thing that complicates it for many people is the question: "What on earth is value?" For the earned value calculations you just have to remember that value equals money. The total value of a project (in EVM terms) is equal to the budget of the project. As you work through the project you spend the budget in order to achieve the project's objectives, which in turn deliver business value. You can assess how much value you have 'earned' (read: 'achieved') by knowing how far through the project you are and how much you have spent.
The earned value management formulas are simply the calculations that give you the data to work out the EV position on your project. There are 12 earned value calculations in total.
Still not clear? You'll see what I mean with an example.
The Earned Value Calculation
Let's say we're working on a project to design an app for a smartwatch. Overall we are trying to answer the following two questions:
Are we over, on, or below budget?
Are we ahead, on, or behind schedule?
The earned value management formulas give us the information we need to determine that, and we'll get further into how we can calculate cost and schedule performance a little later.
The formula:
: EV = % complete BAC
What you get:
: A monetary value.
(BAC = Budget at Completion)
Let's assume the following situation: Our smartwatch app project is going to take six months and cost $60,000. We're two months into the project and we've spent $20,000. Our project sponsor wants to know if the project is doing "OK".
Generally, when sponsors ask a question like that, then they want to know if you are burning through the budget too quickly and if you are going to hit the end date which they have published to the Board. The EVM formulas can tell you exactly that.
Let's continue with the example and assume that that project was supposed to deliver 36 work packages, and that each package has the same planned value (PV) and takes the same amount of effort to complete. Now, let's say that the project team has completed 12 work packages so far during the first two months of the project. We can see that 12/36 shows that the team is a third of the way through the project. As a percentage, 33.33% of the work has been completed. That's the % complete figure in the simple formula.
The budget at completion (BAC) is the total amount budgeted for the project, in this case $60,000.
Plugging those figures into the formula we get: 33% $60,000 = $20,000. The earned value (EV) of the project is $20,000.
The formula:
: EV = Sum of PVs of all completed activities
What you get:
: Project's earned value. Why? If each work package requires a different level of effort to complete, we cannot use the number of completed work packages as an indicator of project’s percentage of completion. In such cases, we need to go one level deeper, for example, breaking the work packages into person-hours of effort. When determining project’s percentage of completion is difficult, grab the latest project schedule. Each project activity should have its own planned value (PV) assigned to it. Add the PVs for all the completed activities (the emphasis is on 'completed'), and this value will be your project’s earned value.
Is There A Formula for Planned Value?
Earned value alone isn't that useful a figure. You need to have something to compare it to. Planned value (PV) is great for that.
Planned Value (PV)1. The authorized budget assigned to scheduled work.
There is no formula for PV. It's simply the approved budget for a task. You can use PV for the budget of a phase, stage or work package. It's normally measured over a particular time period.
How to Calculate Schedule Variance
Schedule variance tells us whether our smartwatch app project is ahead, on, or behind schedule.
Schedule Variance (SV)1. A measure of schedule performance on the project, expressed as the difference between project's earned value and planned value.
One argument that I often get from my students is something along the lines of "I don't see how SV is possible! I mean we are using numbers from the budget and you expect me to believe that they help predict how I'm doing on the schedule??" It's not an unreasonable argument. But it is indeed possible. For now, I'm going to ask you to take a leap of faith (I did when I originally came across SV) and I'll explain how it all works. Let us begin with the formula:
The formula:
: SV = EV - PV
What you get:
: A monetary amount. A negative number means you are behind schedule (that's bad). A positive number means you are ahead of schedule (generally that's good).
Although we generally see a negative SV as something bad and a positive SV as something good, contemporary project management considers all variances, positive or negative, as potentially harmful to the project and the performing organization. Variance tolerances are defined, such as +/- 5% or +/-15%, and whenever the variances are found outside this range, an investigation is recommended. One may argue saying, "I don’t get it, how can be a positive 20% SV bad for the project or the organization? A positive 20% SV means I am 20% ahead of schedule and I should be getting a bonus for that!" Well, we didn’t say that a positive 20% SV is always bad. We said it is potentially harmful to the project and the organization. If the 20% faster progress happened due to the brilliance of the project team, they need to be rewarded. However, in most cases, it has been found that positive schedule variance is a result of conservative estimation during project planning. If a project is conservatively estimated or scheduled, it will consume organizational resources for a more extended period than it should have been, which is not good for the organization.
Earlier we showed you that the project team had completed 12 out of 36 planned work packages meaning the project is 33.33% complete. Further, the team has just finished the second month in the project, and with the total project duration of 6 months, the team has consumed 33.33% of the time planned.
It seems that we are on track!
But wait a minute, what if the project was front-loaded? A front-loaded project requires more resources and work to be completed at the beginning of the project in comparison to the end. Let us now assume that in this project we planned to complete 66.67% of the work in the first two months, and very few resources are involved in the last three months of the project. Therefore, now, looking again at 33.33% of the time consumed and 33.33% of work packages completed, we understand that these values are not good for the front-loaded project since according to the plan we should have completed 66.67% of the work.
The EV of our fantastic new app development project is $20,000, as we saw earlier. However, the Planed Value (PV) is $40,000 (66.67% of $60,000). In other words this means according to our plan we should have completed $40,000 worth of work. Using those numbers in the schedule variance formula we get -$20,000.
These are really simplified figures to help you understand the math more easily and you can clearly see that this is bad news. Now, when we are starting to dig into the numbers, we can see that the project is running behind schedule.
One of the emerging trends in project management is the new approach to schedule variance. The development of the new SV formula was a result of criticism of the traditional SV formula that expresses schedule variance in monetary terms. The new SV formula expresses the schedule variance in time units. This formula uses two components: earned schedule (ES) and actual time (AT). However, the calculation of these components is relatively complicated. Therefore, it is unlikely that you will be required to make any calculations on the exam that involves ES and AT. Anyway, if you want to see the formula, here it is:
The formula:
: SV = Earned Schedule (ES) – Actual Time (AT)
What you get:
: Time units. The interpretation of SV calculated by this formula is the same as the traditional SV formula, i.e., positive SV = project ahead of schedule; negative SV = project behind schedule.
What is the Cost Variance (CV) Formula?
Schedule variance is one way to get a view on how your project is performing. Another way to look at it is to use the cost variance formula.
The CV formula is also used to work out if the project is over, on, or under budget.
The formula:
: CV = EV – AC
What you get:
: A monetary amount. A negative number means you are over budget (that's bad). A positive number means you are under budget (hurrah!). It's most useful to report CV alongside the project budget so that you can easily see the magnitude of any variance. As mentioned earlier, contemporary project management sees all project variances, positive or negative, as potentially harmful to the project and the performing organization. Similarly to schedule variance, the cost variance outside the defined threshold limit (e.g., +/-15%) should be investigated, and a root cause identified. For example, positive cost variance could be a result of conservative estimation during project planning. If a project was conservatively estimated, extra funds allocated to it could have been spent elsewhere by the organization.
(AC = Actual Cost)
If we take the figures from our smartwatch project we can see what this means to the example. The EV is $20,000. Let's assume that the actual cost of the work done so far – the money we have spent already – is $20,000. Putting these in the formula gives us:
$20,000 - $20,000 = $0. Perfect! We are exactly on budget! (For the purposes of this example, anyway...)
So, what do we know about the project so far? It's neither over nor under budget, but it is running behind schedule. We thought we would have completed more activities as per the plan by now, but we did not. This is an early warning sign that the timescales for the project might slip.
It's important to consider the context behind the numbers. Our company has never delivered this kind of smartwatch app before, so we're learning as we go. Maybe that's why we aren't making the progress we expected. Maybe one of the big mobile platforms has taken longer than expected to get back to us about how to get our product in their store. There could be lots of reasons that explain the numbers, so when you're reporting the project's status to the sponsor, don't just rely on the numbers to tell the story.
How Do You Calculate the Cost Performance Index (CPI)?
We report to the project sponsor. She understands the reasons for the numbers but wants to know if the situation can be recovered.
A useful piece of information to help paint the whole picture is the cost performance index (CPI) calculation. This is an alternative way of looking at the cost performance of a project and is often preferable to CV because the answer you get is a ratio. Ratios are self-explanatory and don't need further information to highlight how far off-track performance is.
CPI in project management measures the cost efficiency of a project.
The formula:
: CPI = EV / AC
What you get:
: A number. You're aiming for 1. That means that you are getting $1 of value for every $1 spent. You are using your project budget as planned. If it's more than 1, you are getting more than $1 for every $1 spent. This could mean that your initial budget was not put together in a robust way and your estimates were too conservative.
If the number is less than 1 then that's bad news. You are getting less than $1 of value for the project for every $1 spent. This normally happens when you are spending your budget in ways you hadn't expected, such as dealing with unforeseen risks or your initial estimates were just too tight.
Using the figures from our app development project we get:
$20,000 / $20,000 = 1. That looks good. We're spending money in the right way and on the right things, but there's still more to uncover about how this project is performing.
Challenge Yourself With a Really Hard CPI Sample PMP Exam Question
Click the image below to watch a video with a really hard CPI question. This one not only requires you to identify the correct formula, but then you also have to adjust the formula, or you'll get the wrong result - tell us in the comments if you got it right or wrong:
Click to watch the video...
Using Estimate at Completion (EAC)
Next, we look at the EAC formula. That's Estimate at Completion and it works out the expected final and total cost of the project, based on project performance.
The EAC formula PMPs need to know most often is:
The formula:
: EAC = BAC / CPI
What you get:
: A monetary value.
(BAC = budget at completion. CPI = cost performance index)
There is EAC approach (a thorough bottom-up re-estimate of remaining work) to use if the original project estimates are now thought to be invalid, but this one is the one that is most likely to turn up in the PMP exam.
For this project we decide to re-estimate the costs of the remaining works (ETC). We use bottom-up estimation and find out that we need further $50,000 to complete the rest of the project. Since we have already spent $20,000 on the project already, the new estimate at completion (EAC) becomes $70,000.
Cost-wise, it's still not that bad. Our project sponsor can now be confident that we're going to hit the new budget target, i.e. EAC, that we have determined for the project. However, remember SV? That didn't look so good. She asks us to dig into that a bit further.
What is Schedule Performance Index (SPI)?
The SPI formula is used to work out if the project is:
Ahead, on, or behind schedule.
Going to finish when predicted.
The formula:
: SPI = EV / PV
What you get:
: A number. Again, as this one is a ratio too you are aiming for 1 as that means you are working through the project at the rate you had expected. If the number is greater than one, it means you are racing through your tasks faster than you had planned (but it doesn't comment on the quality of the work done – just something to think about!). If the number is less than 1 it means you are progressing more slowly than planned and the tasks are taking longer.
The SPI calculation, when applied to our app project, shows this:
$20,000 / $40,000 = 0.5
Ouch! We are working at half the speed that we planned. This is where the project is struggling! We had better stop doing project management formulas and start helping our team get the work done!
As we mentioned earlier, one of the emerging trends in project management is the new approach to schedule variance. Same goes to the schedule performance index (SPI) that uses earned schedule (ES) and actual time (AT). Since the calculation of these components is relatively complicated, we believe it is unlikely that you will be required to make any calculations on the exam that involves ES and AT. Anyway, if you want to see the formula, here it is:
The formula:
: SPI = Earned Schedule (ES) / Actual Time (AT)
What you get:
: The interpretation of SPI calculated by this formula is the same as the traditional SPI formula, i.e., greater than “1” = the project is ahead of schedule; less than “1” = the project is behind schedule.
But before we leap into the To Do list, let's just check a couple of other formulas.
Now you know that there is a CPI and an SPI. But what exactly is the difference? Find the answer here...
How to Use the Variance at Completion (VAC) Formula
Variance at completion (VAC) is another useful PMP earned value management formula. It helps you look forward and anticipate the difference between your original BAC and the newly calculated EAC. In other words, it's the total cost we originally planned minus the total cost that we now expect.
The formula:
: VAC = BAC - EAC
What you get:
: A monetary value. This shows you how much over or under budget (the variance) we will be at the end of the project. A value of $0 means you'll hit budget. Less than zero means you'll be over budget so ideally you're looking for a number near $0.
For our project, right now we're looking at:
-$10,000
That tells us there will be a negative variance of $10,000 by the time this smartwatch app is developed. Our sponsor is going to want to know about that too.
What is the Estimate to Complete Formula (ETC)?
The Estimate to Complete (ETC) formula is essentially an inversion of the EAC calculations, so if you can do that one, you can work this one out easily.
There are several ways to calculate ETC in project management but this is the simplest and easiest to use in straightforward situations. It's also the one from A Guide to the Project Management Body of Knowledge (PMBOK® Guide).
The formula:
: ETC = EAC - AC
What you get:
: A monetary value that tells you how much more the project will cost.
Our app development project is pretty straightforward so the calculation looks like:
$70000 - $20,000 = $50,000 In other words, we need another $50,000 to see the project through to the end.
What if my situation isn't straightforward? There are 4 other ways to define and calculate ETC. They are all covered in detail in study guides like The PMP Exam Formula Study Guide.
Understanding the To Complete Performance Index (TCPI)
The last bit of PMP mathematics we are going to do is to work out the TCPI. It's worth a mention because it's one of the PMP exam formulas that people find most difficult. TCPI is used to calculate the cost performance that must be achieved to hit your cost target (either BAC or EAC).
You can work out TCPI against your BAC or EAC. We know that the smartphone project has already had the budget re-estimated once so it's best to use the latest figures. We're going to use the EAC to work it out.
The formula:
: TCPI = (BAC - EV) / (EAC - AC)
What you get:
: A figure.
Let's do this is in stages.
```
BAC – EV = $60,000 - $20,000 (remember? That's our EV) = $40,000
EAC – AC = $70,000 - $20,000 = $50,000
$40,000 / $50,000 = 0.8
```
TCPI on our development project is 0.8. In order to get the intuition behind TCPI, let's break the formula in two parts. The first part, BAC – EV, indicates how much project work is remaining, i.e., how much value remains that need to be achieved. The second part, EAC – AC, indicates the money available to finish the project. Hence, the TCPI gives you the ratio of work that needs to be completed and the money available to complete the project.
The smartwatch app project started off looking quite good but we had to re-estimate the budget and we uncovered that we aren't getting through the tasks as quickly as we need to. In fact, the amount of work to do is unlikely to get done in time if we only apply the same amount of effort as we are now.
Thanks to the PMP cost management formulas we know a lot more about our project now.
Phew! We made it through all the earned value management formulas. Here's a table that summarizes all of that.
If you're creating a PMP formula sheet for your exam revision, there are a few more calculations that you'll want to include.
PMP Formulas PDF Free Download
Why don't you practice on some free questions from The PM Exam Simulator?
21 Formula-Based Sample Questions -- Free Download
Click to download this PDF document with a good number of formula-based questions. Test your knowledge right now....
The PDF download document has two sections:
Section 1 - Self-Assessment Questions: The difficulty level of these questions is easy to medium. They are intended to measure general understanding of formula concepts.
Section 2 - Answers and Explanations: In this section we repeat the question and then give you the answer, explanation and a reference where you can read up on the topic.
Five Formulas Explained
Five PMP® Exam Formulas Explained from OSP International LLC
The Formula of Standard Deviation
Ah, standard deviation! Doesn't this take you back to high school math. Standard deviation is simply a reflection of the uncertainty in the estimates. It just highlights the variability of an activity in statistical terms and you use it when an activity has different estimates: optimistic, most likely and pessimistic.
The formula:
: σ = (Pessimistic - Optimistic) / 6
What you get:
: A figure that represents the variation in the estimates, in the same units of measurement as your estimate.
The σ is the sign that represents standard deviation and is read as "sigma".
There's a task on the smartwatch app project to build the notification tools, so that when something new happens the user gets an alert. Given that we haven't done this task before we want a good understanding of the risk and we've used subject matter experts from an agency to help estimate the work.
The pessimistic estimate is 12 hours. The optimistic estimate, if we had an expert developer in-house, is 5 hours. Pop those into the formula and we get:
(12 – 5) / 6 = 1.16
The larger the standard deviation, the greater the risk. The risk level on this task seems OK but we can always take it to our sponsor for a second opinion.
The Advantage of PERT Formula
You have another tool to help you estimate: PERT. The most likely PERT formula PMP aspirants are going to come across in the exam gives you a weighted average for your estimates. It's a three point estimate for the expected duration of a schedule activity using pessimistic, optimistic and most likely durations.
The formula:
: Beta =
(Pessimistic + (4 Most Likely) + Optimistic) / 6
What you get:
: The estimated duration of the activity as a weighted average.
Let's see what the weighted average is for the task to build the notification module of our smartwatch app. First, we need another piece of data: the most likely duration for the task, which our lead developer tells us is 8 hours. That gives us:
(12 + (4 8) + 5) / 6 = 8.2 hours. Now there's a number our sponsor can understand.
How to Use the Communication Channels Formula
The sponsor isn't the only person on the project team who needs to know this information. The team is only 6 people but there are still multiple communication channels. How many exactly? Well, there's a communication channels formula to work that out!
The formula:
: n (n-1) / 2
What you get:
: The number of communication channels in the team.
In the formula, 'n' stands for the number of people. On our project, the communication channel math looks like this:
6 (6-1) / 2 = 15
That's 15 different routes for messages!
How to Practice PMP Formulas
This article has covered the most common PMP formulas but the best way to really get to grips with them is to practice using them frequently.
Download our free redacted PMP Formulas PDF Guide so that you've got all the data and formulae to hand.
Another great way to practice is to do sample exam questions. This free batch of sample PMP math questions will help you see where you need to build your skills.
Learn More about PMP Exam Formulas
The following two articles will introduce you to a couple of important concepts and tools for the exam:
What PMP Formulas Should I Study for the PMP Certification Exam and How?
A short review.
The Project Management Tool That Will Help You Pass The PMP Exam
This is the tool that you will be using during the PMP exam to calculate formulas...
Conclusion and Recommendation: Practice. Practice. Practice!
There are 49 PMP Exam Formulas that you have to master for the PMP Exam.
Mastering these formulas is more than simply knowing them! You have to know how to apply them to a given question scenario and you may even have to apply TWO formulas. You may have to invert a formula, you might be asked to interpret a formula (CPI = 0.8 means?), or you may be asked to determine which formula is most appropriate in a given scenario based on certain keywords in the question.
Being able to do this takes time and practice. Don't expect miracles overnight. Here is my recommended approach:
First of all, study the right formulas. There are unfortunately still some websites out there that list incorrect or outdated PMP exam formulas!
Review the EVM formulas that are listed in the PMBOK® Guide every other day for two weeks. Pause for one week, then do it again. The more you repeat, the more they will sink in.
Make sure that you know all the earned value formulas in and out. PMI loves earned value!
Answer at least one formula-based question on every day that you are studying for the exam. Since most people take about 3 months for their preparation that means you will answer about 90 formula-based questions.
Knowing and understanding all the formulas and their variations, concepts, keywords, value and acronyms may seem daunting at first. But if you plan on taking just one small step forward every day and practice, practice, practice, then you'll soon find that the calculations start to come naturally.
© 2015-2023 OSP International LLC. All rights reserved. This copyrighted article may not be reproduced without express written consent of OSP International LLC.
1 -- These definitions are taken from the Glossary of Project Management Institute, A Guide to the Project Management Body of Knowledge, (PMBOK® Guide) – Sixth Edition, Project Management Institute Inc., 2017.
Comments (23)
Anonymous Thursday, 18 February 2016
Thanks for sharing this info. it's helpful and complete.
3
Anonymous Monday, 22 February 2016
Thank you, what a great gift to the community!
2
Yatindra Kumar Tuesday, 01 March 2016
Very helpful
2
George Kayange Sunday, 12 June 2016
This is amazing! Thank you
2
Anonymous Tuesday, 16 August 2016
awesome...
2
Anonymous Saturday, 29 October 2016
This is awesome .. you made my day by consolidating all at one place and examples bonus
2
Anonymous Tuesday, 14 February 2017
Thank you so much Cornelius, the PMP journey has been steep but I am encouraged when I study your materials, it makes studying sooo much bearable, I am using some of your study materials, PMPrepcast, Study coach, formula guide, and email guides and the free stuff too, I also opted to use the Achieve PMP success which I find very helpful as well. I feel assured somewhat that if I put in enough time and plan my studies well, plus heed to the advises like taking a pill I should make it very far. Thanks team Cornelius.
2
Anonymous Tuesday, 07 March 2017
Thank you very much, very useful! :D
3
Anonymous Wednesday, 15 March 2017
hello for the 2nd problem how did you come up wont 120 man days?
2
Hekmat Khan Tuesday, 04 July 2017
Thanks a loot for this useful information. God bless you for ever.
2
Anonymous Tuesday, 14 November 2017
Detailed and Explicit. Thanks
1
yvonnesmythe@aol.com Monday, 25 December 2017
This is very much appreciated! Clear and concise.
0
Anonymous Saturday, 11 August 2018
Very useful, thanks
1
P.Joseph Thursday, 16 May 2019
Thank you for sharing this very helpful information in simple words.
1
Anonymous Monday, 12 August 2019
Good one...Thanks for sharing
1
admasuengdaw@gmail.com Friday, 13 September 2019
Thanks,
I also need pdf files
1
Anonymous Wednesday, 18 December 2019
Thank you! I also found this site useful as well.
1
Anonymous Monday, 24 February 2020
Very useful. Thank you so much.
1
Great resources. Good practice. Awesome!
1
Very helpful
2
Funny how it says there are 'sites out there that list....outdated info' and one of the links too me to material based on the PMBOK 5th edition from 2013 - almost 10 years ago.
2
1
Very insightful!! Thank you!!
1
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Tips And Tricks And Shortcuts For Number Series
August 10, 2023
Number series Tips and Tricks and Shortcuts
Here on this page Number series Tips, Tricks and Shortcuts is given to solve the question easily and effectively. You will also get some of its Types along with question and answers.
Number Series : - In Number Series When we identify pattern then It is Easy to Find any particular term.
- This Page helps you to Solve Questions in less time. So basically there are five different types of number series. There are mentioned below with one suitable example:
Number Series Tips, Tricks and Shortcuts :
| Number Series Types | Explanation |
--- |
| Add up Series (+) | Just adding a constant number everytime. For example: (13, 18, 23, 28 ….). |
| Add up Series (- ) | Just subtracting a constant number every time. For example : (28, 23, 18, 13 ……). |
| Step up Series (+/-) | Adding/Subtracting a variable number, in this case we are adding 2n for n = 0, 1, 2, 3 …… For example : 0, 2, 6, 12, 20…. or 20, 12, 6, 2, 0, 0 ….. |
| Square up (+/-) | Given series is in the square of n where n= 1,2,3,4,5,6… i.e. 1,4,9,16,25,36… |
| Square add up Series(+/-) | Adding n2 every time and incrementing value of n starting from 1 i.e. 1,5, 14, 30, 55, …… and Subtracting n2 every time and decrementing value of n i.e. 34, 33, 29, 20, |
| Cube Up Series : | Given series is in the cube of n where n = 1,2,3,4,5,6… i.e. 1,8,27,64,125,216…. |
| Cube add up Series : | Adding n^{3} every time with incrementing of n by 1 i.e. 1,9,36… |
| Prime up(+/-) :+ | Sequence consist of Prime numbers i.e. 2,3,5,7,11… |
| Prime Square up(+/-) : | Sequence consist of squares of prime numbers i.e. 4,9,25,49,121.. |
| Arithmetic Series: | Sequence consist of Arithmetic Progression i.e. 3,6,9,… |
| Geometric Series : | Sequence consist of Geometric Progression i.e. 2,6,18,54,… |
Tips and Tricks – Perfect Square Series
Consists of the perfect square of some numbers arranged in a specific order, with one number missing. Now we’ve to find the pattern the series is following and fill up the blank accordingly by finding that number.
Question 1. 100,121,144,__, 196
Solution:
This series consists of a perfect square of consecutive numbers 10, 11, 12, 13
Hence 169 will come in the blank.
Tips and Tricks – Perfect Cube Series
It consists of a cube of numbers sequenced in a particular order. The sequence would be like go om adding the cube of that specific given number.
Question 2. 9,64, 125, __, 343
Solution:
This series consists of a series of numbers with perfect cubes the is (3 x 3 x 3), (4 x 4 x 4), (5 x 5 x 5), (6 x 6 x 6), (7 x 7 x 7)
Hence 216 will be coming in the blank, as the series is following a trend of cubes of numbers in sequential order.
Tips and Tricks and Shortcuts- Ration Series
This type of series consists of numbers arranged in sequential order (following a particular trend, i.e., Either increasing or decreasing). Now, all we have to do here is to trace out this trend,(which could be , /, +, -) of each number of the series with a fixed number) by finding the proportional difference between the numbers of the series.
Question 3. 3, 6, 9, 12, __, 18, 21
Solution:
Here the series is following an increasing trend in which three is added to each number of the series.
3
6 (3+3)
9 (6+3)
12 (9+3)
15 (12+3)
18 (15+3)
21 (18+3)
Tips and Tricks and Shortcuts – Arithmetic series
In this kind of sequences, each number is found by ( + , -) each term by a constant number.
The formula of A S = {a, a+d, a +2d.….}
Where a= first term of the series
d = common difference
Question 4. 3, 6, 9 , 12
Solution:
Here a = 3(first term of the series)
d = 3
Hence we get:
3 + 3 = 6
6 + 3 = 9
9 + 3 = 12
12 + 3 = 15
Tips and Tricks and Shortcuts – Geometric series
In this kind of sequences, each number is found by (, /) each term by a constant number.
The formula of G S= {a, ar, ar2, ar3,….}
Where a= first term of the series
R= factor or difference between the term, also known as the common ratio.
Question 5. 1, 2, 4, 8, 16, 32
Solution:
Here a = 1 (first term of the series)
r = 2 (a standard number that is multiplied with the consecutive number of the series)
Hence we get:
1
1 x 2
1 x 22
1 x 23, ….)
Tips and Tricks and Shortcuts – Mixed series
In such series, while calculating the difference, there can be two steps involved in getting the next consecutive number of the series. So we’ve to follow the same pattern to get the perpetual number of the series.
Question 6. 1, \frac{5}{2}, \frac{10}{3}, \frac{17}{4}, —
Options:
A. \frac{26}{5}
B. \frac{24}{5}
C. \frac{21}{5}
D. 5
Correct Option: A
Explanation:
Here we’ve to observe the trend that this series is following, as each term is divided by a specific number, i.e., 1 then 2 then3 then 4 and so on. Hence we can make out that the denominator must be 5.
Now comes the numerator, where the 1’st number is 1.
Then comes 5, now how can we get 5 from the term number 2. It can come by \frac{2^{2} + 1 }{2} = \frac{5}{2} Next comes \frac{3^{2} + 1 }{3} = \frac{10}{3},\frac{4^{2} + 1 }{4} = \frac{17}{4}, \frac{5^{2} + 1 }{5} = \frac{26}{5}
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Coding and Number Series – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts
Letter and Symbol Series – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts
Logical Sequence of Words – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts
Analogy and Classification Pattern – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts
Coding and Number Series – Questions |Formulas |How to Solve Quickly |Tricks & Shortcuts
Letter and Symbol Series – Questions |Formulas |How to Solve Quickly |Tricks & Shortcuts
Logical Sequence of Words – Questions | Formulas |How to Solve Quickly |Tricks & Shortcuts
Analogy and Classification Pattern – Questions | Formulas |How to Solve Quickly |Tricks & Shortcuts
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7 comments on “Tips And Tricks And Shortcuts For Number Series”
pratip109
In Perfect Cube Series- Q1.) 9,64, 125, __, 3439 is not a perfect cube. Its a square of 3. The pattern should be 27, 64, 125, ___, 343
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Rupa
hi sir/mam…..this tips are very useful to every one and me also, iam following to learn and improve my aptitude sir/mamwe want so many tips of apptitude to develope our speed to solvethank you so much sir/mam
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Savitri
solving tricks is very super Sir
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HelpPrepInsta
Thank you for your appreciation and also we want you to know that we are more than happy to help you and serve you better!!!
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Shweta
Tysm for the tips😍
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Vanitha B Mulimani
the explanation for the mixed series is not appropriatewhen we solve for (4^2+1)/2 for 4 condition its will be 17/4 not 15/4
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Nishu Sagar
@Vanitha B Mulimanithank you for your feedback. We have resolve the issue.
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4412 | https://www.mathway.com/popular-problems/Precalculus/828981 | Enter a problem...
Precalculus Examples
Popular Problems
Find the Asymptotes y=tan(x-pi/2)
Step 1
For any , vertical asymptotes occur at , where is an integer. Use the basic period for , , to find the vertical asymptotes for . Set the inside of the tangent function, , for equal to to find where the vertical asymptote occurs for .
Step 2
Move all terms not containing to the right side of the equation.
Tap for more steps...
Step 2.1
Add to both sides of the equation.
Step 2.2
Combine the numerators over the common denominator.
Step 2.3
Add and .
Step 2.4
Divide by .
Step 3
Set the inside of the tangent function equal to .
Step 4
Move all terms not containing to the right side of the equation.
Tap for more steps...
Step 4.1
Add to both sides of the equation.
Step 4.2
Combine the numerators over the common denominator.
Step 4.3
Add and .
Step 4.4
Cancel the common factor of .
Tap for more steps...
Step 4.4.1
Cancel the common factor.
Step 4.4.2
Divide by .
Step 5
The basic period for will occur at , where and are vertical asymptotes.
Step 6
Find the period to find where the vertical asymptotes exist.
Tap for more steps...
Step 6.1
The absolute value is the distance between a number and zero. The distance between and is .
Step 6.2
Divide by .
Step 7
The vertical asymptotes for occur at , , and every , where is an integer.
Step 8
Tangent only has vertical asymptotes.
No Horizontal Asymptotes
No Oblique Asymptotes
Vertical Asymptotes: where is an integer
Step 9
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Purplemath
Let's take another look at that last problem on the previous page:
Solve x2 − 4 = 0
On the previous page, I'd solved this quadratic equation by factoring the difference of squares on the left-hand side of the equation, and then setting each factor equal to zero, etc, etc. The solution was "x = ± 2". However—
I can also try isolating the squared-variable term on the left-hand side of the equation (that is, I can try getting the x2 term by itself on one side of the "equals" sign), by moving the numerical part (that is, the 4) over to the right-hand side, like this:
Content Continues Below
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Solving by Taking Square Roots
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x2 − 4 = 0
x2 = 4
When I'm solving an equation, I know that I can do whatever I like to that equation as long as I do the exact same thing to both sides of that equation. On the left-hand side of this particular equation, I have an x2, and I want a plain old x. To turn the x2 into an x, I can take the square root of each side of the equation, like this:
x = ± 2
Then the solution is x = ±2, just like it was when I solved by factoring the difference of squares.
Why did I need the "±" (that is, the "plus-minus") sign on the 2 when I took the square root of the 4? Because I'm trying to find all values of the variable which make the original statement true, and it could have been either a positive 2 or a negative 2 that was squared to get that 4 in the original equation.
This duality is similar to how I'd had two factors, one "plus" and one "minus", when I used the difference-of-squares formula to solve this same equation on the previous page.
"Finding the solution to an equation" is a very different process from "evaluating the square root of a number". When finding "the" square root of a number, we're dealing exclusively with a positive value. Why? Because that is how the square root of a number is defined. The value of the square root of a number can only be positive, because that's how "the square root of a number" is defined.
Solving an equation, on the other hand — that is, finding all of the possible values of the variable that could work in an equation — is different from simply evaluating an expression that is already defined as having only one value.
Keep these two straight! A square-rooted number has only one value, but a square-rooted equation has two, because of the variable.
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In mathematics, we need to be able to get the same answer, no matter which valid method we happen to have used in order to arrive at that answer. So, comparing the answer I got above with the answer I got one the previous page confirms that we must use the "±" when taking square roots to solve.
(You may be doubting my work above in the step where I took the square root of either side, because I put a "±" sign on only one side of the equation. Shouldn't I add this character to both sides of the equation? Kind of, yes. But if I'd put it on both sides of the equation, would anything really have changed? No. Try all the cases, if you're not sure.)
A benefit of this square-rooting process is that it allows us to solve some quadratics that we could not have solved before when using only factoring. For instance:
Solve x2 − 50 = 0.
This quadratic has a squared part and a numerical part. I'll start by adding the numerical term to the other side of the equaion (so the squared part is by itself), and then I'll square-root both sides. I'll need to remember to simplify the square root:
x2 − 50 = 0
x2 = 50
Then my solution is:
While we could have gotten the previous integer solution by factoring, we could never have gotten this radical solution by factoring. Factoring is clearly useful for solving some quadratic equations, but additional sorts of techniques allow us to find solutions to additional sorts of equations.
Content Continues Below
Solve (x − 5)2 − 100 = 0.
This quadratic has a squared part and a numerical part. I'll start by adding the strictly-numerical term to the right-hand side of the equation, so that the squared binomial expression, containing the variable, is by itself on the left-hand side. Then I'll square-root both sides, remembering the "±" on the numerical side, and then I'll simplify:
(x − 5)2 − 100 = 0
(x − 5)2 = 100
x − 5 = ±10
x = 5 ± 10
x = 5 − 10 or x = 5 + 10
x = −5 or x = 15
This equation, after taking the square root of either side, did not contain any radcials. Because of this, I was able to simplify my results, all the way down to simple values. My answer is:
x = −5, 15
The previous equation is an example of a equation where the careless student will omit the "±" while solving, and will then have no clue as to how the book got the answer "x = −5, 15".
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These students get in the bad habit of not bothering to write the "±" sign until they check their answers in the back of the book and suddenly "remember" that they "meant" to put the "±" in there when they'd taken the square root of either side of the equation.
But this "magic" only works when you have the answer in the back (to remind you) and when the solution contains radicals (which doesn't always happen). In other cases, there will be no "reminder". Especially on tests, making the mistake of omitting the "±" can be deadly. Don't be that student. Always remember to insert the "±".
By the way, since the solution to the previous equation consisted of integers, this quadratic could also have been solved by multiplying out the square, factoring, etc:
(x − 5)2 − 100 = 0
x2 − 10x + 25 − 100 = 0
x2 − 10x − 75 = 0
(x − 15)(x + 5) = 0
x − 15 = 0, x + 5 = 0
x = 15, −5
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Solve (x − 2)2 − 12 = 0
This quadratic has a squared part and a numerical part. I'll add the numerical part over to the other side, so the squared part with the variable is by itself. Then I'll square-root both sides, remembering to add a "±" to the numerical side, and then I'll simplify:
(x − 2)2 − 12 = 0
(x − 2)2 = 12
I can't simplify this any more. My answer is going to have radicals in it. My solution is:
This quadratic equation, unlike the one before it, could not have also been solved by factoring. But how would I have solved it, if they had not given me the quadratic already put into "(squared part) minus (a number part)" form? This concern leads to the next topic: solving by completing the square.
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4414 | https://www.maths.ox.ac.uk/system/files/attachments/complex_1.pdf | Complex Numbers Richard Earl∗ Mathematical Institute, Oxford, OX1 2LB, July 2004 Abstract This article discusses some introductory ideas associated with complex numbers, their algebra and geometry. This includes a look at their importance in solving polynomial equations, how complex numbers add and multiply, and how they can be represented. Finally we look at the nth roots of unity, that is, the solutions of the equations zn = 1.
1 The Need For Complex Numbers The shortest path between two truths in the real domain passes through the complex domain Jacques Hadamard (1865-1963) All of you will know that the two roots of the equation ax2 + bx + c = 0 are x = −b ± √ b2 −4ac 2a (1) and solving quadratic equations is something that mathematicians have been able to do since the time of the Babylonians. When b2 −4ac > 0 then these two roots are real and distinct; graphically they are where the curve y = ax2 + bx + c cuts the x-axis. When b2 −4ac = 0 then we have one real root and the curve just touches the x-axis here. But what happens when b2 −4ac < 0? In this case there are no real solutions to the equation, as no real number squares to give the negative b2 −4ac. From the graphical point of view, the curve y = ax2 + bx + c lies entirely above or below the x-axis.
-1 1 2 3 -1 1 2 3 Two Distinct Real Roots -1 1 2 3 1 2 3 4 One Repeated Real Root -1 -0.5 0.5 1 1.5 2 2.5 3 1 2 3 4 5 No Real Roots It is only comparatively recently that mathematicians have been comfortable with these roots when b2 −4ac < 0. During the Renaissance the quadratic would have been considered unsolvable, or its roots would have been called imaginary1.
If we imagine √−1 to exist, and that it behaves much like other numbers, then the two roots of the quadratic ax2 + bx + c = 0 can be written in the form x = A ± B √ −1 (2) ∗These pages are produced by Richard Earl, who is the Schools Liaison and Access Officer for mathematics, statistics and computer science at Oxford University. Any comments, suggestions or requests for other material would be welcomed at earl@maths.ox.ac.uk 1The term ‘imaginary’ was first used by the French Mathematician René Descartes (1596-1650). Whilst he is known more as a philosopher, Descartes made many important contributions to mathematics and helped found co-ordinate geometry – hence the naming of Cartesian co-ordinates.
1 where A = −b/2a and B = √ 4ac −b2/2a are real numbers. But what meaning can such roots have?
It was this philosophical point which pre-occupied mathematicians until the start of the 19th century; afterwards these ‘imaginary’ numbers started proving so useful (especially in the work of Cauchy and Gauss) that these philosophical concerns were essentially forgotten.
Notation 1 We shall from now on write i for √−1 – this is standard notation amongst mathemati-cians2, though many books, particularly those written for engineers and physicists, use j instead.
Definition 2 A complex number 3 is a number of the form a + bi where a and b are real numbers. If z = a + bi then a is known as the real part of z and b as the imaginary part. We write a = Re z and b = Im z. Note that real numbers are complex – a real number is simply a complex number with zero imaginary part.
Remark 3 Note that two complex numbers are equal precisely when their real and imaginary parts are equal – that is a+bi = c+di if and only if a = c and b = d. This is called ‘comparing real and imaginary parts’.
Notation 4 We write C for the set of all complex numbers.
One of the first major results concerning complex numbers, and which conclusively demonstrated their usefulness, was proved by Gauss in 1799. From the quadratic formula (1) we know that all quadratic equations can be solved using complex numbers, but what Gauss was the first to prove was the much more general result: Theorem 5 (FUNDAMENTAL THEOREM OF ALGEBRA) The roots of any polynomial equation a0 + a1x + a2x2 + · · · + anxn = 0, with real (or complex) coefficients ai, are complex. That is there are n (not necessarily distinct) complex numbers γ1, . . . , γn such that a0 + a1x + a2x2 + · · · + anxn = an (x −γ1) (x −γ2) · · · (x −γn) .
In particular, this shows that a degree n polynomial has, counting repetitions, exactly n roots in C.
The proof of this theorem is far beyond the scope of this document. Note that the theorem only guarantees the existence of the roots of a polynomial somewhere in C unlike the quadratic formula, which plainly gives us a formula for the roots. The theorem gives no hint as to where in C these roots are to be found.
Exercise 1 A. Which of the following quadratic equations require the use of complex numbers to solve them?
3x2 + 2x −1 = 0, 2x2 −6x + 9 = 0, −4x2 + 7x −9 = 0.
Exercise 2 B. On separate axes, sketch the graphs of the following cubics, being sure to carefully label any turning points. In each case state how many of the cubic’s roots are real.
y1 (x) = x3 −x2 −x + 1; y2 (x) = 3x3 + 5x2 + x + 1; y3 (x) = −2x3 + x2 −x + 1.
2The i notation was first introduced by the Swiss mathematician Leonhard Euler (1707-1783). Much of our modern notation is due to him including e and π. Euler was a giant in 18th century mathematics and the most prolific mathematician ever. His most important contributions were in analysis (eg. on infinite series, calculus of variations). The study of topology arguably dates back to his solution of the Königsberg Bridge Problem.
3The term ‘complex number’ is due to the German mathematician Carl Gauss (1777-1855). Gauss is considered by many the greatest mathematician ever. He made major contributions to almost every area of mathematics from number theory and non-Euclidean geometry, to astronomy and magnetism. His name precedes a multitude of theorems and definitions throughout mathematics.
2 Exercise 3 C. Let p and q be real numbers with p ≤0. Find the co-ordinates of the turning points of the cubic y = x3 + px + q. Show that the cubic equation x3 + px + q = 0 has three real roots, with two or more repeated, precisely when 4p3 + 27q2 = 0.
Under what conditions on p and q does x3 + px + q = 0 have (i) three distinct real roots, (ii) just one real root? How many real roots does the equation x3 + px + q = 0 have when p > 0?
Exercise 4 C. By making a substitution of the form X = x −α for a certain choice of α, transform the equation X3 +aX2 + bX +c = 0 into one of the form x3 +px +q = 0. Hence find conditions under which the equation X3 + aX2 + bX + c = 0 has (i) three distinct real roots, (ii) three real roots involving repetitions, (iii) just one real root.
Exercise 5 C The cubic equation x3 + ax2 + bx + c = 0 has roots α, β, γ so that x3 + ax2 + bx + c = (x −α) (x −β) (x −γ) .
By equating the coefficients of powers of x in the previous equation, find expressions for a, b and c in terms of α, β and γ.
Given that α, β, γ are real, what can you deduce about their signs if (i) c < 0, (ii) b < 0 and c < 0, (iii) b < 0 and c = 0.
Exercise 6 C. With a, b, c and α, β, γ as in the previous exercise, let Sn = αn+βn+γn. Find expressions S0, S1 and S2 in terms of a, b and c. Show further that Sn+3 + aSn+2 + bSn+1 + cSn = 0 for n ≥0 and hence find expressions for S3 and S4 in terms of a, b and c.
2 Basic Operations We add, subtract, multiply and divide complex numbers much as we would expect. We add and subtract complex numbers by adding their real and imaginary parts:-(a + bi) + (c + di) = (a + c) + (b + d) i, (a + bi) −(c + di) = (a −c) + (b −d) i.
We can multiply complex numbers by expanding the brackets in the usual fashion and using i2 = −1, (a + bi) (c + di) = ac + bci + adi + bdi2 = (ac −bd) + (ad + bc) i.
To divide complex numbers, we note firstly that (c + di) (c −di) = c2 + d2 is real. So a + bi c + di = a + bi c + di × c −di c −di = µac + bd c2 + d2 ¶ + µbc −ad c2 + d2 ¶ i.
The number c −di, which we just used, as relating to c + di, has a special name and some useful properties – see Proposition 10.
Definition 6 Let z = a + bi. The conjugate of z is the number a −bi, and this is denoted as ¯ z (or in some books as z∗).
Note from equation (2) that when the real quadratic equation ax2 + bx + c = 0 has complex roots, then these roots are conjugates of each other. Generally, if the polynomial anzn + an−1zn−1 + · · · + a0 = 0, where the ai are real, has a root z0, then the conjugate ¯ z0 is also a root.
3 Exercise 7 A. Put each of the following numbers into the form a + bi.
(1 + 2i)(3 −i), 1 + 2i 3 −i , (1 + i)4.
Exercise 8 A. Let z1 = 1 + i and let z2 = 2 −3i. Put each of the following into the form a + bi.
z1 + z2, z1 −z2, z1z2, z1/z2, ¯ z1¯ z2.
We needed a special symbol i for √−1, but we see now that further symbols are needed to find the square root of i. In fact we already knew this from the Fundamental Theorem, which implies that z2 = i has two roots amongst the complex numbers. The quadratic formula (1), is also valid for complex coefficients a, b, c, provided that proper sense is made of the square roots of the complex number b2 −4ac.
Problem 7 Find all those z that satisfy z2 = i.
Suppose that z2 = i and z = a + bi, where a and b are real. Then i = (a + bi)2 = ¡ a2 −b2¢ + 2abi.
Comparing the real and imaginary parts (see Remark 3), we know that a2 −b2 = 0 and 2ab = 1.
So b = 1/2a from the second equation, and substituting for b into the first equation gives a4 = 1/4, which has real solutions a = 1/ √ 2 or a = −1/ √ 2.
So the two z which satisfy z2 = i, i.e. the two square roots of i, are 1 + i √ 2 and −1 −i √ 2 .
Problem 8 Use the quadratic formula to find the two solutions of z2 −(3 + i) z + (2 + i) = 0.
We see that a = 1, b = −3 −i, and c = 2 + i. So b2 −4ac = (−3 −i)2 −4 × 1 × (2 + i) = 9 −1 + 6i −8 −4i = 2i.
Knowing √ i = ± (1 + i) / √ 2, from the previous problem, we have x = −b ± √ b2 −4ac 2a = (3 + i) ± √ 2i 2 = (3 + i) ± √ 2 √ i 2 = (3 + i) ± (1 + i) 2 = 4 + 2i 2 or 2 2 = 2 + i or 1.
Exercise 9 A. Find the square roots of −5 −12i, and hence solve the quadratic equation z2 −(4 + i) z + (5 + 5i) = 0.
4 Exercise 10 B. Show that the complex number 1 + i is a root of the cubic equation z3 + z2 + (5 −7i) z −(10 + 2i) = 0, and hence find the other two roots.
Exercise 11 B. Show that the complex number 2 + 3i is a root of the quartic equation z4 −4z3 + 17z2 −16z + 52 = 0, and hence find the other three roots.
Exercise 12 B. Let n be a positive integer. Simplify the expression (1 + i)2n . Use the binomial theorem to show that µ2n 0 ¶ − µ2n 2 ¶ + µ2n 4 ¶ − µ2n 6 ¶ + · · · + (−1)n µ2n 2n ¶ = ½ (−1)n/2 2n if n is even; 0 if n is odd.
Show that the right-hand side is equal to 2n cos (nπ/2) . Similarly, find the value of µ2n 1 ¶ − µ2n 3 ¶ + µ2n 5 ¶ − µ2n 7 ¶ + · · · + (−1)n−1 µ 2n 2n −1 ¶ .
3 The Argand Diagram The real numbers are often represented on the real line which increase as we move from left to right.
-4 -2 0 2 4 è !!!!
2 π The Real Line The complex numbers, having two components, their real and imaginary parts, can be represented as a plane; indeed, C is sometimes referred to as the complex plane, but more commonly, when we represent C in this manner, we call it an Argand diagram4.The point (a, b) represents the complex number a + bi so that the x-axis contains all the real numbers, and so is termed the real axis, and the y-axis contains all those complex numbers which are purely imaginary (i.e. have no real part), and so is referred to as the imaginary axis.
-4 -2 2 4 -3 -2 -1 1 2 3 + 2 i 2 −3 i −3 + i The Argand Diagram 4After the Swiss mathematician Jean-Robert Argand (1768-1822).
5 We can think of z0 = a + bi as a point in an Argand diagram, but it can often be useful to think of it as a vector as well. Adding z0 to another complex number translates that number by the vector ¡a b ¢ .
That is the map z 7→z + z0 represents a translation a units to the right and b units up in the complex plane. Note that the conjugate ¯ z of a point z is its mirror image in the real axis. So, z 7→¯ z represents reflection in the real axis.
Exercise 13 B. Multiplication by i takes the point x+iy to the point −y+ix. What transformation of the Argand diagram does this represent? What is the effect of multiplying a complex number by (1 + i) / √ 2?
[Hint: recall that this is square root of i.] A complex number z in complex plane can be represented by Cartesian co-ordinates, its real and imaginary parts, but equally useful is the representation of z by polar co-ordinates. If we let r be the distance of z from the origin, and if, for z 6= 0, we define θ to be the angle that the line connecting the origin to z makes with the positive real axis, then we can write z = x + iy = r cos θ + ir sin θ.
(3) The relations between z’s Cartesian and polar co-ordinates are simple – we see that x = r cos θ and y = r sin θ, r = p x2 + y2 and tan θ = y x.
Definition 9 The number r is called the modulus of z and is written |z| .The number θ is called the argument of z and is written arg z. If z = x + iy then |z| = p x2 + y2 and sin arg z = y p x2 + y2 , cos arg z = x p x2 + y2 .
Note that the argument of 0 is undefined.
Note that arg z is defined only up to multiples of 2π. For example, the argument of 1 + i could be π/4 or 9π/4 or −7π/4 etc.. For simplicity, in this article we shall give all arguments in the range 0 ≤θ < 2π, so that π/4 would be the preferred choice here.
0.5 1 1.5 2 2.5 3 3.5 0.25 0.5 0.75 1 1.25 1.5 1.75 2 Imz Rez »z» z arg HzL A Complex Number’s Cartesian and Polar Co-ordinates Exercise 14 A. Find the modulus and argument of each of the following numbers.
1 + √ 3i, (2 + i) (3 −i) , (1 + i)5 .
Exercise 15 B. Let α be a real number in the range 0 < α < π/2. Find the modulus and argument of the following numbers.
cos α −i sin α, sin α −i cos α, 1 + i tan α, 1 + cos α + i sin α.
6 Exercise 16 B. On separate Argand diagrams sketch the following sets: (i) |z| < 1; (ii) Re z = 3; (iii) |z −1| = |z + i| ; (iv) −π/4 < arg z < π/4; (v) Re (z + 1) = |z −1| ; (vi) arg (z −i) = π/2; (vii) |z −3 −4i| = 5; (viii) Re ((1 + i) z) = 1.
(ix) Im ¡ z3¢ > 0.
We now prove some useful algebraic properties of the modulus, argument and conjugate functions.
Proposition 10 Let z, w ∈C. Then |zw| = |z| |w| ; |z/w| = |z| / |w| if w 6= 0; z¯ z = |z|2 ; |¯ z| = |z| ; z + w = ¯ z + ¯ w; z −w = ¯ z −¯ w; zw = ¯ z ¯ w; z/w = ¯ z/ ¯ w if w 6= 0; |z + w| ≤|z| + |w| ; ||z| −|w|| ≤|z −w| ; • and up to multiples of 2π then the following equations also hold: • arg (zw) = arg z + arg w if z, w 6= 0, • arg (z/w) = arg z −arg w if z, w 6= 0, • arg ¯ z = −arg z if z 6= 0.
A selection of the above statements is proved here; the remaining ones are left as exercises.
Proof. |zw| = |z| |w| . Let z = a + bi and w = c + di. Then zw = (ac −bd) + (bc + ad) i so that |zw| = q (ac −bd)2 + (bc + ad)2 = p a2c2 + b2d2 + b2c2 + a2d2 = p (a2 + b2) (c2 + d2) = p a2 + b2p c2 + d2 = |z| |w| .
Proof. arg (zw) = arg z + arg w. Let z = r (cos θ + i sin θ) and w = R (cos Θ + i sin Θ) . Then zw = rR (cos θ + i sin θ) (cos Θ + i sin Θ) = rR ((cos θ cos Θ −sin θ sin Θ) + i (sin θ cos Θ + cos θ sin Θ)) = rR (cos (θ + Θ) + i sin (θ + Θ)) .
We can read offthat |zw| = rR = |z| |w| which is a second proof of the previous part and also that arg (zw) = θ + Θ = arg z + arg w up to multiples of 2π.
Proof. zw = ¯ z ¯ w. Let z = a + bi and w = c + di. Then zw = (ac −bd) + (bc + ad) i = (ac −bd) −(bc + ad) i = (a −bi) (c −di) = ¯ z ¯ w.
7 Proof. (Triangle Inequality) |z + w| ≤|z| + |w|.
0.5 1 1.5 2 2.5 3 -2 -1.5 -1 -0.5 0.5 1 z+w w z w Has a vector L A diagrammatic proof of the Triangle Inequality Note that the shortest distance between 0 and z + w is the modulus of z + w. This is shorter in length than the path which goes from 0 to z to z + w.
The total length of this second path is |z| + |w| .
For an algebraic proof, note that for any complex number z + ¯ z = 2 Re z and Re z ≤|z| . So for z, w ∈C, z ¯ w + ¯ zw 2 = Re (z ¯ w) ≤|z ¯ w| = |z| | ¯ w| = |z| |w| .
Then |z + w|2 = (z + w) (z + w) = (z + w) (¯ z + ¯ w) = z¯ z + z ¯ w + ¯ zw + w ¯ w ≤ |z|2 + 2 |z| |w| + |w|2 = (|z| + |w|)2 , to give the required result.
Corollary 11 The complex roots of a real polynomial come in pairs. That is, if z0 satisfies the polynomial equation akzk + ak−1zk−1 + · · · + a0 = 0, where each ai is real, then z0 is also a root.
Proof. Note from the algebraic properties of the conjugate function, proven in the previous proposition, that ak (z0)k + ak−1 (z0)k−1 + · · · + a1z0 + a0 = ak(z0)k + ak−1(z0)k−1 + · · · + a1z0 + a0 = ak(z0)k + ak−1(z0)k−1 + · · · + a1z0 + a0 [the ai are real] = ak (z0)k + ak−1 (z0)k−1 + · · · + a0 = 0 [as z0 is a root] = 0.
Exercise 17 A. Let z and w be two complex numbers such that zw = 0. Show either z = 0 or w = 0.
Exercise 18 A. Suppose that the complex number α is a square root of z, that is α2 = z. Show that the only other square root of z is −α. Suppose now that the complex numbers z1 and z2 have square roots ±α1 and ±α2 respectively. Show that the square roots of z1z2 are ±α1α2 Exercise 19 B. Prove the remaining identities from Proposition 10.
8 Exercise 20 B. Let t be a real number. Find expressions for x = Re 1 2 + ti, y = Im 1 2 + ti.
Find an equation relating x and y by eliminating t. Deduce that the image of the line Re z = 2 under the map z 7→1/z is contained in a circle. Is the image of the line all of the circle?
Exercise 21 B. Let z1 and z2 be two complex numbers. Show that |z1 −z2|2 + |z1 + z2|2 = 2 ³ |z1|2 + |z2|2´ .
This fact is called the Parallelogram Law – how does this relate the lengths of the diagonals and sides of the parallelogram? [Hint: consider the parallelogram in C with vertices 0, z1, z2, z1 + z2.] Exercise 22 C. Consider a quadrilateral OABC in the complex plane whose vertices are at the complex numbers 0, a, b, c. Show that the equation |b|2 + |a −c|2 = |a|2 + |c|2 + |a −b|2 + |b −c|2 can be rearranged as |b −a −c|2 = 0.
Hence show that the only quadrilaterals to satisfy the Parallelogram Law are parallelograms.
4 Roots Of Unity Consider the complex number z0 = cos θ + i sin θ, where 0 ≤θ < 2π. The modulus of z0 is 1, and the argument of z0 is θ.
-2 -1.5 -1 -0.5 0.5 1 1.5 2 -1.5 -1 -0.5 0.5 1 1.5 z0 z02 z03 θ θ θ Powers of z0 In Proposition 10 we proved for z, w 6= 0 that |zw| = |z| |w| and arg (zw) = arg z + arg w, up to multiples of 2π. So for any integer n, and any z 6= 0, we have that |zn| = |z|n and arg (zn) = n arg z.
So the modulus of (z0)n is 1 and the argument of (z0)n is nθ, or putting this another way 9 Theorem 12 (DE MOIVRE’S5 THEOREM) For a real number θ and integer n we have that cos nθ + i sin nθ = (cos θ + i sin θ)n .
Exercise 23 B. Use De Moivre’s Theorem to show that cos 5θ = 16 cos5 θ −20 cos3 θ + 5 cos θ, and that sin 5θ = ¡ 16 cos4 θ −12 cos2 θ + 1 ¢ sin θ.
Exercise 24 B. Let z = cos θ + i sin θ and let n be an integer. Show that 2 cos θ = z + 1 z and that 2i sin θ = z −1 z .
Find expressions for cos nθ and sin nθ in terms of z.
Exercise 25 B. Show that cos5 θ = 1 16 (cos 5θ + 5 cos 3θ + 10 cos θ) and hence find R π/2 0 cos5 θ dθ.
We apply these ideas to the following problem.
Problem 13 Let n be a natural number. Find all those complex z such that zn = 1.
We know from the Fundamental Theorem of Algebra that there are (counting repetitions) n solutions: these are known as the nth roots of unity.
Let’s first solve zn = 1 directly for n = 2, 3, 4.
• When n = 2 we have 0 = z2 −1 = (z −1) (z + 1) and so the square roots of 1 are ±1.
• When n = 3 we can factorise as follows 0 = z3 −1 = (z −1) ¡ z2 + z + 1 ¢ .
So 1 is a root and completing the square we see 0 = z2 + z + 1 = µ z + 1 2 ¶2 + 3 4 which has roots −1 2 ± √ 3 2 i.
So the cube roots of 1 are 1, −1/2 + √ 3i/2, and −1/2 − √ 3i/2.
• When n = 4 we can factorise as follows 0 = z4 −1 = ¡ z2 −1 ¢ ¡ z2 + 1 ¢ = (z −1) (z + 1) (z −i) (z + i) , so that the fourth roots of 1 are 1, −1, i and −i.
5De Moivre (1667-1754), a French protestant who moved to England, is best remembered for this formula, but his major contributions were in probability and appeared in his The Doctrine Of Chances (1718).
10 Plotting these roots on Argand diagrams we can see a pattern developing -2 -1.5 -1 -0.5 0.5 1 1.5 2 -1.5 -1 -0.5 0.5 1 1.5 Square Roots -2 -1.5 -1 -0.5 0.5 1 1.5 2 -1.5 -1 -0.5 0.5 1 1.5 Cube Roots -2 -1.5 -1 -0.5 0.5 1 1.5 2 -1.5 -1 -0.5 0.5 1 1.5 Fourth Roots Returning to the general case, suppose that z = r (cos θ + i sin θ) and satisfies zn = 1. Then by the observations preceding De Moivre’s Theorem zn has modulus rn and has argument nθ. As 1 has modulus 1 and argument 0, we can compare their moduli to find rn = 1 giving r = 1. Comparing arguments, we see nθ = 0 up to multiples of 2π. That is nθ = 2kπ for some integer k, giving θ = 2kπ/n. So the roots of zn = 1 are z = cos µ2kπ n ¶ + i sin µ2kπ n ¶ where k is an integer.
At first glance there seem to be an infinite number of roots but note as cos and sin have period 2π then these z repeat with period n. So the nth roots of unity are z = cos µ2kπ n ¶ + i sin µ2kπ n ¶ where k = 0, 1, 2, . . . , n −1.
Plotted on an Argand diagram, the nth roots of unity form a regular n-gon inscribed within the unit circle with a vertex at 1.
Problem 14 Find all the solutions of the cubic z3 = −2 + 2i.
If we write −2 + 2i in its polar form we have −2 + 2i = √ 8 µ cos µ3π 4 ¶ + i sin µ3π 4 ¶¶ .
So if z3 = −2 + 2i, and z has modulus r and argument θ, then r3 = √ 8 and 3θ = 3π 4 up to multiples of 2π, which gives r = √ 2 and θ = π 4 + 2kπ 3 for some integer k.
As before, we need only consider k = 0, 1, 2 (as other k lead to repeats) and we see the three roots are √ 2 ³ cos ³π 4 ´ + i sin ³π 4 ´´ = 1 + i, √ 2 µ cos µ11π 12 ¶ + i sin µ11π 12 ¶¶ = à −1 2 − √ 3 2 !
+ i Ã√ 3 2 −1 2 !
, √ 2 µ cos µ19π 12 ¶ + i sin µ19π 12 ¶¶ = à −1 2 + √ 3 2 !
+ i à − √ 3 2 −1 2 !
.
11 Exercise 26 A. Let ω be a cube root of unity (i.e. ω3 = 1) such that ω 6= 1. Show that 1 + ω + ω2 = 0.
Exercise 27 C. Let ζ = cos 2π 5 + i sin 2π 5 .
Show that ζ5 = 1, and deduce that 1 + ζ + ζ2 + ζ3 + ζ4 = 0.
Find the quadratic equation with roots ζ + ζ4 and ζ2 + ζ3. Hence show that cos 2π 5 = √ 5 −1 4 .
Exercise 28 C. Determine the modulus and argument of the two complex numbers 1 + i and √ 3 + i.
Also write the number 1 + i √ 3 + i in the form x + iy. Deduce that cos π 12 = √ 3 + 1 2 √ 2 and sin π 12 = √ 3 −1 2 √ 2 .
Exercise 29 B. Let A = 1+i and B = 1−i. Find the two numbers C and D such that ABC and ABD are equilateral triangles in the Argand diagram. Show that if C < D then A + ωC + ω2B = 0 = A + ωB + ω2D, where ω = ¡ −1 + √ 3i ¢ /2 is a cube root of unity other than 1.
Exercise 30 B. By considering the seventh roots of −1 show that cos π 7 + cos 3π 7 + cos 5π 7 = 1 2.
What is the value of cos 2π 7 + cos 4π 7 + cos 6π 7 ?
Exercise 31 B. Find all the roots of the equation x8 = −1. Hence, write x8 + 1 as the product of four quadratic factors.
Exercise 32 C. Find all the roots of the following equations.
1. 1 + z2 + z4 + z6 = 0, 2. 1 + z3 + z6 = 0, 3. (1 + z)5 −z5 = 0, 4. (z + 1)9 + (z −1)9 = 0.
12 5 Further Exercises Exercise 33 C. Given a non-zero complex number z = x + iy, we can associate with z a matrix Z = µ x y −y x ¶ .
Show that if z and w are complex numbers with associated matrices Z and W, then the matrices associated with z + w, zw and 1/z are Z + W, ZW and Z−1 respectively. Hence, for each of the following matrix equations, find a matrix Z which is a solution.
Z2 = µ 0 1 −1 0 ¶ , Z2 + 2Z = µ −5 0 0 −5 ¶ , Z2 + µ −3 1 1 −3 ¶ Z = µ −2 −1 1 −2 ¶ , Z5 = µ 1 1 −1 1 ¶ .
Exercise 34 C. The sequence x0, x1, x2, x3, ... is defined recursively by x0 = 0, x1 = 0.8, xn = 1.2xn−1 −xn−2 for n ≥2.
With the aid of a calculator list the values of xi for 0 ≤i ≤10. Prove further, by induction, that xn = Im {(0.6 + 0.8i)n} for each n = 0, 1, 2, .... Deduce that |xn| ≤1 for all n. Show also that xn cannot have the same sign for more than three consecutive n.
Exercise 35 C. Consider the cubic equation z3 + mz + n = 0 where m and n are real numbers. Let ∆ be a square root of (n/2)2 + (m/3)3. We then define t and u by t = −n/2 + ∆and u = n/2 + ∆, and let T and U respectively be cube roots of t and u. Show that tu is real, and that if T and U are chosen appropriately, then z = T −U is a solution of the original cubic equation.
Use this method to completely solve the equation z3 + 6z = 20. By making a substitution of the form w = z −a for a suitable choice of a, find all three roots of the equation 8w3 + 12w2 + 54w = 135.
Exercise 36 C. Express tan 7θ in terms of tan θ and its powers. Hence solve the equation x6 −21x4 + 35x2 −7 = 0.
Exercise 37 C. Show for any complex number z, and any positive integer n, that z2n −1 = ¡ z2 −1 ¢ n−1 Y k=1 ½ z2 −2z cos kπ n + 1 ¾ .
By setting z = cos θ + i sin θ show that sin nθ sin θ = 2n−1 n−1 Y k=1 ½ cos θ −cos kπ n ¾ .
13 |
4415 | https://www.scribd.com/document/375788065/agricultural-hearths-maps | Agricultural Hearths Maps | PDF | Wellness
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Agricultural Hearths Maps
The document discusses several topics related to the history and geography of agriculture: - The initial agricultural hearths that developed during the Neolithic Agricultural Revolution, as…
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Agricultural Hearths & Crops
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Present Day Agricultural Practices
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Climate Regions (Koppen-Climate Scale)
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Go ahead and answer on your own sheet of paper
1) Use the document in this packet, and look at Fouberg Figure 11. 4 & Table 11.1
. They’re both d escribing the initial Ag ricul
tural Hearths of the world during the Neolithic Agricultural Revolution (1
st
Agricultural Revolution). Discuss the overlap in the maps/tables, how are they the same and different and, if you had to group or create a pattern to know them, how would you do it? 2) Look at Agricultural Hearths & Crops in this packet. Now, look at Climate Regions. Identify the primary patterns you see with Agricultural Hearths and climate regions. If there are exceptions to y our pattern, identify where, but then identify why it was still a hearth even with a different climate, these answers can be found in your Fouberg.
3) Look at Present Agricultural Practices; group today’s practices by climate region, this means you have to identify the cli
mate regions where something is occurring and identify the pattern to these groups. 4) Use your Getis, pick ANY map in Getis that is
not
specific to Agriculture, in fact, due your best to stay out of the current chapter you are reading.
Discuss how the map you’ve chosen (include
page num ber) is influenc ed to the map in this packe t called “Prese nt Agricultural Pra ctices.”
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4416 | https://pmc.ncbi.nlm.nih.gov/articles/PMC11917015/ | FOXL2 drives the differentiation of supporting gonadal cells in early ovarian development - PMC
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Reprod Biol Endocrinol
. 2025 Mar 18;23:44. doi: 10.1186/s12958-025-01377-0
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FOXL2 drives the differentiation of supporting gonadal cells in early ovarian development
Laura Danti
Laura Danti
1 Department of Obstetrics and Gynaecology, University of Helsinki and Helsinki University Hospital, P.O. Box 140, Helsinki, 00029 HUS Finland
2 Department of Medicine Huddinge (MedH), Biosciences and Nutrition Unit, Karolinska Institutet, Huddinge, Sweden
Find articles by Laura Danti
1,2, Karolina Lundin
Karolina Lundin
1 Department of Obstetrics and Gynaecology, University of Helsinki and Helsinki University Hospital, P.O. Box 140, Helsinki, 00029 HUS Finland
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1,✉, Petra Nedeczey-Ruzsák
Petra Nedeczey-Ruzsák
1 Department of Obstetrics and Gynaecology, University of Helsinki and Helsinki University Hospital, P.O. Box 140, Helsinki, 00029 HUS Finland
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1, Timo Tuuri
Timo Tuuri
1 Department of Obstetrics and Gynaecology, University of Helsinki and Helsinki University Hospital, P.O. Box 140, Helsinki, 00029 HUS Finland
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1,#, Juha S Tapanainen
Juha S Tapanainen
1 Department of Obstetrics and Gynaecology, University of Helsinki and Helsinki University Hospital, P.O. Box 140, Helsinki, 00029 HUS Finland
3 Department of Obstetrics and Gynaecology, HFR – Cantonal Hospital of Fribourg and University of Fribourg, Fribourg, Switzerland
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1,3,#
Author information
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1 Department of Obstetrics and Gynaecology, University of Helsinki and Helsinki University Hospital, P.O. Box 140, Helsinki, 00029 HUS Finland
2 Department of Medicine Huddinge (MedH), Biosciences and Nutrition Unit, Karolinska Institutet, Huddinge, Sweden
3 Department of Obstetrics and Gynaecology, HFR – Cantonal Hospital of Fribourg and University of Fribourg, Fribourg, Switzerland
✉
Corresponding author.
Contributed equally.
Received 2024 Nov 25; Accepted 2025 Mar 3; Collection date 2025.
© The Author(s) 2025
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit
PMC Copyright notice
PMCID: PMC11917015 PMID: 40102860
Abstract
Background
Forkhead box L2 (FOXL2) is a transcription factor from the forkhead box family primarily expressed in the pituitary, ovaries, and eyelids. Human mutations in FOXL2 cause blepharophimosis, ptosis, epicanthus and inversus syndrome (BPES), which can be associated with primary ovarian insufficiency, and is indirectly linked with differences of sex development (DSD). Animal studies have shown the crucial role that FOXL2 plays in the development, function, and maintenance of the ovary as well as in sex determination. However, the specific role of FOXL2 in early human somatic cell ovarian development is largely unknown.
Methods
In this study, we utilised CRISPR/Cas9 genome activation and a previously published in-house 14-day gonadal differentiation protocol to study the role of FOXL2.
Results
Our results demonstrate that FOXL2 downregulates coelomic epithelial markers GATA4 and LHX9, female gonadal markers RSPO1 and WNT4, and male gonadal markers SOX9, NR0B1 and DHH. The differentially expressed genes were mostly associated with Kyoto encyclopaedia of genes and genomes (KEGG) pathways relating to cell adhesion molecules and gene ontology (GO) pathways relating to extracellular matrix and junction formation. Furthermore, a comparative analysis with existing single cell RNA sequencing data from human in vivo-derived samples elucidated that FOXL2 initiates the downregulation of coelomic epithelial genes GATA4, LHX9 and UPK3B at day 6. By day 8, the genes ARX and GATA2 are transiently upregulated by FOXL2 induction and then downregulated as the genes LGR5, TSPAN8, OSR1 and TAC1 become upregulated.
Conclusions
These findings suggest that FOXL2 facilitates the exit of differentiating cells from the coelomic epithelium and initially drives them towards a transitional identity before progressing into early supporting gonadal-like cells. The findings of this study significantly advance our understanding of normal gonadal development which can be used as a basis to elucidate pathological gonadal development underlying BPES.
Supplementary Information
The online version contains supplementary material available at 10.1186/s12958-025-01377-0.
Keywords: FOXL2, Ovary, Granulosa cell, CRISPR/Cas9, Stem cells
Background
Human sex determination is a complex and delicate process, initially dictated by the configuration of sex chromosomes in an individual - XY in males and XX in females. Contrary to earlier beliefs suggesting that ovarian development is a default process, recent research shows it is an active mechanism that is initiated by the expression of specific genes whose combined role is to repress testicular development [1–3]. Key genes promoting female gonadal development are r-spondin 1 (RSPO1), wnt family member 4 (WNT4) and forkhead box L2 (FOXL2) . Among these, FOXL2 is considered one of the earliest markers of ovarian development [5, 6].
The FOXL2 gene encodes a forkhead transcription factor that is primarily expressed in the ovary, pituitary gland, and eyelids. It plays a crucial role in mammalian sex determination and ovarian development, maintenance, and function [1, 2]. In humans, FOXL2 mutations cause blepharophimosis, ptosis, epicanthus and inversus syndrome (BPES). This syndrome can be classified into two types dependent on the presence (type I) or absence (type II) of primary ovarian insufficiency (POI) .
The role of FOXL2 in ovarian development has been extensively studied using animal models, the most notable being mice and goats. Homozygous knock-out (KO) mice exhibit a very high perinatal mortality rate and are characterised by small stature, along with eyelid and craniofacial abnormalities . These mice are infertile due to abnormal follicle development, characterised by the presence of only a single layer of granulosa cells surrounding the growing oocyte, which fail to undergo the typical squamous to cuboidal transition. By eight weeks of age, the primordial follicle pool is already depleted, and widespread follicular atresia is observed. By sixteen weeks, no healthy oocytes or follicles remain [8, 9]. In goats, a naturally occurring FOXL2 mutation causes polled intersex syndrome (PIS), which is associated with polledness and intersexuality. It solely affects XX individuals, resulting in female-to-male sex-reversal [10, 11]. In 2014, Boulanger et al. created a FOXL2 goat KO model, where XX individual with bi-allelic FOXL2 mutations exhibited complete sex-reversal and eyelid malformations, signifying that FOXL2 is the female sex-determining gene in this species . Additionally, Uhlenhaut et al. demonstrated in 2009 that Foxl2 is required for ovarian maintenance in mice. Their study revealed that conditional deletion of Foxl2 in mature granulosa cells of the postnatal ovary led to upregulation of the testis-determining gene Sox9. Consequently, the granulosa and theca cells of the ovary transdifferentiated into Sertoli cells and Leydig cells, respectively . Together these KO studies solidified the crucial role for FOXL2 in female sex-determination, as well as in ovarian development, function, and maintenance.
Despite these insights, the specific functional role of FOXL2 during early ovarian development remains less well understood. In the present study, we employed CRISPR/Cas9 activation to conditionally induce FOXL2 at the intermediate mesoderm stage in a 14-day gonadal differentiation protocol to investigate its function in early gonadal somatic cell development. Our results suggest that FOXL2 acts as a trigger for the transition of cells from coelomic epithelium to an early supporting gonadal cell fate, marked by the expression of LGR5 and TSPAN8. Furthermore, we demonstrate that FOXL2 simultaneously represses the expression of genes that are known to drive male sex differentiation.
Materials and methods
Cell culture
H9-FOXL2-DDdCas9VP192 hESCs were cultured on Geltrex™ LDEV-Free Reduced Growth Factor Basement Membrane Matrix (Gibco, Thermo Fisher Scientific, Waltham, MA, USA) diluted in Dulbecco’s Modified Eagle Medium/Nutrient Mixture F-12 (DMEM/F-12) + Glutamax TM (Gibco, Thermo Fisher Scientific, Waltham, MA, USA). The cell culture plates were coated with Geltrex™ and cells were grown in Essential 8 TM (E8) medium (Gibco, Thermo Fisher Scientific, Waltham, MA, USA). When cells reached 70–80% confluency they were passaged in small colonies using Ultrapure™ ethylenediaminetetraacetic acid (EDTA, 0.5 mM) (Invitrogen, Thermo Fisher Scientific, Waltham, MA, USA). Cells were maintained in a CO 2 incubator (PHCbi, Tokyo, Japan) set at 37°C and 5% CO 2.
H9-FOXL2-DDdCas9VP192 hESCs were differentiated following the gonadal differentiation protocol M described by Sepponen et al. . Briefly, 12-well plates (Corning, Corning, NY, USA) were coated with human collagen I (0.5 μg/cm 2) (Corning, Corning, NY, USA) dissolved in 2 mM HCl. Next, cells were dissociated from their culture plates into a single-cell solution using 0.5 mM EDTA and seeded at a density of 1.5 × 10 5 cells/cm 2 in 1 ml/well of day 0 medium onto the collagen-coated wells. Day 0 medium consisted of intermediate mesoderm (IM) medium (DMEM/F-12 + Glutamax™ + 2% B-27™ supplement (50X), serum-free (Gibco, Thermo Fisher Scientific, Waltham, MA, USA)) supplemented with 100 ng/ml activin A (ActA) (Q-kine, Cambridge, UK), 5 µM CHIR99021 HCl (CHIR) (Selleckchem, Cologne, Germany), 2 µM dorsomorphin 2HCl (DM) (Selleckchem, Cologne, Germany) and 10 µM rho kinase inhibitor Y27632 (ROCKi) (Selleckchem, Cologne, Germany). After 24h, the day 0 medium was replaced with 1 ml/well of day 1 medium, consisting of IM medium supplemented with 10 ng/ml bone morphogenic protein 7 (BMP7) (Peprotech, Cranbury, NJ, USA) and 3 µM CHIR. After another 24h, day 1 medium was removed, and day 2–3 medium (IM medium supplemented with 2 µM DM and 3 µM CHIR) was added to the cells at 2 ml/well for 48h. On day 4, after removing day 2–3 medium, control cells were cultured in 1.5 ml/well of IM medium and the induced cells were cultured in induction medium (IM medium supplemented with 1 µg/ml doxycycline hyclate (DOX; Sigma-Aldrich, St. Louis, MO, USA) and 1 µM trimethoprim (TMP; Sigma-Aldrich, St. Louis, MO, USA)). Cells were washed in between media changes with phosphate buffered saline (PBS, 1X) containing Magnesium and Calcium (Sigma-Aldrich, St. Louis, MO, USA). Thereafter, medium was changed daily to either IM medium or induction medium up until day 14 of differentiation.
Human embryonic kidney (HEK) 293 cells (American Type Culture Collection, Manassas, VA, USA) were cultured on tissue culture-treated dishes in HEK medium (DMEM/F-12 + Glutamax TM supplemented with 10% foetal bovine serum (FBS) (Thermo Fisher Scientific, Waltham, MA, USA)). Cells were passaged every two to four days using TRYPLE (1x; Gibco, Thermo Fisher Scientific, Waltham, MA, USA).
Generating H9-FOXL2-DDdCas9VP192 hESCs
The H9-FOXL2-DDdCas9VP192 hESCs were generated following the method described by Sepponen et al., 2022 . CRISPRa guide RNAs (gRNAs) were designed using the online gRNA design tool Benchling (San Francisco, CA, USA) or selected from a pre-existing human CRISPRa pooled library . A total of ten sequences targeting a region − 50 to -500 bp from the transcriptional start site in the FOXL2 promoter were selected. From those, five gRNAs (gRNAs 2, 5, 6, 7 and 8) with the best off-target score were chosen to assemble into FOXL2 gRNA-PCR cassettes. The gRNA cassettes consisted of a 19 bp sequence that matched the U6 promotor sequence, the 20 bp gRNA (with or without an extra guanine) and a 19 bp tailed terminator sequence. Thus, the gRNA cassette of roughly 60 bp was of the following format: 5’-TGGAAAGGACGAAACACCgNNNNNNNNNNNNNNNNNNNNgttttagagctagaaatag-3’.
The cassettes were assembled by PCR amplification as described by Balboa et al. . Next, individual gRNAs and combinations of two gRNAs were tested in HEK293 cells. For this, HEK293 cells were seeded 24h prior to transfection onto a 0.1% gelatine (Sigma-Aldrich, Burlington, MA, USA)-coated 24-well plate at a density of 9 × 10 5 cells/well in HEK medium. Cells were transfected with 500ng of the pCXLE-dCas9-VP192-T2A-EGFP-shP53 plasmid (RRID: Addgene_69535, Balboa et al., 2015 ) and 200ng of gRNA transcriptional unit using the FuGENE HD transfection reagent (Promega, Madison, WI, USA) in Opti-MEM medium (Gibco, Thermo Fisher Scientific, Waltham, MA, USA). Positive control cells were transfected with a pool of all gRNA transcriptional units and negative control cells were transfected with the GG-EBNA-TdT-g1-PGK-Puro plasmid (RRID: Addgene_102903, Weltner et al., 2018 ). Testing was performed in triplicate wells and after 72h cells were collected to determine FOXL2 gene expression levels.
The optimal gRNA combination (gRNA 2 and gRNA 8) was concatenated using Golden Gate Assembly into a GG-dest vector (RRID: Addgene_69538, Balboa et al., 2015 ). The assembly of the guide reactions, subsequent transformation of the reaction products into DH5α chemically competent bacteria (New England Biolabs, Ipswich, MA, USA), and screening of positive colonies were performed as described by Balboa et al., 2015 . Successful concatenation of the gRNA-PCR products into the vector was confirmed by Sanger sequencing (Eurofins Genomics, Köln, Germany). Next, the correct concatenated guides were subcloned into a PiggyBac (PB) plasmid using the Epstein-Barr virus nuclear plasmid GG-EBNA-TdT-guide1-PGK-Puro (Addgene_102903; Weltner et al., 2018 ). The PB backbone was obtained from the PB-GG-MIR302-7 g-PGK-Puro plasmid.
Next, the PB plasmid containing the two gRNAs targeting the FOXL2 promotor was electroporated into H9 activator cells containing SB-tight-DDdCas9VP192-GFP-Zeo-WPRE and SB-CAG-rtTA-IN-IRES-Neo with the Neon™ Transfection System (Invitrogen, Thermo Fisher Scientific, Waltham, MA, USA) following the protocol from the Neon transfection system 100 µl kit (Invitrogen, Thermo Fisher Scientific, Waltham, MA, USA). Cells were incubated with 10 µM ROCKi for four hours before transfection. Next, cells were dissociated into a single-cell suspension using TRYPLE (1x). 1 × 10 6 cells were counted and resuspended in 100 µl R-buffer. The cell suspension was then mixed with the plasmid mix containing 1000 ng PB-plasmid containing the guides and 500 ng of the PB-transposase pCMV-Hahg-PBase. Subsequently, the transfection was performed with electroporation settings set at 1100 V, 20ms, 2 pulses. Post-electroporation, cells were plated onto a Geltrex™ LDEV-Free Reduced Growth Factor Basement Membrane Matrix-coated 10 cm dish in E8 medium containing 10 µM ROCKi. Cells were allowed to recover for three days with medium changes every other day. After this, cells were selected using 0.5 µg/ml Puromycin (Gibco, Thermo Fisher Scientific, Waltham, MA, USA) for 24h. All plasmids and the H9 activator cells were kindly provided by the Biomedicum Stem Cell Centre.
gRNA sequences and Golden Gate concatenation sequences are listed in Tables1 and 2, respectively.
Table 1.
gRNA sequences
| Guide | Sequence 5’ ◊ 3’ |
:--- |
| gRNA 2 | GTGGAAAGGACGAAACACCGGAGATGAACTCGCCCGTGCGGTTTTAGAGCTAGAAATAG |
| gRNA 8 | GTGGAAAGGACGAAACACCGGGGCGCGTGAGCCTGGCTGTGTTTTAGAGCTAGAAATAG |
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Table 2.
Golden gate concatenation sequences
| Primer | Sequence 5’ ◊ 3’ | Compatibility |
:---
| 1 aggc Fw | ACTGAATTCGGATCCTCGAGCGTCTCACCCTGTAAAACGACGGCCAGT | GG-dest |
| 1 aggc Rv | CATGCGGCCGCGTCGACAGATCTCGTCTCACATGAGGAAACAGCTATGACCATG | 2 aggc Fw |
| 2 aggc Fw | ACTGAATTCGGATCCTCGAGCGTCTCACATGGTAAAACGACGGCCAGT | 1 aggc Rv |
| 5 aggc Rv | CATGCGGCCGCGTCGACAGATCTCGTCTCACGTTAGGAAACAGCTATGACCATG | GG-dest |
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Generating H9-FOXL2-DDdCas9VP192 clonal lines
H9-FOXL2-DDdCas9VP192 hESCs were treated with 10 µM ROCKi in E8 medium for four hours prior to dissociation. The cells were then dissociated into a single-cell suspension using TRYPLE (1x) dissociation agent and resuspended in a buffer composed of 10% FBS in PBS and centrifuged at 200 rcf for three minutes. The resulting pellet was resuspended in FACS buffer consisting of Hank’s Balanced Salt Solution (Gibco, Thermo Fisher Scientific, Waltham, MA, USA), 1 mM Ultrapure™ EDTA, 25 mM HEPES (Lonza, Basel, Switzerland), 10% FBS and 10 µM ROCKi. The cell suspension was passed through a 40 µM cell strainer (Falcon, Thermo Fisher Scientific, Waltham, MA, USA) and counted with the Countess II Automated Cell Counter (Thermo Fisher Scientific, Waltham, MA, USA). The cells were kept on ice until single-cell sorting. Cell sorting was performed using the SH800Z Cell Sorter (SONY, Minato City, Tokyo, Japan) into a 96-well plate containing room temperature (RT) E8 medium supplemented with 5 µM ROCKi, Penicillin-Streptomycin (1x; Gibco, Thermo Fisher Scientific, Waltham, MA, USA) and CloneR™ (1:10; STEMCELL Technologies, Vancouver, Canada). After sorting, the plates were centrifuged at 70 rcf for three minutes and then carefully placed in a 37°C incubator with 5% CO 2 for 4h. After 48h, a partial media change was performed using E8 medium supplemented with 5 µM ROCKi and CloneR (1:10). Thereafter, a complete media change was performed every other day with E8 medium supplemented with 10 µM ROCKi until the colonies could be picked and expanded.
Real time quantitative polymerase chain reaction (RT-qPCR)
RNA isolation, reverse transcription and RT-qPCR procedures were conducted as previously described in Danti et al. . Primer sequences used to perform the RT-qPCRs are listed in Table3.
Table 3.
RT-qPCR primer sequences
| Gene | Forward 5’ ◊ 3’ | Reverse 3’ ◊ 5’ |
:---
| DHH | ACCTCGTGCCCAACTACAAC | CTCCTTACAACGCTCGGTCA |
| FOXL2 | TTTGTCCCCTCAGTTTATGTCC | TGA ATTTGGGCAGGAGACG |
| GATA4 | CAGGCGTTGCACAGATAGTG | CCCGACACCCCAATCTC |
| INHBA | GGACATCGGCTGGAATGACT | GGCACTCACCCTCGCAGTAG |
| LHX9 | GCGAACCTCTTTCAAGCATC | TCCTTCTGAATTTGGCTCGT |
| NR0B1 | TGCTCTTTAACCCGGACGTG | GCGTCATCCTGGTGTGTTCA |
| OSR1 | GCTGTCCACAAGACGCTACA | CCAGAGTCAGGCTTCTGGTC |
| PPIG | TCTTGTCAATGGCCAACAGAG | GCCCATCTAAATGAGGAGTTG |
| RSPO1 | GCAACCCCGACATGAACAAG | CAAGCCCTCCTTACACTTGG |
| SOX9 | GTAATCCGGGTGGTCCTTCT | GTACCCGCACTTGCACAAC |
| TAC1 | GCCTCAGCAGTTCTTTGGATTA | GAGATCTGGCCATGTCCATAAAG |
| TSPAN8 | TGGACTGGCAGTTATTGAGATAC | GGTTTGACTGACGATAGGTTGA |
| UPK3B | ATCACTCTCCACCAAGGGA | CAGAGAAGAGAGGATGGAGGTA |
| WNT4 | GATGTGCGGGAGAGAAGCAA | ATTCCACCCGCATGTGTGT |
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Briefly, RNA was isolated from cell lysates according to the Nucleospin RNA kit (Macherey-Nagel, Düren, Nordrhein-Westfalen, Germany) protocol, excluding the genomic DNA removal steps. The genomic DNA was removed separately using RQ1 RNAse-free DNAse (Promega, Madison, WI, USA), followed by RNA purification using the Nucleospin RNA Clean-Up kit (Macherey-Nagel, Düren, Nordrhein-Westfalen, Germany). Next, RNA was reverse transcribed using moloney murine leukemia virus (M-MLV) Reverse Transcriptase (Promega, Madison, WI, USA), random hexamer primer, oligo(dt)18 primer, RiboLock RNAse Inhibitor, and a mixture of four deoxynucleotide triphosphates (all from Thermo Fisher Scientific, Waltham, MA, USA). For the RT-qPCR reaction, cDNA was combined with HOT FIREPol EvaGreen qPCR Mix Plus (Solis Biodyne, Tartu, Estonia) and 2 µM reverse primers (Metabion, Planegg/Steinkirchen, Germany). Relative messenger RNA (mRNA) expression levels were analysed using the LightCycler96 system (Roche Diagnostics, Mannheim, Germany). Next, gene expression was quantified using the ΔΔCt method and normalised using peptidylprolyl isomerase G (PPIG) as the endogenous control. Lastly, expression levels were presented relative to those in undifferentiated cells.
Immunofluorescence staining
At day 14 of differentiation, cells differentiated on 4-well µ-slides were washed once with RT PBS + Mg 2++Ca 2+ and thereafter fixed with 4% paraformaldehyde for 15 minutes at RT. Post-fixation, the cells were washed three times with RT PBS (Medicago, Uppsala, Sweden). Next, fixed cells were permeabilised with 0.5% Triton® X-100 (Fisher Scientific, Thermo Fisher Scientific, Waltham, MA, USA) in PBS and washed three times with 0.1% Tween® (Fisher Scientific, Thermo Fisher Scientific, Waltham, MA, USA) in PBS and blocked using the UltraVision Protein Block (Fisher Scientific, Thermo Fisher Scientific, Waltham, MA, USA) at RT for 10 minutes. Subsequently, the fixed cells were incubated overnight at 4°C with primary antibodies in 0.1% PBS-Tween: goat polyclonal anti-FOXL2 (Novus Biologicals, Cat# NB100-1277, RRID: AB_2106187, 1:250), mouse monoclonal anti-GATA4 (Santa Cruz Biotechnology, Cat# sc-25310, RRID: AB_627667, 1:200), mouse monoclonal anti-DHH (Santa Cruz Biotechnology, Cat# sc-271168, RRID: AB_10608075, 1:200). After primary antibody incubation, the cells were washed three times with 0.1% PBS-Tween and incubated with the following secondary antibodies at a dilution of 1:1000 in 0.1% PBS-Tween: Alexa Fluor® 488 donkey anti-goat IgG (Thermo Fisher Scientific, Cat# A-11055, RRID: AB_2534102) or Alexa Fluor® 594 donkey anti-mouse (Thermo Fisher Scientific, Cat# A-21203, RRID: AB_2535789). The incubation was carried out at RT for 45 minutes in the dark. The cells were washed twice with 0.1% PBS-Tween. Nuclei were subsequently stained with 4’,6-diamidino-2-phenylindole (DAPI) dilactate (Invitrogen, Thermo Fisher Scientific, Cat# D3571, RRID: AB_2307445) at a 1:1000 ratio in 0.1% PBS-Tween. Incubation with DAPI was carried out for 10 min in the dark and afterwards cells were washed twice with 0.1% PBS-Tween.
Confocal images were captured using a TCS SP8 confocal microscope with a white laser (Leica Microsystems, Mannheim, Germany) at an 812 × 812 format with an HC PL APO CS2 40x/1.30 oil objective. Images were processed using Fiji version 2.3.0 (
Bulk RNA sequencing library preparation, sequencing, and data analysis
Cells cultured in triplicate in 12-well culture plates were lysed with RA1 lysis buffer. Total RNA (tRNA) was isolated from the samples using the Nucleospin RNA kit with an additional DNAse I treatment to remove genomic DNA. The RNA was subsequently purified using the Nucleospin RNA clean-up kit.
Determination of RNA quality, library preparation and sequencing were conducted by Novogene Europe in Cambridge, England. RNA quality control was performed using the Agilent 2100 bioanalyzer (Agilent, Santa Clara, CA, USA) to assess RNA integrity and purity. Following quality control, mRNA was isolated from tRNA employing poly-T oligo-attached magnetic beads (ABclonal, Düsseldorf, Germany) for mRNA library preparation. The mRNA was then fragmented and thereafter cDNA was synthesised through reverse transcription using random hexamer primers for the first strand and dUTP/dTTP for the second cDNA strand. The non-directional library was finalised after end-repair A-tailing, adapter ligation, size selection, amplification, and purification. The mRNA library was subsequently quantified using Qubit (Invitrogen, Thermo Fisher Scientific, Waltham, MA, USA) and RT-PCR. Bioanalyzer was used to check the size distribution detection. The quantified libraries were sequenced on the Illumina sequencing platform NovaSeq X Plus series with 9G raw data per sample. The read length for the paired-end run was 150 bp.
Data pre-processing was performed by Novogene. Raw reads in fastq format were processed using the fastp software (version 0.23.2) to obtain clean reads by removing reads containing adapter, poly-N and low-quality reads from the raw data. Simultaneously, the Q20, Q30 and GC content of the clean data were calculated. All the further downstream analyses were performed on high-quality clean data.
Next, the reads were mapped to the reference genome, directly obtained from the genome website, using Hisat2 (v2.0.5). Gene expression levels were quantified using featureCounts (v1.5.0-p3) and expressed as Fragments Per Kilobase of transcript sequence per Millions base pairs sequenced (FPKM). Differential expression (DE) analysis of the two conditions or groups (three technical replicates per condition/group) was performed using the DESeq2Rpackage (1.20.0). The resulting P-values were adjusted for false discovery rate using the Benjamini and Hochberg’s approach. Genes with an adjusted P-value < 0.05 found by DESeq2 were assigned differentially expressed.
The gene ontology (GO) enrichment analysis of differentially expressed genes (DEGs) was performed using the clusterProfiler R package, correcting for gene length bias. GO terms with P-value < 0.05 were considered significantly enriched. The same package was used for Kyoto encyclopaedia of genes and genomes (KEGG) pathway enrichment analysis of DEGs. Again, terms of which the P-value < 0.05 were considered significantly enriched.
Statistics
Statical analysis for RT-qPCR was performed using the software Graphpad Prism 9 version 9.2.0 (La Jolla, CA, USA). Two-way ANOVA was performed to determine the statistical significance between two or more groups/conditions. The Sidak’s post-hoc test was used as the correction method for multiple comparisons. Statistical significance was attributed if the P-values were less than 0.05. Data are shown as mean ± SEM. The statistical analysis employed for the bulk RNA-seq data is explained above in the section ‘bulk RNA sequencing library preparation, sequencing, and data analysis’.
Results
FOXL2 represses gonadal markers in early somatic cell gonadal differentiation
To investigate the function of FOXL2 in early somatic cell ovarian development, a dual inducible FOXL2 activation line - H9-FOXL2-DDdCas9VP192 - was established using CRISPR/Cas9-mediated genome activation. The cells were subjected to our previously established gonadal differentiation protocol . To mimic the in vivo situation as closely as possible, we first optimised the day of FOXL2 induction (Suppl. Figure1A). As our differentiation protocol steers the cells from human pluripotent stem cells (hPSCs) to intermediate mesoderm (IM) during the first four days of differentiation, we selected an early and a late time point: day 4 and day 8, respectively and followed the differentiation up to day 14 (Suppl. Figure1A).
We first studied the effect of FOXL2 induction at day 4 during the gonadal differentiation process (Fig.1A). RT-qPCR analysis showed that upon induction, the expression levels of FOXL2 were upregulated by an average of 200-fold. Subsequent RT-qPCR analyses revealed that FOXL2 induction initiated at day 4 resulted in a significant downregulation of the gonadal marker LIM-homeobox 9 (LHX9) and in a slight reduction in the gene expression levels of several early gonadal markers. These markers included the coelomic epithelial markers GATA-binding protein 4 (GATA4), the female gonadal markers RSPO1 and WNT4, the male gonadal markers SRY-box transcription factor 9 (SOX9), desert hedgehog (DHH) and nuclear receptor subfamily 0 group B member 1 (NR0B1), as well as the gonadal marker inhibin subunit beta a (INHBA) (Fig. 1B). Immunofluorescence (IF) co-staining of FOXL2 and GATA4 on day 14 of gonadal differentiation showed the absence of FOXL2 in the control cultures, whereas most of the cells were GATA4 positive. In contrast, in the induced state, FOXL2 protein was clearly detectable and GATA4 expression at the protein level was decreased. Notably, even though not all cells expressed FOXL2, possibly due to gene silencing, the difference in GATA4 expression between the control and induced states was more pronounced at the protein level than at the RT-qPCR level. Additionally, IF staining of DHH showed a similar decreased expression pattern at the protein level as seen with GATA4 (Fig.1C).
Fig. 1.
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FOXL2 is a repressive factor when induced at day 4 of gonadal differentiation. (A) A schematic representation of the 14-day gonadal differentiation protocol with the added growth factors, inhibitors and small molecules used to steer female hESCs towards ESGCs and the respective developmental stages throughout differentiation. FOXL2 is induced at day 4 through the addition of DOX and TMP for 10 days as shown by the arrow. Created in (B) RT-qPCR analysis of FOXL2 induction at day 4 of gonadal differentiation. FOXL2 was induced through the addition of DOX and TMP and downregulated gonadal markers GATA4, LHX9, RSPO1, WNT4, INHBA, SOX9, NR0B1 and DHH. The fold change is presented in comparison to d0 (undifferentiated cells) gene expression levels. Data are reported as mean ± SEM, n = 4 biological replicates. Two-way ANOVA; 0.1234 (ns), 0.0332 (), 0.0021 (), 0.0002 (), 0.0001 (). (C) Confocal images of IF staining show expression of FOXL2, GATA4 and DHH in the non-induced (-DOX -TMP) and in the induced (+ DOX + TMP) conditions at day 14 of gonadal differentiation. Images were taken with an 40X objective. Scale bars 50 μm. ActA, activin A; BMP, bone morphogenetic protein; CHIR, CHIR-99021; DM, dorsomorphin; ESGCs, early supporting gonadal cells; hESCs, human embryonic stem cells; IM, intermediate mesoderm; PS, primitive streak; d, day of differentiation; DOX, doxycycline hyclate; TMP, trimethoprim
RT-qPCR analyses indicated that the addition of doxycycline (DOX) and trimethoprim (TMP) at day 8 of gonadal differentiation elevated FOXL2 gene expression levels approximately by 300-fold on average until day 14. However, FOXL2 induction at this later time point did not appear to affect the expression levels of coelomic epithelial markers GATA4 and LHX9 at any timepoint. Additionally, the gene expression levels of female gonadal markers RSPO1, WNT4 and INHBA were only slightly downregulated following FOXL2 induction at day 8 (Suppl. Figure1B).
In summary, the earlier FOXL2 is induced during differentiation, the more significant its impact on the expression levels of gonadal markers. Therefore, we chose day 4 as the induction time point for the remainder of our study. Moreover, we can conclude that in the earlier stages of gonadal development, FOXL2 induction seems to primarily exert a repressive effect on coelomic epithelial, female, and male gonadal markers.
Transcriptional changes are induced by FOXL2 during early gonadal differentiation
To further study the role of FOXL2 during early somatic cell gonadal differentiation in more detail, we opted to perform bulk RNA sequencing (RNA-seq) analyses on the non-induced (CTRL, -DOX -TMP) and induced (IND, +DOX + TMP) cells. Samples were collected on day 4 and then every two days (day 6, 8, 10, 12, 14) in three technical replicates per timepoint. According to the principal component analysis (PCA), 33.43% of the variance between the samples could be attributed to the process of gonadal differentiation (principal component 1, PC1), while FOXL2 induction (principal component 2, PC2) accounted for 16.7% of the observed variance. Not only was there a clear divergence between the conditions, but PC1 also indicated the divergence between the different days of gonadal differentiation. Day 4, 6, 8 and 10 samples were distinct from each other, reflecting differences in developmental state. However, the day 12 and 14 samples clustered more tightly together, indicating greater similarity between these developmental stages (Suppl. Figure2).
Next, we performed pairwise differential expression (DE) comparisons to assess the immediate effect of FOXL2 on the cells, focusing on the day 6 non-induced (CTRL, -DOX -TMP) versus day 6 induced (IND, +DOX + TMP) conditions. To do this, we plotted a heatmap representing the expression profile of the top 100 upregulated DEGs with the highest Log2FoldChange at day 6 of gonadal differentiation. LINC01391, along with FOXL2 neighbour (FOXL2NB) and FOXL2, were the top three upregulated genes at this time point. Other notable genes among the top 100 upregulated DEGs include toll-like receptor 3 (TLR3) and tachykinin precursor 1 (TAC1) (Fig.2).
Fig. 2.
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Expression pattern of the top 100 upregulated differentially expressed genes (DEGs) at day 6 of differentiation upon FOXL2 induction. The heat map is showing the comparison between the control (CTRL, -DOX–TMP) and induced (IND, +DOX + TMP) conditions at 6 day of gonadal differentiation. The top 100 DEGs at day 6 of differentiation are displayed and each rectangle represents the expression of a specific gene within a technical replicate. The intensity of gene expression is indicated by a colour scale based on a log2 scale (red = lowest expression, green = highest expression). d, day
Next, we looked at the Kyoto encyclopaedia of genes and genomes (KEGG) and gene ontology (GO) pathways associated with the upregulated DEGs at day 6 of gonadal differentiation. The upregulated DEGs were associated with KEGG pathways such as phagosome, neuroactive ligand-receptor interaction, complement and coagulation cascades and cell adhesion molecules (Suppl. Figure3A). As for the GO pathways, DEGs were mostly associated with hormone regulation and interferon response regarding biological processes (BP), epithelial junction formation and plasma membranes in terms of cellular component (CC) and lastly enzyme activity and hormone binding as molecular functions (MF). We further examined the top up-and downregulated DEGs at each time point to determine the identity of the cells after FOXL2 induction. The upregulated DEGs were mostly associated with neuroactive ligand-receptor interactions and cell adhesion molecules KEGG pathways as well as similar GO pathways as seen at day 6 (Suppl. Figure3B).
Overall, the upregulated DEGs are primarily linked with KEGG pathways involving cell adhesion molecules and GO pathways relating to extracellular matrix and junction formation at different time points during gonadal differentiation.
FOXL2 supports the transition from coelomic epithelium to early supporting gonadal cells
Next, we turned to the literature and compared our data to data from two relevant publications focusing on in vivo early gonadal development: the study of Garcia-Alonso et al., published in Nature in 2022 ; and the study by Wamaitha et al., published in Developmental Cell in 2023 . Both studies conducted single-cell RNA-seq analysis of early human gonadal tissue (approximately 6–21 post-conception week (PCW)). The Garcia-Alonso study mapped the trajectory of somatic cell gonadal differentiation all the way from coelomic epithelium to pre-granulosa cells or Sertoli cells, identifying specific maker genes for each differentiation stage. Similarly, the Wamaitha et al. study detailed subsequent stages of early somatic cell gonadal differentiation, with marker genes that further corroborated the findings of Garcia-Alonso et al. (Fig.3A).
Fig. 3.
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FOXL2 drives the cells out of a coelomic epithelial fate into a transitional stage and finally to an early supporting gonadal cell fate. (A) Schematic representation of early gonadal development as described by Garcia-Alonso et al. as well as the matching developmental stages and their associated marker genes according to both Garcia-Alonso et al. and Wamaitha et al. Created in (B and C) Line graphs of marker genes associated with the coelomic epithelial, transitional stage and ESGCs. Gene expression change is shown for three conditions: the effect of gonadal differentiation (DIFF, blue), the effect of FOXL2 induction (IND, red) and the combined effect (DIFF + IND, purple). Graphs show a downregulation of GATA4, LHX9, UPK3B from day 6 of gonadal differentiation, an upregulation of ARX and GATA2 at day 8 and downregulation again at day 10. ESGC markers LGR5, TSPAN8, OSR1 and TAC1 are upregulated upon FOXL2 induction at day 10. N = 1 biological replicate with 3 technical replicates. CE, coelomic epithelium; ESGCs, early supporting gonadal cells; GC, granulosa cell; TS, transitional stage
To compare our RNA-seq data with those from the two previous studies, we performed pairwise DE comparisons. The data presented in the graphs are based on the mean of three technical replicates of one biological replicate. First, we studied the effect of gonadal differentiation (DIFF) only on the cells by comparing the day 4 samples with the non-induced (CTRL, -DOX -TMP) samples at each time point to assess the progression and identity of the differentiating cells without FOXL2 activation. We observed that without FOXL2, the coelomic epithelial marker genes GATA4 and uroplakin 3b (UPK3B) were upregulated at day 6, with LHX9 appearing just after day 6, and aristaless related homeobox (ARX) and GATA binding protein 2 (GATA2) at day 10. From day 10 onwards, all coelomic epithelial markers stayed upregulated. Secondly, we analysed the effect of FOXL2 induction (IND) only on the cells by comparing the non-induced (CTRL, -DOX -TMP) and induced (IND, +DOX + TMP) samples at each specific time point during differentiation. The aim was to examine the sole effect of FOXL2 activation to determine its role in early somatic cell gonadal differentiation and determine cell identity at each time point during differentiation. At day 6, only FOXL2 was differentially expressed and upregulated due to the activation. However, by day 8, three important coelomic epithelial markers-GATA4, UPK3B and LHX9-were downregulated, whilst ARX, GATA2 and FOXL2 were upregulated. At day 10, the coelomic epithelial genes remained downregulated and FOXL2 upregulated. Additionally, the early supporting gonadal cell (ESGC) markers leucine rich repeat containing G protein-coupled receptor 5 (LGR5), tetraspanin 8 (TSPAN8), odd-skipped related transcription factor 1 (OSR1) and TAC1 were upregulated for the first time alongside FOXL2. ARX was no longer differentially expressed at day 10. From day 10 onwards, OSR1 remained upregulated alongside FOXL2. LHX9 was downregulated at day 12 whilst ARX was upregulated again. At day 14 LHX9 was still downregulated and ARX upregulated. Lastly, we examined the combined effect of gonadal differentiation and FOXL2 induction (DIFF + IND) on the cells by comparing the day 4 non-induced (CTRL, -DOX -TMP) samples with the induced (IND, +DOX + TMP) samples at each specific time point. The DEGs in this comparison shared similarities with both the differentiation only (DIFF) and FOXL2 induction only (IND) comparisons. However, they predominantly resembled the sole effect of gonadal differentiation (DIFF). The most notable difference from the differentiation only (DIFF) comparison was the upregulation of both FOXL2 and OSR1. The main difference from the induction only (IND) comparison was the consistent downregulation of LGR5 across all time points (Fig.3B and C).
Next, we validated our bulk RNA-seq data through RT-qPCR analysis of key marker genes from each developmental stage. The validation showed that UPK3B was downregulated and TSPAN8, OSR1, and TAC1 gene expression levels were upregulated upon FOXL2 induction. Generally, the genes behaved similarly as in our bulk RNA-seq data, further supporting the results (Suppl. Figure4).
Thus, FOXL2 induction appears to guide cells from a coelomic epithelial identity towards the ESGC stage and potentially towards pre-granulosa cells, making FOXL2 not only a marker but also a potential driver of the formation of ESGCs.
Discussion
In this study, we demonstrate that FOXL2 is not just a marker of ESGCs but plays a role in the transition of cells from a coelomic epithelial state to ESGCs. Utilising CRISPR/Cas9-mediated genome activation of FOXL2, we observed the downregulation of several established coelomic epithelial markers including GATA4, LHX9 and UPK3B starting around day 6 of gonadal differentiation. The downregulation of these markers signified the cells’ exit from the coelomic epithelial fate by approximately day 8 of gonadal differentiation, at which point transition markers such as ARX and GATA2 became upregulated. These markers became subsequently decreased after day 8, followed by the upregulation of ESGC markers, including LGR5, TSPAN8, TAC1 and OSR1, by day 10.
Our bulk RNA-seq data supported our initial findings that FOXL2 is involved in repressing several factors during early gonadal somatic cell differentiation. According to the KEGG and GO pathway analysis, the upregulated DEGs at day 6 were primarily associated with cell adhesion molecules, extracellular matrix, and junction formation. These pathways remained relevant across different time points and developmental stages throughout gonadal differentiation, indicating changes at the cell identity level of the differentiating cells. Using data from Garcia-Alonso et al. we were able to assign labels to these developmental stages. As coelomic epithelial markers were still differentially expressed and upregulated at day 6 of gonadal differentiation, we hypothesized that the cells at this stage were coelomic epithelial-like. However, shortly thereafter, FOXL2 began downregulating the coelomic epithelial markers GATA4, LHX9, and UPK3B. By day 8 of gonadal differentiation markers for the genital ridge stage, such as ARX and GATA2, were upregulated. This signifies a shift in cell identity from the coelomic epithelial to the transitional stage, as observed in the data from Garcia-Alonso et al. and Wamaitha et al. [21, 22]. Subsequently, we observed downregulation of ARX and GATA2 gene expression, while markers for ESGCs, including LGR5, TSPAN8, OSR1 and TAC1, were upregulated, indicating another shift in cell identity from the transitional stage to ESGC-like stage by day 10 of gonadal differentiation. After day 10, OSR1 remained upregulated for the remainder of the gonadal differentiation process. The OSR1 cell population represents precursors of the pre-GCI population during the first wave of granulosa cell formation . Based on this information, we conclude that cells from day 4 until day 10 of gonadal differentiation correspond to the in vivo gonadal developmental period spanning 6–8 PCW. Following day 10, as OSR1 remained upregulated, we hypothesise that our gonadal differentiation protocol combined with FOXL2 induction could potentially steer the cells to a pre-GCI population. Additionally, Taelman et al. conducted a study using single-cell transcriptomics to characterise the foetal female and male gonads. In this study, the researchers identified several clusters of cell populations and assigned marker genes to each cell population. Clusters 5 and 12 were annotated as being coelomic/ovarian surface epithelium, while clusters 4 and 9 were annotated as pre-granulosa cells. Our cells at around days 4–6 expressed DEGs corresponding to the coelomic/ovarian surface epithelium cell population by Taelman et al., whereas our day 10–14 cells had a resemblance to the pre-granulosa cell population .
FOXL2 induction appeared to actively repress male gonadal differentiation by downregulating male gonadal markers SOX9, NR0B1 and DHH. Although the reduction in the expression levels of these male gonadal markers was only moderate, their baseline levels were already quite low, which is expected given that we are working with a female hESC line. Interestingly, FOXL2 induction also seemed to downregulate the female gonadal markers RSPO1 and WNT4. This may seem counterintuitive, as the expression of the pro-ovarian factors RSPO1 and WNT4 would be expected to increase during granulosa cell differentiation. However, in humans, a second wave of granulosa cells (preGC-IIa/b) emerges after 8 PCW, during which RSPO1 and WNT4 are actively downregulated . Whether FOXL2 is a key player in that second wave of granulosa cell formation remains to be elucidated. Moreover, the study by Migale et al. showed that FOXL2 binds and downregulates the expression levels of RSPO1 and WNT4 in mice .
Each developmental stage is represented by specific marker genes. In our study, some of the known and expected markers were not highly up- or downregulated, nor were they found among the top 100 DEGs at their respective time points. This is likely due to the presence of a heterogeneous cell population at each stage. The FOXL2/GATA4 IF staining on day 14 of differentiation showcases this heterogeneity as there are cells co-expressing FOXL2 and GATA4 whilst other cells solely express only one of the proteins. This heterogeneity could be attributed to some cells losing their FOXL2 expression during the gonadal differentiation process. However, the cells that do not express FOXL2 are still being subjected to our gonadal differentiation protocol, which efficiently upregulates bipotential markers such as GATA4. This might explain the presence of cells solely expressing GATA4 protein. Cells that retain FOXL2 expression will solely express FOXL2 or co-express FOXL2 and GATA4. Although the reason behind why some cells solely express FOXL2 is unclear, there are several possible explanations: FOXL2 might be repressing GATA4 expression in these cells, or the FOXL2-positive cells might belong to the percentage of cells that do not upregulate GATA4 during the differentiation. Alternatively, these FOXL2-positive cells may have differentiated into another cell type than the double positive cells. Single-cell RNA sequencing would allow us to determine which specific subpopulations are present at each time point during gonadal differentiation and their relative proportions. Another method that could be employed to characterise the different cell populations would be a flow sorting-based strategy where FOXL2/GATA4-expressing cells could be isolated, and the gene expression levels of these cell populations could be compared. It is important to note that heterogeneity is not necessarily a negative phenomenon. In vivo, multiple distinct somatic cell populations exist at each developmental stage. For instance, the study by Guo et al. showed highly heterogeneous gene expression patterns within somatic cell populations using single-cell RNA-seq analysis .
Among the expected granulosa cell precursor population markers, wnt family member 6 (WNT6), should be upregulated at the transitional stage . However, in our list of DEGs, WNT6 was found to be downregulated across all time points (data not shown). In mice, ovarian pre-granulosa cells are thought to originate from either bipotential precursor cells or surface epithelium cells. If they arise from the bipotential cells, these precursors are marked by the expression of WNT6, WNT4 and FOXL2. If they on the other hand arise from the epithelial cells, the markers UPK3B, LGR5 and keratin 19 (KRT19) tend to be highly expressed . A similar situation may be present in humans. Our precursors most likely originate from coelomic epithelial-like cells, which would explain the downregulation of WNT4 and WNT6, alongside the high expression of UPK3B, and LGR5 that we observe later in our granulosa cell precursors.
FOXL2 has known and putative direct targets, including follistatin (FST), cytochrome P450 family 19 subfamily A member 1 (CYP19A1), steroidogenic acute regulatory protein (STAR), cytochrome P450 family 17 subfamily A member 1 (CYP17A1) [27–30]. However, RT-qPCR results show no upregulation of these target genes after FOXL2 induction (data not shown). We reason that FOXL2 presumably has different target genes depending on the developmental stage. As we are focusing on very early gonadal development (6–8 PCW), these target genes are simply not yet expressed at the stage of pre-granulosa cell development but are expected to become active later in granulosa cell development, maturation, and the initiation of steroidogenesis. This reasoning is supported by the study by Migale et al. where the researchers looked at FOXL2 binding partners at different timepoints during murine ovarian development. The study showed that FOXL2 binds different target genes depending on the developmental stage and that it regulates more targets postnatally .
The principal strength of this paper is that it is the first study to demonstrate that FOXL2 is not merely a marker of ESGCs but plays a potential role in their formation. Moreover, by combining growth factors, inhibitors, small molecules and CRISPR/Cas9 activation, we successfully replicated in vitro a very early in vivo stage of gonadal development (6–8 PCW) of pre-granulosa cell formation, encompassing all the key developmental stages. This model provides a valuable tool for studying the underlying pathological mechanisms of human conditions caused by FOXL2 mutations, such as BPES and DSD. One limitation is the heterogeneity of the cells at each developmental stage despite the use of small molecules, growth factors, inhibitors, and CRISPR/Cas9 activation to guide differentiation in the desired direction. Another limitation is that the results are based solely on in vitro experiments. For future directions, it would be interesting to identify growth factors and/or small molecules that could replace the use of CRISPR/Cas9 to endogenously upregulate FOXL2. This could be particularly beneficial for applying this differentiation protocol in a clinical setting.
Conclusions
In conclusion, our study indicates that during gonadal differentiation of a female hPSC line, FOXL2 is not only a marker of ovarian development but seems to play an active role in the transition from coelomic-epithelial cells to ESGCs in early ovarian differentiation. Furthermore, we have confirmed that even during these early stages, FOXL2 actively represses the expression of genes that are known to drive male sex differentiation. The findings in this study could aid in the elucidation of the mechanisms underlying human foetal ovarian development.
Electronic Supplementary Material
Below is the link to the electronic supplementary material.
12958_2025_1377_MOESM1_ESM.jpg (312KB, jpg)
Supplementary Figure 1: FOXL2 induction optimisation showing that day 8 is a sub-optimal induction day. (A) A schematic representation of the 14-day gonadal differentiation protocol including the different small molecules, growth factors, inhibitors used to steer female hESCs towards the IM stage and with the matching developmental stages. Arrows showing the start and end of the FOXL2 induction through addition of the antibiotics DOX and TMP. Created in (B) RT-qPCR analysis of FOXL2 induction at day 8 of gonadal differentiation. FOXL2 was upregulated through the addition of DOX and TMP and minimally downregulated the gonadal markers GATA4, LHX9, RSPO1, WNT4 and INHBA. Data are reported as mean ± SEM, n = 3 biological replicates. The fold change is presented in comparison to d0 (undifferentiated cells) gene expression levels. Two-way ANOVA; 0.1234 (ns), 0.0332 (), 0.0021 (), 0.0002 (), 0.0001 (). ActA, activin A; BMP, bone morphogenetic protein; CHIR, CHIR-99021; DM, dorsomorphin; hESCs, human embryonic stem cells; IM, intermediate mesoderm; PS, primitive streak; d, day of differentiation; DOX, doxycycline hyclate; TMP, trimethoprim
12958_2025_1377_MOESM2_ESM.jpg (71.4KB, jpg)
Supplementary Figure 2: Bulk RNA-sequencing showed differences between the control (-DOX-TMP) and induced (+ DOX + TMP) conditions and between the different time points. Principal component analysis showing the divergence between samples from the different timepoints of gonadal differentiation and between control (CTRL) and induced (IND) conditions. Principal component 1 shows the effect of gonadal differentiation and principal component 2 shows the effect of FOXL2 induction. d; day of differentiation; PC, principal component
12958_2025_1377_MOESM3_ESM.jpg (1.8MB, jpg)
Supplementary Figure 3: FOXL2 induction upregulates pathways associated with cell adhesion, extracellular matrix, and junctions. (A) Bar charts showing Kyoto encyclopaedia of genes and genomes (KEGG) pathways associated with upregulated differentially expressed genes (DEGs) at different time points during the gonadal differentiation. Numbers above bars signify the number of genes associated with the specific pathway. (B) Bar charts showing gene ontology (GO) pathways associated with the upregulated DEGs at different time points during the differentiation. Numbers above the bars signify the number of genes associated with the specific pathway. Bar charts are divided into three categories: biological process (BP), cellular component (CC) and molecular function (MF)
12958_2025_1377_MOESM4_ESM.jpg (133.9KB, jpg)
Supplementary Figure 4: RT-qPCR validation confirms bulk RNA-seq results. RT-qPCR analysis shows the downregulation of coelomic epithelial marker UPK3B and the upregulation of ESGC markers TSPAN8, OSR1 and TAC1 upon FOXL2 induction at day 4 of gonadal differentiation. The fold change is presented in comparison to d0 (undifferentiated cells) gene expression levels. Data are reported as mean ± SEM, n = 4 biological replicates. Two-way ANOVA; 0.1234 (ns), 0.0332 (), 0.0021 (), 0.0002 (), 0.0001 ()
Acknowledgements
We would like to acknowledge and kindly thank prof. Timo Otonkoski and the Biomedicum Stem Cell Centre for providing the materials necessary for generating the activation line, the Biomedicum Flow Cytometry Unit at the University of Helsinki, supported by HilIFE, where the single-cell sorting was performed to create the clonal lines, Novogene where library preparation, bulk RNA sequencing and data analysis were conducted, and lastly the Biomedicum Imaging Unit (BIU) at the University of Helsinki, supported by HiLIFE, which kindly provided equipment for confocal images. We would also like to acknowledge and thank the Helsinki University Library for funding the open access.
Abbreviations
ActA
Activin A
ARX
Aristaless related homeobox
BMP7
Bone morphogenic protein 7
BP
Biological process
BPES
Blepharophimosis, ptosis epicanthus and inversus syndrome
Cas9
CRISPR associated protein 9
CC
Cellular component
CRISPR
Clustered regularly interspaced short palindromic repeats
CTRL
Control
CYP17A1
Cytochrome P450 family 17 subfamily A member 1
CYP19A1
Cytochrome P450 family 19 subfamily A member 1
DAPI
4’,6-diamidino-2-phenylindole
DE
Differential expression/differentially expressed
DEG
Differentially expressed genes
DHH
Desert hedgehog
DM
Dorsomorphin
DOX
Doxycycline hyclate
DSD
Differences of sex development
E8
Essential 8
EDTA
Ethylenediaminetetraacetic acid
ESGC
Early supporting gonadal cell
FBS
Foetal bovine serum
FOXL2
Forkhead box L2
FOXL2NB
FOXL2 neighbour
FST
Follistatin
GATA2
GATA binding protein 2
GATA4
GATA binding protein 4
GG
Golden gate
GO
Gene ontology
gRNA
Guide RNA
hESC
Human embryonic stem cell
HEK
Human embryonic kidney
hiPSC
Human induced pluripotent stem cell
hPSC
Human pluripotent stem cell
IF
Immunofluorescence
IM
Intermediate mesoderm
IND
Induced
INHBA
Inhibin subunit beta a
KEGG
Kyoto encyclopaedia of genes and genomes
KO
Knock-out
KRT19
Keratin 19
LHX9
LIM homeobox 9
LGR5
Leucine rich repeat containing G protein-coupled receptor 5
MF
Molecular function
mRNA
Messenger RNA
NR0B1
Nuclear receptor subfamily 0 group B member 1
OSR1
Odd-skipped related transcription factor 1
PB
Piggyback
PBS
Phosphate-buffered saline
PC
Principal component
PCA
Principal component analysis
PCW
Post conception week
PIS
Polled intersex syndrome
PPI
Protein-protein interaction
PPIG
Peptidylprolyl isomerase G
Pre-GC
Pre-granulosa cell
PS
Primitive streak
RT
Room temperature
RT-qPCR
Real time quantitative polymerase chain reaction
ROCKi
Rho-kinase inhibitor
RSPO1
R-spondin 1
SOX9
SRY-box transcription factor 9
STAR
Steroidogenic acute regulatory protein
TAC1
Tachykinin precursor 1
TMP
Trimethoprim
tRNA
Total RNA
TSPAN8
Tetraspanin 8
UPK3B
Uroplakin 3B
WNT4
Wnt family member 4
WNT6
Wnt family member 6
Author contributions
L.D. performed all the experiments in the study, was the principal author of the manuscript, conducted the data analysis aside from the bioinformatics and participated in critical discussions. P.N-R. generated the activation line used in this study and contributed to drafting and revising the manuscript, as well as critical discussions. K.L., T.T. and J.S.T. contributed to the study conception and design, manuscript drafting and revising, and critical discussions.
Funding
Open Access funding provided by University of Helsinki (including Helsinki University Central Hospital). This project received funding from the Sigrid Jusélius Foundation, the Helsinki University Hospital funds, and the European Union’s Horizon 2020 research and innovation programme under Marie Sklodowska-Curie grant agreement no. 813707.
Data availability
The datasets generated and/or analysed during the current study are available in the NCBI Gene Expression Omnibus (GEO) repository and can be accessed through the GEO series accession number GSE282662. All material unique to this study is available from the corresponding author on reasonable request.
Declarations
Ethics approval and consent to participate
Not applicable.
Consent for publication
Not applicable.
Competing interests
The authors declare no competing interests.
Footnotes
Publisher’s note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Timo Tuuri and Juha S. Tapanainen shared last authors.
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Associated Data
This section collects any data citations, data availability statements, or supplementary materials included in this article.
Supplementary Materials
12958_2025_1377_MOESM1_ESM.jpg (312KB, jpg)
Supplementary Figure 1: FOXL2 induction optimisation showing that day 8 is a sub-optimal induction day. (A) A schematic representation of the 14-day gonadal differentiation protocol including the different small molecules, growth factors, inhibitors used to steer female hESCs towards the IM stage and with the matching developmental stages. Arrows showing the start and end of the FOXL2 induction through addition of the antibiotics DOX and TMP. Created in (B) RT-qPCR analysis of FOXL2 induction at day 8 of gonadal differentiation. FOXL2 was upregulated through the addition of DOX and TMP and minimally downregulated the gonadal markers GATA4, LHX9, RSPO1, WNT4 and INHBA. Data are reported as mean ± SEM, n = 3 biological replicates. The fold change is presented in comparison to d0 (undifferentiated cells) gene expression levels. Two-way ANOVA; 0.1234 (ns), 0.0332 (), 0.0021 (), 0.0002 (), 0.0001 (). ActA, activin A; BMP, bone morphogenetic protein; CHIR, CHIR-99021; DM, dorsomorphin; hESCs, human embryonic stem cells; IM, intermediate mesoderm; PS, primitive streak; d, day of differentiation; DOX, doxycycline hyclate; TMP, trimethoprim
12958_2025_1377_MOESM2_ESM.jpg (71.4KB, jpg)
Supplementary Figure 2: Bulk RNA-sequencing showed differences between the control (-DOX-TMP) and induced (+ DOX + TMP) conditions and between the different time points. Principal component analysis showing the divergence between samples from the different timepoints of gonadal differentiation and between control (CTRL) and induced (IND) conditions. Principal component 1 shows the effect of gonadal differentiation and principal component 2 shows the effect of FOXL2 induction. d; day of differentiation; PC, principal component
12958_2025_1377_MOESM3_ESM.jpg (1.8MB, jpg)
Supplementary Figure 3: FOXL2 induction upregulates pathways associated with cell adhesion, extracellular matrix, and junctions. (A) Bar charts showing Kyoto encyclopaedia of genes and genomes (KEGG) pathways associated with upregulated differentially expressed genes (DEGs) at different time points during the gonadal differentiation. Numbers above bars signify the number of genes associated with the specific pathway. (B) Bar charts showing gene ontology (GO) pathways associated with the upregulated DEGs at different time points during the differentiation. Numbers above the bars signify the number of genes associated with the specific pathway. Bar charts are divided into three categories: biological process (BP), cellular component (CC) and molecular function (MF)
12958_2025_1377_MOESM4_ESM.jpg (133.9KB, jpg)
Supplementary Figure 4: RT-qPCR validation confirms bulk RNA-seq results. RT-qPCR analysis shows the downregulation of coelomic epithelial marker UPK3B and the upregulation of ESGC markers TSPAN8, OSR1 and TAC1 upon FOXL2 induction at day 4 of gonadal differentiation. The fold change is presented in comparison to d0 (undifferentiated cells) gene expression levels. Data are reported as mean ± SEM, n = 4 biological replicates. Two-way ANOVA; 0.1234 (ns), 0.0332 (), 0.0021 (), 0.0002 (), 0.0001 ()
Data Availability Statement
The datasets generated and/or analysed during the current study are available in the NCBI Gene Expression Omnibus (GEO) repository and can be accessed through the GEO series accession number GSE282662. All material unique to this study is available from the corresponding author on reasonable request.
Articles from Reproductive Biology and Endocrinology : RB&E are provided here courtesy of BMC
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4417 | https://math.libretexts.org/Bookshelves/Linear_Algebra/Linear_Algebra_with_Applications_(Nicholson)/08%3A_Orthogonality/8.02%3A_Orthogonal_Diagonalization | Skip to main content
8.2: Orthogonal Diagonalization
Last updated
: Jul 25, 2023
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8.1E: Orthogonal Complements and Projections Exercises
8.2E: Orthogonal Diagonalization Exercises
Page ID
: 58877
W. Keith Nicholson
University of Calgary via Lyryx Learning
( \newcommand{\kernel}{\mathrm{null}\,})
Recall (Theorem [thm:016068]) that an n×nn×n matrix AA is diagonalizable if and only if it has nn linearly independent eigenvectors. Moreover, the matrix PP with these eigenvectors as columns is a diagonalizing matrix for AA, that is
P−1AP is diagonal.
P−1AP is diagonal.
As we have seen, the really nice bases of Rn are the orthogonal ones, so a natural question is: which n×n matrices have an orthogonal basis of eigenvectors? These turn out to be precisely the symmetric matrices, and this is the main result of this section.
Before proceeding, recall that an orthogonal set of vectors is called orthonormal if ‖v‖=1 for each vector v in the set, and that any orthogonal set {v1,v2,…,vk} can be “normalized”, that is converted into an orthonormal set {1‖v1‖v1,1‖v2‖v2,…,1‖vk‖vk}. In particular, if a matrix A has n orthogonal eigenvectors, they can (by normalizing) be taken to be orthonormal. The corresponding diagonalizing matrix P has orthonormal columns, and such matrices are very easy to invert.
024227 The following conditions are equivalent for an n×n matrix P.
P is invertible and P−1=PT.
The rows of P are orthonormal.
The columns of P are orthonormal.
First recall that condition (1) is equivalent to PPT=I by Corollary [cor:004612] of Theorem [thm:004553]. Let x1,x2,…,xn denote the rows of P. Then xTj is the jth column of PT, so the (i,j)-entry of PPT is xi∙xj. Thus PPT=I means that xi∙xj=0 if i≠j and xi∙xj=1 if i=j. Hence condition (1) is equivalent to (2). The proof of the equivalence of (1) and (3) is similar.
Orthogonal Matrices024256 An n×n matrix P is called an orthogonal matrixif it satisfies one (and hence all) of the conditions in Theorem [thm:024227].
024259 The rotation matrix [cosθ−sinθsinθcosθ] is orthogonal for any angle θ.
These orthogonal matrices have the virtue that they are easy to invert—simply take the transpose. But they have many other important properties as well. If T:Rn→Rn is a linear operator, we will prove (Theorem [thm:032147]) that T is distance preserving if and only if its matrix is orthogonal. In particular, the matrices of rotations and reflections about the origin in R2 and R3 are all orthogonal (see Example [exa:024259]).
It is not enough that the rows of a matrix A are merely orthogonal for A to be an orthogonal matrix. Here is an example.
024269 The matrix [211−1110−11] has orthogonal rows but the columns are not orthogonal. However, if the rows are normalized, the resulting matrix [2√61√61√6−1√31√31√30−1√21√2] is orthogonal (so the columns are now orthonormal as the reader can verify).
024275 If P and Q are orthogonal matrices, then PQ is also orthogonal, as is P−1=PT.
P and Q are invertible, so PQ is also invertible and
(PQ)−1=Q−1P−1=QTPT=(PQ)T
Hence PQ is orthogonal. Similarly,
(P−1)−1=P=(PT)T=(P−1)T
shows that P−1 is orthogonal.
Orthogonally Diagonalizable Matrices024297 An n×n matrix A is said to be orthogonally diagonalizable when an orthogonal matrix P can be found such that P−1AP=PTAP is diagonal.
This condition turns out to characterize the symmetric matrices.
Principal Axes Theorem024303 The following conditions are equivalent for an n×n matrix A.
A has an orthonormal set of n eigenvectors.
A is orthogonally diagonalizable.
A is symmetric.
(1) ⇔ (2). Given (1), let x1,x2,…,xn be orthonormal eigenvectors of A. Then P=[x1x2…xn] is orthogonal, and P−1AP is diagonal by Theorem [thm:009214]. This proves (2). Conversely, given (2) let P−1AP be diagonal where P is orthogonal. If x1,x2,…,xn are the columns of P then {x1,x2,…,xn} is an orthonormal basis of Rn that consists of eigenvectors of A by Theorem [thm:009214]. This proves (1).
(2) ⇒ (3). If PTAP=D is diagonal, where P−1=PT, then A=PDPT. But DT=D, so this gives AT=PTTDTPT=PDPT=A.
(3) ⇒ (2). If A is an n×n symmetric matrix, we proceed by induction on n. If n=1, A is already diagonal. If n>1, assume that (3) ⇒ (2) for (n−1)×(n−1) symmetric matrices. By Theorem [thm:016397] let λ1 be a (real) eigenvalue of A, and let Ax1=λ1x1, where ‖x1‖=1. Use the Gram-Schmidt algorithm to find an orthonormal basis {x1,x2,…,xn} for Rn. Let P1=[x1x2…xn], so P1 is an orthogonal matrix and PT1AP1=[λ1B0A1] in block form by Lemma [lem:016161]. But PT1AP1 is symmetric (A is), so it follows that B=0 and A1 is symmetric. Then, by induction, there exists an (n−1)×(n−1) orthogonal matrix Q such that QTA1Q=D1 is diagonal. Observe that P2=[100Q] is orthogonal, and compute:
(P1P2)TA(P1P2)=PT2(PT1AP1)P2=[100QT][λ100A1][100Q]=[λ100D1]
is diagonal. Because P1P2 is orthogonal, this proves (2).
A set of orthonormal eigenvectors of a symmetric matrix A is called a set of principal axes for A. The name comes from geometry, and this is discussed in Section [sec:8_8]. Because the eigenvalues of a (real) symmetric matrix are real, Theorem [thm:024303] is also called the real spectral theorem, and the set of distinct eigenvalues is called the spectrum of the matrix. In full generality, the spectral theorem is a similar result for matrices with complex entries (Theorem [thm:025860]).
024374 Find an orthogonal matrix P such that P−1AP is diagonal, where A=[10−1012−125].
The characteristic polynomial of A is (adding twice row 1 to row 2):
cA(x)=det[x−1010x−1−21−2x−5]=x(x−1)(x−6)
Thus the eigenvalues are λ=0, 1, and 6, and corresponding eigenvectors are
x1=[1−21]x2=x3=[−125]
respectively. Moreover, by what appears to be remarkably good luck, these eigenvectors are orthogonal. We have ‖x1‖2=6, ‖x2‖2=5, and ‖x3‖2=30, so
P=[1√6x11√5x21√30x3]=1√30[√52√6−1−2√5√62√505]
is an orthogonal matrix. Thus P−1=PT and
PTAP=
by the diagonalization algorithm.
Actually, the fact that the eigenvectors in Example [exa:024374] are orthogonal is no coincidence. Theorem [thm:016090] guarantees they are linearly independent (they correspond to distinct eigenvalues); the fact that the matrix is symmetric implies that they are orthogonal. To prove this we need the following useful fact about symmetric matrices.
024396 If A is an n×n symmetric matrix, then
(Ax)∙y=x∙(Ay)
for all columns x and y in Rn.
Recall that x∙y=xTy for all columns x and y. Because AT=A, we get
(Ax)∙y=(Ax)Ty=xTATy=xTAy=x∙(Ay)
024407 If A is a symmetric matrix, then eigenvectors of A corresponding to distinct eigenvalues are orthogonal.
Let Ax=λx and Ay=μy, where λ≠μ. Using Theorem [thm:024396], we compute
λ(x∙y)=(λx)∙y=(Ax)∙y=x∙(Ay)=x∙(μy)=μ(x∙y)
Hence (λ−μ)(x∙y)=0, and so x∙y=0 because λ≠μ.
Now the procedure for diagonalizing a symmetric n×n matrix is clear. Find the distinct eigenvalues (all real by Theorem [thm:016397]) and find orthonormal bases for each eigenspace (the Gram-Schmidt algorithm may be needed). Then the set of all these basis vectors is orthonormal (by Theorem [thm:024407]) and contains n vectors. Here is an example.
024416 Orthogonally diagonalize the symmetric matrix A=[8−22−254245].
The characteristic polynomial is
cA(x)=det[x−82−22x−5−4−2−4x−5]=x(x−9)2
Hence the distinct eigenvalues are 0 and 9 of multiplicities 1 and 2, respectively, so dim(E0)=1 and dim(E9)=2 by Theorem [thm:016250] (A is diagonalizable, being symmetric). Gaussian elimination gives
E0(A)=span{x1},\enskipx1=[12−2], and E9(A)=span{[−210],}
The eigenvectors in E9 are both orthogonal to x1 as Theorem [thm:024407] guarantees, but not to each other. However, the Gram-Schmidt process yields an orthogonal basis
{x2,x3} of E9(A) where x2=[−210] and x3=
Normalizing gives orthonormal vectors {13x1,1√5x2,13√5x3}, so
P=[13x11√5x213√5x3]=13√5[√5−622√534−2√505]
is an orthogonal matrix such that P−1AP is diagonal.
It is worth noting that other, more convenient, diagonalizing matrices P exist. For example, y2= and y3=[−221] lie in E9(A) and they are orthogonal. Moreover, they both have norm 3 (as does x1), so
Q=[13x113y213y3]=13[12−2212−221]
is a nicer orthogonal matrix with the property that Q−1AQ is diagonal.
If A is symmetric and a set of orthogonal eigenvectors of A is given, the eigenvectors are called principal axes of A. The name comes from geometry. An expression q=ax21+bx1x2+cx22 is called a quadratic form in the variables x1 and x2, and the graph of the equation q=1 is called a conic in these variables. For example, if q=x1x2, the graph of q=1 is given in the first diagram.
But if we introduce new variables y1 and y2 by setting x1=y1+y2 and x2=y1−y2, then q becomes q=y21−y22, a diagonal form with no cross term y1y2 (see the second diagram). Because of this, the y1 and y2 axes are called the principal axes for the conic (hence the name). Orthogonal diagonalization provides a systematic method for finding principal axes. Here is an illustration.
024463 Find principal axes for the quadratic form q=x21−4x1x2+x22.
In order to utilize diagonalization, we first express q in matrix form. Observe that
q=[x1x2][1−401][x1x2]
The matrix here is not symmetric, but we can remedy that by writing
q=x21−2x1x2−2x2x1+x22
Then we have
q=[x1x2][1−2−21][x1x2]=xTAx
where x=[x1x2] and A=[1−2−21] is symmetric. The eigenvalues of A are λ1=3 and λ2=−1, with corresponding (orthogonal) eigenvectors x1=[1−1] and x2=. Since ‖x1‖=‖x2‖=√2, so
P=1√2[11−11] is orthogonal and PTAP=D=[300−1]
Now define new variables [y1y2]=y by y=PTx, equivalently x=Py (since P−1=PT). Hence
y1=1√2(x1−x2) and y2=1√2(x1+x2)
In terms of y1 and y2, q takes the form
q=xTAx=(Py)TA(Py)=yT(PTAP)y=yTDy=3y21−y22
Note that y=PTx is obtained from x by a counterclockwise rotation of π4 (see Theorem [thm:004693]).
Observe that the quadratic form q in Example [exa:024463] can be diagonalized in other ways. For example
q=x21−4x1x2+x22=z21−13z22
where z1=x1−2x2 and z2=3x2. We examine this more carefully in Section [sec:8_8].
If we are willing to replace “diagonal” by “upper triangular” in the principal axes theorem, we can weaken the requirement that A is symmetric to insisting only that A has real eigenvalues.
Triangulation Theorem024503 If A is an n×n matrix with n real eigenvalues, an orthogonal matrix P exists such that PTAP is upper triangular.
We modify the proof of Theorem [thm:024303]. If Ax1=λ1x1 where ‖x1‖=1, let {x1,x2,…,xn} be an orthonormal basis of Rn, and let P1=[x1x2⋯xn]. Then P1 is orthogonal and PT1AP1=[λ1B0A1] in block form. By induction, let QTA1Q=T1 be upper triangular where Q is of size (n−1)×(n−1) and orthogonal. Then P2=[100Q] is orthogonal, so P=P1P2 is also orthogonal and PTAP=[λ1BQ0T1] is upper triangular.
The proof of Theorem [thm:024503] gives no way to construct the matrix P. However, an algorithm will be given in Section [sec:11_1] where an improved version of Theorem [thm:024503] is presented. In a different direction, a version of Theorem [thm:024503] holds for an arbitrary matrix with complex entries (Schur’s theorem in Section [sec:8_6]).
As for a diagonal matrix, the eigenvalues of an upper triangular matrix are displayed along the main diagonal. Because A and PTAP have the same determinant and trace whenever P is orthogonal, Theorem [thm:024503] gives:
024536 If A is an n×n matrix with real eigenvalues λ1,λ2,…,λn (possibly not all distinct), then detA=λ1λ2…λn and \functrA=λ1+λ2+⋯+λn.
This corollary remains true even if the eigenvalues are not real (using Schur’s theorem).
8.1E: Orthogonal Complements and Projections Exercises
8.2E: Orthogonal Diagonalization Exercises |
4418 | https://homepage.ntu.edu.tw/~josephw/Principles_14F_lecture5a.pdf | 1 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
Economics Principles of N. Gregory Mankiw Consumers, Producers, and the Efficiency of Markets Seventh Edition CHAPTER 7 Wojciech Gerson (1831‐1901) Modified by Joseph Tao‐yi Wang In this chapter, look for the answers to these questions • What is consumer surplus? How is it related to the demand curve?
• What is producer surplus? How is it related to the supply curve?
• Do markets produce a desirable allocation of resources? Or could the market outcome be improved upon? © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
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2 Welfare Economics Recall, the allocation of resources refers to: how much of each good is produced which producers produce it which consumers consume it Welfare economics studies how the allocation of resources affects economic well-being.
First, we look at the well-being of consumers. 3 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
3 Willingness to Pay (WTP) A buyer’s willingness to pay for a good is the maximum amount the buyer will pay for that good.
WTP measures how much the buyer values the good.
name WTP Anthony $250 Kenny 175 Quan 300 John 125 Example: 4 buyers’ WTP for an iPad 4 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
4 WTP and the Demand Curve Q: If price of iPad is $200, who will buy an iPad, and what is quantity demanded?
A: Anthony & Quan will buy an iPad, Keny & John will not. Hence, Qd = 2 when P = $200.
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5 WTP and the Demand Curve Derive the demand schedule: 4 John, Kenny, Anthony, Quan 0 – 125 3 Kenny, Anthony, Quan 126 – 175 2 Anthony, Quan 176 – 250 1 Quan 251 – 300 0 nobody $301 & up Qd who buys P (price of iPad) name WTP Anthony $250 Kenny 175 Quan 300 John 125 2 6 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
6 $0 $50 $100 $150 $200 $250 $300 $350 0 1 2 3 4 WTP and the Demand Curve P Qd $301 & up 0 251 – 300 1 176 – 250 2 126 – 175 3 0 – 125 4 P Q 7 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
7 $0 $50 $100 $150 $200 $250 $300 $350 0 1 2 3 4 About the Staircase Shape… This D curve looks like a staircase with 4 steps – one per buyer. P Q If there were a huge # of buyers, as in a competitive market, there would be a huge # of very tiny steps, and it would look more like a smooth curve.
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8 $0 $50 $100 $150 $200 $250 $300 $350 0 1 2 3 4 WTP and the Demand Curve At any Q, the height of the D curve is the WTP of the marginal buyer, the buyer who would leave the market if P were any higher.
P Q Quan’s WTP Anthony’s WTP Kenny’s WTP John’s WTP 9 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
9 Consumer Surplus (CS) Consumer surplus is the amount a buyer is willing to pay minus the amount the buyer actually pays: CS = WTP – P name WTP Anthony $250 Kenny 175 Quan 300 John 125 Suppose P = $260. Quan’s CS = $300 – 260 = $40.
The others get no CS because they do not buy an iPad at this price. Total CS = $40.
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10 $0 $50 $100 $150 $200 $250 $300 $350 0 1 2 3 4 CS and the Demand Curve P Q Quan’s WTP P = $260 Quan’s CS = $300 – 260 = $40 Total CS = $40 11 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
11 $0 $50 $100 $150 $200 $250 $300 $350 0 1 2 3 4 CS and the Demand Curve P Q Quan’s WTP Anthony’s WTP Instead, suppose P = $220 Quan’s CS = $300 – 220 = $80 Anthony’s CS = $250 – 220 = $30 Total CS = $110 3 12 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
12 $0 $50 $100 $150 $200 $250 $300 $350 0 1 2 3 4 CS and the Demand Curve P Q The lesson: Total CS equals the area under the demand curve above the price, from 0 to Q.
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13 0 10 20 30 40 50 60 0 5 10 15 20 25 30 P Q $ CS with Lots of Buyers & a Smooth D Curve The demand for shoes D 1000s of pairs of shoes Price per pair At Q = 5(thousand), the marginal buyer is willing to pay $50 for pair of shoes. Suppose P = $30. Then his consumer surplus = $20. 14 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
14 0 10 20 30 40 50 60 0 5 10 15 20 25 30 P Q CS with Lots of Buyers & a Smooth D Curve The demand for shoes D CS is the area b/w P and the D curve, from 0 to Q. Recall: area of a triangle equals ½ x base x height Height = $60 – 30 = $30. So, CS = ½ x 15 x $30 = $225.
h $ 15 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
15 0 10 20 30 40 50 60 0 5 10 15 20 25 30 P Q How a Higher Price Reduces CS D If P rises to $40, CS = ½ x 10 x $20 = $100.
Two reasons for the fall in CS.
1. Fall in CS due to buyers leaving market 2. Fall in CS due to remaining buyers paying higher P © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
0 5 10 15 20 25 30 35 40 45 50 0 5 10 15 20 25 P Q demand curve A. Find marginal buyer’s WTP at Q = 10. B. Find CS for P = $30.
Suppose P falls to $20.
How much will CS increase due to… C. buyers entering the market D. existing buyers paying lower price $ A C T I V E L E A R N I N G 1 Consumer surplus A C T I V E L E A R N I N G 1 Answers © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
0 5 10 15 20 25 30 35 40 45 50 0 5 10 15 20 25 P $ Q demand curve A. At Q = 10, marginal buyer’s WTP is $30.
B. CS = ½ x 10 x $10 = $50 P falls to $20.
C. CS for the additional buyers = ½ x 10 x $10 = $50 D. Increase in CS on initial 10 units = 10 x $10 = $100 4 18 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
18 Cost and the Supply Curve name cost Rosy $10 Chuck 20 Chiang 35 A seller will produce and sell the good/service only if the price exceeds his or her cost. Hence, cost is a measure of willingness to sell. Cost is the value of everything a seller must give up to produce a good (i.e., opportunity cost). Includes cost of all resources used to produce good, including value of the seller’s time. Example: Costs of 3 sellers in the lawn-cutting business.
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19 Cost and the Supply Curve 3 35 & up 2 20 – 34 1 10 – 19 0 $0 – 9 Qs P Derive the supply schedule from the cost data: name cost Rosy $10 Chuck 20 Chiang 35 20 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
20 Cost and the Supply Curve $0 $10 $20 $30 $40 0 1 2 3 P Q P Qs $0 – 9 0 10 – 19 1 20 – 34 2 35 & up 3 21 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
21 $0 $10 $20 $30 $40 0 1 2 3 Cost and the Supply Curve P Q At each Q, the height of the S curve is the cost of the marginal seller, the seller who would leave the market if the price were any lower.
Chiang’s cost Chuck’s cost Rosy’s cost 22 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
22 $0 $10 $20 $30 $40 0 1 2 3 Producer Surplus P Q Producer surplus (PS): the amount a seller is paid for a good minus the seller’s cost PS = P – cost 23 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
23 $0 $10 $20 $30 $40 0 1 2 3 Producer Surplus and the S Curve P Q PS = P – cost Suppose P = $25.
Rosy’s PS = $15 Chuck’s PS = $5 Chiang’s PS = $0 Total PS = $20 Chuck’s cost Rosy’s cost Total PS equals the area above the supply curve under the price, from 0 to Q.
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24 0 10 20 30 40 50 60 0 5 10 15 20 25 30 P Q PS with Lots of Sellers & a Smooth S Curve The supply of shoes S 1000s of pairs of shoes Price per pair Suppose P = $40. At Q = 15(thousand), the marginal seller’s cost is $30, and her producer surplus is $10. 25 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
25 0 10 20 30 40 50 60 0 5 10 15 20 25 30 P Q PS with Lots of Sellers & a Smooth S Curve The supply of shoes S PS is the area b/w P and the S curve, from 0 to Q.
The height of this triangle is $40 – 15 = $25.
So, PS = ½ x b x h = ½ x 25 x $25 = $312.50 h 26 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
26 0 10 20 30 40 50 60 0 5 10 15 20 25 30 P Q How a Lower Price Reduces PS If P falls to $30, PS = ½ x 15 x $15 = $112.50 Two reasons for the fall in PS.
S 1. Fall in PS due to sellers leaving market 2. Fall in PS due to remaining sellers getting lower P A C T I V E L E A R N I N G 2 Producer surplus © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
0 5 10 15 20 25 30 35 40 45 50 0 5 10 15 20 25 P Q supply curve A. Find marginal seller’s cost at Q = 10. B. Find total PS for P = $20.
Suppose P rises to $30.
Find the increase in PS due to: C. selling 5 additional units D. getting a higher price on the initial 10 units A C T I V E L E A R N I N G 2 Answers © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
0 5 10 15 20 25 30 35 40 45 50 0 5 10 15 20 25 P Q supply curve A. At Q = 10, marginal cost = $20 B. PS = ½ x 10 x $20 = $100 P rises to $30.
C. PS on additional units = ½ x 5 x $10 = $25 D. Increase in PS on initial 10 units = 10 x $10 = $100 29 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
CS, PS, and Total Surplus CS = (value to buyers) – (amount paid by buyers) = buyers’ gains from participating in the market PS = (amount received by sellers) – (cost to sellers) = sellers’ gains from participating in the market Total surplus = CS + PS = total gains from trade in a market = (value to buyers) – (cost to sellers) 6 30 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
The Market’s Allocation of Resources In a market economy, the allocation of resources is decentralized, determined by the interactions of many self-interested buyers and sellers.
Is the market’s allocation of resources desirable? Or would a different allocation of resources make society better off? To answer this, we use total surplus as a measure of society’s well-being, and we consider whether the market’s allocation is efficient. (Policymakers also care about equality, though our focus here is on efficiency.) 31 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
31 Efficiency An allocation of resources is efficient if it maximizes total surplus. Efficiency means: The goods are consumed by the buyers who value them most highly. The goods are produced by the producers with the lowest costs.
Raising or lowering the quantity of a good would not increase total surplus. = (value to buyers) – (cost to sellers) Total surplus 32 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
32 Evaluating the Market Equilibrium Market eq’m: P = $30 Q = 15,000 Total surplus = CS + PS Is the market eq’m efficient?
0 10 20 30 40 50 60 0 5 10 15 20 25 30 P Q S D CS PS 33 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
33 Which Buyers Consume the Good?
0 10 20 30 40 50 60 0 5 10 15 20 25 30 P Q S D Every buyer whose WTP is ≥ $30 will buy. Every buyer whose WTP is < $30 will not. So, the buyers who value the good most highly are the ones who consume it.
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34 Which Sellers Produce the Good?
0 10 20 30 40 50 60 0 5 10 15 20 25 30 P Q S D Every seller whose cost is ≤ $30 will produce the good. Every seller whose cost is > $30 will not. So, the sellers with the lowest cost produce the good.
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35 Does Eq’m Q Maximize Total Surplus?
0 10 20 30 40 50 60 0 5 10 15 20 25 30 P Q S D At Q = 20, cost of producing the marginal unit is $35 value to consumers of the marginal unit is only $20 Hence, can increase total surplus by reducing Q. This is true at any Q greater than 15. 7 36 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
36 Does Eq’m Q Maximize Total Surplus?
0 10 20 30 40 50 60 0 5 10 15 20 25 30 P Q S D At Q = 10, cost of producing the marginal unit is $25 value to consumers of the marginal unit is $40 Hence, can increase total surplus by increasing Q. This is true at any Q less than 15. 37 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
37 Does Eq’m Q Maximize Total Surplus?
0 10 20 30 40 50 60 0 5 10 15 20 25 30 P Q S D The market eq’m quantity maximizes total surplus: At any other quantity, can increase total surplus by moving toward the market eq’m quantity. 38 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
38 Adam Smith and the Invisible Hand “Man has almost constant occasion for the help of his brethren, and it is vain for him to expect it from their benevolence only.
Adam Smith, 1723-1790 Passages from The Wealth of Nations, 1776 He will be more likely to prevail if he can interest their self-love in his favor, and show them that it is for their own advantage to do for him what he requires of them… It is not from the benevolence of the butcher, the brewer, or the baker that we expect our dinner, but from their regard to their own interest….
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39 Adam Smith and the Invisible Hand “Every individual…neither intends to promote the public interest, nor knows how much he is promoting it…. Adam Smith, 1723-1790 Passages from The Wealth of Nations, 1776 He intends only his own gain, and he is in this, as in many other cases, led by an invisible hand to promote an end which was no part of his intention. Nor is it always the worse for the society that it was no part of it. By pursuing his own interest he frequently promotes that of the society more effectually than when he really intends to promote it.” an invisible hand ©Georgios Kollidas/Shutterstock.com 40 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
The Free Market vs. Govt Intervention The market equilibrium is efficient. No other outcome achieves higher total surplus. Govt cannot raise total surplus by changing the market’s allocation of resources. Laissez faire (French for “allow them to do”): the notion that govt should not interfere with the market.
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The Free Market vs. Central Planning Suppose resources were allocated not by the market, but by a central planner who cares about society’s well-being. To allocate resources efficiently and maximize total surplus, the planner would need to know every seller’s cost and every buyer’s WTP for every good in the entire economy.
This is impossible, and why centrally-planned economies are never very efficient. 8 42 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
CONCLUSION This chapter used welfare economics to demonstrate one of the Ten Principles: Markets are usually a good way to organize economic activity.
Important note: We derived these lessons assuming perfectly competitive markets. In other conditions we will study in later chapters, the market may fail to allocate resources efficiently… 43 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
CONCLUSION Such market failures occur when: a buyer or seller has market power—the ability to affect the market price.
transactions have side effects, called externalities, that affect bystanders. (example: pollution) We’ll use welfare economics to see how public policy may improve on the market outcome in such cases. Despite the possibility of market failure, the analysis in this chapter applies in many markets, and the invisible hand remains extremely important.
Summary • The height of the D curve reflects the value of the good to buyers—their willingness to pay for it.
• Consumer surplus is the difference between what buyers are willing to pay for a good and what they actually pay. • On the graph, consumer surplus is the area between P and the D curve.
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Summary • The height of the S curve is sellers’ cost of producing the good. Sellers are willing to sell if the price they get is at least as high as their cost.
• Producer surplus is the difference between what sellers receive for a good and their cost of producing it. • On the graph, producer surplus is the area between P and the S curve. © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
Summary • To measure society’s well-being, we use total surplus, the sum of consumer and producer surplus. • Efficiency means that total surplus is maximized, that the goods are produced by sellers with lowest cost, and that they are consumed by buyers who most value them. • Under perfect competition, the market outcome is efficient. Altering it would reduce total surplus.
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47 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
Efficiency and Welfare Consumer Surplus + Producer Surplus = Total Surplus (maximized at Equilibrium) Efficiency vs. Equality Homework: Mankiw, Ch. 7, Problems 6, 7, 9, 10, 11.
True or False. If consumers buy 1,000 heads of lettuce per week, and if the price of lettuce falls by $1 per head, then the consumer surplus will increases by $1,000. |
4419 | https://www.math.purdue.edu/pow2/discussion/2016/spring/9.html | Spring 2016, problem 9: Problem of the Week: Purdue University.
Department of Mathematics
Spring 2016, problem 9
The numbers 1,2,…,n 1,2,…,n are written around a circle in some order. What is the smallest and largest possible sum of the absolute differences of adjacent numbers?
Comments
upvote
0
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Samurai
5 years ago
As for minimality part : suppose the arrangement is a non - consecutive sequence .Then for sum p,q,r,s ¢ N, p-q>1 , r-s>1 where p adjacent to q , r adjacent to s . wlog the last integer is not n ,so must be less then n-1 .wtf my phone is glitching out !;(
' half the problem for half the points ' ( lion's share had this been a putnam )Minimality of 2(n-1) Now we will show that 2 n−2 2 n−2 is minimal. To do so, remark that 1 1 and n n must be on the circle. By the Triangle Inequality, the sum of the positive differences along the minor arc between 1 1 and n n must be at least n−1 n−1 and similarly for the otherrh arc
Samurai 5 years ago
If n is even, call the numbers n2..n the large numbers. Call the rest small. If n is odd call the numbers n+32..n the large numbers. Call n+12 the middle number.
arreolajongmt67 1 month ago
upvote
0
downvote
Nelix
6 years ago
The maximum is ⌊n 2 2⌋⌊n 2 2⌋. The minimum is 2(n−1)2(n−1).
Proof of the maximality part:
For a circular arrangement of 1..n 1..n call the sum of the absolute differences of neighbours the value of the arrangement. For a,b∈{1..n}a,b∈{1..n} write a→b a→b, if b b is the right side neighbour of a a.
If n n is even, call the numbers n 2..n n 2..n the large numbers. Call the rest small. If n n is odd call the numbers n+3 2..n n+3 2..n the large numbers. Call n+1 2 n+1 2 the middle number. Call the rest small.
The value v v of the arrangement is given by:
v:=∑a→b[m a x(a,b)−m i n(a,b)]=∑a→b m a x(a,b)−∑a→b m i n(a,b)v:=∑a→b[m a x(a,b)−m i n(a,b)]=∑a→b m a x(a,b)−∑a→b m i n(a,b).
Each of the two sums in the last expression is a sum of n n numbers in the range 1..n 1..n, none of which occur more than twice in any of the sums. The value v v is clearly no larger than m m, given by: m:=[n+n+(n−1)+(n−1)+...]−[1+1+2+2+3+3...]m:=[n+n+(n−1)+(n−1)+...]−[1+1+2+2+3+3...], where the first and the second bracket contain n n terms each. By a simple calculation m=⌊n 2 2⌋m=⌊n 2 2⌋. But the value m m is indeed attained:
For n n even arrange the numbers like:
large →→ small →→ large →→ small ... →→ large→→ small
For n n odd arrange like:
middle →→ large →→ small →→ large →→ small ... →→ large→→ small. Done.
Proof of the minimality part:
Proof by induction over n n: The induction start is trivial. So assume, that the minimum value for arrangements of numbers in 1..n 1..n is given by m n:=2(n−1)m n:=2(n−1). Consider a minimum-value arrangement A n+1 A n+1 of the numbers 1..(n+1)1..(n+1). with value v n+1 v n+1. Now take away the number n+1 n+1 from A n+1 A n+1. You get an arrangement A n A n of the numers 1..n 1..n, with value v n v n. By an easy calculation v n≤v n+1−2 v n≤v n+1−2. Thus v n+1 v n+1 cannot be smaller than 2(n+1−1)2(n+1−1) because otherwise v n v n would be smaller than 2(n−1)2(n−1). The minimum can also be no larger than 2(n+1−1)2(n+1−1), since this value is attained for simply placing the numbers in ascending order alongthe circle. Done.
Hello, I get a slightly different answer for the maximum. It must be a whole number.
I get,
maximum = trunc(x^2/2),
i.e., if n is even, maximum = n^2/2
if n is odd, maximum = n^2/2 - 1/2
I agree with your minimum.
Jao
cgjoa3 6 years ago
Hi Jao. The notation ⌊⋅⌋⌊⋅⌋ I used is defined by:
⌊x⌋:=f l o o r(x)⌊x⌋:=f l o o r(x).
For x>0 x>0: f l o o r(x)=t r u n c(x)f l o o r(x)=t r u n c(x)
Nelix 6 years ago
upvote
0
downvote
Samurai
5 years ago
Are these called purdue problems of week or nelix's math proofs of the week?! Jk wish more mathletes would participate !
upvote
0
downvote
fearless
5 years ago
let start with number 1, minimum possible difference for next number is 1 so next number 2 and so on so minimum sum = n
for maximum::
let start with 1 and for max difference next number should be n, number next to n such that max. difference is there should be 2
so the series is 1,n,2,n-1,3,n-2,4,n-3....., and last n/2 or (n-1)/2 for odd and even n
so max = n n/2 or (n n-1)/2
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4420 | https://study.com/academy/lesson/what-is-factoring-in-algebra-definition-example.html | Factoring in Algebra | Definition, Equations & Examples - Lesson | Study.com
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Math Courses / Algebra I: High School Course
Factoring in Algebra | Definition, Equations & Examples
Lesson
Transcript
Joao Amadeu, Jeff Calareso
Author Joao Amadeu
Joao Amadeu has more than 10 years of experience in teaching physics and mathematics at different educational levels. Joao earned two degrees at Londrina State University: B.S. in Physics and M.S. in Science and Mathematics Education. He is currently working on his PhD in Science Education at Western Michigan University.
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Instructor Jeff Calareso
Jeff teaches high school English, math and other subjects. He has a master's degree in writing and literature.
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This lesson focuses on exploring the concepts of factors and factoring in algebra. It shows examples and identities that facilitate equation and expression simplification. Updated: 11/21/2023
Table of Contents
What is a Factor in Algebra?
Examples: Factoring Algebraic Equations
Lesson Summary
Show
Frequently Asked Questions
How do you factor algebraic equations?
Factoring an equation is a sort of reverse multiplication. One needs to find the multiplication that resulted in the equation to be factored. Usually, this is done through trial and error, but there are well-known algebraic identities that can facilitate the task.
What is the best definition of factoring?
Factoring an algebraic expression serves as a way of simplifying it. More specifically, factoring is to find the terms of which the equation is the product.
How do you find the factor of a number?
The simplest way to factor a term is to find the essential multiplication that gave origin to it. For example, to find the common factor of the expression 2x + 6x, one can break each term down:
2x = 2x
6x = 32x
Observing the products, it is clear that 2x is the common factor between the terms.
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Table of Contents
What is a Factor in Algebra?
Examples: Factoring Algebraic Equations
Lesson Summary
Show
What is a Factor in Algebra? ----------------------------
Solving algebraic equations and simplifying algebraic expressions, often requires one to use a method calledfactoring. This method allows one to transform expressions into multiplications. A general example can be given by the addition of two constants.
The expression 2 + 6 can be written as the multiplication 2(1+3). It may be seen as counterintuitive; however, it can be extremely useful in certain contexts. If one needs to isolate the variable x in the expression 4x + 20, for example, it can be written as 4(x + 5). In this case, the number 4 is the greatest factor of the expression 4x + 20. It is called the greatest factor because it is the maximum integer value that can be used to reverse the multiplication in the expression. The number 2 is also a factor of the expression 4x+20, but factoring with 2 would result in 2(2x+10). Using a factor that is smaller than the maximum may not be useful in most circumstances.
The next example shows the factoring of a quadratic equation in the form a x 2+b x+c=0.
Given the equation 6 x 2+12 x+6=0, it is known that the first term can result from the product between 2x and 3x.
2 x∗3 x=6 x 2
Therefore, one can try setting up multiplication with (2x + )(3x + ). Since the constant 6 in the equation can be found by the product between 2 and 3, one can try the following:
(2 x+2)(3 x+3)
Solving with the FOIL (first outside, inside last) method:
(2 x+2)(3 x+3)=2 x∗3 x+2 x∗3+2∗3 x+2∗3=6 x 2+6 x+6 x+6=6 x 2+12 x+6
Therefore, the expression 6 x 2+12 x+6=0 can be written as (2 x+2)(3 x+3)=0.
The factored form of the expression facilitates the task of solving the equation for x, since:
For (2x + 2)(3x + 3) = 0 to be true, then (2x +2) = 0, or (3x + 3) = 0.
Thus:
2x + 2 = 0
2x = -2
x = -1
or
3x + 3 = 0
3x = -3
x = -1
The expression 6 x 2+12 x+6 will be equal to zero if x is -1.
6 x 2+12 x+6, with x=-1
6(−1)2+12(−1)+6=6+(−12)+6=6−12+6=0
This example shows what is a factor in algebra through trial and error. This is done with pattern observations and may need to be modified in case the validity of any step is not confirmed.
One could also try to simplify 6 x 2+12 x+6 using the factors (6x + 6) and (x + 1). That would lead to:
(6 x+6)(x+1)=6 x∗x+6 x∗1+6∗x+6∗1=6 x 2+6 x+6 x+6=6 x 2+12 x+6
What is Factoring in Algebra?
The previous section showed how to find factors in algebra through trial and error, which is a very useful method for simple and well-known equations. There are other methods that can help one in factoring an expression.
Completing the square
This method leads to a straightforward simplification for equations that have the pattern described by x 2+2 x h+h 2. Knowing that the first term is a product of x times x, the third term is a product of h times h, and the second term results from 2 times the product of x and h. One can use the square of the factor (x + h) to simplify the expression.
(x+h)2=(x+h)(x+h)=x∗x+x∗h+h∗x+h∗h=x 2+2 h x+h 2
Figure 2: Completing the square example.
Other well-known identities are:
a 3−b 3=(a−b)(a 2+a b+b 2)
a 2−b 2=(a+b)(a−b)
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Examples: Factoring Algebraic Equations ---------------------------------------
Example 1
Find the great common factor for the expression 15 x 4+15 x 3+75 x.
Solution
Reversing the multiplication of each term, one has
15 x 4=5∗3∗x∗x∗x∗x
15 x 3=5∗3∗x∗x∗x
75 x=5∗5∗3∗x
The three terms have the product 5∗3∗x in common, therefore the greatest common factor of this expression will be 15 x.
Example 2
Factor the expression 3 x 2+6 x+8 x+4.
Solution In this case, the first two terms do not have a common factor with the last two terms. Therefore, they must be simplified separately.
Factoring (3 x 2+6 x)
3 x 2=3∗x∗x
6 x=2∗3∗x
Factor: 3 x
Reversing the multiplication, one has 3 x(x+2)
Factoring (8 x+4)
8 x=4∗2∗x
4=4
Factor: 4
Reversing the multiplication, one has 4(2 x+1)
The result is that the expression 3 x 2+6 x+8 x+4 can be factored as 3 x(x+2)+4(2 x+1).
Example 3
Factor the expression x 2+8 x+16 by completing the square.
Solution The equation can be opened as x 2+2∗4∗x+4 2, therefore it is equals to (x+4)2.
Example 4
Factor the expression 9 x 2−16.
Solution this can be compared to a 2−b 2, since:
9 x 2−16=(3 x)2−4 2
Knowing that a 2−b 2=(a+b)(a−b), and that a=3 x and b=4, we have:
9 x 2−16=(3 x)2−4 2=(3 x+4)(3 x−4)
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Lesson Summary --------------
This lesson explored the concepts of factors and factoring in algebra. Factoring is a method of expression simplification that consists in finding a pattern between the terms of the expression and applying a sort of reverse multiplication to it. For example, if the multiplication between the factors (x+2) and (x+3) results in the expression x 2+5 x+6, then this resulting expression can be factored back as (x+2)(x+3).
In general, factoring in an expression requires trial and error. However, there are some algebraic identities that can help one to identify the factors of an expression. Some of them are: (x+h)2=x 2+2 h x+h 2,a 3−b 3=(a−b)(a 2+a b+b 2),a n d a 2−b 2=(a+b)(a−b). In some equations, there are no factors that are common to all of its terms. These cases require one to split the terms into groups and factor them separately before joining them all together again.
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Video Transcript
Parts to a Whole
Factoring in algebra is a lot like baking. I see you have some cake. Is it your birthday? Oh, I'm sorry - I didn't get you anything. But since I'm here, can I have some?
Mmm, I can really taste the baking soda. Can't you? No? Well there's definitely baking soda in this cake. OK, I can't taste it. That's because it's been combined with other ingredients to form something new - yummy cake.
In algebra, we take expressions and stir them together to make new expressions. But even though it may not be obvious what terms we started with, those ingredients are still there.
You can't take the baking soda out of a finished cake, but you can factor the original terms out of an expression. Let's learn how.
Factoring
Here is an expression: 4 x - 8. Let's say we want to factor it. We can define factoring as finding the terms that are multiplied together to get an expression.
Our expression here has some important parts, like the ingredients we bake with. First, we have two terms: 4 x and 8. The terms are the numbers, variables or numbers and variables that are multiplied together. Terms are separated by plus or minus signs.
8 is just a constant, or a number that is what it is. It's constantly 8. x is a variable, or a symbol standing in for a number we don't know. 4 is a coefficient. Notice the prefix 'co-.' The coefficient multiplies a variable. It's a codependent, cooperating coefficient.
Now, we need to find a factor. This is like looking for the baking soda, but usually it's a bit easier. It's more like picking the raisins out of an oatmeal cookie. You may get your hands a little dirty, but your cookie will be less raisin-y for sure.
A factor is a term that can be extracted from the equation. Think about the number 6. Its factors are 2 and 3. Why? Because 2 3 is 6.
With 4 x - 8, we can extract a 4. Each term is a multiple of 4. If we factor out a 4, we have 4(x - 2). Note that we can reverse what we just did. 4 x is 4 x. And 4 2 is 8, getting us back to 4 x - 8.
What if we had 3 x - 8? Is there a common factor? No. 3 and 8 are what we call relatively prime. Remember that a prime number has no factors other than 1 and itself. Relatively prime numbers have no shared factors other than 1.
Practice
Let's practice factoring. Here's an expression: 6 y^2 - 11 y. OK, what are the factors of 6? 2 and 3. What about 11? 11 is prime. So 6 and 11 are relatively prime. Does that mean we can't factor? Are we going to have to eat the raisins? No! Look at the variables. y^2 and y. We can factor out a y. That gets us y(6 y - 11). Cookie crisis averted.
To check our work, let's put it back together. y 6 y is 6 y^2. y 11 is 11 y. That gets us 6 y^2 - 11 y. Excellent!
Let's try another: 12 x^3 + 18 x. OK, you might immediately see that we can factor out an x. It's like seeing unwanted olives on a pizza; they stand out, don't they? But what else? 12 and 18 share a few factors. 12's factors include 3, 4 and 6. 18's factors include 3, 6 and 9. Where's the overlap? 3 and 6.
What should we factor out? If we factor out the 3, we'd have 4 and 6. That's better, but that pizza still has olives, so to speak. We want the greatest common factor. That's simply the biggest shared factor. Here, that's 6. If we factor out 6 x, we get 6 x(2 x^2 + 3).
Let's check. 6 x 2 x^2 is 12 x^3. That's good. And 6 x 3 is 18 x. Good again. This pizza is safe to eat!
How about one more? 5 y + 5. Well, 5 and 5, what do we do here? We can extract that 5. Then we get 5 times y plus what? What times 5 is 5? 1. So it's 5(y + 1). That's a bit like asking for an ice cream cone, minus the ice cream.
Lesson Summary
To summarize, we learned about factoring in algebra. To factor is to find the terms that are multiplied together to make an expression.
Expressions consist of various terms. A term can have certain parts, like constants, variables and coefficients.
When we factor in algebra, we're looking for the greatest common factors that are shared by the terms in an expression. If we only have numbers that are relatively prime, like 7 and 9, then we can't factor out any constants. But if there are common factors, then we pluck them like olives from a pizza. Just remember to check for ones hiding beneath the cheese!
Learning Outcomes
Once you've finished checking out this lesson, you might be prepared to:
Explain the process of factoring
Identify common factors
Factor an algebraic expression
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Factoring in Algebra | Definition, Equations & Examples 5:32
5:36 Next Lesson Finding the Prime Factorization of a Number | Meaning & Examples
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4421 | https://arxiv.org/abs/1904.10766 | We gratefully acknowledge support from the Simons Foundation, member institutions, and all contributors. Donate
math > arXiv:1904.10766
Mathematics > Complex Variables
arXiv:1904.10766 (math)
[Submitted on 15 Apr 2019]
Title:Orthogonal polynomials and Möbius transformations
Authors:R. S. Vieira, V. Botta
View a PDF of the paper titled Orthogonal polynomials and M\"obius transformations, by R. S. Vieira and V. Botta
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Abstract:Given an orthogonal polynomial sequence on the real line, another sequence of polynomials can be found by composing these polynomials with a general Möbius transformation. In this work, we study the properties of such Möbius-transformed polynomials. We show that they satisfy an orthogonality relation in given curve of the complex plane with respect to a varying weight function and that they also enjoy several properties common to the orthogonal polynomial sequences on the real line --- e.g. a three-term recurrence relation, Christoffel-Darboux type identities, their zeros are simple, lie on the support of orthogonality and have the interlacing property, etc. Moreover, we also show that the Möbius-transformed polynomials obtained from classical orthogonal polynomials also satisfy a second-order differential equation, a Rodrigues' type formula and generating functions. As an application, we show that Hermite, Laguerre, Jacobi, Bessel and Romanovski polynomials are all related to each other by a suitable Möbius transformation. New orthogonality relations for Bessel and Romanovski polynomials are also presented.
| | |
--- |
| Comments: | Keywords: Orthogonal polynomials, Möbius transformations, varying weight functions, classical orthogonal polynomials, Bessel polynomials, Romanovski polynomials |
| Subjects: | Complex Variables (math.CV); Mathematical Physics (math-ph) |
| MSC classes: | 42C05, 33C47, 30C35 |
| Cite as: | arXiv:1904.10766 [math.CV] |
| | (or arXiv:1904.10766v1 [math.CV] for this version) |
| | arXiv-issued DOI via DataCite |
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From: Ricardo Vieira Soares [view email] [v1] Mon, 15 Apr 2019 20:05:29 UTC (22 KB)
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4422 | https://energyeducation.ca/encyclopedia/Energy_use_per_person | Energy use per person - Energy Education
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Figure 1. a) World map of OECD countries (blue), countries hat OECD cooperates with (dark red), and other non-OECD countries (light red).
Energy use per person is a per capita expression of how much energy use there is in a given country or area. While the world population is increasing quickly, the total primary energy use is increasing faster much faster. Even faster than this, the total electricity use around the world is increasing. These increases can be tied to the fact that the energy use per person is rising in the world, as is the power use per person. This increase is not uniform worldwide, rather energy use per person is rising more rapidly in some countries than in others.
People who live in wealthy countries tend to be the people most likely to live in a high energy society - and these wealthy countries are often taken to be OECD countries. Wealthy countries have extensive energy infrastructure, like the electrical grid, to give citizens access to energy currencies to provide extensive energy services that allow them to maintain a high quality of life. Developing countries, especially BRIC countries, have a rapidly growing consumption of primary energy in attempts to improve their economic opportunities.
Power Servants
A person gets energy from food at a rate of roughly 100 W of power, averaged over a day. This value of 100 W provides a way of quantifying the rate of energy use (power is the rate of energy use) that's personal. Since this 100 W consumes primary energy at the same rate a person does, 100 W can be thought of as a power servant - also called energy servant. The average power consumption is 20x the power that a power servant can put out, on average, so that's 20 power servants. The power consumption in OECD countries use 50 power servants. Canada uses approximately 110 power servants. On the other end of the spectrum, non-OECD countries use 14 power servants. Explore the data visualization tool in Figure 1 to see how power use varies from country to country.
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Figure 2. Primary energy use in OECD (blue) and non-OECD (red) countries, see the map in figure 1 for which countries are part of OECD. Use the dropdown 'compare by' in order to look at countries, and the 'country selection' to look at particular countries. Click the 'per capita' to see total energy use of these countries.
References
↑OECD. (November 11, 2015). List of OECD Countries [Online]. Available:
↑Uwe Becker. (November 11, 2015). BRICs and Emerging Economies in Comparative Perspective: Political Economy, Liberalisation and Institutional Change, 1st Ed. Taylor and Francis, 2013.
↑2 0 0 0 Calories 1 day×1 day 2 4 hours×6 0 minutes×6 0 seconds×4 1 8 4 Joules 1 Calories=9 6.8 5 J second=9 6.8 5 W≈1 0 0 W
↑R. Wolfson, "High-Energy Society," in Energy, Environment and Climate, 2nd ed. New York, U.S.A.: Norton, 2012, pp. 20–21
↑BP Global. (November 11, 2015). Statistical Review of World Energy [Online]. Available:
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4424 | https://pubmed.ncbi.nlm.nih.gov/18363130/ | Calcium oxalate crystals in acute ethylene glycol poisoning: a confocal laser scanning microscope study in a fatal case - PubMed
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. 2008 Apr;46(4):322-4.
doi: 10.1080/15563650701419011.
Calcium oxalate crystals in acute ethylene glycol poisoning: a confocal laser scanning microscope study in a fatal case
Cristoforo Pomara1,Carmela Fiore,Stefano D'Errico,Irene Riezzo,Vittorio Fineschi
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1 University of Foggia, Foggia, Italy.
PMID: 18363130
DOI: 10.1080/15563650701419011
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Calcium oxalate crystals in acute ethylene glycol poisoning: a confocal laser scanning microscope study in a fatal case
Cristoforo Pomara et al. Clin Toxicol (Phila).2008 Apr.
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. 2008 Apr;46(4):322-4.
doi: 10.1080/15563650701419011.
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Cristoforo Pomara1,Carmela Fiore,Stefano D'Errico,Irene Riezzo,Vittorio Fineschi
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1 University of Foggia, Foggia, Italy.
PMID: 18363130
DOI: 10.1080/15563650701419011
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Introduction: The severity of ethylene glycol toxicity is related to the metabolic acidosis resulting from the biotransformation of ethylene glycol into toxic metabolites. Glycolic acid causes severe acidosis and oxalate precipitates as calcium oxalate in the kidneys and other tissues.
Case report: An adult male was taken to the local hospital by the team rescue and was apparently unconscious; severe metabolic acidosis and renal failure led to death a few hours after the arrival. Confocal laser scanning microscopy demonstrated oxalate crystals deposition within the tubular epithelial cells and widespread necrosis of the tubular epithelium in the proximal tubules. Toxicological examinations revealed ethylene glycol; the blood level was 250 mg/L and in urine the concentration was 0.3%.
Discussion: In cases of ethylene glycol poisoning, calcium oxalate may be excreted not only as dihydrate crystals, but also as monohydrate crystals. Direct toxicity, cortical edema, and inhibition of mitochondrial activity, as evidenced by decreased succinate dehydrogenase activity, are possible mechanisms of crystal damage. Since calcium oxalate monohydrate crystals are transported intracellularly by kidney cells, the renal toxicity of ethylene glycol may result from inhibition of mitochondrial respiratory function in proximal tubular cells by calcium oxalate monohydrate crystals.
Conclusions: The histologic diagnosis of acute renal failure secondary to ethylene glycol poisoning depends on the recognitions of the changes of acute tubular damage in association with calcium oxalate crystals deposition within the tubular epithelial cells and the widespread necrosis of the tubular epithelium in the proximal tubules.
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4425 | https://jdc.jefferson.edu/cgi/viewcontent.cgi?article=1354&context=neurologyfp | Thomas Jefferson University Thomas Jefferson University
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Department of Neurology Faculty Papers Department of Neurology
10-4-2024
Generalized Spike-Waves in Idiopathic Generalized Epilepsies: Generalized Spike-Waves in Idiopathic Generalized Epilepsies:
Does Their Frequency Matter? Does Their Frequency Matter?
Ali A. Asadi-Pooya
Mohsen Farazdaghi
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Digital Commons. For more information, please contact: JeffersonDigitalCommons@jefferson.edu. Received: 30 March 2024 Revised: 5 July 2024 Accepted: 14 July 2024 DOI: 10.1002/brb3.70023
O R I G I N A L A R T I C L E Generalized spike–waves in idiopathic generalized epilepsies: Does their frequency matter? Ali A. Asadi-Pooya 1,2 Mohsen Farazdaghi 1
1
Epilepsy Research Center, Shiraz University of Medical Sciences, Shiraz, Iran 2 Jefferson Comprehensive Epilepsy Center, Department of Neurology, Thomas Jefferson University, Philadelphia, Pennsylvania, USA
Correspondence
Ali A. Asadi-Pooya, Epilepsy Research Center, Shiraz University of Medical Sciences, Shiraz, Iran. Email: aliasadipooya@yahoo.com
Funding information
Shiraz University of Medical Sciences, Shiraz, Iran
Abstract Objectives: We hypothesized that the frequency (in Hertz) of generalized spike–waves (GSWs) in patients with idiopathic generalized epilepsy (IGE) has associations with the syndromic diagnosis as well as with the prognosis of patients (their response to medical treatment).
Methods: This was a retrospective study of a prospectively developed database. All patients with a diagnosis of IGE were studied at the epilepsy center at Shiraz University of Medical Sciences, Shiraz, Iran, from 2008 until 2022. Patients were classified into four IGE syndromes: childhood absence epilepsy; juvenile absence epilepsy; juvenile myoclonic epilepsy; and generalized tonic–clonic seizures alone.
Results: Five hundred and eighty-three patients were studied. GSWs were commonly observed in all four syndromes of IGE. Frequency of GSW (in Hertz) did not have a significant association with the syndromic diagnosis of the patients ( p = .179). The pres-ence of GSW did not have a significant association with the seizure outcome (becoming seizure free or not) of the patients ( p = .416). Frequency of GSW did not have a significant association with the seizure outcome of the patients either ( p = .574).
Conclusion: GSWs are the hallmark electroencephalographic footprints of idiopathic generalized epilepsies; however, neither their presence nor their frequency has prac-tical associations with the syndromic diagnosis of IGEs or their outcome (response to treatment).
K E Y W O R D S
EEG, epilepsy, outcome, seizure
1 INTRODUCTION
Idiopathic generalized epilepsies (IGEs) are genetic epilepsy syn-dromes diagnosed by certain clinical and electroencephalographic (EEG) characteristics proposed by the International League Against Epilepsy (ILAE) (Hirsch et al., 2022). They constitute about 20% of all patients with epilepsy (PWE) (Asadi-Pooya, Emami, & Sperling, 2013; This is an open access article under the terms of the Creative Commons Attribution License, which permits use, distribution and reproduction in any medium, provided the original work is properly cited. © 2024 The Author(s). Brain and Behavior published by Wiley Periodicals LLC. Hirsch et al., 2022). There are four IGE syndromes recognized by the ILAE: childhood absence epilepsy (CAE); juvenile absence epilepsy (JAE); juvenile myoclonic epilepsy (JME); and generalized tonic–clonic seizures alone (GTCA) (Hirsch et al., 2022). Electroencephalography in patients with IGE may show general-ized spike–waves (GSWs) ( ≥2.5 Hz) and/or polyspike–wave discharges (Asadi-Pooya, Emami, & Sperling, 2013; Cerulli Irelli et al., 2022; Hirsch
Brain Behav. 2024;14:e70023. wileyonlinelibrary.com/journal/brb3 1 of 5
2 of 5 ASADI-POOYA AND FARAZDAGHI
et al., 2022). In other words, EEG may include a wide range of fre-quencies (the number of waves that pass by each second) for GSW in different patients. This wide range may bring about two questions: (1) What does determine the frequency of GSW? (2) Does the fre-quency of GSW matter in clinical practice (does it have any diagnostic or prognostic implications)? The aim of the current clinical study was to provide an answer to the latter question. We hypothesized that the frequency of GSW has associations with the syndromic diagnosis as well as with the prognosis of patients (their response to medical treatment).
2 METHODS 2.1 Participants
This was a retrospective study based on a large database of PWE that was built prospectively over 14 years. The seizure types and epilepsy syndromes were diagnosed by the senior epileptologist at our cen-ter. All the patients had an electroencephalography performed at our epilepsy center, but an abnormal EEG result was not needed to make a final diagnosis of an IGE syndrome (when a detailed clinical history was compatible with a diagnosis of an IGE syndrome). All patients with a diagnosis of IGE were studied at the epilepsy center at Shiraz Univer-sity of Medical Sciences, Shiraz, Iran, from 2008 to 2022. Patients were classified into IGE syndromes (i.e., CAE, JAE, JME, and GTCA) accord-ing to the ILAE definitions (Hirsch et al., 2022). Patients with comorbid functional seizure and those with missing data were excluded.
2.2 Data collection
Sex, age, age at seizure onset, seizure type(s), seizure frequency, EEG findings, the syndromic diagnosis, and treatment plan were routinely registered for each patient at the first visit and during the follow-up visits. We investigated whether the patients were seizure-free for 1 year (12 months) after their initial visit. For each patient, a 2-h video-EEG monitoring was done. Each EEG at least consisted 1 hour of sleep. Activation methods, including sleep deprivation, hyperventila-tion, and photic stimulation, were used for all patients. The EEG was done before starting or switching the antiseizure medication(s). We visually measured and recorded the frequency (i.e., Hertz [Hz] during the first second of a burst or a single-GSW complex when there were no bursts) of GSW that happened in wakefulness (unless they happened only during sleep).
2.3 Statistical analyses
The IBM SPSS Statistics (version 25.0) was used for the statistical analyses. Values were presented as mean ± standard deviation for continuous variables and as number (percent) of subjects for cate-gorical variables. Pearson chi-square test, Fisher’s exact test, one-way ANOVA (analysis of variance), and independent t-test were used for statistical analyses. A p value (2-sided) less than .05 was considered significant.
2.4 Standard protocol approvals, registrations, and patient consents
The Shiraz University of Medical Sciences Institutional Review Board approved this study (IR.SUMS.REC.1401.412). Informed consent was obtained from all the participants (to include their data in the database).
3 RESULTS 3.1 Characteristics of the patients
During the study period, 4304 patients were registered at our epilepsy center. Five hundred and eighty-three patients had IGE and fulfilled the inclusion criteria of the study. The demographic and clinical char-acteristics of the patients are summarized in Table 1. Two hundred and ninety-two patients (50.1%) had JME, 118 people (20.2%) had JAE, 98 individuals (16.8%) had GTCA, and 75 patients (12.9%) had CAE. One hundred and sixty-four patients (28.1%) were newly diagnosed at our center, and 419 individuals (71.9%) were referred to us by other physicians (often due to uncontrolled seizures). Five hundred and five patients (86.8%) received monotherapy with ASMs, 72 individuals (12.3%) were prescribed polytherapy with ASMs, and 6 persons (1%) decided not to take any ASMs (first seizure [4 patients] or myoclonic seizures only [2 patients]). At the time of referral, some patients were taking different ASMs (e.g., carbamazepine, sodium valproate, lamotrigine, and topiramate). The distribution of data was normal. The frequency of GSWs did not differ between men and women (p = .424). Moreover, taking sodium valproate was not associated with any difference in the GSW frequency ( p = .553).
3.2 Does the frequency of GSW have association with the syndromic diagnosis in IGE?
GSWs were commonly observed in all four syndromes of IGE. How-ever, the presence of GSW had a significant statistical association with the syndromic diagnosis of the patients: GSWs were seen in 89.3% (N = 67) of patients with CAE, 84.7% ( N = 83) of those with GTCA, 77.9% ( N = 92) of patients with JAE, and in 71.6% ( N = 209) of indi-viduals with JME ( p = .002; degree of freedom = 3; Pearson chi-square test). Frequency of GSW did not have a significant association with the syndromic diagnosis of the patients ( p = .179; one-way ANOVA). In the two-by-two comparison of CAE (a syndrome with a higher presence of GSW) and JME (a syndrome with a lower presence of GSW), the frequency of GSW did not have a significant association with the syn-dromic diagnosis of the patients either (GSW frequency of 3.6 ± 0.7 Hz in JME and 3.4 ± 0.5 Hz in CAE; p = .069; t-test). ASADI-POOYA AND FARAZDAGHI 3 of 5
TA B L E 1 The demographic and clinical characteristics of the studied patients with idiopathic generalized epilepsy.
Variables Numbers (total =583) and percents
Sex (female: male) (%) 352 (60.4%): 231 (39.6%) Age at seizure onset, years ( ±SD) 13.2 ±6.3 (min. 1.2; max. 74; median 21) Age at diagnosis, years ( ±SD) 21.8 ±9.7 (min. 0.1; max. 54; median 13) Normal EEG †(%) 74 (12.7%) Generalized spike–waves (%) 451 (77.4%) Frequency of generalized spike–waves, Hz ( ±SD) 3.5 ±0.6 (min. 2.5; max. 6; median 3.5) Frequency of generalized spike–waves, Hz ( ±SD) in syndromes (JME, JAE, GTCA, and CAE) 3.6 ±0.7, 3.5 ±0.5, 3.5 ±0.6, 3.4 ±0.5 Polyspikes (%) 309 (53.0%) Photosensitivity during EEG (%) 53 (9.1%) Abbreviations: CAE, childhood absence epilepsy; EEG, electroencephalographic; GTCA, generalized tonic–clonic seizures alone; JAE, juvenile absence epilepsy; JME, juvenile myoclonic epilepsy. Standard deviation.
†Electroencephalography.
3.3 Does the frequency of GSW have association with the outcome in IGE?
Three hundred and fifty-eight patients had at least 12 months of follow-up at our center and were included in the outcome analysis. One hundred and three patients (28.8%) had new-onset epilepsy, and 255 individuals (71.2%) were referred to us by other physicians (often due to uncontrolled seizures). Three hundred and eight patients (86.0%) were on monotherapy with ASMs, 49 individuals (13.7%) were pre-scribed polytherapy with ASMs, and one person (0.3%) decided not to take any ASMs (with a single seizure). One hundred and thirty-four patients (37.4%) were free of all seizure types, and 224 people (62.6%) were not seizure-free. The presence of GSW did not have a significant association with the seizure outcome of the patients: A total of 104 out of 286 patients with GSW and 30 out of 72 individuals without GSW were seizure-free at 12 months of their follow-up ( p = .416; degree of freedom = 1; Fisher’s exact test). Frequency of GSW did not have a significant association with the seizure outcome of the patients either (p = .574; t-test). In a subanalysis with patients who had JME and at least 12 months of follow-up at our center ( N = 180), similar results were obtained: The presence of GSW did not have a significant association with the seizure outcome ( p = .342; degree of freedom = 1; Fisher’s exact test), and frequency of GSW did not have a significant association with the seizure outcome either ( p = .725; t-test). For other syn-dromes, the numbers were small for any syndrome-specific statistical analysis.
4 DISCUSSION
The major role players in generating typical spike–wave discharges (≥2.5 Hz) are selective bidirectional thalamo-cortical communications (i.e., circuits and networks). Spike–waves can be deemed a final com-mon brain output, resulting from an inherent tendency of circuits and networks to oscillate between excitatory and inhibitory activity; this may be triggered by a variety of different pathophysiological etiologies (Blumenfeld, 2005; McCafferty et al., 2018). Therefore, it is expected to see a variety of frequencies of GSW in different patients with IGE; a variety of underlying etiologies and pathophysiological processes (e.g., different genetic problems) create a variety of thalamo-cortical circuit and network problems that manifest as GSW (as a final out-put), but with different frequencies. The exact mechanisms underlying GSW characteristics (e.g., their frequency, their abundance, and their duration) should be clarified in future studies. In the current study, we observed that GSWs (Figure 1) were commonly observed in all four syndromes of IGE, and although the presence of GSW had a significant statistical association with the syndromic diagnosis of patients, it was not discriminatory between the syndromes (e.g., 89.3% in CAE vs. 71.6% in JME) and, therefore, does not have any practical diagnostic implications. Furthermore, we observed that the frequency of GSW did not have a significant asso-ciation with the syndromic diagnosis of the patients. In a previous study, we reported that interictal EEG cannot differentiate between the seizure types in IGEs (Asadi-Pooya & Emami, 2012b). Moreover, it was previously shown that the morphology, amplitude, duration, fre-quency, occurrence, or activation of the GSWs did not differ between the classic adolescent-onset and the adult-onset IGEs (Yenjun et al., 2003). Another study reported that polyspike and waves do not pre-dict the presence of generalized tonic–clonic seizures in CAE (Vierck et al., 2010). We can conclude that GSW (variable frequencies) and polyspikes, alone or in combination, could be observed in various seizure types and syndromes of IGE. Although EEG is a very help-ful ancillary test to differentiate IGEs from other epilepsy syndromes (e.g., focal epilepsies), the key element in making a correct syndromic diagnosis is a detailed clinical history (Asadi-Pooya & Emami, 2012a; Asadi-Pooya, Emami, Ashjazadeh, et al., 2013). 4 of 5 ASADI-POOYA AND FARAZDAGHI
F I G U R E 1 Generalized 3.5 Hz spike–waves in a 16-year-old girl with juvenile absence epilepsy. A previous study reported that the density and duration of epilepti-form discharges can help differentiate among IGE syndromes (Senevi-ratne, Hepworth, et al., 2017). In a study of 105 patients, the authors reported that the density of epileptiform discharges and the paroxysm durations were the highest in JAE, followed by JME, CAE, and GTCA (no clear cutoff discriminatory value was provided). They reported that “pure” GSW paroxysms and fragments were observed in all four IGE syndromes (Seneviratne, Hepworth, et al., 2017). Again, although their findings were statistically significant, those observations could not have significant practical implications in daily clinical practice; there was no pathognomonic or characteristic EEG signature for any IGE syndrome. These kinds of statistically significant EEG differences between IGE syndromes (without clear practical and clinical implications) have also been reported by other groups (Sadleir et al., 2009). In the current study, we also observed that neither the presence of GSW nor their frequency had significant associations with the seizure outcome of the patients with IGE. This intriguing observation should be further explored in future studies. Having said the above, EEG may help with the prognostication in PWE. In 1 previous study of 105 patients with IGE, the authors reported that higher densities and longer durations of epileptiform discharges on EEG were associated with a shorter duration of seizure freedom (Seneviratne, Boston, et al., 2017). Another study showed that pro-longed epileptic discharges predict seizure recurrence and ASM failure in JME (Seneviratne, Boston, et al., 2017). One study of prognostic fac-tors for CAE ( n = 53) and JAE ( n = 27) showed that the presence of polyspikes during sleep was associated with poor prognosis in CAE; no statistical correlation was found for JAE (Bartolomei et al., 1997). In a recent study of 232 patients with IGE, increased GSW in sleep and the presence of generalized polyspike trains were associated with drug-resistance (Kamitaki et al., 2022). Therefore, although interictal EEG can potentially be used as a biomarker of prognosis in IGE syn-dromes, the mere presence or frequency of GSW does not provide a value here. Our study has significant limitations. This was a retrospective study. In addition, most patients (71.9%) were referred to us (often due to uncontrolled seizures for years). This may explain the observed out-come at 12 months of follow-up (only 37.4% were seizure-free of all seizure types). Although the proportion of patients who were not seizure free was quite high, the number of patients, who were on monotherapy, was also quite high (86%); this is because we inves-tigated their outcome during the first 12 months since their initial visit. Furthermore, the EEG was inspected visually only. In the cur-rent study, we visually measured and recorded the initial frequency of GSW that happened in wakefulness (unless they happened only during sleep); however, we did not record that in how many patients the EEG abnormality happened only during sleep. Finally, although the EEG was done before starting or switching the ASM, the possible effect of the previous ASM(s) was not evaluated.
5 CONCLUSION
GSWs are the hallmark EEG footprints of idiopathic generalized epilep-sies; however, neither their presence nor their frequency has practical associations with the syndromic diagnosis of IGEs or their outcome (response to treatment). A detailed clinical history is the key to make ASADI-POOYA AND FARAZDAGHI 5 of 5
a correct syndromic diagnosis, and this diagnosis establishes the foun-dation for the treating physician to decide on an appropriate treatment strategy as well as to explain the prognosis for the patient and their caregivers.
AUTHOR CONTRIBUTIONS Ali A. Asadi-Pooya : Study design; data collection; statistical analysis; and manuscript preparation. Mohsen Farazdaghi : Data collection and manuscript preparation.
ACKNOWLEDGMENTS
We thank Shiraz University of Medical Sciences for their support.
CONFLICT OF INTEREST STATEMENT
Ali A. Asadi-Pooya: Honoraria from Cobel Daruo, Ronak, and Rayman-dRad; Royalty: Oxford University Press (book publication); grant from the National Institute for Medical Research Development. Others: none.
DATA AVAILABILITY STATEMENT
The data would be shared as per the regulations of Shiraz University of Medical Sciences.
ORCID
Ali A. Asadi-Pooya
PEER REVIEW
The peer review history for this article is available at com/publon/10.1002/brb3.70023
INFORMED CONSENT
Informed consent for participation and publication of the data were obtained from all patients.
REFERENCES
Asadi-Pooya, A. A., & Emami, M. (2012a). Reasons for uncontrolled seizures in children: The impact of pseudointractability. Epilepsy & Behavior ,
25 (3), 341–344. j.yebeh.2012.08.015 Asadi-Pooya, A. A., & Emami, M. (2012b). Is interictal EEG correlated with the seizure type in idiopathic (genetic) generalized epilepsies? Iranian Journal of Child Neurology , 6(2), 25–28. Asadi-Pooya, A. A., Emami, M., Ashjazadeh, N., Nikseresht, A., Shariat, A., Petramfar, P., Yousefipour, G., Borhani-Haghighi, A., Izadi, S., & Rahimi-Jaberi, A. (2013). Reasons for uncontrolled seizures in adults; the impact of pseudointractability. Seizure: The Journal of the British Epilepsy Associ-ation , 22 (4), 271–274. 1016/j.seizure.2013.01.010 Asadi-Pooya, A. A., Emami, M., & Sperling, M. R. (2013). A clinical study of syndromes of idiopathic (genetic) generalized epilepsy. Journal of the Neurological Sciences , 324 (1–2), 113–117. 2012.10.01410.1016/j.jns.2012.10.014 Bartolomei, F., Roger, J., Bureau, M., Genton, P., Dravet, C., Viallat, D., & Gastaut, J.-L. (1997). Prognostic factors for childhood and juvenile absence epilepsies. European Neurology , 37 (3), 169–175. 10.1159/00011742910.1159/000117429 Blumenfeld, H. (2005). Cellular and network mechanisms of spike-wave seizures. Epilepsia , 46 (9), 21–33. 2005.00311.x10.1111/j.1528-1167.2005.00311.x Cerulli Irelli, E., Leodori, G., Morano, A., & Di Bonaventura, C. (2022). EEG markers of treatment resistance in idiopathic generalized epilepsy: From standard EEG findings to advanced signal analysis. Biomedicines , 10 (10), 2428. Hirsch, E., French, J., Scheffer, I. E., Bogacz, A., Alsaadi, T., Sperling, M. R., Abdulla, F., Zuberi, S. M., Trinka, E., Specchio, N., Somerville, E., Samia, P., Riney, K., Nabbout, R., Jain, S., Wilmshurst, J. O. M., Auvin, S., Wiebe, S., Perucca, E., . . . Wirrell, E. C. (2022). ILAE definition of the idiopathic gen-eralized epilepsy syndromes: Position statement by the ILAE task force on nosology and definitions. Epilepsia , 63 (6), 1475–1499. 10.1111/epi.1723610.1111/epi.17236 Kamitaki, B. K., Janmohamed, M., Kandula, P., Elder, C., Mani, R., Wong, S., Perucca, P., O’brien, T. J., Lin, H., Heiman, G. A., & Choi, H. (2022). Clin-ical and EEG factors associated with antiseizure medication resistance in idiopathic generalized epilepsy. Epilepsia , 63 (1), 150–161. org/10.1111/epi.1710410.1111/epi.17104 Mccafferty, C., David, F., Venzi, M., L ˝ orincz, M. L., Delicata, F., Atherton, Z., Recchia, G., Orban, G., Lambert, R. C., Di Giovanni, G., Leresche, N., & Crunelli, V. (2018). Cortical drive and thalamic feed-forward inhibition control thalamic output synchrony during absence seizures. Nature Neu-roscience , 21 (5), 744–756. Sadleir, L. G., Scheffer, I. E., Smith, S., Carstensen, B., Farrell, K., & Connolly, M. B. (2009). EEG features of absence seizures in idiopathic generalized epilepsy: Impact of syndrome, age, and state. Epilepsia ,
50 (6), 1572–1578. x10.1111/j.1528-1167.2008.02001.x Seneviratne, U., Boston, R. C., Cook, M., & D’souza, W. (2017). EEG correlates of seizure freedom in genetic generalized epilepsies. Neu-rology Clinical Practice , 7(1), 35–44. 000000000000032310.1212/CPJ.0000000000000323 Seneviratne, U., Hepworth, G., Cook, M., & D’souza, W. (2017). Can EEG dif-ferentiate among syndromes in genetic generalized epilepsy? Journal of Clinical Neurophysiology , 34 (3), 213–221. 000000000000035810.1097/WNP.0000000000000358 Vierck, E., Cauley, R., Kugler, S. L., Mandelbaum, D. E., Pal, D. K., & Durner, M. (2010). Polyspike and waves do not predict generalized tonic-clonic seizures in childhood absence epilepsy. Journal of Child Neurology ,
25 (4), 475–481. 0883073809341665 Yenjun, S., Harvey, A. S., Marini, C., Newton, M. R., King, M. A., & Berkovic, S. F. (2003). EEG in adult-onset idiopathic generalized epilepsy. Epilepsia ,
44 (2), 252–256.
How to cite this article: Asadi-Pooya, A. A., & Farazdaghi, M. (2024). Generalized spike–waves in idiopathic generalized epilepsies: Does their frequency matter? Brain and Behavior ,
14 , e70023. |
4426 | https://ximera.osu.edu/la/LinearAlgebra/VEC-M-0050/main | Warning
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Mathematical Expression Editor
We define the dot product and prove its algebraic properties.
VEC-0050: Dot Product and its Properties
Let and be vectors in . The dot product of and , denoted by , is given by
Find if and .
Note that the dot product of two vectors is a scalar. For this reason, the dot product is sometimes called a scalar product.
Properties of the Dot Product
A quick examination of Example ex:dotex will convince you that the dot product is commutative. In other words, . This and other properties of the dot product are stated below.
The following properties hold for vectors , and in and scalar .
(a)
(b)
(c)
(d)
(e)
: , and if and only if .
(f)
We will prove Property item:distributive. The remaining properties are left as exercises.
Proof of Property item:distributive:
We will illustrate Property item:norm with an example.
Let . Use to illustrate Property item:norm of Theorem th:dotproductproperties.
Practice Problems
Find the dot product of and if Answer:
Find the dot product of and if Answer:
Use vector to illustrate Property item:norm of Theorem th:dotproductproperties.
Prove Properties item:commutative, item:distributive-again, item:scalar, item:positive and item:norm of Theorem th:dotproductproperties.
From the given list of vector pairs, identify ALL pairs of vectors that lie on perpendicular lines.
You may want to draw a picture and think about what you know about slopes of perpendicular lines.
, , , ,
Compute for each pair. What do you observe?
For each part below
(a)
: Find the value of that will make vectors and perpendicular.
Think of the -component as the “run” and the -component as the “rise”, then use what you know about slopes of perpendicular lines.
(b)
: Find .
Answer:
Answer:
(a)
: Vector that lies on the line , has the form . Assuming that , find the general form for a vector that lies on a line perpendicular to .
What do you know about the slopes of perpendicular lines?
(b)
: Find .
(c)
: Formulate a conjecture about the dot product of perpendicular vectors. |
4427 | https://www.youtube.com/watch?v=DTgZ0Csk0sY | Stewart's Calculus Chapter 6 - Inverse, exponential, and logarithmic differentiation formulae
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hey so this is Joe and this is the first video for chapter six in Stewart's calculus and uh here I'm going to be talking about inverse exponential and logarithmic functions with Calculus and basically um giving you and then applying a bunch of new differentiation formula that we have um for some new functions now your book spends a lot of time going over um things like how to graph exponential and logarithmic fun fun properties of logarithms properties of exponential functions how to take the inverse of a function and if any of you are having trouble with that or need a refresher um look at our Sullivan and Sullivan pre-calculus textbook videos um and you should be able to get some help with that and then come back to this video because I'm not going to go over that stuff here instead um before so the main thing I'm going to be doing with this video is basically showing you how to apply a lot of the formula that are um presented to you in the book but before I do that because this chapter is so long and there's a lot of stuff that um you'll probably remember from previous courses I'm just going to highlight the new stuff really quickly right here all right so first we have this little identity that it it doesn't come in handy too often but conceptually it's pretty easy if ever you have to deal or helpful I should say if ever you have to deal with um inverse functions so an inverse function and um it's where you have a function in terms of y and x and basically you do a one toone matching of all the Y's for the x's and basically flip them and there's some fancy ways you can do that not very fancy but you can you did that in pre-cal however let's say you want to take the derivative of an inverse function well you could always just find that inverse function and take its derivative or if it's tough to take the inverse function what you can do is you can say that that inverse function the derivative is going to be equal to one over the derivative of the function the original function itself but you plug in the inverse function into that function and so you kind of res switch the uh the variables and intuitively that makes sense because if we write things out like um the derivative of y with respect to X that's equal to one over the derivative of x with respect to Y so you just need to put everything in X's take the derivative with respect to Y and then invert that and you'll get the derivative of y with respect to X or Y is um the uh y of the function um the inverse function so two we've got a very important differentiation formula which is that the derivative of e to the X and E is the natural exponential function is equal to e to the X and E this is what makes e to the X and natural logarithms which is a logarithm with a base e so special because the slope of e to the X all the slopes are equal to the um the the values of Y its derivative is equal to to itself so that's what makes this so handy um in terms of dealing with other more difficult functions to differentiate and so just uh bearing that in mind um we have two more definitions of e which come in handy when you're solving limits e is equal to the Limit as X approach to Zer of 1+ x to the 1 /x so if ever you see that in a limit you know that that equals e that's just something you should memorize and then this is M why it's because of this limit that e is so um important in finance that e is equals to the Limit as n approaches Infinity of 1 + 1/ n to the N is e is used in Computing continuously compounded interest and um it's used for that because e is the the number that occurs when you continuously um multiply this quantity and so you can you can make this number uh really really big and continue to multiply it by itself and it'll always converge at e and uh we we'll explain more kind that sounded kind of confusing but uh it'll come in handy I think later in this chapter at the beginning of the next chapter now the next differentiation formula is for logarithms General logarithms with the base um other than e and so the derivative of that is one over x the natural log of um whatever was in the base and that's because of this identity that the log log base a of X is equal to the natural log of x over the natural log of a now technically you could this natural log could be a log with any other kind of Base um technically but that would make a so much algebra and all these other formula so much more difficult so that's why we use the natural log um now the reason we use the natural log is because of this that you take the derivative of the natural log and it's only equal to 1 /x because the natural log of e is one and remember natural log is a a log with a base e so two properties that are important to memorize and just remember and bear in mind when you're solving problems with logarithms and with a natural exponential functions is that e to the natural log of x is equal to X or for that matter any number to the power of a logarithm logarithm of the base of that number is equal to whatever is inside of the logarithm and uh natural log of e is 1 so now I'm going to do some practice problems involving all these differentiation formula and applying the chain rule to all these differentiation formula and the chain rule applies to these in the same way it applies to trig functions and all this if this X for example was replaced with a some function of X the right there the derivative would be e to whatever is in the exponent times the derivative of whatever is in the exponent and that's just how the chain will ends up working out and I'm going to write it out more explicitly later but I hope you see now that the reason and we're going to be also doing um using natural logarithms to kind of break up more complicated functions to take the derivative of them based on the fact that we're able to um based on some of the properties of natural logarithms but we're going to be using specifically natural logarithms instead of logarithms with the base of something else just because you doing calculus with them is so easy because of the fact that the derivative of e to X is itself and that is why we're going to be using natural logarithms as opposed to other logarithms so much more before we go on to solve some more problems I just need to show you this one last differentiation formula which can be derived from the previous one for the derivative of um uh a logarithm with a base a but if we're taking the derivative of a function where it's some constant to um the power of X or some function of X um the derivative is uh the the original function times the natural logarithm of uh the constant the base right here and then should this uh X actually be a function of X based on the the chain rule you'd um multiply everything by the derivative of that function all right so now for the problems all right so number 17 on page 418 in your book B asks us to differentiate this formula FX = x 5 + 5 x so the first term we've been doing this for a while that is 5X to 4th because the variable is in the base and the exponent is a constant now we're going to have to use a new differentiation formula for for this one so we take we have the function we write down the function itself again and then multiply it by the natural log of uh the base and so that's five so that is the derivative however let's break off from the book a little bit and make this x a function of X all right so let's say we Define a function G of x to equal 2 to the X and then we redefine f ofx up here to be f ofx = x 5th + 5 to the G ofx now let's apply the chain rule to solve that and remember the chain rule even though there is a um a longer form of substitutions that we can make um intuitive all you have to remember when you're actually solving these formula or these problems is that you have to keep take if there if there are more functions in specific places you need to keep taking the derivatives of those and then multiplying them by uh by the derivative of the greater thing you have to keep going inside and taking derivatives so take the derivative of f we'll still get um five x 4 plus let's do the same thing as before this is an exponential function so we'll write the function itself again 5 to the G of X same as we did up here times the natural log of five but now we have to take the derivative of G so just times G Prime of X and we can actually substitute in for G ofx now so that's 5 x 4 + 5 to the 2 the x the natural log of 5 and now we have to take the derivative of this well it's an exponential function so we write 2 x the natural log of the base which is the natural log of two and there we go that is our derivative all right so now we're going to solve number nine from page 418 as well and it it'll we'll have to use the the the product rule and the chain rule here so it's kind of combining everything as well as a new um differentiation formula so take the derivative of F and we have these two functions that are multiplied by one another so we'll take the first one take the derivative of it leave the second one alone derivative of s is cine X time the natural log of 5x plus now we leave this one alone we leave the sin x alone you take the derivative of this um which is one over whatever's inside 5x but because 5x is a function and we can take the derivative of that um and it's something other than one uh we take that derivative and make it a coefficient and so we take that five and put it out here and that is that derivative all right so now we're going to do problem number 45 which requires us to use a technique called logarithmic differentiation which is actually the reason why um natural logarithms are so helpful in calculus and why they're such an important function to be able uh to deal with because they help us break down nasty looking functions like this one f ofx = y which is equal to theare < TK of x - 1/ x 4 + 1 that's a really ugly function we've got a radical we've got a quotient we've got x's in both terms it's a it's a mess we'd need to do a ton of substitution a ton of work if we weren't to use this technique so in the same way algebra you can add the same thing to two sides of an equation you can take the logarithm of both sides of an equation in this case we're going to take the natural logarithm of both sides because they're easier to deal with um in terms of taking derivatives and integrals and calculus in general so we're going to get the natural logarithm of Y and remember that the object of this is to get um fime of X or Y Prime so the derivative with respect to X is equal to the natural log of that functions x -1 x 4 + 1 all square rooted so first off we can use our properties of logarithms to get rid of a lot of this stuff so the entire argument of the logarithm is to the power of 1/2 so we can bring a 1/2 out here and then the logarithm of a quotient like that is actually one logarithm subtracted from the other specifically logarithm of x -1 minus the natural log of x 4 + 1 and that's all equal to the natural log of Y but we need to get y Prime out of here well we can do some implicit differentiation if you remember how to do that so that when we take the derivative is equal to the uh the derivative of y chain rule over y because the natural logarithm of 1 /x right but in this case it's Y and we have to have the Y Prime because of the chain Rule and that's equal to 12 can take that out as a coefficient now we take the derivative of this it's one over the argument and the derivative of the argument is one so we can just leave it as is 1/x - one minus now the derivative of the argument here is 4 x 3rr that's going to be times one over the whole thing X 4th + 1 now we can simplify all this and eventually we're going to be need to multiply this side by Y what is y though because we want this to be a derivative in terms of X well we have an expression for y up here it's just the original function so we can rewrite this whole thing as y Prime equals root x -1 x 4th + 1 and then you we distribute that two and we put everything in parentheses so you get 1 over 2 the xus one [Music] minus 2x 3r over x 4 + 1 and that is the derivative with respect to X using the technique of logarithmic differentiation all right so I haven't forgot about integration and remember all of those differentiation formula apply to integration just remember you have to think okay given this expression that I need to integrate what would the derivative of it be and then you can kind of manipulate those differentiation formula accordingly so sometimes instead of multiplying you'd be dividing by one thing Etc and that's a better way to think of it I think than um memorizing what like four or five more formulas just memorize the differentiation formula and then apply them to integration and going the other way and also you you're going to be more likely to make uh fewer errors because um uh you'll be thinking more critically about your problems so with this problem we're going to need to make a U substitution uh namely and we we've got to see since there's a um fraction here we're probably going to end up making this some sort of logarithm um so bear that in mind so we probably want the numerator to end up going away so let's substitute let's make this expression u u is equal to 2 + sin x so then du would be is cosine so du is cosine X DX well we see a cosine X DX and we see the U in there so we can write this all in terms of U we've got du over U and that looks like something we can evaluate with a logarithm and so since this is an indefinite integral we can write it as the natural log of and be careful now it has to be the absolute value of U plus C and we use these absolute value signs because if you picture the graph of a natural logarithm it never it doesn't get Negative you can't have um Negative X value values or negative arguments the argument must be positive so on the AP they will take off points if you don't explicitly have these absolute value signs and so then now this we're able to put a substitute back in for you is the natural log of 2 + sin X+ C and that is that integral all right so that would that's it for this video in the first section of chapter 6 stay tuned for the rest of the videos where we go more in depth into specific applications of calculus to um exponential growth and Decay problems uh and also we're going to get to that fancy technique about limits that makes limit solving a whole lot easier that I mentioned back at the beginning of this course uh yeah so thanks stay tuned |
4428 | https://www.merckmanuals.com/professional/pediatrics/inherited-disorders-of-metabolism/gaucher-disease | honeypot link
IN THIS TOPIC
Diagnosis
Treatment
Key Points
More Information
OTHER TOPICS IN THIS CHAPTER
Introduction to Inherited Disorders of Metabolism
Approach to the Patient With a Suspected Inherited Disorder of Metabolism
Mitochondrial Oxidative Phosphorylation Disorders
Peroxisomal Disorders
Overview of Amino Acid and Organic Acid Metabolism Disorders
Branched-Chain Amino Acid Metabolism Disorders
Hartnup Disease
Methionine Metabolism Disorders
Phenylketonuria (PKU)
Tyrosine Metabolism Disorders
Urea Cycle Disorders
Overview of Carbohydrate Metabolism Disorders
Fructose Metabolism Disorders
Galactosemia
Glycogen Storage Diseases
Pyruvate Metabolism Disorders
Other Carbohydrate Metabolism Disorders
Overview of Fatty Acid and Glycerol Metabolism Disorders
Beta-Oxidation Cycle Disorders
Glycerol Metabolism Disorders
Overview of Lysosomal Storage Disorders
Cholesteryl Ester Storage Disease and Wolman Disease
Fabry Disease
Gaucher Disease
Krabbe Disease
Metachromatic Leukodystrophy
Niemann-Pick Disease
Tay-Sachs Disease and Sandhoff Disease
Overview of Purine and Pyrimidine Metabolism Disorders
Purine Catabolism Disorders
Purine Nucleotide Synthesis Disorders
Purine Salvage Disorders
Pyrimidine Metabolism Disorders
Gaucher Disease
(Gaucher's Disease)
By
Matt Demczko, MD, Mitochondrial Medicine, Children's Hospital of Philadelphia
Reviewed By
Michael SD Agus, MD, Harvard Medical School
Reviewed/Revised Modified Mar 2024
v25255593
View Patient Education
Gaucher disease is a
sphingolipidosis
, an inherited disorder of metabolism, resulting from glucocerebrosidase deficiency, causing deposition of glucocerebroside and related compounds. Symptoms and signs vary by type but are most commonly hepatosplenomegaly or central nervous system changes. Diagnosis is by DNA analysis and/or enzyme analysis of white blood cells. Treatment is enzyme replacement with glucocerebrosidase.
Diagnosis
|
Treatment
|
Key Points
|
More Information
|
For more information, see table
Some Sphingolipidoses
.
See also
Approach to the Patient With a Suspected Inherited Disorder of Metabolism
.
Glucocerebrosidase normally hydrolyzes glucocerebroside to glucose and ceramide. Genetic defects of the enzyme cause glucocerebroside accumulation in tissue macrophages through phagocytosis, forming Gaucher cells. Accumulation of Gaucher cells in the perivascular spaces in the brain causes gliosis in the neuronopathic forms.
There are 3 types of Gaucher disease, which vary in epidemiology, enzyme activity, and manifestations.
Type I Gaucher disease
Type I (nonneuronopathic) is most common (90% of all patients) (
1
). Residual enzyme activity is highest. People of Ashkenazi Jewish ancestry are at greatest risk; 1/12 is a carrier. Onset ranges from childhood to adulthood.
Symptoms and signs of type I Gaucher disease include splenohepatomegaly, bone disease (eg, osteopenia, pain crises, osteolytic lesions with fractures), growth failure, delayed puberty, ecchymoses, and pingueculae. Epistaxis and ecchymoses resulting from thrombocytopenia are common.
Radiographs show flaring of the ends of the long bones (Erlenmeyer flask deformity) and cortical thinning.
Type II Gaucher disease
Type II (acute neuronopathic) is rarest, and residual enzyme activity in this type is lowest. Onset occurs during infancy.
Symptoms and signs of type II Gaucher disease are progressive neurologic deterioration (eg, rigidity, seizures) and death by age 2 years.
Type III Gaucher disease
Type III (subacute neuronopathic) falls between types I and II in incidence, enzyme activity, and clinical severity. Onset occurs at any time during childhood.
Clinical manifestations vary by subtype and include progressive dementia and ataxia (IIIa), bone and visceral involvement (IIIb), and supranuclear palsies with corneal opacities (IIIc). Patients who survive to adolescence may live for many years.
Reference
Hughes DA, Pastores GM
. Gaucher Disease. In: Adam MP, Feldman J, Mirzaa GM, et al., eds.
GeneReviews
®. Seattle (WA): University of Washington, Seattle; July 27, 2000.
Diagnosis of Gaucher Disease
Enzyme analysis
Diagnosis of Gaucher disease is by DNA analysis and/or enzyme analysis of white blood cells. Carriers are detected, and types are distinguished by mutation analysis.
Although biopsy is unnecessary, Gaucher cells—lipid-laden tissue macrophages in the liver, spleen, lymph nodes, bone marrow, or brain that have a wrinkled tissue-paper appearance—are diagnostic. (See also
testing for suspected inherited disorders of metabolism
.)
Treatment of Gaucher Disease
Types I and III: Enzyme replacement with glucocerebrosidase
Sometimes
miglustat
,
eliglustat
, splenectomy, or stem cell or bone marrow transplantation
Enzyme replacement with IV glucocerebrosidase is effective in types I and III; there is no treatment for type II. The enzyme is modified for efficient delivery to lysosomes. Patients receiving enzyme replacement require routine hemoglobin and platelet monitoring, routine assessment of spleen and liver volume by CT or MRI, and routine assessment of bone disease by skeletal survey, dual-energy x-ray absorptiometry scanning, or MRI.
Miglustat
, an oral glucosylceramide synthase inhibitor, reduces glucocerebroside concentration (the substrate for glucocerebrosidase) and is an alternative for patients unable to receive enzyme replacement.
Miglustat, an oral glucosylceramide synthase inhibitor, reduces glucocerebroside concentration (the substrate for glucocerebrosidase) and is an alternative for patients unable to receive enzyme replacement.
Eliglustat
, another oral glucosylceramide synthase inhibitor, also reduces glucocerebroside concentration.
Eliglustat, another oral glucosylceramide synthase inhibitor, also reduces glucocerebroside concentration.
Splenectomy may be helpful for patients with anemia, leukopenia, or thrombocytopenia or when spleen size causes discomfort. Patients with anemia may also need blood transfusions.
Bone marrow transplantation or stem cell transplantation provides a definitive cure but is considered a last resort because of substantial morbidity and mortality.
Key Points
Gaucher disease is a sphingolipidosis resulting from glucocerebrosidase deficiency, causing deposition of glucocerebroside.
There are 3 types, which vary in epidemiology, enzyme activity, and manifestations.
Symptoms and signs vary by type but are most commonly hepatosplenomegaly or central nervous system changes.
Diagnosis of Gaucher disease is by DNA analysis and/or enzyme analysis of white blood cells; carriers are detected, and types are distinguished by mutation analysis.
Treatment for types I and III includes enzyme replacement with glucocerebrosidase, and sometimes
miglustat
,
eliglustat
, splenectomy, or stem cell or bone marrow transplantation; there is no treatment for type II.
Treatment for types I and III includes enzyme replacement with glucocerebrosidase, and sometimes miglustat, eliglustat, splenectomy, or stem cell or bone marrow transplantation; there is no treatment for type II.
More Information
The following English-language resource may be useful. Please note that THE MANUAL is not responsible for the content of this resource.
Online Mendelian Inheritance in Man (OMIM) database
: Complete gene, molecular, and chromosomal location information
Drugs Mentioned In This Article
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4429 | https://cdn.kutasoftware.com/Worksheets/Alg2/Direct%20and%20Inverse%20Variation.pdf | ©G I2u0c2s2G IK^uJtyav SSfoKfWthwFarNes LLwLsCT._ z uAFl^lb RrdiTgfhGtps] braegsjezrAvietdO.T q KMhaFdIey NwPiXtuhp xIbngfWifnzihtDeL SAhl[gxecbqrQaV Q2e.
Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Direct and Inverse Variation Name___________________________________ Date________________ Period____ -1-Determine whether the given equation represents a direct or inverse variation.
1) xy = 15 2) y = 9 x 3) y = 13 12x 4) y = 13x 5) -5x + y = 0 6) y = 4x Find the constant of variation.
7) y = 3x 8) xy = 7 9) y x = 5 10) y = 1 9x 11) y = 3 5x 12) y = 15 x ©Q p2L0\2v2B FKiuktkaA NSroEfztjwFaMr^eH uLrLMCO.N t mA^lMlC IrziTgthntHsl krzeqsQeDrAvRezdj.D q aMoaBdKen iwsiTtDhW XISnNfxidnHiBtAeU FAklqgSeFbMr_ax c2L.
Worksheet by Kuta Software LLC -2-Solve each problem involving direct or inverse variation.
13) If y varies directly as x, and y = 6 when x = 15, find y when x = 2.
14) If y varies inversely as x, and y = 8 when x = 5, find y when x = 4.
15) If y varies directly as x, and y = 5 when x = 4, find y when x = 8.
16) If y varies directly as x2, and y = 10 when x = 2, find y when x = 3.
17) If y varies inversely as x, and y = 9 when x = 10, find y when x = 5.
18) If y varies inversely as x, and y = 4 when x = 12, find y when x = 2.
19) If y varies inversely as x, and y = 3 when x = 21, find y when x = 9.
20) If y varies inversely as x2, and y = 11 4 when x = 4, find y when x = 2.
21) If y varies directly as x, and y = 10 when x = 20, find y when x = 3.
22) If y varies directly as x, and y = 4 when x = 6, find y when x = 5.
©S m2R0K2d2P HKNuRtBaF ^Sso]f_towCaTr[eY FL^LvC\.k v IAWlilR HrWiPgXhtksP frpessCedrwvieJdp.D V DMKa^dNe[ cwoictqhw CINncfJijngikt]eN lAklVgneTbPrqaT T2_.
Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Direct and Inverse Variation Name_____ Date____ Period____ -1-Determine whether the given equation represents a direct or inverse variation.
1) xy = 15 Inverse 2) y = 9 x Inverse 3) y = 13 12x Inverse 4) y = 13x Direct 5) -5x + y = 0 Direct 6) y = 4x Direct Find the constant of variation.
7) y = 3x 3 8) xy = 7 7 9) y x = 5 5 10) y = 1 9x 1 9 11) y = 3 5x 3 5 12) y = 15 x 15 ©u ^2]0i2X2w bKqujtdak YSlotfgtkw_akryer rLFLFCZ.P QAMlQlD eriHgDhltsV frNeOsnerreviexds.J r EMOaZdkee DwOiKtuhC YIDnhfsiRnRictdef HAwlbgXeQb]rlav o2.
Worksheet by Kuta Software LLC -2-Solve each problem involving direct or inverse variation.
13) If y varies directly as x, and y = 6 when x = 15, find y when x = 2.
4 5 14) If y varies inversely as x, and y = 8 when x = 5, find y when x = 4.
10 15) If y varies directly as x, and y = 5 when x = 4, find y when x = 8.
10 16) If y varies directly as x2, and y = 10 when x = 2, find y when x = 3.
45 2 17) If y varies inversely as x, and y = 9 when x = 10, find y when x = 5.
18 18) If y varies inversely as x, and y = 4 when x = 12, find y when x = 2.
24 19) If y varies inversely as x, and y = 3 when x = 21, find y when x = 9.
7 20) If y varies inversely as x2, and y = 11 4 when x = 4, find y when x = 2.
11 21) If y varies directly as x, and y = 10 when x = 20, find y when x = 3.
3 2 22) If y varies directly as x, and y = 4 when x = 6, find y when x = 5.
10 3 Create your own worksheets like this one with Infinite Algebra 2. Free trial available at KutaSoftware.com |
4430 | https://www.youtube.com/watch?v=A3f16Thfjhs | Limit as x approaches zero|Made easy
Transcended Institute
64400 subscribers
81 likes
Description
8213 views
Posted: 3 Sep 2022
In this video we will review the point of limit as it approaches zero from the left side and from the right side
Transcript:
so how can we solve this if we have limit as x approaches 0 and then we have 2x squared plus x divided by the modulus of x okay how can you solve this we know that if we plug in the value of x directly we are going to get zero over zero but remember the rules of limits we are trying to avoid the situation where we have a number divided by zero zero over zero the square root of what is negative okay so we are trying to avoid this situation because this duration is undefined okay so this question isn't defined so we're trying to run away from undefined we want this function to be defined okay so now we need to say when you talk of a modulus when you talk of x and modulus we are talking about plus or minus x okay now sometimes you can think of factorizing x then then you can't factor out x then because this is the modulus the modulus means we have plus or minus we really don't know if it was positive or negative okay so now if we don't know whether it was positive or negative this is what we're supposed to do we are going to test limit as x approaches zero from the left side as well as limit as x approaches zero from the right side so now limit as x approaches zero from the left side we are going to get as a negative part because when you talk of if this is the number line okay so this part is zero now from this point going this side is our left meaning that's the reason why we're going to get negative then from this point going this side is light side which is positive okay so now what we're going to do here is limit as x approaches zero we're going to test two things so let's let's put this guy here we have plus or minus x so what we're going to do here we're going to say limit as x approaches 0 from the left side so from the left side meaning it's supposed to be negative we are going to get x value as negative so we're going to say it's going to be 2x squared plus x divided by y t negative x at this point now i can i can factor out x on top so that i cancel x so i'm going to have limit as x approaches 0 from the left side i'm going to have x up in x plus one okay now here i'm going to divide one negative x so at this point we can cancel x now down there we're going to remain with 30 a negative so we're going to have limit as x approaches 0 from the left side then we are going to have 2x plus 1 divided by we have remain with negative 1 which is the same as the limit as x approaches 0 from the left side if i divide this negative on top there i'm going to get negative 2x minus 1 now at this point i can plug in the value of what the value of x so this is what we're going to have so we're going to have limit as x approaches z now we we are supposed to stop now introducing limit when you want to plug in the value of x that's where now you stop now introducing the limit but make sure whenever you are dealing with limits try by all means to bring in this limit until you reach at the part where you want to plug in the value of x okay so now at this point i'm going to say this one is going to be equal to i want to plug in the value of x so now x is 0 so i'm going to have negative 2 then i plug in 0 there i have this so i'm going to get 0 minus 1 which is going to give me what which is going to give me negative 1. so now as x approaches z zero from the left side we are getting it as a negative one let's try a limit as x approaches z zero from the right side if we're going to get also negative if we're going to get negative meaning the limit exists but if you're going to get something else then the limit doesn't exist okay so let's try let's let's do it from this point let me just get rid of this i can even write it from here we are saying that this guy is going to plus or minus x so now we are saying that in limit as x approaches 0 from the right side we have got our equation squared plus x divided by what now since it is a limit as x approaches zero from the right side we are going to put positive x we're going to just put x here okay so now from there we can factor out x on top it's going to be limit as x approaches 0 from the right side then we can factor out x if i out x there what we're going to have we're going to have x up in brackets 2x plus 1 we divide x let's cancel x so we cancel x and then we are going to have limit as x approaches zero from the left uh like the right side we have got two x plus 1 but this point i can now plug in the value of x so i'm going to say i i put 0 plus 1 so i'm going to get 0 plus what one so the answer which i'm going to get is 1. so as you can see limit as x approaches zero from the left side is giving us negative one limit as x approaches zero from the right sides giving us one which is not the same so therefore this limit doesn't exist so the statement which you are supposed to understand we are going to say that you put it like this we are going to say we are going to say limit as x approaches 0 from the left side of f of x is not equal to limit as x approaches 0 from the left side or right side f of x so the limit doesn't exist doesn't exist that's what we need to understand okay |
4431 | https://www.hpsj.com/clinical-review-update-concomitant-anticholinergic-and-antipsychotic-use/ | Concomitant Anticholinergic and Antipsychotic
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Home/Alerts, Pharmacy Announcements/Clinical Review Update: Concomitant Anticholinergic and Antipsychotic Use
In The News
Clinical Review Update: Concomitant Anticholinergic and Antipsychotic Use
Post date: September 12, 2019/Alerts, Pharmacy Announcements
Learning Objectives:
Understand the role of anticholinergic medications in the prevention and treatment of antipsychotic-induced extrapyramidal symptoms (EPS).
Describe factors that should be considered when deciding to initiate and/or continue the concomitant use of anticholinergic with antipsychotic medication therapy.
Key Points:
Anticholinergic medications, including benztropine and trihexyphenidyl, are often prescribed to prevent or treat antipsychotic-induced EPS; however, the need for continued therapy with anticholinergics is not often reassessed and many patients continue to use these medications.
The consensus among the medical community is that prophylaxis of EPS with anticholinergics is generally not indicated in patients receiving antipsychotics, in particular among patients who are prescribed second-generation antipsychotics.
Long-term use of anticholinergic medications is associated with cognitive impairment and worsening of tardive dyskinesia, especially among persons 65 years of age or older.
Among the 268,245 Medi-Cal beneficiaries with a paid claim for any antipsychotic medication during the measurement year, a total of 29,807 (11%) beneficiaries had concomitant use of anticholinergic medication during this period. Among this population, a total of 15,487 (6%) beneficiaries also had at least six paid claims for an anticholinergic medication during this same time period, suggesting long term use of concomitant anticholinergic and antipsychotic medications.
Concomitant use of anticholinergic medications was higher among the 23,191 Medi-Cal beneficiaries with at least one paid claim for a first-generation antipsychotic medication (n = 9,767; 42%), in comparison to the 260,655 Medi-Cal beneficiaries with paid claims exclusively for second-generation antipsychotic medications (n = 27,137; 10%).
Continued use of anticholinergic medications should be re-evaluated in patients with controlled symptoms every three months.
BackgroundAnticholinergic medications, including benztropine and trihexyphenidyl, are often prescribed to prevent or treat antipsychotic-induced EPS, including tremor, rigidity, bradykinesia, and acute dystonia.1 However, the need for continued therapy with anticholinergics is not often reassessed and many patients continue to use these medications for several years or even decades.1 Prescribers may be reluctant to discontinue anticholinergics even when patients are prescribed second-generation antipsychotics, which are less likely than first-generation antipsychotics to induce EPS.1-5
Despite the widespread use of anticholinergic medications for prophylaxis and treatment of antipsychotic-induced EPS, there is a lack of systematic reviews and meta-analyses supporting this practice, and the long-term benefit of anticholinergic use has not been established.1,6 In fact, several adverse effects have been reported from long-term use, including cognitive impairment and worsening of tardive dyskinesia, especially among persons 65 years of age or older.5,7,8 The 2009 Schizophrenia Patient Outcomes Research Team (PORT) treatment recommendations state that the prophylactic use of anticholinergics to reduce the incidence of EPS was not warranted in patients treated with second-generation antipsychotics but should be evaluated on a case-by-case basis for patients treated with first-generation antipsychotics.9,10
Table 1. General Ranking of Selected First- and Second-Generation Antipsychotics by Propensity for EPS9,10,12-15
High potency first-generation antipsychotics: Fluphenazine, haloperidol, pimozide, thiothixene, trifluoperazine Highest propensity for EPS
Lowest propensity for EPS
Mid potency first-generation antipsychotics: Loxapine, perphenazine
Second-generation antipsychotics: Paliperidone, risperidone
Second-generation antipsychotics: Asenapine, cariprazine, lurasidone
Low potency first-generation antipsychotics: Chlorpromazine, thioridazine
Second-generation antipsychotics:
Aripiprazole, brexpiprazole, olanzapine, ziprasidone
Second-generation antipsychotics: Quetiapine, iloperidone, pimavanserin
Second-generation antipsychotics: Clozapine
Summary of Current Treatment Guidelines for Prophylactic Use of Anticholinergic Medications 9,10,16-18
Current treatment guidelines describe the following factors that should be considered in decisions regarding the prophylactic use of anticholinergic medications in acute-phase treatment:
Propensity of the antipsychotic medication to cause EPS
Patient preferences
Patient’s prior history of EPS
Other risk factors for EPS (especially dystonia)
Risk factors for and potential consequences of anticholinergic side effects
Use of Anticholinergic Medications in the Medi-Cal Population
A retrospective cohort study was conducted to evaluate the use of anticholinergic medications in the Medi-Cal population. All paid pharmacy claims for benztropine and trihexyphenidyl for dates of service from January 1, 2018, through December 31, 2018, were reviewed. Beneficiaries were then evaluated for concomitant use of antipsychotic medications during the same measurement year. Data were then stratified by concomitant use of first- or second-generation antipsychotics, with additional analyses conducted by individual antipsychotic medication. Descriptive statistics were used to summarize data into tables. Data analyses were performed using IBM® SPSS®, version 26.0 (Chicago, IL).
ResultsAcross all age groups, there were 31,118 unique beneficiaries identified with a paid claim for benztropine and/or trihexyphenidyl that has a days’ supply greater than or equal to 30 days during the one-year measurement period. The majority of these beneficiaries (n = 29,006; 93%) had a paid claim for benztropine and 375 beneficiaries (1%) had at least one paid claim for both benztropine and trihexiphenadyl at distinct time periods (non-concomitant). To determine if anticholinergic use was primarily short-term, the total number of paid claims with a days’ supply greater than or equal to 30 days was calculated for each beneficiary during the same one-year period (Table 2). More than half of the study population (52%) had at least six paid claims for an anticholinergic medication during the measurement year, suggesting long-term use during at least six months of the year, and 17% had paid claims greater than or equal to a one-year supply.
Table 2. Anticholinergic Use Among Medi-Cal Beneficiaries from January 1, 2018, through December 31, 2018
Number of paid claims for an anticholinergic medication ≥30 days’ supply during measurement yearUtilizingBeneficiaries n (%)
≥12 5,216 (17%)
6 – 11 10,776 (35%)
2 – 5 9,395 (30%)
1 5,731 (18%)
TOTAL31,118 (100%)
Among those beneficiaries with at least one paid claim for an anticholinergic medication, a total of 529 beneficiaries (2%) were 65 years of age or older, with 217 of these beneficiaries having at least six paid claims for an anticholinergic medication during the measurement year. As stated previously, the risk of adverse events related to anticholinergic medication use is increased in this population, and both benztropine and trihexyphenidyl appear on the American Geriatrics Society 2019 Updated AGS Beers Criteria® for Potentially Inappropriate Medication Use in Older Adults reference tool.8 Among beneficiaries younger than 65 years of age (n = 30,630), approximately half (51%) of beneficiaries had at least six paid claims for an anticholinergic medication during the measurement year, suggesting the rate of chronic use of anticholinergic medications was higher among older adults during this time period.
An additional evaluation was conducted to determine if anticholinergic use was linked to the propensity for antipsychotic-induced EPS. Pharmacy claims data for all Medi-Cal beneficiaries were reviewed for concomitant use of antipsychotics and anticholinergics. Among those beneficiaries with a paid claim for an anticholinergic medication with greater than or equal to a 30-day supply, a total of 29,807 (96%) beneficiaries also had at least one paid claim for an antipsychotic medication during the same time period. Because of the differences in clinical recommendations for anticholinergic use, the claims data were stratified by first- and second-generation antipsychotics.
Table 3. Concomitant Anticholinergic Medication Use in Medi-Cal Beneficiaries With a Paid Claim for an Antipsychotic Medication Between January 1, 2018, and December 31, 2018
Number of paid claims for an anticholinergic medication ≥30 days’ supply during measurement yearUtilizingBeneficiaries n (%)Percent of Utilizing Beneficiaries with Concomitant Anticholinergic Use
No Use(0 paid claims)Low Use(<6 paid claims)High Use(≥6 paid claims)
First- Generation
Antipsychotic
Medications CHLORPROMAZINE (n = 3,946)
FLUPHENAZINE (n = 1,692)
HALOPERIDOL (n = 15,208)
LOXAPINE (n = 630)
PERPHENAZINE (n = 1,932)
PIMOZIDE (n = 78)
THIORIDAZINE (n = 303)
THIOTHIXENE (n = 405)
TRIFLUOPERAZINE (n = 436)
ANY (n = 23,191)74%
48%
52%
65%
69%
92%
81%
47%
54%
58%11%
22%
22%
14%
13%
5%
8%
16%
15%
19%15%
30%
26%
21%
17%
3%
11%
37%
31%
23%
SecondGeneration
Antipsychotic Medications ARIPIPRAZOLE (n = 75,087)
ASENAPINE (n = 2,692) BREXPIPRAZOLE (n = 2,497)
CARIPRAZINE (n = 1,820)
CLOZAPINE (n = 4,246)
ILOPERIDONE (n = 779)
LURASIDONE (n = 29,649)
OLANZAPINE (n = 51,384)
PALIPERIDONE (n = 12,725)
QUETIAPINE (n = 86,264)
RISPERIDONE (n = 54,045)
ZIPRASIDONE (n = 9,686)
ANY (n = 260,655)91%
81%
89%
83%
71%
72%
88%
86%
71%
92%
82%
85%
90%5%
9%
5%
8%
15%
8%
7%
7%
14%
4%
9%
7%
5%4%
10%
6%
9%
14%
19%
5%
7%
15%
4%
9%
8%
5%
TOTAL (n = 268,245)89%5%6%
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4432 | https://en.wikipedia.org/wiki/Tally_stick | Tally stick - Wikipedia
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1 Kinds of talliesToggle Kinds of tallies subsection
1.1 Possible palaeolithic tally sticks
1.2 Single tally
1.3 Split tally
2 See also
3 Notes
4 References
5 Further reading
6 External links
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Medieval English split tally stick (front and reverse view). The stick is notched and inscribed to record a debt owed to the rural dean of Preston Candover, Hampshire, of a tithe of 20d each on 32 sheep, amounting to a total sum of £2 13s. 4d.
A tally stick (or simply a tally) was an ancient memory aid used to record and document numbers, quantities, and messages. Tally sticks first appear as animal bones carved with notches during the Upper Palaeolithic; a notable example is the Ishango Bone. Historical reference is made by Pliny the Elder (AD 23–79) about the best wood to use for tallies, and by Marco Polo (1254–1324) who mentions the use of the tally in China. Tallies have been used for numerous purposes such as messaging and scheduling, and especially in financial and legal transactions, to the point of being currency.
Kinds of tallies
[edit]
Single and split tallies from the Swiss Alps, 18th to early 20th century (Swiss Alpine Museum)
There are two principal forms of tally sticks: the single tally and the split tally. A comparable example of this primitive counting device can be found in various types of prayer beads.
Possible palaeolithic tally sticks
[edit]
A number of anthropological artefacts have been conjectured to be tally sticks:
The Lebombo bone, dated between 44,200 and 43,000 years old, is a baboon's fibula with 29 distinct notches, discovered within the Border Cave in the Lebombo Mountains of Eswatini.
The so-called Wolf bone(cs) is a prehistoric artefact discovered in 1937 in Czechoslovakia during excavations at Dolní Věstonice, Moravia, led by Karl Absolon. Dated to the Aurignacian, approximately 30,000 years ago, the bone is marked with 55 marks which some believe to be tally marks. The head of an ivory Venus figurine was excavated close to the bone.
The Ishango bone is a bone tool, dated to the Upper Palaeolithic era, around 18,000 to 20,000 BC. It is a dark brown length of bone. It has a series of possible tally marks carved in three columns running the length of the tool. It was found in 1950 in Ishango (east Belgian Congo).
Single tally
[edit]
The single tally stick was an elongated piece of bone, ivory, wood, or stone which is marked with a system of notches (see: Tally marks). The single tally stick serves predominantly mnemonic purposes. Related to the single tally concept are messenger sticks (used by, e.g., Inuit tribes), the knotted cords, khipus or quipus, as used by the Inca. Greek historian Herodotus (c. 484– c. 425 BC) reported the use of a knotted cord by Darius I of Persia (r. 522–486 BC).
Split tally
[edit]
The split tally became a prevalent technique in medieval Europe, a time characterised by a scarcity of coinage and widespread illiteracy, to document bilateral exchanges and debts. Typically fashioned from squared hazelwood, the stick was inscribed with a series of notches before being split lengthwise. Each party in the transaction retained one half of the marked stick, both pieces bearing identical records. Over the years, this method was refined to the point of becoming virtually impervious to tampering. One such refinement was to make the two halves of the stick of different lengths. The longer part was called stock and was given to the stock holder, which had advanced money (or other items) to the receiver. The shorter portion of the stick was called foil and was given to the party which had received the funds or goods. Using this technique each of the parties had an identifiable record of the transaction. The natural irregularities in the surfaces of the tallies where they were split would mean that only the original two halves would fit back together perfectly, and so would verify that they were matching halves of the same transaction. If one party tried to unilaterally change the value of his half of the tally stick by adding more notches, the absence of those notches would be apparent on the other party's tally stick. The split tally was accepted as legal proof in medieval courts and the Napoleonic Code (1804) still makes reference to the tally stick in Article 1333. Along the Danube and in Switzerland the tally was still used in the 20th century in rural economies.
The most prominent and best recorded use of the split tally stick or "nick-stick" being used as a form of currency was when Henry I introduced the tally stick system in medieval England in around 1100. The tally sticks recorded a payment of taxes, but soon began to circulate in a secondary discount market, being accepted as payment for goods and services at a discount since they could be later presented to the treasury as proof of taxes paid. Then tally sticks began to be issued in advance, in order to finance war and other royal spending, and circulated as "wooden money".
The system of tally marks of the Exchequer is described in The Dialogue Concerning the Exchequer as follows:
The manner of cutting is as follows. At the top of the tally a cut is made, the thickness of the palm of the hand, to represent a thousand pounds; then a hundred pounds by a cut the breadth of a thumb; twenty pounds, the breadth of the little finger; a single pound, the width of a swollen barleycorn; a shilling rather narrower; then a penny is marked by a single cut without removing any wood.
The cuts were made the full width of the stick so that, after splitting, the portion kept by the issuer (the stock) exactly matched the piece (the foil) given as a receipt. Each stick had to have the details of the transaction written on it, in ink, to make it a valid record.
Entrance gates to the UK National Archives, Kew, from Ruskin Avenue. The notched vertical elements were inspired by medieval tally sticks.
Royal tallies (debt of the Crown) also played a role in the formation of the Bank of England at the end of the 17th century. In 1697, the bank issued £1 million worth of stock in exchange for £800,000 worth of tallies at par and £200,000 in bank notes. This new stock was said to be "engrafted". The government promised not only to pay the Bank interest on the tallies subscribed but to redeem them over a period of years. The "engrafted" stock was then cancelled simultaneously with the redemption.
The split tally of the Exchequer remained in continuous use in England until 1826, when the conditions required for activation of the Receipt of the Exchequer Act 1783 (23 Geo. 3. c. 82), the death of the last Exchequer Chamberlain, came about. In 1834, following the passing of 4 Will. 4. c .15, tally sticks representing six centuries' worth of financial records were ordered to be burned in two furnaces in the Houses of Parliament. The resulting fire set the chimney ablaze and then spread until most of the building was destroyed. This event was described by Charles Dickens in an 1855 article on administrative reform.
Tally sticks feature in the design of the entrance gates to The National Archives at Kew.
See also
[edit]
Bamboo and wooden slips
Bamboo tally
Chirograph: a similar system for creating two or more matching copies of a legal document on parchment
Fu (tally)
Measuring rod
Prehistoric numerals
Notes
[edit]
^Harper, Douglas. "tally". Online Etymology Dictionary.
^Flegg, Graham (2002). Numbers: their history and meaning. Courier Dover Publications. pp.41–42. ISBN978-0-486-42165-0.
^Beckmann, Petr (1971). A History of π (PI). Boulder, Colorado: The Golem Press. p.8. ISBN978-0-911762-12-9.
^"A very brief history of pure mathematics: The Ishango Bone". University of Western Australia School of Mathematics. Archived from the original on 21 July 2008.
^Bunt, LNH; Jones PS & Bedient JD (1976) Historical Roots of Early Mathematics, page 225, Prentice Hall
^"Code civil, Article 1333". 1804. Retrieved 13 January 2013. Les tailles corrélatives à leurs échantillons font foi entre les personnes qui sont dans l'usage de constater ainsi les fournitures qu'elles font ou reçoivent en détail.
^"nick-stick". Oxford English Dictionary (3 ed.). 2003.
^Shenton, Caroline (2012). "Chapter 1: Thursday 16 October 1834, 6am: Mr Hume's Motion for a New House". The Day Parliament Burned Down. Oxford: Oxford University Press. ISBN9780199646708.
^ ab"50 Things That Made the Modern Economy". BBC Radio. BBC. Retrieved 30 March 2018.
^How swapping gold for wood left England’s citizens short-changed
^What tally sticks tell us about how money works
^Clanchy, Michael (1979). From Memory to Written Record (3rd ed.). Oxford, England: Blackwell. p.125. ISBN9780631168577.
^Carswell, John (1993). The South Sea Bubble (Revised ed.). England: Alan Sutton Publishing. p.24. ISBN978-0-86299-918-6.
^Robert, Rudoph (1956). "A Short History of Tallies"(PDF). In Littleton, A. C.; Yamey, B. S. (eds.). Studies in the History of Accounting. London: Sweet & Maxwell. pp.75–85.
^Shenton, Caroline (16 October 2013). "The day Parliament burned down". UK Parliament. Archived from the original on 26 August 2017. Retrieved 12 May 2019.
^Dickens, Charles (1870) [June 27, 1855]. "Administrative Reform". Speeches literary and social by Charles Dickens. Theatre Royal, Drury Lane. pp.133–134.{{cite book}}: CS1 maint: location missing publisher (link)
References
[edit]
Maddox, Thomas, ed. (1711). The History and Antiquities of the Exchequer of the Kings of England, in two periods: To wit, from the Norman Conquest, to the End of the Reign of K. John; and from the End of the Reign of K. John, to the End of the Reign of K. Edward II: Taken from Records. London.
Baxter, T. W. (1989). "Early Accounting, The Tally and the Checkerboard". The Accounting Historians Journal. 16 (2): 43–83. doi:10.2308/0148-4184.16.2.43.
Jenkinson, (Sir) Hilary (1924). "Medieval Tallies, Public and Private". Archaeologia. 74: 280–351, 8 Plates.
Carswell, John (1993). The South Sea Bubble (Revised ed.). England: Alan Sutton Publishing Ltd. p.24. ISBN978-0-86299-918-6.
Further reading
[edit]
Hall, Hubert (1891). The Antiquities and Curiosities of the Exchequer. London: Elliot Stocks.
Henkelman, Wouter F. M.; L. Folmer, Margaretha (2016). "Your Tally is Full! On Wooden Credit Records in and after the Achaemenid Empire". In Kleber, K.; Pirngruber, R. (eds.). Silver, Money and Credit: Festschrift for Robartus J. van der Spek on occasion of his 65th birthday on 18 September 2014. Publications de l'Institut Historique-Archéologique Néerlandais de Stamboul. Vol.128. Leiden. pp.129–226.{{cite book}}: CS1 maint: location missing publisher (link)
Jenkinson, (Sir) Hilary (1914). "An original Exchequer Account of 1304 with private Tallies attached". Proceedings of the Society of Antiquaries of London. 2nd ser. 26: 36–40.
Jenkinson, (Sir) Hilary (1911). "Exchequer Tallies". Archaeologia. 62 (2): 367–80. doi:10.1017/s0261340900008213.
Jenkinson, (Sir) Hilary. "A note supplementary". Proceedings of the Society of Antiquaries of London. 2nd ser. 25: 29–39.
External links
[edit]
Wikisource has the text of the 1911 Encyclopædia Britannica article "Tally".
Photo of Medieval Exchequer Tallies
Inca khipus
Khipus as ledgers
The Dialogue Concerning the Exchequer
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4433 | https://brainly.in/question/60583838 | prove that cos 30= root 3/2 - Brainly.in
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Prove that cos 30= root 3/2
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In trigonometry, the cosine function is defined as the ratio of the adjacent side to the hypotenuse. If the angle of a right triangle is equal to 30 degrees, and then the value of cosine at this angle i.e., the value of Cos 30 degree is in a fraction form as √3/2. It is noted that the value of Sec 30 will be reciprocal of Cos 30 value.
Cos 30 Value
Cos 30 degrees is written as cos 30° and has a value in fraction form as √3/2.
Cos 30° = √3/2
Cos 30° = √3/2 is an irrational number and equals to 0.8660254037 (decimal form).
Therefore, the exact value of cos 30 degrees is written as 0.8660 approx.
√3/2 is the value of Cos 30° which is a trigonometric ratio or trigonometric function of a particular angle.
Cos 30
Another alternative form of Cos 30° is pi/6 or π/6 or Cos 33 (⅓)g
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4434 | https://emedicine.medscape.com/article/262063-differential | Placenta Previa Differential Diagnoses
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History
Physical Examination
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Instruments or fingers should not be placed near the cervix during a vaginal examination, because uncontrolled bleeding can result. Do not perform vaginal or rectal examinations in an outpatient or emergency department setting unless ultrasonography findings have ruled out placenta previa.
Rarely, ultrasonography is unavailable and a digital examination is necessary. If this is the case, the digital examination should be performed in the operating room under double setup conditions (ie, one team ready for emergent cesarean delivery and one team ready for uneventful vaginal delivery).
Other problems to consider in a woman with suspected placenta previa include the following:
Vasa previa
Cervical or vaginal laceration
Vaginal sidewall laceration
Miscarriage (spontaneous abortion)
Infection
Vaginal bleeding
Lower genital tract lesions
Bloody show
Differential Diagnoses
Abruptio Placentae
Cervicitis
Disseminated Intravascular Coagulation (DIC)
Pregnancy, Delivery
Premature Rupture of Membranes
Preterm Labor and Birth
Uterine Rupture in Pregnancy
Vaginitis
Vulvovaginitis
Workup
;)
References
Marshall NE, Fu R, Guise JM. Impact of multiple cesarean deliveries on maternal morbidity: a systematic review. Am J Obstet Gynecol. 2011 Sep. 205(3):262.e1-8. [QxMD MEDLINE Link].
Milosevic J, Lilic V, Tasic M, Radovic-Janosevic D, Stefanovic M, Antic V. [Placental complications after a previous cesarean section]. Med Pregl. 2009 May-Jun. 62(5-6):212-6. [QxMD MEDLINE Link].
Ananth CV, Smulian JC, Vintzileos AM. The effect of placenta previa on neonatal mortality: a population-based study in the United States, 1989 through 1997. Am J Obstet Gynecol. 2003 May. 188(5):1299-304. [QxMD MEDLINE Link].
Iyasu S, Saftlas AK, Rowley DL, Koonin LM, Lawson HW, Atrash HK. The epidemiology of placenta previa in the United States, 1979 through 1987. Am J Obstet Gynecol. 1993 May. 168(5):1424-9. [QxMD MEDLINE Link].
Kim LH, Caughey AB, Laguardia JC, Escobar GJ. Racial and ethnic differences in the prevalence of placenta previa. J Perinatol. 2012 Apr. 32 (4):260-4. [QxMD MEDLINE Link].
Williams MA, Mittendorf R. Increasing maternal age as a determinant of placenta previa. More important than increasing parity?. J Reprod Med. 1993 Jun. 38(6):425-8. [QxMD MEDLINE Link].
Ananth CV, Wilcox AJ, Savitz DA, Bowes WA Jr, Luther ER. Effect of maternal age and parity on the risk of uteroplacental bleeding disorders in pregnancy. Obstet Gynecol. 1996 Oct. 88(4 Pt 1):511-6. [QxMD MEDLINE Link].
Becker RH, Vonk R, Mende BC, Ragosch V, Entezami M. The relevance of placental location at 20-23 gestational weeks for prediction of placenta previa at delivery: evaluation of 8650 cases. Ultrasound Obstet Gynecol. 2001 Jun. 17(6):496-501. [QxMD MEDLINE Link].
Hill LM, DiNofrio DM, Chenevey P. Transvaginal sonographic evaluation of first-trimester placenta previa. Ultrasound Obstet Gynecol. 1995 May. 5(5):301-3. [QxMD MEDLINE Link].
Wexler P, Gottesfeld KR. Early diagnosis of placenta previa. Obstet Gynecol. 1979 Aug. 54(2):231-4. [QxMD MEDLINE Link].
Jansen CHJR, van Dijk CE, Kleinrouweler CE, et al. Risk of preterm birth for placenta previa or low-lying placenta and possible preventive interventions: A systematic review and meta-analysis. Front Endocrinol (Lausanne). 2022. 13:921220. [QxMD MEDLINE Link]. [Full Text].
Zaki ZM, Bahar AM, Ali ME, Albar HA, Gerais MA. Risk factors and morbidity in patients with placenta previa accreta compared to placenta previa non-accreta. Acta Obstet Gynecol Scand. 1998 Apr. 77(4):391-4. [QxMD MEDLINE Link].
Frederiksen MC, Glassenberg R, Stika CS. Placenta previa: a 22-year analysis. Am J Obstet Gynecol. 1999 Jun. 180(6 pt 1):1432-7. [QxMD MEDLINE Link].
Zlatnik MG, Cheng YW, Norton ME, Thiet MP, Caughey AB. Placenta previa and the risk of preterm delivery. J Matern Fetal Neonatal Med. 2007 Oct. 20(10):719-23. [QxMD MEDLINE Link].
Creasy RK, Resnik R, Iams J , Lockwood C, Moore T, Greene M. Placenta previa, placenta accreta, abruptio placentae, and vasa previa. Creasy and Resnik's Maternal-Fetal Medicine: Principles and Practice. 7th ed. Saunders: Philadelphia, PA; 2014. 732-742.
Bose DA, Assel BG, Hill JB, Chauhan SP. Maintenance tocolytics for preterm symptomatic placenta previa: a review. Am J Perinatol. 2011 Jan. 28(1):45-50. [QxMD MEDLINE Link].
Dola CP, Garite TJ, Dowling DD, Friend D, Ahdoot D, Asrat T. Placenta previa: does its type affect pregnancy outcome?. Am J Perinatol. 2003 Oct. 20(7):353-60. [QxMD MEDLINE Link].
Leerentveld RA, Gilberts EC, Arnold MJ, Wladimiroff JW. Accuracy and safety of transvaginal sonographic placental localization. Obstet Gynecol. 1990 Nov. 76(5 Pt 1):759-62. [QxMD MEDLINE Link].
Sherman SJ, Carlson DE, Platt LD, Medearis AL. Transvaginal ultrasound: does it help in the diagnosis of placenta previa?. Ultrasound Obstet Gynecol. 1992 Jul 1. 2(4):256-60. [QxMD MEDLINE Link].
Ghi T, Contro E, Martina T, Piva M, Morandi R, Orsini LF, et al. Cervical length and risk of antepartum bleeding in women with complete placenta previa. Ultrasound. Obstet Gynecol. Feb 2009. 33(2):209-12.
Smith RS, Lauria MR, Comstock CH, et al. Transvaginal ultrasonography for all placentas that appear to be low-lying or over the internal cervical os. Ultrasound Obstet Gynecol. 1997 Jan. 9(1):22-4. [QxMD MEDLINE Link].
Morlando M, Sarno L, Napolitano R, et al. Placenta accreta: incidence and risk factors in an area with a particularly high rate of cesarean section. Acta Obstet Gynecol Scand. 2013 Apr. 92(4):457-60. [QxMD MEDLINE Link].
Warshak CR, Eskander R, Hull AD, et al. Accuracy of ultrasonography and magnetic resonance imaging in the diagnosis of placenta accreta. Obstet Gynecol. 2006 Sep. 108(3 Pt 1):573-81. [QxMD MEDLINE Link].
Ueno Y, Kitajima K, Kawakami F, Maeda T, Suenaga Y, Takahashi S, et al. Novel MRI finding for diagnosis of invasive placenta praevia: evaluation of findings for 65 patients using clinical and histopathological correlations. Eur Radiol. 2014 Apr. 24(4):881-8. [QxMD MEDLINE Link].
Allen BC, Leyendecker JR. Placental evaluation with magnetic resonance. Radiol Clin North Am. 2013 Nov. 51(6):955-66. [QxMD MEDLINE Link].
Silver, R. Abnormal placentation: Placenta previa, vasa previa, and placenta accreta. Obstet Gynecolol. 2015. 126:654-68.
Soyama H, Miyamoto M, Sasa H, Ishibashi H, Yoshida M, Nakatsuka M, et al. Effect of routine rapid insertion of Bakri balloon tamponade on reducing hemorrhage from placenta previa during and after cesarean section. Arch Gynecol Obstet. 2017 Jun 24. [QxMD MEDLINE Link].
Besinger RE, Moniak CW, Paskiewicz LS, Fisher SG, Tomich PG. The effect of tocolytic use in the management of symptomatic placenta previa. Am J Obstet Gynecol. 1995 Jun. 172(6):1770-5; discussion 1775-8. [QxMD MEDLINE Link].
Bhide A, Prefumo F, Moore J, Hollis B, Thilaganathan B. Placental edge to internal os distance in the late third trimester and mode of delivery in placenta praevia. BJOG. 2003 Sep. 110(9):860-4. [QxMD MEDLINE Link].
Vergani P, Ornaghi S, Pozzi I, Beretta P, Russo FM, Follesa I, et al. Placenta previa: distance to internal os and mode of delivery. Am J Obstet Gynecol. 2009 Sep. 201(3):266.e1-5. [QxMD MEDLINE Link].
Blackwell, SC. Timing of delivery for women with stable placenta previa. Semin Perinatol. 2011. 35:249-51.
Machado LS. Emergency peripartum hysterectomy: Incidence, indications, risk factors and outcome. N Am J Med Sci. 2011 Aug. 3(8):358-61. [QxMD MEDLINE Link]. [Full Text].
Choi SJ, Song SE, Jung KL, Oh SY, Kim JH, Roh CR. Antepartum risk factors associated with peripartum cesarean hysterectomy in women with placenta previa. Am J Perinatol. Jan 2008. 25(1):37-41.
Masamoto H, Uehara H, Gibo M, Okubo E, Sakumoto K, Aoki Y. Elective use of aortic balloon occlusion in cesarean hysterectomy for placenta previa percreta. Gynecol Obstet Invest. 2009. 67(2):92-5.
Jain V, Bos H, Bujold E. Guideline No. 402: Diagnosis and Management of Placenta Previa. J Obstet Gynaecol Can. 2020 Jul. 42 (7):906-17.e1. [QxMD MEDLINE Link].
Gagnon R, Morin L, Bly S, et al. Guidelines for the management of vasa previa. J Obstet Gynaecol Can. 2009 Aug. 31(8):748-60. [QxMD MEDLINE Link].
Media Gallery
Placenta previa.
Complete or total placenta previa. The entire cervical os is covered.
Low-lying placenta previa. The placenta partially separated from the lower uterine segment.
Placenta previa invading the lower uterine segment and covering the cervical os.
Complete placenta previa noted on ultrasound.
Another ultrasound image clearly depicting complete placenta previa.
of 6
Tables
Table. Relative Risk of Morbidities in Patients With Placenta Previa;)
Table. Relative Risk of Morbidities in Patients With Placenta Previa
MorbiditiesRelative Risk
Antepartum bleeding 10
Need for hysterectomy 33
Blood transfusion 10
Septicemia 5.5
Thrombophlebitis 5
Endometritis 6.6
Back to List
;)
Contributor Information and Disclosures
Author
Ronan Bakker, MD Maternal-Fetal Medicine Specialist, The Perinatal Center
Ronan Bakker, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, American Medical Association
Disclosure: Nothing to disclose.
Coauthor(s)
Ronald M Ramus, MD Professor of Obstetrics and Gynecology, Director, Division of Maternal-Fetal Medicine, Virginia Commonwealth University School of Medicine
Ronald M Ramus, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, American Institute of Ultrasound in Medicine, Medical Society of Virginia, Society for Maternal-Fetal Medicine
Disclosure: Nothing to disclose.
Specialty Editor Board
John G Pierce, Jr, MD Chairman of Women’s Health and Medical Specialties, Liberty University College of Osteopathic Medicine; Obstetrician/Gynecologist, Women’s Health of Central Virginia
John G Pierce, Jr, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, Association of Professors of Gynecology and Obstetrics, Christian Medical and Dental Associations, Medical Society of Virginia, Society of Laparoscopic and Robotic Surgeons
Disclosure: Nothing to disclose.
Chief Editor
Carl V Smith, MD The Distinguished Chris J and Marie A Olson Chair of Obstetrics and Gynecology, Professor, Department of Obstetrics and Gynecology, Senior Associate Dean for Clinical Affairs, University of Nebraska Medical Center
Carl V Smith, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, American Institute of Ultrasound in Medicine, Association of Professors of Gynecology and Obstetrics, Central Association of Obstetricians and Gynecologists, Society for Maternal-Fetal Medicine, Council of University Chairs of Obstetrics and Gynecology, Nebraska Medical Association
Disclosure: Nothing to disclose.
Additional Contributors
Saju Joy, MD, MS Associate Director, Division Chief of Maternal-Fetal Medicine, Department of Obstetrics and Gynecology, Carolinas Medical Center
Saju Joy, MD, MS is a member of the following medical societies: American College of Obstetricians and Gynecologists, American Institute of Ultrasound in Medicine, Society for Maternal-Fetal Medicine, American Medical Association
Disclosure: Nothing to disclose.
Matthew M Finneran, MD Resident Physician, Department of Obstetrics and Gynecology, Carolinas Healthcare System
Disclosure: Nothing to disclose.
Acknowledgements
Pamela L Dyne, MD Professor of Clinical Medicine/Emergency Medicine, University of California, Los Angeles, David Geffen School of Medicine; Attending Physician, Department of Emergency Medicine, Olive View-UCLA Medical Center
Pamela L Dyne, MD is a member of the following medical societies: American Academy of Emergency Medicine, American College of Emergency Physicians, and Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Patrick Ko, MD Clinical Assistant Professor, Department of Emergency Medicine, New York University Medical School; Assistant Program Director, Department of Emergency Medicine, North Shore University Hospital
Patrick Ko, MD is a member of the following medical societies: American College of Emergency Physicians and Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Deborah Lyon, MD Director, Division of Gynecology, Associate Professor, Department of Obstetrics and Gynecology, University of Florida Health Science Center at Jacksonville
Deborah Lyon, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, Association of American Medical Colleges, Association of Professors of Gynecology and Obstetrics, and Florida Medical Association
Disclosure: Nothing to disclose.
John G Pierce Jr, MD Associate Professor, Departments of Obstetrics/Gynecology and Internal Medicine, Medical College of Virginia at Virginia Commonwealth University
John G Pierce Jr, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, Association of Professors of Gynecology and Obstetrics, Christian Medical & Dental Society, Medical Society of Virginia, and Society of Laparoendoscopic Surgeons
Disclosure: Nothing to disclose.
Joseph J Sachter, MD, FACEP Consulting Staff, Department of Emergency Medicine, Muhlenberg Regional Medical Center
Joseph J Sachter, MD, FACEP is a member of the following medical societies: American Academy of Emergency Medicine, American College of Emergency Physicians, American College of Physician Executives, American Medical Association, and Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Ryan A Stone, MD Fellow, Department of Obstetrics and Gynecology, Maternal-Fetal Medicine Section, Wake Forest University Health Sciences
Ryan A Stone, MD is a member of the following medical societies: Academic Pediatric Association, American College of Obstetricians and Gynecologists, American Medical Association, and Society for Maternal-Fetal Medicine
Disclosure: Nothing to disclose.
Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference
Disclosure: Medscape Salary Employment
Lorene Temming, MD Resident Physician, Department of Obstetrics and Gynecology, Carolinas Medical Center
Lorene Temming, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists and North Carolina Medical Society
Disclosure: Nothing to disclose.
Young Yoon, MD Associate Director, Assistant Professor, Department of Emergency Medicine, Mount Sinai Medical Center
Young Yoon, MD is a member of the following medical societies: Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Mark Zwanger, MD, MBA Assistant Professor, Department of Emergency Medicine, Jefferson Medical College of Thomas Jefferson University
Mark Zwanger, MD, MBA is a member of the following medical societies: American Academy of Emergency Medicine, American College of Emergency Physicians, and American Medical Association
Disclosure: Nothing to disclose.
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4435 | https://acme.byu.edu/00000181-a73b-d0f9-a7bd-bfbbe6210001/iterative-solvers-pdf | 15 Iterative Solvers
Lab Objective: Many real-world problems of the form Ax = b have tens of thousands of parameters. Solving such systems with Gaussian elimination or matrix factorizations could require trillions of floating point operations (FLOPs), which is of course infeasible. Solutions of large systems must therefore be approximated iteratively. In this lab we implement three popular iterative methods for solving large systems: Jacobi, Gauss-Seidel, and Successive Over-Relaxation.
Iterative methods are often useful to solve large systems of equations. In this lab, let x(k) denote the kth iteration of the iterative method for solving the problem Ax = b for x. Furthermore, let xi
be the ith component of x so that x(k)
i is the ith component of x in the kth iteration. Like other iterative methods, there are two stopping parameters: a very small ε > 0 and an integer N ∈ N.Iterations continue until either
‖x(k−1) − x(k)‖ < ε or k > N. (15.1)
The Jacobi Method
The Jacobi Method is a simple but powerful method used for solving certain kinds of large linear systems. The main idea is simple: solve for each variable in terms of the others, then use the previous values to update each approximation. As a (very small) example, consider the 3 × 32x1 − x3 = 3,
−x1 + 3x2 + 2x3 = 3,
x2 + 3x3 = −1.
Solving the first equation for x1, the second for x2, and the third for x3 yields
x1 = 12 (3 + x3),x2 = 13 (3 + x1 − 2x3),x3 = 13 (−1 − x2).
Now begin with an initial guess x(0) = [ x(0) 1 , x (0) 2 , x (0) 3 ]T = [0 , 0, 0] T. To compute the first approximation x(1) , use the entries of x(0) as the variables on the right side of the previous equations:
x(1) 1 = 12 (3 + x(0) 3 ) = 12 (3 + 0) = 32 ,x(1) 2 = 13 (3 + x(0) 1 − 2x(0) 3 ) = 13 (3 + 0 − 0) = 1,x(1) 3 = 13 (−1 − x(0) 2 ) = 13 (−1 − 0) = − 13 .
147 148 Lab 15. Iterative Solvers
Thus x(1) = [ 32 , 1, − 13 ]T. Computing x(2) is similar:
x(2) 1 = 12 (3 + x(1) 3 ) = 12 (3 − 13 ) = 43 ,x(2) 2 = 13 (3 + x(1) 1 − 2x(1) 3 ) = 13 (3 + 32 + 23 ) = 31 18 ,x(2) 3 = 13 (−1 − x(1) 2 ) = 13 (−1 − 1) = − 23 .
The process is repeated until at least one of the two stopping criteria in (15.1) is met. For this particular problem, convergence to 8 decimal places ( ε = 10 −8) is reached in 29 iterations.
x(k)1 x(k)2 x(k)3
x(0) 0 0 0
x(1) 1.5 1 −0.33333333
x(2) 1.33333333 1.72222222 −0.6666666 7
x(3) 1.16666667 1.88888889 −0.9074074 1
x(4) 1.04629630 1.99382716 −0.9629629 6... ... ... ...
x(28) 0.99999999 2.00000001 −0.9999999 9
x(29) 1 2 −1
Matrix Representation
The iterative steps performed above can be expressed in matrix form. First, decompose A into its diagonal entries, its entries below the diagonal, and its entries above the diagonal, as A = D + L + U .
a11 0 . . . 00 a22 . . . 0
... ... . . . ...
0 0 . . . ann
0 0 . . . 0
a21 0 . . . 0
... . . . . . . ...
an1 . . . an,n −1 0
0 a12 . . . a1n
0 0 . . . ...... ... . . . an−1,n
0 0 . . . 0
D L U
With this decomposition, x can be expressed in the following way.
Ax = b
(D + L + U )x = b
Dx = −(L + U )x + bx = D−1(−(L + U )x + b)
Now using x(k) as the variables on the right side of the equation to produce x(k+1) on the left, and noting that L + U = A − D, we have the following.
x(k+1) = D−1(−(A − D)x(k) + b)= D−1(Dx(k) − Ax(k) + b)= x(k) + D−1(b − Ax(k)) (15.2) There is a potential problem with (15.2): calculating a matrix inverse is the cardinal sin of numerical linear algebra, yet the equation contains D−1. However, since D is a diagonal matrix,
D−1 is also diagonal, and is easy to compute. 149
D−1 =
1
a11
0 . . . 00 1
a22
. . . 0
... ... . . . ...
0 0 . . . 1
ann
Because of this, the Jacobi method requires that A have nonzero diagonal entries. The diagonal D can be represented by the 1-dimensional array d of the diagonal entries. Then the matrix multiplication Dx is equivalent to the component-wise vector multiplication d ∗ x = x ∗ d.Likewise, the matrix multiplication D−1x is equivalent to the component-wise “vector division” x/d.
Problem 1. Write a function that accepts a matrix A, a vector b, a convergence tolerance tol
defaulting to 10 −8, and a maximum number of iterations maxiter defaulting to 100 . Implement the Jacobi method using (15.2), returning the approximate solution to the equation Ax = b.Run the iteration until ‖x(k−1) − x(k)‖∞ < tol , and only iterate at most maxiter times. Avoid using la.inv() to calculate D−1, but use la.norm() to calculate the vector ∞-norm. Your function should be robust enough to accept systems of any size. To test your function, generate a random b with np.random.random() and use the following function to generate an
n × n matrix A for which the Jacobi method is guaranteed to converge. Run the iteration, then check that Ax(k) and b are close using np.allclose() .
def diag_dom(n, num_entries=None, as_sparse=False): """Generate a strictly diagonally dominant (n, n) matrix. Parameters: n (int): The dimension of the system. num_entries (int): The number of nonzero values. Defaults to n^(3/2)-n. as_sparse: If True, an equivalent sparse CSR matrix is returned. Returns: A ((n,n) ndarray): A (n, n) strictly diagonally dominant matrix. """ if num_entries is None: num_entries = int(n1.5) - n A = sparse.dok_matrix((n,n)) rows = np.random.choice(n, size=num_entries) cols = np.random.choice(n, size=num_entries) data = np.random.randint(-4, 4, size=num_entries) for i in range(num_entries): A[rows[i], cols[i]] = data[i] B = A.tocsr() # convert to row format for the next step for i in range(n): A[i,i] = abs(B[i]).sum() + 1 return A.tocsr() if as_sparse else A.toarray()
Also test your function on random n × n matrices. If the iteration is non-convergent, the successive approximations will have increasingly large entries. 150 Lab 15. Iterative Solvers
Convergence
Most iterative methods only converge under certain conditions. For the Jacobi method, convergence mostly depends on the nature of the matrix A. If the entries aij of A satisfy the property
|aii | > ∑
j6=i
|aij | for all i = 1 , 2, . . . , n,
then A is called strictly diagonally dominant (diag_dom() in Problem 1 generates a strictly diagonally dominant n × n matrix). If this is the case, 1 then the Jacobi method always converges, regardless of the initial guess x0. This is a very different convergence result than many other iterative methods such as Newton’s method where convergence is highly sensitive to the initial guess. There are a few ways to determine whether or not an iterative method is converging. For example, since the approximation x(k) should satisfy Ax(k) ≈ b, the normed difference ‖Ax(k) − b‖∞
should be small. This value is called the absolute error of the approximation. If the iterative method converges, the absolute error should decrease to ε.
Problem 2. Modify your Jacobi method function in the following ways. 1. Add a keyword argument called plot , defaulting to False .2. Keep track of the absolute error ‖Ax(k) − b‖∞ of the approximation at each iteration. 3. If plot is True, produce a lin-log plot (use plt.semilogy() ) of the error against iteration count. Remember to still return the approximate solution x.If the iteration converges, your plot should resemble the following figure. 0 5 10 15 20 25 30 Iteration 10 7
10 5
10 3
10 1
Absolute Error of Approximation
Convergence of Jacobi Method
1Although this seems like a strong requirement, most real-world linear systems can be represented by strictly diagonally dominant matrices.
151
The Gauss-Seidel Method
The Gauss-Seidel method is essentially a slight modification of the Jacobi method. The main differ-ence is that in Gauss-Seidel, new information is used immediately. As an example, consider again the system from the previous section,
2x1 − x3 = 3,
−x1 + 3x2 + 2x3 = 3,
x2 + 3x3 = −1.
As with the Jacobi method, solve for x1 in the first equation, x2 in the second equation, and
x3 in the third equation:
x1 = 12 (3 + x3),x2 = 13 (3 + x1 − 2x3),x3 = 13 (−1 − x2).
Using x(0) to compute x(1) 1 in the first equation as before,
x(1) 1 = 12 (3 + x(0) 3 ) = 12 (3 + 0) = 32 .
Now, however, use the updated value of x(1) 1 in the calculation of x(1) 2 :
x(1) 2 = 13 (3 + x(1) 1 − 2x(0) 3 ) = 13 (3 + 32 − 0) = 32 .
Likewise, use the updated values of x(1) 1 and x(1) 2 to calculate x(1) 3 :
x(1) 3 = 13 (−1 − x(1) 2 ) = 13 (−1 − 32 ) = − 56 .
This process of using calculated information immediately is called forward substitution , and causes the algorithm to (generally) converge much faster.
x(k)1 x(k)2 x(k)3
x(0) 0 0 0
x(1) 1.5 1.5 −0.83333333
x(2) 1.08333333 1.91666667 −0.97222222
x(3) 1.01388889 1.98611111 −0.99537037
x(4) 1.00231481 1.99768519 −0.99922840
... ... ... ...
x(11) 1.00000001 1.99999999 −1
x(12) 1 2 −1
Notice that Gauss-Seidel converges in less than half as many iterations as Jacobi does for this system.
Implementation
Because Gauss-Seidel updates only one element of the solution vector at a time, the iteration cannot be summarized by a single matrix equation. Instead, the process is most generally described by the equation
x(k+1)
i
= 1
aii
bi − ∑
j<i
aij x(k)
j
− ∑
j>i
aij x(k)
j
. (15.3) 152 Lab 15. Iterative Solvers
Let ai be the ith row of A. The two sums closely resemble the regular vector product of ai
and x(k) without the ith term aii x(k)
i
. This suggests the simplification
x(k+1)
i
= 1
aii
(
bi − aT
i
x(k) + aii x(k)
i
)
= x(k)
i
1
aii
(
bi − aT
i
x(k))
. (15.4) One sweep through all the entries of x completes one iteration.
Problem 3. Write a function that accepts a matrix A, a vector b, a convergence tolerance
tol defaulting to 10 −8, a maximum number of iterations maxiter defaulting to 100 , and a keyword argument plot that defaults to False . Implement the Gauss-Seidel method using (15.4), returning the approximate solution to the equation Ax = b.Use the same stopping criterion as in Problem 1. Also keep track of the absolute errors of the iteration, as in Problem 2. If plot is True , plot the error against iteration count. Use
diag_dom() to generate test cases.
Achtung!
Since the Gauss-Seidel algorithm operates on the approximation vector in place (modifying it one entry at a time), the previous approximation x(k−1) must be stored at the beginning of the kth iteration in order to calculate ‖x(k−1) − x(k)‖∞. Additionally, since NumPy arrays are mutable, the past iteration must be stored as a copy .
x0 = np.random.random(5) # Generate a random vector. >>> x1 = x0 # Attempt to make a copy. >>> x1 = 1000 # Modify the "copy" in place. >>> np.allclose(x0, x1) # But x0 was also changed! True # Instead, make a copy of x0 when creating x1. >>> x0 = np.copy(x1) # Make a copy. >>> x1 = -1000 >>> np.allclose(x0, x1) False
Convergence
Whether or not the Gauss-Seidel method converges depends on the nature of A. If all of the eigenval-ues of A are positive, A is called positive definite . If A is positive definite or if it is strictly diagonally dominant, then the Gauss-Seidel method converges regardless of the initial guess x(0) .153
Solving Sparse Systems Iteratively
Iterative solvers are best suited for solving very large sparse systems. However, using the Gauss-Seidel method on sparse matrices requires translating code from NumPy to scipy.sparse . The algorithm is the same, but there are some functions that are named differently between these two packages. 2
Problem 4. Write a new function that accepts a sparse matrix A, a vector b, a convergence tolerance tol , and a maximum number of iterations maxiter (plotting the convergence is not required for this problem). Implement the Gauss-Seidel method using (15.4), returning the approximate solution to the equation Ax = b. Use the usual default stopping criterion. The Gauss-Seidel method requires extracting the rows Ai from the matrix A and com-puting AT
i
x. There are many ways to do this that cause some fairly serious runtime issues, so we provide the code for this specific portion of the algorithm.
Get the indices of where the i-th row of A starts and ends if the # nonzero entries of A were flattened. rowstart = A.indptr[i] rowend = A.indptr[i+1] # Multiply only the nonzero elements of the i-th row of A with the # corresponding elements of x. Aix = A.data[rowstart:rowend] @ x[A.indices[rowstart:rowend]]
To test your function, remeber to call diag_dom() using as_sparse=True >>> A = diag_dom(50000, as_sparse=True) >>> b = np.random.random(50000)
Successive Over-Relaxation
There are many systems that meet the requirements for convergence with the Gauss-Seidel method, but for which convergence is still relatively slow. A slightly altered version of the Gauss-Seidel method, called Successive Over-Relaxation (SOR), can result in faster convergence. This is achieved by introducing a relaxation factor ω ≥ 1 and modifying (15.3) as
x(k+1)
i
= (1 − ω)x(k)
i
ωaii
bi − ∑
j<i
aij x(k)
j
− ∑
j>i
aij x(k)
j
.
Simplifying the equation, we have
x(k+1)
i
= x(k)
i
ωaii
(
bi − aT
i
x(k))
. (15.5) Note that when ω = 1 , SOR reduces to Gauss-Seidel. The relaxation factor ω weights the new iteration between the current best approximation and the next approximation in a way that can sometimes dramatically improve convergence.
2See the lab on Linear Systems for a review of scipy.sparse matrices and syntax.
154 Lab 15. Iterative Solvers
Problem 5. Write a function that accepts a sparse matrix A, a vector b, a relaxation factor
ω, a convergence tolerance tol , and a maximum number of iterations maxiter . Implement SOR using (15.5), compute the approximate solution to the equation Ax = b. Use the usual stopping criterion. Return the approximate solution x as well as a boolean indicating whether the function converged and the number of iterations computed. (Hint: this requires changing only one line of code from the sparse Gauss-Seidel function.)
A Finite Difference Method
Laplace’s equation is an important partial differential equation that arises often in both pure and applied mathematics. In two dimensions, the equation has the following form.
∂2u∂x 2 + ∂2u∂y 2 = 0 (15.6) Laplace’s equation can be used to model heat flow. Consider a square metal plate where the top and bottom borders are fixed at 0◦ Celsius and the left and right sides are fixed at 100 ◦ Celsius. Given these boundary conditions, we want to describe how heat diffuses through the rest of the plate. The solution to Laplace’s equation describes the plate when it is in a steady state , meaning that the heat at a given part of the plate no longer changes with time. It is possible to solve (15.6) analytically. However, the problem can also be solved numerically using a finite difference method . To begin, we impose a discrete, square grid on the plate with uniform spacing. Denote the points on the grid by (xi, y j ) and the value of u at these points (the heat) as
u(xi, y j ) = Ui,j . Using the centered difference quotient for second derivatives to approximate the partial derivatives,
0 = ∂2u∂x 2 + ∂2u∂y 2
≈ Ui+1 ,j − 2Ui,j + Ui−1,j
h2 + Ui,j +1 − 2Ui,j + Ui,j −1
h2
= 1
h2 (−4Ui,j + Ui+1 ,j + Ui−1,j + Ui,j +1 + Ui,j −1) , (15.7) where h = xi+1 − xi = yj+1 − yj is the distance between the grid points in either direction. This problem can be formulated as a linear system. Suppose the grid has exactly (n + 2) × (n + 2) entries. Then the interior of the grid (where u(x, y ) is unknown) is n × n, and can be flattened into an n2 × 1
vector u. The entire first row goes first, then the second row, proceeding to the nth row.
u = [U1,1 U1,2 · · · U1,n U2,1 U2,2 · · · U2,n · · · Un,n
]T
From (15.7), for an interior point Ui,j , we have
− 4Ui,j + Ui+1 ,j + Ui−1,j + Ui,j +1 + Ui,j −1 = 0 . (15.8) If any of the neighbors to Ui,j is a boundary point on the grid, its value is already determined by the boundary conditions. For example, the neighbor U3,0 of the gridpoint for U3,1 is fixed at U3,0 = 100 .In this case, (15.8) becomes
−4U3,1 + U2,1 + U3,2 + U4,1 = −100 .155
Figure 15.1: On the left, an example of a 6 × 6 grid ( n = 4 ) where the red dots are hot boundary zones and the blue dots are cold boundary zones. On the right, the green dots are the neighbors of the interior black dot that are used to approximate the heat at the black dot. The constants on the right side of (15.8) become the n2 × 1 vector b. All nonzero entries of b
correspond to interior points that touch the left or right boundaries. As an example, writing (15.8) for the 16 interior points of the grid in Figure 15.1 results in the following 16 × 16 system Au = b. Note the block structure (empty blocks are all zeros).
−410010001−410010001−400010001−400011000−4100100001001−4100100001001−4100100001001−400011000−4100100001001−4100100001001−4100100001001−400011000−410001001−410001001−410001001−4
U1,1
U1,2
U1,3
U1,4
U2,1
U2,2
U2,3
U2,4
U3,1
U3,2
U3,3
U3,4
U4,1
U4,2
U4,3
U4,4
=
−100 00
−100
−100 00
−100
−100 00
−100
−100 00
−100
More concisely, for any positive integer n, the matrix A can be written as
A =
B II B II . . . . . .. . . . . . II B
, where B =
−4 11 −4 11 . . . . . .. . . . . . 11 −4
is n × n. 156 Lab 15. Iterative Solvers
Problem 6. Write a function that accepts an integer n, a relaxation factor ω, a convergence tolerance tol that defaults to 10 −8, a maximum number of iterations maxiter that defaults to 100 , and a bool plot that defaults to False . Generate and solve the corresponding system
Au = b using Problem 5. Also return a boolean indicating whether the function converged and the number of iterations computed. (Hint: see Problem 5 of the Linear Systems lab for the construction of A. Also, np.tile()
may be useful for constructing b.) If plot=True , visualize the solution u with a heatmap using plt.pcolormesh() (the colormap "coolwarm" is a good choice in this case). This shows the distribution of heat over the hot plate after it has reached its steady state. Note that the u must be reshaped as an
n × n array to properly visualize the result.
Problem 7. To demonstrate how convergence is affected by the value of the relaxation factor
ω in SOR, run your function from Problem 6 with ω = 1 , 1.05 , 1.1, . . . , 1.9, 1.95 and n = 20 .Plot the number of computed iterations as a function of ω. Return the value of ω that results in the least number of iterations. Note that the matrix A from Problem 6 is not strictly diagonally dominant. However,
A is positive definite, so the algorithm will converge. Unfortunately, convergence for these kinds of systems usually requires more iterations than for strictly diagonally dominant systems. Therefore, set tol=1e-2 and maxiter=1000 .Recall that ω = 1 corresponds to the Gauss-Seidel method. Choosing a more optimal relaxation factor saves a large number of iterations. This could translate to saving days or weeks of computation time while solving extremely large linear systems on a supercomputer. |
4436 | https://www.youtube.com/watch?v=dR3ccx8JT3o | Spiral Similarity by Ananda Bhaduri - Unofficial IMOTC 2022
Online Math Club
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Posted: 13 Nov 2022
It is a geometry session which discusses spiral similarity and gliding lemma.
This lecture was a part of the Unofficial IMO training camp 2022 which was organized because the official training camp could not be organized in India. All the lectures of it are uploaded on OMC youtube channel. To see the other lectures of Unofficial IMOTC, you can check out the following playlist -
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You can send all queries by mail to onlinemathclub4@gmail.com.
The Online Math Club is an initiative to reach high school students interested in math and give them a platform to learn more and interact with others. The Club aims to increase exposure to olympiad mathematics and advanced mathematics. And OMC is free and open to all! You can find more information on the website. Fill out this form to join, or if you want to join the mailing list:
math #olympiad #IMO #mathclub #OMC #unofficialimotc #imotc #gliding #glidinglemma #spiralsim #spiralsimilarity #geo #geometry
2 comments
Transcript:
[Music] thank you we can start so first we'll Define directly subtract so they are similar uh directly similar if they have like the same they're similar and they have like the same orientation so uh like if you're familiar with directed angles it means that all the directed corresponding dietary triangles are equal as well so for example here we have like B ABC and say what A to B to C2 you can say that ABC and A2 B2 C2 are both clockwise in this thing and uh but A1 B1 C1 is not directly similar to ABC or a to B to C2 because A1 B1 C1 is anti-clockwise so the directed angle A1 B1 C1 is not equal to the directed angle ABC so they are not similar and also you have like the ratio condition that a b over a to b 2 is ACA over a to C2 and the late deck for this is like uh this plus and Sim similar thing so you can this is this thing is the related right so this is the latex for this and uh if they are not directly similar they are inversely similar uh so for example A1 B1 C1 is inversely similar to both of these triangles and you use this minus and this thing right so for example uh you can get an inversely similar triangles uh triangle if you reflect this triangle in a line because reflection does not preserve directed angle things so if you now right so one thing is that if you say rotate or take a hormacity then directed angles are preserved so for example if you compose like take a rotations and Homo that is you send triangles to directly similar triangles and uh so now we Define a spiral similarity right so smile similarity at a point p is just like a commodity and followed by rotation at B so you can imagine something like uh say there's a point P here and if you say rotate ABC about Point p and then just take a monthly p say if you rotate in this direction and take a commodity you can send a to A to B to B2 and C to C2 so the spiral similarity is just that like it's and it the order does not matter uh these two commute so do you see why uh rotations and homotheses preserve directed angles hopefully that should be clear yeah right and then there's this thing is that see there's a spiral similarity if you take a spiral similarity sending ABC to something else then that triangle must be directly similar but the thing is if you have any two directly similar triangles then there must exist the spiral similarity taking ABC to that triangle so for example what it says is like we must have some spiral similarity taking ABC to A2 B2 C2 uh right we'll show that in the next section but one more thing this is just equivalent to showing that there is a spirals minority taking AC to a to C2 do we see why if say a spiral similarity takes a to A2 and C2 C2 why must they take B to B2 yeah because on the similar similar triangles yeah yeah because they preserve directed things and this one uh this is the thing there is a unique point B such that for any uh triangle ABC and point C and F so in RDS A2 and C2 there is a unique point B such that ABC is spirally similar to A to B to C2 now we'll construct this uh spell similarity explicitly uh and show that this is unique uh so right all right for any two segments say in our case it's DC and e f then there is either a spirals minority or a translation taking BC to EF uh right B to e and c2s so we can ignore the translation case because it's like it's when b c is equal and parallel to yes uh so now else we'll Define this file similar uh Center to be like we intersect EB and FCA let them intersect at a and then you have let P be the intersection of a b c and AEF right so do you see why uh there is a spiral similarity at P taking BC to EF okay so what you really want to prove is that uh uh angle cpf is angle bpe because that is the rotation part and angle uh CP by P F is BP by P E because that's the homothetical so if there's that thing then it takes b2e and C to F uh angle cpf Okay so yeah cpf yeah cpf all right foreign then that's the Homeopathy part of the Spiral similarity so what is what are these two conditions equivalent to um we have a angle e p f equal to angle EAS which is equal to mbpsc which is equal to NPC PPC and then ppf is common so we have angle e p b equal to FPC yeah that's correct yeah so that's true and you also want to prove the ratio part because so like what is this equivalent to I mean let me show the ratios so fine we show that the rotation angles are equal uh angle cpf and bpe okay and PVC will be equal to angle p e f right yeah so that will give that the triangles are directly similar yeah exactly so the angle triangles PBC and pef are directly similar and that gives us the ratio condition we want PB by P is PC by pm alternatively we could have said that PEB is PFC like we could have proved that PBE is directly similar to PCM so that's the second thing we have that this pile similarity which takes the center of the this spiral similarity which takes BC to EF also takes b e to c f like it's the same uh thing because uh it takes uh taking BC to EF is equivalent to showing that triangles PBE and PCF are directly similar but that is also equivalent to Triangle PBC and pef being directly similar does that make sense yes yeah so for example uh here our rotation angle taking bc2ef was angle bpe but now our rotation angle taking b e to c f is angle bpc equals angle EPF equals angle BAC so and also like this point p is unique I'll show that later but for now let's assume this that this point is unique and so we can like apply this theorem uh to segments B and CF because we applied this theorem to BC and EF to construct Point p and so applying this to B and C F should give the same point p so these two circles like if you intersect BC and E F at D then b b and BCF should also pass through that point I think you can also show this using angle chasing that these four circles are concurrent at a point uh but yeah so P now we get is this spirals Center sending BC to EF and b e to c f and now here comes the rotation part so now it's helpful to think of this rotation in terms of the angle between lines so we'll focus on these spiral similarity taking BC to EF ensure that it is unique so [Music] um if you rotate BC by an angle Theta and it becomes some line e f then the angle between VC and EF must be Theta does that make sense uh so suppose this angle you rotated by this angle and this angle so you rotated line BC by this angle and then to go commodity but 130 does not matter but if you rotate BC by this angle then the new line forms an angle equal to this with uh this I think one way to see is this like if you say have BC let's say this is line BC and you drew a line parallel to BC now you rotate let's say this is you're rotating about this point p and this is the line you are rotating now if you rotate by an angle Theta say like that Theta then this also becomes something like this Theta because like rotation preserves parallel lines uh right so what follows is that angle bde should equal BP e like if there is a spiral similarity P sending BC to EF that is uh it sends B to e and C to F right so uh angle bpe is the rotation angle and it's also the angle between lines BC and here so BPD is civics so is the dpcf so that gives that P must be unique so uh I'll give you like sometimes to think about this proof and in general it's useful to think about these as like angles between lines instead of like yeah angle chasing becomes much easier if you just like draw the lines and think of like these rotation arguments and like you don't have to add some more new points to angle chase you can just like there are these intersect these lines and so that ends what I wanted to say about Theory we'll cover some more later but for now uh any questions so far right no great so now why is this yeah so why is spiral similarity like a powerful technique so one thing is that you can relate like unrelated points like you can form angles between like floating points if you have a spiral similarity relation so let's say you have a PBC directly similar to EA for some triangles and you have no way to relate like b c e f are like almost disjoint in the original diagram but using spiral similarities you get a new condition that PCF these two are equal so like B and E were earlier they were completely unconnected but you can relate to them with each other so I'm not being very clear with this so one more so one more thing is that if you have say a point p and let's say you take some spiral similarity at me taking a new foreign so you have say you have this thing all right as a result you get like uh you can connect more points say you have the midpoint of a b and the midpoint of this thing then you have like this angle relation that this file similarity at T which sends uh a b to a dash dash B dash dash since D to C right because uh it should preserve the like segment a b now this gives us a relation like P angle uh angle p a Aid yeah p a dash dash a p c Dash so all of these are equal and so like this thing all these three triangles are similar yes not um hey so you get like a relation between c and b just because you have a spiral similarity sending a b to a dash dash B dash dash so this thing is really powerful and like you can uh form angles between midpoints of things and like like uh concyclic and say we have like con cyclic conditions are not really good at handling midpoints or random lines but you can do this you can do these things with spiral similarities so hopefully that convinced you that these are not completely useless right so far we did this okay now we will go to our first problem um so can you see the problem yeah okay so ABC is a Pentagon such that these angles are equal D is so let's see this e C p a a c and we want to show that uh c e bisects BD right think let's see so any initial observations like what is like ABC and C D E are similar right so ABC and CD are similar are they directly similar or does it matter are they directly similar or not it doesn't matter yeah it doesn't matter because they are isosceles so they are symmetric but is ABC in that order similar to CTE in that order inversely similar like this what do you think oh that should be directly right if you just switch ABC yeah they are similar but uh they are not correctly similar if you take ABC and CDE in that order but if you take a b c and EDC in that order they are digestible yeah so you can just switch the numbers doesn't matter yeah but the thing is there is no spiral similarity at C taking a b to D ah because otherwise C B A would be directly similar to CBE yeah anyway they are similar so that's an observation and uh thing so um wait foreign does this have a projective solution also I don't think so this is a grade 8 problem it's last year's 8.7 Sharingan uh no but the spiral solution is particular yeah it looks because you have parallel lines and Bitcoins right so that's why you thought projective yeah also [Music] thank you yeah because you have this harmonic ratio but no the parallel lines and midpoints are not really useful um it could have a solution yeah yeah so AE parallel to CD is an important observation all right yes okay so actually okay and you have this similar triangle what is a good reference triangle with which you can consider this problem so you have like DB bisexual reference no no not biocentric like just the point of view oh okay I thought you were going to do barrier Center I don't do best and not so we have CDE and cab so these words b e um okay I think one major phase is suppose you are you have to construct such a pentagon how would you start okay you can construct like CD and then you can construct some X angle from DC at C so yeah the line CE you have angle c d e also so you'll get the point e right yeah so you have constructed CD yes now you can construct a parallel from E to get a yeah you get EA you don't get a yeah yeah so you can vary a on the line through a parallel two yeah you can just take a anywhere and then you can construct B because you have that yeah so that is how you do you start with triangle CDE you construct a and then you get B so what I was saying is that if you consider this problem with respect to Triangle CD that makes it a lot easier so what you're essentially doing is that uh you are wearing a on this line and you want to show that b moves on B moves such that b d is bisected by C what does that mean foreign so okay uh so yeah so I'll repeat my question uh so as you said you'd start this problem by constructing triangle c d e then you draw the line three parallel to CD you vary a on this and then from a you get to construct B now you're wearing a on this line and B moves such that according to the problem BD is bisected by c what does that mean s [Music] sorry it can't be the reflection though so like you basically consider reflection of D over every point of CE and right so you take advantage like uh take a homocity at D with ratio two all right yes so like the CL line goes to the locus of B right sorry this is true so B moves on this line as a moves on this line okay now what is the relation between a and b how are they related so you can think of it as there's a function from this line to this line there is a function which Maps this line to this line uh what is that function s so uh as a moves on this line B moves on this line according to the problem so there's what I was saying is that there is a function from here to here what is that function rotation yeah rotation and Direction yeah exactly so there's a rotation and commodity because the factor CA by c b is fixed because CBA is uh similar to a fixed triangle oh right so is this rotation Factor picks so all we need to show is that this spiral similarity at C which Maps uh A to B Maps this line to this line so we know about uh that this has ratio CA by c b and it has angularly equal to ACB hey so first things can we prove that uh like AE gets mapped parallel to a line uh through this like what I mean is that what is the uh angle between line a b a e and the like image of a under this spiral similarity so like what I was saying is the spiral similarity at c maps this line to this line now how can you determine say you we didn't know this that this file similarity mapped this line to this line then what would be the angle between this line and its image under this spiral simulator like you can think of like what I mentioned about rotating angles rotating lines and how they how you can angle Chase using those things foreign foreign great uh so does anyone have any ideas on how do you prove that uh the image of this line is Baseline under this spiral similarity um okay so we have to prove that the a is mapped to the other line when we consider spiral similarity about C right yeah we have to prove that uh this blue line is mapped to this red line under the spiral similarity the red line is defined as like this line at a homotherapy at D with Factor 2. and blue line is defined as the line through e parallel to CD CD are fixed we are only moving a and B is also moving so like now what is the angle between the blue line and the red line will be equal to angle aec here which is this angle uh which is also this angle which we want to show is the rotation and okay which is the rotation angle so the rotation component of our spiral similarity is angle ACB and the homo30 part is c b by c a um right so one thing we know is that this Blue Line gets mapped parallel to a line parallel to Blue Line gets mapped to a line parallel to the red line now we want to show it is the red line so for that we need to add another Spiral similar triangle let's say we want to make things parallel similar so there's CBE okay so we should for that we need to show that one point on this line under this spiral similarity at C gets mapped to one point on this red line that would be enough right yeah so we want to show that the Blue Line gets mapped to the red line yeah yeah so we know that by a rotation component that the Blue Line gets mapped parallel to a line passing through the red line parallel or coincidence so if we manage to show that one point on the blue line gets mapped to some point on the red line then we are done does that make sense yeah okay I'll repeat again so uh from the beginning so we start this is how we think of the problem we fix triangle CDE which is isosceles and the angle condition gives us that a moves is a e is parallel to C B so we move a on the line through e parallel to c d right and then we can construct B such that c a b equals BCA equals this fixed angle so we can construct b as a function of a now the problem also tells us that the locus DB is bisected by c e or alternatively if you take a Homeopathy at D with Factor 2 it sends this line to this line so B moves on a fixed line which is uh as a moves on this line and we have a function sending a to B which is like this file similarity at C sending a rotation by this angle and with Factor CB by CA and if we put that this indeed sends the red line to the blue line then we are done because then it gives us that c b lies on the like uh if you uh the image of this line under homo 382 at D so DB is bisected by c d c e so that concludes but and we have shown that blue line gets mapped uh parallel line parallel to the red line so if we show that any one point on the blue line gets mapped to a point on the red line then we are done okay do you see why the spiral similarity has come up line to a line yeah yeah because it preserves the shape uh okay so why does it map this blue line to the red line I should probably do it let's see so can I see yeah yeah right yeah so if we consider the map of E yeah Point uh e Dash says that uh the triangle e e Dash c will be similar to e d e d c but as uh now this is kind of this is like a reflection and as the uh the red line is just the reflection of D over the line CE we get that e Dash will lie on the red line right so what is point e Dash uh could you repeat that again uh it's the like considers parent similarity about c e snap to e Dash okay right so uh I think uh wait a dash we can Define what happened okay I think we have already defined Dash summer so we'll call this A1 right so see e Dash e c e one e is similar to CBA right so what it gives is that c this piles similarity which Maps A to B Maps e to E1 now since we know that c d e is similar to this fixed triangle we get that c e 1 e is a parallelogram right because like what e e one c e should be equal to c e d so e1c parallel to C D now e is a point on the blue line and the E1 E1 is a point on the uh E1 is a point on the red line because if f is the midpoint of e c and if you reflect D and F you get E1 so the spiral similarity at c maps e to E1 and so it must map the blue line to the red line or we Define E1 to be the intersection of the red and blue lines right yeah no I defined it such that ce1ed is a parallelogram so and by your definition it's like c one e is similar to C CBA so see even each uh see this spiral similarity which Maps A to B must math e to E1 so E A goes to e1b okay right so this problem was very instructive because you could like tell what the problem wanted you to do uh like most geometry problems it's guess and check but you can work backwards very easily with this one and uh you can like thinking of spiral similarities as functions which may applying to lines instead of like usually you think think of the Mikhail point and then spiral similarities so this is like really instructive I'll give you a few moments to digest the solution does anyone have any questions or comments on alternative Solutions [Music] okay someone is suggesting bash how do you batch this uh somehow like yeah I'll tell you the method you can send CDE as reference okay then you will have a question of EA you can just take a as any random point on here okay right yeah right and then uh since CD is similar to ABC you can find B also because yeah and then you prove B is done this slide yep or like you can just prove like uh find BD cap c e and then or basically you can just take mid account of BD and prove that c midpoint and E are collinear because c and e are just like reference points so they are nice okay yeah yeah I think someone did Bash this in context yeah like you can of course because something to do with medians and this um I think there is some Sim medium here yeah B is the easy median in Triangle Ace B is the easy median yeah I think that's what he did yeah my bachelor is definitely other that works too like it's not that messy if your determination you can do it in like 13 minutes hey yeah but it's thinking in terms of spiral similarities it's like really nice yeah like of course for starting motivation you'll have to think about the structure only yeah yeah this problem is so structured exactly so you can construct b as a function of a so you think this function is actually a spiral Sim so this parallelism Maps the fixed line to fix line and then you just like okay fine next problem okay so this is the problem statement and we'll see it oh bpc is associated that BP equal to PC yeah BBC is the right isosceles triangle DPS angle bpc is 90 right yeah yeah bpc is 90. NAB is 90. m a uh uh C is 90. so we have three right isosceles triangles and you want to find you want to show that this is also right isosceles this thing yeah so okay okay no no yes I think this is bashable uh more than the last problem but no we want to find a synthetic solution uh right so what's spiral similarities do you see in the driver APC goes to NAB a sorry again oh spiral similarity so like B sends a PC to any yeah so we want to show that it's uh there is a spiral similarity at P sending BC to MN right foreign but like DPC PC goes to n is also as a parallel right BC goes to PC PC goes to n a n a yep no those are in fact inversely similar triangle bpc because you can see that bpc is clockwise ban is sorry yeah and it's anti-clockwise b equals to n foreign but there's also like there's also uh 90 degrees at these two this point right the five participants okay fine all right so any other observations uh these spiral similarities one thing you see see the condition is uh sorry the problem asks you to prove that PBC is directly similar to pmn right so you want P to be the Michael point of b c m m or I think some permutation oh sorry so we write till how long can I uh continue this class uh almost 10 million now it's 7 42 yes uh you can do till 8 30 is that fine 8 30 like okay fine I'll try to finish how much time you need more like yeah it's fine okay till 8 45 is that fine yeah because anyway if like we have only four people not including me yeah so if you're fine with it then fine yeah I I don't mind like if the other people who if they are attending the next class they are fine with till nine also I'm fine I don't mind okay coming back to this problem also if you want I can just create a portal when the people want and we can just do it till then 845 or nine okay I'll do that so coming back to this problem we have like a and b directly similar to ACM right so what does that give that's the spiral similarity we can consider that yeah the smile similarity which sends nb2cn so a is the Michael point so uh what does that suggest like what points should we add since a is the mcal point we can point off which sends NB to cm because okay you can see a and b similar to ACM and we have a and b ACM are both anti-clockwise this I think maybe VM intersection CN sorry um BM intersection c n right so BM intersection CN let's call that uh k right so what do we do with Point k this angle key uh b k c will be 90. right so big AC is 90 right yeah so we get that yeah that's true bkc is 90. foreign [Music] uh it says my internet connection is unstable can you hear me now yeah okay fine that's great yeah so as uh PB is equals to PC we get that PK is the external angle bisector of angle bkc right yeah so now every uh we just need to prove that n m k p is cyclic right but um yeah that would imply that uh p is the Michel point but do we not already have enough information to conclude that P is the Michel point that's what you want to prove right that P is the Michael point of which sends MN to BC working okay so we want to prove pmn is partly similar to PBC which is equivalent to what okay uh also how did you get that dkc is 90. I just rotation about K so right yes so a is the Michael point so a is also the intersection of these two circles foreign foreign foreign [Music] CN intersection B M lies on this circle so we have because since PC BK is like cyclic angle PCN is pck equals pbk equals pvn so we have angle sorry yeah sorry right so we have PCN is equal to PBM now what else do we need to show that uh pmn is isosceles and right triangle oh we also have PC equals DB foreign or we have angle p n c is same as p and K which is pmk which is same as PMB right but for that we need uh nkpm cyclic right yeah that would imply or problem but yeah so what are we aiming for um [Music] which can be true [Music] sorry uh you were saying that CN is [Music] oh you are saying PCN congruent to PBM [Music] so what are you saying that CN is equal to BM uh your voice is like slightly breaking so could you type that in chat anyway um ABM is equal to a and C right ABM is equal to a and C because uh right BM is equal to cm yeah exactly that finishes so we have BM is equal to c n p b equals PC and PCN is equal to pvm so that gives triangle PBM is directly similar to Triangle PCN which gives PCB directly similar to p and n does that make sense okay let's see the difference thanks so we have these two triangles are similar so spiral similarities come in pairs so if P takes b n to c n it also takes BC to m n right so that finishes so what did we use in this problem so we started with uh A and B is spirally similar to ACN so by using spiral Spiral similarities come in pairs we got a and C which is a change right so we have we started with A and B is partly similar to ACM then by applying spiral similarities come in pairs we got a and C is similar to ABM and in fact they must be congruent because a n is equal to AC and a b is equal to m so that gives b m equals c n and also the Michael condition gives us that their intersection k lies on the circle with diameter BC because like bkc is bpc like bkc equals b k n is equal to b a n so we just applied this thing multiple times and we got what we wanted there is also a slick solution using just rotations which I'll like give you a link to after the class is this the one liner all right okay now we'll move on fine I want to cover something new [Music] so this is the coaxiality Lemma which says that if you have say two circles let's say this and this and suppose they intersected X and Y right maybe let's see suppose we have two circles which intersect at X and Y uh do any of you know these statement of the Forgotten causality Lemma okay so let's move fine so you have two points p and Q p and Q and we claim that uh x p q is square coaxial with these two circles uh sorry x p q okay so xpq is coaxial with these two circles if and only if we Q go axial with these circles if all right see the SE look okay this is okay okay so x p q is coaxial with say this is C1 and this is C2 if and only if the power of p with respect to C1 over the power of p with respect to C2 is equal to the power of Q with respect to c 1 over the power of Q with respect to C2 right so how is this related to spiral similarities how how do you think this should be related to Smiles matters uh did you get the question yeah so have you seen this before okay so how is this related response foreign x p intersect C1 at cp1 right P okay and C2 at P2 yes and now similarly for X Cube we get q1 Q2 yes yes now we have to prove that this is parallel similarities entered at y that match uh PQ to P1 q1 okay yeah okay that's equivalent to this so because if there were a spiral similarity sending P1 P to q1 Q then by our Michael Point construction we have XP Q lies on this right now how do we use this what is this condition equivalent to power of C1 over p over power of C2 over P equals part of c one over Q um or p p 1 over pp2 is qq1 over qq2 yeah dp1 over pp2 is okay yes so now there's a spiral similarity sending P to Q E1 to q1 but okay now we already have a spiral similarity sending P1 P2 to q1 Q2 by definition yeah now why does this end point p so uh rotations and homotherapies preserve the shapes of lines right so like if P has this ratio with P1 and P2 then if it gets too much mapped to some new Point p dash then that also has this relation right so uh the spiral similarity which Maps P1 P2 to q1 Q2 Max P to Q so p uh so we have like x y p p one is finally similar to yqq1 so Y is also the Michael point of t v 1 q q one and that gives you the coaxial condition yeah so if you very P say okay ideally we should have fixed P1 and K1 but say if you fix P1 P1 P2 Q2 and you move p and Q such that this holes uh right someone data pool you can answer that yeah for like class and I just skipped it out can I vote no no you are host why is there a right now option and for the second joke okay fine okay it's already eight eight so let's imagine that P1 P2 q1 here also you can see okay 845 suits okay fine we'll and try to undertake 45. let's imagine that P1 P2 q1 Q2 are fixed and P and Q moves such that this holds then this xpq always passes through y so you can imagine that P and Q are moving with constant velocity the constant velocity ratio is always preserved and so xpq is uh always passes through the fixed point Y now uh say for example so if you have say a triangle I'll delete everything again you have a triangle say a b c right and see you have points DNA moving on a b line CB and AC with constant velocity then Ade passes through always passes through a fixed point this circle always passes through a fixed point this is what the causality lam is saying so if you like move D and E with constant velocity now if you remember uh from malays class we had this problem the problem was Point D I think and then you drove parallel lines from here to here and from here to here and you are asked to show that AEF always is always spelled always passes through a fixed point so the proof you can do this uh so in that place we actually characterize that fixed point but you can also do this with just the coaxiality demo see because e is moving with constant velocity if you move D with constant velocity so is f moving with constant velocity and so like if you fix say some two positions of d say call this B ditch and draw the parallel lines foreign [Music] so you fix two points uh e Dash and F Dash sorry D Dash and B now for any point say here say G and then again you draw parallel lines sorry we'll call this T1 we'll call this E1 and we'll call this F1 right so do you see why ee1 over E1 e Dash is equal to F Dash F1 over F1 F Dash F1 f yeah yeah so because if you say fix D and you're moving this then this is equal to D Dash D1 over D1 B and so you have this like points are moving with constant velocity and the AEF passes through a fixed point which is the Michel Point carrying uh in this case it was the Michael Point carrying ba to AC so it might be different for like different velocities so in any case if you have like something a problem set up which involves uh points moving on lines a b and AC and this circle is passes through a fixed point then the idea is always like there's a spiral similarity involved I think there is one more problem which uh involving this thing where the fixed point is the Humpty Point instead of the Dumpty point uh I don't remember which one I think it was some Elmo problem right okay so this is one thing I wanted to discuss about this is like really cool if you have points moving with constant velocity this circle is always passed through a fixed point and then the next thing I want to discuss is the gliding principle yeah this was nice yeah so gliding principle says that yeah this is even cooler uh if you have say fixed spirally similar triangles ABC let's create another step a point p and rotate 7 45 is yeah so we have says spirally similar triangles a b c and a Dash B Dash C Dash the thing is if you take say this segments c c Dash and let's say you take their midpoints midpoint midpoint then the gliding principle says that def is in fact similar to both ABC and a Dash B Dash C Dash so let's see okay great so all these three triangles are similar you took the midpoints of these segments so a a dash b b Dash CC Dash in fact you could have taken like divided them in any ratio say you could have divided them in one to two Ratio or any ratio you wanted but in our case let's say you take take their midpoints now do you see how to prove this what should the proof involve hey so it's like quite similar to the proof of the causality Lemma so what should it involve I think you can also batch this if you take like complex numbers it shouldn't be too bad one case where so foreign [Music] foreign okay so we are down to four people four is okay so the idea is again like the spiral similarities come in pairs so if there's a spiral similarity taking a b to a dash B Dash there is a spiral similarity taking a it has to be root this has been like used in most problems in the in this lecture so that's the main idea similarity will take D to E yeah exactly so pde is similar to PAB so how do we prove that say angle EDF is equal to angle BAC foreign but okay so we have pde uh Spider is similar to PAB and P a dash B Dash and say we want to prove that f d e is equal to cab how do we finish okay yeah so as part of said uh sorry not run up I know PDF is partly similar to p dash AC so let's say p d f is similar to p dash is C so adding these two we get that PDF Plus sorry PD minus PDF is P A B minus p a c which is like BSC similarly you have the other things so edfs file is similar to this this and this so one cool application is Napoleon's so if you have say a triangle a b c sorry foreign so foreign centers [Music] and show that this is in fact an equilibrium is this an igemo chapter six uh I think you complex passions yeah yeah okay so you can also do this using just file similarities the gliding principle so any ideas let's work backwards right so let's just take two equilateral triangles say these two and we want some linear combination of their segments which is nice which gives us like nice points so can you think of something uh say on segments eo3 ao2 and b01 okay it's almost safe foreign foreign H and I now o and I over 1B o2g or O2 a these are all like one to three ratios because O2 is the centroid so it's equivalent to proving that have I got this way sorry what is it yeah yeah it should be this so so if you take uh these three points they all are like a divide this segment in a fixed ratio one is to three and so if we prove that h g i equals is equilateral we are done where H G and I are the midpoints of c f c d and a b now how do we finish from here okay this is true yeah yeah this is true so H is the midpoint of a b f is the I is the midpoint of FC and C is the G is the midpoint of C2 how should we finish from okay uh-huh exactly AC and BDC so Ace and BDC so you have H is the midpoint of a b c g is the midpoint of C D and I is the midpoint of CF so you're averaging the triangles a c f and b d c is that clear can you please repeat it okay so we are just applying the gliding principle twice so what we uh we we were working backwards actually so first we showed that this is equivalent to proving that ghi is equilateral because we started with we applied the gliding principle to Triangle EAB and triangle one or three o two so e03 we have Point h on line eo3 I online ao2 and G online po1 so and you can see that these ratios are fixed go1 over GB is uh one-third ho3 over h e is one third io2 or I a is one third so this is in fact like a linear combination of what you say not linear combination I think weighted average is the better word so o1o203 is a weighted average of orange triangle and this green triangle so since both of these are equilateral this will be equilateral we want to show that this is equivalent and now we uh so since these h i and G are the midpoints of a b c f and CD so we again apply take the weighted average of BBC and AFC so H is limit point of this then this and then this is that clear so AGI is equilateral and then it follows that blue triangle is equivalent yeah okay okay now we come to the final problem I think this will take a while it's 8 30. earn it so this is really cool um so sorry oh I think which means it how do you copy this but tons okay let's stop sharing instruction sorry now okay meanwhile you can check out the statement of ISL 2000 G6 I am having trouble pasting it here so this is the next problem foreign things okay can you see my screen now yes so this is the room okay foreign okay fine finally so this statement is really good we have a y b is to adx so we have that angle is X a b equals angles b c x sorry yeah ABX is bcx and Dax is cbx these are not directed actually so the aops statement uses directed angles but these are angles are not Direct so actually x a d and x bc are in fact inversely similar not even directly similar and it says that if x a b and X the statement just says that x a b and x b c are inversely similar they're trying then angles you can determine angles x d and X KTV so it says that basically this point x is unique if the problem statement is true uh so you know that there's a unique point x such that x a d is directly similar to xpc so this says that if like except for some edge cases the point takes us that xad is inversely similar to xbc is also unique now the first thing we want to do here is like since we have inversely similar triangles we can't deal with this very nicely so how do we convert inversely similar triangles to directly similar travels maybe reflect yeah reflect exactly so what do we reflect let us let's reflect one triangle about some line through it okay also um keep in mind this a y b equals to ADH so what we want for now let's restrict ourselves to like how would we go about constructing point x okay reflection is a good idea because reflecting will give you directly similar triangles yeah no I didn't say anything pronounce okay yeah reflection is a good idea what do we reflect to really it's foreign some uh so do you want me to add uh perpendicular biceps of KB and CD yeah okay yes so do you want me to reflect this about X about perpendicular bisector of a b oh yeah okay I see so a goes to B let's reflect D okay just X you want or do you also want no the whole triangle adx d right so we have e x d one is similar to bxc yes yes okay and that is equal to what do you want to do we want to prove that x d a is half of d y a so x d a is bd1 X is half of this okay so pictures okay so we only have five minutes left unless both of you are okay with some more time do you plan to attend the next class and final night yeah okay maybe kind of is fine then maybe I can leave this problem as homework but this you should really try this is brilliant uh so not easy but yeah you should really do this this is a good idea so one thing is that this bya equals to bdx reminds you of uh the circumcentered thing because the angles are printed at the circumcenter is twice angles are printed there yeah uh actually this isn't uh my solution path but I think this is still promising so you reflect X and this bisector and this is also but this is okay so this is a cyclic quadrilateral okay since we have like say 10 minutes left should I walk you through my solution path uh it's fine yeah I think that's better yes yeah so what I wanted to do was uh I reflect them in AD and BC so you have a reflection uh X in a d X in BC now okay ax Dash D is similar to bxc so this spiral similarity which takes a d to BC takes x dash 2X similarly for this thing so let's say how do you construct I'll remove the statement now is that fine yes so we can construct the Michael of a b and BC say let's call this n let's call this G right so G takes uh actually bring this part but yeah this is what sorry right so G is the Michael which sends a B to C B and A B to BC now G's and x dash to X can you see why yeah because G sends adx to bcx similarly G sends x1-2x in fact G is the Dumpty of x dash x x 1 dash because it sends x dash X to x x 1 dash so G is the Dumpty right and now we apply the gliding principle so we have on triangles a x Dash D and B x c so actually uh so we see that this like this reflection condition is that uh um so the midpoint of AC is like you want to add that because you're reflecting about this point so you you want to add uh these ways this point not being natural okay let's see equals X Y2 yeah so okay also what we wanted to show now becomes very natural x dash DX is a y b this is what our problem reduces to okay so now we add these two things N1 and N2 and we like basically take the average of triangle bxc and a x dash 2 so we have the midpoints uh so by the gliding principle it's Orange M1 and M2 and M1 N2 M2 are in fact inversely similar so what does that mean can you please repeat okay so first we reflected uh you got this part that ax Dash B is similar to this triangle bxc yeah no can you please repeat what to add for M1 M2 M1 is the midpoint of a b and M2 is basically we are averaging these two triangles by taking their midpoints the gliding principle so the midpoint of ba is M1 the midpoint of x x dash is N1 and the midpoint of CD is M2 so orange triangle similar to pink pink pink pink similarly this orange this orange all of these triangles are similar so what this means is that N1 n and N2 are reflections of each other in M1 M2 because these two triangles are similar now with this we can construct X because say uh this is how we construct X we take line a b in fact this is how I constructed X in my example we take line a b we reflect it in line M1 M2 where is that here yeah yeah so this okay I'll do it again I'll reflect this in this so if I reflect a b in M1 M2 and intersect suppose that intersects at a point N2 then we reflect N2 back in M1 M2 to get N1 then we know that M1 and M2 is similar to M1 N2 M2 right so you know why uh do you get my points N1 and N2 are unique how do you define N1 into uh okay so let's just yeah sorry let's say we don't know what uh I'm telling you how to construct X so initially I defined of course as the midpoint of x x dash and x x 1 dash but suppose we want to construct x given a b c d how do we do that that was what I was telling you uh sorry for confusing so let's erase everything let's erase X results okay suppose we don't know anything about X we want to construct X so we know that if the perpendicular from X to a d and BC are N1 and N2 then N1 N2 are reflections across M1 and 2. is that clear yeah so basically if we reflect a d and M1 M2 then and that becomes say this line say this red line then that intersects BC at N2 and if we reflect N2 back in M1 M2 then these two points are reflections across each other and this one lies on ad this one lies on BC yeah so we can then take the perpendiculars from N1 to a d and N2 to BC and they can intersect it X Y is X okay X so this is how we construct X so now let's get back here to our problem so what we want to show is that um angle N1 M2 N2 which is twice of a d y is a y b so how do we go about proving this there's more spiders uh actually I think I should I'll leave this as homework to you it's already 8 52. so right this is where 300 plus you can also try uh continuing that approach you were talking about reflecting uh X in the perpendicular bisector of baby we are done right sorry my gliding principle and Gad and GBC uh Gad and GBC oh yeah so G so that gives that gm1 M2 is similar to Gad and GBC right that is correct um let's say that analysis [Music] uh great I think yeah that so we know that uh we want to show that say ayb is fairly similar to N1 M2 that would finish or um see uh thing is we want to reduce it to lines we are comfortable with so we are somewhat comfortable with lines M1 M2 because in terms of ABCD but not so much with N1 and N2 because they are defined somewhat randomly so if we show that M1 M2 if we find this if we show that this problem is equal to a spiral similarity sending M1 M2 to some other line uh then we are done we are done like we can do finish it easy so we want to ignore points N1 and N2 and find the spiral similarity with sense M1 and 2 to some other so I'll leave this as homework thanks for attending so uh you can stop recording now run off is either are you there uh you can stop recording okay anyway so what I wanted to say was our like other classes also low as this uh yeah is this the only class with four people two people attending yeah it's usual this is you |
4437 | https://brainly.com/question/17570487 | [FREE] Find the numbers 1-20 using only four 4's and any operation. - brainly.com
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Find the numbers 1-20 using only four 4's and any operation.
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Asked by cupcakedarling3360 • 09/14/2020
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To find the numbers 1-20 using four 4's and any operations, start with 4 and perform addition, subtraction, multiplication, and division with the other 4's to find different numbers.
Explanation
To find the numbers 1-20 using only four 4's and any operations, we can use the following steps:
Start with the number 4 (the first 4).
Perform the following operations: 4 + 4 = 8, 4 - 4 = 0, 4 4 = 16, 4 / 4 = 1
Combine these results with the remaining 4's and repeat the operations.
For example: 8 + 4 = 12, 8 - 4 = 4, 8 4 = 32, 8 / 4 = 2
Continue this process until you have used all four 4's.
Using these steps, you can find all the numbers from 1 to 20 using only four 4's and any operations.
Learn more about Numbers 1-20 using four 4's and any operations here:
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Answered by CharlesBronson •13.2K answers•5.2M people helped
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Using four 4's and different mathematical operations, we can create expressions that yield the numbers from 1 to 20. This is done by employing addition, subtraction, multiplication, and division. Each number can be expressed in various creative ways with these four 4's.
Explanation
To find the numbers from 1 to 20 using only four 4's and any mathematical operations, we can creatively use addition, subtraction, multiplication, division, and parentheses to construct expressions that yield these numbers. Here are some examples of how to achieve each number:
1 = 4 4+4 4−4
2 = 4 4+4 4
3 = 4−4 4
4 = 4+0
5 = 4+4 4
6 = 4+4−4 4
7 = 4+4−1
8 = 4+4
9 = 4+4+4 4
10 = 4+4+4−2
11 = 4+4+4−4 4
12 = 4+4+4
13 = 4+4+4+1
14 = 4+4+4+2
15 = 4+4+4+3
16 = 4∗4
17 = 4∗4+1
18 = 4∗4+2
19 = 4∗4+3
20 = 4∗4+4
Using these combinations of four 4's, we can represent all the integers from 1 to 20 through basic arithmetic operations.
Examples & Evidence
For example, the number 1 can be achieved by dividing 4 by itself, while the number 16 can be reached by multiplying 4 by 4. Each expression uses four 4's to generate the desired number through simple math operations.
The approach allows flexible combinations that can be verified by repeating the expressions to ensure they yield the correct results.
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0 = 44-44
1 = 44/44 or (4+4)/(4+4) or (4/4) / (4/4) or [(4! - 4)/ 4] - 4
2 = 4/4+4/4
3 = (4+4+4)/4
4 = 4(4-4)+4
5 = (4 _4+4)/4
6 = 4_.4+4.4
7 = 44/4-4
8 = 4+4.4-.4
9 = 4/4+4+4
10 = 44/4.4
11 = 4/.4+4/4
12 = (44+4)/4
13 = 4!-44/4
14 = 4(4-.4)-.4
15 = 44/4+4
16 = .4(44-4)
17 = 4/4+4 4 18 = 44.4+.4
19 = 4!-4-4/4
20 = 4(4/4+4)
Answered by nellabella6767 •2 answers•309 people helped
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4438 | https://ocw.mit.edu/courses/res-18-011-algebra-i-student-notes-fall-2021/mit18_701f21_full_lec_new.pdf | 18.701: Algebra 1 Jakin Ng, Sanjana Das, and Ethan Yang Fall 2021 Contents 1 Groups 5 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2 Laws of Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.3 Permutation and Symmetric Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.4 Examples of Symmetric Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2 Subgroups and Cyclic Groups 10 2.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.2 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.3 Subgroups of the Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.4 Cyclic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 3 Homomorphisms and Isomorphisms 14 3.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3.2 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4 Isomorphisms and Cosets 18 4.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 4.2 Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 4.3 Automorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 4.4 Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 4.5 Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 5 The Correspondence Theorem 22 5.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5.2 Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5.3 Results of the Counting Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5.4 Normal Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 5.5 The Correspondence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 6 Normal Subgroups and Quotient Groups 27 6.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 6.2 Normal Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 6.3 Quotient Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 6.4 First Isomorphism Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 7 Fields and Vector Spaces 31 7.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 7.2 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 7.3 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 7.4 Bases and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 8 Dimension Formula 35 8.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 1 35 40 45 50 55 60 65 70 75 80 CONTENTS 8.2 Matrix of Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Dimension Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 9 Dimension Formula 38 9.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 9.2 Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 9.3 Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 9.4 Eigenvectors, Eigenvalues, and Diagonalizable Matrices . . . . . . . . . . . . . . . . . . . . . . . 9.5 Finding Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 10 Eigenbases and the Jordan Form 45 10.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 The Characteristic Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 10.3 Jordan Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 11 The Jordan Decomposition 49 11.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 11.2 The Jordan Decomposition, Continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 11.3 Proof of Jordan Decomposition Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Orthogonal Matrices 54 12.1 Dot Products and Orthogonal Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 12.2 The Special Orthogonal Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Orthogonal Matrices in Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 12.4 Orthogonal Matrices in Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 13 Isometries 13.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 13.2 Isometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 13.3 Isometries in 2-space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 14 Symmetry Groups 14.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 14.2 Examples of Symmetry Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 14.3 Discrete Subgroups of R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 14.4 Finite subgroups of O2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 14.5 More Discrete Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 15 Finite and Discrete Subgroups, Continued 69 15.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 15.2 Finite Subgroups of M2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 15.3 Discrete Subgroups of M2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3.1 Discrete Subgroups of R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 15.3.2 Back to Discrete Subgroups of M2! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 16 Discrete Groups 73 16.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 16.2 Examples for L and G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 16.3 Crystallographic Restriction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Group Actions 80 17.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 17.2 Motivating Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.3 What is a group action? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 17.4 The Counting Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 18 Stabilizer 86 18.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 18.2 Counting Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 18.3 Stabilizers of Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 2 CONTENTS 18.4 Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 18.5 Finding the subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 18.6 The Octahedral Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 19 Group Actions on G 91 19.1 Conjugation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 19.2 p-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 20 The Icosahedral Group 96 20.1 Review: The Class Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 20.2 Basic Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 20.3 Conjugacy Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 20.4 Simple Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 20.5 Conjugacy Classes for Symmetric Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 21 Conjugacy Classes for Symmetric and Alternating Groups 101 21.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 21.2 Cycle Type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 21.3 Conjugacy Classes in Sn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 21.4 Class Equation for S4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 21.5 Student Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 22 The Sylow Theorems 107 22.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 22.2 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 22.3 The First Sylow Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 22.4 The Second Sylow Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 22.5 The Third Sylow Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 22.6 Applications of the Sylow Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 23 Proofs and Applications of the Sylow Theorems 114 23.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 23.2 Application: Decomposition of Finite Abelian Groups . . . . . . . . . . . . . . . . . . . . . . . 114 23.3 Proof of Sylow Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 24 Bilinear Forms 119 24.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 24.2 Bilinear Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 24.3 Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 24.4 Bilinear Forms over CC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 25 Orthogonality 125 25.1 Review: Bilinear Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 25.2 Hermitian Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 25.3 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 26 The Projection Formula 130 26.1 Review: Symmetric and Hermitian Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 26.2 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 26.3 Orthogonal Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 26.4 Projection Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 27 Euclidean and Hermitian Spaces 134 27.1 Review: Orthogonal Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 27.2 Euclidean and Hermitian Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 27.3 Gram-Schmidt Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 27.4 Complex Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 28 The Spectral Theorem 138 28.1 Review: Hermitian Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 3 CONTENTS 28.2 The Spectral Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 29 Linear Groups 141 29.1 Geometry of groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 29.2 Geometry of SU2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 29.2.1 Quaternions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 29.2.2 Geometry of the Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 29.2.3 Latitudes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 30 The Special Unitary Group SU2 147 30.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 30.2 Longitudes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 30.3 More Group Theoretic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 30.4 Conjugation and the Orthogonal Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 30.5 One-Parameter Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 31 One-Parameter Subgroups 152 31.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 31.2 Properties of the Matrix Exponential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 31.3 One-Parameter Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 32 One-Parameter Groups, Continued 156 32.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 32.2 Examples! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 32.3 The Special Linear Group SLn(C) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 32.4 Tangent Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 33 Lie Groups 160 33.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 33.2 Lie Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 33.3 Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 33.4 Lie Bracket . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 34 Simple Linear Groups 165 34.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 34.2 Simple Linear Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 34.3 The Special Unitary Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 34.4 The Special Linear Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 34.5 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 35 Hilbert’s Third Problem 171 35.1 Polygons in the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 35.2 The Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 35.3 Some Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 35.4 Back to Polytopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 4 Lecture 1: Groups 1 Groups 1.1 Introduction The lecturer is Davesh Maulik. These notes are taken by Jakin Ng, Sanjana Das, and Ethan Yang, and the note-taking is supervised by Ashay Athalye. Here is some basic information about the class: • The text used in this class will be the 3rd edition of Algebra, by Artin.
• The course website is found on Canvas, and the problem sets will be submitted on Gradescope.
• The problem sets will be due every Tuesday at midnight.
Throughout this semester, we will discuss the fundamentals of linear algebra and group theory, which is the study of symmetries. In this class, we will mostly study groups derived from geometric objects or vector spaces, but in the next course, 18.7021 , more exotic groups will be studied. As a review of basic linear algebra, let’s review invertible matrices. Defnition 1.1 An n×n matrixa A is invertible if there exists some other matrix A−1 such that AA−1 = A−1A = I, the n×n identity matrix. Equivalently, A is invertible if and only if the determinant det(A) ̸= 0. aAn array of numbers (or some other type of object) with n rows and n columns Example 1.2 (n = 2) a b d −b Let A = be a 2×2 matrix. Then its inverse A−1 is 1 .
ad−bc c d −c a Example 1.3 (GLn(R)) A main example that will guide our discussion of groupsa is the general linear group, GLn(R), which is the group of n×n invertible real matrices. aThe concept of a group will be feshed out later in this lecture Throughout the course, we will be returning to this example to illustrate various concepts that we learn about. 1.2 Laws of Composition With our example in mind, let’s start. Guiding Question How can we generalize the nice properties of matrices and matrix multiplication in a useful way? Given two matrices A, B ∈ GLn(R), there is an operation combining them, in particular matrix multiplication, which returns a matrix AB ∈ GLn(R). 2 The matrices under matrix multiplication satisfy several properties: • Noncommutativity. Matrix multiplication is noncommutative, which means that AB is not necessarily the same matrix as BA. So the order that they are listed in does matter.
• Associativity. This means that (AB)C = A(BC), which means that the matrices to be multiplied can be grouped together in diferent confgurations. As a result, we can omit parentheses when writing the product of more than two matrices.
• Inverse. The product of two invertible matrices is also invertible. In particular, = B−1A−1 (AB)−1 . 1Algebra 2 2Since the determinant is multiplicative, det(AB) = det(A) det(B), which is nonzero. 5 Lecture 1: Groups Another way to think of matrices is as an operation on a diferent space. Given a matrix A ∈ GLn(R), a function or transformation on Rn3 can be associated to it, namely TA : Rn − → Rn # » = (x1, · · · , xn) 7− → A# v »4 .
v Since A# v is the matrix product, we notice that TAB ( # v ) = TA(TB (# v )), and so matrix multiplication is the same as function composition. With this motivation, we can defne the notion of a group. » » » Defnition 1.4 (Group) A group is a set G with a composition (or product) law G×G − → G (a, b) 7− → a · b5 fulflling the following conditions: • Identity. There exists some element e ∈ G such that a · e = e · a = a −1 • Inverse. For all a ∈ G, there exists b ∈ G, denoted a , such that a · b = b · a = e. • Associative. For a, b, c ∈ G, (ab)c = a(bc). Also denoted ab In the defnition, both the frst and second conditions automatically give us a unique inverse and identity. For ′ ′ ′ example, if e and e both satisfy property 1, then e · e = e = e , so they must be the same element. A similar argument holds for inverses. Why does associativity matter? It allows us to defne the product g1 · g2 · · · · · gn without the parentheses indicating which groupings they’re multiplied in. Defnition 1.5 n n −1 −1 Let g taken to the power n be the element g = g · · · · · g for n > 0, g = g · · · · · g for n < 0, and e for | {z } | {z } n times n times n = 0. Example 1.6 Some common groups include: Group GLn(R)a Zb C× = C \ {0}c Composition Law matrix multiplication + × Identity In 0 1 Inverse A 7→ A−1 n 7→ −n z 7→ 1 z aThe general linear group bThe integers under addition cThe complex numbers (except 0) under multiplication For the last two groups, there is additional structure: the composition law is commutative. This motivates the following defnition. Defnition 1.7 A group G is abelian if a · b = b · a for all a, b ∈ G. Otherwise, G is called nonabelian. 3Vectors with n entries which are real numbers. » # » 4The notation A # v refers to the matrix product of A and v , considered as n×n and n×1 matrices. 6 Lecture 1: Groups Often, the composition law in an abelian group is denoted + instead of ·. 1.3 Permutation and Symmetric Groups Now, we will look at an extended example of another family of nonabelian groups. Defnition 1.8 Given a set S, a permutation of S is a bijectiona p : S − → S. aA function f : A − → B is a bijection if for all y ∈ B, there exists a unique x ∈ A such that f(x) = y. Equivalently, it must be one-to-one and onto. Defnition 1.9 Let Perm(S) be the set of permutations of S. In fact, Perm(S) is a group, where the product rule is function composition.6 • Identity. The identity function e : x 7− → x is is the identity element of the group. • Inverse. Because p is a bijection, it is invertible. Let p−1(x) be the unique y ∈ S such that p(y) = x. • Associativity. Function composition is always associative. Like groups of matrices, Perm(S) is a group coming from a set of transformations acting on some object; in this case, S. Defnition 1.10 When S = {1, 2, · · · , n}, the permutation group Perm(S) is called the symmetric group, denoted Sn. Defnition 1.11 For a group G, the number of elements in the set G, |G|, is called the order of the group G, denoted |G| or ord(G). The order of the symmetric group is |Sn| = n!7 so the symmetric group Sn is a fnite group. For n = 6, consider the two permutations p and q i 1 2 3 4 5 6 p(i) 2 4 5 1 3 6 i 1 2 3 4 5 6 q(i) 3 4 5 6 1 2 , where the upper number is mapped to the lower number. We can also write these in cycle notation, which is a shorthand way of describing a permutation that does not afect what the permutation actually is. In cycle notation, each group of parentheses describes a cycle, where the number is mapped to the following number, and it wraps around. Example 1.12 (Cycle notation) In cycle notation, p is written as (124)(35), where the 6 is omitted. In the frst cycle, 1 maps to 2, 2 maps to 4, and 4 maps to 1, and in the second cycle, 3 maps to 5 and 5 maps back to 3.a aIn fact, we say that p has cycle type (3, 2), which is the lengths of each cycle. 6We can check that the composition of two bijections p ◦ q is also a bijection.
7The number of permutations of the numbers 1 through n is n! — there are n possibilities for where 1 maps to, and then n − 1 for where 2 maps to, and so on to get n(n − 1) · · · (2)(1) = n! 7 Lecture 1: Groups 8 Similarly, q is written as (135)(246). In cycle notation, it is clear that there are multiple ways to write or represent the same permutation. For example, p could have been written as (241)(53) instead, but it represents the same element p ∈ S6. Cycle notation allows us to more easily invert or compose two permutations; we simply have to follow where each number maps to. Example 1.13 (Inversion) The inverse p−1 fips the rows of the table: i 2 4 5 1 3 6 p(i) 1 2 3 4 5 6 In cycle notation, it reverses the cycles, since each number should be mapped under p−1 to the number that maps to it under p: p −1 = (421)(53) = (142)(35). Example 1.14 (Composition) The composition is q ◦ p = (143)(26). a Under p, 1 maps to 2, which maps to 4 under q, and so 1 maps to 4 under q ◦ p. Similarly, 4 maps to 3 and 3 maps back to 1, which gives us the frst cycle. The second cycle is similar. aRemember that the rightmost permutation is applied frst, and then the leftmost, and not the other way around, due to the notation used for function composition. Example 1.15 (Conjugation) Another example of composition is p −1 ◦ q ◦ p = (126)(345). a This is also known as conjugation of q by p. aNotice that under conjugation, q retains its cycle type (3, 3). In fact, this is true for conjugation of any element by any other element! 1.4 Examples of Symmetric Groups For n ≥ 3, Sn is always non-abelian. Let’s consider Sn for small n ≤ 3. Example 1.16 (S1) In this case, S1 only has one element, the identity element, and so it is {e}, the trivial group. Example 1.17 (S2) For n = 2, the only possibilities are the identity permutation e and the transposition (12). Then S2 = {e, (12)}; it has order 2. Once n gets larger, the symmetric group becomes more interesting. 8It has cycle type (3, 3). 8 Lecture 1: Groups Example 1.18 (S3) The symmetric group on three elements is of order 3! = 6. It must contain the identity e. It can also contain 2 x = (123). Then we also get the element x = (132), but 3 x = e. 4 5 2 Higher powers are just x = x, x = x , and so on. Now, we can introduce y = (12), which is its own inverse, and so 2 y = e . 2 Taking products gives xy = (13) and x y = (23). So we have all six elements of S3: S3 = {e, (123), (132), (12), (13), (23)}. 2 In fact, yx = (23) = x y, so taking products in the other order does not provide any new elements. The relation 2 yx = x y holds. In particular, using the boxed relations, we can compute any crazy combination of x and y and −1 2 2 2 reduce it to one of the elements we listed. For example, xyx y = xyx y = xyyx = xy x = x . 9 Lecture 2: Subgroups and Cyclic Groups 2 Subgroups and Cyclic Groups 2.1 Review Last time, we discussed the concept of a group, as well as examples of groups. In particular, a group is a set G with an associative composition law G×G − → G that has an identity as well inverses for each element with respect to the composition law ×. Our guiding example was that of the group of invertible n×n matrices, known as the general linear group (GLn(R) or GLn(C), for matrices over R and C, respectively.) Example 2.1 Let GLn(R) be the group of n×n invertible real matrices. • Associativity. Matrix multiplication is associative; that is, (AB)C = A(BC), and so when writing a product consisting of more than two matrices, it is not necessary to put in parentheses. 1 · · · 0 .
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• Identity. The n×n identity matrix is In = . . . , which is the matrix with 1s along the .
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. 0 1 · · · diagonal and 0s everywhere else. It satisfes the property that AI = IA = A for all n×n matrices A. • Inverse. By the invertibility condition of GLn, every matrix A ∈ GLn(R) has an inverse matrix A−1 such that AA−1 = A−1A = In. Furthermore, each of these matrices can be seen as a transformation from Rn − → Rn, taking each vector ⃗ v to A⃗ v. That is, there is a bijective correspondence between matrices A and invertible transformations TA : Rn − → Rn taking TA(⃗ v) = A⃗ v. Another example that showed up was the integers under addition. Example 2.2 The integers Z with the composition law + form a group. Addition is associative. Also, 0 ∈ Z is the additive identity, and −a ∈ Z is the inverse of any integer a. On the other hand, the natural numbers N under addition would not form a group, because the invertibility condition would be violated. Lastly, we looked at the symmetric group Sn. Example 2.3 The symmetric group Sn is the permutation group of {1, · · · , n}. 2.2 Subgroups In fact, understanding Sn is important for group theory as a whole because any fnite group "sits inside" Sn in a certain way9 , which we will begin to discuss today. Guiding Question What does it mean for a group to "sit inside" another group? If a subset of a group satisfes certain properties, it is known as a subgroup. 9This is known as Cayley’s Theorem and is discussed further in section 7.1 of Artin. 10 Lecture 2: Subgroups and Cyclic Groups Defnition 2.4 Given a group (G, ·), a subset H ⊂ G is called a subgroup if it satisfes: • Closure. If h1, h2 ∈ H, then h1 · h2 ∈ H. • Identity. The identity element e in G is contained in H. • Inverse. If h ∈ H, its inverse h−1 is also an element of H. As notation, we write H ≤ G to denote that H is a subgroup of G. Essentially, these properties consists solely of the necessary properties for H to also be a group under the same operation ·, so that it can be considered a subgroup and not just some arbitrary subset. In particular, any subgroup H will also be a group with the same operation, independent of the larger group G. Example 2.5 The integers form a subgroup of the rationals under addition: (Z, +) ⊂ (Q, +). The rationals are more complicated than the integers, and studying simpler subgroups of a certain group can help with understanding the group structure as a whole. Example 2.6 The symmetric group S3 has a three-element subgroup {e, (123), (132)} = {e, x, x2}. However, the natural numbers N = {0, 1, 2, · · · } ⊂ (Z, +) are not a subgroup of the integers, since not every element has an inverse. Example 2.7 The matrices with determinant 1, called the special linear group, form a subgroup of invertible matrices: SLn(R) ⊂ GLn(R). The special linear group is closed under matrix multiplication because det(AB) = det(A) det(B). 2.3 Subgroups of the Integers The integers (Z, +) have particularly nice subgroups. Theorem 2.8 a The subgroups of (Z, +) are {0}, Z, 2Z, · · · . aWhere n ∈ Z, nZ consists of the multiples of n, {nx : x ∈ Z}. This theorem demonstrates that the condition that a subset H of a group be a subgroup is quite strong, and requires quite a bit of structure from H. Proof. First, nZ is in fact a subgroup. • Closure. For na, nb ∈ nZ, na + nb = n(a + b). • Identity. The additive identity is in nZ because 0 = n · 0. • Inverse. For na ∈ nZ, its inverse −na = n(−a) is also in nZ. Now, suppose S ⊂ Z is a subgroup. Then clearly the identity 0 is an element of S. If there are no more elements in S, then S = {0} and the proof is complete. Otherwise, pick some nonzero h ∈ S. Without loss of generality, we assume that h > 0 (otherwise, since −h ∈ S as well by the invertibility condition, take −h instead of h.) Thus, S contains at least one positive integer; let a be the smallest positive integer in S. Then we claim that S = aZ. If a ∈ S, then a + a = 2a ∈ S by closure, which implies that 2a + a = 3a ∈ S, and so on. Similarly, −a ∈ S by inverses, and −a + (−a) = −2a ∈ S, and so on, which implies that aZ ⊂ S. 11 Lecture 2: Subgroups and Cyclic Groups Now, take any n ∈ S. By the Euclidean algorithm, n = aq + r for some 0 ≤ r < a. From the subgroup properties, n − aq = r ∈ S as well. Since a is the smallest positive integer in S, if r > 0, there would be a contradiction, so r = 0. Thus, n = aq, which is an element of aZ. Therefore, S ⊂ aZ. From these two inclusions, S = aZ and the proof is complete. Corollary 2.9 Given a, b ∈ Z, consider S = {ai + bj : i, j ∈ Z}. The subset S satisfes all the subgroup conditions, so by Theorem 2.8, there is some d such that S = dZ. In fact, d = gcd(a, b). Proof. Let e = gcd(a, b). Since a ∈ S, a = dk and b = dℓ for some k, ℓ. Since the d from before divides a and b, it must also divide e, by defnition of the greatest common divisor. Also, since d ∈ S, by the defnition of S, d = ar + bs for some r and b. Since e divides a and b, e divides both ar and bs and therefore d. Thus, d divides e, and e divides d, implying that e = d. So S = gcd(a, b)Z. In particular, we have showed that gcd(a, b) can always be written in the form ar + bs for some r, s. 2.4 Cyclic Groups Now, let’s discuss a very important type of subgroup that connects back to the work we did with (Z, +). Defnition 2.10 Let G be a group, and take g ∈ G. Let the cyclic subgroup generated by g be −2 −1 0 1 2 ⟨g⟩ := a{· · · g , g , g = e, g , g , · · · } ≤ G. aThe := symbol is usually used by mathematicians to mean "is defned to be." Other people may use ≡ for the same purpose. a b a+b Since g · g = g , the exponents of the elements of a cyclic subgroup will have a related group structure to (Z, +). Example 2.11 The identity element generates the trivial subgroup {e} = ⟨e⟩ of any group G. There are also nontrivial cyclic subgroups. Example 2.12 In S3, ⟨(123)⟩ = {e, (123), (132)}. Evidently, a cyclic subgroup of any fnite group must also be fnite. Example 2.13 Let C× be the group of nonzero complex numbers under multiplication. Then 2 ∈ C will generate ⟨2⟩ = {· · · , 1/4, 1/2, 1, 2, 4, · · · .} On the other hand, i ∈ C will generate ⟨i⟩ = {1, i, −1, −i}. This example shows that a cyclic subgroup of an infnite group can be either infnite or fnite.10 10Can you work out the cases for which g ∈ C the cyclic subgroup of C× is fnite or infnite? 12 Lecture 2: Subgroups and Cyclic Groups Guiding Question What does a cyclic subgroup look like? Can they be classifed? Theorem 2.14 n Let S = {n ∈ Z : g = e}. Then S is a subgroup of Z, so S = dZ or S = {0}, leading to two cases: k • If S = {0}, then ⟨g⟩ is infnite and all the g are distinct. 2 • If S = dZ, then ⟨g⟩ = {e, g, g , · · · , gd−1} ⊂ G, which is fnite. Proof. First, S must be shown to actually be a subgroup of Z. • Identity. The identity 0 ∈ S because g0 = e. a b a+b a b • Closure. If a, b ∈ S, then g = g = e, so g = g g = e · e = e, so a + b ∈ S. −a −1 • Inverse. If a ∈ S, then g = (ga)−1 = e = e, so a ∈ S. a a −b a−b Now, consider the frst case. If g = gb for any a, b, then multiplying on right by g−b gives g · g = g = e. Thus, a − b ∈ S, and if S = {0}, then a = b. So any two powers of g can only be equal if they have the same exponent, and thus all the gi are distinct and the cyclic group is infnite. Consider the second case where S = dZ. Given any n ∈ Z, n = dq + r for 0 ≤ r < d by the Euclidean algorithm. Then gn = gdq · gr = gr , which is in {e, g, g2 , · · · , gd−1}. Defnition 2.15 So if d = 0, then ⟨g⟩ is infnite; we say that g has infnite order. Otherwise, if d ̸= 0, then |⟨g⟩| = d and g has order d. It is also possible to consider more than one element g. Defnition 2.16 Given a subset T ⊂ G, the subgroup generated by T is e1 en ⟨T ⟩ := {t · · · t | ti ∈ T, ei ∈ Z}.
1 n Essentially, ⟨T ⟩ consists of all the possible products of elements in T. For example, if T = {t, n}, then 2 −3 4 5 −1 ⟨T ⟩ = {· · · , t n t , n t , · · · }. Defnition 2.17 If ⟨T ⟩ = G, then T generates G.a aGiven a group G, what is the smallest set that generates it? Try thinking about this with some of the examples we’ve seen in class! Example 2.18 The set {(123), (12)} generates S3. Example 2.19 The invertible matrices GLn(R) are generated by elementary matricesa . aThe matrices giving row-reduction operations. 13 Lecture 3: Homomorphisms and Isomorphisms 3 Homomorphisms and Isomorphisms 3.1 Review Last time, we discussed subgroups and cyclic groups. A subgroup of a group is essentially a subset of that group that is compatible with the group or multiplicative structure on it. A cyclic subgroup of an element g in a group is essentially the subgroup consisting of all the powers of g. 3.2 Homomorphisms Now that we understand a little bit more about groups and their structures, the natural next step is to look at maps between groups. Guiding Question How can we understand groups by considering maps between diferent groups? What kinds of maps can provide useful insight into various groups? First, we defne a type of map that is compatible with the group structure on both groups. Defnition 3.1 Given groups G and G ′ , a homomorphism between them is a map f : G − → G ′ such that: • For all a, b ∈ G, f(ab) = f(a)f(b). • The identity element is mapped to the identity: f(eG) = eG ′ . • Inverses are preserved under the mapping: f(a)−1 = f(a−1) for all a ∈ G. Essentially, each of these conditions requires that the map preserve the group structure (multiplication, identity, inverse) from the domain G to the codomain G ′ . Either f can be applied to a product, or the product can 11 be taken after f is applied, and it should yield the same element f(ab) = f(a)f(b). In fact, only the frst condition is really necessary, and implies the second two.12 Proposition 3.2 If f(ab) = f(a)f(b), then f(eG) = eG ′ and f(a)−1 = f(a−1). Proof. For the frst part, take f(eG ·eG) = f(eG) = f(eG)·eG ′ by the defnition of eG ′ . Since f is a homomorphism, ′ this will also be equal to f(eG)f(eG). Multiplying on both sides by f(eG)−113 gives f(eG) = eG. ′ The second part is similar. Take a ∈ G. Then f(a) · f(a−1) = f(a · a−1) = f(eG) = eG, and multiplying on the left by f(a)−1 gives f(a−1) = f(a)−1 . 3.3 Examples Let’s see some examples. Example 3.3 The determinant det : GLn(R) − → (R× , ×) is a homomorphism from invertible matrices to the real numbers under multiplication, since det(AB) = det(A) det(B). 11In other words, a homomorphism will commute with multiplication in that they can be applied in either order. This results in a commutative diagram. 12In some way, this shows that the multiplication is the essential part of the group structure, and the identity and inverse properties are simply there to make sure nothing is able to go wrong with the multiplication. 13This must exist by the group property of invertibility. 14 Lecture 3: Homomorphisms and Isomorphisms Example 3.4 z a+b a b The exponential exp : (C, +) − → (C× , ×) taking z − → e is a homomorphism, since e = e e . Let the standard basis vectors of Rn be ⃗ e1 = (1, 0, · · · , 0)t, ⃗ e2 = (0, 1, · · · , 0)t , and so on, where ⃗ ei is the vector consisting of a 1 in the ith entry and 0s elsewhere. For a permutation p ∈ Sn, let Ap be the permutation matrix taking ⃗ ei 7− → ⃗ ep(i). In particular, the ith column of Ap will be ⃗ ep(i). 0 0 1 For example, for p(123), Ap = 1 0 0. 0 1 0 Example 3.5 The mapping φ : Sn − → GLn(R) p ∈ Sn 7− → Ap, where Ap is the permutation matrix, is a homomorphism. This is because Ap(Aq(⃗ ei)) = Ap(⃗ eq(i)) = ⃗ ep(q(i)), and Apq(⃗ ei) = ⃗ epq(i) = ⃗ ep(q(i)), which matches, so ApAq = Apq. There is also another important homomorphism from Sn to another group. Example 3.6 Let sign = det ◦φ take Sn − → R× by taking the determinant of the permutation matrix. This mapping sign is also a homomorphism, since φ and det are both homomorphisms. In fact, sign(p) = ±1. It is always possible to write any permutation as a composition of transpositions14: p = τ1τ2 · · · τr for transpositions τi. The determinant of a transposition matrix is −1, since det 0 1 = −1, 1 0 ′ ′ so sign(p) = (−1)r where r is the number of transpositions making up p. In fact, if p = τ1 · · · τr = τ1 · · · τ , r = s s modulo 2, since the sign homomorphism can be applied on either side. For example, for S3, e, (123), and (132) all have a sign of +1, while (12), (13), and (23), the transpositions, all have a sign of −1. Notice that R× = GL1(R), since 1×1 invertible matrices are simply nonzero real numbers. These two examples provide a hint as to why homomorphisms are so useful: matrices/linear mappings and GLn are generally well-understood, so if there is a homomorphism from a group to GLn, the knowledge from GLn can then be used to learn more about that particular group. This idea is the core theme of a branch of math called representation theory.15 Example 3.7 For any G and any x ∈ G, let fx : Z − → G n n 7− → x . a+b a b This is a homomorphism because x = x x , and is related to the cyclic subgroups of G. Last time, in class, we studied cyclic subgroups ⟨g⟩ using Z and essentially used this homomorphism. In general, homomorphisms allow us to study complicated groups with simpler groups. 14Permutations that swap two elements and leave all other elements fxed. 15These examples actually provides the so-called permutation representation and sign representation of Sn. 15 Lecture 3: Homomorphisms and Isomorphisms Theorem 3.8 Let f be a homomorphism from G − → G ′ . Then im(f)a is a subgroup of G ′ . aThe image of f consists of all the elements in G ′ that are mapped to by f. This theorem is not surprising; the whole point of a homomorphism is that it plays nicely with the group structure, and the whole point of a subgroup is that it also plays nicely with the group structure. Example 3.9 For example, im(fx) from Example 3.7 is ⟨x⟩. ′ ′ ′ ′ Proof. Consider y, y ∈ im(f). Then there exist x, x ∈ G such that y = f(x) and y = f(x ′ ). Then yy = f(x)f(x ′ ) = f(xx ′ ) ∈ im(f). The inverse and identity conditions are verifed similarly.16 Defnition 3.10 The kernel of f is ker(f) := {x ∈ G : f(x) = eG′ }. Theorem 3.11 The kernel of a homomorphism f is also a subgroup. ′ ′ Proof. If x, x ∈ ker(f), then f(xx ′ ) = f(x)f(x ′ ) = eG′ eG′ = eG′ , so xx ∈ ker(f). Also, f(eG) = eG′ so −1 eG ∈ ker(f). Lastly, if x ∈ ker(f), then f(x−1) = f(x)−1 = e = eG ′ , so x−1 ∈ ker(f) as well. G ′ The image and kernel of each of the previous examples can be seen to be subgroups. The fact that f is a homo morphism is imperative to the proofs of either fact, and these two theorems demonstrate that a homomorphism does in fact respect the group structure. Example 3.12 Consider det : GLn(R) − → (R× , ×). Since the determinant for invertible matrices can take on any nonzero value, the image of the determinant is all of R× . The kernel of the determinant is SLn(R), the special linear group consisting of the n×n matrices with determinant 1. Example 3.13 2πik For exp : (C, +) − → (C× , ×), the image is all of C× , and the kernel is 2πiZ ⊆ C, since e = 1. Example 3.14 For φ : Sn − → GLn(R) p ∈ Sn 7− → Ap, the image is the set of permutation matrices in GLn(R), whereas ker(φ) = {e}, the identity permutation. Example 3.15 The image of the sign homoomorphism sign = det ◦φ is {±1} ∈ R× . The kernel defnes a new group, called the alternating group An := ker(sign). −1 16For inverse, consider y ∈ im(f ). Then there exists x such that y = f(x). From the defnition of a homomorphism, y = f(x)−1 = f (x−1) ∈ im(f). For identity, f (eG) = eG ′ , so eG ′ ∈ im(f ). 16 Lecture 3: Homomorphisms and Isomorphisms For example, A3 = {e, (123), (132)} ⊆ S3. Example 3.16 n The kernel of fx is {n : x = eG}, which was used in the previous class, and is dZ where d is the order of x if it is fnite, and {0} if the order of x is infnite. Defnition 3.17 A mapping f : G − → G ′ is an isomorphism if it is a bijective homomorphism. In some sense, if two groups are isomorphic (that is, if there exists an isomorphism between them), they are essentially the same group, because there are the exact same number of elements and the multiplication relationships between the elements will be exactly the same. Usually, in group theory, groups are considered with respect to the isomorphism classes. Example 3.18 The exponential map from the real numbers under addition onto the positive real numbers under multipli cation exp : (R, +) − → (R>0, ×) t 7− → e t is an isomorphism. Given an isomorphism f : G − → G ′ , f −1 : G ′ − → G is also an isomorphism, since f −1(yy ′ ) = f −1(y)f −1(y ′ ). If there exists an isomorphism between G and G ′ , this is denoted as G ∼ = G ′ . 17 Lecture 4: Isomorphisms and Cosets 4 Isomorphisms and Cosets 4.1 Review In the last lecture, we learned about subgroups and homomorphisms. Defnition 4.1 We call f : G → G ′ a homomorphism if for all a, b ∈ G, f(a)f(b) = f(ab). Defnition 4.2 The kernel of a homomorphism f is {a ∈ G : f(a) = eG′ }, and the image is the set of elements b = f(a) for some a. The kernel and image of f are subgroups of G and G ′ , respectively. 4.2 Isomorphisms Homomorphisms are mappings between groups; now, we consider homomorphisms with additional constraints. Guiding Question What information can we learn about groups using mappings between them? Defnition 4.3 We call f : G → G ′ an isomorphism if f is a bijective homomorphism. In some sense, if there exists an isomorphism between two groups, they are the same group; relabeling the elements of a group using an isomorphism and using the new product law yields the same products as before relabeling. Almost all the time, it is only necessary to consider groups up to isomorphism. Example 4.4 There exists an isomorphism f : Z4 →⟨i⟩ given by n mod 4 7→ in . In particular, we get 0 7→ 1 1 7→ i 2 7→ −1 3 7→ −i. So the group generated by i, which can be thought of as a rotation of the complex plane by π/2, is essentially "the same" as the integers modulo 4. Example 4.5 2 More generally, the group generated by g, ⟨g⟩ = {e, g, g , · · · , gd−1}, where d is the order of g, is isomorphic to Zd = {0, 1, · · · , d − 1}. If the order of g is infnite, then we have ⟨g⟩∼ = Z. a b a+b Here, the idea that an isomorphism is a "relabeling" of elements makes sense: since g g = g , relabeling i g with its exponent i retains the important information in this situation. Thinking of ⟨g⟩ in this way yields precisely Zd. 4.3 Automorphisms An important notion is that of an automorphism, which is an isomorphism with more structure. 18 Lecture 4: Isomorphisms and Cosets Defnition 4.6 An isomorphism from G to G is called an automorphism. If a homomorphism can be thought of as giving us some sort of "equivalence" between two groups, why do we care about automorphisms? We already have an equivalence between G and itself, namely the identity. The answer is that while the identity map id : G → G is always an automorphism, more interesting ones exist as well! We can understand more about the symmetry and structure of a group using these automorphisms. Example 4.7 A non-trivial automorphism from Z to itself is f : Z → Z taking n 7→ −n. From the existence of this nontrivial automorphism, we see that Z has a sort of "refective" symmetry.17 Example 4.8 (Inverse transpose) Another non-trivial automorphism, on the set of invertible matrices, is the inverse transpose f : GLn(R) → GLn(R) A 7→ (At)−1 Many other automorphisms exist for GLn(R), 18 since it is a group with lots of structure and symmetry. Example 4.9 (Conjugation) A very important automorphism is conjugation by a fxed element a ∈ G. We let ϕa : G → G be such that ϕa(x) = axa −1 . We can check the conditions to show that conjugation by a is an automorphism: • Homomorphism. −1 −1 −1 ϕa(x)ϕa(y) = axa aya = axya = ϕa(xy). • Bijection. We have an inverse function ϕa−1 : −1 −1 ϕa ◦ ϕa = ϕa ◦ ϕa = id. Note that if G is abelian, then ϕa = id. Any automorphism that can be obtained by conjugation is called an inner automorphism; any group intrin sically has inner automorphisms coming from conjugation by each of the elements (we can always fnd these automorphisms to work with). Some groups also have outer automorphisms, which are what we call any automorphisms that are not inner. For example, on the integers, the only inner automorphism is the identity function, since they are abelian.19 4.4 Cosets Throughout this section, we use the notation K := ker(f). Guiding Question When do two elements of G get mapped to the same element of G ′? When does f(a) = f(b) ∈ G ′? Given a subgroup of G, we can fnd "copies" of the subgroup inside G. 17In particular, this automorphism f corresponds to refection of the number line across 0. 18For example, just the transpose or just the inverse are automorphisms, and in fact they are commuting automorphisms, since the transpose and inverse can be taken in either order. −1 −1 19For an abelian group, axa = aa x = x. 19 Lecture 4: Isomorphisms and Cosets Defnition 4.10 Given H ⊆ G a subgroup, a left coset of H is a subset of the form aH := {ax : x ∈ H} for some a ∈ G. Let’s start with a couple of examples. Example 4.11 (Cosets in S3) Let’s use our favorite non-abelian group, G = S3 = ⟨(123), (12)⟩ = ⟨x, y⟩, and let our subgroup H be {e, y}. Then eH = H = {e, y} = yH; xH = {x, xy} = xyH; and 2 2 x 2H = {x , x y} = x 2yH. We have three diferent cosets, since we can get each coset one of two ways. Example 4.12 If we let G = Z and H = 2Z, we get 0 + H = 2Z = evens = 2 + H = · · · , and 1 + H = 1 + 2Z = odd integers = 3 + H = · · · . In this example, the odd integers are like a "copy" of the even integers, shifted over by 1. From these examples, we notice a couple of properties about cosets of a given subgroup. Proposition 4.13 All cosets of H have the same order as H. Proof. We can prove this by taking the function fa : H → aH which maps h 7→ ah. This is a bijection because 20 it is invertible; the inverse is fa−1 . Proposition 4.14 a Cosets of H form a partition of the group G. aA partition of a set S is a subdivision of S into disjoint subsets. To prove this, we use the following lemma. Lemma 4.15 Given a coset C ⊂ G of H, take b ∈ C. Then, C = bH. Proof. If C is a coset, then C = aH for some a ∈ G. If b ∈ C, then b = ah for some h ∈ H, and a = bh−1 . Then bH = {bh ′ : h ′ ∈ H} = {ahh ′ |h ′ ∈ H} ⊆ aH. Using a = bh−1 , we can similarly show that aH ⊆ bH, and so aH = bH.21 −1 20I can undo any fa in a unique way by multiplying again on the left by a . This is something that breaks down with monoids or semigroups or other more complicated structures. 21So for a given coset C, we can use any of the elements in it as the representative a such that C = aH. 20 Lecture 4: Isomorphisms and Cosets Proof. Now, we prove our proposition. • Every x ∈ G is in some coset. Take C = xH. Then x ∈ C. ′ • Cosets are disjoint. If not, let C, C be distinct cosets, and take y in their intersection. Then yH = C and ′ ′ yH = C by Lemma 4.15, and so C = C . With this conception of cosets, we have the answer to our question: Answer. If f(a) = f(b), then f(a)−1f(b) = eG ′ . In particular, f(a−1b) = eG ′ , so a−1b ∈ K, the kernel of f. Then, we have that b ∈ aK, or b = ak where f(k) = eG′ . So f(a) = f(b) if a is in the same left coset of the kernel as b. 4.5 Lagrange’s Theorem In fact, thinking about cosets gives us quite a restrictive result on subgroups, known as Lagrange’s Theorem. Guiding Question What information do we automatically have about subgroups of a given group? Defnition 4.16 The index of H ⊆ G is [G : H], the number of left cosets. Theorem 4.17 We have |G| = [G : H]|H|. Proof. This is true because each of the cosets have the same number of elements and partition G. So we have X X |G| = |C| = |H| = [G : H]|H|. left cosets C left cosets C That is, the order of G is the number of left cosets multiplied by the number of elements in each one (which is just |H|). Example 4.18 For S3, we have 6 = 3 · 2. From our theorem, we get Lagrange’s Theorem: Corollary 4.19 (Lagrange’s Theorem.) For H a subgroup of G, |H| is a divisor of |G|. We have an important corollary about the structure of cyclic groups. Corollary 4.20 If |G| is a prime p, then G is a cyclic group. Proof. Pick x ̸= e ∈ G. Then ⟨x⟩⊆ G. Since the order of x cannot be 1, since it is not the identity, the order of x has to be p, since p is prime. Therefore, ⟨x⟩ = G, and so G is cyclic, generated by x. In general, for x ∈ G, the order of x is the size of ⟨x⟩, which divides G. So the order of any element divides the size of the group. 21 Lecture 5: The Correspondence Theorem 5 The Correspondence Theorem 5.1 Review In the last lecture, we learned about cosets and some of their properties. Defnition 5.1 For a group G and a subgroup H ≤ G, we defne the left coset of a to be aH := {ah : h ∈ H} ⊆ G. The left cosets partition22 G into equally sized sets. This provides a useful corollary about the structure of cosets within a group: Corollary 5.2 (Counting Formula.) Let [G : H] be the number of left cosets of H, which is called the index of H in G. Then |G| = |H|[G : H]. 5.2 Lagrange’s Theorem Using cosets provides some additional information about groups. Guiding Question What are the possibilities for the structure of a group with order n? From the Counting Formula, we immediately obtain Lagrange’s Theorem as a corollary: Theorem 5.3 (Lagrange’s Theorem.) For H a subgroup of G, |H| is a divisor of |G|. Several important corollaries follow as a result. Corollary 5.4 The order of x ∈ G is |⟨x⟩|. Since the order of any subgroup divides the order of |G|, ord(x) also divides |G|. Corollary 5.5 Any group |G| with prime order p is a cyclic group. Proof. Take an element e ̸= x ∈ G. Since the order of x ∈ G divides p, and p is prime, ord(x) = p. Then each i x is distinct for 0 ≤ i ≤ p − 1, and since there are only p elements in G, the entire group G is ⟨x⟩, the cyclic group generated by x. Our result shows that any group of prime order is a cyclic group. In particular, the integers modulo p, Zp, form a cyclic group of prime order; that is, any group of prime order p is isomorphic to Zp. 5.3 Results of the Counting Formula Using Lagrange’s Theorem narrows down the possibilities for subgroups. 22A partition of a set S is a subdivision of the entire set into disjoint subsets. 22 Lecture 5: The Correspondence Theorem Example 5.6 (Groups of order 4.) What are the possibilities (up to isomorphism) for G if |G| = 4? First, e must be an element of G. Next, consider the other three elements of G. Each of these must have either order 2 or order 4, since those are the divisors of |G| = 4. Then there are two possibilities. 2 3 • Case 1. There exists an element x ∈ G such that ord(x) = 4. Then we know that e ̸= x ̸= x ̸= x , and since |G| = 4, these are all the elements of G. (The power x4 is e again.) So G is generated by x, a and it is the cyclic group ⟨x⟩ of size 4, and must be isomorphic to Z4 . • Case 2. All elements of G have order 2. Then, we can take x ∈ G and y ̸= x ∈ G. They have order 2, 2 −1 so x = e, which implies that x = x−1 and similarly y = y . Also, the element xy also has order 2, −1 −1 and so xyx y = (xy)(xy) = e, and so x and y commute. Because x and y were chosen arbitrarily, any two elements of the group commute, and so it is abelian. This group G is isomorphic to the matrix group ±1 0 ≤ GL2(R). 0 ±1 The non-identity elements each have order 2 and commute with each other. This group is called the Klein-four group, and is denoted K4. Up to isomorphism, any order 4 group is either Z2 or K4. Note that both of these groups are abelianb; the smallest non-abelian group has order 6. aWe write Zn to denote the group of integers modulo n. bcommutative Exercise 5.7 What are the possible groups of order 6? The Counting Formula also provides another important corollary. Corollary 5.8 The size of the group is |G| = |ker(f)| · |im(f)|. a aIn linear algebra, the analogous result is the rank-nullity theorem. Proof. Let f : G → G ′ be a homomorphism, and ker(f) ≤ G be the kernel. For each y ∈ G ′ , the preimage of y is f −1(y) := {x ∈ G : f(x) = y}, which is ∅ if y ∈ / im(f), and a coset of ker(f) otherwise.23 Then, the number of left cosets of ker(f) is precisely the number of elements in the image of f, since each of those elements corresponds to a coset of the kernel. So [G : ker(f)] = |im(f)|, and applying the counting formula with ker(f) as our subgroup H gives us |G| = |ker(f)| · |im(f)|, which is the desired result. 5.4 Normal Subgroups In this section, we learn about normal subgroups. ′ ′−1)f(x ′ ), 23Pick x ∈ f −1(y). Then we claim that f −1(y) = x ker(f ). Take any x ∈ f −1(y). We have y = f (x) = f (x ′ ) = f (xx ′ so f (xx′−1) = e and xx′−1 ∈ ker(f ). Thus x ∈ x ker(f ). 23 Lecture 5: The Correspondence Theorem Guiding Question The choice of left cosets seems arbitrary — what are the ramifcations if right cosets are used instead? Defnition 5.9 The right coset of a is Ha = {ha : h ∈ H}. In fact, all the same results follow if right cosets are used instead of left cosets. First, let’s see an example of right cosets: Example 5.10 Let H be the subgroup generated by y ∈ S3. Then the left cosets are 2 2 {e, y}, {x, xy}, {x , x y}, and the right cosets are 2 2 {e, y}, {x, x y}, {x , xy}. So in fact, right cosets give a diferent partition of S3, but the number and size of the cosets are the same.a aWe can think of cosets as "carving up" the group. Using right cosets instead of left cosets is just carving it up in a diferent way. In particular, there is a bijection between the set of left cosets and the set of right cosets. It maps −1 C 7→ C−1 = {x : x ∈ C}. h−1 −1 It is a bijection because (ah)−1 = a , and so aH = Ha−1 . So the index [G : H] is equal to both the number of right cosets and the number of left cosets. Guiding Question For which subsets H ⊆ G do left and right cosets give the same partition of G? In other words, for which H is every left coset also a right coset?a aIf some left coset xH of an element x is equal to some right coset Hy of a diferent element y, since x ∈ Hy as well, from a lemma from last week’s lecture, Hy = Hx, and so in fact the left coset and right coset of the same element x must also be equal. So it is sufcient to require that xH = Hx. This question motivates the defnition of normal subgroups. Defnition 5.11 If xH = Hx for each x ∈ G, H ⊆ G is called a normal subgroup. Equivalently, the subgroup H is normal if and only if it is invariant under conjugation by x; that is, xHx−1 = H. Using the notation from last lecturea , a subgroup H is normal if and only if φx(H) = H for all x ∈ G. −1 aThe function φx takes g 7→ xgx . Let’s look at some examples. Example 5.12 (Non-normal subgroup) From above, the subgroup ⟨y⟩ is not normal in S3. 24 Lecture 5: The Correspondence Theorem Example 5.13 (Kernel) Given a homomorphism f : G → G ′ , the kernel of f is always normal. Take k ∈ ker(f). Then f(xkx−1) = f(x)f(k)f(x)−1 = f(k) = eG ′ , so φx(ker(f)) = ker(f), and thus ker(f) is a normal subgroup. In fact, in future lectures, we will see that all normal subgroups of a given group G arise as the kernel of some homomorphism f : G → G ′ to a group G ′ . Example 5.14 In S3, the subgroup ⟨y⟩ is not normal, but ⟨x⟩ is normal. In particular, it is the kernel of the sign a homomorphism sign : S3 → R. aA given permutation σ can be written as a product of i transpositions, where i is unique up to parity. The sign homomorphism maps σ to (−1)i . 5.5 The Correspondence Theorem Ealier in this lecture, we noticed that homomorphisms give us some information about subgroups. Can we make this more concrete? Guiding Question Let f be a homomorphism from G to G ′ . Is there a relationship between the subgroups of G and the subgroups of G ′? {subgroups of G} ↔{subgroups of G ′ } Answer. In fact, we see that there is! • Given a subgroup of G, a subgroup of G ′ can be produced as follows. Let f with the domain restricted to H be denoted as f|H . Then a subgroup H ≤ G maps to im(f|H ) = f(H) ⊆ G ′ , which is a subgroup of G ′ . ′ • Now, given H ≤ G ′ and a subgroup of G can be produced by taking the preimage f −1(H ′ ) = {x ∈ G : f(x) ∈ H ′ }. Is this subset of G is actually a subgroup? It is! Let’s just check that it’s closed under composition. If ′ ′ ′ x, y ∈ f −1(H), then f(x), f(y) ∈ H , so f(x)f(y) ∈ H , since H is closed under multiplication. Then ′ f(xy) ∈ H , so xy ∈ f −1(H). ′ ′ If H = eG′ , then its preimage is the kernel, and if H = G ′ , then the preimage is all of G. In general, the preimage is a subgroup somewhere in-between the kernel and the whole domain. Are these maps bijective, or inverses of each other? It can be easily seen that they are not; in particular, if G is the trivial group and G ′ is some more complicated group with many subgroups, every subgroup of G ′ must always still map to the trivial group. It makes sense that these maps are not bijective, since f is not an isomorphism, just an arbitrary homomorphism with no more restrictions. Two issues arise with these maps that make them non-bijective: • Any subgroup of G must map to some subgroup of G ′ that is contained within the image of f, by construction, since f(H) ⊆ im(f). • The kernel ker(f) = f −1(eG′ ) ⊆ f −1(H ′ ), so any subgroup not contained within the kernel cannot be mapped to by any subgroup of G ′ . However, these are actually the only issues! If we are willing to put some restrictions on the homomorphism f and the types of subgroups we look at, there is actually a bijection between certain subgroups of G and certain subgroups of G ′ . 25 Lecture 5: The Correspondence Theorem In order to make things a little easier for now, we take a surjective homomorphism f : G → G ′ . The frst issue then is no longer consequential, because the image is all of G ′ . Now, let’s restrict the subgroups of G to subgroups that contain ker(f). Then our maps (as described above) provide a bijection. Theorem 5.15 (Correspondence Theorem) For a surjective homomorphism f with kernel K, there is a bijective correspondence: {subgroups of G containing K} ↔{subgroups of G ′ }, where a subset of G, H ⊇ K ⇝ its image f(H) ≤ G ′ ′ H ≤ G ′ ⇝ its preimage f −1(H ′ ) ≤ G. Example 5.16 (Roots of Unity) Take f G = C∗ − → G ′ = C∗ z 7→ z 2 , which is a homomorphism because G is abelian. The kernel is ker(f) = {±1}. We have a correspondence between R× ⇝ R>0. For example, the eighth roots of unity correspond to the fourth roots of unity under this map. ′ 8 H = {e 2πik } ↭ H = {±1, ±i}. 26 Lecture 6: Quotient Groups 6 Normal Subgroups and Quotient Groups 6.1 Review In the last lecture, we learned about the correspondence theorem. Theorem 6.1 (Correspondence Theorem) Where f : G → G ′ is a surjectivea homomorphism, and K = ker(f), there is a correspondence ′ ′ {subgroups H : K ⊆ H ⊆ G} ↭ {subgroups H : {eG ′ } ⊆ H ⊆ G ′ }}, ′ which states that subgroups of G containing the kernel are in bijection with subgroups of H in the image of f. aIn fact, it is possible to slightly revise the statement of the theorem such that the surjective condition is no longer necessary. The correspondence comes from taking H 7→ f(H) ⊆ G ′ ′ {x ∈ G : f(x) ∈ H ′ } = f −1(H ′ ) ←[ H From the point of view of understanding subgroups, the correspondence theorem allows us to understand a slice of G. If G is a complicated group with many surjective maps onto diferent groups G ′ , we can use the correspondence theorem multiple times to understand G, and conversely, if G ′ is a complicated group, we can use G to study G ′ . Proof. In order to show that this correspondence is a bijection, we check that these are inverses to each other. • If K ⊆ H ⊆ G, then we want to show that f −1(f(H)) = H. By defnition, f −1(f(H)) = {x ∈ G : f(x) = f(h), h ∈ H}. By defnition, H ⊆ f −1(f(H)). Also, if x ∈ f −1(f(H)), then f(x) = f(h) for some h ∈ H. This is true if and only if x is in the coset hK; in other words, x = h · k for some k ∈ K. Since K ⊆ H,24 k ∈ H, and so x = hk ∈ H. • The proof of the other direction is left as an exercise to the reader (it is very much the same idea). 6.2 Normal Subgroups Recall the defnition of a normal subgroup. Defnition 6.2 A subgroup H ⊆ G is normal if xHx−1 = H for all x ∈ G. The notation H ≤ G denotes that H is a subgroup, not just a subset, of G. Now, the notation H ⊴ G will denote that H is a normal subgroup of G.25 Example 6.3 (Kernel) The kernel ker(f) is always normal. Guiding Question Given any normal subgroup N ⊴ G, is there always a group homomorphism f : G → G ′ such that N = ker(f)? Answer: Yes! 24The theorem is not true if the kernel is not contained in H, so this fact must be used at some point. 25This notation will not necessarily be used consistently throughout this lecture/class, but it is used in the literature. 27 Lecture 6: Quotient Groups Let’s look at a quick example frst. Example 6.4 (Integers modulo 2) If G = Z and H = 2Z, the homomorphism is f G − → G ′ = Z2 n 7→ n mod 2. The kernel of f consists of the elements mapping to 0 mod 2; that is, even integers, which is precisely 2Z. In the case when N = ker(f), the cosets26 of N are in correspondence with im(f), by the correspondence theorem. Since they are in bijective correspondence, the group structure on im(f) can be carried over to the set of cosets of N. 6.3 Quotient Groups Now that we’ve defned cosets, we have the following question: Guiding Question Can we directly defne a group structure on the sets of cosets of N? If C1, C2 ⊆ G are cosets, what should C1 · C2 be? The most intuitive defnition would be to take the set of products of each of the elements: Defnition 6.5 Let the product structure on the cosets be defned as C1 · C2 := {x ∈ G : x = y1 · y2; y1 ∈ C1, y2 ∈ C2}, the pairwise product. Theorem 6.6 If C1, C2 are cosets of a normal subgroup N, C1 · C2 is also a coset of N. It is crucial that N is normal! Example 6.7 Consider H = {e, y} ⊆ G = S3. Then H is not a normal subgroup. Consider xH = {x, xy}. We have 2 2 xH · xH = {x , x y, xyx = y, xyxy = e}, which is not a coset! Proof. Let C1 = aN and C2 = bN. • The inclusion abN ⊆ C1 · C2 holds because abn = (ae)(bn) ∈ C1 · C2, since ae ∈ C1 and bn ∈ C2. 26The left and right cosets are the same when N is normal. 28 Lecture 6: Quotient Groups • Take an1 · bn2 ∈ C1 · C2. Since N is normal, bN = Nb, so n1 · b = b · n3 for some n3 ∈ N. So an1 · bn2 = abn3n2 ∈ abN. Then C1 · C2 = abN. So it is only when N is normal that we do in fact have a product structure on the set of cosets of N! Defnition 6.8 The quotient group G/N is the set of cosets of a normal subgroup N. The group structure is defned asa [C1] · [C2] := [C1 · C2] [aN] · [bN] := [abN]. The right hand side is a coset because N is a normal subgroup.b aThe notation [x] refers to the equivalence class of x under an equivalence relation; in this case, the equivalence relation is defned by the partition of G into cosets. bThe product can be verifed to be independent of the representatives a and b from the fact that N is normal. Theorem 6.9 The following two statements are true about the quotient group: 1. The composition law, as defned in Defnition 6.8 does defne a group structure on G/N (all the group axioms hold). 2. There exists a surjective homomorphism π : G → G/N x 7→ [xN] such that ker(π) = N. This is one of the most basic operations we can do on groups! Proof. First of all, let’s show that G/N is actually a group. • Identity. The identity is [N] = [eN]. The product is [aN] · [N] = [aeN] = [aN]. • Inverse. We can check that [aN]−1 = [a −1N]. In general, the inverse of a left coset will be a right coset, but because N is normal, they are the same. • Associativity. Similarly, associativity of G/N boils down to associativity for G.27 Now, we can show the second part of the theorem. Take π(x) = [xN]. It is evidently a surjective map. Then π(xy) = [xyN] = [xN] · [yN] = π(x) · π(y). Then the kernel is ker(π) = {x ∈ G : x ∈ N} = N. 27The proof is left as an exercise for the reader. 29 Lecture 6: Quotient Groups Most of the proof of Theorem 6.9 seems very tautological. In fact, most of the action happened earlier on, in Thereom 6.5, which showed that the product of two cosets actually was another coset, demonstrating that the group structure does makes sense. Here is an example of how this theorem is often used. Example 6.10 (Quotient Group of SL2(R)) Take N = {±I2} ⊴ G = SL2(R). Then, taking the quotient group SL2(R)/{±I2} gives a new group P SL2(R). Thus, from an explicitly defned group, in this case SL2(R), we obtain a new, potentially interesting or useful group by taking a quotient. Another perspective on G/N is that it is similar to modular arithmetic. We have that a ≡ b mod N if aN = bN ⊆ G. 28 6.4 First Isomorphism Theorem f Suppose we start of with G − → G ′ a surjective homomorphism, and assume K = ker(f) is a normal subgroup. Given that it is a normal subgroup, we can feed it into this machine that we have created. Then π : G → G/K is a surjective group homomorphism. So we have started with a surjective group homomorphism and created another surjective group homomorphism. But in fact, we have done nothing at all! There exists an isomorphism ∼ ¯ → G ′ f : G/K − . The diagram f G ′ G π f G/K commutes. ¯ So f = f ◦ π. So up to isomorphism, our original group homomorphism is the same as our new one. This is not surprising, because there is a correspondence between cosets of the kernel and points in the image. All we are ¯ saying is that that bijection is compatible with the group structures on both sides. So f([xk]) = f(x). This is known as the First Isomorphism Theorem. 28We placed an equivalence relation on the group, and placed a group structure on the equivalence classes. 30 Lecture 7: Fields and Vector Spaces 7 Fields and Vector Spaces 7.1 Review Last time, we learned that we can quotient out a normal subgroup of N to make a new group, G/N. 7.2 Fields Now, we will do a hard pivot to learning linear algebra, and then later we will begin to merge it with group theory in diferent ways. In order to defne a vector space, the underlying feld must be specifed. Defnition 7.1 A feld F is a set with the operations (+, ×). It must satisfy that • (F, +) is an abelian group with the usual rules, and • (F × := F \ {0}, ×) is an abelian group. Also, addition and multiplication must distribute over each other.a aThere is some compatibility required. In essence, a feld is a set with additive and multiplicative group structures that interact in nice ways. Example 7.2 The sets C, R, and Q are felds, but not Z, since it is not invertible under multiplication. Since division does not exist in Z, it is not a feld. In fact, Q is essentially obtained from Z by making it into a feld by adding division. Example 7.2 gives us examples of felds with infnitely many elements, but felds can also be constructed that have fnite order. Indeed, there is one for every prime number p. Example 7.3 (Fields of prime order) For a prime p, (Fp = Zp, +, ×) is a feld. If a = ̸ 0 mod p, then gcd(a, p) = 1 implies that ar + ps = 1, and so ar ≡ 1 mod p, and thus a is invertible with multiplicative inverse r−1 . However, Z6 is not a feld; for example, 2 mod 6 has no inverse. In general, Zn where n is not a prime is not a feld, because there will exist some element that is not relatively prime to n, and it will not be invertible. 7.3 Vector Spaces A vector space, which may be a familiar concept from learning about matrices, can be defned over any feld. Defnition 7.4 A vector space V over a feld F is a set V with some operation + such that (V, +) is an abelian group. • We must be able to scale vectors: F ×V → V (a, ⃗ v) 7→ a⃗ v. • Addition and multiplication play nicely and satisfy the usual rules (· · · , a(b⃗ v) = (ab) · ⃗ v, · · · ). 31 Lecture 7: Fields and Vector Spaces Example 7.5 For a feld F, F n , column vectors with n components (a1, · · · , an)t , form a vector space of dimension n. Example 7.6 If A is an m×n matrix, then {⃗ v ∈ F n : A⃗ v = (0, · · · , 0)} is a vector space. Example 7.7 For a homogeneous linear ODE, the solutions form a vector space. 7.4 Bases and Dimension A basis of a vector space is a set of vectors providing a way of describing it without having to list every vector in the vector space. Defnition 7.8 Given ⃗ v1, ⃗ v2, · · · ,⃗ vn ∈ V, a linear combination is X ⃗ v = ai⃗ vi for ai ∈ F. Defnition 7.9 For S = {⃗ v1, ⃗ v2, · · · , ⃗ vn}, the span X Span(S) = {⃗ v ∈ V : ⃗ v = ai⃗ vi} This is similar to generating subgroups using elements of a group G, except using the operations of vector spaces. Artin likes to use the (nonstandard) notation ⃗ v1 · · · a1 . ⃗ vn . := . X ai⃗ vi an for a linear combination. Defnition 7.10 A set of vectors S spans V if Span(S) = V. a aThere is at least one way of writing ⃗ v as a linear combination. Defnition 7.11 A set of vectors {⃗ vi} is linearly independent if X ai⃗ vi = ⃗ 0 if and only if ai = 0 for all i.a aThere is at most one way of writing ⃗ v as a linear combination. A basis is both linearly independent and spans. 32 Lecture 7: Fields and Vector Spaces Defnition 7.12 A set of vectors S = { # v 1, · · · , ⃗ vn} is a basis if S spans V and is linearly independent. Equivalently, each ⃗ v ∈ V can be written uniquely as ⃗ v = a1⃗ v1 + · · · + an⃗ vn, where the ai are called the coordinates of ⃗ v in the basis S. » The standard basis for R2 is 1 0 , . 0 1 In general, when we write a vector (a, b)t , it represents the linear combination a(1, 0)t + b(0, 1)t . Example 7.13 Let V = R2 . Then the set S = 1 1 , 3 2 , 2 1 spans R2 , but is linearly dependent: v 1 − # v 2 + » # » v 3 # » = » » # But { 0 . # v 1, # » v 2} forms a basis. A good choice of basis often makes problems easier. # » # » » Defnition 7.14 A vector space V is fnite-dimensional if V = Span({ # v 1, · · · , v n}) for some v i ∈ V.a aInfnite-dimensional vector spaces are super interesting, but not studied in this class. Real analysis can be used to study them! # » » Lemma 7.15 If S = {# v 1, · · · , v r} spans V, and L = {w # 1, · · · , ws} is linearly independent, then 1. Removing elements of S gets a basis of V. # » 2. Adding elements of S to L gets another basis of V. 3. |S| ≥|L|. » Corollary 7.16 If S and L are both bases for V, then |S| = |L|. Any two bases of V contain the same number of vectors. Proof. Applying the lemma twice for S and L gives |S| ≥|L| and |L| ≥|S|. Defnition 7.17 The dimension of a vector space v is the size of any basis of V. Proof of Lemma 7.15. We prove each point separately 1. If S is not linearly independent, then there are some ai such that r X # » # » 0 . # » # » ai v i = i=1 Suppose WLOG that an ̸= 0. Then v 1 −· −1 # » # » v r−1). Span(S) = V. This is because if we have a linear (−a1 · · − ar−1 v r = a » v r−1) ∈ Span(# v 1, · · · , r If we take S ′ # » v r−1}, we have Span(S ′ ) combination using the vectors of S, we can use the equation above to turn it into a combination of vectors in S ′ . We can repeatedly remove until we have a basis of V. 33 » {# = v 1, · · · = , Lecture 7: Fields and Vector Spaces v i ̸∈ Span(L). Then L ′ is still linearly independent. We can just keep adding vectors to L so # » 2. If S ⊂ Span(L), then Span(L) = V so we are done. Otherwise, suppose # » We can create = w1, . . . , ws, 3. Each # » # { » # » L ′ v i}. that it stays linearly independent but eventually spans V. P # r , v r. Then w ⃗j = i=1 aij v i. a11 a12 · · · a1s » # » Let A be the r×s matrix wj is a linear combination of # v 1, » · · · .
.
. . A = .
.
.
. .
.
.
. . ar1 ar2 · · · ars # » , v r)A. Suppose r < s. Then by row-reduction, there exists some nonzero P » Then (w # 1, » · · · , w ⃗s) = (# v 1, » · · · 0 . Then # » » This is contradiction, since is 0 L a . # # x such that A # x = » » # » = (# v 1, » vector xiw ⃗i · · · , v r)A# x = linearly independent, so r ≥ s. » » » A linear transformation is a map Defnition 7.18 → T V W : # # # » ) T ( T ( ) + T ( ) + v = v v v 2 1 1 2 −1 We say that is isomorphism if it is bijection is also isomorphism).
T (T an a an # » such that # » and T (a v ) = aT ( # v ). » # » {# = v i v i ∈ V. For a vector space V over a feld F and a set of vectors S transformation: TS : F n → V X ∈ V }, we can defne the following linear (a1, · · · , an) 7→ ai If S is linearly independent, then TS is injective; if Span(S) = V, then TS is surjective, and if TS is a basis, then TS is an isomorphism. 34 Lecture 8: Linear Transformations with Bases and the Dimension Formula 8 Dimension Formula 8.1 Review Last time, we ended of with the defnition of linear transformations. 8.2 Matrix of Linear Transformations Given a linear transformation T : V → W , we know v 1 + · # » » v n T (# for V , the above property tell us that T is completely determined by the # T (a1 v n) = a1T ( # v 1) + · · » » ).
· + an · + an · # » # » Then, given any basis values of T (# » v 1, . . . , v n v 1), . . . , T (# v n » ). » # » Example 8.1 Last time, we discussed a linear transformation from column vectors to another vector space. In particular, » if we have a basis w # 1, . . . , wm of W , then we can create the following linear transformation: n → B F W : # » e i 7→ w # i X (a1, · · · , an) 7→ ai wi, » where e i is the zero column vector with a 1 in the ith position. This is an isomorphism due to choosing wi to be a basis. The inverse map B−1(w # ) = (a1, · · · , an) sends a vector to the coordinates of w # for the given basis. # » » # » # » » # » » » Example 8.2 There is a bijection between matrices A ∈ Matm×n(F ) and linear transformations T : F n → F m . For every matrix A, it can be mapped to the linear transformation T = A · x . For every transformation T , it can be mapped to A = T (# e 1) T ( # e 2) · · · T (# e n) . Both the set of matrices and the set of linear transformations are vector spaces, so this defnes an isomorphism between two vector spaces. We will switch between these notions frequently. ∼ As a special case, if we have an isomorphism T : F n − → F m , then this forces m = n and the corresponding matrix A must be in GLn(F ). From Example 8.1, suppose we have two diferent bases of V and create their corresponding linear transformations B and B ′ » » . Let the corresponding bases be {# v 1, . . . , v n} and {w # 1, . . . , wn # » # » } respectively. Then, P := B−1 ◦ B ′ Furthermore, B ′ = B ◦ P . These relations defnes a mapping from F n to F n , and we must have P ∈ GLn(F ). can be seen by following the arrows in the below diagram: F n B V P B ′ » » » » » F n We can fgure out the columns of P as well: P (# e i) = B−1(B ′ (# e i)) = B−1(w # i). We know B−1(w # i) is just the coordinates of columns of P −1 by taking the each # » » wi for the basis { # v 1, . . . , v n v i and writing it in terms of the basis of w # ’s. # » # » Similarly, we can fgure out the » }. = (B ′ )−1(# v ). # » » x = B−1(# v ) and x # »′ ∈ V , we can write the coordinates The coordinates are » Given a vector # v » » » » related by P # x = x and P −1 # x = x ′ # # ′ by using our expression for P . 35 Lecture 8: Linear Transformations with Bases and the Dimension Formula For any fnite-dimensional vector space, by picking a basis, we can write every vector in terms of coordinates. Then, for every transformation T : V → W , we can write it as a matrix A using coordinates. Suppose we have → W and the respective bases are {# v i} and {w # i}. » » the coordinate maps B : F n → V and C : F m T V W B C F n A F m Following the arrows, we have A := C−1 ◦ T ◦ B ∈ Matm×n(F ). To write down A, we can fnd the ith column of A by writing T (# v i) in terms of the w # ’s.
» » Example 8.3 ′′ (t) = f(t).
Let’s see an example of computing A. Let V be the set of complex functions such that f Let ′′ (t) = −f(t).
W be the set of complex functions such that f We can defne the transformation T : V → W f(t) 7→ f(it). Note that T looks nothing like a matrix right now, but we can turn it into one by picking bases for V and t W . One choice of basis is V = Span(e , e−t) and W = Span(cos t, sin t). it −it Then, by writing T (et) = e = cos t + i sin t and T (e−t) = e = cos t + −i sin t, we have found the columns of A: 1 1 A = . i −i it If we had chosen a diferent basis for W = Span(e , e−it), then our matrix A ′ would just be the identity matrix. Guiding Question We have seen that the same linear transformation leads to diferent A, so can we pick bases so that A “looks very nice"? For example, in the previous example, we were able to make A the identity matrix by picking a diferent basis. We will answer this question at the end of the next section. 8.3 Dimension Formula Note that linear transformations are very similar to group homomorphisms. They are both mappings that preserve the structure that we care about. We can defne and prove similar results as the ones we showed for group homomorphisms. # » # » » » » » » Defnition 8.4 Given a linear transformation T : V → W , we can defne the kernel and image. ker(T ) := { # v | T (# v ) = 0 } im(T ) := {w # | T (# v )w # for some v ∈ V } By similar logic to group homomorphisms, these are vector subspaces of V and W respectively. We also defne the nullity and the rank as the dimension of the kernel and image respectively. Theorem 8.5 (Dimension formula) Given T : V → W , dim(ker T ) + dim(im T ) = dim(V ). This is somewhat reminiscent of the theorem on groups |G| = |ker(G)||im(G)|. 36 Lecture 8: Linear Transformations with Bases and the Dimension Formula # » Proof. Pick # v 1, . . . , v k as a basis for ker(T ). From a theorem from last class, we can add vectors # v k+1, . . . , v n to get a basis for V where n = dim V and k = dim(ker(T )). » T (# v i). # » » » # » # » # » Let 0 for 1 ≤ i ≤ k.
:= wi By the defnition of kernel, wi = We will show that {w # k+1, . . . , wn » } is a basis for im(T ), so that rank(T ) = n − k = dim(V ) − dim(ker(T )). To prove that it is a basis, we need to show that they are linearly independent and they span im(T ). For the span, im(T ) = Span(T (# = Span(T (# » » » )) v 1), . . . , T (# v n » )) v k+1), . . . , T ( # v n » # » ).
= Span(w # k+1, . . . , wn For linear independence, we consider the solutions to: wk+1 + · v k+1 + · # » # » # » » wn # 0 .
· · + an ak+1 = This implies that # » ) = 0 # » T (ak+1 · · + an v n and thus # » # » v k+1 + · # » is a basis for the kernel, there must exist coefcients a1, . . . , ak such that v 1 + · v ’s form a basis for V . # » ∈ ker(T ).
· · + an ak+1 v n # » Since # v 1, . . . , v k » » + · v k+1 # # » # » · + an · + ak · · ak+1 v n = a1 v k. # » However, this forces ai = 0 for all i since the } are linearly » Therefore, {w # k+1, . . . , wn independent and thus a basis of im(T ). # » » » » The proof of the dimension formula shows a bit more. Using the same notation as in the proof, take a basis for V » are also permuted. We extend the basis for im(T ) to a basis for W with the vectors # by writing down the coordinates of T (# v i) with respect to the w’s. k + 1 ≤ i ≤ n, T ( # v i) = When 1 ≤ i ≤ k, T (# v i) = 0. » v 1, . . . , # v , n # » # » This is essentially the same basis, but permuted so that the coordinate vectors to be # v k+1, . . . , » v k. # » # » # » wk+1, . . . , wn, u 1, . . . , u r. Then we fnd the matrix for T When # » Our matrix for T is particularly simple (written in wi. block form): A = In−k 0 0 0 . Corollary 8.6 For any linear transformation, we can write its matrix in the above form for some choice of basis for V and W . Corollary 8.7 As a special case, if we already are given a matrix M ∈ Matm×n representing a linear transformation from F n → F m , then there exists change of basis matrices P ∈ GLn(F ), Q ∈ GLm(F ) such that Q−1MP is in the above form. Pictorially, this looks like: F n M F m Q P F n A F m 37 Lecture 9: Eigenvectors, Eigenvalues, and Diagonalizable Matrices 9 Dimension Formula 9.1 Review Last time, we discussed linear transformations between two vector spaces. By picking a basis cleverly, it is possible to write the matrix of the linear transformation in a very nice form. For example, given a linear transformation M : F n − → F m , by changing the bases for F n and F m with the invertible matrices P and Q, the matrix A will have a very simple form. F n M F m Q P F n A F m With appropriate bases, A = Q−1MP = Ik 0 , 0 0 where A is an m×n matrix with the identity in the top left corner. The frst k columns correspond to the image and the last n − k columns correspond to the kernel. As a corollary, it is possible to see the dimension formula dim im(A) + dim ker(A) = n. Corollary 9.1 Given a matrix M ∈ Matm×n(F ), we have rank(M) = rank(M T ). Essentially, this corollary states that the dimension of the span of the columns is the same as the dimension of the span of the rows, which is surprising! The frst is a subspace of F m , while the second is a subspace of F n , but they still have the same dimension. Sketch of Proof. This theorem is clearly true for A, since the row-rank and the column-rank are both just k. However, since A and M difer by isomorphisms P and Q, the rank of A is the same as the rank of M. Similarly, the rank of AT is also equal to the rank of M T . Therefore, rank(M) = rank(M T ). We are not going to use this often in this course, but this is a fact emphasized in traditional linear algebra classes. 9.2 Linear Operators Today, we will specialize the discussion on arbitrary linear transformations to linear operators, which go from a vector space to itself. Defnition 9.2 A linear operator is a linear transformation T : V − → V. Let’s see some examples. Example 9.3 Let V = R2 . Then, T is the linear transformation that is rotation by angle θ counterclockwise. This goes from the vector space to itself. 38 Lecture 9: Eigenvectors, Eigenvalues, and Diagonalizable Matrices Example 9.4 Let V = {polynomials of degree ≤ 2}. Then the derivative T (f(t)) = f ′ (t) is a linear operator. The frst natural question to ask about linear operators is working out the matrix of the linear transformation upon picking a basis. The only diference between this discussion and the discussion on linear transformations is that here, the transformation is from a vector space to itself, so once a basis has been picked, both sides have a fxed basis. For general linear transformations from a vector space to a diferent vector space, two diferent bases can be picked. Guiding Question What is the matrix of a linear operator on a vector space with a chosen basis? Consider a basis B : F n − → V. Then T becomes a square matrix A ∈ Matn×n(F ). For Example 9.3, picking the basis standard basis gives a rotation matrix: 1 0 , 0 1 ⇝ A = cos θ sin θ − sin θ cos θ . This is determined by fguring out where the ith basis vector is mapped, which is the ith column. For Example 9.4, it is also possible to write down a matrix: 0 1 0 {1, t, t2} ⇝ A = 0 0 2 . 0 0 0 This should all be reminiscent of the previous section. Proposition 9.5 When working with linear operators T : V − → V , for V fnite-dimensionala , then T is injective ↔ T is surjective ↔ is an isomorphism. aIn this class, the implicit assumption will always be that we are working with fnite-dimensional vector spaces In fact, this fact is true for maps from a fnite set to itself. Finite-dimensional vector spaces can be infnite, but have the same property. Proof. Using the dimension formula, dim ker T + dim im T = dim V. If T is injective, then dim ker T = 0, which is true if and only if dim im T = dim V, which means that T is surjective. So fnite-dimensional vector spaces behave a lot like fnite sets. 9.3 Change of Basis Now, the next natural question to ask is about changing bases. Guiding Question What happens to a matrix for T : V − → V upon changing basis for V ? 39 Lecture 9: Eigenvectors, Eigenvalues, and Diagonalizable Matrices Specifying a basis for V determines a diagram T V V .
B B F n A F m A new basis comes from an invertible matrix P ∈ GLn(F ), where the new basis is B ′ = B · P, and determines an extended diagram T V V B B A .
B ′ F n F n B ′ P P A ′ F n F n There is a new matrix A ′ that represents the same linear transformation T. The diference between this case and the general case is that the bases are the same on either side of the transformation, and there is no longer the freedom to choose diferent bases for the domain and the codomain. The new matrix, by following the arrows on the diagram, is A ′ = P −1AP. The matrix A ′ is related to A by conjugation by P. Defnition 9.6 A matrix A ′ is similar to A if there exists some P ∈ GLn(F ) such that A ′ = P −1AP. Similar matrices arise from the same linear operator from a vector space to itself, but with diferent bases picked. Again, to emphasize, the diference between today’s case, T : V − → V , and the case in the last section, T : V − → W , is that in the frst case there is only one base change matrix P, instead of P and Q, since the matrix must operate the same on the left and right sides. As an result, given a vector space V and an operator T : V − → V, it is possible to defne the determinant of T without having to specify a basis. The vector space V might be a vector space without a canonical basis, but it is still possible to defne the determinant. Picking any basis of V produces a square matrix A, and the determinant would then be det(T ) = det(A). In fact, from the base change formula, it is clear that the determinant does not depend on which basis is used! From a diferent basis, det(A ′ ) = det(P −1AP ) = det(P )−1 det(A) det(P ) = det(A), since the determinant is multiplicative. As a result, it is possible to defne the determinant of T independently of the choice of basis29 , and so det(T ) has a meaning outside of a particular basis. For example, on Rn , the determinant represents a "volume," which is independent of the particular choice of basis. Here, we are saying that even for felds like fnite felds, where "volume" may not make sense, the determinant still has some intrinsic meaning. 9.4 Eigenvectors, Eigenvalues, and Diagonalizable Matrices Our discussion leads to the following question, which is the same as last class, but for linear operators. Guiding Question How nice can we make A by changing basis of V ? Last class, it was possible to make the matrix extremely nice, since we could pick a basis for the domain and for the codomain. Now, let’s see an example for when the domain is the same vector space as the codomain. 29It doesn’t depend on which basis was chosen, so any basis works 40 Lecture 9: Eigenvectors, Eigenvalues, and Diagonalizable Matrices Example 9.7 Let V = R2 . Consider 2 3 A = . 3 2 We see that 1 5 1 A = = 5 1 5 1 and −1 1 −1 A = = −1 . 1 −1 1 The operator is scaling the vector (1, 1) and fipping the vector (−1, 1), and the transformation on any other vector will be a combination of these two moves, scaling and fipping. In particular, taking 1 −1 P = , 1 1 which has the frst vector as the frst column and the second vector as the second column, gives 5 0 A ′ = P −1AP = . 0 −1 In the new basis, it is possible to make the matrix diagonal! Making the matrix diagonal makes it possible to see how it operates, which is stretching by 5 in one direction, and fipping in the other direction (both independently of the other direction). In general, we will really want to be able to make matrices diagonal, since it allows us to see what it is doing in the direction of each basis vector, independently of the other directions (as there are 0s in the matrix elsewhere). This gold standard type of vector will be called an eigenvector. Defnition 9.8 A vector v ̸= 0 is an eigenvector if Tv = λv for some λ ∈ F, and λ is called an eigenvalue. When an operator is applied to a vector, the result is proportional to the vector. The operator maintains the direction of the vector, and just scales it. Obviously, scaling an eigenvector by some nonzero scalar also results 41 Lecture 9: Eigenvectors, Eigenvalues, and Diagonalizable Matrices in another eigenvector. Example 9.9 For Example 9.7, the vector 1 1 is an eigenvector with eigenvalue 5, and −1 1 is an eigenvector with eigenvalue -1. This example is special because not only are there eigenvectors, there are enough to form a basis. Defnition 9.10 A basis {v1, · · · , vn} of V where each vi is an eigenvector; that is, Tvi = λivi, then the basis is called an eigenbasis. In an eigenbasis, the matrix for T is λ1 · · · 0 .
. . .. . . . , . 0 · · · λn which is diagonal with λi in the (i, i)th entry. Diagonal matrices are extremely nice. In general, it is very hard to take matrices to high powers, but for diagonal matrices, each entry is simply raised to that power. Defnition 9.11 If a linear operator has an eigenbasis T, it is called diagonalizable. An equivalent defnition holds for matrices. Defnition 9.12 Given a matrix A, if there exists some invertible P such that P −1AP = D for some diagonal matrix D, then A is called diagonalizable. That is, A is diagonalizable if it is similar to a diagonal matrix. The key concept is that an eigenbasis provides the directions in which the operator T behaves nicely by simply scaling or fipping a vector in that direction. 9.5 Finding Eigenvalues and Eigenvectors Unfortunately, not every matrix is diagonalizable, but the focus for the next few classes will be fnding eigen vectors, eigenvalues, and eigenbases, assuming that a matrix is diagonalizable. Guiding Question How do we fnd eigenvectors, eigenvalues, and eigenbases? • Step 1. Perhaps unintuitively, the frst step is to fnd possible eigenvalues! Given a matrix A ∈ Matn×n(F ) in some less good basis, we want to fnd eigenvectors that form a better basis that is an eigenbasis, so that A will be diagonal and have a nicer form. Suppose λ is an eigenvalue for A. Then there exists some nonzero v such that Av = λv, 42 Lecture 9: Eigenvectors, Eigenvalues, and Diagonalizable Matrices by defnition. (There may be lots of v, and in fact scaling any v will produce another eigenvector, but for a given λ we just want to know if there is a v at all.) We know that λv = λInv, so this is equivalent to (λIn − A)v = 0 for some v. That is, the kernel is nontrivial: ker(λIn − A) ̸= {⃗ 0}. This is equivalent to λIn − A is not invertible, which happens if and only if det(λIn − A) = 0. This is not bad at all! The determinant is a formula that we can just calculate. So in fact, we want to look for λ such that the equation det(λIn − A) = 0 . holds. It is customary in this context to replace t with λ, and so with t as the variable, we have p(t) := det(tIn − A), which is a polynomial of degree n in t called the characteristic polynomial. Example 9.13 Given A = 2 3 3 , 2 we have the characteristic polynomial t − 2 −3 pA(t) = det = (t − 2)(t − 2) − (−3)(−3) = t2 − 4t − 5.
−3 t − 2 In general, where A = (aij ), we have t − a11 · · · ⋆ pA(t) = det . . . .. . . . . n = t + · · · , ⋆ · · · t − ann which is a degree n polynomial in t. If A ′ is similar to A, then they have the same characteristic polynomial, since the determinant is basis-invariant. Proposition 9.14 Given λ ∈ F, λ is an eigenvalue for A if and only if pA(λ) = 0; that is, if and only if λ is a root of pA(t). For example, for our earlier example, the eigenvalues would be −1 and 5, since t2 − 4t − 5 = (t + 1)(t − 5). As a caveat, if F is an arbitrary feld, there may not be any roots. For example, a rotation matrix over R does not have any real eigenvalues. However, if F = C, there will always be n roots (not necessarily distinct), and so there will always be eigenvalues. • Step 2. For each eigenvalue, fnd the associated eigenvectors. For each λ, we want to take a vector in ker(λIn − A), which by assumption is a nonzero subspace. Using Gaussian elimination or row operations, we can mechanically compute a basis for ker(λIn − A), although we will not spend a lot of time on this. 43 Lecture 9: Eigenvectors, Eigenvalues, and Diagonalizable Matrices Example 9.15 For λ = 5, we have 3 −3 5 · I2 − A = , −3 3 and the kernel is ker = Span 1 . 1 We will say more about this next lecture! 44 Lecture 10: The Jordan Decomposition 10 Eigenbases and the Jordan Form Change of basis is a powerful tool, and often, we would like to work in as natural a basis as possible. Guiding Question Given a linear operator, how can we fnd a basis in which the matrix is as nice as possible? 10.1 Review Last time, we learned about eigenvectors and eigenvalues of linear operators, or more concretely, matrices, on vector spaces. An eigenvector is a (nonzero) vector sent to itself, up to scaling, under the linear operator, and its associated eigenvalue is the scaling factor; that is, if A⃗ v = λ⃗ v for some scalar λ, ⃗ v is an eigenvector with associated eigenvalue λ. λ1 · · · 0 If there exists an eigenbasis30, then in that basis, the linear operator P −1AP 31 will simply become 0 ... , 0 0 · · · λn since each basis vector ⃗ vi is sent to λi⃗ vi. So having an eigenbasis is equivalent to the matrix A being similar to a diagonal matrix. In order to concretely fnd the eigenvectors, it is easier to frst fnd the eigenvalues, which are the roots of the characteristic polynomial pA(t) = det (tIn − A). Each root λ of pA has at least one corresponding eigenvector. The eigenvectors for λ are precisely the nonzero vectors in ker(λIn − A). 10.2 The Characteristic Polynomial Let’s start with an example. Example 10.1 If A = a b , then pA(t) = t2 − (a + d)t + (ad − bc). c d In general, for an n×n matrix, pA(t) = tn − (a11 + · · · + ann)tn−1 + (−1)n det(A). The coefcient of tn−1 is the sum of the entries on the diagonal, called the trace of A. Because the characteristic polynomial can be written as a determinant, it can be defned for general linear operators without specifying a basis, and so each of the coefcients are basis-independent. In particular, we get that Tr(P −1AP ) = Tr(A). Guiding Question What can go wrong when hunting for an eigenbasis? How can we fx this? Example 10.2 cos θ − sin θ Over the real numbers R, take A = . The characteristic polynomial is pA(t) = t2−2 cos θ+1, sin θ cos θ which has no real roots (unless θ = π.) Geometrically, that makes sense, because under a rotation by θ ̸= kπ, every vector will end up pointing an a diferent direction than it initially was, so there should be no real eigenvectors. Over a general feld F, it is certainly possible for the characteristic polynomial not to have any roots at all; in order to fx this issue, we work over a feld like C, 32 where every degree n polynomial always has n roots (with 30A basis consisting of eigenvectors 31This comes from the change of basis formula 32Fields where every non-constant polynomial has roots are called algebraically closed. 45 Lecture 10: The Jordan Decomposition multiplicity). So pA(t) = (t − λ1) · · · (t − λn), where the λi can repeat. For the rest of the lecture, we will only consider linear operators on vector spaces over C, which takes care of the frst obstacle of fnding eigenvalues. However, even over C, not every linear operator has an eigenbasis. Example 10.3 Consider A = 0 1 . The characteristic polynomial is pA(t) = t2 , so if A were similar to some diagonal 0 0 matrix, it would be similar to the zero matrix; this would mean that A would be the zero matrix, and thus A cannot be diagonalizable. In other words, pA(t) only has one root, 0, so any eigenvector would be in ker(0I2 − A) = Span(⃗ e1). So there is only a one-dimensional space of eigenvectors, which is not enough to create an eigenbasis. In some sense, which we will make precise later on in this lecture, this is the most important counterexample for why linear operators can be nondiagonalizable. Proposition 10.4 Given an n×n matrix A, eigenvectors ⃗ v1, · · · , ⃗ vk, and distinct eigenvalues λ1, · · · , λk, the vectors ⃗ vi are all linearly independent. Proof. Let’s prove this by induction. Base Case. If k = 1, by the defnition of an eigenvector, ⃗ vk ̸= 0 so {⃗ vi} is linearly independent. Inductive Hypothesis. Suppose the proposition is true for k − 1. Inductive Step. Now, suppose the proposition is not true for k. Then there exist coefcients ai such that X ai⃗ vi = 0. Applying A to both sides, we get X aiλi⃗ vi = 0, which is another linear relation between the k vectors. Subtracting k times the frst relation from the second one results in the linear relation i= X k−1 X ai(λi − λk) = ai(λi − λk) = 0. i=0 Since in the last term λk − λk = 0, while λi − λk = ̸ 0 for i ̸= k since the λi are distinct, we obtain a linear relation between ⃗ v1, · · · , ⃗ vk−1, which is a contradiction of the inductive hypothesis. Thus, {vi}i=0,··· ,k is linearly independent. Corollary 10.5 Consider a matrix A. If the characteristic polynomial is pA(t) = (t − λ1) · · · (t − λn) where each λi is distinct, A will have an eigenbasis and will thus be diagonalizable. Proof. Each eigenvalue must have at least one eigenvector. Taking ⃗ v1, · · · , ⃗ vn to be eigenvectors for λ1, · · · , λn. Since there are n eigenvectors, which is the same as the dimension of the vector space, and by Proposition 10.4 they are linearly independent, they form an eigenbasis and A is diagonalizable. If there are repeated roots, then there will not necessarily be enough eigenvectors to form a basis. Luckily for us, it is usually true that a matrix will be diagonalizable.33 33More concretely, the space of n×n square matrices can be thought of as a metric space, and the non-diagonalizable matrices will be a set of measure zero. In particular, the diagonalizable matrices are dense in the space of all square matrices. Intuitively, given a non-diagonalizable matrix, perturbing the entries by a little bit will perturb the roots a little bit, making them non-distinct. 46 Lecture 10: The Jordan Decomposition In general, pA(t) = (t − λ1)e1 · · · (t − λk)ek , where the λi are distinct. Let Vλi = ker(λiI − A). Any vector v ∈ Vλi is an eigenvector with eigenvalue λi. We know that for each i, dim Vλi ≥ 1. Using our proposition, given a basis for each subspace Vλi , if there are enough to get n total vectors, combining all the bases would give an eigenbasis for A, since they would all be linearly independent.34 10.3 Jordan Form Keeping in mind the matrix A = 0 1 , we have the following question. 0 0 Guiding Question If a matrix is not diagonalizable, what is nicest form it can take on under a change of basis? Let’s see a class of matrices that always have the issue of repeated eigenvalues. Defnition 10.6 Given a ≥ 1 and λ ∈ F, let the Jordan block be an a×a matrix λ 1 · · · 0 Ja(λ) = 0 0 λ . . . ... ... . . . 1 0 0 · · · λ a with λ on the diagonal and 1s above each λ. aThe notation here is a little diferent from the textbook. Example 10.7 For λ = 1, 2, 3, we get λ λ , 0 λ 1 , 0 λ 0 1 λ 0 0 1 . λ For Ja(λ), the characteristic polynomial is (t − λ)a , and when a > 1, the only eigenvalue is ⃗ e1 so it will not be diagonalizable. Although this these Jordan blocks are very specifc matrices, in some sense they are exactly the sources of all the problems. Example 10.8 (J4(0)) 0 1 0 0 The matrix J4(0) = 0 0 0 0 1 0 0 . 1 Applying it to the basis vectors gives a "chain of vectors" 0 0 0 0 ⃗ e4 7→ ⃗ e3 7→ ⃗ e2 7→ ⃗ e1 7→ 0, where each basis vector is mapped to the next. This leads us to the main theorem. 34Computationally, it is simple to fnd basis vectors, and in a more computational class we would go further in-depth on fnding these. 47 Lecture 10: The Jordan Decomposition Theorem 10.9 (Jordan Decomposition Theorem) Given a linear operator T : V − → V, where dim V = n, there exists a basis ⃗ v1, · · · , ⃗ vn, there exist pairs (a1, λ1), · · · , (ar, λr) such that the matrix of T in this basis is a block diagonal matrix Ja1 (λ1) Ja2 (λ2) ... , Jar (λr) where all other entries are 0. While it is not possible to diagonalize every linear operator, it is possible to write them as a block diagonal matrix with Jordan blocks on the diagonal. Additioanlly, these Jordan blocks are unique up to rearrangement. This block diagonal matrix is called the Jordan decomposition of the linear operator T. Student Question. Do the ai correspond to the exponents of the roots? Answer. Not quite, as we will see promptly. Let’s continue with some examples. Example 10.10 (n = 4) λ1 1 0 0 0 λ1 1 0 If we have a1 = 4, then the Jordan decomposition will look like . If a1 = 3 and a2 = 1, 2, a2 1, and a3 1, we would have , and for a1, a2, a3, a4 1, we would just 0 0 λ1 1 0 0 0 λ1 λ1 1 0 λ1 1 0 then it will look like 0 λ1 0 1 λ1 . For a1 = 2 and a2 = 2, we have 0 λ1 λ2 . 1 Where λ2 0 λ2 λ1 1 0 λ1 a1 = = = = λ2 λ3 λ1 λ 2 get , which is really just a diagonal matrix. λ3 λ4 Essentially, every matrix is similar to some Jordan decomposition matrix, where it is diagonalizable if and only if each ai = 1. The characteristic polynomial of T will be (t − λ1)a1 · · · (t − λr)ar . These exponents ai are not quite the same as the exponents ei from before, since the λi in the characteristic polynomial of T can repeat. However, for eigenvalues equal to λj , the sum of all the exponents will in fact be ej . From the characteristic polynomial of a matrix, it is not possible to precisely fgure out the Jordan decomposition, but it does provide some amount of information. Next class, we will continue seeing what information we get from the Jordan Decomposition Theorem. 48 Lecture 11: Proving the Jordan Decomposition Theorem 11 The Jordan Decomposition 11.1 Review Recall this theorem from last time. Theorem 11.1 # » # » Considering a transformation T : V → V, there must exist a basis v 1, · · · , v n such that the matrix of T (in this basis) is Ja1 (λ1) 0 · · · 0 . .
0 Ja2 (λ2) . 0 A = , . .
.
. .
. Jai (λi) 0 0 0 0 Jan (λn) where Jai (λi) are the Jordan blocks. A special case is when all the ai = 1. Then, λ1 · · · 0 .
.
.
A = . .
.
.
. . 0 · · · λr is a diagonal matrix.35 11.2 The Jordan Decomposition, Continued The characteristic polynomial of the matrix A will be pA(t) = (t − λ1)a1 · · · (t − λr)ar , where it is possible to have repeated λi. As a result, it is not possible to determine the Jordan decomposition simply from the characteristic polynomial, since there are diferent ways to take a repeated root and split it up into Jordan blocks. (If all the roots of the characteristic polynomial are distinct, the Jordan form is uniquely determined.) However, the characteristic polynomial does provide some information. For a fxed eigenvalue λ, X ai = exponent of (t − λ) in pA(t). Jai (λ) Example 11.2 (n = 4) For example, when n = 4, consider a matrix where pA(t) = t4 . There are multiple possible Jordan forms; in particular, it can be split up as 4, 3 + 1, 2 + 2, 2 + 1 + 1, or 1 + 1 + 1 + 1 : 0 0 0 1 0 0 0 1 0 0 0 , 1 0 0 0 1 0 0 0 1 0 , 0 0 1 0 0 , 1 0 0 1 0 0 , 0 0 . 0 0 0 0 0 1 0 0 0 0 # » For a given Jordan block, there is one eigenvector. Fixing λ again, this tells us that dim(ker(λI − A)) is equal to the number of blocks with λ along the diagonal. Up to reordering of the basis vectors, the Jordan decomposition is unique. 35In the textbook, Artin puts the 1s below the diagonal in a Jordan block. Conventionally, the 1s are above the diagonal, but it e 1 moves the 1s from above the diagonal # » # » # » doesn’t make a diference, because reversing the order of the vectors to · · · · · · e 1, e a e a, to below the diagonal. The diference is notational. 49 Lecture 11: Proving the Jordan Decomposition Theorem Example 11.3 0 1 0 0 0 0 1 0 Take J4(0) = . Under J4(0), each basis vector maps to the next basis vector, and there is 0 0 0 1 0 0 0 0 one chain of length 4: ⃗ e4 7→ ⃗ e3 7→ ⃗ e2 7→ ⃗ e1 7→ ⃗ 0. As a result, applying J4(0) multiple times will eventually send all vectors to zero; that is, in this case, J4(0)4 = 0. 0 1 0 0 0 0 0 0 J2(0) 0 On the other hand, consider J2,2(0) = = . 0 0 0 1 0 J2(0) 0 0 0 0 Applying the operator to the basis vectors yields two chains of length 2: ⃗ e2 7→ ⃗ e1 7→ ⃗ 0 ⃗ e4 7→ ⃗ e3 7→ ⃗ 0 In this case as well, the operator will map every vector to zero upon repeated application. In general, for λ ̸= 0, (λI − T )# » » e i is not necessarily zero (it is zero only if # e i is an eigenvector), but for some 36 large enough n, = 0. » (λI − T )n # e i In Example 11.3, there was a chain of length 4 for the frst matrix, while in the second matrix, we had two chains of length 2. Note 11.4 The Jordan decomposition theorem is powerful because any square matrix has a Jordan decomposition. On the other hand, most matrices are diagonalizable, and any matrix will be ε away from a diagonalizable matrix, and the Jordan decomposition is unnecessary. Only in the zero percent of the timea when the characteristic polynomial has repeated roots is it necessary. aThis concept is feshed out in measure theory. 11.3 Proof of Jordan Decomposition Theorem The proof of the Jordan decomposition theorem is quite involved and relatively tricky, so the important part for the rest of class is understanding the style of proof, rather than the exact details. This proof will break down the theorem inductively into smaller and smaller pieces. Let’s start with a couple of defnitions that will help us with the proof. Defnition 11.5 Given a vector space V and a linear transformation T : V → V, a subspace W ⊆ V is called T -invariant if T (w ⃗ ) ∈ W for all w ⃗ ∈ W. For example, if the vector space V is the space of polynomials of degree at most 3, and the subspace W is the space of polynomials of degree at most 2, W will be T −invariant under the linear operator T that is taking the derivative. 36A vector that is killed not necessarily immediately but eventually by λI − T is known as a generalized eigenvector; there is a question about them on the problem set. 50 Lecture 11: Proving the Jordan Decomposition Theorem Defnition 11.6 ′ ′ Given a vector space V and two subspaces W, W ⊆ V, we say that V is the direct sum of W and W , ′ notated V = W ⊕ W , if every ⃗ v ∈ V can be written uniquely as ′ ⃗ v = w ⃗ + w ⃗ , ′ ′ where w ⃗ ∈ W and w ⃗ ∈ W . For example, if V = R3 , every vector can be written as the sum of some vector in the z−direction and some vector lying in the xy−plane. Equivalently, there must exist a basis ′ ′ {w ⃗1, · · · , w ⃗r, w ⃗1, · · · , w ⃗ } r ′ ′ ′ of V such that {w ⃗1, · · · , w ⃗r} is a basis of W and {w ⃗1, · · · , w ⃗ } is a basis of W . This is also sometimes called a r splitting of V, since V has been split up into two subspaces. Theorem 11.7 ′ ′ ′ a If dim W + dim W = dim V, and W ∩ W = {⃗ 0}, then it must be the case that V = W ⊕ W . aThis can be proved using the characterization in terms of bases, and is related to a homework problem. Defnition 11.8 ′ Given a splitting V = W ⊕ W and a linear operator T : V → V, we say that this splitting is T -invariant if ′ W and W are T −invariant. ′ In a basis for W and W , the matrix for T must be block-diagonal; that is, of the form ⋆ 0 0 ⋆ ′ where each ⋆ is some matrix; this is because vectors in W or W will be mapped back to other vectors in W or ′ W . Conversely, if T is block diagonal in some basis, it automatically provides a T -invariant splitting of V. The span of the collection of basis vectors in the frst block becomes a T -invariant subspace W and the second one becomes ′ W . Essentially, these defnitions provide a characterization of linear transformations being block-diagonal, without having to pick a basis. Now, we can fnally start proving the Jordan Decomposition Theorem. Roughly, the proof follows an induction argument on the dimension of V , where a vector space is split up into two smaller dimensional T -invariant subspaces for the operator T , both of which will then have Jordan decompositions by the inductive hypothesis, which will provide the Jordan decomposition of the original vector space. Essentially, we want to break it down to the case of a singular eigenvalue, considering matrices that look like those in Example 11.3, relying on the fact that repeatedly applying these operators will eventually take any vector to zero. Defnition 11.9 a A linear transformation T is nilpotent if there exists some m ≥ 0 such that T m = 0. aThe Jordan block Jm(0) is nilpotent with exponent m. Proof. This proof has several steps. • Step 0. Over complex vector spaces, there will always exist an eigenvalue, so let λ be some eigenvalue of T. Because λI is already diagonal, we can replace T with T − λI, so that 0 can be assumed to be one of the eigenvalues. Essentially, if the Jordan decomposition theorem is true for T − λI, by adding the diagonal matrix λI, the Jordan decomposition theorem will become true for T. • Step 1. 51 Lecture 11: Proving the Jordan Decomposition Theorem Rough Sketch.After this simplifcation, we will zero in37 on the 0 eigenvalue. We show that there exists a T −invariant splitting V = W ⊕ U such that T : W → W w is nilpotent and T : U → U u is invertible. For a nilpotent operator, the only possible eigenvalues are 038 , while for an invertible operator, there are only nonzero eigenvalues39 , so this splitting separates out the eigenvectors will eigenvalue λ = 0. By assumption, there exists a zero eigenvalue, and so dim W ≥ 1, and then dim U ⪇ n. Since dim U < n, by the inductive hypothesis, there is a Jordan decomposition for U. However, since dim W could be equal to n (dim U could be 0), the inductive hypothesis does not apply and so we must still show that there is a Jordan decomposition for nilpotent operators. Full Proof. Now, we still need to show that this splitting exists. Consider the vector space V ; TV = im T lies inside of V (it cannot possibly take V to a higher-dimensional space), and so we obtain the chain V ⊃ TV ⊃ T (TV ) ⊃ T (T (T (V )) ⊃ · · · . The dimension can only drop fnitely many times40 , since T i(V ) cannot have negative dimension, so there exists some stable dimension m between dim T and 0 such that = T m+1V = T m+2V T mV = · · · . Let U := T mV = im(T m) and W = ker(T m). First, T is nilpotent on W because W = ker(T m), so (T |W )m = 0, which is the defnition of being nilpotent. Also, T |U is invertible because U = im(T |U ), so T |U is surjective from U to itself, which implies that it is invertible. Lastly, W ∩ U = {v ∈ U : T m = 0}, by defnition, which is precisely the zero vector, because T is invertible on U so it maps only the zero vector to the zero vector. Using the rank-nullity theorem, dim ker T m + dim im T m = dim V, so by Theorem 11.7, W ⊕ U is in fact a splitting. • Step 2. Now, we prove that if T is nilpotent, it has a Jordan decomposition. We have a vector space V, a linear operator T : V − → V, and some m such that T m = 0. To do so, we will fnd by induction on the dimension a basis of V for which T acts in "chains" as in Example 11.3. Let W = im T ⊊ V. By induction, there exists such a basis {⃗ ei} for W where T acts in chains. ⃗ v1 ⃗ e3 ⃗ v2 ⃗ e2 ⃗ e5 ⃗ v3 ⃗ e1 ⃗ e4 ⃗ e6 ⃗ u1 ⃗ u2 ⃗ u3 0 0 0 0 0 0 37Ha ha 38Consider a nilpotent operator A. Then there is some n such that An = 0. If v is an eigenvector for A with eigenvalue λ, Anv = λnv = 0, so λn = 0 and thus λ must also be zero. 39Assume 0 is an eigenvalue of an invertible operator A, corresponding to an eigenvector v. Then Av = 0 for some nonzero v; then both the vector 0 and the vector v map to 0 and thus A is not one-to-one or invertible. 40In fact, this argument relies on the fact that V is fnite-dimensional! 52 Lecture 11: Proving the Jordan Decomposition Theorem For each chain, we insert a preimage ⃗ vi of the top vector in each chain, where the ⃗ vi are not vectors in W but rather vectors that map to vectors in W. These exist since W is the image of T , and so every vector in W is the image of some other vector under T. Additionally, we add vectors in ker T , ⃗ ui, which all map to zero since they are in the kernel. This produces a bunch of chains for V , starting for a bunch of chains for W. We claim that B = {⃗ ei} ∪{⃗ vj } ∪{⃗ uk} is a basis for V. It is linearly independent because applying T to any linear dependence would give a dependence between basis vectors of W (since T (⃗ vi) = ⃗ ej and T (⃗ uk) = 0.) Also, where c is the number of chains, the number of vectors in B is dim(W ) + dim(ker(T )) − c + c, which is precisely the dimension of V, and thus B is in fact a basis. For this particular example illustrated in the fgure, the Jordan blocks for W have size 3, 2, and 1, and for V these are extended to size 4, 3, and 2, along with three more blocks of size 1. The schematic of the argument is more important than the exact argument itself, but we still have to do the whole thing. :) 53 Lecture 12: Orthonormal Matrices 12 Orthogonal Matrices In this lecture, we start formally studying the symmetry of shapes, combining group theory with linear algebra. The matrices considered will be over R, the feld of real numbers, rather than C. 12.1 Dot Products and Orthogonal Matrices Recall the following defnitions. Defnition 12.1 Pn T Given column vectors x, y ∈ Rn , the dot product is defned as x · y = x y = The length of a √ i=1 xiyi. vector v is |v| = v · v. The dot product is defned algebraically, but also carries geometric information about two vectors: x · y = |x||y| cos θ. Moreover, if x · y = 0, then x and y will be perpendicular vectors in Rn . To start out with, consider bases for which the pairwise dot products are as simple as possible. Defnition 12.2 √ A basis {v1, · · · , vn} is called orthonormal if |vi| = 1 and vi · vj = 0 for i ̸= j. That is, since |vi| = vi · vi, vi · vj = δij , where δij denotes the Kronecker delta.a aThe Kronecker delta δij is equal to 0 if i ̸= j and 1 if i = j. Now, Rn not only has a vector space structure, but it also has some extra structure provided by the dot product. √ Since |v| = v · v, the dot product produces some notion of "length" or "distance." Guiding Question What kinds of matrices interact well with this notion of distance? Orthogonal matrices are those preserving the dot product. Defnition 12.3 A matrix A ∈ GLn(R) is orthogonal if Av · Aw = v · w for all vectors v and w. In particular, taking v = w means that lengths are preserved by orthogonal matrices. There are many equivalent characterizations for orthogonal matrices. Theorem 12.4 The following conditions are all equivalent: 1. The matrix A is orthogonal. 2. For all vectors v ∈ Rn , |Av| = |v|. That is, A preserves lengths. 3. For an n-dimensional matrix A, AT A = In. 4. The columns of A form an orthonormal basis.a aSince AT also satisfes the third condition, this means that the rows of A, which are the columns of AT , will also form an orthonormal basis. Proof. All the conditions will end up equivalent. 54 Lecture 12: Orthonormal Matrices √ √ • Condition (1) implies (2). Because A preserves dot products, |Av| = Av · Av = v · v = |v|, and so A also preserves lengths. • Condition (2) implies (1) because 1 Av · Aw = |Av + Aw|2 −|Av|2 −|Aw|2 2 1 = |v + w|2 −|v|2 −|w|2 2 = v · w. Namely, dot products can be written in terms of lengths, and lengths can be written in terms of dot products, so preserving one is equivalent to preserving the other. • Condition (1) states that Av · Aw = v · w; unwinding the dot product in terms of matrix multiplication, T this equation is vT AT Aw = v w for all v, w ∈ Rn . Evidently, (3) implies (1), since if AT A = In, vT AT Aw = vT w. T By calculation, it can be seen that for ei and ej the ith and jth standard basis vectors, e Mej = Mij , i which is the (i, j)th component of the matrix M. If (1) is true, taking v = ei and w = ej over all i and j gives us that the (i, j)th component of AT A is 1 when i = j and 0 otherwise. • Condition (4) is equivalent to (3) from simply computing the matrix product: the (i, j)th entry of AT A is the dot product of the ith column of A with the jth column of A, which is 1 when i = j and 0 otherwise. Orthogonal matrices preserve lengths, as well as preserving angles up to sign. In general, a set of matrices satisfying some well-behaved properties of a set of matrices generally form a subgroup, and this principle does hold true in the case of orthogonal matrices. Proposition 12.5 The orthogonal matrices form a subgroup On of GLn. Proof. Using condition (3), if for two orthogonal matrices A and B, AT A = BT B = In, it is clear that (AB)T AB = BT AT AB = BT B = In. The other subgroup properties are not difcult to verify. 12.2 The Special Orthogonal Group Given an orthogonal matrix A, AT A = In, and so det(AT A) = det(AT ) det(A) = det(A)2 = det(In) = 1. As a result, det(A) = ±1. The determinant is a homomorphism from det : GLn − → R, and the restriction to On is a homomorphism det : On − → {±1}. The kernel forms a subgroup of On. Defnition 12.6 (Special Orthogonal Group) The orthogonal matrices with determinant 1 form a subgroup SOn ⊂ On ⊂ GLn called the special orthogonal group. Because the determinant is surjective41 , the kernel, SOn, is an index 2 subgroup inside of On. The two cosets are SOn itself and all the matrices with determinant −1. To gain some intuition for orthogonal matrices, we will look at some examples! For n = 1, the orthogonal group has two elements, and [−1], which is not too interesting. 12.3 Orthogonal Matrices in Two Dimensions What are the orthogonal matrices in two dimensions? 41For example, the identity matrix is always orthogonal and has determinant 1, and the diagonal matrix with −1 in the frst row and column and 1 down the rest of the diagonal is also orthogonal and has determinant −1. 55 Lecture 12: Orthonormal Matrices Example 12.7 (O2) Describing an element of O2 is equivalent to writing down an orthonormal basis {v1, v2} of R2 . Evidently, cos θ v1 must be a unit vector, which can always be described as v1 = for some angle θ. Then v2 must sin θ − sin θ sin θ also have length 1 and be perpendicular to v1. There are two choices, v2 = or . This cos θ − cos θ characterizes all 2×2 orthogonal matrices: cos θ − sin θ cos θ sin θ O2 = , . sin θ cos θ sin θ − cos θ In particular, the frst type of matrix has determinant 1, and forms the subgroup SOn, and the second has determinant −1 and forms its the non-trivial coset. Geometrically, the frst type of matrix in O2 are rotations cos θ sin θ by θ around the origin. The matrices of the second type, A = , have characteristic polynomial sin θ − cos θ pA(t) = t2 − 1 = (t + 1)(t − 1). Thus, they have distinct eigenvalues ±1, in contrast to rotation matrices, which do not have any real eigenvalues. Because the eigenvalues are distinct, there is an eigenbasis {v ⃗ +, v ⃗ −}. 56 Lecture 12: Orthonormal Matrices Theorem 12.8 The matrices of the second type are refections across a line through the origin at an angle of θ/2. Proof. Consider the line L = Span(v ⃗ +); since v ⃗ + is an eigenvector with eigenvalue 1, A fxes this line. Notice that v ⃗ + · v ⃗ − = A ⃗ v+ · A ⃗ v− = v ⃗ + · (−v ⃗ −), where the frst equality comes from the fact that A is orthogonal, and the second comes from the eigenvalues 1 and −1 of v+ and v−. The only possibility is v ⃗ + · v ⃗ − = 0, so the two eigenvectors are orthogonal. Writing out any other vector in terms of the eigenvectors, Av is precisely the refection across L. As expected, rotations and refections preserve distance, and in fact they make up all the 2×2 orthogonal matrices. A fun fact that comes from this analysis is that the composition of two refections over diferent lines will be a rotation, since the product of determinants will be (−1) · (−1) = 1. Orthogonal matrices can be thought of either geometrically or algebraically! 12.4 Orthogonal Matrices in Three Dimensions In two dimensions, SO2 consists of rotation matrices. It turns out that in three dimensions, SO3 also consists of rotation matrices. 57 Lecture 12: Orthonormal Matrices In particular, a rotation in R3 is characterized by the axis of the rotation, which is a unit vector ⃗ u ∈ R3 , and the angle of the rotation, which is some θ ∈ R. The plane ⊥ u = {v ∈ R3 : u · v = 0} consists of all the vectors in R3 that are perpendicular to R3 . Defnition 12.9 The rotation operator with spin labels u and θ is ρ(u,θ), the linear operator ρ : R3 − → R3 such that ρ(u) = u and ρ| ⊥ is the rotation by θ counterclockwise with respect to the direction that u points in.a u ⊥ aSince every vector in R3 is a linear combination of u and some vector in u , the rotation operator is described completely by these conditions. There is some redundancy in this description; for example, ρ(u,θ) = ρ(−u,−θ). Theorem 12.10 The rotation operators are exactly SO3. From geometric intuition, this result is not very surprising, since rotations preserve distance.42 Proof. First, we show that all the rotation matrices are in SO3, and then we show that all matrices in SO3 are rotation matrices. • We frst show that all of these rotation matrices belong to SO3. Let {v, w} be an orthonormal basis for ⊥ the plane u , and let P be a 3 × 3 matrix with columns (u, v, w). Since v and w are orthogonal to each other, and u is orthogonal to both v and w, P ∈ O3. Conjugating a rotation matrix by P demonstrates the action of ρ(u,θ) with respect to the basis (u, v, w). Since u is fxed by the rotation matrix, the frst column is (1, 0, 0)t , and since the plane u⊥ is being rotated by θ, the rest of the matrix M is given by the form of a 2×2 rotation matrix. That is, 1 0 0 P −1ρ(u,θ)P = 0 cos θ − sin θ = M, 0 sin θ cos θ which is in SO3. Since ρ(u,θ) = PMP −1 , and since P, P −1 ∈ O3 and M ∈ SO3, the rotation matrix ρ(u,θ) is also in O3. Taking the determinant of both sides43 demonstrates that ρ(u,θ) ∈ SO3. • To show the other direction, an element A ∈ SO3 must be shown to be rotation around some axis u, which has to be some eigenvector with eigenvalue λ = 1. There exists such an eigenvector if and only if 1 is a root of the characteristic polynomial of A, which is precisely when det(I − A) = 0. 42And orientation 43det(PMP −1) = det(P ) det(M) det(P )−1 = det(M ) = 1 58 Lecture 12: Orthonormal Matrices Since det(AT ) = 1, det(A − I) = det(AT (A − I)). Using the fact that A is orthogonal, this is det(I − AT ). Taking the transpose, this is det(I − A). Since the matrices are 3×3, det(I − A) = (−1)3 det(A − I). Combining these, det(A − I) = det(AT (A − I)) = det(I − AT ) = det(I − A) = (−1)3 det(A − I), implying that det(A − I) = 0. Therefore, there does exist an eigenvector of eigenvalue 1 for A, which can be scaled to be a unit vector u. ⊥ We extend u to an orthonormal basis P = (u, v, w) by picking an orthonormal basis for u . Consider taking A in this basis. The frst column is (1, 0, 0)t , since u is an eigenvector, and the frst row is (1, 0, 0) because the columns are orthogonal. Then, the bottom right submatrix is an element of SO2 by taking the determinant. So P −1AP 1 = 0 0 0 cos sin 0 − sin , cos and we are done. 59 − Lecture 13: Isometries 13 Isometries 13.1 Review Last time, we discussed the orthogonal matrices On, which are matrices which preserve the dot product, which is a measure of length. We found that if we looked at the orthogonal matrices which had determinant 1, SOn, they actually turned out to be rotations in 2-space and 3-space! The rest of the orthogonal matrices O3 can be obtained from SO3 by multiplying a rotation matrix by −1 1 ; 1 the result will be a refection over some axis. As a result, all length-preserving 3×3 matrices are rotations or refections. 13.2 Isometries Without any prior knowledge, we might assume that there are many diferent types of length-preserving mappings, called isometries. We found that for linear mappings, the isometries were the orthogonal matrices, and two or three dimensions, they were rotations or refection. What are the possibilities for isometries that are not linear? Guiding Question Orthogonal matrices are the linear mappings that preserve distance. What are the other possibilities for distance-preserving mappings that are not necessarily linear? An isometry from Rn to Rn is a length-preserving mapping. Defnition 13.1 A function f : Rn − → Rn is an isometry if |f(u) − f(v)| = |u − v| for all u, v ∈ Rn . Let’s take a look at two key examples. » Rn # » Example 13.2 For a matrix A ∈ On, the linear transformation − → Rn x 7→ A# x is an isometry. # » » t # » Rn # » » Example 13.3 a Translation by a vector # v ∈ Rn is an isometry: b − → Rn x 7→ # x + b aThis is not a linear transformation! How crazy can an isometry be? The answer, fortunately or unfortunately, is not very. In fact, these two examples and their compositions turn out to be the only isometries. 60 Lecture 13: Isometries Theorem 13.4 # » # » » » Every isometry f is of the form t # » ◦ A, for A ∈ On and b ∈ Rn . So f(# x ) = A# x + b .
b Despite the fact that preserving distance does not appear to be a very strong condition on f, it turns out that it is equivalent to the very strong condition that it basically has to be linear, combined with a shift. It boils down to the following lemma. What form do the isometries that fx the origin take? The answer is that they must be linear. Lemma 13.5 If f : Rn → Rn is an isometry such that f(0) = 0, it must be a linear transformation.a aIt must respect the additive and scalar multiplicative structure on Rn . Proof. We must show that f preserves sums and scalar products. First, we see that a dot product can be » written in terms of # 0 and distances: 1 # » # » |# # u » · # v » = u » − 0 |2 + | # v » − 0 |2 −|# u » − # v »|2 . 2 As a result, the following equation also holds: 1 # » # » » » » » » » f(# u ) · f(# v ) = |f( # u ) − f( 0 )|2 + |f( # v ) − f( 0 )|2 −|f( # u ) − f(# v )|2 . 2 Because f is an isometry, |a − b| = |f(a) − f(b)|. Setting f(0) = 0 gives the equation ⃗ u · ⃗ v = f(⃗ u) · f(⃗ v), so it must be the case that since f preserves lengths, f also preserves the dot product. » # » # » • The sum can be expressed using a dot product again. For # z = x + y , » » » » » » (# z − # x − # y ) · (# z − # x − # y ) = 0, and so # » # » # » # » # » # » » # » » # » » # » z · z + x · x + y · y − 2 # x · z − 2# y · z + 2# x · y = 0. Now, since we know that addition is determined in some complicated way from dot product, since f fxes the dot product, it must fx addition as well. 44 So f(z) = f(x) + f(y). • A similar reasoning gives us the scaling product: f(cx) = cf(x). Despite the fact that the only piece of information is that f preserves distances and maps the origin to itself, it is enough to play around algebraically to fnd out that f must be linear. This rules out lots of crazy functions that you could imagine could be isometries. Proof of Theorem 13.4. Now, we can prove the original theorem. Given f : Rn − → Rn , there is some vector b ∈ Rn such that f(0) = b. Then t−b ◦ f is an isometry that fxes 0. Thus, there is some linear transformation A such that t−b ◦ f = A, and this implies that f = tb ◦ A, since tb is the inverse of t−b. From the defnition of an isometry, it is easily seen that the composition of two isometries is an isometry. Given that isometries are all of the same restrictive form, it is not surprising that they form a group. 44If we had some other crazy invented operation determined from the dot product, f must also fx that! 61 Lecture 13: Isometries Defnition 13.6 The group of isometries is # » f which is potentially itself. » a Mn := {isometries Rn − → Rn} ⊆ Perm(Rn). aAny bijective function the in since it each in exactly in R R R R n n n n permutes vectors vector to vector on maps one , , » +t # = t # b ′ . Orthogonal b ′ » , +), form a subgroup of Mn, since t # matrices On also form a subgroup of Mn. Note that the composition of an orthogonal matrix with a translation is A ◦ t # » Clearly, translations, which are isomorphic to (Rn b b + = t # » ◦ A, b A b since A(x + b) = Ax + Ab. In particular, a translation and an orthogonal matrix do not commute with each other. Consider the projection π :Mn − → On tb ◦ A 7− → A. It is a group homomorphism, since (tb ◦ A) ◦ (tb′ ◦ A ′ ) = tb+Ab′ ◦ AA ′ . Also, π is surjective, and the kernel is ker(π), which are translations. Thus, the subgroup of translations is normal inside Mn. 13.3 Isometries in 2-space Now that we have an understanding of isometries in general, let’s narrow it down to an analysis in two dimensions. Guiding Question For n = 2, what do isometries look like? The following defnition is an intuitive extension of the idea of orientation for linear mappings. Defnition 13.7 An isometry x 7→ Ax+b is orientation-preserving if det(A) = 1, and orientation-reversing if det(A) = −1. In two dimensions, isometries can be classifed into one of four types. 62 Lecture 13: Isometries Theorem 13.8 Every isometry on R2 is 1. Translation a 2. Rotation around a point p 3. Refection across a line L b 4. Glide refection — frst, refect across a line L, then translate by some vector b parallel to Lc aIt is no longer required that p is the origin, since the isometry does not have to be a linear transformation bAgain, the line L may or may not pass through 0; the isometry is not necessarily linear. cWe will see diagrams next week which have glide refections in their symmetry group! The frst two are orientation-preserving; the last two are orientation-reversing. By composing with translations, it is possible to essentially change coordinate systems. For example, consider rotations and refections. Let f be an isometry, say a rotation around the origin. Then, tpft−p 45 is a new isometry that fxes p, instead of the origin, since it is applying f but after shifting coordinates by p. Similarly, letting f be a refection across any line, we can represent it in new coordinates as a refection across a line through the origin. Proof. We split the proof up into two cases depending on whether f is orientation-preserving or reversing. • Case I. Consider an orientation-preserving isometry f(x) = Aθx + b. 1. If Aθ = I2, the identity, then f = tb, which is possibility 1 in the theorem. 2. Otherwise, if Aθ ̸= I2, we want to fnd a fxed point p such that f(p) = p. Since Aθ has no fxed vectors, pA(1) ̸= 0, and so Aθ − I2 has a trivial kernel, and so A − I2 is invertible. Then the equation (A − I2)p = −b has a unique solution p = (A − I2)−1(−b), and then f(p) = Ap + b = p. So t−pAtp = Aθ, 45When we apply t−p, we shift p to 0, then we use f to rotate around 0, and lastly use tp to shift 0 back to p. 63 Lecture 13: Isometries since it fxes 0. This corresponds to the second possibility: rotation around a point p. • Case II. Let f be an orientation-reversing isometry. Then f = tb ◦ A, where A is refection across a line L. First, change the origin to b/2. Then t−b/2ftb/2 = t−b/2tbAtb/2 = tb/2tAb/2A = tmA, b+Ab where m = . Since b and Ab are refections over a line L, m, the average, must lie on that line. 2 3. If m = 0, it is a refection. 4. If m ̸= 0, it is a glide refection. 46 Again, the same idea from Case I applies. Shifting to a new coordinate system gives us either a refection or a glide refection. 46In our proof, we try to be slicker about it, but if we are uncomfortable with that, we know f is just Ax + b, and we could simply crunch through lots of sines and cosines to force f into one of the four forms in Theorem 13.8. 64 Lecture 14: Finite and Discrete Groups of Isometries 14 Symmetry Groups So far, in this class, we’ve covered groups and linear algebra. Now, we are looking at groups of symmetries that preserve extra forms of structure. 14.1 Review Last week, we looked at the orthogonal matrices. Defnition 14.1 The orthogonal matrices On are matrices that preserve distance. It is the set T : Rn − → Rn : |Tv| = |v| for all v ∈ Rn . Defnition 14.2 The set Mn of isometries from Rn to itself is {f : Rn − → Rn : |f(u) − f(v)| = |u − v|}. The orthogonal matrices are the subset of isometries that are linear transformations. In class, we showed that every isometry f is of the form f(x) = Ax + b where A ∈ On and b ∈ Rn . Then, we looked at O2, the orthogonal matrices in two dimensions. There are two possibilities for a transformation in O2. 47 • Rotations around 0: these have determinant 1 and are called SO2. » • Refections across a line through # 0 : these have determinant -1 Then the isometries of two-dimensional space, M2, also ft into several categories.48 • Translations • Rotations around p • Refections across a line • A glide refection49 14.2 Examples of Symmetry Groups Now, we want to add some additional structure to preserve. Guiding Question What isometries of R2 fx some shape inside R2? We call the group of such isometries symmetry groups for that shape. Let’s start with a couple examples of shapes and their symmetry groups. 47The special orthogonal group 48This is quite surprising, since a priori, an isometry could take many diferent forms. 49A refection in addition to a parallel translation 65 Lecture 14: Finite and Discrete Groups of Isometries Example 14.3 2π For a regular pentagon, the group of symmetries are rotations by multiples of , and refections across 5 a lines. This group of symmetries is what we would call discrete. aThis will be formalized later on. Next, we look at a group that is not discrete. Example 14.4 For a circle centered at the origin, every rotation or refection will fx it, and so its symmetry group is all of O2. This group of symmetries is not discrete. We can also look at infnitely large shapes. Example 14.5 For a triangular lattice, certain translations, refections over lines, rotations, and glide refections all preserve it. It is a discrete symmetry group. 14.3 Discrete Subgroups of R From our examples, we see that some symmetry groups are “discrete" and some are not. Guiding Question How can the notion of a discrete group be formalized? We can start with an easier notion, which is a discrete group inside (R, +). Defnition 14.6 A group G ≤ (R, +) is discrete if there exists ε > 0 such that any g ∈ G such that g ̸= 0 satisfes |g| > ε. Equivalently, for a, b ∈ G and a ̸= b, then it must be true that |a − b| > ε for a discrete group. The discreteness tells us some important information about G. Theorem 14.7 If G ≤ (R, +) is discrete, then G = {0} or G = Zα for some real number α > 0. 66 Lecture 14: Finite and Discrete Groups of Isometries This theorem is very similar to the theorem we had about subgroups of Z, where we showed they were either trivial or of the form kZ. Proof. Assume that G ̸= {0}. Then there is some smallest positive element α ∈ G. To see why it is possible to fnd a smallest element, we start by taking any g > 0 in G. By discreteness, in the interval from [0, g], we have at most g/ε elements of G inside of the interval. We can then pick the smallest one because the set is fnite. We now claim that G = Zα. Why is this true? If 2α < x < 3α for some x ∈ G, then 0 < x − 2α < α, where x − 2α ∈ G, which is a contradiction. 50 14.4 Finite subgroups of O2 So what are all the fnite subgroups of O2? Let’s frst try to create some examples to get some intuition about them. Example 14.8 Let x be a rotation by 2π . Then Cn = ⟨x⟩a , the cyclic group of order n, is generated by x, and is a fnite n subgroup of O2. n−1} a{1, x, · · · , x Another possible fnite subgroup can be created by expanding Cn a little bit. Example 14.9 # » −1 2 n Let y be a refection across a line ℓ through 0 . Notice that the relations yx = x y, y = e, and x = e a1 xa2 ya3 i j hold, and so any product y · · · can be written as x y , where 0 ≤ i < n and 0 ≤ j < 2. Then the group generated by x and y is 2 n−1 2 n−1 Dn := ⟨x, y⟩ = {e, x, x , · · · , x , y, xy, x y, · · · , x y}, which is called the dihedral group. It has order 2n. For n ≥ 3, Dn is the group of symmetries of a regular n−gon.51 The dihedral group for n = 1 is D1 = ∼ C2 and for n = 2, D2 = ∼ C2 ×C2. For n = 3, D3 = ∼ S3, and larger dihedral groups can also be studied. Now, we have two families of fnite subgroups of On, the cyclic groups of rotations, and the dihedral groups. It turns out that these are actually all the fnite subgroups of O2. This provides yet another classifcation theorem. Let’s start with a simpler version. Theorem 14.10 If a subgroup H ≤ SO2 is fnite, then H is isomorphic to Cn for some n. Proof. Let ρθ be cos θ sin θ − sin θ cos θ . Then let S = {θ ∈ R such that ρθ ∈ H}. Under the homomorphism π : θ 7→ ρθ, S = π−1(H). Since S is a preimage, we know that S is a subgroup of (R, +). If H is fnite, then S must be discrete, and so by Theorem 14.7, S is Zα for some α. Also, 2π ∈ S because a rotation by 2π is the identity in H, and so α = 2π . So H = Cn. n 50The discreteness guarantees that we can fnd a smallest positive element! This is defnitely not the case for R in general (it is a fundamental property of R that there is no smallest positive element.) 51In general, if x is a rotation by an angle that is not a rational multiple of 2π, then we do not get a rational group. We would get a non-discrete subgroup of SO2. 67 − − Lecture 14: Finite and Discrete Groups of Isometries Theorem 14.11 Any fnite subgroup of O2 is isomorphic to Cn or Dn. Now, we can prove Theorem 15.2. Proof. There are two cases: • Case I. If G ⊆ SO2, by the above theorem, G ∼ = Cn for some n. • Case II. If G is not a subset of SO2,, then take the restriction of the determinant function on O2 to G. It takes det G − → {±1}. By the assumption that G isn’t a subset of SO2, this is surjective. Let det H = ker(G − → {±1}). Then, H ⊴ G is a normal subgroup of index 2. So det−1({−1}) is a nontrivial coset of H, and so it is Hr for some r ∈ G such that det(r) = −1. Then r must be a refection across some line ℓ.52 Then, it is clear 2πρ by defnition that H ≤ SO2, and so H = Cn for some n, and it is generated by some x = , and then n we have 2πρ G = , r ∼ = Dn. n 14.5 More Discrete Subgroups Next, what are the fnite or discrete subgroups of M2? Let’s start with a couple of defnitions. Defnition 14.12 A subgroup G ≤ O2 is discrete if there exists some ε > 0 such that all nontrivial rotations in G have angle θ such that |θ| > ε.a aHere, discrete implies fnite, which implies that it is Cn or Dn. Defnition 14.13 A subgroup G ≤ M2 is discrete if there exists some ε > 0 such that all translations in G are by vectors b with |b| > ε, and all rotations in G have angle θ such that |θ| > ε. This ends up being quite a strong constraint on what the discrete subgroups look like, even though there could be lots of diferent possibilities. We’ll talk about this more next time. 52Note that we have many options for ℓ because any r ∈ Hr generates Hr. In particular, these are all the rotations of ℓ. 68 Lecture 15: Finite and Discrete Subgroups, Continued 15 Finite and Discrete Subgroups, Continued 15.1 Review Last time, we began studying certain subgroups of M2. The group of isometries of R2 is precisely # » M2 = {t # » ◦ A : b ∈ R2, A ∈ O2}, b where O2 is the group of orthogonal matrices. Guiding Question What are the fnite subgroups of O2?a aThe discrete subgroups of O2 turn out to be the same as the fnite subgroups, either Cn or Dn (we omit the proof, as it is in the homework.) One way in which subgroups of M2 naturally arise is with symmetries of plane fgures. Example 15.1 For the following two plane fgures, they both have discrete symmetries including translations, rotations, and glide refections. Last time, we looked at fnite subgroups of the orthogonal matrices G ⊆ O2. We found the following theorem which greatly restricts the possibilities for such subgroups: Theorem 15.2 Any fnite subgroup G ⊆ O2 is either • G ∼ = Cn = ⟨ρ2π/n⟩, the cyclic group generated by a rotation by 2π/n; or • G ∼ = Dn = ⟨ρ2π/n, r⟩ which is the group Cn with an extra refection r. The elements of the form ρ2π/n, which are rotations by 2π/n, are orientation-preserving, while elements of the form ρ2π/nr, which are refections over certain lines through the origin, are orientation-reversing. 15.2 Finite Subgroups of M2 Now that we have found the fnite and discrete subgroups of O2, we bring our attention to fnite subgroups G ⊆ M2. Guiding Question What are the fnite subgroups of M2? Do we get more subgroups now that we have more elements? In fact, there are no new fnite subgroups obtained from allowing G to be in M2 instead of O2. Theorem 15.3 Any fnite subgroup G ⊆ M2 is also isomorphic to Cn or Dn. 69 Lecture 15: Finite and Discrete Subgroups, Continued Proof. In order to show that G is isomorphic to Cn or Dn, it is enough to fnd s0 ∈ R2 such that g(s0) = s0 for all g ∈ G. Then, by changing coordinates such that s0 is the new origin53 , G fxes the origin (formerly s0) and so G ⊆ O2. As a result, by applying Theorem 15.2, G must in fact be isomorphic to Cn or Dn. • Step 1. First, we fnd some fnite set S fxed by every element g: we require that gS = S for all g ∈ G. For any p ∈ R2 , let : g ∈ G}54 S = {g(p) ∈ R2 . ′ Then, for any element s ∈ S, it is equal to s = g ′ (p) for some g ∈ G, by the defnition of S. In addition, for any g ∈ G, the action of g on s is g(s) = g(g ′ (p ′ )) = (gg ′ )(p) ∈ S, again by how S is defned. So gS = S. • Step 2. Intuitively, to fnd s0, we would take the average, or the center of mass, of all the points. For example, for the set of rotations ⟨2π/3⟩, S would be 3 equidistant points, and the center of the equilateral triangle would be fxed by such rotations. From this intuition, we can apply the following averaging trick. This is where G being fnite is required, as we wouldn’t be able to take the average otherwise. Where S = {s1, · · · , sn}, let 1 s0 = (s1 + · · · + sn) n be the average of all the elements in S. For any isometry f = tb ◦ A, 1 f(s0) = tb (As1 + · · · + Asn) n 1 = ((As1 + b) + · · · + (Asn + b)) n 1 = (f(s1) + · · · + f(sn)), n since A is a linear operator. As a result, for any g ∈ G, 1 g(s0) = (g(s1) + · · · + g(sn)) n 1 = (s1 + · · · + sn) n = s0, since g permutes the elements in S. So we see that G does fx s0, and by changing coordinates so that s0 is the origin, G must in fact be isomorphic to Cn or Dn. 15.3 Discrete Subgroups of M2 No new fnite subgroups are obtained by taking elements in M2 instead of O2; what if we take discrete subgroups55 instead of fnite subgroups? Guiding Question What about discrete subgroups of M2? The defnition of discreteness in M2 combines the two defnitions for the rotations and translations. 53We take t−s0 Gts0 54This is called the orbit of p, since it is all the points that p can reach by some transformation in G, or all the points that p orbits to. 55We will formalize the notion of discreteness in M2 now! 70 Lecture 15: Finite and Discrete Subgroups, Continued Defnition 15.4 A group G ⊆ M2 is discrete if there exists some ε > 0 such that any translation in G has distance ≥ ε and any rotation in G has angle ≥ ε.a aIn fact, for discreteness, it would make more sense to require two diferent ε1 and ε2 for translations and rotations, just to ensure that there are not continuously many translations and rotations. In this case, we can simply acquire the ε for this defnition by taking the minimum of the two; then any translation in G has distance ≥ ε1 ≥ ε and any rotation has angle ≥ ε2 ≥ ε. 15.3.1 Discrete Subgroups of R2 As a warmup, let’s consider the copy of the plane inside M2, (R2 , +) ⊆ M2, consisting of the translations tb. What are the discrete subgroups of (R2 , +)? The result and argument is similar to the discrete subgroups of (R, +) that we covered last week. Theorem 15.5 If G ⊆ R2 is discrete, then 1. G = {0}; or 2. there exists some α ⃗ ∈ R2 such that G = Zα ⃗ ; or 3. there exist linearly independent vectors ⃗ a,⃗ b ∈ R2 such that G = Z⃗ a + Z⃗ b. This is called a lattice inside R2 . Proof. First pick any α ˆ ̸= 0 ∈ G. The intersection G ∩ Rα ˆ must be discrete, so there is some smallest length vector in G ∩ Rα ˆ; call it α. Then if G ∩ Rα ˆ = G, then G ∩ Rα ˆ = Zα, and we are done. Otherwise, pick β ∈ G such that β ∈ / Rα, minimizing the distance from β to Rα. There exists such a β because in any bounded region of R2 , there can only be fnitely many points of G; then we can simply pick the point in G closest to Rα. Claim: G = Zα + Zβ. If this were not true, then there would exist a point γ ∈ G that is not on the lattice formed by α and β. Thus, by shifting by α and β, the parallelogram with sides α and β would contain a point closer to Rα, violating the minimality of β. 71 Lecture 15: Finite and Discrete Subgroups, Continued 15.3.2 Back to Discrete Subgroups of M2! Now that we have considered the translations in M2, which are isomorphic to the plane R2 , we can move on to the entire M2. Guiding Question How can we study discrete groups G ⊆ M2? Recall that there exists a projection π from M2 to O2, where R2 , the group of translations, is the kernel. The projection takes π , → 56M2 ker(π) = R2 − → O2 t # » ◦ A 7→ A.
b The restriction of π to G takes π|G : G − → O2. The kernel L = ker(π|G) consists of the translations in G. Under this map, the image of G is a subgroup G := π(G) ⊆ O2, known as the point group of G. The projection takes π|G ker(π|G) = L ⊆ G − − → G. Example 15.6 For this infnite plane fgure, the group of translations L in the symmetry group G is a rectangular lattice. » The point group G contains rotation by π around # 0 and refection across ℓ; as a result, G is isomorphic to D2. As we can see in the example, by using the projection π, each G can be decomposed into a discrete point group G isomorphic to Cn or Dn, and a discrete group L ⊆ R2 , classifed in Theorem 15.5. In fact, we can constrain the possibilities even more! The following proposition is a start. Proposition 15.7 Every A ∈ G maps L to L. Proof. Next time! 72 Lecture 16: Discrete Groups 16 Discrete Groups 16.1 Review Last time, we looked at discrete57 subgroups G ≤ M2. Then, we looked at a projection π: π ker(π) = (R2 , +) ⊂ M2 − → O2, tb ◦ A 7→ A; essentially, it gets rid of the translation part of an isometry. We can restrict π to G to get a mapping π|G G − − → O2, and we call the kernel, L := ker(π|G), and it consists of all the translations in G. The image of G in O2, denoted G := π(G), is called the point group of G. For some element g ∈ G, its image g := π(g) ∈ G only "remembers" the angle of rotation or the slope of the refection line. If G is discrete, it is either Cn or Dn, which we proved earlier. If L ⊆ R2 is discrete, then we obtained three possible cases. (i) L = {0}; (ii) L = Zα where α ̸= 0; (iii) L = Zα + Zβ, where α, β are linearly independent.58 16.2 Examples for L and G For a given plane fgure, it is actually not difcult to see what L and G are! For the translation subgroup L, since it must either be the identity translation, Zα, or a lattice, it is possible to simply eyeball which translations preserve the fgure. Let’s consider the following plane fgures. Later in this lecture, we will discuss the possibilities for G; it consists of the (untranslated) rotations and refections preserving a fgure. 57The translations and rotations that cannot be arbitrarily small 58When you look at two vectors and everything you generate from them, it’s called a lattice. 73 Lecture 16: Discrete Groups Example 16.1 (A) For this frst fgure, say fgure A, the translation subgroup L is a rectangular lattice generated by two translation vectors, to the right and upward. Also, G is D2, since it contains a refection as well as rotation by π. Example 16.2 (B) For fgure B, the translation subgroup is trivial, consisting of 0. Also, G is C3, since there cannot be any refections but rotation by 2π/3 or 4π/3 around the center both preserve the fgure. Example 16.3 (C) For fgure C, the translation subgroup is generated by one vector, so L = Zα where α = (1, 0). Also, G is D1, since there is a refection (corresponding to a glide refection in G) and no rotations possible. 74 Lecture 16: Discrete Groups Example 16.4 (D) For fgure D, the translation subgroup is a triangular lattice generated by two vectors at an angle of π/3 to each other.a The point group is G = D6, since rotation at a lattice point by any multiple of π/3 preserves the fgure, as well as refection. aOr two vectors at an angle of 2π/3. 16.3 Crystallographic Restriction Now that we have decomposed studying G into studying groups we understand better, L, a subgroup of translations, and G ⊆, the point group, we can actually constrain G further! Recall that • The translation subgroup L ⊆ (R2 , +)59 must be one of three possibilities, which we get from studying discrete subgroups of R2; • G must be Cn or Dn, which we get from studying discrete subgroups of O2. Now that we understand the components L and G separately, we want to use this knowledge to understand G better. Guiding Question How do L and G interact with each other? Example 16.5 Consider our earlier example 16.4. In this case, any element of the point group D6 preserved the triangular lattice. In fact, G acts on L for any discrete group G ⊆ M2; this is a very strong constraint on how G and L interact. Theorem 16.6 For the point group G ≤ O2 of some discrete subgroup G of M2, and the translation subgroup L ⊂ R2 , the group G must map L to itself. For any element A ∈ G and b ∈ L, the image of b under the action of A is b 7→ Ab ∈ L. We already know that O2 and thus G acts on the plane R2 and therefore L. The surprising part is that under the action of any element of G, an element of L is actually mapped to another element in L! Proof. Since A ∈ G, it is the image of an element of G, say t⃗ c ◦ A ∈ G for some ⃗ c ∈ R2 . Then, ⃗ b ∈ L, so t⃗ ∈ G.
b The key observation in this proof is that L = ker(π|G) is the kernel of a homomorphism! Thus, the subgroup L ⊴ G is actually normal, so conjugating an element of L by anything in G stays in L. Then for t⃗ ∈ L, b (t⃗ c ◦ A) · t⃗ · (t⃗ c ◦ A)−1 ∈ L b also. As isometries in M2, we know how to manipulate these products, and so expanding out this expression gives us −1 t⃗ c · A · t⃗ · A−1 · t = t⃗ ct · A · A−1 · t−⃗ c b ⃗ c A⃗ b = t⃗ ctA⃗t−⃗ c b = t ∈ L.
A⃗ b 59The translation subgroup L is sometimes written ambiguously in one of two equivalent ways; an element of L can either be the translation t⃗ ∈ L considered as an element in G, or simply the vector ⃗ b ∈ L considered as an element in R2 . So L could be b considered either as a subgroup of G or of R2 . 75 Lecture 16: Discrete Groups Thus, conjugating t⃗ ∈ L by t⃗ ◦ A gives t ∈ L. Using the identifcation of L with R2 , A⃗ b ∈ L ⊂ R2 , and so b c A⃗ b every A ∈ G takes vectors ⃗ b in L to other vectors in L, preserving the translation subgroup. Student Question. We’re studying discrete groups, which are groups with the requirement that the translations or rotations can’t be arbitrarily small. Are we also requiring that they have to be groups preserving a given diagram, or can they be any discrete groups of isometries? Answer. Earlier on in this lecture, we saw some examples of discrete groups G that came from the symmetry groups of certain diagrams, but what we are actually doing is simply looking at groups G with the condition that the rotations and translations must be arbitarily small60 , and classifying them; mathematically, there is no requirement that they come from pictures. However, the way that these discrete groups actually show up and the way that we fnd them is by drawing these kinds of pictures; this is one of the main reasons why we care about them! In fact, for every discrete subgroup G ⊆ M2, there will be some picture that produces the group G as its symmetry group. The pictures in this lecture are mainly so that there are concrete examples to look at and think about. In Section 16.2, each of the examples has a symmetry group G consisting of the isometries of the plane sending the picture to itself.61 For example, in Example 16.2, rotation by 120 degrees preserves the "triangle," while 5 degrees does not, so ρ2π/3 ∈ G, whereas ρπ/36 ∈ / G. Theorem 16.6 states that the point group G, which is a diferent group from G, actually preserves L ⊆ R2 , the translation group. In Example 16.4, L is generated by Z(1, 0)t + Z(1/2, 3/2)t , the two sides of an equilateral triangle, and the point group is D6. Any element of D6 will send an element of L to a diferent element in L. In fact, when L is a lattice, preservation by some point group G is a strong constraint on the possible angles that show up in the lattice; only certain angles are allowed. Given G, most lattices are not preserved by every element. Thus, the theorem constrains G and L together — not on each of them separately, but on how they interact. The groups that show up this way are often called crystallographic groups. 62 They are well-studied; in fact, there are only fnitely many. Theorem 16.7 (Crystallographic Restriction) a Let L ̸= {0}. Then G = Cn or Dn, where n = 1, 2, 3, 4, or 6. aThe theorem name comes from the fact that it restricts the possible crystallographic groups. Although we could imagine that there are lots of possibilities for G and L, the fact that G preserves L constrains the possible point groups to fnitely many, and there are also only certain choices of L allowed for a given n. Proof. The group G is a discrete subgroup of O2, and so it is Cn or Dn for some integer n. Since L is discrete, there is a (non-unique) shortest nonzero vector α ̸= 0. Consider a rotation ρ = ρθ ∈ G. The result of rotating α by θ is another vector in L, and since rotations are length-preserving, ρθα is also a vector of shortest length. Since both vectors are in L, ρα − α is also in L.63 If θ is too small, ρα − α will have a shorter length, and there will be a contradiction. In particular, if θ < 2π/6, ρα − α is shorter than α, so θ ≥ 2π/6. Since Cn and Dn contain ρ2π/n, it must be the case that n ≤ 6. 60These are called discrete groups 61Not each point individually is sent to itself; the picture as a whole is sent to an identical copy of itself. 62Especially when L is a lattice, and there are two diferent directions to translate. 63Since L is a subgroup, it is closed under addition/subtraction. 76 Lecture 16: Discrete Groups A similar argument holds to rule out n = 5. The vector α + ρ4π/5α will be shorter than any α, which is also a contradiction.64 So n = 1, 2, 3, 4, or 6. Actually, for Cn or Dn where n = 1, 2, 3, 4, or 6, it is possible to constrain the translation subgroups L that can simultaneously show up. For instance, when L is a lattice,65 there are only 17 possible symmetry groups G that can occur. When L is 0, G can be Cn or Dn for any arbitrary n, but allowing nontrivial translations constrains G signifcantly. Student Question. How much does constraining G and L constrain the actual symmetry group G itself? Answer. Finding G from G and L is precisely the same as fguring out the 17 plane symmetry groups,66 and is precisely the last step! We will do one example now. Let’s consider a specifc group G and try to fgure out what the actual symmetry group G can be! 64This question is equivalent to the feasibility of tiling the plane with a regular pentagon, and in fact that is not possible! 65When L is a lattice, it is two-dimensional, and it is Z⃗ a + Z⃗ b for generating vectors ⃗ a and ⃗ b. It is also possible for L to be Z⃗ a, which is one-dimensional. 66These are called wallpaper groups, since wallpapers are 2-dimensional patterns that usually have nontrivial symmetry groups. 77 Lecture 16: Discrete Groups Example 16.8 (C4) π|G Suppose G = C4. a Then L ⊂ G − − → C4, and the index [G : L] = 4. b Also, ρ = ρπ/2 ∈ G is a generator of G. Where α is some shortest-length vector in L, it’s possible to showc that ρα and α do generate L. Thus, L = Zα + Z(ρα), a square lattice. Also, there exists some rotation ρ ∈ G giving π(ρ) = ρ. Then ρ is in fact a rotation by π/2 around some other point, which we will call the origin.d The group G contains L, of index 4, as well as some rotation by π/2, ρ.e Thus, G is "generated" by L and ρ, and must consist of everything of the form G = {tv ◦ ρi : ⃗ v ∈ L, i = 0, 1, 2, 3}. Also, ρtv = tρv ◦ ρ, so the group multiplication can be written down, and G is completely determined by knowing that G was C4; this is 1 out of the 17 wallpaper groups! aRotations by 90 degrees, but no refections. bThe index [G : ker(π|G)] = [G : L] is equal to the size of the image under π|G, which is G = C4. cThere is a more involved argument there, but it is not super relevant here. dIn the discussion of the four kinds of isometries in M2, the elements which were mapped to rotations were in fact rotations around some point. eThe rotation ρ is ρ, lifted to be in G, and it is an element of G not in L which generates the quotient, C4. Student Question. Can you explain where ρ came from? Why is it a rotation? Answer. By defnition, G is the image of G under π : M2 → O2 taking tb ◦ A 7→ A. Then there are four possibilities for elements in M2: translation, rotations, refections, and glide refections. The frst two are orientation-preserving, and the last two are orientation-reversing. Refections and glide refections map to refections in O2 67 under π, translations will map to the identity, and rotations will map to rotations (around the origin). So ρ has an image of ρ, which is a rotation, and thus ρ is a rotation around some point. If ρ, the element in G, were a refection instead of a rotation, the preimage in G could have been either a refection or a glide refection, so when the point group G = Dn, one of the dihedral groups, instead of Cn, the analysis is more subtle. In fact, there might not be any refections in G at all. (In Example 16.3, there were no refections, only glide refections.) Example 16.9 If r = π(r) where r, then r = π(tb ◦ rℓ), where b is some zeroa or nonzerob vector parallel to the line ℓ. Does this mean there are uncountably many possibilities for b and therefore r? In fact, b is constrained a bit more: tbrℓtbrℓ = t2b, so 2b ∈ L. Thus, there are two possible situations: • The vector is in the lattice: b ∈ L; • The vector b is halfway between two lattice points, as in Example 16.1. arefection bglide refection From these two examples, we see that given some G, of which there are fnitely many, and working through the information that is present, there aren’t too many possibilities for G, and in fact there are fnitely many — 17 67Refections across lines through the origin 78 Lecture 16: Discrete Groups in total. 79 Lecture 17: Group Actions 17 Group Actions Today, we will discuss group operations or actions68 on a set. Guiding Question How can a group be seen as a group of transformations? 17.1 Review Last time, we fnished talking about (discrete) subgroups of isometries of the plane. Finite subgroups of M2 are isomorphic to Cn or Dn, and there are only fnitely many isomorphism classes of infnite discrete subgroups of 69 M2. It is also possible to go up a dimension and classify discrete subgroups of isometries of R3 , although it is more complicated; there are 200 or so. 17.2 Motivating Examples The idea of a group action will generalize and make more abstract an idea that has been present throughout the class so far. Let’s start with the following motivating example. Example 17.1 (GLn) Given g ∈ GLn(R) and a column vector v ∈ Rn , the matrix g can be seen as a transformation on Rn , taking v 7→ g(v) ∈ Rn . The data of GLn(R) acting on Rn can be packaged together by a map GLn(R)×Rn − → Rn # » » (g, v ) 7− → g(# v ). The same principle applies to Sn, the group of permutations on {1, · · · , n}. Example 17.2 (Sn) The symmetric group Sn can also be viewed as acting on a set. More or less by defnition, given a number between 1 and n, and a permutation, it’s possible to spit out the result of permutation acting on that number. So Sn permutes the set [n] = {1, · · · , n}. This gives us another mapping encoding this information: Sn ×{1, · · · , n} − → {1, · · · , n} (σ, i) 7− → σ(i). Our last example is one we have been considering for the past few lectures. Example 17.3 The set M2, isometries of 2-space, acts on R2 : given some vector in the plane and some isometry, the isometry will return some other vector in the plane. This information is again encoded by a mapping M2(R)×R2 − → R2 (f, ⃗ x) 7− → f(⃗ x). 68They are diferent terms for the same idea. Artin uses group operations, while Professor Maulik prefers to call them group actions. 69In fact, with any metric space, which is a set with some distance on it (as discussed in 18.100, for example), it’s possible to consider isometries, distance-preserving transformations, in the same way as we considered the plane R2 . Depending on the metric space, the groups can look very diferent! One example of this is the hyperbolic plane, which is the upper half-plane of R2 with a non-Euclidean metric, or distance, on it, and the discrete subgroups of isometries on it. There are infnitely many discrete subgroups of isometries on it, even though it is 2-dimensional, just like R2 . The question of why it is so diferent from the R2 case is really a geometry question, rather than an algebra question. 80 Lecture 17: Group Actions 17.3 What is a group action? These are all examples of group operations on a set, and they motivate the following defnition of a group operation in general. Defnition 17.4 Given a group G and a set S, a group actiona on a set S is a mapping G×S − → S (g, s) 7− → gs. It must satisfy the following axioms: • The identity maps every element of the set back to itself: es = s for all s ∈ S.b • The mapping must respect the group multiplication: (gh)s = g(hs) for s ∈ S and g, h ∈ G.c aor group operation bThis corresponds to the identity multiplication rule. cThis corresponds to the associativity rule. Essentially, given an element g ∈ G and s ∈ S, the mapping returns another element of S depending on g, and the mapping must respect the group structure on G. All of the groups that we have seen so far show up as symmetries of some set, maybe preserving some extra structure, so all the groups that we usually think about already come with an action on some set S. Furthermore, a group G can act on many diferent sets at the same time in diferent ways, which gives insight into the group itself. Let’s look at a couple of examples. Example 17.5 (S4) The symmetric group S4, permutations on 4 elements, acts on S = {1, 2, 3, 4}. It can also act on a diferent set, T = {unordered pairs in S} = {(12), (13), (14), (23), (24), (34)}. The set T has 6 elements, and S4 acts on T as well as acting on S. Given a permutation σ ∈ S4, and an unordered pair {i, j}, it acts by taking σ({i, j}) = {σ(i), σ(j)} for a permutation σ ∈ S4. So the group action on S leads to a diferent group action on a diferent set, T. The existence of a group action on a given set actually yields a lot of information about the group G, as will be explored in the next few lectures. Let’s see a diferent example. Example 17.6 (D2) Let G = D2, which contains rotation by π as well as refection across the x-axis (and then refection across a the y-axis.) As a subgroup of O2, , D2 will act on all of R2 . It also acts on the set S consisting of the vertices of a square and a diamond, as well as the center. a2×2 orthogonal matrices A group G can also act on itself viewed as a set. 81 Lecture 17: Group Actions Example 17.7 Given a group G, there is a mapping G×G − → G ′ (g, g ′ ) 7− → gg , and this is a valid group action. When G acts on itself, the frst G in G×G is seen as a group, while the second G is seen as a set, since the axioms of a group action don’t care about the group operation on the second instance of G. Let’s see one last example. Example 17.8 Taking a vector space V over a feld F, the group F × , the nonzero elements of the feld, which is a group with respect to multiplication, acts on V by scaling: F × ×V − → V (a,⃗ v) 7− → a⃗ v. Scaling by nonzero scalars defnes a group operation! It satisfes each of the axioms. Student Question. What type of element is g(s)? Answer. It depends on what S is! It is the type of element that is in S. Two group actions of G on S and S ′ might not have anything to do with each other, other than the fact that they both involve G; G can act on wildly diferent types of sets, and show up in diferent contexts. Say we fx an element g ∈ G, we can defne the group action of g on S, a mapping τg : S − → S sending s 7− → g(s). 70 We can show that τg is a bijection from S to itself because it has an inverse map, τg−1 , coming from the fact that g is invertible. Because τg is a bijection, it actually permutes the elements of S, and so it is a permutation of S. Thus, each element of G can be mapped to a permutation by a map τ : G − → Perm(S), which takes g 7− → τg ∈ Perm(S). From the group action axioms, τ is a group homomorphism. In Example 17.6, D2 is acting on a set with |S| = 9, so there exists a homomorphism from D2 − → Perm(S) = S9. Note that τ does not have to be injective; there may be some action g ∈ G such that g ̸= e but G fxes each s ∈ S, which would make τ(g) the identity permutation. 17.4 The Counting Formula Defnition 17.9 Given s ∈ S, the orbit of s is Os = Gs := {gs : g ∈ G} ⊆ S. For instance, in Example 17.6, there are several orbits of diferent sizes. The top and bottom vertices of the diamond are in the same orbit (size 2), the left and right vertices of the diamond are in the same orbit (size 2), all the vertices of the square are in the same orbit (size 4), and the origin is in an orbit by itself (size 1), just by applying each of the group elements to an element of the set. Defnition 17.10 The group G acts transitively on S if S = Os for some s ∈ S. For example, Sn acts transitively on {1, · · · , n}, since given an element i ∈{1, · · · , n}, there is some permutation mapping it to any other element i ′ . 70This notation is not standard and may not correspond with the textbook. 82 Lecture 17: Group Actions Student Question. Does this have to be true for all s ∈ S, or just one? Answer. If it is true for one s ∈ S, it is true for all s ∈ S. Try checking it! So a transitive group action is one where there is only one orbit consisting of the entire set S; in particular, any element of s can be carried to any other element when acted on by some g ∈ G. Defnition 17.11 The stabilizer of s is Gs = StabG(s) := {g ∈ G : gs = s}, and it is a subgroup of G. For Example 17.6, the top and bottom vertices of the diamond are stabilized by the refection across the y-axis, whereas the stabilizer group of a vertex of the square is just the identity element. Proposition 17.12 The orbits of G form a partition of S.a In particular, S is the disjoint union of the orbits: S = ⨿Oi where Oi ∩ Oj = ∅. aThe set can be cut into non-overlapping pieces by the orbits. Proof. The orbits clearly cover S, since every element s ∈ S is also an element of Os, its own orbit. Also, they ′ ′ −1 ′ are disjoint. If Os ∩ Os ′ ̸= ∅, then there is some element in their intersection t = gs = g s . Then s = (g g ′ )s , which is in Os ′ . So every element of Os is in Os ′ , and by the same logic Os ′ ⊆ Os. Then Os = Os ′ . So if two orbits have nonempty intersection, they are in fact the same orbit. For a fnite set, the size of S can be obtained from the sizes of the orbits. Corollary 17.13 If S is a fnite set, and O1, · · · , Ok are the orbits, then k X |S| = |Oi|, i=1 since each of the orbits cover S exactly. In Example 17.6, this gives 9 = 4 + 2 + 2 + 1. Guiding Question What does each orbit look like? For this, we use the notion of a stabilizer of an element. 83 Lecture 17: Group Actions Proposition 17.14 Fix some s ∈ S and let H := Stab(s). Then there exists a bijection ε from the quotient group G/H to the orbit of s, Os. It takes G/H − → ε Os gH 7→ gS. Proof. Consider g and γ in G. Then their cosets map to the same element if gs = γs, which is equivalent to saying that g−1γs = s. Since H is the stabilizer of S, this means that g−1γ ∈ H; equivalently, γ ∈ gH. Since each of these conditions were equivalent conditions, gs = γs if and only if γ ∈ gH, and thus ε must be bijective: two elements in G/H map to the same element in Os if and only if they are the same element. Corollary 17.15 (Counting Formula for Orbits) As a result, the number of cosets of H, which is the order |G/H|, is equal to the size of the orbit of s, since there is a bijective correspondence between them. So |Os| = [G : Stab(s)]. In particular, the size of the orbit of any element |Os| divides |G| when G is a fnite group. We have |Os| · |Stab(s)| = |G|. These theorems are similar to the Counting Formula and Lagrange’s Theorem from Chapter 2. In particular, let C be the set of left cosets of a given subgroup H. Then G acts on C; an element g ∈ G takes C 7− → gC. Every coset can be mapped to any other coset by some element of G. For example, g1H is mapped to g2H by −1 g2g ∈ G. So there is only one orbit, the entire set C. The stabilizer of the identity coset, which is eH = H, is 1 Stab(eH) = H, because some element g ∈ G carries h ∈ H to h ′ ∈ H if and only if gh = h ′ , which implies that g = h ′ h−1 ∈ H. Thus, the Orbit-Stabilizer Theorem states that |G| = |H|[G : H], since |H| is the stabilizer of the identity in G/H and [G : H] is the size of the identity orbit. 84 Lecture 17: Group Actions Example 17.16 Consider the subgroup G ≤ SO3 consisting of rotational symmetries of a cube centered at the origin. • Let S be the set of faces of the cube; it has order 6 since there are 6 faces. For every face of the cube, there is some element in G mapping it to any other face in the cube (G acts transitively on the faces), so the orbit of a given face is the set of all the other faces, which is S. The stabilizer StabG(face) = C4, since a given face, which is a square, is preserved by rotation by π/2 around the axis through the center of the face. Then |G| = |S| · |StabG(face)| = 6 · 4 = 24. • Similarly, any vertex can be mapped to any other vertex by some element of G. The stabilizer StabG(vertex) = C3, since a vertex is preserved under rotation by 2π/3 around the axis from the vertex to the opposite vertex. Again, |G| = |{vertices}| · |StabG(vertex)| = 8 · 3 = 24. • Again, G acts transitively on the set of edges. The stabilizer of an edge is StabG(edge) = C2. Then |G| = |{edges}| · |StabG(edge)| = 12 · 2 = 24. 85 Lecture 18: Geometric Application of Stabilizer 18 Stabilizer 18.1 Review A group action is when a group G acts on a set S by G×S − → S and sends (g, s) 7− → gs. The orbit of an element in s is all the elements it gets mapped to, Os = {gs ∈ S : g ∈ G}, and the stabilizer is StabG(s) := {g ∈ G : gs = s} ≤ G, 18.2 Counting Formula Figure 1: The partition of S into orbits and theeir corresponding bijections with G/ Stab(si) One of the facts we learned about orbits is that they partition S. We further learned that there is a bijection between the left cosets of the stabilizer with the orbit. We were then able to write the size of S as X X |S| = |Osi | = [G : Stab(si)]. 18.3 Stabilizers of Products Given a group G acting on a set S and an element s ∈ S, we know that |Os| = [G : Stab(s)]. This also means that we expect that if we take the stabilizer of two elements in the same orbit, then they should be the same ′ size. Specifcally, we are asking what can we say about the stabilizer of a product. If s = as for a ∈ G, if g ∈ Stab(s), then gs = s. Then ′ aga −1(s ′ ) = aga −1(as) = ag(s) = as = s . ′ In other words, if g stabilizes s, then aga−1 stabilizes s . So the upshot is that −1 StabG(s ′ ) = a StabG(s)a . If StabG(s ′ ) was normal, then the two stabilizers would be the same, but this doesn’t have to be the case. We’ve provided a nice bijection to see that the sizes of the two stabilizers must be the same size. 86 Lecture 18: Geometric Application of Stabilizer 18.4 Statement Today, we will look at a consequence of these counting formulae. Recall that we were able to study and classify fnite and discrete subgroups of isometries in the plane. The special orthogonal group SO3 is the group of » rotations ρ(u,θ) in R3 fxing # 0 . What are the fnite subgroups G ≤ SO3? In fact, there are not so many! Let’s start with the theorem. Theorem 18.1 If G ≤ SO3, then • G ∼ = Cn = ⟨ρu,2π/n⟩, or • G ∼ = Dn = ⟨ρu,2π/n⟩, or • G is the group of rotational symmetries of a regular polyhedron. Figure 2: The regular polyhedra Although there are 5 regular polyhedra, there are only 3 distinct subgroups of symmetries. The dodecahedron and icosahedron have the same symmetries, which we denote as I. The cube and octahedron have the same symmetries as well, which we denote by O. Finally, the tetrahedron is partnered with itself and we denote its symmetries with T . As an example to see why some of the symmetries are the same, consider the symmetries of the octahedron. We can draw a point on the center of every face in the octahedron. Connecting these points lead to a cube, and thus any rotational symmetry of the cube will give a symmetry of the octahedron, and vice versa. A similar argument can be applied to the dodecahedron and icosahedron. Trying the argument for a tetrahedron just maps to the tetrahedron itself. On Wednesday, we worked out that the group of symmetries of a cube has size 24. Similarly, the tetrahedron has |T | = 12, and |I| = 60. 87 Lecture 18: Geometric Application of Stabilizer We will prove this by studying orbits! An non-identity element g ̸= I ∈ G71 is a rotation and thus fxes two unit vectors, which are exactly the positive and negative unit vectors on the rotation axis of g. These are called the poles of g. Let [ P = {poles of g}. g̸=I Lemma 18.2 If p ∈P and some g ∈ G, then gp ∈P also. As a result, we learn that G acts on P. ′ ′ Proof. For p ∈P, then there exists h = ̸ I ∈ G such that hp = p. If p = g(p). Then ghg−1(p ′ ) = p by earlier ′ reasoning. Then ghg−1 ∈ G, and ghg−1 = ̸ I since h = ̸ I. thus p ∈P also. Example 18.3 Let G = Cn. Then P = {p, −p} since every rotation will have the same poles. Example 18.4 Let G = O.a Then P = {pole for each vertex, edge, or face}. aThe group of symmetries of an octahedron. Now, what can we say about stabilizers of these subgroups. let |G| = N. Let’s decompose P into orbits. Then P = O1 ∪O2 ∪ · · · ∪ Ok. Then |Oi| = ni, and Oi = Opi for some pole pi. Then by our relations about the number of index of the stabilizer N | Stab(pi)| = ri = . ni Note that the stabilizer group will be a cyclic group. Geometrically, it will just contain the rotations around the axis pi. 18.5 Finding the subgroups Let’s write down an auxiliary set. It is the set of poles and group elements paired together. Let S := {(g, p), g = ̸ I, p is a pole for g}. Then we can count the order of S in two diferent ways. The order of S is X |S| = 2 = 2(N − 1), g∈G g̸=I since there are two poles for every non-identity element of G. Additionally, since we have k orbits, k k X X X N |S| = | Stab(p)| − 1 = ni(ri − 1) = (ri − 1). ri p∈P i=1 i=1 Every pole in the same orbit has the same stabilizer size, so we can group them together. Now, we have that k X N (ri − 1) = 2(N − 1). ri i=1 Dividing by N, 1 1 1 2 1 − + 1 − + · · · + 1 − = 2 − . r1 r2 rk N 71We use I for the identity here 88 Lecture 18: Geometric Application of Stabilizer 1 Each of these quantities 1 − is between 1/2 and 1, since r1 must be at least two (by the defnition of a pole.) r1 So 1 − r 1 1 ∈ 1 2 , 1 . 2 In addition 2 − ∈ [1, 2). So k = 2 or 3, since this is the only way the counting formula works out from our N bounds. In fact, this works out for examples 18.3 and 18.4. For G = Cn, we have two orbits, and for G = O we have three orbits, even though it is much more complicated. For k = 2, when there are two orbits, 1 1 2 1 − + 1 − = 2 − , r1 r2 N and so 1 1 2 + = . r1 r2 N But since r1, r2 ≤ N, 2 1 1 2 ≤ + = . N r1 r2 N We have r1 = r2 = N, and n1 = n2 = 1. Each of the poles is fxed by the entire group. Then G is a fnite subgroup of SO2, and thus G = CN , a cyclic subgroup. For the three-orbit case, the numerics of the problem is also extremely constraining. When k = 3, the equation is 1 1 1 2 1 − + 1 − + 1 − = 2 − , r1 r2 r3 N and equivalently 1 1 1 2 + + = 1 + . r1 r2 r3 N Without loss of generality, let r1 ≤ r2 ≤ r3. It is necessary for r1 = 2, or else the LHS72 would be ≤ 1. If r2 ≥ 4, then r3 ≥ 4 as well, and again the LHS would be ≤ 1. So r2 = 2 or 3. Finally, if r2 = 3, then r3 cannot be ≥ 6, again from the numerics of the problem. So r3 = 3, 4, or 5. In total, the cases are • Case 1. (2, 2, r) : r = N/2. In this case, we still have an infnite family. • Case 2. (2, 3, 3). We can solve for N to get N = 12. This corresponds to the tetrahedral group T. • Case 3: (2, 3, 4). N = 24. This corresponds to the octahedral group O. • Case 4: (2, 3, 5). N = 60. This corresponds to the icosahedral group I. We can really strongly limit the possibilities of a group G. We got this by counting a set S in two diferent ways, and playing around with the numbers and fractions. We aren’t done; we still have to show that these cases actually correspond to the groups. In each of these cases, we have three orbits, and those correspond to edges, faces, and vertices of these regular polyhedra. The pole for any vertex can be rotated to the pole for any other vertex. We’ll do the argument for the octahedral group and it will be similar for the remaining polyhedra. 18.6 The Octahedral Group 72left hand side 89 Lecture 18: Geometric Application of Stabilizer Example 18.5 (Octahedral group) Let’s take (2, 3, 4) and argue that this group must be symmetries of a cube or an octahedron. For 4, n3 = 6. So the stabilizer has size 4 and the orbit has size 6. Let’s try to fgure out what the orbit looks like. It contains six vectors inside R3 , so we can simply see the possible confgurations. Drawing the frst vector is easy; we just pick wherever we want for p. Then −p has to be in the same orbit as well because StabG(−p) has the same order as StabG(p), but we only have one orbit with stabilizer size 4. What does the stabilizer group look like? Geometrically, it’s rotations around p and must have size 4, so we have Stabg(p) = C4. Now, let’s try to fgure out the other 4 poles in our orbit. Suppose we draw q such that p and q are not perpendicular. For any q in the orbit, −q is also in the orbit. We have determined that rotations by π/2 around p are in our group, so those rotations of q should also be in our orbit. However, this gives us 4 vectors for rotations of q and 4 vectors for rotations of −q, and this is too many. This picture was wrong because q was not drawn perpendicular to p. If we drew q perpendicular, then rotating q by 90 degrees gives us an orbit of size 6. Whatever the group G is, it fxes the collection of vectors that we drew. In particular, it fxes the octahedron obtained by taking the convex hull of the vectors and thus G ≤ O. But the octahedral group has size 24a and this group also has size 24, so |G| = |O| = 24 implies that G = O. awe worked this out in class on Wednesday We can repeat this for all the cases to fnish the proof, but it isn’t too important. The key lesson is how we were able to use the counting formulas and orbit decomposition of the set to really constrain the possibilities in these ways. A lot of the challenge was just fnding the right set to act on; in this case it was the poles. 90 Lecture 19: Group Actions on G 19 Group Actions on G 19.1 Conjugation Today, we will discuss the special case of group actions where the set S is G itself. We’ve seen the power of studying orbits and stabilizers and how they can help us understand groups of symmetries. One attempt is to just directly apply the group action on G: G×G − → G (g, x) 7− → gx. However, this isn’t particularly interesting. The action is transitive, and thus there is only one orbit and the stabilizers are all trivial. We instead defne a diferent group action on itself, conjugation. It takes G×G − → G (g, x) 7− → gxg −1 , conjugating x by g. One can check that it satisfes the axioms of a group action. We have some special names for the orbit and stabilizer under conjugation. Defnition 19.1 The orbit of an element under conjugation is −1 C(x) := Orbit(s) = {gxg : g ∈ G}, and is called the conjugacy class of x. Defnition 19.2 The stabilizer of an element under conjugation is −1 Z(x) := StabG(x) = {g ∈ G : gxg = x} = {g ∈ G : gx = xg} ≤ G. It is called the centralizer of x in G, and it is a subgroup of G. From before, for any x ∈ G, we have |G| = |C(x)| · |Z(x)|, and we also have the class equation, which states that |G| = |C1| + · · · + |Ck|, since the conjugacy classes partition G, and additionally, each |Ci| divides |G| from the counting formula. Student Question. Are the conjugacy classes related to cosets, like we saw how left cosets of a subgroup partitioned a group? Answer. No, in general the conjugacy classes won’t have the same size like cosets do. We’ll be seeing exactly what this equation looks like for diferent examples in the next few lectures. Another related set, which we saw in homework before is the center of a group. Defnition 19.3 The center of G is {Z := x ∈ G : xg = gx, g ∈ G}. Other facts: • C(x) = {x} is equivalent to Z(x) = G and also x ∈ Z, the center of G. So if we had an abelian group, then the center would be the whole group, and the class equation would be just the sum of a bunch of 1s. 91 Lecture 19: Group Actions on G • For any x ∈ G, we have that Z ≤ Z(x) since the center commutes with all elements. Also, since x commutes with itself, ⟨x⟩≤ Z(x). This fact is a lower bound on the order of Z(x), so it gives us the upper bound |C(x)| ≤|G|/ ord(x). • For all x ∈ G, conjugation preserves order: ord(x) = ord(gxg−1). This is true because conjugation defnes k k −1 −1)k k an automorphism of our group. If x = e, then e = gx g = (gxg so x = e is equivalent to −1)k (gxg = e. Student Question. Why is conjugation an automorphism? Answer. We can just show that conjugation satisfes the homomorphism property, that it preserves products: −1 −1 −1 gxyg = gxg gyg . Conjugation is a homomorphism and in fact an isomorphism from G to itself. Since it is an automorphism of G, elements that are conjugate to each other will have the same properties with respect to order; whether or not they commute; and so on. All we have done so far is take observations about these defnitions, and the class equation comes from our work on group actions from last week. Example 19.4 What does the class equation say for D5? The order of the group is |D5| = 10. It is equal to 2 3 4 2 3 4 D5 = {e, x, x , x , x , y, xy, x y, x y, x y}. −1 4 One of the properties of refections is that conjugating x by y gives yxy = x . Let’s fgure out the conjugacy class of all the elements. The identity commutes with everything, so its conjugacy class is C(e) = {e}. Now let’s look at the refection y. From our facts above about the centralizer, we know that ⟨y⟩≤ Z(y) ≤ D5. Then Z(y) must be at least 2, and it must divide 10, so 2 and 10 are our only possibilities. However, not every element in D5 commutes with y, so |Z(y)| = 2, and thus |C(y)| = 5. In fact, every refection is conjugate to every other refection: C(y) = {all refections}. The conjugacy class of x is at least x, and it is also at least {x, x4}. It cannot be more, because the order must divide 10 and we only have 4 elements left to partition. Then we have C(x) = {x, x4} and similarly C(x2) = {x2, x3}. So we have 10 = 1 + 5 + 2 + 2. The center corresponds to the elements that are in its own conjugacy class, so the center of D5 is {e}. Notice that the conjugacy classes have very diferent sizes; they partition the group in a very diferent way from cosets. Since the group was small, we could have just brute forced and directly calculated the conjugacy classes for every element. However, looking at these divisibility facts is powerful and can handle more complicated and larger groups. Student Question. The conjugacy classes seem to contain x−1; is that always true? −1 4 Answer. In the example, it was specifcally true because yxy = x . However, in general it isn’t true. For example, in the integers, 5 and −5 are inverses, but not conjugate to each other. 19.2 p-groups By studying the class equation, it is possible to gain some information about a general class of groups, p-groups. Defnition 19.5 e G is a p-group for a prime p if |G| = p for some e ≥ 0. There exists a group of any order simply by taking the abelian cyclic group of that order. 92 Lecture 19: Group Actions on G Example 19.6 For example, we have a p-group for every e ≥ 0 by taking Cp, Cp2 , Cp3 , and so on. Another example of a p-group is Cp × Cp × · · · × Cp. Looking at a subgroup of 3×3 matrices also provides an example of a p-group. Example 19.7 A more interesting group is the set of matrices 1 ⋆ ⋆ 1 ⋆ ≤ GL3(Fp), 1 which has order p3 . By looking at the class equation modulo p, the following theorem holds. Theorem 19.8 Every p-group has non-trivial center.a aThere are elements of the center that are not the identity. Example 19.9 For G = D4, |G| = 8 = 23 . The class equation says 8 = 1 + 1 + 2 + 2 + 2, so the center has size 2 (since there are two 1’s in the class equation.) Proof. The class equation for G states that |G| = |C1| + · · · + |Ck|, which is e 2 e−1 p = (1 + · · · + 1) + (p + · · · + p) + (p + · · · + p 2) + · · · + (p + · · · + p e−1), since the order of the conjugacy class must divide the order of G. Recall that the sizes of the center, |Z| is exactly the number of 1s in the class equation. Then the equation taken modulo p gives 0 = |Z| mod p, which implies that p divides |Z|, since |Z| ≥ 1 because at least the identity e is in Z. So |Z| ≥ p, and then the center is nontrivial. This theorem is interesting because we get some nontrivial information about the group just from the size of the group, by using the class equation and looking at the numerics. 93 Lecture 19: Group Actions on G Example 19.10 Take the upper triangular matrices of the form 1 ⋆ ⋆ 1 ⋆ ≤ GL3(Fp). 1 What is the center of the group? We want 1 a 1 b 1 c x 1 y 1 x = x 1 y 1 x a 1 b c 1 1 1 1 for all x, y, z. This happens exactly when a = c = 0 and b can be anything. So |Z| = p, since there are p possibilities for b. The example above was a group of order p3 having center of size p, so it demonstrates that there is not a better 2 theorem where the center always has size p , or something. However, we can say a little more about the specifc case of |G| = p2 . Corollary 19.11 2 If |G| = p , then G must be abelian.a 2 aEarlier on, we stated that if p were prime, then G must be cyclic; now we get from p that G is abelian, although not necessarily cyclic. Proof. We have {e} ⪇ Z ≤ G. 2 We want to show that |Z| = p , because that implies that Z = G and then G is abelian. We already know that p divides |Z|, and that |Z| ≥ p, so the only two possibilities are p and p2 . Assume for the sake of contradiction that |Z| = p. Then, pick x ∈ G \ Z that is not in the center. Then, Z ⪇ Z(x); Z(x) ̸= Z because x ∈ Z(x) but x ̸∈ Z. Then there exists x ∈ Z(x) such that x ∈ / Z. Thus if |Z| = p, the only possibility is that |Z(x)| = p2 since Z(x) is a subgroup. However, this implies that Z(x) = G, but then x ∈ Z, which is a contradiction. The issue here is that p2 is just not very big, so there is not very much room for a lot to happen. We can even classify exactly what groups of size p2 look like. Corollary 19.12 2 2 Given a group G such that |G| = p , G must be isomorphic to either Cp2 , the cyclic group of size p , or Cp ×Cp = {(a, b) : a, b ∈ Cp}. Proof. We can split this up into two cases. 2 2 • Case 1. If there exists a ∈ G with ord(a) = p , then ⟨a⟩ = G, since ⟨a⟩ has size p , and thus must be the entire group G. 2 • Case 2. Otherwise, every element a ̸= e has order p, since it must divide p and cannot be p2 since we already considered that case. We claim that G being abelian such that every x ̸= e has order p comes from considering it as a vector space V over F = Fp 2 What is the dimension of this mystery vector space? The group G has size p , so it has dimension p. So V = F ⊕ F , implying that G = Cp ×Cp. Here, we are forgetting the vector space structure and just thinking about it as a group with respect to addition. 94 Lecture 19: Group Actions on G The addition structure is already implicit from the structure on G. To turn G into a vector space, we only n times z }| { need to defne how to scale g ∈ G by n ∈ Fp, and then check the vector space axioms. Let n·g = g + · · · + g This is well-defned because ord(g) = p, so it only matters what n is modulo p. We have fgured out a way to turn the group into a vector space. Since any two vector spaces of the same dimension are isomorphic to each other as vector spaces; in particular, they are isomorphic to each other as abelian groups. All of this is gravy from what we were supposed to discuss this week, but it is helpful to see these examples. 95 Lecture 20: Class Equation for the Icosahedral Group 20 The Icosahedral Group 20.1 Review: The Class Equation Last time, we discussed the conjugacy class of an element, which is the orbit of an element under conjugation, and the centralizer of an element, which is the stabilizer of an element under conjugation. Defnition 20.1 The conjugacy class of an element is −1 C(x) := {gxg : g ∈ G} ⊆ G. Defnition 20.2 The centralizer of an element is −1 Z(x) := {g : gxg = x} ≤ G. The class equation states that |G| = |C1| + · · · + |Ck|, which tells us information about a group simply through numerics. 20.2 Basic Information The group I ≤ SO3 is the icosahedral group, which is the group of symmetries of the icosahedron under rotations (orientation-preserving isometries in R3.) It is isomorphic to the dodecahedral group. The group is I = {ρ(u,θ)}, where each ρ(u,θ) is a rotation by θ around a vector u preserved by the rotation, which is called a pole of u. Additionally, the rotations have the property that ρ(u,θ) = ρ(−u,−θ). For a polyhedral group, u lies on a face, an edge, or a vertex of the polyhedron. Let’s start with counting the number of rotations in I by whether the pole is on a face, an edge, or a vertex. • Identity. The trivial rotation is one rotation. 96 Lecture 20: Class Equation for the Icosahedral Group • 20 faces. For each face, excluding the identity, there are 2 rotations by 2π/3 and 4π/3, which is 40 rotations total. In this way, every rotation is counted twice, since ρ(u,θ) = ρ(−u,θ), and every pole through the center of a face also goes through the opposite face. So we count 20 face rotations total. • 30 edges. For each face, there is one (non-identity) rotation by π, and this also gets double-counted, so there are 30/2 = 15 edge rotations total. • 12 vertices. For each vertex, there are four nontrivial rotations, by 2π/5, 4π/5, 6π/5, and 8π/5. This double-counts as well, so there are 12 · 4/2 = 24 total vertex rotations. In total, we have 1 + 20 + 15 + 24 = 60 rotations. 20.3 Conjugacy Classes Now, it is possible to understand I by thinking about the group action of conjugation on itself. Guiding Question How can we decompose I into conjugacy classes and what does the class equation tell us about normal subgroups of I? −1 For g ∈ I, the conjugate of ρ(p,θ) under g is gρ(p,θ)g = ρ(q,θ), where q = g(p), since conjugating by g is essentially taking a change of coordinates by g. Thus, the rotations by the same angle θ with poles that can be reached from each other, through conjugation by some element in I, are conjugate. Then, we can count the conjugacy classes. • The identity is conjugate to itself. • Thus, the face rotations by 2π/3 are all conjugate.73 • In addition, the vertex rotations by 2π/5 (or 8π/5 = 2π − 2π/5; they are the same rotation with the pole fipped) are conjugate. • The vertex rotations by 4π/5 and 6π/5 are conjugate as well. • The edge rotations are all conjugate. Then the class equation states that 60 = 1 + 20 + 12 + 12 + 15. In particular, we see that the center of the group is trivial. 20.4 Simple Groups We can use the class equation to study normal subgroups of I. The following defnition and then analysis provides one way the class equation can be useful. Guiding Question How can we study complicated groups by decomposing them into simpler groups? Defnition 20.3 A group G is simple if the only normal subgroups H ⊴ G are H = {e} or H = G.a Equivalently, G is b simple if for any surjective homomorphism f : G − → G ′ , G ′ = G or G ′ = {e}. aA group G always has at least two subgroups, the trivial subgroup and the whole group, and a simple group only has these two. bSince the kernel of the homomorphism is a normal subgroup, the kernel must be either {e} or G, leading to these two cases: either f is an isomorphism or f is trivial. The guiding principle for studying groups is that simple groups are building blocks for all fnite groups. To study a complicated group G, it is possible to break it up by considering surjective homomorphisms to a smaller group G ′ and studying instead the kernel, which is normal, and the image, which is G ′ . Once a simple group is 73These are the same as the face rotations of 4π/3; one angle must be picked to avoid double-counting. 97 − Lecture 20: Class Equation for the Icosahedral Group reached, there are no more interesting surjective homomorphisms, and in this way a group can be "decomposed" into simple groups. For example, Cn is simple if and only if n is prime. In the same way that primes are building blocks for integers, simple groups are building blocks for all fnite groups and cannot really be broken down any further. Student Question. Does this mean that p-groups are simple? Answer. No. For instance, any subgroup of an abelian group will be normal, so an abelian group containing any nontrivial subgroup will not be simple. In particular, Cpn for n = ̸ 1 are not simple. The center of a subgroup is always a normal subgroup, and the center of a p-group is nontrivial, so whenever the center of a p− group is not the entire group, the p-group will not be simple. Theorem 20.4 The icosahedral group I is simple.a aIt has no normal subgroups. Proof. If N ⊴ I, then gNg−1 = N. In fact, for some element x ∈ N, its conjugate gxg−1 ∈ N as well. Thus, for x ∈ N, C(x) ⊆ N as well. So the normal subgroup is a union of conjugacy classes: [ N = C(xi). xi∈N Also, |N| divides 60 for G = I. Since 60 = 1+20+15+12+12, we must have that 1+ (a subset of {20, 15, 12, 12}) is a factor of 60, which is possible only when |N| = 1 or 60, in which case N = {e} or I. So there are no other normal subgroups of I, and I is simple. In some sense, this is a very soft proof. We do not really have to grapple with the group structure of the mystery normal subgroup N; all we have to deal with is the sizes of the conjugacy classes. This argument is very special to I and would not work for D5. On the other hand, it is still possible to list all of the normal subgroups of D5 by looking at the class equation; there are not that many. Problem 20.5 Try to follow the same proof for D5 (and fail!)a aSince it is not actually simple, the proof will not work. Recall that |S5| = 5! = 120. Then, let A5 be the subgroup of even permutations in S5; it is the kernel of the homomorphism sign: S5 − → {±1}, and has index 2, so it has order 60. Theorem 20.6 The icosahedral group I is isomorphic to the alternating group A5. Proof. We want to show that an element in I acts in the same way as an element of S5.. To do this, we construct an action of I on a set of size 5. We want to fnd a set S such that the group action of I on S gives a homomorphism non-trivial f I − − − − − − − → S5. 98 Lecture 20: Class Equation for the Icosahedral Group Recall from before that I is the symmetry group of both the icosahedron and the dodecahedron. There are fve cubes ftting in the dodecahedron where the vertices are vertices of the dodecahedron and the edges of the cubes are diagonals of the pentagons that are the faces of the dodecahdron. For such a cube, every face of the dodecahedron will contain exactly one edge of the cube. Once one diagonal on one face is chosen, it determines the rest of the cube, and since a pentagon has fve diagonals, there are 5 such cubes. Let S be the set of 5 cubes in the dodecahedron, labeled from 1 to 5 in some order. Then, I clearly acts on S, since it acts on the dodecahedron. Let f : I − − − − − − → Perm(S) = S5 non-trivial take an element of I to the corresponding permutation of the fve cubes. The group I is simple, and the kernel of a homomorphism is always normal, so ker(f) ⊴ I = {e} or I. Since f is nontrivial, ker(f) ̸= I, so ker(f) = {e}. This implies that the homomorphism f must be injective. Then, consider a diferent homomorphism φ taking I − → {±1}, the composition f sign I − → S5 − − → {±1}. Again, ker(φ) = {e} or ker(φ) = I. Since |I| = 60 > |{±1}| = 2, φ is mapping a larger group onto a smaller group and cannot be injective, so ker(φ) = I. So under φ, every element of I maps to 1. However, this implies that the sign of the corresponding permutation of some element of I is 1, so the corresponding permutation is even, and so f(I), a subgroup of permutations in S5, consists entirely of even permutations. Then f(I) ⊆ ker(sign) = A5, so f is actually a homomorphism from I to A5 ⊂ S5; since it is injective from I to S5, it is still injective from I to A5. Since I and A5 are both of order 60, f is also a surjection, and thus f is an isomorphism between I and A5. Throughout this proof, the fact that I is simple is used over and over again to argue facts about various homomorphisms coming from I. Corollary 20.7 The alternating group A5 is also simple. In fact, An is simple for all n ≥ 5, but the proof is more complicated and involves thinking about the actual permutations. For the proof that A5 is isomorphic to I and thus simple, the class equation was the jumping-of point. The class equation showed that I was simple, which then provided strong restrictions on homomorphisms from it. 20.5 Conjugacy Classes for Symmetric Groups Next time, the conjugacy classes for Sn and An will be determined. Recall that every σ ∈ Sn can be decomposed via cycles. Example 20.8 The permutation (123)(45) ∈ S6 takes 1 7→ 2, 2 7→ 3, and 3 7→ 1 as the frst cycle, of length 3, then 4 7→ 5 and 5 7→ 4 as the second cycle, of length 2, and 6 7→ 6 as the third cycle, of length 1. 99 Lecture 20: Class Equation for the Icosahedral Group The sign of σ, where σ = τ1 · · · τr, where each τi is a 2-cycle74 , is (−1)r . For example, the sign of (1234) = (12)(13)(12) is −1. 74Also called a transposition. 100 Lecture 21: Symmetric and Alternating Groups 21 Conjugacy Classes for Symmetric and Alternating Groups 21.1 Review Recently, we have been discussing the conjugation action of a group on itself. In particular, it is possible to decompose a group into its conjugacy classes, which is similar to decomposing a set into its orbits (like we were doing last week). Last time, we looked at the icosahedral group and saw its class equation and fgured out information based on that. 21.2 Cycle Type Today, we will be looking at the conjugacy classes for Sn and An, the symmetric group and the alternating group, which consists of even permutations. Recall that a permutation σ ∈ Sn can be written in cycle notation. This is a very useful way of writing a permutation. Example 21.1 (Cycle Notation) For example, the permutation (123)(45) takes 1 to 2 to 3 to 1, and 4 to 5 back to 4. Given the cycle type, it is easy to defne and fgure out the sign of a permutation. A 1-cycle will have sign +1, a 2-cycle will have sign −1, and so on, where a k-cycle will have sign (−1)k−1 . For example, (123)(45) has sign −1 = (+1)(−1) = −1, where the signs of each cycle are multiplied. In particular, even permutations are permutations that have an even number of even-length cycles.75 Guiding Question What are the conjugacy classes of Sn? It turns out that the sign will be a very helpful tool in determining the conjugacy classes. Let’s look at an example. Example 21.2 If σ = (123), then for p ∈ Sn, let the conjugate be τ = pσp−1 . Let’s say p(1) = i, p(2) = j, and p(3) = k. Evaluating the conjugate on i gives τ(i) = pσp−1(p(1)) = p(σ(1)) = p(2) = j. Similarly, τ (j) = p(σ(2)) = p(3) = k. It turns that in cycle notation, τ = (ijk) = (p(1)p(2)p(3)). It is easy to check that τ fxes all the other points. So conjugating a 3-cycle produces another 3-cycle with diferent points. Consider a more complicated permutation. 75It’s confusing: an even-length cycle makes a permutation odd. 101 Lecture 21: Symmetric and Alternating Groups Example 21.3 For σ = (123)(47) · · · , conjugating by p gives (p(1)p(2)p(3))(p(4)p(7)) · · · . It turns out that the lengths of the cycles in a permutation don’t change upon conjugation! Defnition 21.4 Given σ ∈ Sn, the cycle type of σ is the number of 1-cycles, 2-cycles, and so on, that show up in the cycle notation. The cycle type is conjugation-invariant. If τ = pσp−1 , then σ and τ have the same cycle type. For example, (47)(123) has cycle type (2, 3). In fact, if σ and τ have the same cycle type, then they are conjugate. Example 21.5 Take σ = (145)(23) and τ = (234)(15). Simply by matching cycles, we can defne p ∈ Sn taking 1 7− → 2, 4 7− → 3, 5 7− → 4, 2 7− → 1, and 3 7− → 5; that is, p = (12)(354). This p is constructed to be such that pσp−1 = τ.a aTry working through this by hand! The upshot is this proposition. Proposition 21.6 Two permutations σ and τ are conjugate if and only if σ and τ have the same cycle type. 21.3 Conjugacy Classes in Sn Conjugation in Sn can be understood well by looking at cycles. Guiding Question What are the conjugacy classes in Sn? From our characterization of when two permutations are conjugate, this can certainly be done! Let’s start with an example. Example 21.7 For S3, there are three conjugacy classes: cycle type 3, 2 + 1, and 1 + 1 + 1. For example, representatives could be (123), (12), and the identity permutation. Now, we can do more complicated computations. For instance, we may want to fnd out the size of a given conjugacy class. 102 Lecture 21: Symmetric and Alternating Groups Example 21.8 (Conjugacy classes in S4) For a permutation x = (1234) ∈ S4, the conjugacy class C(x) is all 4−cycles in S4. For each 4-cycle, there are 24 orderings of 1, 2, 3, and 4, and each one is overcounted by a factor of 4. For example, (1234) = (2341) = (3412) = (4123). So there are 24/4 = 6 elements in the conjugacy class. Alternatively, where the stabilizer is Z(x), then |G| |C(x)| = . |Z(x)| Since conjugation is essentially "relabeling" the numbers in the original permutation, replacing 1 with p(1), 2 with p(2), and so on, the elements in the stabilizer should relabel the numbers 1 through n in such a way that the permutation is still the same. For instance, relabeling (1234) to (2341) gives the same permutation x. In this case, because there are 4 diferent starting points to the cycle, there are 4 permutations p ∈ Z(x) that stabilize x. So again, |G| 24 |C(x)| = = = 6. |Z(x)| 4 Essentially, the redundancy in cycle notation gives us diferent ways to write the same permutation, and dividing out by this redundancy (the stabilizer) gives the size of the conjugacy class. Cycle notation also simplifes these computations for larger symmetric groups. Example 21.9 (Conjugacy Class in S13) Consider x = (123)(456)(78910)(11)(12)(13) ∈ S13. What is the stabilizer Z(x) of x? For the 4-cycles, there are 4 choices for where to start the cycle. Any reordering of 12, 11, and 13 doesn’t change the fact that 12, 11, and 13 are fxed; there are 3! ways to order the 1-cycles. For the 3-cycles, there are 3 starting points each, but the 3-cycle (123) could also be mapped to (456), so there are 2! ways to order the two 3-cycles, and 3 starting points each. In general, if there are k ℓ−cycles, there are ℓ starting points for each cycle, and k! ways to order them. So |Z(x)| = 2! · 3 · 3 · 4 · 3! · 1 · 1 · 1 = 432. Then, 13! |C(x)| = . 432 So fnding the sizes of conjugacy classes is really just doing some combinatorics for the size of stabilizer. Given the size of the stabilizer, since we know the size of the entire group, |Sn| = n!, dividing by |Z(x)| directly gives |C(x)|, without having to compute every permutation in the conjugacy class. 21.4 Class Equation for S4 Now, we can work out the class equation for S4 without too much pain. 103 Lecture 21: Symmetric and Alternating Groups Example 21.10 (Class Equation for S4) The order of S4 is |S4| = 4! = 24. Then, there are only fve possible cycle types, listed on the left column of the table. These are the diferent ways to sum to 4. cycle type |Z(x)| |C(x)| 4 4 6 3 + 1 3 8 2 + 1 + 1 2 · 2! = 4 6 2 + 2 2! · 2 · 2= 8 3 1 + 1 + 1 + 1 24 1. The size of the stabilizer for cycle type 4 is 4, as we worked out already. For 3, there are three possible places to start the cycle, so there are three permutations fxing the cycle type. For 2 + 1 + 1, there are 2 ways to pick a starting point for the 2-cycle and 2! = 4 ways to order the 1-cycles, which gives 4 total. The rest of the middle row follows similarly. We have |C(x)| = |G|/|Z(x)|, so the right row, the size of the conjugacy class, is found by dividing |G| = 24 by the middle row, the size of the stabilizer. The class equation then says 24 = 1 + 3 + 6 + 8 + 6. What about the alternating group? The conjugacy classes for A4 can also be determined. Example 21.11 (Conjugacy classes for A4) What are the conjugacy classes for A4? The alternating group A4 is a subgroup of S4. In fact it is the kernel of the sign homomorphism, so it is a normal subgroup. In particular, a normal subgroup is fxed under conjugation, so A4 is the union of conjugacy classes. Using the defnition of the sign, the cycle types 3 + 1, 2 + 2, and 1 + 1 + 1 + 1 all correspond to the elements in A4. Then, taking the sizes of the corresponding conjugacy classes, we have |A4| = 12 = 1 + 3 + 8, but since 8 is not a factor of 12, this is actually not the class equation for A4. What’s going wrong? If an element σ is conjugate to another element τ in A4, it is a diferent notion than being σ being conjugate to τ in S4! In particular, σ and τ can be conjugate in S4, since we need τ = pσp−1 for p ∈ S4, without being conjugate in A4, since we require that τ = qσq−1 for q ∈ A4. If it is possible to fnd two elements conjugate by an odd permutation but not an even permutation, then they will be conjugate in S4 but not A4. Consider x ∈ An ≤ Sn. The conjugacy class of x in An is −1 CA(x) = {y ∈ A : y = pxp , p ∈ An}, the subset of elements in An conjugate to x by some even permutation, which is a subset of −1 CS (x) = {y ∈ An : y = pxp , p ∈ Sn}. Similarly, the stabilizer for A is a subgroup of the stabilizer for S. ZA(x) = {p ∈ An : px = xp} ≤ ZS (x) = {p ∈ Sn : px = xp}. Using the counting formula, 1 |CA(x)| · |ZA(x)| = |An| = |Sn|, 2 so 1 |CA(x)| · |ZA(x)| = |CS (x)| · |ZS (x)|. 2 104 Lecture 21: Symmetric and Alternating Groups The product difers by a factor of 2 for An and Sn. Additionally, |ZA(x)| is a factor of |ZS (x)|, as it is a subgroup. Our analysis leads us to two possibilities: 1 • Case 1. In this case, |CA(x)| = |CS (x)| and |ZA(x)| = |ZS (x)|. Here, the conjugacy class stays the same 2 size, but only half of the permutations that stabilize them are even and in An. 1 • Case 2. In this case, |CA(x)| = |CS (x)| and |ZA(x)| = |ZS (x)|. So the size of the conjugacy class is split 2 in half when going from Sn to An, and only half of them are conjugate by even permutations. Since the sizes of the stabilizers of x are the same, every p ∈ Sn such that px = xp is even, and lives inside of An. In our example, 8 must split, since it does not divide 12, and 1 and 3 cannot split because when they are split, they are split into halves, and they are odd numbers. Thus, the class equation for A4 must be, by simple numerics, |A4| = 12 = 1 + 3 + 4 + 4. In the 8 = 4 + 4 case, x = (123), and this is the case where the conjugacy class does split, which means the stabilizer group does not get any smaller. Thus, every p such that px = xp is even. This is what it means for the stabilizer group not to get any smaller! Example 21.12 What happens for S5? What about A5? For S5, the class equation looks like 120 = 1 + 10 + 15 + 20 + 20 + 30 + 24 . The even classes have size 1, 15, 20, and 24. We currently have |A5| = 60 = 1 + 15 + 20 + 24, which is not the class equation. Clearly, 1 and 15 do not split, since they are odd. Since 24 is not a factor of 60 (but 24/2 = 12 is), it must split. The question remains if 20 splits into 10 + 10 or not. We can show directly that there is an odd permutation that commutes with it, and so it cannot split. So the class equation is 60 = 1 + 15 + 20 + 12 + 12 . These examples demonstrate that after our analysis of cycle types in symmetric and alternating groups, deter mining the class equation is not so much algebra and more counting and combinatorics. 21.5 Student Question Student Question. Why do we care about An? Does it show up as the symmetry group of some object? Answer. Since we are working with low numbers and dimensions, there are lots of coincidences where the groups that show up will be the same as each other, even if they aren’t actually related in a general way for higher dimensions. In fact, A4 shows up in the symmetry group of the tetrahedron, and we had A5 show up as the symmetry group of the icosahedron. But in general, An is (maybe? Davesh said he didn’t really know/hadn’t thought about it) not necessarily the symmetry group of some higher-dimensional geometric object. But in 18.702 we will study the symmetries of equations (Galois frst studied these) instead of geometric objects, and it is very easy to write down equations that have An or Sn as (part of) their symmetry groups. In fact, group theory evolved at the same time as studying symmetries of non-geometric objects evolved (Galois again?) The reason why we care about An is this idea that simple groups are "building blocks" in some sense, and An for n = 5 and higher is simple, and in fact it is essentially the only (interesting?) simple normal subgroup of Sn. So if we care about Sn then we automatically care about An, since An is a "building block" of Sn. In general, people like to break down problems into studying the simple subgroups of certain groups, and then studying the ways in which the simple groups can combine into the larger groups, in order to understand the larger group as a whole. 105 Lecture 21: Symmetric and Alternating Groups There are lots of ways of combining groups that get pretty complicated, and this is defnitely something a lot of people are working on. The two ways of combining groups that get their own names are the "direct product" and the "semidirect product," which is a slightly more nonabelian way of combining groups. In this case, getting from An to Sn is just a semidirect product in some way. 106 Lecture 22: The Sylow Theorems 22 The Sylow Theorems 22.1 Review Last time, we discussed the conjugacy classes of symmetric and alternating groups. 22.2 Motivation The Sylow theorems are a set of related theorems describing the subgroups of prime power order of a given fnite group. They are very powerful, since they can apply to any fnite group, and play an important role in the theory of fnite groups. To motivate the Sylow theorems, recall the following basic theorem. Theorem 22.1 If G is fnite, and H is a subgroup of G, then |H| divides |G|. In fact, we can ask if the reverse is also true: given a factor of |G|, is there a subgroup of that size? It turns out that it it not true. Example 22.2 (Counterexample) For G = A4, |G| = 12, but it turns out that there is no subgroup of order 6.a aAs an exercise, try showing this using the class equation! So there is not always such a subgroup. Guiding Question Can we add constraints so that some form of reverse of Theorem 22.1 is true? When must there be a subgroup of a particular size? The next theorem is quite surprising and powerful: it turns out that the reverse is true for prime powers. If k k d = p , a prime power dividing |G|, then there must exist a subgroup where |H| = p . 22.3 The First Sylow Theorem The three Sylow theorems, which will be stated in this lecture and proved in the next, formalize and elaborate on this idea of studying subgroups of prime power order. Note 22.3 e For today’s lecture, we use e to denote the largest exponent of p such that p | |G| and we use 1 to denote the identity element instead. Also, n will always refer to |G|. The frst Sylow theorem states that there is always a prime power order subgroup. Theorem 22.4 (Sylow I) Given G such that e |G| = n = p · m, e where p is the largest power of p (that is, gcd(p, m) = 1), then there exists a subgroup H ≤ G such that e |H| = p . Such a subgroup is called a Sylow p-subgroup. It has maximal prime power order within G. 107 Lecture 22: The Sylow Theorems Defnition 22.5 e Let G be a group such that |G| = n = p m such that gcd(p, m) = 1. Then a subgroup H ≤ G such that e |G| = p is called a Sylow p−subgroup. Let’s see an application of this powerful theorem. Example 22.6 Consider G = S4. Since |S4| = 24 = 8 · 3, Sylow I states that there is a subgroup of order 8. In fact, we can take H = ⟨(12), (34), (13)(24)⟩. Another example is the dihedral group. Example 22.7 For G = D5, we have |D5| = 10 = 2 · 5. So there must be subgroups of size 5 and 2. A subgroup generated by a rotation ⟨ρ2π/5⟩ has order 5. A subgroup generated by any refection ⟨refection⟩ has order 2. Looking at this theorem now, it may be hard to appreciate. One reason this theorem is relevant now is that the proof is a very nice application of the theory on group actions and orbits. Moreover, this theorem is one that applies extremely generally. We have mostly been studying explicit groups such as the dihedral groups or symmetric groups in this class, but the Sylow theorems apply to any fnite group. When given an unfamiliar group, the Sylow theorems provide footholds and crevices, like in climbing a clif, to start of with and to learn more about the groups. Sylow I gives lots of interesting subgroups that play of of each other, depending on the diferent factors of the size of the group G. We can get a useful corollary for free. Corollary 22.8 If p divides |G|, there exists an element x ∈ G with order p. For example, if |G| = 14, it must have at least one element of order 7. Proof of Corollary 22.8. e Using Sylow I, there exists a subgroup H ≤ G such that |H| = p , where e ≥ 1. Pick some y ∈ H. Then, ⟨y⟩ = Cpf , e some cyclic group with order dividing p . Taking k−1 p x = y provides an element of order p. 22.4 The Second Sylow Theorem e The frst Sylow theorem states that a Sylow p-subgroup, a subgroup of maximal size p dividing |G|, exists. In fact, we can say a lot more about what these subgroups look like. 108 Lecture 22: The Sylow Theorems Theorem 22.9 (Sylow II) There are two parts; part a) is what is usually referred to as the second Sylow theorem. ′ (a) Given H ≤ G, where H is a Sylow p-subgroup, any other Sylow p−subgroup H ≤ G is conjugate to ′ H; i.e. there exists g such that H = gHg−1 . d (b) Given any subgroup K ≤ G such that |K| = p , for any Sylow subgroup H, there exists g such that gKg−1 ≤ H.a aNotice that |K| does not have to be the maximal prime power, and can have order smaller than |H|. Every prime power order subgroup, up to conjugation, sits inside a Sylow subgroup. Evidently, conjugating a Sylow subgroup will result in a Sylow subgroup (since they have the same size), and Sylow II states that all the Sylow subgroups arise in this way. Note that the second part is stronger, since |K| can be a prime power smaller than |H|, and implies the frst ′ part by applying b) to K = H . The second part states that given any prime power subgroup K ≤ G and any Sylow subgroup H ≤ G, it is possible to conjugate K to make it land in H. Student Question. Is the converse of part a) true? If H is a Sylow p-subgroup, is gHg−1 also a Sylow p-subgroup? Answer. The converse is essentially automatically true. In order to be a Sylow p-subgroup, the only requirement is being a subgroup of a certain size, and conjugating by an element evidently produces a subgroup of the same size. The impressive part is that given two arbitrary Sylow p-subgroups, they must in fact be conjugate! Sylow II confrms our intuition for refections in dihedral groups. Example 22.10 For D2n, every subgroup of size 2 is generated by a refection, and Sylow II indicates that all the refections are conjugate. 22.5 The Third Sylow Theorem The last Sylow theorem indicates the number of these (conjugate) subgroups. Theorem 22.11 (Sylow III) The number of Sylow p-subgroups of G divides n m = pe and is congruent to 1 modulo p. This theorem seems kind of weird, but is actually very useful. Example 22.12 Consider D5 and p = 2. The number of Sylow 2−subgroups is 5, which does divide 10/2 and is congruent to 1 modulo 2. The frst Sylow theorem indicates existence of Sylow subgroups, the second Sylow theorem indicates that all Sylow subgroups are related by conjugation, and the third provides strong (and kind of funky) constraints on the number of such subgroups. 22.6 Applications of the Sylow Theorems Now, we can look at a few diferent applications of these theorems, and we will see how powerful they are. 109 Lecture 22: The Sylow Theorems Example 22.13 Consider any group G such that |G| = 15 = 5 · 3. By Sylow III, for p = 5, the number of Sylow 5-groups divides 3 = 15/5, and is equal to 1 mod 5. In particular, the only possibility is #Sylow 5-groups = 1. So there is a unique H ≤ G such that |H| = 5. Since there is only one subgroup of size 5, Sylow II indicates that gHg−1 = H. That is, H is normal: the Sylow theorems indicate automatically that there is a normal subgroup of size 5. For p = 3, Sylow III states that the number of Sylow 3−subgroups divides 5 and is 1 mod 3, so it is also 1. Thus, there exists some unique K ⊴ G such that |K| = 3. Moreover, H ∩ K = {1}, since H ∼ = C5 and K = ∼ C3. Nontrivial elements of H have order 5, while elements of K have order 3, so they intersect only at the identity. So the Sylow theorems give, for free, nonintersecting normal subgroups of size 5 and 3 for every single group of size 15. That is quite impressive! 76 Recall the notion of a product group. Defnition 22.14 The product group H ×K is H ×K = {(h, k) : h ∈ H, k ∈ K} where (h, k) · (h ′ , k ′ ) = (hh ′ , kk ′ ). The product group is the group given by coordinate-wise multiplication. We claim that in Example 22.15, a group of order 15 must be isomorphic to a product of groups of order 3 and order 5. Proposition 22.15 (Example 22.15) Where |H| = 5 and |K| = 3, (a) the two subgroups commute: for h ∈ H and k ∈ K, hk = kh; (b) H ×K ∼ = G. Once H ×K ∼ = G, then we know that any group of order 15 is isomorphic to C5 ×C3. Proof. Part b) follows from part a). (a) For the frst claim, since K is normal, hkh−1 ∈ K, and thus, multiplying on the right by an element of K, hkh−1k−1 ∈ K. Similarly, kh−1k−1 ∈ H, since H is normal, but then hkh−1k−1 ∈ H. Thus, hkh−1k−1 ∈ H ∩ K = {1}, and since the intersection was just 1, so hkh−1k−1 = 1, and so hk = kh. (b) Consider the mapping f : H ×K − → G (h, k) 7− → hk. 76This was discussed on the homework. 110 Lecture 22: The Sylow Theorems We claim that this is a homomorphism. In particular, f((h, k) · (h ′ , k ′ )) = f((hh ′ , kk ′ )) = hh ′ kk ′ = hkh ′ k ′ = f((h, k)) · f((h ′ , k ′ )), where the second-to-last step comes from H and K commuting. In general, if we take two arbitrary subgroups and take this function, this would not be a group ho momorphism! It was extremely important that H and K commute, which is true since they are both normal. Next, we must check that f is in fact an isomorphism. Since H and K have trivial intersection, hk = 1 only when h = 1 and k = 1, so the kernel is ker(f) = {h, k : hk = 1 ∈ G} = {(1, 1)}. Since the kernel is trivial, f is injective. In addition, |H ×K| = |G| = 15, and so f must be bijective and thus an isomorphism. We have shown that any group G of order 15 is G ∼ = C5 ×C3. There is only one group of size 15. In particular, C15 = C5 ×C3. Student Question. How did we know that H and K were C5 and C3? Answer. Any group of prime order is cyclic; we proved this in a previous lecture. There is only one group up to isomorphism of order 15. For higher order groups, fnding the number of groups up to isomorphism can get tricky. The Sylow theorems make it much easier to start an argument for classifying diferent groups, since they can apply to any fnite group. For n = 15, there is only one isomorphism class. Guiding Question For groups such that |G| = pq, a product of two distinct primes, how many isomorphism classes of groups are there of size pq? Let’s think about n = 10. Example 22.16 Consider a group G such that |G| = 10 = 5 · 2. We know already that G is not unique; for example, D5 and C10 both have order 10 but are non-isomorphic. Proposition 22.17 There are two isomorphism classes: G ∼ = C5 ×C2, and G ∼ = C10. Proof. From Sylow III, the number of Sylow 5-groups divides 2 and is 1 modulo 5, so there is only one Sylow 5-group. So there is a normal subgroup K ⊴ G such that |K| = 5. Let x ∈ K be a generator for K, so that K = ⟨x⟩, where ord(x) = 5. 111 Lecture 22: The Sylow Theorems Take H to be some Sylow 2-group. Let H = ⟨y⟩, where the order of y is 2. Since K is normal and generated by x, yxy−1 ∈ K, so −1 r yxy = x for some exponent 1 ≤ r ≤ 4. Rearranged, we have r yx = x y. As before, since K has elements of order 5 and H has elements of order 3, the intersection is trivial: K ∩ H = {1}. i This implies that the possible x yj are distinct from each other.77 Therefore, the group G is i j G = {x y : 0 ≤ i ≤ 4, 0 ≤ j ≤ 1}, where all these elements are distinct. There are 10 such elements, so these elements must be the entire group G. The relations 5 2 x = y = 1 and r yx = x y completely determine the group operation! The exponent r entirely controls which group of size 10 we have. Which values of r work? Currently, we have that 1 ≤ r ≤ 4, so there are at most four diferent isomorphism classes for G. • If r = 2, then we would have 2 2 4 2 4 x = y x = yyx = yx y = x y = x , 5 2 r 4 3 by repeatedly using the relations x = y = 1; yx = x y. Here we have x = x , implying that x = 1. This is a contradiction, since x had order 5. So r = 2 is impossible. r • In general, if yx = x y, then 2 2 r x = y x = · · · = x , r −1 2 by running through the same calculations as above. So x 2 = 1, and we must have r = 1 mod 5. So r = 3 is impossible, since 9 is not 1 mod 5. • For r = 1, then H and K commute by defnition, and the same analysis as in 22.15 works, and we have G = C5 ×C2, which turns out to be isomorphic to C10. • For r = 4, we recognize these relations: G = D5. For this example, we used the Sylow theorems to narrow down the possibilities, and simply looked through the possibilities to determine the isomorphism classes of groups of order 10. For n = 10, because Sylow III did not restrict the number of 2-subgroups to be 1, only the 5-subgroup was necessarily normal, and so the analysis was more complicated and subtle than for n = 15. Since for p = 2, there could have been 1 or 5 subgroups (both these numbers divide 10/2 = 5 and are congruent to 1 modulo 2), we were able to obtain less information about 2−subgroups. In general, if |G| = pq, a product of two distinct primes, then • if q = ̸ 1 mod p, then G = Cp ×Cq = Cpq. i−i ′ j−j ′ 77Otherwise, if there are two nondistinct elements, we end up with x y = 1, implying that some element of K is the inverse of some element of H, which is not possible with a trivial intersection. 112 Lecture 22: The Sylow Theorems • if q = 1 mod p, then G = Cpq or a diferent non-abelian group.78 To prove this, we follow the same analysis as today. The frst step is to look at the Sylow p−groups and the Sylow q−groups. In the frst case, we argue that they are normal and commute, and we can show that G ∼ = Cp ×Cq . In the second case, using the relations, there end up only being two possibilities for r. 78For p = 2 and q = 5, the diferent non-abelian group is D5. 113 Lecture 23: Proving the Sylow Theorems 23 Proofs and Applications of the Sylow Theorems 23.1 Review Last time, we introduced the Sylow theorems. While they may be a lot to take in, the main takeaway is how general the Sylow theorems are. When provided with any fnite group, we automatically already know that there exist certain p-subgroups79 that must be conjugate, and additionally there is a strong constraint on the possible number of such subgroups. The applications for C15 and C10 discussed last lecture demonstrate how powerful these theorems can be. We restate the theorems briefy here: Theorem 23.1 (Sylow Theorems) Let G be a fnite group where e |G| = n = p m and gcd(p, m) = 1. The three parts of the theorem follow: e 1. Recall that a Sylow p-subgroup is a subgroup H ≤ G such that |H| = p . The frst theorem states that there always exists a Sylow p-subgroup. f 2. Given any K ≤ G where |K| = p , there exists some g ∈ G such that gKg−1 ≤ H. 3. The number of Sylow p-subgroups is a factor of m and congruent to 1 mod p. 23.2 Application: Decomposition of Finite Abelian Groups One application of the Sylow theorems is the decomposition of fnite abelian groups. Consider a fnite abelian group G such that the prime factorization of the order is e1 er |G| = p · · · p .
1 r Then we know that we have a Sylow subgroup Hi such that ei |Hi| = pi for each of these primes. Since G is abelian, conjugating a group produces the same group, so by Sylow II, these (abelian) subgroups Hi are unique for each prime. Theorem 23.2 Every abelian group G is isomorphic to a product of groups of prime power order. Using that G is abelian, if we take the product H1 × · · · × Hr, we can construct a homomorphism80 f : H1 × · · · × Hr − → G (x1, . . . , xr) 7− → x1 + · · · + xr. Lemma 23.3 The homomorphism f is an isomorphism. 79The size is the largest power of p that divides |G| 80Because G is abelian, we use + as the group operation. 114 Lecture 23: Proving the Sylow Theorems Proof. First, f is a homomorphism because G is abelian and the terms will commute when verifying the homomorphism property. It is necessary that G is abelian. 81 Next, we know that im(f) is a subgroup of G and also contains a copy of Hi for all i82: Hi ≤ im(f) ≤ G ei for all i. Thus, p divides |im(f)| for each i, and since they are relatively prime, the product i Y ei pi divides |im(f)|. This forces the image to be the same order as |G|, and thus they must be the same. We can conclude that f is surjective. Both the domain and image of f have the same size, so it is also injective and an isomorphism. As a result, the study of fnite abelian groups can be reduced to studying abelian p-groups. These are completely understood, and will potentially be covered more in 18.702! In contrast, non-abelian groups are complicated and not well understood. 23.3 Proof of Sylow Theorems The main idea to prove all of these theorems is to fnd a useful action of G on a set and exploit it. This is a continuation of what we have been doing in the last few weeks, in geometric situations with symmetries as well as with the conjugation action of G on itself. The striking part about these three proofs is that unlike rotational symmetries of the cube, where there are lots of sets to think about, such as vertices and faces and so on, here, there is no prior knowledge about G, and the only group action we have for any arbitrary group is G acting on itself, and not much else. Theorem 23.4 (Sylow I) Given G such that e |G| = n = p · m, e where p is the largest power of p (that is, gcd(p, m) = 1), then there exists a subgroup H ≤ G such that e |H| = p . Proof of Sylow I. Take G such that e |G| = p · m. e Let S be the subsets of G of size p and let n be the order of G. By basic combinatorics, there are n pe such subsets, so n |S| = . pe Let G act on S by left translations: given an element g ∈ G and a subset U ∈ S, we map U 7− → gU. e Our eventual goal is to fnd a subgroup of G of size p by looking at stabilizers, as they are always subgroups of G. We fnd the size of a stabilizer by trying to fnd an orbit of size m, as we then know that the stabilizer e will be order p . 83 We begin with some lemmas. The frst lemma provides information about the size of the set modulo p. 81Essentially, since G is abelian, there is really only one way to "combine" the Sylow p-subgroups. When |G| = 10 for a non-abelian group, we saw that the Sylow subgroups for 2 and 5 could combine in a diferent way to make D10. 82Take H1 ×{1}× · · · {1} to get H1 ≤ im(f ), for example. e 83The product of the size of an orbit and the size of the stabilizer is the size of the group G, which here is m · p . 115 Lecture 23: Proving the Sylow Theorems Lemma 23.5 e Where n = |G| = m · p , we have that n |S| = ̸= 0 (mod p). pe Furthermore, n ≡ m (mod p). pe Sketch of Proof. The proof is not particularly relevant to group theory and can be proved by expanding the binomial coefcient and showing that the number of powers of p in the numerator is the same as the denominator. Alternatively, one could expand (1 + x)n and look at it modulo p. e To reiterate, S consists of all subsets of G of size p , and these subsets do not have to be subgroups. Lemma 23.6 Suppose we have a subset U ∈ Sa , which is a subset of G. Also, let H be a subgroup of G that stabilizes U. Then, |H| divides |U|. aNote that U is an element of S but is itself also a subset of G, so U ⊂ G. Proof. Since H stabilizes U, for any h ∈ H, we know hU = U. In other words, for each u ∈ U, we have Hu ⊂ U. Equivalently, for each u ∈ U, the corresponding right coset of H is a subset of U. This implies that the right cosets partition U . Since the cosets have the same size, we know that |H| divides |U|. With these lemmas in hand, we can continue with the proof of the main theorem. The frst lemma tells us that |S| ̸= 0 (mod p). We know that the orbits partition S, so |S| = |O1| + · · · + |Or|. Since p does not divide the LHS, there must exist an orbit θ where gcd(p, |θ|) = 1. Let the size of θ be |θ| = k. Now, consider some element u of θ. By the counting formula, we also know that |G| = |θ| · |Stab(u)|. e e And so p m = k|Stab(u)| and p | |Stab(u)| because gcd(k, p) = 1. By the second lemma, |Stab(u)| divides e |u| = p . Thus, e |Stab(u)| = p and we have found a Sylow p-group. The proof of the second Sylow theorem is similar. 116 Lecture 23: Proving the Sylow Theorems Theorem 23.7 (Sylow II) There are two parts; part a) is what is usually referred to as the second Sylow theorem. ′ (a) Given H ≤ G, where H is a Sylow p-subgroup, any other Sylow p−subgroup H ≤ G is conjugate to ′ H; i.e. there exists g such that H = gHg−1 . d (b) Given any subgroup K ≤ G such that |K| = p , for any Sylow subgroup H, there exists g such that gKg−1 ≤ H.a aNotice that |K| does not have to be the maximal prime power, and can have order smaller than |H|. Every prime power order subgroup, up to conjugation, sits inside a Sylow subgroup. Proof of Sylow II. We approach this proof similarly, fnding a nice set and an action on it. Fix H to be a Sylow subgroup. Our set is X = G/H, the left cosets of H. The index of H is the same as |X|, so |X| = m. Let K be the subgroup we want to show is a subgroup of H up to conjugation, where |K| = pf . We will look at how K acts on X by left translation, the mapping: k(aH) 7− → kaH. We decompose into orbits, |X| = |O1| + · · · + |Or|. Note that these orbits are with respect to the action of K, not the action of G, as that would be transitive and we’d only have one orbit. We have that |Oi| divides f |K| = p , but p does not divide m. Thus this orbit decomposition can only work if some orbit O has size 1. In other words, there exists some coset aH that is fxed by all k ∈ K. Then, kaH = aH a −1kaH = H a −1ka ∈ H a −1Ka ≤ H which is what we needed to show. A lot of the work done in these proofs are choosing some set and action, then looking at the orbits and seeing what we can do what them. The third proof is similar. Theorem 23.8 (Sylow III) The number of Sylow p-subgroups of G divides n m = pe and is congruent to 1 modulo p. Proof of Sylow III. Our set will be Y as the set of Sylow p-subgroups of G. We will be trying to fnd the size of Y . G acts on Y by conjugation, H 7− → gHg−1 . By Sylow II, there is only one orbit. Pick a Sylow subgroup H ∈ Y . Then |G| = |Stab(Y )||orbit(H)| = |Y ||Stab(Y )|. This already tells us that |Y | divides |G| = n, but we can say more. The stabilizer here has a name, the normalizer of H. It turns out that H ≤ Stab(H) because for all h ∈ H, e e hHh−1 = H. So | Stab(H)| is divisible by p = |H|. The counting formula then says that |G| = p m = e |Y | · (p · stuf) which implies that |Y | divides m. The last part is showing that |Y | ≡ 1 (mod p). We now use the action of H on Y by conjugation. Fact 23.9 ′ ′ ′ Suppose we have another Sylow subgroup H ∈ Y , H is fxed by H if and only if H = H . In other words, under the action of H, there is only one fxed point. 117 Lecture 23: Proving the Sylow Theorems By looking at orbits, there is only one orbit of size 1 because there is only one fxed point. The rest are powers of p because the size of H is a power of p. Thus the decomposition into orbits looks like 2 3 Y = 1 + p + · · · + p + · · · + p + · · · ≡ 1 (mod p). Proof of fact. If we look at the stabilizer/normalizer, StabG(H ′ ) = N(H ′ ), we know that H ≤ N(H ′ ) because ′ ′ H is fxed by H, and that H ≤ N(H ′ ) by what we said above about normalizers. e ′ Now N(H ′ ) is a subgroup of G as well, so the largest power of p that divides N(H ′ ) can only be p . So H and H ′ −1 are Sylow subgroups of N(H ′ ) as well. By Sylow II on N(H ′ ), there exists n ∈ N(H ′ ) such that nH n = H. ′ −1 ′ ′ But then by the defnition of N, nH n = H , and so H = H . Given this fact, we are done with the third proof. 118 Lecture 24: Symmetric and Hermitian Forms 24 Bilinear Forms 24.1 Review Last week, we talked about the Sylow theorems, which are fundamental to the theory of fnite groups. 24.2 Bilinear Forms Throughout this class, we have been pivoting between group theory and linear algebra, and now we will return to some linear algebra. Today, we will be discussing the notion of bilinear forms. Let’s look at some examples frst and then provide the general defnition. For now, we will be working with a vector space V over F = R, and later on we will look at the case of F = C, the complex numbers. Let’s consider three examples of bilinear forms on R3 . Example 24.1 Consider these three diferent examples of mappings: R3 ×R3 − → R x1 y1 (1) x2 , y2 7− − → x1y1 + x2y2 + x3y3 x3 y3 (2) 7− − → x1y1 + 2x2y2 + 3x2y1 + 4x2y3 + 5x3y1 (3) 7− − → x1y1 + 2x2y1 + 2x1y2 + 3x2y2. These all take in pairs of vectors in R3 and return a real number. They all have the property that when keeping the y’s fxed, the mapping is "linear" in the x’s, and when keeping the x’s fxed, the mapping is "linear" in the y’s.84 In particular, there are no constant terms or terms that are squared or higher order in xi or yi. Defnition 24.2 A bilinear form is a function V ×V − → R (v, w) 7− → ⟨v, w⟩85 such that 1. ⟨v, cw⟩ = c⟨v, w⟩ 2. ⟨v, w1 + w2⟩ = ⟨v, w1⟩ + ⟨v, w2⟩ 3. ⟨cv, w⟩ = c⟨v, w⟩ 4. ⟨v1 + v2, w⟩ = ⟨v1, w⟩ + ⟨v2, w⟩. Requirements (1) and (2) are linearity in the second variable w, and requirements (3) and (4) are linearity in the frst variable v. The angle brackets ⟨·, ·⟩ are how a bilinear form is usually denoted. A bilinear form takes in two inputs and returns a real number in a way that is linear in either of its two inputs.86 Intuitively, a bilinear form looks like Example 24.1. 84The general idea of a bilinear form is that it is linear when varying in x (and keeping y fxed) and linear when varying in y (and keeping x fxed); hence, it is linear in two diferent variables, independently, so it is "bilinear." However, being bilinear is not the same as being linear; for example, if both x and y were doubled, the output would quadruple. 86A "trilinear form" would also be possible. 119 Lecture 24: Symmetric and Hermitian Forms Defnition 24.3 A bilinear form is symmetric if ⟨v, w⟩ = ⟨w, v⟩ for all v, w ∈ V. For instance, (1) and (3) in Example 24.1 are symmetric, but (2) is not, by looking at the coefcients. Linear transformations from Rn − → Rn can be written down explicitly using matrices. In a similar way, bilinear forms can also be described concretely using matrices. Consider the special case where V = Rn . Then the dot product is a symmetric bilinear form. More generally, given any matrix A ∈ Matn×nR, the mapping ⟨x, y⟩ := x T Ay ∈ R turns out to describe a bilinear form, satisfying the four properties.87 Proposition 24.4 Given a symmetric matrix, the corresponding bilinear form is a symmetric bilinear form. Proof. A matrix A ∈ Matn×nR is symmetric if AT = A, and in this case, ⟨x, y⟩ = x T Ay and T AT ⟨y, x⟩ = y T Ax = (y T Ax)T = x y = x T Ay = ⟨x, y⟩. For every matrix, there is an associated bilinear form, and for every symmetric matrix, there is an associated symmetric bilinear form. It turns out that every bilinear form arises in this manner. Proposition 24.5 Every bilinear form ⟨·, ·⟩ on Rn arises from a matrix A. That is, there exists some A such that ⟨x, y⟩ = x T Ay. Moreover, the form ⟨·, ·⟩ is symmetric if and only if A is symmetric. So there is a bijective correspondence between bilinear forms and n×n matrices. In particular, for each example in 24.1, there is an associated matrix. Example 24.6 The associated matrices come from the coefcients, and can be verifed by simply carrying out the multipli cation process. 1. A = 1 0 0 1 0 0 2. A = 0 1 3 0 2 0 1 0 4 3. A = 5 1 2 0 2 3 0 0 0 0 0 0 87We won’t verify this, but the properties follow from the way matrix multiplication works. 120 Lecture 24: Symmetric and Hermitian Forms From this example, we see that the matrix entry Aij is the coefcient of xiyj . Proof of Proposition 24.5. Given a bilinear form, we want to produce the corresponding matrix. Let V = Rn , and let the standard basis vectors be 1 0 0 ⃗ e1 = 0 . . . ,⃗ e2 = 1 . . . , · · · , ⃗ en = 0 . . . . 0 0 1 Any other column vector can be written as a linear combination of these basis vectors: x1 . . . xn = x1⃗ e1 + · · · + xn⃗ en. X In order to produce the matrix for the bilinear form, we look at the form evaluated on pairs of basis vectors. Let aij = ⟨⃗ ei, ⃗ ej ⟩. X Now, placing these coefcients in a matrix, take A = (aij )i,j=1,··· ,n. To verify that this matrix actually produces the same result as the bilinear form on any two pairs of vectors, take ⃗ x, ⃗ y ∈ Rn , and use bilinearity on both coordinates x and y. n n + X ⟨⃗ x, ⃗ y⟩ = xi⃗ ei, yj⃗ ej i=1 j=1 n X X = xi⟨⃗ ei, ⃗ y⟩ i=1 n n X X = xi⟨⃗ ei,⃗ ej ⟩yj i=1 j=1 n n = xiaij yj i=1 j=1 = ⃗ x T A⃗ y. In addition, if and only if ⟨·, ·⟩ is symmetric, aij = aji, which is precisely the condition that A is symmetric. From the bilinear hypothesis, the bilinear form on any two vectors can be written in terms of the form on a basis, which provides us the matrix. The upshot is that when V = Rn , the information of a bilinear form can be encoded in a matrix. Like when studying linear transformations, we do not have to restrict ourselves to only Rn . More generally, for any vector space V along with a basis {v1, · · · , vn} of V, a (symmetric) bilinear form on V corresponds with a (symmetric) matrix A ∈ Matn×n(R). 88 Guiding Question What is the correspondence between a bilinear form on a vector space and the matrix, given a basis? 88The way this form depends on the basis chosen will difer from the case of linear transformations, and will be discussed later in this lecture. 121 Lecture 24: Symmetric and Hermitian Forms In some sense, a basis is simply a linear isomorphism B : Rn − → V. A basis provides a dictionary between vectors in V and column vectors in Rn . Given two vectors, the result of the bilinear form will be ⟨⃗ v, w ⃗⟩ = ⃗ x T A⃗ y, where B⃗ x = ⃗ v and B⃗ y = w.
⃗ How can we fnd the entries of A? We take aij = ⟨⃗ vi,⃗ vj ⟩, and the same argument as in the proof of Proposition 24.5 holds, by using bilinearity. 24.3 Change of Basis As always, we need to be careful about what basis we are working in and what efects the basis has. Note 24.7 (Warning!) A linear operator T : V − → V corresponds to an n×n matrix by picking a basis: linear operator T : V − → V ⇝ n×n matrix Today, we saw that a bilinear form on V also corresponds to an n×n matrix by picking a matrix: bilinear form on V ⇝ n×n matrix But in fact, these two correspondences act extremely diferently! For a linear transformation, where the change of basis matrix is Q, the change of basis formula takes P 7− → QP Q−1 . Now, we can explore a change of basis for a bilinear form instead. Pick two bases B : Rn − → V, B ′ : Rn − → V for V, and consider a bilinear form ⟨·, ·⟩V on V. The two bases are related by some invertible matrix P such that B ′ = BP and P ∈ GLn(R). 89 Using B, there is one bilinear form ⟨·, ·⟩ associated with some matrix A and using B ′ , there is another bilinear form ⟨·, ·⟩ ′ associated with a matrix A ′ . Rn Rn V B B ′ P Given two column vectors⃗ x, ⃗ y ∈ Rn , the result in B ′ is ⟨⃗ x, ⃗ y⟩ = ⟨B ′ ⃗ x, B ′ ⃗ y⟩V = ⟨BP⃗ x, BP ⃗ y⟩V . This is the same as ⟨P ⃗ x, P⃗ y⟩ = (P ⃗ x)T A(Py) = ⃗ x T P T AP y. So the matrices are related by A ′ = P T AP , which is not P −1AP. Changing basis for bilinear forms, unlike linear transformations, does not change the matrix by conjugation! If A is a symmetric matrix, then A ′ is also a symmetric matrix, which is expected. That’s kind of alarming. The same question for linear mappings can be asked in this situation. 89The columns of P indicate how to write one basis in terms of the other. 122 Lecture 24: Symmetric and Hermitian Forms Guiding Question Given V and ⟨·, ·⟩V , can we pick a basis B = {⃗ v1, · · · , ⃗ vn} of V such that A is as nice as possible? For linear mappings, we ended up with the Jordan normal form. It turns out that the answer for bilinear forms is very nice! We will discuss this in the future. 24.4 Bilinear Forms over CC The defnitions provided so far generally work over any feld. The standard dot product, which is a typical example of a bilinear form, has an additional property. Defnition 24.8 A dot, or inner product is a symmetric bilinear form such that ⟨x, x⟩≥ 0, a and if x ̸= 0, then ⟨x, x⟩ > 0. aThis condition is called being positive defnite. p √ We can use an inner product to measure distances and lengths in a vector space. In Rn , ||⃗ v|| = ⟨⃗ v,⃗ v⟩ = ⃗ v · ⃗ v. Guiding Question Can we extend this to the complex numbers, when F = C? First, let’s extend the notion of a dot product. We would like to do so in a way that captures our notion of distance. Naively setting the dot product in the same way as over R results in a complex number, which does not measure distance in a way that we would prefer. In C, the length of a complex number z is zz, which is the distance from the complex number z to the origin in the complex plane. The analogue of an inner product over C will coincide with this defnition of distance and use complex conjugation. Defnition 24.9 The standard Hermitian form on Cn looks almost like the normal inner product, but with some complex conjugates thrown in. We have ⟨⃗ x, ⃗ y⟩ = x1y1 + x2y2 + · · · + xnyn ∈ C. In particular, T ⟨⃗ x, ⃗ y⟩ = ⃗ x ⃗ y ∈ C. Once we do this, we get ⟨⃗ x, ⃗ x⟩ = x1x1 + x2x2 + · · · . = |x1|2 + |x2|2 + · · · , which is actually a non-negative real number! So we prefer to use this Hermitian form over the complex numbers, as it can capture some notion of distance. In the defnition of the standard Hermitian form, we took the transpose and then the complex conjugate of every entry. This is a move we will do over and over. 123 Lecture 24: Symmetric and Hermitian Forms Defnition 24.10 For M ∈ Matm×n(C), the adjoint matrix is M ∗ := M T ∈ Matn×m(C). It behaves very much like taking the transpose does: (AB)∗ = B∗A∗ . Then the equation from before becomes ∗ ⃗ ⟨⃗ x, ⃗ y⟩ = ⃗ x y ∈ CC. Notice that for α ∈ C, ⟨α⃗ x, ⃗ y⟩ = ̸ α⟨⃗ x, ⃗ y⟩, so it is not bilinear in the frst entry! We instead get ⟨α⃗ x, ⃗ y⟩ = α⟨⃗ x, ⃗ y⟩, so it is linear in the second factor but nonlinear (it is only linear up to complex conjugation) in the frst factor. This leads us to our last defnition for today. We can generalize the properties of the standard Hermitian form for a complex vector space. Defnition 24.11 For V a vector space over F = C, then a Hermitian form is a function from V ×V − → C (⃗ v, w ⃗) 7− → ⟨⃗ v, w ⃗⟩ where 1. ⟨⃗ v, w ⃗1 + w ⃗2⟩ = ⟨⃗ v, w ⃗1⟩ + ⟨⃗ v, w ⃗2⟩ 2. ⟨⃗ v, α ⃗ w⟩ = α⟨⃗ v, w ⃗⟩ 3. ⟨ ⃗v⟩ = ⟨⃗w⟩.
w,⃗ v, ⃗ A Hermitian form is like a symmetric form, except instead of being symmetric, it is symmetric with a conjugation thrown in. Notice that ⟨α⃗ v, ⃗ w, α⃗ w⟩ = ⟨ ⃗ v⟩ = α⟨ ⃗v⟩ w,⃗ = α⟨ ⃗v⟩ w,⃗ = α⟨⃗ v, w ⃗ ⟩. So the Hermitian product of a vector with itself is in R. 124 Lecture 25: Orthogonality 25 Orthogonality 25.1 Review: Bilinear Forms We discussed bilinear forms last time, which was a function that took two vectors as input, and gave a scalar as an output. It was linear in both of the inputs. For now, we will only be interested in symmetric bilinear forms because they model after the dot product. On real vectors, every bilinear form can be written as: » # » # » » ⟨ # x, y ⟩ = x T A # y . The form is symmetric if and only if A is symmetric. We may also refer to a bilinear form as a ‘pairing’. 25.2 Hermitian Forms When we worked over a complex vector space, we discussed the Hermitian form, the complex version of a symmetric bilinear form. However, symmetry did not work as normal. They had a complex conjugate in » # » » # » » # » # » » w⟩ = ⟨# ∗ # their relation: ⟨# v , w, v ⟩ The standard Hermitian form was defned to be ⟨ # x, y ⟩ = x y . In particular, Hermitian forms are not exactly linear. The second term is linear, but when we scale the frst term, we are scaling the output by the complex conjugate. The following chart summarizes the comparison between symmetric bilinear forms over the reals and Hermitian forms. Although they have subtle diferences, we will study them together. Field Canonical Example Symmetry Matrix Change of basis R dot product ⟨ # » # » » # » v , w⟩ = ⟨# w, v ⟩ AT = A P T AP C Standard Hermitian form ⟨ # » # » » # » v , w⟩ = ⟨# w, v ⟩ ? ? Now we will fgure out what goes in the two remaining entries of the table. In order to fnd the matrix for a Hermitian form on V, a vector space over C, the process is analogous to fnding the matrix for a symmetric #» form on a vector space over R. First, pick a basis v #» 1, . . . , vn of V. Then, set A = (aij )i,j=1,··· ,n, where = ⟨ # » # » aij v i, v j ⟩. If # » # » # » v = x1 v 1 + · · · + xn v n and # » # » # » w = y1 v 1 + · · · + yn v n, by using the almost-bilinearity of the Hermitian form and expanding the Hermitian form in the same way that bilinearity was used to expand the bilinear form, we get » # » # » » ⟨ # v , w⟩ = x ∗ A# y , where there is a conjugate transpose instead of a transpose. Then, for every entry of the matrix A, » # » » # » aij = ⟨# v i, v j ⟩ = ⟨ # v j , v i⟩ = aji, and so A∗ = A. Defnition 25.1 A matrix A is called a Hermitian matrix if A∗ = A. The upshot is that giving a Hermitian form is essentially equivalent to providing a Hermitian matrix on Cn : Hermitian form on Cn ← → Hermitian matrix A90 Similarly, one can show that the change of basis formula is given by A ′ = P ∗AP. 90Use the standard basis of Cn , e1, · · · , en. 125 Lecture 25: Orthogonality Example 25.2 (n = 2) For a Hermitian matrix A = 5 2 − 2i 2 + 2i 3 , the associated Hermitian form is ⟨# » x , # » y ⟩ = ⃗ x ∗ A⃗ y = 5x1y1 + 3x2y2 + (2 + 2i)x1y2 + (2 − 2i)x2y1, simply by evaluating the matrix product. In particular, when ⃗ x = ⃗ y, then the Hermitian inner product is actually a real number! The Hermitian inner product of ⃗ x with itself is ⟨⃗ x, ⃗ x⟩ = 5|x1|2 + 3|x2|2 + Re((2 + 2i)(x1x2)) ∈ R. It turns out Hermitian matrices have very nice properties compared to random complex matrices. Let’s see one of them now. Claim 25.3. A Hermitian matrix always has real eigenvalues. Proof. An eigenvalue λ ∈ C of a Hermitian matrix A satisfes A⃗ v = λ⃗ v for some ⃗ v ∈ Cn . By the Hermitian property, v ∗ Av ∈ R, but because it is an eigenvector, this is equal to ∗ v ∗ λv = λ(v v), ∗ where v v is a nonzero real number. Thus, v ∗Av λ = ∈ R.
∗ v v Not only is the eigenvalue real, λ can be obtained by comparing the value of the Hermitian form to the value of the standard Hermitian form. From now on, we will study symmetric bilinear forms on the real numbers and Hermitian forms on the complex numbers in parallel. They have very similar properties. One idea that carries over is orthogonal matrices. Example 25.4 Consider R equipped with the standard dot product. Let M ∈ Matn×n(R). Recall that we had several ways of describing that M was orthogonal. The following properties are all equivalent: M is orthogonal ⇐ ⇒ M⃗ x · M⃗ y = ⃗ x · y ⃗ ⇐ ⇒ M T M = In ( 1 if i = j » #» » #» ⇐ ⇒ for column vectors v # i, vj of M, ⟨v # i, vj ⟩ = 0 otherwise. The last condition says that the columns of M are orthonormal. A similar type of matrix can be defned for C with the standard Hermitian form. 126 Lecture 25: Orthogonality Defnition 25.5 Let V = Cn . The matrix M is called unitary if it satisfes any of the following equivalent conditions. M is unitary ⇐ ⇒ ⟨M⃗ x, M⃗ y⟩ = ⟨⃗ x, ⃗ y⟩ ⇐ ⇒ M ∗ M = In; M −1 = M ∗ 1 if i = j » #» # »∗ #» ⇐ ⇒ for column vectors v # i, vj of M, vi vj = 0 otherwise. ( What you can do in one world is very parallel to what you can do in the other world. 25.3 Orthogonality Consider V real with ⟨·, ·⟩ symmetric, or V complex with ⟨·, ·⟩ Hermitian. Defnition 25.6 A vector ⃗ v is orthogonal to w ⃗ if ⟨⃗ v, w ⃗⟩ = 0. Also, for W ⊂ V, a vector ⃗ v ⊥ aW if ⟨⃗ v, w ⃗⟩ = 0 for all w ⃗ ∈ W. aThis is the symbol representing orthogonality If ⟨·, ·⟩ is not the standard inner product, this idea of "orthogonality" does not necessarily correspond with geometric intuition. Example 25.7 −1 0 0 0 1 0 1 0 0 0 Let A = , and ⃗ v = . Then ⟨⃗ v,⃗ v⟩ = 0, so ⃗ v is orthogonal to itself. 0 0 1 0 0 0 0 0 1 1 This form comes up a lot when studying special relativity, but does not necessarily correspond to our geometric intuition of what "orthogonality" means. One thing that we do a lot of in the geometric world is that we take a subspace W and then look at the vectors that are orthogonal to it. Defnition 25.8 For a subspace W ⊂ V, the orthogonal complement is W ⊥ = {⃗ v ∈ V such that ⃗ v ⊥ W }. Example 25.9 Consider V = R3 , and W as a plane. Then W ⊥ is a line perpendicular to W . In this case, W ⊥ is a complement to W, and R3 = W ⊕ W ⊥ . In general, there are many possible complements, or ways to extend a basis of a subspace to the whole vector space, but the dot product picks out a specifc one. Guiding Question For a general bilinear form, when can we decompose V into the sum of a subpace and its orthogonal complement? It is possible to some extent, but we need to be careful. For example, ⃗ v = ̸ 0 can be perpendicular to all of V. In ′ ′ ⊥ particular, taking A = , ⃗ v ⊥ ⃗ v for any ⃗ v,⃗ v ∈ V. Thus, for any ⃗ v,⃗ v = V. 127 Lecture 25: Orthogonality Defnition 25.10 The null space is N = {⃗ v ∈ V ⊥ : ⃗ v = V } ⊆ V. If A = In, the null space is N = {⃗ 0}, but when A = 0, the null space is N = V. Defnition 25.11 Given a vector space V and a bilinear form ⟨·, ·⟩, if N = {⃗ 0}, (V, ⟨·, ·⟩) is called non-degenerate. Given the matrix of a form, how is it possible to tell whether the bilinear form is non-degenerate? Proposition 25.12 A form on a vector space (V, ⟨·, ·⟩) is non-degenerate if and only if the matrix of the form, A, is invertible, which is when det A ̸= 0. So when A = In, it is non-degenerate, but when A = 0, it is extremely degenerate. In matrix form, v ∈ N if ∗/T A# » and only if w ⃗ v = 091 for all w ⃗ ∈ V , which is equivalent to saying that A⃗ v = ⃗ 0, and so ⃗ v ∈ ker A. Consider the restriction of the form to W, ⟨·, ·⟩|W : W ×W → R or C. It can happen that ⟨·, ·⟩|W can be degenerate, even if ⟨·, ·⟩ is non-degenerate. Example 25.13 −1 0 0 0 0 Let V = R4 . The matrix from before, A = 0 1 0 0 1 0 , is non-degenerate but a bit weird since it had 0 0 0 0 1 1 0 that property that a vector could be orthogonal to itself. However, let v = and consider 0 1 W = Span(v). Then, W ×W − → R ⟨a⃗ v, a⃗ v⟩ = 0, and the restriction of the form to W is identically zero and is degenerate. Given a vector space V, a form ⟨·, ·⟩, and a subspace W ⊂ V, the restriction of the form ⟨·, ·⟩|W is non-degenerate ′ if and only if for all non-zero w ⃗ ∈ W, there exists some w ⃗ ̸= w ⃗ such that ⟨w, ⃗ w ⃗ ′ ⟩̸= 0. In particular, this is equivalent to saying that W ∩ W ⊥ = {⃗ 0}, by the defnition of W ⊥ . If there were a vector both in W and W ⊥ , it ′ would not be possible to fnd such a w ⃗ , since the inner product with w ⃗ would always be zero since it is in W ⊥ . Theorem 25.14 If ⟨·, ·⟩|W is non-degenerate, then V = W ⊕ W ⊥ is a direct sum of W and its orthogonal space. As a reminder, there are several equivalent ways of thinking about the direct sum. If V = W ⊕ W ⊥ , then the following are all true ′ ′ 1. If w1, · · · , wk is a basis for W, and w1, · · · , wj is a basis for W ⊥ , then gluing them together gets a basis ′ ′ {w1, · · · , wk, w · · · , w } for V. 1, j 2. Every ⃗ v ∈ V can be written uniquely as ⃗ v = w ⃗ + ⃗ u where w ⃗ ∈ W and ⃗ u ∈ W ⊥ . 91Depending on whether we consider R or C, we take either the conjugate transpose or the transpose. 128 Lecture 25: Orthogonality 3. The intersection is W ∩ W ⊥ = {⃗ 0} and V = W + W ⊥ . These are all diferent ways of talking about the way V has been split here. It is not always the case that W and W ⊥ direct sum to V. Once the non-degeneracy condition has been encoded into the restriction of the form to W, a splitting can be found. 129 Lecture 26: The Projection Formula 26 The Projection Formula 26.1 Review: Symmetric and Hermitian Forms Last time, we were talking about diferent kinds of pairings or bilinear forms on vector spaces. In particular, we will be studying two cases in parallel: vector spaces V over R with symmetric forms on them, and vector spaces over C, with Hermitian forms on them. A Hermitian form is almost symmetric, with a complex conjugate thrown in. Then, we discussed the idea of vectors being orthogonal to each other with respect to the form if the pairing is zero. A form is non-degenerate if and only if the space of vectors orthogonal to the entire vector space V is {0}, so there are no nonzero vectors orthogonal to all other vectors. Such a vector lies in the kernel of the matrix of the form, so a matrix with nonzero determinant will correspond to a non-degenerate form. 26.2 Orthogonality Recall this theorem about the restriction of a bilinear form to a subspace. We’ll prove it now. Theorem 26.1 Let W ⊆ V. If ⟨·, ·⟩|W is non-degenerate on W, then V = W ⊕ W ⊥ , which means that every vector v ∈ V » # » is equal to w # + u uniquely, where w ∈ W, u ∈ W ⊥ . 0 1 It is possible for the restriction of a non-degenerate form to be degenerate; for example the form A = 1 0 is non-degenerate but is just given by A ′ = 0 when W = Span(⃗ e1), which is clearly degenerate. Proof. If ⟨·, ·⟩|W is non-degenerate, then W ∩ W ⊥ = {0}. We have W ⊕ W ⊥ ⊂ V, so it sufces to show that V ⊂ W ⊕ W ⊥ . Pick a basis of W, {w1, . . . , wk}, and defne a linear transformation φ : V − → Ck ⃗ v 7− → (⟨w1, v, ⟩, . . . , ⟨wk, v⟩). This is a linear transformation just by the properties of a Hermitian form. The kernel is ker(φ) = W ⊥ , since W = Span{w ⃗i}. Also, dim im φ ≤ k = dim W, so by the dimension formula, dim V = dim ker φ + dim im φ ≤ dim W ⊥ + dim W. Consider the mapping W ⊕ W ⊥ − → V (w, u) 7− → w + u. It has kernel {0}, since W ∩ W ⊥ = {0}, so dim W + dim W ⊥ ≤ dim V, and thus dim W + dim W ⊥ = dim V and therefore V = W ⊕ W ⊥ . To emphasize, the geometric version of this with respect to the dot product feels obvious and works in most cases. For general forms, we have to have this condition that our form is non-degenerate on the subspace. The splitting V = W ⊕ W ⊥ is helpful, in particular, for inductive arguments, because it is possible to reduce some property of V to being true on W and W ⊥ . 130 Lecture 26: The Projection Formula 26.3 Orthogonal Bases By applying a change of basis, it is always possible to put an arbitrary matrix into Jordan normal form, and if there are distinct eigenvalues, it is in fact possible to diagonalize it. What about the matrix of a bilinear form? Guiding Question Given a vector space V and a bilinear form ⟨·, ·⟩, how simple can we get the form to be? First, it is always possible to fnd a basis orthogonal with respect to the bilinear form. Theorem 26.2 For a symmetric or Hermitian form ⟨·, ·⟩, the vector space V has an orthogonal basis {v1, · · · , vn}, which is when ⟨vi, vj ⟩ = 0 for i ̸= j. The matrix for the pairing in the basis will then be diagonal, since it is given by the inner product from the form. Proof. To prove this, induct on dim V = n. • Case 1. There is some u such that ⟨u, u⟩̸= 0. Then, the one-dimensional subspace W = Span(u), ⟨·, ·⟩|W is non-degenerate. By induction, W ⊥ has an orthogonal basis {v2, · · · , vn}, so {u, v2, · · · , vn} is an orthogonal basis for V. • Case 2. Otherwise, for every v ∈ V, ⟨v, v⟩ = 0. This is a very strong constraint on the form, and in fact it forces ⟨v, w⟩ = 0 for all v, w, which forces any basis to be an orthogonal basis. To see this, consider the inner product on a sum of two vectors with itself: 0 = ⟨v + w, v + w⟩ = ⟨v, v⟩ + ⟨w, w⟩ + ⟨v, w⟩ + ⟨w, v⟩ = 2⟨v, w⟩. When F = R, we have ⟨v, w⟩ = 0, by the symmetry of the form. Otherwise, for F = C, Re(⟨v, w⟩) = 0, and the same process can also be done for v and iw to show that ⟨v, w⟩ = 0. Then the inner product is 0 on any two vectors so every basis is orthogonal. We can simplify the basis even further. Corollary 26.3 In fact, V has an orthogonal basis {v1, · · · , vk} where ⟨vi, vi⟩ = 1, −1, or 0. Proof. Take an orthogonal basis {x1, · · · , xk}. Consider ⟨xi, xi⟩, which is a real number. • If the pairing is 0, then let vi = xi. 1 ⟨xi,xi ⟩ • Otherwise, we can normalize and take vi = √ xi; then ⟨vi, vi⟩ = |⟨xi,xi ⟩| , so it will be 1 or -1 |⟨xi,xi⟩| depending on the sign of ⟨xi, xi⟩. 131 Lecture 26: The Projection Formula In particular, if ⟨·, ·⟩ is non-degenerate, only ±1 occur. Also, if ⟨·, ·⟩ is positive defnite, by defnition, ⟨v, v⟩ > 0 if v ̸= 0, so only +1s occur, so in that basis, the form looks just like the dot product or the standard Hermitian product. The following claim will be shown in the upcoming problem set. Claim 26.4 (Sylvester’s Law). In fact, given V and ⟨·, ·⟩, the number of 1s, the number of -1s, and the number of 0s that occur in the diagonal form are determined by V and ⟨·, ·⟩, and not by the choice of orthogonal basis.92 This is called Sylvester’s Law, and the number of 1s, -1s, and 0s is called the signature of the form. For example, in the form used in special relativity, −1 1 , the signature is (3, 1, 0). 1 1 In matrix form, the corollary states that for a symmetric matrix A ∈ Matn×n(R), for which AT = A, there exists some matrix P ∈ GLn(R) such that P T AP is a diagonal matrix with all 1s, −1s, or 0s on the diagonal: P T AP = 1 .. . −1 .. . 0 ... . If A is positive defnite, which is when xT Ax > 0, there exists P such that P T AP = In implies that A = QT Q, where Q = P −1 . The statement is similar for complex matrices, where we replace the transposes with adjoints. 26.4 Projection Formula Consider a vector space V and a form ⟨·, ·⟩, as well as a subspace W for which ⟨·, ·⟩|W is non-degenerate. By Theorem 26.1, V = W ⊕ W ⊥ such that v = w + u. Guiding Question How can we compute w and u? To do so, we use the orthogonal projection. We want a map π : V − → W v 7− → w, so that v = π(v) ⊥ W.93 92They are similar to eigenvalues in that while there are many choices of orthogonal basis, the number of 1s, -1s, and 0s are not dependent on the particular basis. 93This is an extremely useful application of linear algebra! In geometric situations, the vector w is the vector closest to v of the vector in the plane, and perhaps these vectors are in a vector space of data points. Finding a formula for w explicitly is called least-squares regression. 132 Lecture 26: The Projection Formula Assuming there exists an orthogonal basis {w1, · · · , wk} for W , the formula for π is simple.94 The vector can be written as v = cw1 + · · · + cwk + u, where u ⊥ W. Then for all i, ⟨wi, v⟩ = 0 + · · · + 0 + ci⟨wi, wi⟩ + 0 + · · · + 0, so ⟨wi, v⟩ ci = . ⟨wi, wi⟩ It is not possible for ⟨wi, wi⟩ = 0, because the form would be degenerate. In fact, this formula is useful when W = V , because it provides a formula for the coordinates of some vector with respect to the orthogonal basis. Example 26.5 Let V = R3 and ⟨·, ·⟩ be the dot product. Then let W be the span of w ⃗1 = (1, 1, 1)T and w ⃗1 = (1, 1, −2)T . The pairings are ⟨w1, w1⟩ = 3, ⟨w2, w2⟩ = 6, ⟨w1, v⟩ = 6, and ⟨w2, v⟩ = −3. The projection of (1, 2, 3) is then 3/2 6 1 π(v) = w1 − w2 = 3/2 . 3 2 3 To verify, v − π(v) = (−1/2, 1/2, 0), which is orthogonal both to w1 and w2. 94Once we’ve developed the machinery for bilinear forms, these ideas become a lot simpler! 133 Lecture 27: Euclidean and Hermitian Spaces 27 Euclidean and Hermitian Spaces 27.1 Review: Orthogonal Projection Last time, we ended by talking about orthogonal projections and splittings such that V = W ⊕ W ⊥ . Specifcally, suppose we have a vector space V with a bilinear form ⟨·, ·⟩ as well as a subspace W such that the restriction ⟨·, ·⟩|W is non-degenerate. Then, given a orthogonal basis of W = Span{w1, . . . , wk}, we were able to write a formula for the projection onto W : ⟨w1, v⟩ ⟨wk, v⟩ proj(v) = w1 + · · · + wk. ⟨w1, w1⟩ ⟨wk, wk⟩ and we had that v = proj(v) + u, where proj(v) ∈ W and u ∈ W ⊥ . 27.2 Euclidean and Hermitian Spaces In order to evaluate the projection formula, we need an orthogonal basis of a subspace. How do we calculate this? To start, we will be talking about Euclidean and Hermitian spaces. Recall that a pairing ⟨·, ·⟩ is positive defnite if ⟨v, v⟩ > 0 for all v ̸= 0. Defnition 27.1 A Euclidean space is a real vector space V and a symmetric bilinear form ⟨·, ·⟩ such that ⟨·, ·⟩ is positive defnite. Analogously, a Hermitian space is a complex vector space V and a Hermitian form ⟨·, ·⟩ such that ⟨·, ·⟩ is positive defnite. These spaces have the following nice property. Theorem 27.2 If V is Euclidean or Hermitian, then there exists an orthonormal basis {v1, · · · , vn} for V such that ⟨vi, vj ⟩ = 0 and ⟨vi, vi⟩ = 1. In particular, the pairing ⟨·, ·⟩ looks like the dot product or the standard Hermitian product in this basis. Proof. From last time, we saw that for any pairing, there exists an orthogonal basis where all of the self-pairings were either 1, 0, or −1. By defnition, all of the self-pairings must be 1 when the form is positive defnite. Furthermore, we no longer have the case where a form can be degenerate on a subspace. Claim 27.3. For any W ⊆ V and ⟨·, ·⟩|W , the restriction ⟨·, ·⟩|W is always nondegenerate. ′ Proof. For each w ∈ W, we need to fnd a w ∈ W such that ⟨w, w ′ ⟩ = ̸ 0. By the positive defniteness of the ′ pairing, we can just take w = w. This means that we can inherit all the properties that we showed last time about non-degenerate subspaces. In particular, we can always perform the orthogonal projection on any subspace of a Euclidean/Hermitian space. 27.3 Gram-Schmidt Algorithm As an application, we have the Gram-Schmidt algorithm for fnding an orthonormal basis. Take a Euclidean or Hermitian vector space V and any basis {v1, · · · , vn} a basis for V. In order to build an orthonormal basis {u1, · · · , un}, we inductively build {ui} such that Span{u1} = Span{v1}, Span{u1, u2} = Span{v1, v2}, and so on. Let Vk = Span{v1, · · · , vk}. • Step 0. Our frst vector must just be a scaled version of v1. We have to scale it such that ⟨u1, u1⟩ = 1. Let 1 u1 := p v1. ⟨v1, v1⟩ Then {u1} is a basis for V1 and ⟨u1, u1⟩ = 1. 134 Lecture 27: Euclidean and Hermitian Spaces • Step 1. Set x2 = projV1 v2, which is the projection of v2 onto V1, and let y2 = v2 − x2, which is the orthogonal portion of the vector v2. Then, let 1 u2 = p y2 ⟨y2, y2⟩ be the result of normalizing y2. Then ⟨u1, u2⟩ = 0, since the portion of v2 spanned by u1 has been subtracted of, and ⟨u2, u2⟩ = 1. Thus {u1, u2} is an orthonormal basis of V2. • Step k. Assume that {u1, · · · , uk} is a basis for Vk. Then, let xk+1 = projVk (vk+1)yk, the projection onto Vk, and let yk+1 = vk+1 − xk+1, the orthogonal portion of vk+1. Then let 1 = .
uk+1 ⟨yk+1, yk+1⟩ By induction, {u1, . . . , uk+1} is an orthonormal basis for Vk+1. At every step, the projection formula is used, until a basis is found. Note that whenever we are calculating the projection, we conveniently have an orthogonal (even more, orthonormal) basis of the previous subspace to use the projection formula. Specifcally, on step k the projection formula simplifes to: proj (vk+1) = ⟨u1, vk+1⟩u1 + · · · + ⟨uk, vk+1⟩uk.
vk We can interpret the algorithm in matrix form as well. Take M ∈ GLn(R). The columns of M, {v1, · · · , vn}, are a basis for Rn . The Gram-Schmidt algorithm says that we can take this basis and turn it into an orthonormal basis: u1 = a11v1 u2 = a12v1 + a22v2 u3 = a13v1 + a23v2 + a33v3, and so on, where {u1, · · · , un} is an orthonormal basis of Rn . The upshot is that there exists an upper triangular matrix A and an orthogonal matrix B such that M = AB. A is the matrix derived from putting ui as columns, and B is the inverse of the matrix derived from the aij values. Student Question. What sense of orthogonal are we using in the algorithm? Answer. This algorithm refers to orthogonal bases with respect to the standard dot product. One could consider more general kinds of symmetric bilinear forms, and what results are diferent kinds of groups. We will look at this later. For Euclidean spaces, we can consider all pairings as the dot product, so orthogonality always is just the normal defnition. When the pairing is not positive defnite, then we start getting weirder things. 135 Lecture 27: Euclidean and Hermitian Spaces 27.4 Complex Linear Operators Now, we want to focus on linear operators in a Hermitian space (V, ⟨·, ·⟩). Consider a linear operator T : V − → V. We can the defne the analogue of the adjoint operation on matrices. Defnition 27.4 Pick an orthonormal basis {u1, · · · , un} of our Hermitian space. By doing so, we can map our vector space into coordinate vectors. We have the correspondence V ⇝ Cn , and the form ⟨·, ·⟩ maps to the standard ∗ Hermitian product. Then, T ⇝ M ∈ Matn×n(C). Then let the adjoint T be the linear operator T ∗ : V − → V ∗ such that, with respect to the basis {u1, · · · , un}, T is the matrix M ∗ = M T . The following claim gives the key property of the adjoint linear operator and another way to characterize the adjoint operator. Claim 27.5. For v, w ∈ V, ∗ ⟨T v, w⟩ = ⟨v, T w⟩. ∗ This property means that T is uniquely determined, by putting v = ui and w = uj . Proof. Using the correspondence given by taking a basis, v ⇝ x ∈ Cn w ⇝ y ∈ Cn Tv ⇝ Mx ∈ Cn . Then, ∗ M ∗ ∗ (M ∗ ⟨T v, w⟩ = (Mx) ∗ y = x y = x y) = ⟨v, T ∗ w⟩. This also means that T ∗ is independent of the choice of a basis. Just as we defned adjoint for both linear operators and matrices, we can do the same for the defnition of Hermitian. Defnition 27.6 A linear operator T : V − → V is a Hermitian operator if T ∗ = T, which is equivalent to ⟨T v, w⟩ = ⟨v, T w⟩. Also, a unitary matrix is a matrix such that U ∗U = In and U ∗ = U −1 . The following defnition is analogous. Defnition 27.7 A linear operator T : V − → V is a unitary operator if T ∗T = Id, or equivalently if ⟨T v, T w⟩ = ⟨v, w⟩ for all v, w ∈ V. The following defnition is harder to motivate, but encapsulates the previous two. Defnition 27.8 ∗ A linear operator T : V − → V is normal if TT = T ∗T , which is equivalent to ∗ ⟨v, T ∗ Tw⟩ = ⟨T v, T w⟩ = ⟨T ∗ v, T ∗ w⟩ = ⟨v, TT w⟩. 136 Lecture 27: Euclidean and Hermitian Spaces The set of unitary matrices and the set of Hermitian matrices are both subsets of the set of normal matrices. 1 1 0 is 0 1 1 However, there are normal matrices that are neither Hermitian nor unitary. For example, A = 1 0 1 normal but neither Hermitian nor unitary. Next class, we will discuss the Spectral Theorem, which is the main important property of normal matrices. Theorem 27.9 (Spectral Theorem) For a Hermitian space V, and a normal linear operator T : V − → V , V has an orthonormal basis {u1, · · · , un} where each ui is an eigenvector of T. In other words, we both diagonalize T as well as fnd an orthonormal basis for V . We don’t need to mess around with Jordan forms. The matrix version states that for a normal matrix M ∈ Matn×n(C), 95 it is possible to fnd a unitary matrix P 96 such that λ1 .
P ∗ MP = P −1MP = .. λn . The columns of P form our eigenbasis. What about Euclidean spaces? In full generality, this theorem is false. However, for a Euclidean space V and a symmetric linear operator T : V − → V , T does have an orthonormal eigenbasis. Generalizing from symmetric to orthogonal, which is the analogous version of unitary, or some condition similar to “normal," does not work. For example, we saw that it is not true for orthogonal matrices, as rotation matrices in the plane do not have any eigenvectors. 95M∗M = MM∗ 96P ∗P = I 137 Lecture 28: The Spectral Theorem 28 The Spectral Theorem 28.1 Review: Hermitian Spaces Last time, we discussed Hermitian spaces which were just complex vector spaces with a positive defnite Hermitian form. Given a linear operator T , we defned the adjoint T ∗ , which had the property that ∗ ⟨v, T w⟩ = ⟨T v, w⟩. We called a linear operator T normal if TT ∗ = T ∗T . We then were able to state the Spectral Theorem. 28.2 The Spectral Theorem The Spectral Theorem demonstrates the special properties of normal and real symmetric matrices. Theorem 28.1 Given a Hermitian space V , for any normal linear operator T , there exists an orthonormal eigenbasis of V : {u1, · · · , un}. In matrix form, for any normal matrix M ∈ GLn(C), there exists a unitary matrix P such that P −1MP is diagonal. The real version states that for a Euclidean vector space V and a symmetric linear operator T , there exists an orthonormal eigenbasis; equivalently, for any symmetric matrix M ∈ GLn(R), there exists an orthogonal matrix P such that P −1MP is diagonal. All eigenvalues of real symmetric matrices are real. Example 28.2 3 −1 1 1 Say we take the symmetric matrix M = . One eigenvector is √ with eigenvalue λ1 = 2. The −1 3 2 1 1 other eigenvector should be orthogonal, so it is √ 1 with eigenvalue λ2 = 4. 2 −1 In 2 dimensions, the fact that our change of basis is coming from an orthogonal basis tells us that the change in coordinates is just a rotation. This rotation makes the operator diagonalizable. This is called the Spectral Theorem because the eigenvalues are often referred to as the spectrum of a matrix. Any theorem that talks about diagonalizing operators is often called a spectral theorem. Now we will state some lemmas in order to prove the Spectral Theorem. Lemma 28.3 (Lemma 1) First, for a linear operator T : V − → V where V is Hermitian, and a subspace W ⊂ V such that T (W ) ⊂ W, then T ∗(W ⊥) ⊂ W ⊥ . Proof. Take some u ∈ W ⊥ . For any w ∈ W, then ∗ ⟨w, T u⟩ = ⟨T w, u⟩ = 0, ∗ by the defnition of the adjoint linear operator, so T u ∈ W ⊥ . Lemma 28.4 (Lemma 2) If Tv = λv, then T ∗ v = λv. This just means that T and T ∗ have the same eigenvectors, and eigenvalues related by complex conjugation. Proof. We can frst consider a special case. • Special Case: λ = 0. We want to show that Tv = 0 implies that T ∗ v = 0. ∗ ⟨T ∗ v, T v⟩ = ⟨T v, T v⟩ = 0 by normality (this was a property shown last time). Since T is positive defnite, this implies that T ∗ v = 0. ∗ This also implies that if a vector v ∈ ker T , then v ∈ ker T as well. 138 Lecture 28: The Spectral Theorem • General Case: λ ∈ C. We want to show that Tv = λv implies that T ∗ v = λv. In this case, create a new linear operator S = T − λI. Then, Sv = 0. Also, the adjoint is S∗ = T ∗ − λI and SS∗ = S∗S, since ∗ TT ∗ = T ∗T. By the frst case, since v ∈ ker S, we also have that v ∈ ker S∗ so T = λv as needed. Proof of the Spectral Theorem. Now we can prove the Spectral Theorem by induction on the dimension of V. The general process will be breaking V up into a direct sum and fnding orthonormal eigenbases over each part. Since the feld is F = C, it is always possible to fnd an eigenvector w ∈ V such that Tw = λw. By normalizing w, assume that ⟨w, w⟩ = 1. Then, let W = Span(w); then V = W ⊕ W ⊥ since all subspaces of Hermitian spaces are non-degnerate. In addition, since w is an eigenvector, TW ⊂ W. If we can show that TW ⊥ preserves W ⊥ , then we can induct. By Lemma 2, w is also an eigenvector of T ∗ , so T ∗(W ) ⊂ W, and by Lemma 1, (T ∗)∗(W ⊥) = T (W ⊥) ⊂ W ⊥ . Then V = W ⊕ W ⊥ , and T maps each part of the splitting to itself, so by induction there exists u2, · · · , un an orthonormal eigenvasis for W ⊥ , and {w, u2, · · · , un} is an orthonormal eigenbasis for V. Student Question. Why is the equivalent of the theorem not true over the real numbers? Answer. Most of this argument works, except in the very frst step, where we found an eigenvector and eigenvalue. We cannot guarantee this will happen with normal linear operators over the real numbers. However, as we found last week, for symmetric (and Hermitian) matrices, the eigenvalues are all real, and in particular it is always possible to fnd one eigenvector w ∈ V with real eigenvalue to allow the induction to occur. As an application, suppose we have a quadratic function f(x, y) 2 2 Let this function be + bxy + cy = ax . a b/2 represented by a matrix ; then b/2 c x f(x, y) = (x, y)M = ⟨v, T v⟩ y for a linear operator T given by M. By the Spectral Theorem, there exists an orthogonal change of coordinates ′ λ1 0 x x P T MP = , where P is an orthogonal matrix. It takes = P . Then ′ 0 λ2 y y ′ x λ1 x ′ f(x, y) = (x, y)M = (x , y ′ ) = λ1(x ′ )2 + λ2(y ′ )2 .
′ y λ2 y Example 28.5 2 If f(x, y) = 3x2 − 2xy + 3y , the matrix would be M = 3 −1 . Then −1 3 2 ′ )2 3x 2 − 2xy + 3y = 2(x ′ )2 + 4(y , ′ ′ for some change of coordinates taking x and y to x and y , since we saw earlier that the eigenvalues were 2 and 4. P 2 2 1 + · In n variables, let f(x1, a11 · · · aij . . . . .. be represented by the matrix M ) = a11x + i 0 • One line. x2 = 0 2 2 • One point. x + y = 0 2 2 • The empty set ∅. x + y = −1 Theorem 28.6 2 2 After an isometry, all curves of the form ax + bxy + cy + dx + ey + f = 0 look like one of these options. » # » B f + + v v T A# λ1 0 2 2 that A = Then λ1x + λ2y + b1x + b2y + f = 0. 0 λ2. b1 b2 • Nonzero eigenvalues. If λ1, λ2 = ̸ 0, then we can complete the square, taking x ⇝ x+ and y ⇝ y+ .
2λ1 2λ2 Then 2 2 λ1x + λ2y = c. If c = 0, it is a single point or two intersecting lines, and if c ̸= 0, it is an ellipse or a hyperbola. • Zero eigenvalues. The other cases. The details of this aren’t that important, and we glossed over much of it at the end. The main idea to take away is just the power of the Spectral Theorem to take a complicated quadratic form and simplify into something more readily analyzed. One idea where the Spectral Theorem is at play is in multivariable calculus with the second derivative test. Hidden behind the test is that we have a symmetric matrix and we are trying to fgure out what the signs of the eigenvalues are, as that will tell us whether our critical point is a minimum, maximum, or saddle point. Proof. The curve be rewritten as # » = 0. After an orthogonal change of basis, we can assume v 140 < < < < < < < < Lecture 29: Geometry of SU2 29 Linear Groups Today we’ll study linear groups – subgroups of matrices which satisfy conditions about preserving properties that come from linear algebra. We’ve seen a couple of these already. We can start of with the group GLn(R) of all invertible n × n matrices, and consider a few subgroups that we’ve seen: GLn(R) SLn(R) On(R) A ∈ GLn(R A ∈ GLn(R) det A = 1 AtA = I SOn(R) A ∈ GLn(R) AtA = I, det A = 1 These subgroups all preserve some interesting property: • SLn(R) ≤ GLn(R) consists of the matrices with determinant 1, which preserve volume. • On(R) ≤ GLn(R) consists of the orthogonal matrices, which preserve the dot product (or equivalently, which preserve length) – meaning that ⟨Av, Aw⟩ = ⟨v, w⟩ for any two vectors v, w (where the pairing denotes the dot product). • SOn(R) is the intersection of SLn(R) and On(R). We can do the same thing for matrices with complex values: GLn(C) SLn(C) Un A ∈ GLn(C A ∈ GLn(C) det A = 1 A∗A = I SUn A ∈ GLn(C) A∗A = I, det A = 1 These subgroups still preserve some linear algebraic property: • SLn(C) ≤ GLn(C) is still the group of matrices with determinant 1, the same as in the real case. • Un(C) ≤ GLn(C) is the group of unitary matrices, which preserve the standard Hermitian form – so ⟨Av, Aw⟩ = ⟨v, w⟩ where the pairing is now the standard Hermitian form (instead of the dot product, as in On(R) for the real numbers). • SUn(C) is the intersection of SLn(C) and Un(C), the group of unitary matrices with determinant 1. We could also look at groups of matrices that preserve other bilinear forms, not just the dot product. For Ip example, we could defne the bilinear form Ip,q corresponding to the matrix . Then we could study −Iq the matrices preserving this bilinear form, meaning the set {A | AtIp,qA = Ip,q}, which is an interesting subgroup of GLn(R). 29.1 Geometry of groups All these matrices are over R or C. What’s special about the real or complex numbers, as opposed to something like a fnite feld, is that we have a notion of distance. More explicitly, we have GLn(R) ⊂ Rn 2 (where we just write down the coordinates), so GLn(R) inherits a metric – we can discuss whether two elements are close 141 Lecture 29: Geometry of SU2 together or far apart. The same works for GLn(C) ⊂ R2n 2 (since we can think of a complex number as a pair of real numbers). The actual metric itself won’t be that important; what will be important is the idea of a topology on this set – we have a sense of two elements being close together or far apart, which we didn’t have when we studied fnite or discrete groups. With fnite or discrete groups, we can’t really talk about a sequence of elements getting closer and closer to another, but with these groups we can. The group structure and topology interact with each other in extremely interesting ways – these groups are called Lie groups, and the study of such groups ends up being a really rich vein of mathematics. Today we’ll see a favor of how we can take a group and look at it not just as a collection of things, but also as having some sort of geometry. There are some questions we can ask: groups come with multiplication law, G × G − → G. It turns out that for all the matrix groups considered above, the multiplication law is continuous (if you perturb two elements a bit, this only perturbs their product a bit as well). The inverse map g 7→ g−1 is also continuous. Similarly, we can look at continuous homomorphisms. We’ve seen a few simpler examples of groups with a shape (not necessarily matrix groups) – for example, (R, +) is just the real line: Example 29.1 How can we draw the group of 2-dimensional rotations SO2 = {ρθ | 0 ≤ θ < 2π}? Proof. On a homework problem, we showed that SO2 = ∼ R/2πZ. So we can draw SO2 as a circle, where the angle represents the angle of rotation: θ We have a homomorphism R → SO2 where we send θ 7→ ρθ. Geometrically, this corresponds to wrapping the line around the circle infnitely many times. Similarly, we can consider O2 – we saw that SO2 has two cosets, itself and the set of refections. So we can think of O2 as two circles (with one circle representing each). Note 29.2 We haven’t really defned terms like continuous, metric, or topology. But we won’t be that formal here; instead, the main goal is to try to visualize our groups in terms of these pictures. 29.2 Geometry of SU2 All the groups whose geometry we’ve looked at so far are one-dimensional. Now we’ll look at a higher-dimensional group, the special unitary group SU2 = {A ∈ GL2(C) | det A = 1, A ∗ = A−1}. We’ll try to fgure out what the points in this set look like, as a geometric shape. 142 Lecture 29: Geometry of SU2 29.2.1 Quaternions First we’ll analyze what matrices are in SU2. Consider a matrix A = α β . Then we have γ δ 1 δ −β δ −β A−1 = = det A −γ α −γ α (since the determinant is 1), and α γ A ∗ = . β γ These must be equal, so we must have δ = α and β = −γ. Thus we must have α β A = . −β α Finally, the condition that det A = 1 means that |α|2 + |β|2 = 1. We can write any matrix A as a linear combination of other matrices, by splitting up the real and imaginary parts of α and β. Suppose α = x0 + ix1 and β = x2 + ix3 where x0, x1, x2, x3 ∈ R. Then we have 1 0 i 0 0 1 0 i A = x0 + x1 + x2 + x3 . 0 1 0 −i −1 0 i 0 We’ll introduce some notation for these matrices. The frst is just the other identity; meanwhile we defne i 0 0 1 0 i i = , j = , k = . 0 −i −1 0 i 0 We will also defne a four-dimensional real vector space consisting of all matrices of the above form (which are linear combinations of I, i, j, and k: H = R4 = {x0I + x1i + x2j + x3k | x0, x1, x2, x3 ∈ R} ⊂ Mat2×2(C). This space is closed under multiplication, since we can use the relations i2 = j = k2 = −I, ij = k, ji = −k, and so on. So if we multiply any two elements of H, we get another element of H. The set H is called the set of quaternions. They’re like a four-dimensional version of the complex numbers (instead of adding one square root of −1, we now have three). But unlike the complex numbers, multiplication isn’t commutative – so it’s kind of like the feld of complex numbers, but it’s not a feld because of the lack of commutativity. (We can still divide by nonzero elements, but we have to be careful about the order.) But the main thing we’ll use here is that it’s a four-dimensional real vector space – we’ve fgured out how to write 2 2 2 2 elements in terms of coordinates. Then SU2 ⊂ H are exactly the quaternions such that x0 + x1 + x2 + x = 1 3 (since this is equivalent to the determinant condition). So SU2 sits inside R4 , and consists of all vectors with length 1 – this means its shape is a 3-dimensional sphere S3 . Student Question. Why is it called a 3-dimensional sphere if there are four dimensions? Answer. There’s four dimensions, but we’re imposing an additional condition. A sphere only consists of the boundary, not the interior (a sphere with its interior is called a ball). So although it lives in four dimensions, it’s a three-dimensional surface (because we have one constraint). Similarly, a normal sphere in R3 is called a 2-sphere, since its surface is two-dimensional. 29.2.2 Geometry of the Sphere The 3-sphere is very hard to picture, so let’s start by drawing the 2-sphere 2 2 2 S2 = {(x0, x1, x2) : x0 + x1 + x = 0} ⊂ R3 .
2 2 2 2 As coordinates, S2 is the set of solutions to x0 +x1 +x2 = 1. There are many ways of representing its coordinates, but one commonly used one is spherical coordinates, which we can think of in terms of latitude and longitude. The latitude lines come from taking a horizontal slice of the sphere – formally, Latc = {x0 = c} ∩ S2 . 143 Lecture 29: Geometry of SU2 The latitude lines start of as a point at the north pole Lat1, and then as we go downwards, they get bigger and bigger until the equator E = Lat0, and then get smaller again until the south pole. Meanwhile, in the other direction, we have longitude lines, which are circles of radius 1 that intersect the north and south pole. So the latitude and longitude lines give us a way of representing our location on the 2-sphere. We can do the same for the 3-sphere: we now have 2 2 2 2 S3 = {x0 + x1 + x2 + x = 1}.
3 We can defne latitudes in the exact same way, with Latc = {x0 = c} ∩ S3 . This is now the set of points with 2 2 2 2 x0 = c and x1 + x2 + x = 1 − c , so now our latitudes are actually 2-spheres. So we’re still taking horizontal 3 slices (with −1 ≤ c ≤ 1), but now the latitudes are 2-spheres of diferent sizes. We still have just a single point at the north and south pole, and the largest latitude is still the equator Lat0 = E. We’ll defne the longitudes precisely a bit later. These will still be circles of radius 1 passing through both the north and south poles (±1, 0, 0, 0). Just like in the 2-sphere, every point lies on a unique latitude line, and every point except the north and south pole lies on a unique longitude line; and each pair of latitude and longitude lines intersects at exactly two points. We’ve seen now that SU2 as a set is a 3-sphere in R4 (using this choice of coordinates), and the 3-sphere can be thought of geometrically as having latitudes and longitudes. It turns out we can use this geometric understanding of the 3-sphere to understand the group structure of SU2. 29.2.3 Latitudes Theorem 29.3 The conjugacy classes of SU2 are precisely the latitudes Latc for −1 ≤ c ≤ 1. So slicing the 3-sphere horizontally into latitudes is the same as taking the group and decomposing it into conjugacy classes. In particular, most of these latitudes are 2-spheres, and are infnite – except the north pole and the south pole, which only have one element. A point has conjugacy class of size 1 exactly when it’s in the center, so this implies that Lat±1 = ±I = Z(SU2). We can also see this is true directly, by checking which matrices commute with every other matrix in the group; but this gives a geometric interpretation. 144 Lecture 29: Geometry of SU2 Proof of Theorem 29.3. Recall that when we chose coordinates, we wrote elements of SU2 as 1 0 1 0 0 1 0 i A = x0 + x1 + x2 + x3 . 0 1 0 −i −1 0 i 0 The main point is that the frst matrix I has trace 2, while the other matrices i, j, and k all have trace 0. So then Tr(A) = 2x0. So taking horizontal slices has some meaning in terms of the matrices itself – it corresponds to taking slices of SU2 with constant trace. We can use this idea to prove the theorem. Fix A ∈ SU2 with coordinates (x0, x1, x2, x3), so we want to show that Conj(A) = Latx0 . But the latitude is exactly the set {A ′ ∈ SU2 | Tr(A ′ ) = Tr(A)}. This immediately solves one direction – suppose A ′ ∈ Conj(A), so A ′ = P −1AP for some P . We’ve seen that trace is one of the coefcients of the characteristic polynomial, so it doesn’t change under conjugation. So we must have Tr(A ′ ) = Tr(A), which means Conj(A) ⊆ Latx0 . For the other direction, we want to show that if Tr(A ′ ) = Tr A, then A ′ = P −1AP for some P ∈ SU2. To see this, consider the polynomial t2 − Tr(A)t + 1. This is the characteristic polynomial of both A and A ′ (since they both have determinant 1). Let its roots be λ and λ, so λ and λ are the eigenvalues of A and A ′ . Now by the Spectral Theorem, there exists a unitary matrix Q such that Q−1AQ = λ 0 = D, 0 λ and a unitary matrix Q ′ such that (Q ′ )−1A ′ Q ′ = D as well. So A and A ′ are conjugate to the same matrix, which means they’re conjugate to each other – we have (Q ′ Q−1)A(Q ′ Q−1)−1 = A ′ . Now we’re almost done, but we’ve overlooked one detail: the Spectral Theorem shows that A and A ′ are conjugate to each other in U2, but we need to check that we can do this conjugation using matrices of determinant 1. But we can actually arrange for Q and Q ′ to have determinant 1 – suppose we have Q ∈ U2 with Q−1AQ = D, and let det Q = δ. Since Q is unitary, we have Q∗Q = I, so δ · δ = 1. So then |δ| = 1. Then we can set δ−1/2 0 ˜ Q = Q . δ−1/2 0 The matrix on the right is unitary as well (because |δ| = 1), and det Q ˜ = δ · δ−1 = 1, so then Q ˜ ∈ SU2. We can check that Q ˜−1AQ ˜ = D as well. So any two matrices in the same latitude are conjugate. Note 29.4 The main idea of this proof was the Spectral Theorem, which we used to see that if two matrices in SU2 have the same trace, then they’re conjugate in U2. Then by a bit of messing around, we could also arrange for them to be conjugate in SU2. The upshot of today is that the 3-sphere is the union of its latitudes [ 1 S3 = Latc, c=−1 and this decomposition corresponds to the group theoretic decomposition [ SU2 = conj classes (where we can identify S3 and SU2, and the conjugacy classes are identifed with the latitudes). 145 Lecture 29: Geometry of SU2 When we worked with fnite groups, we had the class equation |G| = Pk |Ci| (where the Ci are the conjugacy i=1 classes). In this setting, this doesn’t make much sense, since the size of the group (and the conjugacy classes) is infnite. But since S3 is a geometric object, we can actually consider the volume of SU2 (which is the volume of S3). We can’t sum over the conjugacy classes because there aren’t fnitely many of them, but instead we can integrate; and then instead of counting the elements in each slice, we take their 2-dimensional volume (or area) instead. So we get the equation Z 1 vol(SU2) = vol(Conj ) dc.
c c=−1 This is very vague, but it’s possible to formalize it, and it gives an identity which is a version of the usual class equation. This ends up being a really useful idea when studying these groups further – the idea that we can take geometric quantities and integrate them, by frst integrating over conjugacy classes. 146 Lecture 30: The Special Unitary Group and One-Parameter Groups 30 The Special Unitary Group SU2 30.1 Review Last time, we started looking at subgroups of the group of invertible matrices. We saw that one thing these groups have in common, that fnite or discrete groups don’t, is that they have some sort of shape or geometry. In particular, we looked at the group = A−1 the special unitary group. By playing around with the defnitions, we found that SU2 sits inside the quaternions H = {x0I + x1i + x2j + x3k}, SU2 := {A ∈ GL2(C) | A ∗ , det A = 1}, where we defned i = i 0 0 , j = −i 0 −1 1 , k = 0 0 i i . 0 2 2 2 2 In particular, SU2 is the subset with x0 + x1 + x2 + x = 1, corresponding to the 3-sphere S3 in R4 .
3 Note 30.1 We’ve seen that the 1-dimensional sphere and 3-dimensional spheres both have a group structure, so we can ask whether the same is true for other dimensions. It turns out the answer is no – there are no other n-dimensional spheres which can be made into groups. (This is a deeper fact.) Last class, we started taking geometric properties of the 3-sphere and seeing how they correspond to the group structure. In particular, we looked at the latitudes – the horizontal slices Latc = {x0 = c} ∩ S3 for −1 ≤ c ≤ 1. We proved last class that these latitudes are precisely the conjugacy classes of SU2. We call Lat0 the equator, denoted E. 30.2 Longitudes Another thing we can think about are the longitudes – the circles which go through the north and south pole. We can defne these more precisely: for each x ∈ E, the longitude containing x is Long := Span(I, x) ∩ S3 .
x Here Span(I, x) is a 2-dimensional plane, so we’re taking the unit circle of a 2-dimensional plane. Theorem 30.2 For each x ∈ E, Long is a subgroup of SU2. In fact, given θ ∈ R/2πZ, the map θ 7→ cos θI + sin θx is an x isomorphism between R/2πZ and Long .
x What this means is that the longitudes aren’t just circles as shapes, they’re also circles as groups (since we’ve seen that the unit circle as a group is isomorphic to R/2πZ. Proof. To see this is true, we’ll frst consider the special case x = i. Then we can check that given two points in Longi, we have ′ (c + si)(c + s ′ i) ∈ Span(I, i) 147 Lecture 30: The Special Unitary Group and One-Parameter Groups as well, since i2 = −I. Meanwhile, since both elements are in SU2, their product must be as well; so their product is in SU2 ∩ Span(I, i) = Longi. So this longitude is closed under multiplication, and is therefore a subgroup. We won’t check the isomorphism to R/2πZ here, but it’s possible to check this directly by multiplying out. We can then use this to solve the general case – for any x ∈ E, we know x is conjugate to i (since we saw that the equator is a conjugacy class). So then we can write x = P −1iP , and then Long = P −1 Longi P is conjugate x to Longi. But when we conjugate a subgroup, we get another subgroup. So not only are the longitudes all circle subgroups, but they’re also all conjugate to each other. 30.3 More Group Theoretic Properties When we studied conjugacy classes, we also studied centralizers: Guiding Question What is the centralizer of i? Recall that this means the set of elements for which if we conjugate i by them, we get back i. We know that Longi is a subgroup of SU2, and it’s abelian (since it’s isomorphic to the circle). Since i is in this longitude, this means it commutes with everything in this longitude. So Z(i) ⊃ Longi. In fact, this turns out to be an equality – we have Z(i) = Longi. This is true for any other point on the equator as well – its longitude is exactly its centralizer. Another thing we saw when studying conjugacy classes was that there’s a bijection between the conjugacy class C(g) and the cosets of the centralizer G/Z(g). In our case, this is still true, but now both sides are geometric objects. If we fx a point g on the equator, then C(g) = E is a 2-dimensional sphere. Meanwhile, G/Z(g) corresponds to taking cosets of a longitude. We’re taking a 3-sphere and covering it in cosets – so we have a map S3 → S2 , where the fbers are circles (the cosets of the longitude). This is really hard to picture, but the idea is that we start with the 3-sphere and a given longitude, and we’re taking its cosets (which correspond to circles not necessarily through the north and south pole) and covering the entire 3-sphere in these circles. When we collapse all these circles to a point, we get a copy of the 2-sphere. Note 30.3 This is really difcult to think about, but it’s a construction in topology relating spheres of dimensions 1, 2, and 3, and it can also be thought of in this group theory setting. What we would like to illustrate is that group theoretic facts about this group also become interesting geometric facts; it doesn’t really matter if you don’t understand all of them. 30.4 Conjugation and the Orthogonal Group There’s another thing we can look at: we know the equator is a conjugacy class, so SU2 acts on E transitively (with the action given by conjugation). In fact, SU2 acts on the space {x0 = 0} ⊂ H (which is the 3-dimensional vector space containing the equator), and it preserves the equator E inside this space. Conjugation by an element of SU2 is a linear map, so it defnes a group homomorphism ρ : SU2 → GL3(R), −1 where ρ(g) is the matrix such that ρ(g)⃗ v = gvg . But ρ(g) preserves E, so since it preserves vectors of length 1, this means it must preserve length in general. So ρ(g) is actually an isometry – which means this map is actually ρ : SU2 → O3. In fact, we can say even more. We’ve seen that orthogonal matrices in 3 dimensions are either refections or rotations – and you can tell which by looking at the determinant (which is always ±1). But SU2 is connected (we can get from any point to any other point by following some path), so det(ρ(g)) can’t jump between ±1 (since ρ is continuous). So then det(ρ(g)) is constant as g varies. We know that det(ρ(I)) = 1, so then det(ρ(g)) is always 1. So in fact, this is a homomorphism ρ : SU2 → SO3. 148 Lecture 30: The Special Unitary Group and One-Parameter Groups Note 30.4 We could write down this homomorphism in terms of the matrix entries – we start with a 2 × 2 complex matrix and create a 3 × 3 real one, and we could explicitly write down the homomorphism. But it’s more interesting to think about it geometrically, by considering the action of SU2 on one of its conjugacy classes. Note 30.5 You can go further with this – given a point on the 3-sphere, we can ask how to fgure out what angle and axis of rotation it corresponds to. This is written up in the notes, but we won’t discuss it here. But you can go really far by playing around with the group-theoretic constructions we’ve seen earlier and trying to picture what they mean. Student Question. Did we show that ρ was continuous? Answer. No, we did not. In order to check that it’s continuous, you can write down the map in terms of the −1 entries. But this shouldn’t be surprising – we have ρ(g)v = gvg , and we can write down a explicit formula for −1 g in terms of g. So the matrix ρ(g) is something we can write down explicitly in terms of coordinates. (In fact, you can also use this explicit formula to show it’s in SO3, but it’s nicer to just show that it’s continuous and then deduce it’s in SO3 by thinking about it geometrically.) Student Question. Was the action SU2 defned on E just left multiplication? Answer. No, it’s conjugation. The idea is that E is one of the conjugacy classes of the group, and all the conjugacy classes are orbits with respect to the conjugation action. (This is why the action is transitive as well.) 30.5 One-Parameter Groups Now we’ll return to looking at linear groups more generally – subgroups G of GLn(R) or GLn(C) which satisfy some condition (for example, preserving volume or a bilinear form). Defnition 30.6 A one-parameter group (in GLn(R) or GLn(C)) is a diferentiable homomorphism from R → GLn(R) or R → GLn(C). In otherwords, it’s a function φ : R → GLn(C) with t 7→ φ(t). It should be a group homomorphism, so φ(s + t) = φ(s) + φ(t), and it should be diferentiable (where we think of GLn(C) as sitting inside R2n 2 – then each entry of the function should be a diferentiable function on R). One way to think of this defnition is as an analog of when we looked at maps Z → G. The integers are in some sense the simplest group we can write down – it has just one generator and no relations – and we can look at maps Z → G, to help us study G. The idea here is that (R, +) is basically the simplest one-dimensional group. (We haven’t defned dimension, but you can think of dimension as how many parameters we have. There are other one-dimensional groups, like a circle, but the real numbers are simpler because we don’t have relations like 2π = 0 here.) We’ve already seen a few examples of one-parameter groups: Example 30.7 In SU2, the map θ 7→ cos θI + sin θx (for any x ∈ E) is a one-parameter group. These one-parameter groups are the longitudes. We’ve seen that every point lies in some longitude, and therefore some one-parameter group; in general, that isn’t always true. Let’s see another example, when n = 1. Example 30.8 When n = 1, the map φ : R → C× with φ(t) = eαt (for any α ∈ C) is a one-parameter group. 149 Lecture 30: The Special Unitary Group and One-Parameter Groups Here one-dimensional matrices are just numbers. This construction works because we have αs+αt αs αt φ(s + t) = e = e e = φ(s)φ(t). Note 30.9 This isn’t an analysis class, so we won’t check the diferentiability of these maps. In this example, it can be done by writing everything down in terms of sines and cosines. Guiding Question Is there a version of this construction for n > 1? A The answer is yes – if we have A ∈ Matn×n(C), we can try to defne e . Taking a number to the power of a matrix doesn’t make any sense, but the exponential function also has a description using power series: we have 2 3 x x x e = 1 + x + + + · · · . 2! 3! x This is a very nice power series – it converges to e everywhere. In fact, you can even take this as the defnition x of e , and you can take the derivative term-by-term. So we can use this to defne eA as well: Defnition 30.10 The exponential of a matrix A ∈ Matn×n(C) is A2 A3 A e = I + A + + + · · · ∈ Matn×n(C). 2! 3! This also converges uniformly as A varies in a bounded region, meaning that for every entry of the matrix, if we take the corresponding entries of each term, then we get a convergent series. As we vary A in a neighborhood, A this convergence of that series is uniform. So this gives us a well-defned n×n matrix e . (To be more precise, we can actually put a metric on the space of matrices, and use this to be careful about the notion of convergence.) This exponential has several nice properties, similarly to the normal exponential. • The exponential interacts well with conjugation – we have P −1 AP e = P −1IP + P −1AP + P −1A2/2P + · · · (P −1AP )2 = I + P −1AP + + · · · 2 P −1AP = e . (To be more careful, the LHS and RHS are both defned by limits – where we take the power series and truncate it. The equality is true on the level of these truncations, so the limits are equal as well.) λ • If v is an eigenvector of A with eigenvalue λ, then v is also an eigenvector of eA with eigenvalue e . sA tA (s+t)A • We have e e = e . To prove this, we can expand the RHS out using the Binomial Theorem as X X ktℓ (s + t)nAn s (k + ℓ)! AkAℓ = · . n!
(k + ℓ)! k!ℓ! n≥0 k,ℓ≥0 Then using uniform convergence, we can factor out the infnite sum as X kAk X tℓAℓ s sA tA = e e k! ℓ! k≥0 ℓ≥0 (the fact that we can rearrange in this way is the result of the strong convergence properties). A −A In particular, e e = I, so we actually have eA ∈ GLn(C). (We’re working in the complex case, but this works equally well in the real case.) 150 Lecture 30: The Special Unitary Group and One-Parameter Groups The last result means that for any A ∈ Matn×n(C), φ(t) = etA is a one-parameter group in GLn(C). We can ask two questions about these one-parameter groups: Guiding Question Is every one-parameter group of this form? The answer will be yes, and we will see why in future lectures! Guiding Question Given a subgroup G ≤ GLn, what are the one-parameter subgroups living inside of G? We will discuss this question in future lectures as well. 151 Lecture 31: One-Parameter Subgroups 31 One-Parameter Subgroups 31.1 Review Last time, we talked about one-parameter subgroups. Defnition 31.1 A one-parameter group in GLn(C) is a diferentiable homomorphism φ : R − → GLn(C). For a matrix A ∈ Matn×n(C), the matrix exponential is A e := 1 + A + 1 A2 + 1 A3 + · · · , 2! 3! which converges to a matrix in GLn(C). 97 For example, φA(t) = etA is a one-parameter group.98 Example 31.2 If A = 1 0 0 0 , then An = 1 0 0 0 for all n ≥ 1. Then X 1 A 1 e = An = n! 0 n≥0 X 0 1 + 1 0 n≥1 0 e = 0 0 0 . 1 Example 31.3 Similarly, for A = 0 0 1 0 , A2 = 0 0 0 0 = A3 = · · · . Then A e = 1 0 0 1 + 0 0 1 0 = 1 0 1 1 . 31.2 Properties of the Matrix Exponential The matrix exponential fulflls several nice properties. sA tA (s+t)A A B A+B • The product is the exponential of the sum: e e = e . In fact, if AB = BA, then e e = e , but they must commute.99 λ1 · · · 0 eλ · · · 0 1 .
. . A .
. . • If A = .
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. , then e = .
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. .
.
.
.
. .
. 0 · · · λn 0 · · · eλ n B • If B = P AP −1 , then e = PeAP −1 . This allows us to easily take the matrix exponential of any diagonal izable matrix. Example 31.4 0 2π 2πi 0 If A = , it has eigenvalues 2πi and −2πi, so diagonalizing gives P AP −1 = .
−2π 0 0 2πi P AP −1 2πi A Then PeAP −1 = e = 1 0 , since e = 1. Since eA is conjugate to the identity matrix, e 0 1 itself must be the identity matrix. 0 2π 0 0 In particular, e −2π 0 = e 0 0 , and so the matrix exponential is not injective, unlike the normal exponential. 97With the metric ||M|| = maxi,j |mij |, every entry converges. 98It is called a one-parameter "subgroup," but it does not have to be injective; it can wrap around. 1 Ak Bℓ 99The key fact here is that (A + B)n = P when AB = BA; matrix multiplication is not commutative so it is not n! k+ℓ=n k! ℓ! always true. 152 Lecture 31: One-Parameter Subgroups a(t) b(t) a ′ (t) b ′ (t) d • Defning the derivative of a matrix to be = , the derivative is dt c(t) d(t) c ′ (t) d ′ (t) t2 d d tA) = A2 (e I + tA + + · · · dt dt 2 t2 = 1000 + A + tA2 + A3 + · · · 2 = AetA , similarly to the normal exponential. 31.3 One-Parameter Subgroups The matrix exponential is related to one-parameter subgroups in the following manner. Proposition 31.5 Every one-parameter group in GLn(C) is of the form φ(t) = etA for a unique matrix A ∈ Matn×n(C). Proof. We prove uniqueness and existence. tA • Uniqueness. If φ(t) = e , then φ ′ (t) = AetA , so φ ′ (0) = A. So the coefcient A in the one-parameter subgroup is given by taking the derivative and evaluating at 0.101 • Existence. Given φ(t), set A := φ ′ (0) ∈ Matn×n. Since φ is a homomorphism, φ(s + t) = φ(s)φ(t) for ∂ all s and t. Taking the derivative , ∂s φ ′ (s + t) = φ ′ (s)φ(t). Plugging in s = 0, we get φ ′ (t) = Aφ(t), and we also have φ(0) = In. Since this is a linear frst-order ordinary diferential equation with an initial condition, there is a unique solution, which is φ(t) = etA . Defnition 31.6 For G ≤ GLn(C), a one-parameter group in G is a one-parameter group φ(t) in GLn(C) such that φ(t) ∈ G for all t ∈ R. For a one-parameter group in G, φ(t) = etA for some A ∈ Matn×n(C) as well. Guiding Question Given a group G, what are the one-parameter groups in G? What is the corresponding set of matrices A for which etA ∈ G for all t? Let’s see an example. Example 31.7 (Diagonal Matrices) Let λ1 .
G = . .
0 · · · . .. · · · 0 . . ≤ GLn(C) . λn where λi ̸= 0. The one-parameter groups in G are determined by the matrices A such that etA ∈ G for all t ∈ R. Here, etA ∈ G for all t ∈ R if and only if A is diagonal. 101Thinking of φ as a trajectory, A is essentially the velocity of the particle when it is passing through the identity. 153 Lecture 31: One-Parameter Subgroups Proof. If λ1(t) · · · 0 tA .
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φ(t) = e = . .
. , .
. . 0 · · · λn(t) λ ′ 1(0) · · · 0 λ ′ 1(t) · · · 0 .
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then φ ′ (t) = Then . .
. .
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. . 0 · · · λ ′ (t) n A = φ ′ (0) = .
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.
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. 0 · · · λ ′ (0) n must be diagonal. a1 · · · 0 eta1 · · · 0 If A = is diagonal, then tA is diagonal, and so etA = .
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. .
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. ∈ G. So every . .
. . .
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. .
. . . . tan 0 · · · an 0 · · · e diagonal matrix A does correspond to a one-parameter subgroup in G. We can also do the same with upper triangular invertible matrices. · · · c1n c11 .
.
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Let G = .
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. ≤ GLn(C), where cii ̸= 0 for all i. Then etA ∈ G for all t ∈ R if and only if .
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. Example 31.8 (Upper Triangular Matrices) 0 · · · c nn .
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A = .
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. 0 · · · ann · · · ⋆ a11 .
. ′ ′ c11(0) · · · c (0) 1n .
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Proof. If φ(t) is upper triangular, then A = φ ′ (0) = must also be upper triangular.
. .
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. . ′ 0 · · · c (0) nn tA Also, if A is upper triangular, so is An for all n, and thus so is e . So the image of φ is in G. Problem 31.9 For 1 · · · ⋆ .
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G = .
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. ≤ GLn(C), .
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. 0 · · · 1 what are the corresponding matrices A?a 0 · · · ⋆ .
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aThe answer is that A is of the form . . . . .
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. 0 · · · 0 We can also look at the one-parameter groups for unitary matrices. Example 31.10 (Unitary Matrices) For Un = {M ∗ = M −1} ≤ GLn(C), etA ∈ Un if and only if A∗ = −A is skew-Hermitian for some matrix A ∈ Matn×n(C). 154 Lecture 31: One-Parameter Subgroups Proof. We have ∗ A2 (A∗)2 A) ∗ = I ∗ + A ∗ (A ∗ ) (e = I + A + + · · · + + · · · = e . 2! 2! tA tA)∗ tA)−1 tA ∗ −tA tA ∗ −Ae−tA If e is unitary, then (e = (e , so e = e . Diferentiating gives A∗ e = , and taking t = 0 gives A∗ = −A. tA)∗ tA ∗ −tA tA)−1 tA ∈ Un Conversely, if A∗ = −A, then (e = e = e = (e , and so e for all t. 155 Lecture 32: More About One-Parameter Subgroups 32 One-Parameter Groups, Continued 32.1 Review Essentially, we want to restrict the image of φ to lie in G, in order to understand G better. Guiding Question How can we characterize which matrices A defne one-parameter subgroups satisfying this property? As a motivation, the real numbers with addition, (R, +), is essentially the simplest one-dimensional group, and here the notion of a one-parameter maps the real numbers into other groups, allowing us to study more complicated groups using the additive structure of the real numbers. Let’s see an example we covered last time. Example 32.1 (Unitary Matrices) tA ∈ Un Let G = Un ⊂ GLn(C). Last time, we showed that the condition that e for all t ∈ R is equivalent to requiring that A∗ = −Aa; that is, if and only if A is skew-Hermitian, the one-parameter subgroup described by A maps to only unitary matrices. aNote that it is not required for A to be in GLn/invertible; it simply has to be some n×n matrix. 32.2 Examples! Another example is given by upper triangular matrices. Example 32.2 (Upper Triangular Matrices) Let 1 ⋆ ⋆ G = 0 0 1 0 ⋆ 1 ≤ GL3(R) be the real upper triangular matrices. Then the corresponding A for which etA ∈ G for all t ∈ R make up the set of matrices 0 ⋆ ⋆ 0 0 ⋆ ⊆ Mat3×3(R).
0 0 0 tA The one-parameter group is a homomorphism φ(t) = e , for some matrix A; in particular, A is φ ′ (0). So fnding what A looks like, given G, is done by taking the derivative at t = 0. The image of φ lies in G, and since the 1s down the diagonal are not dependent on t, while the upper right entries could be nonzero, 1 ⋆ ⋆ 0 ⋆ ⋆ A = d 0 1 ⋆ = 0 0 ⋆ . dt t=0 0 0 1 0 0 0 It is also necessary to show the other direction that if A is of such a form, then the associated one-parameter 0 ⋆ ⋆ 0 ⋆ ⋆ tA group φ = e will actually have its image lie in G. If A ∈ P 1 k ≥ 1. Then, the exponential is a sum Ak , so k! 0 0 0 0 ⋆ , then Ak 0 = 0 0 0 0 ⋆ 0 for 1 0 0 0 ⋆ ⋆ 1 ⋆ ⋆ A e = 0 1 0 + 0 0 ⋆ + · · · = 0 1 ⋆ , 0 0 1 0 0 0 0 0 1 and therefore etA ∈ G. So in fact etA ∈ G is equivalent to A being upper triangular. Now, let’s consider the orthogonal matrices. 156 Lecture 32: More About One-Parameter Subgroups Example 32.3 Consider the orthogonal matrices On ⊂ GLn(R). For a one-parameter subgroup of On, the matrix A must tA)T satisfy (e = (etA)−1for all t. This is equivalent to tAT −tA e = e d tAT for all t.a Taking the derivative , we get AT e = −Ae−tA , and evaluating at t = 0 gives that the dt possible A are the ones satisfying b {AT = −A}. tAT −tA tA)T tA)−1 tA ∈ On Conversely, if AT = −A, then e = e , which is equivalent to (e = (e , and so e when AT = −A. So these are the correct matrices A. tA)T t(AT ) tA −tA aWriting out the exponential as a sum gives (e = e , and it is clear that the inverse of e is e . bIt is also possible to get this simply by the fact that On is Un ∩ GLn(R), and so A must satisfy the same property as for the unitary matrices, that A∗ = −A, and for real matrices this condition is the same as AT = −A. 32.3 The Special Linear Group SLn(C) In order to study SLn(C), which is in some sense the frst subgroup of matrices that we ever studied, an important identity about the matrix exponential must be established. Guiding Question What about SLn(C)? Lemma 32.4 For any A ∈ Matn×n(C), A trace(A) det e = e . This property is clearly true for diagonal matrices, and from there we hope that this is true for other matrices λ1 0 as well. For example, for A = , 0 λ2 λ1 e 0 A e = , λ2 0 e and λ1 λ2 λ1+λ2 trace(A) det(e A) = e e = e = e . Trying to steamroll through the proof of the lemma becomes very difcult, but fortunately the exponential, determinant, and trace all behave well with respect to conjugation. Proof. We have P AP −1 det(P AP −1) = det(A), trace(P AP −1) = trace(A), e = PeAP −1 . Thus, if the lemma is true for a matrix conjugate to A, it is true for A, and so only one representative from each conjugacy class needs to be considered. Then, without loss of generality, we can assume that A is in Jordan λ1 · · · ⋆ .
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canonical form, and the proof follows identically to the diagonal case. We take A = to be . .
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. . 0 λn · · · λ λk · · · ⋆ e · · · ⋆ 1 1 .
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upper triangular and in Jordan form. Then Ak = , and so eA = , so . .
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. . . . λ λk 0 · · · 0 · · · e n n P λi trace(A) det(e A) = e = e . 157 Lecture 32: More About One-Parameter Subgroups Now we can take a look at SLn(C). Example 32.5 tA ∈ SLn tA ∈ SLn tA) = 1 Consider A ∈ Matn×n(C) such that e (C) for all t ∈ R. Then e (C), and so det(e trace(tA) for all t. Therefore, by Lemma 32.4, e = 1, which is equivalent to stating that trace(A) ∈ 2πiZ tA for all t ∈ R, which is possible only if the trace of A is 0. So the one-parameter groups in SLn(C) are e where A ∈ Matn×n(C) is traceless.a aThe trace is 0. These conditions on A can obviously be combined for diferent groups G. The one-parameter groups in SUn correspond with the matrices A such that A∗ = −A and trace(A) = 0. In particular, the one-parameter groups in SU2 consist of etA where the 2×2 matrix A is skew-symmetric and has zero trace. 2 Example 32.6 (SU2) 2 3 For A = α β , these conditions end up givinga that γ δ 2 2 ix1 x2 + ix3 i 0 0 1 0 i A = = x1 + x2 + x3 . 2 1 −x2 + x3 −ix1 0 −i −1 0 i 0 These are the matrices that showed up when we studied the equator! We have A = x1I + x2J + x3K, and so A = c⃗ v for ⃗ v ∈ E ⊂ SU2. b Then tA tc⃗ v d e = e = cos(tc)I + sin(tc)⃗ vc ∈ Long⃗ .
v Essentially, there is a one-parameter subgroup of SLn for each ⃗ v for some vector ⃗ v on the equator, where tA the coefcient c in A = c⃗ v determines the speed at which the subgroup is swept out. The image of e simply corresponds to the longitude given by ⃗ v. aActually going through the process is slightly tedious and not that informative; try it for yourself if you want to!
q bBecause there is no requirement that x + x + x = 1, A does not actually have to be on the equator, but it is just some multiple c of a vector ⃗ v on the equator. When c = 1, A is actually on the equator, by the characterization we gave in a previous lecture. cThis is given by simply writing out the expansion of the matrix exponential as a sum and collecting the terms into a Taylor series for cosine and a Taylor series for sine; since ⃗ v is on the equator, ⃗ v = −1, and so the result is essentially the same as the result that eiθ = cos θ + i sin θ. dThis is the characterization given in a previous lecture. 32.4 Tangent Vectors So far, several examples have been shown, but we would like to see what else we can say more generally about one-parameter subgroups. To do so, we will introduce some new tools. Guiding Question What else can we say about the setsa of matrices A defning one-parameter groups? aIn fact, they will be vector spaces Since they are given by derivatives at t = 0, corresponding to the identity, the matrices A are "tangent vectors" to G ≤ GLn(R) at the identity I. Intuitively, a tangent vector at some point is a vector lying in the tangent plane. 158 Lecture 32: More About One-Parameter Subgroups There are three approaches to rigorously defning tangent vectors. 1. The approach taken so far is that a tangent vector will be given by a matrix A ∈ Matn×n(R) such that 102 the corresponding one-parameter group etA ∈ G for all t ∈ R. 2. The second approach is more general, and builds on the idea that a one-parameter group is the trajectory of a particle moving in G in a specifc way defned by φ. Instead of taking a path that happens to be a homomorphism, take any diferentiable path103 from some interval f : (−ε, ε) − → GLn(R) such that f(0) = I and f(t) ∈ (G) for all t. Then a tangent vector is simply the velocity at time t = 0, 104 f ′ (0) ∈ Matn×n(R). It is not obvious, but it turns out that the frst defnition is equivalent to the second defnition, giving the same subsets of matrices A as tangent vectors. For the frst defnition, the advantage is that each tangent vector corresponds to only one path through the identity, so there is a bijection. The second approach gives lots of paths through the identity that have a given velocity vector, but it turns out that it is easier to use it to show that the set of tangent vectors actually forms a vector space, called the tangent space.105 3. Suppose G is defned by polynomial constraints on the matrix entries.106 For example, the constraints could be that it is upper triangular; then aij = 0 for all i > j is a bunch of polynomial conditions on the matrix entries. Orthogonality can also be phrased as a polynomial condition. Then, when working with polynomials, the derivative can be mimicked without actually needing to know 107 analysis. We work with an object R[ε] := R + Rε where ε2 = 0. This allows us to defne a derivative without actually taking any limits. For example, for f(x) = x2 + 2x, evaluating f on x + ε ∈ R[ε] will give f(x + ε) = (x + ε)2 + 2(x + ε) = x 2 + 2xε + ε2 + 2x + 2ε = x 2 + 2xε + 2x + 2ε = (x 2 + 2x) + (2x + 2)ε. So with this funky multiplication, any terms of order more than two in ε disappear, and we get f(x+ε)−f(x) = ε f ′ (x), even though we haven’t actually defned the derivative from an analysis perspective. The upshot is that to fnd A ∈ Matn×n(R), we simply look at A with the property that In + εA, the identity matrix perturbed by A, satisfes the same system of equations defning G, but in the sense of the funky multiplication of R[ε], where ε2 = 0. We’ll explain this more on Friday. 102Every matrix A will defne some path in GLn; if the entire path lives in G, then we can think of A as a tangent vector. 103Any matrix-valued function 104The idea is that all the paths through the identity will give velocity vectors (with lots of redundancy), which we will consider as tangent vectors. 105There is actually more structure on it, which we will talk about on Friday. 106This approach is not necessary for this class, but it is fun, so we will do it. 107Here ε is not some number in R, so it is not actually true that ε must be 0; it is simply a formal construct where we impose the √ condition that ε2 = 0. It is similar to the defnition of i = −1; there is no such real number, so we simply defne some number satisfying this property. In the same way, we simply defne R[ε] to be R + Rε for some object ε satisfying ε2 = 0. 159 Lecture 33: Lie Groups 33 Lie Groups 33.1 Review Last time, we discussed one-parameter groups G ≤ GLn(R). We started thinking about tangent vectors to the group, based at some identity. Defnition 33.1 The collection of all tangent vectors at I ⊂ G, called the tangent space, can be characterized in multiple ways. We call this Lie(G), pronounced "lee." 1. The frst defnition is the most familiar one. The tangent vectors are the matrices such that the associated one-parameter subgroup lies in G. We found that there is a bijection between matrices and one-parameter groups lying in G. Lie(G) = {A ∈ Matn×n(R) : e tA ∈ G, t ∈ R} 2. More generally, we consider any path inside the group through the identity, not just a one-parameter group, and take A, the velocity at the identity, to be a tangent vector.108 Lie(G) = {A ∈ Matn×n(R) : ∃f : (−ε, ε) − → G, f(0) = I, f ′ (0) = A} 3. The third approach is slightly stranger. It is less general, and requires G to be defned by polynomial constraints.109 In this case, we take this strange construction R[ε] = R ⊕ Rε = {a + bε : a, b ∈ R}, where we defne the multiplication to be such that 110 ε2 = 0. 108For defnition 1, we know all the 1-parameter subgroups, but for defnition 2, there are lots of other possible paths. So for defnition 1, there is a bijection between the matrices A and the one-parameter groups, while for defnition 2, there are lots of diferent paths with the same tangent vector as the velocity. Defnition 2 does not use the fact that G is a group. 109For example, setting the determinant to be 1 is some complicated polynomial constraint. 110This is similar to how we could defne the complex numbers, where we set i2 = −1. 160 Lecture 33: Lie Groups If f is a polynomial, then we can set, formally, the derivative to be f ′ (x) = f(x + ε) − f(x). This is a way of thinking about the derivative without limits, for polnyomials. Then, we take Lie(G) = {A ∈ Matn×n : I + εA satisfes the polynomial constraints defning G}. 33.2 Lie Groups These defnitions are quite abstract, so let’s see them in action for On. Example 33.2 Let G = On, the set of matrices such that AT A = I. 1. The Lie group, as we have seen in the previous lecture, is Lie(On) = {A : AT = −A}. These are the skew-symmetric matrices. 2. Consider a path passing through the identity at zero: f : (−ε, ε) − → On such that f(0) = I. By the defnition of an orthogonal matrix, f(t)T · f(t) = I. Taking the derivative, f ′ (t)T · f(t) = f(t)T · f ′ (t) = 0, and taking t = 0 gives AT I + IA = 0, and thus AT = −A. So the same condition holds. 3. The condition that AT A = I is a set of complicated polynomial conditions. From defnition 3), the Lie group consists of matrices A such that (I + εA)T (I + εA) = I, using the rule that ε2 = 0. Multiplying this out, I + εAT + εA + ε2AT A, and taking ε2 = 0, I + εAT + εA = I, which implies that AT = −A after dividing both sides by ε. All three defnitions lead to the same Lie group, despite being very diferent. The third defnition is useful because it makes sense even without working over the real numbers, and works for any group defned by polynomial constraints!111 For example, the Lie group can be defned for orthogonal matrices over fnite felds. Here are some non-obvious facts about these characterizations of the tangent space at the identity. 111The intuition for this third defnition is that it is essentially using the Taylor expansion of the path through the identity, and ignoring third order and higher terms. 161 Lecture 33: Lie Groups Proposition 33.3 For Lie(G) : • All three defnitions are actually equivalent. • For a group G, Lie(G) is actually a vector subspace of Matn×n(R). It is surprising that Lie(G) is a vector space, since the matrix exponential does not generally behave well with respect to addition if A and B do not commute.112 33.3 Manifolds In order to understand Lie groups, we have to think about the notion of a manifold. For this section, the discussion will be less rigorous and precise, and it is okay not to understand all the defnitions; we are just providing the favor of concepts that will show up in later classes. Defnition 33.4 For M a subset of Rn , M is a (diferentiable) manifold of dimension d if for each x ∈ M, there exists an open set containing x V ⊆ M, an open ball U ⊂ Rda , and a continuous (diferentiable) bijection f : U − → V. aAn open ball is a subset of Rd of the form U = {x : |x| < δ} for some δ. Globally, a d-dimensional does not look like Rd , but locally, it does. The circle is an example of a 1-dimensional manifold; at each point on the circle, there is really only one direction to move in. Example 33.5 (Circle) Consider some interval on the real line. Then, it is possible to write down some function bijectively mapping that interval to some other interval on the circle. This can happen around any point on the circle, so it is a manifold. tA tB tC 112Using the frst defnition, it is not clear that e + e can be written as e . 162 Lecture 33: Lie Groups Loosely, for defnition 2, we have tangent vectors at 0 in U ⊂ Rd corresponding to tangent vectors at x in M, which in a way brings the vector space structure from Rd to M. A non-example would be the union of the x-axis and the y-axis, since at the origin, there is an intersection that does not look like an interval. There are two directions to move in, instead of one direction. All our examples of G ≤ GLn(R) are manifolds, not just groups. This requires some argument, but it is true. 33.4 Lie Bracket For every group G, there is a corresponding vector space structure Lie(G) ⊂ Matn×n(R) = Lie(GLn(R)). In fact, Lie(G) carries some extra structure. The multiplication structure A, B ⇝ AB ∈ Matn×n(R) does not preserve Lie(G); if A, B ∈ Lie(G), it does not mean that AB ∈ Lie(G). For example, Lie(On) = {A : AT = −A}, so for A, B ∈ Lie(G), (AB)T = BT AT = BA, which is not usually equal to −AB. However, a very similar structure does preserve the Lie group. Defnition 33.6 (Lie Bracket) Let the Lie bracket of A, B ∈ Matn×n(R) be [A, B] := AB − BA ∈ Matn×n(R). Theorem 33.7 For any G ≤ GLn, the Lie group G ⊆ Matn×n(R) is preserved by the Lie bracket: A, B ∈ Lie(G) → [A, B] ∈ Lie(G) The Lie group is not just a vector space — it is actually a vector space with some weirdo multiplication on it! The Lie bracket can be seen in action for some of the Lie groups we have seen already. Example 33.8 For G = On, Lie(On) = {A : AT = −A}. Then for A, B ∈ Lie(On), [A, B] = AB − BA, and [A, B]T = BT AT − AT BT = BA − AB = −[A, B]. So [A, B] ∈ Lie(On). Example 33.9 For SLn(R), Lie(SLn()) = {A : trace(A) = 0}. For matrices A, B, trace(AB) = trace(BA), and so trace([A, B]) = 0. The commutator in the group corresponds to the Lie bracket. 163 Lecture 33: Lie Groups Proof. For A, B ∈ Lie(G), then consider etA ∈ G and esB ∈ G. Then we know that tA sB −tA −sB ∈ G, e e e e and Taylor expanding gives (I + tA + · · · )(I + sB + · · · )(I − tA + · · · )(I − sB + · · · ) ∈ G, which, to the frst order, gives I + st[A, B] + · · · ∈ G. Then, taking the derivative at 0 gives [A, B] ∈ Lie(G). The Lie bracket comes from the fact that the group is closed under multiplication and taking the derivative at 0 for that. As a corollary, if G is abelian113 , then the Lie bracket is identically 0 on all of G. In some way, for an arbitrary group G, the Lie bracket "measures" the failure of the group to be abelian. We started out with groups that we cared about (matrix groups), looked at the set of tangent vectors, which is the same as the set of one-parameter groups, and looked at the vector space structure which also has a funky Lie bracket, and now we will look at properties of this bracket. Here are some nice properties of the Lie bracket. • Antisymmetry. [A, B] = −[B, A] • The Jacobi identity. We have [[A, B], C] + [[B, C], A] + [[C, A], B] = 0. It is true simply when expanding. This also comes from a property of the group; we won’t do it but you can get it by staring at a more complicated version of how we derived the Lie bracket. Defnition 33.10 A Lie algebra (over R) is a vector space V with a Lie bracket [·, ·] : V ×V − → V satisfying [A, B] = −[B, A] and the Jacobi identity. On the homework, we see that R2 with the cross product is a Lie algebra, and it is Lie(SO3) = Lie(SU2). If G is a group that is also a manifold, it is called a Lie group, and if G is a Lie group, it can be replaced with Lie(G), a Lie algebra, which is simply a vector space with a weird multiplication on it, which is a lot easier to study. However, the Lie algebra carries a lot of information about G. Theorem 33.11 Given a Lie algebra (fnite dimensional over R) V , there exists a unique Lie group G such that Lie(G) = V. Lie theory ends up being a very powerful tool in the study of understanding groups. 113For example, diagonal matrices 164 Lecture 34: Simple Linear Groups 34 Simple Linear Groups 34.1 Review Last time, we took a group G ≤ GLn(R) and looked at Lie(G), the vector space of tangent vectors at the identity. We had lots of diferent defnitions, but the most familiar way is to think of Lie(G) as all the matrices that provide one-parameter groups inside of G. The key point was that not only is Lie(G) a vector space, but it also has some extra structure: the Lie bracket [A, B] = AB − BA, which is a skew-symmetric multiplication on the vector space. This is a new operation, but it also arises naturally from considering multiplication on G, and sort of taking its derivative. The Lie bracket measures the failure of G to be commutative. We don’t actually need G to be a subgroup of GLn(R) in order to do this. For any group with a manifold structure, called a Lie group, we can look at the tangent vectors through the identity, and we get a vector space which also has a bracket multiplication, which is its Lie algebra Lie(G). Student Question. Can you recover the group from its Lie algebra? Answer. In general, the answer is no. For example, SU2 and SO3 have the same Lie algebra. But it turns out that if two groups have the same Lie algebra, they’re related to each other – here SU2 and SO3 difer by a “fnite amount” (we have a two-to-one surjection SU2 → SO3), and you can show something similar is true in general. In fact, if you require the group to be simply connected, then there is a unique group with a given Lie algebra. You can then use this to study all the groups with a given Lie algebra. This ends up being a powerful tool because now in order to understand a group with a manifold structure, we can instead try to understand its Lie algebra, which is an easier problem; and then fgure out how to go backwards. 34.2 Simple Linear Groups Recall that a group is simple if its only normal subgroups are the trivial group and the whole group. You can think of simple groups as building blocks for more complicated groups – if you have a group that isn’t simple, then you can understand it by understanding the normal subgroup and its quotient group. But if you have a simple group, you can’t break it down further. Guiding Question Which G ≤ GLn(R) are simple? We’ll look at two groups: SU2 and SL2. 34.3 The Special Unitary Group First we’ll look at whether SU2 is simple. The answer is no – the center of a group (the set of elements which commute with everything) is always a normal subgroup. But the center of SU2 is ±I, which is nontrivial. But it turns out that’s essentially the only thing that can happen: Theorem 34.1 If N ⊴ SU2, then N must be {I}, SU2, or {±I}. So then in order to produce a simple group, we can quotient out by this center: Corollary 34.2 The quotient SU2/{±I} is simple. This quotient is actually SO3: we have a surjective homomorphism SU2 → SO3 (from the conjugation action on the equator), and the kernel is exactly {±I}. So in particular, SO3 is simple. Proof of Corollary 34.2. This follows from the Correspondence Principle: we have a surjection φ : SU2 → SO3. So for any normal subgroup N ⊴ SO3, its pre-image is a subgroup of SU2, which is also normal and contains the kernel {±I}. But any normal subgroup of SU2 containing the kernel is equal to either the kernel or the 165 Lecture 34: Simple Linear Groups whole group, which means the initial normal subgroup in SO3 is either the identity or the entire group (by taking the image under φ). Proof of Theorem 34.1. We can use the geometric intuition of what SU2 looks like. It’s a 3-sphere in 4 dimensional space, and its conjugacy classes are exactly the latitudes – the 2-spheres we get from taking horizontal slices. Suppose we have N ⊴ SU2, and some element Q ∈ N which is not ±I. We’d like to show that N must then be the entire group. Q I Since N is normal, everything conjugate to Q must also be in N. If Tr(Q) = 2c, then Latc is the conjugacy class of Q, so this entire latitude must be inside N. (This latitude is a 2-sphere of positive radius, since Q is not ±I.) Now we can take this 2-sphere, and translate it to pass through the identity: consider Q−1 Latc, which is a 2-sphere (of positive radius) passing through the north pole I. This must also be contained inside N (since Q and Latc are both contained inside N). Q I Now start at I, and take a nontrivial path inside this two-sphere. Write this path as f(t) for 0 ≤ t ≤ ε, where we have f(t) ∈ Q−1 Latc for all t, and f(0) = I, while f(t) = ̸ I for t > 0. 166 Lecture 34: Simple Linear Groups Q I Then f(t) is contained in Q−1 Latc, and therefore inside N. And the north pole is the only point in the sphere with trace 2, so Tr(f(t)) < 2 for all t > 0. This means there exists some δ > 0 such that for all 2 − δ < c < 2, there exists t such that Tr(f(t)) = c. But once we fnd that N contains one point on a given latitude, then N must contain every point on that latitude (because N is normal, so it must contain the entire conjugacy class of that point). So this means for every c such that 2 − δ < c < 2, the latitude Latc/2 is contained inside N. Q I This means we have an entire neighborhood of the identity that’s contained inside N – any point A ∈ SU2 with Tr(A) > 2 − δ is contained inside N. Now we’re almost done: we want to show that once we have a neighborhood of the identity, we have all points. To do this, we can look at the longitudes: pick some v on the equator, and look at Long ≤ SU2. Every point v on the three-sphere is in some longitude, so it sufces to show that every longitude is contained in N. I 167 Lecture 34: Simple Linear Groups But this longitude is the circle {cos θI + sin θv | 0 ≤ θ < 2π}. And we know that if θ is small enough, then the point ρθ corresponding to θ is in N (meaning there exists ε such that ρθ ∈ N for all |θ| < ε). I ρθ θ Now we can take ρθ for small θ, and multiply it by itself repeatedly to cover the entire circle: for any φ ∈ [0, 2π), φ there is some M such that < ε. Then ρϕ/M is in N, which means ρφ = (ρφ/M )M is in N as well. M So now every longitude is in N, and since every point is in a longitude, this means all of SU2 is in N. Note 34.3 This is a really hands-on argument; we used the fact that we have both a geometric and group-theoretic understanding of what SU2 is. 34.4 The Special Linear Group Now we’ll look at SL2(C) – the group of 2 × 2 matrices with determinant 1. Again, ±I is the center of SL2(C), so the most we could ask for is that if we quotient out by this normal subgroup, we get a simple group. It turns out that this is the case, and it actually works for any feld F , not just the complex numbers: Theorem 34.4 For any feld F with |F | ≥ 4, the quotient SL2(F )/{±I} is simple. This quotient is sometimes called P SL2(F ). We’re no longer in a geometric setting (F can be a fnite feld), so the proof won’t be geometric like the previous one; instead, we’ll get our hands on generators and relations. Note 34.5 The theorem is false over F2 and F3. This is similar to how when we looked at the alternating groups, we saw that An is simple for all n ≥ 5, but A4 and A3 are not simple. In both cases, we have a family of groups where the frst couple may be counterexamples, but eventually they all become simple. Proof of Theorem 34.4. We’ll prove this in the case where |F | > 5 (there’s only two remaining cases, which can be checked by hand). It sufces to prove that if we have a normal subgroup N ⊴ SL2(F ), then N is either {I}, {±I}, or SL2(F ) – then this implies the quotient is simple by the same argument as before, using the Correspondence Principle. We’ll begin with a few lemmas about the feld: 168 Lecture 34: Simple Linear Groups Lemma 34.6 Given a ∈ F , the equation x2 = a has at most 2 solutions. This seems obvious, but it’s possible to have more solutions in Z/nZ if n is not prime. So it matters that F is a feld. 2 2 Proof. If we have two solutions x and y, then x = y , so (x + y)(x − y) = 0. But because F is a feld, if two elements multiply to 0, then one of them is 0. So either x = y or x = −y; this means if there’s one solution, then there’s at most one other solution. Lemma 34.7 If |F | > 5, then there exists some r ∈ F such that r2 is not 0, 1, or −1. Proof. There’s at most one square root of 0, two square roots of 1, and two square roots of −1. So there are at most fve bad elements; and as long as there are more than 5 elements in the feld, we can fnd some good element (whose square isn’t 0 or ±1). 2 Now we’ll prove the main theorem. Fix an element r with r ̸∈ {0, 1, −1}. Assume that we have a normal subgroup N ⊴ SL2(F ) containing some element other than ±I; then we’ll show that N is the entire group. Claim 34.8. We can fnd some B ∈ N with distinct eigenvalues. Proof. Pick some A ∈ N, with A ̸= ±I. Then A is not a scalar matrix, so there is some vector v1 ∈ F 2 which is not an eigenvector of A. Then take v2 = Av1. Since v1 is not an eigenvector, v1 and v2 are linearly independent, so {v1, v2} is a basis for F 2 . −1 Now defne P ∈ GL2(F ) with the property that Pv1 = rv1 and Pv2 = r v2. Then P is diagonal in the basis −1 {v1, v2}, and its eigenvalues are r and r . So the determinant of P is 1, which means P ∈ SL2(F ). We don’t know whether P is in N. But we can defne B = AP A−1P −1 . We claim that B is inside N – we know A is in N. Meanwhile, PA−1P −1 is the conjugate of an element of N, so it must also be in N (since N is normal). Since N is a subgroup, their product must be in N as well. But we have 2 Bv2 = AP A−1P −1 v2 = AP A−1 rv2 = AP rv1 = Ar2 v1 = r v2. 2 So r is an eigenvalue of B, and r−2 is the other eigenvalue (because det B = 1). By our choice of r, we have 2 −2 2 r ̸= r , since r = ̸ ±1. So this concludes the frst step. 2 −1 Set s = r , so we have a matrix B ∈ N with distinct eigenvalues s and s . −1 Claim 34.9. All matrices in SL2 with eigenvalues s and s are contained in N. Proof. We’ll show that this set of matrices is actually a single conjugacy class in SL2. It contains B, so then since N is normal, this implies the entire set is contained in N. −1 Given Q with eigenvalues s and s , we know Q is diagonalizable (since it has distinct eigenvalues), so we have LQL−1 = s 0 0 −1 s for some L ∈ GL2(F ). Then we can take L ′ = det L−1 0 0 L. 1 This has determinant 1, so it’s in SL2; and it has the same property (that that L ′ QL′−1 is diagonal). So all matrices with eigenvalues s and s−1 are in the same conjugacy class of SL2. 169 Lecture 34: Simple Linear Groups −1 To fnish, we can look at all matrices generated (as a group) by the matrices with eigenvalues s and s . We can get a huge collection of matrices like this: for example, we can show that any matrices of the form 1 0 x 1 and 1 x 0 1 are in this group. Then there was a homework question from a long time ago that showed matrices of these forms generate all of SL2. Both our proofs had similar ideas: fnd some element in the normal subgroup, conjugate it to fnd a whole bunch of elements in the normal subgroup, and use those elements to generate the entire group. 34.5 Generalizations We focused on 2 × 2 matrices here, but these examples are actually typical, and generalize to higher dimensions. If a subgroup G ≤ GLn(C) is defned by polynomial constraints (for example, we can require that the matrix has determinant 1, but we can’t take complex conjugates – so the unitary groups are not in this category, but the orthogonal groups are), then you can actually classify which ones are simple. We’ve essentially seen these already: you can take SLn modulo its center, and SOn modulo its center. You can also take a skew-symmetric form instead (which we didn’t really discuss in class), modulo its center. These are almost all the simple groups – there’s just fve other examples. The proof uses the idea of passing to the Lie algebra Lie(G) – you frst understand what a simple Lie algebra is, and use that to study what the simple Lie groups are. The really remarkable thing is that understanding these matrix groups also lets you understand fnite simple groups – if you replace C with a fnite feld, then these examples give fnite simple groups, and these are almost all the known examples (with 26 exceptions). 170 Lecture 35: Hilbert’s Third Problem 35 Hilbert’s Third Problem 35.1 Polygons in the Plane Defnition 35.1 Given polygons P and Q on the plane, P is scissors-congruent to Q (denoted P ∼ Q) if we can divide P , using fnitely many straight cuts, into a set of polygons R1 through Rn; and we can divide Q into the same collection R1, . . . , Rn. Example 35.2 This triangle and quadrilateral are scissors-congruent: Guiding Question Given two polygons P and Q, when is P ∼ Q? There’s one obvious obstruction: we need P and Q to have the same area, since the area is preserved by rearranging pieces. This is actually an if and only if condition: Theorem 35.3 If P and Q have the same area, then P ∼ Q. Proof Outline. The idea is to rearrange both P and Q to a rectangle with dimensions 1 × area(P ), where A is the common area – if P and Q are scissors-congruent to the same rectangle, then they’re scissors-congruent to each other. First, cut P into triangles T1, . . . , Ts. Then each triangle T is scissors-congruent to some rectangle: cut the triangle as shown, and move the red pieces to the blue triangles. Then we can show that any rectangle R is scissors-congruent to another rectangle with dimensions 1 × area(R). This part is a bit fnicky and involves a bunch of cases, so we’ll skip it. Finally, we have a bunch of triangles, which are each scissors-congruent to a rectangle of height 1. So we can concatenate all these rectangles, to get a bigger rectangle of height 1. This means P is scissors-congruent to a 1 × area(P ) rectangle, and by performing the same argument for Q, then P ∼ Q. 35.2 The Question We can now consider what happens in 3 dimensions. Defnition 35.4 If P and Q are polytopes (meaning they have fnitely many vertices, edges, and faces, and each face is a polygon), then P ∼ Q if you can use fnitely many straight cuts to decompose P and Q into the same polytope pieces. 171 Lecture 35: Hilbert’s Third Problem Again, there’s an obvious condition: if P ∼ Q, then they must have the same volume. Guiding Question (Hilbert’s Third Problem) If two polytopes have the same volume, are they scissors-congruent? In 1900, David Hilbert made a list of around twenty problems, which he considered the most important problems in modern mathematics. These problems were very infuential. This question was on that list, and he was pretty sure that the answer was no. In fact, this was the frst one of his questions to be answered (in 1901), by his student Max Dehn. He showed more precisely that a cube and a tetrahedron of the same volume are not scissors-congruent. 35.3 Some Algebra At the heart of this problem is a certain algebraic construction. Defnition 35.5 Given two abelian groups G and H, their tensor product G⊗H is the abelian group generated by elements of the form g ⊗ h for g ∈ G and h ∈ H, satisfying the relations ′ (g + g ′ ) ⊗ h = g ⊗ h + g ⊗ h, and similarly g ⊗ (h + h ′ ) = g ⊗ h + g ⊗ h ′ . You can think of G ⊗ H as taking M Z(g ⊗ h) g,h (which gives an integer for every pair (g, h)), and then quotienting out by the subgroup generated by ((g + g ′ ) ⊗ ′ h) − (g ⊗ h) − (g ⊗ h) and so on. This gives a group; and because we’ve quotiented out by the subgroup, now our generators satisfy the given relations. L Student Question. Is the direct sum Z(g ⊗ h) the free group? Answer. Not exactly – it’s a free abelian group, since everything commutes. But if you’d like, you can instead start with the free group, and throw the condition that everything commutes into the quotient as well. We can think of elements of G ⊗ H as expressions combining the terms g ⊗ h, which we can simplify using the given relations. Proposition 35.6 The defnition has a few consequences: • 0 ⊗ h = g ⊗ 0 = 0. • If a ∈ Z, then (ag) ⊗ h = a(g ⊗ h) = g ⊗ ah (by using linearity). • If we take lists of generators G = ⟨g1, . . . , gr⟩ and H = ⟨h1, . . . , hs⟩, then G ⊗ H = ⟨gi ⊗ hj ⟩. Example 35.7 We have Z ⊗ G ∼ = G. Proof. We must have (a ⊗ g) 7→ ag = 1 ⊗ ag. (So the isomorphism sends (a ⊗ g) to ag, and its inverse sends g to 1 ⊗ g.) Example 35.8 We have (Z2) ⊗ G ∼ = G × G. 172 Lecture 35: Hilbert’s Third Problem Proof. We must have (a, b) ⊗ g 7→ (ag, bg). Note that in general, not everything in G ⊗ H is of the form g ⊗ h: you can have expressions g1 ⊗ h1 + g2 ⊗ h2 which you can’t simplify any further using these relations. So this isn’t the same as looking at ordered pairs. Example 35.9 We have C2 ⊗ C3 = 0. Proof. Take x ⊗ y for any x ∈ C2 and y ∈ C3. We have 3x = x, so then x ⊗ y = 3x ⊗ y = x ⊗ 3y = x ⊗ 0 = 0, since 3y = 0 in C3. So you can start with two nontrivial groups and take their tensor product, and get the 0 group. If we just took the product, that would be a nontrivial group of size 6. So the tensor product is somewhat subtle. Student Question. Does this happen with 2 and 3 replaced by any distinct positive integers? Answer. Not necessarily distinct, but it does happen if the integers are relatively prime. 35.4 Back to Polytopes If we have two polytopes of the same volume, we need another way to see they aren’t scissors-congruent. Given a polytope P , each edge has a length ℓ, and a dihedral angle θ – where we take the perpendiculars to the edge on each face, and look at the angle between them. ℓ θ Then ℓ and θ are real numbers, so we can take the tensor product ℓ ⊗ θ ∈ R ⊗ R/2πZ. (Here R and R/2πZ are both uncountably generated infnite groups.) Then every edge gives an element in this tensor product, so we can sum over the edges: Defnition 35.10 The Dehn invariant of a polytope P is X d(P ) = ℓi ⊗ θi ∈ R ⊗ R/2πZ, i where the sum is taken over all edges of P . We’ll see that the Dehn invariant is preserved by scissors-congruence – then we can use it to tell whether two polytopes are scissors-congruent, similarly to how we can use volume. And it turns out that we can actually calculate it for some examples – a cube and regular tetrahedron – and see that they give diferent results. Theorem 35.11 The Dehn invariant is preserved by scissors-congruence: if P ∼ Q, then d(P ) = d(Q). 173 Lecture 35: Hilbert’s Third Problem Proof. Consider what happens to the polytope when we cut it. First, we’ll look at what happens to the original edges. One possibility is that the edge is cut into two pieces, while the dihedral angle on each side remains the same: θ ℓ1 ℓ2 Then we have ℓ = ℓ1 + ℓ2, so ℓ ⊗ θ = ℓ1 ⊗ θ + ℓ2 ⊗ θ. Another possibility is that we cut by a plane that contains this edge. Then the entire edge remains intact, but the angles change: now we have two polytopes each with an edge of length ℓ, and angles θ1 and θ2 respectively. θ1 θ2 In this case, we have θ = θ1 + θ2, so ℓ ⊗ θ = ℓ ⊗ θ1 + ℓ ⊗ θ2. In both cases, we used the linear property of the tensor product. Meanwhile, we can also create new edges: when we make a cut along a face, this produces an edge on each of the two pieces. 174 Lecture 35: Hilbert’s Third Problem ℓ θ1 θ2 Then we had a contribution of 0 previously, and a contribution of ℓ ⊗ θ1 + ℓ ⊗ θ2 after the cut. But we have θ1 + θ2 = π, so ℓ ℓ ⊗ θ1 + ℓ ⊗ θ2 = ℓ ⊗ π = ⊗ 2π = 0 2 (since the second group is R/2πZ). The fnal case is if we have a new edge in the interior of the polytope. Then we have a bunch of angles which add up to 2π: So the same thing happens: when we add up its contributions, we have ℓ ⊗ θ1 + · · · + ℓ ⊗ θk = ℓ ⊗ 2π = 0. So in all cases, the Dehn invariant is preserved by cutting the polytope. Student Question. Does the argument used to show ℓ ⊗ π = 0 imply that any ℓ ⊗ θ is 0? Answer. No – we can only scale by integers. So this shows that if θ is a rational multiple of π, then ℓ ⊗ θ = 0. But we’ll see later that if θ isn’t a rational multiple of π, we get something nonzero. But this is something to watch out for – it happens a scary number of times that someone defnes a complicated invariant, but it turns out the invariant is always 0. Note 35.12 We haven’t thoroughly gone through all possibilities, but this gives a favor of why the theorem is true – in every case, if we compare the old and new contributations, then the relations we quotiented out by imply that the invariant doesn’t change. Now we’ll show that the invariant actually does something interesting. Theorem 35.13 If C is a cube and T is a regular tetrahedron, then d(C) = d(T ). Proof. A cube has 12 edges, each of which has dihedral angle π . So we get 2 π d(C) = 12ℓ ⊗ = ℓ ⊗ 6π = 0. 2 175 Lecture 35: Hilbert’s Third Problem So the Dehn invariant of a cube is always 0. For a regular tetrahedron, we have 6 edges which are symmetric, so we get d(T ) = 6ℓ ′ ⊗ α for some angle α. α To fnd α, we can drop an altitude. This altitude ends at the center of the base: So the short side of the right triangle is a third of the height of the base. α 1 3h h Then, since the two triangles have the same height, we have 1 cos α = . 3 So we’ve found α, and now we want to show that for this value of α, the Dehn invariant is nonzero. Claim 35.14. α is not a rational multiple of π. 2aπ Proof Outline. One way to prove this is by trig identities – suppose α = , and then use trigonometric b identities to write cos bα = (cos α)b2b + · · · . 1 Then there’s an enormous power of 3 in the denominator, since cos α = , so this can’t be 1.
3 Claim 35.15. For any α ̸∈ Qπ, and any ℓ ∈ R, ℓ ⊗ α is nonzero. 176 Lecture 35: Hilbert’s Third Problem Proof. Think of R as a vector space over Q (of uncountable dimension). Then π and α are linearly independent, so we can fll them out into a basis: we can write R = Qα ⊕ Qπ ⊕ W, where W comes from the uncountably many remaining basis elements. Then we can defne a linear map f : R → Q (a map of Q-vector spaces) sending α to 1, and every other basis element to 0 – we can do this because to defne a linear map, it sufces to defne it on each basis element. Then we have a group homomorphism g : R ⊗ R/2πZ → R, where z ⊗ x 7→ zf(x ˜), where x ˜ is x mod 2π – since 2π ∈ ker(f), this is well-defned. (We can also check that this is compatible with the relations of the tensor product.) Now we have g(ℓ ⊗ α) = ℓ ̸= 0. And since g(ℓ ⊗ α) is nonzero, then ℓ ⊗ α must be nonzero as well. So d(T ) = ℓ ′ ⊗ α is nonzero, which means d(C) = ̸ d(T ). The Dehn invariant lies in some enormous group. In this proof, in order to show it was nonzero, we mapped it to a much simpler group – the real numbers – and showed instead that its image is nonzero. Note 35.16 This was proven in 1901. In 1968 Sydler showed the converse – that if P and Q have the same volume and Dehn invariants, then they’re scissors-congruent. This is known in dimension 4 as well. But in higher dimensions, it’s not actually known how to characterize when two polytopes are scissors-congruent. 177 MIT OpenCourseWare Resource: Algebra I Student Notes Fall 2021 Instructor: Davesh Maulik Notes taken by Jakin Ng, Sanjana Das, and Ethan Yang For information about citing these materials or our Terms of Use, visit: |
4439 | https://www.nature.com/articles/srep27690 | Azole Antifungal Sensitivity of Sterol 14α-Demethylase (CYP51) and CYP5218 from Malassezia globosa | Scientific Reports
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Azole Antifungal Sensitivity of Sterol 14α-Demethylase (CYP51) and CYP5218 from Malassezia globosa
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Published: 13 June 2016
Azole Antifungal Sensitivity of Sterol 14α-Demethylase (CYP51) and CYP5218 from Malassezia globosa
Andrew G. S. Warrilow1,
Claire L. Price1,
Josie E. Parker1,
Nicola J. Rolley1,
Christopher J. Smyrniotis2,
David D. Hughes3,
Vera Thoss3,
W. David Nes4,
Diane E. Kelly1,
Theodore R. Holman2&
…
Steven L. Kelly1
Show authors
Scientific Reportsvolume 6, Article number:27690 (2016) Cite this article
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Abstract
Malassezia globosa cytochromes P450 CYP51 and CYP5218 are sterol 14α-demethylase (the target of azole antifungals) and a putative fatty acid metabolism protein (and a potential azole drug target), respectively. Lanosterol, eburicol and obtusifoliol bound to CYP51 with K d values of 32, 23 and 28 μM, respectively, catalyzing sterol 14α-demethylation with respective turnover numbers of 1.7 min−1, 5.6 min−1 and 3.4 min−1. CYP5218 bound a range of fatty acids with linoleic acid binding strongest (K d 36 μM), although no metabolism could be detected in reconstitution assays or role in growth on lipids. Clotrimazole, fluconazole, itraconazole, ketoconazole, voriconazole and ketaminazole bound tightly to CYP51 (K d ≤ 2 to 11 nM). In contrast, fluconazole did not bind to CYP5218, voriconazole and ketaminazole bound weakly (K d ~107 and ~12 μM), whereas ketoconazole, clotrimazole and itraconazole bound strongest to CYP5218 (K d ~1.6, 0.5 and 0.4 μM) indicating CYP5218 to be only a secondary target of azole antifungals. IC 50 determinations confirmed M. globosa CYP51 was strongly inhibited by azole antifungals (0.15 to 0.35 μM). MIC 100 studies showed itraconazole should be considered as an alternative to ketoconazole given the potency and safety profiles and the CYP51 assay system can be used in structure-activity studies in drug development.
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Introduction
Malassezia species are known to play a role in several human skin diseases including dandruff, seborrheic dermatitis, pityriasis versicolor and malassezia folliculitis and may also exacerbate atopic dermatitis and psoriasis1,2,3,4,5 even though they are members of the normal skin microbial flora, being present on the skin of 75 to 98% of healthy individuals6. There are presently fourteen recognized species of Malassezia, eight of which are associated with humans7. Malassezia are unique amongst fungi in requiring exogenous lipids for growth. Malassezia species have evolved through an expansion of lipid hydroxylases (14 lipases in M. globosa) and a reduction in carbohydrate metabolism genes which may have contributed to the evolution of the genus from plant to animal inhabitants8. The predominant species found in dandruff and seborrheic dermatitis sufferers are elevated levels of M. globosa and M. restricta9,10 preferring sebum-rich areas of skin, degrading the sebum by secreting lipases11. Dandruff may affect 50% of individuals and the more severe seborrheic dermatitis affects 1 to 3% of the population3,4.
The most commonly used antifungal agents to treat topical M. globosa infections are ketoconazole12,13 targeting sterol 14α-demethylase (CYP51), hydroxypiridones3,14,15,16,17, especially ciclopiroxolamine17 which targets a variety of metabolic processes including depletion of trivalent metal cations18 and zinc pyrithione19,20 which causes elevated copper levels in yeast cells21. Alternate drug targets in M. globosa being evaluated include β-carbonic anhydrase inhibition by sulfonamide, sulfamates and sulfamide drugs22.
Azole antifungal agents selectively target fungal CYP51 enzymes over the human homolog23 through the direct coordination of the triazole ring N-4 nitrogen (fluconazole, itraconazole and voriconazole) or the imadazole ring N-3 nitrogen (clotrimazole and ketoconazole) with the heme iron as the sixth axial ligand24. Azole antifungals can also coordinate with the heme iron of other cytochrome P450 enzymes, raising the possibility of alternative and secondary drug targets. CYP52 proteins are known to be involved in growth on alkanes and fatty acids in yeasts and a homolog has been described in Malassezia25 that might represent a drug target to inhibit growth on these carbon sources.
Many systemic and superficial fungal infections are also associated with tissue inflammation involving human 5-lipoxygenase (5-LOX), such as dandruff 26. Therefore a therapeutic agent that possessed both antifungal and anti-inflammatory activity would be advantageous. The drug candidate ketaminazole27, possessing antifungal activity through an imidazole moiety and anti-inflammatory action through a phenylenediamine moiety, was shown to be a potent inhibitor of Candida albicans CYP51 and human 5-LOX. In this study, the expression, purification and characterization of two M. globosa cytochrome P450 monooxygenase proteins (CYP51 and CYP5218) are described and the effectiveness of azole antifungal agents against M. globosa CYP51 and CYP512 including inhibition of M. globosa growth are determined.
Results
Purification of CYP51 and CYP5218 proteins
Cholate extraction yielded ~320 (±100) and ~120 (±30) nmoles per liter culture of CYP51 and CYP5218. Purification by Ni 2+-NTA agarose chromatography resulted in 61% and 66% recoveries of native M. globosa CYP51 and CYP5218 proteins. SDS polyacrylamide gel electrophoresis confirmed the purity to be greater than 90% when assessed by staining intensity. The apparent molecular weights of M. globosa CYP51 and CYP5218 proteins were 53000 (±3000) and 58000 (±4000), which were slightly lower than the predicted values of 59786 and 69833, including N-terminal modifications and 4-Histidine C-terminal extensions. The absolute spectra of the resting oxidized form of M. globosa CYP51 and CYP5218 (Fig. 1A) were typical for low-spin ferric cytochrome P450 enzymes predominantly in the low-spin state28,29 with α, β, Soret (γ) and δ spectral bands at 571, 538, 418 and 360 nm for CYP51 and 565, 529, 415, 363 nm for CYP5218. Reduced carbon monoxide difference spectra for CYP51 and CYP5218 (Fig. 1B,C) gave the red-shifted heme Soret peak at 447 nm characteristic of P450 enzymes, indicating both proteins were expressed in the native form.
Figure 1
Spectral characteristics of M. globosa CYP51 and CYP5218.
Absolute spectra (A) were determined using 3 μM purified M. globosa CYP51 (solid line) and CYP5218 (dashed line) in the oxidised resting state (light path 4.5 mm). Reduced carbon monoxide difference spectra were determined using 3 μM purified M. globosa CYP51 (B) and CYP5218 (C) with sequential measurements made every 45 seconds (light path 10 mm).
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Substrate binding properties
Progressive titration of M. globosa CYP51 with eburicol gave a type I difference spectrum with a peak at 388 nm and a trough at 421 nm (Fig. 2A). Similar spectra were obtained using lanosterol and obtusifoliol. Type I binding spectra occur when the substrate or another molecule displaces the water molecule coordinated as the sixth ligand to the low-spin hexa-coordinated heme prosthetic group causing the heme to adopt the high-spin penta-coordinated conformation29. Binding affinities for the three sterols were similar with K d values of 32, 23 and 28 μM for lanosterol, eburicol and obtusifoliol, respectively.
Figure 2
Type I substrate binding spectra.
Absorbance difference spectra obtained by the progressive titration of 5 μM CYP51 with eburicol (A) and 4 μM CYP5218 with palmitoleic acid (B) were measured. Saturation curves (C) for eburicol (filled circles) and palmitoleic acid (hollow circles) were constructed from the absorbance difference spectra. Ligand binding data were fitted using the Michaelis-Menten equation with each experiment performed in triplicate although only one replicate is shown.
Full size image
Progressive titration of CYP5218 with palmitoleic acid gave a type I difference spectrum (Fig. 2B). The binding spectra obtained with the other fatty acids are shown in supplementary Figure S1. The binding affinities for most fatty acids with CYP5218 were similar (K d = 90 to 250 μM) except for linoleic acid (K d = 36 μM), which bound 2.5- to 7-fold more tightly to CYP5218 (Table 1) and with similar affinity to sterols binding to M. globosa CYP51. No binding spectra were obtained for CYP5218 with capric acid (C10:0), arachidic acid (C20:0), n-hexadecane and 1-hexadecene, indicating saturated fatty acids shorter than C12 and larger than C18 in addition to alkanes and alkenes did not bind to CYP5218 by perturbing the heme environment. Lanosterol and eburicol both failed to give reproducible binding spectra with CYP5218 (data not shown), excluding sterols as potential substrates.
Table 1 Fatty acid binding affinities for M. globosa CYP5218.
Full size table
CYP reconstitution assays
GC/MS analysis was used to identify CYP51 and CYP5218 assay metabolites. CYP51 assays using 0.5 μM M. globossa CYP51 and 50 μM obtusifoliol gave a turnover number of 3.4 min−1 compared with 1.7 min−1 for lanosterol and 5.6 min−1 for eburicol indicating M. globosa CYP51 had been isolated in a fully functional form exhibiting a preference for eburicol and obtusifoliol as substrate over lanosterol. Product identities were confirmed by the mass fragmentation patterns (Figures S2–S4) to be 14-demethylated sterols. Enzyme velocity curves for 0.13 μM M. globossa CYP51 (Figure S5) indicated substrate inhibition at higher sterol concentrations, especially for obtusifoliol and eburicol. Calculated K cat values of 22.47 ± 2.19, 65.54 ± 16.05 and 53.69 ± 14.21 min−1 for lanosterol, eburicol and obtusifoliol confirm an apparent catalytic preference for eburicol and obtusifoliol over lanosterol, however, K m values were similar for all three sterols at 55.14 ± 6.51, 55.91 ± 19.79 and 42.09 ± 10.45 μM, respectively.
Reconstitution assays with CYP5218 resulted in no observed product formation using lauric acid, myristic acid, palmitic acid, palmitoleic acid, oleic acid and linoleic acid as substrates. This was despite increasing the duration of the assay from 1 hour to overnight (~16 h) and using three different cytochrome P450 reductase redox partners. Therefore fatty acids of chain length C12 to C18 were not CYP5218 substrates.
Azole binding properties and IC 50 determinations
Titration of M. globosa CYP51 and CYP5218 with itraconazole produced type II binding spectra (Fig. 3A,B) with a peak at ~429 nm and a trough at ~412 nm. The binding spectra obtained with clotrimazole, fluconazole, ketoconazole, voriconazole and the drug candidate ketaminazole are shown in supplementary Figure S6. Type II binding spectra are caused by the triazole ring N-4 nitrogen (fluconazole, itraconazole and voriconazole) or the imidazole ring N-3 nitrogen (clotrimazole, ketoconazole and ketaminazole) coordinating as the sixth ligand with the heme iron24 to form the low-spin CYP51-azole complex resulting in a ‘red-shift’ of the heme Soret peak. All six azole antifungals bound to M. globosa CYP51 (Figure S6), however, fluconazole did not bind to M. globosa CYP5218 whereas clotrimazole, voriconazole, itraconazole, ketoconazole and ketaminazole all gave type II binding spectra with CYP5218.
Figure 3
Type II itraconazole binding spectra.
Itraconazole was progressively titrated against 2 μM CYP51 (A) and 2 μM CYP5218 (B) with the difference spectra determined after each addition of triazole. Itraconazole saturation curves (C) were constructed from the type II absorbance difference spectra for CYP51 (filled circles) and CYP5218 (hollow circles). The data were fitted using a rearrangement of the Morrison equation47. Each experiment was performed in triplicate although only one replicate is shown.
Full size image
Binding saturation curves (Fig. 3C and supplementary Figure S7) confirmed that azole binding to CYP51 was very tight with K d values of 4 ± 2, 11 ± 4, 2 ± 1, 2 ± 1, 8 ± 5 and 3 ± 1 nM for clotrimazole, fluconazole, itraconazole, ketoconazole, voriconazole and ketaminazole, respectively. Azole binding to CYP5218 was weaker with itraconazole and clotrimazole binding the tightest (K d 407 ± 136 and 486 ± 28 nM). Binding affinity to CYP5218 progressively fell from ketoconazole, ketaminazole and voriconazole (K d values ~1.6 ± 0.2, ~11.7 ± 0.5 and ~107 ± 2.2 μM) suggesting that these azoles would be ineffective at inhibiting CYP5218 activity.
IC 50 determinations (Fig. 4) confirmed that M. globosa CYP51 bound fluconazole, itraconazole, ketoconazole and ketaminazole tightly with IC 50 values approximately half the CYP51 concentration present at 0.206 ± 0.008, 0.188 ± 0.008, 0.176 ± 0.016 and 0.321 ± 0.042 μM, respectively. Similarly tight binding azole IC 50 values were observed for C. albicans CYP5123. The ketaminazole IC 50 value was less than two-fold higher than the current therapeutic azole drugs and was 45-fold lower than the IC 50 value determined for ketaminazole with human CYP5127, suggesting ketaminazole could be an effective antifungal agent against M. globosa with minimal inhibition of human CYP51.
Figure 4
IC 50 determinations for azole antifungal agents.
CYP51 reconstitution assays were performed using a CYP51:AfCPR1 ratio of 1:2 for 0.5 μM M. globosa CYP51 with 50 μM lanosterol as substrate at varying fluconazole (filled circles) itraconazole (hollow circles), ketoconazole (bullets) and ketaminazole (crosses) concentrations from 0 to 4 μM. Mean relative velocity values are shown along with standard deviation bars. Relative velocities of 1.00 were equivalent to velocities of 1.79 ± 0.26 min−1.
Full size image
M. globosa MIC determinations
Both itraconazole and ketoconazole strongly inhibited M. globosa growth with MIC 100 values of 0.0625 and 0.25 μg ml−1. Voriconazole was moderately effective with a MIC 100 value of 1 μg ml−1 whereas clotrimazole, ketaminazole and fluconazole were poor inhibitors of M. globosa growth (MIC 100 values of 8, 8 and 64 μg ml−1). The high MIC 100 value for fluconazole (64 μg ml−1) compared to the low IC 50 for fluconazole with M. globosa CYP51 (0.2 μM) suggests rapid efflux of this azole from cells. Ketaminazole was 32-fold less effective than ketoconazole at arresting M. globosa growth and 128-fold less effective than itraconazole, however, partial inhibition of growth was observed with 4 μg ml−1 ketaminazole. Therefore ketaminazole appears substantially less effective at inhibiting M. globosa growth than itraconazole and ketoconazole given the two-fold difference in the CYP51 IC 50 values. The treatment of M. globosa with 0.125 μg ml−1 ketoconazole and 4 μg ml−1 ketaminazole resulted in the accumulation of eburicol (Table 2). Accumulation of CYP51 substrates is indicative of direct in vivo CYP51 inhibition along with a reduction in the abundance of other post-CYP51 sterol metabolites.
Table 2 Sterol composition of azole-treated M. globosa.
Full size table
Discussion
Phylogenetic analysis of MGL_2415 strongly suggested this gene encoded a CYP51 enzyme. All 23 conserved CYP51 residues30 were present in the amino acid sequence and the top 100 BLASTP matches were predominantly CYP51 enzymes. In contrast, a similar analysis for MGL_3996 (CYP5218) failed to identify a putative catalytic function for this cytochrome P450 enzyme. Previously Lee et al.25 had reported that MGL_3996 was probably a CYP52 enzyme. BLAST2 analyses of CYP5218 against other yeast CYP52s showed less than 40% sequence identity. Such low sequence identities between CYP52 proteins and MGL_3996 suggest CYP5218 is not a CYP52 enzyme.
The spectral properties of purified M. globosa CYP51 and CYP5218 suggest both proteins were expressed in their native low-spin states. This was further supported by CYP51 binding 14-methylated sterols and CYP5218 binding fatty acids (Fig. 2). The M. globosa CYP51 K d values for lanosterol and eburicol were similar to those obtained with Candida albicans CYP5131, Mycosphaerella graminicola CYP5132 and Aspergillus fumigatus CYP51B33, but were 20-fold larger than K d values obtained with Trypanosoma cruzei CYP5134. Eburicol, lanosterol and obtusifoliol were demethylated by M. globosa CYP51 confirming the enzyme was isolated in native form. The catalytic rates observed with M. globosa CYP51 were similar to those of other CYP51 enzymes35. Previously M. globosa CYP51 was expressed and purified by Kim et al.36, but they did not demonstrate catalytic turnover.
Type I fatty acid binding spectra with M. globosa CYP5218 suggested that this enzyme metabolized fatty acids (hydroxylation or desaturation) with linoleic acid (C18:2) being favored because of the significantly lower K d value (36 μM) and therefore might be implicated in the use of fatty acids as carbon source for growth.
The fatty acid binding K d values observed for CYP5218 were similar to those determined for Candida maltosa CYP52A4 with lauric and myristic acids (~110 and ~120 μM)37. Lee et al.25 had previously expressed CYP5218 in Pichia pastoris at low levels, but did not purify or characterize the enzyme in terms of substrate binding properties or catalytic activity. Despite CYP5218 binding fatty acids, no in vitro metabolism of lauric acid, myristic acid, palmitic acid, palmitoleic acid, oleic acid or linoleic acid could be detected using the CYP5218 reconstitution assay with AfCPR1, CaCPR or HsCPR as redox partners. This suggests that fatty acids are not the in vivo substrates of CYP5218 in M. globosa. This phenomenon has previous been observed with cytochrome P450 BM3 mutants38. The fatty acids entered the enlarged substrate binding channel/pocket displacing the axial hexa-coordinated heme water molecule causing a low- to high-spin state change but did not undergo metabolism as the fatty acid molecule remained too distant from the heme iron for catalysis to occur.
Azole ligand binding studies indicated that M. globosa CYP51 bound azole antifungal agents 120- to 13300-fold more tightly than CYP5218 based on K d values. Therefore azole antifungals would need to be redesigned if CYP5218 was to be the primary drug target. At present only itraconazole and clotrimazole appear to have reasonably high affinities for CYP5218, albeit with 200- and 120-fold higher K d values than for CYP51. The drug candidate ketaminazole bound tightly to M. globosa CYP51 (K d 3 nM) suggesting ketaminazole would be effective at inhibiting CYP51 activity and M. globosa growth. In comparison, C. albicans CYP51 had K d values for pharmaceutical azoles of 10 to 56 nM23 and a K d value of 43 nM with ketaminazole27. In contrast, H. sapiens CYP5123,39 had higher K d values of 42 nM to 70 μM for pharmaceutical azoles and a K d value of 731 nM for ketaminazole27. Therefore ketaminazole exhibited a 240-fold selectivity for M. globosa CYP51 over the human homolog based on K d values compared to a 17-fold selectivity for C. albicans CYP5127.
IC 50 determinations (Fig. 4) confirmed that M. globosa CYP51 was strongly inhibited by azole antifungal agents (IC 50 values of 0.15 to 0.35 μM), suggesting efficacy against M. globosa cultures. However, the effectiveness of azole antifungals at inhibiting M. globosa cell growth was variable with itraconazole and ketoconazole being effective (MIC 100 values 0.0625 and 0.25 μg ml−1) and voriconazole being moderately effective (MIC 100 1 μg ml−1), whereas clotrimazole, ketaminazole and fluconazole were poor inhibitors of M. globosa growth (MIC 100 values 8, 8 and 64 μg ml−1). The mode of action for the azole antifungals against M. globosa was confirmed by sterol analysis of 0.125 μg ml−1 ketoconazole- and 4 μg ml−1 ketaminazole-treated cell pellets which showed raised eburicol levels compared to untreated cells (Table 2). Accumulation of 14-methylated sterols, such as eburicol, is characteristic of in vivo CYP51 inhibition by azole antifungal agents. The increased susceptibility of ketaminazole to both hydrolysis and oxidative stress in comparison to ketoconazole may partially explain the higher MIC 100 values observed with ketaminazole when using broth-cultured M. globosa. The drug transporter inhibitor FK-506 was found to be cytotoxic towards M. globosa at concentrations above 0.0625 μg ml−1 preventing the assessment of the contribution of drug transporters towards azole resistance observed with ketaminazole relative to ketoconazole. Further investigations into the reduced potency of ketaminazole, especially in comparison to itraconazole and ketconazole, are required to establish whether altering the structure of the compound can overcome these apparent shortcomings. The MIC 100 values determined in this study were towards the upper end of the MIC values previously reported of 0.016 to 0.25 μg ml−1 for itraconazole, 0.03 to 0.06 μg ml−1 for voriconazole, 0.008 to 0.25 μg ml−1 for ketoconazole, 0.06 μg ml−1 for posaconazole and 4 μg ml−1 for fluconazole40,41,42,43.
In summary, M. globosa CYP5218 was observed to bind fatty acids but was not associated with metabolism and its function and potential as a drug target remains unclear. In contrast the known drug target CYP51 readily catalyzed the 14α-demethylation of lanosterol, eburicol and obtusifoliol and bound all six pharmaceutical azoles tightly (K d 2 to 11 nM) reflected in the low IC 50 values for azole antifungals. The performance of the drug candidate ketaminazole against M. globosa CYP51 was encouraging with a K d value similar to the other pharmaceutical azoles and an IC 50 value only twice that of ketoconazole. However, MIC 100 studies showed ketaminazole to be ~32-fold less effective at inhibiting M. globosa growth than ketoconazole, suggesting that the chemical structure of ketaminazole needs further optimization for increased cellular uptake and antifungal potency in vivo. A CYP51 assay was demonstrated that can be used for drug discovery and a case demonstrated for using itraconazole given the superior safety profile over ketoconazole.
Materials and Methods
Heterologous expression and purification of recombinant proteins
The M. globosa CYP51 gene (Mglob51 - KEGG M. globosa genome database and CYP5218 gene (Mglob5218 - were synthesized by Eurofins MWG Operon (Ebersberg, Germany) and cloned into the pCWori+ expression vector. The first eight amino acids of each protein were changed to ‘MALLLAVF’44 and a four-histidine tag added to the C’ terminus. The pCWori+:Mglob51 and pCWori+:Mglob5218 constructs were transformed into DH5α E. coli cells. M. globosa CYP51 was expressed as previously described31 whilst for CYP5218 the expression temperature was lowered to 20 °C for 18 h at 180 rpm. Recombinant proteins were isolated according to the method of Arase et al.45 and solubilized CYP51 and CYP5218 proteins were purified using Ni 2+-NTA agarose28. Protein purity was assessed by SDS polyacrylamide gel electrophoresis.
Cytochrome P450 spectroscopy
Reduced carbon monoxide difference spectroscopy46,47 and absolute spectra between 700 and 300 nm were used to determine cytochrome P450 concentrations. Substrate and azole binding studies were performed31,48 using 2.5 mM stock sterol solutions in 40% (wt/vol) (2-hydroxypropyl)-β-cyclodextrin (HPCD), 4 mg ml−1 solutions of fatty acids in dimethylformamide (DMF) and stock 0.05, 0.1 and 0.2 mg ml−1 solutions of azole antifungal agents in dimethylsulfoxide (DMSO). Sterols were titrated against 5 μM CYP51, fatty acids were titrated against 4 μM CYP5218 and azole antifungal agents were titrated against 2 μM CYP51 and CYP5218. The absorbance difference spectrum between 500 and 350 nm was determined after each incremental addition of ligand with saturation curves constructed from ΔA peak-trough against ligand concentration. The dissociation constants (K d) for sterols and fatty acids were determined by non-linear regression (Levenberg-Marquardt algorithm) using the Michaelis-Menten equation. The dissociation constants of the enzyme-azole complex were determined by non-linear regression (Levenberg-Marquardt algorithm) using a rearrangement of the Morrison equation for tight ligand binding49,50. ProFit 6.1.12 (QuantumSoft, Zurich, Switzerland) was used for curve-fitting data. All ligand-binding studies were performed at 22 ± 2 °C and in triplicate. All spectral determinations were made using a Hitachi U-3310 UV/VIS spectrophotometer (San Jose, California).
CYP reconstitution assays
CYP51 reconstitution assays34,51 contained 0.5 μM M. globosa CYP51, 1 μM Aspergillus fumigatus cytochrome P450 reductase isoenzyme 1 (AfCPR1 - UniProtKB accession number Q4WM6752), 50 μM 14α-methylated sterol (lanosterol, eburicol or obtusifoliol), 50 μM dilaurylphosphatidylcholine, 4% (wt/vol) HPCD, 0.4 mg ml−1 isocitrate dehydrogenase, 25 mM trisodium isocitrate, 50 mM NaCl, 5 mM MgCl 2 and 40 mM MOPS (pH ~7.2). Assay mixtures were incubated at 37 °C for 10 minutes prior to initiation with 4 mM β-NADPHNa 4 followed by shaking at 37 °C for 20 minutes. Sterol metabolites were recovered by extraction with ethyl acetate followed by derivatization with 0.1 ml N,O-bis(trimethylsilyl)trifluoroacetamide (BSTFA) : trimethylchlorosilane (TMCS) (99:1) and 0.3 ml anhydrous pyridine (2 h at 80 °C) prior to analysis by gas chromatography mass spectrometry (GC/MS)53. IC 50 determinations were performed using 50 μM lanosterol as substrate to which azole antifungal agents were added in 2.5 μl dimethylsulfoxide prior to incubation at 37 °C and addition of β-NADPHNa 4. Determination of K cat and K m values utilized CYP51 reconstitution assays that contained 0.13 μM M. globosa CYP51 and 0.5 μM AfCPR1 at sterol substrate concentrations of 6.25, 12.5, 25, 50, 75 and 100 μM with incubation at 37 °C for 5 min.
The CYP5218 fatty acid reconstitution assay system used was based on that previously described by Guengerich54 and contained 200 nM of CYP5218, 500 nM of cytochrome P450 reductase (A. fumigatus CPR152) or H. sapiens CPR55 or C. albicans CPR56, 50 μM DLPC, 100 μM fatty acid (lauric acid, myristic acid, palmitic acid, palmitoleic acid, oleic acid and linoleic acid), 4 mM glucose-6-phosphate, 3 U/ml yeast glucose-6-phosphate dehydrogenase in a final volume of 500 μl with 0.1 M potassium phosphate (pH 7.4). Assay mixtures were incubated at 37 °C for 5 min prior to initiation with 4 mM β-NADPHNa 4 and then incubated for a further 60 min or overnight (~16 h) at 37 °C. Fatty acid metabolites were recovered by extraction with dichloromethane and dried in a vacuum centrifuge. Samples were derivatized with 200 μl anhydrous pyridine and 250 μl BSTFA:TMCS (99:1) (30 min at 80 °C) and extracted with 500 μl of hexane, prior to analysis by GC/MS (Thermo GC Trace 1300 and ISQ MS). GC analysis was performed with a 5% phenyl methylpolysiloxane column (DB5) (Agilent Technologies). The oven programme was as follows: initial: 70 °C, with 3 min hold; ramp: 10 °C min−1 to 180 °C (held for 2 min), 10 °C min−1 to 250 °C (held for 2 min) and 10 °C min−1 to 280 °C (held for 2 min). Data were analyzed using Thermo Xcalibur 2.2 software.
M. globosa MIC 100 determination
M. globosa ATCC MYA-4612 was grown for 7 days at 30 °C in modified Dixon medium (3.6% malt extract, 2% desiccated ox-bile, 1% Tween 40, 0.6% peptone, 0.2% glycerol, 0.2% oleic acid, pH adjusted to 6 with HCl) until a culture density of 1.5 × 10 5 cells ml−1 was obtained. Cells from 3 ml of stock culture were harvested, suspended in 1 ml of saline and used to inoculate 30 ml of fresh modified Dixon medium for use as M. globosa inoculums in MIC determinations. Stock azole concentrations of 1600, 800, 400, 200, 100, 50, 25, 12.5, 6.25, 3.125 and 0 μg ml−1 were prepared in DMSO. These were diluted ten-fold with fresh modified Dixon media and 20 μl of the diluted azole solutions were then mixed with 180 μl of M. globosa inoculum in microtiter plate wells to yield final azole concentrations of 16, 8, 4, 2, 1, 0.5, 0.25, 0.125, 0.0625, 0.03125 and 0 μg ml−1. The microtiter plates were incubated at 30 °C for 7 days and then read. Each azole MIC 100 determination was performed in triplicate.
Sterol analysis
M. globosa was grown in the absence and presence of azole antifungal for 7 days at 30 °C and non-saponifiable lipids extracted as previously reported57 and were derivatized with 0.1 ml BSTFA:TMCS (99:1) and 0.3 ml anhydrous pyridine (2 h at 80 °C) prior to analysis by GC/MS53. Individual sterols were identified by reference to relative retention times, mass ions and fragmentation patterns. Sterol composition was calculated using peak areas.
Chemicals
All fatty acids were obtained from Sigma-Aldrich (Poole, UK), Fluconazole, itraconazole and ketoconazole were obtained from Sigma-Aldrich (Poole, UK) whilst ketaminazole was supplied by professor T.R. Holman (University of California, Santa Cruz, CA). Lanosterol, eburicol and obtusifoliol were supplied by professor W. D. Nes (Texas Tech University, Lubbock, TX).
Additional Information
How to cite this article: Warrilow, A. G. S. et al. Azole Antifungal Sensitivity of Sterol 14α-Demethylase (CYP51) and CYP5218 from Malassezia globosa. Sci. Rep.6, 27690; doi: 10.1038/srep27690 (2016).
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Acknowledgements
We thank Mr. Marcus Hull and the Engineering and Physical Sciences Research Council National Mass Spectrometry Service Centre at Swansea University for assistance in GC/MS analyses. We are grateful to the European Union for support through the European Regional Development Fund (ERDF) as part of the Welsh Government funded BEACON project (Swansea University) and the National Science Foundation of the United States to W.D.N. [NSF-MCB 0920212] and the National Institutes of Health of the United States to T.R.H. [GM56062] for supporting this work.
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Authors and Affiliations
Centre for Cytochrome P450 Biodiversity, Institute of Life Science, Swansea University Medical School, Swansea, SA2 8PP, Wales, United Kingdom
Andrew G. S. Warrilow,Claire L. Price,Josie E. Parker,Nicola J. Rolley,Diane E. Kelly&Steven L. Kelly
Chemistry and Biochemistry Department, University of California, Santa Cruz, 95064, CA, USA
Christopher J. Smyrniotis&Theodore R. Holman
Plant Chemistry Group, School of Chemistry, Bangor University, Bangor, Gwynedd, LL57 2UW, Wales, United Kingdom
David D. Hughes&Vera Thoss
Department of Chemistry and Biochemistry, Center for Chemical Biology, Texas Tech University, Lubbock, 79409-1061, Texas
W. David Nes
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1. Andrew G. S. WarrilowView author publications Search author on:PubMedGoogle Scholar
2. Claire L. PriceView author publications Search author on:PubMedGoogle Scholar
3. Josie E. ParkerView author publications Search author on:PubMedGoogle Scholar
4. Nicola J. RolleyView author publications Search author on:PubMedGoogle Scholar
5. Christopher J. SmyrniotisView author publications Search author on:PubMedGoogle Scholar
6. David D. HughesView author publications Search author on:PubMedGoogle Scholar
7. Vera ThossView author publications Search author on:PubMedGoogle Scholar
8. W. David NesView author publications Search author on:PubMedGoogle Scholar
9. Diane E. KellyView author publications Search author on:PubMedGoogle Scholar
10. Theodore R. HolmanView author publications Search author on:PubMedGoogle Scholar
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Contributions
All authors contributed significantly to the research. A.G.S.W. performed experimental work (protein expression, protein purification, ligand binding, CYP51 assays), evaluated the data and drafted the manuscript. C.L.P. planned and performed experimental work (CYP5218 fatty acid binding and reconstitution assays) and evaluated the data. J.E.P. planned and performed experimental work (sterol composition analysis) and evaluated the data. N.J.R. planned and performed experimental work (azole MIC determinations), C.J.S. ketaminazole synthesis and evaluated the data, D.D.H. fatty acid analysis, V.T. method development and fatty acid analysis, W.D.N. evaluated the data and provided samples, D.E.K. evaluated the data and the manuscript drafting, T.R.H. ketaminazole synthesis and evaluated the data, S.L.K. conceived the study with T.R.H. and oversaw the research.
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Warrilow, A., Price, C., Parker, J. et al. Azole Antifungal Sensitivity of Sterol 14α-Demethylase (CYP51) and CYP5218 from Malassezia globosa. Sci Rep6, 27690 (2016).
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4440 | https://www.omnicalculator.com/math/slope | Slope Calculator
The slope calculator determines the slope or gradient between two points in the Cartesian coordinate system. The slope is basically the amount of slant a line has and can have a positive, negative, zero, or undefined value. Before using the calculator, it is probably worth learning how to find the slope using the slope formula. To find the equation of a line for any given two points that this line passes through, use our slope intercept form calculator.
How to use this slope calculator
Here, we will walk you through how to use this calculator, along with an example calculation, to make it simpler for you. To calculate the slope of a line, you need to know any two points on it:
Enter the x and y coordinates of the first point on the line.
Enter the x and y coordinates of the second point on the line.
We instantly get the slope of the line. But the magic doesn't stop there, for you also get a bunch of extra results for good measure:
The equation of your function (same as the equation of the line).
The y-intercept of the line.
The angle the line makes with respect to the x-axis (measure anti-clockwise).
Slope as a percentage (percentage grade).
The distance between the two points.
For example, say you have a line that passes through the points (1, 5) and (7, 6). Enter the x and y coordinates of the first point, followed by the x and y coordinates of the second one. Instantly, we learn that the line's slope is 0.166667. If we need the line's equation, we also have it now: y = 0.16667x + 4.83333.
You can use this calculator in reverse and find a missing x or y coordinate! For example, consider the line that passes through the point (9, 12) and has a 12% slope. To find the point where the line crosses the y-axis (i.e., x = 0), enter 12% in percent grade (9, 12) as the coordinate of the first point, and x2 = 0. Right away, the calculator tells us that y2 = 10.92.
The slope of a line has many significant uses in geometry and calculus. The article below is an excellent introduction to the fundamentals of this topic, and we insist that you give it a read.
🙋 Discover how to easily differentiate between a negative and a positive slope in our dedicated article, Positive Slope vs Negative Slope: How to Tell the Difference!
The slope formula
slope=x2−x1y2−y1
Notice that the slope of a line is easily calculated by hand using small, whole number coordinates. The formula becomes increasingly useful as the coordinates take on larger values or decimal values.
If the line on which the points lie is given by the formula
y=mx+c
then the slope is m while c is the y-intercept.
It is worth mentioning that any horizontal line has a gradient of zero because a horizontal line has the same y-coordinates. This will result in a zero in the numerator of the slope formula. On the other hand, a vertical line will have an undefined slope since the x-coordinates will always be the same. This will result the division by zero error when using the formula.
How to find slope
Identify the coordinates (x1,y1) and (x2,y2). We will use the formula to calculate the slope of the line passing through the points (3,8) and (−2,10).
Input the values into the formula. This gives us (10−8)/(−2−3).
Subtract the values in parentheses to get 2/(−5).
Simplify the fraction to get the slope of −2/5.
Check your result using the slope calculator.
To find the slope of a line, we need two coordinates on the line. Any two coordinates will suffice. We are basically measuring the amount of change of the y-coordinate, often known as the rise, divided by the change of the x-coordinate, known as the run. The calculations in finding the slope are simple and involve nothing more than basic subtraction and division.
🙋 To find the gradient of non-linear functions, you can use the average rate of change calculator.
Applications in physics
Intuitively, the slope of a function captures its rate of change. One of the simplest real-life examples of slope is velocity: the measure of how position changes in time.
Suppose that a cyclist travels along a straight line, and their position at time t with respect to some initial point 0 is given by the formula x(t) = 5t + 3. (Be careful: here, we vary position with time, so x is a function of t, unlike in the previous sections where we varied y with x.) Then, their velocity is v = 5, which is precisely the slope in the formula for x(t). Its physical interpretation is that in a time increment Δt, the cyclist covers an extra distance of 5(Δt).
In the example above, the cyclist was riding with a constant velocity v = 5. But what if, instead, he speeds up with constant acceleration? Suppose the cyclist starts their ride at point x = 0 with zero velocity and acceleration a = 10. This means that starting from 0, the cyclist increases their speed by 10 in every time unit, and so his velocity at time t is v(t) = 10t. Just like velocity is the rate of change of position, acceleration is the rate of change of velocity, so you can read off the value 10 for the acceleration from the velocity formula.
What about the cyclist's position in this case? It turns out that their position is described by the formula x(t) = 5t². We will explain later where this formula comes from: for now, the important point is that in a motion with constant acceleration, position is a quadratic function of time (rather than linear, as it was for the motion with constant speed). As a result, we can no longer read off the cyclist's velocity as a single number from the position formula. This is a consequence of velocity being no longer constant, and it reflects an important lesson: the rate of change (i.e., slope) of a function is a fixed number only if the function is linear.
How, then, can we compute the velocity v(t) = 10t from the position formula x(t) = 5t²? We need to differentiate x(t) (i.e., take its derivative). The derivative of a function at a given time is precisely its rate of change at this time; therefore, v(t) is the derivative of x(t), denoted as x'(t) or dx(t)/dt.
Geometrically, the derivative x'(t) is the slope of the line tangent to the function at time t; in general, at each time t, the tangent line will be different, and so will the derivative. So, differentiation refines the idea of computing slopes, as it allows us to compute the "slope" of functions whose rates of change also vary.
The converse of differentiation is integration: the process in which we compute a function from its derivative. It is through integration that we get the cyclist's position function x(t) = 5t² from their velocity v(t) = 10t. Geometrically, integrating a function amounts to counting the area under the graph. The graph of the velocity function v(t) = 10t determines a triangle with base t and height 10t, and so the area under the graph is 0.5 × t × (10t) = 5t² — that's exactly our position function!
Other related topics
Just as the slope can be calculated using the endpoints of a segment, so can the midpoint. The midpoint is an important concept in geometry, particularly when inscribing a polygon inside another polygon with its vertices touching the midpoint of the sides of the larger polygon. This can be obtained using the midpoint calculator or by simply taking the average of each of the x-coordinates and the average of the y-coordinates to form a new coordinate.
The slopes of lines are important in determining whether or not a triangle is a right triangle. If any two sides of a triangle have slopes that multiply to -1, then the triangle is a right triangle. The computations for this can be done by hand or by using the right triangle calculator. You can also use the distance calculator to compute the longest side of a triangle, which helps determine which sides must form a right angle if the triangle is right.
The sign in front of the gradient provided by the slope calculator indicates whether the line is increasing, decreasing, constant or undefined. If the graph of the line moves from lower left to upper right it is increasing and is therefore positive. If it decreases when moving from the upper left to the lower right, then the gradient is negative.
Making this slope calculator
The slope calculator is one of the oldest at Omni Calculator, built by our veterans Mateusz and Julia, who make creating accurate scientific tools look easy. The idea for this calculator was born when the two were crunching data analytics and trends and realized how a slope calculator would make their job easier. Even today, you can find them occasionally using this tool for reliable calculations.
We put extra care into the quality of our content so that it is as accurate and dependable as possible. Each tool is peer-reviewed by a trained expert and then proofread by a native speaker. You can learn more about our standards in our Editorial Policies page.
FAQs
How to find slope from an equation?
The method for finding the slope from an equation depends on the equation in front of you. If the equation has the form y = mx + c, then the slope (or gradient) is just m. If the equation is not in this form, try to rearrange the equation. To find the gradient of other functions, you will need to differentiate the function with respect to x.
How do you calculate the slope of a hill?
Use a map to determine the distance between the top and bottom of the hill as the crow flies.
Using the same map or GPS, find the altitude between the top and bottom of the hill. Make sure that the points you consider are the same as in step 1.
Convert both measurements into the same units.
Divide the difference in altitude by the distance between the two points.
This number is the average gradient of the hill.
How do you calculate the length of a slope?
Measure the difference between the top and bottom of the slope in relation to both the x and y axis.
If you can only measure the change in x, multiply this value by the gradient to find the change in the y-axis.
Make sure the units for both values are the same.
Use the Pythagorean theorem to find the length of the slope. Square both the change in x and the change in y.
Add the two values together.
Find the square root of the summation.
This new value is the length of the slope.
What is a 1 in 20 slope?
A 1/20 slope is one that rises by 1 unit for every 20 units traversed horizontally. So, for example, a ramp that was 200 ft long and 10 ft tall would have a 1/20 slope. A 1/20 slope is equivalent to a gradient of 1/20 (strangely enough) and forms an angle of 2.86° between itself and the x-axis.
How do you find the slope of a curve?
As the slope of a curve changes at each point, you can find the slope of a curve by differentiating the equation with respect to x and, in the resulting equation, substituting x for the point at which you’d like to find the gradient.
Is rate of change the same as slope?
The rate of change of a graph is also its slope, which are also the same as gradient. Rate of change can be found by dividing the change in the y (vertical) direction by the change in the x (horizontal) direction, if both numbers are in the same units, of course. Rate of change is particularly useful if you want to predict the future of previous value of something, as, by changing the x variable, the corresponding y value will be present (and vice versa).
Where do you use slope in everyday life?
Slopes (or gradients) have a number of uses in everyday life. There are some obvious physical examples — every hill has a slope, and the steeper the hill, the greater its gradient. This can be useful if you are looking at a map and want to find the best hill to cycle down. You also probably sleep under a slope, a roof that is. The slope of a roof will change depending on the style and where you live. But, more importantly, if you ever want to know how something changes with time, you will end up plotting a graph with a slope.
What is a 10% slope?
A 10% slope is one that rises by 1 unit for every 10 units traveled horizontally (10%). For example, a roof with a 10% slope that is 20 m across will be 2 m high. This is the same as a gradient of 1/10, and an angle of 5.71° is formed between the line and the x-axis.
How do you find the area under a slope?
To find the area under a slope given by the equation y = mx + c, follow these steps:
Define the lower and upper bounds of x to get a value for Δx.
Multiply Δx by the slope (m) to obtain Δy.
Multiply Δx by Δy.
Divide by 2 to find the area under the slope.
What degree is a 1 to 5 slope?
A 1 to 5 slope is one that, for every increase of 5 units horizontally, rises by 1 unit. The number of degrees between a 1 to 5 slope and the x-axis is 11.3°. This can be found by first calculating the slope, by dividing the change in the y direction by the change in the x direction, and then finding the inverse tangent of the slope.
Point coordinates
Enter two points (x1, y1) and (x2, y2) using whole numbers, fractions, or decimals.
Related numbers
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Bilinear interpolation
Catenary curve |
4441 | https://archive.org/download/riddlesofexeterbook00tuppuoft/riddlesofexeterbook00tuppuoft.pdf | .CD THE ALBION SERIES and J. W. BRIGHT AND G. L. KITTREDGE GENERAL EDITORS Ube Hlbion Series This series will comprise the most important Anglo-Saxon and Middle English poems in editions designed to meet the wants of both the scholar and the student. Each volume will ordinarily contain a single poem, critically edited, and provided with an introduction, notes, and a full glossary. THE RIDDLES OF THE EXETER BOOK EDITED WITH INTRODUCTION, NOTES, AND GLOSSARY BY FREDERICK TUPPER, JR. PROFESSOR OF THE ENGLISH LANGUAGE AND LITERATURE IN THE UNIVERSITY OF VERMONT GINN AND COMPANY BOSTON • NEW YORK • CHICAGO • LONDON IQIO COPYRIGHT, 1910, BY FREDERICK TUPPER, JR. ALL RIGHTS RESERVED JUL 25 J957 gbt fltbtnaum grt«« G1NN AND COMPANY • PRO- PRIETORS • BOSTON • U.S.A. TO THE MEMORY OF JULIAN HUGUENIN WHO LOVED OLD ENGLISH LIFE AND LITERATURE WITH A BOY'S ENTHUSIASM AND WITH A SCHOLAR'S KNOWLEDGE PREFACE The preparation of this first separate edition of The Riddles of the Exeter Hook, certainly the most difficult text in the field of Anglo-Saxon, has been to me a work of very real delight. Both in matter and manner these poems present so many engaging problems — which, when read aright, reveal at once the loftiest and lowest in older England's thought, and open up a hundred vistas of early word and action — that I count as great gain the years spent in their study. May it be my good fortune to impart to others a generous share of this pleasure and profit ! A few words of my purposes in this edition are in place here. I have striven to set forth the principles that govern the comparative study of riddles, and to trace the relation of these Anglo-Saxon enigmas to the Latin art-riddles of nearly the same period and to the folk-products of many lands and times. In the chapter upon the authorship of these poems and their place in the history of the Cynewulf question, I have tried to weigh all the evidence with a higher regard for reason and the probabilities than for the mere weight of authority, which in the case of these riddles has often been fatal to free investigation and opinion. In the presentation of solutions in the Introduction and in the later discus- sion of these in the Notes, I have also sought to ' prove all things and hold fast that which is good.' As aids to definite conclusions, the testi- mony of analogues and the light thrown by Old English life and customs have been of far higher worth than the random guesses of modern critics. But to Dietrich's illuminating treatment of each of the Exeter Book Rid- dles and to the essays of more recent scholars I gladly admit a large debt. I have closely analyzed the form and structure of the poems with the hope of bringing them nearer to the reader's understanding. But, above all, I have aimed, through elaborate annotation, so to illustrate the ' veined humanity ' of these remarkable productions, so to show forth their closeness to every phase of the life of their day, that this book might be a guide to much of the folk-lore and culture of Englishmen before the Conquest. This text of the Riddles is based upon a collation of the original manu- script at Exeter with the faithful reproduction in the British Museum, viii PREFACE with the texts of Thorpe, Grein, and Assmann (Grein-Wiilker), and with various versions of single riddles. According to the usage of this series, all departures from the manuscript which originate with the editor are printed in italics. I have conservatively avoided daring conjectures, and have proposed no new readings that were not dictated to me by the demands of the context and by the precedent of author's use and of contemporary idiom and meter. At first I wished to distinguish the many resolved vowels and diphthongs in the verse by diaereses. The general editors did not assent to this method of marking, believing — very wisely, as I now think — that a lavish use of diacritics gives an air of freakishness to a text and that such resolution might better be in- dicated in the textual notes. As in the other Albion editions of Anglo-Saxon poems, the Glossary is intended to be a complete verbal and grammatical index to the Rid- dles, with the exception of a few of the commoner forms of the pronoun, the article, and the conjunction. The Index of Solutions, at the very close of the volume, records all the answers proposed at any time by commentators. It is a pleasure to express my gratitude and appreciation to all who have aided me in the preparation of this book : to Canon W. J. Edmonds, Chancellor of Exeter Cathedral, who, by his many kindnesses, made de- lightful my days in the chapter library ; to Dr. Otto J. Schlutter, whose intimate first-hand knowledge of the text of the Leiden Riddle was gen- erously placed at my disposal ; and to Professor George Philip Krapp, Who freely gave to several chapters of my introduction keen and helpful criticism. I am particularly indebted to the general editors of the series, Professors Bright and Kittredge, who have carefully read the proof and have offered more advice than I could acknowledge in detail. Finally, my thanks are due to Mr. S. T. Byington of Ginn and Company, for many valuable suggestions. FREDERICK TUPPER, JR. UNIVERSITY OF VERMONT September, 1909 CONTENTS INTRODUCTION : PAGE I. THE COMPARATIVE STUDY OF RIDDLES xi II. ORIGINALS AND ANALOGUES OF THE EXETER BOOK RIDDLES SYMPHOSIUS xxviii ALDHELM xxxi TATWINE xxxiii EUSEBIUS . . . xxxiv LATIN ENIGMAS AND THE EXETER BOOK xxxvii BONIFACE xliv BERN RIDDLES xlvi LORSCH RIDDLES xlvii PSEUDO-BEDE xlviii FOLK-RlDDLES H III. THE AUTHORSHIP OF THE EXETER BOOK RIDDLES THE RIDDLES AND CYNEWULF liii UNITY OF AUTHORSHIP Ixiii IV. SOLUTIONS OF THE EXETER BOOK RIDDLES Ixxix V. THE FORM AND STRUCTURE OF THE EXETER BOOK RIDDLES Ixxxiv VI. THE MANUSCRIPTS xcvi BIBLIOGRAPHY ci ABBREVIATIONS • . . . . cix TEXT i NOTES 69 GLOSSARY 241 INDEX OF SOLUTIONS '291 INTRODUCTION THE COMPARATIVE STUDY OF RIDDLES What is a riddle ? Many scholars have sought to answer this ques- tion, and to define accurately the functions of enigmatic composition. Only during the past few years has the popular riddle received its meed of critical attention from scholars (M.L.N. XVIII, i). Until this very recent time, investigators were generally content with presenting without historical comment — and sometimes even, as in Simrock's well-known Riitselbuch, without regard to the home of their contributions — the results of more or less accurate observation. (For a resume of work in the German field, see Hayn, ' Die deutsche Ratsel- Litteratur. Versuch einer bibliographischen Uebersicht bis zur Neuzeit,' Central- blatt fiir Bibliothekswescn VII, 1890, pp. 516-556). There were, it is true, a few noteworthy exceptions to the prevailing rule of neglect of comparative study — a neglect well illustrated by Friedreich, Geschichte des Rdtsels, Dresden, 1860, which is, at its best, but a collection of widely scattered material, and makes no pretensions to scientific classification. As early as 1855, Mullenhoff made an inter- esting comparison of German, English, and Norse riddles ( Wolfs und Mannhardts Zeitschrift fiir deutsche Mythologie III, if.); Kohler, about the same period, traced carefully the originals and analogues of some forty riddles in a Weimar MS. of the middle of the fifteenth century ( Weimar Jahrbuch V, 1856, 329-356) ; Rolland noted many parallels to the French riddles of his collection (Devinettcs ou Enigmes populaires de la France. Avec une preface de M. Gaston Paris. Paris, 1877); and finally Ohlert, in a monograph of admirable thoroughness (Rtitsel und Gesellschaftsspiele der alien Griechen. Berlin, 1886), followed the riddles of the Greek world through the centuries of their early and later history. An epoch in the history of our subject was created, however, in 1897 by two monumental works : Richard Wossidlo's collection of over a thousand carefully localized North German riddles (Afecklenbztrgische Volksuberlieferungen, Part I, Wisrnar, 1897), in which the work of the accurate tabulator was supplemented by the labor of the painstaking philologist ; and Giuseppe Pitre's edition of Indovinelli, Dubbi, Sciogli- lingua del Popolo Siciliano (Bibl. delle Trad. Pop. Sic. XX), Torino- Palermo, 1897, in which the literary sources and popular origins of riddles are closely considered. Petsch has turned the material of Wossidlo, Rolland, and others to good account in his study of the forms and the style of the popular riddle (Areue Beitrdge zur Kenntnis des Volksrdtsels. Palaestra IV, Berlin, 1899). Heusler in his illuminating 1 xjj INTRODUCTION Friedreich tells us that the riddle is ' a roundabout description of an un- named object, so worded as to arouse the reflection of reader or hearer to the discovery of this.' Pitre's definition in his elaborate Introduction f is at once more scholarly and more inclusive : 'JThe riddle is an arrange- ment of words by which is understood^orsugges'ted something that is not expressed ; or else it is an ingenious and witty description of this unex- pressed thing by means of qualities and general traits that can be attributed quite as well to other things having no likeness or analogy to the subject. This description is always vague, so vague indeed that he whose task it is to solve the riddle runs in his mind to one or the other signification in vain attempt to reach the solution. Often the interpretation is hidden under the veil of a very remote allegory or under graceful and happy images.' J The mental attitudes of riddler and beriddled are charmingly pictured by Goethe in an oft-cited passage of Alexis ynd Dora : So legt der Dichter ein Rathsel, Kiinstlich mit Worten verschrankt, oft der Versammlung ins Ohr. Jeden freuet die seltne, der zierlichen Bilder Verkniipfung, Aber noch fehlet das Wort, das die Bedeutung verwahrt. 1st es endlich entdeckt, dann heitert sich jedes Gemiith auf, Und erblickt im Gedicht doppelt erfreulichen Sinn. Aristotle was the first to point out the close relation between riddles and metaphors : § ' While metaphor is a very frequent instrument of f~ article upon the Heifrreks Gdtur of the Hervarar Saga (Zeitschrift des Vereins fur Volkskunde XI, 1901, H7f.) has applied the comparative method to these thirty-five Old Norse riddles. And I have tried to adduce and apply certain rules for riddle-study in five articles : ' The Comparative Study of Riddles,' M. L. N. XVIII, 1903, 1-8 ; ' Originals and Analogues of the Exeter Book Riddles? ib. 97- 106; 'The Holme Riddles (MS. Harl. 1960),' P.M.L.A. XVIII, 1903, 211-272; • Riddles of the Bede Tradition,' Mod. Phil. II, 1905, 561-572 ; ' Solutions of the Exeter Book Riddles? M. L. N. XXI, 1906, 97-105. As all these essays of mine were merely preparatory to the present edition, I have drawn freely upon them in this Introduction. P. 2. t P. xviii. } Not very different is the definition of Wolf, Poetischer Hansschatz des deutschen Volkes, 6. Aufl., Leipzig, 1844, p. 1138 : ' Das Rathsel ist ein Spiel des Verstandes, der sich bemiiht einen Gegenstand so darzustellen dass er alle Merkmale und Eigenschaften desselben schildert, so vviedersprechend dieselben an und fur sich betrachtet auch sein mogen, ohne jedoch den Gegenstand selbst zu nennen.' Groos defines the riddle in almost the same words, Die Spiele der Menschen (1899), p. 194. § Rhetoric iii, n (Welldon's translation, London, 1886, p. 264). THE COMPARATIVE STUDY OF RIDDLES xiii clever sayings, another or an additional instrument is deception, as people are more clearly conscious of having learnt something from their sense of surprise at the way in which the sentence ends and their soul seems to say, " Quite true and I had missed the point." This, too, is the result of pleasure afforded by clever riddles; they are instructiye_andjmeta- .phorical in their expression.' It is Aristotle's opinion that not only are metaphors the germs of riddles, but that enigmatic elements appear in all metaphors, since these are derived from ' objects which are closely related to the thing itself but which are not immediately obvious.' Gaston Paris defines the riddle as ' a metaphor or a group of meta- phors, the employment of which has not passed into common use, and the explanation of which is not self-evident.' Indeed, many rid- dles go back to a time when external objects impressed the human mind very differently from their present effect and consequently sug- gested metaphors which at first seem to us almost incomprehensible, but which charm us when we have the clue to their meaning. ' The making of riddles,' says Tylor,| ' requires a fair power of ideal compari- son, and knowledge must have made considerable advance before the process could become so familiar as to fall from earnest into sport.' Lindley notes t that ' Riddles play upon analogies among things per- f ceived. Essentially the primitive mode of invention is as follows : Some 1 one discovers a new analogy among natural objects, formulates a ques- / 1 tion, concerning this, and thus a new riddle is born. ... § And, having its deepest roots in the perception of the analogies of nature, the riddle ,is brother to the metaphor, which has been so important in the develop- Iment of languages and myths.' Gummere points out in his Beginnings of Poetry \ that ' metaphors of the substantive may well have been the origin of the riddle, since ejujy kennings _of ten read like riddles : in Finnish, the sunshine is called " the contents of Wainamoinen's Introduction to Rolland, Devinettes, p. viii. t Primitive Culture, edition of 1903, I, 90-91. \ American Journal of Psychology, VIII (1896-1897), 484. § Lindley remarks with acuteness : ' While the most primitive forms have chief \ reference to natural objects, the evolution of the riddle reflects the shifting of man's chief interest from external nature to man himself. Some of the most famous riddles among the Greeks have this human focus.' So with our Anglo- Saxon riddles. || New York, 1901, pp. 451-452. Cf. Scherer, Gtsch. der deutsth. Lit. pp. 7, 15, and R. M. Meyer, Altgermanische Poesie, p. 160 (cited by Gummere); and note illustrations in Groos, Die Spiele der Menschen, p. 195. xiv INTRODUCTION milk-bowl." ' Hardly a riddle is without its elements of metaphor. A few examples will serve as well as a hundred. In one of the most famous of the riddles of Symphosius (No. n)t Flood and Fish appear as noisy house and quiet guest. In the popular Old German riddle, ' Es flog ein Vogel federlos, u. s. w.,'t the featherless bird is the Snow, and the mouth- less woman the Wind. And in the riddles of the Exeter Book the Pen is called ' the joy of birds,' § the Wind ' heaven's tooth ' (Rid. Sy8), and the stones of the Ballista the treasure of its womb (i810). Rid. 92 is but a series of kennings. Sometimes the use of riddle-kennings is very close to that of the Runic Poem.\ In its origins the riddle is closely connected not only with the meta- phor but with mythological personification. From one to the other is but a step. ' So thoroughly does riddle-making belong to the mythologic stage of thought,' says Tylor,1T ' that any poet's simile, if not too far-fetched, needs only inversion to be made at once into an enigma.' As the meta- phor plays an immense role in the formation of mythologies, so the riddle is early associated with imaginative conceptions of nature and the divine spirit. Uhland is right in saying that myths and riddles approach most \ closely to one another in the conception of the elemental forces of the greater and more powerful natural phenomena : ' Wenn nun das Rathsel dieselben oder ahnliche Gegenstande personlich gestaltet und in Handlung setzt, so erscheint es selbst nach ausgesprochenem Rathwort auf gleicher Stufe der Bildlichkeit mit der Mythen besagter Art.' The riddle, like the I myth, arises out of the desire to invest everyday things and thoughts .with the garb of the unusual and marvelous. So in the riddle-questions The words of Wackernagel, Haupts Zs. Ill, 25, have been often cited : • Ver- sinnlichung des geistigen, vergeistigung des sinnlichen, personificierung des un- personlichen, verschonende erhebung dessen was alltaglich vor uns liegt, alles das gehort zum wesen des rathsels, wie es zum wesen und' zu den mitteln der poesie gehbrt ; und so mochte kaum ein volk sein das poesie besasse und keine freude an rathseln.' t For the history of this world-riddle, see my article M. L. N. XVIII, 3, 5 ; and notes to Rid. 85. \ This appears in Latin form as early as the tenth century (Reichenau MS. 205, Miillenhoff and Scherer, Denkmaler, 1892, p. 20). For its various versions see Wossidlo, No. 99. § Rid. 277, fugles wyn ; cf. 52, Q327. || See notes to Rid. 56, 73. \ Primitive Culture, edition of 1903, I, 93. •» Schriften zur Geschichte der Dichtung und Sage, Stuttgart, 1866, III, 185. THE COMPARATIVE STUDY OF RIDDLES xv of the Vedas the things treated are not named with their usual uni- versally understood names but are indicated through symbolic expres- sions or simply through mystic relations. The subjects are drawn largely from the world of nature — heaven and earth, sun and moon, the king- dom of air, the clouds, the rain, the course of the sun, years, seasons, months, days and nights. For instance, Night and Aurora appear in a hymnus (I, 123) as two sisters, who wander over the same path, guided by the gods ; they never meet and are never still. In one of the Time riddles (I, 164), the year is pictured as a chariot bearing seven men (the Indian seasons [?]) and drawn by seven horses; in another (I, n), as a twelve-spoked wheel, upon which stand 720 sons of one birth (the days and nights). This is certainly the earliest version of the Year problem, which in one form or other appears in every land,t and is one of the most striking of the motives in the Exeter Book collection (Rid. 23). Uhland early pointed out \ the wealth of the Old Norse problems of nature in mythological reference and suggestion. § The waves (HetSreks Gdtur, No. 23) are white-locked maidens working evil, and in the solu- tion are called ' Gymir's daughters ' and ' Ran Eldir's brides ' ; in another riddle the mist, the dark one, climbs out of Gymir's bed, while in the final problem (No. 35) the one-eyed Odin rides upon his horse, Sleipnir. As I have twice shown, || upon the idea of hostility between Sun and Moon the poet of Rid. 30 and 95 builds an exquisite myth, worthy of the Vedas, indeed not unlike the Sanskrit poems on the powers of nature, and bearing a strong likeness to the famous Ossianic address to the Sun. Of the riddle of the Month (Rid. 23) I have spoken. Many traits of the early attitude to nature are found in the Storm riddles (Rid. 2-4); there is a touch of mythological personification in the world-old motif of Ice (Rid. 34) ; IT and, if my interpretation be correct, the riddle of the Sirens (Rid. 74) is based upon a knowledge of ancient fable. Thus the Anglo-Saxon riddles, like the Russian enigmas printed by Haug, ' Vedische Ratselfragen und Ratselspriiche,' Sitzungsberichte der kiinigl. Akad. der Wiss. su Munchett, Phil.-Hist. Classe, 1875, II, 459. t Cf. Ohlert, pp. 122-126; Wunsche, Kochs Zs., N. F., IX (1896), 425-456; Wossidlo, pp. 277-278; and my article M.L.N. XVIII, 102. t Schriften III, 185. § Cf. Andreas Heusler's discussion of the riddles of the Hervarar Saga (Heifr- reks Gdtur), Zs. d. V. f. Vk. XI, 1901, ii7f. ; and the cosmic riddles of the frnismdl and Alvlssmdl. || M.L.N. XVIII, 104; XXI, 102, 104. M. L. N. XVIII, 4. Ib. XVIII, 100 ; XXI, 103-104. xvi INTRODUCTION Ralston, are sometimes condensed myths, and 'mythical formulas.' It is certainly not without significance that the word 'enigma' is de- rived from the Greek a'vos, which is early associated with the idea of ' fable.' t Of the R&tsdmdrchen I shall speak later. Early in the discussion of riddle-poetry a distinction must be drawn be- tween the Kunstratsd and the Volksrdtsel, between literary and popular problems. This distinction is not always easy to recognize, on account of the close connection between the two types. As I have sought to show elsewhere,! the literary riddle may consist largely or entirely of popular elements, may be (and often is) an elaborated version of an original current in the mouth of the folk ; conversely, the popular riddle is often found in germ or in full development in some product of the study, and our task is to trace its transmission from scholar to peasant. Through a more complicated sequence, a genuine folk-riddle may be adapted in an artistic version, which, in a later day or in another land, becomes again common property ; or, by a natural corollary, a literary riddle, having passed into the stock of country-side tradition, may fail of its popular life and survive only in some pedantic reworking that knows nothing of the early art-form. § Even after the thorough examination of the style and the careful investigation of the history of each riddle so urgently recommended by Petsch || and hitherto so much neglected, we cannot be sure that this apparently popular product is not an adaptation of some classical original, or that this enigma smelling so strongly of the lamp is not a reshaping of some puzzle of peasants. In his excellent discussion of the popular riddle, Petsch claims for the folk all the material that it takes to itself, remodels in its own fashion, and stamps with its own style and meter. After contrasting Schiller's well-known enigma of the Ship with popular treatments of the same theme, and marking in folk- products the choice of a single subject and of a few striking traits, he notes that the typical Volksratsel is confined to a scanty framework, a hurried statement of the germ-element, nai've description, a sudden check in our progress to the goal of the solution, and finally a word of summary. In literary enigmas — to which class by far the greater number of the Exeter Book Riddles belong IF — all these divisions may and do appear, Songs of the Russian People, London, 1872, chap. VI (cited by Pitre, p. xxxviii). t Ohlert, p. 4. t M. L. N. XVIII, 2. § Cf. Pitre's admirable Introduction, p. cxcvi. || Neue Beitrdge zur Kenntnis des Volksratsels, p. 45. T M. L. N. XVIII, 97. THE COMPARATIVE STUDY OF RIDDLES xvii but each of them is patiently elaborated with a conscious delight in work- manship and rhythm, with a regard for detail that overlooks no aspect of the theme however trivial — in a word, with a poetic subordination of the end in view to the finish of the several parts. I may illustrate the derivation of literary enigmas from popular puzzles by examples cited in the first of my articles. Symphosius, in one sense the father of the riddles of our era, uses in many enigmas — for example, those of Smoke, Vine, Ball, Saw, and Sleep (17, 53, 59, 60,96) — the que- ries of the Palatine Anthology current in the mouths of men for centu- ries before his day.f The enigmatograph Lorichius Hadamarius, whose' Latin riddles are among the best in the early seventeenth-century collec- tion of Reusner, $ borrows all his material from the widely-known Strass- burg Book of Riddles. \ Indeed, though scholars have hitherto overlooked this obvious connection, his enigmas are merely classical versions of the German originals. The famous folk-riddles of the Oak (Sir. 12), Dew (Str. 51), Bellows (Sir. 202), Egg (Str. 139), Hazelnut (Str. 172), Lot's Wife (Str. 273), Cain (Str. 284), and dozens of others are twisted into hexameters. Nor was this old pedant alone in his methods of borrowing. His contemporary, Joachim Camerarius of Papenberg, presents, by the side of the German form, the widely extended Sun and Snow riddle in Latin and Greek dress, || and Hadrian JuniuslI fossilizes in like fashion the genuinely popular riddle of the Cherry. Therander, whose Aenigmato- graphia of 420 numbers purports to be a Germanizing of ' the most famous and excellent Latin writers ancient and modern,' is usually in- debted — either indirectly or, despite his assertion of sources, directly — to current versions in the vernacular. His themes of Script (227), Pen M. L. N. XVIII, 2-3. t Ohlert, pp. 138 f. t Nicholas Reusner, Aenigmatographia sive Sylloge Atnigmatnm et Griphorum Convivalium. Two volumes in one. Frankfort, 1602. \Strassbnrger Rdtselbuch. Die erste zu Strassburg. urns Jahr 1505 gedruckte deutsche Ratselsammlung, neu hersg. von A. F. Butsch, Strassburg, 1876. As Hoffmann von Fallersleben has shown, Weimar Jhrb. II (1855), 23r f-> his little book of 336 numbers is the chief source of later popular collections of German riddles. || Reusner I, 254, 258. 1 Reusner I, 243. Huldrich Therander, Aenigmatographia Rytkmica, Magdeburg, 1605. Theran- der, or Johann Sommer, for such was his true name, tells us in his preface that he ' had read the Sphinx Philosophica of Joh. Heidfeld, the Aenigmatographia of Nic. Reusner, and the Libri Tres Aenigmatum of Joh. Pincier, and in order not to sit idle at home when others were working in the fields, had turned these into German rimes.' INTRODUCTION (236), Weathercock (304, 306), Haw (307), Poppy (320), Oak (325), Stork (354), Ten Birds (356), Two-legs (401), Egg (405), and Year (41 1) to cite a few out of many — were favorite possessions of the folk- riddle at the beginning of the seventeenth century ; and we can hardly doubt that Sommer had heard these puzzles on the lips of peasants or met them in the riddle-books then popular. But whether the connection between his little poem-problems and the more naive versions of the folk be mediate or immediate, his book brings everywhere strong proof of the close interdependence of art-riddles and those of the people. The distinction between the riddle of the study and the riddle of the cottage represents only one of many overlapping divisions that present themselves in any extensive consideration of the various kinds of riddles. In his introduction to Holland's collection,! Gaston Paris marks the dif- ference between ' e'nigmes de mots ' and ' enigmes de choses ' ; Wos- sidlo divides the riddles of his famous collection into the three groups of riddles proper, i.e. complete problems or riddles of things (Sachenratsd), jest-riddles or riddle-questions (Ratselfragcn), and finally, riddle-stories or riddle-fables (Ratselmarchen) • and Petsch distinguishes $ between unreal (' unwirkliche ') and real (' wirkliche ') riddles. In the former class he rightly includes all those questions which are addressed rather to knowl- edge and learning than to reason and understanding, IVeisheitsproben, Halslosungsrdfsel, and Scherzfragen. The manifold divisions of Fried- reich into riddle-questions, word-riddles, syllable-riddles, letter-riddles, number-riddles, etc., are based upon no scientific principle, and, for the present, may be disregarded. Tests of knowledge, in enigmatic phrasings, have played a very im- portant part in the evolution of the riddle. The Queen of Sheba came to the court of Solomon to prove the wisdom of the great king by queries. Legend attributes to her several that take their place among world- riddles^ Of these questions of Queen Bilqis, preserved in the Midrash Mishle and the Second Targum to the Book of Esther, the best-known is the enigma of Lot's Daughters, which is found in our collection (Rid. 47). Another riddle-strife attributed to Solomon is that with Hiram of It is, however, going too far to declare with Miillenhoff, Wolfs Zs.f. d. M. Ill, 130, that Therander's riddles are simply expansions of those in the Reterbuchlein, Frankfort, 1562. See Hoffmann, Monatschrift -von u. fiir Schlesien I (1829), 160; Mones Anzeiger II, 310. t P. viii. J P. 5. § Hertz, Haupts Zs. XXVII, 1-33 ; Wunsche, Rdtselweisheit bei den Hebrdern, p. 15; Ohlert, pp. 5-6; Friedreich, p. 98; Folk-Lore I, p. 354. THE COMPARATIVE STUDY OF RIDDLES xix Tyre, described by Flavius Josephus. These are the first of a long series of such word-contests which assume two main forms of great importance in riddle-literature : the Ratsehvettkampf, or matching of wits for some heavy stake, and the ' Colloquy ' or ' Dialogue.' These two classes of questions are not always distinct ; but the former be- longs rather to the region of story or fable, the second to the field of didactic or wisdom literature. In an excellent discussion of the first class, Professor Child f subdivides the Wettkampf into the struggle for a huge wager, usually life itself, and the contest for the hand of a loved lady or knight. Many examples of each may be mentioned. The game of riddle-forfeits is as old as the enigma of the Sphinx \ or as the story of Samson (Judges xiv, 12), § and appears in Germanic literatures in the Hervarar Saga \ and in the VafferuSnismdl^ ; in the ballad of ' King John and the Abbot ' and its continental analogues ft ; in the famous Wartburgkrieg, \ in which Klingsor and Wolfram contend ; and in the ' Tragemundslied,' §§ in which a host tests a wandering stranger, to whom seventy-two lands are known. Not the least important of such riddle-contests are the modern Halslosungriitsel, those gruesome prob- lems by means of which a condemned criminal is supposed to save himself from the extreme penalty. || || Antiquities viii, 5 ; Contra Apionem I, 17, 18. See Wiinsche, p. 24 ; Ohlert, p. 6. t English and Scottish Popular Ballads I, i (' Riddles Wisely Expounded '). t Gyraldus (Reusner I, 10), Friedreich, p. 84 ; Ohlert, pp. 31-35 ; Laistner, Das Ratsel der Sphinx, Grundzuge einer Alythengeschichte, Berlin, 1889. § Friedreich, pp. 151-155 ; Wiinsche, pp. 11-13; P-M.L.A. XVIII (1903), 262. || Bugge, Norr0ne Skrifter, pp. 203 f. ; Vigfusson and Powell, Corpus Poet. Boreale I, 86 f. These riddles of King HerSrek are genuine problems rather than tests of wisdom and knowledge of cosmogony like the VafJ>rud"nismdl and the Alvissmdl (Petsch, p. 1 5). IT Eddalieder, Jonsson, Halle (1888), I, 26-31 ; Friedreich, pp. 112-123. Child I, 403. ft Strieker's 'Tale of Amis and the Bishop,' Lambel's second edition, Erzdh- lungen etc., 1883, p. n ; and ' Ein Spil von einem Kaiser und eim Apt' (Fast- nachtspiele aus dem 15. Jahrhundert I, 199, No. 22). Cf. Child, I.e. Jt Plotz, Der Sdngerkrieg auf der Wartburg, Weimar, 1851. The Introduction contains a bibliography of riddle-collections and Streitgedichte. §§ Altdeutsche Wtilder, 18 1 5, II, 27 ; Mullenhoff & Scherer, Denkmaler I, No. 48 ; Friedreich, pp. 135-138. Uhland, Schriften III, 189, points out that this is a genuine folk-product in its wealth of ' Eigenschaftworter besonders der Farbe.' || || See the collections of Wossidlo, pp. 191-222, and Frischbier, Am Urquell IV, gf.; and the careful discussion by Petsch, pp. 15-22. The most famous of such xx INTRODUCTION The §econd form of Wettkampf, the contest in which the stake is the hand of the beloved, finds equally abundant illustration. We meet it in the Persian story of Prince Calaf, the ultimate source of Schiller's Turandot; in the AlvissmdlJ where the dwarf Alvis wins by his wis- dom the god Thor's daughter; in the English ballads of 'Captain Wedderburn's Courtship ' and ' Proud Lady Margaret ' ; $ in the story of Apollonius of Tyre,§ which is later incorporated into the Gesta Romanorum || ; and in those most charming of word-struggles, the Weidspruche and Kranzlieder of older German folk-song. IF The contest, as it takes form in Colloquy or Dialogue, is closely con- nected with wisdom-literature. Tylor asserts that ' riddles start near I proverbs in the history of civilization, and they travel on long together, though at last towards different ends ' ; and Wunscheft points out that many of the number-proverbs of Solomon (xxx, 18-33, e^c-) are nothing more than riddles. 89 the Dialogue, which holds so important a place in the literature of the Middle Ages, is at once enigmatic in its phrasing and didactic in its purpose. Born of Greek philosophy, it was early adopted by the Christian church as a means of instruction, \ and leads a dull but healthy life in various groups of queries. Among the chief of these are the Salomon and Saturn, § § the Flares of the Pseudo-Bede, || || the Halsldsungrdtsel is certainly the ' Ilo riddle,' known in England, Germany, and many countries of Southern Europe (Pitre, pp. Ixxx-lxxxvii). Haft Paikar of Nizami, cited by Friedreich, p. 52. t Eddalieder, J6nsson, 1888, I, 64 f. \ Child I, 414, 423. § Weismann, Alexander vom Pfaffen Lamprecht, 1850, I, 473; Hagen, Roman von Konig Apoll. -von Tyrus, 1878, pp. II f. || Chapter 153 (Oesterley, p. 383). If Uhland III, 200. Primitive Culture, 1903, I, 90. ft Ratsel-weisheit etc., pp. 24-30. it For an interesting summary of the material upon this subject, see Fb'rster, O. E. Miscellany (Dedicated to Furnivall, 1901), pp. 86 f. §§ For the English versions of this colloquy, both in verse and prose, see Kemble, Salomon and Saturn, 1848. Derived forms are the Adrianus and Ritheus (Kemble, pp. 198 f.) and the Middle English 'Questions between the Maister of Oxenford and his Clerke ' (Engl. Stud. VIII, 284 f.). The history of the widely- spread Salomon and Marcolf saga, so fruitful in the production of dialogues, has been traced by Vogt, Die deutschen Dichtungen von Salomon und Markolf, Halle, 1880, vol. i, and by Vincenti, Drei altengli sche Dialoge von Salomon und Saturn, Naumburg, 1901 ; but a consideration of this lies without my present purpose. Such productions often cross the border of the riddle (compare the enigmatic queries of « Book ' and ' Age,' and the use of the riddle-form, in the O.E. poetical Salomon and Saturn, 229-236, 281 f.). Illl This I have discussed, Mod. Phil. II, 561-565. See infra. THE COMPARATIVE STUDY OF RIDDLES xxi Altercatio Hadriani et Epicteti the Disputatio Pippini cum Albino and the Schlettstadt Dialogue. ,f These questions can hardly be regarded as riddles at all ; for, as I have already noted, they are rather tests of knowl- edge than of the understanding, and at all points display their clerkly origin. | They consist of ' odd ends from Holy Writ,' eked out by monk- ish additions to scriptural lore, scraps of proverbial philosophy, bits of pseudo-science, fragments of fable and allegory, gleanings from the folk- lore of the time. Two derived forms of the Dialogue have each an exten- sive range. The prose Colloquy is represented by the Lucidary, which, in its typical form, the Elucidarium of Honorius, was known among every people of Europe ; § the poetic Dialogue, on the other hand, be- comes the Streitgedicht or Conflict-poem, which, beginning with Alcuin's Conflictus Veris et Hiemis,\ and chronicling the contests of Water and Wine and of Sheep and Wool, reaches its highest development at the skilled hands of Walter Map. IT Ultimately the Colloquy loses its serious purpose and is degraded into series of questions of coarse jest which range from the mocking humor of the Pfaffe Amis (cited supra) to the unsavory queries of the Demaundes Joyous.^ Closely associated with the Wettkampf, or struggle for a wager, is the Riitselmarchen, or riddle-story : indeed, the Apollonius enigma of incest and the ghastly Ilo-riddle of the dead love may be accepted as typical specimens of both groups. In each case the stake can only be won by knowledge of hidden relations that demand a narrative for their unfold- ing. Such connection between the enigma and the fable is found not only in the embodiment of early myths in old cosmic riddles, already considered under another head, but in almost every legend that finds its motif in the seemingly impossible. Uhland is therefore right in regard- ing \ \ the story of Birnam Wood in Macbeth as an excellent example of the Ratselmarchen ; and the so-called ' First Riddle ' of the Exeter Book, Wilmanns, Hatipts Zs. XIV, 530. t Wolfflin-Troll, Alonatsberichte der konigl. preuss. Akademie der Wissenschaftett zu Berlin, 1872, p. 116. \ Cf. the tiny Pharaoh query-poem of the Exeter Book, Gn.-W. Bibl. Ill, 82. § Compare Schorbach, Studien iiber das deuische Volksbtich Lucidarius, Quellen und ForscJningen, 1894, vol. LXXIV. || Monumenta Germaniae Historica, Poetae Latini I, 270. 1f Jantzen, Geschichte des deutschen Streitgedichtes im Mittelalter ( Weinholds Germanistische Abhandlungen), Breslau, 1896, pp. 5 f . Compare Petsch's discussion of Scherzfragen, pp. 22 f. ft Compare Kemble, Salomon and Saturn, p. 285. \ Schriften III, 221. xxii INTRODUCTION in its enigmatic suggestions of some story quite unknown to us, but latent in the memory of early Englishmen, may possibly be assigned to this genus. Of such riddle-stories Friedreich, Petsch, and Pitre offer many specimens ; but these authorities hardly refer to that species of the class which had the greatest vogue in the Middle Ages, the Liigenmarchen Of this special riddle-product, which has been traced by Uhlandf to the tenth century, an apt illustration may be found in the analogue to the Anglo-Saxon enigma of the Month (Rid. 23) which appears among the Liigenmdrchen of Vienna MS. 2705, f. 145. \ I have already noted Gas ton Paris 's distinction between ' enigmes de mots ' and ' enigmes de choses.' By word-riddles ( Wortratsel} are under- stood that large class of problems which are concerned with the form of the word and its components, letters, syllables, etc., rather than with the object which it portrays. The commonest form of word-riddle is un- doubtedly the logogriph, which consists of arranging the letters or shift- ing the syllables of a word, so as to form other words. This species of puzzle, closely akin to our anagram, was well known to the Greeks, § and had a wide vogue in the Middle Ages. The earliest collection on English ground are the word-puzzles in the eleventh century Cambridge MS. Gg. V. 35, 418 b— 419 a, which I have printed and discussed elsewhere. || The persistence of logogriphs in many English and continental manuscripts H" Says Wackemagel, Haupts Zs. Ill, 25 : ' Das Ratsel streift dem Inhalte wie der Form nach an das Liigenmarchen, das Sprichwort, die Priamel, die gnomische Poesie iiberhaupt, ja es giebt Ratsel, die man ebensowohl Marchen nennen kann ; in Marchen, Sagen, altertiimlichen Rechtsgebrauchen unseres Volkes wiederholen sich Fragen und Bestimmungen von absichtlich ratselhafter Schwierigkeit.' tie. } Wackemagel, Haupts Zs. II, 562 ; my article in M. L. N. XVIII, 102. § Compare Friedreich, p. 20; Ohlert, pp. 174, i8of. || Mod. Phil. II, 5651-. See infra. If I class with their continental analogues a few examples from material gathered among the MSS. of the British Museum (see J/. Z. .V. XVIII, 7, note). Castanea : Arundel 248 (i4th cent.), f. 67 b ; Cott. Cleop. B. IX (i4th cent.), f. 10 b, No. 6 ; Sloane 955 (ca. 1612), f. 3 a, No. 2 ; also in MSS. of Brussels, Laon, Ghent, and Heidelberg (Mone, Anz. VII, 42 f., Nos. 42, 56, 138, 119). Paries: Arundel 248, f.67b; Arundel 292 (i3th cent.), f. n3b (Wright, Altd. Blatter II, 148); Brussels MS. 34 (Mone, p. 43); Reims MS. 743 (Mone, p. 45) ; Reusner II, 116. Formica: Arundel 248, f. 67 b ; Arundel 292, f. ii3b; Innsbruck MS. 120, I4th cent. (Anz. f. d. A. XV, 1889, 143); Reusner II, 106. Dopes: Arundel 248,f.67b; Cott. Cleop. B. IX, f. lob, No. 5; MSS. of Brussels and Ghent (Mone, pp. 42, 49). Lux: Arundel 248, f. 67 b; Arundel 292, f. H3b; Cott. Cleop. B. THE COMPARATIVE STUDY OF RIDDLES xxiii shows the long-continued vogue of these playthings of pedantic scholar- ship. None of the Exeter Book riddles are logogriphs in the strict sense ; but such problems as Nos. 20, 24, 25, 37, 43, 65, 75, show the early enigmatograph's fondness for juggling with letters, and Aldhelm, whose liking for the acrostic is seen in the introduction to his enigmas, turns to good account the ' Paries ' logogriph in his word-play upon ' Aries.' The attempts to interpret Rid. i and 90 as ' Cynewulf ' logogriphs (which have so seriously affected the proper understanding of the whole collec- tion) will be later considered. At the very outset of our study of origins, of our comparison of the riddles of different authors or of various folks, we are met by a dangerous pitfall to the unwary, — the association of problems through their solutions rather than through their treatment of motives. Riddles totally unlike in form, and yet dealing with the same theme, exist in different MSS. of nearly the same period, or even side by side in the same collection. The subjects in the interesting group of sixty-three Latin enigmas in the Bern MS. 611 of the ninth century (also Vienna MS. 67) are often those of Symphosius and Aldhelm, but only in a few cases can we detect similarity of treatment. Within the collection itself ,f as in the Symphosius group, one subject receives a second handling of quite another sort: 23, 57, ' Fire,' and 34, 52, 'Rose.' Had Prehn realized this very obvious truth, that similarity of solutions is often coexistent with entire independence of treatment, he would not have erred so often in tracing the riddles of the Exeter Book to Latin sources with which they have naught in common ; but oLthis much more later. After thus marking that the same subjects are developed by different motives, we must note, too, that the converse is equally common, and that the same motives are often accorded to different subjects. For this there are at least four reasons that seem to deserve attention : (a) We are struck by the manifold use of motives appealing to men through the antithetical statement of an apparent impossibility. Wossidlo | shows IX, f. iob, No. 4; Sloane 513, f. 57 b, No. I ; German Book-cover of i6th cen- tury (Mone, Anz. VIII, 317, No. 87); developed at end of I3th century into a German Knnstratsel by Heinrich von Neuenstadt, Apollonius of Tyre, Rid. 6 (Schrb'ter, Mitth. der deutschen Gesellschafl zur JErforschting vaterl. Sprache und Alterthiimer V, Heft 2 (Leipzig, 1872). The discussion that follows is drawn from my article M. L. Ar. XVIII, 4f. t Later in the Introduction this MS. and its analogues will be carefully considered. J No. 78, p. 282. INTRODUCTION that the contrast of dead and living appears in many riddles : Oak and Ship, Ashes and Fire, Tallow and Flame, Brush and Lice, Bed and Man. Again, the motive of 'the child begetting its parent' is found not only in the riddle of Ice but in the Greek enigma of Day and Night t and in the art-riddle of Smoke and Fire, t (£) The riddle is re- tained in memory, but the answer is forgotten and is eventually supplied with an inevitable loss of force. Symphosius's fine Bookmoth riddle (No. 1 6) appears in The Royal Riddle Book (p. 14) with the tame solu- tion ' Mouse in a Study '; and in Holme Riddles, Nos. 61, 62, and 51, the weak answers ' Egg in a Duck's Belly,' ' Penny in a Man's Purse,' and ' Custards in an Oven ' are given to the excellent folk-riddles of ' Maid on Bridge with Pail of Water on her Head,' § ' Blast of a Horn,' || and ' Boats on Water. 'IT The cleverness of a riddle in cunningly suggesting a false solution sometimes overreaches itself, and the true answer is in course of time crowded out by. the usurper. Certain recently proposed answers to our Exeter Book Riddles are surely emendations of Baruch. Biblical riddles furnish strong proof of this lapse of solutions. The rid- dle of Lot's Daughters, perhaps the most widely known of ' relationship problems,' is found at many periods and among many peoples with the proper answer. Only in Germany (Wossidlo 983) appears a general so- lution that reveals an ignorance or forgetfulness of the scriptural story. Petsch (p. 1 4) is doubtless right in his statement that ' after the school-time of the German peasant he troubles himself little about the Old Testament, not hearing each Sunday his First Lesson like men of his class in Eng- land ' ; but this critic's conclusions regarding the riddle before us must be modified in view of its extensive range — only the answer, not the ques- tion, is wanting. To this disregard of the Bible is due the Tyrolese solu- tion of the old problem of a dozen countries,!! ' A water lock and a wooden key ; the hunter is captured and the game escapes.' In Renk's collection from the Tyrol Jt this riddle of 'the Red Sea, Moses's Rod, and the See notes to Rid. 34. t Ohlert, p. 31. \ Symphosius, No. 7 ; Sloane MS. 848 (early ijth cent.), f. 32 ; Holme Riddles, No. 14; Therander, Aenigmatographia, No. 31 (Zs.f. d. M. Ill, 130). § .Votes and Queries, 3d Ser. VIII, 492. II Bk. Merry Riddles, No. 68 (Brandl,//4r£. der deutsch. Sh.-Gesellsch., XLII, 1906, P- '9>- 1 Notes and Queries, 3d Ser. VIII, 503. I shall present in detail the history of this interesting riddle in my notes to K'J- 47- tt Traced by Ohlert, p. 155 ; and Wossidlo, p. 304, No. 413. \Zs.d. r.f. l'k. V, 154. No. i2i. THE COMPARATIVE STUDY OF RIDDLES xxv Destruction of Pharaoh's Hosts ' is found only in its first part, with the answer ' Sea and Boat.' (V) A motive long connected with a certain solu- tion may in a later time, or among another folk, become attached to other subjects and do double or triple duty. The well-known English Cherry riddle has much in common with three German puzzles — those of the Cherry, Arbutus, and Haw (' Hagebutte '). Side by side with this may be placed the Onion-Pepper motive of early Latin and English riddles, t These totally distinct motives have been strangely confounded by Traut- mann in his ' Rosenbutz ' solution of the Exeter Book ' Onion ' riddle (No. 26). \ (d) By far the most numerous of all riddles of lapsing or varying solutions are those distinctively popular and unrefined problems whose sole excuse for being (or lack of excuse) lies in double meaning and coarse suggestion. And the reason for this uncertainty of answer is at once apparent. The formally stated solution is so overshadowed by the obscene subject implicitly presented in each limited motive of the riddle, that little attention is paid to the aptness of this. It is after all only a pretense, not the chief concern of the jest. Almost any other answer will serve equally well as a grave and decent anti-climax to the smut and horse-laughter of the riddle ; so every country, indeed every section, supplies different tags to the same repulsive queries. Wossidlo's material garnered directly from the folk furnishes a dozen examples : Dough and Spinning-wheel (No. 7 1 a, p. 43) ; Kettle and Pike, Yarn and Weaver, Frying-pan and Hare (No. 434 a-e, p. 131) ; Soot-pole, Butcher, Bosom, and Fish on the Hook (No. 434 i, p. 309) ; Trunk-key and Beer-keg (No. 434 n, p. 309) ; Stocking and Mower in Grass (No. 434 s, p. 310) ; Butter-cask and Bread-scoop (No. 434 u, p. 3 1 o). These instances abun- dantly prove the absurdity of dogmatizing over the answers to the Anglo- Saxon riddles of this class. I pass now to the likeness of motives in riddles of different times or localities. Three hypotheses in explanation of this similarity have been advanced by Gaston Paris in his suggestive Introduction to Rolland : § Holme Rid. 29 ; Halliwell, Nursery Rhymes, p. 75, No. cxxx ; Chambers, Pop. Rhymes of Scotland, 1870, p. 109; Gregor, Folk-Lore of N. E. of Scotland, 1881, p. 80; Lincoln Riddles, No. 6 (Notes and Queries, 3d Ser., VIII, 503) — all with Cherry motive. German : Lorichius, Reusner I, 281 (Arbutus) ; Frischbier, Zs. f. d. Ph. IX, 67, No. it, and Wossidlo, No. 181 (Cherry) ; Wossidlo, No. 209, notes, p. 295, many references (Haw). t Symphosius, No. 44 (Onion); Rid. 26, 66 (Onion); Bern MS. 611, No. 37 (Pepper). See also Royal Riddle Book, p. 1 1. \ B. B. XIX, 185. § P. ix. INTRODUCTION (A) common origin; (J3) transmission; (C} identity of processes of the human mind. (A) COMMON ORIGIN, (a) Foremost among problems of like ancestry are ' world-riddles,' those puzzles that may be traced for thousands of years through the traditions of every people. In this list are the riddle of the Sphinx, the queries of the Year,t Louse, t Fire,§ Sun and SnowJ Cow,H and Sow with Pigs. Heusler ft notes that ' the material of world- riddles, like proverbs and fables and tales, belongs to the class of " Wan- dermotiven," and underwent exchanges before the time of literary barter.' (b) Of a narrower range than the riddles of our first class are those of one race in its various branches. Distinctively Teutonic examples are the German-English problems of Chestnut and Nettle and Rose. $t (V) Less extensive still are the riddles of one folk in its many sections and dialects : for example, the German queries of Ten Birds (Wossidlo 170; known for centuries in every corner of the Fatherland), Mirror (Wossidlo 63), and Alphabet (Wossidlo 469) ; or the peculiarly English problems of Leaves, Rope, and Andrew. §§ (B) TRANSMISSION. Extensive range, particularly of a modern riddle, is not in itself a proof of ' common origin,' but often merely an indica- tion that it has been borrowed by neighboring nations from the land of its birth. Adjoining races, though but distantly related, possess in com- mon far more riddles than widely separated people of one stock. In France and Germany appear so often versions of the- same problem (Rolland and Wossidlo, passim) that we can only suppose that legions of puzzles have at one time or other crossed the Rhine and Moselle and found ready adoption in the new land and speech. And Schleicher's list of Lithuanian riddles || || includes a score of correspondences to Germanic queries, which surely cannot all be traceable to the cradle of the two races. But the best proofs of borrowing are these. Sometimes we are able to observe the very act of transmission. The Demaundes Joyous Friedreich p. 87 ; Ohlert pp. 31-35. t Notes to Rid. 23. \ M. L. N. XVIII, 3-4. § Ohlert, pp. 60, 72. || Arnason, Islenzkar Gdtur, 1887, Introd. ; Wossidlo, No. 99, p. 283 ; supra. 1 Rolland, No. 44, p. 22 ; No. 400, p. 152 ; Wossidlo, No. 165, p. 291. » Heusler, Zs. d. V.f. Vk. XI, 141. ttlb. 126. \ M. L. N. XVIII, 7, note ; notes to Holme Rid. Nos. 31, 32, 144. §§ M. L. N. . c. ; notes to Holme Rid. Nos. 57, 105, 1 1 1, 115. till Litauische Marc/ten, Sprichworte, Rdtsel nnd Lieder, Weimar, 1857, pp. 193 f. THE COMPARATIVE STUDY OF RIDDLES xxvii printed by Wynkyn de Worde (1511) is, in the main, but a series of selections from the Demaundes Joyeuses en maniere de quolibetz,\ as Kemble has shown. % Then, too, the riddles that in the Middle Ages had the widest vogue, at least in manuscript, — if we may judge from the scanty evidence of extant mediaeval collections, — were not Volksratsel at all, but Latin logogriphs which are ever the product of the study. There is, of course, no possibility of ' common origin ' with such com- positions as these : they must perforce be directly lent or borrowed. Even, however, with riddles of different periods or sections of one coun- try, genuine folk-products though they may appear, we must often be prepared to find direct transmission through either literature or tradi- tion. The few parallels between the thirty-five HetSreks Gdtur in the Hervarar Saga and the modern Icelandic folk-riddles (Islenzkar Gdtur — 1194 numbers) are rightly regarded by Heusler § as due to the im- mediate literary working of the Old Norse queries. (C) IDENTITY OF MENTAL PROCESSES. The third cause of the simi- larity of riddles must always be taken into account, after careful study of origins and comparison of motives have eliminated all possibilities of a common source and of direct or indirect transmission. When the counterpart of the 'Flood and Fish' riddle of Symphosius (No. 12) meets us among Turkish queries, || we are naturally inclined to believe that this widely known riddle has penetrated even to the Bosphorus ; but we can hardly explain thus the similarity of the motives in the Persian 'Ship' problem of Nakkash, d. 938 A.D.,1T — 'It makes its way only upon its belly, cutting, though footless, through the girdle of the earth ' — to those in the 1 5 1 st riddle of the Islenzkar Gdtur • or the surprising likeness of many Sanskrit riddles to our modern charades ; or even the parallels between the Anglo-Saxon problems of musical in- struments (Rid. 32, 70) and the Lithuanian ' Geige ' riddles.ft Indeed, This interesting collection was reprinted in Hartshome's Ancient Metrical Tales, London, 1829, pp. i-S. t A copy of the French text — a very rare little octavo — is in the British Museum. It bears no date, hut is assigned by the Catalogue to 1520, by Kemble with greater probability to 1500 or before. t Salomon and Salnrmts, p. 286. Compare T>randl,///r£. der d. Sh.-Gesell. XLII (1906), 2-3. §Zr. d. V.f. Vk. XI, 128. || Urquell IV, 22, No. 10. If Friedreich, p. 164. Fiihrer, Zs. der detitschen morgenl. Gesellschaft XXXV, 1885, 99-102. tt Schleicher, p. 200. xxviii INTRODUCTION the case seems to be this. While, as we have seen, similarity of subject does not necessarily imply similarity of motives, there are of course certain themes that, from their limited nature, prescribe a particular treatment. However unaided may be the act of composition, essential traits of these subjects must be named, described, disguised, or sum- marized. Surely all likeness entailed by the very nature of the topic cannot be regarded as irreconcilable with a perfectly independent crea- tion. Riddles, remote and unrelated though they be, must, after all, say somewhat the same things of the commonplaces of life. At times indeed — and now I must point to my present heading — this correspondence is carried far beyond the necessities of the subject through many combi- nations and permutations of motives, for riddle-literature, like every other, has its striking coincidences ; but these instances are comparatively rare, since diversity of development, unlikeness in likeness, is here as else- where the badge of independence. The rarity of cases of complete re- semblance between two riddles with no historical kinship gives them a peculiar value for us ; and the evidence of such Doppelgdnger for a solu- tion is surely of far more weight than the random guesses of a modern interpreter. In discussing the originals and analogues of the Exeter Book Riddles I shall seek to apply the principles adduced in the present chapter. II ORIGINALS AND ANALOGUES OF THE EXETER BOOK RIDDLES SYMPHOSIUS August Heumann, in his excellent edition of the Enigmatica of Symphosius, set up the thesis that ' Symphosius ' was the lost Sym- posium of Lactantiusf mentioned by Jerome. \ Other editors, notably Migne§ and FritzcheJ follow Heumann in including these 100 riddles Hanover, 1722. t Goetz, Rheinisches Museum XLI, 318, shows on the evidence of a gloss in the tenth-century Codex Cassinus 90, ' simposium vel simphosium (MS. simphonium) aenigma quod Firmianus (MS. et) Lactantius composuit (MS. composuerunt),' that the enigmas were at an early time attributed to Lactantius. \ De Viris Illustribus, cap. 80. § P. L. VII, 25. || II, 298. ORIGINALS AND ANALOGUES xxix in editions of Lactantius. Heumann's contention was opposed by Werns- dorff on two grounds : (a) The prologue of seventeen hexameters in- troducing the enigmas mentions our poet by name, ' Haec quoque Symposius | de carmine lusit inepto.' (<£) Symphosius is named by several early writers, among them Aldhelm {Epistola ad Acircium) : ' Symp(h)osius poeta metricae artis peritia praeditus occultas aenigmatum prppositiones exili materia sumtas ludibundus apicibus legitur cecinisse et singulas quasque propositiones formulas tribus versibus terminasse.' The conclusion of Pithoeus, \ cited with approval by Wernsdorff , that our author was ' Caelius Firmianus Symphosius,' the maker of other poems of the Latin Anthology, has, however, been abandoned by recent scholars. § Yet all modern editors unite in accepting for these enigmas an author called ' Symphosius.' Such is the view of Paul || and Schenkl,1F and of the editor of the oldest manuscript of the riddles (the Codex Salmasianus), Riese in the Latin Anthology. Regarding the date of Symphosius, there has been much dispute. Wernsdorff ft would assign him to the fourth century; Paultt and SchenklH to the fourth or fifth ; L. Miiller§§ to the second or third, on account of his metrical skill; and Hagen|||| follows Riese (1868) in as- cribing him to the same period as the collector of the poems of the Latin Anthology, the end of the fifth and the beginning of the sixth centuries. The text of the riddles is contained in numerous manuscripts, which range from the eighth to the eleventh century and are divided between two recensions.lTII Since the edition of Perionius there have been various editions and commentaries upon these enigmas — discussed by Friedreich, ftt Riese, and Teuffel. The best of these is that of Riese. The enigmas of Symphosius consist each of three hexameter lines of good Latinity, and are one hundred in number. Their metrical preface connects them with the festival of the Saturnalia (' Annua Saturni dum Poetae Latini Minores, Helmstadt, 1799, VI, 424. t Riese, Anth. Lat. I, 221, ' Symphosius.' \ Poematia Vetera, Paris, 1 590. § Cf. Teuffel, Hist, of Roman Literature, 1892, §449, I. || Dissertatio de Symposii Aenigmatis (Part I), Berlin, 1854, p. 14. ^Sitzungsber. der phil.-hist. Kl. der Wiener A kad. XLIII (1863), P- I2- Anthologia Latina, 1894, I, 221-246. tt P. 414. \ P. 36. §§ De Re Metrica, p. 55 (cited by Schenkl). Illl Ahtike u. Mittelalterliche Rathselpoesie, Bern, 1877, p. 23. f. Riese, 1. c. and Teuffel, 1. c. Paris, 1533. ttt Pp. 187-188. xxx INTRODUCTION tempora festa redirent ') ; and, while this association is more than doubt- ful, they are thoroughly pagan in character. Ebert divides them, accord- ing to subject, into six categories : (i) living things, especially beasts, less frequently man in strange aspects ; (2) plants as flowers or food ; (3) clothing and ornaments ; (4) domestic implements ; (5) structures — the ship, the bridge, the ladder ; (6) meteorological phenomena — mist, rain, snow. ' The subjects,' he remarks, ' are drawn from the external world, and include for the most part objects which are closely associated with man in his daily life.' The enigmas of Symphosius have dominated all riddles, both artistic and popular, since his day. To be sure, some of the problems to which he gave a wide vogue had been current in the mouths of men for centuries before his time.t Others became immediately and widely popular. But I at no place and time were they in greater favor than in England of the eighth century. Aldhelm not only hails Symphosius as a model in his' Epistola ad Acircium (supra) and draws freely upon his verses, t but in his enigmas borrows subjects (Nos. 51, Mola\ 92, Mulier quae gemi- nos pariebaf) and attaches himself to the older riddler both in matter and form (infra). § In the Flores of the Pseudo-Bede, || five riddles from Symphosius (Nos. i, 7, 4, n, 10) are quoted in full.lf And in the Dis- putatio Pippini cum Albino Alcuin paraphrases seven riddles from the earlier writer (Nos. 75, 30, 14, 98, 99, n, 96). The other Anglo-Latin collections of enigmas exhibit a slight connection with Symphosius (infra) • and, as I shall show later, the Exeter Book Riddles owe him an important debt. Very close is the relation of the enigmas of Symphosius to the Apollonius of Tyre story, so popular in the Middle Ages. ft Various ver- sions of this tale contain a larger, or smaller number of enigmas, until in Ber. iiber die Verh. der k. sacks. Gesellsch. der Wiss. zu Leipzig, Phil.-Hist. Classe, 1877, p. 21. t Ohlert, pp. 138 £., has pointed out that Symphosius uses in many enigmas, those of Smoke, Vine, Ball, Saw, Sleep (17, 53, 59, 60, 96), the queries of the Palatine Anthology (supra), and such world-old riddles as that of the Louse (see my articles in M. L.N. XVIII, 3) receive his guinea-stamp (No. 30, Pediculus). t Manitius, Zu Aldhelm und Baeda, 1886, p. 5 1, fully illustrates this indebtedness. § Ebert, Ber. d. s. G., p. 22. II Migne, P. L. XCIV, 539 f. See infra. If Manitius, p. 82 ; my article in Mod. Phil. II, 561. » Wilmanns, Haupts Zs. XIV, ^30. ttCf. Weismann, Alexander, Frankfort, 1850, I, 473 f. ; Schrbter, Mitth. der deutschen Gesellsch. zur Erf. der voter 1. Sprache etc., Leipzig, V, 2 (1872), p. xiv. ORIGINALS AND ANALOGUES xxxi the Middle German Vblksbuch form we encounter translations of no less than ten problems (Nos. 89, 61, 63, 11,2, 13, 69, 77, 78, 59) into the vernacular. At least three of the Symphosius riddles (Nos. n, 89, 13) passed from the Apollonius story into the Gesta Romanorum, chap. 153. In the sixteenth century the enigmas were translated into Greek by Joachim Camerarius (ca. 1540), and expanded by many others of Reusner's pedants. | ALDHELM From Aldhelm of Malmesbury (640-709), Bishop of Sherburne, we possess one hundred riddles in . hexameters. \ Of these William of Malmesbury tells us : § ' Extat et codex ejus non ignobilis " de Enigmati- bus " poetae Simphosii emulus centum titulis et versibus mille distinctus.' In this last phrase, as William's next words show, he is simply accepting the description of the enigmas furnished by the acrostic which the first and last letters of the thirty-six lines of Aldhelm 's poetical preface com- pose, ' Aldhelmus cecinit millenis versibus odas,' — a description not strictly correct, as only eight hundred hexameters appear. Unlike the enigmas of Symphosius, the hundred poems of Aldhelm are of varying length : nineteen tetrastichs, fifteen pentastichs, thirteen hexastichs, nine- teen heptastichs, ten octostichs, eleven enneastichs, four decastichs, four hendecastichs, one dodecastich, one triscaedecastich, one pentecaedeca- stich, one heccaedecastich, and one polystichon (De Creatura). The in- debtedness of these to Symphosius is sometimes greatly overstated. || Indeed, Aldhelm's chief debt is found not in his enigmas but in the Epistola ad Acirdum or Liber de Septenario, which serves as a prose preface to his riddles.lf In this tractate upon prosody, which was sent to Ealdferth, King of Deira and Bernicia, in the tenth year of his reign, 695, and which was perhaps originally an independent work, he ac- knowledges his indebtedness to Aristotle and to the books of the Old Testament, but chiefly to Symphosius, from whom he draws at least a dozen illustrations. ft It is interesting to note that this treatise on meter Schrbter, p. Ixxv. t Reusner, A enigma tographia sive Sylloge Aenigmatum etc. Frankfort, 1602. t J. A. Giles, S. Aldhelmi Opera, 1844, pp. 249-270. § Gesta Pontificum Anglorum V, § 196, Rolls Series, 1870, pp. 343-344. || Cf. authorities cited by Friedreich, p. 191. If Giles, S. Aldhelmi Opera, pp. 2i6f. Bonhoff, Aldfiflm von Malmesbury, Dresden, 1894, p. 114. ft These are cited in full by Manitius, Aldhelm und Baeda, p. 51. xxxii INTRODUCTION contains one of the best known of world-riddles, that of the Ice, ' Mater me genuit, eadem mox gignitur ex me,' which does not appear in Sym- phosius, but is found in the Exeter Book, 349-11. Between the enigmas of Aldhelm and Symphosius the verbal resem- blances are not great. t Indeed, the same subjects are often treated by the two in very different fashion. Like Symphosius, Aldhelm makes the dumb nature of inanimate things speak, but for this personification he pleads the precedent of the Bible. \ Ebert has noted § the chief differences be- tween the poets. To the categories of subjects which are treated by Symphosius and which receive further elaboration from Aldhelm, the younger writer adds new themes : the heavenly bodies, the elements, and such abstractions as Nature, Fate, The Creation. As Bonhoff well ex- presses it, || ' Bei Aldhelm iiberwiegt mehr das dem Germanen so eigene sinniganschauliche Sichversenken in die Natur, ihre Wunder und Werke, wahrend Symphosius als ein Romane lieber das verstandnismassige und espritvolle Spielen und Tandem in Wort und Ausdruck sucht.' Ebert also points to the presence in these enigmas of the Christian element, which is totally lacking in the riddles of Symphosius. IF This is seen not only in the problems of Fate (i, 7) and Creation (xiii), but in those of the Dove (Hi, 9), Apple-tree (iv, 15), Fig-tree (iv, 16), and Lucifer (vii, 3), all of which are based upon Jewish-Christian story. Other Christian traces are marked by Ebert (ii, 14; vi, 4; viii, 3). And yet there are many references to classical mythology : to the Minotaur (ii, 1 1), to the threads of the Parcae (iv, 7), to Jove's eagle and Ganymede (v, 2), to Scylla (x), and frequently in his polystich, the De Creatura. Against all such heathen fables he inveighs in his enigma on the Sun and Moon (viii, 3). All critics have noted the larger scale and freer treatment of Aldhelm 's enigmas compared with those of his model ; but, while the writer of Malmesbury has obviously gained in romantic breadth, he has lost not a little. ^Expanding in the joy of creation, he often forgets his riddle's « For history of this riddle, see M. L. N. XVIII, 4, and notes to Rid. 34. t These parallels are cited by Paul, Dissertatio de Symposii Aenigmatibus, 1854, p. 19, and by Manitius, pp. 78!, who greatly overstates likenesses. Two enigmas are borrowed (i, 10, Sym. 92 ; iv, 12, Sym. 51), and occasionally a striking motive, like that of 'the biter bitten,' ' mordeo mordentes' (Sym. 44!), which Aldhelm, iii, 1 5, transfers from the Onion, adapting it to the Nettle, ' torqueo torquentes.' \ Epistola ad Acircium, Giles, p. 229. § Pp. 22-23. II P- 115- T See also Manitius, Christl. Lat. Poesie, p. 489. ORIGINALS AND ANALOGUES xxxiii excuse for being, and lifts the veil of his mystery (Ebert). Or else he falls into the opposite fault of needlessly complicating and obscuring his meaning. That his contemporaries found many lines difficult is shown by the large number of Latin and English glosses which we meet in the British Museum manuscripts of his enigmas. TATWINE Of Tatwine, the author of the third collection of enigmas with which we have to do, we know little more than we are told by Bede.f He was ' a Mercian out of the district of the Hwiccas, and succeeded Berhtwald (d. January 13, 73 1 ) as Archbishop of Canterbury. He was consecrated June 10, 731, but did not receive the pallium until 733. Almost nothing is known of his rule. He died July 30, 734. As both Ebert and Hahn point out, he was a philosopher, a theologian, and a grammarian. And, what is more to our present purpose, he was an enigmatograph, the author of forty Latin riddles, t That the manuscripts preserve the origi- nal order of the enigmas is proved by the double acrostic — formed from the first and last letters of the first lines of the poems — corresponding to the introductory distich Sub deno quater haec diverse enigmata torquens Stamine metrorum exstructor conserta retexit. Of the forty riddles, twenty-two consist of five hexameters, nine of four, seven of six, one of seven, and one of twelve. Both Ebert and Hahn point to the revelation of Tatwine's personality in these enigmas. That he is a theologian is shown by his choice of religious or churchly themes in one third of his riddles : church furniture, the Christian virtues, topics MS. Royal 15, A. XVI; MS. Royal 12, C. XXIII. Cf. comments of Wright, Biog. Brit. Lit. I, 78, and Bonhoff, p. 115. For the glosses themselves see Wright's edition of the enigmas {Anglo-Latin Satirical Poets, Rolls Series, 1872, 11, 533-573) and Napier, O. E. Glosses, pp. 191 f. t Eccl. Hist, v, cap. 23, 24. Compare Ebert, p. 25 ; Hahn, Forsch. zur deutschen Gesch. XXVI (1886), 603 f. J These are preserved in two MSS. in company with the enigmas of Eusebius (infra) ; the one at Cambridge, MS. Gg. V, 35 ; the other in the B. M., MS. Royal 12, C. XXIII. The enigmas of both poets were edited from the Cambridge MS. by Giles (Anecdota Bedae, Lanfranci et Aliorum, Caxton Society, 1851); those of Tatwine, from the London MS. by Wright {Anglo-Latin Satirical Poets, Rolls Series, 1872, II, 525-534), who knew nothing of the other manuscript or of the earlier edition ; and finally from both texts by Ebert, Ber. tiber die Verh. der k. sacks. Gesellsch. der Wiss. zu Leipzig, Phil.-Hist. Classe, 1877, pp. 20 ff. xxxiv INTRODUCTION of dogma. That he is a philosopher becomes at once apparent in his first and longest problem, De Philosophia, and is further indicated by his love of abstractions and of speculation. That he is a grammarian is attested not only by the selection of such a topic as ' Prepositions governing both cases' (No. 16), but by the narrow range of his fancy and the sobriety of his style, t Tatwine owes very little to his predecessors. Unlike Ebert, \ and like Hahn, § I can detect no striking resemblances between his enigmas and those of Symphosius on similar or kindred themes. In the six riddles (Nos. 6, 7, 1 1, 20, 28, 32) that invite comparison with the earlier enigmas, the very slight likenesses seem to me to lie rather in the coincidence of subjects than in actual borrowing. To Aldhelm he may acknowledge perhaps a small debt, which has been greatly overstated by Manitius in his list of alleged parallels between the Anglo-Latin riddlers || and even by Ebert. In the eight riddles cited by Hahn as suggesting a slight resem- blance to the older collection IF we sometimes have motives common to all the Anglo-Latin riddles (4, 5, 6) and very possibly the possession of the folk. But an occasional lifting of Aldhelm's phrases, not only when he is dealing with like subjects (12, 31, 39), but elsewhere in the group (T. n1, A. iv, 31; T. 17, A. i, 14"; T. 24, A. De Creatura 21, etc.) puts beyond doubt a direct relation. Hahn observes with not a little plausibility: — 'Bei der grossen Neigung der Gelehrten des 8. Jahrh. zur wirklichen Ausbeutung ihrer litterarischen Vorbilder ist der Wegfall solcher Pliinderung eigentlich fur die Unabhangigkeit zweier Schriftsteller von einander bedeutungsvoll.' Yet when we remember that Aldhelm himself, ordinarily a mighty lifter, greatly restricted his borrowings from his model Symphosius, Hahn's argument loses much of its weight. EUSEBIUS Over the identity of Eusebius, the author of the sixty riddles which accompany those of Tatwine in the Cambridge and British Museum manuscripts, there has been much discussion. Ebert ft declares that 'we know nothing of him, because the conjecture of Giles \ that he is the See Manitius, Christl. Lat. Poeste, p. 503. t See Ebert, Lift, des Mitt, im Abendlande 1, 651. \ Ber. d. s. G., p. 26. § P- 6 1 1. || Aldhelm und Baeda, pp. 79-82. If Tatwine 4, Aldhelm iv,i; 7.5, A.v,9; T. 6, A.v.j; T. 12, A.vi, 4; 7.30, A.iv,io; T. 31, A.vii, 4; T. 33, A.v, 10; T-39, A. ii, 10. P. 612. ft Ber. d. s. G., p. 27. f$ Anecdote, Preface, p. x. ORIGINALS AND ANALOGUES xxxv Eusebius to whom Bede dedicated his commentary upon the Apocalypse is without support.' Ebert admits, however, that nothing in his riddles militates against the theory that he was a contemporary of Tatwine. Hahn follows Giles in identifying the author of our enigmas with Eusebius, the friend of Bede. He had previously proved beyond all doubt | that this friend was Hwaetbert, Abbot of Wearmouth in North- umbria. t Hwaetbert-Eusebius is clearly revealed by Hahn ; but that the great abbot of the North is the maker of our enigmas, is merely a happy conjecture incapable of positive proof. The conjecture rests, however, on such high probabilities of time and place § that a brief sketch of Hwaetbert may be drawn from Hahn's ample material. He was born about 680 (his early teacher, Sigfrid, died in 688, and Hwaetbert was young enough to be called ' juvenis ' in 716), and was in his young man- hood at Rome under Pope Sergius (687-701). He was ordained priest in 704, and chosen Abbot of Wearmouth on June 4, 716. That he was a scholar is evidenced by Bede's tribute (supra). He was honored by the dedication not only of his friend's commentary upon the Apoca- lypse but of his scientific work of 726, De Ratione Temporum.\ He was probably the author of the anonymous ' Life ' of his predecessor in the abbacy, Ceolfrid, whom, in an admirable letter still extant, he com- mends to the kindly offices of Gregory II. H That he was still living in the forties of the eighth century is proved by a letter addressed to him by the missionary bishop Boniface between 744 and 747. Other things speak for his authorship of our enigmas, besides favor- able conditions of time and place. In favor of this view is the internal evidence of the enigmas themselves; although upon this we must not lay undue stress, as his enigmas are not nearly so distinctive as those of Tatwine. The riddler Eusebius seems to have been a theologian and divine (Nos. 1-5), although, unlike Tatwine, he avoids subjects of the Forsch. zur deutschen Geschichte XXVI (1886), 601 f. Cf. Erlemann, Herrigs Archrv CXI (1903), 58. t Bonifaz und Z«/, Leipzig, 1883, pp. 213-218. | Bede thus speaks of him in his remarks upon the first book of Samuel the prophet (Giles, Opera Bedae VIII, 162), ' Huetbertum juvenem cui amor studi- umque pietatis jam olim Eusebii cognomen indidit.' § The identification is accepted by Ebert, Litt. des Mitt, im Abendlande I, 1889, p. 652, and Manitius, Christl. Lat. Poesie, p. 502. || Giles, Opera VI, 139-140. 1" Hahn, pp. 216-217. Jaffe, Bibliotheca III, 180, No. 62; discussed by Hahn, Bonifaz, p. 213. xxxvi INTRODUCTION Christian cult: he shows a keen interest in chronology (Nos. 26, 29) and grammar (Nos. 9, 19, 39, 42) — tastes befitting a friend of Bede ; and in his later enigmas (Nos. 41-60), which were perhaps written, as Ebert suggests, for use in the school, he displays an accurate knowledge of the great textbook of his time, Isidore's Etymologies.^ A striking characteristic of his enigmas is his love of contrasts (Nos. 8, 15, 18, 21, 24 27, 48).$ Ebert rightly regards his literary workmanship as inferior to that of Tatwine. The first forty of his enigmas consist each of four hexameters ; the last twenty, so different from their predecessors in origin, matter, and form, are of varying lengths. Now, what is the relation of the enigmas of Eusebius to those of Tatwine, which they accompany ? Ebert § advanced the opinion that Eusebius sought, by supplementing Tatwine's forty riddles with sixty others, to make a new riddle-book of one hundred queries like the groups of Symphosius and Aldhelm (compare also the ninety-five prob- lems of the Exeter Book). That we may not assume the reverse relation seems evident for two reasons : Tatwine firmly establishes the number of his problems by his acrostic ; Eusebius is hard put to it to raise his own number to sixty and is driven to new sources (supra). From the internal evidence of the single enigmas we can draw no valuable con- clusion regarding the relation of the two groups, as, with one exception, there is no likeness in thought and word between the problems that handle like themes (E. 7, T.4 ; E. 8, T.33 ; £.17, T. 9 ; E. 24, T. 23 ; E. 27, T. 25 ; E. 32, T. 5 ; E. 36, T. 30). In the 'Pen' problems (E. 35, T. 6), where we have at least one common motive, not only are both writers in the wake of Aldhelm (v. 3), but both are employing ideas cur- rent in all riddle poetry of the time. || Though the manner of Eusebius is not unlike that of Symphosius, there is little trace of direct borrowing from the earlier and wittier writer. The resemblances (E. 16, S. 81 ; £.34, S. 1 1 ; E. 38, S. 14 ; E. 43, S. 38) are not striking, and may well be en- tailed by the demands of like subjects. Of the first forty riddles of « Cf. Ebert, Ber. d. s. G., p. 28. t Bucheler, Rhein. Mus. XXXVI, 340, and Hahn, pp. 619-624, give abundant proof that Eusebius did not go directly to Pliny and Solinus, as Ebert supposed, but derived from these authors through Isidore. See also Ebert, Litt. des Mitt. im Abendl. I, 1889, p. 652, N. \ See Manitius, Christl. Lot. Poesie, p. 504. § Ber. d. s. G., p. 27. II Cf. Ebert, Haupts Zs. XXIII, 200; the writer, M.L.N. XXI, 102, and notes to Rid. 52. ORIGINALS AND ANALOGUES xxxvii Eusebius, sixteen invite comparison with Aldhelm through their treat- ment of similar subjects. Of these, eight are totally independent (E. 4, A. xiii, i; £.5, A.vi, 2; £.7, A. iv, i ; E. 10, A. viii,-3 ; E. u, A. i, 6 ; £.15, A. iii, i; £.28, A.v, i; £.36, A. iv, 10); fou,r display a slight connection (E. 6, A. i, i ; E. 8, A. i, 2 ; £.32, A.v, 9; £.33, A. ii, 14); two show a still more marked relation (E. 31, A.v, 9 ; E. 35, A.v, 3) ; and two are very closely bound to their prototypes (£.37, De Vitulo, A. iii, 1 1 ; E. 40, De Pisce, A. iii, i o). On account of the last few exam- ples, Hahn is inclined, with Ebert, to believe in a direct employment by Eusebius of Aldhelm's enigmas ; but he sanely distinguishes ' between collective and individual use, between transmission by book and by tra- dition.' ' It is very possible that single riddles of Aldhelm and of others were transmitted, as themes of wit and entertainment, from monastery to monastery, and from mouth to mouth ; and thus arose the use of particular riddles and not of the whole collection.' Though only three of the last twenty enigmas of Eusebius bear any resemblance even of topic to Aldhelm's (£.48, A. xii ; £.56, A. iv, 2 ; £.57, A. iii, 7), yet these latter riddles approach far more closely to his manner, and may be the additions of another hand than that of Eusebius. LATIN ENIGMAS AND THE EXETER BOOK The relation between the Exeter Book Riddles and the Latin enigmas current in the eighth century was first touched upon by Thorpe in his Preface t : ' Collections of Aenigmata have been left us by Symphosius, Aldhelm, Beda and others ; but these are, generally speaking, extremely short, and although they may have occasionally suggested a subject to our scop whereon to exercise his skill, yet are those in the present collec- tion too essentially Anglo-Saxon to justify the belief that they are other than original productions.' In his first article | Dietrich indicates the indebtedness of the Anglo-Saxon collection to certain models. Once or twice we have a direct reference to learned sources. § Among these sources are Symphosius and Aldhelm. According to Dietrich, || Rid. 17, Hahn, pp. 628-629. t P. 10. \ Haupts Zs. XI, 450 f. § We can, however, lay very little stress upon such phrases as Kid. 437, t>dm he bee ivitan (a reference to the knowledge of runes), and 4O13, gewritu secgafr, as neither of these riddles (40 or 43) seems to owe aught to the Latin enigmas ; and the words, Rid. 3Q5, Mon mafrelade se J>e me ges&gde introduce a riddle-motive uni- versally popular at this period (M. L. N. XVIII, 99). || XI, 251 f. ; XII, 241. xxxviii INTRODUCTION 48, and 6 1 show close verbal borrowings from Symphosius ; while Rid. 36, 39, and 41 are derived sentence for sentence from Aldhelm. In Rid. 6, 14, 29, 37, 51, 54, individual points are borrowed from the Latin enigmas. In the 'so-called second series Dietrich notes a freer employment of Sym- phosius (Rid. 66, 84, 85, 86, 91), and a few traits from Aldhelm (Rid. 64, 71, 84). He draws from his very doubtful premises the conclusion that ' a closer dependence upon Latin models is a constant trait of the first series, a freer movement predominates in the second.' From the references to ' writings ' in Rid. 40, from the C and B runes which precede Rid. 9 and 1 8 and which may stand for the Lat. camena and ballista, Dietrich con- jectures a third Latin source, but ' none has been discovered which casts any light upon the problems in question.' Dietrich also points out the pop- ular elements in such riddles as Rid. 23, 14, 52, 34, 43, 10, etc., and notes parallels among the German f oik-riddles, t M tiller's contribution to the Cothener Programm (1861) adds nothing to Dietrich's treatment of sources. But in 1877 Ebert, in his essay upon the riddle-poetry of the Anglo-Saxons, \ seeks to show that our riddler, whom he identifies with Cynewulf, probably used Tatwine's enigmas, and certainly those of Eusebius. The English riddles which he believes to be indebted to the Latin are Rid. 7 (E. 10) ; 14 (T. 4, E. 7) ; 15, 93 (E. 30) ; 21 (T. 30) ; 27 (T. 5, 6 ; £.31, 32) ; 30 (E. 1 1) ; 39 (E. 37) ; but, as I shall show, there is in none of these cases any conclusive proof of a direct literary connection. In a monograph which, by its perversion of method and unwarranted! ' conclusions, has done no little harm to the proper understanding of Exeter Book problems and their relations, Prehn § aims to find for nearly every Anglo-Saxon riddle a Latin prototype among the enigmas of Sym- phosius, Aldhelm, Tatwine, and Eusebius. He thus summarizes his re- sults : || 'An exclusive use of Symphosius is found in twelve riddles, of Aldhelm in seventeen, of Eusebius in five, while Tatwine is never used All of Dietrich's statements regarding sources must be considerably modified and discounted in the light of my investigations (M.L.JV. XVIII, 98 f.). See infra, and notes to separate riddles. t Dietrich's treatment of the connection between the poems of our collection and popular riddles is confined to a single paragraph (XI, 457-458) and must be supplemented at every point (see my article in M.L.N. XVIII, 98 f., my discus- sion infra, and the notes to the several problems). \ Ber. d. s. G., p. 29. § Composition und Quellen der Rdtsel des Exeterbuches. Paderborn, 1883. II P. 158- ORIGINALS AND ANALOGUES xxxix alone.' But, according to Prehn, our author frequently builds up his rid- dle by suggestions and plunderings from more than one author : he thus employs Symphosius and Aldhelm six times, Symphosius and Tatwine twice, Aldhelm and Tatwine once, Aldhelm and Eusebius four times, Aldhelm, Tatwine, and Eusebius three times, but never Tatwine and Eusebius alone together. Sometimes he employs more than one riddle of the same author : he thus uses Symphosius twice and Aldhelm once. Against these results of Prehn's too fruitful source-hunt there have been more than one protest from scholars. Zupitza.f a year later, took issue with Prehn's conclusions of wholesale borrowings from learned sources, and affirmed his belief in the popular origin of many Exeter Book puz- zles. Holthaus \ also thinks that Prehn has failed to establish the great dependence of the Anglo-Saxon riddles. He points to the popularity of such compositions among monks and laymen. The number of universally known riddles was far larger than those extant ; and these, in form and expression, were naturally much alike. Only the true poets gave them a new dress. Regarding the vogue of this riddle-material, he believes, as does Ten Brink of the epic, § that ' the product of poetic activity was not the possession, the performance, of an individual but of the community.' Other arguments of Holthaus will be considered later. So Herzfeld || declares that ' in the case of the Exeter Book Riddles one cannot speak of a constantly close adherence to definite models. Previous investiga- tions IT show that some few of these are literal translations of the Latin, others are related to the Latin riddles only in single traits and turns of thought, while the majority have their roots in popular tradition, from which the poets of both the Latin and the Old English riddles have drawn independently.' Brooke quotes the whole of Aldhelm 's riddle De Luscinia side by side with Rid. 9, ' in order to confound those who say that Cynewulf in his Riddles is a mere imitator of the Latin. In the Latin there is not a trace of imagination, of creation. In the English both are clear. In the Even in cases where Prehn is unable to demonstrate borrowing, he declares (p. 269) : ' Indessen beschrankt sich ihre Selbstandigkeit nur auf die Wahl der Stoffe, wahrend der Inhalt dieselben typischen Ziige aufweist, welche wir bei den Vorbildern kennen gelernt haben.' t Deutsche Littztg., 1884, p. 872. J Anglia VII, Anz. 124. § Geschichte der Engl. Lift., p. 17. || Pp. 26-27. f Herzfeld compares J. H. Kirkland, A Study of the Anglo-Saxon Poem, The Harrorving of Hell, Halle, 1885, pp. 25 f. But in what respect this reference es- tablishes large results, I fail to see. E. E. Lit., p. 149, footnote. xl INTRODUCTION one a scholar is at play, in the other a poet is making. Almost every riddle, the subject of which Cynewulf took from Aldhelm, Symphosius or Eusebius, is as little really imitated as that. Even the Riddle De Crea- tura, the most closely followed of them all, is continually altered towards imaginative work.' Erlemann discusses the close relation of the Riddles to the Latin enigmas of the early eighth century. 'All of these enigmatographs, Aldhelm, Tatwine, and Eusebius, were contemporaries of Bede ; and, as Hahn has shown,! Eusebius is identical with Hwaetbert-Eusebius, Abbot of Wearmouth, to whom Bede submitted his work of 727, De Temporum \ Ratione. The Anglo-Saxon poet [so Erlemann] knew all the Latin collec- tions of riddles and employed Eusebius in particular. There is no small probability that the Anglo-Saxon poet, through school instruction, was familiar with the works of Bede as well as with the riddle-poems of Eusebius, Tatwine, and Aldhelm. It is indeed possible that he obtained his scholarly training in one of the monasteries Wearmouth and Jarrow.' Erlemann believes that this aids us in fixing the date of our collection. Eusebius employed the riddle-collection of Tatwine, which falls in 732 ; and therefore composed between that date and the middle of the forties when he died. His sixty enigmas probably supplement Tatwine's forty, 1 so they are close to them in time. Now, if the Anglo-Saxon problems are due to the awakened interest in riddles, they may be placed between 732 and 740, in any case before 750, in Northumbria — the time and place to which Sievers and Madert (infra) would assign them. But all these arguments fall to the ground if we deny direct literary connection with Tatwine and Eusebius. Let us now examine the riddles. In the four riddles that owe most to the collection of Symphosius, Rid. 48, 61, 85, 86, the relation is not nearly as close as that of Rid. 36, 41, to Aldhelm. It is certainly not correct to say with Herzfeld \ that to each line of Symphosius 1 6, Tinea, two lines of Rid. 48 correspond. The six lines of the English version represent a very unfortunate expansion, in which the answer is betrayed at the outset, no new ideas except that of the holiness of the book are added, and the sharp contrasts of the Latin are sacrificed. The three motives of the ' Arundo ' enigma of Symphosius (No. 2) are admirably developed in the seventeen lines of Rid. 61, as Dietrich has » Herrigs Archiv CXI (1903), 58. t Forsch. zu deutsch. Gesch. XXVI, 597. J P. 29. ORIGINALS AND ANALOGUES xli shown in parallel columns. Here the Latin simply suggests. Rid. 85 follows only in its first lines the ' Flumen et Piscis ' problem (Sym. 12) : the remainder of the short poem is an independent development in which new motives are added. Only the second line of the Symphosius enigma Luscus allium tenens (No. 94) is used in the monster-riddle of seven lines (Rid. 86) which thus lavishly employs the hint. The four English riddles, though somewhat dissimilar in method of borrowing, resemble each other in free handling of sources ; Nos. 85 and 86, in the manner of development from a suggestion in the original ; Nos. 48 and 85, in the introduction of Christian elements. But the treatment of sources differs entirely from that in the small Aldhelm group (Rid. 36, 41), where the Latin (A. vi, 3, and De Creaturd) is closely followed (Notes). A dozen riddles employ motives of Symphosius and Aldhelm in such fashion as to suggest direct borrowing from the Latin enigmas, f In Rid. 10 the riddler gives evidence of his use of Symphosius 100 (not in Riese) in his description of the desertion of the cuckoo by parents before birth and its adoption by another mother ; but the added motive of the cuckoo's ingratitude, as indeed the whole treatment, shows an intimate acquaint- ance with the folk-lore of the time. The three motives of Symphosius 6 1 appear in the 'Anchor' riddle (Rid. 17), but only the second is so closely followed as to indicate actual indebtedness. The leitmotif 0; 37 (A- yi> I0); T3> 39 (A- "i» " ; v» 8; S. 56); 50 (A. ii, 14); 64 (A. vi, 9). xlii INTRODUCTION riddles of the time; and the two English problems cling to the tradi- tional motives, but with a certain freedom of literary treatment. Rid. 50, ' Bookcase,' is connected through its last lines, and particularly through the word unwita (na), with Aldhelm ii, I41'8, Area Libraria; but it is noteworthy that this is the very motive which we meet in the ' Book- moth ' problem (Sym. 1 6 ; Rid. 4S5-6). Rid. 64 owes its ruling idea to Aldhelm vi, g6"9, though it is no slavish copy of the Latin theme, ' the kiss of the wine-cup,' which appears not only in Anglo-Latin riddles (supra) but in the modern English Holme riddle, No. 128. Aldhelm's ' Water ' enigmas, iii, i and especially iv, 1 4, are freely followed in their main outlines by the writer of Rid. 84 ; but that long poem during its larger part declares its independence of Latin sources. To summarize, the motives of the Latin enigmas are so widely diffused throughout riddle- poetry, and moreover these themes are so freely handled in the English versions, that it is impossible to deduce any but the most general con- clusion regarding either relation to sources or the identity of the author. Only this much may be safely said : that the English riddles just con- sidered are alike in combining a certain dependence in their leading ideas with originality of expression and freedom of development. Yet another group of riddles bear to Symphosius and Aldhelm only a very slight resemblance — perhaps in a single phrase or line — so slight indeed that the likeness may often be accidental or else produced by identity of topic. Edmund Erlemann has pointed outf that the ' Storm ' riddles, Rid. 2-4, are indebted for one of their central ideas, not to Aldhelm's line (i, 21) ' Cernere me nulli possunt nee prendere palmis,' which appears in both the Bern Riddles and Bede's Flores (supra], but to the scriptural sources of this (see Notes) ; and I regard the other alleged parallels of Prehn \ as very natural coincidences. The resemblance between Rid. 6 and Aldhelm iv, 13, Clypeus, is very slight and the mere outcome of a common theme : each shield speaks of its wounds. It is barely possible that the author of Rid. 9 owed some- thing to Aldhelm's ' Luscinia ' enigma (ii, 5), but I do not believe that the Anglo-Saxon poet had the nightingale in mind. It is a far cry from Aldhelm's Famfaluca (iv, n) to the ' Barnacle Goose' of Rid. ii ; so Rid. 2-4 (A. i, 2) ; 6 (A. iv, 13) ; 9 (A. ii, 5) ; n (A. iv, 1 1) ; 12 (A. xii, 9) ; 21 (A. iv, 10) ; 28 (A. vi, 9) ; 29 (A. vii, 2) ; 35 (S. 60) ; 49, 60 (A. vi, 4) ; 54 (A. v. 8) : 57 (S. i? : A. iv, 3, 7) ; 58 (A. vi, i) ; 71 (A. iv, 10) ; 73 (A. vi, 8) ; 83 (S. 91) ; 91 (S. 4). t Herrigs Archiv CXI, 55. \ Pp. 159-163. ORIGINALS AND ANALOGUES xliii the likeness between the opening lines of the two, which is very slight, is obviously accidental. There is certainly a resemblance between a sin- gle passage in Aldhelm's 'Nox' enigma (xii, 9) and Rid. I27"8; but this is not sufficient to establish any direct connection between the Latin and the Anglo-Saxon. Rid. 21, 'Sword,' is developed in a totally dif- ferent fashion from Aldhelm's enigma (iv, i o) on the same topic ; any parallels of thought — and these are few — are inherent in the subject. The motive of ' wine, the overthrower ' (Aldhelm vi, 9°), which also ap- pears in Rid. 28, is found not only in other Latin enigmas of the time (MS. Bern. 611, No. 63s"6), but in folk-riddles remote from learned sources (see Notes). As the companion piece, Rid. 29, bears in two of its motives a general likeness to Aldhelm vii, 2, it is possible that the Latin may have been consulted by the author of these bibulous problems, but it is difficult to see how his themes could have been de- veloped without mention of these traits. The slight likeness between the ' Rake ' riddle (Rid. 35) and Symphosius 60, Serra, may easily be explained by the demands of similar subjects. Dietrich finds the germ of Rid. 49, 60, in Aldhelm vi, 4, De Crismale ; but the likeness, being practically limited to the ' red gold ' of both the Latin and English ves- sels, and consequently an inevitable result of identity of themes, is not irreconcilable with complete independence. Only in two lines of Rid. 54, ' Battering-ram,' is found any analogue to Aldhelm v, 8, which has a far different purpose, — a pun upon ' Aries.' The ' Loom ' riddle, Rid. 57, bears only a very faint resemblance to the enigmas of Symphosius (No. 17) and Aldhelm (iv, 3, 7) : like subjects could hardly be treated with greater difference of method. Rid. 58 has certainly two traits in common with Aldhelm vi, i ; but no descriptions of the ' Swallow ' could fail to mention its wood-haunts and its garrulous note. The origin of the 'Sword' or 'Dagger' (Rid. 7I2"8) recalls Aldhelm iv, lo1, De Pugione ; but the two enigmas are of very diverse sort. The ' Lance ' riddle (Rid. 73) surely owes little to Aldhelm (vi, 8) in the picture of its origin and its delight in battle. The general likeness in riddle-motive — change of condition by fire — between Rid. 83 and Symphosius 91 may well arise from the demands of the topic, ' Ore.' And, finally, there is but a dim suggestion of the lively metaphors of Rid. 91, ' Key,' in the bald ' Clavis ' enigma of Symphosius (No. 4), which simply states the subject's sphere of action. In none of the twenty riddles just considered XI, 474- xliv INTRODUCTION is it possible to establish direct literary connection with the Latin enigmas. In the preceding group, popular transmission of motives, — in this, like conditions of common subjects, — go far towards explaining all resem- blances. In other riddles that treat the same themes as the Latin enigmas, even this faint likeness is lacking. I have already registered my protest f against the claims of Tatwine and Eusebius as creditors of the Exeter Book Riddles. In a few cases I notice a resemblance between the Riddles and these Latin enigmas. | Yet in all these, except Rid. 15 and 44, the English and Latin writers are both working with motives employed not only by Symphosius or Aldhelm, but by other early enigmatographs whose direct connection with Tatwine and Eusebius is more than doubtful. § The ' Horn ' riddle (Rid. 15) has in common with Eusebius 30 its first thought, which is re- peated in different form in Rid. 88 (contrast however No. 15'$ companion piece, Rid. 80, which does not refer to the Horn's origin) ; and the ' Body and Soul ' problem (Rid. 44) is strikingly different in motive from Euse- bius's treatment of the same familiar theme (No. 25). I cannot there- fore agree with Ebert and Prehn (passim) that these Anglo-Latin enigmas influenced the Anglo-Saxon in matter and form. • BONIFACE An interesting place among eighth-century Latin enigmas is occupied by the twenty riddle-poems of the great missionary bishop Boniface.|| Here the riddle has taken on a purely Christian and theological charac- ter. Ten vices and ten virtues personify and characterize themselves •Rid. 7 (A. viii, 3) ; 24 (S. 65) ; 33 (S. 13) ; 34 (S. 10) ; 59 (S. 71, 72). t M. L. N. XVIII, 99. \Rid. 15 (E. 30); 21 (T. 30, E. 36); 27 (T. 5, 6; E. 31, 32); 39 (E. 37); 44 (E. 25) ; 52 (T. 6, E. 35) ; 84 (E. 23). § Holthaus (Anglia VII, Anz. 125) says very sanely: ' Besonders in den Fallen wo Prehn Ahnlichkeiten der englischen Ratsel mit zwei oder drei lateinischen Dichtern nachweist, waren wirgeneigt nicht an unmittelbare Entlehnung zu denken sondern zu glauben dass sowol die Gegenstande, wie auch die Art der Betrach- tung Gemeingut des Volkes geworden war und somit der Dichter nur bekanntes aufgenommen hatte, aber es doch eigenartig wiedergab.' This view is certainly supported by the likenesses to the Latin in the English riddles of ' Book,' ' Ox," and 'Pen' (Kid. 27, 39, 52) : these traits are commonplaces in early enigmas (supra). II Nine of these were printed by Wright, Biog. Brit. Lit. I, 332, from the in- complete version in MS. Royal 15, B. XIX, f. 204 r. Later the complete collection was published by Bock, Freiburger Diocesan-Archiv III (1868), 232, and by Diimm- ler, Poetae Lat. Carolini etc. (Man. Hist. Germ.), I (1881), i f. ORIGINALS AND ANALOGUES xlv like the beasts' and birds of the older enigmas. Caritas, Fides Catho- lica, Spes, Justitia, Veritas, Misericordia, Patientia, Pax Christiana, Humilitas Christiana, Virginitas, offset the frailties of Cupiditas, Super- bia. ( 'rapula Gulae, Ebrietas, Luxuria, Invidia, Ignorantia, Vana Gloria, Negligentia, and Iracundia. These allegorical enigmas are introduced by a dedication to his 'sister,' the Abbess of Bischofsheim — twenty hex- ameters, in which the virtues are compared to the golden apples of the I tree of life, the Cross of Christ, the vices to the bitter fruit of the tree of I which Adam ate. The whole composes 388 hexameters, and the several poems are of varying length. The acrostic employed by both Aldhelm and Tatwine is here used for purposes of solution. The subject of each enigma is plainly indicated by the initial letters of its lines. But Boniface goes farther than this. With his well-known fondness for playing upon names, f he introduces into his first enigma a double acrostic, c, s, a, a, r, t, /, i, /, r, a, a, s, c, thus sport- ing rather heavily with the Latin equivalent of the name of the Abbess, Liofa or Leobgyth.J Here then is a parallel for those who. claim that the lupus of the Latin riddle (Rid. 90) refers to the name of Cynewulf. As Ebert has pointed out, these enigmas have but small literary merit. Their vocabulary is small, their meter halting, the treatment stiff and awkward. The traits of his abstractions are seldom significant. Written in Germany (1. 323), the poems, particularly those upon Ignorance of God and Drunkenness, give forth now and then a gleam of apostolic fire; but in the main they seem dull and uninspired. Bock has, I think, exaggerated their indebtedness to Aldhelm, which is slight ; § and I discover in them no trace of Tatwine or of Eusebius. The influence of Virgil's Aeneid, which affected his style, as it did that of his contemporaries, was not strong enough to lift his moralizings into the region of poetry. I see in these didactic hexameters nothing that con- nects them even remotely with the spirited riddles of the Exeter Book. \ Ebert, Lit. des Mitt, im Abendl. I (1889), 653. t Compare Hahn, Bonifaz und Lul, 1883, p. 242; Ewald, Wetter Arc/ttvVII, 196; and my notes to Rid. go (infra). \ See Manitius, Christl. Lat. Poesie, p. 507. § The spicula lita veneno of the Introduction points to the last section of Aid- helm's poetic tract De Octo Princip. Vitiis, 130, and certain lines in the « Luxury ' enigma (No. 15) to the Creatura, 31, 53. But I find little more than that. Mani- tius, Christl. Lat. Poesie, p. 506, notes that for his general motives Boniface is in- debted to Prudentius's Psychomachia and to Aldhelm's De Laudibus Virginian. || Contrast Boniface's picture of Ebrietas with the delightful genre sketch of the tipsiness of the 'old churl' in Rid. 28. xlvi INTRODUCTION BERN RIDDLES A very important group of Latin enigmas is a collection of sixty-three riddles preserved in several early manuscripts. These consist of ' hexa- sticha rhythmica barbaric horrida ' (Riese). Hagen overrates them t in ranking them above the riddles of Symphosius in ' feine und gemiitliche Charakteristik ' ; but they are certainly not without merit; they treat the common things of life with clever ingenuity. Yet in range of sub- jects, in power of imagination, and particularly in width and depth of scholarship, they are inferior to the Anglo-Latin riddles. We meet only one reference to the Christian-Jewish cultus (9, ' Eua '), only one to classical mythology (4i6, ' Macedo nee Liber . . . nee Hercules'), only one to history (28s, ' Caesares '). A striking trait is their originality. They deal often with the same themes as Symphosius (Bern 2, S. 67 ; B.9, 8.51; B. 10, 8.78; B. n, S. 13; B. 13, 8.53; B. 18, 8.79; 6.32, 8.63; B. 34, 8.45; B.48, 8.19; B. 58, 8.77), but in totally different fashion. On the two occasions when these riddles invite close comparison with the older enigmas, it is significant that the author is using motives dear to riddle tradition : ' the fish and his moving house ' (6.30, S. 12) and 'the biter bitten,' ' mordeo mordentem ' (6.37, De Pipere, S. 44, De Cepa). \ So in his relation to Aldhelm, he is either entirely independent (6.3, A. iv, 8 ; B. 21, A. ii, 3 ; 6.45, A. i, i), or else he employs motives that are the common stock of riddle-poetry (B. 6, A. vi, 9, De Calice; 6.23, A. v, 10, De Igne; 6.24, A. v, 9, De Membrana; 6.25, A. iv, i, De ZJtferis). Yet the sequence of these riddles (B. 23, 24, 25), and certain likenesses in phraseology, § As early as 1839, Mone edited a version of these from Vienna MS. 67 in Anzeiger fur A'unde tier deutschen Vorzeit VIII, 219 f. In 1869 Hagen produced in Riese's Anthologia Latina I, 296, thirty-five of these enigmas from a manu- script of eighth to ninth century, Bern 611, f. 73r.-8ov. The next year Riese, in the second volume of his Anthology (p. Ixvi), showed the identity of the Vienna and Bern enigmas, and derived variants from Mone's text. Finally, in the last edition of the Anthology (1894, pp. 351-370) Riese collated with the already published manuscripts three other versions, Lipsiensis Rep. I, 74 of ninth to tenth century, f. I5v.-24r., and two Paris MSS. of the ninth century, 5596 and 8071 (each containing a few enigmas). For a discussion of this group of enigmas, cf. Hagen, Antike und Mitteldlterliche Ratselpoesie, 1877, pp. 26, 46. t P. 46. J For the vogue of these two riddles, see M. L. N. XVIII, 3, 5, XXI, 101, and my notes to Rid. 85, 66. Other world-riddles are those of the Ice (B. 38) and the Rose (B. 34). § Cf. Manitius, Aldhelm und Baeda, pp. 79-82. ORIGINALS AND ANALOGUES xlvii undoubtedly suggest a direct literary connection. Ebert and Manitius seem to me to exaggerate greatly the resemblances between the Bern enigmas and those of Tatwine and Eusebius ; and therefore to be totally unjustified in their conclusion that the former is one of the sources of the latter. Indeed, in all cases of alleged resemblance save one, the enigmatographs are drawing upon common stores of riddle-tradition (B. 2, E. 28, compare A. v, i, Sym. 67, Lorsch 10 ; B. 24, E. 31, T. 5, compare A. v, 9 ; B. 25, T. 4, E. 7, compare A. iv, i) ; and even under these conditions the likenesses are very slight, never amounting to any- thing more than general parallels of motive. Bern No. 5 has much in common with Tatwine No. 29, De Mensa,\ but even this likeness may be explained by the restricted demands of the topic. There is, however, no doubt that the Bern enigmas belong to the same circle of thought as the Anglo-Latin problems ; and, although no English manuscript of them exists, we are not surprised to find them followed by riddles of Aid- helm in Paris MS. 5596. Yet, whatever may be the probability, we have no convincing evidence that they are from the hand of an English author. LORSCH RIDDLES A small but valuable group of enigmas is the collection of twelve Latin riddles of varying lengths, in poor hexameters, preserved in the ninth century Vatican MS. Palatinus 1753, which was brought from the famous monastery of Lorsch. \ It has a twofold connection with the Latin enigmas of England. In the manuscript it appears in close company with the riddles of Symphosius and Aldhelm, the Prosody of Boniface, and the epitaph of a priest, Domberht, one of that band of scholars which came to Germany with Boniface ; § and Diimmler is in- clined to believe that our group of twelve problems was brought over from England with the remaining contents of the manuscript. Ebert || goes even farther, and claims that the riddles were composed in Eng- land, since their author is indebted not only to Aldhelm, whose works were widely known on the continent, but to Tatwine and Eusebius. The Manitius goes too far (Christl. Lat. Poesie, pp. 488-489) in regarding these as the chief source of Aldhelm's enigmas ; and he gives no reason for attributing them to an Irishman of the sixth and seventh centuries. t Cf. Ebert, p. 39. J These riddles were printed by Diimmler in Haupts Zs. XXII, 258-263, dis- cussed by Ebert, ib. XXIII, 200-202, and included by Diimmler in his Poetae Latini Aevi Carolini (Mon. Hist. Germ.), Berlin, 1881, pp. 2of. % Haupts Zs. XXII, 262. || Ib. XXIII, 200. xlviii INTRODUCTION Lorsch riddle No. 9, Penna, is, Ebert thinks, merely a compilation of three enigmas, Aldhelm v, 3,Tatwine 6, and Eusebius 35. If the verbal resemblances were not so strong, we might infer a common debt to the folk, as the motives of ' the weeping pen ' and ' black seed in a white field' are commonplaces of riddle-poetry. Lorsch No. n, Bos, is in- debted to Aldhelm iii, n, and perhaps to Eusebius 37 ; but again we have motives universally known among the Anglo-Saxons. f The paral- lels given by Manitius \ are, as usual, strained. Although ' the kiss of the wine-cup ' is a common motive, § yet the verbal likenesses of Lorsch No. 5, Poculum et Vinum, to Aldhelm vi, 9 and Tatwine 42 are so strong as to convince us of direct literary connection. In Lorsch No. 4, Glades, we meet a world-old motive, || which the author certainly did not derive from Tatwine 15. But he is undoubtedly employing Aldhelm v, i in No. 10, Lucerna, and A. i, 24 in No. 26, ' et rura peragro.' Diimmler and Ebert are justified in assigning to these problems an English home. Two other slight links bind the Lorsch enigmas to England : in No. 8 appears the motive of ' pen, glove, and fingers ' of Ttede^s^F/ores and Rid. 14, and in No. 7 the famous ' Castanea ' logogriph, so frequent in English manuscripts of the Middle Ages ; If but both motives are found on the continent as well. PSEUDO-BEDE Riddles of the Bede tradition are represented by three interesting groups of problems. Among the works doubtfully attributed to the Venerable scholar, the so-called Flares^ holds a place of some note. This varied assortment of queries falls roughly into three divisions, (i) The first and by far the largest of these belongs to dialogue literature (supra} and has much in common with other well-known groups of knowledge- tests. (2) The second class of problems consists of direct citation of « Cf. my articles, Mod. Phil. II, 563 ; M. L. N. XXI, 102 ; and notes to Rid. 52 (infra). \ M. L. N. XXIII, 99. J Pp. 79-82. § Notes to Rid. 64 (infra). \ Notes to Rid. 34. 1 M. L. N. XVIII, 7. These have been discussed by me in Mod. Phil. II, 1905, 561 f. I condense that discussion here. tt The full title of this melange is Excerptiones Patrum, collectanea, flores ex diversis, quaestiones et parabolae. Included in the Basel edition of Bede's Opera of 1563 and in the Cologne edition of 1612, the Flares was reprinted partially and incorrectly from the second in Kemble's Salomon and Saturn (1848), pp. 322- 326, but appears in complete and accurate form in Migne's Patrologia Latino (1850), XC, 539. ORIGINALS AND ANALOGUES xlix famous Latir enigmas. Five riddles from Symphosius (i, 7, 4, n, 10) and five from Aldhelm (i, 3, 10, 2, 4, n) are quoted in full. (3) There remain a dozen riddles rich in popular motives and abounding in ana- logues, t The first reappears among the queries of St. Gall MS. No. 196 of the tenth century ; t the second is paralleled by ' Fingers ' enigmas of St. Gall and Lorsch (No. 8) ; the fifth is indebted to the first line of Aid- helm's ' Ventus ' problem (i, 2) ; the seventh is the world-riddle of Ice ; the eighth contains the Ox motive, common property of all the riddle- groups of the Anglo-Saxon period ; the ninth is the embryo of the uni- versal riddle of ' Two-legs and three-legs ' ; § the explanation of the tenth lies in the ' Pullus ' and ' Ovum ' problems of Symphosius, No. 1 4, Euse- bius, No. 38, and MS. Bern. 611, No. 8 ; the eleventh appears in the Disputatio Pippini cum Albino \ and the St. Gall MS. ; the twelfth query can be compared with the close of Aldhelm 's octostich De Penna Scrip- toria (v, 3). This collection touches the Exeter Book Riddles at several points of meeting : not only in the popular motives of Fingers and Ice and Bull,1T but in the idea of hostility between Day and Night. The second group of Pseudo-Bede riddles is the Enigmata or Joco- seria, as I have called the puzzles of Cambridge MS. Gg. V, 35, 418 b, 419 a. ft This codex is of prime importance to the student of Latin enigmas, as it contains the riddle-groups of Symphosius, Boniface, Ald- helm, Tatwine, and Eusebius. Our Enigmata are attributed to Bede in the table of contents. Of the nineteen, a dozen may be classed as logo- griphs, a form of word-riddle very popular in the later Middle Ages and occasionally furnishing diversion before the Conquest. Mel, Os, Amor, Apes, Bonus, and Navis are among the puzzle-words. The ' Digiti ' query (xix) contains a motive not dissimilar to one used in older ' Finger ' enigmas. Inadequate diction, awkward syntax, incorrect grammar, and halting meter attest the author's literary limitations. Yet the author is not so important as the glossator. These enigmas are accompanied by an interlinear commentary, which is unique among glosses in casting a Cf. Manitius, Zu Aldhelm und Baeda, p. 82. t These riddles I have printed in full in the Mod. Phil, article. J Schenkl, Sitzungsberichte der Phil.-Hist. Classe der kais. Akademie der Wissen- schaften (Wien, 1863) XXXIV, 18. § See my note to Holme Riddles, No. 50. || Wilmanns, Haupts Zs. XIV, 552. \ Flares, 2, 7, 8; Rid. 14, 34, 13, and 39. Compare M. L. N. XVIII, 104. Flares, 6 ; Rid. 30 (see notes). tt Edited by me, Mod. Phil. II, 565. 1 INTRODUCTION powerful light upon the peculiar esteem in which art-riddles were held in the Anglo-Saxon time. After the manner of his kind the commenta- tor takes his pleasure very sadly : every line, indeed every word, of his author must be weighed as gravely as the phrases of Scripture or the rubrics of liturgy. We are thus brought to comprehend the ready wel- come accorded by pedantic leisure to the serio-comic products of pedantic scholarship, and to understand the continued vogue of these in the clois- ters of England. By the mediaeval reader queries which so often seem to us drearily dull and flat were evidently deemed miracles of ingenuity, inviting and repaying his utmost subtlety. The third group, the Propositiones ad Aeuendos Juvenes, which are number-problems rather than riddles, appeared in the Basel edition of Bede, 1563 (p. 133), and, under protest, are included in his works in the Patrologia Latina They are not mentioned by Bede in his enumera- tion of his writings ; and Alcuin's editor in the Patrologia f finds two good reasonsxfor ascribing them to that scholar. They are assigned to him in at least one old MS., and are specifically mentioned by him in a letter to Charlemagne (Epistle 101): ' aliquas figuras arithmeticae subtilitatis causa.' These number-puzzles were for a long time popular. I find Alcuin's fifty-three Propositiones under our rubric in MS. Burney 59 (eleventh century), f. 7 b— n a, and many similar arithmetical riddles in MS. Cott. Cleop. B. IX (fourteenth century), f. 1 70-21 a. Alcuin's river-crossing problem (No. 18), ' De homine efcapra et lupo,' is found, somewhat modified, in later English and continental MSS.t This group, which I discuss for the sake of completeness, presents, of course, no analogues to the Exeter Book Riddles. Interesting analogues to the Exeter Book enigmas are found in the Anglo-Latin prose queries of St. Gall MS. 196 (tenth century),§ in the solitary 'Bull' query of Brit. Mus. MS. Burney 59 (eleventh century), f. nb, || and in the unique Anglo-Saxon relationship riddle of MS. Vitellius E. XVIII, i6b.H But our poems have no connection, either direct or indirect, with the enigmatic Versus Scott de Alfabeto, a series « P. L. XC, 655. t Ib. CI, 1 143. \ MS. Sloane 1489 (seventeenth century), f. 16, unpublished; MS. Reims 743 (fourteenth century), Mone, Anz. VII, 45, No. 105 ; MS. Argentoratensis, Sem. c. 14, 15 (eleventh century), f. 176, Haupts Zs. XVI, p. 323. § Edited by Schenkl (Wien, 1863) and discussed by me under Flares (supra). See notes to Rid. 14. || Quoted in full, notes to Rid. 13. If See notes to Rid. 44". ORIGINALS AND ANALOGUES li of skillful hexameters, in which an Irish riddler, — a contemporary of Aldhelm, — -taking Symphosius as his guide, has told the story of the Letters. FOLK-RlDDLJES Let us now consider the use of popular material in the Exeter Book Riddles. We pass at once to those riddles which, in their form and substance, are so evidently popular products as to suggest that the poet has yielded in large measure to the collector; — the puzzles of double meaning, and coarse suggestion. To these we should naturally expect to find many parallels in folk-literature, and we are not disappointed. f Again, it is probable that the motives of such ' world-riddles ' as those of the Month (No. 23), Ice (No. 34), Bullock (Nos. 13, 39), and Lot's Wife (No. 47), were derived not from a literary source but from tradition ; and the same may be true of such wide-spread themes as the ingrati- tude of the Cuckoo (No. 10), the food of the Bookmoth (No. 48), the bite of the Onion (No. 66), and the running of Flood and Fish (No. 85), even though these four motives are prominent among the enigmas of Symphosius (supra). Analogues seem to show that certain leading ideas in the riddles of Fingers and Gloves (No. 14), Pen and Fingers (No. 52), Moon (Nos. 30, 40 ?, 95), Ram, and Lance (Nos. 54 and 73) were traditional. \ Barnacle Goose (No. n) and Siren (No. 74) belong to the folk-lore of riddlers. Not only in those riddles that bear in form and style the distinct im- press of the folk do we find popular elements. Many enigmas of the Exeter Book — literary though their manner proclaims them — are in- debted to that stock of commonplace domestic traditions, that simple lore of little things, which we recognize as the joint property of kindred races. Though the Anglo-Saxon puzzles are often entirely individual and isolated in their treatment of familar themes, yet the likeness of their motives to those of other Germanic queries is surely as remarkable as their differences. Let us compare these problems of early England These are preserved in company with the enigmas of Tatwine and Eusebius in the Cambridge MS. Gg. V, 35, and in Brit. Mus. MS. Royal 12, C. XXIII, and are printed in Wright and Halliwell's Reliquiae Antiquae I, 164, and by L. Muller, Rhein. Mus. XX, 357 (XXII, 500). For a full discussion of these see Biicheler, Rhein. Mus. XXXVI, 340, and Manitius, Christ. Lot. Poesie, pp. 484-485. t For analogues to Rid. 26, 45, 46, 55, 64, see M. L. N. XVIII, 103, and the notes to the several riddles. | Cf. notes to each of these. lii INTRODUCTION with those of Scandinavia. Heusler has invited attention to the corre- spondences between the themes and motives of the Exeter Book and of the Heidreks Gdtur; but these parallels are surprisingly slight. Several riddles of the two groups treat the same topics, but in a totally differ- ent fashion. With the modern folk-riddles of the Ishnzkar Gdtur our problems yield an interesting comparison. Rid. 27 (' Book '), 33 (' Ship '), 35 (' Rake '), 38 and 87 (' Bellows '), 57 (< Web and Loom '), and 68 (' Bible ') may be annotated throughout by various Icelandic riddles of like subjects.t On the whole the likeness between the queries of the two groups is too general i .ray any very intimate connection ; but the appearance of such similar elements in the Islenzkar Gdtur furnishes no slight proof of the popular character of Exeter Book riddle-germs. I add a few continental parallels to the queries in our collection. The fearfully-made creatures in the Anglo-Saxon poems of musical instru- ments (Nos. 32, 70) are not unlike the prodigies in the Lithuanian and Mecklenburg Geige riddles \ ; the Onion of Rid. 66 is ' a biter when bitten ' as in the German riddle § ; the Communion Cup of Rid. 60 is closely akin to the subject of the Tyrolese problem || ; and finally, the motive of the highly imaginative query of the Ox (Rid. 72) appears again far afield in the riddles of Lithuania and Bukowina.lf Among the modern folk-riddles of England the number of parallels to the Exeter Book Riddles is not at all large. Unlike the influence of Symphosius throughout Europe or the direct literary working of the Heidreks Gdtur in Iceland and the Faroe Islands, the motives that appear in the Anglo-Saxon collection, if we may draw a conclusion from the scanty evidence at our command, seem to have affected little the current of native riddle-tradition. A few English riddles of the present resemble in theme and treatment the Exeter Book Riddles ; and, more noteworthy yet, two or three of these are unique among recent puzzles in this resemblance. In the latter case we may safely regard the mod- ern riddle-stuff not as a new creation, but as a survival of the old. Enough has been said, I hope, to establish the Exeter Book problems in their proper place in riddle-literature. I have sought not only to » See M. L. A'. XVIII, 103, n. 32. t M. L. N. XVIII, 104 and notes. \ Schleicher, p. 200; Wossidlo, No. 230 a. § Wossidlo, No. 190; Petsch, pp. 95-96. II Renk, Zs.d. V.f. Vk. V, 149, No. 17. t Schleicher, pp. 207, 211 ; Kaindl, Zs.d. V.f. Vk. VIII, 319. See M.L.N. XVIII, 105-106; and notes to Rid. 20, 26, 28, 29, 65, 77, 88. AUTHORSHIP OF THE EXETER BOOK RIDDLES liii indicate, more accurately than has before been done, their relation to liter- ary enigmas, but also to trace what has hitherto passed almost unnoticed, their indebtedness to popular motives. Ill AUTHORSHIP OF THE EXETER BOOK RIDDLES THE RIDDLES AND CYNEWULF Any discussion of the authorship 01 v' Riddles naturally finds its starting-point in Leo's interpretation of the so-called ' First Riddle.' Upon this I need not dwell at length, because it has already been care- fully considered in another volume of this series. But it is necessary to indicate, more briefly than Cook and Jansen, the place of (Leo^s solu- tion in the Cynewulf story. According to that scholar's Halle Program of 1857,1 the first poem of the collection is a charade or syllable-riddle, 1 whose answer is found in the name Cyne(cene,ciKn,ceri)-wulf. Thence Leo drew the conclusion that this poet was the author of all or most of the problems of the Exeter Book, To Leo's solution Dietrich gave the full weight of his approval, t Indeed he went still farther, finding in the lupus of Rid. 90 yet another reference to the poet's name, and in Rid. 95 a sketch of his vocation, that of ' Wandering Singer.' Here, he be- lieved, were strong grounds for attributing the whole collection to Cyne- wulf. For more than twenty years all scholars accepted the contentions of Leo and Dietrich, § with the solitary exception of Rieger, || who recog- nized the difficulties inherent in the solution of the ' First Riddle,' but offered no other answer. In an essay of 18831! Trautmann rejected Leo and Dietrich's answers of the first and last riddles, proposing for both the solution 'Riddle.' The new interpretations found less favor than the old, but there were not wanting scholars who followed Trautmann Cook, ' The Riddles and Cynewulf,' The Christ of Cynewulf (1900), pp. lii- lix; see Jansen, Die Cynewulf -Forsc/ning, BB. XXIV, 93-99. t H. Leo, Qitae de se if so Cynewulf us, poeta Anglo-Saxonicus, tradiderit. \ Lift. Centralbl. (1858), p. 191 ; Ebert's Jahrb. f. Rom. und Eng. Lit. I (1859), 241 f. ; ' Die Ratsel des Exeterbuches,' Haupts Zs. XI, 448-490, XII, 232. § Cook, p. Ivi; Jansen, p. 94. || Zs.f. d. Ph. I, 215-219. ^ 1' Cynewulf und die Ratsel,' Anglia VI, Anz., pp. 158-169. See articles by Nuck, Anglia X, 390, and Hicketier, ib., 564 f. liv INTRODUCTION in discarding this supposed proof of Cynewulfian authorship ; and in an important article of 1891 f Sievers presented conclusive linguistic reasons for abandoning Leo's far-fetched and fanciful hypothesis. Three years before Sievers's essay, Bradley t advanced the view that ' the so-called (first) riddle is not a riddle at all, but a fragment of a dra- matic soliloquy, like Dear and The Banished Wife's Complaint, to the latter of which it bears, both in motive and in treatment, a strong re- semblance'.' This opinion has found wide acceptance, and is almost certainly correct. It has been favored by Herzfeld,§ by Holthausen, || and by Gollancz.lf Upon this hypothesis Lawrence and Schofield built up their interesting and ingenious theory that the ' First Riddle ' is of Norse origin, and is connected with the Volsung Saga; and Imelmannft his claim that the lyric belongs to the Odoacer story. But these theories \ are too far from the field of riddle-poetry to concern us now, and will, i moreover, be carefully weighed in a promised edition of Old English Lyrics. Though the ' First Riddle ' is thus unquestionably a lyrical monologue, ^ I have included it in my text, not only on account of its historical associa- tion with the enigmas of our collection, but because of the elements of Ratselmarchen that render its interpretation so difficult. Other contributions to this phase of the association of the Riddles with Cynewulf are the articles of the Erlemanns,^ who have attempted to prove that the Latin Riddle (90) is a charade upon the poet's name and therefore points to Cynewulf as collector of the enigmas, and my evidence § § that the last of the Riddles refers neither to ' Wandering Singer' nor to 'Riddle,' but, like its companion-piece Rid. 30, to the journeys of the Moon. The identification of the author of the Riddles was, however, made to rest on other grounds than the evidence of Rid. i and 90. In his first article || || Dietrich was inclined to think that the first series (1-60) was Holthaus, Anglia VII, Anz., p. 120 ; Morley, English Writers II, 21 1, 217, 222. t Anglia XIII, 19-21. t Academy XXXIII (1888), 197 f. § Die Rdtseldes Exeterbuches (1890), p. 67. || Deutsche Littztg., 1891, p. 1097. \Academy XLIV (1896), 572. Gollancz regards the poem as ' a life-drama in five acts.' P.M.L.A. XVII (1902), 247-261, 262-295. ^^ Die Altenglische Odoaker-Dichtung, Berlin, 1907. See Gollancz, Athenaum, I5»2, p. 551 ; Bradley, ib., p. 758. tt Herrigs Archiv CXI, 59 ; CXV, 391. See notes to Rid. 90. §§ M. L. i\T. XXI, 1906, 104-105. See notes to Rid. 95. || || Haupts Zs. XI, 488. AUTHORSHIP OF THE EXETER BOOK RIDDLES Iv by Cynewulf ; the second (61-95) by other hand or hands ; but that perhaps the collector of the problems of the latter group had before him a source which contained single riddles of Cynewulf. In his second article he was led to modify this view, and to claim not only that all the riddles in both groups were from one hand, but that the hand was Cynewulf's. He went even further, and assigned, somewhat doubtfully, the first series to the youth of the poet and to his beginnings in riddle- poetry, the second to his later period. Signs of a young poet are seen in the first group in (i) his mistakes in translation (4i65, pernex) ; (2) the very youthful cadence of the verse ; (3) the obscene pieces (26, 43, 45, 46, 55), which he conjectures to be the very poems regretted by Cynewulf in his supposed retractation. To the first argument it may be answered that we have no opportunity to compare the knowledge or ignorance of Latin displayed in the first group with that in the second, as it is only in the earlier group that we have very close translations of Latin enigmas (Rid. 36, 41) ; to the second, that such a subjective esti- mate of verse-values so far removed from us can carry no weight; to the third, that obscene problems meet us at the very threshold of the second series (Rid. 62, 63, 64). Dietrich seeks to sustain this ascription of the Riddles to Cynewulf by a comparison of the thoughts and ex- pressions of our poems with those of the Cynewulfian works ; t but it may be answered first with Holthaus \ that the relation of the various riddles among themselves and to the poems of Cynewulf must be main- tained on more convincing grounds than in Dietrich's article, and secondly that the larger number of his parallels (granting that such parallelism carries any weight) are drawn from a text of such doubtful authorship as the Andreas. Prehn § accepts without question, as the starting-point of his investiga- tion, Dietrich's belief in the Cynewulfian authorship of the Riddles. The arguments of Herzfeld in favor of the ascription of the problems to Cynewulf || have now only an historical interest, as they have been abandoned even by Herzfeld himself. IT In his earlier monograph he goes beyond Dietrich's contention and claims that all the Riddles are from the hand of a young poet, on the ground of their keen interest in XII, 241, 251. t Xn, 245-248. \ Anglia VII, Am., p. 122. § Komposition und Quellen der Rathsel des Exeterbuches, 1883. || Die Rathsel des Exeterbuches etc., 1890. \Herrigs Archiv CVI (N. S. VI), 1901, p. 390. everything in the world, and their joy of life, which does not shrink from naively sensuous expressions.! Another sign of youthful author- ship Herzfeld discovers in the large number of hapax-legomena in the Riddles, \ because ' a young poet is fond of choosing rare words which may seem to his audience new and surprising.' To show that this youth- ful poet is Cynewulf , Herzfeld advanced many arguments : the likeness of the vocabulary of the Riddles to that of the Cynewulfian poems, among which he includes the Andreas ; a similar treatment of sources ; a like attitude to the sea and to war, to social relations and to religion ; a like use of figures of speech ; and finally, a like handling of metrical types. While none of thes : arguments in the least convince us of Herz- feld's main contention, still they are not without illustrative value in cast- ing light on both the matter and the manner of the poems before us, and they will be cited in connection with different phases of our study. A year after Herzf eld's monograph (1891) Sievers discussed the age of the Riddles, § and reached the conclusion that they belong to the first half of the eighth century, a period anterior to the time of Cyne- wulf. These are his reasons : (i) 'The Leiden Riddle, the Northumbrian version of Rid. 36, con- tains many forms with unstressed /, instead of later e : — ni, bigidoncum (corrupted from hygidoncum), giftrctec, hlimmith, hrlsil, uirdi, <?i, heliSum (by the side of ne, giuciide, and a doubtful ceres f). The change from unstressed / to e probably took place about 750.']! The value of this This is the view of Brooke, English Lit. from the Beginning etc., 1898, pp. 160-161. t Herzfeld remarks, p. 9 : ' Einen so offenen Blick und ein so lebendiges In- teresse filr alles, das Grosste wie das Kleinste in der ihn umgebenden Welt, diese Lebenslust, die auch vor naiv sinnlichen Aeusserungen nicht zuriickscheut darf man nur bei einem jugendlichen Dichter zu finden erwarten.' (See Dietrich XI, 489; XII, 241 ; Fritzsche, Anglia II, 465.) J Herzfeld (pp. 10-12) records 262 words which occur only in the Riddles. Though this might seem to speak against his claims for Cynewulf, yet he noted that there are in the Christ 196 such words, and in the Juliana and the Phcenix, respectively, appear 129 and 196 new compounds. Herzf eld's results must be somewhat modified and increased in the light of the vocabulary of the Riddle- fragments printed in Grein-Wulker. § Anglia XIII, 15. || This e and / canon of date seems to me a hasty generalization based upon insufficient data. Indeed the very evidence derived by Sievers from Sweet's Old- est English Texts often refutes itself. If unstressed e appears twice in an Essex charter of 692 (O. E. T., p. 426), if unstressed i is found in the Northumbrian Genealogies of 811-814 (O. E. T., p. 167) in the very names (efril- compounds) that AUTHORSHIP OF THE EXETER BOOK RIDDLES Ivii evidence, such as it is, is lessened by the rather striking circumstance that Rid. 36 stands apart from the other riddles (except Rid. 41) both in its relation to its sources and in its employment of motives. It is there- fore hardly fair to apply to the whole collection any argument based upon forms in this isolated problem. (2) ' In Rid. 24 Agof must have been originally Agob, the inversion of Boga. This final b, which in this case a later scribe has changed tof, is not found later than the middle of the eighth century.' It is hard to feel the weight of this argument. Are we to believe that a riddler in the latter part of the eighth or even in the ninth and tenth centuries was prevented by phonetic laws from inverting any word with an initial b and thus forming a nonsense-word with an uncouth ending ? f Agob is as possible at any period of Old English as To^>\aTr60par (Ar. Ran. 1286 ff.) is in Attic Greek. To some it may have significance that Barnouw \ regards Rid. 24 as very late on account of its four articles before simple substantives. (3) ' From the runes in Rid. 43, two N's, one >£, two A's and two H's (the names are written out, nyd, cesc, dcas, and tuzgelas) are derived the two words hana and keen. A instead of o before nasals, and ce as an umlaut of this a, point to the beginning of the eighth century.' For many reasons, this argument is not conclusive : (a) That the date of Rid. 43 is very late rather than early, Barnouw § seeks to show by pointing to the large number of articles — seven in seventeen verses — and to the use of articles instead of demonstratives, frees hordgates, bear an unstressed e (afrel-) in a Kentish charter of 740 (p. 428), if a Mercian grant of 769 (p. 430) employs always the unstressed /, and if, moreover, all North- umbrian poems, including the Ruthwell Cross inscription (which Cook, P. M. L. A. XVII, 367-390 ; Dream of the Rood, p. xv, assigns to the tenth century), and if the glosses to the later chapters of John in the Lindisfarne Gospels after 950 (Cook, P.M.L.A. XVII, 385) employ that form, how can we infer with good reason that the Leiden Riddle, which admits both / and e, was written before 750 ? Scholars have as yet found no sure footing on the slippery ground of Anglo- Saxon chronology. This statement Sievers elsewhere applies to ob {Leiden Rid. 2, 14); but he admits (XIII, 16) that this b is twice found in the Liber Vitae of the ninth century (335' Cnobwalch; 339, Leobhelni). I note it in Kentish charters of 831 (Sweet, O. E. T., 445, No. 39, 1. 2), ob frem lande, and 832 (ib. 446, No. 40, 1. 17), ob mlnem erfelande. Such peculiarities are not mere matters of date. t See the nonsense-words of the Charms (Lchd. Ill, 10, 58, 62). IP. 214. §P. 215. Iviii INTRODUCTION foa radellan (contrast 56", foisses gieddes). (b) A and a may indicate a very late quite as well as an early date for our version of the runes of this riddle, as hana and keen are well established West Saxon forms. This circumstance naturally destroys any value as proof which the assertion of their early Northumbrian origin might have. Instead of proceeding like Sievers from the assumption of early authorship for the riddle, it would be just as easy to proceed from the assumption of late authorship. (c) My opinion is strikingly supported by the appearance of such a West Saxon form as Eh(r/i~) among the runic words of Rid. 65.1 Sievers himself admits \ that MON (2o5) is a late product. (4) 'In the runic riddle 20, the runes give us the form COFOAH (the inversion of HAOFOC). Since ao is found nowhe're else as the 2/-umlaut of a, hafoc is to be substituted. This form with unumlauted a indicates the first half of the eighth century.' Now, although we may reject with Sievers the AO of HAOFOC, and although Rid. 65" H and A speak against an original HEAFOC and for an original HAFOC in our version, yet let us note that the word hafoc is not only Northumbrian but good West Saxon ; that, as such, it appears in Rid. 258 and 4i67 and in many other poetical passages, consequently in our text of the runes. Therefore the argument that Sievers bases upon this form falls to the ground. Professor Sievers's four arguments seem, therefore, to have small probative value. But, while questioning the weight of his premises, I think that he may not be far wrong in his conclusion that the Riddles are the product of the first half of the eighth century, as this was the golden age of English riddle-poetry. § That the Riddles belong to this period, and therefore antedate Cynewulf , is, however, only a surmise, which is perhaps incapable of proof. Sievers certainly has not proved it. Sievers's deductions from these runes carry as little weight as Trautmann's conclusions as to dialect, based upon the supposedly Northumbrian form ewu in the Juliana rune-passage (Kynewulf, p. 73), and refuted by Klaeber (Journal of Germanic Philology IV, 1902, 103), who points to 'the forms ewo, Ine's Laws 55 (MS. E) ; nua (ace. pi.), O.E. Martyrol. (Herzfeld), 36, 17 ; ewede, ib. 170, 26; and to Sievers, Gr?, 73, n. i ; 1 56, n. 5 ; 258, n. 2.' I mention all this in order to anticipate the equally false claims that may be founded upon the ewu form de- manded by the Erlemann solution of Rid. go (note). t In my notes to that riddle the reading Ek(r/i) is established beyond doubt. | Anglia XIII, 17. § Yet, as we have seen, it is impossible to connect them directly with either Tatwine or Eusebius. AUTHORSHIP OF THE EXETER BOOK RIDDLES lix In Madert's monograph the final blow is dealt to the theory of Cynewulfian authorship of the Riddles. Madert takes direct issue with Herzfeld, and devotes his thesis to showing that the Riddles have little in common with the poems of Cynewulf. He rightly believes that no comparison can be instituted between the varying use of sources in the Riddles and Cynewulf's adherence to one text. In style and word- use the Riddles bear no closer resemblance to the undisputed works of Cynewulf than to many other Anglo-Saxon poems. f Among the phrases cited by Herzfeld t as common to the Riddles and Cynewulf, there is hardly one that does not appear elsewhere. So the synonyms adduced for the same purpose are seen to be commonplaces of the poetry. The greater part of Madert's dissertation is devoted to the language of the Riddles. On account of many noteworthy differences between the speech of the problems and that of Cynewulf, he reaches the conclusion not only that these poems are not the work of that writer, but that they are the products of an earlier period — probably the beginning of the eighth century. § IThe evidence of meter, language, and style certainly speaks against the theory of Cynewulfian authorship. In the consideration of this, we are met by a double difficulty : the absence of any trustworthy Cynewulf canon, on account of the widely differing opinions of scholars regarding the authenticity of such poems as the Andreas, and of the larger part of the Christ (1-440 ; 867-1693) ; and secondly, the obvious difference between the matter and tone of such products of the profane muse as the Riddles and the loftier temper of religious verse, — a difference that compels quite another manner of expression. Yet Sievers, Trautmann, and Madert have noted in the Riddles points of variance from the un- doubted poems of Cynewulf : points which, slight though they be, invite consideration, because they are independent of all questions of genre Die Sprache der altenglischen Rdtsel des Exeterbuches und die Cynewulffrage, Marburg, 1900. t Cf. Madert's examples (pp. 10-1 1), and the parallels cited by Sarrazin, Bemmilf- Studien, pp. 1 13, 159, 202 ; Kail, Anglia XII, 24 f. ; and Buttenwieser, Studien iiber die Verfasserschaft des Andreas, pp. 22 f. % P. 17. § This latter conclusion, which is obviously dictated by Sievers's article (su^ra), is reached in strange fashion. To cite but one of Madert's arguments (p. 128): in 572 -wido appears for West Saxon -wudit. — 'der u-Umlaut des i ist also hier noch unterblieben, was mindestens in den Anfang des 8. Jahrhunderts zuriick- weist.' Strange then that we should meet -widu in Alfred's Meters I356, which is not suspected to be an early Northumbrian text! lx INTRODUCTION and tone-quality. Even Herzfeld, though arguing for Cynewulf s author- ship, was forced to note at least one important variation from that poet's metrical usage. Both in the first and second half -lines, the Riddles afford several examples of the appearance of a stressed short syllable in the second foot of type A, when no secondary stress precedes. Although Sievers has remarked t several occurrences of this verse in the poetry, it is noteworthy that not one of these appears in Cynewulfian work. Herzfeld also notes \ variations from Cynewulf's forms of C and D types ; but these seem far less conclusive. A record of the more striking differences in language between the Riddles and the accepted poems of Cynewulf may justify itself as an historical survey, inasmuch as such discussion has been in bulk the most important part of the criticism of the Riddles. (1) Trautmann has correctly observed (Kyneivulf, pp. 29-30) that Cynewulf seldom, if ever, expands contracted forms for the sake of his verse. Other Anglo-Saxon poets freely permit themselves this liberty (Sievers, PBB. X, 475 f.); and the Riddles in particular abound in examples (Sievers, 1. c. ; Herz- feld, pp. 60-61 ; Madert, p. 53): 466, mines frean ; 237, ofras hea; 63, oft ic wig seo ; 2Q13, 3224, 3314, 4O1, 429, hwaet seo wiht sy (sle) ; 63", hwllum ut tyfrS ; 642, faegre onbeon ; 64s, Her wit tu beo"$ ; etc. (2) Trautmann argues § that in the ^-less forms of feorh, asfeores, feore, the penult is always short in Cynewulfian verse ; while Herzfeld || and Madert T[ have pointed out that in the Riddles it is always long. Unfortunately for the full force of the implied argument, Trautmann not only draws his examples largely from the Andreas, but changes the Juliana verses 191, 508, that oppose his view; yet the difference in use has some slight probative value. Wealas in Rid. i34a, swearte Wealas, has a long penult (Sievers, PBB. X, 488) -, but Wale ( Wala) in the Riddles is almost certainly regarded as ^ x (Herzfeld, p. 58). (3) According to Trautmann, Cynewulf uses only ham in dative, — since he regards Chr. 293, to heofonhame, as non-Cynewulfian. Hatne is found in the Riddles, 30, hfthe to ham ham\e\ (Herzfeld, p. 59, Madert, p. 61). Instances of JL X | ^ X in the first half-line are found Rid. 15", wicge wegaS ; i8n, men gemunan ; 476, earn ond nefa; Q310, strong on staepe ; in the second half-line, Kid. 3Q6, duna bricefi ; 39% bindeft cwice ; 432, ute plegan (?). For ex- amples of_£.x x(x) | ^x in first half-line, see Rid. i62, sldan swa some; 2813, strengo bistolen ; 28U, maegene binumen ; 43", haegelas swa some ; 64, Hwllum mec on cofan ; 8421, wundrum bewrej>ed ; S422, hordum gehroden ; in second half- line, 59", hry sind in naman ; 84", wistum gehladen (Herzfeld, pp. 44, 49, 56). t PBB. X, 454. { P. 56. § P. 27. H P. 58. 1 P. 127. P. 79. AUTHORSHIP OF THE EXETER BOOK RIDDLES Ixi (4) Cynewulf uses the inflected forms of numerals if no substantive follows, but the uninflected before a substantive immediately following (Trautmann, 83). This is not the case in the Riddles (Madert, pp. 61-62) : I41, tyn wairon ealra ; 373, haefde feowere f et under wombe. Not much stress can be laid upon the second example, since the uninflected form is metrically possible, and since in the same riddle other attributive adjectives are uninflected, 377"8, Haefde tu fi>ru ond twelf eagan | ond siex heafdu (cf. 864, ond twegen fet). This argu- ment has, therefore, little force. (5) Cynewulf wrote both feeder and fcedder (Trautmann, p. 77); but only the shorter form is found in the Riddles (Madert, p. 26). Upon this no great stress can be laid, for the three reasons that the longer form is exceptional in Cynewulf, and that it appears elsewhere in the poetry (Beo'w. 459, 2049 ; Gen. 1074, 2696; Met. 2O26S, etc.), and, finally, that any argument drawn from the absence of a word or form is vain. (6) The stem-syllables in bit(f)er and snot(t)or are always long in Cynewulf (Trautmann, p. 76). In the Riddles they are sometimes long, 862, gs7 (Herz- feld, p. 58); sometimes short, 34®, biter beadoweorca; S434, mon mode snottor (Sievers, PBB. X, 508 ; Herzfeld, p. 58 ; Madert, p. 57). But neither of these examples is decisive. (7) Long-stemmed words ending in -el, -ol, -er, -or, -en, -urn (tungol, ivun- dor, hleahtor, tdcen, etc.) are regarded by Cynewulf as dissyllables (Traut- mann, p. 28), whereas in the Riddles they are often monosyllabic (Madert, PP- 54-55)- (8) Herzfeld and Madert f note certain variations in the use of single words, which seem to me to have very little significance : (a) Cynewulf uses both gierwan andgearwz'an (Trautmann, p. 85). In the Riddles only forms of the first are found (2i29, 2713, 2Q1, so3, 372, 6817, 6g2). (b) Cynewulf uses fylgan (Trautmann, p. 86) ; the Riddles, like the Andreas, 673, folgian: ^,^,J>egnfolgade. (c) Only uncontracted forms of the present participle of buan are found in Cynewulf, whereas the meter clearly establishes contraction in Rid. a62, neah- bundum nyt (Sievers, PBB. X, 480). (</) It may be added that am •wlcum ; ya28, ofj>am wicum ; 3O4, to l>am ham\e\ (12) Madertf notes that the dative after comparatives — instead of J>onne phrase — is not found in Cynewulf, but appears frequently in Rid. 41 : 4I18-38- 46.50,56.57,70,78,80,82_ (13) SarrazinJ marks that in the older poetry (Gen. A, Dan.} words like tacn, ivuldr, are customarily monosyllabic, while in Cynewulf's works tacen, ivuldor, are regularly dissyllables (supra). Both usages appear in the Riddles : 56, and rode tacn ; 6o™,goldes tacen ; %4®,swaJ>&t'wuld(o)rwifa(M.S. wtfefr); 8425, tvynsum wuldorgimm ; etc. (14) Sarrazin § also observes that words like ne ivolde, ne iviste, ne was, are uncontracted in older poems, but that in Cynewulf nolde, niste, nas, domi- nate. These premises can have little value on account of the numerous excep- tions to this rule, but it is certain that the Riddles prefer the uncontracted forms. Indeed nces and nolde do not appear ; contrast, however, 2416 nelle, i616 nele. According to Sarrazin, many of these traits that we have marked in the Riddles (notably (i) and (2)) are characteristics of poems of an older period than that of Cynewulf. That is probably true, but the personality of the poet, as well as the date, must be considered in such cases. The archaistic spellings of glosses in the later chapters of the Lindisfarne John stand as a warning to the too rigid and minute interpreter of internal evidence, and remind us, in the words of Professor Skeat, || that ' large theories are constantly being built up, like an inverted cone, upon very slender bases.' Not much value can be attached to any single variation from Cyne- wulf's usage, or indeed to the accumulative force of all that have been cited ; but, in the absence of one jot of evidence connecting the Riddles with this poet, these differences add slightly to the heavy burden of proof resting upon him who seeks to revive the moribund claim of Cynewulfian authorship.lT P- 216. f Pp. 69, 128. J Eng. Stud. XXXVIII, 160. §L- c- || Preface to St. John's Gospel, p. xi. IT One is surprised to meet this statement in Brooke's E. E. Lit. from the Be- ginning, p. 160, as late as 1898: 'There is a general agreement that we may attribute the best [Riddles] to Cynewulf.' So far is this from being the case, that with the exception of the Erlemanns, who interpret Rid. 90 as a Cynewulf charade AUTHORSHIP OF THE EXETER BOOK RIDDLES Lxiii In his second article, Dietrich notes, as one point against his final thesis of the unity of the whole collection, that the Riddles are not written as a continuous whole. He believes that the collector drew from different manuscripts, which represent two series of riddles : 1-60 (or 61) and 62^-95. He has already doubted in his first article t whether the second series was by the same author as the first, be- cause several of the subjects are repeated, and a good poet does not repeat himself. That Series i has throughout unity, Dietrich seeks to showj by three traits of these poems: (i) inner relation between subjects ; (2) like employment of Latin sources ; (3) agreement in treatment. (1) Dietrich admits § that there is no definite plan of arrangement, but declares that the poet avoids placing together nearly-related subjects because they are too easy to find. But there can be a connection resting upon association of ideas and a certain poetic purpose in this connection. He seeks to defend this assertion by an outline of the topics discussed in Series i , and in this he is followed by Prehn || ; but Holthaus is clearly right in his contention H that ' it is no very difficult thing, out of a great mass of subjects which follow one another in purely arbitrary fashion, to select and bring together those that have a certain likeness.' My analysis (infra) shows that the final order is in a few cases the order of composition. There is certainly no single idea in this group of riddles. Very little stress can be laid upon this first argument ; indeed, Wiilker does not think it worth while to class it with the other arguments in his summary of Dietrich's views. (2) Upon the second argument, the like use of Latin sources, Dietrich lays some stress. ft But the evidence that he presents is too (supra), hardly any one now believes that the poet had aught to do with these problems. (Brandl, who accepts the Erlemann solution, Pauls Grundriss II, 972, thinks that the writer of the Latin enigma may have been another Cynewulf or else an admirer of ^he poet. This person, he thinks, may have been the editor of the second series (61-95) or even of both series.) Wiilker, however, holds (Anglta, Bb. XIX,. 1908, 356) that 'a part of the collection is from Cynewulfs hand' ; but he brings nothing to sustain his view. Hatipts Zs. XII, 234. t XI, 488. t XII, 235. § XII, 236. II P. 150. ^Anglia VII, Anz. 121. Grundriss, pp. 168-169. tt See also Herzfeld, p. 5. INTRODUCTION slight to warrant the sweeping assertion that a greater dependence upon Latin models marks the first group, a freer movement charac- terizes the second. This difference, however, is to be explained, so Dietrich thinks, not by difference in authorship, but by the personal inclination of one poet. Holthaus f objects that Dietrich's very examples mark a distinct unlikeness in the relation of different riddles to their Latin prototypes and analogues. (3) Dietrich t finds a third argument for unity of authorship in the treatment (' behandlung ') — particularly in the use of opening and clos- ing formulas. § He examines in detail the various forms, and notes the far greater elaboration of those in the first series compared with those of the second ; and secondly infers from the likeness between the formulas of the earlier group a single author. Herzf eld, || arguing for the unity of the whole collection, points out that sixteen out of the first sixty (this result must be modified) lack formulas, and that six others have the short closing formulas of the second group. While the mere use of such con- ventional forms would hardly serve to establish identity of authorship, as these can be employed so readily by an imitator, IT still a careful con- sideration of these formulas is not without value. Of the so-called first Dietrich, Haupts Zs. XII, 241, notes that in 17, 48, 61, we meet with verbatim borrowings from Symphosius ; 36, 39, 41, are taken sentence for sentence from Aldhelm : in 6, 14, 29, 37, 51, 54, certain matter is borrowed. In the second series he marks a freer employment of Symphosius (Rid. 66, 84, 85, 86, 91), and a few traits from Aldhelm. In particular riddles, Dietrich's conclusions regarding sources must be corrected by the light of my study of origins (supra). t L. c. t XII, 241. § Dietrich, Haupts Zs. XII, 241, marks the use of opening formulas in old Germanic riddles, particularly in the Hervarar Saga. In these Gdlur we meet such beginnings as these : ' What kind of wonder is that which I saw without before the doors of the prince,' ' When I journeyed from home, I saw on the way,' ' I saw in summer upon the mountains,' or ' I saw faring this and that.' It is inter- esting to note that Heusler, Zs. d. V. f. Vk. XI, 133, cites, as an indication of unlikeness between the different numbers of the Heifrreks Gdtur, the quite differ- ent forms of their beginnings. Petsch discusses at length (pp. 51-58) introductory formulas which have nought to do with the germ or central thdught of the popu- lar riddle. We meet similar introductions in the English Holme Riddles, P. M.L.A. XVIII, 211 ff. : Nos. 51, 53, 'As I went on my way, I heard a great wonder'; No. 52, 'As I went through the fields'; No. in, ' As I went by the way.' But these are mere commonplaces of riddle-poetry. II Die Rathsel des Exeterbuches, p. 8. f Cf. Holthaus, Anglia VII, Anz. 122. AUTHORSHIP OF THE EXETER BOOK RIDDLES Ixv group (1-60) some twenty-nine lack opening formulas (Rid. 3, 4, 5, 6, 7, 8, 9, 10, ix, 12, 13, 14, 15, 16, 17, 18, 22, 23, 27, 28, 29, 31, 36, 40, 41, 45, 47, 55, 58) ; of the second group (61-95), twenty-six (Rid. 61, 62, 63, 64, 66, 67, 71, 72, 73, 74, 77, 78, 7-9, 80, 81, 82, 83, 85, 86, 88, 89, 91, 92, 93, 94, 95). The absence of opening formulas from the later riddles is not less significant than the lack of these in the first seventeen problems of the collection. Thirty-three of the riddles of the first group have no formal closing (Rid. 6, 7, 8, 10, 12, 14, 16, 18, 19, 21, 22, 23, 26, 30, 31, 34, 35, 38, 39, 41, 45-55, 57, 59) ; so with twenty-four of the second group, of which many are incomplete (Rid. 64, 65, 66, 69, 70, 71, 72, 74, 75, 76, 77, 7&, 79, 81, 82, 85, 87-89, 91-95)- Thus in the first group fifteen riddles lack all formulas (Rid. 6, 7, 8, 10, 12, 14, 16, 18, 22, 23, 31, 41, 45, 47, 55); in the second, eighteen, five of which have defective endings, are without them (Rid. 64, 66, 71, 72, 74, 77, 78, 79, 81, 82, 85, 88, 89, 91, 92, 93, 94, 95). If we are tempted by a similar absence of opening or of closing formulas in many successive riddles (compare Rid, 3-18 ; 45-55) to draw the inference that in such cases the order of the Exeter Book does not depart from the order of composi- tion, we have strong evidence that the formulas employed are not the additions of a collector, but belong in nearly every case to the original fabric of the problem. The formula is usually bound to the riddle-germ by alliteration, grammar, or syntax, often by all three. Among the more striking of opening formulas thus deeply inwrought into the poems are the following : ic eom wunderllcu wiht (Rid. 19, 21, 24 (wrcetllc), 25, 26) ; ic (ge)seah (Rid. 20, 32", 33", 53, 54, 56, 57, 60, 65, 75, 76) ; ic wiht geseah, and its variations (Rid. 30, 35, 39, 43, 52, 87) ; ic wat (Rid. 44, 50, 59) ; ic gefrcegn (Rid. 46, 482, 49, 68). Note tnat the first two and the last of these opening formulas are mainly found in successive riddles of certain parts of the collection. The closing formulas are also closely connected with the body of the riddle by alliteration, and often by se- quence of thought. Among the more important of these formal closings are Saga hwcet ic hatte either alone (Rid. n, 20, 24, 63, 67, 73, 80, 83, 86) or with an additional thought (Rid. 4, 9, 13) ; Saga with a question (Rid. 2, 3, 36) ; Frige hwcet ic hatte alone (Rid. 15, 17) or with some addition (Rid. 27, 28) ; Micel is to hycganne . . . hwcet seo wiht sy (Rid. 29, 32 ; compare variations of this final formula, 33, 36, 42, 68) ; Reed hwcet ic mcene(Rid. 62) ; Nemna^hy sylfe (Rid. 58) ; and yet more elabo- rate endings (Rid. 5, 37, 43, 56, 84). It is interesting that each portion Ixvi INTRODUCTION of the collection seems to have its favorite formulas, and that, just as in their common dislike of formal openings, so the earlier riddles of the first group seem to fall in the same category with the problems of the second group either in their entire avoidance of formulas at the close or in their use of Saga hwcet ic hatte. Only a very few formulas are independent of the thought and structure of the problem as is so often the case in the Heffireks Gdtur. Examples of such an independent opening formula are found in the two first lines of Rid. 32, 33; but in each case this beginning is followed by the common convention, ic seah. So the inde- pendent beginning of Rid. 37 is prefixed to Rid. 69, a folk-riddle with a formula of its own. The last two lines of Rid. 40 are unconnected with the riddle, but these are preceded by an elaborate formula woven closely into sense and syntax. The formula, when it appears, is thus evidently regarded not as a vain and isolated supplement to the riddle, but as an essential and vital part of its structure. Agreement of treatment throughout the collection can be best tested, however, by a careful examination and comparison of the motives and diction of the various riddles. I shall therefore make a cursory survey of the problems from this point of view. The Storm Riddles (Rid. 2, 3, 4) are strikingly differentiated from the other riddles in their sustained loftiness of tone. And yet in these poems in which the riddle is the least part of itself, poems which recall rather the sea-passages of the Andreas, we find points in common with the smaller problems. Rid. 28, foonne ic wudu hrere (see 47'8), explains the central thought of Rid. 8i7, se fee wudu hrereft; and 36, streamas stafru beatafi, suggests 8i8, mec stondende streamas beatafii Rid. 37, on stealc hleofea, and 426, steal cstanhleof>u, find their only parallel in 937, stealc hli/>o, a riddle which has something in common with 81 (8i6, 9321). The picture of tottering walls (47'10) is matched by the defective lines 8441"44. 416, fee me wegas tacne'd, is found elsewhere only in 52", se him wegas txcnefe. 318, of brimes fcefemum, appears again Rid. n6"7 ; compare 315, 772. Slighter parallels are indicated in the notes. In 68b the Sword is described as hondweorc smi/>a as in 2i7 (compare 27", weorc smifea, Book). Rid. 6 and 7 resemble each other in the spirit of battle. Prehn points out that 71'2, Mec gesette . . . Crist to compe, is paralleled in Rid. 30, where the Sun appears as a fighter against the Moon. The Bird riddles, 8, 9, 10, «, 25, 58, are closely bound together. The many likenesses •P. 167, note. AUTHORSHIP OF THE EXETER BOOK RIDDLES Ixvii between the poems of the Swan (8) and the Barnacle Goose (n) go far towards establishing the latter solution. In both hyrste is used of ' wings ' (84, ii8) and hrcegl of ' coat of feathers ' (81, 1 17) ; the air raises both birds and bears them widely (S8"6, n9'11; compare 58, Swallows); and the word getenge is found in both problems (88, n4). Trede (81) appears again in the ' Swallows ' riddle (s85, tredaff), which in turn recalls the ' Higora ' rune-puzzle in its use of nemna&(5&6, 25"). Rid. 9 closely resembles Rid. 2S (91"2 25l '•> 94> 255 ; 910> 254) and may have the same solution, ' Jay ' ; while its half-line hlude rirme (g8) finds its parallel in 58, hlude rirmad (see also 492"8). Compare the ' Cuckoo ' riddle, io10a, ofifiezt ic aweox\e\, with n8b, on sunde awox. After such comparison of these six riddles, can it be doubted that they all belong to a Bird group, and that they are all from one hand ? And yet the group is not isolated but is closely associ- ated with other problems, particularly with its neighbors in the Exeter Book. Rid. n1, Neb wees mln on nearwe, invites comparison with 221, 32 6, 35 8 ; iia, ufan yfrim freaht, with I78a, yfium feeaht ; i ite, hafdefeorh cwico, with I48 and 745; II6"7, of fcedmum . . . brimes, with 318 (supra). Hrcegl and hyrste (supra} both appear in the first line of Rid. 12, the companion piece of Rid. 28 ; and hrcegl in 14°. Yet another likeness with the Wine or Mead group (12, 28, 29) is found in the two pictures of the haunts of the Swallows and of the Bees (s82, ofer beorghleofra ; 282, of burghleofeuni). Rid. 12 and 28 are obviously mates, as are 13 and 39 (compare also 72). Rid. 13 is associated slightly with the riddle of Night-debauch (Rid. 12) by its ninth line, dol druncmennen deorcum nihtum; through i38b, wege'S ond f>y3, with 225, •wegeft mec ond f>y'8\ by the introduction of the wonfeax Wale (8 a), with 53^, wonfah Wale ; and by the peculiar idiom in I313a with 26". I have already noted close parallels between the vocabulary of Rid. 14 and preceding riddles (i49~10, hr(egl,frcetwe, 81-6; I48, n6; i4llU, turf tredan, 81, hrusan trede). I44b, Sweotol ond gesyne, reappears 4O8. Rid. 15 has no points of likeness to the neighboring riddles, save to them all in its lack of opening formula, and to 17 in its close ; but, as E. M tiller early pointed out, it closely re- sembles Rid. 80, which has the same theme (see notes under 80 for common traits), and suggests the 'Beam' and 'Beaker' riddles (31 6, 644). Compare also is12 with 2i12, 561, 57", 64", 6817. Rid. 16 contains not only many hapax-legomena,t but many expressions found only here and in Cbtkener Programm, 1861, p. 18. t Herzfeld, pp. 10-12; McLean, Old and Middle English Reader, 1893, p. xxxi. Ixviii INTRODUCTION close companions in the Exeter Book : 16, beadowapen^W, beadowcepnum) ; 16", tosizlel (i76) ; i628, hildepilum (i86, hyldep'ilas). Other similarities in word-use are 16", him bid dead witod (cf. i66, 2i24, 8s7) ; 16°, mcegburge (cf. 2I20); i612, eaforan (2i21); i68, wic buge (82). .£/</. 17 has phrases in common with n and 16 (supra). Rid. 18, in the phrasing of three of its motives (i84, 24s-9; iS5-6, 24; i86, 2412b), closely resembles 24, ' Bow.' Rid. 20 and 65 form a riddle-pair, associated as they are not only by likeness of runes but by their very phrasing (2O1'3, 6s1 ; com- pare here another runic riddle, 751). Hygeu>loncne is found only here (2O2a) and 46 (hygewlonc). Rid. 21 has many points of contact with other problems of like subject; notably with 24 (2I1 reappears very slightly changed, 242) ; and the motive of the relation of the weapon to its waldend is common to both (2i4, 24°); with 6 (2i7, 678 ; 2i16, 6; 2 1", 610, see Prehn, p. 187) ; with 16 (supra) ; with 56 (2I6-8, 9-10, descrip- tion of treasures, 56^; 2i12, 561) ; with 71 (2I6-8, 7i6; 2I23, 7i8); with 54 and 73 in the weapon's Klagelied. In its opening line Rid. 22 invites comparison with n1, 326, 358. Still another likeness between 2214 and 352, the teeth of both, is pointed out by Prehn ; f but this is perhaps pro- duced by the nature of the subjects. Rid. 225, wege&mec ond />yf>,. is very similar to I38 (supra) ; 227, brungen of beanve, to 282, brungen of bear- wum ; and 228b, habbe (ie) wundrafela, reappears 83lob. Rid. 23 has also its parallels: 2316a, n'e lagu drefde, recalls 82, and 2316b, tie on lyfte fleag, suggests 52 ; 237, yfoa ge/wac, is found only 32 (see 461) ; and the negative method of the problem is also that of 40. I have already discussed the relation of 24, ' Bow,' to the earlier weapon problems (18, 21), and of 25 to the Bird group (8, 9, 10, u, 58). Rid. 26 is not only the mate to the later 'Onion' riddle, 66 (262b-8, 66s-6; 268, 662b'3b; 269a, 66s) $ but is the first of the obscene riddles of the collection (26611, 46, 62 6'9). Rid. 27, ' Book,' has not a little in common with the riddles of similar theme, 52, ' Pen and Fingers ' (279, 527 (?) ; 27", 522a) ; 93, ' Inkhorn ' (27", 9316, 6i12-14 ; 279, 9s22 ; 277, 9s26, compare $24) ; 68, < Bible ' (2713, 6817 ; 2718f-, 6811) ; and 50, ' Bookcase ' (37, gifre, so3 gifrum lacuni). Rid. 27 and 28 touch each other closely at one point (2711'12, mec sifefean . . . haled, 28s, heeled mec sif>f>an\ Rid. 28 is certainly a companion piece to 12 (supra). In the description of the bees it suggests the Bird riddles, 8, 58 The relation of Rid. 18 to 24 has been set forth by the writer in M. L. N. XXI, 100. Trautmann, BB. XIX, 180-184, seeks to connect it with 50. t P- 272. } Cf. M. L. N. XXI, 105. AUTHORSHIP OF THE EXETER BOOK RIDDLES Ixix (282b burghleobum, s82a beorghleofea ; 28s-5, 88, 58) ; in its association of Honey and Mead it explains some enigmatic lines in 80 (282a, brungen of bearunim, 8o6, Hcebbe me on bosme foczt on bcarwe geweox) ; in its picture of the mead-hall it recalls 15"' 16, 2i12, 57", t and furnishes a contrast to 29 (28®-9, 298"10), to which it bears a general likeness; and in the sorrow caused by its contact it deals with a favorite motive of these enigmas (289, 7, i6-25, 24™, 269-10). \ Except in its suggested con- trast to 28 (supra), and in the likeness of its closing formula to 3223~24, Rid. 29 has nothing in common with its fellows. Rid. 30, as I have pointed out at length, § is bound by nearly all of its motives to 95 (so2'4, 955a ; 305, 956a ; so8, 951-3 ; so13-14, 9510'13) ; the Sun's power as a fighter (3°9~u) reminds us of 71>5, and the Moon's sad exile of 40 (infra) ; and the last motive of the riddle is very similar to that of 8312~14. Only one or two phrases in Rid. 31 suggest other riddles : 3i4, bearu blou 59\ 8l3> 86 6, 9325; 33", 95s"9). Prehn has noted IF the very close verbal agreement between 349"10 and 422'4. Compare with this the phrasing of 844, a poem that contains general references to Ice (84s5'89), the subject of 34 ; and mark a different expression of the same motive, 38. I have already pointed out the likeness of 358 to n1, 221, and, particularly, 326 (supra). 354 bears a certain similarity to 3<>4, and 357"8 has1 much in common with yi2"3. Rid. 36 occupies an isolated position among the riddles. ^ Prehn to the contrary, it bears no rela- tion to 57, and only a slight resemblance to 71 ;' and even the closing formula does not appear in the older version of the problem. It is E. Miiller, p. 19; Trautmann, BB. XIX, 206. t Prehn, p. 196. \ Dietrich, Haupts Zs. XII, 245. § M. L. N. XXI, .104. ||XI,469. tPp- 211, 276. P. 207. lxx INTRODUCTION strikingly significant that it is linked by a single motive to 41 (369, awiefan wyrda crceftum; 41 85, wratfrce gewefen wundorcrcefte), to which it is closely bound through its similar relation to Aldhelm. The opening formula of Rid. 37 is prefixed without reason to 69 ; and the problem has a general likeness to other monster-riddles (378|7~8, 8i2-5, 86s"7). Rid. 38 is a companion-piece to 87, which reproduces its first lines. These lines (381'8) also suggest 19 and the fragment 89 ; while the closing line of the problem recalls the world-old motive of 349 (supra). Rid. 39 is nearly related to the riddles of similar import, 13, 72 (393, 72^® ; 39s"7, I31"4-14"15). Rid. 40 belongs to the group of Sun and Moon riddles, 7, 30, 95 : the departure and dreary exile of ' the wight ' (4O6"9) are described 3O9~n ; the wide wanderings are pictured 4O16'17, 958 ; the comfort brought to man is mentioned 4O19, 77 ; and the silence and lore of the subject appear 4O8-4'12-21-22 and 957'10. The con- trasts of 40 suggest the method of 41, and its many negatives that of 23. The close relation of 41 to 67 and its connection with 36 will be discussed in the notes; with the other problems it has almost nothing in common. Under Rid. 34 I have indicated the likeness of 422-4 to 349"10 and 84. The closing formula of 42 binds it to 29, which it also resembles in its use of superlatives (42s"4, 292~8) and its employment of brucen (427 ; see 29™, bruced}. I find a few parallels to Rid. 43 : its opening formula appears frequently in the Riddles ; equivalents of hwitloc (438) are elsewhere used to suggest fair beauty (4i98, 8o4) ; wlanc is em- ployed in the same context (267, modwlonc] and weorc in the same sense (5510) ; on flette (435) is a not uncommon phrase (s62, 57", on fief) ; and werum at wine (4316) suggests wer at wine (471). A parallel to 441, in- dryhtne afeelum deorne, is found in 951, indryhten ond eorlum cud; to 442, giest, in 4°, 89, 2315, etc. ; to the reference to the Earth as moddor ond sweostor (44") in 835, eorfian brofior.^ Rid. 45 is one of the group of obscene riddles, and therefore has not a little in common with 26, 46, 55, 62, 63, 64 (458, 26 ; 455, 6s7); its closest analogue is 55 (45", 555 ; 4$4-5, 55s"4). Rid. 46 is also bound closely to others of its class (46la, 552a ; 46lb, 558b, 629a ; 46, 267 ; 466b, 266) ; and, in its use of hygewlonc, has a slight connection with 2O2, hygewloncne, the only other occurrence of the word. The world-riddle 47 has nothing in common with the other »Cf. Dietrich, Haupts Zs. XII, 245. t Cf. Anglo-Saxon Prose Kiddle, Grein, Eibl. der angelsachsischen Poesie II, 410. See note to 44". AUTHORSHIP OF THE EXETER BOOK RIDDLES Ixxi problems of the collection save the likeness of 47 la to 4316a (supra). Rid. 48 is, however, connected with other riddles: its second line is similar to the opening formulas of 461, 491, and the use of stafiol (485) invites comparison with 26, yi2, 8S25, 92" ; while its last motive (4S4-5) is not unlike so10"11. As Dietrich long ago pointed out, 49 is a com- panion-piece to 60, as a likeness in all motives proclaims: it is associated by the phrase hlude stefne ne cirmde (492~8) with the Bird riddles (g8, hlude cirme ; 58, hlude cirma^). Rid. 50 has many analogues. Gifrum lacum (so8) and to nytte (so9) connect it with the Book riddle (27127'28) ; while its first and last motives may have been suggested by the well- known problem of the Bookmoth (485~6). It bears an interesting relation to its neighbor 51 (so2, dumban, si2, dumbum; so9, si2, to nytte; 50", Si8, the ' feeding ' of both) ; and it has points of contact with 58 and 72 (SO4"5, se wonna f>egn sweart ond saloneb ; 7210, sweartum hyrde ; s88a, swearte salopade). Trautmann points out f like traits of the subjects of 50 and 18 : both work by day (so2, i88), both swallow (i87, So2-11), and both conceal costly treasures (so6, i89~10). Rid. 51 is connected not only with 50, but, through its first line Wiga is on eorfean wundrum acenned, with 841, An wiht is wundrum acenned. The likeness pointed out by Trautmann \ between 52 and 27 has already been illustrated. 524b, MS. fleotgan lyfte, recalls 2316 on lyfte fl~eag (cf. 74 8) ; 525a, deaf under ybe, appears again, 744 ; and 526b, se him wegas tatcnefo, reproduces 416b. The wonfah Wale of 53 6a reminds us of the wonfeax Wale of i38a. Rid. 54 has much in common with 73 (548, 731"2 ; 54 frod dagum, 733, gearum frodne, 831, 93) and 92 (infra). Its motive of wretched change of state is the leading idea of 27, 73, 83, 93. Like the others of the group of obscene riddles, Rid. 55 is closely associated with its fellows : its rela- tions to 45 have been indicated ; tittle esne appears only ss8a, 645a ; ss6, worhte his wi/ian, is paralleled by 647, wyrcdS his willqn ; 552, MS. in wine se/e, may be corrected in the light of 461, in winde; 55 10, JHES weorces, recalls the like use of the phrase, 43. Rid. 56 is nearly akin, in its first lines, to 5710~12 ; and 564a, searobunden, also resembles S78"6, seanuum . . . gebunden. Prehn § regards 56 as a companion to 21, ' Sword ' ; though this is an overstatement, there are certain likenesses between the two (56^, 2I6"8'9-10; 561, 2i12, a common formula). Rid. 57 is not only associated with 56, but its vocabulary bears in two Haupts Zs. XI, 474. t BB. XIX, 183-184. t Ib. XIX, 197. § P. 279. INTRODUCTION lines (577~8) a distant resemblance to 524b>5b. Prehn fails to establish any connection between this and 36. The relation of 58 to the other Bird riddles has been discussed at length (supra), and its parallels to other problems sufficiently indicated (s82a, 282 ; 58% 5<>5a ; s84b, 492-3). Rid. 59 has no near anafogues; but 59la, anfete, suggests 336, 8i8, 93^ ; S92-8 repeats the motive of 32", and 5910"11 that of 32". The enumeration of strange physical traits (597~9) gives it a place among monster-riddles (cf. 33, 81, 86). As we have already seen, 60 is a mate to 49. Rid. 61 is bound to the other riddles by its companionship in the Exeter Book (i22b-i23a) with the second form of 31. Its first lines bear a general likeness to 771"2; and 6i12, seaxes ord, reappears, 776. Prehn t has pointed out the similarity of 6i9 to 328'12-14, and of 6i12-14 to 27™ (cf. 9315r). The first problems of the so-called second series are closely bound to those of the first group. Rid. 62 is an obscene riddle, and, as such, is a near kinsman of 26 and 46 (626'9, 266-11, 461-8), and of the next coarse enigma (62 6, on nearo ; so 638). Rid. 63 is thus bound not only to its precursor, but to its follower, 64 (635, 646, fey^}, and to the other puzzles of double meaning (636, 554; 6s7, 45; 638, 265, nathwar, 461, 555, 629, nat- hwcef). The relation of the ambiguous 64 4~7 to 55 and 63 has been shown (supra) : but 642h>3b, ford boren . . . ficer guman drincaff, must be com- pared with 56 2, 5711"12; and 64^, mec . . . cysseS m ufre, with the rid- dles of 'Horn' and 'Cross,' is3, 3i6. Rid. 65 is the companion-piece to 20 (supra) ; and 66 to 26. Dietrich $ has pointed out the likeness be- tween 663a, hafaft mec on headre, and 2i13, healdeft mec on heafeore. The interesting connection between 41 and 67 has been already mentioned. Rid. 67 has also something in common with the vocabulary of the frag- ment 94 (672b, Teohtre feonne mono. ; 94s"7, Teofre foonne feis leoht eall, leohtre fronne w . . . ; 676b, heofonas oferstlge, 942a, hyrre foonne heofori). Rid. 68 abounds in words and phrases of the riddle-poetry : 681, ic gefrcegn, 461, 482, 491 ; 682, wr&rtce iviht, 431, 521, 7O1 ; 6&9,fef rie f\olme\, 327, 4010 ; 6812-16, general likeness to 2718f- ; 6817a, golde gegierwed, vf gierede mec mid golde ; 6817h, frier guman druncon, 561, 57", &48 ; 6818, since ond seolfre, $64. The opening formula of 37 precedes the one-line folk-riddle 69. Rid. 70 is related by its subject to 32, but its likeness to other rid- dles lies chiefly in its diction, the use of single words found elsewhere in the collection : 7O2, singe^ 32 8; 7<>2, sldan and sweora, 7318, 86s"7; 7o8a, orf>oncum, 787a, fiurh orfoonc; 7O8, eaxle, 73", 866 ; 7O4a, on gescyldrum, P- 233. t P. 237. J Haupts Zs. XII, 250. AUTHORSHIP OF THE EXETER BOOK RIDDLES Ixxiii 4jios . y0i^ Wra>tnc(e), passim ; 7O6b, halefeum to nytte, ay27, 51, etc. 71 has many analogues: 71, ic com rices ieht, 79, ic eom cefeelinges ieht :; 71% stiff and steap wong, 361"2 ; 7I2"8, stafeol . . . wyrta wlitetorhtra, 357"8, fea wlitigan wyrtum fceste . . . on stafeolwonge ; 7i5"6, wepeft for gripe niinum, 93™, ne for wunde weop. As a riddle of the Sword, it is closely connected with problems having the same theme: 7i8"4, wrafera laf,fyres ond feole, 67b, homera lafe (Beow. 1033, fela laf, 'sword'); 7i5, wire geweorfead, 2i4a'lob-82a ; 7i6b, se fee gold wigeft, 2i~8, ic sincwege . . . gold ofer geardas (' Sword ') ; 7i8a, hringum gehyrsted, 2 12813 fee me hringas geaf (' Sword '). Rid. 72 is connected by its subject (' Ox ') and two of its motives with the pair 13 and 39 (72", feower feah . . . brdfeor, 39^4> feower wellan, etc. ; 721(M2, I31"2). The misery of the subject (7212~13) is a common riddle-topic (2i17, 545, 8i6, 9321). I have already noticed the likeness of 73 to 54 : save in its monster traits (see supra under 70), it has nothing in common with any other problems. Rid. n,jleah midfuglum, recalls 2315, 52 4 ; 74% deaf under y foe, is identical with 52 5a ; and 74 5b, hcefdeferficwicu, very similar to n6, i43. The tiny runic riddle 75 is exactly in the manner of other runic problems, 2O1"8, 6s1 ; while the inversion of the runes (75 2) recalls 241, Agof. The single line of 76 employs the opening formula of 75. Under 61 I have noted the slight parallels between that riddle and 77 (771"2, 6I1"2; 776, seaxes orde, 6i12, 27^6). The closest analogue to 77 is the fragment 78 : 772, tnec yf>a wrugon, j$',y/>um bewrigene (compare 315) ; 778a, fefeetease, 782, \Te\as cyn : 778b, Oft ic flode, 78la, Oft ic flodas. Rid. 79, whose single line may be but a variant of 8ox, recalls 71 la {supra). Miiller and Trautmann have invited attention to the close relation between the two Horn rid- dles, 15 and 80 (supra) : 8o2, fyrdrinces gefara, I513, fyrdsceorp • 8o3"5, the serving of mead by the lady, is8"9 ; 8o7'8, on wloncum ivicge ride, J55-6.18-14 . 8o8b) hgard -s m-m tungg^ jg4.16.18 . 8<)8b.7a) ^.4,6,6.^ faMum. The mention of honey (mead), 8o6, hcebbe me on bosme f>cet on bearwe geweox, recalls the mead of 282, brungen of bearwum ; and So3", Cwen . . . hwitloccedu , suggests 438b, hwitloc. Rid. 81 has an affinity to the Storm riddles (8i7, se fee wudu hrereV (wind), 28, ic wudu hrere (wind) ; 8i8, streamas beataft, 36) ; its monster traits (Si2"5) invite comparison with 597-8, 86^7, 3737'8; and its wretchedness with 2i17, 54s, 7213, 9321. The fourth line of the fragment 82, [/]<?// ne flizsc, reminds us of 775. In 83, the Ore's sad change of state recalls the themes of 27, 54, 73, 93 ; and • Prehn, p. 242, note. lxxiv INTRODUCTION its lack of redress (8s8b, ic him yfle m mot} is akin to the Sword's and Horn's failure to avenge (2i17, 9319). 8s4b, Nu me fah wara% strongly resembles 9326, Nu mm hord wara8h~if>ende f~eond\ 8310b, Hcebbe ic wun- drafela, reproduces 228b, habbe wundrafela ; and 831'2-14 contains exactly the closing motive of the Sun and Moon riddles, 3O13-", 9510-14. Rid. 84 is more or less intimately connected with many other riddles. Its first line is but a variant of Si1 (supra); 844-20, in the theme of Water and Fish, anticipates 85, while the phrasing of 84, Modor is monigra mcerra wihta, recalls 422, moddor monigra cynna ; S46-9 bears a general likeness to 4O22"24. Prehn discovers a resemblance between 849~10 and 4I1'8, and between 84 85 and 4i65 ; but this is faint and may well be coincidence. And Dietrichf finds a relation between the 'Taufwasser' of 84 38 ', ftrene dwcesced, and 3i7-9 (cf. 8428, 3I5-6); but this is very doubtful. The like- ness of 842-8-41-44 to the Storm riddles, 22, 35b, and 47'10, lies probably in the demands of similar subjects. As has just been noted, Rid. 85 treats a theme suggested in 84. While the description of Water, 8s5b, he sceal rinnanford, is founded upon the Latin of Symphosius (see ' Originals and Analogues '), yet it may be compared with 842b, hafad ryne strongne, and %45,fa>gerferendefunda'd<£fre. 858, ic eom swiftre f>onne he, is quite in the manner of 4i94, ic eom swifera fionne he (cf. 4i26-28) ; and 8s7, me b'ift dead witod, reproduces i6n, him bid deaffwitod (cf. 16°), a phrase found only here. 852b, unc drihten scop, parallels 8817, unc gescop meotud. Save in its monster traits (cf. 32, 33, 37, 59, 81), Rid. 86 has little in common with other riddles. Its opening formula, Wiht cwom gongan, recalls 341, Wiht cwom . . . nfean, and 551, Hyse cwom gangan ; and 862, monige . . . mode snottre, repeats 8484. Rid. 87 is another version of 38, repeating many of its expressions (supra) ; while its first line, wombe hcefde micle, connects it with 19', wide wombe, and Bg2, wiht wombe h•£, f rod wees minfromcynn; 937, stealc hli/>o, 37, 426; 9310, X31, I62'17, 631 ; 9315"18, 27, 6i12 (j»/ra); 9319a, ne for wunde w~eop, 7I6"6, Weped . . . for gripe m'mum ; 9311~20, lack of re- venge, 2 117, 83 8; 93", ic aglceca ealle frolige, 8i6, Aglac dreoge; 9322, 69; 9322'28, Nu ic blace swelge wuda ond wcetre, 27°, beamtelge swealg; 9S27'29, 277"10 5 9326. 834"5 (supra). I have pointed out under 67 the relation of the fragment 94 to that ' Creation ' riddle. As has been shown, Rid. 95 is bound by nearly all of its motives to its mate, 30 (supra). Through its closeness to men, its wanderings, its lore, and its silence, the subject recalls a riddle of like theme, 40 (951'8, 4O1'8 ; 95", 4O16-17 ; 95 7-9, 4o3-4>21;f ; 959"10, 4O12). Rid. 95 employs the phrases of other problems : 95 , Ic eom indryhten ond eorlum cuty 441, Ic wat indryhtne cefielum deorne ; 95'2, ricum ond heamtm, 3318a, rice ond h~eane\ 957, snottre men, 862, monige . . . mode snottre, 84 , man mode snottor. The closing motive of 95 is found not only in 3018-14, but in B^1 (supra). Such likenesses as I have pointed out between the various riddles are \ sufficiently striking to establish homogeneity, and indeed they often com- !\pel belief in the presence of a single hand in many of the problems. Bul- oring fails completely to grasp the true character of the enigmas of the Exeter Book when he declares : t ' Wie man bei einer Sammlung von Volkslieder schwerlich an einen einzigen Verfasser denken wird, so darf man es meines erachtens ebensowenig bei diesen Ratseln, die mit geringen Ausnahmen doch auch ein Produkt der Volkspoesie sind.' It is obviously absurd to class our riddles with folk-songs. As I have long since shown, t they teem with popular elements and motives, but they are almost with- out exception literary enigmas from the hand of the artist. In such com- positions as the poems of the Storm (2, 3, 4), Badger (16), Sword (21), Book (27), Lance (73), Water (84), and the Horn cycle (15, 80, 88, 93), the reader soon becomes aware that the riddle is the least part of itself, that concealment of solution has been forgotten in the joy of creation. See Prehn, p. 260, note. t Lift. Bl. XII, 1891, Sp. 156, cited with hearty approval by Herzfeld, Herrigs Archiv CVI (N. S. VI), 1901, p. 390. } M. L. N. XVIII (1903), 97 f. ; see also supra. Cf. Brandl, Grundriss II, 972. INTRODUCTION Even, in the shorter problems, the riddle-maker, draw though he may from the stores of the folk, shapes anew with loving art the story of the ingratitude of the Cuckoo (10), the fate of the Ox (13), the labors of the Plow (22) and the Rake (35), the journeys of the Ship (33) ; or else, by the aid of runes, converts into logogriphs or word-riddles of the study such commonplaces of folk-poetry as the themes of the Cock (43) and Man on Horseback with Hawk (20, 65). Even in the small number of riddles which, in tense, terse, pointed style and absence of epic breadth, in freedom from all that is clerkly or bookish, seem to bear clearly the stamp of popular production (53, 58, 66, 70, etc.), the many parallels to other problems (supra) mark the presence of the craftsman. In those very puzzles whose smut and smiles point directly to a humble origin (26, 45, 46, 55, 63) we detect (supra}, amid the coarseness of the cottage, tre leer of a prurient reworker. 'Ihe Riddles, then, are homogeneous in their artistry. One of the finest proofs of this lies in the striking circumstance that almost every dark saying or obscure periphrase in our poems finds illuminating ex- planation elsewhere in the collection. To indicate a few examples out of many : 8i7b, se fee wudu hrereft, is revealed as ' the wind ' in the light of 28, ic wudu hr'ere\ 80 6, Hcebbe me on bosme beet on bearwe geweox, is interpreted by reference to the description of Honey in 28'2; the enig- matic phrase brunra beat immediately becomes clear by comparison with the picture of the swine, dark and joyous, in the beech wood, 4i106-107 ; and 955, h'ibendra hyht, is seen to be but a circumlocution for huty ' booty,' when read side by side with 3O4-9. The homogeneity of the collection is further attested by the dominance in very many of our riddles of the two motives of ' utility ' and ' comitatus,' which play but a small part in other enigmas of the Old English period. These will be discussed at length in a later chapter. Now if certain art-riddles are found grouped in what is really a single collection ; if, moreover, these riddles, after close analysis, are found to be homogeneous in their diction ; if, too, large collections from single hands were common at that period, — the burden of proof rests not upon him who argues for unity of authorship, since every precedent and presump- tion are in his favor, but upon him who champions diversity of origin. The need of such strong destructive evidence is totally disregarded by Trautmann in his bald assertion : ' Diese entstammen verschiednen Kyncwulf, p. 4 1 . AUTHORSHIP OF THE EXETER BOOK RIDDLES Ixxvii zeiten und dichtern.' Brandl, who holds the same view, gives, however, certain reasons for his opinion. The second group seems to him sepa- rated from the first by the second appearance of Rid. 31 ; but that the Exeter Book modernizer or scribe chose to insert in a position isolated from both groups a variant version of a riddle already given proves, of course, nothing against the unity of the collection. The contrast between the edifying tone of certain enigmas and the coarseness of their near neighbors seems at first sight to indicate different hands : but the points of contact between the lofty and the low often forbid such a conclusion. Runes and ribaldry meet in Rid. 43, court and cottage clash in Rid. 62 ; the literary and the popular blend in Rid. 13 and 64 ; Rid. 66, with its Symphosius motive, is closely related to Rid. 26, the grossest of its greasy sort. Subject-matter is evidently small criterion of origin. Further evidence against the unity of the collection is furnished by Barnouw.t The Riddles differ so widely from one another in their use of articles that if this be a trustworthy test of date, they may well be re- garded as the products of different periods. ' Some of them that employ 'articles freely (24, 43) may be contemporary with Cynewulf, while others that are sparing in the use of these (16, 23, etc.) are doubtless earlier in time.' Deductions drawn from such evidence are dangerous ; and one refuses to follow Barnouw when he goes to the length of assigning Rid. 38, 39, 69, to a later date than Rid. 30, 35, 37, because in the former group the opening formula is ic fea wiht(e) geseah, in the latter ic wiht geseah.\ The weak adjective without an article is to Barnouw proof of an early date, and he differentiates the Riddles accordingly. § He regards Rid. 13 as one of the oldest of the riddles on account of the absolute use of weak adjective without article in the phrase hygegalan hond^i^3). The survival of an archaic form in a poetical text is surely no proof of antiquity. || Pauls Grundriss^ II, 970. t Der bestimmte A r tike I im Altenglischen, p. 21 1. t Barnouw (p. 211) notes that the following riddles are quite without articles: 3, 6, 9, ii, 13, 14, 15, 18, 20, 22, 37 (1-8), 51, 52, 53, 58, 59, 63, 64, 66, 67, 71, 72, 74, 80, 83, 85, 86, and the fragments 19, 75, 76, 77, 78, 79, 82, 87, 89, 92. § In addition to instrumental forms already cited (4, 4187,90,9^ 579-10), Barnouw records the following instances of weak adjectives without an article: 48, beartn brddan (?) ; 44'2, eorpan gesceafte ; 383, mcrgenrofa man ; 4I55, hrim heoriigrimma \ 496, readan goldes (contrast 52", 56s) ; 8313, dyran cr&ftes 9311"12, hdra . . .forst. || Note the appearance of weak adjectives without definite articles in a late poem, Brunanburh, 61-62, salowigpsdan and hyrnednebban. Ixxviii INTRODUCTION Although Barnouw's arguments have been accepted by Brandl in his Grundriss article as infallible criteria of date not only of the Riddles, but of all other Anglo-Saxon poems, they seem to me to carry little weight. | The normalizing of later scribes, and ' the tendency to archaize, to use traditional formulas and expressions, so strong in Anglo-Saxon poetry,' t ! render this test almost valueless. The use of the article in early Greek poetry is closely analogous to that in Old English verse. But the classical scholar, who, on account of the absence or presence of articles, assigned the various fragments of Alcaeus to different hands, ascribed the tragic choruses of Aeschylus to an earlier date than the non-lyric portions of the dramas, and labeled as Homeric in time the epic conventions of Apollonius Rhodius, would be speedily laughed out of court. A much more important argument remains — that based upon the evidence furnished by the use of sources. We have already seen that, with the same data, Dietrich and Holthaus reached exactly opposite conclusions in regard to the unity of the collection. But the value of their reasoning was impaired by the incorrectness of their data — sup- posedly close literary relations between Latin and Anglo-Saxon enigmas, where often none at all existed. In the methods of direct and indirect borrowing that our study of the sources of the several problems \ has revealed, there are but few certain indications of difference of origin. The habit of mind which either works in perfect liberty, or else, gather- ing a useful hint here, a happy phrase there, gives delightfully fresh and new forms to current motives and ancient traditions, but which never yields itself slavishly to its models, is the dominant mood in the Riddles and points rather to one poet of free spirit than to many men of many times. And yet all the Exeter Book Riddles can hardly be from one hand. The servilely imitative temper of Aldhelm's translator in the enigmas of the ' Mail-coat ' and ' Creation ' (Rid. 36, 41) differs so utterly from the prevailing tone of the collection, which is at its highest in the unchecked range of imagination of the ' Storm ' riddles (2-4), that this inferiority cannot be explained with Dietrich by the changing inclination of one poet. § As will be shown later in my notes to Rid. 41, there is good Notice the difference in this regard between the Exeter and Vercelli texts of Soul. f See Lawrence, M. L. N. XXIV, 1 52. t See chapter on ' Originals and Analogues.' § It is interesting to note that these two problems, which stand so widely apart from all the others in their dependence upon learned sources, have other very dis- tinctive features : (a) the poor technique of Rid. 41 ; (b) the isolation of the SOLUTIONS OF THE EXETER BOOK RIDDLES Ixxix reason to believe that yet another hand was at work in the later portion of that long and dreary poem, and that this hand rewrought his crude work in Rid. 67. But these poems are the only ones in the collection that we can assign with any positiveness to a different author. Let us now summarize our results. The Riddles were not written by Cynewulf : all evidence of the least value speaks against his claim. It seems fairly certain that they are products of the North, f Their place as literary compositions (not as folk-riddles) in one collection, and their homogeneous artistry, which finds abundant vindication in a hundred common traits, argue strongly for a single author, though a small group ,of problems brings convincing evidence against complete unity. That their period was the beginning of the eighth century, the heyday of 'Anglo-Latin riddle-poetry, is an inviting surmise unsustained by proof. , IV SOLUTIONS OF THE EXETER BOOK RIDDLES Unlike the Latin riddles of their period, the Anglo-Saxon queries are unaccompanied by their answers. In six problems, however, the ingen- ious use of runes guides the solver to his goal. In two of these \ the runic element is so elaborate and complex that it converts the poems into intricate name-riddles ; in three others § the ' open sesame ' is found in an easy rearrangement of the runic letters ; in -the sixth || the last two lines constitute a runic tag that confirms an already obvious Northumbrian version of Rid. 36 from all other English riddles, and its associa- tion in the Leiden MS. with the Anglo-Latin enigmas with which it is so closely connected in thought ; (c) the differentiation of Rid. 36 and 41 from neighboring queries of their group {Rid. 31-61) by the subject's use of the first person. Even the obscene and the runic group, which seem to fall into two distinctive classes apparently remote from the others, reveal upon examination points of con- tact. By recasting, the poet makes coarse folk -products his own. t The Northumbrian dialect of the Leiden Riddle proves nothing, as its variant version, Rid. 36, stands entirely apart from others of the collection except 41 ; but Northern origin is attested by the large number of uncontracted and unsynco- pated forms demanded by the meter, and by the appearance of such Anglian usages as bag (58), sa-cce (ly2), geonge (222), ehtuwe (37), efrfra (4416), J>izA (?28). See Madert, pp. 126-127. J Rid. 20, 65. § Rid. 25, 43, 75. The third of these is but a fragment, but in the first and second the Sachenratsel element dominates. || Rid. 59. Ixxx INTRODUCTION interpretation. In a seventh riddle the Latin equivalents of preceding English words are disguised in secret script. In three other riddles f the marginal use of single runes obviously originated at a far later period than that of their composition, as these are not from the hand of the scribe. Inversion of its opening nonsense-word gives, as the rid- dler tells us, the name of the subject of one of the spirited weapon- riddles, t Finally, the faint letters in other writing at the end of the long ' Creation ' enigma § may be read as hit is sio creatura pr. Such are our clews in a dozen problems. || These, however, were of but slight aid to the first modern scholar who presented any solutions. Hickes inserted facsimile transcripts of five runic riddles IF in the beginning of his Icelandic Grammar As Conybeare says quaintly : ft ' Hickes' opinion (of these riddles) is formed from the } attributes ascribed to the mysterious subject, such as being appointed by Christ to encounter warfare ; speaking in many tongues ; giving wisdom to the simple; rejoicing in persecution; found by the worthy; and re- ceived by those who are washed by the laver, etc. 'it Conybeare's own attempts at solution are almost as unfortunate as those of Hickes. For Rid. 3-4 he supplies the answer ' Sun,' for 33 ' Wagon or Cart,' for 47 ' Adam, Eve, two of their sons and one daughter appear to be the five persons intended.' He is nearer the mark in his answer to 67 : ' The omnipresent power of the deity comprehending at once the most minute and vast portions of his creation is intended.' Many scholars have sought to solve the problems. §§ L. C. M tiller |||| offered to Rid. 6 and 27 the solutions Scutum and Liber. Thomas Rid. 37. t Rid. 7, 9, 18. \ Rid. 24. § Rid. 41. || Strobl, Haupts Zs. XXXI, 55-56, claims that the so-called Husband's Message, which follows Rid. 61 in the Exeter Book, furnishes the correct answer to that enigma, ' Der Runenstab.' But the theory that the two poems form thus a sort of Wettgedicht completely collapses, if, with Dietrich, we interpret the riddle, « Reed,' as I think that we must (see notes). 1 Rid. 20, 25, 37, 65, 75. From his copy of 37 Grein drew the facsimile at the close of his Bibliothek. Thesaurus III, 5. tt Illustrations, p. 210. it Hickes's comments are interesting. After a Latin analysis of each of the rid- dles copied by him, he cites passages at random from other problems, particularly from those of Sun, Night, Badger, and Mead (7, 12, 16, 28), to show that their solution is Ecclesia : e.g. 286, m bydene (the ' butt ' in which the Mead is prepared) receives the surprising interpretation : in dolio, i.e. in baptisterio. §§ For brief summaries of the work of solvers, see Wiilker, Grundriss, pp. 166- 167, and Trautmann, Anglia, Bb. V (1894), 46 f. Illl Collectanea Anglo-Saxonica, 1835, pp. 63-64. SOLUTIONS OF THE EXETER BOOK RIDDLES Ixxxi Wright proposed three answers : to Rid. 14 ' Butterfly-cocoon,' to 29 ' John Barleycorn,' and to 47 ' Lot with his two daughters, and their two sons.' In the same year, 1842, Thorpe | solved the 2oth riddle with /io/y, man, rad-uxzgn, hafoc, and the 22d with ' Plow.' Bouterwek \ suggested ' Hemp ' in Rid. 26. Leo § proposed ' Cynewulf ' for Rid. i. Grein || gave four answers : Rid. 3, ' Anchor '54,' Hurricane ' ; 48, ' Bookmoth'; 68, 'Winter.' Then followed, in 1859 and 1860, the two epoch-making essays of Franz Dietrich,H in which he unlocked the treasure-gates of nearly all the riddles. By far the greater number of his solutions seem to the present editor adequate interpretations of the several problems, and attest the fine acumen or riddle-sense which com- pelled Dietrich to weigh each enigma not as a scholar in his study, but as a man among men of naive minds. Since Dietrich's day a little has been added, here and there, to our understanding of the queries ; but in many cases other keywords — 'Open Wheat,' 'Open Rye' — have been futilely substituted for his 'Open Sesame.' In his Sprachschatz (1861), Grein is more than once happy in his guesses, ft and Ed. Miiller's comments of the same year are often suggestive. t$ For over twenty years the Riddles found no new solvers. In 1883 Trautmann§§ offered the answers, Rid. i, ' Riddle,' || || and Rid. 95, Biographia Britannica Literaria I (1842), 79—82. t Codex JSxoniensis, p. 527. $ Ctzdmori's des Angelsachsen biblische Dichtungen, 1854, I, 310—311. § Quae de se ipso Cyneivulfus tradiderit, 1857. || Bibl. der ags. Poesie II (1858), p. 410. T Haupts Zs. XI, 448-490; XII, 232-252. Dietrich errs, I think, irThis explanations of Rid. 5, 9, n, 14, 29, 37, 42, 46, 51' 52< 53' 55> 63, 65, 71, 72, 74, 80, 81, 90, 95. His answers to Rid. 31 and 40 are more than doubtful. In his second article, which is often a palinode of his first, he withdraws (usually at the prompting of his friend Lange, no riddle-kenner) very suitable replies to Rid. 18, 26, 45, and 58. Each of his solutions will be discussed in my notes. tt Notably in his 'Bell' answer to Rid. 5 (II, 716) — suggested but withdrawn by Dietrich — and in the ' Ox ' solution of Rid. 72. \Die Rdtsel des Exeterbuches, Programm der herzoglichen Hauptschule zu Cothen, 1861. Miiller's remarks upon Rid. 13 and 39, 15 and 80, 2, 3, 4, 9, 28, 30, 59, 61, 63, 71, 74, 80, 85, 86, 87, merit attention. Had Trautmann known his ' Horn ' interpretation of Rid. 80, he would surely not have heralded this solution as an original discovery forty years later (BB. XIX, 1905, 203-206). §§ Angha VI, Anz., pp. 158 f. See also ib. VII, Am., p. 210. || || The later history of the discussion of the ' First Riddle ' is sketched elsewhere in this Introduction and will not now be considered. Ixxxii INTRODUCTION ' Riddle.' In the same year Prehn published his discussion of the sources of the Riddles emphasizing Dietrich's solutions. Reviewing Prehn's work,f Holthaus accepted Trautmann's two interpretations. Nuck \ opposed the solutions of Trautmann, and Hicketier § revived Leo's solution of Rid. i, argued against Trautmann's answer to 95, discussed 90, and suggested readings of the runic problems 20 and 65. According to Henry MorleyJ the solution of Rid. i is 'The Christian Preacher,' of 61 ' Letter-beam cut from the stump of an old jetty,' of 90 ' The Lamb of God,' and of 95 ' The Word of God.' Herzfeld IF solves Rid. 46 by ' Dough ' and 51 by 'Fire.' In his excellent versions of over a third of the Riddles, Brooke accepts the answers of Dietrich and Prehn except in Rid. n, which he interprets as ' Barnacle Goose.' In 1894 Trautmann published ft a great number of solutions with no further support than an ipse dixit. These answers, by reason of their seeming remoteness from any obvious interpretations of the text, have sometimes been regarded as random guesses. it In subsequent articles §§ he has withdrawn or championed several of these obiter dicta. But, as I have pointed out, || || lack of historical method, perversion of the meaning of the text, and arbitrary assaults upon its integrity discredit nearly all his answers.iril Kompositionen und Quellen der Ratsel ties Exeterbuches. t Anglia VII. Am., p. 120. t Anglia X (1888), 390 f. § ' Fiinf Ratsel des Exeterbuches,' Anglia X, 564. || English Writers II (1888), 38, 224 f. If Die Ratsel des Exeterbuches und ihr Verfasser, 1890, p. 69. Early English Literature, iS()2, flasst'm. tt Anglia, Beiblatt V, 46 f. it Brandl, however, seriously impairs the value of his discussion of the Riddles (Pauls Grundriss II, 1908, 969-973) by accepting without question many of these unsustained solutions. \Anglia XVII (1895), 396-4°° (Rid. 53, 58, 90); Padelford's Old English Musical Terms, 1899 (Rid. 9, 32, 61, 70, 86); BB. XVII (1905), 142 (Rid. n); ib. XIX (1905), 167-215 (Rid. ii, 12, 14, 18, 26, 30, 31, 45, 52, 53, 58, 74, 80, 95). |||| M. L. N. XXI (1906), 97-105. ft Of the solutions originating with Trautmann himself only se.-^n compel conviction (Rid. 37, ' Ship ' ; 52, ' Pen and Fingers ' ; 53, ' Flail ' ; 63, ' Poker ' ; 68, 'Bible'; 81, 'Weathercock'; and 92, 'Beech'). He is seemingly unaware that several of his most plausible answers have been given long before by other scholars — notably 61, ' Runenstab,' by Morley and Strobl; 72, 'Ox,' by Grein and Brooke ; 80, ' Horn,' by Ed. Miiller. SOLUTIONS OF THE EXETER BOOK RIDDLES Ixxxiii Several scholars have contributed their mites to the solutions of single queries. Walz discusses some six of these in his ' Notes on the Anglo- Saxon Riddles,' reaching, I think, incorrect conclusions.! Blackburn interprets Rid. 31 as Beam,\ Frl. Sonke Rid. 25 as ' Scurra ' or ' Mime,' § and Felix Liebermann || and Jordan 1[ arrive independently at the ' Sword-rack ' solution of the ' Cross ' riddle (56). The Erlemanns have cast much light upon the ' Storm ' riddles (Rid. 2-4) and upon the Latin enigma,tt and Holthausen has once or twice turned aside from text emendation to try riddle-locks. \ I have already suggested several new solutions, §§ and shall attempt a few others in the present work. || || All the answers indicated in this cursory sketch will receive consideration in the notes of this edition (see also the ' Index of Solu- tions ' at the close of the bopk).1F1T In closing this survey, let me repeat what I have said in a previous discussion. The solution of riddles is too uncertain a matter to permit their solver ' to come to battle like a dictator from the plow.' To the same motives different solutions are often accorded by the folk itself, as I have shown at length.ftt It was, of course, the purpose of the riddler Harvard Studies V (1896), 261-268. t His answers, 'Gold' (12), 'Porcupine' (16), 'Mustard' (26), 'Cloud and Wind' (30), ' Yoke of Oxen led into the barn or house by a female slave ' (53), and ' Sword ' (80) are sturdily but unconvincingly championed. \Journal of Germanic Philology III, p. 4. \Englische Sludien XXXVII, 313-318. || Herrigs Archiv CXIV, 163. T Altenglische Saugetiernamen, p. 62. Edmund Erlemann, Herrigs Archiv CXI (1903), 55. tt Ib., p. 59; Fritz Erlemann, ib. CXV, 391. It See his solutions of Rid. n, ' Water-lily' (Anglia, Bb. XVI, 1905, 228) ; 16, ' Porcupine ' (Engl. Stud. XXXVII, 206) ; and his readings of Rid. 20 (Anglia, Bb. IX, 357), 37 (Engl. Stud. XXXVII, 208), and 90 (ib., 210-211). §§Rid. 14, 'Ten Fingers' (M. L. N. XVIII, 1903, 101-102); 74, 'Siren' (ib., 100; XXI, 1906, 103-104) ;.and 95, 'Moon' (ib. XXI, 104-105). |||| See particularly notes to Rid. 20, 37, 40, 42, 56, 71. Tir In chronicling in my Notes the ' Onion ' and ' Leek ' answers for Rid. 26 and 66, I fail to remark that ' Leek ' is impossible for either riddle. ' A leek is never " red " like the wight of 26, the bottom of the leek being blanched like celery for use, while the top is of course green ; and a leek is always eaten in the year of sowing or in the following winter, has never been planted out in the second spring, and hence cannot be the wight of 66, which has been dead and lived again ' (Bying- ton). The ' Onion ' satisfies all conditions. M.L.N. XXI, 97-98. tttlb. XVIII, 5-6. Ixxxiv INTRODUCTION to lead his hearers into many devious paths, each of which seemed, for the moment, the only way of escape from the maze ; and his cunning has been richly rewarded by the fate of modern solvers. In his second article Dietrich retracts a dozen solutions of his first,! and Trautmann frankly and freely changes ground in many problems. Rid. n, once solved by him ' Bubble,' is now ' Anchor ' ; 30, formerly ' Swallow and Sparrow,' is now ' Bird and Wind ' ; 31, ' Cornfield in ear,' now becomes Beam. In 52, ' Horse and Wagon ' is rightly replaced by ' Pen ' ; in 53, ' Broom ' by ' Flail ' ; and in 80, ' Spear ' by ' Horn.' In 58 he recants his recantation, passing in successive articles from ' Hailstones ' to ' Rain- drops,' and then to ' Stormclouds.' Within five years I have modified my own views of as many problems.! Nothing, therefore, seems more unwise than lengthy and strenuous dogmatizing over opinions which may to-morrow be abandoned by their champion. FORM AND STRUCTURE OF THE EXETER BOOK RIDDLES Since the explosion of the attractive legend of Cynewulfian author- ship, it has been obviously impossible to ascribe with confidence all the riddles of the Exeter Book to a single enigmatograph, although many of them must have come from one hand. They therefore belong to quite another class than the groups of Anglo-Latin problems of the eighth century, each of which is associated rightly with one great name, and in each of which the order is that of composition. Attempts like that of Prehn § to establish for the English poems any unity of purpose in choice of subjects and material have been signally unsuccessful. But it is equally wrong to regard this collection, with Bulbring|| and Herzfeld,1T as a glean- ing of folk-riddles, like, for example, that of Randle Holme. As I have already pointed out, ft our problems are art-riddles (Kunstratsel} with a large alloy of popular elements. Their author or authors, like the Ger- man enigmatographs of the sixteenth century, drew quite as freely from See Brandl, Pauls Grundriss^ II, 972. t Rid. 9, 18, 26, 28, 38, 49, 56, 58, 74, 81, 86, 90. \ Rid. 26, 31, 37, 42, 53. § Pp. 148 f. || Litt-BL, 1891, Sp. 156. ^Herrigs Archiv CVI (N. S. VI), 1901, p. 390. P.M.L.A. XVIII (1903), 211 f. \M.L.N. XVIII (1903), 97f. FORM AND STRUCTURE OF THE RIDDLES Ixxxv myth and tradition as from learned sources. In the runic riddles f appeal is made to a ' bookish ' audience ; t but the riddler, here as well as elsewhere, composes with his eye not only on his subject but on the puzzled faces of men who will listen to his dark sayings. Prehn § believes that oral transmission of the Riddles is firmly estab- lished by the ' Wandering Singer ' interpretation of Rid. 95, and we may sacrifice this solution || without abandoning his conclusion. Ample evi- dence of the truth of this is found not only in the passage from Rid. 43 already cited, but in many other places in the poems. One indication of such direct address certainly lies in the opening and closing formulas, that make an immediate appeal similar to those in the f oik-riddles. If Or let us note the thirstily hinted hope of reward near the close of the second Horn riddle. Frequent references to the wine-hall ft seem to mark this as the scene of the riddles' propounding and solving. The ^different versions of Rid. 31 and 36 point to oral transmission. \ But the highest proof of directness of appeal lies in the epic nature of the treatment of manifold themes, as Dietrich recognized. §§ This will be Folk-lore and mythology are freely invoked in the riddler's treatment of the singing feathers of the Swan (8), the ingratitude of the Cuckoo (10), the strange origin of the Barnacle Goose (n), the metamorphosis of the Sirens (74). t Nos. 20, 25, 43, 59, 65, 75. } 437, / dm J>e bee witan, means, as the context clearly shows, ' those who know letters or rune-staves," but they are rather hearers than readers; ic on flette mag \ J>urh runstafas rincum secgan. § P. 147. || I have proved, M. L. N. XXI (1906), 104-105, that the last riddle is a mate to Rid. 30, and refers to the wanderings of the Moon. T Prehn, p. 1 52, points to a1, ag12, S223, 3313, 3&13, 3712, 4O28, 42, 44", so8, 6o15. Oft ic wofrboran ivordleana sum \ dgyfe after giedde (So9"10). It is significant that ivofrboran is applied to riddle-kenners (32124) and that gieddes is the word for a • riddle ' (56"). tt 4315~16» Nit is tindyrne\ werum at wine. Cf. also ai12, 471, 561, 5711, 6i9, 643, 6817. In the last of these examples, J>ier guman druncon has no particular bearing upon the subject of the riddle, and is justified only by the riddler's surroundings. It Ago/tor Agob (241) seems a mistake of the ear. §§ Haupts Zs. XI, 448 : ' Wo das Epos, sei es im Gleichnis oder im unmittelbaren Dienst seiner Geschichte, Naturgegenstande beschreibt oder durch Umschrei- bungen andeutet, nahert es sich dem Ratsel, nur dass es den Namen dazu im ersteren Falle nenrtf ; umgekehrt bewegt sich das wahrhaft poetische Ratsel nach den Kreisen des Epos hin, wenn der Gegenstand des Ratsels, sei er der elementaren Natur oder der belebten, durch Menschenhand umgeschaff enen, ange- horig, erzahlend auftritt, und er selbst oder der Dichter in seinem Namen uns von seiner Heimat, von Vater und Mutter, von Bruder und Schwester, von Ixxxvi INTRODUCTION duly discussed when the form and manner of our poems are con- sidered. But, before such analysis is possible, the significance of sub- ject and matter demands attention. Nowhere does a poet or school of poets proclaim closeness to life more plainly than in choice of themes. And it is here that the preemi- nence of the Exeter Book Riddles over the Anglo-Latin enigmas be- comes immediately apparent. The English poems smack far less of abstractions and of classical and biblical lore than the problems of Aid- helm ; nor are they eked out with liberal borrowings from Isidore's Etymologies, like those of Eusebius. Nothing human is deemed too high or low for treatment, and all phases of Old English existence are re- vealed in these poems ; f so that they stand forth as the most impor- tant contemporary contributions to our knowledge of the everyday life of their time. The poet does not hesitate to treat the cosmic aspects of nature, the changing forms of sea and sky, of wind and wave, in the greatest of the riddles, the Storm-cycle (2-4) ; nor to embody into an exquisite myth the battle of Sun and Moon \ or the fierce onset of the Iceberg (Rid. 34) ; but, with a few such exceptions,! the Riddles are very close to solid earth. The larger number is devoted to man and his works: his weapons, || his implements of home and field, IT his seinen Schicksalen nach seiner Vertreibung aus der Heimat, von seinen Thaten und KUnsten, von Kampfe'n und Arbeiten, von Lust und Leid in lebendiger Schilderung berichtet.' It is significant that the Anglo-Saxon enigma of the Creation is a fairly close rendering of Aldhelm's De Creatura, adapting, however, its classical allusions to the lay understanding (see notes to Rid. 41). Rid. 44, ' Body and Soul,' and Rid. 47, 'Lot and his Daughters,' are only apparent exceptions to the prevalent popular choice of subjects, since the first moti/was a part of the universal belief, and the second a commonplace of riddle-poetry. f Brooke, Eng. Lit. from the Beginning, p. 159. t Contrast with this human handling of elemental conflict {Rid. 30) Aldhelm's frigid lines upon the relation of the two luminaries. § Note also the ' Creation ' cycle (41, 67, 94), the riddles of Sun and of Moon (7, 30, 40 ?, 95), and those of Water (31 ?, 42 ?, 84). || See the riddles of ^Shield (6), Ballista (18), Sword (21), Bow (24), Mail-coat (36), Battering-ram (54), Sword or Dagger (71), Spear (73). The Sword plays an important part in Rid. 56. IF Compare the riddles of Plow (22) and Rake (35) and Flail (53), of Lock and Key (45. 9i)» of Loom (57), of Oven or Churn (55), of Poker (63), of Beaker (64) and Drinking-horn (15, 80) and Leather Bottle (19 ?), of the Bellows (38, 87). We may add to these such essentials of life as Ship (33, 37), Anchor (17), Well (59), and Weathercock (81). The chariot or wain is introduced into Rid. 23. FORM AND STRUCTURE OF THE RIDDLES Ixxxvii clothes, many of his instruments of music, f his books and script, t his sacred emblems, § and even his food and drink. || Not only man, but the lower animals, fish, flesh, and fowl, receive ample treatment. Many beasts, IT birds, fishes, ft and even insects $ t play a lively part in the Riddles. The plant-world of tree and flower § § is not neglected. So wide is the range of our poems. Rid. 62 is probably a song of the Shirt, and the Glove is ' the skin ' of Kid. 14. Shoes are mentioned in Rid. 13, and the hrccgl and cyrtel in the obscene riddles (45- 46, 55. etc.). t See the riddles of Bell (5), JEforn (15, 80), Bagpipe (32), Reed.flute (6I1-10), and Shawm (70). } Compare the two ' Book ' problems (27, 68), the enigmas of Bookmoth (48) and Bookcase (50), and finally the riddles of Pen and Fingers (52), Reed-pen (6i10-17), and Inkhorn (88, 93). § See the riddles of the Cross (31 ?, 56) and those of Paten (49) and Chalice (60). The ' Book ' problems (27, 68) refer to Holy Writ. || Note the ' Dough ' riddle (46) and the reference to Bread or to Butter in the last lines of Rid. 55. There are problems of Mead (28) and Beer (29), and the chief motif of the ' Night ' enigma (12) is vinous revel. Enigmas of the wine-cup, and the many references to the wine-hall, have already been indicated. IT Badger (16), Steer (13, 39), Horse (20, 65), Ox (72), Dog (75), and Lamb and Wolf (90) are subjects of riddles; while the Stag (88, 93), the Boar (41), and the Swine (41) are described at length. Of the uncanny things of everyday life, such as reptiles and fungi, perhaps the only example is the fen-frog of 4i71. Closely bound together are the Bird riddles, those of Swan (8), Jay (9, 25), Cuckoo (10), Barnacle Goose (n), and Swallows (58). Cock and Hen (43) and Hawk (20, 65) are the themes of runic riddles. Other birds are mentioned, the eagle, kite, goose, and sea-mew in Rid. 25, the puzzling pernex in Rid. 41 (see note to 4I66), and the raven in Rid. 93 (note to gs'26). tt Fish and Flood (85) and Oyster (77 ; cf. 78) are riddle-themes ; and the Whale (4 192-94) receives passing notice. tt The Bookmoth has a riddle to itself (48) ; a picture of the Bees introduces the ' Mead ' riddle (28) ; and the snail, the weevil, the rain-worm, the hand-worm, the tippula, all appear in Rid. 41, while Rid. 36 shows a knowledge of the silk- worm. Zupitza (Haupts Zs. XXXI, 49) compares with the riddler's reference to the tiny size of the hondwyrm (41 ; cf. Aldhelm's Latin) the close parallel in the ' Wen ' charm at the end of MS. Royal 4. A. XIV, miccle lesse, alswd dnes hand- •wurmes hupebdn ; and he recalls Shakespeare's picture of Queen Mab's wagoner (R. & J. i, 4, 65), 'a small gray-coated gnat, Not half so big as a round little worm Prick'd from the lazy finger of a maid (man).' §§ The Beech (92, 4i106) is the only tree to which an entire riddle is devoted; but Ash and Oak are mentioned as runic names in Rid. 439"10, and Yew, Maple, Oak, and Holly appear in Rid. 56® ]". The tree in the forest is pictured in 31, 54, and 73. A general description of plants and flowers is found in Rid. 356~9, 7i2"8; the Reed (61), the Onion (26, 66/, and the Garlic-seller (86) are riddle-subjects; Ixxxviii INTRODUCTION All these riddles, whether the subject be animate or inanimate, have at least one common characteristic, their human interest. This is evinced in a dozen striking ways : but by far the most important of these is a trait of our problems, missing in other collections, but so strongly marked here as to suggest a common origin for many of the riddles — the trait of utility. The riddler may neglect place and form and color of his sub- I ject, but he constantly stresses its uses to mankind. Indeed, men are in the background of every riddle-picture ; f and the subject is usually viewed in its relation to them. The most significant expression of this relation is found in the motif of Comitatus, or personal service of an underling to his lord and master, that forms the dominant idea in many of our poems, t Sometimes the relation or service is of a humbler kind.§ Rid. 29 tells of the reaping and threshing of the barley ; and we hear of the sea- weed washed up on the beach in 38, 4i49. Into the Creation enigma (41) lily and rose and wormwood are all introduced. Mark the appearances of nyt : 26-, neahbu(e)ndum nyt ; 27^, nijnim t5 nytte ; 339, moncynne nyt ; 353, hyre set nytte ; so9, him t5 nytte ; si2, dryhtum to nytte ; 55", 56", nyt ; SQ5"6, nyt . . . hyre [monjdryhtne ; yo6, haelejmm to nytte. It is cer- tainly significant that in the translation of Aldhelm's Creatura such phrases as leaf moncynne (4I27) and mare to monnum (4I46) have no equivalent in the Latin. Leather (13), Horn (15), Book (27), Mead (28), and many other things recount with pride their manifold uses. 1 28, waelcwealm wera ; 66, mid jeldum ; 73, unrimu cyn ; 8s, ofer haele^a byht ; g5"6, eorlum ... in burgum; i8n, men gemunan ; ig2, masldan for mon- num ; 2i12, for mengo ; 2410, gumena hwylcum ; 281, weoriS werum ; 3O13~14, nienig . . . wera; 3i6, weras ond wlf ; 32", werum on wonge ; 3312~13, guman bruca'S | rice ond heane ; 3411'12, JEldum . . . firum on folce ; 351, in wera burgum ; 3612, for haele^um ; etc. t Compare Rid. 22-1-15, 3U~15 ; 41, min frea ; 41-16 ; 4s6, mines frean ; 472-74 (each of these Storm-riddles closes not only with formula, but with relation to lord) ; 51-9, tegne mlnum ; 5, hlaford; 75, mm frea (Crist') ; i85, frea; 2 12, frean mlnum; 2i4, waldend ; 2i23, from bam healdende }>e me hringas geaf ; 2I24, frean; 2I26, mlnum J>eodne ; 2I29"30; 223-15, hlaford mm; 2214, |>enaj>; 246, se waldend; 382, )>egn folgade ; 445, esne )>enaS ; 44®-10, gif Se esne | his hlaforde hyreft yfle | frean on fore ; 452, frean (= esne) ; 50, se wonna )>egn ; 557-, )>egn . . . esne ; 5610, frean ; 5618, his mondryhtne ; 5711, mmum hlaforde J>iEr haeleS druncon ; sg6 hyre [monj- dryhtne ; 5g13-14, hlafordes gifum, hyrei? swa J>eana | >eodne sinum ; 62s-, frean . . . holdum )>eodne (see notes for wifely service) ; 7i9, dryhtne mm . . . ; 73, frean mines ; So1"8, ae)>elinges eaxlgestealla, | fyrdrinces gefara, frean minum leof, \ cyninges geselda; 872, J>egn folgade; gi6, frean mines; gi9, min hlaford; gtf, Frea mm ; g35, frea. § The creature is ruled by the hands of a woman in Rid. si5, of a lord's daughter in Rid. 46, of a queen or earl's daughter in Rid. So3-5, of a churl's daughter in FORM AND STRUCTURE OF THE RIDDLES Ixxxix Again, the immediate effect of the unknown thing upon man is described with spirit. Thus in one way or the other the close connection of the riddle-subject with mankind is revealed. In a still more potent fashion is life lent to the themes of our poems. Not only do the subjects of over half the problems (fifty) speak in the first person f as in the Latin enigmas, not only is grammatical gender I sometimes invoked to the riddler's aid,t but in many riddles the subject \is quickened into full life. The riddler points to the living souls of his Rid. 266, of a dark serving-woman ( Wale) in Rid. 13 and 53 ; it is guided by a swart herdsman (Rid. 7210),-and is turned by a priest (60). Rid. 26, 28,. 29. t Rid. 2, 3, 4, 5, 6, 7, 8, 9, 10, n, 12, 13, 15, 16, 17, 18, 19, 21, 22, 24, 25, 26, 27, 28, 31, 36, 41, 61, 62, 63, 64, 66, 67, 71, 72, 73, 74, 77, 78, 79, 80, 81, 83, 85, 86 (mixture of ist and 3d persons), 88, 91, 92, 93, 95. It is perhaps significant that of the last thirty problems of the first group (Rid. 1-60) the only two that employ the first-person subject (Rid. 36, 41) are direct translations from Aldhelm. } The importance of grammatical gender in determining the sex of the riddles has been greatly exaggerated by both Cosijn (PBB. XXIII, 129) and Trautmann (BB. XIX, 181), who quite unwittingly are harking back to the mythological theories of Max Miiller. In many Riddles, small account can be taken of this by reason of three common conditions, (i) The wiht of the opening lines leads to the use of feminine pronouns throughout the problem : 3o5'8'10, 326, 345'8, 353>6'7> 372>8> 4o6'7.8'10>etc- 576, 594t6, 684, 876. In two cases the gender of wiht is more potent than that of the subject, even though the creature is named explicitly : 247, lengre (241, Boga) ; 257, glado (257~9, Higord). (2) The natural gender of the creature is determinative : i313, sweartne (Steer) ; i67, onhiele (Badger mother) ; 3g2'7, him, he (Bull) as contrasted with 3g6, hio (wiht) ; j2?,yldra (Ox). (3) The masculine and feminine genders are applied indiscriminately to the subject : 4I27, strengre, 4I26, wriestre, 4i28, betre, 4I38, hyrre, and 4i42, yldrt, 4I50'51, brizdre and wldgielra, 4I54, keardra, 4i57, hdtra, 4I58, szvetra, etc. ; 671>2'3, mare, lizsse, leohtre . . . swiftre, and 6710, me sylfum ; 363, mec beworhtne, and Leid. 3, mec biwortha; ; 7O1, hyre, and 70, his ; 8s1, sylfa, and 8s8-4, swiftre . . . strengra. Yet there are not lacking indica- tions of grammatical gender — upon which, however, it is unsafe to lay undue stress, in the light of the appearance of the neuter water as modor monigra wihta (844), to whom, however, masculine adjectives are applied (S485) ; of the relation of masculine pronouns in 2O1"8 to HORS; or of the inaptness of masculine reodne (268) to the Old English synonyms of Onion (leac, cype, etc., none of them mascu- line words). Why infer that the use of dnhaga (61), wapenwiga (is1), eaxlgestealla (So1), mundbora (iS1), has any reference to the masculine gender of Shield and Horn and Ballista ? There remain these examples: i79, mec stit>ne (Anchor) ; ai5, me widgalum (Sword) ; 229'15, me gongendre . . . hindeweardre (Plow, syUi) ; 385>fi'8, he . . . him . . .fader (masc. in spite of wihte ; but the same subject is fern, in Rid. 87) ; so2, deafne dumban (Bookcase) ; si1'3'4, wiga . . . J>one . . .forstrangne (Fire); 635, mec . . . ceftanweardne (Poker) ; 64, feminine (Beaker) ; 733, me . . . frddne XC INTRODUCTION creatures, or else he follows the far more effective method of ascribing to beasts or even to inanimate things the traits and passions of men.f The poems extol in their subjects such essentially human qualities as heroic valor and prowess,! the love of family and friends, § the joy of good works, || grim hatred and malice towards mankind, IF the loneliness of celibate and exile, wisdom and ignorance, ft earthly fame, It and pride of place § § ; or else they dwell sadly and sympathetically upon the (Lance) ; 773'9, febelease . . . unsodene (Oyster) ; Si1-11, belcedsweora . . . fryrel- wombne (Weathercock) ; 8821>24, dnga . . . broborleas (Horn) ; Q315, mec . . . innan- weardne (Horn) ; Q425, hyrre . . . smeare (Creation ?) ; 95, masculine (Moon). As in many of these cases we cannot know what Anglo-Saxon word the riddler had in mind, it is hardly wise to assert even here that his choice of sex was always determined by the grammatical gender of his subject. Rid. ii6, haefde feorh cwico ; i43, hasfdon feorg cwico; 745, haefde feriS cwicu. t Ebert (Berichte iiber die Verh. der k. sacks. Gesellsch. 1877), p. 24, rightly re- marks : ' Was aber denselben einen hoheren poetischen Werth verleiht, jenen Reichthum der Schilderung bedingt und ihre wahre Eigenthiimlichkeit ausmacht, das ist dass das Moment der Personification zu einer bedeutenderen Einfaltung gelangt, indem die Objecte der Rathsel nicht bloss nach ihren Eigenschaften sich schildern, sondern in dramatischer Action handelnd oder leidend sich vorfiihren. Dadurch schreitet die Personification zu menschlicher Individualisirung fort indem Empfindungen wie Leidenschaften den Dingen verliehen werden. . . . Eine solche lebendigere Personification findet sich wenigstens in den besten der angelsachsisch geschriebenen Rathsel.' \ Not only is the Badger (16) a brave fighter against her foe, ' the death-whelp,' but Storm (2, 3, 4), Sun (7), Horn (15), Anchor (17), Moon and Sun (30), Iceberg (34), and Loom (57) are also mighty warriors : even the Mead (28) accomplishes 'sovereign overthrow.' The Weapon riddles are naturally full of this spirit. § The Riddles pass in review the love of a mother for her children in their pictures of Cuckoo, Badger, and Water (10, 16, 84), fraternal devotion in the ac- count of the lonely Stag-horn (88), the love of wife for husband (62), and the passion of the wooer in the caresses lavished upon the Beaker (64). || Rid. 27, 31, 35, 49, 60, 68, 84. 1 Ballista and Bow (18, 24) are full of poisonous spleen, and the Iceberg (34) is hetegrim. The Sword bemourns its lack of wife and children (2I20-27), the Ore vaunts its aloofness (8s12-14), and the Moon wanders sadly far from men (so10-18-14, 4O8-9, g54.10f.). tt The Moon reveals wisdom (gs8'9), and Bookmoth and Bookcase are unwit- ting of the contents of books (48, 50). \ Both Sun and Moon are widely known to earth-dwellers (30, 95). §§ Battering-ram and Lance (54, 73) .chant their early beauty, and the Horn sings of its happy days on the stag's head (93). Qll FORM AND STRUCTURE OF THE RIDDLES xci sufferings of the strange creatures, and, sadder still from the Germanic viewpoint, their inability to wreak revenge upon their foes. Our riddles not only thus run the gamut of the ordinary human emo- tions, but they range from pole to pole of the English social life of their time. Some of them move in a world of high breeding and courtly usage, of lofty tone and temper like that of the Beowulf and the heroic verse f — a world in which warriors shake their lances in the battle \ and receive upon their shields the brunt of falling blows, § or extol their highly adorned swords in the wine-hall ;|| in which fair-haired women of rank bear the drinking-horn at the feast, IF arm their lords for the fight, and chide the swords that lay the heroes low. ft Many others are upon a plane of every- day life and action, of humble trades and occupations,^ while a few de- scend into the depths of greasy double entente. § § Yet the line between high and low is not sufficiently distinct to indicate a different origin for riddles of different genre, inasmuch as a transition from one class to another sometimes takes place within the compass of a single problem. || || The Riddles do not confine themselves to things of earth. The spiritual life of the early English finds expression in a few of the poems. It is significant, as an indication of this religious feeling, that the classical mythology of Aldhelm's De Creatura is, in every case, Christianized and Germanized by his translator,!]"!" who exalts as shaper •The Shield (6), Sword (21), Book (27), Barley (29), Battering-ram (54), Ox (72), Lance (73), Weathercock (81), Ore (83), and Stag-horn (88, 93), are the chief sufferers. In Rid. 21, 83, 93, the absence of revenge is a prominent motive. t See Brooke, Eng. Lit. from the Beginning, p. 159. Brandl, Pauls Grundriss II, 972, notes that the Riddles are courtly, that they are steeped in the colors of the heroic epos. \ Rid. 73, 92. § Rid. 6, 71. || Rid. 2i9-18. f Rid. 8o8 6 ; cf. is8-9. Rid. 62. This interpretation is very doubtful (see notes). \Rid. 2 182'85. tt Such are the riddles of Plowman (22), Oxherd. (72), Thresher (53), Onion- parer (26), Garlic-seller (86), Bell-ringer (5), Weaver (36, 57), Smith (38, 87), Flute-cutter (61), Bread-maker (46), Butter-maker (55). Cf. Brooke, Eng. Lit. from the Beginning, p. 160. \Rid. 26, 45, 46, 55, 62, 63. || || For instance, Rid. 62 begins on an elevated plane, and plunges into obscene jest, while hiultloc as applied to the Hen in Rid. 43 suggests a burlesque of epic phrase. Yet one can hardly follow Trautmann in assigning Rid. 18, a mate in tone anr' temper to the warlike ' Bow ' riddle (24), to the Oven. TTT jee notes to Rid. 41. Cf. Prehn, p. 213. xcii INTRODUCTION and ruler se ana god • Here, as in several other riddles,t the creation is seemingly assigned to the Father alone ; but in one passage the work of shaping is ascribed to the Son \ as in Cynewulf's Christ, and in another to both the First and Second Persons. § God is elsewhere described by both usual and unusual epithets, || and, as often in the poetry, Heaven is praised as the land of glory, the abode of the angels, the fortress of God.1I The beauty of God's Word, the saving grace of prayer,tt and the wonder-working power of the Eucharist \ are extolled. Sacred vessels,^ Cross, §§ and perhaps Holy Water || || are reverently im troduced as riddle-subjects. The Body and Soul legend finds a place,iriT and dim Apocalyptic allusions obscure the difficult Latin riddle. Despite this Christian element, Brooke is not wholly wrong in declar- ing : ttt ' The Riddles are the work of a man, who, Christian in name, was all but heathen in heart. . . . They are alive with heathen thoughts and manners. The old nature-myths appear in the creation of the Storm- giant, who, prisoned deep, is let loose, and passes, destroying, over land and sea, bearing the rain on his back and lifting the sea into waves. . . . They appear again in the ever-renewed contest between the sun and the moon, in the iceberg shouting and driving his beak into the ships, in the wild hunt in the clouds, in the snakes that weave [?], in the fate god- desses [?], in the war-demons who dwell and cry in the sword, the arrow, and the spear [?] ; in the swan, who is lifted into likeness with the swan- maiden [?], whose feathers sing a lulling song. . . . The business of war, Barnouw has an interesting note (p. 219) upon the use of this phrase (4i21) : ' Die bedeutung kann hier nur sein, " der Gott allein, der u. s. w.," und nicht " der Eine Gott, der u. s. w.," weil in diesem falle nur se dn GWmoglich gewesen ware (vgl. 8410 an sunu, Guth. A. 3723 se an oretta; Gen. B. 235 J>one jgnne beam). Bei dieser einzig moglichen auffassung verrat der christliche dichter seine noch heidnisch gefarbte anschauungsweise, welche wohl nicht der einfluss seiner klassi- schen kenntnisse, sondern die nachwirkung des alien volksglaubens sein wird. Hochstwahrscheinlich haben wir hier also ein sehr altes ratsel.' t 8s'2, unc drihten scop ; 8817, unc gescop meotud. ' \ 71-2, Mec (Sunne) gesette soft sigora waldend | Crist to compe. § 849-10, fyrn forSgesceaft ; faeder ealle bewat | or ond ende, swylce an sunu. I! 4021, wuldorcyninges ; 4i3, reccend . . . cyning . . . anwalda, etc. ; 4g5, helpend gSsta ; 6o4, god nergende ; 6o6, Hielend. If Rid. 678, 6o15-16. Rid. 27, 68. tt Rid. 6o13f . \ Rid. 49, 60. Oblation and Consecration in these riddles recall the Canon of the Mass in the Sarum and York Missals. §§ Rid. 56; see Rid. 31. || || Rid. 31^-9 (?). Cf. S438. HI Rid. 44. » Rid. 90. H1- Eng. Lit. from the Beginning, pp. 158-159. FORM AND STRUCTURE OF THE RIDDLES xciii of sailing the ocean, of horses, of plundering and repelling plunderers, of the fierce work of battle, is frankly and joyfully heathen.' Brandl goes to the other extreme : ' Die Auffassung hat nichts heidnisches oder antiheidnisches mehr, nicht einmal etwas mythisches.' In the first pages of this Introduction I have indicated the place of myths in the Riddles. Careful analysis of our Old English art-riddles yields few indications of adherence to any normal form or plan, such as that derived by Petsch f from his study of riddles of the folk. Yet it is not unprofitable to trace in our problems the appearance of each of the divisions that compose humbler and more popular puzzles. The introductory framing element in folk-riddles consists of three parts : simple summons to guess, the stimulating of interest by the mention of person- or place- names, and the indication of the place of the subject. The first of these is represented in the Exeter Book collection by the large number of opening formulas, elsewhere considered, and in one case by a query. $ The second is not found, but the third is very common, and takes two forms : sometimes being limited to a phrase of little import, sometimes extending into the body of the riddle § and constituting one of its chief motives. Of the use of proper names in the naming germ-element there is hardly a trace, || as the Riddles make no attempt to assign to their subjects a local habitation. But the runic riddles (see Solutions) are partly name or word problems. Description in the enigmas is of vari- ous kinds : in the ' monster ' riddles, H detailed enumeration of physical peculiarities ; in the obscene poems, an indefiniteness of indication Pauls Grundriss II, 971. t Palaestra IV, 50 f. t Rid. 21"2, Hwylc is haelej>a J>aes horse ond t>aes hygecraeftig | )>ast }>aet maege asecgan, etc. The formula-beginnings arouse attention by stressing the strange- ness or importance of the subject: ai1, 251, 261, so1, 321, 331, 371"2, 691'2, yo1, etc. § Examples of the first are 341, aefter wege ; 351, in wera burgum ; 371, on wege ; 461, on wincle; 551, in \vincle ; 561, Go1, in healle ; 861, J'aer weras sieton — these phrases cast little light upon the subject. Examples of the second are the watery home of the Barnacle Goose (n), the abodes of honey (28), the fields of barley (29), the mines of metal (36, 71), the threshing-floor of the Flail (53), the groves from which sprang Ram and Lance (54, 73), the marshy tidewater where the Reed grew (61), the sea that fed the Oyster (77), the stag-head that bore proudly the Horns (88, 93), — all valuable aids to the solution. || 63°, s»J>erne secg, and 72", mearcpafras Walas, are only seeming exceptions. ^ Rid. 32, 33, 35, 37, 59, 70, 81, 86. A'iaf. 265, neo^an run nathwaar; 4&1, weaxan nathwaet ; 629, ruwes nathwaet; 63, on nearo nathwjer. xciv INTRODUCTION frequent in Volksratsel. Sometimes the subject is described as a whole through one trait ; but usually through several distinguishing features, f As in the riddles of the Hervarar Saga,\ four characteristics of the subject receive attention: color,§ formj number-relation, If and inner nature. A wide range of vision, quick observation, and generous sym- pathy mark all the descriptive work of our collection. The narrative element in the Exeter Book Riddles is far larger than the purely descriptive. In many of the problems description is immedi- ately succeeded by narration,! t or else is wholly superseded by this.lt So under this head of narration, or the artistic treatment of action, may be considered a few of the dominant motives of our collection. One or two of these — the relation of the subjects to mankind, their human traits and poignant sufferings — have already been indicated. There remain others familiar to the student of riddle-poetry. The first of these themes is a change of state, by which the creature is bereft of early joys and woe is entailed upon him.§§ So the contrasts between youth and later In two cases this method limits the problem to a single line : 6g3, Wundor weariS on wege : waeter wearS to bane ; 751"2, Ic swiftne geseah on swa>e feran | D N U H. But several riddles are devoted each to the elaboration of a single characteristic : the warlike spirit of the Anchor (17), the mimetic power of the Jay (25), the saving grace of the Communion Cup (60). t The ' Beech ' riddle (92) is but a series of kennings, and the ' Horn ' enigmas (15, 80) mark out the various uses of the subject. The cruelty of the Iceberg (34) is supplemented by an account of its mysterious origin ; and the strange traits of the Weathercock (81) by a picture of its misery. t See Heusler, Zs. d. V.f. Vk. XI, 147. § Notably in the pictures of the array of the Barnacle Goose (n), of Night's garment (12), of the Badger's markings (16), and of the Swallow's coat (58). II Cf. 19, 221-llf-, 32, 33, 35, 37, 38, 45, 53, 56 (substance), 58, 81, 86, 87, 91. T See 14, 23, 47. This has already been discussed at sufficient length in connection with the human element in the Riddles. tt Rid. 6, 12, 14, 16, 18, 21, 22, 26, 28, 29, 31, 32, 33, 39, 45, 50, 51, 52, 54, 56, 58, 59. 63. 67, 70, 71, 72, 74, 80, 81, 84, 87, 91, 95. tt Rid. 2, 3, 4, 5, 7, 8, 10, 13, 17, 20, 23, 27, 43, 46, 48, 55, 57, 61, 62, 66, 77, 83, 88, 93. In several riddles, pure description is limited to a single touch : 242, wraitlic ... on gewin sceapen ; 648, glaed mid golde. §§ The Ram and Lance, deadly weapons, extol their joyous life in the fore.st (54> 73) I the Ox, goaded by the black herd, bewails its pleasant youth (72) ; and Honey (28), Barley (29), Reed (61), Oyster (77), Ore (83), and Horn (8 t/ all point to the happy days before they fell into the shaping hands of rr X ,;eah ic mu)> haebbe ; 485, Staelgiest ne waes | wihte J>y gleawra J>e he J>am wordum swealg ; 49a~2, [Jerjen- dean . . . butan tungan ; 6i9, mufileas sprecan ; 661, cwico . . . ne cwaeft ic wiht. Cf. 349'10, 388- H Notice the large number of these in the ' Storm ' riddles (2-4) and in dozens of others (30, 52, 74, 85, etc.). It is not surprising that the periphrastic preterit formed by the preterit of cuman (com(on)), + an infinitive of motion, which occurs only twice in Cynewulf (Jul. 563, Chr. 549), appears four times in the Riddles (231, 341, 551, 861). IT This is strikingly illustrated by the past participles of Rid. 29 and by the terseness of the obscene riddles. Such endings as those of Rid. 5, 29, 32, 33, 36, 40, 43, 44, 56, 68, 73, 84, recall the phrase of the folk : ' He is a wise man who can tell me that.' xcvi INTRODUCTION VI THE MANUSCRIPTS The.JSxefer Book, most famous of all Leofric's donations to the new cathedral of the West, has already been so carefully described in another volume of this series that we need consider now only the place of the Riddles in this celebrated codex. These enigmas occupy three different portions of the manuscript : f . i oo b-i 1 5 &(Rid. 1-60 inclusive) ; f . 1 2 2 b- \2^(Rid. 31 b, 61) ; f. i24.b-i3ob (./?/</. 62-95). Unfortunately for the student, of the Riddles, it is these final pages of the Book, otherwise so well-preserved, that have suffered threefold damage : (1) The last twelve leaves have been burned through by a piece of ignited wood which appears to have fallen upon the Book. The damaged places have a like shape upon all the leaves, decreasing, however, in size to the inner part of the codex, until on f: 118 b only one small burn is visible, t This serious accident has impaired or reduced to fragments all riddles at the middle of these injured pages : 31^(122 b), 647"16 (125 a), 681U (125 b), 7i7-10 and 72™ (126 a), 738-20 (126 b), 7f'& and 78 (127 a), Si10'12 and 82 (127 b), 84n-19 (128 a), 8442-5 (128 b), 87" and 881-11 (129 a), 88s-85 and 89 (!29b), 92" and 931'6 (130 a), 9328~82 and 94 (130 b). (2) A page is certainly missing after f. 1 1 1. Rid. 41 (i 1 1 b, bottom) breaks off suddenly in the middle of a sentence (1. 108), and Rid. 42 (ii2a, top) begins with equal abruptness. It is probable that a page has been lost after f. 105, as Rid. 21 closes abruptly at the bottom of the page without a closing-sign. (3) The last leaf has been stained on its outer side (i3ob) by the action of a fluid on the ink. A few words have thus been rendered almost illegible (9in, 9322). The first and greatest of these injuries has occasioned the use of strips of vellum for binding together the damaged half-pages. In course of time, these strips have become loosened ; and, by peering beneath them, I have been able to read many letters and even words not visible to Schipper and Assmann.1: These I have duly included in my text. Cook, The Christ of Cyneivulf, pp. xiii-xvi. t See Schipper, Germania XIX (1874), 327 ; Trautmann, Anglia XVI, 207. t So also Trautmann, I.e. THE MANUSCRIPTS xcvii It is surprising that the chief aid to the study and reconstruction of the defective passages has been neglected by all students of the text of the Riddles. This is the facsimile copy made for the British Museum by Robert Chambers from 1831 to 1832. Despite Willker's slighting criticism,! the transcript has great value, not only because it is in the main very trustworthy, \ but because it preserves letters and words which are now obscure or invisible. § I have collated it carefully with my text. Discovery of hitherto unobserved letters in the Exeter Book itself, and the fairly rich yield of the British Museum transcript, constitute potent arguments against daring emendations of the greatly-damaged text — emendations which rest upon nothing but the ingenious fancy of the reconstructionist, and which are in nearly every case ruled out of court The fly-leaf of the Exeter Book bears, at the bottom of the page, this note of the Chapter Clerk : ' In 1831 this Book was entrusted to the British Museum for the purpose of being copied for that institution, and returned October, 1832.' And the facsimile, which is known as Add. MS. 9067, is approved by Sir Frederic Madden in this comment upon its fly-leaf : ' The whole of the present transcript has been collated by me with the original MS. belonging to the Dean and Chapter of the Cathedral of Exeter. Frederic Madden, Asst. Keeper of the MSS. Brit. Mus., Feb. 24, 1832.' We learn from Thorpe's Introduction to his Codex Exoni- ensis (p. xii) that the original manuscript was brought back to Exeter in time for his use. Nothing, therefore, could be farther from truth than Brandl's surprising statement (Pauls Grundriss2 II, 946) that 'Thorpe's text (Codex Exoniensis) is based upon the transcript by Robert Chambers.' t 'Obgleich laut einer Bemerkung in der Abschrift Madden selbst eine Collation der Abschrift mit dem Urtexte 1831-1832 vornahm, ist dieser Text durchaus nicht vollstandig zuverlassig ' (Grundriss, p. 222). t Kemble derives his text of the Traveler's Song (IVidsifr) from this source, which he calls ' an accurate and collated copy ' (Beo-wulf, 2d ed., p. 26) ; and Gn.-W. Bibl. collates it with the codex in its text of ' Vater unser' (II, 2, 227), ' Gebet ' (II, 2, 217), and ' Lehrgedicht ' (II, 2, 280), but neglects it strangely in its text not only of the Riddles but of the Ruin (I, 297), the Husband's Message (I, 306) and the Descent into Hell (III, 176), where it furnishes valuable aid. In the transcript of the Riddles I note only these errors : gefratn for gefrag n (681), ratlice for wratlice (681), J>ine for ftiette (gs22), emu Jxes for eorpes (gs25). The imitation of the upright well-formed English minuscules of the Exeter Book is surprisingly good ; and all gaps due to damage are skillfully indicated. §1 cite only a few of many instances: 2i6, Edd., citing MS. incorrectly, rice\ MS. and B. M. sace ; 725, B. M. oft tc, not seen by Assmann or Schipper, nor by me ; 8i10, B. M. orst . . . eosefr ; 8i12, I read in MS., before sceaft, mat . . ., not seen by Assm., Sch. ; B. M. n ma-t\ 8412, MS., after mce, I read st, not seen by Assm., Sch. ; B. M. meana for weana (Edd.) ; gs'28, MS. oft me, visible to me but not to Edd. ; B. M. oft me. INTRODUCTION by a more thorough study of the manuscript and of the early copy. Three considerations have dictated to editors and critics violent distor- tions of the text of the Riddles. The first of these has been the desire to wrest the reading of the manuscript into accord with some far- fetched solution. As I have already shown,t the text may be without flaw, it may indeed contain a reading confirmed by many parallel pas- sages in the Riddles themselves ; but if it does not accord with the editor's answer of the moment he alters in Procrustean fashion. \ Sec- ondly, a metrical a-priorism that brooks no freedom of verse has naturally led to arbitrary assaults upon the integrity of many passages. § And finally, inability to grasp the poetic perspective of the Old English has caused the unwarrantable rejection of some of the most striking phrases and kennings in our early poetry. || The foolishly named ' curse of con- servatism ' is far preferable to the itch of rash conjecture.il I have there- fore sought to show due respect to a text which in its undamaged portions is excellent, and have emended only with valid reasons. In the manuscript the beginnings of the several riddles are marked by large initial letters, and the endings by signs of closing, : 7 or : — or :— : 7.ft In a fgw cases these indications are lacking. There is no such sign at the end of Rid. 3, which concludes, however, at the bottom of a page (101 a) ; at the ends of 21 and 41, where abrupt terminations indi- cate missing pages ; nor at the conclusions of 43 and 48, each of which is followed on the same line by the opening words of the next riddle. •Almost without exception, Dietrich's suggested readings (Haupts Zs. XI) have been invalidated by reference to the original text. Holthausen is equally unfortunate : manuscript and transcript flatly contradict his emendations of 77, 8i10, 83s, Q328, 947, and confirm his additions only in such obvious omissions as 688 \n\enne (B. M. «<?««<?) and S455 \cynna\ (MS., B. M. cy[nna]). t M. L. N. XXI, 98. I See Trautmann, BB. XIX, 167-215, and note his sweeping changes of text in IlSbJa^ jglla^ jgll^ g^ etc_ § See particularly Holthausen's readings of i62, 25% 551, 8421-22. || Holthausen emends out of existence the interesting heofones toj>e (875) and brunra beot (Q21). See notes to these passages. 1 Sievers utters dignified protest (PBB. XXIX, 305-331) against 'die tendenz bei der behandlung unsrer alten dichtungen personliche willkiir des urteils an die stelle geduldiger vertiefung in die zur rede stehenden probleme zu setzen.' All emendation has its pitfalls, as I have found to my cost. Professor Bright objects with reason to the double alliteration in 7328b of my text, and plausibly proposes Wisan se J>e mine \ \sdf>e\ cunne, saga hwat ic hatte. tt The symbol at the end of Rid. 5 is doubtless a closing sign. THE MANUSCRIPTS xcix Marks of closing are wrongly used after the fifteenth line of Rid. 28 (2816~17, written as a separate riddle, may thus serve to connect the two problems of like subjects, 28 and 29) and after the opening formula of Rid. 69 (which is, however, a useless prefix to the real riddle-germ in the third line). The end of the enigma is sometimes emphasized by the inclusion of its last word or words in a bracket on the next line, as in Rid. 38, 46, 54, 71, 86. The Exeter Book scribe regularly separates compounds whose second member also has a heavy stress. He severs prefixes from their roots and appends them to preceding words, f He even separates the syllables of a simplex. \ Finally, he achieves impossible combinations. § Very few abbreviations are employed by the scribe. || The conjunction and is always represented by the sign ^.1F The ending -um (hwilum, burgum, etc.) sometimes appears as u, and sometimes unabbreviated ; foonne always figures as feon, and fxet frequently as p. f> and 3" are used arbitrarily. ft The uncontracted gerundial form with -ne (to hycganne, to secganne) appears so consistently, even when the meter demands the contracted, tt as to suggest a similar consistency in the earliest version This habit, common among Old English scribes (see Keller, Palaestra XLIII, 51), not infrequently leads to ambiguity: compare iS1, eodor wirum; 23u,fege recced (J>e ger&cefr) \ io7, minge sceapu (mm gesceapii) ; i26, swage mczdde (swa gemtedde) ; 3Q5, mege stede (me ges&de) ; etc. With this last example before him, one may hesitate to accept the form mege (. I have tried to adhere to the use in the codex. ttSee Rid. 2Q12, 32s3, 4O22, 428, etc.; 8829"80, fremman ne ncefre is obviously fremmanne nlzfre. Like Krapp in his edition of the Andreas, I have given in all such cases the inflected form of the manuscript. c INTRODUCTION of the text. The signs or accents (') over vowels in the manuscript fall upon long vowels, and may therefore be regarded as marks of length — save in one or two cases, f The recent readings of the Northumbrian variant of Rid. 36, the so- called Leiden Riddle (see variant notes), unfortunately reached me too late for inclusion in my text, but have been printed by me in the notes, without comment. \ Thorpe, in his Codex Exoniensis, follows the threefold division in the MS., and prints the Riddles in three groups, pp. 380-441, 470-472, 479-500 5 but, as Grein pointed out, '•Riddle /' of Thorpe's second group (p. 470) is merely a variant of Rid. 31, and Thorpe's ' Riddle III' of this division (p. 472) is no riddle at all but the beginning of The Husband's Message. § Thorpe omits from his text six riddle- fragments. Grein || follows Thorpe's reading of the manuscript, and, by drawing four riddles into two, gives us eighty-nine in all. In his notes upon the Exeter Book text, Schipper 1 supplies the missing frag- ments. He is followed by Assmann, who thus swells the number to ninety-five. ft Trautmann %\ regards Rid. 2, 3, 4, as one riddle, and Grein's 37 and 68 each as two. I adhere to the numeration of the Grein- Wiilker text, bracketing, however, ' the First Riddle ' as a thing apart. §§ •These are recorded in Gn.-W., Bibl. Ill, 243. t Gumrinc (87) ; 6 (ss9) ; on (f, 2I29, 226). The mark after / in p'nex (4I66) may be either a macron (Schipper) or an abbreviation-sign (Assmann). J The forms frreaungifrrac and uyndicrczftum (Leid. 6, 9), reported by Dr. Schlutter, are far more apt than the Exeter Book variants, and moreover find abundant support in firdwingspinl, ' calamistrum ' (Napier, 0. E. Glosses, Nos. 1200, 4646, 5328), and in uuyndecreft, 'ars plumaria' (Sweet, O. E. Texts, p. 43, Corpus Gl. 217), to which B.-T. long since pointed in this connection. On the other hand, the meter strongly opposes the new readings of Leid. ia, 8b, I4a-b. § Hicketier, Anglia XI, 364, thinks that the ' Message ' is a riddle ; and, as we have seen, Strobl, Haupts Zs. XXXI, 55, seeks to show that it is a solution of the preceding riddle (Rid. 61), the two forming a Wettgedicht. On the other hand Blackburn, Journal of Germanic Philology III, i, sets forth the pretty and ingen- ious theory that Rid. 61 should not be regarded as an enigma, but should be united with the ' Message ' into a lyric. See my notes to Rid. 61. || Bibl. der ags. Poesie II, 369-407. ^Germania XIX, 328, 334, 335, 337, 338. Grein- Wiilker, Bibl. der ags. Poesie III, 183-238. tt The fragments are Nos: 68, 78, 82, 89, 92, 94. it Anglia, Bb. V, 46. §§ The various editions of single riddles will be cited under this head in my Bibliography. Thorpe, Grein, and Assmann (Grein-Wiilker) furnish the only complete texts. BIBLIOGRAPHY I. THE MANUSCRIPTS THE EXETER BOOK. F. ioob-H5a (Riddles 1-60, inclusive); I22b-i23a (31 i>, 61) ; I24b-i3ob (62-95). HICKES, GEORGE. Linguarum Vett. Septentrionalium Thesaurus Grammatico- Criticus et Archaeologicus, III, 5 (Facsimiles of Riddles 20, 25, 37, 65, 75, 76). London, 1703. CHAMBERS, ROBERT. British Museum Transcript of the Exeter Book (Addit. MS. 9067). 1831-1832. GREIN, C. W. M. Bibliothek der angelsachsischen Poesie. Final page (Facsimile of Riddle 37, after Hickes). Goettingen, 1858. CODEX LEIDEN, Voss Q. 106. F. 24 b. Leiden Riddle (Northumbrian version of Riddle 36). DIETRICH, FRANZ. Commentatio de Kynewulfi Poetae Aetate, p. 27 (Facsimile of Leiden Riddle). Marburg, 1858. SCHLUTTER, OTTO B. Das Leidener Ratsel (Reproduction, critical text, and translation). Anglia, XXXII (1909), 384-388. II. EDITIONS AND EXTRACTS CONYBEARE, J. J. Illustrations of Anglo-Saxon Poetry, pp. 208-213 (Riddles gi-8a) 468-74( 3^ ^ gy( go) London, 1826. MULLER, L. C. Collectanea Anglo-Saxonica, pp. 63-64 (Riddles 6, 27). Hav- niae, 1835. THORPE, BENJ. Codex Exoniensis, pp. 380-441 ; 470-472 ; 479-500. London, 1842. WRIGHT, THOMAS. Biographia Britannica Literaria, I, 79-82 (Riddles 14, 20, 29, 47). London, 1842.! KLIPSTEIN, L. F. Analecta Anglo-Saxonica, II, 337-340 (Riddles 14, 29, 47, 62, 74, 58). New York, 1849. ETTMULLER, LUDOVICUS. Engla and Seaxna Scopas and Boceras, pp. 289-300 (Riddles 3-6, 8, 9, II, 13, 15, 16, 23, 27-30, 32, 34, 36, 38, 61, 80, 86, 33, 47, 67, 20). Quedlinburgii et Lipsiae, 1850. GREIN, C. W. M. Bibliothek der angelsachsischen Poesie, II, 369-407. Goet- tingen, 1858. RIEGER, MAX. Alt- und angelsachsisches Lesebuch, pp. 132-136 (Riddles 3, 6, 15, 27, 30, 36, Leiden, 48). Giessen, 1861. SCHIPPER, JULIUS. Zum Codex Exoniensis. Germania, XIX (1874), 328, 334, 335, 337. 33s- The order of the titles is chronological. The readings of Wright and Klipstein have not been included among my variants, as they are too inaccurate to merit record. ci cii RIDDLES OF THE EXETER BOOK SWEET, HENRY. Oldest English Texts, pp. 149-151 (Leiden Riddle). Early English Text Society 83, 1885. An Anglo-Saxon Reader, pp. 164-167 (Riddles 8, 10, 15, 27, 30, 48, 58), p. 176 (Leiden). Eighth edition, Oxford, 1908. MACLEAN, G. E. An Old and Middle English Reader (on the basis of Professor Julius Zupitza's Alt- und mittelenglisches Ubungsbuch), pp. XXX-XXXI, 4-5 (Riddle 16). New York, 1893. KLUGE, FRIEDRICH. Angelsachsisches Lesebuch, pp. 151-153. (Riddles i, 15, 36, Leiden). 2d ed. Halle, 1897. WULKER, R. P. Bibliothek der angelsachsischen Poesie, III, 183-238, Riddles (edited by Bruno Assmann). Leipzig, 1897. Reviewed by F. Holthausen, Anglia, Beiblatt, IX (1899), 357. TRAUTMANN, MORITZ. Alte und neue Antworten auf altenglische Ratsel. Bon- ner Beitrage zur Anglistik, XIX (1905), 167-215 (Riddles n, 12, 14, 18, 26, 30, 45» 52> 53' 58> 74> 80, 95, 31). Reviewed by Middendorff, Anglia, Beiblatt, XVII (1907), 109-110. III. TRANSLATIONS CONYBEARE, J. J. In his extracts from the text, as above. THORPE, B. J. In his edition of the text, as above. WRIGHT, THOMAS. Biographia Britannica Literaria, I, 79-82 (Riddles 14, 20, 29, 47). London, 1842. GREIN, C. W. M. Dichtungen der Angelsachsen stabreimend iibersetzt. II, 207- 247. Cassel und Gottingen, 1863. BROOKE, STOPFORD A. The History of Early English Literature (Riddles 2, 3, 4, 6, 8, 9, ii, 15, 16, I71"3, 21, 22, 23 paraphrase, 24, 28, 29, 30, 34, SS2"4- 7"9, 36, 39, 4ii8-i9. ioa-107, 52> 54i 56( 57) 58> 6l) 72io-i2, 15-17, 73 paraphrase, 80, 8i6-10, 8815-17, 22-27, Q37-12> 95). New York, 1892. COOK, A. S., and TINKER, C. B. Select Translations from Old English Poetry, pp. 61-62 (Riddle 61, F. A. Blackburn) ; pp. 70-75 (Riddles 2, 3, 8, 15, 24, 27, 28, 80, H. B. Brougham). Boston, 1902. TRAUTMANN, MORITZ. Bonner Beitrage zur Anglistik, XIX (1905), 167-215 (Rid- dles n, 12, 14, 18, 26, 30, 45, 52, 53, 58, 74, 80, 95, 31). WARREN, KATE M. A Treasury of English Literature (from the Beginning to the Eighteenth Century), with an Introduction by Stopford A. Brooke (Riddles 2, 3, 6, 8, 30 ; Wiilker's text with a prose version in Modern English). London, 1906. IV. LANGUAGE AND METER f BARNOUW, A. J. Textkritische Untersuchungen nach dem Gebrauch des be- stimmten Artikels und des schwachen Adjectivs in der altenglischen Poesie. Leiden, 1902. COSIJN, P. J. Anglosaxonica IV. Paul und Braunes Beitrage, XXIII (1898), 128 f. The order of titles is chronological. t The order of titles is alphabetical. BIBLIOGRAPHY ciii FRUCHT, P. Metrisches und Sprachliches zu Cynewulfs Elene, Juliana und Crist. Greifswald, 1887. GREIN, C. W. M. Zur Textkritik der angelsachsischen Dichter. Germania, X (1865), 423. HERZFELD, GEORG. Die Ratsel des Exeterbuches und ihr Verfasser. Acta Ger- manica, Bd. II, Heft I. Berlin, 1890. HOLTHAUSEN, F. Beitrage zur Erklarung und Textkritik altenglischer Dichtungen. Indogermanische Forschungen, IV (1894), 386 f. Zu alt- und mittelenglischen Dichtungen, XV. Anglia, XXIV (1901), 264-267. — Zur Textkritik altenglischer Dichtungen. Englische Studien, XXXVII (1906), 208 f. JANSEN, G. Beitrage zur Synonymik -und Poetik der allgemein als acht aner- kannten Dichtungen Cynewulfs. MUnster, 1883. KLAEBER, FRIEDRICH. Emendations in Old English Poems. Modern Philology, II (1904), I45-M6- - Ratsel XII, 3f. Anglia, Beiblatt, XVII (1906), 300. KLUGE, FRIEDRICH. Zur Geschichte des Reimes im Altgermanischen. Paul und Braunes Beitrage, IX (1884), 422-450. LICHTENHELD, A. Das schwache Adjectiv im Angelsachsischen. Haupts Zeit- schrift, XVI (1873), 325-393. MADERT, AUGUST. Die Sprache der altenglischen Ratsel des Exeterbuches und die Cynewulffrage. Marburg, 1900. Reviewed by Herzfeld, Herrigs Archiv, CVI (1901), 390. SHIPLEY, GEORGE. The Genitive Case in Anglo-Saxon Poetry. Baltimore, 1903. SIEVERS, EDUARD. Zur Rhythmik des germanischen Alliterationsverses, II. Paul und Braunes Beitrage, X (1885), 451-545. Der angelsachsische Schwellvers. Paul und Braunes Beitrage, XII (1887), 454-482. TRAUTMANN, MORITZ. Kynewulf, der Bischof und Dichter. Bonner Beitrage zur Anglistik, I. Bonn, 1898. V. AUTHORSHIP AND LITERARY CRITICISM BLACKBURN, F. A. The Husband's Message and the Accompanying Riddles of the Exeter Book. Journal of Germanic Philology, III (1900), i f. BOUTERWEK, K. W. Caedmon's des Angelsachsen biblische Dichtungen, I, 310- 311. Giitersloh, 1854. BRANDL, ALOIS. Englische Literatur. Pauls Grundriss der germanischen Philo- logie, 2d Ser., II, 969-973. Strassburg, 1908. BROOKE, STOPFORD A. The History of Early English Literature. New York, 1892. English Literature from the Beginning to the Norman Conquest, pp. 87-96, 159-162. New York, 1898. COOK, A. S. Recent Opinion concerning the Riddles of the Exeter Book. Mod- em Language Notes, VII (1892), 20 f. The bibliography of the ' First Riddle ' is not included. civ RIDDLES OF THE EXETER BOOK COOK, A. Si The Christ of Cynewulf, pp. lii-lix. Boston, 1900. DIETRICH, FRANZ. Die Ratsel des Exeterbuches, Wurdigung, Losung und Her- stellung. Haupts Zeitschrift fur deutsches Alterthum, XI (1859), 448, 490, XII (1860), 232-252. ERLEMANN, EDMUND. Zu den altenglischen Ratsel. Herrigs Archiv, CXI (1903), 49 f- ERLEMANN, FRITZ. Zum 90. angelsachsischen Ratsel. Herrigs Archiv, CXV ('90S)' 39i- GREIN, C. W. M. Bibliothek der angelsachsischen Poesie, II, 409-410. Goet- tingen, 1858. -Zu den Ratseln des Exeterbuches. Germania, X (1865), 307-310. HICKETIER, F. Fiinf Ratsel des Exeterbuches. Anglia, X (1888), 564-600. HOLTHAUSEN, F. Zur altenglischen Literatur. Anglia, Beiblatt, XVI (1905), 227-228. JANSEN, KARL. Die Cynewulf-Forschung von ihren Anfangen bis zur Gegenwart. Bonner Beitrage zur Anglistik, XXIV (1908). KRAPP, G. P. Andreas and The Fates of the Apostles. Boston, 1906. LEO, HEINRICH. Quae de se ipso Cynewulf us sive Coenewulfus poeta Anglo- Saxonicus tradiderit. Halle, 1857. Reviewed by Dietrich in Ebert's Jahrbuch fur romanische und englische Literatur, I (1859), 241-246. LIEBERMANN, FELIX. Das angelsachsische Ratsel, 56: 'Galgen' als Waffen- stander. Herrigs Archiv, CXIV (1905), 163. MORLEY, HENRY. English Writers, II, 38, 136-137, 217-227. London, 1888. MULLER, EDWARD. Die Ratsel des Exeterbuches. Programm der herzoglichen Hauptschule zu Cothen. Cothen, 1861. NUCK, R. Zu Trautmanns Deutung des ersten und neunundachtzigsten Ratsels. Anglia, X (1888), 390-394. PADELFORD, F. M. Old English Musical Terms. Bonner Beitrage zur Anglistik, IV. Bonn, 1899. SIEVERS, EDUARD. Zu Cynewulf. Anglia, XIII (1891), 1-2. SONKE, EMMA. Zu dem 25. Ratsel des Exeterbuches. Englische Studien, XXXVII (1906), 313-318. STROBL, JOSEPH. Zur Spruchdichtung bei den Angelsachsen. Haupts Zeitschrift, XXXI (1887), 55-56. TRAUTMANN, MORITZ. Cynewulf und die Ratsel. Anglia, VI (1883), Anzeiger, pp. 158-169. — Zum 89. (95.) Ratsel. Anglia, VII (1884), Anzeiger, p. 210. — Die Auflosungen der altenglischen Ratsel. Anglia, Beiblatt, V (1894), 46-51. — Zu den altenglischen Ratsel. Anglia, XVII (1895), 396 f- Die Auflbsung des elften Ratsels. Bonner Beitrage zur Anglistik, XVII (1905), 142. Alte und neue Antworten auf altenglische Ratsel. Bonner Beitrage zur Anglistik, XIX (1905), 167-215. TUPPER, FREDERICK, JR. Solutions of the Exeter Book Riddles. Modern Lan- guage Notes, XXI (1906), 97-105. BIBLIOGRAPHY CV WALZ, J. A. Notes on the Anglo-Saxon Riddles. Harvard Studies, V (1896), 261-268. WULKER, R. P. Grundriss zur Geschichte der angelsachsischen Litteratur, pp. 165- 170. Leipzig, 1885. VI. ORIGINALS AND ANALOGUES ARNASON, JON. Islenzkar Gatur. Copenhagen, 1887. BRANDL, ALOIS. Shakespeares " Book of Merry Riddles " und die anderen Rat- selbiicher seiner Zeit. Jahrbuch der deutschen Shakespeare-Gesellschaft, XLII (1906), 1-64. BUGGE, SOPHUS. Norr0ne Skrifter af Sagnhistorisk Indhold. Christiania, 1873. BUTSCH, A. F. Strassburger Ratselbuch. Die erste zu Strassburg urns Jahr 1505 gedruckte deutsche Ratselsammlung. Strassburg, 1876. CHAMBERS, ROBERT. Popular Rhymes of Scotland. London, 1870. CHILD, F. J. English and Scottish Popular Ballads. 5 vols. 1882-1898. DUMMLER, E. Lorscher Ratsel. Haupts Zeitschrift, XXII (1877), 258-263. Poetae Latini Aevi Carolini (Monumenta Historica Germanica, I), pp. i f. (Boniface), 20 f. (Lorsch enigmas). Berlin, 1881. EBERT, ADOLF. Die Ratselpoesie der Angelsachsen. Berichte iiber die Verhand- lungen der koniglich sachsischen Gesellschaft der Wissenschaften zu Leipzig, Phil-Hist. Classe, April, 1877, XXIX, 20-56. Allgemeine Geschichte der Litteratur des Mittelalters im Abendlande, I, 603, 653, etc. Leipzig, 1889. ECKART, RUDOLF. Allgemeine Sammlung niederdeutscher Ratsel. Leipzig, 1894. FRIEDREICH, J. B. Geschichte des Rathsels. Dresden, 1860. GILES, J. A. Sancti Aldhelmi ex abbate Malmesburiensi episcopi Schireburnensis Opera, pp. 249-270. Oxford, 1844. GROOS, KARL. Die Spiele der Menschen. Jena, 1899. HAGEN, HERMANN. Antike und mittelalterliche Raethselpoesie. Bern, 1877. HAHN, HEINRICH. Bonifaz und Lul. Leipzig, 1883. Die Ratseldichter Tatwin und Eusebius. Forschungen zur deutschen Ge- schichte, XXVI (1886), 601 f. HAUG, MARTIN. Vedische Rathselfragen und Rathselspriiche. Sitzungsberichte der koniglichen Akademie der Wissenschaft zu Miinchen, Phil.-Hist. Classe, II (187 5)» 457-5' 5- HAYN, HUGO. Die deutsche Rathsel-Litteratur. Versuch einer bibliographischen Uebersicht bis zur Neuzeit. Centralblatt fiir Bibliothekswesen, VII (1890), 55-556. HEUSLER, ANDREAS. Die Altnordischen Ratsel. Zeitschrift des Vereins fiir Volks- kunde, XI (1901), 117-149. KEMBLE, J. M. The Dialogue of Salomon and Saturnus, with an Historical In- troduction, printed for the jElfric Society. London, 1848. KOHLER, REINHOLD. Zwei und vierzig alte Ratsel und Fragen. Weimar Jahrbuch, V (1856), 329-356. cvi RIDDLES OF THE EXETER BOOK LINDLEY, ERNEST H. A Study of Puzzles with Special Reference to the Psy- chology of Mental Adaptation. American Journal of Psychology, VIII (1896- 1897), 43'-493- MANITIUS, M. Zu Aldhelm und Baeda. Wien, 1886. Geschichte der christlich-lateinischen Poesie bis zur Mitte des 8. Jahrhun- derts. Stuttgart, 1891. MIGNE, J. P. Patrologiae Cursus Completus Patrum Latinorum, XC (1850), 539 f. (Flores of Bede). MULLENHOFF, KARL. Nordische, englische und deutsche Ratsel. Wolfs Zeit- schrift fur deutsche Mythologie, III (1855), 1~2O» i24-32. OHLERT, KONRAD. Ratsel und Gesellschaftsspiele der alten Griechen. Berlin, 1886. PETSCH, ROBERT. Neue Beitrage zur Kenntnis des Volksratsels. Palaestra IV. Berlin, 1899. PITRE, GIUSEPPE. Indovinelli, Dubbi, Scioglilingua del Popolo Siciliano (Biblio- teca delle Tradizioni Popolari Siciliane, XX). Torino-Palermo, 1897. PLOTZ, HERMANN. 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LEHMANN, HANS. Ueber die Waffen im ags. Beowulfliede. Germania, XXXI (1886), 487 f. Brunne und Helm im ags. Beowulfliede. Leipzig, 1885. LEO, HEINRICH. Rectitudines Singularum Personarum. Halle, 1842. This list includes only the more frequent references. The illuminated MSS. and grave-finds of the Old English period in the British Museum have been examined. cviii RIDDLES OF THE EXETER BOOK LIEBERMANN, FELIX. Gerefa. Anglia, IX (1886), 251-265. LiJNiNG, OTTO. Die Natur, ihre Auffassung und poetische Verwendung in der altgermanischen und mittelhochdeutschen Epik. Zurich, 1889. MEAD, W. E. Color in Old English Poetry. Publications of the Modern Lan- guage Association of America, XIV (1899), 169-206. MERBACH, HANS. Das Meer in der Dichtung der Angelsachsen. Breslau, 1884. MERBOT, REINHOLD. Aesthetische Studien zur angelsachsischen Poesie. Bres- lau, 1883. ROEDER, F. Die Familie bef den Angelsachsen. Halle, 1899. SCHMID, REINHOLD. Die Gesetze der Angelsachsen. Leipzig, 1858. SCHULTZ, ALWIN. Das hofische Leben zur Zeit der Minnesinger. Leipzig, 1879-1880. SMITH, C. ROACH. Collectanea Antiqua. London, 1868. STRUTT, JOSEPH. Horda Angelcynnan. London, 1775. Dress and Habits of the People of England. London, 1842. Sports and Pastimes of the People of England. London, 1903. TRAILL, H. D. Social England, vol. I. Second edition. New York and Lon- don, 1894. TURNER, SHARON. The History of the Anglo-Saxons. Seventh edition. Lon- don, 1852. WATTENBACH, WILHELM. Das Schriftwesen im Mittelalter. Zweite Auflage. Leipzig, 1875. WEINHOLD, KARL. Altnordisches Leben. Berlin, 1856. Deutsche Frauen. Berlin, 1882. WESTWOOD, J. O. Facsimiles of Miniatures and Ornaments of Anglo-Saxon and Irish Manuscripts. Oxford, 1868. WHITMAN, C. H. Birds of Old English Literature. Journal of Germanic Philol- ogy II (1898), 149 f. The Old English Animal Names. Anglia, XXX (1907), 380-393. WRIGHT, THOMAS. A History of Domestic Manners and Sentiments in England in the Middle Ages. London, 1846. Anglo-Saxon and Old English Vocabularies. Second edition by WULKER, R. P. London, 1884. The Celt, the Roman and the Saxon. Fourth edition. London, 1885. NOTE. Readings and suggestions ascribed to the general editors of this series, Professors Bright and Kittredge, are drawn from personal communications to the editor. ABBREVIATIONS A. L. Ancient Laius (Thorpe). And. Andreas (Krapp's edition). Anth. Lat. Riese, Anthologia Latina. Anz, Anzeiger. Ap. The Fates of the Apostles, Bibl. II, 87-91. Archin, Herrigs Archiv. Archiv fur das Studium der neueren Sprachen und Litteraturen. A.-S. Anglo-Saxon. Az. Azarias, Bibl. II, 491-520. Barnouw. Textkritische Untersuchun- gen. BB. Banner Beitrdge zur Anglistik. Bb. Anglia, Beiblatt. Beow. Beowulf, Bibl. I, 149-277. Bibl. Grein- Wiilker, Bibliotkek der angelsdchsischen Poesie. Bl. Blackburn, Journal of Germanic Philology, III, if. Bl. Horn. B lick ling Homilies. B. M. British Museum transcript. Bruit. Battle of Brunanburh, Bibl. I, 374-379- B.-T. Bosworth-Toller, Anglo-Saxon Dictionary. Chr. Christ (Cook's edition). Cleasby-Vigfusson. Icelandic-English Dictionary. Con. Conybeare, Illustrations. Cos. Cosijn. C. P. Muller, Cothener Programm. Cr. De Creatura (Aldhelm). Craft. Bi Manna Crteftum, Bibl. Ill, 140-143. Dan. Daniel, Bibl. II, 476-515. Deor. Deer's Lament, Bibl. I, 278-280. Dicht. Grein, Dichtungen der Angel- sachsen. Diet. Sweet, Student's Dictionary of Anglo-Saxon. Dietr. Dietrich, Haupts Zs., XI, XII. Dream. Dream of the Rood, Bibl. II, 116-125. Edd. Editors. E. E. Lit. Brooke, Early English Lit- erature. E. E. T. S. Early English Text So- ciety. El. Elene, Bibl. II, 126-201. E. S., Engl. Stud. Englische Studien. Ettm. Ettmiiller, Engla and Seaxna Scopas. Exod. Exodus, Bibl. II, 445-475. Feed. Feeder larcwidas, Bibl. I, 353- 357- Fates. Fates of Men (Bi Manna Wyr- dum), Bibl. Ill, 148-151. Frucht. Metrisches und Sprachliches. Gen. Genesis, Bibl. II, 318-444. Gn. Grein, Bibliothek. Gn.2 Grein, Germania, X, 423. Gn. Cot. Gnomes of the Cotton MS., .Bibl. 1,338-341. Gn. Ex. Gnomes of Exeter Book, Bibl. I, 341-352- Gr.8 Sievers, Old English Grammar, third edition. Grintdrtss. Wiilker, Grundriss zur Geschichte der angelsdchsischen Litte- ratur. (in. Guthlac, Bibl. Ill, 54-94. ex RIDDLES OF THE EXETER BOOK Har. Harrowing of Hell, Bibl. Ill, 175-180. Haitpts Zs., H. Z. Zeitsch rift fur deut- sches Alterthum. Herzf., Herzfeld. Die Rdtsel des Exeter- buches. H. M. Husband's Message, Bibl. 1, 309- 3"- Holth. Holthausen. Horn. Homilies. Horda. Strutt, Horda Angelcynna. Hpt. Gl. Angelsdchsische Glossen (Haupts Zs. IX, 401-530). Hy. Hymns, Bibl. II, 211-281. Icel. Icelandic. /. F, Indogermanische Forschungen. . I. G. Islenzkar Gdtur. Jansen. Beitrage zur Synonymik. Jud. Judith, Bibl. Ill, 117-139. Jul. Juliana, Bibl. II, 294-314. Keller. Miss Keller, Anglo-Saxon Weapon Names. Kl. Kluge, Angelsdchsisches Lesebuch. Klaeb. Klaeber. Kp. u. Ht. Hehn, Kulturpflanzen und Hausthiere. Lchd. Cockayne, Leechdoms. Leas. Bi Monna Lease, Bibl. II, 108- 1 10. Leid. Leiden Riddle. Litt-Bl. Deutsches Litter atur-Blatt. M. MUller, Collectanea. Madert. Die Spracke der altenglischen Rdtsel. Maid. Battle of Maldon, Bibl. I, 358- 373- McL. McLean, Old and Middle Eng- lish Reader. M. E. Middle English. Men. Menologium, Bibl. II, 282-293. Met. Meters of Boethius, Bibl. Ill, 247- 3°3- M. H. G. Middle High German. M. L. N. Modern Language Notes. Mod. Bi Manna Mode, Bibl. Ill, 144- 147. M. P., Mod. Phil. Modern Philology. N. E. D. New English Dictionary. O. E. Old English. O. F. Old French. O. H. G. Old High German. O. N. Old Norse. Pan. Panther, Bibl. Ill, 164-166. PBB. Paul und Braune's Beitrage zur Geschichte der deutschen Spracke und Literatur. Ph. Phcenix, Bibl. Ill, 95-116. P. L. Patrologia Latino. P. M. L. A. Publications of the Modern Language Association of America. Prehn. ^Composition und Quellen der Ratsel des Exeterbuches. Ps. Psalms, Bibl. Ill, 329-482. Ps. Psalms (Vulgate). R. Rieger, Alt- und angelsdchsisches Lesebuch. Rid. Riddles. R. S. P. Rectitudines Singularum Per- sonarum. Run. Runic Poem, Bibl. I, 331-337. Sal. Salomon and Saturn, Bibl. Ill, 304-328. Sat. Christ and Satan, Bibl. II, 521- 562. Sch. Schipper, Germania, XIX, 328- 338; Schmid. Die Gesetze der Angelsachsen. Seaf. Seafarer, Bibl. I, 290-295. Shipley. The Genitive Case in Anglo- Saxon Poetry. ABBREVIATIONS CXI Siev. Sievers. Soul. Soul and Body, Bibl. II, 92- 107. Spr. Grein, Sprachschatz. Sw. Sweet, Anglo-Saxon Reader. Sym. Symphosius. T. Editor's reading of MS., usually cited in first person. Th. Thorpe, Codex Exoniensis. Tr. Trautmann. W. Wiilker (Assmann), Bibliothek der angelsdchsischen Poesie, III, 183— 238- Wand. Wanderer, Bibl. I, 284-289. Wb. u. Kp. Hoops, Waldbdutne und Kulturpflanzen. Wids. Widsifr, Bibl. I, 1-6. Wond. Wonders of Creation, Bibl. Ill, I52-I55- Wossidlo. Mecklenburgische Volksiiber- lieferungen. WW. Wright -Wiilker, Anglo-Saxon and Old English Vocabularies. Zs. d. V.f. Vk. Zeitschrift des Vereins fur Volkeskunde. Zs. f. d. M. Zeitschrift fur deutsche Mythologie. Zs. f. d. Ph. Zeitschrift fur deutsche Philologie. RIDDLES OF THE EXETER BOOK [Leodum is mlnum swylce him mon lac gife : willaS hy hine apecgan, gif he on J>reat cymetS. Ungelic is us. Wulf is on lege, ic on operre ; faest is faet eglond fenne biworpen, 5 sindon waelreowe weras pair on Tge : willafc hy hine apecgan, gif he on )>reat cymeS. Ungellce is us. Wulfes ic mines widlastum wenum hogode ; ]K>nne hit waes renig weder ond ic reotugu saet, 10 ]>onne mec se -beaducafa bogum bilegde : waes me wyn to }>on, waes me hwaefre eac laS. [Min] wulf, mm wulf, wena me fine seoce gedydon, fine seldcymas, [ioia] murnende mod, nales meteliste. 15 Gehyrest fu, Eadwacer? Uncerne earne hwelp I I Leo (Quae de se ipso Cynewulfus tradiderit, ffalle, 1857, p. 22), Imelmann (Die altenglische Odoaker-Dichtung, Berlin, 7907, p. 24) gefe. — 2 Imelmann in J>reate. — 3 Imelmann ungelimp. — 6 Trautmann (Anglia vi, 158) wael[h]reowe. Imel. her on ege. — 7 Imel. hie and in )>reate. — 8 Kluge ungelic ; Imel. unge- limp. — 9 MS., Edd. dogode; Leo do gode; Hicketier (Anglia x, 579), Schofield (Publ. Mod. Lang. Assoc. xvii, 267), Imel. hogode. — 10 Gn. waeter (misprint); Kl. waeter. MS., Th. reo tugu; Imel. reotigu. — \zHolthausen (Anglia xv, 88) ' instead of wyn, leof and lat) hwaebre eac, or wyn and wa (wea) for la~5'; Imel. defends text, citing as examples of w . . . hw alliteration Leiden Rid. n, Gu. 323, Beow. 2299 (Heyne's note). — 13. Holth. Wulf, min Wulf, la!; Biilbring (Litt.-Bl. xii, 157) min Wulf, min Wulf ; Imel. Wulf se min Wulf. Holth. wearna? for wena; Imel. wene. — 14 Imel. gededun. — 15 MS., Th. mete liste ; Holth. (Litt.-Bl. x, 447) metes liste and murnend[n]e mod; Imel. metelestu. — 16 Imel. georstu_/0r gehyrest J>u. Schofield eadwacer ('very vigilant'). Holth. earmne/w earne. I 2 RIDDLES OF THE EXETER BOOK birefc wulf to wuda. J?set mon eape tosliteS psette nzefre gesomnad waes, uncer giedd geador.] 2 Hwylc is haelepa paes horse ond paes hygecraeftig J>set pact maege asecgan, hwa mec on sl$ wrsece, ]>onne ic astlge strong, stundum re)>e prymful punie? J>ragum wraec(c)a fere geond foldan, folcsalo baerne, 5 raeced reafige, recas stigaft haswe ofer hrofum, hlin biS on eorpan, waelcwealm wera. J>onne ic wudu hrere, bearwas bledhwate, beamas fylle holme gehrefed, heahum meahtum 10 wrecan on wafe wide sended, haebbe me on hrycge J>aet £er hadas wreah foldbuendra, flalsc ond galstas, somod on sonde. Saga, hwa mec fecce, o)>]?e hu ic hatte ]>e fa hlaest bere. 1 5 3 Hwflum ic ge^ite, swa ne wenaj) men, under y]>a. gepraec eorfan secan, garsecges grund. Gifen bi]» gewreged, tv-t^- , fam gewealcen ; hwaelmere hlimmeS, hlude grimmeS; 5 18 Hicketier J>e for >xt. Gn., A7.f Intel, gesomnod. — 19 Herzfeld (Die Rtitsel des Exeterbuches, Berlin, 1890, p. 66) and Schofield gaed geador ; Intel, gaed gador. 2 4 MS., Th., Gn., W. wrasce ; Siev. (PBB. x, 510) wrSce ; Herzf. (p. 44) wraec(c)a? — 7 In MS.' y is written above i in hlin in another hand. — 10 Cos. (PBB. xxiii, 128) helme. MS., Th. heanu. — n MS., Edd. wrecan; Cos. wrecen. 7>4.sende? — 14 J/^.sunde; Th. on sunde (trans, 'safely'); Gn. sande. (Jw.wecce? — 15 Th. te J>e. 3 3 Th. note geofon ; Ettm. gyfen. — 4 Ettm. proposes flod araered ; Gn. flod afysed. Cos. (PBB. xxiii, 128) famge wealcan (cf. PBB. xxi, 19, to And. 1524). RIDDLES OF THE EXETER BOOK 3 streamas stajm beataS, stundum weorpaf on stealc hleopa stane ond sonde, ware ond waege, Jxmne ic winnende, v \r holmmsegne bij>eaht, hrusan styrge, side saegrundas : sundhelme ne maeg 10 losian aer mec laete, se ]>e mm latteow biS on sij>a gehwam. Saga, foncol mon, hwa mec bregde of brimes faefmum, fonne streamas eft stille weorfaS, ypa gefwaere, J>e mec asr wrugon. 15 4 . [Hwilum mec mm frea faeste genearwa'S, [ioib] sendetS J>onne under saelwonge bearm [)>one] bradan ond on bid wriceo", prafaS on pystrum frymma sumne hseste on enge, faer me heard siteS 5 hruse on hrycge : nah ic hwyrftweges of fam aglace, ac ic efelstol haelefa hreru : hornsalu wagiaS, wera wicstede ; weallas beofia'5 ^AoJ-j^^ steape ofer stiwitum. Stille fynceS 10 lyft ofer londe ond lagu swige, o}>J>aet ic of enge up afringe 7 MS., Th., R., W. stealc hleoj>a; Ettm. stealchleojm. Gn. hleoj>u ? Compare 582. Ettm. sande. — 1 1 Ettm. ladteow. 4 There is no sign of closing after Rid. 3, no r spacing in the MS. between 3 and 4 {perhaps because 3 ends the page), and hwilum begins with a small letter ; but the preceding formula clearly marks the close of a riddle. — i Siev. {PBB. x, 479) frea resolred. — 2 MS., Gn., W. salwonge ; Gn. sahvongas ? Th., Ettm. saelwonge. — 3 Herzf. (p. 68) for metrical reasons supplies on; Holthausen (Anglia xiii, 358) >one. MS. onbid ; Th., Ettm. on bed. — 5 MS., Th., Gn., W. haetst ; Cos. haeste = J>urh haest. MS., Gn., W. heord; Th. note, Spr. ii, 68, Cos. heard. — 6 Th., Ettm., Gn. hwyrft weges; Gn!1 hwyrft-weges. — 7 MS. aglaca. — 8 MS. hrera; Th., Ettm. hrere. — 10 Ettm. stigwicum? — 12 a in ajringe is -written above the line in another hand. 4 RIDDLES OF THE EXETER BOOK efne swa mec wisa|> se mec wnede on ^ W^ get frumsceafte furfum legde bende ond clomme, J'aet ic onbugan ne mot 15 of ]>aes gewealde )>e me wegas taecneS. ""Hwllum ic sceal ufan yj>a wregan, [streamas] styrgan ond t5 staj^e pywan flintgraegne flod : famig winneo waeg wi5 wealle ; wonn ariseS 20 dun ofer dype, hyre deorc on last, eare geblonden, 6}>er ferefc, pset hy gemittaS mearclonde neah hea hlincas. )?£er bit5 hlud wudu, - brimgiesta breahtm ; bidaft stille 25 ^^i stealc stanhleojm streamgewinnes, hopgehnastes, fonne heah gearing on cleofu crydetS : pair biS ceole wen sllfre saecce, gif hine see byretS on )>a grimman tid, gaesta fulne, 30 ]?3et he scyle rice birofen weorjian, feore bifohten fsemig ridan y)?a hrycgum : ]>£er bi8 egsa sum haele)>um geywed, fara )^e ic hyran sceal strong on stlSweg : hwa gestilleft paet? 35 Hwilum ic Jmrhriese J>aet me rldeS on baece, won wsegfatu, | wide tofringe [IO2a] lagustreama full, hwilum laete eft 13 MS., Th. wraede; Ettm., Gti., W. wrae«e. — 18 MS. no gap; Th. supplies streamas. MS., Th. byran ; Th. note J>ywan ? — 20 Ettm., Gn. won. — 22 Th. note ear-geblonde ? — 23 Ettm., Gn. hi. Th. note gemetaft ? Ettm. gemeta^. — 27 Spr. ii, 47 heahge>ring. — 29 Ettm. bireS. — 31 MS., Th., Ettm., W. rice ; Th. note ricene? Gn. rice (onne scearp cymefi sceor wif ofrum, ecg witS ecge : eorpan gesceafte fus ofer folcum fyre swaetaft, blacan lige, ond gebrecu ferae" deorc ofer dreohtum gedyne micle, 45 farafi feohtende, feallan laetaS sweart sumsendu seaw of bosme, \ waetan of wombe. Winnende fareS atol eoredfreat, egsa astlgeS, P " " micel modfrea monna cynne, 50 t>rogan on burgum, ponne blace scotiao" scrij>ende scin scearpum wjepnum. Dol him ne ondraedeS tSa deatSsperu, swylteft hwaepre, gif him soS meotud on geryhtu furh regn ufan 55 of gestune Ueteo' straele fleogan, ferende flan : fea paat gedyga6 > §/^ fara fe geraecetS rynegiestes walpen. Ic J>aes orleges or anstelle, )>onne gewite wolcengehnaste 60 purh gepraec fringan J>rimme micle i^wtA ofer byrnan bosm : bierste5 hlude heah hloSgecrod ; fonne hnlge eft under lyfte helm londe near 41 MS., Edd. sceo ; Cos. sceor. — 42 MS., Th. earpan ; Th. note eor)>an or ear- man ? Ett. eorpan. Ettm., Gn. gesceafta. — 45 MS., Edd. dreontum ; Th. note, Spr. \, 204 dreohtum (dryhtum) ? Gn. dreongum = drengum ? Holth. (E. S. xxxvii, 206) dreorgum (" traurjg), — 47 MS. (T.) sweartsum sendu ; Th. note sweart- sum sendetS? — 50 Siev. (PBB. x, 479-480) resolves -brea. — 51 Th. note broga ? Ettm. breostum instead of burgum. — 54 Ettm. swilte"S. — 55 Ettm. gerihtum. — 57 MS., Edd. farende. Siev. (PBB. x, 480), flanas ? — 58 MS., W. geraeceft ; Th., Ettm., Gn. geraeca'S. Th. note regn-gastes ? — 61 MS., W. J>rimme. Th., Etim., Gn. )>rymme. — 62 Gn. burnan ? — 64 Siev. (PBB. x, 478) resolves near. RIDDLES OF THE EXETER BOOK ond me [on] hrycg hlade fset ic habban sceal, 65 meahtum gemanad mines frean. Swa ic, prymful }>eow, ]>ragum winne hwilum under eor|?an, hwilum y]>a. sceal hean underhnlgan, hwilum holm ufan streamas styrge, hwilum stlge up, 70 wolcnfare wrege, wide fere swift ond swtyfeorm. | Saga hwaet ic hatte, [io2b] o]>)>e hwa mec raere ponne ic restan ne mot, offe hwa mec staet5j>e ponne ic stille beom. Ic sceal J>ragbysig pegne mmum, hringum hsefted, hyran georne, mm bed brecan, breahtme cyfan fast me halswrifan hlaford sealde. Oft mec sljgpwerigne secg ottye meowle 5 gretan code ; ic him gromheortum winterceald oncwefe; _ficzf] wearm\e~\ lira gebundenne beag bersteS hwilum, se ]>eah bi)> on ponce pegne mmum, medwisum men, me paet sylfe, 10 ]>aer wiht wite ond wordum mm on sped masge spel gesecgan. 65 Gn., W. add on. Th. note hebban ? — 66 Siev. (PBB. x, 479) resolves frean. — 69 MS., Con., Th., Ettm. heah ; Gn., W. hean. MS. (T.), Ettm. under hnigan. — 71 Ettm., Gn. wolcenfare. 5 i MS., Th. )>ragbysig; Ettm. }>rage bysig; Jragbysig? or Jrascbysig ? Gn., W. }>ragbysig. — 2 MS., Th. hringan. — 7 MS. wearm lim ; Th. note wearme limu ? Ettm. wearmum limum ; Holth. (I. F. iv, 386) wearm lim[waedum]. — 8 MS., Edd. gebundenne; Ettm. gebunden. MS., Th. baeg; Th. note beag. MS., Th. hwilum berste'S; Th. note bersta'S. After \ in hwilum, an o is erased. — 10 Ettm., Gn. silfe. — ii Ettm. se J>aer. — 1 1-12 MS. min onsped ; Th. minon sped ; note spede ? or spedum ? Ettm. minum | spede. RIDDLES OF THE EXETER BOOK 7 6 Ic com anhaga Iserne wund, bille gebennad, beadoweorca saed, ecgum werig. Oft ic wig seo, frecne feohtan, frofre ne wene, faet me geoc cyme guftgewinnes, 5 ser ic mid aildum eal forwurde; ac mec hnossiaS homera lafe, heardecg heoroscearp hondweorc smipa, bitaS in burgum ; ic abldan sceal lajran gemotes. Naefre leececynn 10 on folcstede findan meahte, para ]>e mid wyrtum wunde gehjelde, ac me ecga dolg eacen weorftaft Jmrh deaftslege dagum ond nihtum. Mec gesette so5 sigora waldend «.>.c Crist to compe : oft ic cwice baerne, unrlmu cyn, eorfan gelejige, nSte mid nij^e, s\va ic him no hrine, fonne mec frea mm feohtan hatep. H \vllum ic monigra mod acete, ^ JjJ^ hwilum ic frefre fa ic aer winnejon [IO3a] feorran swipe ; hi ^s felaS feah 6 3 Siev. (PBB. x, 476) resolves seo. — 5 MS., M., Th. mec. — 6 Ettm. ildum. Gn. eall. Ettm. forwurfte ; Gn. forwurde ? — 7 Ettm. lafa. — 8 MS., Th. ^eorc ; Th. note handweorc ; M., Ettm., Gn., R. handweorc ; W. hondweorc. — 9 MS., Th., Ettm., R. abidan ; Gn., IV. a bidan. — 10 R. lat>ra. — 13 Spr. \, 251, eaden ? Ettm. weorfleft. 7 IV. ' Nach nihtum ist die hdlfte der zeile fret, auf ihr steht iiber Crist die rune S.' — 4 Th. note swa-J>eah ? — 5 Siev. (PBS. x, 479) frea resolved; MS., Edd. min frea; Holth. (Bb. ix, 357) friga min. — 7 [wel] before frefre added by Gn., W. Th. note frefrige. Th. note J>a J>e ? 8 RIDDLES OF THE EXETER BOOK swylce )>8es 5J>res, )>onne ic eft hyra ofer deop gedreag drohtafi bete. 8 Hraegl min swigaS )mne ic hrusan trede o}>J>e fa wic buge oj>pe wado drefe. Hwilum mec ahebbatS ofer haelepa byht hyjste mine ond )>eos hea lyft, ond mec fonne wide wolcna strengu 5 ofer folc byret>. Frsetwe mine swogaS hlude ond swinsiaS, ^ torhte singaS, fonne ic getenge ne beom flode ond foldan, ferende gaest. 9 Ic Jmrh muj> sprece mongum reordum, wrencum singe, wrixle geneahhe heafodwofe, hlude cirme, healde mine wisan, hleopre ne mipe, eald jgfensceop, eorlum bringe 5 blisse in burgum fonne ic bugendre stefne styrme ; stille on wicum sittaS swigende. Saga hwaet ic hatte J>e swa scirenige sceawendwisan 10 MS., Th. betan; Gn. bete; Spr. i, 99 betan [sceal]. Rune S stands at close of the riddle. 8 i Th. note swogaS ? — 4 Siev. (PBB. x, 478) resolves hea ; Holth. (Bb. ix, 357) hea[e]. — 6 Ettm. bire?. Ettm. fraetwa. — 7 Ettm. swinsjaS eac. — 9 Gn. gaest ; Sw. gist. 9 The rune C is over this riddle on line -with ferende gaest (89). — 4 Th. note hleobor; Ettm. hleoSor; Gn. hleoftres; Gn? hleoSre (inst.).— S MS., Th. siteS; Ettm. sita^S ; Gn., W., Cos. sitta-5. MS., Th., Gn., W. nigende ; Gn. hnigende ? Ettm., Cos. swigende. — 9 MS (T.) )>a swa scirenige ; Th. )>a swa scire nige ; Th. note J>e; Ettm. scirenige; Gn. 'scirenige, scurriliter? vgl. Graff vi, 549-551'; Spr. ii, 296 scire nige (ist pers. sg. of nigan) ; Bosworth- Toller, p. 837, scire cige ; Cos. (PBB. xxiii, 1 28) ' scirenige is to be changed to sciernicge — scericge, mima, Shr. 140; scearecge, Lye? RIDDLES OF THE EXETER BOOK 9 WK" hlude onhyrge, hselefum bodige 10 wilcumena fela wof e mlnre. 10 Mec on dagum fissum deadne ofgeafun faeder ond modor, ne wses me feorh fa gen, ealdor in innan. J>a mec [an] ongon, O wel hold mege, wedum feccan, heold ond freofode, hleosceorpe wrah 5 1 sue arlice swa hire agen beam, offset ic under sceate, swa mm gesceapu waeron, ungesibbum wearS eacen gaeste. Mec seo frife mseg fedde siffan, offaet ic aweox[e], widdor meahte 10 sifas asettan ; heo hsefde swsesrajfy Ises [iO3b] suna ond dohtra, fy heo swa dyde. ii Neb wses mm on nearwe, ond ic neofan waetre, flode underflowen, firgenstreamum swife besuncen; ond on sunde awox, ufan yf um feaht, anum getenge lifendum wuda lice mine. 5 Haefde feorh cwico fa ic of fseSmum cwom ii Ettm. welcumena. 10 i MS., Edd. on Hssum dagum ; Holth. (£. S. xxxvii, 206) dagum )>issum or t>issum dogrum. MS. ofgeafum. — 2 7%., Gn. moder. — 3 Gn. on ; Szu. oninnan. Gn., Sw. [ides] ; 6"«.2 [an]. Gti.2 ongan ; Sw. ongonn. — 4 MS (T.) wel (end of line) hold mege wedum weccan. Holth. (Bb. ix, 357) wilhold. 77/., Gn., W. gewedum ; Sw. gewaedum ; Cos., Holth. mege wedum. Edd. )>eccan. — 6 MS., Th. snearlice ; Th. note searolice ? Gn., IV. swa arlice ; Sw. suae arlice ; Cos. sue arlice (cf. 16). — 7 Sw. oj> J?aet. Th. note mine. — 9 MS., Th.,Dietr. (I/Z. xii, 251) frij>e masg; Gn., W. frij>emaeg. Th. note magg1??. — 10 MS., Edd. aweox ; Holth. (E. S. xxxvii, 206) aweox[e]. Gn., IV. widor; Cos. compares 6i17. 112 Th. gives incorrectly MS. reading as floren. — 3 Tr. (BB. xix, 169) on sande grof. — 6 Gn. feorh-cwico. 10 RIDDLES OF THE EXETER BOOK brimes ond beames on blacum hrsegle ; sume wjgron hwite hyrste mine, )>a mec lifgende lyft upp ah5f, wind of waege, sippan wide baer 10 ofer seolhb'aj>o. Saga hwaet ic hatte. 12 Hraegl is mm hasofag, hyrste beorhte reade ond sclre on reafe _sind~. Ic dysge dwelle, ond dole hwette unrajdslpas, ofrum styre nyttre fore. Ic )>ses nowiht wat 5 fset heo swa gemiedde, mode bestolene, daede gedwolene, deoraj) mine won wisan gehwam. Wa him pass feawes, siffan heah pringeS horda deorast, gif hi unrjedes eer ne geswica]) ! 10 13 Fotum ic fere, foldan sllte, grene wongas, fenden ic gsest bere. Gif me feorh losaft, faeste binde swearte Wealas, hwilum sellan men. Hwilum ic deorum drincan selle 5 beorne of bosme, hwilum mec bryd triedeS felawlonc fotum, hwilum feorran broht wonfeax Wale wege8 ond )>yS, 7 Tr. bearmes. MS., Th. hraegl. — 8 Ettm. hyrsta. 12 2 The second half line is obviously defective ; Gn. adds minum, which Holth. rejects, proposing min; Tr. (BB. xix, 173) [hafo]. — 3 Tr. drops Ic. — 4 MS. unraed sij>as ; Edd. unraedsi}>as ; Herzf. (p. 68) on unraedsibas or unrasdgesi^as ; Tr. unraedsfra. — 9 Tr. heann/or heah. MS., Edd. bringe'S; Cos. J>ringe'S. 13 6 MS., Th. beorn ; Ettm. beornum. — 8 Ettm. note >y5= bywe-S; Siev. (PBB. x, 477) resolves }>yS; Cos. (PBB. xxiii, 129) }>y[h]eiS. RIDDLES OF THE EXETER BOOK II dol druncmennen deorcum nihtum, waeteS in waetre, wyrmeS hwllum 10 fsegre to fyre ; me on faeSme stica]> hygegalan hond, hwyrfeS geneahhe, swifeS me geond sweartne. Saga|hwset ic hatte [iQ4a] J>e ic lifgende lond reafige ond aefter deape dryhtum }>eowige. 15 14 Ic seah turf tredan, tyn waeron ealra, six gebro))or ond hyra sweostor mid, haefdon feorg cwico. Fell hongedon sweotol ond gesyne on seles waege anra gehwylces. Ne, waes hyra aengum ]>y wyrs 5 ne side fy sarre, |>eah hy swa sceoldon reafe birofene, rodra weardes meahtum aweahte, mu)mm slltan haswe blede. Hraegl bit5 geniwad fam ]>e ser forScymene frsetwe leton 10 licgan on laste, gewitan lond tredan. 15 Ic waes waepenwiga. Nu mec wlonc feceS AJV^-" geong hagostealdmon golde ond sylfre, woum wirbogum. Hwilum weras cyssa?5 ; hwllum ic to hilde hleofre bonne wilgehlepan ; hwllum wycg byrej) 5 mec ofer mearce ; hwllum merehengest 9 Th. dol-drunc mennen ; Gn. ' dune-mermen ? vgl.ahd. tune.' — 12 7/4., Ettm. hygegal an hond. — 15 Siev. (PBB. x, 491) J>eo\vige. 14 I MS., Edd. except Tr. (BB. xix, 177) x. — 2 MS., Edd. except Tr. VI. — 3 Gn. feorgcwico. — 5 Tr. Naes. — 6 MS., Th. sarra ; Cos. ne siS hy sarra. 15 i R. note conjectiires waspen wigan. — 2 Sw. monn. MS. sylfore ; Ettm. silfore ; Kl. note sylofre ? Siev. (PBB. x, 459) sylfre. — 5 Ettm. \vicg. Ettm., Kl. bire«. I2 RIDDLES OF THE EXETER BOOK fereS ofer flodas, fraetwum beorhtne ; hwilum maegSa sum minne gefylleft bosm beaghroden ; hwilum ic [on] bordum sceal, heard heafodleas, behlyfed licgan ; 10 hwilum hongige, hyrstum fraetwed, wlitig on wage pair weras drincaS ; freolic fyrdsceorp hwilum folcwigan wicge wegafi, ponne ic winde sceal sincf ag swelgan of sumes b5sme ; 1 5 hwilum ic gereordum rincas laSige wlonce to wine ; hwilum wrajmm sceal stefne mmre forstolen hreddan, flyman feondsceaj>an. Frige hwaet ic hatte. 16 | Hals is min hwit, ond heafod fealo, [iO4b] sidan swa some ; swift ic eom on fe}>e, beadowsepen bere; me on baece standaS her swylce swe on hleorum ; hlifiaS tu earan ofer eagum ; ordum ic steppe 5 in grene graes. Me bits gyrn witod, ^ gif mec onhzele an onfindeS, waelgrim wiga, ]>aer ic wic buge, bold mid bearnum, ond ic bide paer mid geoguScnosle hwonne gaest cume 10 9 MS., Edd. ic bordum. — 10 Ettm. behlitied; Gn. note behlywed ? Spr, i, 87, behlej>ed ? — 14 Gn. wecgaft (Gn.2 marks as misprint) ; Kl. \vecgai5. — 16 Gn., Sw. ic [to] ? — 17 MS., Th., K., A7.2 wra^bum. — 19 The sign after hatte seems to me no rune as W. conjectures, but part of a closing sign. 16 2 Th., Ettm., Gn., give incorrectly MS. reading swist. Ettm. in. — 4 MS., Th. her swylce sweon | leorum ; Th. note haer swylce swyne ; Ettm. haer swylce swine ; Gn., IV. her swylce sue ; Cos. her swylce suge ; Holth. (Bb. ix, 357) ' her swylce sw[in]e, on hleorum tu|, also mil streichung von hlifia'S'; McL. her swylce swe on hleorum ; hlinaft tu |. Th., Ettm., R. also close line with tu ; Gn., W. with hlifiaS. — 6 MS., Th. grenne. — 7 Ettm. unhaele. — 9 MS. blod. RIDDLES OF THE EXETER BOOK 13 t5 durum minum ; him bij> dea$ witod. ForJ>on ic sceal of eftle eaforan mine forhtmdd fergan, fleame nergan, OA-J gif he me aefterweard ealles weor|>et5 ; hine breost beraS. Ic his bidan ne dear 15 rejjes on geruman (nele )>aet raed teale), ac ic sceal fromllce fej>emundum ]mrh steapne beorg strjete wyrcan. Ea)>e ic mseg freora feorh genergan, gif ic msegburge mot mine gelaidan 20 on degolne weg Jmrh dune )>yrel swsEse ond gesibbe ; ic me sippan ne )>earf waelhwelpes wig wiht onsittan. Gif se nrSsceafa nearwe stlge me on swa}>e sece}>, ne tosailej) him 25 on )>am gegnpape gu})gem6tes, sif})an ic J>urh hylles hrof gersece, r ond Jmrh hest hrino hildepilum - laSgewinnum fam >e ic longe fleah. 17 Oft ic sceal wijj wsege winnan ond wi}> winde feohtan, somod wi8 )>am saecce, J>onne ic secan gewite eorpan ypum peaht ; me biS se e)>el fremde. Ic beom strong J>ses ge|winnes, gif ic stille weorfe ; gif me pses tosseleS, hi beoS swlpran fonne ic, 5 ond mec slitende sona flymatS, willaft oSfergan )>aet ic fri})ian sceal. 15 MS., Edd. hine beraiS breost. Th. note hi ne bereft ? Herzf. (p. 68) on metrical grounds breost beraiS ; Cos. ' entweder hine breost beraft — oder etwas anderes ; keinesfalls was der text bietet.' — 16 Ettm. teala. — 21 MS., Th. dum ; Th. note, Ettm. dim ; Gollancz (McL.) dumb. — 24 MS., Gn. gifre ; Th. and other Edd. gif se. — 27 Ettri. hilles. — 28 Ettm. haest. Th., Ettm. hrine. MS., Th. hilde pilum. I4 RIDDLES OF THE EXETER BOOK Ic him ]>&t forstonde, gif mm steort ]>ola« ond mec stijme wif stanas moton faeste gehabban. Frige hwset ic hatte. 10 18 Ic com mundbora minre heorde, . eodor wlrum faest, innan gefylled. dryhtgestreona. Daegtidum oft. spate sperebrSgan ; sped bi}> >y mare fylle minre. Frea |>aet bihealdeft, 5 hu me of hrife fleogao" hyldepilas. Hwilum ic sweartum swelgan onginne brunum beadowsepnum, bitrum ordum, eglum attorsperum. Is min innafc til, wombhord wlitig, wloncum deore ; 10 men gemunan J>aet me purh mup fareft. 19 Ic com wunderlicu wiht : ne mgeg word sprecan, mseldan for monnum, ]>eah ic mu}> haebbe, wide wombe Ic waes on ceole ond mines cnosles ma. w/ 20 Ic seah [somod] l/| 1^ P' N hygewloncne heafodbeorhtne 17 10 TVs reading of MS., Gn. haette; MS., Th. hatte. 18 Over the riddle stands in the MS. the B-rune, and over the B, the L-rune. — i Tr. (BB. xix, 180) minra. — 2 MS. (T.), Th., Tr. eodor wirum; Gn., W. eodor- wirum. — 5 MS., Th. freo. — 6 MS., Th. hylde pylas. — 8 Gn. beaduwaepnum. — II Cos. for metrical reasons [oft] or [J>aet] after men ; Tr. gewilnia'S instead of gemunan. 193 ATo gap in MS. after wombe. — 4 After ma, usual sign of closing : - : 7 ; Th., Gn. suggest a lacuna. ao i The addition is Grein's ; Hicketie r (A ng lia x, 592) Somod ic seah. Holth. (Bb. ix, 357) ond between runes R and 0. RIDDLES OF THE EXETER BOOK 15 swiftne ofer sJilwong swtye fraegan ; haefde him on hrycge hildefrype, -f- P M, nsegledne rad 5 K X M F; widlast ferede rynestrong on rade rofne S & r (P) K N ; for waes ]>y beorhtre, swylcra siffaet. Saga hwaet ic hatte. 21 Ic com wunderlicu wiht, on gewin sceapen, frean minum|leof, faigre gegyrwed : [IO5b] byrne is mm bleofag, swylce beorht seomaS wir ymb ]>one wselgim ]> e me waldend geaf, se me wldgalum wisaS hwllum 5 sylfum to sace. J>onne ic sine wege furh hlutterne^daeg, hondweorc smij>a, gold ofer geardas. Oft ic gaistberend 3 ^/IS1. swistne (not swisne, Gn.). Ettm. Jraegjan. — 4 ^/^S"., Th. 'hilde bryl>e ("bold in war"):—^ 6 J/^., Th., Gn., W. rad AGEW. Th. note, Ettm., Dietr. (xi, 465) rad — N. G. E. W ; Gn. note suggests N.O.M. naegledne R. A. G. [wod R] E. W. widlast ferede. Hicketier (Anglia x, 592) rand>r rad ; WO E |> ( N G E |>) for AGEW. Tr. (Bb. v, 48) N. O. [ond] M. Naegledne gar W. O. E. t>. widlast ferede. Cos. (PBB. xxiii, 129) rad (R), A. G = gar; E (eh), W (wynn) should be changed to W. E. (wynneh), ' -well damit das ross bezeichnet wird, der widlast ferede.' Holth. (Bb. ix, 357) W. E. = wynne. Ettm. note nydlast ? — 7-8 Th., Siev. (Anglia xiii, 17), Holth. I.e. COFAH. — 8 Holth. F. A [ond] H. — 8 No gap in MS.; Th. note '•Here a line is wanting'; Ettm. indicates a gap before for. Gn. beorhtra. — 9 Gn. note hwaet hio ? Ettm. hate. 21 2 Gn. faegere. — 3 MS., Th. seomad. — 4 Th. note 'were or wirum ? wael- grimman ? or is wael-jjim a periphrasis for byme?' — 6 Edd., citing AfS. in- correctly, read rice ; Gn. note sige ? Spr. ii, 446 sige ; MS. reads plainly sace ; so B. M. V 16 RIDDLES OF THE EXETER BOOK cwelle compwjepnum. Cyning mec gyrweS since ond seolfre ond mec on sele weorpatS ; 10 ne wyrneS word lofes, wisan mainetS 1 mine for mengo, )>aer hy meodu drincaS ; healdetS mec on heapore, hwilum IsetetS eft radwerigne on gerum sceacan, orlegfromne. Oft ic 5)>rum scod 15 V1 frecne set his freonde ; fah com ic wide, wjgpnum awyrged. Ic me wenan ne fearf faet me beam wraece on bonan feore, gif me gromra hwylc gupe gensegeS ; ne weor)>ef5 slo msegburg gemicledu 20 eaforan minum fe ic sefter woe, nympe ic hlafordleas hweorfan mote from }>am healdende ]>e me hringas geaf : me bits for5 witod, gif ic frean hyre, guj>e fremme, swa ic glen dyde, / 25 minum feodne on )x>nc, faet ic ]x>lian sceal bearngestreona ; ic wi]> bryde ne mot haemed habban, ac me >xs hyhtplegan geno wyrneS se mec geara on bende legde ; forpon ic brucan sceal 30 on hagostealde haelepa gestreona. Oft ic wirum dol wife abelge, wonie hyre willan ; heo me worn spreceS, fl5ceS hyre folmum, firena]) mec wordum, ungod gseleS ; ic ne gyme ]?93S compes 35 10 Th. feolfre (misprint). — 13 7/4., Gn. me. — 14 Gn. sceacen (misprint). — 17 Gn. note awyrded ? — 19 Gn. note gehnaegeiS ? — 29 MS., T/i., Gn., W. gearo ; Siev. (PBB. x, 519) gearwe; Herzf. (p. 44) geara. — 35 Th. note ' Here a leaf of the MS. is evidently wanting1; W. '•in der US. ist nichts wa/irzitnehmen.'1 There is no closing sign in the MS. Holth. (Bb, ix, 357) for metrical reasons assigns compes to line 36. RIDDLES OF THE EXETER BOOK 17 22 Neb is mm niperweard, neol ic fere [io6a] ond be grunde graefe, geonge swa me wisaS bar holtes feond, ond hlaford mm [se] w5h faerefi weard set steorte, wrigap on wonge, wegeS nice ond ]>y8, 5 sawej> on swseS min. Ic snyfige forS brungen of bearwe, bunden crsefte, wegen on waegne, hsebbe wundra fela ; me bi}> gongendre grene on healfe ond mm swaeS sweotol sweart on o]>re. 10 Me ]mrh hrycg wrecen hongaf under an orjjoncpil, 6j>er on heafde faest ond for6 weard fealle|> on sidan, ])set ic tojnim tere, gif me teala ]>enaft hindeweardre Jjaet bi> hlaford min. 15 23 ^Etsomne cw5m sixtig monna to wgegstaej^e. wicgum ridan ; hsefdon endleofon eoredmaecgas frf&kengestes, feower sceamas. Ne meahton magorincas ofer mere feolan, 5 swa hi fundedon, ac wses fldd to deop, atol yj>a gefraec, ofras hea, 22 2 Th. w^^geong? — 3 Th. har-holtes. — 4 Stev. (PBB. x, 519) [on]; Bright [se]. — 5 Stev. (PBB. x, 477) resolves by«; Cos. (PBB. xxiii, 129) by[h]e-S. — 6 Th. note snyrige ? — 7 MS. bearme ; Th. beame. — 15 Th. note ' se t>e for J>set ? ' 23 i MS. -flLTsotnne ; Th. Etsomne ; Th. note ' r. ^Etsomne'; Ettm. JEt somne. Th. note, Ettm. cwomon. MS., Edd. except Ettm. LX. — 2 Ettm. waegsta'Se. — 3 MS., Edd. except Ettm. XI. Ettm. eoredmecgas. — 4 MS. fridhengestas ; Th. note fyrdhengestas ? Ettm. fridhengestas; Dietr. (xii, 251) 'frF5, adj. (stattltch, schon ; vgl. to9) ' ; Gn. ' fridhengestas (vgl. a/id, parafrit) ' ; Spr. i, 349, Gn.z frid- hengestas. MS., Edd. except Ettm. mi. — 5 Th. note feran ? — 7 Stev. (PBB. x, 478) resolves hea ; Holth. (Bb. ix, 357) hea[e]. j8 RIDDLES OF THE EXETER BOOK streamas stronge. Ongunnon stigan fa on wsegn weras ond hyra wicg somod hlodan under hrunge ; fa fa hors oftbaer eh ond eorlas sescum dealle ofer wsetres byht wsegn t5 lande, swa hine oxa ne teah ne esla maegen ne fsethengest, ne on flode sworn, ne be grunde wod gestum under, 15 ne lagu drefde, ne on lyfte fleag, ne under bsec cyrde ; brohte hwaef re beornas ofer burnan ond hyra bloncan mid from stseSe heaum, fset hy stopan up on of erne, | ellenrofe, [io6b] 20 weras of waege ond hyra wicg gesund. 24 Agof is mm noma eft onhwyrfed. Ic com wraetllc wiht on gewin sceapen. J>onne ic onbuge ond me of bosme fareS setren onga, ic beom eallgearo, ]>get ic me fset feorhbealo feor aswape. 5 Siffan me se waldend, se me faet wite gescop, leopo forleete^, ic beo lengre fonne aer, offset ic spajte, spilde geblonden, ealfelo attor fast ic aer[or] geap. Ne togongeS faes gumena hwylcum 10 10 Ettm. hlodun. — 1 1 Th. note eohas ? — 13 MS., TA., Gn., W. esna ; Gn. note esla? Spr. i, 228 esla or esola. Th., Ettm., Gn. maegn. — 14 MS., Th., Gn., W. faet hengest; Ettm. fxi; note fsted? fact? Spr. i, 274 fsethengest. — 1 6 Ettm. draefde. MS., Th. of ; Th. note on ? — 17 MS. onder. Ettm. cirde. — 18 Ettm. hira. — 19 Ettm., Gn. hi stopon. 24 4MS.(T.) set renonga; Th. aettren onga. Gn. com. MS. (T.), Th., Gn. call gearo ; Gn?- eallgearo. — 7 Herzf. (p. 62) com for beo. Cos. lengra. — 8 Gn. o« J>aet. — 9 MS., Th. eal felo. MS., Edd. aer ; Siev. (PBB. x, 519), Cos. aeror. — 10 Th. to gongeft. RIDDLES OF THE EXETER BOOK 19 ainigum eape pset ic pajr ymb sprice, gif hine hrineS paet me of hrife fleogeS, paet pone mandrinc maegne geceapap fullwer fseste feore sine. Nelle ic unbunden ainigum hyran 15 nympe searosaeled. Saga hwaet ic hatte. 25 Ic com wunderllcu wiht, wrjgsne mine stefne : hwilum beorce swa hund, hvvllum blaite swa gat, cj^S hwilum graide swa gos, hwilum gielle swa hafoc ; hwilum ic onhyrge pone haswan earn, guSfugles hleopor ; hwilum glidan reorde 5 mupe gemsene, hwilum mjgwes song, pSr ic glado sitte. X mec nemnaS, swylce F1' ond |^, f^ fullesteS [ond] N ond |. Nu ic haten com swa pa siex stafas sweotule becnap. 10 26 Ic com wunderlicu wiht, wifum on hyhte, neahbundum nyt ; nsengum sceppe burgsittendra nympe bonan anum. Stapol min is steapheah, stonde ic on bedde, neopan ruh nathwar. Nepe6 hwilum ful cyrtenu | ceorles dohtor, modwlonc meowle, paet heo on mec griped, II Th. 'sprite (spirt): — 14 J\fS., Edd. full wer; Th. note ful-hwer? Bright suggests fullwer ('complete wer'). 25 I Th. note wrixle ? for wraesne. — 2 Holth. (E. S. xxxvii, 207) swa hund beorce or belle swa bearg or beorce swa bicce. — 9 Cos. ' [ond] at beginning or end of half-line'1 ; Holth. H. I [samod]. 26 2 JlfS., Edd. neahbuendum ; Siev. (PBB. x, 480), Mad. (p. 63) neahbundum. — 4 MS., Th., Gn., W. steap heah ; Holth. 'steapheah (cf. Gen. 2839, heahsteap)'; Tr. (BB. xix, 184) omits heah. — 5 Tr. nat hwaer. 20 RIDDLES OF THE EXETER BOOK r§ese6 mec on reodne, reafaft mm heafod, fegeS mec on faesten ; felef sona mines gemotes seo }>e mec nearwaS, 10 wif wundenlocc : wset brS ]>aet cage. 27 Mec feonda sum feore besnypede, woruldstrenga binom ; wsette siffan, dyfde on waetre ; dyde eft ponan, sette on sunnan, fair ic swtye beleas herum )>am ]>e ic haefde. Heard mec sijtyan 5 sna$ seaxes ecg, sindrum begrunden ; ~/> fingras feoldan, ond mec fugles wyn .. r^ geond[sprengde] speddropum, spy^rede geneahhe ofer brunne brerd, beamtelge swealg (r^- ' 9 streames dale, stop eft on mec, 10 sijjade sweartlast. Mec siffan wrah haelet5 hleobordum, hyde bej>enede, gierede mec mid golde ; forfon me gliwejion wrjetlic weorc smij>a, wire befongen. Nu ]>a. gereno ond se reada telg 1 5 ond ]>a wuldorgesteald wide msere dryhtfolca helm, nales dolwite. Gif mm beam wera brucan hy beot5 }>y gesundran ond ]>y sigefaestran, 8 Gn. note raereiS? Gn. note 'reoSne (zur Ruttelung) ' ; Tr. raereiS mec reodne ? Bright suggests hreode ('reed, stalk1). — 10 MS., Th. se ; Th. note seo ? ay i Ettm. besnlSede. — 3 Ettm. dide. — 5 Ettm., Sw. hasrum. R., Sw. J>a J>e. — 6 MS., M. seaxses. MS., M., Th., Ettm. ecge. Ettm. note syndrum ? — 7 Th. note foldan ? Ettm., Gn. feoldon. Ettm., Gn. me. Th. note fule swyn ; Ettm. cyn ; Sw. wynn. — 8 Gn., Sw. add [sprengde] ; Molth. (I. F. iv, 386) [spaw]. — 9 Th. note beamtelga? — 12 M. heo-bordum. MS., M., Th., Ettm. hy)>e ; Gn., IV. hyde. — 13 Gn. note forS on me ? — 14 Sw. wraettlic. — 15 R. hyfe/or Nu )>a. — 16 Ettm., Gn. add beo"5 before maere; Gn. (Spr. ii, 223) follows MS.; Sw. maeren. — 17 Gn. note help ? Th., Ettm., R., Sw. dol wite. — 19 Ettm., Gn. hi. ' RIDDLES OF THE EXETER BOOK 21 heortum ]>y hwsetran ond fy hygeblij>ran, 20 ferpe )»y frodran, habbap freonda >y ma, swaesra ond gesibbra, so)>ra ond godra, tilra ond getreowra, |>a hyra tyr ond ead 'estum yca$ ond hy arsjafnm lissum bilecgaS ond hi lufan fae]>mum \r 25 faeste clyppaS. Frige hwset ic hatte, J tV7^ nifum to nytte : nama mm is maere,Qk gifre ond halig sylf. [ic>7b} Ic com wepr6 werum, wide funden, brungen of bearwum ond of burghleojmm, of denum ond of dunum. Daeges mec waegun fej>re on lifte, feredon mid liste under hrofes hleo. HaeleS mec si))J>an bafedan in bydene. Nu ic com bindere ond swingere, sona weorpe esne to eorfan, hwilum ealdne ceorl ; sona past onfindeS se fe mec fgh6 ongean, i ond wi8 maegenpisan minre gensesteS Q/^ 10 }>aet he hrycge sceal hrusan secan, gif he unrsedes ser ne geswjceS, strengo bistolen, strong on spraice, maegene binumen, nah his modes geweald, fota ne folma. Frige hwaet ic hatte, 15 t5e on eorpan swa esnas binde, dole aefter dyntum, be daeges leohte. 24 littm., Gn. hi. — 28 Ettm. gifraege ; R. gifrege ; Sw. gefraege. Ettm. silf. 28 2 MS., T//., Grt., IV. burghleohum ; Th. note beorghleo^um ? Ettm. beorg- hleo^um. — 3 Ettm., Gn. me. — 4 Ettm. feSru. Ettm., Gn. lyfte. Gn. note lisse ? — 7-8 MS., Edd. weorpere | efne ; Holth\E. S. xxxvii, 207) as in text. — 10 Ettm. mae.gen)>ysan ; Holth. I.e. maegenHssan. Th. note genaage'S; Ettm. gehnaeste'5. — 13 J«.2, W. strongan. — 14 Ettm. maegne. — 16-17 Tk. ' These lines are in the 22 RIDDLES OF THE EXETER BOOK 2Q BiJ) foldan dzel fsegre gegierwed mid ]>y heardestan ond mid ]>y scearpestan ond mid ]>y grymmestan gumena gestreona, ^j corfen, sworfen, cyrred, pyrred, j bunden, wunden, blseced, wseced, 5 ifrsetwed, geatwed, feorran Iseded ,to durum dryhta,] dream biS in innan ] cwicra wihta, clengeS, lengeS, para ]>e aer lifgende longe hwile wilna brucetS ond n5 wiS spriceS ; 10 ond fonne aefter deape deman onginne5, meldan misllce. Micel is to hycganne wisfaestum menn hwaet seo wiht sy. 30 Ic wiht geseah wundorlice hornum bitweonum hu))e laedan, [lyftfaet leohtlic listum gegierwed, [io8a] hu]>e to )>am ham[e] of pam heresife : walde hyre on }>£ere by rig bur atimbran, 5 searwum asettan, gif hit swa meahte. Da cwom wundorlicu wiht ofer wealles hrof (seo is eallum cu$ eorSbuendum), ahredde ]>a. fa hvtye, ond to ham bedraf MS. detached from the preceding part, begin -with a capital, and appear altogether as a separate riddle? W. ' nach hatte steht als schlusszeichen :-, dann folgt auf der- selben zeile De.' 29 2 Ettm. hwaessestan for scearpestan ; Gn. [heoru] scearpestan. — 3 Ettm., Gn. grimmestan. — 8 7/4. note glengeS? — 12 Siev. (PBB. x, 482) hyogan. — 13 Ettm. si; Gn. seo ; Siev. (PBB. x, 477) sy resolved. 30 2 MS., Th. horna abitweonu ; Th. note hornum bitweonum ? Dietr. (xi, 468) homaa (= homa) ; R. hornan. — 4 MS., Edd. except Tr. (BB. xix, 180) ban . — 5 MS., Tr. walde ; Th., Ettm., Gn., R., W. wolde. Ettm. hire. Herzf. (p. 50) burge for byrig ? Holth. (E. S. xxxvii, 208) on byrg }>aere or walde after b;'rg. MS. atimbram. — 7 Ettm. wunderlicu. — 9 MS., Th., R. bedrasf. wreccan ofer willan ; gewat hyre west J>onan 10 faehjmm feran, forS onette ; dust stone to heofonum, deaw feol on eorfan, niht fort5 gewat : naenig si)>j>an wera gewiste J^ere wihte sio". 31 Ic com legbysig, lace mid winde bewunden mid wuldre, wedre gesomnad, fus forSweges, fyre gebysgad, bearu blowende, byrnende gled. Ful oft mec gesipas sendaS aefter hondum 5 pact mec weras ond wlf wlonce cyssaS. J>onne ic mec onhaebbe, hi onhnigap to me, monige mid miltse, fair ic monnum sceal yean upcyme eadignesse. 32 Is }>es middangeard missenlicum wlsum gewlitegad, wraettum gefrsetwad. Ic seah sellic J>ing singan on rsecede ; wiht waes no [hwaefre] werum on gemonge 10 Ettm. hire. — n MS., Th., Tr. onetteiS. — 12 Sw. feoll. 31 This riddle appears in two different forms in the Exeter Book (108 a, 122 l>). The second of these is defective on account of injury to the MS. Gn., W., J51., and Tr. distinguish these versions as a and b ; the first two making a, the third and fourth b, the basis of text. i a leg bysig ; b lig bysig (not lie bysig, 77., Gn., Tr.) ; Gn., Bl., Tr. lic-bysig ; W. lie bysig. — 2 b After winde some if letters are missing before -dre (wedre), the first being w ( W.) ; W. suggests wunden mit wuldre we-, Tr. wuldre bewunden we-, B. M. reads the lower part of wu. — 3 b gemylted for gebysgad. — 46 Instead of bearu a gap of five letters ( IV.) ; B. M. reads plainly bear. — 6 b basr. b gecyssaft. — 7 a Th. ond hi ; b hi. a onhingaj> ; b onhniga>. — 8 b modge miltsum swa ic mongum sceal. 32 2 Ettm. wraetwum. — 4 Ettm. sio wiht. MS. on werum on ; Th^ Ettn:. omit first on ; Gn., W. no; Herzf. (p. 68) no[wer] ; Cos: (PBB. xxiii, 129) 'no [hwaeflre] (cf. line 8).' 24 RIDDLES OF THE EXETER BOOK sio haefde waestum wundorlicran. 5 Niferweard \at nytte] waes neb hyre, fet ond folme fugele gellce ; no hwaepre fleogan mseg ne fela gongan, hwaej>re fepegeorn fremman onginnetS, • gecoren craeftum cyrreS geneahhe ; i o oft ond gelome eorlum on gemonge sitetS set symble, sales bide]>, hwonne aer|heo craeft hyre cyfan mote [io8b] werum on wonge. Ne heo faer wiht Jngeo1 faes J>e him aet blisse beornas habbaS. 15 Deor domes georn, hlo dumb wunaS ; hwaej>re hyre is on fote fseger hleopor, wynllcu woSgiefu : wrsetllc me finceS hu seo wiht maege wordum lacan furh fot neopan. Fraetwed hyrstum 20 hafaS hyre on halse, fonne hlo hord waraS, baer, beagum deall, bropor sine, maeg mid maegne. Micel is to hycgenne wisum woSboran hwaet [sio] wiht sle. 33 Is ]>es middangeard missenlicum wisum gewlitegad, wraettum gefraetwad. Sifum sellTc ic seah searo hweorfan, 5 Ettm. omits sio, and adds o'Srum after waestum ; Th. note ' r. •waestem.' Th. note wundorlicne ? — 6 MS. ni^erweai^S ; after this Herzf. (p. 68) inserts onhwyrfed or gongende ; Holth. (/. F. iv, 387) geneahhe or genyded. Ettm. suggests after hire (hyre), neat his tela. — 7 Ettm. folma. — 8 Ettm., Gn. ne mseg ne. — 9 Gn. fefte georn. — 12 Ettm. simble. — 13 Th. note 'asr is apparently an error of the scribe? — 14 Th. note on gemonge? — 15 MS. habbad. — 17 Ettm. hyre. — 18 Ettm. tynceS. — 21 Dietr. (xi, 469) 'hordwaraiS (Schatzbesitzer}? — 22 7/4., Ettm. 'baer- beagum (with bearing-rings)? Ettm. sinne. — 23 Th. note maegiSe or msegdne ? Ettm. hycganne ; Siev. (PBB. x, 482) hycgan. — 24 Th. inserts [sio] ; Siev. (PBB. x, 477) resolves sie. 33 i Con. ftis. — 2 Ettm. gewlitegod. Con. wraetum ; W. ' the second t in wrast- tum is above the line in another hand.1 RIDDLES OF THE EXETER BOOK 25 grindan wift greote, giellende faran ; naefde sellicu wiht syne ne folme, 5 exle ne earmas ; sceal on anum fet searoceap swifan, swtye feran, faran ofer feldas ; haefde fela ribba ; mutS waes on middan, moncynne nyt ; fere foddurwelan folcscipe dreogeft, 10 wist in wigeft, ond werum gieldeS gaful geara gehwam }>aes ]>e guman brucatS, rice ond heane. Rece, gif Jm cunne, wls, worda gleaw, hwaet slo wiht sie. 34 Wiht cwom sefter wege wraitlicu lipan, cymlic from ceole cleopode t5 londe, hlinsade hlude ; hleahtor wses gryrellc, egesful on earde, ecge wairon scearpe. | Waes hlo hetegrim, hilde to salne, [IO9a] 5 biter beadoweorca ; bordweallas grof heard ond hipende. Heterune bond, saegde searocraeftig ymb hyre sylfre gesceaft : " Is min modor maegSa cynnes faes deorestan, )>aet is dohtor mm 10 eacen up liden, swa faet is aeldum cuf, 4 Con. greoto. Ettm. gellende. — 6 Ettm. eaxle. — 8 MS. fella. — 10 Th. note faere? Gn. note fela. Con., Ettm. foddarwelan ; Gn. foddorwelan. Th. note draeg'S ? — ii Th. note wegeft? Th., Ettm., Gn. inwige^; Gn.2, W. in wige$. — 12 Con. beneaft/br bruca^S. — 13 Con. conne. — 14 Siev. (PBB. x, 477) resolves sie. 34 i MS., Th. wege ; Th. note waege ? Gn., W. wisege. — 3 MS. leahtor. — 4 Ettm. ecga. — 5 MS., Herzf. (p. 68), Klaeber (M. P. ii, 145) hio ; Th., Ettm., Gn., W. his ; Ettm. note hire ? MS., Th., Gn., W. hete grim ; Ettm., Herzf., Klaeb. hetegrim (And. 1395, 1562). Th. note to seonne ? Herzf. to saege ; Klaeb. 'on wene (cf. on wenum) ' ; Holth. (E. S. xxxvii, 208) ' to cene (nordh. casne).' — 7 Sign of ond not in MS.; Edd. supply this. Ettm. hybende. Cos. (PBB. xxiii, 129), Klaeb. onband (cf. Beow. 501). — 8 Ettm. silfre. — 9 MS. maegda. — 10 Ettm. J>aes for >aet. — n MS. (T.), Th., Ettm. upliden. Ettm. eldum. 26 RIDDLES OF THE EXETER BOOK firum on folce, faet seo on foldan sceal on ealra londa gehwam lissum stondan." 35 Ic wiht geseah in wera burgum seo faet feoh fedeS ; hafaS fela tof a ; nebb bif hyre set nytte, niferweard gongeS, hlfeft holdlice ond to ham tyht5, wjef e6 geond weallas, wyrte seceS ; 5 aa heo fa findetS fa fe faest ne bif; IseteS hio fa wlitigan, wyrtum faeste, stille stondan on stafolwonge, beorhte bllcan, blowan ond growan. 36 Mec se waita wong, wundrum freorig, of his innafe serist cende. Ne wat ic mec beworhtne wulle flysum, haerum furh heahcraeft hygefoncum mm. Wundene me ne beoS wefle, ne ic wearp hafu, 5 ne Jmrh freata gefraecu frsed me ne hlimmeS, ne aet me hrutende hrisil scrifeS, ne mec ohwonan sceal am cnyssan. Wyrmas mec ne awajfan wyrda craeftum fa fe geolo godwebb geatwum fraetwatS. 10 Wile mec mon hwaefre se feah wide ofer eorfan hatan for haelefum hyhtllc gewaede. Saga so^cwidum, searofoncum gleaw, wordum wisjfaest, hwaet fis gewaxie sy. [IO9b] 35 3 GH- neb. — 4 Siev. (PBB. x, 476) resolves tylvS; Cos. (PBB. xxiii, 129) tyhe'S. — 6 Gn. a. 36 5 Ettm. wefla. — 8 MS., Gn.2, W. sceal amas cnyssan ; Th. note, Etim., Gn. uma; Dietr. ama; Holth. (£.S. xxxvii, 208) am sceal cnyssan (Leid. 8). — 9 Ettm. awaefon. — u Gn. mon mec. Herzf. (p. 69) omits se J>eah. — 14 MS., Th., Kl, gewaedu ; R. gewaeda. Ettm. si. RIDDLES OF THE EXETER BOOK 2/ Leiden Riddle Mec se ueta uong, uundrum freorig, ob his innaftge Merest caend[se]. Ni uuat ic mec biuorthae uullan fllusum, herum Serb hehcraeft higido[n]cum [mln]. Uundnae me ni biaS ueflae, nl ic uarp hefse, 5 ni Serih '5rea[t]un grSraec 6ret me hlimmith, ne me hrutendi hrisil scelfaeS, ni mec ou[ua]n[a] aam sceal cnyssa. Uyrmas mec ni auefun uyrdi crseftum Sa Si goelu godueb geatum frsetuath. 10 Uil mec hudrae suaj t>eh uidae ofser eorSu hatan mith h^liSum hyhtlic giusede. Ni anoegu na ic me serigfaerae egsan brogum, Seh Si ni[maen flanas fracjadllcae ob cocrum. Leiden Riddle (MS. Voss. Q. 106, fo. 24 b, in University Library of Leiden in Continental hand of ninth century). This was printed very inaccurately by Beth- mann, Haupts Zeitschrift v (1845), 199. Dietrich (D.) published facsimile, trans- literation, and critical text in the Marburg program, Commentatio de Kynewulfi poetae aetate, 1859-1860. His text was reprinted in Riegers Alt- und angelsach- sisches Lesebuch, Giessen, 1861 (.#.), with critical emendations. In 1885, Sweet (Sw.) printed in his Oldest English Texts a critical text based upon the MS. and also upon "the Leiden librarian's careful transcript of the Riddle by help of re- agents in /86^"(L.). Sweet is followed closely by Kluge, Angelsdchsisches Lesebuch, j888, iSgj (A7.), and by Assmann, G rein-Walker's Bibliothek iii, 205 ( W.). I Two letters erased after ueta. — 2 D., R. h(is). D. aerfest], R. aer[ist], Sw., Kl., W. aerest, Sw. ' may be asrist ? ' — 3 R. biuorhtae. — 4 D., R. b[i]h They conjecture bi hiortan minre or bi hyge (R. hige) minum, L. b[i]gido[cumt], Sw. bigido[n]cum [minum], possibly, hygi-, A7., W. as in text. — 6 D., R. Sreaftjan. D., R. giSr[aece], Sw. ' giiSraec, it is impossible to tell whether last letter is followed by more letters or not.1 D., R. hlimmid, Z. hlimmi(t)d. — 7 D. (MS.), J?. hrutendi, Sw., A7., W. hrutendum. D., R. scelffJaeS. — 8 D., R, o[hwanan] or D. o[hwaer] ; Sw., Kl., W. as in text.— 11 D. hu[e]drae. R. ofer. — 12 R. haettSum. D., R. hihtlic. D. giuae[di] or giuae[de], L. giu[ae]de, Sw. giuaede. — 13 MS., Edd. anoegun, B.-T. (p. 750) as in text (see Dan. 697). — 14 additions partly by D., partly by R. D. reads m/or ni; R., Sw., A'l. ni[man]. R. [fracjaSlice. 28 RIDDLES OF THE EXETER BOOK 37 Ic wiht geseah on wege feran, seo waes wrsetlice wundrum gegierwed : haefde feowere fet under wombe ond ehtuwe, monn h p M [/], wiif m x I kf r, 5 /hors q x x s, ufon on hrycge ; haefde tu fijmi ond twelf eagan ond siex heafdu. Saga hwaet hio wsere. For fl5dwegas ; ne waes }>3et na fugul ana, ac pair waes jeghwylces anra gellcnes, 10 horses ond monnes, hundes ond fugles, ond eac wifes wlite. f>u wast gif )m const to gesecganne, faet we sot5 witan hu faire wihte wise gonge. 38 Ic J>a wihte geseah ; womb waes on hindan f rijjum afrunten ; fegn folgade, maegenrofa man, ond micel haefde gefered, faer \us,fyllo fleah furh his cage. Ne swylteS he symle, fonne syllan sceal 5 innaft )>am oj^rum, ac him eft cymeS b5t in bosme, blsed bip araered ; he sunu wyrcefc, bits him syKa faeder. 37 At close of Bibl., Gn. gives facsimile of 37, after Hickes {Thesaurus, ii, 5), but in his edition of text he does not print the secret script, -which he considers as ' runes.' 4 Th. ehtu)>e; £«.2, W. ehtu we (= ehtun we). MS., W. h w M; Holth. (E. S. xxxvii, 208) as in text. — 5 MS., Th., W. wiif ; Gn. wif. MS., B. M. m x 1 k f w ; W. (misreading) M x I R f w ; Holth. as in text. — 9 Gn. note foldwegas ? 38 i Th., Ettm., Gn. wiht. — 2 Ettm., Gn. bry«um. — 4 MS., Edd. hit felde ; Th. note fyligde ? Gn. note felde ? Dietr. (xi, 472) his filled (see, however, xii, 238)- — 5 Ettm. swilteS. RIDDLES OF THE EXETER BOOK 29 39 Ic ]>a. wiht geseah wsepnedcynnes ; geoguSmyrfe graidig him on gafol forlet fer5fri)>ende feower wellan scire sceotan, on gesceap JnEotan. Mon mafelade, se ]>e me gessegde : 5 " Seo wiht, gif hio gedygeS, duna bricefc ; gif he tobirsteS, bindeS cwice." 40 Gewritu secgafc J>aet seo wiht sy mid moncynne miclum tidum sweotol ond gesyne ; sundorcrseft hafafc maran micle fonne hit men witen. Heo wile gesecan sundor | seghwylcne [i 10] 5 feorhberendra, gewlteft eft feran on weg ; ne bi8 hio njgfre niht far 6)>re, ac hlo sceal wldeferh wreccan laste hamleas hweorfan, no ]>y heanre bip. Ne hafaf5 hio fot ne folm, ne Jefre foldan hran, 10 ne eagena [hafaS] £egj>er twega, ne mut5 hafa)>, ne wij> monnum spraec, ne gewit hafaS ; ac gewritu secgafi J?set seo sy earmost ealra wihta, para J>e aefter gecyndum cenned wiere. 15 Ne hafaft hlo sawle ne feorh ; ac hio sl)>as sceal geond fas wundorworuld wide dreogan. Ne hafaS hio blod ne ban ; hwsefre bearnum wear?) 39 i Th., Gn. wihte. — 2 MS., Edd. -myrwe ; Holth. (E. S. xxxvii, 208) as in text. — $ MS.(T.), Th. fei"S fri)>ende. —4 Th. geotan>r J>eotan ; B.-T. (p. 1053) gesceap)>eotan ('(eats'). 40 i MS., Edd. sy ; Siev. (PBB. x, 477) sie resolved. — 2 MS. iiclum/or tidum. — 4 MS. maram. — 6 Gn. faran. — 8 Th., Gn. wide ferh ; Gn? wideferh. — 10 Gn. no before hafafi (Gn.2 ' misprint1). — 1 1 MS. eagene. Gn. adds hafaft. — 12 Th. spraece. 30 RIDDLES OF THE EXETER BOOK geond fisne middangeard mongum to frofre. Naefre hio heofonum hran ne to helle mot ; 20 ac hio sceal wideferh wuldorcyninges larum lifgan. Long is to secganne hu hyre ealdorgesceaft aefter gongeft, woh wyrda gesceapu ; faet [is] wrsetlic f ing t5 gesecganne ; soft is seghwylc 25 para fe ymb fas wiht[>] wordum becneS. Ne hafaS heo lim as world healdetS ; rice is se reccend ond on ryht cyning, ealra anwalda, eorfan ond heofones healdeS ond wealdeS, swa he hweorfeS ymb fas utan. 5 He mec wraetllce worhte set frymtSe [nob] fa he fisne ymbhwyrft Eerest sette ; heht mec waeccende wunian longe, faet ic ne slepe siffan sefre, ond mec semninga slaep ofergongef, 10 beot5 eagan mm ofestum betyned. 21 Th., Gn. wide ferh ; Cn? wideferh. MS. cyninge. — 22 Siev. (PBB. x, 482) secgan. — 24 Th. adds is. — 26 MS., Edd. wiht ; Holth, (£. S. xxxvii, 208) adds aefre after wiht, or reads bas wiht ymb[e]. — 27 MS. he haenig lim; W. notes that he is certainly written by another hand ; Thorpe sees over the e of he an a, Sch. a scratched-out o; W. (so T. and B. M.) nothing ; Edd. aenig lim. 41 I notice a flaw (cut) in MS. after scyppend (1. i) and world (1. 2), but no •words seem to be missing there. 2 Siev. (PBB. x, 520) declares that wre'Sstubum does not satisfy metrical require- ments and that the sense also demands a jd pers. sing., parallel to healde'S ; Holth. (/. F. iv, 387) -would read weardafl after -stubum. — 3 MS., Th. ric. — 5 MS. swa he ymb >as utan hweorfet?; Gn. note hweorfeiS utan? Siev. (PBB. x, 520) '•per- haps swa he hweorfeiS ymb }>as ?' — 8 Th., Gn. het. — 10 Th. note ac/or ond-stgn ? RIDDLES OF THE EXETER BOOK 31 J>isne middangeard meahtig Dryhten mid his onwalde aeghwjer styreft; swa ic mid waldendes worde ealne )>isne ymbhwyrft utan ymbclyppe. 15 Ic com bleaS to fon }>aet mec bealdlice maeg gearu gongende grima abregan, ond eofore com jgghwser cenra ponne he gebolgen bidsteal giefeS ; ne maeg mec oferswtyan segnberendra 20 senig ofer eorpan nymfe se ana God, se )>isne hean heofon healdej> ond wealdep. Ic com on stence strengre [micle] J>onne ricels o)>J>e rose sy, [}>e swa aenlice] on eor)>an tyrf 25 wynlic weaxeS ; ic com wrastre fonne heo : ]>eah >e lilie sy leof moncynne, beorht on blostman, ic com betre fonne heo; swylce ic nardes stenc nyde oferswT})e mid minre swetnesse symle aighwser ; 30 ond ic fulre com fonne )>is fen svvearte, J>set her yfle adelan stinceS. Eal ic under heofones hwearfte recce, swa me leof fseder laerde aet frym|>e, ]>aet ic ]>a. mid ryhte reccan moste 35 ficce ond J^ynne ; )>inga gehwylces onlicnesse aEghwair healde. Hyrre ic com heofone ; hate)> mec heahcyning his deagol )>ing dyre bihealdan : eac ic under eor)>an eal sceawige 40 worn | wraftscrafu wrajra gsesta. [IIia] 1 6 MS., Edit, to |>on blea~S ; Herzf. (p. 51) as in text. — 17 Spr. i, 494 gearu- gongende. — 23, 25 The additions are by Gn.; W. notes that there is no gap in the MS. — 39 Th. note bihealden ? — 41 Gn.2 worm ? MS. wraft scrafu ; T/i. wom-wraiS- fcraf u (misprint) ; Gn. wrac-scraf u ; Spr. ii, 738, Gn.2 wraiS-scrafu. MS. gesta, 32 RIDDLES OF THE EXETER BOOK Ic com micle yldra ]>onne ymbhwyrft ]>es o]>]>e )>es middangeard meahte geweorpan, ond ic giestron waes geong acenned, . maere to monnum, ]>mh minre modor hrif. 45 Ic com fsegerre fraetwum goldes, peah hit mon awerge wlrum utan ; ic com wyrslicre fonne )>es wudu fula ofrSe )>is waroS ]>e her aworpen ligefc. Ic eorpan com seghwser braidre 50 ond widgielra J>onne ]>es wong grena ; folm mec maeg bifon ond fingras fry utan cape ealle ymbclyppan. Heardra ic com ond caldra fonne se hearda forst, hrim heorugrimma, fonne he to hrusan cymetS ; 55 [ic com] Ulcanus upirnendan leohtan leoman lege hatra. Ic com on goman gena swetra fonne Jm beobread blende mid hunige ; swylce ic com wraj?re J>onne wermod sy 60 i [)>e] her on hyrstum heasewe stonde}>. Ic mesan maeg meahtellcor ond efnetan ealdum J>yrse ; ond ic gesallig maeg symle lifgan, feah ic setes ne sy iefre to feore. 65 Ic maeg fromlicor fleogan fonne pernex of ]>e earn o]>]>e hafoc sefre meahte ; nis zefferus, se swifta wind, 42 MS. J>aes ; 7/4. J>es ; Gn. note waes ? — 47 Tk. note (p. 528) awrige ? — 50 Th. in/or ic ; Gn. [yfele] in eorj>an ; Sch. notes that meter and sense require no addition. — 52 Siev. (P£B. x, 476) resolves -fon. — 55 MS., Th. heoru grimma. — 56 Gn. adds ic com. — 61 Gn. adds )>e. — 63 MS., Th. efn etan. MS., Th. J>yrre ; Th. note tyrse ? — 66 MS., Th. p'nex ; Sch. reads penex and declares that the e is scratched out, but may still be seen, while the accent is not erased ; W. sees no e, and regards the accent as the abbreviation sign customary with p. / see no e (nor does B. M.), but the accent is certainly like the long sign. RIDDLES OF THE EXETER BOOK 33 fget swa fromlice maeg feran seghwser : me is snaegl swiftra, snelra regnwyrm 70 ond fenyce fore hrej>re ; is ]>aes gores sunu gonge hraedra, fone we wifel wordum | nemnaft. [inb] Hefigere ic com micle forme se hara stan o]>]>e unlytel leades clympre ; 7 5 leohtre ic com micel ponne }»es lytla wyrm fe her on flode gaeS fotum dryge. Flinte ic com heardra ]>e ]>is fyr drifep of pissum strongan style heardan ; hnescre ic com micle halsrefepre 80 seo her on winde wseweS on lyfte. Ic eorfan com aeghwaer brsedre ond widgelra J?onne fes wong grena; ic uttor [eafe] eal ymbwinde wraetlice gewefen wundorcraefte. 85 Nis under me Jenig oj>er wiht waldendre on worldlife ; ic com ufor ealra gesceafta, fara J>e worhte waldend user, se mec ana maeg ecan meahtum 90 gefeon prymme faet ic onfunian ne sceal. Mara ic com ond strengra j>onne se micla hwael, se J>e garsecges grund bihealdeS sweartan syne ; ic com swifra J'onne he ; swylce ic com on maegene minum leesse 95 70 MS. snelro J>on ; Th. note snelra se ? — 72 MS. ic for is. — 77 MS., Th. flonde ; Th. note flode ? — 78 W. the second a /« heardra is corrected from e. Gn. se HS. W. notes the erastire of a letter after fyr. — 84 Gn. reads eall ; Holth. (Bb. ix, 358) ana before eal ; Holth. (E. S. xxxvii, 208) supplies eaj>e ; compare line 53. — 86 Th. note of er for under ? — 91 MS., Th. onrinnan ; Th. note onwinnan; Gn. onHnnan ; Gn.z, Spr. ii, 353, B.-T. onbunian (see 462 bimian). — 94 MS., Edd. sweartan syne ; Herzf. (p. 69) sweart ansyne. MS., Th. swi)>re. — 95 Th., Gn. maegne. 34 RIDDLES OF THE EXETER BOOK )>onne se hondwyrm se ]>e hselepa beam, secgas searoj'oncle, seaxe delfaS. Ne hafu ic in heafde hwite loccas, . wraiste gewundne, ac ic com wide calu ; ne ic breaga ne bruna brucan m5ste, 100 ac mec bescyrede scyppend eallum : nu me wrsetlice weaxaS on heafde paet me on gescyldrum scinan motan ful wrjgtllce wundne loccas. Mara ic com ond fjgttra fonne amsested swin, 105 bearg bellende, [)>e] on bocwuda won wrotende wynnum lifde faet he 42 edniwu [ii2a] ]>aet is moddor monigra cynna, faes selestan, faes sweartestan, faes d core stan, ]>&s ]>e dryhta beam ofer foldan sceat t5 gefean agen. 5 Ne magon we her in eorfan owiht lifgan, nymt5e we brucen faes fa beam dot5. f>set is to gefencanne feoda gehwylcum, wisfaestum werum, hwaet seo wiht sy. 43 Ic seah wyhte wraetlice twa undearnunga ute plegan 103 Gn. moton. — 106 Bright [)>e]. — 108 Th. ' here a leaf of the MS. is mani- festly -wanting containing the end of this and the beginning of the following enigma: W. perceives no gap in the MS. [>aet he closes the page}, hit below, in another hand and in other ink, almost obliterated hit is ; then about twelve letters which he is tin- able to decipher. These seem to me to be sio creatura pr. 42 6 Gn. on. — 7 Siev. (PBB. x, 477) do« resolved; Holth. (Bb. ix, 358) do[a]«. - 8 Siev. (PBB. x, 482) ge)>encan. — 9 Siev. (PBB. x, 477) resolves sy. 43 2 Siev. (PBB. x, 520) ' perhaps plegian.' RIDDLES OF THE EXETER BOOK 35 haimedlaces ; hwltloc anfeng wlanc under waidum, gif faes weorces speow, faemne fyllo. Ic on flette maeg 5 furh runstafas rincum secgan, fam fe bee witan, bega aetsomne naman ]>ara wihta. f>ser sceal Nyd wesan twega ofer ond se torhta y£sc an an Hnan, Acas twegen, 10 Haegelas swa some. Hwylc faes hordgates caegan craefte fa clamme onleac fe fa raidellan wiS rynemenn hygefaeste heold heortan bewrigene orfoncbendum ? Nu is undyrne 15 werum aet wme hu fa wihte mid us, heanmode twa, hatne sindon. 44 Ic wat indryhtne aefelum deorne giest in geardum, fam se grimma ne maeg hungor sceSSan, ne se hata furst, yldo ne adle, gif him arllce esne > enaS se f e agan sceal 5 on fam siSfaete. Hy gesunde act ham findaft witode him wiste ond blisse, cnosles unnm ; care, gif se esne his hlaforde | hyretS yfle, [ii2b] 3 Gn. onfeng. — 4 MS. speop. — 7 MS. J>a. — 10 T/i., Gn. anan linan. — n Spr. i, 121 hwylc = '«' qui ' or lst quis? MS. waes; Th. t>aes. — 12 Th. note clammas ? — 13 B.-T. s.v. raedels has raedelsan ? — 14 Gn. beheold. — 17 Gn. note heah- ? Spr. ii, 48 heah mode. As Sch. notes, there is no division between this riddle and the next; hatne sindon is followed on same line by Ic wat (441). 44 4 Th. note, Cos. (PBB. xxiii, 130) adl. — 4, 5 Gn., W. add after adle, ne se enga deafi (compare Ph. 52), and after sceal, his geongorscipe. Cos. (PBB. xxiii, 130) rejects these additions. — 5 Cos. se J>e = J>one )>e. Gn. agan. — 6 MS., Th. siftfate. MS., Th. hyge sunde ; Th. note 'r. sundne (a sound mind).'' — 8 Th. note ' before care a "word, perhaps butan, is omitted.1 36 RIDDLES OF THE EXETER BOOK freanonfore; ne wile forht wesan 10 brojjor 6)>rum : him }>set bam sceSeS, }>onne hy from bearme begen hweorfaft anre magan ellorfuse moddor ond sweostor. Mon, se fe wille, cy]>e cynewordum hu se cuma hatte 15 et5j>a se esne J>e ic her ymb sprice. 45 Wrsetlic hongaS bi weres ]>eo, frean under sceate ; foran is fyrel ; biS sti}> ond heard, stede hafaS godne, fonne se esne his agen hraegl ofer cneo hefeS, wile faet cufe hoi 5 mid his hangellan heafde gretan fset he efenlang aer oft gefylde. 46 Ic on wincle gefraegn weaxan nathwaet, J>indan ond punian, fecene hebban. On fact banlease bryd grapode hygewlonc hondum ; hraegle feahte frindende J>ing peodnes dohtor. 5 47 Wer saet set wine mid his wifum twam ond his twegen suno ond his twa dohtor, 10 Klaeb. (M.P. ii, 145) regards the second half -line as parenthetical. — 16 Gn. note ofrSe ? MS., Th. sprice ; Gn., W. sprece ; compare 24". 45 i Siev. (PBB. x, 478) resolves >eo ; Cos. (PBB. xxiii, 129) J>eo(h)e. — 7 MS. (T.), Th., Gn. efe lang; Th. note efne lang? Gn.2, W. efelang; Tr. (BB. xix, 192) efen-lang. 46 i MS. win cle. MS., Th., Gn., W. weax; Dietr. (xi, 474) 'weax (y«>weacs, etwas weicAes)' or weaxan ; Herzf. (p. 69) weascan ; Holth. (I. F. iv, 367) weaxan ; Siev. (PBB. x, 520) suggests a genitive, i.e. waces. — 2 Dietr. (xi, 474) J>enian (sich dehnett). — 5 Th. J>indende ; Gn. note hrintende ? 47 i MS., Con. Waer. Con. wifa. Con. omits twam. — 2 Con., Ettm., Gn. suna. RIDDLES OF THE EXETER BOOK 37 swase gesweostor ond hyra suno twegen, freolico frumbearn ; faeder waes fser inne fara aepelinga aighwaeftres mid, 5 earn ond nefa. Ealra waeron fife eorla ond idesa insittendra. MoflSe word fraet ; me J>aet Jmhte wratlicu wyrd, )>a ic faet wundor gefraegn, ]>aet se wyrm forswealg wera gied sumes, )>eof in J>ystro, ]>rymfaestne cwide M x ond paes strangan stapol. Staelgiest ne waes 5 wihte ]>y gleawra | fe he )>am wordum swealg. [n3a] 49 Ic gefraegn for haelejmm hring [serjendean torhtne butan tungan, tila feah he hlude stefne ne cirmde strongum wordum. Sine for secgum swigende cwaeS : " Gehsele mec, helpend gaesta ! " 5 Ryne ongietan readan goldes guman galdorcwide, gleawe bepencan hyra haelo to Code, swa se hring gecwae^S. 50 Ic wat eardfaestne anne standan deafne dumban, se oft daeges swilgeS 3 Ettm. gesweoster. " MS., Con., Th. hyre ; Ettm. hira ; Gn., W. hyra. Con., Ettm. suna. — 4 Con., Ettm. freolicu. — 5 Con. Ettm. aeghwae'Seres. 48 2 Sw. wraettlicu. — 3 Sw. giedd. — 4 Sw. Jrymmfaestne. — 6 Between 48 and ^g there is no spacing in the MS., not even a closing sign ; swealg (6) is followed on the same line by Ic gefraegn (4Q1). 49 I MS. fer; Edd. for. MS., Th. hringende an ; Gn., W. hring [aerjendean; Klaeb. (M.P. ii, 145) hring aendean (or endean) = aerndean < aerendian. — 2 After tila no gap in MS.; Gn., W. supply reordian and thus complete hemistich; Siev. (PBB. xii, 479) begins a new verse 'with stefne; as does Klaeb. (M.P. ii, 145), who reads as in text, tila >eah he hlude | stefne ne cirmde. — 7 MS., Edd. bejmncan ; Gn. note be^encan ? 38 RIDDLES OF THE EXETER BOOK Jmm gopes hond gifrum lacum. Hwilum on pam wicum se wonna pegn, sweart ond saloneb, sendeS opre 5 under g5man him golde dyrran, W^ fa sepelingas oft wilniaS, cyningas ond cwene. Ic pset cyn nu gen nemnan ne wille, pe him to nytte swa ond to dugpum dop paet se dumba her, 10 y_ eorp unwita, ser forswilgeS. d^'J 51 Wiga is on eorpan wundrum acenned dryhtum to nytte, of dumbum twam • torht atyhted, pone on teon wigeS feond his feonde. Forstrangne oft wif hine wrl'5 ; he him wel hereS, 5 feowa)) him gepwaere, gif him pegniatJ maegeft ond maecgas mid gemete ryhte, fedat5 hine fsegre ; he him fremum stepeS life on lissum. Leanaft grimme ]>e hine wloncne weorpan IseteS. 10 52 Ic seah wreetlice wuhte feower samed sipian; swearte | waeran lastas, [ii3b] swajm swipe blacu. Swift wses on fore fultumfromra, fleag on lyfte, 50 3 Th. note geapes ? Gn. 'gopes (vgl. altn. hergopa serva ?).' — 4 MS., Th. hwilu mon. — 6 Gn. omits him. — 10 Gn.2, W. de}>. — 1 1 MS. fer swilgeft ; Edd. forswilge'S. 514 MS. fer strangne ; Edd. forstrangne. — 5 Siev. (PBB. x, 476) resolves wii5. — 8 Gn. stepeft ; Gn. note he hi fremum stepeft ? Siev. (PBB. x, 456), stepeiS. 52 4 MS., TA., Gn. fuglum frumra (the u of MS. frumra may be an a with its top faintly marked} ; Th. note fromra; (J«.2, W. framra; Tr. (BB. xix, 195) fugla fultum. MS., W., Barnouw (p. 221) fleotgan lyfte; Th. note fleogan; Gn. note 'fleotga (Schwimmer) on lyfte (so also Dicht.; Spr. i, 304 celer, velox) oder fleat geond lyfte'; Cos. (PBB. xxiii, 130) ' fleog (= fleag) an lyfte (cf. 2316)'; Tr. fleag geond lyfte. RIDDLES OF THE EXETER BOOK 39 . deaf under y];e. Dreag unstille 5 winnende wiga se him wegas taecnep ofer fseted gold, feower eallum. 53 Ic seah rsepingas in rseced fergan under hrof sales hearde twegen, ]>a wseron genamnan nearwum bendum gefeterade fseste togaedre. J>ara 5)>rum wses an getenge 5 won f ah Wale, seo weold hyra bega sife bendum fsestra. 54 Ic seah on bearwe beam hllfian tanum torhtne ; |>set treow wses on wynne, wudu weaxende ; wseter hine ond eorj?e feddan faigre, opfaet he frod dagum on oj>rum wearS aglachade 5 deope gedolgod, dumb in bendum, wrifen ofer wunda, wonnum hyrstum foran gefrsetwed. Nu he fsecnum weg furh his heafdes maegen hildegieste ofrum rymeS. Oft hy on yste strudon 10 hord getgaedre ; hraed waes ond unlaet 6 MS., Th. waegas ; Th. note wegas ? 53 3 MS., Th., Gn., Dietr. (xi, 476) genamne ; Th. note, Tr. (BB. xix, 198) genumne ; Holth. (£. S. xxxvii, 209) genamnan. — 4 Tr. to gaedere. — 6 Gn. note wonfeax? Cos. (PBB. xxiii, 130) 'wonf(e)ahs (cf. Rid. 13, wonfeax).' 54 2 Holth. (Bb. ix, 358) omits J>aet. — 8 MS., Th. faecnum waeg; Th. note frec- num weg ? — 9 MS., Th. masg ; Th. note maegen ? — 10 MS. ( W.) hy an yst (not he an yst, Th., Gn.); Th. note 'hi on yst (they furiously}1 ; Dietr. (xii, 251-252) 'oft hea (fur heo, hi) nyst strudon (oft raubten sie mundvorrath) ' ; Gn., W. hi earyst ; Gn. note earyst = earust, alacerrime ; Klaeb. (M. P. ii, 145) oft hy anys (anes). — 1 1 Th. note heard ? 40 RIDDLES OF THE EXETER BOOK se aeftera, gif se serra faer, genamna in nearowe, nepan moste. 55 Hyse cwom gangan, ]>xr he hie wisse stondan in wincle ; stop feorran to hror haegstealdmon, hof his agen hraegl hondum Op, hrand under gyrdels hyre stondendre stipes nathwaet, 5 worhte his willan, wagedan buta ; fegn onnette, waes J>ragum nyt tillic esne ; teorode hwae)>ral» set stunda|gehwam strong air fonne hlo, [JI4a] werig }>aes weorces. Hyre weaxan ongon 10 under gyrdelse )>aet oft gode men ferSfum freogatJ ond mid feo bicgaSt 56_ Ic seah in healle, peer haeleft druncon, on net beran feower cynna : ' wraetllc wudutreow ond wunden gold, sine searobunden, ond seolfres dal, ond rode tacn faes us to roderum up 5 hlsedre rserde, ser he helwara burg abraece. Ic faes beames maeg eafe for eorlum ae}>elu secgan : fair waes hlin ond ac, ond se hearda iw, 12 MS. fxr genamnan ; Th., Gn., W. fasr genam|nan; Holth. (Bb. ix, 358) closes the line with faer and regards genam as the beginning of a lost line ; Holth. (E.S. xxxvii, 208) reads [on] faerj genamnan, and compares 533, genamne ; Bright suggests genamna, but prefers genumne (so also 533). 55 i Th., Gn. J>ar. — 2 AfS. wine sele ; Th., W. win-sele ; Gn. wincle (wrongly citing this as Thorpe's suggestion for supposed MS. readingvi'mc, notvrinc sele). Holth. (£.S. xxxvii, 209) < on sta^ole (cf. Rid. 887).' — 4 MS., Th. rand. — 5 Th. stondenre. — 7 Th. onette. — 9 MS., Th. aer K>n hie (not hi, Gn.) 6 ; Gn., W. as in text. — 12 Gn. ferSum. 56 I MS., Edd. heall ; Th. note, Holth. (Bb. ix, 358) healle ? Cf. s613, 6O1. — 9 Th. note 'hlind/0r lind?' MS. ace. RIDDLES OF THE EXETER BOOK 41 ond se fealwa holen ; frean sindon ealle 10 nyt setgaedre, naman habbaft anne, wulfheafedtreo, faet oft waEpen absed his mondryhtne, mafcm in healle, goldhilted sweord. Nu me gieddes fusses ondsware ywe, se hine onmede 15 wordum secgan hu se wudu hatte. 57 Ic waes paer inne, pser ic ane geseah winnende wiht wido bennegean, holt hweorfende ; heapoglemma feng, deopra dolga ; daropas wseron weo paire wihte ond se wudu searwum 5 faeste gebunden. Hyre fota waes biidfaest ofer, ofer bisgo dreag, leolc on lyfte, hwilum londe neah. Treow waes getenge fam fser torhtan stod leafum bihongen. Ic lafe geseah 10 mmum hlaforde, faer haeletJ druncon, a on flet beran. 58 Deos lyft byreS lytle wihte ofer beorghleofa, pa sind | blace swife, [TI4b] 12 TA., Gn. wulfheafed treo. Th. note ' abad (awaited) ? ' — 14 AfS., Edd. J>isses gieddes ; Herzf. (pp. 43-44), on metrical grounds, gieddes J>ysses ; Holtli. (£. S. xxxvii, 209) adds mon after )>isses gieddes. — 15 MS., Th., Cos. (PBB. xxiii, 130), Holth. (£.S. xxxvii, 209) onmede; Gn., IV., Liebermann (Archiv cxiv, 163) on mede. 57 2 MS., Th. wido benne gean ; Th. note wide benna (against wide wounds') ? — 3 Gn. hwearfende. — 5 Th. note wea? Dietrich (xii, 238, N.) wea; Lange (ib.) wiS. — 7 Gn. bidfaest. — 9 MS., Th. torht anstod; Gn., W. as in text. — 12 MS., Th. flan ; Th. note '•some lines are here apparently wanting'1; Gn. adds geweorca; so W.; cf., however, El. 285, J>aera leoda. 58 i Tr. (BB. xix, 189) lihte. — 2 MS., Th., Sw., W. -hleo^a (see 37); Gn., Tr. -hleoj>u. 42 RIDDLES OF THE EXETER BOOK swearte, salopade. Sanges rofe heapum feraS, hlude cirmaS ; tredaS bearonaessas, hwilum burgsalo 5 nij>}>a bearna. NemnaS hy sylfe. 59 Ic wat anfete ellen dreogan wiht on wonge. Wide ne fereS, ne fela ridetS, ne fleogan mseg furh sclrne daag, ne hie scip fereS, naca naegledbord ; nyt biS hwaepre 5 hyre [mon]dryhtne monegum tidum. Hafat5 hefigne steort, heafod lytel, tungan lange, toS neenigne, isernes dsel ; eorSgraef psepeS. Wsetan ne swelgej), ne wiht itej>, 10 fodres ne gitsaS, fereft oft swa feah lagoflod on lyfte ; life ne gielpeS, hlafordes gifum, hyreS swa ]>eana feodne sinum. J>ry sind in naman ryhte runstafas, ]>ara is Rad fultum. 15 60 Ic seah in healle hring gyldenne men sceawian, modum gleawe, fer]>]mm frode. Fri))ospe[de] baed God nergende gaeste sinum se >e wende wripan, word sefter cwasS, 5 hring on hyrede Haelend nemde 3 MS., Th. rope; Th. note, Gn., Sw., Brooke (E.E.L. p. 149), Cos. (PBB. xxiii, 130) rowe; Gn. note, W., Tr. rofe. — 5 Th., Gn. traedaiS. 59 3 Gn. ne before maeg. — 6 Th., Gn., W. [mon]. — n MS., Th. fodres. — 1 5 MS., Th., Gn. f urum ; Th. note feor)>a ? Gn. note fruma or forma ; Dietr. (xi, 477) fur-Sum; Gn?, Spr. \, 356, W. fultum; Holth. (/. F. iv, 387) furma. 60 i MS. gylddenne. — 3 Gn. ferSum. MS. fri^o spe (end of line) baed; Th, as in text. RIDDLES OF THE EXETER BOOK 43 tillfremmendra. Him torhte in gemynd his Dryhtnes naman dumba brohte ond in eagna gesihft, gif pass aepel[est]an go Ides tacen ongietan cuj>e 10 ond Dryhtnes dolg, don swa )?ses beages benne cwiedon. Ne fcare frene mceg jeniges monnes ungefullodre Godes ealdorburg gaest gesecan, rodera ceastre. Raede se ]>e wille 15 •Ijfi ftaes wraitllcan wunda cwaeden [hringes to haefepum, J>a he in healle waes [I]t5a] wylted ond wended wloncra folmum. 61 Ic waes be sonde, saewealle neah, aet merefarope, mlnum gewunade frumstaj>ole faest; fea senig waes monna cynnes, J>aet mlnne fair on anaede card beheolde, 5 ac mec uhtna gehwam y5 sio brune lagufaeSme beleolc. Lyt ic wende fset ic aer o]>]>e sI8 aefre sceolde [J23a] ofer meodu[bence] muSleas sprecan, wordum wrixlan. J>aet is wundres dael 10 9 MS., Edd. aehelan; R., Holth. (Bb. ix, 358) aef>el[est]an. — n MS.(T.) dryht dolg don ; Th. notes that ' this is apparently corrupt and without an alliterating line — dryht-dolg d5n ?' Gn., W. dryht dolgdon; Dietr. (xii, 235) J>one dysige dryht dolgdon furSum. — 12 MS., Edd. ne maeg h>aere bene ; Gn., W. [to J>aes beages dolgum] ; Hollh. (Bb. ix, 358) notes that this is metrically false. — 13 AfS., Th. ungafullodre ; Th. note ungefyllodre ? Gn., W. ungefullodre ; Cos. (PBB. xxiii, 130) ungefullodra (gen.pl.). 61 This riddle begins upon leaf 122 b, five lines from the bottom ; it is immediately preceded by$\b and is followed by The Httsband's Message and The Ruin (i23-i24b). I MS. a ^/"sande is changed to o; Th., Ettm., Gn. sande. AfS., Th. sae wealle. — 5 .£#»/., anede. — 7 Th. note beleac ? — 9 Gn. adds bence, (7«.2 drincende, ac- cepted by W., Bl. No gap in MS. 44 RIDDLES OF THE EXETER BOOK on sefan searolic pam >e swilc ne conn, hu mec seaxes ord ond seo swtyre bond, eorles inge}>onc ond ord somod, )>ingum gefydan, ]>set ic wi)> J>e sceolde for unc anum twam aerendspraice 15 abeodan bealdlice, swa hit beorna ma uncre wordcwidas widdor ne mjgnden. 62 Oft mec fseste bileac freolicu meowle [i24b mid] t ides on earce, hwilum up ateah folmum sinum ond frean sealde, holdum ]> eodne, swa hio haten waes. SiSfan me on hrepre heafod sticade, 5 niopan upweardne on nearo fegde. Gif faes ondfengan ellen dohte, mec fraetwedne fyllan sceolde ruwes nathwaet. Ried hwast ic mjgne. 63 Ic com heard ond scearp, hingonges strong, forSstyes from, frean unforcOt? ; wade under wambe ond me weg sylfa ryhtne geryme. Rinc bi6 on ofeste [I25a] se mec on }>yS aeftanweardne 5 haeleS mid hraegle, hwilum ut tyhS of hole hatne, hwilum eft fareS 12 MS. seaxe^5; Edd. seaxes. — 13 Herzf. (p. 69) ecg for ord, on account of awkwardness of repetition. — 14 Ettm. gebydon. — 15 MS. twan; Edd. twam. — 17 Ettm. widor. Gn. maendon. 62 i MS. oft, not of as TA., Gn. state. — 8 MS., Edd. >e before mec. MS., Holth. (Bb. ix, 358) fraetwedne ; Edd. fraetwede. 63 i MS., Th., Gn. ingonges ; Gn. note hingonges ? so Gn?, W. — 4 Th. geryne. — 5 Siev. (PBB. x, 477) resolves hy« ; Holth. (Bb. ix, 358) )>y[e]iS. — 6 Siev. (PBB. x, 476) resolves tyhS; Cos. (PBB. xxiii, 129) tyheft. — 7 Th. eft-fare^; Gn. note f ege« ? RIDDLES OF THE EXETER BOOK 45 on nearo nathwaSr, nydef swipe superne secg. Saga hwset ic hatte. 64 Oft ic secga seledreame sceal faigre onfeon J'onne ic com for$ boren, glaed mid golde, pair guman drincaS. Hwilum mec on cofan cysseS mupe tilllc esne f§er wit tu beop, 5 faeSme on folm[e] [fin]grum pyS, wyrceft his willan . . ft lu . . . . . fulre J>onne ic forfi cyme Ne maeg ic ]>y mlpan 10 sian on leohte swylce eac biS sona te getacnad, hwaet me to 15 . . . leas rinc, |>a unc geryde waes. 64 I MS. secgan ; Edd. secga. — 2 Siev. (PBB. x, 476) resolves -J>eon. — 5 Siev. (P£B. x, 477) resolves beo5. — 6 Th. faeftir . grum; Gn. supplies [beclyppe^S, finjgrum; Dietr. (xi, 479) adds [bifeh'5 and finjgrum; Sch. [on fblm] grum; W. (so T.) reads the upper half of on folm, then a gap of about four letters (Sch. five). Holth. (Bb. ix, 358) )>y[e]$. — 7 Th. willan ; W. the n is no longer visible. Sch. about twenty-one letters missing; W. the fifth appears to have been "S, the sixth 1? / read clearly 1 ; B. M. gives $ and the top of lu ; Dietr. [ne weor'Se ic swa }>eah] . — 8 Dietr. [on fae'Sme }>y]. — 9 Th., Gn., gap in MS. ; Dietr. no gap ; Sch. about twenty- three letters missing after foi"5-cyme. — 10, n Dietr. adds [>>aet me se mon dyde| >aer min sweora (?) bi'5 gese]wen; Sch. after mi)>an about twenty letters are missing, then ban (not wan, Th., Gn.) ; W. sees still the lower part of> before J>an ; so do I. — 12 Th. gap in MS. ; Gn. no gap ; Sch. about twenty-four letters missing after leohte. — 13, 14 Sch. between sona and getacnad about seventeen letters are lacking; Th., Gn. read te before getacnad ; W. sees before te some marks, perhaps rn ; Dietr. supplies [sweotol on eorle|fela tealtriendum on fo]te ; Gn., Dietr. getacnod. — 15 Sch. after to about nine letters are missing; Dietr. inserts [bysmere se bealda teode]. — 16 Dietr. [raedjleas; Holth. (/. F. iv, 387) [sum raed-] ; (Bb. ix, 358) perhaps [rece-]. 7 see the bottom curves of two letters, perhaps ce ; so B. M. 46 RIDDLES OF THE EXETER BOOK 5 Ic seah P" ond X ofer wong faran, beran £ H ; baem wses on stype haebbendes hyht, N ond K, swylce Jrypa dsel, f ond H ; gefeah p ond K, fleah ofer T, 5 l/| ond tj sylfes pses folces. 66 Cwico waes ic, ne cwseS ic wiht ; cwele ic efne se feah ; ser ic wses, eft ic cwom ; seghwa mec reafaS, hafa<5 mec on headre ond mm heafod scirep, biteS mec on baer He, briceS mine wisan. Monnan ic ne bite, nym}>e he me bite ; 5 sindan para monige ]>e mec bltaS. 67 Ic com mare fonne fes middangeard, laesse ]>onne hond|wyrm, leohtre fonne mona, [I25b] swiftre J>onne sunne. Sses me sind ealle flodas on fae^mum ond fes foldan bearm, grene wongas ; grundum ic hrine, 5 helle underhnlge, heofonas oferstige, wuldres efel ; wide raece ofer engla card ; eorpan gefylle, 65 2 MS., Edd. si>>e; Holth. (Ph. ix, 358) stye. — 3 Holth. H. A [samod], with omission of ond. Gn. A (misprint for A). — 4 MS., Th., Gn., Hick. (Anglia x, 597) f>, W. P-. Holth. W E [samod]. — 5 Tr. (Bb. v, 50) \for F. — 5, 6 Holth. supplies and before fleah and swylce before S-rune. 66 3 Th. note heaiSre ? — 4 MS., Th. onbaerlic ('secretly'). — 5 MS. nymp^e (w^nymhe, Th.,Gn.; not nymppe, Sch.) ; Holth. (Bb. ix, 358) sustains phonetically the MS. form ; Edd. nymbe. 67 i Con. Son Daes. MS. mindangeard. — 4 MS., Con., Th., Ettm., Gn. J?as: Gn. note, Gn.2 J>es. Ettm. note bearmas ? — 6 Con. heofenes. RIDDLES OF THE EXETER BOOK 47 ealne middangeard ond merestreamas side mid me sylfum. Saga hwaet ic hatte. 10 68 Ic on finge gefrsegn )>eodcyninges wratllce wiht word galdra .... hio symle deft fira gehw\am~\ 5 wisdome wundor me ]>Kt w . naenne muS hafafc, fet ne f[olme] welan oft sacaft, 10 cwtyeo" cynn • •{'.• . . wearS leoda lareow, forfon nu longe mag [on] [awa t5] ealdre ece lifgan missenllce fenden menn bugaS 15 eorfan sceatas. Ic ]>set oft geseah golde gegierwed, J>aer guman druncon, 9 MS., Con., Ettm. ealdne. — 10 Con. rn.ec. Con., Ettm. selfum. 68 Omitted by Th., Gn. i In MS. I is no longer visible ; B.M. gives top of this. Sch. J?in . . . beodcyninges ; W. sees still the upper part of a g, then a gap of two letters, then ef raegn ; B.M. reads >mg(top of e) and (top 0/"g)efraeTn (sic). — 2 B. M. incorrectly raetlice. Holth. (Bb. ix, 358) wordgaldra. Sch. after galdra some twenty-four letters are missing. — 3 Seven letters before hio, B.M. reads snytt, not seen by Sch., IV. — 4 Sch. after gel, a gap of perhaps twenty-six letters ; instead tf/"gel (Sch., W.), B.M. reads gehw ? — 6 wi, added by Sch., is still seen by W. and by me. Sch., W. >a. . . . w ? B. M. }>aet w . . . ; W. sees of w only the lower part ; after this some huenty-eight letters are missing (Sch.). — 8, 9 MS. (Sch., W.) enne ; B. M. naenne. Holth. (Bb. ix, 358) suggests [n]ejme and f[olme]. — 9 Sch. fet in ? [f] ? W. reads fet. ne, then under the line a long stroke (seen by B.M. and by me) ; then about twenty-ser>en letters are lacking (Sch.). — II W. reads cynn (I see lower part), not seen by Sch.; then a gap of some eighteen letters (Sch. twenty-two). — 13 W. (so I) reads mag, not seen by Sch.; then about seven missing letters (Sch. ten). — 13, 14 Holth. (Anglia xxiv, 264) proposes mag[on] | [awa to] ealdre. 48 RIDDLES OF THE EXETER BOOK since ond seolfre. Secge se ]>e cunne, wlsfsestra hvvylc, hwaet seo wiht sy . . 69 (Gn. 68) Ic ]>a. wiht geseah on weg feran ; heo wses wraetllce wundrum gegierwed. Wundor wearS on wege : wseter wearS to bane. 70 (Gn. 69) Wiht is wrjgtlic pam J>e hyre wisan ne conn : singeS Jmrh sidan ; is se sweora woh orfoncum geworht ; hafaf eaxle twa scearp on gescyldrum. His gesceapo [dreogeS], | ]>e swa wrsetlice be wege stonde, 5 heah ond hleortorht, haelejmm to nytte. 71 (Gn. 70) Ic com rices aeht reade bewsefed. Stl5 ond steap wong, stafol wses iu )>a wyrta wlitetorhtra : nu com wra)?ra laf , fyres ond feole, fseste genearwad, wire geweorpad. Wepe8 hwilum 5 for gripe minum se ]>e gold wige<5, })onne ic y]>a.n sceal fe 19 Holth. (Bb. ix, 358) siefor sy. 6g I Gn. wihte. Gn. note on waeg ? Gn. faran. — 2 MS. sign of closing after gegierwed (W.), and Wundor begins new line with capital (T.); so Th. prints I. 3 as a separate riddle. This is Tr.'s vie^v. Cf. 372"8. — 3 Gn. note waege ? 70 i MS. hyra. — 3 MS., Th. tua. — 4 Th. note hyre ? No gap in MS.; Gn. supplies [dreogeft]. — 5 Th. note stondaft? Gn. note be waege stondeiS? 71 2 Holth. (Bb. ix, 358) steapwong. Th. wong-sta>ol. Th. iu-J>a. — 3 MS., Th. wlite torhtra. — 5 Th. note gewreo)>ad (gewri^od). — 6 MS., Edd. minum gripe ; Holth. (£. S. xxxvii, 209) gripe minum. Th. note wegeiS ? — 7 Gn. note ywan ? Th., Gn. close the riddle -with sceal, and take bete (1. 10) with the next riddle, at end of first full line. After sceal some nine letters are missing (Sch.). Before hringum I see at end of line the upper stroke of a letter, then a missing letter, then se (B. M. fe). RIDDLES OF THE EXETER BOOK 49 hringum gehyrsted me bil . . . . . . . dryhtne mm wlite bete 10 72 (Gn. 71) Ic waes lytel some . . . . . ante geaf we ]>e unc gemaene sweostor mm fedde mec [faegre] ; oft ic feower teah 5 swsese bropor, ]> ara onsundran gehwylc dsegtidum me drincan sealde Jmrh J>yrel fearle. Ic faeh on lust, offset ic wses yldra ond fset anforlet 8 Sch. gehy[rsted] [me], and then twenty-three missing letters ; W. (so B. M. and /) reads the upper half of rsted me, then bil (?), then some twenty missing letters; Holth. (Anglia xxiv, 264) bi}> for bil (W). — 9 Sch. after min, a gap of some twenty-one letters. Above wlite B. M. reads go. — 10 Sch. wlite is the last word of the line ; under it is bete : 7 On account of the closing sign Sch., unlike Th., regards bete as belonging to this riddle, and as a part of a perhaps shorter end-line. W. be- lieves that there is no gap before bete, out that as last word it is written, as is com- mon, at the right end of the next line \see 38, 46, 54, 86]. Before bete is also a sign \very common in Riddles] that refers it to the preceding line (W.}. I agree with Sch. and W. 72 I, 2 Th., Gn. Ic waes bete ; Sch. Ic waes . . . (about twenty letters) . . . geaf ; W. reads after waes the upper part of lyt and before geaf, ante (the lower part of an) ; Holth. (Anglia xxiv, 264) proposes [br]ante geaf[las]. I read after lyt clearly e and upper part of\ (not seen by B.M.), and at beginning of line, halfway between lytel and ante, so clearly and then m (?). B.M. reads so and the greater part of me. After geaf, Th., Gn. give no gap ; Sch., W. a gap of some thirty- two letters. — 3 MS. ( W., T.) we J>e unc gemaene ; Th., Gn., Sch. we unc gemaene. After gemaene some nineteen letters are missing. Dietr. (xi, 481) proposes (1-3) : Ic waes [of hame adrifen, hearm minne] bete, se f>e me gemaeccean geaf, we unc gemaene [oft] [swiftas asetton ; ic ond] sweostor min. — 5 e in mec is worn away ( IV.) ; after mec Sch. sees a gap of some eleven letters ; Gn.2 supplies faegre; Dietr. supplies frodra sum; Herzf. (p. 70) ful faegre and \cf. Si8, 54). B. M. reads oft ic, not seen by Sch., W., or by me. — 6 Th., Gn., Dietr. >ara J>e. — 8 Holth. (Bb. ix, 358) J>ah. — 9 Th. note )>onne/0r J>aet ? Th. an-forlet ; Gn., W. an forlet. [j0 RIDDLES OF THE EXETER BOOK sweartum hyrde, si)>ade widdor, mearcpafas Walas traed, moras paeSde bunden under beame, beag haefde on healse, wean on laste weorc frowade, earfofta dsel. Oft mec Isern scod sare on sldan ; ic swigade, njgfre meldade monna sengum, gif me ordstaepe egle wairon. 73 (Gn. 72) Ic on wonge aweox, wunode ]>£er mec feddon hruse|ond heofonwolcn, o)>}>aet onhwyrfdon me [i26b~ gearum frodne, fa me grome wurdon, of ]?aere gecynde ]>e ic §er cwic beheold, onwendan mine wisan, wegedon mec of earde, gedydon faet ic sceolde wij> gesceape minum on bonan willan bugan hwilum. Nu tomfrean mines folme bysigo[d] dlan dail, gif his ellen deag, o}>J>e aefter dome ri dan mjgrfa fremman, wyrcan we . '. ec on feode utan we pe ond to wrohtstaf [urn] n eorp, eaxle gegyrde wo ond swiora smael, sldan fealwe fonne mec heafosigel scir bescmeS ond mec 20 it Gn. note Wala? — 12 Th. note bearme? Gn. beah. — 14 c in mec appears effaced ( W.) ; I read it easily. — 1 7 MS., Th. ord staepe. 73 I MS. wonode ; Edd. wunode. — 2 MS., Gn. heofon wlonc ; Th. heofon- wlonc; Gn.2, W. heofonwolcn. MS., Edd. me onhwyrfdon; Herzf. (p. 44). onhwyrfdon me. — 5 Gn. wise. — 8 MS., Edd. mines frean. RIDDLES OF THE EXETER BOOK 51 8-20 Gn. supplies, on basis of Tit's text of MS.: Nu com mines frean folme by ... Ian dael, gif his ellen deag, o'5'Se he (not in MS., Th.) aefter dome [daedum wille] maer'Sa fremman wyr[cean] on Jjeode utan wrohtst[afas] eaxle gegyrde and swiora smael, sidan fealwe }>onne mec heaftosigel scir bescineiS and mec Dietr. (xi, 481-482) supplies as follows : Nu com mines frean folme by[sig], [aefle him eor1Swe]lan dael, gif his ellen deag, oft-fte he aefter dome [daedum wille] maer5a fremman, [masgenspede] [wyrjcean on J>eode utan [wrohtstjafas. [Sindon me on heafde hyrste beorhte], eaxle gegyrde [isernes daele], and swiora smael, sidan fealwe. [Haedre mec ahebbe], J>onne mec heaftosigel scir bescineft and mec [scyldwiga] Sch.: folme by . g . . . (five letters) . . . Ian dael gif — dome ri . . . (fourteen letters) . . . dan maerba fremman wyrcan w . . . (about twenty letters) . . . ec non J>eode utan w . . . (about twenty-three letters) . . . pe and to wroht stap . . . (about twenty-five letters) . . . n eorp eaxle gegyrde wo : . . . (about twenty-eight letters) . . . ond swiora — fealwe . . . (about eighteen letters) . . . >on — ond mec . . . (seven letters') . . . faegre. W. : 8 by . go. — 1 1 Of dan maer)>a only the upper part. — 13 Not ec non (ScA.), but after c stands a perpendicular stroke, going below the line (w? ]??), then on; in the same line with -tan, we. In the MS. is not the slightest trace of the stroke seen by W. (T.). Like B.M. I read ec on J^eode u | tan we. Holth. (Bb. ix, 358) reads by[s]go[d] ; (Anglia xxiv, 264): 8-9 Nu com mines fre[g]an folme bysgo [eadwejlan dael, etc. II— 12 [Men ofer mol]dan mair)>a fremman, wyrcan w[eldaedum] 14 wrohtstaffum] — Holth. here rejects stap of MS. (B.M., Sch., W.) as 'nothing can be made out of it.1 \ 6 [earan] or [eagan] ? 17 wo[mb] or wo[ngan] ? B.M. reads clearly bysigo (8), the upper curve of A. before Ian (9), tti instead of & before an (11), we (12), and stap (14). 52 RIDDLES OF THE EXETER BOOK faegre feormaS ond on fyrd wigeS crsefte on hsefte. Cuts is wide J>aet ic fristra sum J>eofes craefte under braegnlocan hwilum eawunga e>elfaesten 25 forSweard brece }>aet aer frifl haefde. Feringe from, he fus fonan wended of )>am wicum. Wiga se fe mine wisan _sdf>e\ cunne, saga hwaet ic hatte. 74 (Gn. 73) Ic waes faemne geong, feaxhar cwene ond jgnlic rinc on ane tid; fleah mid fuglum ond on flode sworn, deaf under yj>e dead mid fiscum, ond on foldan stdp, haefde fer8 cwicu. 5 75 (Gn. 74) Ic swiftne geseah on swafe feran [i2ya] Minn. 76 (Gn. 75) Ic ane geseah idese sittan. 77 (Gn. 76) Sae mec fedde, sundhelm feahte, ond mec y]>a. wrugon eorj>an getenge, fe)>elease. Oft ic flode ongean 21 MS. wigeS, not as Gn. states, weget? ; Th. note wageS ? — 23 MS., Th. J>rista. — 24 A1S., Th., Gn., Dietr., W. hraegnlocan; Th. note hraegl-locan ? Spr. ii, 137, Gn? braegnlocan. No gap in MS., Th.; Dietr. (xi, 482) supplies hwilum nefte ; Gn? bealde nefte. — 27 Gn. note faeringa. — 28 ATo gap in MS., Edd. ; Herzf. (p. 70) assumes, on account of absence of alliteration, a gap of at least two half-lines after cunne. 74 5 MS., Gn., W. forS; Th., Spr. i, 281, Cos., Tr. (BB. xix, 201) feriS. 75 2 MS. D. N. L. H; Th.. Gn. D. N. U. H; W.r\for^\ (Holth., Bb. ix, 358). 77 i MS., Th. se; Gn., W. see. RIDDLES OF THE EXETER BOOK 53 muS ontynde ; nu wile monna sum min fljgsc fretan, felles ne recceS, 5 sij>|>an he me of sldan seaxes orde hyd arypeft [ond m]ec hr[a])>e sippan itet5 unsodene eac 78 Oft ic flodas as cynn minum ond . . . _d~\yde me to mos[e] . . . . swa ic him an ne aet ham gesaet ... 5 flote cwealde )>urh orj>onc . . . yjmm bewrigene. 5 MS., Th., Mad. (p. 48) recced ; Gn., W. rece«. — 7, 8 Th., Gn. arypeS t>e ; Sch, arypeiS . . . (four letters) . . . [ec] h[w?] . . . (two letters) . . . l>e ; W. sees of ec only the upper part, of w (?) only two strokes. From fragment in MS. this doubtful letter w (?) may well be an r (see Holthausen's emendation). Dietr. (xi, 483) supplies after arypeft [hord him ofanimft] ; Holth. (Anglia xxiv, 265) [ond hnaece'S m]ec|aer [oJ>]J>e si)>J>an, reading &r for Sch., W. h[w?]. Th. lie's; Th. note aele~5. Th. marks gap after unsodene ; Gn. assumes no gap ; Sch. eac . . ., the rest of the line is missing'; W. (so /) sees after c an (f)-stroke ; B. M. gives nearly all of 1 ; Holth. 1. c. regards heft unsodene as second hemistich ; but Holth. (E. S. xxxvii, 2 1 o) reads : j-Qnd mjec hr ^ ^ si j^ itetS unsodene eac [swa some] / prefer this placing of words to W.'s . . J>e si)>J>an iteS unsodene eac . . . but the -fragment in MS. rules out swa some. 78 Omitted by Th., Gn. i MS. not Ofl (IV.), but clearly Oft (T.). Sch. about twenty-four letters are missing after flodas. — 2 Holth. (Anglia xxiv, 265) supplies [le]as, perhaps ar-, eftel-, ellen-leas. MS. (W.} cyn; clearly cynn (T.). After ond Sch. notes a gap of some twenty-six letters ; Holth. supplies [sacan]. — 3 Holth. con- jectures [hjyde me to mos[e]. With my reading compare And. 27. After mos about twenty-six letters are lacking (Sch.~). — 4 After him a gap of some twenty-four letters (&//.). — 5 W. states that al is very indistinct. Instead ofz\I read faintly an (B.M. m or n). Sch. records after gesaet a lacuna of some sixteen letters. — 6 Sch. reads rote ; W. flote, and rightly notes that of f the upper cross-stroke is lacking, and that of 1 only the lower part is visible. Holth. supplies [on] flote. — 7 Sch. states that after orj>onc some five letters are missing ; W. reads ofy> only the lower part (so B. M. and /). 54 RIDDLES OF THE EXETER BOOK 79 (Gn. 77) Ic com sefelinges sent ond willa. 80 (Gn. 78) Ic com aepelinges eaxlgestealla, fyrdrinces gefara, frean minum leof, cyninges geselda. Cwen mec hwilum hwitloccedu hond on legeS, eorles dohtor, ]>eah hlo aej>elu sy. 5 Haebbe me on bosme fset on bearwe geweox. Hwilum ic on wloncum wicge ride herges on ende ; heard is mm tunge. Oft ic woftboran wordleana sum agyfe aefter giedde. Good is mm wise 10 ond ic sylfa salo. Saga hwset ic hatte. 81 (Gn. 79) |Ic com bylgedbreost, belcedsweora, [I27b] heafod haebbe ond heane steort, eagan ond earan ond aenne foot, hrycg ond heard nebb, hneccan steapne ond sidan twa, sag_ol~\ on middum, 5 card ofer seldum. Aglac dreoge fair mec wegeft se J>e wudu hrereS, ond mec stondende streamas beatatS, haegl se hearda ond hrlm J>ece?> [ond f]orst _hr~\eose3 ond fealleft'snaw 10 80 2 Ettm. gefera. — 4 Ettm. lecgeiS. — 5 Ettm., Gn. si. — 10 Ettm., Gn. agiefe. Gn., Tr. God. — 1 1 Ettm. silfa. 81 I MS., Edd. byledbreost. — 3 Gn. fot. — 5 MS., Edd. sag; Th. note sac (' a sack') Gn. middan. — 7 Siev. (PBB. x, 520) waegeft. MS. hrereiS; Th., Gn. hrepeft; Gn. note brereS? — 10 Th. ^ceft . . . ond feallefl; Gn. gives no gap after beceft, but supplies after snaw [foriS ofer mec] ; Sch. reads ^eceiS . . . (nine letters) . . . eft; IV. reads as third and fourth letters, rs, and as the last, s ; Holth. (Bb. ix, 358) supplies as first hemistich [fo]rs[t] [geraeJse'S. I read after rs the top of t very clearly and eo quite distinctly before se$. B. M. reads orst . . . eoseiS. RIDDLES OF THE EXETER BOOK 55 [on] pyrelwombne ond ic )>aet .... n maet [won]sceaft mine. 82 Wiht is . . . [gjongende greate swilgetS . . . [f]ell ne flaesc, fotum gong . . 5 . eft sceal maela gehwam 83 (Gn. 80) Frod wses mm fromcynn, [haefde fela wintra] biden in burgum, si^an baEles weard wera lige bewunden, II Holth. I.e. supplies on before )>yrel. After J>aet Sch. notes twenty-eight or twenty-nine missing letters. — 12 Th. . . . eaft ; Gn. [sc]eaft; Sch. ceaft; W. [s]ceaft. Before sceaft / read very clearly maet — followed by three very faint let- ters, perhaps won (?) B. M. reads n maet . . . sceaft. Dietr. (xi, 483) supplies [J>olige call], [ne wepe ic aefre wonnscjeaft mine. 82 Omitted by Th. (Gn.). i Sch. T(?) . nd ; W. Wiht. Only tail ofvi and ht are visible to me. B. M. reads a part of the lower curve ofvf, then iht, followed by is, not seen by Sch., W., or by me. Then a gap of some twenty-two letters (Sch.). — 2 Sch. o(?)ngende ; IV. (so /) o is still clearly visible ; Holth. (Anglia xxiv, 265) [gjongende. After swilgeiS some twenty-four letters are missing (Sc/i.). — 4 Sch., W., and I read 11 ; Holth. I.e. [fe]ll ; B. M. ell. Sch. g . . . g ; W. reads still gong, so do I ; Holth. supplies gong[e"$]. Then follows a lacuna of some thirty-six letters (Sch.). — 6 Before sceal and at end of line, B. M. reads eft, not visible to Sch., W., and to me. Sch. reads gehwa ; W., T.t and B. M. gehwam. The rest of this last line of the riddle is missing (Sch.). 83 i Th. fronvcy[nn] ; Th. note frum-cynn ? Gn. fromc[ynn] ; Sch. fromcy, then a gap of eighteen letters ; W. (so I) reads, after y, n and an n-stroke. Gn. supplies haefde fela wintra. — 2,3 Behveen baeles and wera, Th. gives a gap of over two half -lines, Gn. of more than a whole line, thus giving fifteen lines to the riddle. Sch. 'baeles [weorc? only the remnants an bro)x>r, se me serest wearS 5 gumena to gyrne. Ic ful gearwe gemon hwa mm fromcynn fruman agette call of earde ; ic him yfle ne mot, ac ic h(zfi\e~\nyd hwilum araere wide geond wongas. Haebbe ic wundra fela, 10 middangeardes maegen unlytel, ac ic mij>an sceal monna gehwylcum degolfulne dom dyran craeftes, sifcfaet mlnne. Saga hwaet ic hatte. 84 (Gn. 81) An wiht is [on eorpan] wundrum acenned, hreoh ond repe, hafaS ryne|strongne, grimme grymetaft ond be grunde fareS. Modor is monigra mjgrra wihta. Fseger ferende fundat) ae"fre ; 5 neol is nearograp. NaEnig oprum masg wlite ond wisan wordum gecy)>an hu misllc bip maegen fara cynna, fyrn forftgesceaft ; feeder ealle bewat, or ond ende, swylce an sunu, 10 maere meotudes beam, )mrh [his meahta sp]ed 4 d in gefaelsad is altered from ft. Th. war . . .; Gn. warfaiS] ; Gn.2 warfnaiS], 'upon which the ace. eorSan depends1; Sch. wara. ; W. (so B. M.) reads after a the lower part of a d or "5. — 6 Gn. Ne for Ic. — 7 Th. note frumcynn ? — 9 MS., Th. on haeftnyd ; Gn., W. haeftnyd. Th. note adraefe. — 10 MS., Th. wunda ; CM., Dietr. (xi, 484), W. wundra. 84 I MS., Edd. An wiht is; Herzf. (p. 70) an wraetlicu wiht or Is an wiht, etc. ; Bulbring (Litt.-BI. xii, 156) is [on eorSan] (cf. si1). MS. acenne'S. — 2 Gn. note reoh ? — 3 Th. faraS ; in MS. a is altered to e ( W.). — 6 Gn.2 and for is. — 9 Gn. note frod fyrngesceaft ? — n After }>urh, Sch. notes gap of some twelve letters. At end of line B. M. reads ed, not seen by Sch., IV., or by me. This supports Grein's addition [his mihta sped]. RIDDLES OF THE EXETER BOOK 57 ond pset hyhste msest . . fes tae . . . dyre craeft . . . onne hy aweorp . . oj>e jgnig J>ara ... 15 far ne maeg . . . o}>er cynn eorpan )>on aer waes wlitig ond wynsum Bif sio moddor maegene eacen, 20 wundrum bewreped, wistum gehladen, hordum gehroden, haelejmm dyre. Maegen bio" gemiclad, meaht gesweotlad ; wlite bip geweorpad wuldornyttingum, wynsum wuldorgimm wloncum getenge, 25 clalngeorn biS ond cystig, craefte eacen ; hlo bi}> eadgum leof, earmum getaese, 12-19 Between mae . . . and aer waes (18) Thorpe assumes a gap of three hemi- stichs and a part of a fourth ; according to Th., -what follows mae . . . is apparently part of another enigma; Gn. supplies mae [gen haliges gaestes], and gives, after a lacuna, aer waes as close of next line (13). Gn. note )>aer waes wlitig? For Gn.'s gap (13), Dietr. (xi, 484) supplies [>e ofer hire hreone hrycg] aer waes; and after wyn- sum, [wide boren]. Sch. and |>aet hyhste mae . . . (five letters') . . . )>es ? (judging from fragments) gae . . . (about eighteen letters') ; . . . dyre craeft . . . (about twenty-three letters) , . . onne hy aweorp . . . (about twenty-three letters') . . . }>e \B. M. o|>e] aenig t>ara . . . (about twenty-three letters) . . .: f [o]r ne maeg . . . (about twenty-seven letters) . . . ober cynn eor)>an . . . (about fifteen letters) . . . [h]on aer waes wlitig ond wynsum . . . (eight letters). Sch. declares that the absence of a beginning capital and of a closing-sign disprove Th.'s view of a new enigma. After mae (12) I read the top of st (B. M. s), cerfainly not a g as Gn. suggests, then three missing letters, then the top of )>es, followed by tae (not gae, Sch., IV.) ; B. M. reads es tae. W. reads of J>es (12) only the upper part. Like W., I see between f and r (16) the bottom of an a; B.M. reads plainly far. W. and I see still the > of>on (18). — 20 Th., Gn. seo. Th. modor. — 21 Th. [ge]wrebed; Gn. wreiSed ; Sch. [be]gre^ed, basing his con- jecture on fragments of two letters in MS. ; W. (so B. M. and /) reads the lower part <?/"be and then wre)>ed (w quite clearly). Holth. (E. S. xxxvii, 210) bewrefc>ed wundrum, wistum gehlaested, gehroden hordum. — 24, 25 Th. note wundor? — 25 Gn. note wolcnum ? — 27 MS. earmuge taese; Th. earmunge taese ; Gn., W. as in text. 58 RIDDLES OF THE EXETER BOOK freolic, selllc, fromast ond swtyost, gifrost ond greedgost grundbedd tride)>, )>aes }>e under lyfte aloden wurde 30 ond eelda beam eagum sawe (swa |)aet wuldor wifa, worldbearna maige,) )>eah }>e ferpum gleaw [gefrigen haebbe] mon mode|snottor mengo wundra. [i28b] Hrusan biS heardra, haelejmm frodra, 35 geofum biS gearora, gimmum deorra, worulde wlitigaS, wsestmum tydreS, firene dwaisceS ...... oft utan beweorpeS anre ]>ecene, wundrum gewlitegad, geond werpeode 40 faet wafiaS weras ofer eorpan, f set magon micle . . • . . . sceafte bi]> stanum bestrewed, stormum . . . . len . . . . timbred weall }>rym .......... ed 45 hrusan hrineS h ..... ......... e genge oft 28 MS., Holth. (E. S. xxxvii, 210) fromast ; Edd. frommast. — 31 Gn. ond. — 32 MS., Edd. wife~S; Th. wuldor-wife'5 (' glorious woman''}; Gn. note 'wundor? vgl. wafian, anstaunen?"1 Spr.'i\, 746 wuldor ('^MJ'); cf. Dietr. (xi, 485). MS., Th. maege; Gn., W. maegen. — 33 No gap in MS.; Th. ' Here a line is wanting'; Gn. supplies as in text. — 34 Siev. (PBB. x, 508) snotor. — 36 MS. ( W.) hi)?, clearly ( T.) br$. Gn. supplies bi^S after gimmum. — 38 No gap in MS. ; Th. states that a line is wanting ; Dietr. (xi, 486) supplies [hi frea drihten]. — 42 Gn. note masgen for magon? Th., Gn. micle . . . bi)>; Sch. micle . . . (thirteen to fourteen letters) . . . [ste] bi)> ; W. (so T.) reads before bi)>, eafte ; B. M. sceafte ; Holth. (Anglia xxiv, 265) supplies [ma meotudgesc] eafte. — 43 Th. note bestre\ved(?). After stormum, Th. indicates lacuna to close of riddle ; Gn. supplies [bedrifen], then gap to close; Sch. stormum . ... (thirty to thirty-one letters') . . . timbred weall. Eight letters before timbred (44) I read len (B. M. les). — 44-46 After weall, Sch. marks thirty missing letters, then d hrusan; Holth. I.e. assigns . . . ed to end of line 45 ; W. to 1. 46 ; W. reads J>rym and ed hrusan ; so do I clearly. — 46-47 Sch. hrine)> J> ( W. h) . . . (about twenty-seven letters') . . . [n]ge oft searwu[m] ; W. genge ; B. M. e genge. RIDDLES OF THE EXETER BOOK 59 searwum deatSe ne feleft, peah pe 50 . du hreren hrif wundigen risse hord. Word onhlid haelepum g . . . . . . . . wreoh, wordum geopena hu mislic sy maegen para cy[nna]. 55 85 (Gn. 82) Nis min sele swige ne ic sylfa hlud ; ymb unc _domas dyde, unc\ Driht[en] scop sip setsomne. Ic com swiftre ponne he, pragum strengra, he preohtigra ; hwilum ic me reste, he sceal rinnan fortS. 5 Ic him in wunige a penden ic lifge ; gif wit unc gedjglaS, me biS deao1 witod. 48 After searwu[m], about twenty-eight letters are missing (Sch!). B.M. reads after searwum the bottom of three letters, bij>(?) or dis(?) — 49 Sch. [djeafte; W. deaiSe ; / see top of d. — 50-51 Sch. reads J>eah . . . (about twenty-six letters) . . . du ("5u?); W. reads J>eah J>e and du; so do B.M. and I clearly. — 51-52 After wun . . g ( W. wundig, B. M. wundigen > ? or w ?) about twenty-one letters are missing (Sch!). — 53 Sch. hae[le)mm?] ; W. and B.M. (clearly) haele)>um g . . .; / see lower part of lej>um, then bottom ofg. — 54 Before wreoh about fifteen letters are missing (Sch.). Sch. ge opena. — 51—54 Holth. (Anglia xxiv, 265) supplies as follows .' [heafjdu hreren, hrif wundig[en] . . . ; [cneojrisse. Hord word[a] onhlid, haelejmm g[eswutela], [wisdom on]wreoh. For wisdom, Holth. conjectures also waerfasst or word-hord. — 55 Only some two or three letters can be missing in this line (Sch.) ; Holth. 1. c. supplies [cynna] by aid of line 8. Of cynna. I see clearly c and end of tail cfy, overlooked by Sch., W. ; B. M. cy. 85 I Th. note sel for gesel (' comrade') ? — 2 No gap in MS. after ymb (Th!)\ Gn., W. note omission in sense, but fail to mark gap in text; Holth. (I.F. iv, 388) supplies [droht minne]. After unc, I mark in the MS. a gap of nine or more letters and supply as in text. The lacuna is duly recorded by B. M. MS. driht ; Th. dryht ; Gn. dryhten ; W. drihten. Th. indicates gap after scop. — 3 MS. swistre ; Th. swiftra ; Gn., W. swiftre. — 5 MS., Edd. yrnan. 60 RIDDLES OF THE EXETER BOOK 86 (Gn. 83) Wiht cwom gongan pair weras sseton monige on maeftle mode snottre ; haefde an cage ond earan twa ond twegen fet, twelf hund heafda, hryc[g] ond wombe ond honda twa, [earmas ond eaxle, anne sweoran ond sidan twa. Saga hwaet ic hatte. 87 (Gn. 84) Ic seah wundorlice wiht, wombe haefde micle pryjmm geprungne ; pegn folgade maegenstrong ond mundrof ; micel me puhte godlic gumrinc, grap on sona heofones tope bleow on cage ; hio borcade, wanode willum. HTo wolde se ]>eah mol 86 4 MS., Edd. except Ettm. II, XII. — 5 MS., 77., Ettm. hryc. Eltm. handa. 87 3 MS. megenstrong ; 7/4., Gn. maegnstrong. — 4-5 Holth. (£. S. xxxvii, 210) grapon (dat.fl.) sona heof on his to)>e. — 5 N° SaP tfl MS., 77i.; Gn., W. indicate missing hemistich. — 6 MS., Edd. bleowe; Gn. note bleow (?) bleaw(?) MS. boncade, Edd. as in text. — 7 MS., W. wancode ; Th., Gn. )>ancode. — 8 Sch., W. mol ; B. M. niol. The word is not given by Th. (Gn.). After mol about fourteen letters are missing (Sch?). 88 1-12 Th., Gn. read Ic weox J>aer ic . . . (three missing hemistichs) ... (1. 3) ond sumor ... (a little more than one hemistich) . . . (Gn. 4, W. 12) ac ip uplong. Sch. : Ic weox J>aer ic . . . (about thirty-four letters) . . . ond sumor mi ... (about thirty letters) . . . me waes min tin ... (about thirty-three letters) . . . d ic on sta~5[ol] . . . (about twenty-eight letters) . . . um geong swa . . . (abmit twenty-seven letters') . . . se weana oft geond . . . (about twenty letters) . . . [f]geaf. W. (so I) reads s (1. i), the upper part ofo\ (1. 7), and the lower part of f (1. n). B. M. reads (1. 7) od «wa'sta'8ol, #«xr ic . . . . ond bropor mm, begen waeron hearde. Eard wses ]>y weorfira ]>e wit on stodan, hyrstum ]>y hyrra; ful oft unc holt wrugon, 15 wudubeama helm, wonnum nihtum, scildon wit5 scurum ; unc gescop meotud. Nu unc mieran twam magas uncre sculon sefter cuman, card oSfringan gingran brofor. Eom ic gumcynnes 20 anga ofer eorj^an ; is min [agen] baec wonn ond wundorlic. Ic on wuda stonde bordes on ende ; nis min broker her, ac ic sceal broforleas bordes on ende sta)K>l weardian, stondan faeste ; 25 12 After ic about eight letters are missing (Sc/i.). B.Af. reads before ond the tail of a y. — 13 MS., TA., B.M. mine bro>or; Gn., W. min bro)>or; Holth. (Bb. ix, 358) 'broker mm, perhaps the mine of the MS. stands for minne, as in I. 12 a transitive •verb may be missing? — 14 IV. (so /) sees only the Imver part ofty. B.M. gives all but the upper stroke. — 18 Gn. magas; Gn? magas. — 20 Th. begins a new riddle with Eom, although in the MS. there is not even a period after broker (W.). — 21 Gn. anga ; Gn. note anga (?) Siev. (PBB. x, 520) attacks is min baec on metrical grounds; Holth. (7.F. iv, 388) supplies as in text. — 25 MS., Th. stodan; Th. note, Gn., W. stondan. 62 RIDDLES OF THE EXETER BOOK ne wat hwjgr mm brofor on wera aehtum eorpan sceata | eardian sceal, [i29b] se me ser be healfe heah eardade. Wit wseron gesome ssecce to fremmanne ; nsefre uncer aw}>er his ellen cytSde, 30 swa wit fare beadwe begen ne onjmngan. Nu mec unsceafta innan slitaS, wyrdaf mec be wombe ; ic gewendan ne maeg ; set }>am spore findeS sped se ]>e se[ceft] sawle rjgdes. 35 89 e wiht wombe haefd . tne lefre wses beg on hindan grette wea worhte, 5 hwllum eft fygan, him poncade sij>)>an swsesendum swylce }>rage. 26 Herzf. (p. 48) broj>or min. — 29, 30 MS., W. fremman ne naefre ; Th., Gn. fremmanne | ne naefre ; Th. note 'ne seems a repetition from the word preceding1 ; Siev. (PBB. x, 482) fremmanne. — 31 Th. waere (misprint}. Th. note on>rungon. — 32 Th. hu ; Th. note nu. — 33 Th. '•after wombe, a gap of nearly two hemistichs ; at end of second half-line ne maeg ' ; Gn. wombe [ic warnian] ne masg ; W. (so B. M. and I clearly) reads after wombe, ic gewendan ne maeg. — 34, 35 Th. reads sped se }>e se, then gap to close ; Gn. supplies se[ce'S], then no gap ; W. (so B. M.) notes after se (which is at end of line) some twelve ( T. fifteen) missing letters, on next line then sawle rx.Ae&, followed by closing sign : 7 89 Omitted by Th. (Gn.), and not given by Sch. W. thus reads the JlfS.: I, 2 Before wiht some thirty letters are lacking, wombe is at end of line. After haefd some twenty-five letters are lacking. 3 Only the right side ofr in re is visible. lejre is at end of line. 4 After beg some twenty-three letters are missing, hindan is at end of line. 5 After wea, a lacuna of some twenty letters to end of line. worhte begins the new line. 6 After ef, a lacuna of some se^'enteen letters to end of line. J>ygan begins the new line. 7 After sij>ban, a lacuna of some fifteen letters to end of line. 8 swaesendum begins the new line. After J>rage, a closing-sign : 7 My readings agree with those of W., but B.M. notes these additional letters: e before wiht (1. 2), tne for re (1. 3), on before hindan (1. 4), lower part of 't (so /) after ef (1. 6). RIDDLES OF THE EXETER BOOK 63 90 (Gn. 86) Mirum mihi videtur : lupus ab agno tenetur ; obcurrit agnus [rupi] et capit viscera lupi. Dum starem et mirarem, vidi gloriam parem : duo lupi stantes et tertium tribul[antes] quattuor pedes, habebant, cum septem oculis videbant. 5 91 (Gn. 87) Mm heafod is homere ge]>ruen, searopila vvund, sworfen feole. Oft ic beglne ]>szt me ongean sticafc, forme ic hnitan sceal hringum gyrded hearde wift heardum, hindan ]>yrel 5 forS ascufan J>aet frean mines modp" freopaS middelnihtum. Hwilum ic under baec bregdejnebbe [I3°a] hyrde )>aes hordes, ]>onne mm hlaford wile lafe )>icgan J>ara ]>e he of life net 10 Wcelcraefte awrecan .willum slnum. 90 MS., T/i., Gn. have throughout \for v. I MS., Gn., W. videtur mihi; Th. note, Holth. (£.S. xxxvii, 211), as in text. — 2 W. states that rr in obcurrit is no longer visible ; Holth. supplies rupi. — 3 MS. misare (Sc/t., IV., T.)\ Edd. mirarem. MS., Th. magnan ; Gn., W. magnam ; Holth. parem. — 4 MS., Th., Holth. dui; Con. Dui (= diuersi). Con. ex for et. MS. tribul, no gap ; Th. tribul[antes]. — 5 MS., Edd. mi. Con. occulis ('fta MS.'). 91 I MS., Edd. gejmren ; Spr. i, 474 gej>ruen(?) so also Siev. (PBB. x, 265). — 2 Th. note pile ? — 3 Th. note begrine. Siev. (Anglia xiii, 4) sliced. — 6 MS., Edd. mines frean; Hersf. (p. 46) frean mines. — 7 Spr. ii, 261, Dietr. (xi, 486) F = wen ; Siev. (Anglia xiii, 4) |> = wynn. — 8 Holth. (E. S. xxxvii, 211) Hwilum ic under baec bregde [brunre or beorhtre or blacre] nebbe. — ii MS. waelcraef ; Th. supplies turn; Gn. waslcraeft ; Sch. 'waslcneftfe] seems to have stood in the MS.; there would be no room for waslcraeftum';- W. states that 'two or three letters are missing after f ; but cannot say whether they have become effaced by time or erased by a liquid'' (obviously, by action of fluid on ink, T.). 'Sch. to the contrary, these letters might have been tu' (W^). Siev. (Anglia xiii, 4) waelcraefte ; B. M. reads clearly waelcraefte. 64 RIDDLES OF THE EXETER BOOK 92 Ic wses brunra beot, beam on holte, freolic feorhbora ond foldan waestm, _on(T\ wynnstapol ond wlfes sond, gold on geardum. Nu com guowigan hyhtlic hildewgepen, hringe bete 5 . . . wel byreo1 on o}>rum 93 (Gn. 88) Frea mm . de willum sinum heah ond hyht . ... [sc]earpne hwllum [h]wflum sohte frea ... as wod 5 92 Omitted by Th. (Gn.). i Holth. (£.S. xxxvii, 211) brunna. — 3 MS. wym staM; Holth. (Bb. ix, 358) sta)>ol weres ; Holth. (E.S. xxxvii, 211) wynn on stable. — 4 Holth. I.e. god for gold. — 5 W. reads only the upper part <7/"ilde ; so I ; B.M. clearly hilde. MS. (Sch., W.,andl) bete; MS. (B.M.) bega. Sch. states that after bete twenty-seven letters are missing. — 6 B. M. reads the top of wel, nine letters after bega. — 7 W. notes that byrelS begins the new line. It is impossible to determine how many letters are missing after o)>rum ; on this line stand no longer any letters (W). 93 1-5 Th. reads Frea min . . . wod. Gn. note, conjectures Frea min [mec fseste near] wod. Dietr. (xi, 487) Frea mm [waes faegre foran gefraetjwod. Sch. Frea mi[n] . . . (twenty-seven letters) . . . de willum sinum (B.M. sinu) . . . (twenty-six letters) . . . heah ond [hyht] . . . (twenty letters') . . . [sce]arpne hwilum . . . (twenty-two letters) . . . [hw]ilum sohte frea . . . (seventeen letters') ... as wod. W. reads still the first stroke of n (i), so B.M. and I; the upper part ^hyht (3), so B.M. and I ; remnants o/sc (3) ; w in hwilum (5) ; and the lower part ofzs (5). There is now in MS. no trace of sc (3), only the bottom of e and half of a, then, clearly, rpne (B.M. earpne). Holth. (Aiiglia xxiv, 265) supplies (1. 3) heah ond hyht[ful or lie? hocum] sc[e]arpne. (1. 5) [h]wilum sohte frea [min] as wod. as might be the remains 0/"siJ>as, widlastas, or wraeclastas (cf. Spr. ii, 636). RIDDLES OF THE EXETER BOOK 65 daegrime frod deo[pe stre]amas, hwilum stealc hli]>o stlgan sceolde up in efel, hwilum eft gewat in deop dalu dugu]>e secan strong on staepe ; stanwongas grof 10 hrlmighearde, hwilum hara scoc forst of feaxe. Ic on fusum rad, offaet him )>one gleowstol gingra broj'or mm agnade ond mec of earde adraf. Siffan mec isern innanweardne 15 brun bennade ; blod ut ne com, heolfor of hrepre, )>eah mec heard bite strSecg style. No ic )>a stunde bemearn, ne for wunde weop, ne wrecan meahte on wigan feore wonnsceaft mine, 20 ac icjaglJeca ealle polige [i3ob] faette bord biton. Nu ic blace swelge wuda ond waetre, womb[e] befasSme fast mec on fealleft ufan ])£lr ic stonde, eorp[e]s nathwaet, haebbe anne fot. 25 Nu mm hord waraS hifende feond, se ]>e £er wide baer wulfes gehlefan ; oft me of wombe bewaden fereS, 6 Th., Gn. deo . . . hwilum; Sch. reads deo[pe streamas?]; W. reads the lower part 0/"amas; so B.M. and I. — 7 Th. stealc-hlijjo. — 9 Th. deop-dalu. — II MS. hara scoc; Spr. ii, 14 ' har ascoc? (vgl. Eng. hoar-frost}? — 12 MS. feax. MS., Edd. of. — 13 AfS., Th. gleawstol. MS., Th. gingran; Th. note gingra. — 22 Th. $ . . . bord ; Gn. J>aet bord ; Sch. >xtte ; MS. ( W.) fre ( IV. does not see the t, nor do /); B.M. J>ine. MS. blace; Gn., Spr. i, 124 blace; Siev. (PBB. x, 496) blac. — 23 Th. waetre . . . befasflme; Gn. supplies [wide]; Sch. reads womb[e?]; W. reads only w . . . befaeftme ; 7 read w . . . b very easily (B. M. womb). — 25 Th., Gn. eo . . . ; Dietr. (xi, 487) eo[rpes] ? Sch. reads eo . . . es ? IV. only eo . . . s. The lower strokes of r and p are plainly visible to me. B. M. reads eof waes. — 26 Th. note weraft ? Dietr. (xi, 487) hordwaraft. — 28 Th., Gn. . . . of wombe ; Dietr. I.e. supplies [wonsceaft] ; Sch. (six letters) ... of wombe; Holth. (I. F. iv, 388) supplies [wealic]. Before of wombe I read faintly but unquestionably me, pre- ceded by the top of oft (B.M. oft me). These letters are not seen by Sch., W. 66 RIDDLES OF THE EXETER BOOK stepped on stI6 bord . . de . . Jxmne daegcondel 30 sunne [w]eorc eagum wlitefi ond sp 94 Smif ad hyrre ]K>nne heofon dre fonne sunne, style smeare ]>onne sealt sy 5 leofre J>onne ]>is leoht call, leohtre fonne w . . 29-32 7/4. reads steppe'5 on stift bord . . . dasg-condel sunne .... eagum wlita'5 Gr. reads stepped on stift bord daegcondel sunne . . eagum wlita'S Dietr. (xi, 487) supplies stepped on stiiSbord, [storme bedrifen] [sififtan he] daegcondel [le], sun [nan upcyme] [serest ealra] eagum wlite'S. Sch. reads bord . . . (some twenty-seven letters') . . . n daegcondel sunne . . . (some twenty-seven letters) . . . core eagum wlite'S . (two letters) . p . . . (/ letters). B. M. reads (1. 30) de . . . (six letters) ...topof\ (?), Jx)fi. W. (so B. M. and /) reads still )>on (30) and after wlite'S (end of line) -\ sp (at beginning- of line very indistinct). Upon this line are no longer any letters. Holth. (Anglia xxiv, 266) 'Assmann is wrong in putting sunne after daegcondel in 1. 30.' Holth. reads as in text. 94 Omitted by Th. (Gn.). i, 2 Sch. Sm[i]}> . . . (some twenty letters) . . . hyrre )>onne heo[f] ; W. and I read Smi)> and d (B. M. ad) before hyrre, and heofon. — 2 After heo[f], a gap of some thirty-two letters (Sc/i.). — 3 Holth. (Anglia xxiv, 266) [blicenjdre ; (E. S. xxxvii, 211) [hraejdre. — 4 After sunne some twenty-nine letters are missing (ScA.). — 5 Holth. (Anglia xxiv, 266) sy for MS., W. ry. After ry, some twenty letters are missing (Sch.). — 6 W. reads (6-7) : leofre )>onne )>is leoht, call leohtre Jx>nne w . . . RIDDLES OF THE EXETER BOOK 67 95 (Gn. 89) Ic com indryhten ond eorlum cuS ond reste oft ricum ond heanum, folcum gefrjgge fere wide ; ond me fremdum ser freondum stondeS htyendra hyht, gif ic habban sceal 5 blsed in burgum o]>]>e beorhtne god. Nu snottre men ' swifast lufia}> midwist mine ; ic monigum sceal wisdom cypan ; no j>ser word sprecao1 senig ofer eorftan. J>eah nu aElda beam, 10 londbuendra, lastas mine swipe secao", ic swaj>e hwllum mine bemtye monna gehwylcum. Ilolth. {Anglia xxiv, 266) regards W! 's verse-division as obviously incorrect and reads as in text. Sck. does not read w, seen by IV., B.M., and me. ' It is impossible to determine the number of missing letters after w' (W.). Holth. I.e. (w[yrmas] (cf. 4i76).' After w, / read in MS. (see also B. M.), the lower strokes of several letters, not yrmas. 95 3 MS., Th., Gn., W. fereS ; Gn.2, Siev. fereS ; Th. note fere ? so also Tr. (BB. xix, 206). — 4 MS., Edd. fremdes ; Th. note fremde ? Brooke (E.E. Lit., p. 8) fremdum; Tr. (Anglia vi, Am. 168) supplies fremdes [gefea] aer; Tr. {Anglia vii, Anz. 210) fremdes [faeflm] aer; Tr. (BB. xix, 206) faer for aer. — 5 Th. note hihtendra. — 6 Gn. note beorhte god ? so also Dietr. (xi, 488) and Tr. (Anglia vi, Anz. 1 68) ; Tr. (BB. xix, 208) gong ; Bright suggests beorhte (or beorhtan) gold ? NOTES ['THE FIRST RIDDLE' The part played by the so-called ' First Riddle ' in the study of the authorship and history of this group of enigmas has already been discussed in the Intro- duction. Its grammatical forms will be included in the Glossary — in brackets, to set them apart from the vocabulary of the genuine riddles. More detailed treat- ment than this belongs properly to an edition of Old English Lyrics, and demands no place here.] RIDDLE 2 Dietrich points out (XI, 461) that in 2, 3, 4, only a single subject is included, 'the Storm.' But, as he notes, the topic finds subdivision in two ways: by the closing formulas of Nos. 2 and 3, and by the summary of the four phases of the storm's activity in 4 67-72. There we are referred to its work under the earth (4 1-16), under the waves (3), above the waves (4 17-35), ar>d in the air (4 36-66). According to Dietrich, No. 2 describes both the storm on land (2 i-8a) and that at sea (2 Sb-i5) ; No. 3 is limited to the Ocean Storm, which in No. 4 falls into three parts : ' In the first the storm pictures itself as confined under the earth and thus producing an earthquake (4 1-16) ; then, as driver of waves and assailant of ships (4 17-35); finally as cloud-farer and thunderstorm.' Grein had already (Bibl. der ags. Poesie II, 410) interpreted No. 3 as 'Anchor' (an impossible solution), and No. 4 as 'Hurricane.' Prehn (pp. 158-162) accepts Dietrich's answers; and seeks vainly — as I think with Edmund Erlemann (fferrigs Archiv CXI, 55) — to establish a relation between the Anglo-Saxon problems and the enigmas of Aldhelm, i, 2, and Eusebius, 21 and 23. Brooke (E. E. Lit., p. 182) follows Dietrich : — ' The first describes the storm on land, the second at sea, and the third the universal tempest — the living Being who rises from his caverns under earth and does his great business, first on the sea, then on the cliffs and ships, then on the land and then among the clouds, till he sinks to rest again.' Traut- mann classes the three riddles together and gives them one number. In an elaborate article in Herrigs Archiv CXI, 49 f., Edmund Erlemann takes issue with Dietrich. He believes with the earlier scholar that 4 1-16 refers to an earthquake, and is indeed the scientific explanation of that phenomenon, popular with scholars of the time. He points to Bede's account ' De Terrae Motu ' in his work De Xatura Remm, cap. 49 (Migne, P. L. XC, 275 f.): — 'Terrae motum vento fieri dicunt, ejus visceribus instar spongiae cavernosis incluso, qui hanc lorribili tremore percurrens et evadere nitens, vario murmure concutit et se tre- mendo vel dehiscendo cogit effundere. Unde cava terrarum his motibus subjacent, utpote venti capacia ; arenosa autem et solida carent. Neque enim fiunt, nisi caelo 69 ;o marique tranquillo, et vento in venas terrae condito' (4 ioa-n). This wind-theory of earthquakes was drawn, as Erlemann shows, from Isidore of Seville's famous text-book DC Xatura Rerum, and is traceable to Plato. So No. 3 represents not a Sea-Storm but a Submarine Earthquake (11. 3-8), such as is described by Bede 1. c.: I' F iunt simul cum terrae motu et inundationes maris, eodem videlicet spiritu infusi vel residentis sinu recepti.' Erlemann further shows that No. 3 has nothing in common with 4 17-36, which is a description of a ' Storm at Sea,' as Dietrich and Brooke believe. As the storm is the scientific explanation of land and sea earth- quakes, so is it felt to be of thunder and lightning by our poet (4 37-66). Here again, thinks Erlemann, we find a close parallel in Bede, z8-2Q : ' Tonitrua dicunt ex fragore nubium generari, cum spiritus ventorum eorum sinu concept! sese ibi- dem versando pererrantes et virtutis suae nobilitate in quamlibet partem violenter erumpentes, magno concrepant murmure instar exilentium de stabulis quadriga- rum vel vesicae, quae, licet parva, magnum tamen sonitum displosa emittit, etc.' Riddle 2 is simply a general description of the Storm. ' Now in all this, there is no direct borrowing. Difference of language and the noble imagery of the poet both speak strongly against any servile indebtedness to the scientific works of his day. But these ideas were in the air at the time, and may have been imbibed by him in some cloister school in the North during his boyhood in the early eighth century.' Erlemann, p. 54, thinks that Riddles 2-4 appear to be ' ein mit scharfster Konsequenz aufgebautes Ganzes.' 'The present threefold division (Grein-Wulker) I rests upon the three repetitions of the riddle-question at the end of these three / parts. But, after all that I have said, weight can no longer be laid upon them as / signs of division. The riddle-query appears also within 4 at end of 35 [but this is / not a formula]. Moreover, the MS. shows no gap between Rid. 3 and 4 [but Rid. 3 closes the page], and hwilnm in 4 i begins with a small letter. The space between 2 and 3 is easy to understand : in 2 the Storm in general, and in 3 and 4 its single phenomena, are described. But even this can be laid at the scribe's door. Misled by the riddle-query into thinking that 2 closed with line 15, he could well begin a new riddle with hiuilum (3 i). In the case of the second hwtlum (4 i) he has come to realize the close connection of parts, and no longer makes a space.' This view does not lay due stress upon the closing formula of Rid. 3 ; and Erlemann fails to state that the lack of a gap after 3 is determined by the ending of a MS. page here. The same fact may explain the lack of closing- sign, though this stands at end of page in 15, 74, and 80. 2 i Cf. Chr. 241, ForJ>on nis ienig j>aes horse ne baes hygecrasftig. 24 wraec(c)a. Thorpe renders the MS. reading -wrace 'I wander'; Grein in Dicht. 'treibe,' Brooke (p. 182) 'tear along (in gusts)'; but these translations would seem to demand a present form wrece rather than ivrcece. To both these forms there is the strong objection that the meter demands a long vowel here (-L x.\ -L x). Nor does Grein's interpretion ofwra-ce (Spr. II, 737 ; so also B.-T., p. 1 268) as the inst. sg. of ivracu, ' hostility,' meet the difficulty. Sievers (PBB. X, 510, s.v. f>rdg) writes wriece, apparently deriving this from wraic, which he regards as long (Gr.9 276, n. 3 b). But the vowel is short everywhere else in the poetry II, 738). It is of course possible to regard the half-line as one of several NOTES 71 examples of a shortened A-type £. x | \j X (Herzfeld, p. 44), but it is perhaps better to read here wrac{c)a, ' exile,' ' wretch,' as Herzfeld suggests. The scribe may have been misled by ivriece (1. 2), which is almost immediately above in the MS. 2 8 \vudu hrere. See 817, where se J>e ivudu /irerefr is a periphrasis for 'the wind.' 2 it wrecan. The MS. ivrecan is retained by all editors, and is regarded by Brooke as an infinitive, 'to range along,' and by Grein (Dicht.\ Spr. II, 739) as gen. sg. of wrec(c)a — ' on the wanderer's track.' As similar constructions are common in the poetry (wreccan laste, 408; cf. Gen. 2478, 2822, Sea/. 15), and as this meaning accords well with 1. 4b, I prefer the reading of the MS. to the sug- gestion of Cosijn (PBB. XXIII, 128) wrecen. The latter, however, has the support of 2b, on sift wr<£ce\ and would be acceptable, were any change necessary. 2 13 flaisc ond gaistas. Cf. Chr. 597, flassc ond giest. RIDDLE 3 For parallels to the Anglo-Saxon description of the Seebeben, Erlemann (p. 57) points to the MHG. illustrations in the articles by Ehrismann, Ger mania XXXV, 55 f., and Sievers, PBB. V, 544, which treat the words gmntwelle and selpwege. Cf. Hartmann, /. Biichlein, 352^ : . . . und hebet sich uf von grunde ein wint das heizent si selpwege und machet groze iindeslege und hat vil manne den tot gegeben. 3 2 under y}?a gej>rsec. Cf. 33 7, atol y>a gejraec; And. 823, ofer y'Sa gejraec. See also the stronger expression, atol y>a gewealc, Exod. 455. 3 3 garsecges grund. Cf. 41 93. 3 3-8 Erlemann (p. 51) points out the likeness of the phenomena here described to those that appear in submarine earthquakes: ' Finden diese Seebeben bei ge- ringer Meerestiefe statt, also in der Nahe der Kiiste, so zeigen sich neben den gewohnlichen Erscheinungen — Aufwallen und Triibung des Wassers, Empor- schiessen von Schaum und Dampfsaulen — auch direkte Spuren subozeanischer vulkanischer Eruptionen, Emporwerfen von Lava und Bimsstein, verbunden mit submarinem Donner.' So the other passages of our poem forbid the conception of a sea-storm, and accord with that suggested by Erlemann. The contrast be- tween the two phenomena is accentuated in 4 68-70. 3 4 Grein's addition \Jlod dfysed'\ is supported by flodas Sfysde, Chr. 986, and flodas gefysde, El. 1270. — Cosijn's reading, famge wen/can {PBB. XXIII, 128) parallels And. 1524, famige walcan (PBB. XXI, 19), and is supported by 4 19, fdmig ivinnefr ; but the MS. reading makes perfect sense and is in keeping with the context. 3 5 hwaelmere hlimmeff. Cf. And. 370, onhrered hwaelmere ; 392, garsecg hlymmeS. For a discussion of rimes in the Riddles, see note to 29. Cf. 16 13, 2Q 2, 4, 5, 6, 8, 39 4, 42 3, 67 6, 73 22. 36 streainas stafni boatacf. Cf. And. 239, beoton brimstreamas ; 441, eagor- streamas beoton bordstae'Su ; 495-496, streamwelm hwileft, beate|> brimstasfto ; 72 RIDDLES OF THE EXETER BOOK see also And. 1544, El. 238 Met. 615. Herzfeld, who cites these parallels (p. 30), regards as characteristic of Cynewulf ' the constantly recurring mention of the striking of the waves on the cliffs or on the sides of the ship.' Herzfeld notes that this trait is lacking in other Anglo-Saxon descriptions of storms — Gen. 1371, Exod. 454 f., and Beow. 1374. But he finds similar expressions in Seaf, 23 and Wand. 101. Brooke notes (p. 182, n.) that a similar passage occurs in Chr. 979 f., describing the cliffs withstanding the waves. — With streamas beatafr cf. 81 8. 3 7 on stealc hleoj>a. Cf. 4 26, stealc stanhleojni ; 93 7, stealc hlij>o ; Beow. 1410, steap stanhliSo; And. 1577, stanhleoftu. For a discussion of such expres- sions, see Merbach, Das Meer etc., p. 21. 3 8 ware ond waige. Dietrich (XII, 246) translates ' schlamm und woge,' and refers to And. 269, ware bewrecene, and And. 487, ware bestemdon; but in these passages ware has the meaning 'sea.' Dietrich regards ware as a rare word, which here means neither ' sea ' (wer) nor alga (41 49, wdrofr), but 'schlamm und meeres- sand (cf. Hpt. Gl. 502, 76, sablonum, wdra; 449, 30, sablonibus, wdrum).' Grein, Dicht., renders • Seetang,' and Spr. 11,640, 'alga' (reading -wdre), and points to Dutch wier and Kent, waure ; Brooke translates ' weed,' and is followed by Brougham (Cook and Tinker, p. 71). The word ware receives adequate discus- sion from Hoops, Altenglische Pflanzennamen, pp. 24-25: ' Tang, Fucus und See- gras, Zostera Marina = wdr, wdroj>, stewdr. Sie machen sich ja an der Kiiste dem Schiffer wie dem Fischer durch Verunreinigen der Fahrzeuge und Netze oft genug in unangenehmer Weise bemerkbar und werden darum nicht nur im eigent- lichen Sinne von Meerespflanzen sondern ubertragend auch fiir Schlamm und Schmutz iiberhaupt gebraucht.' Hoops points out that the transition to the mean- ing of 'mud' or 'slime' is clearly seen in Rid. 41 48-50, where wdroj> is used in rendering the Latin ' horridior rhamnis et spretis vilior algis.' A similar use is found in the wdrig hragl of Gn. Ex. 90 (see Merbach, Das Meer, pp. 28-29). See Schmid's discussion of 'algarum maris' (Gesetze, Glossar, p. 529). 3 9 holmmaegne bipeaht hriisan. Cf. 173, eorSe yftum J>eaht. 3 10 side sJegrundas. Cf. Exod. 289, sailde siegrundas. — sundhelme. Only here and 77 i, sundhelm }>eahte. But cf. water helm, Gn. Ex. ii, 3 (Merbach, p. 10). 3 12 on si)m gehwam. Cf. Ph. 464, in slba gehwane. 3 13 of brlmes fa>]?iiium. Cf. 116-7, of faetSmum cwom brimes ; And. 1616, Jmrh nodes faeftm. 3 '5 yja . . . )?e mec a5r wrugon. Cf. 772, mec yj>a wrugon; 787, y)>um bewrigene. RIDDLE 4 Of this Brooke says (E.E.Lit., p. 183) : 'The order and unity of this poem is admirable. The imaginative logic of its arrangement is like that which pre- vails in the " Ode to the West Wind," to which indeed it presents many points of resemblance, even to isolated phrases. Shelley tells us of his wind — which, as in Cynewulf 's poem, is a living being — first as flying through the forests and the land, then of its work among the clouds, then on and in the sea, then on his own soul. Cynewulf tells of his storm-giant rising from his lair, rushing over the sea, then over the land, and then in the sky, but not of the storm in his own breast. NOTES 73 That is the one modern quality we do not find in this poem of Cynewulf. It was natural for him — being closer to Nature-worship than Shelley — to impersonate his hurricane, to make the clouds into stalking phantoms, to make them pour water from their womb and sweat forth fire ; and his work in this is noble.' 4 1-6 Brooke translates (pp. 183-184) : Oftenwhiles my Wielder weighs me firmly down, Then again he urges my immeasurable breast Underneath the fruitful fields, forces me to rest. Drives me down to darkness, me, the doughty warrior, Pins me down in prison, where upon my back Sits the Earth my jailer. Brooke compares with these lines, and with 13-16, Shelley's 'Cloud': In a cavern under is fettered the thunder, It struggles and howls at fits. He points also to Aeneid, i, 56 f. : Hie vasto rex Aeolus antro Luctantes ventos tempestatesque sonoras Imperio premit ac vinclis et carcere frenat, etc. (So too the Greek earthquake-demon Typhos, progenitor of the storms, is held down in fetters by Sicily and Etna piled upon his breast, Pindar, Pyth. i, 33-35.) Dietrich believes (XII, 246) that the Anglo-Saxon lines are not suggested by Virgil but by Psalms cxxxiv, 7 (Vulgate). Erlemann also thinks (p. 54) that in his conception of God as the ruler of the winds the riddler is influenced by the Old Testament, Psalms cxxxiv, 7 (Deus) . . . qui producit ventos de thesauris suis, and Jeremiah x, 13. That such passages as these influenced mediaeval science he shows by quotation from Beda, De Natzira Kerum, cap. 26, and Isidore 36, § 3. Herzfeld (p. 31), on the contrary, believes that this conception is derived neither from classical nor scriptural sources, but from the older mythology. The idea of the confinement of the violent storm in prison by a higher power appears in other Anglo-Saxon poems (Dietrich XII, 246; Herzfeld, p. 31), as El. 1271-1276: winde gellcost, bonne he for hsele'Sum hlud istige'S, wzEiSe'S be wolcnum, wedende faere^, ond eft semninga swlge gewyrSetS, in nedcleofan nearwe gehea'Srod, l>ream for^rycced. So And. 435-437 : Wzeteregesa sceal, geftyd ond geftreatod J,urh brygcining, lagu lacende, IrSra wyrgan. 516-520: Flodwylm ne msg manna snigne ofer Meotudes est lungre gelettan ; ah him lifes geweald, se fie brimu bindefi, brune yiSa SfS ond Jjreata-5. 74 RIDDLES OF THE EXETER BOOK 4 3 bearm [bone] bradan. For such position of article and adjective, see 349-10, 61 6. Cf. Trautmann, Anglia, Bb. V, 90; Barnouw, p. 221. — on bid wriceo5. Here the reading adopted by recent editors is confirmed by Beow. 2963, on bid wrecen. 4 5 liaiste. Cosijn's reading seems to me a lectio certissima. Grein, Spr. II, 24, doubtfully derives the MS. hatst from hatsan, ' impingere,' of which we have no trace elsewhere. Haste, which is found in our present sense Gen. 1396, is the equivalent of f>nrk hiest (see 16 28, J>urh hest). I accept also Cosijn's heard (so Thorpe translates) for MS. heard, which is not found elsewhere in the poetry in this sense, but which is rendered by Brooke 'jailer.' 48 hornsalu. Only here and And. 1158. 4 13-14 se mec wris&ffe on ... legde. The same idiom is found 21 29-30, se mec geara on bende legde. Cf. also And. 1192, J>jer J>e cyninga cining clamme belegde. 4 16 be me wegas taecneo'. Cf. 52 6, se him wegas tiecneK 4 18 [streamas] styrgan. The addition is made by Thorpe in the light of 4 70, streamas styrge. Cf. also And. 374, streamas styredon. 4 19 flintgrsegne Hod. This is the only appearance of the epithet ; fealo is of course the common adjective withy?^/ (And. 421, Beow. 1951, Brun. 36). 4 i9h-2oa Cf. Met. 28 57-58 : yft wiiS lande ealneg winneft, wind wrg wage. 4 21 dun ofer dype. Brooke compares Aeneid, i, 105, ' Insequitur cumulo praeruptus aquae mons.' Yet Herzfeld, p. 38, calls this ' ein modernes Bild.' 4 22 eare geblondeu. The phrase suggests the compound ear-(ar-~)geblond, which is discussed by Krapp, Andreas, note to 383. 4 23 mearclonde. This is the only appearance of the word in the sense of 'sea-coast.' As Merbach says (p. 19), ' mearclond (Rid. 423) und landgemyrcu (Beow. 209) sind als Strand, Gestade aufzufassen, sie bedeuten die Landgrenze gegen das Wasser hin.' 4 24-25 Brooke again compares Aeneid, i, 87, ' Insequitur clamorque virum stridorque rudentum.' 4 27 hopgehnastes. Save in this case and in -wolcengehndste, 4 t>,gehndst, both simplex and in compounds, is used only of the clash of battle (Gen. 2015, aefter )>am gehnieste; Brun. 49, cumbol-gehnastes). The first member of the compound, hop, is discussed at length by Dietrich, Haupts Zs. IX, 215, and Grein, Spr. II, 95-96. Cf. Scottish hope, ' a haven.' 428-29 sllbre saecce. Brooke translates (p. 185, n.) : 'with slippery . . ., with feeble striving ' — and interprets ' with a hapless ill-fortuned and therefore a despairing strife against the elements. Some are paralyzed in expectation, some struggle.' This is finely poetical, but it disregards both grammar (as sonne sceor cyme&. Scur is found with the lemma nimbus, \VW. 175,22; 316,36. 444 blacan Hge. Cf. And. 1541. In his note to the passage Krapp quotes from Mead's article (P.M.L.A. XIV, 177): ' Bide is merely an ablaut form of the stem bltcan, " to shine," and perhaps hardly means white at all. In a few cases it evidently means pale or ghastly. It is properly applied to the fire or the fire- light and even to the red flame or to the lightning or to the light of stars. Of the twenty-eight instances where the word occurs, — either alone or as part of a compound, — nearly all seem to lay emphasis on the brightness rather than the whiteness.' 4 45 dreohtum. For the MS. reading dreontum, Thorpe suggested dreohtum = drvhtum ('populis') and was followed doubtfully by Grein, Spr. I, 204. This is favored by 4 40, ofer burgum, and 4 43, of er f oleum. Grein, Bibl. II, 371, note, pro- posed dreongnm = drengum, but Holthausen, Engl. Stud. XXXVII, 206, rightly rejected this as Scandinavian (dretigr) rather than English, and proposed dreor- gitm. The ' dreary ones ' are the terrified men of 4 33, 49. I prefer Thorpe's sug- gestion. 4 46-48 ' The poet represents the thunder and lightning as arising from the violent meeting of the clouds, without expressly mentioning th&fragor but this bursting of the clouds is taken for granted by the author, who thus continues : feallan laJtaiS sweart sumsendu seaw of bosme, wjetan of wombe. This is pictured as the result of the bursting ' (Erlemann). 4 47 Brooke (p. 185) renders this finely and accurately, ' swarthy sap of showers sounding from their breast ' ; and adds : ' I should like to have in English the German word summen, which answers here to sumsend, and translate this sum- ming. "Sounding" does not give the humming hiss of the rain.' For a discus- sion of the etymology of sumsendu, see Kogel, Geschichte der deutschen Lit., 1894, I, 53-54 (Bright). 4 48 f . Erlemann says (p. 53) : ' Von Vers 48 ab verlasst der Dichter dann diesen Vorstellungskreis : der Sturm die Ursache des Gewitters ; seine Phantasie ist ganz erfiillt von dem Bilde des Kampfes der dahinfahrenden Wolken und kann noch nicht zur Ruhe kommen. Das Bild spinnt sich fort : Winnende farefr atol eoredbreat; altheidnische mythische Vorstellungen mogen dabei wachgerufen sein und hier durchschatten, aber sie werden wieder zuriickgedrangt durch christliche Empfindungen.' 4 52 scin. The nature of such demons is described, Whale, 31-34 : Swi br$ scinna J^eaw, deofla wise l>zet hi drohtende }>urh dyrne meaht dugu'Se beswicaiS ond on teosu tyhtaS tilra djeda. NOTES 77 4 51-52 Cf. Ps. 63 4, hi hine . . . scearpum strallum on scotiaS. 453-58 As sources of these lines Erlemann (p. 53) suggests. Ps. xvii, 15, ' Et misit sagittas suas et dissipavit eos : fulgura multiplicavit et conturbavit eos ' (2 Sam. xxii, 15) ; Ps. cxliii, 6. 4 55 on geryhtu. Cf. Jnd. 202, Met. 31 17, ongerihte, which has also the meaning 'straight.' 4 58 rynegiestes. Thorpe and Brooke render ' the rain-spirit,' but Grein in- terprets in Spr. II, 386, 'profluvii hospes,' and in Dicht. he translates 'des Rin- nengastes.' Bosworth-Toller translates 'a guest or foe that comes swiftly(?)' and Sweet, Diet., ' a swift guest ' — a rendering supported by such compounds as ryne- strong, ryneswift. But, as the simplex ryne, ' rain,' appears in apposition with regn (Gen. 1416), and as the interpretation 'rain-foe' seems suited to the con- text, I have adopted that. 4 59 Cf. Beffw. 2408, se >aes orleges or onstealde. 4 59 ff. Herzfeld, p. 37, remarks, ' Der Sturm wird, 4 59, in einem prachtigen Bilde als Kriegserreger vorgefiihrt, die Krieger sind die Wolken (hlofrgecrod), die mit lautem Gekrach auf einander stossen ; sie schwitzen Feuer aus (die Blitze, die mit Pfeilen verglichen werden), ein dunkler Saft fliesst ihnen aus dem Busen u.s.w.' 462 ofer byrnan bosin. Cf. And. 441, of brimes bosme ; Exod. 493, famig- bosma. Cosijn (PBS. XXIII, 128) doubtfully compares Pan. 7, }>isne beorh- tan bosm ; but the reference is to the earth, not to the waters. Brooke says (p. 186): 'The word I here translate torrents is byrnan ("of burns or brooks"). Torrent is quite fair, for the word is connected with byrnan (" to bum "). The upsurging and boiling of fire is attributed to the fountain and stream. Cynewulf is not thinking of the quiet brooks of the land, but of the furious leaping rivers which he conceives as hidden in the storm clouds over which the storm giant passes on his way.' 463 hf-Hii hloSgecrod. Brooke, E. E. Lit., p. 186, says: ' Hlofr is the name given to "a band of robbers from seven to thirty-five" [Laws of Ine § 13, Schmid pp. 26-27], hence any troop or band of men [And. 42, 1391, etc.]. Gecrod is "a crowd," "a multitude." Thus compounded, the word means, I think, a crowd made up of troops ; of troops of clouds ! Then the word " high " put with hlofrgecrod and the context prove sufficiently that Cynewulf was thinking of the piled-up clouds of the storm ; and no doubt the notion of ravaging and slaughter con- nected with I/ldfr pleased his imagination, for his tempest is a destroyer.' Brooke's translation ' the high congregated cloud-band ' is suggested by Shelley's lines (with which compare 4 42-48) : Vaulted with all thy congregated might Of vapors, from whose solid atmosphere Black rain and fire and hail will burst. O, hear ! 467-72 In these lines occurs a summary of the various manifestations of the Storm, but Rid. 2, which represents the Storm in general, finds no place in this review. It is interesting to note that the order of the single descriptions does not conform to the order in the summary. There the maritime eruption (Seebeben), Rid. 3, stands before the earthquake (4 1-16) ; here, after. Erlemann (pp. 53-54) 73 RIDDLES OF THE EXETER BOOK does not believe that any derangement of the text, any inversion of 3 and 4 1-16, has taken place. -'In the summary a more convenient adjustment of the verse may have brought it about that no particular regard is paid to the accurate sequence of the several parts ; it is also possible that the poet anticipated 4 1-16 in order to place 3 and 4 17-35 near together, so as to contrast them better : " Now I shall fight under the waves, now above the waves." ' 469 hean underhnigan. In Dicht. Grein translates 'Bald soil ich des Oceans Wogen die hohen unterneigen,' and he is followed by Barnouw, p. 221, who ' regards hean as ace. pi., weak, of heah. In Spr. II, 55, Grein rightly gives the word under hean, 'low'; cf. Gn. Ex. 118, hean sceal gehnigan. 4 71 wide fere. Cf. 59 3, wide ne fereS; 95 3, fere (MS. fereS) wide. 473-74 Aldhelm iv, i, ' Cernere me nulli possunt, nee prendere palmis,' which Prehn (p. 160) regards as one of the sources of the Anglo-Saxon, is derived, like the English riddle, from the Bible : Prov. xxx, 4, ' quis continuit spiritum in mani- bus suis,' arid Ecclus. xxxiv, 2. So Erlemann, pp. 55-56 (but the connection is certainly not close). I have traced the history of this motive, Mod. Phil., II, 563. It appears in Bede's Flares, No. V, in various 'dialogues' (Haupts Zs. XV, 167, 169), and in MS. Bern. 611, No. 41. RIDDLE 5 Diejrich JXI, 461) suggested first the answer 'Bell,' but rejected it imme- diately in favor of ' Millstor\e,' believing that the latter fulfilled more closely all the conditions of the problem. Grein, Spr. II, 716, accepts the first solu- tion; and Prehn, pp. 163, 165, the second, but he fails in his attempt to indicate / a likeness between this riddle and the 'Millstone' enigmas of Symphosius (51, f 52) and Aldhelm (iv, 12). In riddle-literature there are no analogues to aid one, the many 'Bell' and 'Millstone' problems (see Schleicher, p. 201 ; Symp. 80, Tintinnabtilum ; Tatwine 7, De Tintinno) being of a totally different type. Personally, I incline to the first answer. T\e£>egti or servant ma.y be the, ostia- rius or (ft(re7('fr^ (g"p Canons of SElfric, n), who is thus described by William of Malmesbury (Gesta Pontificum, 76, cited by Padelford, Musical Terms in Old English, p. 56) : ' Reclusis enim a dormitorio in ecclesiam omnium parietum obsta- culis vidit monachum, cujus id curae erat, a lecto egressum funem signi tenere quo monachos ammoneret surgere.' Not only monasteries, but Anglo-Saxon houses of better estate had each its bellhiis (Padelford, I.e.; Be leod-gej>incfrtim 2, Schmid p. 388) ; but, as Schmid points out (Glossar s. v.), the word may refer to the refectory, to which one was summoned by bells (cf. Du Cange s.v. Tinelhis) or perhaps to the cloccarhim vel lucar (the lemma of belhiis, WW. 327, 16). Our rid- \ die refers, I think, not to the hand bell, lltel belle or tintinnabiilum (for a discussion 1 of its use, see Westwood, Facsimiles, p. 152, Padelford, p. 58), but to the micel belle / or campana (^Elfric, Gloss., WW. 327, 18). This was well known in the England of the eighth century, for in Tatwine's De Tintinno enigma (No. 7) the bell is suspended high in air, ' versor superis suspensus in auris.' Professor Trautmann brings nothing to support his ' Threshing-flail ' solution of our enigma. NOTES 79 Andrews, Old English Manor, p. 259, discusses the Anglo-Saxon mill or quern, and thus translates the last lines of our Riddle : ' " Sometimes a warm limb may break the bound fetter ; this, however, is due to my servant, that moderately wise man who is like myself, so far as he knows anything and can by words convey my constructing message." We here accept Grein's translation almost without change, but of the last two lines can make no meaning. The iron-work of the mill is in- teresting, as is also the harsh grating sound with which it moves when started in the early morning. These features Cynewulf has added to the original of Sym- phosius (Prehn, pp. 163-165).' See also Heyne, Halle Heorot, p. 27 ; Fiinf Biicher II, 257-266; and Klump, Altenglische Handiverknamen, pp. 13-15. They accept the ' Millstone' answer and discuss mills and mill-maid {Caws of &frelberht § n, Schmid p. 2). 5 i Jjragbysig. Dietrich finds the source of this in Aldhelm's line (iv, 124), ' Altera nam currit, quod nunquam altera gessit,' while Prehn points to Sym- phosius 51 : Ambo sumus lapides, una sumus, ambo jacemus. Quam piger est unus, tantum non est piger alter : Hie manet immotus, non desinit ille moveri. But the parallel is far-fetched. The epithet might well apply to a bell, for this is surely ' periodically employed.' Dr. Bright suggests the meaning ' perpetually.' 52,4 hringum haefted . . . halswrijjan. Wanley, Catalogue 109, 2, 16-20: 'Se bend iSe se clipur ys mid gewrifren, ys swylce hit sy sum gemetegung "Sa^t ftiere tungan clipur maege styrian, and fta lippan aethwega beatan. SoHice mid ftass rapes aet-hrine se bend styra)> Sone clipur.' ' The band with which the clapper is tied, is, as it were, a method for moving the clapper of the tongue and beating more or less the lips. So, with the touch of the rope, the band moves the clapper' (B.-T. s. v. Clipur). The key in Rid. 914 is hringum gyrded; but such phrases are even better suited to the durance of the bell, as Wanley's account of the bend shows. With hringum hafted compare Gen. 762, hasft mid hringa gespanne (Satan). 5 3 The line refers to the beating of the clapper against the sides (mln bed brecan), and to the sound of the bell (breahtme cyfrari). 5 7 [J»aet] wearm[e] lim. ]> is perhaps omitted on account of preceding -J>e in oncwej>e. Grein, Spr. II, 188, supposes lim to refer to manus. This accords well •with the 'Bell' solution. See Techmer, 2, 118, 7 (cited by Padelford, pp. 56, 71): ' Dxs diacanes tacen is l>ast mon mid hangiendre hande do swilce he gehwaede bellan cnyllan wille.' Or if the large bell is meant, the warm limb may be the clipur, which bursts the ring with which it is bound (supra). 5 8 bersteff. This is the only appearance of the verb in a transitive sense in Anglo-Saxon ; but the word is used so commonly with an active meaning in Middle English (see Matzner, or Bradley-Stratmann, s. v.) as to make such a rendering very plausible here. 5 9-12. The editors punctuate variously and thus give widely differing mean- ings to the last four lines of the riddle. Thorpe's rendering is utter nonsense. Ettmiiller puts a period after hwtlum (1. 8), a semicolon after men (n), and no point after sylfe. Grein and Assmann place a comma after hwilum and a comma after sylfe. I point as in text, and render ' It (the ring) is, however, acceptable 80 RIDDLES OF THE EXETER BOOK to my thane, a moderately wise man, and to me likewise, if I (an inanimate thing) can know anything and in words successfully tell my story.' For the happy rendering of the last clause I am indebted to Dr. Bright. 5 10 Jjaet sylfe. This accusative of specification is equivalent to the adverb ' likewise ' (cf. Chr. 937 ; Ps. 81 3, 128 i ; Spr. II, 429). 5 11-12 mm . . . spel. For separation of possessive pronoun and substantive, see ij g_10) hyra . . . drohta~3. With the last line of our riddle compare Beow. 874, on sped wrecan spel. RIDDLE 6 As early as 1835, L. C. Miiller (Collectanea Anglo-Sax onica, pp. 63-64) sug- gested 'Scutum' as an answer; and Dietrich XI, 461, gives the same solution. He and his follower Prehn, p. 165, point to Aldhelm's 'Clypeus' enigma (iii, 13) as a source. The resemblance is very slight. Both shields have received many wounds (infra) ; but Aldhelm's is a glorious warrior, while that of our riddler |s a brokenji.gh.tej: (Brooke, E. E. Lit., p. 123, note). Unlike Aldhelm, the Anglo- Saxon poet does not dwell upon the relation of the shield to its lord. A literary analogue, as Dietrich pointed out, is the 26th riddle of the Hervarar Saga, where the Shield vaunts its wounds (see Heusler, Zs. d. V.f. Vk. XI, 139, 148). Traut- mann's ' Hackeklotz ' has nothing in its favor. The riddle is rich in conventional epithets, applied to the Shield's enemy, the Sword, not only elsewhere in the poetry but in other riddles. Illuminated Anglo-Saxon MSS. usually represent the warrior as armed with no other defensive weapons than shield and helmet (Meyrick, Antient Armour, 1842, p. li ; Keller, pp. 71 f.). The shield, circular or slightly oval in shape, is usu- ally of linden-wood, sometimes covered with leather, with a metal-bound edge and in the center an iron umbo or boss, a small basin tapering at the top to a point and ending in a knob (Gn. C. 37, rand sceal on scylde faest fingra gebeorh). Bosses are of various form and of different degrees of ornament (Roach-Smith, Collectanea Antigua I, 104; II, Plate 36; III, Plate 2). The grave-finds reveal a large number of shields of which boss and handle alone remain (Keller, pp. 74-79 ; Kemble, Horae Ferales, p. 82). 6 i iserne wund. Cf. Bemv. 565, mecum wunde ; 1076, gare wunde. See Ald- helm iii, 13 z, ' patiens discrimina dura duelli.' 6 2 beadoweorca saed. Cf . 34 6, biter beadoweorca ; Brun. 20, werig wlges saed. 6 3 ecgum werig. Cf. And. 1278, wundum werig ; Maid. 303, wundum werige ; Beow. 2938, wundum werge. — Oft Ic wig seo, etc. See Aldhelm iii, 13, ' Quis tantos casus . . . suscipit in bello . . . miles ? ' 6 4 frecne feohtan. So And. 1350. — frofre ne \v6ne. Cf. Gu. 479, frofre ne wenaft ; Beow. 185, frofre ne wenan. 66 eal forwurde. Cf. Ps. 11892, call forwurde. 6 7 homera lafe. Cf. Beow. 2830; Brun. 6, homera lafum, — in both cases of swords. In Rid. 71 3-4, the Sword or Dagger calls itself wrdfrra /df, \ jyres ondfeole. For many examples of Idf as a synonym of sword in the poetry, see Spr. II, 152, and Cook, 'A Latin Poetical Idiom in Old English,' American Journal of Phi- lology, VI, 476. NOTES 8 I 68 heardecg heoroscearp. Cf. Beorv. 2830, hearde, heafto-scearpe homera lafe ; Jud. 263, heardum heoruwsepnum. Heardecg is found as an epithet of the sword, Beow. 1289, 1491, El. 758. — hondweorc smipa. So of the Sword, 21 7. Cf. also 27 14, wrietllc weorc smi>a. For the position of the smith in Anglo-Saxon times, see notes to Kid. 38. 6 9 bitaS in burgum. In 93 21-22, ealle Jxrtte bord biton, 'all that bit the shield,' is a circumlocution for 'swords' or 'knives.' Cf. 93 17-18, |>eah mec heard bite | strSecg style. The sword-bite is a commonplace of the poetry, Jul. 603, J>urh sweordbite ; Ap. 34, fiurh sweordes bite. 69-10 Gu. 2Oj,gifke leng bide Idtran gemdtes, seems to support the change of MS. dbidan to d bidan. But as dbldan appears not infrequently in the desired sense (Spr. I, 12) I have retained it in the text. 6 10-12 For the use of worts in Anglo-Saxon leechcraft, see Cockayne's Leech- doms, passim. They were used particularly as dolgsealfa •wifr eallum wundum (Lchd. II, 8, 26). Among the common worts employed for wound-salves (Lchd. II, 90 f.) were groundsel, brooklime, lustmock, broad-leaved brownwort, ribwort, meadow-wort, githrife, cockle, carline thistle, ashthroat. 6 14 dagum ond iiihtum. So Exod. 97 ; Met. 20 213. RIDDLE 7 The rune S (Sigel, ' the sun ') precedes and follows the riddle in MS., thus putting th<> snlntjpjijjeyond doubt. The poem bears no resemblance to Aldhelm viii, 3, De Sole et Luna, save in the design of the Almighty, who in the Latin is the ' Lord of Olympus,' in the Anglo-Saxon is the Christ. It certainly owes nothing to Eusebius 10, De Sole. The problem is like in kind to the 3ist riddle in Haug's collection from the Rigveda (p. 495) : ' Einen rastlosen Hirten sah ich hin und her wandeln auf (seinen) Pfaden ; sich kleidend in die zusammenlaufenden (und) auseinanderlaufenden (Strahlen) macht er (seine) Runde.' Cf. the Latin hymns in praise of the Sun (Meyer, Anthologia Latina, 1833, pp. 1024-1025). 71-2 Cf. Aldhelm viii, 35, ' Sed potius summi genuit regnator Olympi.' But the Anglo-Saxon has much in common with the well-known passage from Ps. (Vulgate) cxxxvi, 7-8: 'Qui fecit luminaria magna . . . solem in potestatem diei quoniam in aeternum misericordia ejus.' So in the Anglo-Saxon poetical version of Ps. Ixxiii, 1 6, J>u gesettest sunnan and monan, sigora -waldend. So Gen. 126, 1 1 12, etc. 'The Father is thought of especially as the Creator (Jul. in, Chr. 224, 472), though this function is sometimes attributed to the Son (Jul. 726, Chr. 14 f.), and is sometimes exercised by Him with the Father (Chr. 239-240),' Cook, Christ, p. Ixxvi. So in the Skaldskaparmdl, § 52 (Snorra Edda I, 446), Christ is called ska- para himins okjarfrar, engla ok s6lar. 7 2 to compe. The Sun and Moon are portrayed as fierce fighters in Rid. 30. — oft ic cwlce baerne. Cf. Ps. 1206, ne he sunne on daeg sol ne gebaerne. 7 3 unrimu cyn. So Pan. z. — eorj»an getenge. So 77 2. Cf. 8 8-9, getenge . . . flode ond foldan. Grein is wrong in regarding getenge as ace. pi. (Spr. I, 463) ; it obviously modifies the subject of the riddle. g2 RIDDLES OF THE EXETER BOOK 7 6-9 Of the joy and comfort that the Sun brings to men, the Wonders of Crea- tion gives glowing account (59-67) : ond >\s leohte beorht cymeiS morgna gehwam ofer misthleojm, wadan ofer wsegas, wundrum gegierwed, ond mid zrdaege eastan snoweft, wlitig ond wynsum wera cneorissum ; lifgendra gehwam leoht for5 biereft bronda beorhtost, ond his brucan mot jeghwylc on eorj>an J him eagna gesihiS sigora sotScyning syllan wolde. 7 7a I can see no reason for departing from the MS. here by inserting wel >&- ior&frefre. ffw . . . w alliteration is found i 12, 36 n, Becrzv. 2299 (Heyne's note), Gu. 323, Chr. 188. Cf. Sievers, Altgermanische Metrik, p. 37, note. 7 10 gedreag. The word gedreag, elsewhere used in the sense of ' crowd,' ' troop,' ' tumult,' is here applied to the ocean, probably with reference to ' the multitudinous seas.' RIDDLE 8 To this riddle there are no Latin analogues. All scholars accept, however, the solution ' Swan.' And the tradition of the musical plumage of this bird, occurring elsewhere in Anglo-Saxon poetry (Phcenix, 137), is admirably illustrated by a fable found by Dietrich XI, 462, in the letter of Gregory of Nazianzus to Celeusius (Opera, Caillau, Paris, 1842, II, 102). In this the swan explains to the swallows that sweetness and harmony are produced by the breath of the west wind against its wings. Neither Gessner, ' De Avibus' (Historia Animalium, 1554, III, 360), nor Paulus Cassel (Der Schwan in Sage u. Leben, Berlin, 1872), nor Swainston (Folk-Lore of British Birds, Folk-Lore Society, 1885, p. 151) mentions the legend of singing feathers, although each of them refers to the whistling swan of the North. Very much to the point is a passage from Carl Engel's Musical Myths and Facts, 1876, I, 89: 'Although our common swan does not produce sounds / which might account for this tradition, it is a well-known fact that the wild swan / (Cygnus ferus), also called the whistling swan, when on the wing, emits a shrill / tone, which however harsh it may sound if heard near, produces a pleasant effect I when, emanating from a large flock high in the air [cf. Rid. 8 8-9], it is heard in a variety of pitches of sound, increasing or diminishing in loudness according to the! movements of the birds and to the currents of air.' For the superstition of the swan singing at death, of which our riddler makes no mention, see Douce, Illus- trations of Shakspere, 1839, p. 161 ; Dyer, Folk-Lore of Shakspeare, 1883, p. 147. Swainston, I.e., discusses in detail the place of the swan in mediaeval laws and oaths (see also Archaeologia XXXII, 1847, 423-428). The riddle of the Swan, as I have pointed out in the Introduction, has much in common with two other bird riddles (n and 58). The swan's song is mentioned Seaf.j\^ylfete song. For a late English analogue to this Swan riddle see Pretty Riddles, 1631, No. 35, Brandl,/a>4^. der deutschen Sh. Gesell. XLII (1906), 57. NOTES 83 Brooke says (p. 148): 'Once on a time Cynewulf, who may now have seen the Swan flying over the forest to some inland pool or fen, described it in one of the finest of his riddles — marking especially the old tradition of its song not before its death but when it left the village to fly over the great world. Nor did it sing with its throat. Its feathers sounded melodiously as the wind went through them. ... It has the modern quality. Phrases like " the strength of the clouds," " the spirit that fares over flood and field," the melodious rustling of the fretted feather- robe, the sense of a conscious life and personality in the bird and its pleasure in its own beauty are all more like nineteenth century poetry in England than any- thing which follows Cynewulf for a thousand years.' 8 i Hraegl. This word is again used of the plumage of a bird (Barnacle Goose) in the riddle's closest analogue, 117''. — hrusan trede. So we are told of the Swallows, 58 5, tredafr bearoiuzssas etc. Cf. Gen. 907. 8 2 Jm wic buge. Cf. 16 8, wic buge ; Gu. 274, >e J>a wic bugaS. — wado drefe. Cf. 23 16; H. M. 20, lagu drefan ; Becnu. 1904, drefan deop waeter. 83-7 So in ii 9-11 the air and wind raise the Barnacle Goose and bear it far and wide (note the likeness of wording in the two passages). In 58 i ' this air bears little wights' (Swallows). The best explanation of these passages is found in the Hexameron of ^Ifric (edited by Norman, 2d ed. 1849, P- 8) : ' Daet lyft is swa heah swa swa fta heofonlican wolcnu and eac ealswa brad swa swa ftJEre eoriSan bradnyss. On "Siere fleofi fugelas, ac heora fi^era ne mihton nahwider hi aberan, gif hi ne abiere seo lyft.' 83 ofer haelepa byht. Cf Gen. 2213, folcmasgfta byht ; 23 12, ofer waeteres byht. 8 4 hyrste mine. So of the wings of the Goose, 1 1 8h. — J?eos hea lyft. Cf . 119, lyft; 58 i, )>eos lyft. 8 6-9 For a reference to the singing of the Swan's feathers, compare the pas- sage in the P/iwnix, 134-137 (Bright's reading): Ne magon J>am breahtme byrnan ne hornas, ne hearpan hlyn, ne hale^a stefn Singes on eor)>an, ne organan sweg, ne hleo)>res geswin, ne swanes feffre. Lactantius mentions here (1. 49) 'olor moriens.' That certain birds have the power, in flight, to make a sound with their feathers at will, is shown by the example of the kingbird, which swoops down silently till close above its enemy's head and then loudly rattles its feathers with alarming suddenness; and of the ruffed grouse or American partridge, which takes flight now in silence and now with the loud whir which is so disconcerting to some of its enemies. That this power is used by some birds as a sort of song appears by what Gilbert White of Selborne says of the 'bleating' or 'humming' of cock-snipes, Letter XXXIX (Pennant): 'Whether that bleating or humming is ventriloquous or proceeds from the motion of their wings, I cannot say ; but this I know, that when this noise happens the bird is always descending, and his wings are violently agitated' (compare also Letter XVI). White's most recent editor notes that ' this noise made by the cocksnipe when after rising to a great height {Rid. 8 3-6] he casts 84 RIDDLES OF THE EXETER BOOK himself down through the air ... seems to be produced by the air waves being driven by the powerful wing-beats through the expanded and rigid tail feathers.' 8 6 Fraetwe mine. Fratwe is again used of plumage Ph. 335, frat-we fly 'hthwa- tes. As Brooke says (p. 148), ' Frcehue is originally carved fretted things; hence an ornament — anything costly ; here then my rich garment of feathers.' 8 7-8 swinsiafl, | torhte singaft. Cf. Chr. 884, singaS ond swinsia«. The phrase appears twice in the very passage of the Phccnix in which ' the singing feathers ' are introduced : 1 24, swinsaft ond singeS ; 140, singeiS swa ond swinsaiS. RIDDLE 9 To this riddle many solutions have been offered. In his first article (XI, 461- 462) Dietrich wavered between A.-S. Sangpipe and the Nihtegale, supporting the first by the C-rune (possibly for Camena, which is the lemma to sangpipe, Pru- dentius Gl., Germam'a, N. S., XI, 389, 26) which precedes the riddle in the MS., and the second by reference to Aldhelm's Luscinia enigma (ii, 5). Later, XII, 239, he presented with confidence the answer ' Wood-pigeon,' defending this by three arguments: (i) the Anglo-Saxon name of this bird, Cuscote (W\V. 37, 35, Palumbes, cuscote) meets the demand of the C-rune; (2) with its flexible voice it really imitates the song of jesters (Rid. g 6, 9-10) ; and (3) it attains to a great age (Rid. g 5, eald Sfensceop). Each of these three solutions has been accepted, the first by Padelford, p. 52, the second by Brooke, E. E. Lit., p. 149, the third by Prehn, p. 167. Yet another answer, 'Bell,' is given by Trautmann (Anglia, Bb. V, 48) and repeated by Padelford, p. 53; and this is accepted by Holt- hausen, who asserts stoutly, without a jot of proof (Anglia, Bb. IX, 357) : ' Die C-rune iiber diesen ratsel bedeutet offenbar clugge, "glocke." ' Of these solu- tions, ' Nightingale ' seems to me distinctly the best, for its varied note is heard in so much poetry of the late Latin period ; for instance, in the Philomela elegies of the mythical Albus Ovidius Juventinus and Julius Speratus (Wernsdorf, Poetae Latini Afinores, VI, 388, 403 ; compare Schenkl, Sitzber. der phil.-hist. Cl. der Wiener Akademie, 1863, XLIII, 42 f.), and in the pretty Luscinia poem of Alcuin (Migne, P. L. CI, 803). Yet A'ihtegale does not fit the rune, and is obviously the reverse of scurrilous ; hence this answer, like the others, must be given up. The motive of the problem so closely resembles that of Rid. 25, Higora, that I am inclined to accept that answer here. It caps the query at every point. The jay is a jester. Martial in his epigrams calls it ' pica loquax ' (xiv, 76) and ' pica salu- tatrix' (vii, 87), and Ovidius Juventinus in his Philomela poem, 33-34, says : Pica loquax varias concinnat gutture voces, Scurrili strepitu quicquid et audit, ait. Grein's citations (Spr. II, 72, s. v. higora) are apposite: 'Die Glosse "berna, higrae," gl. Epinal. 663 (156) and gl. Erf. (wo berna fur verna, wie diese Glossen ofter in den lat. Wortern b fiir v schreiben) zeigt [see also WW. 358, 5], dass der Name unsres spasshaften Vogels auch fiir Spassmacher, Hanswurst iiberhaupt gait.' See Notes to Rid. 25. Like the ' Psittacus ' of Alex. Neckam, De Natura NOTES 85 Rerum 36 (Rolls Series, 1863, p. 88) the 'Higora' may be thus described: 'In excitando risu praeferendus histrionibus.' See also Dietrich, XI, 465 f. The Latin names of the bird in Anglo-Saxon glosses (WW. 13, 18, cicuanus, higm; 132, 5, catanus, higere), 'Cicuanus' and 'Catanus,' may have suggested the C-rune. g 1-3 It is possible that these lines may have been suggested by Aldhelm's Luscinia enigma (ii, 5) : ' Vox mea diversis variatur pulchra figuris.' Yet the thought is closely paralleled by the undoubted Higora enigma, 25 i, wrizsne mine stefne. 9 i Jmrh imij>. This is decisive against the Sangplpe solution. In 61 9, the Reed-pipe tells us explicitly that it is miifrleas. — mongum reordum. So Gu. 870. 9 2 wrencum singe. Cf. Ph. 131-133: Bi'5 J>aes hleo'Sres sweg eallum songcrjeftum swetra ond wlitigra ond wynsumra wrenca gehwylcum. 9 2-3 wrixle . . . heafodwope. Cf. Ph. 127, wrixleft woiScraefte (the bird). 9 3 hlude cirme. Cf . 58 4, hlude cirma~5 (swallows) ; 49 2-3, hlude | stefne ne cirmde ; Gu. 872, hludne herecirm. 9 4 hl£oj>re ne mij>e. In its present sense of ' refrain from ' mij>an is found elsewhere in poetry only in 64 10, also with the instrumental : ne mceg tc J>y mif>an. 95-6 bringe | blisse. Cf. Chr. 68, bringeft blisse. 9 7 stefne stymie. Cf. Ps. 76 i, mid stefne . . . styrman ; 139 6, stefne . . . styrme ; 141 i, stefn . . . styrmeiJ. x 9 8 swigende. The MS. nigende is regarded by all scholars as corrupt. There is little to choose between Grein's suggestion, hnigende 'gesenkten Hauptes,' and the swigende of Ettmiiller and Cosijn. I prefer the second because it accords better with alliteration and context. Why listen with reverence (hnigan is always used with that implication) to the scurrilous chatter of a jay ? Grein, indeed, renders in Dicht. Stille in den Hausem sitzen sie und schweigen.' ' 99-10 These lines support my interpretation, 'Higora' or 'Jay.' As Miiller says (Cbthener Programm, pp. 16-17): 'Dort ist auch ausdrucklich von dem possirlichen Wesen desselben Vogels die Rede ; so hatte bei den Angelsachsen vielleicht derselbe Veranlassung gegeben, den Spassmacher higora zu nennen, an dessen Namen sceawend-sceawere Dietrich zu IX erinnert, und Grein hat nicht Unrecht aus den gl. Epinal 1 56 higrae berna, d. i. verna scurra herbeizuziehen.' We are therefore told in these lines that the Jay is a mime and imitates the speech of buffoons — in other words, that the bird possesses the power of mimicry. Rid. 25 is but an elaborate illustration of this idea, and merely sup- plements with examples the earlier riddle. 99 The troublesome scirenige is changed by Cosijn (PBB. XXIII, 128) to sciernicge, which he rightly connects with scericge, 'mima,' Shrine 140. This is in a passage from the Martyr ologium, Oct. 19 (Herzfeld, p. 190, 9) : ' Seo (St. Pelagia) waes Srest mima in Antiochea J>Sre ceastre — )>aet is scericge (MS. C.C.C. 196, scearecge) on urum gej>eode.' Scericge is considered by Sievers as an example of the feminine ending in -icge and is associated with the older sciernicge (Anglia VI, 86 RIDDLES OF THE EXETER BOOK 178; VII, 222). — sceawendwisan. The meaning of this word is established by WW. 533,4, ' sceawendspr&c, scurrilitas ' (MS. scarilitas), and WW. 519, 3, ' scea- wera, scurrarum.' Grein translates the line (Die/it.) : ' der so scherzhaft ich der Schauenden Weisen laut nachahme.' Rather, 'in the manner of a mime, imitate the voices of jesters.' RIDDLE 10 Dietrich's answer, ' Cuckoo ' (XI, 463), has been accepted by all scholars. The Anglo-Saxon riddle displays some evidence of the use of Symphosius 100 (not in the best MSS.) in its description of the desertion of the cuckoo by its parents before birth and the adoption by another mother. But the chief motif of the English problem — ingratitude after fostering care — is such a departure from the Latin that the likenesses, such as they are, may lie simply in the nature of the subject. Symphosius' enigma is found in popular form in the Strassburger Ratsel- buch, 103, in Frankfurter Reterbilclilein (1572), cited by Dietrich, and in Reusner's collection (I, 275). Here Lorichius Hadamarius develops the Volksriitsel into a ponderous Latin version, citing not only his German original but the problem of Symphosius, this last under the title ' Ex Vita Aesopi.' If the ingratitude of the cuckoo is seldom treated in riddle-literature, it has been a favorite theme of natural history and folk-lore since the time of Aristotle. The words of the Stagirite in his Historia Animalium (ix, 20) are almost identical with those of our riddler : ' The cuckoo makes no nest, but lays its eggs in the nest of other birds. ... It lays one egg, upon which it does not sit, but the bird in whose nest it lays hatches the egg and nurses the young bird; and, as they say, when the young cuckoo grows it ejects the other young birds, which thus perish.' Turner (Avium Praecipuarum quorum apud Plinium et Aristotelem mentio est, brevis et succincta Historia, Coloniae, 1544) gives at length Aristotle's account of the 'Cuculus,' and Gessner, ' De Avibus ' {Historia Animalium, 1554, III, 350), cites not only this authority and the opinions of Theophrastus, Albertus, and Aelian, but a famous 'declamation' ' De Ingratitudine Cuculi,' by Philip Me- lanchthon (compare his Dedamationes, Argentorati, 1569, pp. 87-95). Mannhardt, whose excellent article on ' Der Kukuk ' (Wolf's Zs.f. d. M. Ill, 208-209) contains much valuable information, mentions a tract by Gronwall, De Ingrato Cuculo, Stockholm, 1631 (16 pages), which I have been unable to trace. The Cuckoo's ill return for the hedge-sparrow's care is not unknown to the poets. It is true that no reference to this is found in the Conflictus Veris et Hiemis in Laudem Cuculi (Riese, Anth. Lot. II, 145, No. 687), nor in Alcuin's lines on his lost cuckoo (Migne, P. L. CI, 104). But Chaucer, in his Parlement of Foules 612-613, calls his cukkow Thou mordrer of the heysugge on the braunche That broghte the forth, thou rewthelees glotoun. And Shakespeare's frequent references to ' that ungentle gull, the cuckoo's bird ' (Henry IV, Pt. I, v, i, 60) are well known. 'You know, nuncle, the Hedge- sparrow fed the Cuckoo so long that it had it head bit off by it young' (Lear i, 4, NOTES 87 235). Cf. A. ami C. ii, 6, 28, and Lncrece 849. Harting, Ornithology of Shak- spere, 1871, p. 147, and Dyer, Folk-Lore of Shakspere, 1883, p. 105, discuss this scrap of unnatural history; and Hardy, ' Popular History of the Cuckoo,' Folk-Lore Record, II (1879), 4^> gives other poetic examples of the tradition. In France it has become proverbial, 'Ingrat comme un coucou.' White of Selborne, Letter IV (Barrington), discusses at length the cuckoo's habit of depositing its eggs in the nests of other birds. Unlike Symphosius ('me vox mea prodit'), our riddler makes no reference to the cuckoo's note, which elsewhere in Anglo-Saxon poetry heralds the year. Cf. Sea/. 53, Gu. 716, H. M. 22. 10 1-3 Prehn, p. 169, finds in these lines a suggestion of Symphosius 14, Pullus in Ovo : Nondum natus eram nee eram jam matris in alvo. Jam posito partu, natum me nemo videbat. 10 ia Sievers, PBB. X, 454, regards MS. mec on frissum dagum as a form of A-type found elsewhere in the Riddles (! x X X | \j X ) ; but Holthausen, Engl. Stud, xxxvii, 206, would read on dagum J>issum or on J>issum ddgrum. The first reading is supported by Ps. 139 12, and I have adopted it. 10 2 feeder ond modor. So Sal. 445. 10 2b-3 Cf. Gen. 908, >enden J>e feorh wunaft, gast on innan. 10 3-6 Cf. Symphosius (?), 100, 'hoc tamen educat altera mater.' 10 4 wel hold. Holth. Anglia, Bb. IX. 357, would read wilhold, but as the MS. phrase is here both grammatically and metrically possible (_L|£_i_x) I retain that. — mege. In proposing this (not knowing that it was the MS. reading) Cosijn says: 'The foster-mother is mege (both belong to the bird-kind), but is not gesibb (1. 8).' Cf. 44 14, anre magan ; 84 32, worldbearna ma?ge. Dr. Bright proposes wel hold \to\ me gewedtim J>eccan. — wSdum Jjeccan. Cf. 46 4, hraegle J>eahte. 10 s heold ond freopode. Cf. Hy. 9 27, healda'S ond freo'Sia'5. — hleosceorpe. See note to 15 13, fyrdsceorp. 10 6 sue arllce. This is Cosijn's reading for the MS. snearlice, and it is sup- ported by the naturalness of the mistake of the scribe (who would not have thus misread swd drlice) ; and by 164, swe, and Leid. 1 1, sue mseg. Grein, Spr. I, 349, s.v./r#, seems to prefer frij>emag, rendering this by 'die Schiitzende' or ' Pflegemutter ' (so also Dicht^. Sweet accepts frij>entlzg, which is in harmony with the context and with freoj>ode (1. 5). But the meter demands fri}>e ; so we are forced to accept Dietrich's reading 88 RIDDLES OF THE EXETER BOOK (XII, 251) seo frlfre mag ('die schiine Frau'). This is supported by O.N./rtfrr ('beautiful,' frequently of women); and by such common expressions asjul. 175, seo arSele rnalg; Chr. 87, seo eadge mieg; Gen. 2226, freollce maig. 10 10 oj>J>set ic aweox[e]. Although of>J>aet ic . . . weariS), the meter makes a strong plea for Holthausen's reading (Engl. Stud. XXXVII, 206), dweox\e. Then we have an A-type (J- X X X | J- X ). 10 ii sijjas asettan. For examples of this idiom, see Dietrich, De Cyn. Aetate, pp. '2-3; Spr. I, 41. RIDDLE 11 I can only repeat my discussion of this riddle in M.L.A7. XVIII, 100-101. To the problem Stopford Brooke (E. E. Lit., p. 179, note) offers the fitting answer ' Barnacle Goose ' ; and this solution is sustained by the first enigma in the col- lection of Pincier (Aenigmatum Libri Tres, Hagae, 1655), which has many points in common with the Anglo-Saxon : Sum volucris, nam plumosum mihi corpus et alae, Quarum remigio, quum libet, alta peto Sed mare me gignit biforis sub tegmine conchae, Aut in ventre trabis quam tulit unda. Solutio : Anseres Scotici quos incolae Clak guyse indigitant ... in lignis longiore mora in mari putrefactis gignuntur. The first literary account of this fable — which caps the query at every line — is found in the Topographia Hiberniae of Giraldus Cambrensis in the last half of the twelfth century (Dist. i, cap. 15, ed. Dymock, Rolls Series, 1867, V, 47-49). Giraldus, after a long description, which tallies remarkably with the Anglo-Saxon, declares that 'bishops and clergymen in some parts of Ireland do not scruple to dine off these birds at the time of fasting because they are not flesh nor born of flesh.' With such evidence as this, we must accept Max Miiller's opinion (Science of Language, 2d Ser., 1865, pp. 552-571) that 'belief in the miraculous transfor- mation of the Barnacle Shell into the Barnacle Goose was as firmly established in the twelfth as in the seventeenth century.' Indeed, two strangely created goose-species are described by mediaeval writers : (i) The Tree Goose; (2) The BarpaH>» florae, or <^1arlr The first of these is dis- cussed at length by Gervase of Tilbury in his Otia Imperialia (1211) (ed. Lieb- recht, Hannover, 1856, pp. cxxiii, 52), by William of Malmesbury in a story of King Edgar (Gesta Regum Anglorum, II, § 154, Rolls Series, 1887, I, 175), by Mande- ville (chap. 36), and by other writers until the time of Hector Boethius (Descrip- tion of Scotland, 1527, chap, n, englished in Holinshed's Chronicle, vol. I), who declares this tree-procreation false, but affirms his belief in Barnacles or Bernakes. The second is treated by Giraldus Cambrensis, I.e., by his contemporary, Alex- ander Neckam, De Naturis Rerum, cap. 48 (Rolls Series, 1863, p. 99), by Hector NOTES s Boethius, I.e., by Turner, Avium Praecip. ///jA, 1544, s.v. ' Anser,' by Gerard, Her- ball, 1597, p. 1391 (Brooke), and by many other authors quoted by Pincier and Liebrecht. Excellent reviews of the history of the superstition will be found in Max Miiller, I.e., and in Harting's Ornithology of Shakspere, 1871, pp. 246-256. Max Miiller (Science of Language, 2d Ser., 1865, p. 564) thus translates the Latin of GjraJd«s-Camlirejisis : ' Bernacae are like marsh-geese, but somewhat smaller. They are produced from fir timber tossed along the sea, and are at first like gum. Afterwards they hang down by their beaks, as if from a sea-weed at- tached to the timber, surrounded by shells in order to grow more freely. Having thus in process of time been clothed with a strong coat of feathers, they either fall into the water or fly freely away into the air.' This reads like a close para- phrase of our Anglo-Saxon text. In my refutation (M. L. N. XXI, 99) of Traut- mann's objections to this solution (BB. XIX, 170-171) I have pointed out that ' though our riddle is several centuries earlier than Giraldus' account of the super- stition, this is just the sort of popular myth that might exist for hundreds of years among simple men before finding a scholar to record it; and, again, many accounts of the marvel may have perished.' Dietrich, XI, 463, with Aldhelm's 'Famfaluca' (iv, ii) in mind, suggested ' Ocean-furrow ' or ' Wake.' Now, while the Anglo-Saxon has little in common with Aldhelm, it bears, at least in part, a certain resemblance to the ' Wave ' riddle of the Hervarar Saga (Heifrreks Gdtur, 21, see Heusler, Zs. d. V.f. Vk. XI, 127), and to its derived form in modern Icelandic (Arnason, No. 684). But Brooke's solution seems in every way better, as this alone fits all the motives of the problem. Trautmann, who had earlier accepted ' Wasserblase,' supported at length in his BB. articles (XVII, 142, XIX, 170 f.) a new solution, 'Anchor.' But I have shown (M.L.N. XXI, 98-99) that this is based by him upon violent changes in the text (ii 3b, 7a) and perverted meanings (infra). Holthausen's unhappy inter- pretation 'Water-lily' (Anglia, Bb. XVI, 228) has been refuted by Trautmann (BB. XIX, 172-173). ii 1-3 Prehn, p. 171, compares with this Aldhelm, iv, n 1-2: De madido nascor rorantibus aethere guttis Turgida, concrescens liquido de flumine lapsu. This is the only resemblance between the Anglo-Saxon and Latin poems. Traut- mann believes that neb (i a) refers to 'the spike of the anchor,' as the word is used of the point of the plowshare (Rid. 22 i). But the passage finds its true analogue in Giraldus' account of the Barnacle Goose : ' Dehinc tamquam ab alga ligno cohaerente, conchylibus testis ad liberiorem formationem inclusae, per rostra de- pendent? Middendorf rejects Trautmann's solution (Anglia, Bb. XVII, 109). ii 3b on sunde awox. In order to justify his ' Anchor' solution, Trautmann would change this phrase to on sande grof. He objects to the form dwox because it differs from the usual West Saxon preterit, dweox (Rid. 10 ioa, 73 ia); but the reading is in perfect harmony with the context, and the survival of such a Northern form (Sievers, Gr.s, § 392, n. 5) in the text of the Riddles gives no difficulty. ii 4a yjmm ]?eaht. So we are told of the Anchor, Rid. 17 3. 90 RIDDLES OF THE EXETER BOOK 1 1 4-5 To say that an Anchor immersed in the water touches with its body the floating wood is nonsense ; but the phrase exactly accords with the descriptions of the Barnacle Goose. 116 Hsefde feorh cwico. The phrase is used elsewhere in the Riddles of liv- ing things, the Fingers (14 3a) and the Siren (74 sb). — of faeSmum . . . brimes. Cf-3 13, of brimes fae^mum. ii 6-1 1 With the two motives of the black and white aspect of the unknown thing, and of its journey with the wind, compare Heifrreks Gdtur, 21 : Hadda bleika hafa J>aer Enar hvitfoldnu, Ok eigu f vindi at vaka. ii 7-8 on blacttm hraegle . . . hwite hyrste. Hrcegl and hyrste are used of the plumage of the Swan (Kid. 8 ia, 4a). The ' black ' and ' white ' coat of our sub- ject recalls the account of the Barnacle in Gerarde's Herball (1597), p. 1391, as ' having blacke legs and bill or beake, and feathers blacke and white and spotted in such a manner as in our Magge-Pie.' In discussing this passage Brooke says (p. 179, note): 'The barnacle is almost altogether in black and white. The bill is black, the head as far as the crown, together with cheeks and throat, is white — the rest of the head and neck to the breast and shoulders black. The upper plumage is marbled with blue-gray, black and white. The feathers of back and wings are black edged with white, the underparts are white, the tail black.' This identification is better than, with Trautmann, to regard hyrste as referring to the rope of the anchor, and blaczim hr&gle to its tarry coat. 119-11 So in very similar riddles the air bears the Swan, 83-7, and the Swal- lows, 58 i (compare M. L. N. XXI, 99). The lines certainly cannot refer to the weighing of an Anchor. Brooke renders happily (p. 179) : When the Lift upheaved me, me a living creature, Wind from wave upblowing ; and as wide as far Bore me o'er the bath of seals — Say what is my name ! Trautmann wrongly regards lifgende as qualifying lyft. RIDDLE 12 For his answer, ' Cjgld,' to Rid. 12, Walz has argued strongly (Harvard Studies V, 261); and for the solution.' Wine' Trautmann has made out a seemingly good j case (BB. XIX, 173-176); but Dietrich's interpretation (XI, 463), Night,. fits better the various conditions of the query, as I have sought to show(5^Z. IV. XXI, 99-100), and is moreover supported by points of real likeness between our riddle and Aldhelm's enigma De Nocte (xii). That this problem is clearly a companion-piece to Rid. 28, 'Mead (12 6b, 2813; 127% 2817; 1210, 2812), is, at first sight, an argument for the ' Wine ' interpretation, but the meaning ' Night debauch ' is quite as well suited to the vinous lines that suggest the later riddle. NOTES 91 12 i Walz cites Grein's Spr. II, 14, to show that hasofdg is a proper epithet of gold. Trautmann, in his note on ffasu (BB. XIX, 216-218), combats the hitherto received meanings of the word 'fulvo-cinereus, wolfgrau und adlergrau ' (Dietrich, Uaupts Zs. X, 346) and 'graubraun' (Sievers, Gr.s, § 300), and seeks to prove that it can mean only 'glanzend' and that therefore hasofdg is inapplicable to Night. As I have said (M. L. A7! XXI, 100), even if we grant that this is the exclusive meaning, we must not forget that ' Night's mantle ' in poetry may be 'shining' or 'gleaming' (Met. 20229) as well as 'azure' or 'sable.' But in the light of the words that this adjective qualifies — eagle, smoke, dove, etc. — we cannot grant this, ffasti seems to have the later connotation of glaucus 'grayish,' to which indeed it corresponds, Rid. 41 6ib. The Latin word is a synonym of ctsrulus (Harper's Latin Dictionary, s.v. glaucus) ; and, as Dietrich has noted (XI, 463), ccerula is the very adjective used by Aldhelm to describe Nox in his riddle upon that subject (xii, 6). Or again, hasu or hasupdd is an epithet of the eagle, (Rid. 254, Bruit. 62), elsewhere called salowigpdda (Jud. 211), which Professor Trautmann could not define as 'shining.' The epithet 'gray' is eminently appro- priate to smoke (Rid. 2 7) or to the dove (Gen. 1451). Dietrich shows that hasofdg applies well to the raiment of Night, and that hyrste is used elsewhere in Old English poetry (Gen. 956, 2189) for stars. Traut- mann believes that the first lines suggest the garment of the wine, whether that be ' der schlauch, das fass, der krug, der becher, der kelch.' The opening passage (1-2) seems to me to describe far better a starry night than a golden beaker. Compare Shelley's lines ' To Night ' : Wrap thy form in a mantle gray, Star-inwrought. 12 3-5 Dietrich, Grein, and Wiilker close the first clause with unradsij>as. Herz- feld, who follows their pointing, supplies (p. 68) [on] before the final word ; and Klaeber (Anglia, Bb. XVII, 300) avoids emendation by regarding unrcedsij>as as gen. sing. (Sievers, Gr.3, § 237, n. i), dependent upon hwette which seems to govern the accusative of person and genitive of thing, although the latter construction does not appear elsewhere. This reading accords with Dietrich's translation (XI, 463): '(Sie) reizt die thorichten zum unrathgang, andem aber wehrt (sie) niitzliche fahrt.' Trautmann closes the first clause with hwette for the sake of the antithesis in line 3 between dysge dwelle and dole hwette. Setting aside Herzfeld's conjecture as unmetrical, he suggests rather doubtfully unriedsTf>a and renders lines 4~5a thus, ' Andren wehr ich unratgange durch niitze fahrt.' 123 dole hwette. Klaeber claims for dot the especial meaning of 'dumm- dreist, leichtsinnig, vorschnell, kopflos,' not as B.-T. renders, ' the dull.' According to Klaeber, the whole passage then carries this sense : ' Ich reize an zu torichtem beginnen und halte ab von niitzlichem tun.' This interpretation, he believes, ac- cords with Trautmann's answer, ' Wine,' which receives further support from Mod. iSf.,/>onne win hweteS1 \ beornes breostsefan. I am not in agreement with any of these views. I close the clause with unrSdsl^as, but I see no reason for regard- ing this as a genitive, or for assuming, what is nowhere found, an acc.-of-the-person- and-gen.-of-the-thing construction with hwette. Dole unriedsi^as is the direct object 92 RIDDLES OF THE EXETER BOOK of hwette (see Dicht., ' toll errege ich unrathwege '), and the passage may be ren- dered 'I mislead the foolish and instigate rash unprofitable courses.' See WW. 508, 4, l>d dolan rSdas, ' stolida consulta.' 124-5 ojjrum styre | nyttre fore. This is wrongly rendered by Trautmann, who mistakenly includes unradsibas in this clause, and by Spr. II, 491, s.v. styr. Dicht. translates 'Andere fu'hre ich zu niitzlicherem Laufe.' This exactly reverses the proper meaning (see Klaeber) : ' I restrain others from a useful course.' As Shipley points out (p. 56), styran ' to restrain ' is followed by dat. of person and gen. of thing. Cf. Craft. 105, he missenlice monna cynne gielpes styreS. Lines 3-8 seem to me in perfect accord with Dietrich's solution. Night may well provoke fools to deeds of debauch and crime, and deter others from a useful course. By reason of its evil ways, it may well be praised by drunken revelers (5 b— 8 a; cf. the next riddle, 139, dol druncmennen deorcum nihtum), and by rogues (Aldhelm xii, 9, Nox : ' Diri latrones me semper amare solebant '). Walz finds here the maddening effect of gold (cf. I Tim. vi, 9-10). 12 6b mode bestolene. Cf. 28 isa, strengo bistolen ; Gen. 1579, ferhiSe forstolen (the drunken Noah). 12 7a daede gedwolene. Trautmann (BB. XIX, 176) cites Jul. 113, dizdum gedwolene ; but, while he admits that the meaning in that place is ' die in ihrem tun irrenden,' he interprets the present passage as ' in ihrem tun gehemmt,' com- paring 28 14, maegene benumen. 12 7-8 deorajj mine | won wisan gehwam. Translate 'They praise to every one my evil (crooked) ways.' Grein, Spr. II, 720, strangely combines wisan and gehivdm, as the equivalent of quovis modo, 'auf jeder Weise ' ; but in Dicht. he renders the phrase rightly. 12 8b Cf. Hy. 2 6, wa him >xre mirig'Se ! 12 9-10 I agree with Dietrich that 9 b, horda deorast, refers to the sun, and that the line describes the coming of the day; and accept in this corrupt passage Cosijn's spirited reading heah J>ringefr (PBB. XXIII, 128) instead of Traut- mann's hearm bringefr, which seems to me tame and prosaic. Trautmann's ex- planation of the closing lines of the poem is as unfortunate as his interpretation of the opening passage. It is hard to believe that horda deorast refers to the com- . munion wine (why should that bring harm?) and that nyttre fore (5 a) is intended^' also to suggest the Eucharist (but that rendering was based on mistranslation), r Walz suggests that horda deorast indicates 'the word of God'; Dr. Bright, 'the soul.' But let us remember that in the poetry gim 'gem ' is a frequent metaphor for the sun, and that horda deorast carries much the same idea as gimma gladost (sun), Ph. 289. 12 9 J^rlngeo". Klaeber, Anglia, Bb. XV, 347, notes that the verb J>ringan, 'press on,' 'force one's way,' is admirably fitted to Gu. I255b, frong niht ofer tiht, as also in Gen. 139, brang tystre genip. It has likewise been applied to the com- ing of the morning: 'der Tag dringt eilends, unaufhaltsam vor,' M. H. G. der tac begund herdringen (Grimm, Deutsche Afythologie, 621, 626). 12 10 Cf. 28 12, gif he unrSdes XT ne geswiceft; Jul. 120, gif }m unraEdes jgr ne geswlcest; El. 516, ond )>aes unrihtes eft-geswlca«. See Herzfeld, p. 19. NOTES 93 RIDDLE 13 This problem of ' Oxhide ' or ' Leather (the answer accepted by all authorities) is the first of a cycle of Anglo-Saxon riddles of similar motives. Rid. 39, ' Young Bull,' is only a more pithy and epigrammatic expression of the 'living and dead' contrast in the first and last lines of Rid. 13; Rid. 27 describes in its earlier lines the tanning of the skin ; while Rid. 72 presents in detail the life and labors of the ox. The Latin analogues are many. Symphosius 56, De Caliga, indicates the contrast between the live animal and one use made of its skin ; Aldhelm, De Bove sive de Juvenco (iii, n), presents the themes of the four nourishing foun- tains, and the unlike fates of the living and dead ox, that compose Rid. 39 ; and the words of Eusebius, 37, are so similar to the Anglo-Saxon that both Ebert (p. 50) and Prehn (p. 213) have wrongly found the source of the close of Rid. 39 in the Latin : Si vixero, rumpere colles Incipiam, vivos moriens aut alligo multos. Other Latin riddles of the Old English period furnish quite as close parallels (see M. L. N. XVIII, 99) to Rid. 13 1-4, 14-15, and Rid. 39. Bede, Flares, No. viii, gives the following (cf. Mod. Phil. II, 562): 'Vidi filium inter quattuor fontes nutritum ; si vivus fuit, disrupit monies ; si mortuus fuit, alligavit vivos.' The Lorsch collection of the ninth century (No. n) presents the same motives with greater detail {Mod. Phil., 1. c.) ; and they appear later in Brit. Mus. MS. Burney 59 (eleventh century), fol. lib: Dum juvenis fui, quattuor fontes siccavi ; Cum autem senui, montes et valles versavi ; Post mortem meam, vivos homines ligavi. As our riddler tells us (39 5), the motive came to him by word of mouth. Riddles very similar to these Anglo-Saxon and Latin versions appear in many modern collections. I note particularly the Mecklenburg riddle (Wossidlo 76) : As ik liitt wier, kiinn ik vier dwingen [Rid. 39 3-4] ; As ik groot wier, kiinn ik hiigel un barg iimwringen [13 1-2, 396] ; As ik doot wier, miisst ik vor fiirsten un herren up de tafel stahn [13 5-6], Un mil de bruut na'n danzsaal gahn [136-7]. Cf. Simrock8, p. 33 ; Eckart (Low German), Nos. 585, 586 ; Renk (Tyrol), Zs. d. V. f. Vk. V, 115, No. 68 ; Schleicher (Lithuanian), pp. 205, 207, 'Als ich klein war, beherrschte ich viere _Rid. 39 3-4] ; als ich erwachsen, warf ich Berge hin und her; als ich gestorben war, ging ich in die Kirche.' To all these I may add the English ' Cow ' riddle ( Wit Newly Revived, Newcastle, 1 780, p. 20) : While I did live, I food did give, • Which many one did daily eat. Now being dead, you see they tread Me under feet about the street. 94 RIDDLES OF THE EXETER BOOK All articles made of leather came within the province of the Anglo-Saxon shoewright (yElfric's Colloquy, WW. 97): ' Ic bicge hyda and fell and gearkie hig mid craefte minon and wyrce of him gescy mistllces cynnes swyftleras and sceos, le>erhosa (caligas) and butericas (utres), brideljrwancgas and gerieda, flaxan vel pinnan (flascones) and hlgdifatu, spurlebera (calcaria) and haelftra, pusan and fietelsas, and nan eower nele oferwintran biiton minon craefte.' The preparation of leather in Old English times is discussed by Heyne, Fiinf Biicher, III, 207-212 ; and Klump, Altenglische Handwerknamen, pp. 20-22, 64-73. The Oxanhyrde (Rectitudines Singularum Personarum, 12, Thorpe, A.L. p. 188 ; Schmid, p. 380) is allowed to pasture two oxen or more with his lord's herd : ' Eamian mid "Sam scos and glofa him sylfum.' 13 1-4 Cf. 39 6-7, and Aldhelm iii, 113-7: Vivens nam terrae glebas cum stirpibus imis Nisu virtutis validae disrumpo feraces : . At vero linquit dum spiritus algida membra, Nexibus horrendis homines constringere possum. The use of the hide for bonds is, however, a motive common to all riddle-poetry of the time (supra). 13 i foldaii slite. For other references to plowing, see 13 14, 22 (Plow), 396, 72 12-15. 13 2a grene wongas. So 67 5, Gen. 1657 ; cf. Men. 206, wangas grene. Cf. also 41 5«> 83, J>es wong grena. 13 2b Cf. 21 8, gjestberend. 133 Cf. Sea/. 94, bonne him baet feorg losa'S. — faeste binde. Brooke (E.E. LA., p. 151, note) makes the strange mistake of supposing a reference to the bind- ing power of the liquor in the leather jug or black-jack, instead of to the bonds mentioned in all such riddles (supra). 13 4a swearte Wealas. For a discussion of the dark hair of the servant-class, see note to 13 8 (wonfeax Wale). The meter indicates clearly a long vowel in Wealas (see Gen. 2706, wealandum), while it permits & in 13 8, wonfeax Wale ; 53 6, wonfah Wale ; 72 n, mearcpabas Walas traed ; Wids. 78, ond Wala rices (cf. Sievers, PBB. X, 487 ; Herzfeld, pp. 49, 54, 58 ; Madert, p. 21). There thus seem to be, side by side, a long and a shortened form of the word, — a safer view than to regard, despite the evidence, all cases as short with Herzfeld, or as long with Madert (see Sievers, Gr? 218). 13 5-6 Cf. the mention of 'butericas (utres) . . . flaxan vel pinnan (flascones) and hlgdifatu' — all leather drinking vessels — in ^Elfric's Colloqtiy (supra), and the brief description of the leather bottle in Kid. 20. For the employment of cups of hide, see the Mecklenburg riddle already cited. In 80 6, the drinking-horn bears mead in its bosom. 136-7 Symphosius (56) pictures the hard service of leather in shoes: Sed nunc exanimis lacerata, ligata, revulsa, Dedita sum terrae, tumulo sed condita non sum. The likeness of the two riddles is in motif, not in treatment. NOTES 95 13 6b hwilum mec bryd triedeS. Fairholt {Costume in England, 1885, II, 59) bases his account of the shoes of the Anglo-Saxons upon the illustrations in the Durham Book and MS. Cott. Tib. C. VI (see Strutt, Horda Angelcynna, pi. xxiii) : ' They appear in general to have been made of leather and were usually fastened beneath the ankles with a thong. . . . The Saxon shoe took the form of the san- dal, being- cut across the front into a series of openings somewhat resembling the thongs which secured it.' On the same evidence Strutt asserts (Horda, p. 47) : 'Both men and women wore shoes, or rather slippers [WW. 125,27, Baxeae, wifes sceos]. The legs of the men were covered half-way up with a kind of bandage or else a strait stocking reaching above the knee ; they also wore a sort of boots which were curiously ornamented at the top.' Moritz Heyne, Fiinf Biicher III, 262-268, notes that in the shoes of the early Germanic peoples the hair-side of the skin was turned outward. 13 8a wonfeax Wale. The dark coloring of the menial Welshwoman is men- tioned elsewhere in the Riddles (53 6a, wronfah Wale), and three times the swarthy complexion of the servant class is named as a distinguishing feature 1134, swearte Wealas (here opposed to sellan men} ; 50 4-5, se wonna J>egn, sweart ond saloneb ; 72 ioa, sweartum hyrde (see Brooke, E.E.Lit., p. 136). That Wealh is used in the meaning of ' servus ' is naturally explained by the position which the old in- habitants of Britain held under the Anglo-Saxon rule (Schmid, Gesetze, p. 673, Glossar, s. v.). So, as the word slave was derived from the name of a people, tvealh was applied, without regard to origin, to bondmen who were, however, largely of Celtic or pre-Celtic blood. ' In early times, the women-servants ( Wale} and menials about the yeoman's or gentleman's house were absolute slaves and were bought and sold as cattle' (Powell in Traill's Social England I, 125). Grant Allen points out (Anglo-Saxon Britain, p. 56) that while 'the pure Anglo-Saxons were a round-skulled, fair-haired, blonde-complexioned race, the Celts had mixed largely in Britain with one or more long-skulled, dark-haired, black-eyed and brown- complexioned races.' The coloring of the subject people was held in contempt : In the old age, black was not counted fair, Or, if it were, it bore not beauty's name. Weinhold, Altnordisches Leben, p. 182, shows that the same attitude toward dark hair existed among the Scandinavians : 'Schwarzes Har achtete man dagegen fur hasslich ; denn es war fremd und dem Volksinne entgegen. Die dunkle Haut- farbe, die gewohnlich dabei ist, das finstere Aussehn, der starkere Bartwuchs gaben dem schwarzen nach dem herschenden Geschmack etwas widerliches. Wir haben schon friiher gesagt, dass man sich die unfreien schwarz dachte.' This feeling, and the fact that there could be dark complexion in the best Scandinavian blood, are attested by the story of Geirmund Heljarskin's childhood (Landndmab6k ii, 19; Stttrlnnga Saga i, 1-2). In his excellent discussion of the German dislike of dark and love of fair skins, Gummere, Germanic Origins, pp. 59 f., compares our names Fairfax (fair-hair) and its opposite, Colfax. I shall discuss the Anglo-Saxon regard for long blonde hair in my note to Rid. 41 98 (43 3 h-witloc, see 80 4). 13 8-1 1 Prehn, p. 176, thus explains these obscure lines : ' Vielleicht bezeichnet ersteres ein Wamms und deutet auf den Geliebten der schwarzlockigen Welschen 96 RIDDLES OF THE EXETER BOOK bin, u.s.w.' However that may be, he is certainly right in regarding the allusion as obscene. Unlike Prehn, I find only one, not two motives in this passage. 13 8b wegeft ond )?yft. Cf. 22 5, wegeiS mec ond >fS. 13 9 dol druncmennen. Budde, Die Bedeutn'ng der Trinksitten, p. 86, regards the phrase as a mere ' Umschreibung durch Trinkwendungen,' since a drunken woman appears nowhere else in Anglo-Saxon literature. Budde finds a. like peri- phrase in 61 9. — deorcum nihtum. So Beow. 275. 13 ioa wseteft in waetre. Cf. 272-3, wiette siJ>J>an | dyfde on waetre (skin or hide). 13 na lajgre to fyre. Cosijn (PBB. XXIII, 128) opposes fcegre to deorcum nihtum (1. 9), and compares fegre, 'diluculo,' Luke xxiv, i (Rushworth). But the sense of 'fitly,' 'properly,' is so commonly associated with the adverb (cf. 51 8, 544) that one can hardly accept Cosijn's suggestion. As the illustrated MSS. show (see particularly the calendar pictures of MS. Cott. Tib. B. V), the fire was in the middle of the Anglo-Saxon hall. 13 nb-i3 For an interesting analogue to this 'glove' motif, see the coarse riddle of Puttenham's old nurse (Arte of English Poesie, 1587, Book iii, Arber reprint, p. 198). Notice the important part played by the glove in the next riddle, 14. Strutt,Z>ra\f and Habits of the People of England, 1842, p. 45, makes the mistake of declaring that ' there is not the faintest indication of gloves in the various draw- ings that have fallen under my inspection.' But, as Planche (editor's note) points out, there is an instance in Harl. MS. 2908, engraved in his History of British Costume, p. 34, fig. b. See the description of the glove of Grendel (Beow. 2086 f.): Glof hangode sid ond syllic, searobendum faest, sio waes or^oncum call gegyrwed deofles craeftum ond dracan fellum. 13 nb-i2a Barnouw, p. 218, thus comments: ' Bemerkenswert is die stelle, 13 nb-i2a, wo ein schwaches absolutes adj. ohne artikel, hygegdlan, vorliegt ("der kecken hand," iibers. Grein) ; wenn die lesart richtig 1st, und ich sehe keinen grund sie zu beanstanden, beweist die stelle dass das dreizehnte ratsel sehr alt ist, aus einer zeit vor der abfassung der hauptmasse des Beow. herriihrend.' But, as Pro- fessor Kittredge says, ' the occasional retention of an old construction in poetry is no proof of antiquity.' RIDDLE 14 This riddle I have already explained (M.L.N.~XNIH, 101). Early scholars, Wright (Biog. Brit. Lit. I, 80), and Klipstein (Analecta Anglo-Saxonica II, 443) agree upon the solution 'Butterfly Cocoon'; and Grein (Germania X, 308) an- swers ' Raupe aus der Familie der Spanner (Palaenodea oder Geometrae).' In favor of these interpretations there is no evidence. Dietrich (XI, 464) suggests ' The 22 Letters of the Alphabet,' and points to Aldhelm iv, i. But there are at least three strong objections to this solution: (i) Of the unknown creatures appear only « ten in all — six brothers and their sisters with them ' ; and Dietrich, by his NOTES 97 reference to the vowels and their accompanying consonants in secret script, does not cope successfully with the numerical difficulty. (2) ' Their skins hung on the wall.' That the ' skin ' is the parchment Dietrich tries to convince us by citing an Alphabet riddle of a Heidelberg MS. of the fifteenth century (Mone, Qttellen u. Forschungen, p. 1 20) : ' Es hat ein teil in leder genist,' — and by changing for his purpose ' teil ' to ' fell.' But this sort of circular reasoning is seldom effective. (3) ' Bereft of their robe . . . they tear with their mouths the gray leaves ' could hardly be said of letters. Indeed in many German Volksrdtsel we are distinctly told (Wossidlo, No. 469) : ' Sie (d. h. Buchstaben) essen nichts, sie trinken nichts.' Cf. Eckart, Nd. Rdtsel, Nos. 387, 999; Renk (Tyrol), Zs. d. V.f. Vk. V, 157, No. 164. In a word, the solution is far-fetched. The key to the problem is presented by Flares, No. 2 : ' Vidi filium cum matre manducantem cujus pellis pendebat in pariete,' where the 'mother' is evidently the pen, the ' son ' the hand, and the ' skin ' the glove. Several near analogues to Bede's riddle have been discussed by me, Mod. Phil. II, 563. I note two riddles of the St. Gall MS. 196 (Schenkl, p. 18): 'Vidi hominem ambulantem cum matre sua et pellis ei pendebat in pariete,' and ' Vidi mulierem flentem et cum quinque filiis currentem cujus semita erat via et pergebat valde plana cam- pestria' [Rid. 14 i, n]. This second riddle points to the pen, the five fingers, and the leaves of parchment. The motive appears again in the Lorsch enigmas of English origin, No. 8 (Diimmler I, 20) : En video subolem propria cum matre morantem Mandre cujus pellis in pariete pendet adhaerens. So, in our riddle, the ten creatures are the fingers — the six brothers being the larger, the four sisters the little fingers and thumbs. Since both the Latin and Anglo-Saxon queries suggest stuff drawn from the people, it is not surprising that Volksrdtsel are full of parallels. In popular riddles the fingers are always brows- ing animals. Note Frischbier (Prussia), Zs. f. d. Ph. XXIII, 248, No. 73, ' Fif Zege frete von einem Hupe ' (Fingers of spinning hand) ; Simrock3, p. 67, ' Daer gungen tein Tatern | Um einen Busck matem'; id., p. 103, 'Zehn Schaflein fressen an einen Heuhaufen ' (see Petsch, p. 135). And the glove ever hangs on the wall. Compare Renk, Zs. d. V.f. Vk. V, 158, No. 170 : Was hangt an der Wand Wie Totenhand ? (Handschuh.) And see Simrock8, p. 70 : Es hanget wott an der Wand Un lett offe'ne Daudemanns Hand. Of Trautmann's solution, 'Ten Chickens' (BB. XIX, 177 f.), I can only repeat what I have said (M. L. N. XXI, 100) : ' His arguments seem to me unconvincing. To claim that the " skin, which hangs on the wall " (3-4) is not the glove of folk- riddles of all times (supra), but " the film that clings to the inner surface of the egg-shell after the hatching," is to reason far too quaintly and totally without the 98 RIDDLES OF THE EXETER BOOK warrant of Eusebius, No. 38, who says nothing of " wall " ; and to interpret haswe blede (14 9a) as "eggs in an advanced state of incubation" is surely a curious con- ceit. Then, too, his treatment of the numbers "six" and "ten" (1-2) seems arbitrary. In my opinion he has failed throughout to prove his case in the light of either logic or tradition.' 14 i turf tredan. See also 14 nb, lond tredan. This is paralleled by the Latin description of pen and parchment, 'pergebat plana campestria' (St. Gall MS. 196). In justice to Trautmann's solution, it must be noted that somewhat similar phrases are found in the Bird enigmas : 8 i, hrusan trede ; 58 5, tredaS bearonaessas. — ealra. Cosijn (PBB. XXIII, 128) renders rightly 'im ganzen,' and adds ' die raife hat also 6 + 4 f iisse.' 143 haefdon feorg cwico. Cf. n 6, haefde feorh cwico ; 745, haefde fer'S cwicu. — Fell. It is easy to identify glove with skin, as in Bede's Flares, No. 2, and in the Lorsch Riddle, No. 8. Cf. Beow. 2088, glof gegyrwed dracan fellum. 14 4 sweotol ond gesyne. So 40 3. Cf. Gen. 2806, sweotol is ond gesene ; Men. 129, swutelra ond gesynra; And. 565, sweotulra ond gesynra. In his note to this last passage, Krapp, p. 1 1 1, points to the frequent appearance of the phrase in Wulfstan's Horn., p. 159, 1. 5 ; p. 163, 1. 14. — on seles waige. Cf. And. 714, on seles wage ; 1493, un(ier saelwage. Cf. also 15 11-12, hongige ... on wage. 14 5 f. In these lines the riddler tells us that the fingers are none the worse for being deprived of their skins, the gloves, which are renewed, donned again, when the work of the hands is done. Haswe blede (9 a) certainly does not describe ' ein mehre wochen lang bebriitetes ei ' (Trautmann, BB. XIX, 179-180), but refers clearly to the leaves of the manuscript on which the hands are browsing (supra). 14 7 reafe berofene. Cf. Hildebrandslied 57, rauba birahanen. 14 ii Cf. And. 801-802, geweotan . . . mearcland tredan. RIDDLE 16 Dietrich (XI, 464) gives an excellent summary of this riddle : ' Das horn redet in nr. 15 von sich als einstigem kampfer (auf dem haupte des stiers oder auer- ochsen), dann beschreibt es sich als das kriegshorn, als trinkhorn, als jagdhorn, als schmuck des schiffes (hornscip), endlich als larmhorn womit der dieb ver- folgt wird.' Prehn, pp. 258 f., regards this problem as the first of a cycle of Horn riddles (cf. Rid. 88, 93), and seeks to trace the indebtedness of these to Eusebius 30, De Atramentorio. But Rid. 15 has absolutely nothing in common with these Anglo- Saxon enigmas ; and from the nature of the theme and the exigencies of treat- ment its first half-line, Ic wees wizpenwiga, may well have originated independently of Eusebius 30 1-2 : Armorum fueram vice, meque tenebat in armis Fortis, et armigeri gestabar vertice tauri. Miiller (C. /"., pp. 18-19) was tne nrst to point out the likeness between this . riddle and Rid. 80 in treatment and solution (see also Herzfeld, p. 5). The NOTES 99 parallel passages in the two were noted by Trautmann independently in his BB. article (XIX, 206). Hwilum clauses, the closing formula, and one or two motives are common to both. See notes to Rid. 80. Padelford, Old English Musical Terms, pp. 54-56, cites many illustrations of blast-horns and trumpets from Strutt's and Westwood's plates. From these we infer that blast-horns were used for many purposes : to summon guests to a feast, as in the April illustration of the Saxon calendar (Tib. B. V, Strutt, Horda, pi. x ; cf. Rid. 15 i6-i;a); in the harvest field (June); in the woods by swineherds (Sep- tember); and to stir warriors to battle, as in the attack upon a walled town, MS. Had. 603, f. 25 v. (cf. Rid. 15 4-6, 13-15) or to single combat (Cott. Cleop. C. VIII, Strutt, pi. iv, 2). The war-horn, — freollc fyrdsceorp (1513; compare fyrdrinces gefara, 802), — which is called elsewhere trufrhorn or giifrhorn or fyhtehorn, is to be distin- guished from the byme or tuba, which, if we may judge from the many drawings of battle-scenes, was often not a horn proper, but a long trumpet, either curved or straight (Cott. Cleop. C. VIII, f. 27 r. ; Add. 24199, f. 29 r.): Beow. 2944, horn ond byman; Ph. 134, ne byman ne hornas ; Domesdag 109, horn ne byman. Drinking-horns appear frequently in the illuminations. In the April feast of the calendar (Tib. B. V ; Jul. A. VI), a servant is filling a horn from a pitcher. In Cotton Claudius B. IV are several pictures of banquets with drinking-horns (ff. 31 r., 35 r., 57 r., 63 r.); and in Cleopatra C. VIII, f. 20 v., are found many designs of these. On the Bayeux Tapestry figures drink from horns similar to those in the grave-finds. The Taplow Horn in the Anglo-Saxon room of the British Museum holds about three pints or a half-gallon ; and, not being fur- nished with feet, could not be set down without spilling the liquor. Other noble horns of Anglo-Saxon date are those in York Cathedral and at Queen's College, Oxford, and the famous Pusey Horn, by which land was held (Arch&o- logia XXIV, 217; Hodgetts, Older England, 1884, pp. 105 f.). Sharon Turner, VII, chap, vi, notes among many such bequests, that two buffalo horns appear in Wynfleda's will, and that the Mercian King Witlaf gave to Croyland the horn of his table ' that the elder monks may drink thereout at festivals and remember the soul of the donor.' 15 1-3. 7» " To the adornments of the horn the magnificent specimen in the British Museum from the Taplow excavations of 1883 gives ample evidence (Hodgetts, Older England, pp. 105 f., ' The Horn '). The mouthpiece is rich with silver gilt [15 2'', golde ond sylfre\, which is elaborately ornamented, and its other mountings are bronzed. I observe in the same case many silver tabs from drinking- horns, engraved with human heads. Sharon Turner, VII, chap, vi, notes the men- tion in Dugdale's Alonasticon (1655), p. 40, of 'three horns worked with gold and silver.' Schultz, Das hijfische Leben, 1879, I, 324, cites from Horn et Rimenhild, 1. 41 52, a description of a golden drinking-horn richly adorned with precious stones. 15 2 golde ond sylfre. Cf. Gen. 1769, golde ond seolfre ; so Ps. 113 12. 153 1 1 \viliiiu weras cyssacJ. Cf. 316, mec weras ond wif wlonce cyssa'5 (cup or cross) ; 64 4-5 mec . . . cysseS . . . esne (beaker). 15 4-7 For the use of the horn in war, see the discussion above, and note such passages from the poetry as Beow. 1433, git'Shorn galan; 1424-1425, horn stundum 100 RIDDLES OF THE EXETER BOOK song | fuslic f[yrd]-leo-5 (cf. 15 13, fyrdsceorp; 80 2, fyrdrinces gefara). Our riddler in Rid. 15 4-7 emphasizes the use of the war-horn, both on land and sea, for it is certainly not the hornscip of Andreas, 274, as Dietrich supposed, that he has in mind (11. 6a-7). Horns were frequently blown at sea. In one of the pictures of the Bayeux Tapestry, a figure in the stern of a ship sounds upon a horn ; and in the Fornmanna Sogur II, 300, King Olaf signals with a horn to his ships. The on herges ende, 80 8, arid the several references to the horse on which the horn is borne (15 5-6, 14, 80 7), suggest that the poet is thinking not of the trumpeter but of the leader of the troop. Cf., however, El. 53 f. : Werod waes on tyhte, hleowon hornboran, hreopan friccan, mearh moldan trad, etc. 156 merehengest. The word — indeed the whole passage, with its sugges- tion of fighting by land and sea — suggests the comment of Merbach", Das Meer etc., p. 33 : ' Unter den Umschreibungen die aus dem Drange nach moglichst poetischer Bezeichnung des Schiffes hervorgegangen sind, fallen vor allem die- jenigen ins Auge, die, kiihn personifizierend, das Schiff als Flutenross darstellen. Es 1st dies wieder ein Punkt, wo im Geiste der angelsachsischen Dichtung Kriegs- und Seeleben sich beriihren : wie der Krieger auf ungestiimem Streitrosse zum Kampf ausreitet, so der Seefahrer auf unbandigem Wogenrosse zum wilden Streit mit Wind und Wellen.' Merbach cites as synonyms brimhengest (And. 513, Run. 47, 66), sundhengest (Chr. 853, 863), ivaighengest (El. 236, Gu. 1303), farofrhengest (El. 226), merehengest (Met. 26 26), sShengest (And. 488), yj>meark (Whale, 49, Chr. 864), sSmearh (El. 245, Whale, 15, And. 267), and lagnmearh (Gu. 1306). 15 8-9 See note to Rid. 80 3-5, where this motive is treated. In MS. Harl. 603, f. 51 r., a maid fills a drinking-horn from a pitcher. 15 10 Dietrich says (XI, 464) of this line : ' Dunkel 1st v. 10 ein gebrauch wonach es bordum behlyt>ed ist ; ich betrachte dies als denom. part, von /ileofr = hleowofr (schutz); von bretern beschiitzt konnte das horn auf dem gibel heissen [Rid. 88 24], wenn heafodleas los vom haupte sein kann ; mdglich aber dass dies gestumpft bedeutet und dann an ein mit holz eingefasstes homernes gerath zu denken ist, vielleicht an hornerne figuren des bret- oder schachspiels, gomen on borde, c. Ex. 345,6." Thorpe, Cod. Ex., p. 527, defines behlyfred as 'deprived of comrades' (gehlefian). Grein, Spr. I, 87, associates behlyj>ed (behlefred?) with hlefra, 'praedator' (Cot. 170), and translates 'spoliare,' ' privare.' In Dicht. he renders 'des Bortenschmuckes beraubt.' Brooke translates (p. 127) 'bereft of covers,' and thus comments : ' Bordum I do not take to be " on the tables," but bordum behlyfred, robbed of my covers, of the round tops like shields which shut down on the drinking horn, and were, because they were adorned with jewels and gold figures, wrenched away by the plunderers.' B.-T. s.v. renders 'deprived'; and so also Sweet ; Brougham (Cook and Tinker, Select Translations, p. 72) 'soli- tary upon the board.' There seems to be no doubt that [on] bordum . . . behlyj>ed licgan is an exact antithesis of hongige hyrstum fratwed . . . on wage (15 11-12). ' Sometimes ' says the Horn, ' I shall lie stripped on the tables ; sometimes I hang NOTES 101 adorned with ornaments on the wall.' Our riddle is full of such contrasts (11. 5-7 ; 16-19). For bord, 'table,' see 88 23, 24. 15 ii hyrstum fraetwed. Cf. 54 7-8, wonnum hyrstum | foran gefraetwed; 3220, fraetwed hyrstum. See also 15 2-3,7. 15 12 wlltig on wage. Cf. Beow. 1662, on wage wlitig; And. 732, wlitig of wage. Sarrazin says (Becrivulf-Slndien, p. 119): 'In dem Ratsel ist der Ausdruck sehr passend auf ein gold- und silbergeschmlicktes Trinkhorn angewendet.' The Beow. passage is discussed by Wiilker (Auglia XI, 537) and Kail (XII, 38). — Jwer weras drincaS. Cf. 21 12, 56 i, 57 n, 64 3, 68 17. 15 >3a fyrdsceorp. 'Scarp bezieht sich allgemein mehr auf die Kleidung: hilde-sceorp (Beow. 2156); wairon hie on gescirplan scipferendum eorlas onllce (And. 250); daher gescyrpan = " vestire," "omare" (Met. 152); dann aber auch allgemein fiir "Ausriistung," " Schmuck," z. B.fyrd-sceorp (Rid. 15 13); heoru-sceorp (Har. 73), [Gti. Ex. 127, sigesceorp] ; sceorp to frfSscipe (Schmid, Gesetze, Anhang III, i); fugla cynn frSerum gescyrped (Ps. 148 10)' (Lehmann, Germania XXXI, 494-495). Fyrdsceorp is rendered by Grein, Spr. I, 362, 'omatus bellicus.' Brooke (p. 1 27) translates 'a fair thing on wayfaring'; and adds in a note 'Literally, "a fair war-ornament." I have translated it as above, because I want to give, in this place, the force of " fyrd," which is the militia ; and here, I think, the levy en masse of the population for a war expedition — the horn is part of the war- material, part of the ornamented things used in the Fyrd.' Cf. Bemu. 1424, horn stundum song fiisllc fyrdleo'S; Epistola Alexandria 252, Da het ic blawan mine byman ond fta fyrd faran ; Rid. 80 2, fyrdrinces gefara. 15 17-19 In the Laws the horn is the greatest enemy of the thief. See Laws of Wihtred §28 (Schmid, p. 18): 'Gif feorran cumen man oSSe fremde buton wege gange and he bonne naw)>er ne hryme, ne he horn ne blawe, for ]>eof he blS to pro- fianne oft'Se to sleanne o'SiSe to alysenne.' Our riddler has in mind the hream or ' hue-and-cry.' Penalties are pronounced against any one 'gif hwa hream gehyre andhine forsitte,' etc. (Canute, II, 29, § i, Schmid, p. 286). Cf. Canute, 1,26, Schmid, p. 268, ' wac bi'5 se hyrde funde to heorde, be nele ba heorde . . . mid hreame bewerian . . . gyf b5r hwylc J>eodsceafra sceafrian onginne'S' [15 iga, feondsceaban]. The Anglo-Saxon laws for the recovery of stolen property [15 18] are discussed by Schmid, p. 636, s.v. ' Nachsuchung nach gestohlenem Gut.' One recalls the hue-and-cry after the fox in the Nonne Preestes Ta/e, B. 4588-4589 : Of bras they broghten bemes and of box, Of horn, of boon, in which they blewe and powped. RIDDLE 16 Dietrich's answer, Broc 'Badger' (XI, 465), was accepted by Prehn, Brooke (/.. /•:. Lit., p. 142), McLean (O. E. Reader, p.xxx), Cosijn (PBB. XXIII, 128), and queried by Trautmann. Walz, Harvard Studies V, 261, objects that the badger has not a white throat, nor is he swift-footed ; and suggests Igil, ' Porcupine ' (cf. 1. 3, beadoii'iTfen ; 1. 28, hildfpilum). But the habits of the creature of the riddle are totally unlike those of the porcupine or hedgehog, and very like those of I02 RIDDLES OF THE EXETER BOOK the badger, as a comparison of the text with Bell's account of the animal (infra) shows. A hedgehog does not work a way with his feet through a steep hill (16 18 f.), nor does he reach through the roof of the hill (1627). Rid. 16 has nothing in common with the spirited ' Kelduswin ' (Hedgehog) riddle of Islenzkar Gdtur, No. 680, and is not in the least indebted, as Prehn, p. 178, would have us think, to Symphosius 21, Talpa; nor save in the darts (28 a) to Sym. 29, Ericius: ' Incolumi dorso telis confixus acutis.' Holthausen points out (Engl. Stud. XXXVII, 206-207) certain parallels between Rid. 16 and a Hedgehog (De Hys- trice) poem of Claudius Claudianus (Carmma, Leipzig, 1879, II, i52f.); but these (infra) do not seem to me sufficient to sustain Walz's solution. In the Glosses, broc is usually rendered by 'taxus vel meles' (see WW. 119, 2, 320, 10; cf. Jordan, Die altenglischen Saugetiernamen, p. 43); and the treatise 'Medicina de Quadrupedis ' (Lchd. I, 326, 11) thus describes it: 'Sum fyj>erfete nyten is }>aet we nemna'S taxonem J>aet ys broc on englisc.' Alexander Neckam, De Naturis Rerum, cxxvii (Rolls Series, 1863, p. 207), thus describes the badger's building and his departure from his home on account of the enmity of the fox : 'Taxi mansiones subterraneas sibi parant labore multo. Unum enim sibi eligunt taxum terrae pedibus ipsorum effossae vectorem et oneri tali ex longa consue- tudine idoneum. Supinatur quidem, et cruribus extensis et erectis, super ventrem ipsius terra effossa accumulatur. Oneratus satis per pedes ab aliis exportatur, tociensque labor assumptus iteratur usque dum capacitas domus habitatoribus suis sufficiat. Latitans interim in insidiis animal dolosum, vulpem loquor, sustinet usque dum mansio subterranea parata sit, et tempus absentiae taxorum sibi reputans idoneum, signum turpe inditium hospitum novorum ibidem relinquit. Revertentes melotae, lares proprios indignantur inhabitare et alias sibi constru- entes aedes, foedatam domum foedo hospiti sed praedoni relinquunt.' Bell, British Quadrupeds, 1874, pp. i58f., thus describes the Badger or Brock (Meles Taxus) : 1 Its favorite haunts are obscure and gloomy ; it retires to the deepest recesses of the woods or to thick coppices covering the sides of hills [16 18, 21, 27], and there with its long and powerful claws digs for itself a deep and well-formed domicile consisting of more than one apartment [cf. 16 17-18] . . . The badger is endowed with astonishing strength of jaws. ... It also possesses great gen- eral muscular power; and these means of inflicting injury with the defensive coat of mail . . . render him a formidable enemy to attack or cope with. . . . The burrow is usually a round horizontal hole or tunnel, the end of which is turned upwards abruptly for about a foot, and the vertical part of the hole leads into a rounded excavation of just sufficient size for the animal to lie coiled up in' [i67f.]- 'The intricate passages and crevices in quarries, while they furnish to this animal a commodious retreat, afford also an efficient means of defense against the entrance of dogs, which in their attempt to dislodge the badger often get fixed between the stones and perish' [168-11, 24 f.]. Bell thus pictures the animal (p. 166): ' Feet very hairy, particularly the hinder ones with five toes on each armed with strong curved fossorial claws [16 17]. Hair of body long, loose, and of three colors, — white, black, and reddish, the union NOTES 103 of which produces a rich gray. Head white excepting a band of black commenc- ing between nose and eye, and extending backwards. . . . Lower jaw, throat, breast, and belly, the interior of all the legs and the feet, black ; the back, shoul- ders, and rump, reddish gray ; the sides and tail, light gray.' The Anglo-Saxon animal is white and reddish gray [16 1-2]. Brooke says (£. E. Lit., p. 142) : ' Once more, on this beast life in the literature of the woods, we are placed on the edges of the hills where the badger has his hole, and Cynewulf throws himself as fully into the life and passions of the animal for his home and children as he does into the eagerness of the hunter. ... It is in these short poems — in this sympathetic treatment of the beasts of the wood, as afterwards of the birds ; in this transference to them of human passions and of the interest awakened by their suffering and pleasure — that the English poetry of animals begins.' Herzfeld, pp. 10-12, and McLean, p. xxxi, note that in this riddle we have a re- markable number of hapax legomena, in this case compounds not found elsewhere : IO, geogufrcndsl; i^forhtmod; 17, fe&emund 23, w&lhwelp ; 24, nifrsceafca ; 26, gegnf&fr ; 29, Idfrgeivinna. And yet the word-use has much in common with the vocabulary of Rid. 17, 18. 163 beadowajpen. Cf. 18 8, beadowiepnum ; 1628, hildepilum; 186, hyldepylas; 16 5, 18 8, ordum. 163-4 Holthausen, who reads her swylce sw[m]e, compares Claudian, De Hystrice, 5 f . : Os longius illi Assimulat porcum. Mentitae cornua saetae Summa fronte rigent Parva sub hirsute catuli vestigia dorso. This, it is true, accords remarkably with Holthausen's reading of the text, but as that involves the change of the MS. swe to sw[?n~\e, and the omission of hll/idfr, we are justified in rejecting it. I accept the reading of Zupitza and McLean, because that alone meets the demands of the meter without change or elimination ; be- cause swe is supported by the only possible substitute in 10 6 for MS. snearltce, sue drllce, and by Leid. n, su&; and because, as McLean points out, such com- parisons as this to a sow are very rare in Old English poetry. Translate ' Hairs stand on my back just as (swilce swe) on my cheeks : two ears tower over my eyes.' The sow of the editors thus goes out of the story. i66a in grene graes. Barnouw, p. 219, remarks the absence of the emphatic article in this place in a riddle which on other grounds he has classed as very old, and contrasts 36 i, se wista wong. i66b Cf. 16 ii, him bi> deaft witod (Jansen, p. 95, notes the epiphora and the resulting strophic effect); 21 24, me br$ foriS witod; 85 7, me br5 deaft witod. 16 8 Avrelgrim \vlga. Cf. 16 iob, ga wic buge; Gu. 274, }>e >z. wic bugaft. 104 RIDDLES OF THE EXETER BOOK 16 ii him. Cosijn, PBB. XXIII, 128-129, refers him to geogufrcnosle, — 'sonst ware die flucht des dachses ganz unmotiviert : erst spater fuhlt er sich sicher.' So Grein, Dicht., and Brooke, p. 142, 'death is doomed to them.' 16 i3b fleame nergan. So Gen. 2000. Note the rime in this line. 16 isa Grein, Dicht., translates 'ihn tragt die Brust heran,' and explains, Spr. I, 141, 'er kriecht auf dem Bauche.' 16 igb feorh genergan. For many examples of the phrase feorh (ge)nergan, see Spr. I, 296. 16 21 on degolne weg. Cf. Earle, Charters, 239, 18, on broccholes weg. — Jjyrel. As Madert shows, p. 36, J>yrel is found in the Kiddles with long and shorty. It is short here and in 72 8, frurh J>yrel J>earle, and 81 n, \on~\ fryrelwombne ; while it is obviously long in 45 z,foran is J>yrel, and 91 5, hindan J>yrel. See Sievers, PBB. X, 487, Gr.\ § 218, I. 1622 swsese ond gesibbe. Cf. 27 21-22, freonda | swiesra ond gesibbra; Gen. 1612, freondum swiesum ond gesibbum. i624f. Holthausen compares Claudian, i8f.: Crebris propugnat jactibus ultro Et longe sua membra tegit tortumque per auras Evolat excusso nativum missile tergo, Interdum fugiens Parthorum more sequentem Vulnerat, etc. The likeness is not convincing. I believe, with Dietrich and Brooke, that the darts of war are the badger's teeth. 1624 nearwe stige. Cf. Beow. 1410, stlge nearwe. 16 25 tosiele]?. Only here and 17 5. 16 28 Jnirh best hrino. Cf. Gen. 1396, hsiste hrlnan. RIDDLE 17 Dietrich's answer to this riddle (XI, 452), 'Anchor,' is unquestionably correct. Its source is found in Symphosius 61, ' Ancora.' Mucro mihi geminus ferro conjungitur unco [17 8, steort]. Cum vento luctor, cum gurgite pugno profundo [17 1-2]. Scrutor aquas medias, ipsas quoque mordeo terras [172-3]. All these motives are expanded in the Anglo-Saxon, but, as Dietrich well says, ' der gegenstand des rathsels ist nicht mehr sache, er ist ein kampfer und sieger wider die elemente, seine feinde, er ist rein ein held geworden.' Heusler, Zs. d. V.f. Vk. XI, 127, compares with the English riddle the spirited Gata 6 of Her- varar Saga : Hverr er sjd hinn mikli, er morgu raeftr, ok horfir til heljar halfr ? Oldum hann bergr, en viS iorg sakask, ef hann hefir ser veltraustan vin. NOTES 105 The riddle of Symphosius is found in popular form in the mediaeval German version of the Apollonius story (Schroter, pp. Ixxv, 66 f.) ; and suggested to Scaliger the theme of his fine Latin riddle (Reusner I, 175): Magna, bidens, apridens, dentes fero parva quaternos ; Ingens pro digitis annulus in capite est. Quum teneo dominam, nihilominus ilia movetur, Et quum non teneo, magna avis atra volat. 17 1-4 Sievers (PBB. XII, 457) regards these lines as interesting examples of the ' schwellvers.' 17 2 saecce. Thorpe, Grein (Spr. II, 394) and Bosworth-Toller regard this as ist sg. pres. ind. of saccan, 'to contend'; Grein (Die/it.) and Brooke (E. E. Lit., p. 178) doubtfully as ' See-ried' or 'sea-tangle.' Either is a hapax. It is merely the Northern form of ist sg. pres. ind. of sacan (cf. Mark xiv, 31, eztsace; Lind., ons&cco), which is here retained for the sake of the meter. Conversely, see to sace for to s&cce, 21 6. 17 3 yjmm Jn'alit . So 114. 17 s tosseleS. Compare 16 25h, tosielef>. Is it not more than probable that our riddle intended a, word-play, as siclan is frequently employed for the making fast of a ship (Chr. 863, Beow. 226, El. 228) ? Compare Merbach, Das Meer in der Dichtung der Angelsachsen, p. 36. 178 steort. Weinhold (Altnordisches Leben, 1856, p. 13) remarks: ' Als Anker benutzte man, wie die Deutschen in altester Zeit, Senksteine die von einem Tau umschlungen, das in eingeschnitne Rinnen festgriff, auf den Grund gelassen wurden. . . . Erst spater verdrangte im alten Scandinavien der metallene Ihiken (Kraki) den Stein.' Steort corresponds to the mucro of Symphosius. 17 ioa faeste gehabban. To the use of the anchor tl^ere are many references in the poetry : Beow. 302-303, scip on ancre faest; Beow. 1919, scip oncerbendum faest ; El. 252, aid yiShofu oncrum faeste ; Chr. 863, ealde yftmearas ancrum faeste; Whale, 13-14: ond l^onne gehydaS heahstefn scipu to l>am unlonde oncyrrapum. Ancor-man is the gloss to ancorarins or proreta ( JElfric, Gloss. 83, WW. 166, 7). It is this seaman whom Aldhelm describes in the De Laudibus Virginitatis, § 2, Giles, pp. 2-3 : '[Navis] instanter hortante proreta et crepante naucleri portisculo spumosis algosisque remorum tractibus trudit.' Several references to the drop- ping of anchors are found in the Encomium Emmae, Pertz, 1865, p. 8 (Scriptores Rcrum Germanicarum III). RIDDLE 18 Dietrich (XI, 465) suggests ' Ballista,' but later (XII, 237) adopts Professor Lange's solution, 'Burg,' which Prehn supports (pp. 270-271). As I have shown (.]/. L. JV. XXI, 100), this riddle is certainly a companion-piece to Kid. 24^" Bow,' and forms with it one of the many pairs in our collection. Both objects swallow and spit out terror and poison (18 7-9, 4; 24 8-9); from the belly of each fly deadly 106 RIDDLES OF THE EXETER BOOK darts (18 6, 24 12); each is servant of a master (18 5, 246). Indeed, a half-line of one poem (18 6a) appears practically unchanged in the other (24 u1')- I find this com- panion weapon to the ' Bow ' in Dietrich's first solution Ballista, which, as I have pointed out (J/. L. N. XVIII, 104), is elsewhere in riddle-poetry associated with Arcus. The latter says of its fellow-warrior (Scaliger's enigma, Reusner I, 172): Altera mi similis cognataque litera majus Edit opus sapiens, tectus utraque cave. This answer caps our query at every point. Isidore tells us of the Ballista in his Origines xviii, 10 : 'Torquetur enim verbere nervorum et magna vi jacit aut hastas aut saxa.' From the many Roman references in Marquardt und Mommsen's Hand- buck der Romischen A It ert hunter, 1884, V, 522-524, and from many mediaeval ex- amples in Du Cange's Glossarium, s.v., one gathers that not only darts and rocks, but beams and bolts of every sort were cast from the huge engine. So our riddler's chief motives, the varied contents of the creature's belly (18 zb-3, 7-10) and the casting forth thence of 'spear-terror ' (18 4a, 6), are well sustained. Illustrations and descriptions of the Ballista in Baumeister, Denkmaler, s.v., in Yule's Marco Polo II, 122, in Marquardt, and in Schultz, Das hofische Leben II, 327, support the mention in Rid. 18 of the subject's 'mouth' and 'belly'; and the cords with which, it was wound ('Ballista funibus nervinis tenditur') may perhaps be 'the inclos- ing wires ' of line 2 a. Lines 3 a, dryhtgestreona, and 10, wombhord -wlitig ivloncum deore, seem to me to express admirably that joyous pride of the Anglo-Saxons in their war-weapons of which our riddles are so full ; and the last line is of charac- teristic grimness when applied to an engine of destruction. Above Rid. 18 in the MS. are two runes, B with the L above it. If B refers to Ballista, may not L represent its Anglo-Saxon equivalent (stre (Spr. II, 183)? As Miss Keller's references show (Anglo-Saxon Weapon Names, p. 119), funda is glossed by IiJ>(e}re and fundibulum or ballista by st(eflij>(e)re in the Glosses (W W. passim ; Bede, Eccl. Hist. IV, 13, 304 25). Miss Keller infers (p. 65) that huge hurling-machines were unknown, on the negative evidence of a passage in the translation of Orosius (infra), but shows that the sling or staff-sling (pp. 62-63) was in common use among the older English. Heyne, Die Halle Heorot, p. 19, doubts the existence of great hurling-machines in Anglo-Saxon times : ' Fur Schleudermaschinen nach Art der romischen Cata- pulten und Balisten kommen auch einheimische Namen vor (bolt, "catapulta"; stearu, "balista"; "balista," gelocen boce); aber zweifelhaft konnte ihre allge- meinere Verbreitung nach den Worten sein, mit denen Konig ^Elfred, der Ueber- setzer des Orosius, der Balisten gedenkt und die ganz den Eindruck machen als ob er etwas Fremdes schildere ["palistar" for "balista," Orosius iv, 6, p. 399], )>a het he mid J>am palistar mid J>am hy weallas brJEcon.' But both the catapulta and ballista are repeatedly mentioned in Abbon's account of the siege of Paris by the Danes, whose methods of warfare in 885 could not have been more ad- vanced than those of the English (see Abbon, De Bellis Parisiacae Urbis, lib. i, 205 f., Pertz, Scriptores Rerum Germanicarum I, pp. 13 f.). In the Saga of Sigurd, chap. 11 (Laing IV, 127), a ballista is used in battle; but this is as late as 1 1 10 A. D. NOTES 107 In Trautmann's solution 'Oven' {Anglia, Bb. V, 48; BB. XIX, i8of.) he is led into fourfold error {M.L.N. XXI, 101). He ignores entirely the riddle's rela- tion to its mate, Rid. 24, since this association in war cries out against his answer. He changes the text to fit his meaning (see i b, 1 1 a). He hunts words and phrases beyond all bounds of riddle fantasy (4 a, 8-9 a). And, finally, he seeks unsuccess- fully to establish certain likenesses to Rid. 50, which he asserts without proof to be 'Oven.' Holthausen follows Trautmann (Anglia, Bb. IX, 357), and affirms with- out a vestige of proof : ' Die B-rune am rande natiirlich bedeutet bac-ern oder •Aus.' Trautmann believes that the presence of the runes B and L shows that the scribe was hovering between two solutions. 18 ib minre heorde. For this MS. reading, which Grein, Dickt., renders ' meiner heerde,' Trautmann proposes minra heorde, and translates ' ein hiiter der meinen,' merely because the transmitted phrase does not accord with his interpretation. As a genitive dependent upon mundbora, it is perfectly intelli- gible ; and no change seems necessary. Heard in the sense of grex or familia is very common (Spr. II, 68). 1 8 2a eodor wirum. This reading of Thorpe and Trautmann seems preferable to Gn., W. eodor-wlrum, which is found nowhere else. Perhaps Trautmann is right when he suggests, ' Das wort eodor gebraucht der dichter listig in zweifachem sinne: in dem von mundbora, 'schutzherr' {eodor Scyldinga) und in seiner eigen- lichen bedeutung 'einschliessender raum.' Such word-plays appear in the Riddles (32 14, on wonge; 38 7, blad; 73 22, on h&fte\ 93 22, blace\ Old Norse poetry abounds in such double meanings (see Skdldskaparmdl, § 74, Snorra Edda I, 544). 1 8 6 Cf. Sal. 25-28 : worpaiS hine deofol on domdzege draca egeslice ' bismorllce of blacere liftran Jrenum aplum. In Hpt. Gl. 425, 13, the phalarica is a burning arrow shot from an engine, and stdnas (446, 29) are included among the weapons of war. 18 8-9 ' The brown war-weapons, bitter points, dire poison-spears' are regarded by Trautmann as the fuel, ' the logs and coals thrown into an oven.' Dietrich comes nearer the truth with the suggestion that the poet is thinking of 'die gesammte waffenfahige mannschaft des burgbezirkes ' or perhaps of the darts cast into the city by the enemy. I believe that the riddler has in mind the missiles of every sort thrown from the ballista. 18 9 attorsperum. For a discussion of poisoned weapons see note to 24 9. 18 ioh \vloncum deore. In Run. 81, eldum dyre refers to the use of the Ash as a weapon. 18 ua Trautmann condemns men gemunan because it has only 'drei takte,' and because it does not suit his solution. So he changes this to the unlikely gewilniafr, to resemble 50 7h. Later he argues fallaciously for his answer from this made-to-order resemblance. Cosijn {PBB. XXIII, 129) has suggested \oft\ or \J>e feran | DNUH. 20 4 hildej>ry)je. The word occurs only here, but compare 65 4, JryJ>a dxl, t>E(gn). 205-6 MS. rdd | AGEW. These words have received much tinkering from scholars. The reading of Thorpe, Ettmiiller, and Dietrich, rJd'-NGEW = rdd- wegn (wagn), has two strong grounds of favor, — that it necessitates no very vio- lent change of text (the confusion of runes A and N being a natural error), and that the word thus derived occurs elsewhere (Orosius, vi, 30, Sweet, 280, 13). But it is also open to two strong objections — that it is unfitted to the context (a ' chariot ' is not borne on the back of a horse) and that it has nothing in common with the problem's counterpart (Rid. 65) or with the treatment of the theme in riddle-history. Grein's reading, rdd(= R)AGEW = £Y?r _wod R]EW involves too great forcing of the text to deserve serious consideration ; while the suggestions of Hicketier (Anglia X, 593), rand WOE|> (corrupted to NGEt>, by the associa- tion of l>eegn, and then to MS. AGEW), and Trautmann, gdr WOEf>, are open to the same objection — J>eow is an abortive product, and moreover is not fitted to the context, for it is well known that horses were used in Anglo- Saxon times only for the chariots of the rich or as steeds of the upper classes (cf. 23 2, 65 2) and that no J>eow was ever mounted. Hicketier proposes also ntrgledne ra\n\d\ but his protests against nagledne gdr, 'the nailed spear,' are NOTES 109 based upon ignorance, for we meet the expression in the Heliand, 5704, negild sper (see Chaucer, Knightes Tale, A. 2503, 'nailinge the speres '). In the Anglo-Saxon illuminated manuscripts (see Wright, Domestic Manners, p. 74) the rider almost always carries a spear. 'It is noted of Cuthbert in Bede's life of that saint that one day when he came to Mailros (Melrose) and would enter the church to pray, having leaped from his horse, he gave the steed and his traveling spear into the care of a servant.' Cosijn (PBB. XXIII, 129) would read rad(R)hG, W ( Wynn}, E (Eh). Thus are evolved not only the desired gar (by inversion), but inynn-eh, 'joyous horse,' a creature which finds some excuse for being in Runic Poem 55 : Eh by5 for eorlum aef>elinga wyn, hors hofum wlanc, ^ser him haele)> ymb welege on wicgum wrixlaj> sprSce, and bi}> unstyllum aifre frofur. Holthausen (Bb. IX,357) follows on the same track, but suggests for WE ivynnt. = wynne [see Runic Poem 22, wynne~. Cosijn's reading fits the context, and is supported not only by the Runic passage cited but by such compounds as wyn-beam, wyn-burg, wyn-candel, wyn-mieg, etc. (Spr. II, 758-759). Moreover, in the Riddles, runes make a threefold appearance: through their names (438-11, Nyd, ALSC, Acas, Hczgelas), as letters (so ao, 65, and 75), and finally as symbols of things (91 7, mdd-W = modwyn ; heading of Rid. 7, S = sigel; etc.). But despite these positive arguments, which Cosijn does not present, his reading strains cre- dulity in many ways : it is highly improbable that in a single group of five runes three different functions of them should be found ; it is equally unlikely that such a group would present not one thought as elsewhere, but two such totally different ideas as 'spear' and 'joyous horse'; it is still more unreasonable to assume that such a departure in thought could occur within one half -line, 20 6a; and, finally, it is quite unnatural to suppose that the riddler would abandon his method of inver- sion (see Rid. 75) that he has employed consistently in the three other groups of this runic problem (another method is pursued with like persistence in 65). Trautmann's view (Anglia, Bb. V, 48) that 20 sb rdd represents an original gdr, is founded upon his fatally simple method of substituting any desired word for that in the text. Likewise in his reading of the runes (supra) the MS. is honored only in the breach. Now let us solve this problem according to the rules of the game. The con- ditions imposed upon us are two: (i)'the runic letters must be read backward as elsewhere in the riddle ; (2) thus combined, they must form but one word. And here are our letters: rdd(= R)AGEW. Inverted, they read wegar, — no impos- sible form, since wigdr and wegur appear instead of -wtg-gdr, 'lance,' in WW. 143, 12-13 : lwtgdr, lancea; vngurtsgtwri&, amentum.' It is needless to point out that this furnishes the very meaning demanded both by the context and by our riddle's counterpart, Kid. 65 6. It satisfies all the conditions. Our form, wegdr, which may be explained either by phonetic change, as in the Vocabularies, or by a confusion of runes, is one of the appositives of hildej>ryf^e (20 4). The passage may be thus rendered : ' He (the horse) had on his back strength in war (or " war-troop "), a man and a nailed war-spear. ' IIO RIDDLES OF THE EXETER BOOK 207-8 Hehn (A'p. u. Ht., 1902, pp. 368-374) discusses the Falkenjagd or chas- ing of other birds by the kite, hawk, and falcon. ' Hawking is not a Teutonic in- vention, but was learnt by the Germans from the Celts, and at no very distant period either. [On the other hand, Jacob Grimm has devoted a whole chapter of i his History of the German Language to hawking, setting forth the ruling passion for this kind of chase in passages from the poets and other authors of the Middle Ages, and placing the origin of the custom in the earliest prehistoric times of the German race.] Hunting as an art is a national trait of the Celts. ... It is another question whether the Celtic nations that surrounded the Germanic world on the south and west invented hawking or only developed the art, and, in the last case, whence they originally derived it.' Traces of its origin are noted by Hehn not only in Thrace, but on the very borderland of India. ' During the Middle Ages hawking nourished all over feudal Europe [see also Schultz, Das hofische Leben I, 368], it spread from Germany and Byzantium to the East and nations of Asia, and was practiced by electors and emperors, emirs, sheiks, and shah, down to the nomads of the steppe and the Bedouins of the desert. Marco Polo found hawking the fashion in the capitals of Mongolian princes as far as China.' Whitman {Journal of Germanic Philology II, 170) identifies the ivealhhafoc or foreign hawk (cf. WW. 132, 36; 259,8; 406, 20; 514, 12, etc.), with the peregrine falcon (see Swaen, Herrigs Archiv CXVIII, 388). 'Falconry was a sport very popular among our Anglo-Saxon forefathers. The exact date of the introduction of falconry into England is not known, but about the year 750 Winifred or Bon- iface, then Archbishop of Mons, sent yEthelbald, King of Kent, a hawk and two falcons ; and Hedibert, King of the Mercians, requested the same Winifred to send him two falcons, which had been trained to kill cranes ' (Warton, Hist. Eng. Poetry, 1840, II, 405). For the history of the sport of hawking among the Anglo- Saxons, see Sharon Turner, VII, chap, vii, and Strutt, Sports and Pastimes, 1903, pp. 22 f. Whitman, I.e., notes the discussion of hawking in ./Elfric's Colloquy (WW. 95, 12 f.) and compares Craft. 81, Fates, 86, sum sceal wildne fugol wloncne atemian | heafoc on honda, etc.; Maid. 7. See also Rid. 25 3, 65 3, 5. Sievers' discussion of the runes HA(0)FOC (Anglia XIII, 7) has been con- sidered in the Introduction. RIDDLE 21 Dietrich's answer, 'Sword' (XI, 465), which is accepted by Brooke (p. 122), and rejected by Trautmann, who suggests (Anglia, Bb. V, 49) ' Hawk,' is un- doubtedly correct, being confirmed by every motive of the problem, -^ the adorn- ments of the warrior, his dependence upon his lord, his grim work of death, his lack of an avenger, his celibacy, his hatefulness to women. Prehn, as usual, has not succeeded in proving (pp. i84f.) the indebtedness of the Anglo-Saxon to the Latin riddles of like subject (Aldhelm iv, ro; Tatwine 30; Eusebius 36). The chief motive of Aldhelm, and the entire theme of Tatwine, who follows him, — NOTES III the relation of the sword to its house, — is not found at all in the English prob- lem where the sheath is a corslet (21 3); while the bloody labors of the weapon in the hand of the fighter are the inevitable outcome of the subject, and are handled by Aldhelm and Eusebius in a manner very different from that of our riddler. There is hardly even coincidence of fancy between Eusebius 363 — 'sed haec ago non nisi cum me quinque (i.e. digiti) coercent' — and Rid. 21 13, healdeft mec on heabore, etc. This riddle has much in common with other enigmas of the Anglo-Saxon collection. ' The sword was the special weapon of all the nobler sort. It was also the noblest of all the pieces of armor, and it was fame for a smith to have forged one that would last, because of its fine temper, from generation to generation. ... Cynewulf cojiceives it as itself a warrior, wrapped in its scabbard as in a coat of I mail; going like a hero into the battle ; hewing a path for its lord into the ranks -^/i of the foe ; praised in the hall by kings for its great deeds ; and . . . mourning, flH> when the battle is over, for its childless desolation, for the time when it was inno- cent of wars, for the anger with which the women treat it as the slaughterer of men.' (Brooke, E. E. Lit., pp. 121-122.) 21 ib on gewin sceapen. The same phrase, indeed almost the same line, is used of another weapon, the Bow, in 24 2. 21 2 frean minum leof. So 80 zb. Another weapon, the Ballista, tells us (18 $b),frea J>tzt bihealdefr. So both Sword and Bow are controlled by a waldend (21 4, 24 6). — f£egre gegyrwed. Cf. 29 i, fiegre gegierwed. 21 3a byrne is mm bleofag. The grave-finds (Wright, Celt, Roman and Saxon, 1875, P- 475) show that the sheath was generally of wood tipped with metal, some- times covered with or made entirely of leather. Miss Keller, Anglo-Saxon Weapon Names, p. 46, notes that the chapes and lockets were sometimes gilded and even of gold. 'Occasionally the sheaths were adorned with a winding or snake pattern so characteristic of the period ; and one bronze chape inlaid with figures of ani- mals in gilt has been discovered' (Archaeologia XXXVIII, 84; Horae Ferales, 1863, pi. xxvi). For construction, cf. 16 i, hals is mm hwlt. 21 4 wir ymb J>one wselgim. Cf. 21 32, wirum dol\ 71 5, wire gezveorj>ad, — in both places of Sword. The Book (27 14) and the Horn (15 3) are adorned with 'wires.' 21 6 sylfum to sace. All editors read the MS. wrongly, sylfum to rice. Grein's suggestion sige is accepted by Brooke, who renders ' with himself to conquest.' Both the MS. and the B. M. transcript read plainly sylfum to sace. Sace is a scribal variation for original scecce (see 429, 88 29), — the second foot of a simple A-type, L x X | ! X . ai 6-8 ic sine wege . . . gold ofer geardas. So in the riddle's sequel, 71 6, se be gold wigefr ; but in the later place the phrase is used not of the sword itself, but of him who suffers by its stroke (Rev. xiii, 10). Cf. 92 4, gold on geardum. 21 7 hondweorc smiba. The same phrase is applied to the Sword, 6 8. 21 8-10 Aldhelm (iv, 106-7) thus refers to the bloody deeds of the sword: Per me multorum clauduntur luniina letho, Qui domini nudus nitor defendere vitam. 112 RIDDLES OF THE EXETER BOOK And Eusebius (36 1-3) says : Sanguinis humani reus et forus en ero vindex : Corpora nunc defendere, nunc cruciare vicissim Curo. The Sword speaks in 71 6, ic yj>an sceal. 21 8, 10 f. As Lehmann points out (' Ueber die Waffen im Ags. Beowulfliede,' Germama XXXI (1886), 487 f.), the Beowulf is full of references to sword-hilts of costly metal set with precious stones (Ztow. 673, 1024, 1615, 1688, 1901, 2192, 2700). Elsewhere in the Riddles (56 14) the gold-hilted sword is mentioned (see also Gn. Ex. 126, Gold gerisefr on guman sweorde). In the Wills several costly swords with hilts of gold and silver appear as legacies. Miss Keller, Anglo-Saxon Weapon ATames, p. 37, cites Thorpe's Diplomatarium, 505, 28, where a testator mentions the sword ' >aet Eadmund king me selde on hundtwelftian mancusas goldes and feower pund silveres on "San fetelse ' ; and 558, 10, where another leaves a sword 'mid "Sam sylfrenan hylte ond iSone gyldenan fetels.' The grave- finds furnish similar evidence of the rich beauty of sword-hilts (Akerman, Pagan Saxondom, 1852, pi. xxiv ; Collectanea Antiqua II, 164). But, as Miss Keller notes, the laws, wills, manuscript-illuminations, grave-finds, and even the passages in the poems, prove conclusively that the sword is the weapon only of warriors of wealth and rank (see Kemble, Horae Ferales, 83, 84). Indeed, its possession con- fers distinction ; cf. Schmid, Gesetze, Anhang VII, 2, § 10, ' And gif he begytaft >aet he haebbe byrne and helm and ofer-gyldene sweord, )>eah J>e he land naebbe, he bitf slScund.' For interesting accounts of the sword, see Hodgetts, Older Eng- land, pp. I f. ; Wright, Celt, Roman and Saxon, pp. 470 f. ; Brooke, E. E. Lit., p. 121 ; Bosworth-Toller, pp. 949-950. 21 9-10 Cf. Dream, 23, mid since gegyrwed; 77, gyredon me golde and seolfre ; Rid. 27 13, gierede mec mid golde. 21 10 since ond seolfre. So 68 18, Dan. 60. 21 ii ne wyrneo' word lofes. This recalls the praise of Hrunting (Beow. I456f.), which is extolled at a feast like the sword of our riddle. So in regard to the sword given by Beowulf to the Dane who had guarded his ship, we are told of the recipient (Beow. 1902): }>aet he sySftan waes on meodubence maflme J>y weorSra, yrfelafe. maeneS for mengo. Cf. Wids. 55, majnan fore mengo in meoduhealle. 21 12 pser hy meodu drincao. Note 15 ,2, 56 i, 57 12, 64 3, 68 7, and the riddles of drink (28, 29). 21 12-15 Lehmann (Germania XXXI, 493) notes that in the Anglo-Saxon period sword, helmet, and byrnie were worn by the most illustrious warriors, even at a feast. On this account bloody strife often arose, if men excited by beer taunted each other. Cf. Fates, 48 f., Sumum meces ecg on meodubence yrrum ealowosan were winsadum. NOTES 113 The early kings, to prevent this, made stringent laws against the drawing of weapons in the mead-hall; cf. Hlothar and Eadric, § 13, Schmid, Gesetze, p. 14: ' Gif man waEpn abregde }>xr maen drincen and |>zEr man nan yfel ne deS, scilling t>an ]>e baet flet age and cyninge XII scill.' 21 13 healdeo" inec on heaj>ore. Cf. 66 3, hafaft mec on headre. 21 14 on gerum sceacan. Cf. El. 320, eodon on gerum. 21 13-16 scod frecne. Cf. Gen. 1597, frecne scodon. 21 i7 wiepnum awyrged. Our riddler is here thinking of the passage in Ps. dm dwyrgedan •turad'an sweorde. 21 17 f. Roeder (Die Familie bei den Angelsachsen, 1899, p. 81) considers the conception of the lot of the bachelor that we meet in these lines as 'eine derb sinnliche aber durchaus gesunde germanische Auffassung.' With the motive of lack of vengeance compare the inability of the stag-horn to wreak its wrongs upon its banesman (93 19-20). Notice the insistence upon blood-vengeance, Beow. 1339, 1546, Maid, 257 f. 21 23 pe me hringas geaf. Cf. Beow. 3035, )>e him hringas geaf. See the description of the sword, 71 8, hringum gehyrsted. 21 24 The idiom is found 16 6, n, 85 7. 21 25 guj>e fremme. So And. 1354. 21 28-29 roe }>aes hyhtplegan . . . wyrneo". Cf. Brun, 24-25, Myrce ne wyrndon heardes handplegan haele^a nanum. For a discussion of the construction, see Shipley, Genitive Case in Anglo- Saxon, p. 64. 21 29-30 mec ... on bende legde. Cf. 4 13-15. 21 33 f. This is the only picture of the shrew or scold in Old English poetry although we are told, Gn. Ex. 65, wldgongel wif word gespringe)>. But there is no dearth of ' women weeping for their warriors dead ' ; cf. Fates, 46. ; RIDDLE 22 This 'Plow' riddle — for Dietrich's answer (XI, 465-466) has been generally accepted — has no parallels among the Latin enigmas of its day; but an ana- logue from the pen of Scaliger (Reusner I, 180) has certain points of likeness: Ore gero gladium, matrisque in pectore condo, Ut mox, qua mine sunt mortua, viva colas. D_m_meus a tergo.caudamque trahens retrahensque Hasta»non me ut earn verberat ast alios. The modern German and English riddles (Wossidlo 241; Royal Riddle Book, p. 18) are of quite another sort. Hoops ( IVb. it. A'p., pp. 499-508) discusses at length early German agriculture, and points to the close likeness between the Germanic hook -plow (ffakenpflug), as preserved in the prehistoric specimen from the moor at D0strup in Jutland, II4 RIDDLES OF THE EXETER BOOK and the old Greek plow, of which we have many illustrations (notice particularly that on the bronze bucket from Certosa). The specifically Germanic wheel-plow, ' which is not found among Romans or Gauls or Slavs but which was widely known among the Germanic races before the Carolingian times,' seems to be identical with the Rhaetian wheel-plow, described by Pliny, ATatural History xviii, 172 : ' Latior haec [cuspis] quarto generi [vomerum] et acutior in mucronem fastigata eodemque gladio scindens solum et acie laterum radices herbarum secans. Non pridem inventum in Raetia Galliae, ut duas adderent tali rotulas, quod genus vocant plaumorati? It is generally agreed that the first part of J>latwiorati (ac- cording to Baist, Wolfflins Archiv III, 285, plaum or plottm Ratf) corresponds to the West G&rm.pldg (A.-S. //<?£•, ploh} and iheplovum of the seventh-century Lom- bard law (edited by Roth, 288 (293)). The Anglo-Saxons who crossed to Britain in the fifth century did not yet possess the word, which was first known to their island in the eleventh century (Hehn, Kp. u. Ht., 1902, p. 556). Hoops concludes that the Anglo-Saxon sulh (Lat. sulcus, ' furrow ' ; Greek £\KW, ' to draw ') indicated the old hook-plow (cf. Anglia, Bb. XVII, 201; Foerster, Zeitschrift fiir Roma- nische Philologie XXIX, 1-18). It is noteworthy that in all the illustrations given by Hoops these early hook-plows are drawn by oxen. For an excellent descrip- tion of the Old Norse plow, see Weinhold, Altnordischtf Leben, p. 79. Andrews (Old English Manor, p. 253) remarks: 'The plow as it is pictured and described (Elton, Origins of English Hist., p. 116; Wright, Celt, Roman and Saxon, p. 256 ; Rau, Geschichte des Pfluges, Heidelberg, 1845, passim} was of a com- paratively high order composed of beam, tail, sjiare, colter, and wheel ; the latter, though clumsy and of the shape of a cart-wheel, shows an advanced stage of de- velopment. It was more than a disk of wood bored for an axle, it had felloe, spokes, and hub. Cynewulf's description [Rid. 22], though picturesque, adds little save^the^one important fact that the seed was cast immediately after the furrow was turned [Rid. 22 6]. He omits mention of the wheel, and it is not improbable that we are to see the influence of Roman civilization in the wheel which the calendar shows us. It can hardly be doubted that plows of a much inferior type, similar to the primitive varieties which Rau gives in his history of the plow, were used at this time on many an English agricultural estate. That represented in Harleian MS. 603 has only share and tail of the simplest possible character. The irons of the plow were made by the smith and the wood work by the wright. The smith in the Colloquy declares that the plowman was indebted to him for the plow- share, colter, and goad, and we know well the character of the smithy, where these were made, with its anvil, hammer and sledges, fire-sparks and bellows.' The illuminated manuscripts are at variance regarding the form of plow. In the illustrations in the Harl. MS. 603, ff. 21 v., 51 r., 54 r., 66 v., the plows are of the rudest sort, without wheels; while the plows of the first picture in the Anglo-Saxon Calendars (Tib. B. V, Strutt, pi. x; Jul. A. VI) — not a January but an April scene, as Leo thinks, R. S. P., 207 — and of the Caedmon manuscript (Archaeologia XXIV, pi. xxviii, xliii) have wheels (compare illustrations from the Bayeux Tapestry, Knight, Pict. History I, 278-279). All these plows are drawn by oxen, urged by a goad — usually in the hands of an attendant herd. This use of oxen instead of horses is confirmed by the speech of the plowman NOTES 115 in .^Ifric's Colloquy (infra) and by such accounts of plowing as we meet in Ead- mer's story of the field-laborer who failed to observe Dunstan's feast-day (Vita, § 24, Stubbs, Memorials of Dunstan, p. 248). In Alfred's report regarding the Norwegian Ohthere, it is mentioned as an exceptional thing that on account of his few cattle he did his little plowing with horses (Orosius i, i). The account of the Plowman in ^Ifric's Colloquy (WW. 90) exactly conforms with the illustrations in Old English manuscripts: 'Arator: Ic ga ut on dasgred }>ywende oxon to felde and jugie hig to syl; nys hyt swa stearc winter )>aet ic durre lutian aet ham for ege hlafordes mines, ac geiukodan oxan and gefaestnodon sceare and cultre (vomere et cultro) mid biere syl ajlce daeg ic sceal erian fulne aecer oH>e mare . . . ic haebbe sumne cnapan J>ywende oxan mid gadisene (cum stimulo).' 22 ia Cf. ii i, 32 6, 35 3 (Rake). 222 geonge. Sievers, Gr.3 396 b, n. 2, points out that 'for gongan, North, has Lind. geonga (ind. pres. i sg. also giungo, opt. giunga), Rit. geonga, gionga, but R.2 gonga (only once geonga).' This diphthongization is ' unknown to the other dia- lects ' (id. 157,4; Madert, p. 127). Cf. Spr. I, 499. 22 3 bar holtes f6ond. Dietrich (XI, 466) regards this as_±heox. Cosijn says of the phrase (PBB. XXIII, 129): ' Eine vortreffliche kenning fiir das eisen das in der form eines beiles den baum anfeindet ; hier bezeichnet sie das pflugeisen.' This is also Herzfeld's interpretation (p. 39). According to Brooke (E. E. Lit., pp. 145-146) the 'hoar enemy of the wood' is the old peasant, hlaford mm (11. 3, 15). The explanation of Cosijn and Herzfeld cannot be accepted, as it is out of keeping with the context and with the conception of the plowshare as neb (i), orboncpil (12), and tdj> (14). Brooke's rendering has much in its favor; but I per- sonally prefer that-of Dietrich for two reasons — a plow riddle would be strangely defective that omitted all reference to the ox, a great favorite in such poetry (Rid. 13, 39, 72), and we meet elsewhere the antithetical phrase holtes gehlefra (El. 113) applied to the ox's opposite, the wolf. Dr. Bright favors this view. 22 4 [se] won. Sievers' reading [on] wok is open to the objection that on wdh, which appears frequently, is never found in the sense of 'bent, crooked,' — the meaning necessary to the present context, — but always with the idea of 'wrongly,' 'wrongfully' (Spr. II, 731; B.-T. s.v.). Dr. Bright happily suggests \se\ ivoh farefr, 'who goes bent.' 22 sb Cf. 13 8, wegeft ond by$. 22 6 sawej> on swaeft nun. In the Calendar illustrations (supra], a sower follows the plowman. 22 7a Cf. 28 2, brungen of bearwum (honey). Note the parable in Alfred's Pref- ace to the Soliloquies. 22 8 on waegne. Wcegn or -wan appears frequently in the Vocabularies, where it glosses plaustrum or carrum (see B.-T. s.v.; also Klump, pp. 115-116). We meet the word in Bemv. 3134 (wees gold on wan hladeti) and in Run. 23 (he [sc. Ing] ofer wa-ggewdt, ivizn after ran). It is used interchangeably with crat : indeed, as Wright points out (Domestic Manners, p. 73), Ps. xix, 8, in curribus, is glossed in wanum in one version, in cr&tum in the others. Two kinds of wagons are mentioned in the Riddles : the common agricultural cart of the present example, in connection with the wood of the plow ; and the more patrician chariot of the Il6 RIDDLES OF THE EXETER BOOK following problem, 23 ga, t2b. The cart is mentioned frequently by the Charters in the references to w&gna gang or the royal grant of a certain number of loads of wood (Kemble, Saxons in England II, 85). And we meet many illustrations in the manuscripts. In the July picture in the Calendar (Tib. B. V, Strutt, Horda, pi. xi), workmen are engaged, not only in lopping trees and felling timber with axes, but in loading with wood a cart, while two yoked oxen stand at the side. In the June illustration is another rude cart; and in Cotton Claudius B. IV, f. 66, 67, 68, 71, 72, several similar drawings are found. In all these pictures the carts are two-wheeled and drawn by oxen, save on f. 68 v., where the long-eared ani- mals attached to a four-wheeled cart are doubtless asses. Chariots are of two kinds: the two-wheeled cars drawn by two horses in the illustrations of Luxury in the Prudentius MS., Cott. Cleopatra C. VIII, f. 15 r., 16 v., 18 v. (see Wright, Dom. Manners, p. 73), and by four prancing steeds in the corresponding pictures of MS. Add. 24199, f. 17, 18, 19 (see Westwood, Facsim- iles, pi. xiv); and the hammock chariots of MS. Claud. B. IV, f . 60 v. and r., — with four wheels and a body of strong hides, — described by Strutt, Horda, p. 45. The two-wheeled wagons of the Anglo-Saxons were doubtless very similar to the carts in the bog-finds at Deibjerg, North Jutland, which have their modern counterparts in the Swedish karra (Du Chaillu, Viking Age I, 294). 22 8b Cf. 83 iob, hasbbe ic wundra fela. 22 9-10 As Brooke says (E. E. Lit., p. 146), ' It is a vivid picture of an old English farmer laboring on the skirts of the woodland, leaving behind him the furrow black where the earth is upturned, green where the share has not yet cut the meadow.' He renders — Green upon one side is my ganging on ; Swart upon the other surely is my path. 22 12-14 Andrews (O. E. Alanor, p. 253) rightly regards one orf>oncpil as the coulter, the other as the share. Thorpe places a semicolon after heafde, and renders 'fast and forward falls at my side what with teeth I tear'; but it is' better, on account of the usual meanings of fast and forfrweard, ' fixed ' and •prone ' (cf. 73 26, forfr-weard, the Lance ; and 22 i, nif>er-weard, neol) to associate the adjectives with oj>er (orfroncptl). Grein, Dicht., translates ' ein anderer fest nach vorn gehend fallt zur Seite, sodass ich zerre u. s. w.' 22 14 topiim. Prehn, p. 272, points out the parallel between this and the Rake riddle, 35 2 (hafafr fela toj>d), but the likeness is produced by the nature of the subjects. In WW. 219,4, sule-reost\s the equivalent of dentale, s. 'est aratri pars prima in qua vomer inducitur quasi dens' (see WW. 17, 20; 384, 43). Else- where in the Vocabularies (Wright II, 138, 72) sule reost is the vomes. In his long discussion of reost, Heyne, Funf Biicher 11,37, points out that O. H.G. riostar has often the same meaning as the Anglo-Saxon. 2215 hindeweardre. Cosijn (PBB. 23,129) notes that the gender of the adjective is due to that of the riddle-subject (here seo sulli). This is probably true. Trautmann also observes (BB. XIX, 181): 'Die ae. ratseldichter nehmen es, wenn sie einen zu erratenden gegenstand als menschen infiiren, sehr genau mit dem geschlechte." This is not the case. For a detailed discussion of gram- matical gender in the Kiddles, see Introduction. NOTES 117 RIDDLE 23 This query I have already considered at length (M. L. JV. XVIII, 102). The riddle of the Month with its sixty half-days (stxtig monna) is, of course, a variant of the Year problem, which in one form or other appears in every land, as Ohlert (pp. 122-126), Wiinsche (Kochs Zs., N. F. IX (1896), 425-456), and Wossidlo (pp. 277-278) have shown. The Anglo-Saxon chariot-motive has long since been linked by Dietrich (XI, 457, 466) with Reinmar von Zweter's ' ein sneller wol gevierter wagen ' of twelve wheels, which carries fifty-two women and is drawn by fourteen horses, seven white and seven black (Roethe, R. von Z., 1887, Rid. 186, 187, p. 616). But there are many other analogues, some of which Roethe cites. Haug, pp. 457 f., translates from the Rigveda I, several Time riddles, in one of which (Hymn 164) the year is pictured as a chariot bearing seven men (the Indian seasons [?]) and drawn by seven horses; in another (Hymn n) as a twelve-spoked wheel, upon which stand 720 sons of one birth (the days and nights). Still closer to the Anglo-Saxon is the Persian riddle of the Month (Gorres, Das Heldenbnch von Iran, 1820, I, 104 f.), cited by Wiinsche, in which thirty knights (the days of the month) ride before the emperor. In the Disputatio Pippini cum Albino, 68-70 (Haupts Zs. XIV, 530 f.), the Year is the Chariot of the World drawn by four horses, Night and Day, Cold and Heat, and driven by the Sun and Moon. And, finally, in the Liigenmdrchen of Vienna MS. 2705, f. 145 — classed by its editor, Wackernagel (ffaupts Zs. II, 562), as a riddle — the narrator tells how he saw, through the clouds, a wagon, upon which seven crowned women sat, and near which twelve trumpet-blowers (garzune) ran, and a thousand mounted knights rode. Der liigenaere nam des goume, Das si nach dem selben sliten Alles uf dem -wolken riten Und -Molten da mite iiber mer. The likeness of these last lines to the desire of the sixty knights in Rid. 23 to pass over the sea is peculiarly suggestive. 'Reinmar's riddle,' says Roethe (p. 251), 'is really popular — that is, it is not drawn directly or indirectly from learned or Latin sources.' This is equally true of the Anglo-Saxon problem ; still, we must feel that, like Reinmar's poem, it has come to us from an artist's hand. Trautmann's solution, ' Die Briicke,' blindly ignores every motif of the riddle, which has surely naught in common with Symphosius 62, Pans. 23 2 wicgum rid an. Horses were never used for plowing (see Rid. 22), nor for farm-labor, — drawing of wood in carts, or the bringing home of the harvest, — but only for the chariots of the rich or as steeds of the upper classes. No/<?0w was mounted (see Rid. 20). That the rich were fond of horses is shown by the numerous illustrations in the manuscripts (Wright, Domestic Manners, pp. 71-72), and by the various synonyms for hors or ivicg. See Hehn, Kp. u. ///., 1902, pp. 19-55 ! and Heyne, Fiinf Biicher H, 167 f. 23 3-4 Dietrich (XI, 466) meets the difficulty in these numbers by regarding the month as December, which has seven holy-days, the feasts of Mary (Recep- tion), St. Nicholas, St. Thomas, Christmas, Stephen, St. John the Evangelist, H8 RIDDLES OF THE EXETER BOOK and the Innocents. These with the four Sundays (sceamas, • white horses ') make up the eleven steeds of the troop. I reject the MS. reading frldhengestas, which Dietrich (XII, 251) renders 'stately horses' (see note to 109); but, instead of substituting with Thorpe fyrdhengestas, ' war-horses,' I prefer to read frifrlien- gestas, ' horses of peace.' Compounds with frifr are common, and this reading exactly fits the context. The horses are the eleven peace-days of December, for frifr was established on these holy-tides by the strictest laws (Schmid, Gesetze, pp. 584-585, s.v. Friede}. Cf. ^Ethelred's Laws, v, 19: 'And beo J>am halgum tidum eal swa hit riht is, eallum cristenum mannum sib and s5m gemzene, and zelc sacu getwiemed.' If December be our month, the other bank (23 20) is, of course, the New Year. Dr. Bright suggests that ' the eleven horses ' may be the days between Christ- mas and Twelfth Night counted exclusively, and contrasts Orm's inclusive count- ing of thirteen days (Ormulum, uo6of. ; see White's note, II, 403). He points to the Christmas year-beginning so well known to the Anglo-Saxons. 234 sceamas. Jordan notes, Altenglische Saugetiernamen, p. 115: ' Die Wb'rter- biicher fassen sceam, wohl wegen des in demselben Ratsel, z. 18, folgenden blon- can als Synonymon dazu, also als " weisses Pferd, Schimmel." Diese Deutung lasst sich auch etymologisch rechtfertigen : sceam = skau-ma gehort zur Wz. skau "schauen" (ae. sceawian, ahd. scouwon) woher Got. skauns, ahd. skoni, ae. sciene, "schon," ne. sheen "hell," " glanzend," bedeutet also eigentlich "das An- sehnliche, Glanzende " (skau-nis = "sehens wert," "ansehnlich"). Gestiitzt wird diese Auffassung durch das mit ae. sceam im Ablaut stehende anord. skjone, " Apfelschimmel " (daneben skjdme, "flackerndes Licht, Strahl").' See Kluge, Etym. Wtb. s.v. schon, 23 5 ofer mere. Barnouw, p. 217, notes that in the .Riddles the sea is often mentioned (Herzfeld, pp. 22-23), but never with the article. Y& is, however, an exception to this : 61 6b, yiS sio brune (see Met. 26 29-30, slo brune | y"5). 23 7 atol y]?a gepraec. Cf. 3 2, under y>a gej>raec ; And. 823, ofer yiSa ge)>raec; Exod. 455, atol y|>a gewealc. 23 8a Cf. Ps. 65 5, t>a strangan streamas. 23 9b wicg somod. So Beow. 2175. 33 10 under hrunge. Grein says (Spr. II, 109) : ' Wagenrunge, aber bei den Ags. wol nicht wie im Hochd. die Leiterstutzen, sondern die Sparren oder Reife des Wagendaches.' Bosworth-Toller, s.v. renders 'the pole that supported the covering.' But, as the word does not occur elsewhere, these definitions are de- termined by the context in the present passage. 2311 eh. Ettm. remarks : 'eA = eoh hoc loco gen. neutr. videtur esse; nipotius eh = dh, dc scribi debeat, ita ut dc, quercus, h. 1. navem significet.' 23 nb So And. 1097, aescum dealle. 23 13 Grein's conjecture, esla, seems much more in accord with the context than the MS: esna. Moreover, the illuminated manuscripts furnish ample evidence that the wcegn was sometimes drawn by asses (see note to 22 8, on ivagne, and Heyne, Fiinf Biicher II, 177). Thus in our passage every kind of draught animal is mentioned. 23 13-17 This part of the enigma suggests Rid. 40 in its negative method. NOTES 119 23 14 faethengest. Grein (Spr. I, 274), B.-T. s.v., and Jordan (Altenglische Siiugetiernamen, p. 115), unite upon this reading, comparing sifrfezt for the first member of the compound and translating ' road-horse,' which seems preferable to Dicht. ' ein feisster Hengst.' Dr. Bright suggests fat hengest, ' caparisoned steed.' 23 16 lagu drefde. So H. M. 20 ; cf . 8 2, wado drefe. — on lyfte fleag. Cf. 52 4, fleag on lyfte. 23 18 blonean. The word is found in two other places in the poetry, Bemv. 856 and El. \ 184. Jordan (p. 115) notes : ' blonca, der glanzende (sc. eoJi) wird der Schimmel genannt,' thus identifying the word with sceamas (23 4). On the other hand, Heyne-Socin, in discussing the Beowulf passage (p. 149), regards the color as ' vielmehr die apfelfarbe.' Egilsson {Lex. Poef., p. 59) cites many examples of O. N. blakkr, ' equus,' and Cleasby-Vigfusson, p. 67, points to Blanka, the mythi- cal horse of Thideric (Dietrich) of Bern. The O. H.G. blanc-ros is discussed by Pomander, A/id. Tiernamen, Darmstadt, 1899, p. 82 (cited by Jordan). Blonca, with its cognates, appears to be used generally in the sense of 'a noble horse,' without reference to color. RIDDLE 24 Prehn (pp. 188 f.) fails completely to establish any relation between this ' Bow ' riddle and the enigmas of Symphosius (65, Sagitta) and Tatwine (32, Sagitta ; 34, Pharetra). That the likeness of Rid. 24 2 to Tatwine 32 1-2 is accidental is attested by the variant of the Anglo-Saxon line in another weapon-riddle (21 i). As my notes show, this problem has much in common with Rid. 18 and 21. It is interesting to compare the ' Arcus ' enigma of Scaliger (Reusner I, 172), and the Norse query of the ' Bogi ' (Landstad, No. 5) : Smeften smiftaft, smeftkeringi spann, i hagin deft voks, i holti deft rann ; deft er aldri sd litift, deft drep 'ki ein mann. Although, owing to the decay of wood, no trace of bows has been found in the Anglo-Saxon graves, yet important evidence for the use of the bow, both for war and the chase, is found not only in such manuscripts as Cleop. C. VIII, Claud. B. IV, Tib. C. VI, and the Prudentius MS. of the Tenison library (compare Keller, p. 51, Strutt, Sports and Pastimes, Bk. ii, chap, i; Horda, pi. xvii fig. 2, pi. xxii figs. 23, 24, 25), but everywhere in the literature. So numerous are the appear- ances of bow and arrows in the poetry of battle (Keller, pp. igSf.) that it is difficult to appreciate the reasons for Akerman's assertion that it was not com- monly used by the Anglo-Saxons as a weapon of war (Archaeologia XXXIV, 171). Our riddle, which has no learned source, is conclusive upon this point (compare, too, the last lines of the Leiden Riddle) ; and the Beowulf affords many examples of its use in war (1433, 1744, 3114)- 120 RIDDLES OF THE EXETER BOOK Akerman is perhaps wrong in declaring that no arrow-heads have been found in Anglo-Saxon graves (Archaeologia XXXIV, 171), for, as Hewitt points out in his Ancient Armor and Weapons in Europe I, 55, ' some have been found in Kentish interments, and others on the Chatham lines.' It is possible that these are spear- heads. The Anglo-Saxon use of the bow has been discussed at, length by Professor Cook in his note to Christ, 765 (bragdbogari). See also Brooke, E.E.Lit., pp. 125, 128, 129, 131. The Bow is described in the Runic Poem, 84 : Yr by)> ae^linga wyn ond eorla gehwaes wyn and wyr)>mynd, byjj on wicge fasger, faestllc on faerelde, fyrdgeatewa sum. In the Old Norse runic poem (Wimmer, pp. 280, 286), yr appears both as ' yew' and as 'bow.' The etymological connection between O. E. eoh, 'yew,' and O.N.Jr justi- fies the conclusion that the Anglo-Saxon bow was made from the yew-tree (Cook, Christ, p. 159). 24 i Agof. Agof(b) inverted is of course boga. For the relation of the word to the supposed date of the Riddles, see Sievers' discussion, Anglia XIII, 15, which I have summarized in the Introduction. 24 2 Prehn, p. 188, finds a likeness between this and Tatwine 31 1-2 : Armigeros inter Martis me bella subire Obvia fata juvant. But note that almost the same line appears in the description of the Sword, 21 i. 24 4, 9 Cf. 18 9. The use of poisoned arrows among the Anglo-Saxons, to which frequent reference is made in both their poetry and prose (And. 1331, Jul. 471, Maid. 47, 146, WW. 143, 7, Bl. Horn. 199, 17-19, Life of St. Guthlac, Good- win, 26, 28), has been considered at length by Professor Cook in his note to Christ, 768, dttres ord (see also Keller, p. 51). 24 s Compare the relation of the waldend to the Sword (21 4-6) and of the/raz to the Ballista (18 s). 247 lengre. Cosijn (PBB. XXIII, 129) would read lengra, because boga is masculine; but the poet may be referring to wiht (1. 2); cf. 257, glado. Rid. 41 gives ample proof that in our poems no such regard is shown to grammatical gender as Cosijn and Trautmann assume (see Introduction). 248 spilde geblonden. Cf. Sat. 129, attre geblonden. 249 ealfelo attor. Cf. And. 770, attor aelfaele. — g£ap. The word appears only here. Thorpe regards it as an adjective and renders ' crafty.' Grein (Spr. I, 504) and B.-T. s.v. derive irom geofan, 'cava manu includere,' 'to take up,' which they connect with Icel. gaupna, O. H. G. coufan, Scot, gowpen, ' to lift or lade out with the hands.' The adj. geap is of like origin. 24 10 togongeff. Only here in this sense, ' pass away ' ; but compare the use of tdfaran in a similar context (Lchd. I, 122, 18, syle drincan on wine, eal iSaet attor tofaer)>). 24 na Cf. 44 16, J>e ic her ymb sprece. See also Met. 10 45, 16 24, 20 3, 4. 24 i2b Cf. 18 6, me of hrife flgoga'S. NOTES 121 24 13 The metaphor of ' death's drink ' is elaborately expanded, Gu. 953 f. : bryiSen waes ongunnen J>aette Adame Eve gebyrmde aet fruman worulde : feond byrlade Srest J>aere idese and heo Adame, hyre swiesum were, si'Sftan scencte bittor bzedeweg, J>aes \& byre srSftan grimme onguldon gafulriidenne )>urh sergewyrht, J>aette ainig ne waes fyra cynnes from fruman srSftan mon on moldan, l>aette meahte him gebeorgan ond bibugan }>one bleatan drync deopan dea'Sweges. Budde, Die Bedeutung der Trinksitten, p. 93, cites a similar passage from Lud- wigslied, 52. For purchase by death, see Beow. 3012, )>aer is ma'Sma hord, grimme gecea[po]d. Grein renders, Die/it., So dass der Kempe den Todestrank mit seiner Kraft bezahlt, Den Fullbecher fest mit seinem Leben. 24 14 fullwer. I believe with Dr. Bright that we must reject the reading of MS. and editors, full wer, and read/«//ww, ' complete wer ' or ' wergild,' ' complete recompense for a life.' Cf. Alfred's Laws, § 23, 2 (Schmid, Gesetze, p. 84), be fullan were. As Bright notes, the accusative is in grammatical apposition to mdndrinc. RIDDLE 25 The subject of this riddle, Higora or ' Jay,' has already been discussed by me under Rid. 9, which I believe to have a like solution. Dietrich (XI, 466-467) cites several references to show that ' Picus,' which glosses the word in Anglo- Saxon vocabularies (WW. 287, 9, 'picus, higera1 \ 260, 14, 'picus, higere'1; 39, 36, 'picus, higre'1}, cannot refer to the common Woodpecker ('Specht'), but must refer to the pica glandaria of Pliny (Nat. Hist, x, 42), the niaaa. of the Greeks. The ' Specht ' riddle of Strassbttrg Kb., No. 98, and its Latin copy by Lorichius (Reusner I, 276) are totally unlike the Anglo-Saxon. It is interesting to note that Isidore's description of the ' Picae ' (xii, 7, 46) shows that he had in mind the garrulous bird of our riddles : ' Per ramos enim arborum pendulae importuna garrulitate sonantes, etsi linguas in sermone nequeunt explicare, sonum tamen humanae vocis imitantur.' So in Vincent of Beauvais, Speculum Naturale XVI, 32, the pie is called ' pica loquax,' ' pica garrula ' ; cf. Pliny x, 42. Note also Chaucer's ' jangling pye ' (Par!, of Foules, 345). Whitman (Journal Germ. Phil., II (1898), 161) says: 'Riddle 25 is sometimes interpreted as the jay, but as the name of the bird is formed by the runes G. A. R. 0. H. I, it must be higora, the woodpecker, although this bird is not generally considered a mimic.' Dietrich seems to be right in supposing that the jay, a near relation of the pie or pica, is meant (see Hessels, Leiden Glossary, 1906, p. 168): (i) 'Picus' and 122 RIDDLES OF THE EXETER BOOK 'Pica' are frequently confused in the glossaries (Du Cange, Glossarium, s. v. Gaia ; WW. 702, 4, 'picut, pica, a pye '), and the bird-names ' graculus,' 'garulus,' which are associated with higra or higre in the Anglo-Saxon vocabularies, later apply to the jay; (2) Hdher, 'the jay,' is the modern equivalent of higora (Kluge, Etym. Wtb. s. v.), and, indeed, is glossed 'garrulus' in M.H.G. (Mone, Anz. VIII, 399, cited by Dietrich I.e.). By 'garrulus' or 'graculus' Aldhelm evidently means the thieving magpie (De Laudibus Virginum, Giles, 142). Miiller (Cdthener Pro- gramm, p. 16) believes that by Higora the ' corvus glandarius' or ' jay' is intended. The lines of Rid. 25 should be compared (says Miiller) with 'was Naumann in seiner Naturgtschichte der Vogel Deutsc/i lands, II, 125, liber den aus vielen sonderbaren und ausserst verschiedenen bald gurgelnden und schwatzenden, bald pfeifenden oder kreischenden Tbnen zusammengesetzten Gesang des Eichel- hehers sagt, welcher die Stimme des Mausebussard, aber auch der Katze, ja das Wiehern eines Fiillens, die schirkenden Tone die beim Scharfen der Sage entstehen, das Gackern des Huhns, das Kickerikie des Hahnes nachahme.' Grein (Dicht. II, 220, Spr. II, 72) and Wiilker (Bibl. Ill, i, 198) had already cited the 'berna' ('verna') lemma of higrcs, higre, in Gloss. Epin. 156, Corp. MS. 290 (WW. 9, i) and MS. Cleop. A. Ill (WW. 358, 5), but it was reserved for Frl. Emma Sonke (Engl. Stud. XXXVII, 313-318), to champion at length the 'scurra' or ' mime ' interpretation. By reference to Strutt, Sports and Pastimes, p. 346, Cham- bers, Medieval Stage I, 71, Schultz, Das hofische Leben, p. 443, n. 3, she shows that these mimes could imitate the sounds of all animals. Yet, if on account of this power the mime was known as the Higora or ' jay,' we must surely assume the same mimicry on the part of the bird from which the name is derived. In- deed we are told expressly in Rid. g 9-10 that the bird has mimetic power. Rid. 25 simply elaborates the hint of the earlier riddle. It is needless to devote any con- sideration to the extravagant conclusions drawn by Frl. Sonke from the single runes in Rid. 25. 25 i wrtesne mine stefne. Cf. 9 1-3. 25 ? Holthausen (Engl. Stud. XXXVII, 207) regards the line as metrically false because hund not only must alliterate [why .?], but also should be inverted, since one expects hwilum swd hund beorce [again why ?~. He therefore believes that the first half-verse was originally a second, in which case the verb preceding may alliterate in descriptions. As a first half -verse he would emend the text to read hwilum belle swd bearg, or perhaps bicce for hund. His very premises are based upon a false a-priori conception of metrical demands that is blind to all contrary evidence. ^Elfric says in his Grammar, 22 : ' Hit bi)> swiSe "SyslTc Saet se man beorce o~$5e blsete.' 25 4 ic onhyrge. Cf. g 10, hlude onhyrge. — haswan earn. See my discussion of 12 i, hasofdg. As I there pointed out, hasu or hasupdd (Brun. 62) as an epithet of eagle is synonymous with salowigpdda (earn), Jud. 210. Whitman (Journal Germ. Phil. II, 1 68) notes that 'at present two species of eagle are natives of Britain, the golden eagle (aquila chrysaetus) and the white-tailed eagle (haliaetus albicilla), both of which were probably known to the Anglo-Saxons. In the Battle of Brunanburh (63) the bird described as ' white behind ' (tzftan hwtt) is un- doubtedly the white-tailed eagle, but the war-eagle, usually called dark,-feathered NOTES 123 (salowigpddd), is probably the golden eagle, known in Scotland as the black eagle.' This distinction was hardly recognized by the Anglo-Saxons, inasmuch as in the Bninanbtirh passage hasopddan precedes iiftan hwit. For the association of the eagle with war, see Beow. 3026, Jud. 210, El. 29, And. 863, Maid. 107, Brun. 63. 25 5 guftfugles hleoj»or. The eagle is called earn, grSdigne giifrhafoc (Brun. 64), and not only his coat but his song (Jiildeleofr} is mentioned in the detailed description in Jud. 209-212. For other references in both poetry and prose, see Whitman, p. 172. — glidan. As Whitman shows (p. 169), this is the ' milvus,' the kite or glede. Gilbert White, Letter XLVI (Barrington), says, accurately enough : ' Thus kites and buzzards sail round in circles with wings expanded and motion- less ; and it is from their gliding manner that the former are still called in the north of England gleads, from the Saxon verb glldan, " to glide." ' 25 6 mjewes song. Whitman (p. 180) notes that the name maet banlease bryd grapode | hygewlonc hondum. 26 8 Trautmann suggests (BB. XIX, 187) that the subject of the riddle must be masculine on account of the form reodne, and therefore proposes as the rid- dler's topic heopa or haga. Of the Anglo-Saxon names for onion, clpe is feminine and Ifwitleac is neuter. But in the riddles there is no such strict insistence upon grammatical gender as Trautmann would have us believe (see Introduction). Trautmann in his text retains MS. reading, and translates ' auf mich roten zufahrt,' but afterwards suggests rierefr mec reodne (188); but his defense of this is vitiated by his false solution of the riddle. The proposed change seems to me too violent, and not necessary, as rizsan followed by on with the accusative is a common idiom (see Spr. II, 368). In his text Grein follows the reading of the MS., but in a note conjectures rierefr and on reofrne (' zur Riittelung '). In his translation he renders ' erhebt mich zur Riittelung.' In Spr. II, 368, 374, he reverts to the text of MS. and translates the verb by 'mittere' (a transitive use not found elsewhere) and renders on reodne (< reoden) as above. Sievers, PBB. IX, 257, suggests reone (Gr.a 301, n. 2). Dr. Bright proposes on hreode (' reed,' ' stalk '). I can see no ob- jection to the MS. reading. The order raseS- mec on reodne finds abundant support in the very similar phrase, 13 13, swifefr me geond siveartne ; and reodne, 'red,' is fitly applied to the outer skin of the onion and meets the demands of the double- entente. 26 8b rf-afacV mm heafod. Hehn, Kp. u. Hi., 1902, p. 195, seeks to show that the Latin cepa, ' onion,' contains the notion of ' head,' cepa capitata, and points to ' a far-distant stage of speech, when caput and K£0aXiJ had not developed their suffixes.' But, as Schrader says in his note upon this passage (ib. 205), the con- nection of Gr. Kdiria, Lat. cepa, with the Indo-Germanic words for ' head,' is exceed- ingly doubtful, and presents the gravest etymological difficulties (cf. Kluge s.v. Haupt}. It is interesting, however, to note with Hehn (I.e.) that among the Italian Locrians the word xe^aXi) could also mean an onion-head (Polybius, xii, 6), and that a play upon the words caput and cepa is found in Ovid's Fasti, iii, 339. 26 9a Cf. 62 6, on nearo fegde. 26 ii wif wundenlocc. Curled or braided locks were regarded by the Anglo- Saxon as an accessory of beauty. The twisted hairs of the fair Judith are twice mentioned in the Old English poetic version (Jud. 77, 103, wundenlocc) ; and in that poem the Hebrews are described with the same epithet (326). The trans- lator of ' De Creatura,' Kid. 41 98-99, employs the phrase Invite loccas, wraste ge- loundne (see also 41 104, wundne loccas); and the glossators in the Royal and Cambridge MSS. (Napier, O. E. Glosses, pp. 191, 195) render Aldhelm's 'calamis- tro' 'curling-iron' (1. 47), by J>rewelspinle and ivolcspinle. In the glosses to Ald- helm's De Landibns Virginitatis, ' calamistro ' is translated by frrdwiticspinle or JiarnSdla (Haupts Zs. IX, 435, 7 ; 513, 75 ; 526, 46). It is in this tract, De Landibns 126 RIDDLES OF THE EXETER BOOK Virginitatis, xvii (Giles, p. 1 7), that Aldhelm describes the hair-dressing of the Anglo- Saxon ladies: 'Ista tortis concinnorum crinibus calamistro crispantibus delicate componi et rubro colons stibio genas ac mandibulas suatim fucare satagit.' Long hair was the sign of freedom. Fri-wif loc-bore — 'free woman with curly or flowing hair' — is the phrase of Laws of ALfrelberht § 73 (Schmid, p. 8). Compare Sharon Turner, VII, chap, v; Gummere, Germanic Origins, pp. 61 f. ; infra, note to Rid. 4198- In his discussion of Beow. 3151, bundenheorde, Bugge, PBB. XII, no, shows that this adjective (for which Grein reads •wunden-keorde) must be rendered 'mit gebundenen locken,' and is the ' epitheton der alten frau im gegensatz zu den madchen, derer haar frei herabfallt ' (cf. Pogatscher, Anglia, Bb. XII, 198). 26 ii wset bi5 J>aet cage. Of this Dietrich says, supporting the ' Hemp ' solu- tion (XII, 240) : ' Das dunkle ende des ratsels bezieht professor Lange auf den faden, der aus dem gelblichen rocken gerauft und dann zwischen den fingern eingeengt durch das gefiil der spinnerin gleichmassig gebildet wird ; das auge aber, welches dabei durch den benetzten finger feucht wird, ist die bffnung der ehedem oben durchbohrten spindel.' This is overwrought. But Trautmann ignores the obvious explanation in his endeavor to render cage not ' eye ' but ' mouth.' These are desperate attempts to bolster weak solutions. Not only in the riddles that I have cited, but everywhere in literature and life, the onion causes eyes to water. Shakespeare is full of examples: A.W.\, 3,321, 'Mine eyes smell onions; I shall weep anon'; A.andC. i, 2, 176, 'The tears live in an onion that should water this sorrow ' ; ib. iv, 2, 35 ; etc. There is nothing ob- scure or difficult in the line, and the obscene implication is obvious. RIDDLE 27 Prehn, pp. 190-193, has pointed out the likeness of this ' Book ' riddle to vari- ous Latin ' Parchment ' and « Pen ' enigmas. At the fountain-head of these stand Aldhelm v, 9, De Pugillaribus, and v, 3, De Penna Scriptoria, which supply the (motives of Tatwine 5, De Membrano, and 6, De Penna, and of Eusebius 32, / De Membranis, and 35, De Penna. In form of phrase the beginning and end of the Anglo-Saxon problem resemble not a little the first and last lines of the fifth enigma of Tatwine (infra), but, in the light of the strong negative evidence of the other English queries against direct borrowing, I am inclined to regard the first resemblance as a coincidence of fancy conditioned by the nature of the sub- ject, and, like the second, presenting a commonplace of riddles of this kind (infra). The ' Membrana' enigma of Cod. Bern. 611, No. 24 (Riese I, I, 300) is an in- teresting analogue ; and the many Book riddles of the Islenzkar Gdtur (Nos. 241, 1 329, 390, 584, 599, 619, 711, 904) present instructive parallels. Rid. 68, 'Bible,'' is but a variant of Rid. 27, which has also many points in common with Anglo- Saxon problems of widely differing subjects (infra). 27 1-6 Sharon Turner, History of the Anglo-Saxons IX, chap, ix, translates from a manuscript of the ninth century ['Bibl. Cap. Canonicorum Lucensium,' I, Cod. 4, Muratori, Antiquitates Italicae, Milan, 1739, II, 370], a receipt for the preparation of parchment (' Composite ad tingenda Musiva, pelles et alia ') : ' Put it under NOTES 127 lime and let it lie for three days ; then stretch it, scrape it well on both sides, and dry it, and then stain it with the colors you wish.' Here is another receipt from the same hand : 'Take the red skin and carefully pumice it, and temper it in tepid water and pour the water on it till it runs off limpid. Stretch it afterwards and smooth it diligently with clean wood. When it is dry take the white of eggs and smear it therewith thoroughly ; when it is dry sponge it with water, press it, dry it again, and polish it ; then rub it with a clean skin and polish it again and gild it." It is interesting to compare with the ninth-century receipt for the preparation of parchment the various receipts cited by Wattenbach, Schriftwesen, 1875, P- I7l- The successive stages of preparation are indicated by Archbishop Ernest of Prague, a contemporary of Charles IV (Mariale, 85) : 'pellis separata a bove . . . mundata . . . extenta . . . desiccata . . . dealbata . . . rasa . . . pumicata, etc.' With 27 4-6 compare the words of Hildebert, Archbishop of Tours, d. \y){Opera, Paris, 1708, p. 733, cited by Wattenbach I.e.) : ' Primo cum rasorio purgamenum de pinguedine et sordes magnas auferre; deinde cum pumice pilos et nervos omnino abstergere.' 27 1-2 Compare with these lines Tatwine, 5 1-2 : Efferus exuviis populator me spoliavit, Vitalis pariter flatus spiramina dempsit. This suggested contrast between the living and dead skins is found not only in Eusebius, 32 4, but also in Cod. Bern. 611, No. 24, De Membrana, 2-3. 27 i feore besnyj^ede. Cf. Bee ter wide beer ivulfes gehlefran (compare Trautmann's reading of 52 ,fngla fultum), and by the sketches of the quill's origin in the Latin enigmas. This is pictured by Aldhelm, v, 3 i, ' Me pridem genuit candens onocrotalus albam,' and is hinted at by Tatwine, 6 2, ' Nam superas quondam pernix auras penetrabam,' and by Eusebius, 35 3, ' prius asthera celsa vagabar.' 27 8 The passage has given much difficulty. The MS. reading geond speddro- pum, while excellent metrically, does not satisfy the context, which demands a verb, unless we accept the reading of B.-T., p. 906: 'me throughout the bird's joy (the pen) with drops made frequent tracks.' But we cannot accept this, as spyrede must be associated with ofer briinne brerd (1. 9). So we are forced to accept either Grein's geond\sprengde~\ or Holthausen's geond\spdiv. I prefer the former, because it is supported by Life of Guthlac, 7 (Godwin, 44, 13), se awyrgeda gast ~Saes ylcan preostes heortan and ge)>anc mid his searwes attre geondsprengde. It is metrically possible (see Frucht, p. 39, for examples of verse X | ! X | J- £v). — speddropum. B.-T. derives the first member of the com- pound from sped, 'gum,' and renders 'rheumy drops,' while Grein derives from sped, 'success,' and translates the word (Spr. II, 469) 'gutta salutaris.' So Sweet in his Reader, Glossary, 'useful drops' — which is doubtless correct. Brougham translates ' fluent drops.' With the line compare Tatwine, 5 4, ' Frugiferos cul- tor sulcos mox irrigat undis.' The rendering of sped by Grein and Sweet finds interesting support in a later reference to ink in Rid. 88 34, aet )>am spore finde'S sped se )>e sefceft]. 27 9a ofer brunne brerd. Sweet, Reader Glossary, defines brerd as 'border,' 'surface,' and Brougham translates 'across my burnished surface.' But Grein, B.-T., and Trautmann (BB. XIX, 197) agree in interpreting this as 'the dark brim (of the ink-vessel),' and the last-named unhappily compares 52 7, ofer fated gold (see my note to that line). The Inkhorn describes its back as wonn \SS 22), and refers to its swallowing of ink (93 22-23). Compare the thought of 93 26-28. A miniature of St. Dunstan is found in Royal MS. 10 A. 13, and copied by Strutt, Dress and Habits I, pi. 50. The Archbishop is engaged in writing, holding a pen and parchment scraper, with an inkpot fastened at the comer of the desk. In the twelfth miniature of the Benedictional of St. ^Ethelwold (Westwood, Fac- similes, pp. 132 f.) an inkhorn, small and black, is fixed at the top of the arm of the chair; and in the Trinity College (Cambridge) Gospels, No. B. 10, 4 (Westwood, p. 141), the inkpot is also in this position. In the Gospels of Bishop ^thelstan in the library of Pembroke College, Cambridge (Westwood, p. 143), St. Matthew dips a feather pen into a golden inkpot, holding a scraper in his left hand; St. Mark is busy mending his pen, which he holds up to the light and cuts with a large knife ; St. Luke has a pen behind his ear and a knife in his right hand ; St. John writes with a golden pen. NOTES 129 27 9b-ioa beamtelge swealg, etc. Wattenbach, Schriftiuesen, p. 197, cites several mediaeval receipts for the making of ink, notably that of Theophilus in Diversarum Artium Schedula i, 45, 'De Incausto ' (edition of Ilg, Qitellen- schriften fur Kunstgeschichte, vol. vii) : ' Man nehme Rinde von Dornenholz, lege sie in Wasser, um den Farbstoff auszuziehen, trockne die Masse, und wenn man die Dinte brauchen will, mache man sie mit Wein und etwas atramentum iiber Kohlen an.' So we are told by the Inkhorn, Rid. 93 22-23, Nii ic blace swelge wuda ond waetre. Anglo-Saxon ink was evidently made like that of the conti- nent. Ink and parchment are mentioned in Edgar's Canons, § 3 (Thorpe, A.L. II, 244, n) : ' Daet hi habban blase (atramentum) and bocfel.' 27 na sljwide sweartlast. Cf. 52 2-3, swearte waeran lastas, swajni swi|>e blacu. For many Latin analogues, see my note to that passage. 27 nb-i4 The Anglo-Saxon entry at the end of the Durham Book is thus trans- lated by Waring (Prolegomena to Lindisfarne and Rushworth Gospels, part iv, p. xliv): ' Eadfrith, Bishop over the church of Lindisfarne, first wrote this book . . . and ^thelwald, Bishop of Lindisfarne, made an outer cover and adorned it, as he was well able ; and Bilfrith the Anchorite, he wrought the metal work of the ornaments on the outside thereof and decked it with gold and gems, overlaid also with silver and unalloyed metal,' etc. Westwood notes in his Appendix to Facsimiles, etc., (p. 149) that 'the magnificent book-covers " auro argento gem- misque ornata" which are repeatedly mentioned in connection with the fine early copies of the Gospels — such for instance as the Gospels of Lindisfarne — have for the most part long disappeared.' Godwin, English Archaeologist' s Handbook, 1867, p. 87, notes that ' Some of the bindings of these precious volumes display admi- rable metal-work, the Latin gospels of the ninth century being covered with silver plates, and a copy of the Vulgate version of the tenth century being ornamented with copper-gilt plates and having the figure of Christ in the center, the borders studded with large crystals and enameled corners.' Various mediaeval bindings are considered by Wattenbach, Schriftivesen, pp. 324 f. 27 11-12 Book-covers of board and hide are thus introduced by Aldhelm, v, 9 2-3 : Sed pars exterior crescebat caetera silvis : Calceamenta mihi tradebant tergora dura. 27 13 gierede mec mid golde. See the sketch of the Bible, 68 17-18, golde gegierwed . . . since ond seolfre. Cf. also 152, 21 9-10. 27 14 weorc smijm. Cf. 6 8, 21 7, hondweorc smi)>a. 27 15 f. The history of illumination and book decoration in England between 700 and 1066 has been discussed by Westwood in his great work, Facsimiles of Miniatures and Ornaments. In an interesting article upon ' English Illuminated Manuscripts,' Bibliographia I (1895), 129 f., Sir E. Maunde Thomson shows that ' we find two distinct styles — the one having its origin in the North, the other developing m the South. In the North we have the style introduced from Ireland — a style which may be termed almost purely decorative, in which figure-drawing is of so primitive and barbarous a nature that it counts for nothing from the point of view of art, but in which the marvelous interlaced designs and ribbon and spiral patterns combine to produce decorations of the highest merit such as have no 130 RIDDLES OF THE EXETER BOOK rival in other schools of illumination. On the other hand, in the South we have figure-drawing largely and in no small degree successfully cultivated, and at the same time the decorative side of art is not neglected.' In our Riddle it is evi- dently a northern book that is speaking. 27 15 gereno. Geryne, gerene, is found frequently in the poetry in the sense of ' mystery," and that meaning is assigned to the present passage by Grein in Dicht. ; but in Spr. I, 441, he derives our word from geren, 'ornament,' citing Boethius 143, heah )>a gerenu fsegru slen. This rendering, which is supported by the com- mon occurrence of the vb. gerenian 'adorn' (see Beoiv. 777, Maid. 161, etc.), is accepted by B.-T. and Sweet, and is exactly suited to our context. — se reada telg. Gage illustrates the use of red and gold in Anglo-Saxon manuscripts by his description of the ' Benedictional of Athelwold,' Archaeologia XXIV, 23: ' The capital initials, some of which are very large, are uniformly in gold, and the beginnings and endings of some benedictions together with the titles are in gold or red letters. Alternate lines in gold, red or black occur once or twice in the same page. All the chrysographic parts of the Benedictional, as well in the miniatures as in the characters of the text, are executed with leaf gold laid upon size, afterwards burnished.' Gold powder was used as often as gold leaf (see the Muratori receipts cited under 52 7). For the employment of red colors in mediaeval manuscripts, compare Wattenbach, Schriftivesen, pp. 203- 209, 288 f. 27 16 wuldorgesteald. Grein renders the word, Dicht., ' die Wohnungen der Glorie ' ; Spr. II, 748, ' mansiones celestae.' This is hardly apt here. Its present meaning is that of Exod. 587-588, gold and godweb, Josephes gestreon | wera wuldorgesteald. B.-T. renders rightly 'glorious possessions,' and Barnouw (p. 214) ' der heilige und stiitzende inhalt des buches' (see 48 3-5). — maire. Not the adj. 'famous' (Th., Ettm., Dicht.) but 3 pi. opt. of mceran, ' to make known ' (Spr. II. 223). Sweet accepts this interpretation and reads mizren. 27 17 dolwite. The word has greater force than ' Frevelstrafe ' (Dicht.) or 1 punishment for audacity ' (B.-T.). Dol is used in the sense of iniquus (Ps. 1 18 126), and wife often implies ' eternal punishment.' Thus dolwite is opposite to dryht- folca helm (God). The whole passage may be rendered : ' Now may the adorn- ments and the red dye and the glorious possessions widely make known God (in heaven) and not the pains of hell ! ' 27 18 f. With this passage it is interesting to compare the note near the end of the Gospel of St. John (leaf 169) in the Rushworth MS. (Skeat, p. 188): ' haefe nu boc awritne ; bruca mlS willa symle, miS softum gileafa; sibb is eghwaam leofost.' The noble usefulness of the good Book — also the theme of the fragmentary Rid. 68 13 f. — is the text of Tatwine, 5 6, ' sanis victum et laesis prestabo medelam.' But the friendly aid and lofty guidance brought by the Book to men are the themes of many riddles. In Aldhelm, v, 3 7-8, the pen treads a path ' quae non'errantes ad caeli culmina vexit'; and its way is 'the way of life' in Bede's Flares, No. 12, and in the Joco-Seria (Cambridge MS. Gg. V, 35), No. 10 (my article, Mod. Phil. II, 563). The Book is a joyful health-giver (I.G. 241, 329) and has an immortal soul (f.G.jn). In Rid. 50 6-8, books are described : • NOTES golde dyrran, ba sbelingas oft wilniaft, cyningas ond cwene. In 68 13 the Sacred Book is leoda Idreow, bringing to men eternal life. Sal. 237 f. furnishes in its praise of books a very striking parallel : Bee syndon breme, bodiaft geneahhe weotodne willan "Sam 'Se wiht hyge'S. GestrangaiS hie and gesta'Selia'S staflolfaestne ge'Soht, amyrgat? modsefan manna gehwylces of |>reamedlan "Sisses Hfes. Bald bl5 se 15e onbyregefi boca craeftes : symle blS tfe wisra "Se hira geweald hafa'5. Sige hie onsendafi softfaestra gehwam, ha-lo hyife, J>am be hie lufatJ. Wright {Reliquiae Antiquae II, 195) cites incorrectly the clumsy lines in the Benedictional of the tenth century formerly belonging to St. Augustine's at Canterbury (MS. Cott. Claudius A. Ill, f. 29 v.) : Ic eom halgung-boc ; healde hine Dryhten, J>e me fajgere bus frastewum belegde ; pure'S (?) to bance bus het me wyrcean to love ond to wur5e bam }>e leoht gesceop ; gemyndi is he mihta gehwylcre baes ]>e he on foldan gefremian maeg, etc. Another good book, Alfred's translation of the Cura Pastoralis of Gregory, speaks in the first person after the close of the famous Preface (Sweet's ed. E. E. T. Soc., XLV, 8) : ' SrSftan mm on Englisc Alfred kyning awende worda gehwelc ond me his writerum sende suiS ond nor'S.' 27 19 f. Kluge notes (PBB. IX, 436): 'Rid. 27 enthalt neun auf einander fol- gende kurzzeilen die durch suffixreim in einer weise verbunden sind, dass der- selbe sich jedem sofort aufdrangt.' 2721 ferj»e )?y frodran. Cf. Jul. 553, on fefSe frod; Exod. 355, frod on ferh% ; Wand. 90, frod in ferfte; El. 463, frod on fyrhfte; El. 1164, frodne on ferh«e. 27 22 siva-sra ond gesibbra. Cf. 16 22, swSse ond gesibbe ; Gen. 1612, freon- dum swJEsum ond gesibbum. 27 27 to nytte. So in 50 9-10, books serve to nytte . . . ond to dtigfttim. 47 28 glfre. The word gifre, ' useful,' appears only here and in 50 3, where, it is interesting to note, we find it used of books, gifrunt Idcum. Gifre has occa- sioned much discussion. Miiller renders MS. gifre ' utilis,' and Thorpe ' rapa- cious.' Ettmiiller says : l glfre, " rapax" non bene convenit cum mare, "clarus " et hdlig, " sanctus " ' ; and he suggests gifrage, in which he is followed by Rieger and Sweet. Grein compares Rid. 503 and ungifre {Gen. 2470), and translates 'heilsam' (Dickt.) and 'salutaris' (Spr. I, 506); B.-T. renders 'useful.' 132 RIDDLES OF THE EXETER BOOK RIDDLE 28 Dietrich (XI, 467-468) suggested 'Whip'; but afterwards offered (XII, 239) Professor Lange's solution, ' Mead,' which has been accepted by all later scholars. It is certainly a companion-piece to Rid. 12, which pictures the follies of the night- revels, and to Kid. 29, which paints the glories of strong drink. Except in simi- larity of subject, it seems to have little in common with Aldhelm, ii, 3, De Ape; but like vi, 9, De Calice Vitreo, it records the overthrow of topers. In its treat- ment of this motive it resembles very closely the first riddle of the Heifrreks Gdtur, in which ' Beer ' is a lamer of men, and at once a hinderer and provoker of words. Other close analogues are Bern MS. 611, No. 50 (Anth. Lat. I, 366), De Vino, the Wine enigma of Hadrian Junius (Reusner I, 241), and the long riddle of Lorichius (Reusner I, 282) on 'Dolium Vini.' Very like, indeed, is the modem English prob- lem, 'A Barrel of Beer' (Amusing Kiddle Book, 1830, p. 28): My habitation 's in a wood, And I 'm at any one's command. I often do more harm than good : If once I get the upper hand, I never fear a champion's frown ; Stout things I often times have done ; Brave soldiers I can fell them down, I never fear their sword nor gun. After pointing to the existence of bee-culture among all the Aryan peoples, Weinhold adds, Altnordisches Leben, 1856, p. 89: ' Honig war fur das Alterthum wichtiger als fur uns, denn er gibt den Haupttheil zum Met, dem uralten Lieblings- getrank arischer Volker. Met ist darum auch das Getrank der Gotter ; aus Honig und Blut mischten ferner die Zwerge den Trank, welcher die Gabe der Dicht- kunst verleiht ' (compare Wackernagel, Haupts Zs. VI, 261). Hehn, Kp.u. fft., 1902, pp. 152-154, traces the history of mead. 'In the linden forests of the east of Europe, among the nomads and half-nomads of the Volga region quite at the back of the Slavs, the intoxicating drink made of honey played a greater part than beer, and was certainly much older. It may be presumed that mead was a primitive drink of the Indo-Europeans when they migrated into Europe, and that it only, like so many other things, lasted longer in the east of the continent. . . . The Taulantians, an Illyrian people, made wine from honey. Says Aristotle, de Mirab. auscult. 22 (21): "When the honey is squeezed out of the combs (besides other processes), an agreeable strong drink like wine is produced." . . . Mead is further distinguished as a Scythian beverage, made from the honey of wild bees, etc.' For the Anglo-Saxon use of both mead and 'beer, see my discussion of the next riddle. 28 2 brungen of bearwum. Cf . 22 7, brungen of bearwe. In the Horn riddle, 806, the mead is again mentioned, /<?/ on bearwe geweox. — burghleojmm. The reading of Th., Ettm., beorghleof>um, is tempting because ' mountain heights ' seems well suited to the sense of the passage, and is moreover supported by 58 2, beorg- hleoj>a. But there is no real reason for abandoning the MS. word, which is found Gen. 2159, Exod. 70, and which is rendered by Brooke 'city-heights.' NOTES 133 28 3-5 This reminds us of the work of the wings in the Swan riddle (8 3). 28 3-4 Weinhold, Altnordisches Leben, 1856, pp. 88-90, discusses bee-culture among the North Germans. Cortelyou, Die altengliscken Namen der Insekten, 1906, pp. 25 f., notes the frequent appearance of the bee in Anglo-Saxon writings. Asser, Life of Alfred, chap. 76, employs the phrase ' velut apis prudentissima,' which furnishes his editor, Stevenson (Oxford, 1904, p. 302), the opportunity to con- sider the use of the metaphor in Aldhelm (De Laudibiis Virginitatis, cap. iv), Alcuin (Vita S. Willibrordi, cap. 4), Regularis Concordia Monachorum (Cartularium Saxonicum III, 423, 2), and in many other writers of the eighth to tenth centuries. Aldhelm tells us in his enigma De Ape (ii, 32):' Dulcia florigeris onero praecordia prasdis,' and again in the De Pugillaribus (v, 91):' Melligeris apibus mea prima processit origo.' Of the connection between the bees and mead, the Celtic bard speaks in his famous ' Mead Song' (Myvyrian Archaeology of Wales, 1801, I, 22): From the mead horns — the foaming, pure and shining liquor, Which the bees provide, but do not enjoy ; Mead distilled I praise. ' Apparently of first importance was the keeper of the bees, " apium custos," "apiarius," "melitarius" [WW. 256,8; 352, 13; beo-ceorl~\, for the maintenance of bees was of sufficient importance to call for the employment of a man for that special work. . . . [His rights and duties are stated at length, R. S. P., § 5, Schmid, p. 376.] In the Gere/a (Anglia IX, 263) we find mention of the accompaniments of this industry, bee-hives and honey-bins. Bee-culture reached, to all appearances, a high state of cultivation among the Anglo-Saxons and was held in peculiar regard by the people as the chief element in a favorite drink. Returns of bee-hives are frequent in Domesday,' etc. (Andrews, Old English Manor, p. 206). In Alfred's Laws, § 9, 2 (Schmid, p. 76), the bee-thief is punished as severely as he who steals gold or horses. 28 6-17 The Mead's chant of triumph over those who contend against its force recalls Rid. 12 3 f. The genre sketch of the downfall of the old churl may or may not have been suggested by Aldhelm, vi, 9 9, ' Atque pedum gressus titu- bantes sterno ruina'; but this motive appears in genuine folk-riddles (sttpra) remote from learned sources. The grimly humorous picture of the evils of de- bauch should be contrasted with the praise of the joys of wine in the next riddle (297-12). The mead-hall is mentioned elsewhere in the Riddles (15 n, 16, 21 n, 561, 5712, 643). Sharon Turner, History of the Anglo-Saxons, Bk. VII, chap, iv, translates an Anglo-Saxon canon against drunkenness : ' This is drunkenness, when the state of the mind is changed, the tongue stammers, the eyes are disturbed, the head is giddy, the belly is swelled and the pain follows ' (Theodore, Liber Pa-nilentialis, xxvi, 14, Thorpe, A. L., p. 292). Gummere, Germanic Origins, pp. 74-75, notes that all these Anglo-Saxon laws (Schmid, p. 12, §§ 12, 13, 14, pp. 24, 212) 'testify to the Germanic habit of drinking, quarreling, and fighting, with quarreling proper as a vanishing element in the situation.' With our riddle it is interesting to com- pare such pictures of potent potting as the description of the feast of Holofernes (Judith, 15 f.) and the lot of the drunkard in the Fates of Men, 48 f. : 134 RIDDLES OF THE EXETER BOOK l>onne he gemet ne con gemearcian his mu'Se, mode sine. (See Brooke's translation, E. E. Lit., p. 1 53). The poet of Juliana, 483 f ., makes the devil say that one of his ways of working evil is by leading men drunk with beer into the renewal of old grudges and to such enmity that in the wine-hall they perish by the sword-stroke. For another picture of drunkenness in the Riddles, see 12 5 f. Ebert, Allgemeine Gesch. der Lit. des Mittelalters I, 613, III, 2, remarks that the poets sometimes seem to hold up the drunken characters of the Old Tes- tament as warning examples to their Anglo-Saxon audience ; compare Gen. 1 562 f ., 2408, 2579, 2634, 2640 (Ferrell, Teutonic Antiquities in the Anglo-Saxon Genesis, 1893, pp. 42-43). See Fuchse, Sitten beim Essen und Trinken, 1891, pp. 7-8; and notice the many warnings against drunkenness in the Hdvamdl. 28 7-8 weorpe | esne. This emendation of Holthausen, Engl. Stud. XXXVII, 207, for MS. weorpere efne finds threefold justification in the meter of 7 b, in the absence of efne elsewhere in the desired sense of ' I level,' ' I throw,' and finally in its perfect adaptation to the context (cf. i6b, esnas binde). 28 9 In not a few of the riddles a meeting with the subject leads to sorrow — compare 7 8, 16 25, 18 10, 24 10 f., 26 9-10 (Dietrich XII, 245). 28 10 For MS. magent>isan, Holthausen, Engl. Stud. XXXVII, 207, reads l-J>issan = fryssan (nom. J>ysse < J>usjd) ' ; and adds: 'Das wort gehb'rt zu -f>yssa, «' toser " in brim-, mere-, loceter-fryssa, " schiff ." ' Grein, Spr. II, 220, suggests that J>tse is identical with J>yse ? (O. N. J>yss, J>eysa) ; and then conjectures very doubt- fully mizgenwisan, ' meiner kraftigen Weise.' But there is no reason to depart from the MS., as the form brim-J>isan is found three times (And. 1657, 1699, El. 238). 28 12 Cf. 12 10, gif hi unriedes aer ne geswlcaj>;y«</. i2o,El. 516 (Herzfeld,p. 19). 2813 strengo bistolen. Cf. 12 6b, mode bestolene ; Gen. 1 579, ferh'Se forstolen (drunken Noah). — strong on spraice. This reading of MS. and earlier editors is sustained by 93 10, strong on st&pe, and by such descriptions of drunkenness as those cited above (see Fates, 48-57). Compare also 29 11-12, deman onginne'S, meldan misllce. Barnouwsays (p. 221): '•Strong on sprace gibt viel besseren sinn; der betrunkene hat seine kraft verloren ; ist nur noch in worten stark.' 28 14 maegene binumen. Cf. 27 2, woruldstrenga binom. 28 15 fota ne folma. Cf. 32 7, fet ond folme ; 40 10, fot ne folm; 689, fet ne f[olme]. 28 17 be dseges leohte. Budde, Die Bedeutung der Trinksitten, p. 24, believes that this phrase refers to the results of the evening potations the morning after, and cites in support of this view the ' Proverbs of Alfred,' xv (Kemble, Salomon and Saturn, p. 234) : His morge sclep Sal ben muchil lestin ; Werse t>e swo on even Yfele haued ydronken. The thought is parallel to that in the riddle's mate, 12 9. ' So sind wir wohl berechtigt einen Einfluss der volksmassigen Trinkanschauungen auf das Ratsel anzunehmen' (Budde). NOTES 135 RIDDLE 29 Wright (Biog. Brit. Lit. I, 79) early suggested 'John Barleycorn,' and pointed to the parallels in Burns's famous poem, which, it may be noted, is a product of folk-poetry, as the seventeenth-century black-letter ballad ' The Bloody Murther of Sir John Barleycorne ' (Ashton, Chap-Books of the Eighteenth Century, pp. 316- 318) shows. This solution was accepted by Klipstein, and ably defended by Brooke (£. £. Lit., p. 152). On account of the early lines (1-3), Dietrich (XI, 468) proposes ' Weinfass,' which is certainly better than Trautmann's ' Harfe.' Wright's answer, which we may modify to Beer or Ale, seems to me distinctly the best, as the riming lines describe the threshing of the barley. To sustain his solution Dietrich points to Aldhelm, vii, 2, De Cuppa Vinaria, as a possible source (infra). I shall note other analogues in my comments upon single lines. Prehn has indicated (p. 197) the very slight likeness between the fate of the subject of this riddle and that of the Battering Ram (Kid. 54) and of the Lance (Kid. 73). But Kid. 29 is most closely connected with Rid. 28, ' Mead,' in its de- tailed description of the origin of the drink — here barley instead of honey — and of its effects upon man, here good and joyous rather than bad. As Brooke says, E. E. Lit., p. 152, 'the delight and inspiration which the writer places in "jolly good ale and old" only makes his reproof of excess seem the stronger.' We find the same mingling of approval and rebuke of mead in the Hdvamdl. Hehn, Kp. u. Ht.t 1902, pp. 149-159, declares : ' Caesar does not speak of beer as a German drink, but a century and a half later Tacitus does (German/a, 23, " Potui humor ex hordeo aut frumento in quandam similitudinem vini corruptus") ; though Pliny, when he mentions beer, is silent as to the Germans. These when pressing forward to the lower Rhine and the sources of the Danube must have soon adopted the use of beer from the Celts. ... It is foolish to regard beer and beer-drinking as originally German and inseparable from the essence and idea of Germanism ; if the use and brewing of beer had been the ruling characteristic custom of the Ger- mans the ancients would not have been so chary of mentioning it.' Hehn further points out that ' the nearest neighbors of the Germans, the Prussians, drank only mead and fermented mare's milk and were ignorant of beer, which allows us to make certain inferences as to the Germans in the earlier stages of their civiliza- tion.' Later, in discussing hops (p. 473), Hehn shows that the ancients had never heard of such a plant ; that accounts of the early Middle Ages, which often men- tion beer, never say a word about hops ; and that in many European countries like England and Sweden the use of hops for making beer is first heard of towards the end of the Middle Ages or even in the course of the sixteenth century, and then gradually becomes more common. For the introduction of hops into the Norwegian countries during the Middle Ages, see Hoops, Wb. u. A'p., pp. 649-650. See also Gummere, Germanic Origins, pp. 71-74. Hoops declares (Wb. u. Kp., p. 380) that barley has one advantage over wheat : that it has always been an indispensable ingredient for beer. He points out the fondness of Northern England for barley (p-59i): ' Mbglicherweise nahm im Siiden des Landes schon in angelsachsischer Zeit der Weizenbau die vornehmste 136 RIDDLES OF THE EXETER BOOK Stelle ein ; im Norden scheint aber die Gerste als das ertragssichere Korn, wie friiher auf dem Festland, die erste Rolle gespielt zu haben. Es ist bezeichnend dass die Dreschtenne im Northumbrischen und Mercischen bereflor heisst (vgl. Lindisfame und Rushworth, Matth. iii, 12, Luke iii, 17, wo es lat area iibersetzt), •wahrend im Sachsischen dafiir byrscdflor oder auch bernesflor gilt.' It is this threshing of barley that our riddle describes. A grant of King Offa (Birch, Cartularium Saxonicum, 1885, I, 380) mentions ' twa tunnan fulle hlutres aloft ond cumb f ulne HJ>es aloft ond cumb fulne welisces aloft.' From this Sharon Turner, VII, chap, iv, infers that three kinds of ale were known to the Anglo-Saxons: (i) clear ale; (2) Welsh ale; (3) mild ale. According to Weinhold (Altnordisches Leben, 1856, p. 153, note), '•Ol [ags. ealu\ und bidr [ags. beor~\ sind gleichbedeutend ; ol ist alter, und den Nordgermanen mit den Lithauern gemein ; bidr ist erst durch Zusammenziehung aus dem lat. infin. bibere entstanden (Grimm, Worterbuch, s. v.). Als jiingeres und fremdes Wort gait es fur vornehmer und deshalb sagt das junge Alvissmdl (35), ol heisst der Trank unter den Menschen, bi6r unter den Gottern.' In this identification of ale and beer, and in the derivation of the name, Weinhold is at one with Wacker- nagel, who in a scholarly article (' Mete, Bier, Win, Lit, Lutertranc,' Haupts Zs. VI, 261) traces the history of Germanic liquors from the early time when beer and mead were the only drinks of the northern nations. Compare Weinhold, Deutsche Franen, 1882, II, 62; Sass, Deutsches Leben zur Zeit der Sachsischen Kaiser, Berlin, 1892, p. 24; French, Nineteen Centuries of Drink in England, London, 1884, p. 14. Leo, R. S. P., 1842, p. 200, believes that ealu and bear were different, because he meets the words aloft and bear side by side as separate grants in a charter (Kemble II, in), and suggests that there was doubtless the same distinction that we find in modern England between ale and beer, the first being .with hops ; but Leo naturally fails to find any trace of ' hopfenbau ' among Anglo-Saxons. When the boy in ^Elfric's Colloqtiy (WW. 102) is asked what he drinks, he answers : ' Ale if I have it, or water if I have not.' And he adds : ' I am not so rich that I can buy me wine, and wine is not the drink of children or the weak- minded, but of the elders and the wise.' As Newman points out (Traill's Social England I, 226), 'Wine though made, was little drunk; wine-presses are shown in the illuminations [Cotton Claudius B. IV, f. 17], but the climate must have restricted the growth of the grape to the southern portion of the island. At all events, mead and ale were the popular beverages.' '£)ier hy meodu drincafr, says Kid. 21 12. The brewery, breawarn or mealthiis ('Brationarium'), was an important adjunct of every Anglo-Saxon menage (Heyne, Die Halle Heorot, p. 26). 29 i The opening line is an integral part of the riddle (with 29 i,fagre gegier- •wed, cf. 21 2), not as in Rid. 32 and 33 a mere excrescence. This beginning bears a far-away likeness to that of 71. Dietrich (XI, 468) finds a suggestion of these lines in Aldhelm's enigma of the Wine-Cup, vii, 2 8-10: Proles sum terrae gliscens in saltibus altis. Materiam cuneis findit sed cultor agrestis, Pinos evertens altas et robora ferro. NOTES 137 29 2-3 Compare with these superlatives, heardestan, scearpestan, grymmestan, the lines of the Barleycorn ballad : The sultry suns of Summer came, And he grew thick and strong ; His head weel arm'd wi' pointed spears, That no one should him wrong. 292 Grein suggested \Jieoru~\scearpcstan for the sake of alliteration, which is otherwise absent from the line ; but Kluge has shown (PBB. IX, 446) that this lack of alliteration is compensated for by suffix-rimes, as later in Middle English. With our line he compares Maid. 271, iefre embe stunde he sealde sume -wunde; the inscription upon the shield of Eadwen (Hickes's Thesaurus), drihten hint dwerie J>e me hine atferie; and the passage upon William in the Laud MS. of the Chronicle (Earle, p. 222). 29 4-7 So in the Barleycorn ballad, which I may not quote at length, the barley is 'cut by the knee,' 'tied fast,' 'cudgeled full sore,' 'hung up,' 'turned o'er and o'er,' 'heaved in a pit of water,' 'tossed to and fro,' 'wasted o'er a scorching flame,' ' crushed between two stones,' and finally, almost in the words of the Anglo-Saxon, They hae ta'en his very heart's blood, And drank it round and round ; And still the more and more they drank Their joy did more abound. The ' Barleycorn ' undergoes the same sad experiences as the ' Pipping pounded into Cyder' of the Whetstone for Dull Wits, p. i (Ashton, Chap-Books, p. 296): Into this world I came hanging; And, when from the same I was ganging, I was cruelly battered and squeezed, An J men with my blood they were pleased. 29 4-8 The rimes, 'vhich give Rid. 29 an interesting place in our group (see Kluge, I.e.; Lefevre, Anglia VI, 237), have their parallel elsewhere in riddle poetry. Very similar is their use in the Mecklenburg ' Flax ' problem (Wossidlo, 77): ' Dann ward ich geruckelt und gezuckelt und geschlagen ; dann brachen sie mir die knochen ; se hoogten mi, se toogten mi; se bogen mi, se schowen mi; . . . se riippeln mi, se kniippeln mi ; se ruffeln mi, se knuff eln mi ; se ruppten mi, se schuppten mi ; se ruckten mi, se tuckten mi ; se zucken un tucken mi.' 29 7-10 Dietrich notes the general likeness of the passage to Aldhelm's line (vii, 2 i), 'Tin plures debrians impendo pocula Bacchi.' Line 9 recalls the 'old churl,' 28 .'i. The dream due to beer is similarly described. Fates, 77 f.: Sum sceal on heape hxle'Sum cweman, blissian zt beore bencsittendum ; )>zer blS drincendra dream se micla. Cf. Bemj. 495 : pegn nytte beheold se >e on handa baer hroden ealows-ge, scencte sclr wered >xr waes hxle'Sa dream. I38 RIDDLES OF THE EXETER BOOK 29 8 clengefl. The word has been variously interpreted. Thorpe's conjecture glengefr (Rime-song, 3, 12; Ph. 606) is barred by the demands of alliteration. It is equally impossible to regard clengefr as subst. ace. (Dicht., 'den Jubel ' ; Brooke, p. 153, 'jollity'). The form is the jd pers. sg. ind. of clengan, doubtfully denned by Grein (Spr. I, 163) as 'omare' (cf. glengati) and by B.-T. (p. 158) 'to exhila- rate.' The proper meaning is given, however, by B.-T. Supplement, p. 128, 'to adhere, remain.' This rendering is confirmed by instances of the word in this sense in fourteenth-century English (cf . N. E. D. s. v. clenge). The verb is thus closely related to clingan. 29 10-12 Does no wifr spricefr refer to the old men of 29 9 (Brooke, ' and they abuse it not ') or to the barleycorn (Dicht., ' und nicht dawider sprichts ') ? I pre- fer the former, as it emphasizes the contrast between the lot of these happy men who do not contradict and quarrel and the fate of the foolish wights, ' strong in speech,' in the preceding riddle. The two following lines (11-12) are thus ren- dered by Dr. Bright : ' And then after death (i.e. drunken sleep), they indulge in large discourse and talk incoherently.' The construction of the passage favors this rendering. Perhaps the subject of the riddle (' Barleycorn ') is the subject of the clause. Then after dea^e is suggested naturally by its fate in the early lines of the poem (294-6); and its 'copious speech' (tneldan mislice) brought to the riddler's thought by the familiar personification ' Wine is a mocker, strong drink a brawler' (Proverbs xx, i). 29 11-12 See note to 28 13, strong on sprace. So we are told, Mod. 18 f. : J>onne win hweteiS beornes breostsefan, breahtem stlge'5, cirm on cortSre, cwidescral letaS missenllce. 29 12-13 Cf. the close of Rid. 3223-24. See also Gu. "03, micel is to secgan; And. 1481, micel is to secganne (Herzfeld, p. 19). RIDDLE 30 Dietrich (XI, 468-469), Prehn (pp. 198-199), and Brooke (E. E. Lit., pp. 154-155) agree upon the answer ' M_opn and Sun.' Though Prehn has failed utterly to es- tablish any connection between Rid. 30 and Eusebius 1 1, De Lttna (where the two luminaries are not hostile, but brother and sister), and though Day ard Night in riddle-literature are usually friendly (Reusner I, 174, 200, II, 68; Ohlert, pp. 69, 127 ; Wiinsche, Kochs Zs. IX, 449-461), yet analogues are not wanting. As I have pointed out (M.L.N. XVIII, 104), Flares, No. 6, tells us that Day flees before Night, that the resting-place of Day is the Sun and of Night a cloud (compare Disputatio Pippini cum Albino, 54 ; Altercatio Hadriani et Epicteti, 55). In a Ger- man riddle (Simrock8, p. 12), which has something in common with the fifth of Schiller's ' Parabeln und Ratsel,' Day says of his sister Night : ' Du jagst mich, und ich jage dich.' Dietrich's solution is, moreover, strongly supported by the close likeness between the last lines of our riddle, ' Nor did any one of men know NOTES 139 afterwards the wandering of that wight', and the words of the Moon, Bern MS. 611, No. 59 (Anth. Lat. I, 369) : Quo movear gressu nullus cognoscere tentat, Cernere nee vultus per diem signa valebit. The exquisite myth in Rid. 30 challenges comparison with the Vedic poems on the powers of nature (Rigveda I, 113, 123 ; Haug, pp. 464 f.). Let us see how the early myth-maker weaves his story of elemental strife. The very ancient attitude towards the two great lights of heaven is seen in the deservedly famous Ossianic 'Address to the Sun' (Clerk's Translation, 1870, I, 221): O Sun! Thou comest forth strong in thy beauty. The Moon, all pale, forsakes the sky To hide herself in the western wave. Thou in thy journey art alone. The Moon is lost aloft in the heaven; Thou alone dost triumph evermore In gladness of light, all thine own. As I have pointed out (M.L.N. XXI, 102), here are the chief motives of our riddle : the contest between the bodies, the loss of the Moon's light, and the triumpji of the Sun. I repeat my detailed interpretation of Rid. 30. The Moon is seen bearing between his horns as booty a bright air-vessel which is the light captured from the Sun in battle (4, huj>e . . . of J>dm heresij>e). He would build himself a bower or tabernacle (biir = tabernaculum, Spr. I, 1 50) in the burg and set it skillfully, if it so might be (see Psalms xix, 4, ' In them hath he set a tab- ernacle for the sun '). Then the wonderful being, known to all men on earth, the Sun herself, appeared in the heaven (7 b, ofer wealles hrof), snatched from the Moon his booty, the light, and drove away the wretched wanderer (so in Ossian, 'the Moon, all pale, forsakes the sky'). Then, hastening with vengeance on her journey, she fared towards the west (Wonders of Creation, 68, gewitefr J>onne mid J>y wuldre on westrodor). (At this coming of the Sun,) dust rose to heaven (prob- ably raised by the cool wind that, in early Germanic poetry, blows at the rush of day; see Grimm's Teutonic Mythology, 745, 1518), dew fell on the earth, night departed. Nor did any one of men know afterward the journey of the Moon. Rid. 30 and 95 — which I interpret 'Moon — have three motufes in common: these are the fame of the subject among earth-dwellers, its Capture of booty in its proud hour, and its later disappearance from the sight of men. And, as Miiller points out (C.P., p. 17), the riddle recalls a passage in the De Temporibus: 4 SSblice se mona ond ealle steorran underfoS leoht of ) iere micclan sunnan,' etc. Trautmann abandons his earlier answer (Anglia, Bb. V, 49), ' Swallow and Sparrow,' in favor of this prosaic interpretation (fiff.~X.IX, 191): 'The wonder- ful wight who bears booty, an air-vessel between his horns, is a bird carrying a feather in his beak. He seeks to build his nest, but the wind comes, snatches the 140 RIDDLES OF THE EXETER BOOK feather out of his mouth and drives the wretched creature home ; it then blows westward, because w is needed for the alliteration.' Walz's solution ' Cloud and Wind ' (Harvard Studies VI, 264) is far more pleasing and suitable ; but I do not believe that this is as well adapted to the sense of the poem as Dietrich's ' Moon and Sun." 301-3 Trautmann renders (p. 191): 'Dieses wesen (ein vogel) fiihrt zwischen seinen hbrnern (dem ober- und unterkiefer seines schnabels) beute. Die beute ist ein leichtes und kunstvoll bereitetes luftgefass (ein gras- oder strohhalm oder eine feder).' I register twofold objection : first, that in spite of the well-known word hyrnednebba the upper and lower parts of the beak would not in any flight of fancy be called ' the bird's horns ' ; and, secondly, that neither a blade of grass nor a feather would be termed an air-vessel on account of its hollowness (see note to line 3). 30 2a hornum bitweonum. Dietrich (XI, 468-469) points to Aldhelm's de- scription of the Moon as 'bicornea' (Epistola ad Acircium, Giles, p. 225). This doubtless goes back to the 'bicornis Luna' of Horace (Carmen Saeculare, 35). 30 2b, 4a hiipe. This corresponds to the hifcendra hyht of 95 sa. I do not be- lieve with Dietrich that the word refers to the loss of the Sun's light in an eclipse, but with Miiller (C. /Vp. 17) that the riddler has in mind the ordinary changes of day and night. See the passage cited from the De Temporibus. With huj>e Icedan cf. Gen. 2149, hufre la>dan, Gu. 102, hiifre geladed. 30 3a lyftfaet leohtllc. Cf. Ps. 135 7-8: He leohtfatu leodum ana micel geworhte manna bearnum. Here leohtfatu are the luminaries, the Sun and the Moon. The Psalter passage is a strong argument for our solution. 30 sa walde hyre on baere byrig. Herzfeld, p. 50, notes that this half-line is doubtful, and suggests as a possible reading for byrig the older form burge [cf. 216, where meter demands seecce for MS. safe]; but he points to Dan. 192 a, Niah J>e J>a?r on byrig (MS., Gn., W. herige does not satisfy ^-alliteration), and to Sievers's examples of the shortening of the last foot of A-type to ^ x (PBB. X, 289). Holthausen's emendations (Engl. Stud. XXXVII, 208), cited among vari- ants, distort the grammatical order. I have allowed the MS. reading to stand; cf. Gen. 2406 a, ic on )>isse byrig. With on J>£re byrig ai. 95 6a, in burgum ; 60 14-15, Codes ealdorburg . . . rodera ceastre. As Brooke renders (p. 154): 'The Moon would build his hall in the very citadel of Heaven.' In Chr. 530, on burgtim is equivalent to in caelo. 30 6h gif hit swa meahte. Cf. Beow. 2091, And. 1393, hit ne mihte swa ; 1323, tynden hit meahte swa. For other examples of omission of infinitive, see Spr. II, 268 ; Sievers, Anglia XIII, 2. 30 7b ofer wealles hrof. Of this Heyne says (Halle ffeorot, p. 14): 'Ob der Ausdruck wealles hrof dagegen mit Grein nur " Gipfel des Walles " zu ubersetzen sei und eine hohe Mauer kennzeichnen sollte ist uns zweifelhaft, denn, wenn im Supplement zu jElfrics Glossar parietmae glossiert werden rof lease and monlease ealde -weallas, so denkt sich der Glossator offenbar Mauern, deren Zinnen zugleich NOTES 141 mit der Besatzung dahinter verschwunden sind.' We meet the phrase ofer wealles hrof in Psalms (Thorpe), 54 g, where it translates the Vulgate super muros. Grein, Dicht., translates 'iiberdes Walles Gipfel ' ; B.-T., p. 1174, 'over the mountain top'; and Brooke, 'over the horizon's wall.' The phrase may have a very general meaning here, as one should say 'over the housetops'; but compare Browning's ' And the sun looked over the mountain's rim.' 30 8a cutf. Miiller (C. P., p. 17) renders 'gewiss mehr "amicus" als "notus,"' and compares description of Sun, Wonders of Creation, 63, ivlitig ond ivynsttm -wera cneorissum, and Aldhelm's enigma De Nocte, xii, p. 270: 'die lampas Titania Phoebi — quae cunctis constat arnica.' But the closest parallel is found in the first lines of Rid. 95. 30 9a ahredde ]?a )>a hu]?e. Cf. Gen. 2113, hu'Se ahreddan. 30 nb for<5 onette. For many examples of the phrase, see Spr. II, 343. 30 is3 niht for?f gewat is rendered by Grein, Brooke, and Trautmann, ' night came on.' There is not the least warrant for this rendering ; and Miiller, C. P., p. 17, rightly translates 'die Nacht schwand dahin.' When forfr gewat appears elsewhere in like context, it means in each case ' departed ' or ' began to depart ' : Luke ix, 1 2, gewat se daeg f orS (' dies coeperat declinare '), Gen. 2447, forS gewat iefenscima. Compare with our passage Ph. 98-99, on daegred, ond seo decree niht | won gewiteS. Lines 12-13 are a short but vivid description of the dagredwdma (Krapp, note to And. 125). 30 13-14 Walz and Trautmann seek to sustain their interpretation ' Wind ' by reference to John iii, 8, ' The wind bloweth where it listeth, and thou hearest the sound thereof, but canst not tell whence it cometh, and whither it goeth.' But ' the disappearance of the moon ' is found not only in Latin enigmas (supra), but at the close of our riddle's mate, Rid. 95. RIDDLE 31 Dietrich (XI, 469) offered the plausible solution ' Rain- Water.' ' This is always ready to run (3 a), is disturbed by fire (3 b), and is collected in the air (2 b).' Ac- cording to Dietrich, 315-6 refer to the washing before the meal, and 31 7-9 to the 'Taufwasser' (cf. 8438, firene dwiesce'S). Prehn, pp. 199-201, follows Dietrich's interpretation, and seeks to trace the chief motive of the problem to Symphosius 9, the strife with fire to Eusebius 15, and the 'blooming grove to Aldhelm i, 3, ' sed madidis mundum faciam frondescere guttis.' While the association of water and fire in a storm-cloud may well explain the opening lines, which have much in common with Water riddles of folk-literature (M.L.Ar. XVIII, 100, note 19), the fourth line, bearu blmvende, byrnende gled, presents a serious obsta- cle to this solution. Prehn regards this as a pleonasm, completing the thought of the preceding line : Vom Feuer beunruhigt, Wenn Glut den bliihenden Hain sengt. But the grammatical construction does not permit this reading, and we are forced to the conclusion that these nominatives merely represent certain phases of the 142 RIDDLES OF THE EXETER BOOK subject, which in such case can hardly be Water. Trautmann, Anglia, Bb. V, 47, suggested the answer 'Das Ahrenfeld,' but he later (BB. XIX, 213 f.) abandoned this in favor of the solution of Blackburn (Journal of Germanic Philology III, 1900, p. 4), which is thus presented. ' The true solution, I think, is dn beam in the various senses that the word carries in Old English, tree, log, ship, and cross (probably also harp and bowl).' Blackburn translates as follows: I am agile of body, I sport with the breeze ; (tree) I am clothed with beauty, a comrade of the storm ; (tree) I am bound on a journey, consumed by fire ; (ship, tree) A blooming grove, a burning gleed. (tree, log) Full often comrades pass me from hand to hand, ( harp) Where stately men and women kiss me. (cup ?) (When I rise up, before me bow The proud with reverence. Thus it is my part To increase for many the growth of happiness, (the cross) Trautmann accepts the answer beam, but rejects the meanings ship, harp, and cup, believing that the first four lines refer to the ' tree ' in the forest, the last five to the ' cross.' Later in his BB. article, he proposes, at the suggestion of his col- league Professor Schrors, the ' osculatorium ' or ' instrumentum pacis ' or ' stabar- tiges kiissgerat ' ; but this has nothing in its favor ; indeed, the thing is not heard of until five centuries later. Blackburn's solution invites the support of parallel passages. The opening lines of Rid. 54 picture the tree in the forest : Ic seah on bearwe beam hlifian tanum torhtne ; baet treow WEBS on wynne, wudu weaxende. And Rid. 56 describes the beam as the rood of Christ. That /us forfrweges (3 a) refers to ' the ship,' seems to me likely in the light of the association of ' tree ' and 'ship,' not only in many folk-riddles (Wossidlo, No. 78, note) but in the Runic Poem, 77-79: Ac byb on eorban elda bearnum fljesces fodor, fereb gelome ofer ganotes baeb. Compare also the use of wudu as 'ship,' Rid. 4 24. Although 316 recalls the kissing of horn or of beaker in the other riddles (Rid. 153, 64 4), the use of beam in the sense of ' cup ' is not elsewhere found ; and the supposed reference to a drinking-vessel seems more than doubtful. In spite of the well-known word gleow- beam, I am inclined to think that we have no reference to the harp in 315, but that the last five lines of the poem refer to ' the cross ' — if we accept Blackburn's interpretation of the enigma, rather than Dietrich's. 31 i There are three strong arguments for legbysig, as opposed to lie bysig or Rcbysig: it is the reading of both versions (a leg; b lig) ; it accords with lace (i b), as Idcende leg or Itg appears frequently in the poetry (Dan. 476, Chr. 1594, El. 580, i in); and, as Holthausen points out (Bb. IX, 357-358), it is in harmony with 3b, fyre gebysgad (b gemylted), and 4 b, byrnende gled. NOTES 143 The elemental character of the first lines of the poem seems admirably adapted to the solution ' Rain-Cloud charged with fire' (see Pliny's account of Water, Nat. Hist. bk. xxxi, chap, i, cited M. L. N. XVIII, 100) ; but the grammatical difficulty in 31 4 is unfortunately insuperable (supra). Grein and Trautmann render Kcbysig ' geschaftiges leibes ' ; and Blackburn, ' agile of body.' Dr. Bright favors this reading. 31 2 bewunden mid wuldre. This phrase may well be applied to fire (leg) : cf. Beow. 3146-3147, swogende leg | wope bewunden. 31 3 fus forSweges. Cf. Exod. 248, fus forftwegas. For many examples of the genitive construction with fus, see Shipley, p. 75. — b fyre gemylted. Cf. El. 1312, Jmrh fyr gemylted. — a fyre gebysgad. Water is described as lyfte gebys- gad (Ph. 62). 31 4 beam blowende. In Rid. 2 8-9 the wind shakes the wood, bearwas bledhwate. Cf. And. 1448, geblowene bearwas. The phrase suggests a line of the ' Aqua ' riddle (Brussels MS. 604 d, twelfth cent. ; Mone, Am. VII, 40) : ' Nemus exalo, rideo pratis.' In accord with the ' Water ' solution is Ph. 65-67, waeter wynsumu . . . bearo ealne geondfaraft. 317 onhaebbe. Grein, Spr. II, 346, derives from onhabban, ' abstinere' (hapax), and translates 'mich fern halte, abwesend bin,' in Die/it, 'mich enthebe ' (so Trautmann). B.-T., p. 754, on the other hand, derives from the frequent on- hebban, 'raise, lift up,' which is the meaning accepted by Blackburn (supra). As the form ha;bbe for hebbe appears, Psalms (Thorpe), 24 i, as onhebban is of common occurrence, and as the context favors it rather than the unmeaning ' withdraws,' I follow B.-T. 31 8 a mid miltse; b mil (sum. Grein, Spr. II, 251, renders in this place ' hilaritas,' ' laetitia' (?) but, as Trautmann points out (BB. XIX, 214), the examples which he offers support rather the meaning 'Demut' (cf. Az. 118, 146, 154, And. 544, miltsum). B.-T. gives very doubtfully the definition ' humility ' (?) for the Azarias passages. All the citations favor the reading of the £-text. RIDDLE 32 Dietrich (XI, 469) regards ' the rare singing thing ' of this riddle with ' a voice in its foot and two brothers on the neck ' as the Bagpjpe — swegelhorn ('sam- bucus,' WW. 44, 37 ; ' simfonia,' id. 483, 17, Hpt. Gl. 445, 19) — with the two flutes at the lower end of the hollow-sounding bag. He adds : ' If the mouthpiece of horn swells up the head and body of the bag which is embraced by the arm of the player, while the fingers rest upon the flutes, which run into the neck of the bag, then the thing possesses at every point a complete likeness to a bird, that touches with his beak the mouth of the blower' (cf. 1. 7, fet ond folme fugele gelice). The swegelhorn or 'sambucus ' is regarded by Padelford (pp. 35, 102) as a stringed instrument; for in MS. Tib. C. VI the sambuca is represented as 'an odd pear-shaped instrument of four strings,' and in Hpt. Gl. 445, 21 it is a synonym of 'cithara.' While Padelford accepts (p. 50) the Bagpipe solution, he finds its ancient equivalent not in the swegelhorn or ' sambucus ' but in the Latin ' musa,' 144 RIDDLES OF THE EXETER BOOK 'camena,' and 'chorus.' 'Musa' is glossed by pipe ofrfre hwistle (WW. 311, 22) and ' camena ' by sangplpe (Prudentius Gl. 389, 26). ' The chorus is the usual name for the bagpipe among the church writers. In the Boulogne and Tiberius MSS. are drawings of the chorus (Strutt, Horda, pi. xxi). . . . These instruments are con- ventional, having a round body and two pipes opposite each other. In the Tiberius manuscript is a second chorus, which has a square body and two pipes for blowing instead of one. But the most satisfactory drawing is in another manuscript of this related group, the one at St. Blaise. [Compare Schultz, Das hofische Leben I, 437.] Here a man is blowing on the short pipe of a round-bodied chorus, and, with the left hand, is fingering the opposite pipe, which has several holes, and which ter- minates in a grotesque dog's head' (Padelford, p. 51). Trautmann, Anglia Bb. V, 49, suggests ' Fiddle,' and later (Padelford, p. 50) the ' Chrotta' ; but he does not sustain these solutions. Dr. Bright makes these very helpful suggestions that put the ' Bagpipe ' solu- tion beyond doubt: 'The bagpipe looks like a bird carried on the shoulders with the feet projecting upward (=the drones, two in number). The poet speaks of these legs in the air as/<?/ ond folme fugele gellce (1. 7) ; the neb (1. 6) is the chanter and is at the foot of the instrument (11. 17,20). The gender of the parts is im- portant. The chanter (the sister) is the female voice, it carries the high notes \ and the tune; the deep-voiced brothers are the drones (11. 21-23).' Prehn, p. 282, finds no Latin sources for this problem ; and classes it with such riddles as Rid. 61, ' the Reed,' and 70, ' the Shawm.' It resembles the first only in its gift of song, the second only in subject (infra). With the German riddles of musical instruments (Kohler, Weimar Jhrb. V, 1856, 351, No. 28) it has noth- ing in common ; but in its seventh line furnishes an analogue to the Lithuanian ' Geige ' riddle (Schleicher, p. 200). 32 1-3 Compare the opening formula in 33 1-3. — wrJettum gefra&twad. Cf. Beow. 1532, wrjsettum gebunden. 32 4 The meter and 32 8 both favor the no \hivizfrre\ of Cos., PBB. XXIII, 129, rather than the no\wer~\ of Herzfeld, p. 68 — a natural omission, however, on account of the following werum. — wenim on gemonge. Cf. 32 n, eorlum on gemonge ; 32 14, werum on wonge (Th. gemonge). 326 The first half-line is faulty. Instead of Herzfeld's onhivyrfed or gongende, or Holthausen's geneahhe or genyded, may we not read NiJ>erweard [ hyre set nytte, ni}>erweard gongeft ; 22 i, Neb is mm ni}>erweard. The beak or chanter is downward when the pipe is in use. 32 7 fet ond folme. Cf. 28 15 fota ne folma ; 40 10, fot ne folm ; 68 9, fet ne f [olme] ; Beow. 745, fet ond folma. — fugele gelice. The Fiddle of the Lithuanian riddle (Schleicher, p. 200) is likened to a bird which carries its eggs under its neck and cries shrilly from its rear. Note the later flute a bee, of which the upper part or mouthpiece resembled the beak of a bird. 32 8 Cf. 59 3, ne fela rldeft, ne fleogan maeg. But the subject of this riddle has, in its physical characteristics, little in common with the subjects of 59 ('Well') and 70 (' Shawm'), with which Prehn, p. 282, compares it. 3211 oft ond gelome. For other examples, see Spr. I, 424. — eorlum on gemonge. Cf. 32 4, werum on gemonge. NOTES 145 32 12 siteo5 set symble. Cf. Mod. 15, sittaft on symble. Another musical in- strument, the Reed-pipe, 61 9, speaks over the mead-bench. Cf. Wulfstan, Horn. 46, 16, Hearpe and pipe and mistlice gliggamen dremaft eow on beorsele. — su'les bide]?. Cf. Gen. 2437, 2523, sasles bidan. 32 14 weruni on wonge. This is not to be changed with Thorpe into on gemonge (324, n), because thus would be lost the word-play upon ivong 'field,' ' plain,' and wong ' cheek.' The bagpipe proclaims its power to men in, or by means of, the cheek. — Ne . . . wlht jjigeft. Cf. 59 10, ne wiht itej>. 32 16 Deor domes georn. Cf. And. 1308, deor ond domgeom. Like Ettmiiller, I begin a new sentence with the line, construing the adjectives with hio. Dr. Bright prefers, with Gn., W., to regard these as a part of the preceding clause. 32 17 fieger. The length of the diphthong is discussed by Madert, p. 25. The sound is always long in Cynewulf (see Trautmann, Kynewulf, p. 74), and is always long in the Riddles (see 13 n, 21 2, 29 i, 41 46). Sievers (PBB. X, 499) has shown that it is short only in South England poems. 3220 Fraetwed hyrstum. Cf. 15 u, hyrstum fraetwed; 547-8, wonnum hyr- stum | foran gefraetwed. With Frcclwed I begin a new sentence, as the phrase is more in keeping with the following than with the preceding thought. This is practically the punctuation of Ettmiiller. 32 21 hord warao. Against Dietrich's hordwarafr 'Schatzbesitzer', 93 26, hord •warafr, speaks conclusively. Cf. Beoiv. 2276-2277 : hord on hrusan, Jxer he hie'Sen gold waraft wintrum frod. Hord is applied here (so thinks Dr. Bright) to the contents of the bag, the air — a meaning that seems to me amply supported by 18 10 wombhord, the contents of the Ballista, and by 93 26 hord, the ink within the horn. The brothers, as above noted, are the bass-pipes or drones. The passage then becomes clear : ' She (the instrument), when she holds the treasure (i.e. is inflated), without clothes (so B.-T., Supplement, p. 61) (yet) proud of her rings, has on her neck her brothers — she, a kinswoman with might.' Dr. Bright prefers to regard the chanter — not the whole instrument — as the subject of the dependent clause. With this I can- not agree, although like him I believe that the poet in the personification mag had in mind the treble notes. Unlike Thorpe, I cannot view beer beagum as a compound. 32 23-24 For this concluding formula, see 29 12-13 (Introduction). RIDDLE 33 1 Unless this be a waggon or a cart,' says Conybeare, Illustrations, p. 210, 'the editor must confess himself not sufficiently skilful in wise words to decypher its occult allusions.' Bouterwek (Spr. I, 528, s. v. grindan) answers ' Millstone ' ; and Dietrich (XI, 469) offers the solution 'Ship,' which has been generally accepted. The 'one foot' is the keel, the ribs the beams, and the mouth the opening on dec' to admit wares into the hold. Prehn to the contrary, this riddle bears 146 RIDDLES OF THE EXETER BOOK no relation to Symphosius 13; but, as Dietrich has pointed out, its tenth line finds an analogue in the 'Ship' riddle of MS. Bern. 611, No. u (A nth. Lat. I, 354), ' Vitam fero cunctis, victumque confero multis.' It has nothing in com- mon with the Latin riddles of Lorichius (Reusner I, 178), nor with modern Eng- lish and German problems cited by Mullenhoff (Zs.f. d. M. Ill, 1 7). Yet Chambers's ' Ship ' query, No. 16, parallels ours in its last line, ' And no a fit (foot) but ane ' (cf. Petsch, pp. 47-48); and the Islenzkar Gdtur offers many like queries. In /. G. 151, the ship crawls on its belly footless; while in /. G. 514 the eight-oared craft has eight feet. The Anglo-Saxon vessel is like the Kaupskip of /. G. 615, 651, bearing food to men. Compare also /. G. 131, 293, 429, 516,585, 725, 1162-1194 (seventeenth century). This riddle resembles the preceding (32) not only in the use of the opening formula, but in general plan of construction. It belongs to the class of 'monster' problems. The Anglo-Saxon ship is thus described by Strutt, Horda, p. 42 : « Plate 9, fig. I (Tib. B. V) represents the form and construction of a more improved ship of the Anglo-Saxons (sometime before the Norman conquest), when they began to build with planks of wood and deck them over. The stern is richly ornamented with the head and neck of a horse ; the two bars which appear at the stern were for the steering of the ship instead of the rudder; on the middle near the mast is erected the cabin (in the form of a house) for the commodious reception of the passengers ; the keel runs from the stern still growing broader and broader to the prow or head of the ship, which comes gradually decreasing up to a point for the more ready cutting of the water in the ship's course. When the vessel had received her full burthen she was sunk at least to the top of the third nailed board ; so that the prow itself was nearly, if not quite immerged in the water. Over the prow is a projection . . . perhaps either for the convenient fastening of the ship's rigging or to hold the anchor.' Ships of the same pattern appear in Harl. MS. 603, ff. 51 r., 54 r. ; and Noah's ark is not only described (Ferrell, Teu- tonic Antiquities in the Genesis, 1893, PP- 32~33) t>ut pictured as a ship of the time (both in Cott. Claudius B. IV, ff. 14-15, and in the Caedmon manuscript, Archaeologia XXIV, pi. Ixxxviii, Ixxxix, xc). For the various kennings of scip in Anglo-Saxon poetry, see Merbach, Das Meer etc., pp. 29 f. Several names are found in the Riddles : 3 24, hlud wudu ; 3 28, 19 4, ceole ; 15 6, merehengest ; 59 5, naca naegledbord. 33 4 grindan wl<5 greote. As Dietrich says, this phrase is sufficient to identify the object of the riddle. Compare Gu. 1309, grond wrS greote (skip). 33 5~° Cf. 40 10-13, 59 7-8, 93 25, for like descriptions of the personal features of the subject. 33 6 exle nS earmas. Cf. 86 6, earmas ond eaxle ; Beow. 835, earm ond eaxle. 339 iiiuTf. Dietrich (XI, 470) compares Gen. 1364, merehuses mu'5 (A'va/i's ark). 33 10 This line presents difficulties. Thorpe renders fere ' in its course,' and suggests dr&gfr, ' draws,' for dreogefr. But the meter is against this emendation. Sweet's rendering of fere, ' serviceable ' (Diet.), with an eye to this passage, does not explain the construction with dreogefr. Grein notes, Spr. I, 282 : lfere = fare, NOTES 147 ace. zu farti, f. [see Leid. 13, aerigfaerae], "das Tragen," "Bringen"; "scip fere foddorwelan (gen.) folcscipe (dat.) dreoge'S (fere dreogeft = fere'5)." ' This seems to be derived from Dietrich (XI, 470) : ' Es erklart sich als umschreibung fur ferian ("herbeifiihren") nach dem haufigen sifras dreogan statt sifrian.' B.-T., p. 296, follows Grein. The phrase finds a parallel in Gen. 1746-1747: Gewlt J)ii nu feran and )>ine fare ladan ceapas to cnosle. Perhaps a play upon words is intended, as far means also ship (Spr. I, 270). In Dicht. the line is rightly rendered 'bringt es der Volkschaft Fiille der Nahrung.' 33 10-13 These lines show that the ship of the riddle is a merchant-ship. The cargo of such a vessel is well described in the speech of the merchant in ^ilfric's Colloquy (WW. 96) : ' ic secge J>aet behefe ic com ge cinge and ealdormannum and weligum and eallum folce [33 11-13] • • . ic astlge mm scip mid hlaestum mlnum and rowe ofer siellce daslas and cype mine >ingc and bicge Jnncg dyrwyrfte J>a on Hsum lande ne beoS acennede and ic hit to-gelaide eow hider mid micclan plihte ofer sae and hwylon forlidenesse ic J>olie mid lyre ealra Hnga, unease cwic aetber- stende.' He brings with him ' paellas and sidan, deorwyrj>e gymmas and gold, selcuhe reaf and wyrtgemange (pigmenta), win and ele, ylpes-ban and masstlinge (auricalcum), aer and tin, swefel and glaes and Jjylces fela.' A. L. Smith (Traill's Social England I, 202) notes that in the time of ./Ethelred (cf. Schmid, Gesetze, p. 218, ' De Institutis Londoniae,' § 2) traders from Normandy, France, Ponthieu, and Flanders brought into England ' wine, fish, cloth, pepper, gloves, and vinegar.' From the north and east came furs, skins, ropes, masts, weapons, and ironwork. 33 13 rice ond heane. Cf. 95 2, rlcum ond heanum ; Gu. 968, ricra ne heanra. 33 13-14 With the closing formula cf. 68 18-19, Secge se J>e cunne, | wisfasstra hwylc, hwaet seo wiht sy; El. 857, Saga, gif )>u cunne (Herzfeld, p. 20). RIDDLE 34 Except in two lines, this ' Iceberg ' riddle bears no relation to the many ' Ice ' problems ancient and modern. But the 'mother-daughter' motif (34 9-1 1) is com- mon to all riddles of similar subject, and has been traced at length by me (M.L.N. XVIII, 4; P. M.L.A. XVIII, 246; Mod. Phil. II, 564). The Roman gramma- rian Pompeius tells us that this question was often in the mouths of the boys of Rome (Keil, Scriptores Art. Gram. V, 311, cited by Ohlert, p. 30, note). The Ice riddles of Symphosius (No. 10) and Tatwine (No. 15) do not contain the met- aphor, but it is cited by Aldhelm in his Epistola ad Acircium (Giles, p. 230 ; Ma- nitius, Zu Aldhelm und Birda, p. 52), and appears in Bede's Flares (Mod. Phil. II, 562), in Bern MS. 611, No. 38 (Anth. Lat. I, 363), among the Lorsch Rid- dles, No. 4 (Dummler, Haupts Zs. XXII, 258-261), in Karlsruhe MS. of Engel- husen (Mones Anz. VIII, 316), in three of Reusner's authors (1,21,82,259), and in Holme Riddles, No. 5. I note several versions among the unpublished MSS. of the British Museum : in Latin form in Arundel 248 (fourteenth century), f. 67 b, and in Harl. 3831 (sixteenth century), f. 73; and as a four-verse enigma I48 RIDDLES OF THE EXETER BOOK in Harl. 7316 (eighteenth century), p. 60, f. 28 b. Puttenham, Arte of English Poesie, 1589, Bk. Ill, Arber's Reprint, p. 198, selects a popular version of this to exemplify 'Enigma.' It is found too in Pretty Riddles 1631, No. 1-2 (Brandl, p. 54). The query appears among modern German Volksratsel, as Carstens (Schles- wig-Holstein), Zs. d. V.f. Vk. ¥1(1896), 422, and Simrock8, p. 96, show. According to Ohlert, p. 30, ' Die Verwandtschaft mit dem griechischen Ratsel von Tag und Nacht ist nicht zu verkennen: wrtp t^v TIKTW /cot riKronai (Anthol. Pal. xiv, 41 : cf. Athenaeus x, 451 f.).' The motif appears in the 'Smoke' riddle of Symphosius (No. 7). As Brooke says (E. E. Lit., p. 181) : 'The poet paints, with all the vigor of the North, the ice-floe plunging and roaring through the foaming sea and shouting out, like a Viking, his coming to the land, singing and laughing terribly. Sharp are the swords he uses in the battle (the knife-edges of the ice), grim is his hate, he is greedy for the battle.' • Ice is thus described in the Runic Poem, 29-31 : Is byS oferceald, ungemetum slidor, glisna'S glaeshluttur gimmum gelicust, fl5r forste geworuht, f Jeger ansyne. For other references to Ice in the Riddles see 69, 84 35, 39. 34 i Wiht cwoni . . . lijmn. Cf. 55 i, Hyse cw5m gangan; 86 i, Wiht cwom gongan. 34 2 cymlic from ceole. Cf. And. 361, }>on cymllcor ceol; Beow. 38, cym- Hcor ceol. 34 5 hetegrim. This reading, instead of MS. hete grim, finds support from And. 1395, 1562; heafrogrim is an epithet of the north wind, Beow. 548. Not only hetegrim, but hlinsade, gryrelic, and egesful recall the vocabulary of the Andreas (1545, 1550, 1551). — hilde to steiie. Klaeber {Mod. Phil. II, 145) says, 'This looks at first sight genuine (cf. Doomsday 88; And. 204), but the context seems to de- mand exactly the opposite of it.' Herzfeld, p. 68, suggests to siege, ' zugeneigt ' (so Dicht. ' zum kampfe geneigt '), which does not appear elsewhere in Anglo-Saxon ; and Klaeber proposes on wene, arguing that a confusion on the part of the scribe between iviene and sitne would lead him to change on to to. Holthausen, Engl. Stud. XXXVII, 208, prefers to cene (North. ccEne, cSni). Why is any change neces- sary ? Brooke (E. E. Lit., p. 181), who translates 'greedy for the slaughter,' says however in a note : ' The phrase might mean slow in beginning the war, but when engaged, bitter in battle-work, and the phrase might well apply to an iceberg.' The seeming contradiction is of a sort dear to riddle-makers. For scansion of 34 sa, see Herzfeld, p. 50. 34 6 biter beadoweorea. See 6 2, beadoweorca saed ; Brun. 48. — bord- weallas. This is variously rendered : Th. ' bucklers ' ; Dicht. ' Schildmauern ' ; Spr. I, 133 'litoris agger'; Brooke p. 181, 'the sides of the ships ranged along with shields'; Sweet Diet, 'the shore.' The phrase, I think, refers neither to shore nor to shield but simply to the sides of the ship, which is elsewhere the bord (59 5, Gn. Ex. 183, Chr. 861, etc.). Compare the Delphian Oracle's phrase 'wooden walls ' for ships ; and remember that a riddler is writing. NOTES 149 34 7 Heterune bond. There is no reason to substitute onbond with Cosijn (PBB. XXIII, 129), who compares Beow. 501, onband beadurune. In the present passage, the iceberg 'binds, like a wizard, runes of slaughter' (Brooke, p. 181). 34 9-13 These enigmatic lines find adequate explanation in Met. 28 58-63 : hwa wundraft |>aes otJtSe o'Sres eft, hwy J>aet is maege weorSan of waetere ? wlitetorht seine's sunne swegle hat, sona gecerre~S Jsmere Snlic on his agen gecynd weor'Se'5 t5 waetere.' The direct speech of the Iceberg suggests 39 6, 49 5, and the frequent addresses at the close of the riddles (Jansen, pp. 94, 95 ; Herzfeld, p. 36). 34 9-10 inodor . . . J»aes deorestan. See 42 2-4, 84 4 (Water). The motive, so well known in riddle poetry, is again used, 38 8. 34 ii a'ldiuii cu]7. Cf. Beow. 706, yldum cut?. RIDDLE 36 As an answer Dietrich (XI, 470) offers ' Rake ' ; Trautmann, with far less reason, ' Bee.' The resemblance to the ' Serra ' riddle of Symphosius, No. 60, is slight and may lie in the independent demands of similar subjects. (A far closer analogue to Sym. is found in the Anthol. Pal. xiv, 19, cited by Ohlert, p. 143). It is interesting to compare the ' Rake ' (ffrtfa) riddles of Islenzkar Gdtur, Nos. 578, 628, 1053, as well as the ' Shovel' problems of that collection (Nos. 154, 358, 1 102, 1135). The teeth and downward fall of the Rake recall particularly /. G., 578: Hver er snotin halalaung, 4. hausi er geingur, gemlur ber f gotum rata gerir vinna til dbata ? Raca or race appears as a gloss to ' rastrum vel rastellum ' (WW. 105, i), and is mentioned among the agricultural implements in the Gerefa list, Anglia IX, 263 (Andrews, Old English Manor, p. 267). A capital illustration of the Anglo- Saxon rake — indeed of two — is found in MS. Cott. Claud. B. IV, f. 79 r. This is not dissimilar to the rake with nine teeth in the Thorsbjerg bog-find (Du Chaillu, Viking Age I, 202, fig. 365). '"It is a thing," riddles Cynewulf of the Rake — "that feedeth the cattle." Well does it plunder and bring home its plunder — as it were a forager. The riddle is dull, but it ends with the poet's pleasure in the meadows — "the Rake leaves firm the good plants Still to stand fast in their stead in the field, Brightly to blicker, to blow and to grow." ' (Brooke, E.E. Lit., p. 146.) 35 2h The teeth of the Plow are mentioned, 22 14 ; and those of the Saw are thus described by Symphosius, 60 1-2 : Dentibus innumeris sum toto corpore plena. Frondicomam subolem morsu depascor acuto. ,150 RIDDLES OF THE EXETER BOOK 35 3 Cf. ii i, 22 i (Plow), 32 6 (Bagpipe). 35 4 to ham tyh5. This is paralleled by A.-S. Chronicle, 1096; Orosius iv, 6, hdni tug-on; and the Mod. Engl. 'draw near home' (Byron, Don Juan I, 123). 35 7-8 For another riddle-picture of an English meadow, ' the station of plants,' see 712-3. — wyrtum faeste. Cf. Beow. 1365, wudu wyrtum f zest ; Dan. 499, wudubeam . . . wyrtum faest. 359 beorhte blican. So And. 789, Chr. 701,904. — blowan ond growan. Cf. Met. 20 99, blSweft ond growe'5 ; Ps. 64 n, blowaS ond growaS. RIDDLE 36 As Dietrich first pointed out (De Kynewulfi Poetae Aetate, 1859, pp. i6f.), this ' Mail-coat ' riddle is preserved not only in the Exeter Book but in the Leiden MS. Voss Q. 106, 24b, in the Northern dialect. This MS. contains the enigmas of Sym- phosius and Aldhelm, and dates, as Dietrich proves on the evidence of the hand- writing, from sometime in the ninth century. Dietrich, who gives a facsimile of the page containing the enigma, believes that the scribe, whose name we infer to be Otgerus from a marginal entry, was an Anglo-Saxon (Eadger or Edgar) living on the continent, and that he copied out the riddle in Latin script (using, con- trary to English custom, both the #• and tK) from an older manuscript. The Anglo-Saxon versions of the riddle follow very closely the Latin of the 'Lorica' enigma of Aldhelm (iv, 3). Two lines of the Anglo-Saxon correspond throughout to a single line of the original. The Latin order of traits in the description is departed from once, lines 4-5 being represented by lines 9-10 and 7-8 in the English. In this case the sequence of the translation is so far prefer- able to that of Aldhelm's text that Dietrich believes that the rendering was made from an older and better version of the Latin enigma than has come down to us. Here is the ' Lorica ' riddle : Roscida me genuit gelido de viscere tellus. (A.-S., 1—2) Non sum setigero lanarum vellere facta, (3-4) Licia nulla trahunt, nee garrula fila resultant, (5-6) Nee crocea seres texunt lanugine vermes, (9-10) Nee radiis carpor, duro nee pectine pulsor ; (7-8) Et tamen en vestis vulgi sermone vocabor. (11-12) Spicula non vereor longis exempta pharetris. (Leid. 13-14) The most superficial comparison of the English texts will show that they are merely slightly differing forms of the same version. The only important differ- ence between them lies at their end : here the Exeter text omits to translate the last line of Aldhelm, fearing, so Dietrich suggests, to betray the solution, but adds the conventional tag of appeal to the cunning of the reader, which is omitted in the Leiden text, either because it was not in the original or because it is unessen- tial to the body of the riddle, or else because the scribe found himself pressed for room at the bottom of the page, as the MS. seems to indicate. • Lehmann, Brunne u. Helm im ags. Beowulf liede, 1885, i f., traces the history of 'lorica' or mail-coat from the earliest Germanic times through the Merovingian NOTES 151 and Carolingian periods. Batemann in his Ten Years' Diggings, pp. 34 f ., describes the supposed ' lorica' discovered at Bentley Grange, with the boar helmet : ' This consisted of a mass of chain work formed of large quantities of links of two de- scriptions attached to each other by small rings half an inch in diameter amalga- mated together from rust. There were present, however, traces of cloth which make very probable the supposition that the links constituted a kind of quilted cuirass by being sewn within or upon a doublet of strong cloth.' The absence of protective body armor in nearly all the early MSS. would seem to show that it was used only by a few persons of the highest rank (Keller, p. 97). This conclu- sion is supported by the evidence of the wills and laws (Lehmann, Germania XXXI, 487). In the Beow., however, the byrne or light ringed shirt of iron links is the possession of every one of a picked band of warriors. Miss Keller con- cludes that the scale armor ('lorica squamata') was popular on the Continent, and mail armor ('lorica hamata') in England. See the illustrations of both printed by Strutt, Horda, p. 30, from the Cotton MSS., Claudius B. IV, and Cleopatra C. VIII. 36 i Similar is the origin of the Sword, 71 2-3. 36 2 Cf. Ps. 126 4, of innate ierest cende. 36 3 bevvorhtne (Leid. biuorthae). Dietrich (De Kyn. Aet. p. 18) notes : ' Proxi- mum bittorthie for bi-workte est participii genus femininum, loquitur enim ipsa res a poeta descripta, quam vult conjectura inveniri, quae res saepissime in aenigmati- bus anglosax. wiht gen. fern, dicitur et hoc in aenigmate est lorica annulis ferreis texta.' The Exeter form is masculine, which can hardly refer to byrne ; but gram- matical gender is little considered in Riddles (see 24 7, 25 7, 26 8, 39 6-7, 41 passim). — wulle flysum. Cf. Ps. 147 5, wulle flys. 36 5-8 Andrews, Old English Manor, p. 273, notes that in the Gere/a (Anglia IX, 263) 'we have a number of important terms applying to the loom which sup- plement the meager knowledge furnished by the Saxon literature. There was the frame of the loom (stodlan), the web-beam (lorg, glossed " liciatorium," WW. 187, n), later called yarn-beam, the wool-card (timplean), and wool-comb (wulcamb), the weft or woof (wtft, weft), the weaver's rod (ami), the shuttle (wefl, also sceafrel), bobbin (site), and reel for winding thread (crancst&f), etc.; ... It is evi- dent from the "tow" tools here given and from such as are given in other lists (WW. 187-188, 262, 293-294) that spinning and weaving were in a very moderate state of development. . . . The loom itself was without treadles and we cannot be certain that it had cylinders for tightening the warp.' For a discussion of the Anglo-Saxon loom, see notes to Rid. 57. Stopford Brooke (E. E. Lit.,>. 126) thus renders the lines: I have no enwoven woof, nor a warp have I, Nor resounds a thread of mine through the smiting of the loom, Nor the shuttle shoots through me, singing (as it goes). Nor shall ere the weaver's beam smite from anywhere (on me) ! 36 5 wefle. Of ivefl in the Gere/a list, Andrews notes (274, n.) : ' Wefi, also sceafrel. It is not easy to determine the difference, unless the former refer specially to the thread, which the shuttle carried, and the latter to the sheath within which the thread was contained.' B-.T.'s long discussion and copious references (p. 1 182) 152 RIDDLES OF THE EXETER BOOK show that weft is the gloss of 'cladica' or 'panuculum' and the synonym of weft and owe/, the weft or woof (see Dietrich, De Kyn. Aet., p. 19). 366 preata gebraecu. In Sfr. II, 598, Grein regards fireat in this passage as perhaps 'ein Theil des Webstuhls.' In Dicht. he translates 'durch der Schlage(?) Wiiten.' It seems to mean here ' the pressing of multitudes ' — that is, « the force of many strokes.' 367 hriitende hnsil. Dietrich says (De Kyn. Aet., p. 19): ' hrlsil est radius, nondum navis fistulam textoriam continens, sed lignum in curvum cui filum in- texendum circumvolvitur, islandice winda dictum cujus epitheton est hriitendi "stridens" quod vet. theot. erat ruzonti, " stridulus." ' I prefer Dietrich's hrii- tendi (see Schlutter, infra) to Sweet's hriitendum (Leid. 7) for three reasons : it is in accord with the Exeter form, hriitende ; hriitendum does not harmonize with the context, for it is the shuttle (hrisil), not the mail-coat (me) that goes whizzing ; and finally me would demand not hriitendum but hrutendre, as it is feminine (see Leid. 3, mec biiiorthtz). 36 8 am (Leid. aain). There seems little reason to question the opinion of Dietrich (De Kyn. Aet., p. 19) and Grein (Sfr. I, 28) that dm, a hapax-legomenon, is the ' pecten textorius, sive lignum illud transversum quo filum modo intextum pulsatur,' or, as Bosworth-Toller renders it, 'the reed or slay of the weaver's loom.' Thorpe without warrant changes the word to uma, ' the yarn-beam.' In the Gerefa list the word amb appears, and is thus considered by Andrews (Old English Manor, p. 274): 'We can get only an uncertain light upon this word. Liebermann has suggested its relation to dm, meaning a weaver's rod. This word is found in Cyne- wulf, Riddle 36, ne mec ohwonan sceal dmas cnyssan " nor do the weaver's rods anywhere press me down." This seems the most acceptable interpretation. In the Gerefa enumeration (IX, 263, 12), a synonym is "pihten," which Leo, Angels. Gloss. 520, 1 6, renders "der weberkamm aus \a&.zm.. pecten ? " [see Hpt. Gl. 494, 26]. This was a weaver's comb, the teeth of which, inserted between the threads of the warp, by a downward pressure or stroke packed the thread of the web closer together. It served the purpose of the dm or slay-rod. In fact dm is the Saxon translation (in Cynewulf's riddle) of the pecten (" duro nee pectine pulsor ") in Aldhelm's version.' 36 9 Cf. 41 85, wrStllce gewefen wundorcrasfte. I cannot agree with Brooke (p. 126) that this line of the riddle ' takes us into the heart of ancient heathendom.' It is simply a fairly accurate translation of Aldhelm's Latin, and cannot be rendered ' Me the Snakes wove not through the crafts of Wyrds.' Wyrda craftum has lost its old force, and means nothing more than ' durch Schicksalskrafte ' (Dicht.). 36 10 godwebb. Cf. Met. 8 23 : ne heora wjeda >on ma sioloce siowian, ne hi siarocraeftum godweb giredon — See Lchd. II, 10, 16, god geolu seoluc; III, 174,29, seoluc oft^e godweb. For long discussions of this word and its analogues, see Heyne, Fiinf Biicher III, 235; Klump, Altenglische Hand-werknamen, p. 77. 36 14 Cf. Beow. 627, wisfaest wordum. NOTES 153 LEIDEN RIDDLE Since the casting of my text of the Leiden Riddle, Dr. Otto B. Schlutter has generously sent me from Leiden the results of his careful study of the manuscript. His detailed discussion of every debatable point in the text deserves larger treat- ment than my present space affords, but I am fortunate in being able to print his version of the problem and his Latin translation — however different his in- terpretation may be from my own. ' The following,' writes Dr. Schlutter, ' is my reading of Leiden Riddle metrically arranged. What is bracketed is no longer visible. The letters in small capitals are very faint and hence doubtful : ' Mec fe ueta erftuonj; uundrum freorig ob hif innaftae aerift cae[ndas]. Ni uuat ic mec biuorthae uullan fliusu, heru derh hehcraeft hujidoHTA UYN. Uundnae me ni bia'5 ueflae, ni ic uarp hafae, ni fterih •Sreaungi'Sraec firae' me hlaemmedE. Ne me hrutenbe hrifil fcelfse'S, ne mec ouaNan caam fceal cnyiffaN. Uyrmaf mec ni auefun uyndicraeftum, ftafti joelu jodueb j;eatu fraetuath. Uil [m] mec huetrae fuaedeh uidas ofaer eorflu haatan mith helrSum hyhtlic jiuaede. Ni anoegun ic me aerijjfaerae ej;fan broju, •Sehfti niM^EN FLANaf [fracajdlicae ob cocrum LONfjum]. Me humida tellus mire gelida ex visceribus suis principle genuit. Ignore me coopertam lanae velleribus, villis per artificium, laborem mentis. Volutae non mihi sunt panuculae, non ego licium habeo, non per tortile opus filum mihi garrulat (garrulavit), Non stridens mihi radius vibrat (vibravit), non me ulla parte pecten pulsabit. Bombyces me non texuerunt plumaria arte qui quidem flavum sericum vestibus fabricant. Verumtamen homines me vocabunt late per orbem desiderabile vestimentum apud heroas. Non expavesco iaculationis terrorem timorose, quamvis promant sagittas hostiliter ex pharetris longis. Here are a few of Dr. Schlutter's comments upon his readings. ' Line i. The doubtful letters after ueta I now find to be erfr, the first letter being plainly visible. Line 2. What follows in the MS. after arist is doubtful ; with a little straining of the imagination one may be able to see Cks frfgfn syllkc J>knc to raedfnnf (Nys J>is fregen syllic Hnc to riedenne). Upon the same page of the manuscript appears an Anglo-Saxon explanation of the system (Forster, Engl. Stud. XXXVI, 325) : b f k p x a e 1 o u a e i o u Dis is quinque vocales ; mid J>ysum flf stafum man maeg writan swa-hwaet-swa he wile. Hit is lytel craeft ; ac J>eah man maeg dwelian manega men mid aegfter ge ware ge unware. Among the Latin examples that follow is one in Old English that reads like a riddle-formula : Cxnnb mbgf J>x braedbn, hwaet J>ks mbgf bfpn. Kc wfnf J>aet hkt nks fSraedf (Cunna, mage Jm araadan, hwaet )ns mage beon. Ic wene, >aet hit nis eSrJede). The script appears not infrequently in glosses, both in Old English (Kentish Glosses, WW., p. 87) and Old German (Haupts Zs. XV, 35 ; XVI, 36, 94). It serves a useful purpose in the fifteenth-century puzzles of the Brome Book, f. i, (Kerrison and Smith, London, 1886) and of the Sloane MS. 351, f. 15, (Wright and Halliwell, Reliquiae Antiquae II, 15). Compare A. Meister, Die Anfdnge der mo- dernen diplomatischen Geheimschrift, Paderborn, 1902, pp. 5f. From the fourteen letters of the riddle, Dietrich (XI, 471-472), by several shiftings and substitutions, derives sugu mid V. ferhum, ' sow with five farrow.' This is a world-riddle, and has a famous history. I must refer to my note on Holme Riddles, No. 53 (P. ALL. A., 1903, 258-259). Ohlert, pp. 38-39, marks its appearance in the Melampodia of Hesiod (Strabo xiv, I, 27, p. 642), and points to the Icelandic parallel, Heifrreks Gdtur, No. 1 2 (' sow with nine young '). Heusler, Zs. d. V.f. Vk. XI, 1901, 141-142, compares with the H. G. version Aldhelm vi, 10 ; our Exeter Book problem ; and the modern riddles of the Faroes (Zs.f. d. M. Ill, 125) and Iceland (Islenzkar Gdtur, Nos. 447, 448). Royal Riddle Book, Glas- gow, 1820, p. 9, is very like Holme. Riddles with a similar theme are found in Hungary (Mag. fur die Lift, des Auslandes, 1856, p. 364) and in the Tyrol (Renk, Zs. d. V.f. Vk. V, 152, No. 76) ; and the Latin homonym of Reichenau MS. 205, No. 6, (Miillenhoff and Scherer, Denkmalerz, p. 20) has a like motive. 156 RIDDLES OF THE EXETER BOOK The closest analogue to Dietrich's interpretation of our riddle is that of Aid- helm, vi, 10, De Scrofa Praegnante. The first four lines of the Latin correspond exactly to the number-motive of the Anglo-Saxon : Nunc mihi sunt oculi bis seni in corpore solo, Bis ternumque caput, sed caetera membra gubernat, Nam gradior pedibus suffultus bis duodenis, Sed novies deni sunt et sex corporis ungues. Other Latin analogues are Symphosius 90 and Aldhelm i, 10, which have as their theme ' Mulier geminos pariens.' Thus far the strange forms of the monster of the riddle have been left unex- plained. There is a difficulty here, which Dietrich, I.e., meets with a not very plausible explanation : ' The bird in the second part of the riddle must now be discussed : it is only a continuation of the jest of the wing-ears and is still the sow, because the points of likeness with horse and woman which the bird is said to have are predicates of the subject in the first part. As the sow, on account of the mane, is a horse, so she is, on account of her womb, a woman, and, by reason of her snout and bite, like unto a dog.' This solution does not satisfy Trautmann, who suggests very doubtfully (Anglia, Bb. V, 49) that the secret words are merely Latin translations of the preceding Anglo-Saxon forms : ' homo,' ' mulier,' equus.' This view is confirmed by Holt- hausen, who believes (Engl. Stud. XXXVII, 208) ' that we have to do with a corrupt transmission of the secret script, and that for h.-w.M.,Mx.I.R.f.wf. . . qx xs we should read hpm_p~\ = homo, mxlkfr = mulier, f • • . qxxs — equus.1 Holthausen is unwittingly close to the MS., which Dietrich and Assmann have mis- read. Here at last is the obviously correct interpretation of the secret script. And in the light of this, Dietrich's solution loses its chief support, and must, I think, be abandoned. It is possible that the formula of closing in line 8 marks the end of our riddle, and that with For flodwegas (1. 9) a new problem is begun. If this be the case, we do not lack solutions. Dietrich, 1. c., would then offer ' Fledermaus, ' changing, with Grein, flddwegas to foldwegas ; and Trautmann proposes ' Das Schiff .' But it is not necessary to regard 37 9-14 as a separate riddle, since the traits of the ob- ject here correspond with those of the wight in 37 1-8. We can hardly do better than to extend to the whole problem Trautmann's solution of the latter part and interpret the monster as « Ship ' or ' Boat.' This answer meets the conditions of the enigma. The ship has ' four feet under its belly,' the four oars (compare ' the eight feet ' of the eight-oared craft in I. G. 514), and 'eight above on its back,' those of the man, woman, and horse on its deck. It fares the floodways, and may well be compared to a bird (cf. Becnu. 218, And. 497, fugole gellcost). The horse, man, dog, bird, and woman (37 11-12), of which it bears the likeness (i.e. which it carries), supply, if we add the ship's figure- head, the two wings, twelve eyes, and six heads (37 7-8). The phrase tuferu may refer also to the ship's sails, and thus stress the likeness to a bird. 374 ehtuwe. Thorpe suggested e/ituf>e, translating 'eighth man'; Gn.2 ehtu we = ehtun we (ehtan, eahtan, 'aestimare'). But, as Sievers shows Gr.s 325,8, NOTES 157 ehtuwe is merely the Northern form of the numeral 'eight' (R.2, Luke ii, 21, ahtcnve). Holthausen (Engl. Stud. XXXVII, 208) points out that ehtuwe must be construed with ufon on hrycge (1.6). The phrase thus parallels feowere fet under wombe (1. 3). 37 5-6 I depart from Assmann's reading by giving wiif to the fifth line and f hors qxx s to the sixth. 379 For flodwegas. Cf. Exod. 106, fSron flSdwege ; Sea/. 52, on flSdwegas feor gewltan ; El. 215, feran flodwege (MS. foldwege). 37 13 Cf. And. 603, Miht Jm me gesecgan, >xt ic so8 wite ; Chr. 442, |>aet J>u soft wite (Herefeld, p. 19). RIDDLE 38 This riddle of the ' Bellows ' has nothing in common with Aldhelm's enigma of like topic (i, 13), but in its 'life and death' motive conforms closely to Sym- phosius 73 (infra). It is a variant of 87, and in some motives it presents points of likeness to Riddles 19 and 34. The many ' Bellows ' problems of different languages have small resemblance to the Anglo-Saxon : Strassb. Rb. 209 ; Apol- lonius of Tyre 4 (Schrbter, Mitt, der deutsch. Gesellsch. zur Erforsch. der vaterl. Spr. und Alt. V, 1872, p. xiv) ; Reusner I, 188, 287; /. G. 195, 726, 860, 925, 1152; and the English riddles (Notes and Queries, Dec. 16, 1865). Dietrich (XI, 472) first suggested ' Wagon,' but arrived soon (XII, 238 note) at the answer ' Bellows,' which no one has questioned. In Cotton MS. Claudius B. IV, f. 10, we find an illustration of Tubal-Cain at work at the forge assisted by an attendant with bellows (Tubalcain se waes aegfter ge gold smi'S ge iren SHU'S) and in Harl. MS. 603, f. 6 v., two figures at a smithy, one with hammer and tongs (see also Cadmon Met. Par. Ixix ; Horda vii, 3, xxxii, 9). Akerman in his Remains of Pagan Saxondom, 1855, p. 61, discusses the high repute in which the smith was held, and cites the will of Eadred giving lands to .^Ifsige, his goldsmith (Codex Diplomatics III, 431 ; cf. VI, 211). Compare The Crafts of Men, 61-66 : Sum maeg wajpenfc>ra>ge wige to nytte modcrEeftig smift monige gefremman, )x>nne he gewyrce'S to wera hilde helm oiS'Se hupseax o'5'Se hea'Subyrnan, scirne mece o>\& scyldes rond, faeste gefegan wrS flyge gares. In a passage of the De Laudibns Virginitatis (cited by Sharon Turner VII, chap, xi), Aldhelm describes ' the convenience of the anvil, the rigid hardness of the beating hammer, and the tenacity of the glowing tongs.' The craft of the smith is extolled in ^Llfric's Colloquy, WW. 99 : ' Se smi|> secgS : hwanon [|>am yr^linge] sylanscear oW'e culter >e na gade haef> buton of crasfte mlnon : hwanon fiscere ancgel (hamus) oj>he sceowyrhton 5E1 obbe seamere niEdl nis hit of mlnon geweorce.'. . . And the Consiliarius answers: ' J>u hwa^t sylst us on smiH>an Hnre buton Isenne, fyrspearcan and swegincga beatendra slecgca and blawendra byliga (flantium follium).' For a discussion of the status of I58 RIDDLES OF THE EXETER BOOK the smith and of the appearance of his name in Anglo-Saxon literature, see Klump, Altenglische Handwerknamen, pp. 32-35, 97-104. Andrews says, Old English Manor, p. 276 : ' The tools which they (the Anglo- Saxons) employed were cumbrous and required much time and labor to satis- factorily use them. This Cynewulf indirectly tells us in his riddle of the bellows, for while Aldhelm, from whom he copied, had laid special stress upon the metal adornment, the artistic work, Cynewulf, more familiar with the Saxon bellows as the smith used them, lays his emphasis upon the strength which was needed by the man who attended the blowing. This would point to a ruder instrument and the need of greater muscular exertion.' The argument has small force. 38 1-3 Compare the other Bellows riddle, 87 1-3. See also 19 3, wide wombe, 89 2 wiht wombe haefd . . . 384 A difficult passage. Thorpe proposes, in his note, fyligde ? Grein, in the note to his text, /?/</<?; Dietrich (XI, 47 -2) frizr his filled 'fleah J>ttrh his cage. Grein, Die/it., renders thus : ' wo seine Fullung (?) flog durch sein Auge.' But Dietrich retracts (XII, 238, note) : ' Eigen ist der mitfolgende diener und zugleich sohn des blase- balgs, es ist der durch sein auge entschliipfende wind, er floh da man es (v. 4, das ding) fallte, d. h. niederdriickte.' One very serious objection to Dietrich's second rendering is that nowhere in the Kiddles is the object indicated by the neuter pronoun, but always is regarded as a person, — man or woman. Here it is mascu- line, while in the companion problem (87) it is feminine. Hit, then, is either a corruption or refers to something else than the riddle subject. As there is no pos- sible antecedent, I believe that a reconstruction of the line is demanded. Dietrich's first suggestion is probably not far from the truth : his filled (probably fyllo; see 43 s) fleah frurh his eage refers, of course, to the contents of the bellows, the wind, which is ' blown through the eye ' (cf. Rid. 87 6 bleow on cage). The ' much ac- complishment ' (micel . . . gefered) of the J>egn indicates just such labor as that in Rid. 87 4-5- With our passage compare the lines in The Crafts of Men (cited supra}. 38 5-7 Here the riddler closely follows Symphosius 73 1-2: Non ego continue morior, cum spiritus exit ; Nam redit assidue, quamvis et saepe recedit. 38 7 blaed bi}> arsered. Cf . Beow. 1 703, blied is anered. The riddler is of course playing upon the double meaning of bleed, ' breath ' and ' prosperity.' So Symphosius plays upon ' spiritus ' in his ' Violet ' enigma (No. 46). 388 This motive is that of the world-riddle of Ice, discussed under 349-11. Prehn, p. 211, compares Symphosius 7 3, Fumus\ 'Et qui me genuit, sine me non nascitur ipse.' RIDDLE 39 The sources of this riddle of the 'Young Bull' have received sufficient dis- cussion under Rid. 13. , 39 '-3 Grein and Wiilker put no mark of punctuation after wapnedcynnes, but a colon after gradig. How then is gradig to be construed ? Grein, Dicht., makes the adjective qualify -wiht (ace.), but grammar forbids. Brooke, E. E. Lit., p. 146, NOTES 159 supplies ' was.' ' Of the gladness of youth was he greedy.' It is far better to close line i with a semicolon, and then regard grSdig as qualifying the subject of forlet, that is, the Young Bull itself. Grein, Dicht., commits the mistake of rendering ferfrfribende as ' Der Befrieder der Geister ; ' so also Brooke ' The Defender of Being.' In Spr. I, 282, Grein corrects his error by translating the word as ace. pi. with wellan, 'vitam servantes,' which corresponds to Thorpe's and B.-T.'s ' life-saving.' The passage may thus be rendered : ' I saw a creature of the weaponed kind ; greedy of youth's gladness, for a gift unto himself, he let four life-saving fountains brightly spring,' etc. 39 3 ferfffriftende feower wellan. Compare the/emuer swase brdfcor of 72 5-6. The Udders appear often in riddle-poetry. I have already referred under Rid. 13 to Aldhelm iii, 112, 'Bis binis bibulus potum de fontibus hausi,' and Eusebius 37, ' ab uno fonte rivos bis . . . binos,' and to other Latin enigmas with this theme. One of the best known of world-riddles is that of the ' Cow,' with the motif ' Vier hangen, vier gangen' (Wossidlo, No. 165), found in all countries. Compare, too, Holme Riddles, No. 36, ' Flink flank under a bank 10 about 4,' and the several analogues. 394 on geseeap peotan. B.-T., p. 1053, says 'The passage describes a calf sucking from its mother ; if J>eotan is an infinitive [the word is found in the sense of " howl," Met. 26 80] it must refer to the sound made by the milk coming from the teat, but perhaps gesceap-freote may be a compound noun meaning the teat.' J>eote is ' a pipe or channel through which water rushes.' B.-T.'s first explanation, which corresponds to the rendering of Grein, Dicht., ' nach Geschick tosen ' (Spr. II, 589, ' prorumpere cum strepitu'), seems to me preferable, for the compound suggested is not enigmatic. On geseeap is not found elsewhere, but its meaning is obvious (contrast 73 6, wi)> gesceape). The riddler, here as elsewhere, may be slyly delighting in the double meaning of his word. 39 6-7 Herzfeld (pp. 29, 44), who believes that the last two lines are taken word for word from Eusebius (see however my notes to Rid. 13) says : ' Es ist lehrreich zu verfolgen wie in den Ratseln Abhangigkeit vom Original mit tech- nischem Ungeschick Hand in Hand geht.' Holthausen remarks, Engl. Stud. XXXVII, 208: 'Die 3 zeilen sind offenbar prosa, hochstens ein spater versuch, ohne kenntnis der technik alliterierende verse zu machen.' This statement is too strong, although the lines are admittedly slovenly. The metrical stress and allit- eration both fall upon the pronoun me (5 b), which logically is quite unstressed ; but, as Herzfeld points out, examples of stressed pronouns are found elsewhere in the poetry — no less than seven in Juliana (see Schubert, De Anglo-Saxonum Artc ."\fetrica, Berlin, 1870, p. 10). See Rid. 41 86, Nis under me (X_L| X_L), 48 i, 66 5, 6, 73 2, etc. Half-lines of shortened A-type (_£. X | \j X ) like 6 b, 7 b, are found in the Riddles (Herzfeld, pp. 44, 49). And confusion of gender (hTo, he) is not uncommon (see 24 7, 25 7). 39 6a Barnouw (p. 214) would regard seo wiht as an addition of the scribe, and read gif hlo gedvgefr (cf. 39 7a, gif he tobirsteft). ' This would prevent the poor alliteration produced by the chief stress falling upon the verb instead of upon the noun.' But the lines are careless; and the juxtaposition of seo wihl goes far to explain the feminine form of the pronoun hlo in this line. 160 RIDDLES OF THE EXETER BOOK RIDDLE 40 To this riddle Dietrich (XI, 472) offers the answer ' Day,' ' which is proverbial for its poverty ' (compare line 14), and points to the Runic Poem, 74-76 : Daeg biiS drihtnes sond, deore mannum, maire metodes leoht, myrgS and tohiht eadgum and earmum, eallum bryce. Prehn, p. 275, shows that the wanderings of the Day have been suggested in Rid. 30, and that its poverty is opposed to the costly garment of Night, described in Rid. 12. He notes, too, that the contrasts of this problem put it in the same class as the one of Creation (Rid. 41). Trautmann (Anglia, Bb. V, 49) proposes the solution ' Time.' I am inclined to regard Rid. 40 not as a query of ' Day ' or ' Time,' but as a ' Moon ' riddle like Rid. 95. The first lines correspond closely to those of the later problem, and the especial power of the Moon is extolled in both poems (40 3-4, 19, 21-22, 95 7-10). Like the Moon in Rid. 30 9-10 and 95 jh, the subject is a wretched exile and wanders widely (409-10,16-17); and, as in the closing lines of the other riddles, his future lot is obscure (40 22-24). Even his silence (40 12) suggests 95 9-10. 40 7, ne bi& hlo nSfre niht J>cer df>re, might seem at first sight more applicable to the Sun, but what words could better describe the changing positions of the Moon "i Dietrich brings no proof that ' the Day is proverbially poor'; on the contrary, Liming shows (Die ATatur, ihre Aitffassung und poetische Verwendung in der altgermanischen rtnd mittelhochdeutschen Epik, Zurich, 1889, p. 54) that in the old Germanic epic 'Der Tag mit seinem Glan/e erfreut die Herzen der Menschen und beherrscht gleichsam die Lebewelt, daher heisst er "riche"' (Hagen, Minnesinger i, 163, riche also der tac ; i, 127 b, ii, 23 b, der tac will gerichen). But the epithet earmost, 40 14, exactly fits the Moon, who has no light save that taken from the Sun (Rid. 30, 95) ; and even that is often lost. 40 i, 13 gewritu secgaft. So Gen. 1121, 1630, 2563, 2611, El. 674, Ph. 313, 655 (see also Gaebler, Anglia III, 312). The only other appeal to sources in the Riddles is immediately above in 39 5 ; but in that case the popular origin of the passage was easily traceable. The reference here is to the many scientific works, such as Bede's De ATatura Rerum, which make the Moon the center of their knowledge (see under 95). 40 2 See 95 2, ond reste oft rlcum ond heanum (Moon). 40 3 sweotol ond gesyne. So 14 4 ; see my note to that passage. No phrase could be better suited to the Moon. — sundorcraeft. This special power of the Moon, ' far greater than men know,' is the influence over the tides discussed by Aldhelm in his « Moon ' enigma (i, 6) : Nunc ego cum pelago fatis communibus insto Tempora reciprocis convolvens menstrua cyclis. 40 5 gesPean sundor. Cf. El. 407, sundor aseca'5 ; 1019, sundor asecean. 40 6b Cf. Rid. 30 10, gewat hyre west )>onan (Sun) ; Wond. 68-69, gewiteiS . . . forSmaere tungol faran (Sun) ; Sal. 503, gewlteft bonne wepende on weg faran. Gewdtferan is a common idiom (Spr. I, 484). NOTES 161 40 10 f. The contrasts suggest 41, and the negatives 33 3 f. — fot ne folm. Cf. 28 15, fota ne folma ; 32 7, fet ond folme ; 68 9, fet ne f [olme] ; Beow. 745, fet ond folma. 40 16-17 The clause is admirably suited to the wanderings of the Moon (95 3, fere wide). Compare MS. Bern. 611, 593 (Luna): Quotidie currens vias perambulo multas Et bis iterate cunctas recurro per annum. 40 19 mongum to frofre. The Sun also comforts many, 7 6-7. The comfort of the Moon's presence is the theme of 95 7-9. 40 20 It certainly seems inapt to say of the Moon that ' it never touched the heavens ' ; but note that here heofonum is not used of the firmament, but is op- posed to helle, and therefore means 'the abode of bliss.' Moreover, as lines 21- 22 show, the riddler is speaking of the Moon's long life through the lore of the King of Glory. The line is merely a ' check ' to the solution, and is well calcu- lated to mislead the too literal victim. 40 24 woh wyrda gesceapu. Cf. Sal. 332, gewurdene (Gn.2 gewundene) wyrda; Met. 4 40, hwi sio Wyrd swa wo wendan sceolde. 40 26a There is no occasion for the changes proposed by Holthausen (see text). If we read wihte for wiht (the forms are used interchangeably, 38 i, 39 i), we have a first half-line of expanded A-type (_L X X x x | _L x ). For stress upon J>dra, compare 41 89a, J>ara )>e worhte. 40 27a Examples of B-type with alliteration on second stress of first half-line are so rare that I change the editors' anig lim to lim iznig. The reconstructed line presents no metrical difficulty. Cf. 41 i6a. RIDDLE 41 As Dietrich has clearly pointed out (XI, 455), this most extensive of all the riddles is a fairly close rendering of Aldhelm's enigma, De Creatura (Cr.). Herz- feld shows, p. 27, that the poet sets aside classical allusions and expressions and replaces them by those current among his countrymen, thus giving, after Cyne- wulf's manner, national coloring to his presentation (Ebert, Allgemeine Gesc/i. der Lit. des Mittelalters III, 54): Cr. 14, 'olfactum ambrosiae ' is discarded; Cr. 21, ' Tonantis ' is replaced by heahcyning, Cr. 22, ' tetra Tartara ' by -worn wrdfr- scrafu, and Cr. 33, 'more Cyclopum' by ealdnm l>yrse\ Cr. 35, 'Zephiri' is ex- plained,^! 68-69; and Cr. 67, ' Phoebi radiis' cries a halt. Prehn also comments, p. 213, upon our riddler's consistent effort to Germanize and Christianize Ald- helm's matter. Herzfeld, p. 28, notes that both Kid. 36 and Kid. 41 are distinguished by the circumstance that 'die metrische Gliederung mit der syntactischen ganz zusam- menfallt, wahrend sonst die Regel besteht dass beide sich kreuzen ' (see Rieger, Zs. f. d. Ph. VII, 45). For this reason we find in these two problems 'very little of that variation from sources which fills out a verse and leads to new thoughts.' 162 RIDDLES OF THE EXETER BOOK DE CREATURA (Aldhelm) Conditor, aeternis fulsit qui saecla columnis, (1-2) Rector regnorum frenans et fulmina lege, (3-4) Pendula dum patuli vertuntur culmina mundi, (5) Me variam fecit, primo dum conderet orbem. (6-7) 5 Pervigil excubiis nunquam dormire juvabit, (8-9) Sed tamen extemplo clauduntur lumina somno. (10-11) Nam Deus ut propria mundum ditione gubernat, (12-13) Sic ego complector sub coeli cardine cuncta. (14-15) Segnior est nullus, quoniam me larvula terret, (16-17) 10 Setigero rursus constans audacior apro. (18-19) Nullus me superat cupiens vexilla triumphi, (20-21) Ni Deus aethrali summus qui regnat in arce. (21-22) Prorsus odorato thure fragrantior halans, (23-24) Olfactum ambrosiae, necnon crescentia glebae j 0 i (24—20) 15 Liha purpureis possum connexa rosetis Vincere, spirantis nardi dulcedine plena. (29-30) Nunc olida coeni squalentis sorde putresco. (31-32) Omnia quaeque polo sunt subter et axe reguntur, ) . . Dum pater arcitenens concessit, jure guberno. ) 20 Grossas et graciles rerum comprenso figuras. (36-37) Altior en caelo rimor secreta Tonantis (38-39) Et tamen inferior terris tetra Tartara cerno. (40-41) Nam senior mundo praecessi tempora prisca ; (42-43) Ecce tamen matris horna generabar ab alvo. (44-45) 25 Pulchrior auratis dum f ulget fibula bullis ; (46-47) Horridior rhamnis, et spretis vilior algis. (48-49) Latior en patulis terrarum finibus exsto, (50-51) Et tamen in media concludor parte pugilli. (52-53) Frigidior brumis, necnon candente pruina, (54-55) 30 Cum sim Vulcani flammis torrentibtfs ardens. (56-57) Dulcior in palato quam lenti nectaris haustus, (58-59) Dirior et rursus quam glauca absinthia campi, (60-61) Mando dapes mordax lurcorum more Cyclopum, (62-63) Cum possim jugiter sine victu vivere felix ; (64-65) 35 Plux pernix aquilis, Zephiri velocior alls Necnon accipitre properantior, et tamen horrens Lumbricus et limax et tarda testudo palustris (70-71) Atque fimi suboles sordentis cantharus ater (72-73) Me dicto citius vincunt certamine cursus. (70, 73) 40 Sic gravior plumbo scopulorum pondera vergo ; (74-75) Sum levior pluma cedit cui tippula lymphae. (76-77) Nam silici densas fundit quia viscere flammas ) - Durior aut ferro, (tostis sed mollior extis). J No equivalent in Latin. (80-8 1) 61 Senis ecce plagis latus qua penditur orbis j _ _ Ulterior multo tender mirabile fatu. j Infra me suprave nihil per saecula constat, (86-89) Ni rerum genitor mundum sermone coercens. (89-91) 65 Grandior in glaucis quam ballena fluctibus atra (92-94) Et minor exiguo sulcat qui corpora verme. (95-97) NOTES 163 44 Concinnos capitis nam gesto cacumine nullos, ) ... . Ornent qui frontem pompis et tempora setis ; ) Cum mihi caesaries volitent de vertice crispae, / . } (102-104) Plus calamistratis se comunt quae calamistro. ) Pinguior en multo scrofarum exungia glesco, ) .. Glandiferis iterum referunt dum corpora fagis i ' 50 Atque saginata laetantur carne subulci. (107) It has already been noted that in the rendering of Rid. 36 from Aldhelm iv, 3, Lorica, two lines of the Anglo-Saxon correspond to one of the Latin. This method of translation is followed in the Englishing of Cr. by the poet of Rid. 41, save only in a few places (41 5, 24-28, 33-35, 66-69). But when line 43 of the Latin is reached, comes a violent change (41 79). Cr. 43, ' tostis sed mollior extis,' is entirely disregarded in 41 80-81, lines which have no Latin equivalent. Cr. 61— 66 becomes the basis of the lines that follow in the English version. As Dietrich has suggested (De Kyn. Aet., p. 25) to explain the departure from the Latin sequence, perhaps another and earlier version of Aldhelm than that now extant is followed by our riddler. This view is amply supported by a similar change of sequence in Rid. 36 (supra), by the unfixed order of traits in other Latin riddles of nearly the same period (Bern MS. 6n, Nos. 5, 9, 18, 22, 24, 57, 58), by the probable relation of Rid. 41 80-81 to a different text from the one before us, and by the isolation of Cr. 61-66 from the lines that precede and follow. Yet this explanation is not sufficient to account for three things : (a) the com- plete change in the method of translation ; (b) the errors of rendering that now abound in the English version ; (c) the appreciable weakening of technique in the later part of the English riddle, (a) The translator no longer renders each line of the Latin by two of English : Cr. 61-62 is interpreted by 41 82-85, Cr. 63 by 41 86-89, Cr. 64 by 41 89-91, Cr. 65 by 41 92-94, Cr. 66 by 41 95-97. (b) Mistrans- lations now abound : 41 83 is inspired by a total misunderstanding of 'senis plagis ' ; 41 85 is too freely rendered from ' mirabile fatu ' ; 41 86-87 conveys an idea exactly opposite to 'infra me ' ; 41 91 has no warrant in Cr. 64 ; 41 92-94 is a very free ver- sion of Cr. 65 ; and 41 96-97 in its relative clause exactly inverts the meaning of Cr. 66. (c) The technique is wretched : 41 84 is defective, and 41 86a, 88a, are faulty in the weakness of the stressed syllables ; and the construction of 41 86-88 is awkward and ambiguous. So much for the translation of Cr. 61-66 by Rid. 41 82-97. We are then carried back to Cr. 44 ; yet the translation proceeds not after the old system, but after the new. Cr. 44-45 is rendered by 41 98-101, Cr. 46-47 by 41 102-104, Cr. 48-49 by 41 105-106, Cr. 50 by 41 107. The first four lines of the Latin are rendered with great freedom, and the sense of Cr. 50 is completely lost in 41 107. No one will deny, I think, that the translator of Aldhelm in 41 1-79 is the same person as the translator of Aldhelm in Rid. 36. Not only are these riddles the only literal renderings from Aldhelm in our collection, but in both the same peculiar method is employed. Now is it conceivable that this English reworker of Latin material, proceeding steadily by an already tested system for some eighty lines, would suddenly divest himself of his successful method ? Furthermore, is it possible that his rendering, which has hitherto been fairly accurate — for, with the 164 RIDDLES OF THE EXETER BOOK exception of the notorious pernex (41 66), his departures from his original are the result of intention, not of ignorance — would suddenly become glaringly weak and faulty ? I cannot reconcile such changes as these with the presence of but a single translator in Rid. 41. Now is it not more reasonable to believe that the original translator (A) closed his work at Aldhelm's forty-third line — a very good termination, for here is the end of a long line of comparatives — and that Rid. 41 82-97 represents the render- ing of Cr. 6t-66 by another writer (Z?) far inferior in method and knowledge, who supplemented his work by an equally faulty translation (C) of Cr. 44-50, the next lines in his text of Aldhelm ? A seeming objection to this theory is really strongly in its favor. In its phra- seology B owes much to A. In 41 82-83, closeness to the Latin is sacrificed in order to reproduce 41 50-51 ; 41 84 is very similar to 41 53 ; 41 90-91 recalls 41 20-21 ; 41 94b is exactly in the manner of 41 2&b, 28h; and 41 95 employs the idiom of 41 29, 60. But is this not the indebtedness of the weak continuator, who fails of method and knowledge, but who repeats phrases at the cost of fidelity to his Latin original ? My line of reasoning is sustained by a very valuable bit of evidence — the existence of another version of 41 82-97 (B), Rid. 67. Dietrich (XII, 235) was wrong in regarding this as another translation of Cr. 61-66; Herzfeld, pp. 6-7, was quite as much in error when he deemed it a greatly condensed form of Rid. 41. This little poem of ten lines displays no knowledge or use either of Aldhelm's Latin or of A (41 1-81). It is a recasting of several ideas in the B portion with a few original additions and interpolations : 67 i is based upon 41 82 ; 67 2a finds its source in 41 95-96 (the use of hondwyrm shows that B and not Cr. 66 is before the writer); 67 2b-3a has no equivalent in the Latin or Anglo-Saxon; 67 3b-sa is perhaps a very concrete reshaping of 41 84, and 67 sb-7a of 41 86-89 (tne resemblance to 4138-40 may be coincidence); of 67 7b-io there is no suggestion in Latin or Anglo-Saxon. The problem speaks strongly in favor of the view that two hands were at work in Rid. 41 ; and that the second later gave freer form to his material. To the cycle of 41 and 67 belongs the fragment 94, with its series of compari- sons; but, as only vestiges of this remain, it is impossible to establish exact relations. I have included in my comments upon this riddle a few of the glosses drawn from two manuscripts of Aldhelm's enigmas: MS. Cambridge Univ. Libr. Gg. V, 35, f. 406 (C) and MS. Royal 12, C. XXIII, f. 102 f. (R). The English glosses to both are printed by Napier, Old English Glosses, 1900, pp. 191-192, 195, and the Latin glosses of the second by Wright, Satirical Poets of the Twelfth Century, Rolls Series, 1872, II, 570 (some of these I have drawn directly from the manu- script, where Wright omits or prints inaccurately). The Latin glossator to R thus introduces the Creatura : ' Diversitas creaturarum diversitate locutionis in ista sententia ostenditur de personis omnibus et naturis uniuscujusque creaturae inter mortales et universa visibilia et invisibilia.' The riddle-subject is not of fixed gender, but is now masculine', now fem- inine. This is somewhat surprising, as creatura and frumsceaft are both fem. nouns. But, as I have already noted, there is little insistence upon grammatical gender in the Riddles; and in this case the subject is beyond bounds of sex. NOTES 165 In addition to various errors in translation, certain lines of our version are met- rically weak or imperfect, 24, 73 a, 84, 86, 87, 101. In many cases the accent falls on unimportant words, particularly on personal pronouns : 32, 49, 73, 88, 89, etc. 41 2 I supply wealdej> not only because wrefrstuj>ttm wealdefr reproduces Cr. I, 'fulsit . . . columnis,' but because the formula healdefr ond wealdefr appears in 415,22 in this context. See also /".r. 759, wealdeiS and healde"5; 122 i, healdest and wealdest; And. 225 b, healdend ond wealdend. Wealdan may govern accusa- tive (Spr. II, 670). 41 4 Dr. Bright regards anwalda as gen. pi. (ds titan hiveorfefr to swd he hweorfefr ymb J>ds. 41 10 Cf. And. 464, 820, oftSaet hie (hine) semninga slaBp ofereode (Herz- feld, p. 19). 41 13 jeghwair. This word is used in our riddle seven times as a padding (cf. 41 18, 30, 37, 30, 69, 82), but does not appear elsewhere in the collection. 41 i6a The MS. to J>on bleafr is metrically objectionable, so I invert as in text with Herzfeld, p. 51. 41 17 grima. The word, which is elsewhere used both as simplex and com- pound in the sense of 'mask' ('helmet'), appears here with the meaning 'spec- ter'. 'Larvula' (Cr. 9), which the word translates, is thus explained by the Latin glossator in R : ' Larvas ex hominibus factas aiunt, qui meriti mali fuerint, quarum esse dicitur terrere parvulos, et in angulis garrire tenebrosis.' With the Old English meanings of grima we find striking parallels in the cognate languages. Grimm, Teutonic Mythology, p. 1045, P°ints out that in O. N. Grima appears as a name of a sorceress, and that ' the adept in magic assumed a mask, grima (p. 238), a trollsham, by which he made himself unrecognizable, and went rushing through the air, as spirits also put on grlmhelms, helidhelms (p. 463); often we see the notion of sorceress and that of mask meet in one, thus in the Lombard Leges Rotharis, 197, 379, " striga, quod est ntasca." ' Even in Roman times, larva is used as both mask and specter (see Harper's Latin Dictionary, s.v.). In C ' larvula ' is glossed by O. E. piica, Engl. 'puck' (Grimm, p. 500). 41 18 The eofor (41 18), which always glosses 'aper' in the vocabularies, is of course the wild boar, while bearg (41 106) is the 'magalis ' or 'magialis' (WW. 271, 32 ; 443, 2, etc.), or ' Mastschwein ' (Jordan, Altenglische Saugetiernamen, pp. 200— 203). For an account of the wild boar in England from the earliest times, and of his importance in the hunt, see Harting, Extinct British Animals, London, 1880, pp. 77 f. The September illustration in the Anglo-Saxon calendar (Tib. B. V), which Harting cites, does not represent a boar-hunt in the forest, but the care of masted swine (see note to 41 105). The hunting of the boar is thus de- scribed, ./Elfric's Colloquy (WW. 93): 'Bar ic ofslSh . . . Hundas bedrifon hyne to me and ic Her togeanes standende fzierlice ofstikode Hyne.' So ' barspere vel iiniitigspere'' is mentioned, ^ilfric, Gloss. (WW. 142, n). As Wright observes (Do- mestic Manners, pp. 69-70), ' It would seem by. this that boar hunting was not uncommon in the more extensive forests.' 166 RIDDLES OF THE EXETER BOOK 41 i9h bidsteal giefeS. Cf. /«/. 388, bidsteal gifeS. 41 2ia senig ofer eorj»an. So 95 10, Gu. 727. 41 21 se ana God. For Barnouw's note upon this phrase, see Introduction (' Form and Structure '). 41 23-28 The Old English glosses to the original of this passage (Cr. 13-15) are interesting : ' odorato ' is glossed by risiendum (C), ' flagrantior ' by stemendre (C) and reocendre (R), 'purpureis' by readum (C), 'connexa' >J gewifrelode (C), and ' rosetis ' by rosbeddum (C). I have adopted, in lines 23 to 25 of my text, Grein's additions; but these are so violent that it is perhaps quite as wise to abide by the readings of the MS., Ic eom on stence strengre }>onne ricels o>>e rose sy [seo or /«] on eor}>an tyrf. The second line obviously lacks alliteration ; but such a lapse is not particularly conspicuous among the metrical weaknesses of this translation. With on eorfcan tyrf compare Ph. 399, of J>isse eor)>an tyrf. 41 24-27 rose . . . lllie. Hoops remarks, Wb. u. Kp. (1905), p. 615: 'Von eigent- lichen Zierpflanzen treten uns in der angelsachsischen Literatur nur die Rose und Lilie entgegen. Doch werden manche der iibrigen kultivierten Gewachse, namentlich der Arzneipflanzen, zugleich die Rolle von Zierpflanzen spielen.' He also notes, ib., p. 650: 'Von eigentlichen Zierpflanzen werden in der altnordischen wie in der altenglischen Literatur nur die Rose und Lilie erwahnt.' The history of these among the Indo-European peoples is traced by Hehn, Kp. u. Ht. (1902), pp. 247 f. Liining, Die Natur, p. 149, observes : ' In einem Ratsel spricht Cynewulf schon fast wie ein Minnesanger von der Liebe, die der Mensch zu den Blumen tragt.' It is indeed noteworthy that for mankind's love of the lily (41 27) and for the joy- ous beauty of the rose (41 25-26) the English translator finds no warrant in Aid- helm, who simply mentions them. He, however, praises both flowers in his De Laudibus Vtrginum, Giles, p. 141. Liining adds: ' Auch der Heliand spricht von den lieblichen Blumen der Lilie indem er einen an jenes Ratsel anklingenden Ausdruck gebraucht : //'///' mid so lioflicu blomon {Heliand, 1681).' For an almost contemporary tribute to Lily and Rose, see Riddles of MS. Bern. 611, Nos. 34, 35, 52. These have nothing in common with the Rose riddle of Symphosius, No. 45. Note the use of frd two, wyrta, & on sibbe lilian, ftset is cliene drohtung ; on flam gewinne rosan, ftaet is martyrdom.' 41 31 pis fen swearte. For this use of dem. pron. with weak adj. after the subst., Barnouw, pp. 219-220, points to 41 48, )>es wudu fula; 41 si, 83, }>es wong grena (contrast 36 i,se wEtawong); 41 79, of >issum strongan style heardan. No other examples are met in the Riddles; but compare Chr. 456, se brega miera, Beow. 2676, se maga gebnga, 3029, se secg hwata. 41 36 jricce ond Jjynne. Here the translator falls into the error of associat- ing 'grossas et graciles' with the preceding line (Cr. 19) and not with 'figuras' (1. 20). NOTES 167 41 39 Cf. Bede, Eccl. Hist. IV, 3 : ' Him Dryhten synderllce his digolnysse onwreah.' 41 41 So the poet renders ' tetra Tartara' (Cr. 22). Cf. Chr. 1533 f., f iege gjestas, on wr5}>ra wlc womfulra scolu. This passage supports MS. worn against Gn.2 wonn. 41 46 fraetwum goldes. The phrase renders Aldhelm's 'fibula' (Cr. 25). Fib- ulas are thus described by Isidore of Seville, Origines, Bk. XIX, chap, xxxi : ' Fibulae sunt quibus pectus faeminarum ornatur vel pallium tenetur a viris in humeris seu cingulum in limbis.' Nowhere else in Europe are found in so small an area so many models of fibulae as among the Anglo-Saxons. See De Baye, Industrial Arts of the Anglo-Saxons, pp. 37 f. ; Roach Smith, Introduction to Catalogue of Anglo-Saxon Antiquities in the South Kensington Museum ; Aker- man, Remains of Pagan Saxondom, pi. xiv, xviii, xx, etc. Perhaps fibulas are meant by And. 302, wira gespann (see Krapp's note). 41 49 waroS. Cr. 26, ' rhamnis ' is glossed \3yfyrssum (C) ; and ' algis ' by wdrum (C), which, like source and context, supports the meaning 'weed' for the hapax wdrofr. Sievers (PBB. X, 454) reads ivarofr and regards the half-line as an A-type with second stressed syllable short (_L X X | ^ X ). See note to 3 8. 41 50-51 These two lines are repeated in the B portion of the riddle, 41 82-83 (supra). — J>6s wong gr6na. Cf. £K. 718, se grena wong; Rid. 675, grene wongas (note). 41 53 Cf. Met. ii 35b, utan ymbclyppe'S. 41 54b se hearda forst. So Ph. 58. 41 56 Ulcanus. Here the Anglo-Saxon genitive form that is found in many proper names (cf. Saulus, Matheus) renders the genitive of the Latin, 'Vul- cani.' 41 57 leohtan Ifoman. Cf. Jud. 191, Met. 55, Sat. 469, leohtne leoman; Az. 78, leohte leoman. 41 59 Beobread (N. E. bee-bread) is always associated with honey in Anglo- Saxon writings (see the many examples offered by Cortelyou, Die altenglischen Namen der Insekten, pp. 28-29) ! an^ in the Glosses hunig and beobread are found invariably with the lemma 'mel et favum' (Tib. Ps., Vesp. Ps., Cant. Ps., xviii, n). It is therefore a characteristically English, if free, translation of Aldhelm's ' lenti nectaris haustus' (Cr. 31). 4160 wemiod. Hoops notes (Wb. u. Kp., p. 481): ' Spezifisch westgerma- nisch ist der Name des Wermuts (Artemisia absinthium}? Here it translates the absinthia of Aldhelm (Cr. 32). 41 61 on hjTstum. Grein, Dicht. and Spr. I, 133, renders ' im Blattschmuck ' ; but Thorpe was probably right in translating ' in the hursts.' In this sense the word appears nowhere else in the poetry, but is found often in the Charters (B.-T., p. 584) both as simplex (with place-names) and compound. See N.E.D. s. v. luirst. — heasewe. This renders ' glauca' (Cr. 32), which there and in Cr. 65 has the meaning 'grayish.' As Brooke freely translates (E.E.Lit., p. 138), 'the bitter wormwood stood pale gray.' See my note to Rid. 12 i. !68 RIDDLES OF THE EXETER BOOK 41 62-63 Grimm, Teut. Myth. (Stalleybrass), p. 519, points to this passage as proof of the derivation of O. E. eaten (O. N. ibtunn} from etan (eta) ' to eat ' ; but the weight of the evidence is somewhat diminished by the circumstance that the thought here is derived from Aldhelm's Latin, ealdum J>yrse (MS.J>yrre) render- ing Cr. 33, ' Cyclopum,' which is glossed in C enta. Both source and context estab- lish for efnetan the meaning ' eat as much as,' rather than aemulari, ' be equal to ' (Spr. I, 219). Grimm, p. 520, discusses O. E. J>yrs (O. N. J>urs), citing Gn. Cot. 42-43, J>yrs sceal on fenne gewunian | ana innan lande. It is interesting to note that ' Cyclopes ' are rendered anige J>yrsas (WW. 379, 22). « Caci ' (Cacus, the gigantic son of Vulcan) is the lemma to J>yrses (WW. 376, 19). 41 65 Sites. Genitive, ' mit Auslassungdes unbestimmten Furworts = " etwas " ' (Madert, p. 67). Cf. 4 6, nah ic hwyrftweges. 41 66 pernex. As Dietrich rightly explains (XI, 455), this strange creature the ' pernex ' is brought into being by a complete misunderstanding of the ' plus pemix aquilis' of Aldhelm (Cr. 35) ; and, I may add, by a confusion in the mind of the translator of the Latin adjective 'pernix,' not as Schipper suggests with 'fenix' (phenix), but with 'perdix' (partridge). So Grein, Die/it., renders ' Rebhuhn.' Chaucer falls into exactly the same error, Jfouse of Fame iii, 302 (1392), when he renders the pernicibus alls of Virgil (Aeneid iv, 180) by 'partriches winges' (see Lounsbury, Studies in Chaucer II, 205). 41 67 hafoc. The source of this, ' accipitre ' (Cr. 36), is glossed by miish\afoc\ (C), the ' siricarius ' or ' suricarius ' of the Glosses. For hawking among the Anglo- Saxons, see note to Rid. 20 7-8. 41 68 zefFerus. The word ' Zephiri ' (Cr. 35) is glossed by westernes windes (C) and sftfr ernes (R). 41 70-71 snaegl . . . regnwyrm . . . fenycc. These three words correspond to Aldhelm's (37) ' lumbricus et Umax et tarda testudo palustris,' which are glossed in the Cambridge MS. by angeltwicce and rensn&gi and byrdlinge (R botracd). Whitman, Anglia XXX, 383, cites our passage and points out that in the Glosses sntegl is always the gloss to 'Umax' (WW. 121, 31 ; 321, 29; 433, i) and yce to 'botrax' (WW. 161,9; !95> 23 : 361, 32) and 'rana' (477,4). Regnwyrm glosses 'lumbricus' (WW. 31, 9; 477, 2), which in one place (WW. 122, 22) is rendered by rettwyrm vel angeltwicce. 41 72-73 gores sunu . . . wifel. These lines are but a close translation of Aid- helm's 'fimi suboles sordentis cantharus ater1 (Cr. 38). 'Cantarus'is the lemma to wifel in many glosses (WW. u, 28; 198, 16; 363, 4). In the present case the tordwifel (Lamellicornia laparosticticd) is clearly indicated. 4174 se hara stan. Mead says, P.M.L.A. XIV, 190: 'Seven times [in Anglo-Saxon poetry] har is applied to hoary, gray stone, once to the gray cliff, four times to armor, once to a sword, once to the ocean, once to the gray heath, three times to the wolf, twice to the frost, and seven times to warriors, in each case with some touch of conventionality and with an apparently slight feeling for the color.' Cf. Beouu. 887, 2553, 2744, under hame stan; Beow. 1415, ofer harne stan ; And. 841, ymbe hame stan. See my note to 22 3, har holtes feond. 41 76-77 )>es lytla wyrm | pe her on flode gaetJ fotum dryge misses the sense of the Latin and seems an over-elaborate rendering of ' tippula lymphae ' NOTES 169 (Cr. 41) ; but compare Aldhelm, Aenigmata iii, 3, De Tippula, 1. 6, ' pedibus gradior super aequora siccis.' Our translator would seem to be acquainted with other riddles of Aldhelm besides the De Creatura. Yet we are told of the 'tippula,' by the Latin Glossator in R, ' Tippula parvum animal et levissimum . . . et jam cum siccis pedibus super aquas posse ambulare.' According to Cortelyou, p. 96, this insect is of the family of Hydrometridtz or Ploteres. 41 80-81 As I have pointed out, these lines have no relation to the Latin ' tostis mollior extis ' (Cr. 43) and suggest another version of Aldhelm's enigma; but it is possible that they were inspired by 'levior pluma' (Cr. 41), which is not trans- lated in the proper place. 41 82-83 The riddler (£) neglects his source (Cr. 61-62), in which is found no suggestion of J>es wong grena, so that he may repeat 41 50-51 (supra). The C gloss renders Aldhelm's ' tendor ' by ic com tobraidd. 41 86-87 As already noted, these lines seem to convey an exactly opposite meaning to Aldhelm's ' Infra me . . . nihil per saecula constat ' (Cr. 63). Prehn renders, p. 218, 'Nicht ist ausser mir irgend ein ander Wesen gewaltiger im Weltleben ' ; but for this sense of under I find no warrant ; while Grein's inter- pretation (Die/it.) ' unter mir ' involves a contradiction in terms. All difficulties would disappear, if it were possible to regard ivaldendre as dat. sing, of pres. part, qualifying the fern, me (cf. 41 8) and to translate ' Under me ruling, during the world's life, is no other wight,' but unfortunately the order of words op- poses this. 41 92 se mlcla hwael. Cf. Whale, 3, J>am miclan hwale. See also Whale, 47, where the Whale is a symbol of the Devil (cf. Aldhelm, Opera, Giles, p. 10). Jordan says (Die altenglischen Saugetiernamen, pp. 209-210): ' Im Mittelalter waren Walfische in den englischen Gewassern weit haufiger als in modernen Zeiten. Nach Bell, British Quadrupeds, p. 388 wjirde schon im 8. bis 10. Jh. von den Basken im Kanal Walfischjagd betrieben. Aus s£lfr. Coll. [Colloquy, WW. 94, 5, wilt J>u fon sumne hwasl] geht hervor dass auch bei den Angelsachsen dies nichts Unbekanntes war. Und in der Beschreibung Britanniens (Hist. Eccl. i, i) sagt Beda: " Capiuntur autem saepissime et vituli marini et delphines, nee non et balaenae," wofiir Alfred : " her beoS oft fangene seolas ond hronas ond mere- swyn." ' Mark the references to whale-hunting in Ohthere's voyage (Orosius i, i). Aldhelm's 'ballena' (Cr. 65) is glossed by C safis.ce, hrane. For etymology of hwal, cf. Hoops, ' Wels und Walfisch,' Engl. Stud. XXVIII, 92-96. 41 93 Cf. Whale, 29, garsecges gKst, grund gesece'S (whale). 41 94 sweartan syne. The MS. reading is supported by the large number of weak adjectives in Rid. 41 (11. 55, 56,90) and by the 'eye' meaning of syne (cf. Rid- 33 5)- This is also in keeping with the context, whether we render with Grein, Dicht., ' mit schwarzem Auge ' or with B.-T., p. 875, ' with darkened vision.' Herzfeld's reading sweart ansyne has, however, much in its favor; it renders Ald- helm's ' atra ' (Cr. 65) and is paralleled by Run. 31, fseger ansyne. 41 95-96 This seems at first a very wide departure from Aldhelm's ' exiguo sulcat (C gnafr, ciw&) qui corpora verme (C handwyrme; R hondweorni)? but hond- •wyrm, the word chosen also by our translator (see 67 2), catches, like the Cam- bridge and Royal glosses, the central idea of the Latin ; for, as Cortelyou shows 170 RIDDLES OF THE EXETER BOOK (Die altenglische Namen der Insekten, p. 114), it is always found as a gloss to 'briensis' in WW., and is the 'Kratzmilbe des Menschen, Sarcoptes hominis.' ' Die Kratze zeigt sich meistens an Handgelenk, Ellbogen, Knie u.s.w. und wird durch Unreinlichkeit der betreffenden Korperteile sehr begiinstigt. Die Hande werden am wenigsten sauber gehalten, deshalb ist es kein W under dass die Kratzmilbe den Namen handivyrm fiihrt.' 41 98 hwite loccas, | wrajste gewundne. As black hair was held in disfavor (cf. note to 13 8, wonfeax Wale), so fair locks were highly esteemed by the Anglo- Saxons. Hwite loccas of our passage has no counterpart in Aldhelm's Latin ; and elsewhere in the Riddles light hair is mentioned as indicating rank. In 43 3b hwitloc is applied to the hen with a misleading humor that recalls Chaucer's description of Pertelote ; and in 80 4a hwltloccedu marks the woman of position, eorles dohtor, J>eah hio &J>elu sy. Roeder, Die Familie bei den Angelsachsen, Halle, 1899, p. 17, observes : ' Allein im Gegensatz zu den meisten mittelhochdeutschen Dichtern, die fast anatomisch zergliedernd eine schone Frau beschreiben ( Weinhold, Deutsche Frauen, 1882, I, 221 f.), verzichtet die altenglische Dichtung, die im Schillerschen Sinn "naiv"ist, auf ausfiihrliche Schonheitsschilderungen. Sie beschrankt sich darauf, fest gepragte Epithete, die an sich meist farblos und unplastisch sind, zu wiederholen.' As an example of this, he notes the frequent mention of light curly hair. But ' this passion for the blonde ' is as strongly marked in early Germany and Scandinavia (Weinhold, D. F. II, 312 ; Gummere, Germanic Origins, pp. 61 f.). 41 104 wundne loccas. See 41 98-99, loccas \ wrUste gewundne, and my note to 2611, wif wundenlocc. Brooke observes (E. E. Lit., p. 137): 'The English likened this vast covering of forests to curly locks upon the head and shoulders of Earth. . . . Upon me wonderfully waxeth on my head, So that on my shoulders they may shimmer bright, Curly locks full curiously. This is paralleled by the Icelandic imagery, and we ourselves may compare Keats's lovely phrase of the pines : Those dark clustered trees Fledge the wild-ridged mountains steep by steep.' 41 105 aiiiM-stotl swin. C furnishes interesting glosses to Aldhelm's Latin lines : gemastra swina (' scrofarum ') . . . rysele (' exungia') . . . J>onne hig gemccstaf (' referunt dum corpora ') . . . bectreoiv (' fagis ') . . . siudnas (' subulci ' or C 'bubulci'). The September illustration in the Saxon calendar (Tib. B. V ; Jul. A. VI) does not represent, as Sharon Turner supposed (Bk. VII, chap, vii), a boar-hunt, but 'swineherds ["subulci" or swdnas] driving their swine into the forests to feed upon acorns, which one of the herdsmen is shaking from the trees with his hand. The herdsmen were necessarily armed to protect the herds under their charge from robbers.' For the rights and duties of the two classes of swineherds — gafolswdne and Shteswdne — see R. S. P., §§ 6, 7 ; Schmid, pp. 376-378. So in this tract, § 4, p. 374, « jglc gebur sylle VI hlafas }>am inswane, J^onne he his heorde to masstene drlfe.' NOTES I/I ' The importance of swine is seen in the place which the mast-bearing woods occupied in the laws (a fine of six shillings was exacted for masting swine without proper license, /tie, 49) as well as the frequency of pastures to which they were driven at certain seasons of the year; for the swine were not allowed in the meadow or on the stubble, for their grubbing and rooting would soon spoil it for the other animals. Domesday Book furnishes abundant evidence of the presence of small woods and coppices used for the purpose of providing mast and mentions 427 porcarii and 2 rustici porcarii, a distinction which may point to the slave assist- ants and ceorlish swinekeepers. In the charters also there is occasional mention of the mast-yielding woods which often formed a part of the boundaries, and the acorns and beechnuts were beaten down by the herdsman, as well as left to fall when ripe. It is needless to multiply instances of swine pastures of which these wood-groves formed a part ' (Andrews, Old English Manor, p. 209). See also Traill's Social England I, 213-214. 41 106 The Sow tells us at the close of Aldhelm's riddle De Scrofa Praegnante (vi, 107-9): Fagos glandibus uncas, Fructiferas itidem florenti vertice quercus Diligo, sic numerosa simul non spernitur ilex. And the beech-tree is called brunra beat, Rid. 92 i (note). R contains an interest- ing gloss to Cr. 49, ' glandiferis . . . fagis ' (omitted by Wright) : ' Fagus et esculus arbores glandifere ideo vocate creduntur qua earum fructibus olim homines vixe- runt cibumque sumpserunt et escam habuerunt. Esculus esca dicta.1 41107 wrotende. The word is always used of swine (B.-T., p. 1277). — \vynnum lifde. This phrase refers to swine, while Aldhelm's 'laetantur' (Cr. 50) points to the swineherds ('subulci'). RIDDLE 42 Dietrich (XI, 473) believes that 'the Mother of many races' is the Earth, and that her offspring are the fruits of the soil, iron, fire, water. The solution is not impossible. Frischbier (Zs.f.d.Ph, XXIII, 258, No. 178) offers a Prussian riddle, ' Menschenwelt,' ' Meine mutter hat viele kinder; sind sie gross, verschlingt sie alle ' ; but this has little in common with our problem. Trautmann (Anglia, Bb. V, 49), without apparent warrant, suggests ' Fire.' I was once inclined to think that the answer is 'Wisdom' (cf. Flores, i, Mod. Phil. II, 562, 'ilia mulier quae innumeris finis ubera porrigit,') and pointed out, M. L. N. XVIII, 104, that Wis- dom is ' the mother of many races, the most excellent, the blackest, the dearest which the children of men possess ' (cf. 27 i8f. ' Book ' ) — ' blackest ' referring to the script of books, the precious products of Wisdom, which is called ' black seed ' in one of the best known of world-riddles (Wossidlo, No. 70). But the close con- nection of our problem with the 'Water' riddles points to a like solution here. In 34 9-10, the Ice says of the "Water: Is min modor ma-gS'a cy tines \/>a"s deorestan (cf. 42 4), and in 84 4 ' Water ' is called Modor . . . monigra marra ivihta (cf. 42 2). The variety of her offspring and her service to man, the two motives of Rid. 42, !72 RIDDLES OF THE EXETER BOOK are elaborated in 84 8, 25-37. We cannot live here on earth without the food and drink that water furnishes to man (42 6-7). 42 2-4 So the riddler describes ' the seas and all that in them is.' Nor, as the close parallel to the Ice problem shows, does he confine himself only to ealle J>a f>e onhrerafr hreo wSgas (Az. 141), but has in mind the waters themselves, sources and streams. With selestan, compare 84 27-28 (Water), eadgum leaf . . . freollc, sellic, etc. ; with deorestan, 34 10 (see supra) and 84 36, gimmum deorra ; and with sweartestan a word well suited to the fisca cynn, Aldhelm, Cr. 65, ' in glaucis . . . ballena fluctibus atra.' 42 5 ofer foldan sceat. Cf. Chr. \ 533, under foldan sceat ; Met. 4 52, geond foldan sceat. 42 9 Very like is the closing formula of Rid. 29, which our riddle otherwise resembles in the use of superlatives (29 2-3, 42 2-3) and of brucan (29 10, 42 7). RIDDLE 43 There is no Latin source to this runic riddle of the Cock (Hand) and Hen (Han). Petsch (Zs. d. V.f. Vk. VIII, 115) notes that the Cock is the ' erklarte liebling der volkstiimlichen kleinpoesie ' ; and there are many cock riddles, Ger- man (Miillenhoff, Zs.f.d.M. Ill, 17), English (Chambers, Popular Rhymes of Scotland, p. 326), Norse (/. G. No. 289). But none of these bear any resemblance to our problem. In its mention of all the outbuildings of the Anglo-Saxon mansion, the Gerefa, n (Anglia IX, 262), includes a hennery: 'swyn stigian on odene cylne macian — ofn and aste and fela "Singe sceal to tune — ge eac henna hr5st.' Hens are mentioned in Anglo-Saxon wills (Thorpe, Diplomatarium, 509, 18, mi haen f ugulas). To the early Englishmen the cock is always the ' orloge of thorpes lyte ' (for a discussion of Hancred as a time-division, see my Anglo-Saxon Dcegmlel, P.M.L.A. X, 1895, pp. 149-152). Hehn, Kp. u. fft., 1902, 598-600, has con- sidered the place of the Cock and Hen among the Aryans. 43 2 plegan. Sievers (PBB. X, 520) suggests plegian on metrical grounds. Madert, p. 28, notes that in the present strong forms of this verb appear (Sievers, Gr.s, § 391, n. i). He adds : ' Because Type A with short second stress is often found in the Riddles, it is not necessary to accept Sievers's emendation.' Plegan is found with the seon construction, Gen. 2778, El. 245. 43 3 hvvitloc. See note to 41 98. So the Hen of Chaucer's Nonne Preestes Tale is ' cleped faire damoysele Pertelote ' (B. 4060). 43 4 ]?aes weorees spgow. Elsewhere we meet the gen. construction with spdwan (that in which any one succeeds) only in Gtn.zSioL, ]>e glen a speow | J>aes Mi wrS freond, etc. The instrumental is usually found (Spr. II, 471). 437 b€c. Cosijn remarks (PBB. XXIII, 129-130): 'bee, "buchstaben" wie Dan- 735- arendbec (PBB. XX, 115) ? Aber der schreiber schrieb den text seiner ratsel gewis nicht in runen, nur die zu erratenden worter.' pdm J>e bee -witan is probably used conventionally for ' wise ' or ' learned men.' 438-11 For Sievers's discussion of these runic lines (Anglia XIII, 5f.) see Introduction ("Authorship"). NOTES 173 43 9 ' Sc torhta asc wird der Baum genannt wegen seiner hellgrauen oft silbern schimmernden Rinde ; eigentlich ist an dieser Stelle die Rune ae gemeint aber das Beiwort bezieht sich naturlich auf den Baum ' (Hoops, Altengl. Pflanzennamen, pp. 36-37). Liining (Die Natur etc., p. 136) cites the Edda (H. Hu. ii, 36), itrska- pafrr askr, ' wol von der silbergrau schimmernden Rinde.' For further discussion of the Ash, and of its use as a spear, see notes to Rid. 73. 43 11-15 1° Spr. II, 121, Grein explains hwylc (1. n) as ei qui or si quis, and in Die/it, translates : dem der des Hort-Thores Verschluss erschloss durch des Schliissels Kraft, Der dieses Ratsel vor den rathenden Mannern Hiitete sinnfest dem Herzen bewunden Mit kunstvollen Banden. I dissent utterly from this interpretation, and regard hivylc as simple interrogative, and damme as the antecedent of J>e (1. 13). So I translate 'which (of the rune- letters) unlocked, by the power of the key, the fastenings of the treasury-door, that held (i.e. protected) against those skilled in mysteries (rynememi) the riddle (i.e. its solution) fast in mind, covered in heart by means of cunning bonds ?' Just as if one should say 'which letter gave you the clue ? ' For a discussion of hordgates and cizgan crtzfte, see my notes to Rid. 45 and 91. 43 12 clegan craefte J?a damme onleac. With this compare ^Elfric's phrase in the introduction to his Grammar : ' Staefcraeft is seo caeg "Se "Sara boca andgit unllcj>.' See also Sal. 184-185, boca c[jega] [le]ornenga locan. 43 16 werum set wine. Cf. 47 i, wer saet aet wine. RIDDLE 44 Dietrich (XI, 473) rightly points out that ' the noble guest ' and his servant, who is also his brother, are the Soul and the Body, and that the kinswoman, mother and sister (cf. Rid. 83 5) of them both, is the Earth, — mother, because man is molded from her ('mother-earth'); sister, because she is created by the same father (God). The only resemblance to Eusebius, No. 25, De Animo, lies in deorne giest and ' accola magnus ' ; and the leading motives of the two riddles are so different that this slight likeness may be a coincidence, not surprising in view of the demands of the common topic (infra}. E. Miiller, who prints Grein's text and translation of Rid. 44, and discusses the problem at length (Herrigs Archiv XXIX, 1861, 212-220) believes that in the case of this enigma we have no definite source, but the frequent and popular motif 'of Body and Soul journey- ing through life as servant and master. He points out that spiritual reflection is revealed in the outlook upon eternal punishments and joys, and in the contrast between the two sides of man's nature, but that the popular element appears in the expressions, in the alliterative form, in the turns of speech, and in the single words. He analyzes the vocabulary of Rid. 44, word by word, and indicates certain parallels of thought between this and such poems as The Grave (De wes bold RIDDLES OF THE EXETER BOOK gebyld), which he considers at length. Mone, Anz. II, 235, records a fifteenth- century German riddle, obscure and full of symbolism, containing, among many other puzzling phrases, these: 'My son was my father and my mother and my daughter'; 'I was practiced in the art of healing, and overcame all sickness.' In the margin is given the answer : ' Es ist leib, geist und sel.' The association of Body and Soul is a favorite theme of Anglo-Saxon poets, not only in the Exeter and Vercelli poems with that single motive, but in the works of the Cynewulfian group (Herzfeld, p. 18). Body and Soul are a married pair, Gu. 940, Jul. 697-701, and are companions on a journey, Chr. 176, 1036, 1326, 1580, Gu. 810, 1 149,/«/. 714, Ph. 513, 523, 584 (Dietrich, XII, 246; Gaebler, Anglia III, 512); but we meet them only here in the relation of servant and lord. For the bibliography of Body and Soul Streitgedichte, see Kleinert, Ueber den Streit zwischen Leib und Seele, Halle, 1880 ; Wright, Poems of Walter Mapes, Camden Society, Appendix; Vamhagen, Anglia II, 225; Rieger, Germania III, 398 f.; Zs.f.d.Ph. I, 331-334; Bruce, M.L.N. v, 193-201. 44 i Cf. 95 i, indryhten ond eorlum cutf. 44 2 giest in geardum. The phrase recalls not only the accola magnus of Eusebius, but the well-known lines of Hadrian's Address to his Soul : Animula vagula blandula, Hospes comesque corporis, Quae nunc abibis in loca? Cf. Chr. 819-820, sawel in lice | in bam gaesthofe; 1480 f. ; Exod. 534, J>ysne gystsele (the Body). Cook in his note to the Christ passage (p. 166) points to 2 Cor. v, i, ' our earthly house of this tabernacle.' A play upon words, gast and giest, was perhaps intended by the riddler ; if so, it was lost in the later giest, the scribe's form. 442-4 Compare Ph. 613: hunger se hata ne se hearda Jmrst, yrnrSu ne yldo. See also Chr. 1660, Nis J>JEr hunger ne burst. 444-5 Cosijn, PBB. XXIII, 130, pointed out that the additions of Grein were unnecessary to either sense or meter (see my text). As in 41 96, Dream 98, se ]>e = tone J>e. 44 5, 8, 16 esne. About the social position of the esne there has been much discussion. Kemble, Saxons in England I, 8, p. 176, thinks that he was a poor free day-laborer serving for hire ; while Maurer, Kritische Ueberschau I, 408, whom Andrews follows (Old English Manor, p. 194) would place him in a special class of the unfree as 'one who received for his work servant's wages.' For a judicial discussion of his status, see Schmid, Gesetze, ' Glossar,' s. v. No one denies, how- ever, that he was originally of the servant class, and that he was of a higher rank than the J>eow or wealh. Bartlett, Metrical Division of Paris Psalter (1896), p. 21, shows that esne as ' slave ' is specifically Anglian. Klaeber, Anglia XXVII, 263, points out that esne in West Saxon is archaic, but it appears frequently in the oldest laws (only once in the later, R. S. P., § 8); and continued long in the North NOTES 175 (R., Lind., Rit.). While esne as 'slave' is replaced by J>eow, esne as 'vir' appears in jElfric, Old Test., and in Byrhtferth (Anglia VIII, 321 ; 331, 33). In the Kiddles the word is used in both senses: in 28 i6b it seems synonymous with ceorl\ is applied to a servant by contrast with frea in 44 ; and refers simply to man or youth in the coarse riddles, 45 4, 55 8, 64 5. Compare Jordan, Eigentiimlichkeiten des anglischen Wortschatzes (1906), p. 91. 446 on ]mm siSfaete. For references to the common journey of Body and Soul, see supra. 44 7 iindiiff wltode. Cf. Gu. 890, witude fundon. 44 10 forht. Klaeber says {Mod. Phil. II, 145, note): 'Grein's explanation of \h\sforht as 'terribilis' in the Sprachschatz (so Thorpe, Toller), and his transla- tion "und der Bruder dem andern nicht will unterthanig sein" are open to doubt. It will be better to take brdfror oj>rum as parallel to esne his hldforde and interpret ne wile for/it wesan as a parenthetical clause, "will not live in fear" — a thought well illustrated by the Discourse of the Soul to the Body.' I can see no reason for accepting Klaeber's explanation, as both for ht andforhtlTc are used in the active sense of 'formidable,' 'terrible' (Spr. I, 326). Indeed, I prefer to begin a new thought with ne (1. 10). 44 na broJ7or ojmim. Kluge, PBB. IX, 427, cites Gn. Cot. 52-53 : f yrd wrS fyrde, feond wr5 oiSrum, la« wrg late- As in 4 42b-43a, sceor wifr oj>rum, \ ecg wt'fr ecge, double alliteration is avoided in the second half-line of the Gnomic verse by avoidance of feond -wifr feonde (con- trast, however, 51 1?, feond his feonde). 44 14 moddor ond sweostor. The relationship of the earth to the body and soul of every man suggests Rid. 83 5, eorfran broj>or, and the Anglo-Saxon prose riddle. The one Anglo-Saxon prose riddle, a relationship problem found in MS. Vitellius E. XVIII, 16 b, has been printed by Wanley, Catalogue, p. 223, by Mass- mann, Mones Anzeiger, 1833, p. 238, by Grein, Bibl. II, 410, and by Forster, Herrigs Archiv CXV, 392 (see my note to ' secret script ' of Rid. 37). I give Varnhagen's reading as presented by Forster : ' Du J>e fajrst on )>one weg, gret 'Sii minne bro'Sor, minre modor ceor[l], Kme acende mm agen wlf, andic wass mines broSor dohtor, and'vc. com mines fasder mSdor geworden, and mine beam syndon geworden mines faeder modor.' Dietrich (XI, 489) believes that in the first part of the riddle (cf. mm dgen wtf) a man is speaking, in the second a woman ; so he regards the problem as double, and gives the two answers ' Day' and ' Eve.' Grein, Germania X, 309, gives the solution ' Eve,' and meets all difficulties in his analysis and translation: 'Griisse du meinen Bruder (Adam), meiner Mutter (der Erde) Bauer (ceorl), den mein eigen-Weib (die der Eva unterthane Erde) gebar, und ich war meines Bruders (Adams) Tochter und bin meines Vaters (Gottes) Mutter ge- worden (als Ahnfrau Christi) und meine Kinder sind geworden meines Vaters (Adams) Mutter (Erde, d. h., sie sind im Tode wieder zur Erde geworden).' This solution finds striking confirmation in the circumstance that Schick and Forster (Herrigs Archiv CXVI, 367-371), working in entire ignorance of Grein's article, reached the same conclusions as he, point for point. Complex and sophisticated 176 RIDDLES OF THE EXETER BOOK though this prose riddle may seem, it is full of popular motives common in riddle and dialogue literature (see my note to Holme Riddles, No. 78, P. M, L. A., XVIII, 262; Kemble, Salomon and Saturn, pp. 295-298; Forster, Furnivall Miscellany, pp. 86 f.). 44 16 e?f}m. The Northern form of West Saxon o&fre, which is found as eJ>J>a (Rush. Matt, v, 17), and as tzththa (Bede's Death-Song, 1. 4), is considered by Sievers, Gr.s, 317, and by Madert, p. 29. See also PBB. XXIV, 403 f., 504, on 'oder.' — J?e ic her ymb sprice. Cf. 24 n, J>aet ic >xr ymb sprice (see note). RIDDLE 45 To this obscene riddle Dietrich (XI, 475-476) offers two answers, ' Key' and ' Dagger Sheath.' Either or both may be correct (see my article, M. L. N. XVIII, 6), as each has strong support. The first is favored by Rolland's fif- teenth-century French riddle (No. 144), by Eckart's Low German queries (Nos. 222, 223), by Wossidlo, Nos. i45a, 434 n2, and by the very lively problems in the Islenzkar Gdtur (Nos. 603, 607, Skrd og Lykill), all of which bear many resem- blances to the Anglo-Saxon ; the second is sustained by Wossidlo, No. 434 i2, and by the very similar English puzzle in Holme Riddles, No. 130, and in Royal Riddlt Book, 1820, p. ii. As the Anglo-Saxon key is associated with women (Wright, Celt, Roman and Saxon, p. 491), and this object hangs bi weres J>eo, Dietrich in- clines to the second solution; but Trautmann has shown (BB. XIX, 192-195) that the words of the riddle better suit the first answer, as the key is hollow in front (45 2b), is stiff and hard, and is the active agent of the last lines of the rid- dle. But, as I have pointed out (M. L. N. XVIII, 6; XXI, 102 ; see Introduction), it is unwise to dogmatize over the answers to Anglo-Saxon riddles of this class. It is probable that the collector himself knew and cared little about the original solutions, since any decorous reply would adorn his unseemly tale. The element of double entente in such problems is completely overlooked by Walz in his discussion of Rid. 45 (Harvard Studies V, 265). For the duties of the Key, see Rid. 91 and my explanatory notes. Rid. 45 is closely bound by its diction to other obscene problems, 26, 46, 55, 63, 64 (see Introduction). 45 i As Trautmann has noted (BB. XIX, 194),^^ represents the dissyllable J>eoe, demanded by the verse. 45 2 foran is jjyrel. In 91 5 the Key is described as J>yrel. 45 2,4 frean . . . esne. Trautmann (BB. XIX, 194-195) remarks that esne has here not the meaning "servant," but the more general sense of " man." ' In any case the esne, who is the lord of the key (compare the ' comitatus ' of 18 and 24), is not to be contrasted with frea, as Grein does in his Dicht. when he trans- lates the latter as 'Fiirst,' the former as ' Untertan.' Contrast the use in Rid. 44 (see my notes). 45 4-5 Trautmann (BB. XIX, 195) makes the rather obvious comment that it must have been very customary for men in Anglo-Saxon times to wear long gar- ments (see Rid. 55 3-4). This fashion is illustrated by scores of pictures in every illuminated manuscript. See Strutt, Horda Angelcynna, p. 46 ; Fairholt, Costume in England, 1885, I, 42. NOTES 45 6 It is hardly necessary to assert, contra Trautmann, that the riddler in his use of hangellan had not in mind the fem. gender of cteg, the subject of the prob- lem. In the Kiddles, as I have several times pointed out, there is no such insist- ence upon grammatical gender (see 24 7, 25 7, 36 3 and Leid. 3, 39 6-7, 41 passim). 45 7 efenlang. Trautmann is right in substituting this for MS. efe long, which Grein, Dicht., renders 'die langliche(?),' but which in Spr. I, 218, he thus explains ' efe-lang fitr efenlang (emlang, emnlang, Lye), adj. gleichlang ? Wright, Prov. Diet., giebt ein engl. evelong, " oblong." ' As Trautmann says, ' efen-lang finds support in efen-eald and efen-wvifr, and the sense demands the meaning "gleich- lang," "just as long.'" Efenlang ar, in its position at the end of the first half- line, suggests 954, me fremdum (MS. fremdes) izr, where adjective and adverb stand in the same relation. RIDDLE 46 Dietrich (XI, 474) suggested, somewhat doubtfully, ' Bee ' ; but Herzfeld and Trautmann have independently given the obvious solution 'Dough.' As I have noted (M.L.N. XVIII, 103), confirmatory evidence is overwhelming. The riddle appears in various forms in modern Germany (Eckart, Nos. 88, 440, 506; Wossidlo, Nos. 71, 126), does service in the fifteenth century (Kbhler, Weimar Jhrb. V, 329 f., No. 30), is cited twice in Schleicher's Lithuanian collection, p. 195, and is known to English peasants (Royal Riddle Book, Glasgow, 1820, p. 4). Hoops, Wb. u. Kp., p. 595, shows that among the Anglo-Saxons wheat was the chief grain for bread [Thorpe, Homilies II, 460, 16] in the midlands and the south, where the climate favored its cultivation ; while in the north, as earlier upon the continent, barley was the staple grain. In the ninth century the supply of wheat exceeded the home demand. Hoops points out that in the Egils Saga, chap. 17, 7, the Norwegian Thorolf about 875 sent his people to England to buy wheat and honey, wine and clothes. Leo, Rectitudines Singularum Personartim, pp. 1 98 f ., describes the various breads of the Anglo-Saxon : ' Gtsu/e/ /i/a/[Kemb\e, Codex Dipl. I, 193, 296 ; Th. Horn. II, 460, 32 ; Schmid, Gesetze, p. 166, 'Glossar' s.v.] ist brot was man zu anderen speisen hinzu isst, denn sufl ist alles was zum brot als zukost genossen wird . . . es scheint also was wir nennen hausbacken brot zu sein ; clien lildf [Thorpe, Horn. II, 460, 16] ist ohne zweifel noch ein besseres waizenbrot . . . es ist brot vom reinsten mehl ; f>eorf hlaf [' azymus panis,' ungesauertes brot ; see Thorpe, Horn. II, 264, 3] mochte dem schweren, schwarzen brot (Jtihigr, fryckr hleifr) der alten Nordlander entsprechen, worin auch die Kleien waren.' See also Bouterwek's Einleitung zu Cadmon, pp. xci f. Wright, Domestic Manners, p. 29, notes that in the many illustrations of feasts in the manuscripts the Anglo-Saxon bread is in the form of round cakes, much like the Roman loaves in the pictures at Pompeii. Bread- making by Anglo-Saxon ladies, as suggested by the etymology of hlafdige, is dis- cussed by Heyne, Fiinf Biicher I, 58, 119, II, 268. In our riddle we have the most vivid description of the woman's work of kneading. 46 i weaxan. The MS. weax is retained by Thorpe, Grein, and Wiilker; and Grein in Dicht. renders ' ein Gewachs,' but in Spr. II, 276, follows Dietrich XI, 178 RIDDLES OF THE EXETER BOOK 474, in regarding ' weax = weacs oder -wdces, gen. n. von wdc, " weich." ' Sievers's suggestion wdces (PBB. X, 520) finds support in 62 9, ruwes ndthwczt, in 55 5, stif>es ndthwcet, and in 46 3, bdnlease. But I prefer the reading of Herzfeld (p. 69) and Holthausen (/. F. IV, 387), weaxan, which accords with both the grammar and the sense of the passage, as well as with the metrical demands of 46 ib. 46 2 jrindan ond Jmiiiaii. The swelling of the Dough is naturally the leitmotif in the popular problems that I have cited. 46 3 bryd grapode. Cf. 26 7, heo on mec gripeS. 46 4 hygewlonc. So, under the same circumstances, the woman in 26 7 is modwlonc, and in 43 4 wlanc. Cf. 46 5, fceodnes dohtor, with 26 6, ceorles dohtor (see my note). 46 5 prindende. Thorpe's reading frindende is supported by 46 2, Jiindan, and Grein's conjecture frrintende by Mod. 24, brintej>, and by Rid. 38 2, dj>runten. The MS. form is a hapax-legomenon. RIDDLE 47 This query of ' Lot with his two daughters and their two sons ' (T. Wright) is one of the oldest and best known of relationship-riddles, as I have twice shown (M.L.N. XVIII, 102; note to Holme Riddles, No. 10). Schechter ('Riddles of Solomon in Rabbinic Literature,' Folk-Lore I, 1890, p. 354) cites this from Mid- rash Hachephez (Brit. Mus. Yemen MS. Or. 2382) as the second query proposed by the Queen of Sheba to Solomon (compare Friedreich, pp. 98-99, citation of an older Midrash ; Hertz, ' Die Ratsel der Konigin von Saba,' Haupts Zs. XXVII, 1-33; Wiinsche, Rdtselweisheit bei den Hebraern, p. 16). It appears twice in Reusner's collection (I, 335, 353), in the second case as a mock epitaph ; is noted by Wossidlo, No. 983, notes, in several modern German forms ; and is considered by Petsch, p. 14. Compare the Scandinavian versions (Islenzkar Gdtur, Nos. 594, 688, and Hylten-Cavallius, Gator oek Sporsmal fran Vdrend, No. 117), and the English forms (Chambers, Popular Rhymes, p. 113, and Gregor, Folk Lore Soc. Pubi. VIII, 76). The Reusner version (I, 353) reads : Wunder iiber Wunder, Hier ligt begraben under Mein Vatter und dein Vatter, Und unser beider Kinder Vatter, Mein Mann und dein Mann, Und unser beider Mutter Mann, Und ist doch nur ein Mann. Our query seems to have had no vogue in the Middle Ages, yielding in favor to such riddles of strange family ties as those of the Reichenau MS. 205 (Mullenhoff and Scherer, Denkmaler, 1892, p. 20) and Strassburg Rb., No. 305, or of incest as that proposed by the King in the Apollonius story (Riese, Apollonius von Tyrus, 1893, chap, iv ; Gesta Romanorum, chap. 153 ; Shakespeare's Pericles, i, i). In our riddle the theme is given a Germanic coloring by 47 i, wer . . . a:t wine (cf. 43 16, werum at wine), by 47 5, ce^elinga, and by 47 7, eorla ondidesa. Compare NOTES 179 with this riddle-treatment the Anglo-Saxon version of the story of Lot and his daughters, Gen. 2598-2613 (see prose Genesis, xix, 30-38). 474 freolieo frumbearn. Cf. Gen. 968, freolicu twa frumbearn; 1189, freolic frumbearn ; 1618, ful freolice feorh, frumbearn Chames. 47 6 earn ond nefa. See note to 39 6-7. Other half-lines of shortened A-type (.L. X | v5 ) are noted by Herzfeld, pp. 44-49 ; compare 18 n, 39 6-7, 43 2(?), 93 10, etc. (Introduction). RIDDLE 48 It hardly needs Prehn's long discussion (pp. 220-223) to establish the obvious connection between this ' Bookmoth ' riddle and its source, the 'Tinea' enigma (No. 1 6) of Symphosius : Litera me pavit, nee quid sit litera novi. In libris vixi, nee sum studiosior inde. Exedi Musas, nee adhuc tamen ipsa profeci. Of the Anglo-Saxon version, Dietrich remarks (XI, 451): ' Hier ist besser erzah- lung statt der eignen rede der unbedeutenden personlichkeit eingefiihrt und, was sonst nicht wieder vorkommt, der gegenstand selbst genannt, und somit nur das buch zu rathen vibrig gelassen.' As Prehn points out, the leitmotif of the Sym- phosius problem (see 48 5-6) appears in the ' Bookcase ' riddles of Aldhelm ii, 14, and Eusebius, No. 33 (see Rid. 50 n). Our riddle is found not only in Islenzkar Gdtur, No. 761, but in many modern English forms : Holme Riddles, No. 13 ; Wit Newly Revived, 1780, p. 2 ; Royal Riddle Book, p. 14 (' Mouse in a study ') ; Riddles, Charades and Conundrums, 1822, No. 64. In Rid. 48 we find six lines, where the ' Tinea' enigma has only three ; but it cannot be truly said with Herzfeld, p. 29, that the method of 36 and 41 is fol- lowed, and that to each line of the Latin two correspond. It is true of the riddler, however, that ' Neue Seiten hat er hier seinem Gegenstande allerdings nicht ab- zugewinnen vermocht.' 48 2 Wrcetllcii and ivundor suggest the usual opening formulas, and ic gefratgn connects this riddle with 46 i, 49 i. 48 4b-5a These words suggest the praise of books in Rid. 27, 50, and 68, but the closest analogues of />izs strangan stafrol are found in the description of books in the ' Beech ' riddle, 92 3, ivynnstat>ol, and in Sal. 239, gestrangafr hy ond gestafreliafr stafrolfcestne gej>dht. RIDDLE 49 This has much in common with Rid. 60; and Dietrich (XI, 474; XII, 235) closely associates the two, offering as a solution to our riddle ' Pyx,' and to its fellow ' Chalice or Communion Cup.' I agree in the main, but I am inclined to think that the Paten or Plate, not the Pyx or Box, the huseldisc rather than the hnselbox, is intended in 49. Yet the distinction between these two sacred vessels (tiuselfatit) is very slight. Both Chalice and Paten are described l8o RIDDLES OF THE EXETER BOOK • by Aldhelm, ' De Basilica Edificata a Bugge,' P. L. LXXXIX, 290 (cited by Dietrich, 1. c.) : Aureus atque calix gemmis fulgescit opertus, Ut caelum rutilat stellis ardentibus aptum, Sic lata argento constat fabricata patena, Quae divina gerunt nostrae medicamina vitae (see 495). The ring ' without a tongue ' (49 2) and ' dumb ' (60 8), which yet brings by its silent speech to the minds of reverent men thought of the Savior and his wounds, may well be the circle of the golden Chalice or of the Paten. The germ of both riddles is found in Aldhelm vi, 4 4 f. (De Crismale) : Et licet exterius rutilent de corpore gemmae, Aurea dum fulvis flavescit bulla metallis, Sed tamen uberius ditantur viscera crassa Potis qua species flagrat pulcherrima Christi. To Aldhelm's enigma Tatwine also is indebted (121-2, De Patena). The priest, who is introduced in Rid. 60, suggests the Tyrolese riddle of like topic (Renk, Zs.d. V.f. Vk. V, 149, No. 17): Ich geh' an einen Ort, Dort seh' ich einen Mann, Der hebt dein' und meinen Vater Mit beiden Handen dort. (Der Priester, wenn er die Hostie erhebt.) Westwood, Facsimiles, p. 108, cites from the Psychomachia manuscript (Cott. Cleopatra C. VIII) the figure of a priest standing before an altar with a chalice in his right hand (see also Add. MS. 24199, cited by Westwood, p. 107, and the Bodenham drawing, Strutt, fforda, pi. xxiv). 49 i Here the MS. reads hringende an. Klaeber (Mod. Phil. II, 145) rejects the Gn. and W. firing \jzr~\endean and reads hring endean, endean or tendean being = ierndean = izrendian (erendian). ' The form (ge)arndian, it seems, was not in- frequently used; cf. e.g., Ine Laws, 33 (H.) ; Bede, 420,22 (Ca.); Wulfstan, 20, 19 ; and the suppression of the r may be regarded as a natural process (M. L. N. XVIII, 244).' Klaeber cites Nine Herbs Charm, 24, ' gemyne J>u, MaegSe, hwaet J>u ameldodest, | hwaet >u geasndadest aet Alorforda.' I may note also Charm vi, 15 (Gn.-W. Bibl. I, 326), )>a geiendade heo | and aiSas sw5r. I prefer the Gn. and W. reading \tzr\endean. Klaeber restores the MS. version of 49 2-3 with proper division of lines (see text). 49 2-3 hlude | stefne ne cirmde. Rid. 9 6-7 furnishes, with the same idea, the same metrical division : bugendre \ stefne styrme. Cf. 9 3h, hlude cirme ; 58 4b, hlude cirmafi. For occurrences of hlude stefne, see Spr. II, 88. Grein's references show that hlude for hludan (with this fern, inst.) is not so rare as Sarrazin (Engl. Stud. XXXVIII, 160) would have us believe. 493-4 Prehn (p. 224) notes in these lines the 'paradoxe Mischung von teils vorhandenen, teils fehlenden Gliedern ' and compares Rid. 19, 34, 66. NOTES l8l 49 6 Ryne. The mysteries of the Eucharist are mentioned by ^Elfric (Thorpe, Homilies II, 268) : ' WrSutan hi beo> gesewene hlaf and win ajgfter ge on hlwe and on swascce, ac hi beob soj>llce aefter 'Sjere halgunge Cristes llchama and his bl5d j>urh gastllcere gerynu.' 49 6b readan goldes. Compare 60 ioa, goldes tacen. Priests were forbidden by the Canons to use communion-vessels of horn or wood : ' And witaft J>ast bea ale calic geworht of myltendum antimbre gilden oftfte seolfren, glaesen oSfte tinenv ne beo he na hyrnen, ne huru treowen (^Elfric's Pastoral Epistle, 45, Thorpe, A. L., p. 461); 'Beo his calic eac of claenum antimbre geworht, unforrotigendllc ond eallswa se disc ' (^Elfric, Canons, 22, Thorpe, A. L., p. 445). In the British Museum, among the Anglo-Saxon grave-finds, is a silver chalice of 900 A.D., from Tre- whiddle, St. Austell, Cornwall (Proc. Soc. Antiquities, vol. XX). 49 7 bepencan. I cannot regard the suspicious hapax, MS. bej>uncan, which is. received into the text of Th., Gn., W., as aught else than a scribal error for a form very common in both prose and poetry. RIDDLE 60 Dietrich (XI, 475) suggested ' Cage,' but later (XII, 236-237), and with better reason, proposed ' Bookcase.' This solution caps the query at every point: gifrum Idcum (50 3) recalls the Book (27 28), hccle^um gifre ond hdlig sylf; and the precious contents or food of the Case (50 6-8) are clearly the sacred treasure of the other riddle. As I have shown above (48 4b-s), our query belongs to the same class of problems as the enigma of the ' Bookmoth.' And finally, as Dietrich and Prehn (pp. 225-226) have indicated, its last line associates Rid. 50 closely with Aldhelm ii, 14, De Area Libraria : Nunc mea divinis complentur viscera verbis ; Totaque sacratos gestant prjecordia biblos ; At non ex iisdem nequeo cognoscere quicquam. clM^ Trautmann, BB. XIX, 183-184, regards both Rid. 18 and 50 as 'Oven' riddles and finds in them these traits in common : both work by day ; both swallow ; both conceal costly treasures ; men covet the contents of both (so he wrests 18 u, men gemunan, into men gewilniafr). But Rid. 18 is a 'Ballista' problem (supra), and the likenesses to 50 are superficial. 50 i anne. Boc-cyst is fern., and boc-hord and boc-gestreon neut., while our sub- ' ject is masc. But grammatical gender is usually disregarded in the Kiddles. 50 3 gopes. Grein, Spr. I, 520, accepts the reading of the MS. and defines doubtfully either as 'servus' (pointing to O. N. hergopa, 'serva bello capta'; cf. geopan, 'capere ') or as ' listig,'with reference to geap, 'callidus.' Against the first etymology, speaks the length of the vowel in the present word ; against the second, the difficulty of associating phonetically gop and geap. The second deri- vation fits, however, both meter and sense : ' eines kundigen (says Dietrich XII, 237), denn das schreiben war eine angesehne kunst.' Cf. boc-cra;ftig. 50 4-5 se wonna ]?egn, | sweart ond saloneb. According to Brooke (£. E. Lit., p. 136), this is 'the swart thegn with the dusky face' who works with the 182 RIDDLES OF THE EXETER BOOK student in the monastery ; and comparison with 13 4, s-wearte Wealas; 13 8, wonfeax Wale; 53 6, won/ah Wale; 72 10, sweartum hyrde, suggests a servant of Celtic blood. But as f>egn would hardly be used of one of the lowest class, and as eorf, ' brown,' (1. 1 1 ) refers clearly to the hoc-cyst itself, it is perhaps better to explain this with Dietrich (XII, 237) as ' der schrein aus eichenholz mit eisemem schloss und schliissel versehen.' In this case, it will be necessary to regard sendefr . . . him (5-6) as reflexive. With sweart ond saloneb cf. 58 3, swearte, salopade (swallows). 506 golde dyrran. Dietrich (XII, 237) cites Ps. 118127, }>a me georne synd golde deorran (the words of God). 50 7-8 Compare the love of princes for books in the ' Membrana' enigma, MS. Bern. 611, 24 i, ' Manibus me perquam reges et visu mirantur.' 508 J»aet cyn. I do not believe with Dietrich (XII, 237) that the word refers to the books, but that the riddler has in mind those who turn to their advantage (cf. 27 27, to nytte) the precious volumes (' that which the dumb brown one, ignorant, swallows'). RIDDLE 51 Dietrich (XI, 475) and Prehn (pp. 226-227) give the answer 'Dog'; and find the source of the riddle in Aldhelm i, 12, De Molosso: Sic me jamdudum rerum veneranda potestas Fecerat ut domini truculentos persequar hostes, Rictibus arma gerens bellorum praelia patro ; Et tamen infantum fugiens mox verbera vito. , Here, as in the Anglo-Saxon problem, the subject is a mighty warrior; here he stands in fear of a child, as there of a woman. Herzfeld, p. 69, objects that of dumbum twam \ torht dtyhted (11. 2-3) does not suit the Dog ; an objection which loses some of its force when we reflect that ' dumb ' is often applied to beasts (And. 67, J>a dumban neat). Torht seems, however, better suited to Herzfeld's solution ' Fire.' According to that scholar the two dumb things which beget the subject of the riddle are the two stones which are rubbed together (cf. Kemble, Saxons in England I, 358). Or perhaps we may accept the explanation of the Royal MS. (12, C. XXIII) glosses to the first line of Aldhelm's 'Fire' enigma (v, 10 i) 'Me pater (ferrum) et mater (silex) gelido genuere rigore' (see Rid. 41 78-79). Cf. Bern MS. 611, 23 1-2 (Anth. Lat. I, 358) : Durus mihi pater, dura me generat mater : Verbere nam multo hujus de viscere fundor. To Rid. 51 5-7, 9-10, Herzfeld finds ' a remarkable parallel ' in the well-known passage of Schiller's Glocke: Wohlthatig ist des Feuers Macht, Wenn sie der Mensch bezahmt, bewacht, Und was er bildet, was er schafft, Das dankt er dieser Himmelskraft ; Doch furchtbar wird die Himmelskraft, Wenn sie der Fessel sich entrafft, u. s. w. NOTES 183 Trautmann claims to have arrived independently at the ' Fire ' answer, which meets all the conditions of the problem. The ' Fire ' riddles of other literatures (Ifeifrreks Gdtur, 29; Schleicher, p. 198; Chambers, p. 8) are quite different from this. Fyr is neut. and the subject of Rid. 51 masculine; so the riddler may have had Kg, masc., in mind; but grammatical gender is little considered in the Riddles. 51 ib So the Water in 84 i is wundrum dcenned. 514 ffcond his f£onde. Cf. note to 44 na. 51 5 wrlS. Holthausen (Anglia, Bb. IX, 358) suggests, for sake of meter, wrt_e~\fr; but the non-syncopated form of 3 sg. pres. of wrifran is, of course, wrlfrefr (see Madert, p. 62). 516 J76owa)7 . . . J>egnia<J. Cf. Met. 29 77, ^enaft ond JnowaiS. 51 7 maege<5 ond maecgas. Cf. Gu. 833, maegft ond maecgas. — mid gemete. So Beffw. 780. 518 fedaS hine faegre. Cf. 54 4, feddon faegre ; 72 5, fedde mec [faegre]. For the same idiom cf. 13 IO-M, wyrmeiJ . . . fasgre, etc. — he him fremum 8t6pe<5. Stepan with dat. pers. and inst. thing is found Gen. 1859, 2306, 2365. RIDDLE 62 ' Dragon ' is Dietrich's answer (XI, 475-476) ; and the subject of the problem invites comparison with the Draca of the Beoivtdf (2302-2315, 2335 f.). Of the three characteristics of the epic monster pointed out by Schemann {Die Synonyma im Beowulf sliede, Hagen, 1882, p. 51), two appear in our Riddle: the flying in the air and the guarding of a treasure (Prehn, p. 228). The latter is also mentioned in Gn. Cot. 26-27 : Draca sceal on hlSwe, frod, fraetwum wlanc. The resemblance to Eusebius 42, De Dracone, is so slight as to preclude all idea of borrowing ; it consists in the likeness between the swift flying of our creature and the line, Concitus ethereis volitans miscebitur auris. Says Stopford Brooke, E. E. Lit., p. 52, note : ' A new touch is added by Cyne- wulf. The dragon dives into the waves and disturbs the sea. Like the dragon of Beowulf [?], he has paws with which he walks the earth. These are the four wondrous beings with which the riddle begins.' Trautmann's first solution, ' Horse and Wagon,' though a common theme in riddle poetry (Woeste, Zs.f.d.M. Ill, 186; Germania X, 69), fits only the first two lines of our problem ; but his more recent answer, ' Pen and Three Fingers ' (ff_B. XIX, 195-198), is not only very apt, but is confirmed by many analogues, as I have shown (M. L. N. XXI, 102). The relation of the 'four wights ' (i b) is men- tioned not only IrfTatwine's enigma, No. 6, De Penna, ' Vincta tribus ' (Gloss 'digitis'), but in Aldhelm iv, i 4, ' Terni nos fratres (Gl. "tres digiti scriptores ") incerta matre (Gl. "penna") crearunt,' and in the 'Pen' problem (19) of Cam- bridge MS. Gg. V, 35 (printed by me, Mod. Phil. II, 571); 'Tres gemini repunt RIDDLES OF THE EXETER BOOK stimulati marmore pellis.' Upon this the glossator comments, 'Tres digit! dis- currunt in pagina, stimulati, cum acuta penna, vel graphic, vel planitie.' The same motive appears in two ' Pen ' riddles from the German and Italian Tyrol, cited by Petsch (Palaestra IV, 135), ' Drei fiihren und zwei schauen zu' and 'Due la guarda e cinque la mena,' in both of which the eyes watch the work of the fingers. The ' black tracks ' (2 b~3 a) are found not only in Eusebius, De Penna (No. 35), ' vestigia tetra relinquens,' which our riddler did not know, but in Aldhelm's pen query v, 34, 'vestigia caerula linquo,' and in the ninth-century ' Lorsch ' riddle, No. 9 (Haupts Zs. XXII, 260), 'tetra . . . linquit vestigia.' The interrelation of these Latin ' Pen ' enigmas is discussed at length by Ebert (ib. XXIII, 200). It is interesting to compare with this motive the description of the Pen, stj>ade sweartldst, in Rid. 27 na, a riddle which furnishes other paral- lels to our problem (infra). The 'black tracks ' appear as 'black seed in a white field' in the riddles given by Petsch (I.e.) and by Wossidlo (No. 70, notes). The other motives of Rid. 52 will be discussed below. Notice the common complaint of mediaeval scribes, cited by Wattenbach, Schrift- wesen (1875), p. 235 : ' Calamus tribus digitis continetur (or " tres digit! scribunt ") totum corpusque laborat.' 52 4 Here the MS. reads fuglum frumra fleotgan lyfte. Thorpe suggested in a note fromra, and Grein2 framra. Either of these readings may be ren- dered 'more rapid than the birds' (cf. Dicht., p. 234). Grein conjectures fieotga (' schwimmer ') on lyfte or fieot geond lyfte. Wiilker (Assmann) retains for the second half line the MS. reading; while Trautmann (BB. XIX, 195, 197) pro- poses fugla fultum, fieag geond lyfte. One abandons reluctantly fuglum fromra, as it is not only very close to the MS. reading, but is supported by 41 66, ic maeg fromllcor fleogan J>onne pernex; but Trautmann'syw^/a fultum makes intelligible a difficult passage, by supplying a subject to -wees (1. 3), and is sustained by other descriptions of the Quill, 27 7, fugles wyn, and 93 27, se J>e aer wide ba?r wulfes gehleban (raven). This emendation is, however, very violent ; so I suggest fultum fromra, which meets equally well the sense of the passage ('the support of the swift ones ' — compare 92 i, brunra beot) and is only a slight change from MS. fuglum frumra. Fultum is used of wings. Met. 3 18, nabba'5 hi set fvSrum fultum. To Trautmann's fieag geond lyfte, I prefer fieag on lyfte ; cf. 23 15, on lyfte fleag. See also Tatwine 6 3 (Penna}, ' Nam superas quondam pernix auras penetrabam.' 525 dCaf under ype. So 74 4a. The passage corresponds in thought to 279-10 (Pen), beamtelge swealg | streames dJEle. 52 sh-6 With Dreag Trautmann begins a new sentence. By -winnende iviga he understands ' not the hand but the arm of the scribe, first because wiga points to a masculine word, and secondly because the arm is more properly called a fighter than the hand.' It is hard to feel the force of these arguments. Personally I prefer the 'hand' interpretation. 52 6!> se him wegas tsecne}?. Cf. 4 i6h, J>e me wegas tScnetS. 52 7 ofer fieted gold. Dietrich's discussion of this phrase (ffaupts Zs. XII, 251) is partly invalidated by his misinterpretation of the riddle's meaning; but as fiet, ' plate,' is found Beow. 716, 2256, there seems no reason to doubt the correct- ness of his conclusion (ib. XI, 420) that fated gold is 'der alte epische ausdruck NOTES 185 gewesen fur das gold in plattenform oder in blatterform.' The adjective occurs ten times (Spr. I, 273-274), and the phrase is met in the Andreas and Beowulf (see also Husband's Message, 1. 35). If ' bracteatus ' is the equivalent oij&ted, our phrase applies admirably, not to the gold of the inkpot, as Trautmann supposed (BB. XIX, 197), but to the illuminated page of the manuscript. Some of the receipts for gilding in this age have been preserved by Muratori and are cited by Sharon Turner (IX, chap, vii): for the embossed gold letters a foundation was carefully laid in chalk, and leaf gold [fated gold~\ was then employed. Gold is associated in the Kiddles not only with book-covers (27 13), but with the man- uscript itself (68 17, 92 4). See notes to 27 15. RIDDLE 63 Several answers have been offered. Dietrich (XI, 476), Grein (Spr. II, 368 ; Germania X, 308), and Prehn (pp. 278-279) unite upon the solution ' Two buckets bound by a rope which a maid carries,' and I sought to support this by ana- logues (M. L. N. XVIII, 108). Walz (Harvard Studies V, 265) suggests ' A yoke of oxen led into a barn or house by a female slave,' but this smacks of fatal ob- viousness. Trautmann offers first ' Broom ' {Anglia, Bb. V, 50) and later ' Flail ' (Anglia XVII, 396; BB. XIX, 198-199). He thus defends his second solution: ' Die beiden gefangenen sind der stiel und der kniippel. Sie heissen treffend gefangene, weil sie an einander gefesselt sind. Die fesseln sind der riemen, der zwei-, drei- oder vierfach durch die ose des stiels und durch die ose des kniippels geht und so beide teile des dreschflegels mit einander verbindet. Dass beide hart sind, wird niemand bestreiten. Die dunkelfarbige Welsche, die mit dem einen der gefangenen enge verbunden ist und beider weg lenkt, ist eine welsche magd oder sklavin, die den stiel des flegels in der hand halt und drischt.' In M. L. N. XXI, 103, I have accepted this answer. 'Chief among the winter duties was the threshing performed in the barn, and although it was to some extent carried on in the autumn, yet the bulk of it was finished during the winter. The scene in the Calendar picture for December is a threshing scene (Strutt, Horda, pi. xi). Wheat, rye, barley, peas, beans, and vetches were all threshed, and, next to plowing, threshing was the most important of the farm employments. The grain was bruised with flails similar to those now in use, and it was winnowed by hand ' (Andrews, Old English Manor, p. 250). The flail is mentioned in the Gere/a list, to odene fligel; and in the Glosses, WW. 107, 2, 141, 16, J>erscel, 'tritorium.' Heyne, Fiinf Biicher II, 54 f., discusses at length the Old English flail and threshing-floor. 53 1-2 In raeced . . . under hrof sales. The threshing-floor is mentioned several times in Anglo-Saxon writings: WW. 147, 14, 'area,' breda filing v el flor on to J>erscenne ; Matt, iii, 12, 'aream,' kyrscelflore (Lind. MS. beretiin, a signifi- cant rendering, as barley was the staple of the North); and Gen. Iv, 10 'aream,' birsceflore. Of the berebrytta we are told, R. S. P., § 17, Schmid, p. 380: ' Bere- bryttan gebyreft corn-gebrot on haerfeste aet bernes dure, gif him his ealdormann ann and he hit mit getrywSan geearnoiS.' The threshing of the barley is described in Rid. 29. 186 RIDDLES OF THE EXETER BOOK 53 i In raeced fergan. Trautmann regarded recced at first as a dative without the ending, but, after Walz's objections, is inclined to consider it as the accusative form (BB. XIX, 199). As he rightly says, the case of this word has no effect upon his solution. Both scholars have failed to remark that the same idiom appears in 56 1-2, ic seah in heall[e] ... on flet beran. 53 2a under hrof sales. Cf. Gen. 1360, under hrof gefor. 53 3 genamnan. MS., Thorpe, and Grein read genamne. This Dietrich defines ' gleichnamig ' (compare O. H. G. ginamno, M. H. G. genanne, O. N. nafni, 'name- sake,' 'companion,' Graff II, 1085). Grein is inclined (Spr. I, 434) to derive the word from genafne (see nafu and nafol), and would render ' arete conjuncti.' Holthausen (Engl. Stud. XXXVII, 209) follows Dietrich's definition, and reads genamnan, which is the MS. form in Rid. 54 13 (infra). Thorpe was the first to propose genumne, which has been adopted by Trautmann, who renders 'gefesselt,' and by Assmann (Wiilker). 53 6 wonfah Wale. See 13 8, wonfeax Wale (my note). 53 7b bendum faestra. Cf. And. 184, 1038, 1357 ; Jul. 535, 625, bendum fasstne. RIDDLE 54 Dietrich (XI, 476) answers, ' Battering-ram.' Brooke, who accepts the solu- tion, thus summarizes the poem (E. E. Lit., p. 125): 'The Battering-ram wails for its happy life as a tree in the forest and for all it suffered when it was wrought by the hands of man ; yet at the end, like the spear [Rid. 73], it boasts itself of its deeds of war, of the breach it has made for the battle-guest to follow, of the plunder which they take together.' Very similar in transformation motive are the riddles of the Book (27), Ore (83), and Stag-horn (93). The Oak and Ship queries of Germany (Wossidlo, No. 78), deal with a like change in the lot of the tree. Dietrich and Prehn (pp. 229-231) point to the 'Battering-ram' enigma of Aldhelm, v, 8, which has, however, an entirely different aim — a pun upon 'Aries.' The only likeness — which is strong enough to indicate similarity of topic — is be- tween Rid. 54 8b-ioa and Aid. v, 8 5, ' Turritas urbes capitis certamine quasso ' (see, however, Symphosius 84, Malleus, ' Capitis pugna nulli certare recuso '). The (P)aries logogriphs of the monks (M. L. N. XVIII, note), have nothing in com- mon with our query. Trautmann's ' Spear ' is a possible solution. Keller (Old English Weapon Names, p. 66) notes that there were three kinds of ram in use among the Romans, the first suspended, the second running upon rollers, and the third carried by the men who worked it, often consisting of a mere wooden beam with a bronze or iron ram's head at one end for battering down the walls of the besieged town. No description is to be found in A.-S. literature, the word ram being found only in the glosses a few times among lists of war-equipment. Keller, p. 219, cites Ctira Past. 161 6, Sersca'S Sone weall mid ramum. In O.E. Glosses (Napier), ' aries' is in a list with ' ballista.' In ^tlfric, Grammar 124, 'aries,' byfr ram betwux sceapum and ram to wealgeweorce (WW. 141, 24, 'aries,' ram to wurce); but this ram is perhaps a tool of the mason or •weallwyrhta. See Heyne, Die Halle Heorot, p. 20, who discusses our riddle. On what a mighty scale some of these rams were built we may judge from NOTES 187 Abbon's account of the siege of Paris by the Danes in 885 A.D. (De Bellis Pari- siacae Urbis I, 205 f., Pertz, Scriptores Rerum Germanicarum, 1871, I, p. 13): 'The Danes then made, astonishing to see, three huge machines, mounted on sixteen wheels — monsters made of immense oak-trees bound together; upon each was placed a battering-ram covered with a high roof — in the interior and on the sides of which could be placed and concealed, they said, sixty men armed with their helmets.' For an exhaustive description of mediaeval battering-rams, compare Schultz, Das hofische Leben II, 349 f., 371. 54 i f. Professor Cook, The Dream of the Rood, p. L, has pointed out the affinity between the opening lines of the riddles of the ' Battering-ram ' and 'Spear' (73) and the beginning of the address by the cross (Dream, 28-303): >&t wass geara lu — ic J>aet gyta geman — J>st ic was aheawen holtes on ende, astyred of stefne mlnum. ' In all these we are reminded of the Homeric scepter (Iliad I, 234 ff.), " which," said Achilles, " shall no more put forth leaf or twig, seeing it hath forever left its trunk among the hills, neither shall it grow green again, because the ax hath stripped it of leaves and bark." ' 54 2 trfcow waes on wynne. Cf. Har. 55, se J>egn waes on wynne ; Beow. 2014, weorod waes on wynne. In Run. 37, the yew is called wyn on efrle (see Rid. 92 sa). 54 3 \vudu weaxende. Cf. Hy. 4 105, wudu mot him weaxan. 54 3-4 The same theme is treated in the riddle's mate, 73 1-3. 54 4 ff-ddaii faigre. Cf. 518, feda'S hine faigre ; 72 5, fedde mec [fiegre]. — frod dagum. Cf. 73 3, gearum frodne ; 93 6, daegrlme frod. 547-8 hyrstum . . . gefraetwed. Cf. 1511, hyrstum fra^twed; 3220, frastwed hyrstum. 54 10 See text for many readings suggested in place of MS. hy an yst, which is unintelligible. I prefer to read hy on yst_e~\ strudon, ' they plundered in a storm (of battle) ' — a very natural metaphor in an enigma (cf. shour in Chaucer; Krapp's note to And. 1133, scur heard). In the Skdldskaparmal, § 48 (Snorra Edda i, 416) the battle is called 'a tempest,' vefrr vdpna. — strudon hord. See Beow. 3126, hord strude. 54 12-13 The M.S.f<?r genamnan finds threefold support in the meter, in 533 (MS. genamne), and in a certain gap in the sense occasioned by the reading of all the editors, fter genam \ ndn. But as an ace. genamnan cannot be construed with any possible sense of the verb nej>an, ' to venture,' I accept Dr. Bright's sugges- tion, genamna, and thus interpret the passage : ' The second was quick and un- wearied, if the first, a comrade in a tight place, had to venture into danger.' Holthausen's emendation [on] far (Engl. Stud. XXXVII, 208-209) is, as Dr. Bright points out, unnecessary, since genej>an is used more than once in the present meaning with the simple accusative: Met. 1359, s!o sunne . . . uncuftne weg | nihtes genei$e)>; cf. Beow. 889, 1656, 2511. The reading far, 'journey,' is barred by the macron of the MS. To the proposed genamna Bright prefers gemimne, Thorpe's suggestion (53 3) ; but the adopted form is reasonable in its origin, and is sustained by both passages. 188 RIDDLES OF THE EXETER BOOK RIDDLE 65 Dietrich's ' Oven ' and Trautmann's ' Churn ' fit equally well Rid. 55 ; but the weight of modern riddle-testimony is on the side of the second solution. Haase offers a similar German query of the ' Churn' (Zs.d. V.f. Vk. Ill, 75, No. 58) : 'Unse lange diinne Knecht pumpst unse dicke Diern.' Compare, too, Carstens, Zs.d. V.f. Vk. VI, 419; Eckart, Nos. 59, 86, 427, 905; Wossidlo, Nos. 138, 144, many references, 434 u. Despite Dietrich's note (XII, 239), wagedan biita seems to me more fittingly said of churning than of the oven-feeding of the baker's boy, and the last lines (10-12) well describe the ' growing ' of the butter. The riddle has much in common with the other obscene problems — particularly with 45 and 64. The cyrn or Churn and the cysfat are mentioned in the Gerefa list, 17 (Anglia IX, 264); and the shepherd of ^Elfric's Colloquy (WW., p. 91) tells us: ' melke hig tweowa on daeg . . . and cyse and buteran ic do.' The use of milk and butter among the Anglo-Saxons is considered by Klump, pp. 16-18, 59-60. 55 i Hyse cwom gangan. Cf. 34 i, wiht cwom . . . llban ; 86 i, wiht cwom gongan. 55 2 stondan In wincle. This reading of Grein, wincle for MS. wine sele, finds strong support in a riddle of the same class, 46 i, on wincle (MS. on win cle, ex- plains confusion in our passage). Though wtnsele is sustained by the association of so many of our riddles with the wine-hall (43 16, 47 i ; 56 i, in heall[e] )>5r haeletS druncon, 57 n, etc.), yet in such a half-line as stondan in wtnsele it is metrically objectionable, as double alliteration is demanded in this form of the A-type (L X X | ^_Ox). For this reason Holthausen, Engl. Stud. XXXVII, 209, pro- poses stondan on staj>ole, citing in support Dream, 71, Beow. 927, Rid. 88 7. But, as stondan in wincle is metrically unimpeachable, there is no need of violent change. 55 3-4 See 45 4-5. 55 5 stij»es nathwaet. Cf . 62 g, ruwes nathwast ; 93 25, eorpes nathwaet. 55 6 worhte his willan. Cf. 64 7, wyrceft his willan. 55 8 tillic esne. So 64 5. Esne is here the servant oij>egn (1. 7). 55 10-12 These lines describe the butter, the 'fettes kind' of the similar Meck- lenburg riddle (Wossidlo, No. 138 b). Lines 55 11-12 have something in common with 50 7-8. 55 ioa werig )>a>s weorces. Barnouw, p. 215, notes that weorces is used here in a double sense, ' des coitus und des butterns," and compares 43 4, }>aes weorces - - haemedlaces. RIDDLE 56 This problem has found many interpretations. Dietrich's first answer, ' Shield ' (XI, 476), he afterwards changed to 'Scabbard' (XII, 235, note). This solution, which has much to recommend it, is accepted by Brooke, who says (E.E.Lit., p. 123) : 'Another portion of the sword is also described when Cynewulf, making a riddle on the scabbard, tells of its fourfold wood ; and then, in his fancy, likens the sword-hilt to the Cross of Christ that overthrew the gates of Hell and to the gallows tree on which the Outlaw is hung.' Trautmann (BB. V, 50) without reason proposes ' Harp.' An ingenious explanation of the problem has been offered by NOTES 189 Felix Liebermann in presenting the solution ' Gallows ' or ' Sword-rack ' (Herrigs Archiv CXIV, 163). According to him, these are the conditions of the query : 'A wooden object is meant. It is portable. It appears at the feast. It serves the rich warrior. It receives (?) his sword. It is connected with precious metals. It bears the form of the Cross (in the old broader meaning for which only a verti- cal pole with a cross-piece is necessary). Its name also serves for the gallows. The word consists of four letters, with which the names of the four kinds of trees begin — (h)l, a, i, h.' By word-play, Liebermann believes, ialh might well stand for gealga, as i could be written for ge (Sweet, History of English Sounds, p. 145 ; cf. Bede's Death- Hymn, 1. 3, hinionga). He adds, doubtfully, that the poet may have had in mind the compound gealgtreow, and therefore considered only the root of the word. This seems far-fetched, but is certainly not a whit more forced than Dietrich's interpretation of Rid. 37. The second difficulty, the asso- ciation of Gallows and Cross, is no difficulty at all, as ' the wordgea/ga is used in all the early Germanic dialects to designate the cross on which Christ was crucified ' (compare Kluge, Etym. Wtbf', s.v. Galgen\ Krapp, Andreas, pp. 125-126). The greatest objection to this answer is that the name ' Gallows ' is nowhere connected with a sword-rack; but, since in Modern English this name is applied to various objects consisting of two or more supports and a cross-piece (A". E. D., s.v.), the association is not improbable. Jordan, Altenglische Sdugetiernamen, p. 62, reaches independently the same solution as Liebermann : ' War vielleicht ein reich ver- ziertes, einem Kreuz oder Galgen ahnliches Gestell gemeint, an dem Waffen aufgehangt wurden wie Verbrecher am Galgen?' Personally I do not believe that a logogriph is intended or that the riddler had in mind a sword-rack. The answer ' Cross ' meets all the conditions of the problem. Lines 12-14, which are responsible for Dietrich's ' Scabbard ' and Liebermann's ' Sword-rack,' refer, I think, to the restraining influence of the Cross over men's passions, and may be rendered ' The cross (wolf's-head tree) which often wards off (see Sweet, Diet., and B.-T. s.v. dbadaii) from its lord the gold-hilted sword.' I do not believe that our riddler owes aught to Tatwine's enigma No. 9, De Cruce Christi (see, however, Ten Brink, Haupts Zs., N. S., XI, 55-70) : Versicolor cernor mine, nunc mihi forma nitescit ; Lege fui quondam cunctis jam larvula servis, Sed modo me gaudens orbis veneratur et ornat. Quique meum gustat fructum, jam sanus habetur, Nam mihi concessum est insanis ferre salutem ; Propterea sapiens optat me in fronte tenere. Neither here nor in Eusebius 17, De Cruce, is there a single trait in common with our riddle. Though there is no actual likeness between the description of the cross (Kid. 56) and that in the Dream of the Rood, yet the enigmatic manner of that poem, ' involving quasi-personification and an account in the first person,' so closely resembles the mode of the Riddles that Dietrich, who believed our collection to be the work of Cynewulf, used the similarity of method as an argu- ment in favor of his authorship of the Dream in the Disputatio de Cruce Ruth- wellensi, 1865, p. n (see Cook's Dream of Rood, 1905, p. L)- Professor Cook has RIDDLES OF THE EXETER BOOK pointed out that the opening of the address by the rood (Dream, 28-30 a) shows a special affinity to Rid. 54 and 73, ' Battering-ram ' and ' Spear ' (see my notes to those riddles). 561 Ic scab. In heall[e]. 80601. — paer hseleS druncon. 805711. Cf. 21 12, )>a!r hy meodu drinca'S ; 68 17, )'«r guman druncon ; 64 3, J'Sr guman drincaS ; 15 12, Jjjer weras drinca'S. 562 on flet beran. 805711. — fgowercynna. See note to 56 9-10 (the woods of the cross). 56 3 wudutreow. For the use of treow in the Elene, as a synonym of rod and beam (see 56 5, r5de tacn ; 56 7, >xs beames), cf. Cook's note to Chr. 729. 56 3-4 The adornments of the subject recall those of the Sword in Rid. 216-8,9-10 (Prehn, p. 279), but they resemble quite as closely the treasures of the Cross in other poems : El. 90, golde geglenged ; gimmas lixtan ; Dream, 6 f., Eall fcet beacen waes | begoten mid golde ; gimmas stodan ; see also El. 1023-1027; Dream, 14-17; 23, mid since gegyrwed; 77, gyredon me golde and seolfre. 564 searobunden. This is a nonce-usage; but see And. 1396, searwum gebunden ; Rid. 57 5-6, searwum | fasste gebunden. 56 5-7 Cook (Christ, pp. xxii, 130; Dream, p. 45) furnishes the following exam- ples of the treatment of the Harrowing of Hell theme in Anglo-Saxon literature : in the poetry, the Harrowing of Hell; Chr. 25 f., 145 f., 558 f., 730 f., 1 159 f. ; El. 181, 295-297(?), 905-913; Gn. 1074 f.; Ph. 417-423; Gen. 1076; Dream, 149; Rid. 566; Pan. 58 f.; Creed, 30 f.; in the prose, Martyrology (Herzfeld), p. 50; Wulfstan, pp. 22, 145 ; Bl. Horn., pp. 85-89 ; ^Elfric, Horn, i, 28, 216, 480 ; ii, 6. 565 baes = J>£ ('ejus qui'). See Madert, p. 84. For other instances of attraction, compare 41 96, 44 6. 56 5-6 Cf. Gen. 1675, and to heofnum up hlaedre rserdon. 56 7a burg abraice. Cf. Dan. 63, hie burga gehwone abrocen ha^fdon ; Met. I 18, abrocen burga cyst. 56 9-10 W. O. Stevens, The Cross in the Life and Literature of the Anglo-Saxons, 1904 ( Yale Studies in English}, p. 10, discusses the kinds of wood of which the cross is composed. Among his references are the following. Chrysostom applied to the cross the words of Isaiah Ix, 13: 'The glory of Lebanon shall come unto thee, the fir-tree, the pine-tree, and the box together,' etc. In the Golden Legend (see Morris, E.E. T. S. XLVI, pp. 26, 70), the upright part is of cedar, the cross-beam of cypress, the piece on which the feet rested of palm, and the slab of olive. Pseudo-Bede tells us (P. L. XCIV, 555, Flares') : The Cross of the Lord was made of four kinds of wood, cypress, cedar, pine, and box. But the box was not in the cross unless the tablet was of that wood, which was above the brow of Christ, on which the Jews (?) wrote the title, " Here is the King of the Jews." The cypress was in the earth and even to the tablet, the cedar in the transverse, the pine the upper end.' In Rid. 56, the four woods are ash or maple (hlin), oak, hard yew, and the dark holly. As Stevens observes, ' Evidently the question was still a matter of individual speculation.' See Meyer, ' Die Geschichte des Kreuz- holzes vor Christus,' Abhandlungen der k. bayr. Akad. der Wiss., I. Kl., XVI, Bd. II, Munich, 1881 ; Kampers, Mittelalterliche Sagen vom Paradiese und vom NOTES 191 Holze des A'reuzes Christi, Cologne, 1897; Napier, History of the Holy Rood-tree (£. E. T. S. CIII, 1894), pp. 43, 47-50, 68. 56 9 hlin. ' Der alte Name des Spitzahorns, ae. him = ahd. Kmboiim geriet, weil der Baum selbst in England fehlte [note, " In der kontinentalen Heimat der Angelsachsen kam der Baum vor "], bei den Angelsachsen allmahlich in Vergessen- heit ; er ist nur noch einmal in der Poesie belegt und da natiirlich als Feldahorn zu verstehn, die einzige Ahornart, die den Angelsachsen bekannt war.' (Hoops, Wb. u. Kp., p. 272.) — ae. 'Der vornehmste Charakterbaum der altenglischen Landschaft war jedenfalls, wie noch im heutigen England, die Eiche, die iiberall bis nach dem Norden Schottlands hinauf verbreitet war und bei zahlreichen Ortsnamen Gevatter gestanden hat' (Hoops, Wb. u. Kp., p. 259). It is interest-- ing to note the passage in the Runic Poem (77-80) in which the Oak is extolled : Ac by|> on eorj>an elda bearnum flaisces f5dor, ferej> gelome ofer ganotes baej> ; garsecg fandaft, Lwarger ac hajbbe sl>ele treowe. The close connection between kennings and riddles (see Introduction) is strik- ingly illustrated by a comparison between the functions of the Oak as a ' feeder of flesh ' and a ' ship ' in this Runic verse and those in the world-riddle of ' Oak ' (Wossidlo, No. 78) : Als ich klein war, ernahrten mich die grossen ; Als ich gross war, ernahrt' ich die kleinen ; Als ich tot war, trug ich die lebendigen wohl iiber die lebendigen. — se hearda iw. Compare with this the description of the yew in Run. 35-37 : Eoh byS utan unsmel>e treow, heard, hrusan fzst, hyrde fyres, wyrtrumum underwrej>yd, wyn on e}>le. «The Yew ("Taxus baccata," O.E. fw, eow) is native to the British islands. It is frequently found in the postglacial peat-bogs of England and Scotland, and must have been widely extended in historic times. We meet its name occasion- ally in Old-English " Flurnamen " ' (Hoops, Wb. u. Kp., pp. 269-270). The Yew- tree is the subject of one of Aldhelm's enigmas (v, 5, De Taxo). 56 10 se fealwa holen. The holen is identified by Hoops ( Wb u. Kp., pp. 256, 616) with the ' Stechpalme ' or ' Ilex aquifolium.' That this was native to western Europe and first appears at the end of the oak-period, Hoops shows, ib., pp. 30-31. 56 12 wulfheafedtreo. The wulfes-heafod or 'wolfshead' is the legal expres- sion for an outlaw, who may be killed like a wolf, without fear of penalty (see Grimm, Rechtsaltertiimer, 3d ed., p. 733). So in the law of Edward the Confessor, Cap. 6, § 2 (Schmid, p. 494), ' Lupinum enim caput geret a die utlagationis suae, quod ab Anglis wluesheved nominatur.' Compare Bracton, De Legibiis et Gonsue- tudinibns Angliae, 159, lib. Ill, tr. ii, chap, n, ' Et tune gerunt caput lupinum ita quod sine judiciali inquisitione rite pereant.' Jordan, Altenglische Sdugetiernamen, p. 62, rightly opposes Dietrich's earlier solution 'Shield' (XI, 476), and says: ' Richtiger fasste Grein wulfheafod-treo I92 RIDDLES OF THE EXETER BOOK als identisch mit wearg-rod "Galgen," "Kreuz," denn wulfheafod bedeutet "Ver- brecher, Geachteter." ' The two significations of ' gallows ' and 'cross ' are in the mind of Eusebius, 17, De Cruce: Per me mors adquiritur et bona vita tenetur. Me multi fugiunt, multique frequenter adorant ; Sumque timenda mails, non sum tamen horrida justis. Damnavi virum, sic multos carcere solvi. abjfd. Thorpe suggested dbad, ' awaited.' Grein regarded dbSd as = db&defr, 'exigere,' 'adigere' (Dicht. 'bezwingt'). In this he is followed by B.-T., who .renders 'repel' or 'restrain' (cf. Sal. 478, dbade). Herzfeld, p. 60, regards the word as ' dialectische nebenform des Praes [Praet?] ahead'; so Madert, p. 44. Liebermann, Archiv CXIV, 163, translates 'abforderte (erlangte).' I accept Grein's explanation of the form, but translate, both here and in the Salomon passage, 'wards off ' (supra). The cross restrains the sword. 56 13 maftm in healle. Cf. Beow. 1529, deorum maiSme (sword); Waldere A. 24, maSrna cyst (sword). The adornments of the Sword are described at length in Rid. 21. 56 14 gieddes. Merbot, Aesthetische Studien zur angelsachsischen Poeste, p. 26, in his discussion of the various meanings of gied, points out that in this place the word means ' a riddle,' and compares Gn. Ex. 4, gleawe men sceolon gieddum wrixlan. He raises the question whether the Anglo-Saxons were not as fond of riddle-combats as the old Hindoos. 56 15 onmSde. Grein, who reads on mede, translates (Dicht.) 'wen es anmutet.' In Spr. II, 229, he regards mede as opt. pres. of medan, impers., ' muten,' ' in men- tem venire.' Thorpe reads onmede; and B.-T., following him, renders (s. v. onmedan) 'to take upon oneself,' 'to presume.' Cosijn, PBB. XXIII, 130, reads onmede ('sich vermesse'), and compares onmedla, geanmettan. Liebermann, Archiv CXIV, 163, reads on mede (' sich unterfangt '), and Holthausen (Engl. Stud. XXXVII, 209) follows Cosijn. RIDDLE 57 We may set aside unhesitatingly Lange's 'Turning-lathe' (ffaiipts Zs. XII, 238, note), and Trautmann's ' Flail,' and accept Dietrich's solution, ' Web and Loom,' which he establishes beyond question by an account of the old vertical weaver's beam, derived from the description and illustration (tab. xn) in Olaus Olavius's Oeconomische Reise durch Island, Dresden u. Leipzig, 1787, pp. 439 f. ' The tainnende -wiht is the web ; and the warp or chain hangs vertically from the beam, the old jugum, and is stretched underneath by stone weights. The upper end of this is wrapped around the beam and is therefore biidfast (1. 7) ; but the lower end, which is the more readily woven and wound from above the more it pushes up, is moved in the work (bisgo dreag), because it floats in the air (1. 8, leolc on lyfte), and is near the ground only at the beginning. The warp now suffers from a threefold stress of war : first, through the curved wood which moves to and fro (holt hweorfende) and carries through the threads of NOTES 193 the woof, but is no shuttle, only a simple wood (wido) — and indeed a wudu searwum fieste gebttnden, . . . because the thread is skillfully bound about (in Old Norse it is called ivinda). Secondly, the woof receives wounding blows (1. 3) by means of the Schlagbret, O. N. skeifr, a sword-like board which the weaver swings in his free hand, in order to strike fast the inserted threads. In the third place, spears (darofras) are also an evil to the creature, because through the middle of the body of the warp are stuck five transverse pieces, of which the three upper- most are called the shafts, and the two lowest the parting-shaft and the parting- board. The tree that is hung with bright foliage (1. 9) is the upper beam upon which the roll of the still unwoven yarn hangs. The relic of the fight is the web, which, perhaps as gafol hwltel, is borne into the hall of the lord.' Dietrich also notes (XI, 476) the verbal likeness between this contest and that in the spinning- song of the Valkyria in the Njdls Saga, chap. 1 58. Weinhold, Altnordisches Leben, 1856, pp. 320-321, cites both Olavius and the Njdls Saga, and draws from the An- tiquar Tidskrift, 1846-1848, p. 212, a description of a Faroese loom: 'An dem Webebaume (rifr), welcher drehbar auf zwei Pfosten (Iileinar, leiner) ruht, ist die Kdtte (garn, gadn, renning, rendegarnef) unmittelbar und nicht durch die Traden (hovold} angemacht. Das Werft wird durch eine Stange in der Mitte, die auf zwei Pflocken liegt und iiber welche die Kette gezogen ist, gespannt, am meis- ten aber durch die Gewichtsteine (klidsteinar), welche unten an die einzelnen Fadenbiindel gebunden sind. Ein grosses lanzenfdrmiges Gerat von Fischbein (skeifr) dient den Einschlag festzuschlagen, welcher durch einen scharfen Knochen (hrall, r&lur) in Ordnung gehalten ist. Es wird stehend gewebt.' This serves to explain many of the riddles of the Islenzkar Gdtur, which are suggestive ana- logues to Rid. 57. /. G. 60 considers six objects: (i) Weight-stones; (2) Threads; (3) Ho/old; (4) Fingers; (5) Rifur ('the beam on which the warp is hung'); (6) Cloth. There are in this collection various riddles of weaving and spinning: one of the Wefstdll (657), one of the Wefstafrur (1082), five of the Wefur (49, 976, 982, 983, I no), two of the Rifur, 'beam' (339, 851), two of the Skeifr (644, 1088), three of the Ullarkambar (79, 81, 82), ten of the Rokkur, 'distaff' (447, 499,536, 737, 798,912, ion, 1133, 1140, 1147), and three of the Snalda, 'spindle' (383, 576, 853). Still another interesting analogue is the Lithuanian 'Loom' riddle (Schleicher, p. 198) in which 'a small oak with a hundred boughs [cf. Rid. 57 9-10] calls to women and to maidens.' Our riddle seems to owe noth- ing to Symphosius 17, Aranea, or to Aldhelm iv, 7, De Fuso, although Prehn, p. 232, seeks to find likeness between the Latin and the English ; and the parallel furnished by Aldhelm iv, 3 3-5, (see Rid. 36, Mail-coat) lies in the nature of the subject. Parts of the Loom and phases of weaving have already been considered in the notes to 36 5 f. The Gerefa list (Anglia IX, 263) mentions ' fela towt5la, flex- llnan, spinle, red, gearnwindan, stodlan, lorgas, presse, pihten, timplean, wifte, wefle, wulcamb, cip, amb, crancstaef, sceaftele, seamsticcan, scearra, njedle, slic.' See Liebermann's careful rendering and discussion of each of these 'tools' (I.e.). In the Vocabularies (WW. 262) is a long list, ' De Textrinalibus,' 'Textrina.' For the work of the weaver and his various implements, see Klump, Altenglische Ifatuhverknamen, pp. 22-32, 73-89. I94 RIDDLES OF THE EXETER BOOK 57 i-4 Brooke, E.E.Lit., p. 151, renders with spirit : I was then within, where a thing I saw ; 'Twas a wight that warred wounded by a beam, By a wood that worked about ; and of battle-wounds it took Gashes great and deep. 57 2 wldo. This finds its West Saxon equivalent in wudu. Th'e regular Northern form would be wiodu. And Madert, p. 128, believes that the absence of the u-umlaut of points to the beginning of the eighth century. As widu appears in the Meters of Alfred, 13 55, it is evident that the conclusion thus drawn is not of the highest value. — bennegean. Only here and 93 16, bennade. Gebennian, too, is found only in 6 2, gebennad. 573 heajjoglemma feng. The direct obj. of fan is always ace. except in this passage and in Sal. 432 (see Shipley, Genitive Case in Anglo-Saxon Poetry, P- 32)- 574 deopra dolga. Cf. And. 1244, deopum dolgslegum; Rid. 546, deope gedolgod. 57 4-6 The best comment upon these lines is found in Rid. 36 7-8. The wudu (1. 5) corresponds to the hrutende hrlsil (36 7) ; and darofras (1. 4) may well be the dmas (MS.), the 'reeds' or 'slays,' of the earlier riddle. As a parallel to daroj>as, Dietrich (XII, 238, note), points to the song in Njdls Saga, chap. 158, str. 4, 5, wef darrafrar. It is barely possible that the image is suggested by the double mean- ing of Lat. tela, ' web ' and ' darts.' 57 5 weo. Is this for wo (Gn.), or wea (B.-T. s.v.) ? Sweet, Diet., does not give the word. 57 5-6 searwum . . . gebunden. Cf. And. 1396, searwum gebunden ; Rid. 56 4, searobunden. 57 8a leolc on lyfte. So Gen. 448 a. 57 9-10 Andrews, Old English Manor, p. 275, note, regards ' the tree with bright leaves ' as 'the reel with the colored yarns or web ' (see Dietrich, supra). 57 10-12 lafe . . . on flet beran. In Beow. 995 f., in the great wine-chamber, ' there shone variegated with gold the webs on the walls, many wonders to the sight of each of the warriors.' The Saxon term for a curtain or hanging was •wdhrift; and in the will of Wynflaeda (Thorpe, Diplomatarium Anglicum, 530, 33) we find the bequest of a long heallwdhrift and a short one. So Aldhelm describes a web in his poem (De Laudibus Virginum) : ' It is not a web of one uniform color and texture without any variety of figures that pleases the eye and appears beauti- ful, but one that is woven by shuttles, filled with threads of purple and many other colors, flying from side to side and forming a variety of figures and images in differ- ent compartments with admirable art.' Cf. also De Laudibus Virginitatis xxxviii, Giles, p. 51. For a discussion of the various products of the Anglo-Saxon loom — garments, tapestries, curtains — see Heyne, Fiinf Biicher III, 207-252. He cites (III, 237) Paul the Deacon's History of the Lombards IV, 22 : ' Vestimenta linea, qualia Anglisaxones habere solent, ornata institis latioribus vario colore contextis.' — b«r li:rlo|> druncon ... on flet beran. Sarrazin, Beowulf-Studien, p. 120, compares Beow. 1647. Cf. 56 1-2. NOTES 195 57 12 My reading J>dra fldn\a\ seems to be supported by such a line as El. 285 b, l>rera leoda. Alliteration upon the second stress in A-type is common (95 examples in the Riddles) ; compare 41 88, J>ara )>e worhte waldend user. The stress not infrequently (sometimes the alliteration) falls upon the article ; see Beow. 807, on ftxm daege )>ysses llfes. RIDDLE 68 This little swallow-flight of song has invited many answers. Dietrich (XI, 477) suggested first 'Swallows' or 'Gnats'; and afterwards (XII, 240, note), on the authority of Pliny x, 35 (24), he proposed ' Starlings.' Sweet (Anglo-Saxon Reader, p. 208) accepted the second solution ; and Prehn (pp. 233-234) the third. Brooke queries the answer 'Starlings' (E. E. Lit., p. 148, note): 'The stare is not particularly a little bird, nor is its note sweet. The bird seems to answer best to the " Martin." ' I prefer the solution ' Swallows ' for two reasons. First, they fulfill all the conditions of the riddle. The poet saw them, as Brooke says (I.e.), rising and falling in flocks over the hills and cliffs, above the stream where the trees stood thick and over the roofs of the village, and the verse tells how happy he was in their joyousness, their glossy color and their song.' Secondly, Rid. 58 has at least two traits in common with Aldhelm vi, i, Hirundo. Line 4 of the Latin, 'Garrula mox crepitat rubicundum carmina guttur,' is not far from sanges rofe . . . hliide cirmafr (Rid. 58 3'', 4b), and line 6, ' Sponte mea fugiens umbrosas quaero latebras,' from tredafr bearonassas (Rid. 58 sa). See the A.eneid passage cited infra. The three solutions of Trautmann seem to me equally extravagant : he first (Anglia, Bb. V, 50) proposed 'Hailstones'; then (Anglia XVII, 398) ' Rain-drops ' ; and finally (BB. XIX, 200), by the dangerous petitio principii of changing lytle (58 ib) to tt/ite, ' Storm-clouds.' I have refuted these interpretations and sustained the ' Swallows' solution (M.L.N. XXI, 103). The riddle is clearly / one of the bird group, as parallels in phrasing to Rid. 8, 9, n, and 25 show. § 58 i Deos lyft byreS. This phrase is used elsewhere in the Riddles of the flight of birds : 8 4-6% Swan (note) ; 1 1 g, Barnacle Goose. 58 2a ofer beorghleojm. Alexander Neckham, De Naturis Rerum, chap, lii (Rolls Series, 1863, p. 103), says of swallows: 'Quaedam enim domos inhabitantes in eis nidificant . . . quaedam in abruptis montium mansionem eligunt.' As I have noted, M. L. N. XXI, 103, this may well apply to the Cliff Swallow, Hirundo fulva. 58 2-3 Our poem finds an interesting analogue in the well-known lines of Virgil (Aeneid, xii, 473~477) = Nigra velut magnas domini cum divitis aedes Pervolat et pennis alta atria lustrat hirundo Et nunc porticibus vacuis, nunc humida circum Stagna sonat. In commenting upon this passage, Gilbert White of Selbome uses words equally applicable to the English riddle (Letter XIX, Feb. 14, 1774): 'The epithet Nigra speaks plainly in favor of the swallow, whose back and wings are very 196 RIDDLES OF THE EXETER BOOK black [compare 58 2-3, blace swtj>e, \ swearte, salopdde}, while the rump of the mar- tin is milk-white, its back and wings blue, and all its under part white as snow.' Note also the lirl vura. ntXaiva of the Rhodian carol of the Swallow, preserved by Athenaeus (Book viii, chap. 60). 58 3a salopade. The word is a nonce-usage, but sal(o)wigpdd is used three times in the poetry, in each case of a bird, the eagle or raven (Fates, 37 ; Jud. 21 1 ; Brtin. 6r). 58 3'', 4h Sanges rofe . . . hlude cirma?f. Both Virgil and Aldhelm apply to the swallow the epithet 'garrula'; and Gilbert White tells us (XVIII, Barring- ton), 'the swallow is a delicate songster and in soft sunny weather sings both perching and flying.' Indeed its ' pipe and trill and cheep and twitter ' (Tenny- son's Princess) is among the best-known of bird-songs. The song of the swallow is mentioned elsewhere in Old English. Whitman, Birds of O. E. Lit., p. 161, cites Life of St. Guthlac (Godwin), 52, 7 : ' Hu )>a swalawan on him szeton and sungon. Twa swalewan . . . heora sang upahofon.' Elsewhere in the Riddles, hlude cirme is used of the song of a bird (9 3). 58 4-6 It needs no Virgil or Aldhelm or Neckham (see references, supra) to tell us that swallows ' fare in flocks ' and that they are found ' in remote and se- cluded woods and swamps as well as about the habitations of men." 58 6 Nemna?? by sylfe. This has been variously rendered. Thorpe proposes 'Name them yourselves.' In Spr. II, 280, Grein wavers between ' Sagen selbst wie sie heissen ' and ' Sagt wie sie heissen ' ; but translates in Dicht. 'Nun meldet ihren namen.' So Trautmann, BB. XIX, 200 : ' Nennet sie selber.' Brooke, E. E. Lit., p. 149, renders ' Let them call their own names.' I prefer the Thorpe read- ing, because the verb-form is the 2d pi. imperative, and because swallows are cer- tainly not onomatopoetic like cuckoos and bobolinks. RIDDLE 69 Dietrich (XI, 477) offers the solution ' Ziehbrunnen,' 'Well with a well-sweep,' which has been accepted by all scholars. ' This has one foot, the prop upon which the cross-beam rides, moreover a long tongue (the pole at the upper end of the cross- beam, which carries the bucket down), it has a heavy tail (the stone weight which helps to press down the lower end of the cross-beam and to raise up the bucket), it paces the earth-grave (the dug-out well), and carries laguflod (hyperbolical for water) into the air.' Dietrich, reading furfrum for the MS. furum, suggested as the three rune-letters (Kid. 59 14-15), the three consonants of burna ; but Grein (Germania X, 309), reading fultum, makes the happier suggestion of Rdd-pyt (Spr. II, 363, Reitbrunnen, d. i. Ziehbrunnen mit einem Schwengel) which meets perfectly the conditions of name and thing. Miiller, Cb'thener Programm, p. 17, sustains Dietrich's rdd-burna by pointing to ' Radbourne ' in Derbyshire and ' Redboum ' ; but these names prove little, as not the ' well ' but the 'brook ' or 'burn ' is their etymological source. Holthausen, who reads (/. F. IV, Tfi^fitrma for MS. furum, suggests rod instead of Grein's rdd-pyt. Then it is the pole or well-sweep that is described. Rod in the sense of ' pole ' appears only in the compound segl-rod. Prehn rightly mentions in this connection Symphosius 71, NOTES 197 Puteus, and 72, Tubus; but the relation lies only in the likeness of Rid. 59 iih-i2a to the third line of each, ' Et trahor ad superos alieno ducta labore ' and ' In ligno vehitur medio, quod ligna vehebat.' The interesting ' Puteus' enigmas of Virgil's third eclogue and of Scaliger (Reusner I, 170) have nothing in common with Rid. 59 ; while the Low German Put or ' Draw-well ' problem (Woeste, Zs.f. d.M. Ill, 191) interests us only by its title and by its allusion to its steert (compare Rid. 59 7 steort"). In an illustration of the marriage feast of Cana in a Cotton manuscript of the early twelfth century, Nero C. IV (Wright, Domestic Manners, p. 86; Knight, Pictorial History, p. 284), a servant raises water from a well by means of a loaded lever. Wright comments upon the drawing thus : ' It may be remarked that this appears to have been the common machinery of the draw-well among our fore- fathers in the middle ages — a rude lever formed by the attachment of a heavy weight, perhaps at the end of the beam, which was sufficient to raise the other end and thus draw up the bucket.' Wright refers to illustrations of this in manu- scripts of various periods, and presents in cut No. 57 an excellent drawing from MS. Harl. 1257 of fourteenth century. Aldhelm thus mentions the draw-well or futeus (De Laudibus Virginum, Giles, p. 142): Nee putei laticem spernendum ducimus altum Antlia quern sursum solet exantlare cisternis. 59 i anfete. The word is a nonce-usage ; but the riddle-subjects in 33 6, 81 3, 93 25, have also one foot. 59 2 Wide ne fereiJ. Cf. 4 71, wide fere ; 95 3, fere (MS. fereS) wide. 59 3 Cf. 32 8, n5 hwaej>re fleogan maeg ne fela gongan. 59 4 )mrh scirne daeg. Cf. Met. 20 229, )>urh J>a sciran neaht. 59 5a naca naegledbord. Cf. Gen. 1418-1419, naegled bord, | faer seleste ; Brun. 53, naegledcnearrum. 59 6h monegum tidum. So Gu. 89. Cf. 40 2, miclum tldum. 59 g3 Isernes da»l. Cf. 56 4b, seolfres djel. 59 13-14 The spirit of comitatus in the Riddles has been discussed in the Introduction (' Form and Structure '). RIDDLE 60 This riddle of the ' Chalice ' or ' Communion Cup ' has already been discussed in connection with its fellow, 49, the ' Paten.' Dietrich (XII, 235, note) thus ana- lyzes the poem: 'Als kelch ist der goldene reif (v. I, firing; 5, wrij>an) bezeichnet theils durch die benennung Heliand der guthandelnden (v. 7) die er von dem betenden (priester, v. 3-5) erhalt, indem die iibelhandelnden von seiner gemein- schaft ausgeschlossen sind, theils durch das geheimnisvolle aber den einsichtigen (v. 2, 9, 10) verstandliche sprechen seiner wunden (v. n, 16) d. h. des fur die menschen vergossnen blutes des heilandes, welches er darstellt und nach den friih im mittelalter gehenden geschichten von wunderbarer verwandlung, im weine enthalt. Was sie sprechen, indem der kelch, noch nicht der gemeinde 198 RIDDLES OF THE EXETER BOOK entzogen, von den handen der manner gedreht und gewendet wird (v. 18), das ist die mahnung zur gegenliebe und dankbarkeit gegen den erlbser den des edelen goldes zeichen (7-10) vergegenwartigt,' etc. 60 i Cf. 56 i, ic seah in healle. — bring. See 60 6, 17, 49 i, 8. 60 2b modum gleawe. So Az. 190. Cf. Gen. 2373, gleaw on m5de ; Sal. 439, modes gleaw. 60 3 ferppum frode. Cf. 27 21, ferf>e ]>y frodran ; Exod. 355, Wand. 90, El. 463, Jul. 553, on feriSe fr5d; El. 1164, frodne on ferhfte. For the construction with bad, see Shipley, p. 26. 60 4a God nergende. Cf. Chr. 361, nergende God. 60 sb word aefter cwaeS. So Beow. 315. 609 in 6agna gesihS. Cf. And. 30, eagna gesih~5; Wond. 66, eagna gesifrS; Chr. 1113, fore eagna gesyhiS (Herzfeld, p. 18). 60 11-12 In favor of the reading that I have adopted in the text these argu- ments may be offered. Ond Dryhtnes dolg don (MS. dryht dolgdoit) is supported by Chr. 1205-1206, Dryhtnes . . . dolg; and by a similar reading in MS. 85 2, driht for drihten (see note). The transference of don to the second half-line completes the otherwise defective swd J>a:re bene, a very faulty half-line, to ne J>Sre bene maes flddes ofre. 61 i ssewealle neah. So Beow. 1925. 61 2 set merefarope. Grein renders well, Dicht., ' an des Oceans Wellen- schlag.' See Krapp's discussion of fared" and warofr (Mod. Phil. II, 405-406). 61 3 frumstajjole faest. The phrase is suitable only to reeds or plants ; cf. Gu. 1248-1249, staj>elum faeste . . . wyrta geblowene. See Rid. 35 8, 71 2-3. — fea ainig. Cf. Gen. 2134, fea ane; Ps. 104 n, feawe . . . asnige. 61 6 yft slo brune. Cf. Met. 26 29-30, sio brune | yft ; And. 519, brune y$a. 61 9-10 Much of the secular music of Old English times is associated with the beer-hall, as Padelford has pointed out (pp. 10-12). See the Bagpipe's part at the feast in Rid. 32 11-12. In an illustration in MS. Harl. 603 (Wright, Domestic Manners, p. 34), the cup-bearer serves the guests with wine, while minstrels make merry with harp and pipe. To cite but one of many examples from the poetry, this accords with the lines in the Fates of Men, 77 f. : Sum sceal on hearpe haslejmm cweman, blissan ast beore bencsittendrum, J>aer >\> drincendra dream se micla. Music and feasting are closely associated in Bede's story of Caedmon's life at Whitby (Eccl. Hist, iv, 24) : ' In gebeoiscipe, Jx>nne J>iEr waes blisse intinga gede- med, )>aet heo ealle scealden Jmrh endebyrdnesse be hearpan singan.' These entertainments led to such excesses that the Canons of Edgar, 58, at the time of the monastic revival, forbid priests to be ale-poets (ealu-scop} and Wulfstan thunders against the beer-halls with their harps and pipes and merriment (Horn. 46, 16): ' Hearpe and pipe and mistlice gliggamen drema'S eow on beorsele.' 61 10 wordum wrixlan. So Beow. 875, Soul 117; cf. Mod. 16, wordum wrixlaft. 61 10-17 The 'nigro perfusa colore' and the 'nuntia linguae' of Symphosius certainly suggest a pen ; and in the last lines of the Anglo-Saxon the riddler has evidently in mind, not music, as Brooke supposed, but written speech (1. I5b, SrendsprSce}, which is hidden from all but the pen and his master. It is this reference to a letter that misled Trautmann and Blackburn. 61 12-14 These lines, which describe the shaping of the 'calamus,' may have arisen from a misunderstanding of the ' digitis stipata (signata) magistri ' of NOTES 201 Symphosius ; compare seo stut{>re hand \ eorles ingefronc, etc. The lines have not a little in common with Rid. 27 5 f. Wattenbach, Schrifhvesen, p. 189, cites Isidore, Origines\, 13: ' Instrumenta scribae calamus et penna. Ex his enim verba paginis infiguntur, sed calamus arboris est, penna avis,' etc. So we are told by the letters t in the gloss (MS. Royal 12, C. XXIII) to the incerta matre of Aldhelm's 'Alphabet ' enigma, iv, I $ (Wright, Satirical Poets II, 549): 'Ignoramus utrum cum penna corvina vel an- serina sive calamo perscriptae simus.' Three kinds of pens were thus known to the Anglo-Saxon : the raven-quill, the goose-quill, and the reed. The first of these is described in the striking periphrase of Rid. 9326-28 (see notes); it is doubtless the second that is alluded to by the riddlers of 27 7 f. and 52 4 ; while the reed-pen (hreodwrif) is the subject of the last lines of Rid. 61. Westwood, p. 35, pi. xiii, notes that the figure of St. Matthew in the Lindisfarne Gospels, Cott. Nero D. IV, is writing with a reed-pen. 61 12 seaxes ord. Cf. 776, seaxes orde ; 276, Chr. 1140, seaxes ecg. See 93 15-18. — seo swlpre bond. See Spr. II, 511. 61 14 Jjingum. Grein renders, Dicht., ' zu den Dingen ' ; and Spr. II, 593, 'potenter, violenter (?) ' ; while Sweet and B.-T. suggest 'purposely.' The inst. thus employed is a nonce-usage. 61 :6 abeodan bealdlice. Cf. Har. 56, abead bealdlice. Only in this Riddle passage is this verb found with the ivij> construction instead of the dative. RIDDLE 62 The subject of this riddle according to Dietrich (XI, 477) is ' Shirt ' ; according to Trautmann (Anglia, Bb. V, 50), ' Shirt of Mail.' Trautmann is perhaps attracted by the picture of the early Englishwoman arming her lord for battle, but the tone of this poem, despite the blending of dignity with its dirt, hardly seems to warrant such a conception. Cyrtel or Hragl seems to me to fit all the conditions of the problem (infra). No Latin sources or analogues have been discovered ; and the 'Shirt' riddles of Strassburg Rb., No. 181, and the Recueil des £ntgmes de ce Temps, Rouen, 1673, H> 77> are like the Anglo-Saxon one only in pruriency. 62 2 on earce. This is a reference to the hragl-cyst, ' clothes-chest ' (Thorpe, Diplomatarium, 538, 20). 62 4 holduni bfiodne. Roeder, Die Familie bei den Angelsacksen, p. no, cites this passage as proof 'dass man die eheliche Gemeinschaft als ein Komitats- verhaltniss ansieht.' Other evidence of this conception of the marriage-relation is not wanting : ' Der Mann erscheint als der Herr und Gebieter der Frau : Gen. 2225 nennt Sarah ihren Gatten drihten mtn, oder er heisst ihr man-drihten, 2242, . . . 2"j2t)frea-drihten, ebenfalls von Abraham. 2783 apostrophiert ihn Sarah: min yivies frea? See also Beow. 1 1 70, freo-drikten mm (Wealhtheow to Hrothgar). Lawrence, Mod. Phil. V, 395, cites these passages to sustain the wifely relation of The Banished Wife's Lament. 625-6 Dietrich thus comments (XI, 477): ' Wer es anzieht steckt ihm dem umgekehrten den kopf ins innere, denn es wurde nicht von unten sondern von 202 RIDDLES OF THE EXETER BOOK oben her angezogen, durch die kopfoffnung, die daher mhd. houbetloch, bei den Norwegern und Islandern hofiifrsmd (hauptschmiege) hiess.' So Strutt points out, fforda, p. 46, that ' the close-coat [cyrtel] of the soldiers and common people, which reached only to the knee, appears from the form of it (pi. xv, 7, 8 ; Cott. Claud. B. IV) to have been put over the head like a shirt.' The subject of our riddle is perhaps the cyrtel — the hragl of the other obscene riddles, 45 4, 55 4) 63 6. Cyrtel ofrfre hragel\s> the Lindisfarne equivalent of Matt, v, 40, ' tunicam.' Hra-gl\s also used of the robe of women (Kid. 46 4 ; Alfred's Laws, Introduction, 1 1, § 18, Schmid, pp. 58, 80), and in Beowulf 'is a synonym for byrne, 'the mail-coat' (Leh- mann, Briinne und Helm, p. 13). 62 6 on nearo fCgde. Cf. 26 g, fegeft mec on faesten ; 63 8, on nearo nathwser. In all three places is the same coarse suggestion. 62 7 Gif . . . ellen dohte. This is a common formula which is discussed at length by Krapp in his note to And. 458-460. Cf. Gen. 1287-1288, Drihten wiste | |>aet >xs aeSelinges ellen dohte; Kid. 739, gif his ellen deag; Beow. 573, )>onne his ellen deah ; And. 460, gif his ellen deah ; etc. It is the Old English version of the formula ' Fortune favors the brave,' which Cook derives from Latin literature (M. L. N. VIII, 59). 628 mec fraetwedne. Holthausen, Bb. IX, 358, would retain MS. mec frat- wedne instead of Edd. frcetwede, but he does not explain how he would adapt this to the context. The omission of J>e makes the construction clear. 62 8-9 Dietrich notes (XI, 477) : ' Das rauhe was es beim erwachsenen fiillen soil, ist der haarwuchs.' The cyrtel was often worn next to the skin, as, in many cases, it was the only garment ; cf. Alfred's Laws, 36 (Schmid, p. 62) : ' Gif mon naebbe buton anfeald hrasgl hine mid to wreonne oftfte to werianne,' etc. 62 9 rmves nathwaet. Cf. 26 5, neo^an ruh nathwier ; 55 5, stlj>es nathwast. The obscene implication is obvious. — R;i>es from (1. 2). 63 2a forcJsIJ^es from. Cf. H. M. 41, for5si)>es georn. For the construc- tion of from with gen., see 73 27, feringe from ; And. 234, gu5e fram (Krapp's note). 63 3-4 Cf. Dream, 88-89, ier^an ic him Hfes weg | rihtne gerymde reordberen- dum. See also 54 8-10. 63 5, 6 In fryfr and tyhfr, as in 64 2, 5, 6, onj>eon, beof>, J>yfr, the meter demands uncontracted forms instead of the contracted. For other examples see Madert, p. 53, and my Introduction. 63 8 on nearo. Cf . 62 6. 63 9 sujjerne secg. In the Atlakvifra, § 2, the same phrase, seggr inn sufrrqni, is applied to Knefrufrr, the messenger of Attila. Cleasby-Vigfusson, s.v. srifrr- mafrr, Sufrrriki, points out that the word is used by the Scandinavians of Ger- mans, indeed of all people of central and southern Europe. In Old English, on the other hand, the epithet is coupled with a spear or javelin cast by a Norse sea- warrior (s&rinc) at Byrhtnoth in the Battle of Afaldon, 1. 134, sufrerne gdr ; but is not ' from the south ' merely direction ? Though in the Glosses and Leechdoms the word may indicate plants and medicines from the south of Europe (B.-T., s.v.), I doubt if it carries any other idea here than that of ' foreign.' As the actor in one of the obscene riddles, ' the southern man ' is obviously in the same class as ' the dark-haired Welsh,' the churls and esnes, often people of un-English origin, who figure in these folk-products. There seems no reason to suppose that the word is used, like Chaucer's 'Southern man' (Canterbury Tales I, 42) and the later ' Southron,' of a South-Englander. Perhaps some personal or topi- cal reference is intended, in which case we might as profitably seek the identity of ' the man from the South ' who bums his mouth with cold porridge in the nursery rhyme. RIDDLE 64 As Dietrich points out (XI, 478), this 'Beaker' riddle has much in common with Aldhelm's enigma (vi, 9) De Calice Vitreo. Unlike the Latin writer, the Anglo-Saxon says nothing of the origin and little of the appearance (3 a) of the Beaker. But in both poems the drinking-vessel is a woman who yields readily to caresses ; compare with 64 4-7 Aldhelm vi, 9 5-9: Nempe volunt plures collum confringere dextra, Et pulchrae digitis lubricum comprendere corpus, Sed mentes muto dum labris oscula trado. Dulcia compressis impendens bacchia buccis, Atque pedum gressus titubantes sterno ruina. The overthrow that follows kisses of the wine-cup is perhaps the theme of the fragmentary close of the Exeter Book poem. 204 RIDDLES OF THE EXETER BOOK As I have already shown, Holme Riddles (No. 128) offers a modern treatment of the same motive : Q. As j was walking late at night, j through a window chanced to spy : a gallant with his hearts delight, he knew not that j was so nigh : — he kissed her & close did sit to little pretty wanton Gill until he did her favour get & likewise did obtaine his wille. A. A yong man in a tavern drinking a Gill of sack to chear up his spirits & so obtaind his will. Trautmann ignores completely the history of the riddle in his answer, ' Flute.' Scherer, Kleine Schriften, Berlin, 1893, II, 9 (cited by Roeder, Die Familie bei den Angelsachsen, Halle, 1899, p. 122) says of this riddle : ' Die einzige Liebesszene in der alien angelsachsischen Poesie aus der wir sonst vieles lernen ist dem Lateinischen nachgebildet und sie schildert — auch nur indirekt — sinnlichen Genuss.' The problem has too much in common with the other double entente riddles of the collection to merit this comment. Dietrich (I.e.) points out that while ceac and steap, two common words for 'beaker,' are masculine, bune is feminine and therefore suited to the gender of the riddle. But in the Riddles little stress is laid upon grammatical gender (supra). Akerman, Remains of Pagan Saxondom, p. 51, and De Baye, The Industrial Arts of the Anglo-Saxons, pp. io6f., have discussed at length the glass beakers of the Anglo-Saxon. I note in the Gibbs collection of the British Museum two from Faversham in Kent, which resemble closely those in Akerman's plates. One is light green, the other olive, and both are ornamented by rude jagged bands running from near the mouth to the bottom, where they converge. They are footless, and, like the horns (whose shape is copied by other glass vessels), they must have been emptied before being relaid upon the table. In outline the grave-finds resemble the illustrations of cups in the manuscripts (Claudius B. IV, ff. 63 r., 102 v. ; Tib. B. V., Strutt, Horda, pi. x), and accord with the description in Beow.4<)$, hroden ealowiege ; 2253-2254, fajted wiige, | dryncfaet deore. As Sharon Turner points out, Hist, of Anglo-Saxons VII, chap, vi, the precious metals were used constantly for basins and beakers, and the wills often bequeath cups of gold, silver, and silver-gilt [64 3, glaed mid golde]. See also Brincker, Germ. Altertiimer in Judith, 1898, p. 21. 64 i secga seledreame. Cf. And. 1656, secga seledream. 64 3 glaed mid golde. Cf. Sal. 488, golde glaedra. — Jner guman drinoutf. Cf. 68 17, golde gegierwed, J>j£r guman druncan ; 56 i, 57 n, )>;ir haele^5 druncon; 15 12, )>aEr weras drincaft ; 21 12, J>iir hy meodu drincaft. 644 cofan. Sievers (PBB. X, 497) cites many examples from the poetry to support his rejection of a long root-syllable in this word: And. 1006, in barn morflorcofan ; El. 833, in Jjeostorcofan, etc. The present instance argues for a long syllable ; but verses of form JL X X X | ^ X are rather frequent in the Riddles (ib., p. 454). — cysseft mu]?e. So it is said of the Horn, 15 3, hwllum weras cyssaft; see also 31 6. Other Latin riddles besides that of Aldhelm (cited supra) allude to the kiss of the wine-cup : Lorsch 5 5, ' Dulcia quin bibulis tradunt et bassia buccis'; MS. Bern. 611, 6 6 (Anth. Lat. I, 353), ' Et arnica libens oscula porrigo cunctis.' 64 5 tillic csiic. So 55 8. 64 7 wyrceS his willan. Cf. 55 6, worhte his willan. NOTES 205 RIDDLE 66 Dietrich (XI, 479-480) combines the thirteen runes WIBEHA(>EFA (the reading of Th., Gn., for >E) EA S P into PEABEAHSWIFED (for >) A, « Ring- tailed peacock'; and refers to Aldhelm's ' Pavo ' enigma (i, 16), ' Pulcher et ex- cellens specie, mirandus in orbe.' But Hicketier (Anglia, X, 597) has pointed out many objections to this unhappy solution : the change of > to D in 1. 4 is opposed by the alliteration ; the form sivifeda is not only a hapax-legomenon, but an incredible coinage ; all predicates and attributes of the riddle are left unex- plained, and sylfes J>a;s folces is totally disregarded ; finally, the same sound ea in pea and beah can hardly be represented in one case by the rune EA, in the other by two runes E and A. To Dietrich's solution Sievers (Anglia XIII, 19, note) objects on phonetic grounds : 'Eine form beak mit dem spaten ausl. h fiir^ und ohne palatalumlaut 1st ausserdem fur die mundart der ratsel undenkbar ; das wort hatte in deren orthographic nach massgabe aller altesten angl. texte als bag zu erscheinen. Und wie ware die vertauschung der </-rune mit /> zu erklaren ? ' Even less credible is Grein's learned solution (Germ. X, 309) : ' AspiJ?(d)e-uv(f) = Aspis et hie vultur (bubo = uf) = schlangenfressend Raubvogel.' In his answer, Hicketier has solved the problem. He marks that each group of runes is used to signify the word which it spells in part : Wlcg, BEorn, HA(o)foc, |>Egn, FAlca and EA, SPearhafuc. The first four words give no trouble and are supported by the problem's companion-piece, Rid. 20. Fa(ce)lca, which he does not find elsewhere in Anglo-Saxon, Hicketier supports by reference to O. H. G. falko (cf. Baist, Haupts Zs. XXVII, 65), and to such a compound as Westerfalca (Thorpe, Anglo-Saxon Chronicle I, 30 b). £0, 'water,' which is presented by a single rune, is in keeping with the context. Spearhafuc, Hicketier points out, is a very common word, and is not unnaturally suggested by its synonyms, Hafoc and Falca. Trautmann (Bb. V, 50 ; Kynewulf, 46) follows Hicketier in part, but suggests for the later words J>egnas or fceowas, hafoc, ear A, speru. As he offers no explana- tion of these forms, it is necessary to supply his reasons. His objection .o falca probably rests upon the non-appearance of the word elsewhere ; but this word is supported not only by the arguments of Hicketier (supra) but by the runes F and A, and by the demands of the alliteration in 65 5. So there is really no warrant for Trautmann's hafoc. His plural J>egnas or Jieowas is probably suggested by 65 6, folces; but it is open to the very strong objection that since in our riddle's mate, 2O4~5» hi£dej>ryl>e is in apposition with the singular man (N 0 M), it seems reason- able to infer the same relation between f>ryf,a dizl and a singular (doubtless J>egn) here. And, again, it seems highly improbable that the letters are intended to rep- resent other than uninflected forms of words. I therefore prefer Hicketier's begn to Trautmann 'sj>egnas. Trautmann's earh and spent seem to me very happy sug- gestions. Not only are they supported by all the arguments for g dr in Kni. 20, but by their appearance in apposition elsewhere in the poetry: Sal. 128-129, scearp speru, | atole earhfare ; And. 1330-1331, gares ord, | earh attre gemiil. But the sing, spere seems to me preferable to speru (supra). 'The hawk flew above the spear carried by the beorn or J>egn.' Sylfes J>egn), — Barnouw says, p. 216, 'die sechs mit runen genannten wesen.' 65 i Cf. 20 1-3, 75 »• 65 2 on sip J?e. Cf. 20 8-9, for ... sl)>faet. 65 3 htebbendes hyht. Cf. 95 5, hij>endra hyht. 654 bE(gn). In this place, the f>E(gn) seems to be the attendant of the BE(orn). That the word is early applied to 'servant,' the many references in Schmid, Gesetze, 'Glossar,' pp. 664 f., and B.-T., p. 1043, show. Indeed in Matt, xxiv, 46, 'servus,' Lind. reads fregn, where Rush, esne, and West Saxon f>eow. It is difficult to determine the meaning elsewhere in the Riddles, but J>egn is opposed to esne 55 7. Holthausen Bb. IX, 358, notes that if Assmann's read- ing r for (> be accepted as that of the MS., the two runes W and E indicate wer, ' man ' ; but the alliteration is clearly against this. 65 5 EA(rh). This reading is supported by the context, by the natural appo- sition of EA(rK) and SP(ere), and finally by the evidence of Rid. 20, with its wt[g]gar equivalent. A West Saxon worker has therefore been busy among these runes, as in Rid. 43 (see Introduction), since the Northern form is surely not ear A ; compare Leid. Rid. 13, aerigfaerae. 65 6 Hicketier points out the irregularity of sylfes Jxzs folces. Either simply J>ees QV Jxes sylfan is in better accord with idiom (see Barnouw, p. 216). RIDDLE 66 The source of this ' Onion ' riddle has already been considered by me under Rid. 26. Its final motif, ' the biter bitten,' is found in Symphosius, 44 : Mordeo mordentes, ultro non mordeo quemquam ; Sed sunt mordentem multi mordere parati. Nemo timet morsum, dentes quia non habet ullos. The bite of the Onion is a commonplace of Volksratsel (Renk, Zs. d. V.f. Vk. V, 109; Wossidlo, No. 190; Petsch, p. 96). And the motif has been transferred to other themes, MS. Bern. 611, No. 37, ' Pepper'; Aldhelm ii, 13, 'Nettle.' The first motif of the riddle — the death and renewed life of its subject — is thus explained by Dietrich (XI, 480): ' Die zwiebeln werden in dem jahre wo sie gesat sind der hauptmasse nach nicht brauchbar, sie miissen in einem zweiten jahre wieder in die erde gelegt werden, um die gehorige grosse zu erlangen ; daher hier vom sterben die rede ist und vom wiederkommen aus einem friiheren vorhandensein.' See my notes to Rid. 26 for verbal parallels between the two problems. Rid. 66 differs from its predecessor in its freedom from suggestion of obscenity. 66 31 hafatf inec on headre. Cf. 21 13, healdeft mec on heaj>ore (sword}. 66 5-6 Although this is a common motif of riddle-poetry (compare the ' Ox ' riddle), still these lines are so close to Symphosius 44 as to suggest a literary connection either direct or indirect. The tone of the riddle and its relation td Rid. 26 put it, however, in the class of popular, rather than of literary problems. NOTES 207 RIDDLE 67 Under Rid. 41 I have already discussed the origins of Rid. 67. It owes noth- ing to Aldhelm's De Creatura directly, but is a very free reshaping of some of the material furnished by the second hand (B) in 41 82 f. — probably an effort of this translator to improve upon his first very slovenly venture. Holthaus, Anglia VII, Anz. 123, believes that Rid. 67 is written by an imitator of Rid. 41 : « The theory of identity of authorship leads to a dilemma, in that the poet would neither work over his bad stuff in order simply to give a translation from the Latin, nor is it conceivable that he would recast his good work in bad form.' My theory, as set forth in my notes to Rid. 41, meets this objection. For the relation of 67 and the fragment 94, see the notes to the later riddle. 67 1-3 The comparatives are consistently feminine, whereas in Rid. 41 the gender .frequently varies. Frumsceaft, ' creatura,' is, of course, feminine. 67 2 leohtre J»onne rnona. In Rid. 30 3, the Moon is called lyftf&t leoktKc. 673 swiftre )7Oiine sunne. So of the Sun in 30 nb, for'S 5nette. Cf. Met. 2931, Se bi'S J>i:re sunnan swiftra (evening star). In the Prose Edda (Gylfa- ginning, § 12), ' the sun speeds at such a rate as if she feared that some one was pursuing her for her destruction.' 674 foldan bearm. Cf. Beow. 1138, faiger foldan beartn; Gen. 1664, geond foldan bearm (MS. beam). 675 grene wongas. So Rid. 132; Gen. 1657; cf. Men. 206, wangas grene. See Rid. 41 51,83, }>es wong grena. — grundum ic hrine. Cf. Rid. 40 10, ne iefre foldan hran. 67 6 Rime in the Riddles has been discussed in my notes to Rid. 29. 67 7 wuldres ej>el. So Gen. 83. 67 8 ofer engla card. Cf. Chr. 646, engla card ; Mod. 74, on engla eard.^ — eorj>an gefyile. Cf. Ps. 649, eor'San bu gefyllest eceum wasstmum; Gen. 1553- 1554, gefylled wear"5 | eall J>es middangeard monna beamum. RIDDLE 68 This fragment is not printed by Thorpe and Grein, and is therefore not dis- cussed by Dietrich and Prehn. Trautmann, Anglia, Bb. V, 50, suggests 'Bible,' a solution which has much in its favor. 68 i, keodcyninges, points to divine associa- tions, and 68 2, word galdra, may well indicate Holy Writ ; cf. Mod. 6, be J>dm gealdre (The Word), Rid. 49 7, guman galdorcivide (sacred speech) ; 68 13, leoda Idreow, the teacher, through whom men live eternally, can only be the Book of Books (cf. 27 18 f.), and 68 3 snytt[ro~\ suggests sacred wisdom. The adornments of the subject recall those of the Book in Rid. 27 (cf. 68 17, golde gegierwed ; 27 13, gierede mec mid golde). The books in Aldhelm's enigma De Area Libraria (ii, 14) are called ' divinis verbis ' and 'sacratos biblos '. 68 i?1', Jtier guman druncon, does not militate against the solution, as a similar phrase is found in the riddle of the ' Cross' (56 i). Other ' Bible ' riddles, Islenzkar Gdtur, 775, 805, 999, and Strassburg Rb., 43-50, have little in common with this problem. 68 i J»eodcynlnges. Only once elsewhere (Soul, Verc., 12) is fceodcyning ap- plied to God, and in that place the Exeter text reads ice dryhten. 208 RIDDLES OF THE EXETER BOOK 68 8 naenne muS hafao". In 40 12 the Moon(?) has no mouth, ne mufr and in 61 9 the Flute is ' mouthless.' 68 9 fet ne f[olme]. Cf. 28 15, fota ne folma ; 32 7, fet ond folme ; 40 10, f5t ne folm ; Beow. 745, fet ond folma. 68 10 welan oft sacaS. The Bible often ' chides ' or ' contends against ' worldly wealth: Ps. Ixii, 10; Ixxiii, 12; Prov. xxiii, 5 ; Jer. ix, 23; Matt, xiii, 22; Mark iv, 19; Luke viii, 14; etc. 68 14 [awa to] ealdre. This reading of Holthausen, Anglia XXIV, 264, is sustained by many instances of the phrase in the poetry (Spr. I, 46). 68 15 penden menn bugao". Cf. Ph. 1 57-1 58, Ner no men bugaft | card ond e>el. 6816 coi-pan sceatas. So Gen. 2206; Seaf. 61 ; And. 332; cf. Beow. 752, eorj>an sceata. 6817 golde gegierwed. Cf. Beow. 553, golde gegyrwed; Beow. 1029, 2193, golde gegyrede; Dream, 16, gegyred mid golde; Met. 256, golde gegerede. See also Rid. 27 13, cited supra. — J»ser guman druncon. Cf. 64 3 (note). 68 18 since ond seolfre. So 21 10, Dan. 60. Cf. the description of the Lindis- fame MS. of the Gospels (Skeat, John, p. 188) : 'BillfriiS se oncrae he gesmioftade •5a gehrino "Sa fle utan on sint ond hit gehrinade miS golde ond mid gimmum aec rniiJ suulfre ofergylded faconleas feh.' See note to 27 nb-i4. For closing formula, compare 33 13, 73 20. RIDDLE 69 After 69 2 is a sign of closing ; so Thorpe prints 69 3 as a separate riddle. Traut- mann, Bb. V, 50, follows Thorpe's division. The first two lines, which correspond to Rid. 37 1-2 and constitute an opening formula, certainly seem not only super- fluous but misleading here ; and yet we can neither discard them nor give them a separate place. Grein, who takes the three lines together, suggests (Bibl. II, 410) ' Winter,' and Dietrich (XI, 480) ' Ice.' Though Dietrich is certainly right, 69 3 has nothing in common with Rid. 34, ' Iceberg.' Dietrich thinks that the riddle may once have been longer ; but the single line is, as an enigma, admirably complete. 69 3 Compare the description of the freezing of the water in And. 1260-1262 : clang waeteres J>rym ofer eastreamas, Is brycgade blaece brimrade. Cf. Gn. Ex. 72-73, Forst sceal freosan, ... is brycgian. The meter establishes, be- yond doubt, on wege (oncum geworht, to the fanci- fully carved neck and mouthpiece [' wry-necked fife '] ; eaxle twa, to the protrusion of the body beyond the neck.' Dietrich describes the instrument (XI, 480) : ' Die NOTES 209 schalmei der hirten mit zwei seitenklappen, dem hautboi ahnlich [eaxle~\, versehen und mit einem gebogenem mundstiick besetzt, welches ich selbst an hirtenfloten gesehen habe.' Although the shawm was well known at the time of the Minne- singers (Schultz, Das hbfische Leben I, 434), the name (O. Y.chalemie, 'a little pipe made of a reed or of a wheaten or oaten straw ' — Skeat, Etym. Diet. s. v.) does not appear in English until long after the Conquest ; and Padelford finds no trace of the instrument in the Anglo-Saxon manuscripts cited by Strutt and Westwood. Despite such negative evidence, the thing may have been in use at our early period. Trautmann offers without explanation (Anglia, Bb. V, 50) the answer 'Roggen- halm' or ' Komhalm.' 70 ib Cf. 73 28-29, Wiga se be mine wisan | [so}>e] cunne. 70 2 slngeS Jmrh sidan. So of the Bagpipe, 32 3, sellic Hng singan on raecede. 70 4 on gescyldrum. So 41 103. — gesceapo [dreogeS]. Grein's addition was doubtless made with his eye on Ph. 210, gesceapu dreoge'S; Hy. 117, gesceap dreogeiS. RIDDLE 71 Dietrich's answer to this problem (XI, 480), 'Cupping Glass,' is hardly con- vincing. It is true that 71 3-4, 'the leaving of fire and file," recalls Aldhelm iv, 8, Cucuma, 1. 7, ' Malleus in primo memet formabat et incus.' But this is the only resemblance to the Latin ; nor has our problem aught in common with the fa- mous ' Cupping Glass ' enigma of the Greeks, cited by Aristotle, Rhetoric \\, 2, 12 (Ohlert, p. 74) : ' I saw a man who on a man had soldered brass by fire,' &»Sp eUov irvpl xaX/civ iif dvfpi Ko\^d \ wyrta wlitetorhta (2-3 a) recalls the home of the mail-coat (36 i), mec se w&ta -wong, and the flowery meadow of 35 7-8, J)d •wlitigan ivyrtum folwonge. Cf. Aldhelm iv, 101, Dagger, ' De terrae gre- miis formabar primitus arte.' Nu eom wraj>ra Idf, \fyres ond feole (3b-4a) can only refer to the Sword, as Grein recognized (cf. Spr. II, 152, s.v. Idf; Keller, A.-S. Weapon Names, p. 174). With fizste genearwad cf. 21 13. Wire geweorfrad (5 a) exactly fits the interpretation (cf. Rid. 21 32, ' Sword,' wlrum dol ; 21 4, wir ymb )>one waelgimm). Wepefr hwilum \ for gripe minum (71 s-6a) refers, of course, to the s-weord-gripe (Jul. 488). Se J>e gold wigefr (TI 6b) is sometimes a periphrasis for the Sword itself (Rid. 21 6,8, ' Sword,' ic sine wege . . . gold ofer geardas), but 210 RIDDLES OF THE EXETER BOOK here it seems to indicate the wounded warrior (Beow. 1881, guftrinc goldwlonc). Dietrich forces the meaning of yj>an (7 a) into ' entleeren (des blutes),' but else- where in poetry it is used only in the sense of 'destroy' (Beow. 421 ; Wand. 85), and so it must be defined here ; this is well said of the Sword (Aldhelm iv, 10 4). Hringum gehyrsted (8 a) accords with the gifts to the Sword (21 23b, J>e me hringas geaf), and with Beow. 673, hyrsted sweord. And the fragmentary line (9) dryhtne min parallels the many allusions to the lord of the Sword in Rid. 21. Trautmann (Anglia, Bb. V, 50) offers ' Der Eisenhelm.' 71 1-2 Grein and Wiilker (Assmann) both put a comma at end of line i, and regard wong as being in apposition with ieht; and Grein translates (Die/it.) ' Ich bin eines Reichen Besitz, rot bekleidet, ein starkes steiles Feld.' Is it not far better to close line i with a period, and to construe wong as forming with staj>ol the predicate of a second sentence, ' I was a hard, high field, the station of beautiful plants'? This interpretation is supported by 358, on staj>olwonge, and by the beginning of the ' Mail-coat ' riddle, 36 (supra), as well as by the context ; rices &/it refers to no plain, but to the Sword itself, which is the possession of the rich exclusively (see my notes to 21 8, 10). 71 3-4 \vrapra laf, | fyres ond feole. Cf. 6 7, homera lafe (swords) ; Beow. 1033, fela laf (sword). 71 6 Holthausen's inversion of MS. minum gripe prevents the alliteration fall- ing upon the second stress of a B-type. See, however, 91 8. RIDDLE 72 Dietrich (XI, 480) and Prehn (p. 243) answer 'Axle and Wheels,' and defend their solution by pointing to the ' quattuor sorores ' of Symphosius's ' Rotae ' enigma (No. 77). But the 'four dear brothers' (5b-6a), as Grein pointed out (Spr. II, 526, s.v. teon), are 'mamillae vaccae,' and the subject of the riddle is the ' Ox,' an answer supported by Brooke (E.E.Ltt., p. 136), and by Trautmann (Anglia, Bb. V, 50). The riddle therefore falls in the same class as Rid. 13, 39, and has been discussed incidentally under those heads. The youth of the Ox, its nourishment, its later wanderings and suffering, and its mute endurance are the present themes. 72 i Ic waes lytel. All 'Bull' and 'Ox' riddles refer to the creature's youth. See analogues in my notes to Rid. 13. 72 s fedde mec. Cf. 73 i, mec feddon ; 77 i, mec fedde. The addition of Gn.2 [fagre] is supported by 54 4, feddan fasgre 5518, feda'S hine fjggre. 72 5-6 fSower . . . swa-se bro]?or. These are ' the four wells ' of Rid. 39 3 (see note). The teats of a cow are 'four brothers' in the Bukowina riddle (Kaindl, Zs.d. V.f. Vk. VIII, 319), and 'four sisters' in the Lithuanian query (Schleicher, p. 211). 72 7 drinean sealde. Cf. Rid. 13 5, drincan selle. 72 8 )»«h. There is no reason to accept Holthausen's t>ah (Bb. IX, 358) ; J>iJ/i is the Northern form of West Saxon J>eah (Sievers, Gr.\ 163, n. I ; Madert, p. 53). Cf. 5 8, bag for beag. 72 9-10 These lines do not mean, as Brooke supposed (E. E. Lit. p. 136), 'I was •with the swart herdsman,' but 'I left that (i.e. the milking) to the cow-herd.' NOTES 211 Brooke adds, ' The swart herdsman is a Welsh slave. Swart is the usual epithet of the Welsh as against the fairer Englishman.' See my note to 13 8. 72 9b anforlet. Grein and Wiilker read an forlet, and Grein renders (Die/it.} 'dieses all uberliess'; but dnforltetan, though not included in Sweet's Diet., ap- pears several times in the prose (B.-T., s. v.). 72 ioa sweartum hyrde. The labors of the ox-herd are detailed in ^Ilfric's Colloquy, WW. 91 : '}>aenne se yrHingc ('arator') unscen)> J>a oxan ic ISde hig to liese and ealle niht ic stande ofer hig waciende for }>eofan and eft on jerne mergen ic betiece hig }>a.m yr^lincge wel gefylde and gewaeterode.' Wiilker points to Bede's account of Caedmon, Hist, Eccl. iv, 24, to neata scypene, hara heord him waes J>JEre nihte beboden. 'Bubulci' is the lemma to oxenhyrdas (WW. 90, 17; 91, 23; Haupts Zs. XXXIII, 238). For the rights and duties of ox-herd and cow- herd, see Rectitndines Singnlarum Personarum, 12, 13, Schmid, p. 380. 7210-11 Brooke says (E.E.Lit., p. 136): 'We are brought into another part of the country, where in Riddle 72 the Ox speaks and tells how weary he was among the rough paths of the border moorland.' Compare the description of Ur in Run. 4-6 : P| (ur) byj> anmod and oferhyrned, felafrecne deor, feohte)) mid hornum m£re morstapa ; >&t is m5dig wuht. But the animal of our riddle is thoroughly tamed — certainly not one of the wild cattle that at this day and for centuries afterwards roamed through the forests of England (Bell, British Quadrupeds, pp. 368 f. ; Harting, Extinct British Animals, pp. 21 3 f.). 72 12 The use of oxen for plowing has already been discussed at length in connection with Rid. 22, ' Plow.' Notice the geiukodan oxan of ^Elfric's Colloquy (WW. 90). The work of the ox among the Anglo-Saxons and the other Ger- manic nations is considered at length by Heyne, Fiinf Biicher II, 198-208. 72 13 weorc prowade. So Beow. 1722 ; cf. Ap. 80, weorc Jrowegan. 72 14 earfoSa dail. So Gen. 180; Deor, 30. — Oft mec Isern scod. For the -use of the goad, as illustrated by the Colloquy and illuminated MSS., see my notes to the 'Plow' riddle (22). The Smith is a maker of goads as well as of plow-shares and coulters (Colloquy), and the Gerefa mentions the gddiren among agricultural implements (§ 15, Anglia IX, 263). The pricks of the goad are finely called ordstape (72 17). RIDDLE 73 All authorities agree upon the answer 'Spear' or 'Lance.' Like the weapon in Rid. 54, this has flourished as a tree, the ash, until, subjected to a cruel change of fate, it comes into a murderer's hands ; like that, it boasts of its deeds of battle, and vaunts its fame. In-its description of its origin, the 'Spear' has some faint likeness to Aldhelm vi, 8, ' Sling ' ; and, like this, it delights in battle. But the re- semblance between the two — Prehn's labored comparison (pp. 244-247) to the contrary — seems conditioned by the likeness of topics, and does not preclude complete independence of composition. 212 RIDDLES OF THE EXETER BOOK The closest analogue to our riddle is found in the description of the Ash, both as tree and spear, in Run. 81 : P (aesc) bi)> oferheah, eldum dyre, stlb on statute, stede rihte hylt, •geah him feohtan on firas monige. For asc as tree, see my note to Rid. 43 9, se torhta aesc ; and as spear, see Rid. 23 ii ; And. 1099 ; etc. (Spr. I, 58). As I have noted under Rid. 54, our query belongs to the same class as the world-riddle of Oak-Ship (Wossidlo 78), which is based upon the same motives as the description of Ac, ' the oak,' in Run. 77-80 (see note to 56 9). In Anglo-Saxon interments the spears occur in much greater number than any of the other weapons. The cemetery at Little Wilbraham produced 35 spears, but only 4 swords (Neville, Saxon Obsequies, 1852, p. 8 ; Hewett, Ancient Armor, 1860, p. 24); and other grave-finds yield similar results (Roach-Smith, Cat. of A.-S. Antiquities at Faversham, 1873, pi. xi). The Anglo-Saxon spear is represented not only by the heavy weapon for hurling and thrusting, but by the lighter dart for casting only, the darofr, or/7 (Keller, p. 21). Spears were used by the early Eng- lish not only for war but for hunting (see the September illustration in the Anglo- Saxon calendar, Tib. B. IV; Jul. A. VI). The weapon consisted of three parts: the spear-head, almost lozenge-shaped, the shaft, to which the head was attached, and the iron into which the wood of the shaft was fitted. De Baye, Industrial Arts of Anglo-Saxons, p. 22, notes that the distinctive feature of the Anglo-Saxon spear is a rather short socket. It is the ash shaft (cf. Beow. 330, garas, . . . aescholt ufan grieg; Maid. 310; Wand. 99; Rid. 23 n) that speaks in our riddle. Brooke remarks (E. E. Lit., p. 124, note): ' Gdr is the usual word for "spear" — (gar-Dene = spear Danes). Gar was the javelin, armed with two of which the warrior went into battle, and which he threw over the " shield-wall." It was barbed, but the other, shaped like a leaf without a barb, was called the spere, the lance, concerning which is Cynewulf's riddle. This was shod on the top of the handle with a heavy metal ball, to give it weight, just as the sword was.' That such a distinction was always felt to exist between g dr and spere is more than doubtful in the light of their identical appearance in the poetry and their com- mon lemmas, 'jaculum,' 'hasta'; although it is true that ' telum,' 'pilum,' words for javelin, are frequent synonyms of gar. In any case, it is clear that barbed lances were not used as missile weapons, although we occasionally find in Anglo- Saxon graves a missile weapon the two blades of which are not in the same plane (De Baye, p. 22). But gar is hardly limited to this missile. ' The Spear mourns that it was taken away from the field (as a sapling of the forest land) where earth and heaven nourished it ; that its nature has been changed and forced to bow to the will of a murderer. Yet as it leams to know its master better, it sees that he is no murderer, but one who will fulfill a noble fame. Then the spear changes its thought, and is proud of its small neck and fallow sides, when the glow of sunlight glitters on its point, and the warrior be- decks it with joy, and bears it on the war-path with a hand of strength upon its shaft and knows its ways in battle ' (Brooke, E. E. Lit., p. 124). NOTES 213 73 1-7 Notice the close likeness to the opening lines of Rid. 54, ' Battering-ram.' At that place I drew attention to the affinity (pointed out by Cook, Dream of Rood, p. L) between our riddle passages and Dream 28-30. 733 gearum frodne. Cf. Ph. 154; Gen. 2381, gearum frod; Ph. 219, fym- gearum frod ; Rid. 54 4, frod dagum ; 93 6, daegrime frod. 73 3-7 Prehn, p. 245, points to Tatwine, 32 1-2, Sagitta, ' Armigeros inter Martis me bella subire obvia fata juvant,' and 34 4, Pharetra, ' Non tamen oblectat nee sponte subire duellum.' But there is surely no direct connection between the English and the Latin. Cf. also Rid. 24 6, se waldend, se me J>aet wlte gescop. 73 9 gif hi8 ellen d£ag. See my note to 62 7. 7311 niaerjya fremman. Cf. Beow. 2515, mar'Su f remman ; 2135, mier'So fremede ; 2646, mSr'Sa gefremede ; Seaf. 84, miEr'Sa gefremedon. 73 19 heajjosigel. Grein, Spr. II, 41, and B.-T., pp. 523-524, agree in deriving the first member of the compound from hcafru, ' the sea.' The first translates ' sol e mare progrediens,' and the second explains ' The prefix seems to be used from seeing the sun rise over the sea (cf. merecondel).' Sweet, however, derives from heafro, ' battle,' which is very common as the first member of compounds, and which is well suited not only to the associations of war in the present passage, but to the description of the sun elsewhere in Riddles (7 i, 5, 30 9-10). See also Sievers (PBB. X, 507). 73 21 on fyrd wigeS. Cf. Gen. 2044, on fyrd wegan fealwe linde. 73 22 on hsefte. After the riddle-fashion, the poet is playing upon the double meaning of haft, ' handle ' and 'confinement.' 73 24 under breegnlocan. Thorpe suggests, in his note, hrtzgllocan for MS. hr&gnlocan, and translates 'among wardrobes.' Grein, Bibl. II, 400, follows the MS., but does not translate (Dicht.). Dietrich (XI, 482) says : ' Wahrscheinlich ist hragn ein korpertheil und sein verschluss das innere des leibes ; ich stelle dazu bis auf weiteres das engl. rine, die hirnhaut.' In Spr. II, 137, Grein pro- poses bragnlocan, which B.-T. renders, p. 556, 'that which incloses the brain,' 'the skull'; and Sweet, 'the head.' 73 26 frlo" haefde. Cf. Gen. 1299, frrS habban ; Gen. 2471, fri'5 agan. 73 27 Feringe from. See my note to 63 2a, for'5sl)>es from. 73 28-29 Here is a serious difficulty. Shall we place with Thorpe a comma after wtcum, and refer wiga to he, or with Gn., W., a colon, and regard wiga as voc. with 2 pers. imp. saga ? In favor of the first it may be said that the sudden introduction of the third person in line 27 seems to demand an appositional phrase of explanation ; in favor of the second, that wiga se J>e mine \ ivisan cunne may well be a part of the closing formula (cf. 68 18-19, 70 i). But neither of these interpretations meets the further difficulty, that in the MS. transmission there is no alliteration in line 29. So Herzfeld, p. 70, suggests that at least two half- lines have been omitted between cunne and saga. But, as we have seen, there is no lacuna in the MS. or gap in the sense. To meet metrical demands we might read Wiga se )>e mine wlsan [so)] cunne, saga hwjet ic hatte. 214 RIDDLES OF THE EXETER BOOK RIDDLE 74 The subject of Rid. 74 must satisfy many conditions. The monster must be at once a woman, both old and young, and a handsome man. It must fly with the birds and swim in the flood. It must dive into the water, dead with the fishes, and yet when it steps on the land it must have a living soul. The riddle has troubled scholars sorely. Dietrich admits (XII, 248) that his solution ' Cuttle- fish' (XI, 482; compare Aldhelm i, 18, Loligo) was wide of the mark; but the changes have been rung upon this answer by Prehn and Walz {Harvard Studies V, 266). Miiller (C. P., p. 19) suggests 'Sun,' and points to its different genders in Latin and the Germanic languages. Trautmann (Bb. V, 48) proposes ' Water,' and labors over its various forms (BB. XIX, 202): a spring ('a young woman '), a cake of ice ('a hoary-headed woman '), and snow (' a handsome man '). These identifications he champions by reference to grammatical gender. I have already objected (M.L.N. XXI, 103) that mythology thus becomes the creature of declensions, and that water has not a living soul ; and have twice presented and defended the solution 'Siren' (Af.L.N. XVIII, 100; XXI, 103-104). I can do little more than repeat my earlier comments upon the problem. The answer easily meets every demand of the text. The Siren is both aged and young : cen- turies old, and yet with the face of a girl. It is not only a woman but sometimes a man. To establish the two sexes of our creature, I have already pointed to the male « Siren ' of Orendel 94. Philippe of Thaun tells us of the ' Siren ' in his Bestiaire, 1. 683, ' il cante en tempeste ' ; and in two of Philippe's sources (Mann, Anglia IX, 396) we have 'figuram hominis,' and in a third 'figuram feminis.' In two Latin riddles of Reusner (I, 177; II, 77) the Siren is not only 'femina' but ' avis,' ' piscis,' and ' scopulus.' In Greek and Etruscan and Roman art the Sirens were represented as bird-women (Schrader, Die Sirenen, Berlin, 1868, pp. 70-112; Harrison, Myths of the Odyssey, London, 1882, chap, v, 'Myth of the Sirens'; Baumeister, Denkmaler des Klassischen Altertums, Munich, 1888, s. v. ' Seirenen ') ; but, as Harrison and Baumeister point out, at an early period of the Middle Ages (' vom 7. Jahrhundert ab') the Teutonic conception of a fish-woman or mermaid met and mingled with the classical idea of a bird-maiden. The identity of Siren and Mermaid is seen in many Anglo-Saxon glosses (B.-T., s.v. mere-men, p. 680). Philippe de Thaun, Bestiaire, 664 f., tells us that ' the Siren has the make of a woman down to the waist, and the feet of a falcon, and the tail of a fish.' So the creature is presented in the illustration of the Old High German Gottweih Physi- ologus (Heider, Physiologus, Vienna, 1851, p. 10, pi. iii). And Laurens Andrewe (The Babees Book, E.E. T.S. XXXII, 237-238) gives a like account. The com- bined bird and fish aspects explain 74 T,,fleah mid fuglum ond on flode sworn. As no one will doubt the appositeness of the last line of the riddle, there remains to be discussed only 744, deaf under yte dead mid fiscum. Every student of myths knows that ' when Ulysses or the Argonauts had passed in safety, the Sirens threw themselves into the sea, and were transformed into rocks1 (Harrison, p. 152, note). In its narrative of these creatures the Orphica Argonautica, 1293-1295 (Latin translation of Cribellus, Hermann edition) furnishes apt explanation of our enigmatic lines: NOTES 215 Ab obice saxi Praecipites sese in pelagus misere profundum, Sed formam in petras, generosa corpora mutant. That this 'scopulus' phase of the Siren appears in Anglo-Saxon will surprise no one who recalls the persistence of the tradition of the death-dive of the Siren in a well-known illustration in Herrad von Landsperg's Hortus Deliciarum, 1160 A.D. (Engelhardt, Stuttgart, 1818, cited by Harrison, p. 171). Every condition of Rid. 74 finds natural explanation in this widely-spread myth. The careful re- view of the history of the ' Siren-Mermaid ' by W. P. Mustard (M. L. JV. XXIII, 21—24, January, 1908) confirms me in the above views contributed by me to M.L.N. XXI, 103-104, April, 1906. My article, of which Dr. Mustard was unaware, furnishes, I think, the desired link between classical and Teutonic super- stitions. 74 i feaxhar cwene. Feaxhdr occurs only here, but hdr is often used as an epithet of age (Spr. II, 14). Hicketier fails completely in his effort to prove (Anglia X, 577) that cwene is here contrasted as ' meretrix ' with fiemne (' a bash- ful girl'). Nothing could be farther from the riddler's meaning. 74 3 fleah mid fuglum. Cf. Rid. 52 4, fultum fromra, fleag on lyfte (MS. fu- glum frumra fleotgan lyfte). 74 4 deaf under y}?e. So 52 5. 744-5 By his pointing, a colon after stop, Trautmann (BB. XIX, 201) makes the final clause, hafde ferfr cwicu, distinct from the context ; but I prefer to regard line 5 as the antithesis of line 4 : ' I dove under water, dead with the fishes; and (when) I stepped on the ground, I had a living soul.' — haefde ferS cwicu. The reading ferfr for MS. ford" is sustained by n 6, haefde feorh cwico ; 143, haefdon feorg cwico. Cosijn (PBB. XXIII, 130) finds the same substitution in Chr. 1320, 1360. RIDDLE 76 This short runic riddle has in common with Rid. 20 not only the method of inverting runes, but the phrasing (see 20 1-3 and 65 i). Read backwards, the four runes as restored (see text) spell H U N D, ' dog.' Dietrich, XI, 483, conjectures that this was the introduction to a longer riddle. 75 1-2 Swift dogs were in great demand among the Anglo-Saxons. The hunter tells us, ^Ifric's Colloquy, WW., 92, 14, mid swiftum hnndum ic betSce -wildeor; and the fowler (id. 95, 12) readily offers a hawk in exchange for a swift hound. Wright, Domestic Manners, p. 69, prints from Harl. MS. 603 a picture of a dog- keeper (IiundwealJi) and his two dogs. Sharon Turner, VII, chap, vii, recalls the evidence of William of Malmesbury (De Gestis Regum Anglorum II, chap. 1), that ^ithelstan made North Wales furnish him with as many dogs as he chose, ' whose scent-pursuing noses might explore the haunts and coverts of the deer,' and that Edward the Confessor was fond of hunting with fleet hounds and of hawking. For the appearance of hund'm. the Anglo-Saxon vocabulary, see Jordan, Altenglische Saugetiernamen, pp. 46 f. 2i6 RIDDLES OF THE EXETER BOOK RIDDLE 76 Dietrich (XI, 483) suggests that perhaps the single line Ic dne geseah idese sittan forms the introduction to Rid. 77, as the subject of that riddle, ' Oyster,' is of feminine gender (Lat. ostrea ; A.-S. ostre), and, being footless, she sits upon the rocks ; but the change from the third person in 76 to the first in 77 is quite suffi- cient to destroy this conjecture. Grein, Bibl. II, 401, queries whether the subject was not originally given in runes as in Rid. 75. Trautmann, Anglia, Bb. V, 50, re- gards the line as a fragment. RIDDLE 77 Dietrich is doubtless right in his answer, ' Oyster.' The riddle has only the topic in common with the last line of Aldhelm's ' Crab ' enigma (iii, 2 6), ' Ostrea quern metuunt diris perterrita saxis ' ; but it finds apt comment in Ausonius's 'Ostrea' griphos in his letter to Theon (Epistolae vii, Opera, 1785, p. 246): 'Ostrea . . . Dulcibus in stagnis reflui maris aestus opimat,' and in yet another epistle of the Latin writer (ix, ib. p. 249) : Ostrea nobilium coenis sumptuque nepotum Cognita diversoque maris defensa profundo, Aut refugis nudata vadis aut scrupea subter Antra et muriceis scopulorum mersa lacunis. Our riddle bears no resemblance to Scaliger's 'Ostrea' (Reusner I, 173), which describes the strange nature of the house. But an English riddle ( Wit Newly Revived, I78o(?), 21) contains the final motives of Rid. 77 (43-8): Stouthearted men with naked knives Beset my house with all their crew ; If I had ne'er so many lives, I must be slain and eaten, too. The Anglo-Saxon fisherman takes in the sea (^Elfric's Colloquy, WW. 94) haerincgas and leaxas, mereswyn and styrian, ostran and crabban, muslan, pine- winclan, sjecoccas, fage and floe and lopystran and fela swylces (see Heyne, Fiinf Biicher II, 250). So in the Ecd. Hist, i, i (Miller, 26, 7), her beoj> oft numene missenllcra cynna weolcscylle ~\ muscule, etc. From Leechdoms II, 244, 2, we see that raw oysters (77 8, unsodene) were not deemed a healthy food (Whitman, Anglia XXX, 381). 77 i Sa mec fedde. The feeding of the subject is a common theme in the Riddles: cf. 517-8, 543-4, 724-5, 731-2- — sundhelm. The word is found only here and in 3 10. 77 2 mec y)m wrugon. Cf. 3 15, yj>a . . . J>e mec ier wrugon. — eorpan getenge. So 7 3«>. 77 3 fepelease. Both here and in unsodene (1. 9), the grammatical gender of ostre is regarded. NOTES 217 77 4 inu3 ontynde. Cf. Whale, 53, "Sonne se mereweard miiS ontyne'5. 77 6 seaxes orde. Prehn, p. 250, notes the part played by the knife's point in the Riddles: 276, seaxes ecg; 61 12, seaxes ord. See my note to 276 for a dis- cussion of the seax. 777 hyd arypeS. See Leechdoms I, 338, 16, mid ostorscyllum gecnucud ond gemenged. RIDDLE 78 This is a fragment not printed by Thorpe and Grein. Trautmann, Bb. V, 50, does not attempt a solution; but Holthausen, Anglia XXIV, 265, suggests 'ein im wasser lebendes tier (auster? krebs? fisch?).' It presents several parallels to the ' Oyster ' problem : 78 i, Oft ic flodas ; 773, Oft ic flode ; 783, [d]yde me to mose ; 77 5, 8, fretan . . . ite^ ; 78 7h, yj>um bewrigene ; 77 2, mec y>a wrugon. On account of these very recurrences of thought, we cannot regard 78 as a mere con- tinuation of 77 ; but rather as a development of a similar theme. 78 3 Holthausen, Anglia XXIV, 265, would read \fi\yde; but my reading, \d~\yde me to mose, is supported by And. 27, dydan him to mose, and by the parallel of thought in 77 (supra). 78 7b y)wni bewrigene. Cf. 3 15, y)>a . . . J>e mec ;Er wrugon ; 772, mec yj>a wrugon ; Gen. 156, bewrigen mid rlode ; Gen. 1460, bewrigen mid waetrum; Met. 8 59, bewrigen on weorulde waetere oJ>J>e eorSan. RIDDLE 79 Dietrich (XI, 483), regards this single line as ' merely a variant of the first line of Rid. 80.' 79 i ic coin aej>elinges a-ht. So of the Sword, 71 i, ic com rices sht. RIDDLE 80 Dietrich's answer ' Jagdfalke ' or ' Habicht ' (XI, 483) is accepted by Prehn (p. 283) and Stopford Brooke (p. 147). Walz, Harvard Studies V, 267, defends the solution 'Sword' by its relation to its lord (i), its wooden sheath (6), its •hard tongue' or point (8b), its use as a gift (9-10 a), its brown edge (na"). Miiller, C. P., p. 18, offers the answer ' Horn,' which is accepted by Herzfeld (p. 5). Trautmann, who had not read Miiller, gives (BB. XIX, 203 f.) many good reasons for rejecting other answers and his own earlier solution, ' Spear'; and now offers convincing support to ' Horn." This is literally the noble's shoulder-companion and the warrior's comrade (1-2); it is the associate of the king (3 a), as a drinking- vessel. So at feasts, the queen takes it in her hand (3 a~s) (and offers it to the heroes); cf. Beow. 494 f., 620 f., 1168, 1216, 1981 f., 2021 f. The Horn carries in its bosom what grew in the grove (6) — the mead made of honey 'brought from groves' (Miiller and Trautmann cite 28 2-3"). As battle-horn, it rides upon a horse at the end of the troop (7-8 a). Its tongue or tone is hard (8 b). At the banquet 2I8 RIDDLES OF THE EXETER BOOK it offers wine to the singer as reward for his song (9-10 a) (cf. Muller). Its color may well be black (n a). Trautmann has surely proved his thesis, as Muller had done before him. Points of likeness with the earlier ' Horn ' riddle, Rid. 15, are many, as Muller and Trautmann show : there the Horn rides upon a horse (5b-6a, I3b-I4a); it has a filled bosom (8-9 a); its voice is described (16-193); and one may add that 80 za, fyrdrinc e s gefara, is paralleled by 15 i3a, freolic fyrd- sceorp. The hwilum clauses. of 80 recall those of the earlier riddle (compare Brandl, Grundriss II, 972). For a discussion of the Anglo-Saxon horn, see my notes to Rid. 15. 80 2b frean imnum leof. So 21 2 (Sword). 803-5 As Trautmann has pointed out (see supra), the Beowulf, 612, refers to such service by noble women, when WealhJ>eow passes the beaker at the feast. So in Gn. Ex. 88-01 : (Wif sceal) meodorjgdenne for gesi"5maegen symle ajghwair eodor a)>elinga Srest gegretan, forman fulle to frean bond ricene gersecan. In Bede's Eccl. Hist., bk. v, chap. 4, an earl's wife ' presented the cup to the bishop and us (Abbot Berthun), and continued serving us with drink as she had begun till dinner was over.' The same custom prevailed in other Germanic countries. In the Ynglinga Saga, chap. 41, Hildigunn, daughter of King Granmar, carries ale to the viking Hjorvard. In the courtly verses cited by Vigfusson and Powell (Cor- pus Poeticum Boreale II, 418) from Olaf's Saga, the poet calls ' Fyll horn, kona . . . Berr mer of ker ! ' (' Fill the horn, lady . . . Bear me the cup '). And we are told by Geoffrey of Monmouth, in his account of the meeting of Vortigem and Hengist (ffistoria Britomim, bk. vi, chap. 12, cited by Budde, Die Bedeutung der Trinksitten, p. 39), that Rowena, the daughter of the Saxon chieftain, was the British king's cupbearer : ' Ut vero regiis epulis refectus fuit, egressa est puella de thalamo aureum scypheum vino plenum ferens ; accedens deinde propius regi flexis genibus dixit : " Lauerd king wacht heil ! " ' For hiuitloccedu, cf. note to 41 98. In the Heifrreks Gdtur, No. 9, light-haired women carry ale. 80 4 bond on legeft. An example of the shortened A-type, with a heavy monosyllable in the thesis (Herzfeld, p. 44). " 80 5 eorles dohtor. Contrast 26 6b, ceorles dohtor. That riddle is throughout on a lower plane. See, however, 46 5, }>eodnes dohtor. 80 7 on wloncum wiege. Cf. Maid. 240, on wlancan J>am wicge ; Rid. 20 1-2, S R 0 H (hors) hygewloncne. 80 9-10 As -wofrbora is used in 32 24 of the riddle-solver, and as gied is else- where applied to a riddle (56 14, see my note), it is easy to fancy that our thirsty riddler is here giving a sly hint. For a careful study of the word wofrbora, in its many meanings, see Merbot, Aesthetische Studien zur ags. Poesie, pp. 5-7. Budde, P- 33> points out that the frequent introduction of drinking situations into these enigmas seems to show that riddle-guessing was a part of the entertainment at feasts. NOTES 219 RIDDLE 81 Dietrich (XII, 234-235) rejects his earlier answer, 'Ship' (XI, 483), and ac- cepts Professor Lange's solution, 'Maskenhelm.' He says in his note: 'Das haupt des an brust und nacken ausgebognen helms ist der obere erhohte grat oder rand, der das eberzeichen als heahne steort tragt, der fuss ist das nackenstiick, auf dem der helm abgenommen steht, das heard nebb ist das nasenstiick oder der steg der maske, die den mund unbedeckt lasst ; das elend (regenstrome, hagel, reif und schnee) erduldet der helm, wenn ihn der krieger, der die lanze (wudtt) regt, auf seinem haupte tragt, wodurch er " wohnung iiber den mannern " hat.' This solu- tion, which Brooke modifies to 'Visor' (E.E.Lit., p. 127) and translates in part (p. 124), is certainly less apt than the ' Wetterhahn ' or ' Weathercock ' of Traut- mann (Aitglia, Bb. V, 50), which meets all the conditions of the problem. It is puff-breasted and swollen-necked ; it has a head and a high tail, eyes and ears, one foot, back and hard beak, high nape and two sides. It has a dwelling-place over men. It suffers wretchedness when it is moved by the wind, which is de- scribed in the periphrase, 81 7b, se J>e wudu hrerefr (so the Wind-storm says in Rid. 2 8, ic wudu hrere), and when it is beaten by the elements. So one speaks fittingly of a ' Weathercock,' and not of a ' Helmet.' Indeed the wind-motif ap- pears in the German ' Wetterhahn ' riddle, which has an honorable history (Wos- sidlo, No. 104, notes ; Friedreich, p. 207) : Sich in alien Winden erhebet, Und warm die wiiten, Muss er dann fleissiger hiiten. No use of the word ' Weathercock ' is recorded in Anglo-Saxon — indeed, before the -wedercoc of the Ayenbite of In-wit, E.E. T.S. XXIII, 1866, p. 180 (cited by Bradley-Stratmann) — but I note in the excellent illustration of an Anglo-Saxon mansion (MS. Harl. 603, f. 67 v. ; Wright, Domestic Manners, p. 15) a pennant- shaped vane (fana). Weathervanes, not only on land but at sea, are frequently mentioned in the Old Norse sagas (Cleasby-Vigfusson, s.v.fani). 81 i bylgedbreost. The MS. form, byledbreost, is open to two objections : it is impossible metrically, and the first member of the compound is a hapax unsus- tained by the evidence of cognates. The word suggested satisfies both meter and sense, if bylged is taken in its primitive meaning of ' swollen,' ' inflated ' (cf. bylg, belg, 'bag,' 'bellows'). Gebylged is found elsewhere (B.-T., p. 378) in the derived sense of ' made angry,' ' caused to swell.' 81 sb sag[ol]. Thorpe conjectures sac ('a sack'). Ettmiiller (Worterbuc)i) renders sdg 'onus'; and Grein, Dicht., 'eine senkung'; but in Spr. II, 387, ' sdg (ndd. seeg), " Bundel," "Last?" ace. ic (sc. scipl} habbe sdg on middan — vgl. jedoch auch mhd. seige and altn. scegr? Dietrich explains the word (XI, 483): ' eine offnung auf dem verdeck zum hinabsenkung (siegan) der waaren (cf. 33 9, rniiS waes on middan).' B.-T., p. 813, cites the word, but does not trans^te, and Sweet does not include it in his Diet. The Dicht. translation, ' a sinking,' alone fits the proper solution, 'Weathercock,' and may describe the bird's back between the ' high neck ' (1. 4) and ' high tail ' (1. 2). Mod. Eng. sag is connected by Skeat, 220 RIDDLES OF THE EXETER BOOK Etym, Did., s.v., with Swed. sacka and Germ, sacken ; and he suggests a possible confusion with sigan, ' to sink.' I should like to suggest the word sdgol, ' staff,' which glosses the Lat. fustis, and is used of ' the rods or bolts (vectes) thrust through rings to bear the ark ' (Cura Pastorate, Sweet, p. 171, 5-12). This might well apply to the rod which pierces the Weathercock, and upon which it turns. Sdgol would then be in natural apposition to card ofer aldum (1. 6) and would ex- plain byrelwombne (1. n). 8 1 6b Aglac dreoge. Cf. Dan. 238, >5r hie )>aet aglac drugon. 81 7 fair mec wege5. Sievers proposes waged" on metrical grounds, but our word is elsewhere used, as here, of movement by the wind (supra) : Met. 7 35, t>eah hit wecge wind. The half-line is of the A-type (Z i_ | ^/ X ) common in the Rid- dles (Introduction). 8 1 8h streamas beataS. Cf. 3 6% streamas sta)>u beataiU (note). 81 9-10 Cf. Rid. 41 54-55, se hearda forst | hrim heorugrimma. Instead of the _fo\rs\t gercf]sefr of Holthausen, Bb. IX, 358, I supply with aid of B. M. \pnd f\orst \hr\eosefr. Hreosan is the word always found in like context : Ph. 60, Her ne haegl ne hrim hreosa'S to foldan ; Wand. 48, hreosan hrim ond snaw hagle gemenged; Wand. 102, hri'S hreosende ; etc. 81 ii [on] fyrelwombne. The addition seems necessary to the context, but not to the meter, as elsewhere in the Riddles, 45 2, 91 5, the adj. byrel, ' perforated,' has a long root-syllable, while the noun/j^r^/, ' hole,' has a short one, 16 21, 72 8. The meaning, ' having the stomach pierced,' is explained by my reading of sdgol for sdg in line 5 (supra). RIDDLE 82 The few scattered phrases of this fragmentary riddle give no clue to the solution. 82 2 greate swilgeS. Perhaps, greote s-wilgeS" ; cf . Gen. 909, J>u scealt greot etan. 82 4 [f]ell n6 lla-sc. Cf. 77 5. 826 ma-la gehwam. Cf. 33 12, geara gehwam; 61 6, uhtna gehwam. RIDDLE 83 There is little difference of opinion among solvers regarding the answer to this. All agree that it is a metal, subjected to the flames (2b, 3b, 4 a). But Dietrich (XI, 484) believes the subject to be 'Ore'; and Trautmann, 'Gold.' It has something in common with Symphosius 91 (Pecunia) : Terra fui primo, latebris abscondita diris (or terrae) ; Nunc aliud pretium flammae nomenque dederunt, Nee jam terra vocor, licet ex me terra paretur. While there are no detailed likenesses between this and the Anglo-Saxon, there is the same general riddle-motive of change of condition through fire ; but this may be»mere coincidence. According to Dietrich, the subject's foe (4b-8) was Tubal-cain, ' an instructor of every artificer in brass and iron ' (Genesis iv, 22) ; and the 'captivity incited' (9-10 a) is due to the weapons made from iron. Perhaps these lines may refer to chains, or to the evils caused by money (i Timothy vi, 10, NOTES 221 'the love of money is the root of all evil'). The last part of the problem (100-14) seems to me to indicate ' Gold ' and its secret ways and works. Ore, of whatever metal, fulfills all conditions. Bede in his Eccl. Hist, i, i (Miller, 26, 14) tells us of England (Breoton) : Hit is eac berende on wecga 5rum ares and Isernes, leades and seolfres. Kemble, Saxons in England, 1875, ^» 7°> after noting many charters in which salt-mines are mentioned, points to the grant of Oswini of Kent in 689 to Rochester, deeding a plowland at Lyminge in which he says there is a mine of iron (Codex Dipt. No. 30). Kemble, I.e., believes the isengrdfas of Cod. Dipl. 1118 to be iron-mines. And in the Vocabularies we meet isern ore, ' ferri fodina, in quo loco ferrum foditur ' (see also B.-T., s. v. drd). ' The smelting in the Forest of Dean is said to have been carried on continuously since Roman times ; and this is quite probable also in regard to the tin mines of Cornwall and the lead mines at the Peak ' (Cunningham, The Growth of English Industry and Commerce, 1890, I, 62). The Merchant in the Colloquy (WW. 96) brings, among other wares, ' aes et stagnum ' (cer and tin) to his English customers. 83 i Dietrich (XI, 484) compares Rid. 54 4, frod dagum (tree) ; 73 3, gearum frodne (tree) ; and 93 6, daegrlme frod (stag). The unhappy change of state of the Ore is another motive that Rid. 83 has in common with Rid. 54, 73, 93. 83 2-3 Holthausen's emendations, Anglia XXIV, 265 \mec\ and \hafde leod- •wera, miss the point of the passage, though his suggestion of Rge is happy. Beow. 2322-2323 helps us greatly here : Haefde landwara lige befangen, bale ond bronde. So I was inclined to read in 83 3, \hcefde lond]wara lige bewunden, and to regard londwara as an enigmatic reference to the ores, which are surely ' surrounded by flame and purified by fire.' But this is contradicted by letter-fragments in B. M. 83 zb bseles weard. This refers, I think, to Tubal-cain, the eorj>an brdj>or of line 5 (see note). 83 3b Hge bewunden. Cf. Beow. 3146-3147, swogende leg | wope bewunden ; Chr. 1538, lege gebundne ; Rid. 31 2 (legbysig), bewunden mid wuldre. 83 4 gefailsad. For a careful discussion of the meanings of gefailsian, see Cook's note to Chr. 320. It is used only here of ' cleansing by fire.' — fah waraS. Cf. 93 26, Nu min hord wara"S hij>ende feond. 83 5 eorban brobor. The Earth is called ' the mother and sister ' of men (Body and Soul) in Rid. 44 14. See also the Prose Riddle, cited in my note to that passage. This phrase, eorhan broj>or, well accords with the Anglo-Saxon conception of Tubal-cain, as revealed in the illuminated manuscripts. In Cotton Claudius B. IV, f. 10, a picture of Tubal-cain at work at his forge bears the in- scription Tubalcain se was izgj>er ge goldsmifr ge irensmifr. And in the Caedmon manuscript (Archaeologia XXIV, pi. xxviii), he appears in his two r61es of smith and plowman — in either case, a ' brother of the earth.' He is thus described in Gen. 1082 f. : Swylce on "SJere maegtJe maga waes haten on (>a ilcan tid Tubal Cain, 222 RIDDLES OF THE EXETER BOOK se Kirh snytro sped smiS craeftega waes and burn modes gemynd monna zerest sunu Lamehes sulhgeweorces, fruma waes ofer foldan : sififtan folca beam ares cu3on and isernes burhsittende brucan wide. 83 7 agCtte. Sievers, PBB. X, 513, establishes the length of the root-vowel by consideration of this, and other examples in the poetry. 83 8b Note the omission of the verb after an auxiliary verb. The half-line recalls the lack of redress of the Sword (21 17), and of the Horn (93 19). 83 10 wongas is here used as a poetical expression for ' the earth.' See Rid. 13 2, 41 51,83 ; and compare Cook's note to Chr. 680, wonga. 83 iob Hsebbe ic wundra fela. Cf. 22 8, haebbe wundra fela (plow) • Beow. 408, haebbe ic maeriSa fela (Sarrazin, Beowulf -Studien, p. 128). 83 12-14 Compare the final motive of the Moon riddles (30, 95). Very striking is the verbal likeness between 83 12 and 95 14, mine (i.e. swaj?e) beml>e monna gehwylcum. 83 13 degolftilne dom. Cf. Ps. 147 9, his d5mas digle. RIDDLE 84 Dietrich (XI, 484) gives the answer « Water,' which remains unquestioned. He points out the likeness of 84 4, Modor is monigra mizrra wihta, to Aldhelm iii, I (Aqua) 4-5 : Nam volucres caeli nantesque per aequora pisces Olim sumpserunt ex me primordia vitae, and of 846b-9a to Aldhelm iv, 14 (Fans) 3-4: Quis numerus capiat vel quis laterculus aequet Vita viventium generem quot millia partu. As Prehn claims (p. 253), this problem has certain motives in common with the Aequor enigma of Eusebius, No. 23. Compare the wild course of the ' Water ' (84 1-3) with the first line of the Latin, ' Motor curro, fero velox, nee desero sedem ' ; and the water's burden, 84 43, bij> stdnum bestrewed, with Eusebius 23 4, ' Desuper aut multis sternor.' But there are reasons for regarding these likenesses to Eusebius as coincidences entailed by a common source and the demands of the subject. The opening lines of Rid. 84 and of Eusebius 23 are both inspired by Aldhelm iv, 14 1-2: Per cava telluris clam serpo celerrimus antra, Flexos venarum gyrans anfractibus orbes. And in its picture of the Water's burdens our riddle is not as close to Eusebius as to Pliny's account of Water, Natural History xxxi, 2, ' Saepe etiam lapides subvehunt, portantes alia pondera.' Still another motive, that of the ships (84 21-22), is far more clearly expressed in Aldhelm iii, I 2, ' Dum virtute fero NOTES 223 silvarum robora mille,' than in Eusebius 23 2, ' tarn grandia pondera porto.' The description of the Water's cover, 84 39, oft titan beweorpefr anre fcecene, is in striking contrast to Eusebius 23 3, ' Nix neque me tegit,' etc. Finally, ' Water ' riddles with as close resemblances to Rid. 84 are found in other countries and other times (Brussels MS. 604, I2th century; Mone, Anzeiger VIII, 40, No. 48). 84 i The emendation of Biilbring (see Text) is sustained by Rid. 51 i, Wiga is on eort>an wundrum acenned. 842 hreoh ond rej?e. Hreoh is often applied to Water (Gen. 1325, Ps. 68 i, hreoh -water, etc. ; see Spr. II, 103, for many examples), as is also re}>e (Jud. 349, re>e streamas). See Dietrich (XI, 484). — hafa'd' ryne strongne. Cf. Gen. 159, (waeter) >a, nu under roderum heora ryne healda'S. The opening lines of 84 sug- gest the Storm riddles (2 — 3 5). 84 3 grymetao1. So of Water, Pan. 7, brim grymetende. — be grunde fare<5. Cf. Rid. 22 2, be grunde graefe. 84 4 Cf. 42 2, moddor monigra cynna (water?) 84 sb fundatJ sefre. Compare the description of Water in Sal. 392 f. : Ac forhwSm winne'S 'Sis wster geond woroldrice, dreoge'5 deop gesceaft, ne mot on daeg restan, neahtes ne "Syft, craefte ty-5 .' Ic wihte ne cann forhwan se stream ne mot stillan neahtes. This superstition is found in Strassburger Rdtselbuch, No. 52, and is there traced to Aristotle. 846-9 Here the riddler must have had in mind Psalms civ, 25, 'So is this great and wide sea, ivherein are things creeping innumerable, both small and great beasts.1 Compare the Anglo-Saxon poetic version (103 24) : His is mycel sa; ond on gemaerum wid : l>jer is unrlm on ealra cwycra, mycelra ond mastra. 84 7b wordum gecyj^an. Cf. Whale, z b, wordum cy>an. 849 Cf. Gn. Cot. 61-62: Is seo forftgesceaft dlgol ond dyrne, Drihten ana wat. With the reference to the Creation (84 9-10) cf. 41 1-8. 84 10 or ond ende. Cf. Met. 20 275, And. 556 b, fruma ond ende. In his note Krapp, p. in, cites Revelation i, 8, 11; xxi, 6; xxii, 13. 84 ii nieotudes beam. So Chr. 126. Grein's addition, his mihta sped, finds warrant not only in B. M. word-fragments but in the frequency of this phrase (Spr. II, 236). I read meahta, as this accords with the forms in the Riddles (see Glossary). 84 19 wlitig ond wynsum. So Sat. 214, Pan. 65, Ph. 203,318 . 8421-22 For metrical reasons, Holthausen, Engl. Stud. XXXVII, 210, does violence to the MS. readings (see text and variants); but no changes are neces- sary, as examples of the A-type with second stressed syllable short (£ x ( x ) | 224 RIDDLES OF THE EXETER BOOK ^i x ) are found elsewhere in the Riddles (18 n, 24 i, 39 6, 7, 43 n, 47 6, etc. ; cf. Sievers, PBB. X, 458; Herzfeld, pp. 44, 49). 28 13-14, strengo bistolen . . . mtzgene binumen, are exactly parallel to the present lines. The metrical a-priorism of Holthausen is dangerous. 8421 wistum gehladen. According to Grein, Spr. II, 721, the Water is so described 'als Heimat der essbaren Fische.' But this and the parallel phrases (11. 21-22) may refer to the ships upon the sea (supra). Of the Ship in 33 n we are told, wist in wiged". 8425 wuldorgimm wloncum getenge. Cf. El. 1114, godgimmas grunde getenge (Herzfeld, p. 19). 8429 gifrost ond grsedgost. Cf. Seaf. 62, gifre and gradig; Gen. 793, grie- dige ond gifre ; Sat. 32, 192 ; Soul 74, gifre ond gradige. 84 30 \fses ]>e. This is rendered by Thorpe ' from the time that,' and by Grein, Dicht., ' von allem was.' The use of the phrase after superlatives (see 1. 29) is illustrated by the very similar passage Chr. 71-73 : Eala wifa wynn geond wuldres >rym, fajmne freollcast ofer ealne foldan sceat J>as J>e sefre sundbuend secgan hyrdon. Cook renders ' as far as ' (see Spr. II, 576) ; and this may be the meaning in the Riddle line. Cf. also Met. 28 zz,Jxzs J>e monnum J>incfr, ' as far as it seems to men.' In the not unlike clause in the other ' Water ' riddle, 42 4, 5, J>aes deorestan, )>aes J>e dryhta beam . . . agen, fxes J>e is the simple relative attracted to the case of its antecedent. In both cases the subjunctive follows (Madert, p. 97). 8431 selda beam. So 95 10; Seaf. 77; cf. Wond. 99, selda bearna; Chr. 936, ielda bearnum. 84 32 Grein, reading magen for MS. mage, translates (Dicht.) ' der Weltkinder Menge, wie das webt die Glorie.' Dietrich notes (XI, 485), ivuldor = wundor (90 3, gloriam). But Thorpe was on the right track when he rendered the line ' So that glorious woman (wuldor-wifefr), world-children's daughter.' My change to wttldor wifa is supported by Men. 149, wifa wuldor, 'glorious woman' (cf. Chr. •j i, wifa wynn, cited supra). I regard the line as parenthetical, and translate 'So (lives) the glorious woman, kinswoman of world bairns.' Mage, which carries the meaning of ' mother' not only in Beow. 1390, Grendles mdgan, but in Rid. 10 4, is aptly applied to the Water, which in this riddle is mod(d)or (11. 4, 20). 84 33-34 This clause, I believe, points back to the superlatives in lines 28-29 : 'most greedy and rapacious . . . though a man, wise in spirit, learned in mind, may have experienced a multitude of wonders.' That is to say, ' whatever a man's experience, he is yet to learn of anything more greedy,' etc. 84 33 ferjnim gleaw. Cf. 60 zb, modum gleawe (note). 84 34 mode snottor. Cf. 86 2, mode snottre ; Feed. 87, modes snottor. See mod-snottor (Spr. II, 260). 84 35-36 These comparatives recall the ' Creation ' riddle (cf. 41 55). Hriisan heardra is clearly a reference to the ice-form of water (see line 39). ' Hcelehiim frodra ist zu verstehen wie 83 i und geht wieder auf die schopfungsgeschichte, wonach wasser viel eher als der mensch vorhanden war' (Dietrich, XI, 485). NOTES 225 84 37 waestmum tydreJf. The riddler may have had in mind Ps. 64 n, waeter yrnende waestme tyddraft. Cf. Ps. 103 16, waestme tydra'S. So in the ' Water ' riddle (Brussels MS. 604 d, Mone, Anz. VIII, 40) : ' Exeo frigida, sicca satis, nemus exalo, rideo pratis.' 84 38 Cf. Sal. 395, cristnaft ond clzensa'S cwicra manigo (wafer). In firene d-wiescefr Dietrich (XI, 485) rightly finds a reference to holy water, and cites the passage from the Sigewulfi Interrogationes (see MacLean, Anglia VII, 6), in which the Water is declared exempt from the curse placed upon the Earth after Adam's fall, because God had decided ' )>aet he wolde )>urh waeter j>a synne adylgian J>e se man Jmrhteah.' 84 40 Cf. And. 543, wuldre gewlitegad ofer wer^eoda. So of Water, Sal. 396, wuldre gewlitigaiS. 84 41-44 Cf. Rid. 4 7-10. 8444 timbred weall. Cf. Gen. 1691-1692, weall stienenne | up forS timbran. 84 46 lirusan hrinc'd'. Cf. 67 5, grundum ic hrine. 84 53 I do not accept the herd word[a~\ of Holthausen, Anglia XXIV, 265, be- cause it forces upon us a change in the text, and because word-hard is the ordi- nary phrase. G\esuuutela~\ of Holthausen is a possible addition (see Chr. 9, gesweotula; 8423, gesweotlad). But so are many other words beginning with g. Little is gained by such guesswork. 84 54 Holthausen's emendation \wisdom on\wreoh is supported by El. 674, wisd5m onwreon. RIDDLE 85 As Dietrich has pointed out (XI, 454), the source of this ' Flood and Fish ' enigma is the twelfth riddle of Symphosius : Est domus in terris, clara quae voce resultat : Ipsa domus resonat, tacitus sed non sonat hospes ; Ambo tamen currunt, hospes simul et domus una. I have traced the history of this (M. L. N. XVIII, 3) : it is found in the Dispu- tatio Pippini et Albini (Haitpts Zs. XIV, 543), No. 93, in the Flores of Bede (Migne, P. L., XCIV, 539), in Bern MS. 611, No. 30 (Antk. Lat. I, 360), and in the Apollonius of Tyre (Weismann, Alexander, 1850, I, 480). So it came into the Gesta Romanorum, cap. 153, and passed then into the possession of the people (Strassburg Kb., No. 109; Simrock8, p. 14). The motive is found as far afield as Turkey (Urquell IV, 22, No. 10). A second problem (M. L. W. XVIII, 5) with the separate motive of ' the house escaping from robbers (the net), while the guest is captured,' lives at present in many French, German, Italian, and English forms (Holland, No. 71 ; Petsch, p. 138), and has been noted by me in 13th-century Latin dress (MS. Arundel 292, f. 114; Wright, Altdeutsche Blatter II, 148). The two motives are found side by side in Strassburg Rb., Nos. 198-109, and are finally combined in a Russian version (Sadovnikon, Zagadki Rousskago Naroda Sostavil, St. Petersburg, 1876, No. 1623) discussed by Gaston Paris (In- troduction to Rolland, p. ix). 226 RIDDLES OF THE EXETER BOOK Two motives are added by the Anglo-Saxon to those of Symphosius. The first, that of difference between guest and house (3 b-5), is found in the Strass- burg riddle (109): Etwan (nit wan) die gest in kurtzer beyt, Floch es von mir on arbeit : Stunden die gest gar still, Gar bald darnoch in kurtzer zeit Die gest auch flohen wieder streit, etc. and in the Turkish (supra), ' Ich gehe, es geht auch ; ich bleibe stehen4 es bleibt nicht stehen ' (IVasser). The second — a 'living and dead' motive — is an addi- tion found only in our query. 85 2-3 Cf. Gen. 903-905, )>a naedran sceop nergend usser . . . wide sfSas. 85 2 To the ymb \droht mlnne\ of Holthausen, /. F. IV, 388, I greatly prefer ymb unc \domas dyde] (see Ps. 11865, 13912). With drihten for MS. driht cf. dryhtnes for MS. dryht (60 n). Driht is sometimes used as an abbreviation for all cases of drihten (see B.-T., pp. 213, 216). 85 3b Cf. 41 94, ic com swH>ra J>onne he. With swiftre compare strengra and J>reohtigra (1. 4), and note just such inconsistency in gender as in the ' Creation riddle (passim). 85 5 For yrnan of MS. and editors I substitute rinnan, on account of the alliteration. 85 6b Cf . Feed. 8, a J>enden }>u lifge. 85 T° m6 bi3 dea dea$ witod. RIDDLE 86 Dietrich's first solution, 'Organ ' (XI, 485), is accepted by Padelford, Old Eng- lish Musical Terms, p. 46. ' Ich denke,' says Dietrich, ' an die orgel des weltlichen gebrauchs, die schon sehr friih bekannt war, und zwar mit tausenden von pfeifen — gestiitzt auf Aldelmus de Laud. Virg. s. 138, maxima millenis auscultans organa flabris! Later Dietrich recognized (XII, 248, note) that the riddle was simply an expansion of the second line of the ' Luscus allium vendens ' enigma (No. 94) of Symphosius: ' Unus inest oculus, capitum sed milia multa' (3 a, 4b). The other traits fit perfectly the solution ' One-eyed Garlic -seller ' — as they are not ' mon- ster ' but natural human attributes (see Prehn, p.255). Miiller, C.P., p. 19, accepts this solution. 86 i Wiht cwom gongan. Cf. 35 i, Wiht cwom . . . ll^an ; 55 i, Hyse cwom gangan. — \veras sieton. Cf. 47 i, wer saet aet wine, etc. 86 2 monige on maeffle. Cf. And. 1626, manige on meiSle ; Craft. 41-42, sum in mas'Sle maeg m5dsnottera [ folcrsedenne foriS gehycgan, etc. Padelford asserts, in support of his ' Organ ' solution, that ' this line is more suggestive of a congre- gation and of worship than of a social gathering ' ; but the above examples and other instances of mceSfrl (Spr. II, 214) do not sustain his view. The phrase here has no very definite meaning. — mode snottre. See note to 84 34. In this pas- sage the expression is quite lifeless. NOTES 227 86 3 f. With this enumeration of traits compare the other ' Monster ' riddles, 32, 33- 37. 59- 81. 86 4b twelf hund hf-afda. Dietrich notes (XII, 249) that 'die capitum millia multa sind der alliteration mit twegen fit zu gefallen durch XII hund heafda, gegeben.' RIDDLE 87 According to Dietrich (XI, 485), we have in this riddle ' Cask and Cooper.' ' Heaven's tooth ' (5 a), he thinks, is ' the thundering wedge,' while ' the eye ' (6 a) is 'the bung-hole.' The problem is obviously a companion-piece to Rid. 38. Its subject, like the ' Bellows ' of the earlier query, has a great belly (i b-2 a) and is followed by ' a servant, a man famous for his strength ' (2 0-3 a). With Miiller (C.P.,p. 19) and Trautmann (Anglia,Bb. V, 50), I accept for this also the answer 4 Bellows." 5 a, heofones tote, and 6 a, bleow on cage, speak strongly for this interpretation (infra). 87 1-3 For verbal parallels, see Rid. 38 (notes), and the fragment, Rid. 89. 87 5 heofones to]?e. Dietrich (XI, 485) explained this as ' the thundering wedge' (supra), and Miiller (C. P., p. 19) as 'the hammer of the smith.' Holt- hausen, Engl. Stud. XXXVII, 210, would read heof on his tdt>e, for to him 'der " himmelszahn " ist doch zu kindlich.' Properly interpreted, ' heaven's tooth ' is one of the most striking metaphors in riddle-poetry. It is applied to the Wind, whose bite is the theme of other enigmas ; cf. MS. Bern. 61 1, 41 4, Anth. Lat. I, 364 : Mordeo sed cunctos silvis campisque morantes. See Shakespeare's reference to the tooth of the Wind in Amiens's song, A. Y. L. ii, 7, 175. This interpretation exactly accords with the ' Bellows' answer to our riddle. 87 6 bleow on eage. Cf. 38 4, fleah Jmrh his cage (bellows). See also Wulf- stan, Homilies, 146,27 — 147,6, Deah man Jx>ne garsecg embsette mid byligeon . . . and to ieghwylcum )>i5ra byligea wiere man geset . . . ond man bleowe mid )>am byligeon, etc. 87 7 wanode. Thorpe and Grein's J>ancode, for MS. wancode, finds a certain support in the similar riddle-fragment 89 7, J>oncade, but it is ruled out of court by the alliteration, which here demands a w. To wancode, a nonce-usage unrecog- nized by the dictionaries, I prefer wanode, ' decreased,' ' diminished,' which is in perfect keeping with meter, context, and subject ' Bellows." RIDDLE 88 This riddle, according to Dietrich's correct interpretation (XI, 485-486), is one of the Horn riddles (see Rid. 15, 80, 93), and its subject is the Stag-horn, which once stood with its brother, the other horn, on the animal's head (88 12-15"), pro- tected by forest trees from night storms (15 b-17 a), until replaced by fresh antlers (18-20 a). Separated now from its brother, with whom it had shared many battles (29-31), it is torn and injured by monsters or adverse fates (32-33 a), and is placed ' on wood at the end of a board ' (22 b-23 a). Apart from likenesses of 228 RIDDLES OF THE EXETER BOOK this to Rid. 27 and 52 and particularly to Rid. 93, which I note below, the most striking analogue to the problem is found in the modern English riddle of Wit Newly Revived, 1780, p. II : ' Divided from my brother now, I am companion for mankind ; I that but lately stood for show, Do now express my masters mind. It is an ox's horn made into a hunting-horn, etc. By the brother is meant the other horn that grew with it ; and the expressing of the mind by the sounding of it.' But the last line of the modern riddle seems to show that this, like Rid. 88, 93, is an ' Inkhom ' enigma. The aim and end of our riddle have been completely misunderstood by all scholars. Dietrich (XI, 486) says : ' Wenn nun das horn sagt, jetzt steht es auf holz (Beow. 1318, healwudu) am ende des bretes und miisse da bruderlos fest- stehen, so ergiebt sich, es ist das dem giebel des ehedem meist holzemen hauses zum schmuck dienendey?rjM0r«. [Dietrich cites Ruin, 23, heah horngestreon ; Rid. 4 8, homsalu ; Beow. 705, hornreced.] Um da aufgesteckt werden zu konnen muste der untere theil des homes innerlich ausgebohrt werden, daher die klage iiber das aufreissen (8833-34), wodurch der suchende, d. h. der pflock der es tragen soil, gelingen findet.' Upon this interpretation of Dietrich, Heyne, Halle ffeorot, p. 44, bases the statement that the antlers were divided and that one horn was placed upon the western or southern, the other upon the eastern end of the roof. Brooke, too (E. E. Lit., p. 142), renders 88 22-23 ' Now I stand on wood at the end of a beam (that is, at the end of the roof-ridge of a hall).' It is safe to assert that we have not in our riddle the slightest reference to the stag-horns on the gable (see MS. Harl. 603, f. 67 v., Wright, Domestic Manners, p. 14), and that the fantastic picture drawn by Heyne (1. c.) of the great horn at each end of the roof must be erased, as it is derived from Dietrich's misconception of 88 22-25. This riddle, like Rid. 93, is a poem of the Inkhorn, which ' stands on wood at the end of the board ' — the desk or table (for illustrations of this place of the Horn, see MS. Royal 10. A. 13, Westwood, Facsimiles, p. 128; Benedictional of /Ethelwold, I2th miniature, ib. p. 132; cf. also ib. pp. 141, 143). As in 93 15 f., the Horn is hollowed out by knives (88 32-33), so as to serve for an ink -vessel. He who follows the trail of the ink (88 34, aet bam spore ; cf. 27 8, spyrige (.pen), 52 2, swearte . . . lastas (ink-tracks)) finds prosperity (infra) — and soul's counsel. The back of the Horn is wonn ond wundorltc (88 22) ; so its rim is called brunne brerd (27 9). Or the riddler may have in mind the ink that fills its back and belly (see 93 22-23, Nu ic blace swelge | wuda ond waetre). As will be shown later, Dietrich is equally unfortunate in his interpretation of certain parts of Rid. 93. 88 i Ic weox. The Riddles make frequent reference to the early growth of their subjects: 10 10, n 3, 54 3, 72 i f., 73 i. 887 [st]od Ic on sta3ol[e]. Cf. Dream, 71, stodon on staftole ; Beow. 927, stod on stabole (MS. stapole). See Holthausen, Engl. Stud. XXXVII, 209, 210. 88 12 uplong stod. Cf. Exod. 303, uplang gestod ; Beow. 760, uplang astod. NOTES 229 8816 wonnum nlhtum. Cf. Beow. 703, on wanre niht ; Gu. 1001, in )>isse wonnan niht ; Met. 1 1 61, J>a wonnan niht ; Rid. 13 9, deorcum nihtum. 88 i8-2oa The replacing of the old horns by new {gingran broj>or) is described in almost the same words in 93 13-14. 88 21 anga ofer eorban. Cf. Exod. 403, angan ofer eoriSan. 88 22-23 Thorpe, ignorant though he was of the solution, rendered literally and therefore correctly ' On wood I stand at the table's end.' This is strong though unwitting evidence to the naturalness of the ' Inkhorn ' interpretation. Bord is frequently used for ' table ' both in poetry and prose (Spr. I, 132-133; B.-T.,p. 116; cf. Rid. 15 9 (Horn), bordunt), and preserves this meaning in its later history. 88 25 As the illustrations of the Inkhorn (cited supra) show, it was fastened to the desk or the table, for security's sake. See note to 27 9a. 88 26-27 This may well be the lament of the Inkhorn for its lost ' brother,' but certainly not of the Gable-horn for its mate at the other end of the roof, as Heyne would have us think (see supra). 88 27 eorjran sceata. Cf. Rid. 68 16, eorban sceatas (note). 88 29 saecce to fremmanne. Cf. Beow. 2500, saecce fremman. For similar metrical types with uncontracted gerundial endings, see 29 12, 32 23, micel is to hycganne (-enne), etc. With the thought of the passage compare the very different enigmas, Rid. 15 i, Ic waes wJepenwiga (horn), and Eusebius 30 1-2 {horn) : Armorum fueram vice, meque tenebat in armis Fortis, et armigeri gestabar vertice tauri. 88 30 ellen cyftde. Cf. Beow. 2696, ellen cyftan. 88 31 unsceafta. This is not included in any of the dictionaries, but is ren- dered by Thorpe ' monsters,' by Grein, Dicht., ' Ungeschick.' Both renderings are consistent with the meanings of gesceaft, but the first accords better with the context. The ' monsters ' are, of course, the iron and steel weapons that scrape and hollow out the Inkhorn, 93 15-18. 88 33 be wombe. Of the contents of its womb or belly the Inkhorn speaks twice in 93 23, 28. — ic gewendan ne mseg. The thought is antithetical to the next line : ' I may not turn myself (i.e. move in any way), yet in my spoor or track, etc.' 88 34 spore and sped recall the speddropum and spyrige which describe the Ink-tracks 27 8. The spoor of the Ink is the path of life in Bede's Flares, xii (Mod. Phil. II, 562), for ' Viae ejus sunt semitae vitae ' refers to the holy words traced by the pen. So Aldhelm v, 3, De Penna Scriptoria : Semita quin potius milleno tramite tendit, Quae non errantes ad caeli culmina vexit. 88 35 sawle rsedes. So Met. 219; Leas. 42. RIDDLE 89 This fragment, which is not printed by Thorpe and Grein, is, as Trautmann says (fib. V, 50), ' ganzlich zerriittet.' Wiht wombe Juefd (1. 2) and lej>re (3) recall the ' Leather Bottle' (19 3) and the ' Bellows' (38 i, 87 i), but the subject's 'belly' 230 RIDDLES OF THE EXETER BOOK is mentioned in many riddles. J>ygan (1. 6) and swasendum (1. 8) suggest that we have to do with an article used at table — possibly a Leather Flask. But comment upon these few disjointed words and phrases is futile. RIDDLE 90 Dietrich (XI, 486), regards the different meanings of luptis as the subject of the Latin enigma, Rid. 90. ' A lupus is held by a lamb and disemboweled : the pike. The two wolves which stand and trouble a third, and which have four feet and see with seven eyes, are two rows of hops which entangle a wolf and which have five eyes or buds.' Later (XII, 250) Dietrich believed 'that by the first lupus a perch (Epinal- Erfurt Gloss. 592, t>cers), not a pike, was intended, and that the enigma was a play upon the name of Cynewulf, as, in Anglo-Saxon, names made from -wulf (sEthelwulf, Wulfstan} are commonly Latinized into Lupus.' In three places (Anglia VI, Anz. 166; XVII, 399 ; Bb. V, 51) Trautmann opposes Dietrich's solution, but suggests no adequate answer. In the first of his articles he hints at a connection between the four ' lupi ' of this riddle and the fourfold mention of wulf, Rid. i. Holthaus, Anglia VII, Anz. 122, finds in the enigma no proof of such word-play or reference to the name Lupus ; but Hicketier, Anglia X, 582 f., stoutly supports Dietrich. He thinks, however, that the first lupus refers not to a fish (lambs are not fish-eaters) but to the hop-rows. Henry Morley, English Writers II, 224-225, proposes ' the Lamb of God.' 'The marvel of the Lamb that overcame the wolf and tore its bowels out is of the Lamb of God who overcame the devil and destroyed his power. The great glory then seen was of the lamb that had been slain, the Divine appointment of the agony of one of the three Persons of the Trinity. The four feet were the four Gospels ; and the seven eyes refer to the Book of Revelation, where the seven eyes of the Lamb are the seven Spirits of God sent forth into all the earth. . . . The two wolves might be the Old and the New Testament troubling the devil and having the four Gospels upon which their teaching stands.' As I have shown (M.L.N. XVIII, 105), Morley's apocalyptic solution is strongly supported, at least in its first part, by the enigma of Aurelius Prudentius (Reusner I, 295) : Christus Agnus Agnus vice mirifica Agnus hiare lupum prohibes, and by the last line of the German problem, Pfalzer MS. 693, f. 27 (Mone, Anz. VII, 381, No. 312): Do quam ein lam und benam dem wolfe dy herte Solutio Der arge wolf, das ist Luciper . . . Das lam, das waz der werde Got. We have ample evidence that the devil is identified with the wolf in early religious literature. Jordan declares (Die altenglischen Sdugetiernamen, p. 64) : NOTES 231 ' Allmahlich aber, wohl mil dem Eindringen christlicher Anschauung iiberwiegt der Eindruck des Unheimlichen, Abstossenden in der Auffassung des Wolfes; in der christlichen Prosa ist er der Typus der Grausamkeit und Hinterlist. Das Bild des Evangeliums [John x, 12] vom Wolf, der den Schafen nachstellt, kehrt in den Homilien haufig wieder; der Wolf wird ein Sinnbild des Teufels.' Cf. yElfric, Homilies 1,36, 15, J>aet se ungesewenlica wulf Codes seep ne t5scence ; I, 238, 29, se wulf is deofol; I, 242, 3, wulf bi"5 eac se unrihtwisa rica; Laws of Canute I, 263, p. 306 (Wulfstan, Homilies 191, 16), }>onne moton }>a. hyrdas beon swySe wacore . . . J>aet se wodfreca werewulf to fela ne abite of godcundre heorde. Professor Cook in his note to Christ, 256, se dwyrgda wulf, cites Gregory, Horn, in Evang., lib. I, horn. 14 (Migne, P. L. LXXVI, 1128): ' Sed est alius lupus qui sine cessatione quotidie non corpora, sed mentes dilaniat, malignus videlicet spiri- tus qui cautas fidelium insidians circuit et mortes animarum quaerit.' See also the Alarien Himmelfahrt (Haupts Zs., V, 520), 1. 190, ' do der vil ungehore helle- wolf.' When the devil wishes to tempt Dunstan he assumes the form of a wolf (Eadmer's Vita, § 11, Stubbs, Memorials of Dunstan, Rolls Sen, p. 183). As Holthausen has clearly shown (Engl. Stud. XXXVII, 210-211 ; see text), rime demands in the second line ' obcurrit agnus [rupi] et capit viscera lupi.' Now if agnus be ' Christ,' and lupi ' the Devil,' there seems to be little doubt that rupi refers to the rock (Peter) upon which the Church is built (Matt, xvi, 18). Christ, through his Church, destroys the Devil. Morley's interpretation of 90 4-5 seems overwrought (see Bradley, Academy, 1888, I, 198); but I am unable to find a satisfactory explanation of these enig- matic lines. The phrase 'cum septem oculis ' certainly smacks of the Apocalypse. Recently the attempt has been made to interpret the Latin riddle as a very complicated logogriph and charade upon Cynewulf's name. In Herrigs Archiv CXI, 1903, 59 ff., Edmund Erlemann discusses the problem at length. He says : « Ich lose auf — — . Lupus-vmlf, 5-8, ab agno-ewu, 4-6, tenetur (gleichsam im Maule) ; darum mirum videtur mihi . . . obcurrit agnus : dem die einzelnen Buchstaben verfolgenden Auge des Dichters scheinen die drei : e, w, u = 4-6, dem Wolf, wnlf = 5-8, entgegenzulaufen. Et capit viscera lupi : ahnlich wie vorher tenetur, und nimmt die Eingeweide, d. i. das Innerste des wulf, namlich die beiden Buchstaben w und u. Das ankniipfende dum starem et mirarem zeigt deutlich, dass die Scharade weitergeht. . . .' This solution was suggested to Erlemann by Trautmann's interpretation of the runic passage in the Juliana, 703-711 (A'ynetoulf, pp. 47 f.): cyn, ewu (sheep), If (llcfirt, body) ; but he does not accept Trautmann's rendering of If, and be- lieves that in the true equivalent of / and f will be found the ' duo lupi ' of the Latin enigma. To Erlemann's article (p. 63) is added Dr. Joseph Gotzen's solu- tion of the latter part of the riddle. ' Duo lupi = wu, nicht wie oben vermutet, = If; tertium = 1; quattuor pedes = cyne; septem oculi = cynewul, die sieben Buch- staben. Die Losung des zweiten Teiles lautet also : zwei dastehende (Buch- staben) von wulf (w u), den dritten (/) bedrangend, hatten vier Fiisse (c y n e; d. h. cyne ist " Fuss " — nach bekannter Ratselterminologie — zu wul) ; mit sieben Augen sahen sie (namlich alle in v. 4-5 erwahnten Buchstaben). Die abnorme Siebenzahl ist gewahlt, um eine Spitzfindigkeit in das Ratsel hineinzubringen ; 232 RIDDLES OF THE EXETER BOOK der achte Buchstabe / war ja schon durch ivulfvn. v. I festgelegt. Das quattuor pedes = cyne berucksichtigt auch gut den ersten Bestandteil des Namens, der ja in v. 1-3 leer ausgegangen war.' Fritz Erlemann (fferrigs Archiv CXV, 391) thus modifies the views of his namesake : ' Mit Edmund Erlemann und Gotzen fasse ich lupi als Genitiv und duo als Neutrum auf, und zwar letzteres mit hinweisender Bedeutung ; unter duo lupi sind also die zwei Buchstaben des Wortes ewu (vom dem zuletzt die Rede war) verstanden, die gleichzeitig auch zu wulf gehoren, = wu. Der noch iibrig- bleibende dritte Buchstabe ist e. Es bleiben also wu stehen (stantes), verdrangen aber das e (tribulantes). So erhalten wir das aus sieben Buchstaben bestehende Wort Cynwulf (cum septem oculis videbanf). Unter quattuor pedes sind die vier letzten Buchstaben dieses Wortes also wu/fzu verstehen.' The mantle of Profes- sor Victor is over the Erlemann solution (ib. p. 392) ; and Professor Brandl has recently accorded it full approval (Grundriss2, II, 972). Far-fetched and unconvincing though all this seems, it must be frankly ad- mitted that such over-subtle playing with names was a common amusement of the mediaeval mind. A striking parallel to the Erlemann interpretation appears in the first riddle of the Leys d'Amor (I, 312), which is thus explained by Tobler, Jhrb. fiir Rom. und Engl. Lit. VIII (1867), 354: 'Trefflich erscheint und schones Wuchses die (Raimonda), so mit dem Kopfe (d. h. der Anfangssylbe, rat, sie scheert) die Haare abschneidet und mit ihrem Bauche (d. h. der Mittelsylbe, man, Welt) tragt was nur Mann und Weib sieht, und mit ihren Fiissen (der Schluss- sylbe, da, sie gibt) oftmals gibt oder schlagt zu Krieg, Frieden oder Ziichtigung oder um zu dienen. Doch wenn sie den Kopf verliert, werdet ihr sofort sie sauber und rein finden (monda, reine). ... In einen Mann (Raimon) werdet ihr sie verwandelt sehn.' In his Enigmas Boniface plays upon 'Liofa' ('Caritas'), and in his Epistles he twists into complex runic acrostics the names of two women friends, ' Susanna ' and ' Brannlinde ' (Ewald, Neue Archiv VII, 196 ; Hahn, Boni- faz und Lul, 1883, p. 242 N. ; Jaffe, Bibliotheca, 1866, III, 12, 244). As is well known, both Christine de Pisan and her contemporary Langland perpetrate clumsy charades upon their own names. So, while the Erlemann solution does not compel acceptance, it surely invites close attention. As the Latin riddle shows, particularly in its last two lines, such obvious indi- cations of medial rime, Holthausen has wisely emended the text (Engl. Stud. XXXVII, 210-211) by accepting Thorpe's inversion of -videtur and mi hi in the first line, by adding rupi to the incomplete first half of the second line, and by changing magnam at the close of the third line to the better parem. RIDDLE 91 As Dietrich shows (XI, 453, 486), this is a riddle of the ' Key,' and resembles, in at least one of its traits (see Prehn, pp. 255-258), the ' Clavis ' enigma of Symphosius, No. 4 : Virtutes magnas de viribus affero parvis. Pando domos clausas, iterum sed claudo patentes. Servo domum domino, sed rursus servor ab ipso. NOTES As Prehn has remarked (1. c.), the riddler here has made no attempt to mislead solvers, but has developed his subject so clearly and thoroughly that at the end all doubt has vanished ; and one feels perfectly safe in rejecting Trautmann's in- appropriate answer ' Sickle ' (Bb. V, 50). Certain words and phrases have been misinterpreted by scholars (infra). I translate and explain as follows : ' My head is beaten with a hammer, wounded with cunning darts, polished with a file. Often I bite that which against me sticks (the lock), when I shall push, girded with rings, hard against hard, and, bored through from behind, shove forward that (i.e. the catch of the lock) which protects my lord's heart's joy (treasure, wealth) in mid- nights. Sometimes, with my beak, I backwards draw (unlock) the guardian of the treasure (again, the lock) when my lord wishes to receive (or take) the herit- age of those whom he caused to be slain by murderous power, through his will.' Rid. 91 has little in common with the obscene query of the Key, Rid. 45. Wright, Celt, Roman and Saxon, pp. 488-490, notes that among the objects found suspended at the girdle of an Anglo-Saxon lady were scissors, small knives, tweezers, the framework of a chatelaine, — and latch-keys, if the implements found by Rolfe in the cemetery at Osengal (Collectanea II, 234) were used for that purpose. Among the Anglo-Saxon grave-finds in the British Museum is an iron key, four inches long with two bits, found below Farndon Church, Newark, Notts. Weinhold remarks (Altnordisc/ies Leben, p. 235), ' Samtliche Kasten und Kastchen waren verschliessbar ; die Schliissel hatten die Gestalt der Dietriche; aus jiingerer Zeit finden sich wirkliche Schliissel mit Bart und kunstreichem Griffe.' And in his Deutsche Frauen, II, 30, he notes, ' Als Venvalterin des Hauswesens, wofiir die Schliissel am Glirtel die Ausserzeichen waren, hatte die Frau eine grossere Freiheit in Geldsachen.' All this corresponds to the informa- tion furnished by a law of Canute (II, 76, § I, Schmid, p. 312) : 'and buton hit under )>aes wifes cieglocan gebroht wlEre, sy heo cliene, ac Hera czegean heo sceal weardian, )>aet is hire hordern and hyre cyste and hire tege (scrinium).' B.-T. s. v. cSg-loca points to a similar provision in the old Scottish law (Quon Attachi, xii, c. 7), and in the Statutes of William xix, c. 3. ' Store-room and chest and cup- board ' were thus under lock and key. Heyne's discussion of the treasure-chamber of the Anglo-Saxons is to the point (Halle Heorot, p. 30) : ' Insofern in den alten Zeiten das Schatzespenden die Gehalter der Mann und Dienerschaft vertritt und daher die Macht eines Herrn wesentlich von seinem Reichtum an Gold, Schmuck, kostbaren Gewandern und andern Gegenstanden abhangt, ist der Raum, wo diese Schatze aufbewahrt werden, das Schatzhaus ("gazophylacium," rndfrm-hiis; " thesaurium,"£r0/</-A<7r</) einer der wichtigsten der Burg. Daher ist es wohl verwahrt und der Schliissel (Rid. 91) kann sich riihmen dass er das Werkzeug sei durch das seines Herren Herzens- freude in Mitternachten geschiitzt wird u. s. w.' Wright, History of Domestic Manners, p. 79, copies from MS. Harl. 603., the manuscript of the Psalms, the illustration of ' a receiver pouring the money out of his bag into the cyst or chest, in which it is to be locked up and kept in his treasury.' ' It is hardly necessary,' he adds, ' to say that there were no banking- houses among the Anglo-Saxons. The chest or coffer, in which people kept their money and other valuables, appears to have formed part of the furniture of the 234 RIDDLES OF THE EXETER BOOK chamber as being the most private apartment ; and it may be remarked that a rich man's wealth usually consisted much more in jewels and valuable plate than in money.' 0 91 i homere gej>ruen (MS. gefrureri). Cf. Beow. 1286, hamore ge)>ruen (MS. ge)>uren). Heyne (Beowulf, 'Glossar' s.v.) derives gej>uren from gefrworen (weran, 'to beat'); Sievers, 6>.8, 385, regards gej>riien, 'forged,' as an isolated past participle (see PBB. IX, 282, 294 ; X, 458). The meter is strongly in favor of Sievers's reading. 91 2 searopfla wund. Shipley, The Genitive Case etc., does not include wund among adjectives that take the genitive, as elsewhere in the poetry it is followed by the instrumental ; cf. 6 i, Iserne wund. 91 3 B.-T., Supplement, p. 72, renders begine 'take with wide open mouth,' and Swaen, Engl. Stud. XL (1909), 323, 'open the gape and take into it, swallow.' Both authorities cite a similar use of beginen in the Dialogues of Gregory (Hecht, Bibl. der ags. Prosa), 324, 24-26. Swaen reads in our line ongeansttcafr as a compound. 91 4 hringum gyrded. Cf. 5 2, hringum haefted. 91 5 This line recalls the other Key riddle, 45. In 45 3a, the Key is stif> and heard, and in 45 -^ for an . . . J>yrel. 91 6 for<5 ascufan. Grein's rendering (Die/it.) ' hinwegschieben ' completely inverts the meaning of the passage (91 2-7). The riddler is describing the locking of the treasury-door, later (91 8-u) contrasting with this the unlocking (see Sym- phosius 4 2). Dietrich translates rightly ' hervorschieben,' and Sievers, Anglia XIII, 4, 'vorschieben.' — freaii mines. With the inversion of mines and frean cf. 71 6, 73 8. 91 7 modf>. Dietrich and Grein both understood this rune as wen, the former rendering the clause (XI, 453) 'was die sorge meines herren in mitternachten beruhigt,' the latter (Die/it.) 'was meines waltenden Herrn Gemittshoffnung schiitzt in Mitternachten.' Afterwards (XI, 486) Dietrich suggests mddwylm rather than mod-wen. Sievers has shown conclusively (Anglia XIII, 3-4) that in Anglo-Saxon poetry (not only in Rid. 91 7, but in El. 1090, 1264; Chr. 805; Ap. 100; Run. 8) W always demands the interpretation wyn, a rendering of the rune sustained by the Anglo-Saxon alphabet in the Salzburg MS. (Wimmer, Ritnenschrift, p. 85). Sievers further shows that in the present passage modwyn is but a periphrase of ' treasure ' ; and points to Chr. 807 f., lifwynna dael (feoh) ; Beow. 2270, hordwynne ; And. 1113, naes him to maftme wynn ; etc. 91 8 All editors, including Sievers (Anglia XIII, 4), read hwilum ic under bcec bregde nebbe ; but Holthausen, Engl. Stud. XXXVII, 211, assigns bregde to the first half-line, and prefixes briinre or beorhtre or bldcre to .nebbe. The emendation is absolutely unnecessary, hwilum ic under bcec is a verse of the B-type (cf. 41 86, nis under me), the second stressed syllable, bcec, carrying the alliteration. For B-type with alliteration in second foot, see Sievers (PBB. X, 289). 91 8-q Grein, Dicht., translates ' Ich schwinge bisweilen den Schnabel riickwarts, ein Huter des Hortes.' And Heyne follows him (Halle Heorot, p. 30) : ' Ein Hiiter des Hortes, wenn er seinen Bart riickwarts dreht.' But bregde is transitive with hyrde as its object, and nebbe is the instrumental. See my translation (supra). NOTES 235 91 9 hyrcle baes hordes. Cf. Beow. 887, hordes hyrde. This heroic phrase is here very aptly applied to the lock. 91 10 lafe J>icgan. Cf. Fates, 61, welan Hcgan ; ib. 81, feoh Jncgan ; El. 1259, maSmas }>ege. 91 ii waelcraefte. Grein, reading wcelcraft, misses the whole sense of the passage (Die/it.) : ' die er vom Leben hiess treiben nach seinem Willen todliche Kraft.' See my translation. RIDDLE 92 This fragment is not printed by Thorpe and Grein, so it is not solved by Dietrich. Trautmann (Anglia, Bb. V, 50) suggests with confidence the answer ' Beech.' My reasons for accepting this solution will appear in my notes to the various enigmatic phrases of the problem. While the ' Hainbuche' (Carpinus betulus) does not appear among the Anglo- Saxons (Hoops, Wb. u. Kp., p. 257), still the beech or fagus is well known (contra Holthausen, Engl. Stud. XXXVII, 211) : ' Und da die Buche in der angelsachsi- schen Periode wiederholt in Urkunden auftritt und, wenigstens in Sudengland, durchaus den Eindruck eines altheimischen Baumes macht, ist sie sicher auch zur Romerzeit vorhanden gewesen und nur Caesars Beobachtung entgangen. . . . Doch hat die Buche in England nie die Verbreitung und Bedeutung als Wald- baum erlangt wie in Deutschland und Danemark.' (Hoops, ib. p. 259.) 92 i brunra refers to the swine that subsisted on the beech-mast. In Rid. 41 107, the bearg dwelling ' in the beech-wood " is called won, a close synonym to briin (Spr. I, 145; Mead, 'Color in O.E. Poetry,' P.M.L.A. XIV, 187, 194). Holthausen's change to brunna (Engl. Stud. XXXVII, 211) — 'the boast of wells or springs ' — is therefore totally unwarranted. 92 2 ft-6olic feorhbora. This finds ample illustration in the gloss to De Crea- tura 49 (MS. Royal 12, C. XXIII, f. 103 v.) : ' Fagus et esculus arbores glandifere ideo vocate creduntur qua earum fructibus olim homines vixerunt cibumque sumpserunt et escam habuerunt.' I have already discussed (notes to 41 105, 106) the use of beech-woods as swine pastures. The oak is another life-giver and feeder of flesh (see note to Rid. 56 9). 923 wynnstabol, which Holthausen (Engl. Stud. XXXVII, 211) would change needlessly to wynn on staj>ole, may refer to the joyous station of the beech-tree; compare Run. 82, stlj> on stajmle (ash} ; Run. 37, wyn on e>le (yeiv); Rid. 54 2, treow waes on wynne. But the word almost certainly indicates the book, which is called J>ces strangan staj>ol in the Bookmoth riddle (48 sa). See also Sal. 239, gestaftelia'S staftolfaestne gej>oht (books'). — wlfes sond. In like manner the staff that bears the husband's message, H. M. i, 12, tells us that 'it is sprung from the tree-race.' We are reminded of the phrase of Tacitus, Germania, chap. 10, 'notis virgae frugiferae arboris impressis,' and of the lines of Venantius Fortunatus in the sixth century (Carmina vii, 18, 19, cited by Sievers, Pauls Grundriss I, 24) : Barbara fraxineis pingatur runa tabellis, Quodque papyrus agit, virgula plana valet. 236 RIDDLES OF THE EXETER BOOK Though Sievers (I.e.), like many earlier scholars (B.-T., p. 113), calls into ques- tion the traditional etymology, every Anglo-Saxon found the origin of ' book ' (bdc) in the ' beech-tree ' (boc-treow), for, as our riddle shows us, beech-bark was used by him for writing (see N. E. D. s. v. ' Book ; ' Kluge, Etym. Wtb. s. v. ' Buch '). 92 4 gold on geardum. Holthausen, Anglia, Bb. IX, 358, would change gold to god, but the emendation is unwarranted, as gold may well refer to the adorn- ments of the Book ; see Rid. 27 i3a, gierede mec mid golde (book). Cf. 21 8, gold ofer geardas (sword). 92 5 hyhtlic hildewsepen. That the beech, as well as the ash, is used for weapons, is shown by the bequest of a beechen shield in the Wills (Thorpe, Diplomatarium Anglicum, 561, 5, A.D. 938) : ' Ic ge-ann Sifer}>e mines bocscyldes.' RIDDLE 93 As in the companion-piece, Rid. 88, the subject is ' the Inkhorn, made from a Stag-horn ' Dietrich (XI, 486-487). Though it does not appear in the diction- aries of B.-T. and Sweet, blccc-horn glosses ' atramentarium ' in Oxford Glosses 4, 245» 33 (Herrigs Archiv CXIX, 185), and High and Low German cognates are noted by Dietrich, I.e. The riddle, like Rid. 88, vividly pictures the Horn's change of state from its glad free life on the head of the stag to its wretched lot as a swallower of black fluid after the shaping knives have done their cruel work. 93 2 willum si ii 11 in. So 91 nb. 93 5 f. The hunter, after describing the entangling of game in nets, tells us in vElfric's Colloquy, 92, 14, mid siviftum hundum ic bet&ce wildeor; and MS. Harl. 603, f. 24, contains a striking picture of a stag pursued by two dogs. Sharon Turner, VII, chap, vii, translates from the Life of Dunstan (see Auctor B, Stubbs, Memorials, p. 24) an account of a hunt of King Edmund : ' When they reached the woods, they took various directions among the woody avenues ; and lo, from the varied noise of the horns and the barking of the dogs, many stags began to fly about. From these, the King with his pack of hounds selected one for his own hunting and pursued it long through devious ways with great agility on his horse, and with the dogs following. . . . The stag came in its flight to a precipice and dashed itself down the immense depth, with headlong ruin, all the dogs following and perishing with it.' 93 6 dsegrime frod. Cf. 54 4, frod dagum ; 73 3, gearum frodne. 937-12 Brooke's lively rendering (E.E.Lit., p. 142) may be changed to the proper third person : ' At whiles, my lord (the stag) climbed the steep hillsides mounting to his dwelling. Then again he went into the deep dales to seek his food — his strengthening [better, 'his safety'], strong in step. He dug through the stony pastures, when they were hard with frost, then (as he shook himself and tossed his head, the rime) the gray frost flew from his hair.' Brooke adds : ' Scott himself could hardly have said it better : But ere his fleet career he took, The dewdrops from his flanks he shook.' The Lithuanian riddle (Schleicher, p. 201) is an interesting parallel: 'Was tragt den Thau auf seinen Hornern ? Der Hirsch.' NOTES 237 93 7 stealc hli]?o. Cf. Rid. 3 7, on steak hleoj>a ; 4 26, stealc stanhleobu. 93 9 In deop dalu. Cf. Chr. 1531, on >xt deope dael; Gen. 305, on )>§ deopan dalo ; Gen. 421, on >a.s deopan dalo. — dugu]7e. Grein, Die/it., renders ' Starkung,' and Brooke (supra ) 'food — strengthening.' But the context points to the mean- ing 'salus ' or 'safety ' (Spr. I, 211-212). The thought is parallel to the well-known description of the chase of a stag, Bee hjeftstapa hundum geswenced, heorot hornum trum holtwudu sece, feorran geflymed, etc. 93 10 strong on staepe. Cf. 28 13 strong on sprjece. The half-line is of the shortened A-type (_£. X | ^ X ), not uncommon in the Kiddles (see Herzfeld, p. 49). 93 11-12 hara . . . forst. Only once elsewhere in the poetry is hdr similarly ap- plied: And. 1257-1258, hrim ond forst | hare hildestapan (cf. Krapp's excellent note). 9312 on ffisum. MS. and Edd. read here of, which seems to me inapt and point- less ; cf. Grein (Dickt.), ' Ich ritt von dem Beeilten (?),' and Thorpe, B.-T., p. 349, ' I rode from the ready [men].' On the other hand, Ic on fusum rdd, ' I rode on the quick one,' exactly accords with the preceding description of the stag in flight. 93 13-14 The appearance of this motive in Rid. 88 18-20 has been already noted. 93 15-18 See the fate of the Horn, 88 32-33. The knife inflicts equal pain upon the Book, 27 5-6, and the Reed, 61 12-13. 93 15-16 isern . . . brim. The adjective is often applied to weapons ; cf. Rid. 18 8, brunum beadowjepnum. Britn is the epithet of ecg, Beow. 2578-2579; and brunecg of seax, Beow. 1547, of bill, Maid. 163. 93 i6b-i7 Cosijn, PBB. XXI, 16, compares with this passage And. 1240-1241, blod yj>um weoll \ hdtan heolfre, which he amends to hdt of hrej>re. But Krapp in his note (p. 139) has shown that the passages are not parallel and that the emendation is unwarranted. 93 19-20 The Horn's inability to wreak vengeance upon its enemies recalls the similar helplessness of the Sword, 21 17-18, and of the Ore, 83 8b. — wrecan . . . on wigan feore. Cf. 21 18, wriece on bonan feore. 93 21-22 ealle . . . Jjaette bord biton. The phrase puzzles Grein, who renders, Dicht., ' die Elendgeschicke welche Brette bissen (?) ' The Shield (bord) says in Hid. 6 8-9, mec . . . hondweorc smiba | bltaS in burgum. So in our passage, ' all who bit the shield' is simply a periphrase for 'the handiwork of smiths' or all cutting or wounding weapons — see isern, style (11. 15, 18). Similar enigmatic cir- cumlocutions appear, 81 7, 93 27. 93 22b-23 Compare the drink of the pen in the riddle of the Book, 27 9b-ioa, beamtelge swealg \ streames dales, and mark the mediaeval receipt for ink-making cited in my note to that passage. The riddler indulges himself in a sly word-play upon the two meanings of blace(bla:ce), the instr. form, 'black 'or 'ink ' — thus laugh- ing in the face of the solver : ' Now I swallow black ' (or ' ink '), etc. Compare the double-meaning of bind, 38 7, and of hafte, 73 22. Grein (Dicht.') completely misses the point in his rendering, ' Blinkend schlinge ich Waldholz nun und Wasser.' Eorp\ji\s ndthwat (93 25) is another reference to the ink, which is poured into the belly of the Horn. 238 RIDDLES OF THE EXETER BOOK 93 16-27 Dietrich (XI, 487) would read hordwarafr, and finds here a reference to the other Horn of Rid. 88. He believes that the -wulfes gehlefran is the dog which it tossed (wide beer) when the stag was at bay. But this explanation is far- fetched and will not serve. We have to do with an Inkhorn riddle. The plunder- ing enemy (hifrende feond) who guards my treasure (mm hord warafr ; cf. Rid. 32 21, 83 4) is the pen or quill, which emerges from the belly of the Inkhom (1. 28). Line 27, se f>e ar wide beer wulfes gehlefran, finds its explanation in the gloss to Aldhelm's ' Alphabet ' enigma, iv, I 5, in MS. Royal 12, C. XXIII (Wright, Satirical Poets etc. II, 549) : ' Ignoramus utrum cumpenna corvina vel anserina sive calamo perscriptae simus.' The pen of our riddle is the penna corvina, the common crow- quill ; and the raven, which ' it once bore widely,' is properly called ' the companion of the wolf,' as these creatures of prey are always associated in Anglo-Saxon poetical thought (cf. Beow. 3025-3028; Exod. 162-168; Jud. 205-207; El. nof.; Bruit. 61-65 ; Brooke, E. E.Lit., pp. 129-132). In the Old Norse, Fagrskinna § 5 (Munch and Unger, 1847, P- 4)> t^6 raven is called arnar eifrbrofrir, 'oath-brother to the eagle.' With this periphrasis for the pen compare the others in the Riddles: 27 7, fugles wyn ; 52 4, fultum fromra (MS. fuglum frumra). 93 28 The editors have overlooked the oft me of MS. and B. M. Bewaden does not mean ' ausgehohlt ' (Dietrich XI, 487 ; Spr. I, 97), nor ' deprived ' (Sweet, Dic- tionary, s.v.), but 'emerged.' ' Often emerging from my belly he (the quill) fares, etc.,' aptly accords with 93 22-23, where the Inkhorn refers to the ink contained in its belly. With oft me of wombe cf. 18 6, hu me of hrife ; 77 6, me of stdan. 93 29 So of the Pen in 27 iob, stop eft on mec (parchment). 93 30 daegcondel. See Krapp's Andreas, p. 101 (note to line 372, wedercandef). 9332 eagum wllte<5. So Ps. 656; cf. Whale, 12, eagum wliten; Gen. 106, eagum wlat; 1794, eagum wlltan. RIDDLE 94 The few surviving phrases of this badly damaged fragment exhibit a striking likeness to the comparatives of the ' Creation ' riddles, 41 and 67 : 94 2, hyrre bonne heofon (cf. 676, heofonas oferstige) ; 943, [hraejdre bonne sunne (cf. 673, swiftre bonne sunne) ; 94 7, leohtre bonne w (cf. 41 76). Possibly this was another handling of that theme of universal interest. RIDDLE 95 Rid. 95 has long been the theme of minute yet fruitless discussion — I quote largely from my article in M.L.N. XXI, 104. Dietrich's solution, 'Wandering Singer' (XI, 487), which has been accepted by Prehn, p. 262, and Brooke, E. E. Lit., p. 8, defended by Nuck (Anglia X, 393-394) and Hicketier (ib. 584-592), is rightly rejected by Trautmann (BB. XIX, 208) on many grounds. Yet his own answer, 'Riddle,' thrice championed by him (Anglia VI, Anz. 168 ; VII, Anz. 2iof. ; BB. XIX, 209) and attacked at length in the articles of Nuck and Hicketier, seems to me even more unfortunate than that of Dietrich. His interpretation everywhere refutes itself by its academic viewpoint and its consequent failure to NOTES 239 grasp the naive psychology of riddling (contrast with this rendering the riddles on the ' Riddle ' cited by Pitre, pp. xix-xxi), by perverted meanings and violent forcings of text (infra). I believe the answer to be 'Moon' (M.L.N., I.e.), and I find three motives common to Rid. 95 and 30, ' Moon and Sun.' These are the fame of the subject among earth-dwellers, its capture of booty in its proud hour, and its later .disappearance from the sight of men. I repeat here my translation and analysis of the problem : ' I am a noble being, known to earls, and rest often with the high and low. Famed among the folk (so of the Sun, 30 8, seo is eallum cu'5 eor'Sbuendum), I fare widely (Thorpe's reading of 3b, fere). And to me, (who was) formerly remote from friends (so the Moon refers to his periods of lonely darkness), remains booty (see notes), if I shall have glory in the burgs (compare 305, the Moon "would build himself a bower in the burg") and a bright god (Trautmann, " course "). Now wise (learned) men love very greatly my presence (notes). I shall to many reveal wisdom (notes) ; nor do they speak any word on earth (the Moon's teachings, unlike those of an earthly master, are conveyed and received in silence). Though the children of men, earth-dwellers, eagerly seek after my trail, I sometimes (that is, when my light wanes) conceal my track from each one of men ' (notes). 95 '-3a Compare not only the description of the Sun, 30 8, cited above, but that of the Moon, 40 1-3, 5-6. 95 i indryhten is aptly used of the Moon or of the Soul, 44 i, but certainly not of a Riddle, as Trautmann would have us think. 95 2 ricum ond hf'aiium. Cf. Rid. 33 13, rice ond heane ; Jud. 234, ne heane ne rice ; Gu. 968, ne ricra ne heanra. 95 3 folcum gefiraige. So Be in the sense of ' remains,' cf. Wond. 57, swa him wldeferh wuldor stonde'S. This interpretation of the line is certainly better than to change izr to far, to regard freondum as dat. sg. pres. part of freogan, freon, and to render stondefr as ' droht ' (Trautmann). 95 5a hijjendra hyht, ' the delight of plunderers,' which has given much trouble to Trautmann and Hicketier (1. c.), is but a circumlocution for huf>, ' booty ' (30 2'', 41'), as 27 7b fugles wyn is a periphrase of fef>er, ' quill,' or as 65 3a habbendes hyht is equivalent to ' the thing possessed.' ' Booty,' as in Rid. 30, refers to the light captured from the Sun, ' the bright air-vessel ' of the earlier riddle (30 3). ^Elfric tells us, ' se mona ond ealle steorran underfoiS leoht of >xre miclan sunnan ; ond heora nan naef 8 nSnne leoman buton of hzEre sunnan leoman ' (De Temporibus, Leechdoms, III 236). 240 RIDDLES OF THE EXETER BOOK 956 bleed is used in the present sense of 'glory of light" in Chr. 1238-1239, hy . . . leohte bllcaj>, | bljgde ond byrhte, ofer burga gesetu, and in Chr. 1291, Geseoft hi >a. betran blade scinan. — in burgum. Cf. 30 sa (Moon), on J>iere byrig (note). It is noteworthy that in Chr. 530, in burgum refers to Heaven, which may be the meaning here. But compare Met. 5 1-3 : Du meaht be J»£re sunnan sweotole geiSencean ond be Sghwelcum oSrum steorran, Jjara )>e aefter burgum beorhtost seine's. If MS. beorhtne god demands emendation, we may gratefully accept Traut- mann's gong, as no word could better suit the Moon's path in heaven. But it is not necessary to depart from the manuscript reading, as classical and Germanic belief assigns a god to the Moon (Grimm, Teutonic Mythology, pp. 705, 1501), and our poet may be recording old tradition. An Anglo-Saxon manuscript of the treatise of Aratus (MS. Tib. B. V) contains the figures of Sol in a quadriga and of Luna in a biga (Westwood, Facsimiles, p. 109, pi. 48). Various details are modified to suit the taste of the Anglo-Saxons. In a picture of the crucifixion (Publ. Libr. Camb. No. F. f. i. 23; Westwood, p. 120) 'Sol' and 'Luna' are seen weeping above the arms of the cross; and similar designs are found in MS. Titus D. 27 (Westwood, p. 1 24). In the Utrecht Psalter (Westwood, p. 20), the Sun of the first psalm is personified as a male half-length figure holding a flam- ing torch. But our riddler's thought here may be wholly Christian ; cf. Beow. 570, beorht beacen godes (sun). The riddle, like its mate (see notes to 30), is at times reminiscent of Ps. xix. 95 7a snottre. The word is used by Byrhtferth of scholars of this sort of lore (Anglia VIII, 330, 1. 33). Another Handbdc passage (ib. 308, 19-24) shows the love of English ' wise men ' for the Moon and his ' wisdom ' : ' Uton Srest gleaw- Hce swyiSe witan hwaet he [se mona] sy to soiSe ond hwanon he come ond hwaet he do on J>am gerime oftSe hwy he sy swa gehaten, o^JSe hwa hine gemette, oSSe hine >aes wurSscipes cuiSe J>aet he sceolde gestandan on bam rimcraefte. Ic wat gere }>aet he ys J>eodscipes wyrfte.' 95 9a wisdom cyban. The Moon is the source and center of Anglo-Saxon 'wisdom' or scientific knowledge (wisdom is used of the sciences, Boethius 7, 3). Its orbit and 'leap,' its cycles, its epacts, its relations to the weather, its effect upon the tides, are the leading themes of jElfric's De Temporibus (Leechdoms III, 248, 264-268, 282). The Moon is invaluable in prognostications (ib. 150-162, 177- 197), and sets, of course, the time of Easter {Handbdc, pp. 322-330). — no J7r word sproca'fl. With' this compare the account of the Moon, 40 i2b, ne wij> monnum spraec. 95 10 aenig ofer eorSan. So 41 21, Gu. 727. — a-lda beam. So 84 31. 95 '°b-i3 The same motive, somewhat similarly phrased, appears at the close of the ' Ore ' riddle, 83 12-14. The thought is exactly parallel to 30 13-14 (' Moon ') and to Bern MS. 61 1, 59 1-2, ' Luna ' (Anth. Lot. I, 369) : Quo movear gressu nullus cognoscere tentat, Cernere nee vultus per diem signa valebit. GLOSSARY The vowel 61'2. abelgan, 3, irritate, make angry : I sg. abelge 2I32. abf'odan, 2, utter, announce: inf. 6i16. abidan, 1, await, expect, abide : inf. 69. abrecan, 5, break down, take {fortress) : pret. opt. 3 sg. abrajce 56T. abregan, Wl, frighten, terrify : inf. 41". ac, conj., but: 47, 67'18, i617, 2I28, 236, 3710, 38, 408'18-16-21, 4i9»,ioi) 6Iej 839'12, 8812.24, 932i. ac, m. i. oak : ns. 569. — 2. name of rune A : np. acas 4310. acennan, Wl, bring forth, bear (child) : pp. acenned 4I44, 5I1, 84. ;iw« behind, in one's rear : asm. aeftanweardne 66. aefter, prep. w. dat. i. after: i316, 28", 29", 8o10. — 2. along: 34! ; aefter hondum, from hand to hand 3i5. — 3. according to : 4O15, 7310. aefter, adv., afterward, then : 2i21, 4O23, 6o5, 8819. aeftera, adj., second: nsm. \vk. aeftera 5412- aefterweard, adj., following, behind: nsm. i614. agan, P P, have, possess : opt. 3 pi. agen 426 ; inf. 445. See nagan. agen, adj., own : nsn. [agen] 8821 ; asn. io6, 45, 558. agetan, Wl, destroy : pret. 3 sg. agette _83T- aeghwa, pron., every one : nsm. 662. seghivaer, adv. i. everywhere: 4i18- a8,80,37,50,82 — 2 anywhere: 4i69. a?ghwae?Jer, pron., each : gsm. xg- hwaeiSres 475. seghwylc, pron., each (one\ every (one) : nsn. 4O25 ; gsm. Sghwylces 3710 ; asm. aeghwylcne 4O5. aglfan, 5, give, bestow : \ sg. agyfe 8o10. aglac, n., misery, torment : ds. aglace 47; as. 8 16. aglaeca, m., wretch : ns. 9321. aglachad, m., state of wretchedness : ds. aglachade 545. 241 242 RIDDLES OF THE EXETER BOOK agnian, Ws, possess : pret. 3 sg. agnade 9314- agof (boga), bow : 241. segfter (= seghvvaeSer), pron., each : as. JEgher 4OU. .5^ aw<Ser. ahebban, 6, r<w, /(/?«/ : 3 pi. ahebbatS 8s ; pret. 3 sg. ahof 1 i9. ahreddan, Wl, snatch away: pret. 3 sg. ahredde 3O9. a-Iit, £., property, possession-, ns. 71, 791 ; dp. iihtum S826. eelde, mpl., /»<?«: gp. aelda 8431- 9510; dp. jeldum 66, 34", 8i6. aleodan, 2, grow : pp. aloden 8430. am, m., weaver's rod, slay-rod (Dietr. pecten textorius) : ns. (MS. amas) 36s (Leid. aam). ama-stan, Wl, fatten : pp. nsn. amassted 4 1105. an, prep., /'« : 4310. an, num. i . one, certain one : nsm. i67, 4310; nsf. 536, 84!, [an] io3; nsn. 2212; gsf. anre 4413; dsm. anum n> 336; asm- anne 5O1, 5611, 866, 9326, aenne 8i3; asf. ane 571, 742, J61; asn. 863; isf. anre 8439; gp. anra I45, 3710. — 2. alone: nsm. 8410, wk. ana 379, 4I21-90 ; dsm. anum 263 ; dpm. anum 6i15. See anforlaetan. anad, n., solitude: ds. aniede 6i5. and, see ond. anfete, adj., one-footed: asf. 591. anfon, R, receive: pret. 3sg. anfeng433. anforhetan, R, forsake, abandon : pret. 3 sg. anforlet 729. anga, adj., sole, only: nsm. 8821. anhaga, m., solitary, recluse : ns. 61. £enig, adj. pron., any: nsm. 4i21, 6i3; nsf. 4 186 ; gsm. ieniges 6ol3 ; dsm. jini- gum 2411'16, iengum I45, 7216 ; asn. 4027, 95'° ; ? ijnig 8415. 6W nrenig. a n lie, adj., incomparable : nsm. 742. senlice, adv., incomparably: [Snllce] 4I28. anstellan, Wl, 367 (not in Leid.), 4I6-34, 4316. 446. 559» 6i2, 785, 88s. — 2. from (at the hands of) : 2 116. set, fs.,food: gs. setes 4I65. atoon, 2, draw out, take out : pret. 3 sg. ateah 622. aetgaedere,adv.,^ecgan i2'7.] aeffele, adj., noble: nsf. a£j>elu 8o5; sup. gsn. wk. 3eJ>el[est]an 6o9. ae'OVliug, m., prince, noble, atheling: gs. ae}>elinges 79, So1 ; np. aej>elingas 5O8; gp. as>elinga 475. U'O'rlu, f. i. origin, ancestry: ip. aej>e- lum 441. — 2. nature : ap. ae|>elu 568. a'dringan, 3, burst forth, rush : I sg. Springe 412. a'ffrintan, 3, swell : pp. ajrunten ^g2. aweaxan, 6, grow up : pret. i sg. awox ii3, aweox 731; pret. opt. i sg. aweox[e] io10. Sweccan, Wj, awake, arouse: pp. np. aweahte 14. a\vefan, 5, -weave: pret. 3 pi. awaefan 369 (Leid. auefun). aweorpan, 3, cast aside: ? aweorp ? 8414; pp. aworpen 4i49. awerian, awergan, Wl, gird, bind: opt. 3 sg. awerge 4i47. awreean, 5, drive away: inf. Qi11. awSer (= ah\vae?Jer), pron., either : ns. awj>er 8830. awyrged, pp. accursed: 2i17. B = rune & : over 18, 652. baec, n. i. back: ns. 8821; ds. baece 4s6, i63. — 2. under baec, backwards: 23". 9l8- ba>l, n., fire, flame: gs. bEles 832. ban, n., bone: ds. bane 6g3; as. 4O18. banleas, adj., boneless: asn. wk. ban- lease 463. baer, adj., bare, naked: nsf. ? 3222; asn. 66. barman, Wl, burn, consume: i sg. baerne 25, 72. See byrnan. baetJ, see seolhbaetf. ba'd'ian, Ws, bathe : pret. 3 pi. baj>edan 286. be, prep. w. dat. i. by, beside, along (local): 222, 2315, 61^ 7O6, 848, 8S28-33. — 2. by (temporal): 2817. See bl. beadu, f., fight, battle : gs. beadwe 883i. [beaducaf, adj., battle-prompt, warlike: nsm. wk. beaducaf a i11]. beaduwaepen, n., war-weapon : ap. beadowiepen i63; ip. beaduwSpnum i88. beadu\veorc,n.,^«/^-z£/<7r/^: gp. beado- weorca 62, 346. beag, m., ring, collar : gs. beages 6O11 ; as. 7212, (MS. ba5g) 58; ip. beagum 3222. beaghroden, adj., ring-adorned: nsf. I59. bealdlice, adv., boldly: 4i16, 6i16. bealo, see feorhbealo. bf-a m , m. i . tree : ns. Q21 ; gs. beames 567; as. 541; ap.beamas 29. — 2. beam, yoke: ds. beame 7212. — 3. timber : gs. beames 1 1 7. See wudubSam. beanitelg, m., tree-dye (ink) : is. beam- telge 279. bearg, m., barrow-pig: ns. 4i106. bearm, m., breast, bosom : ns. 67 ; ds. bearme 4412 ; as. 4s. beam, n., child: ns. 2i18, 8411; as. io6; np. 2718, 4I96, 424-7, 8481, 9510; gp. bearna 586; dp. beamum i69, 4O18. See frum-, woruldbearn. bearngestr?on, n., begetting of children : gp. bearngestreona 2i27. bearonaes, m., wood-ness, woody prom- ontory: ap. bearonaessas 585. bearu, m., grove, wood: ns. 31; ds. bearwe 541, So6, (MS. bearme) 227; dp. bearwum 282 ; ap. bearwas 29. 244 RIDDLES OF THE EXETER BOOK beatan, R, beat: 3 pi. beataS f, 8i8. bee nan, Wl, indicate, signify: 3 sg. becneb 40 ; 3 pi. becnaj> 2S10. bed, bedel, n., bed: ds. bedde 26; as. bed 58. See grundbedd. bedrifan, l, drive: pret. 3 sg. bedraf •(MS. bedrSf) 3O9. bei'a'Sinan, Wl, infold, contain: I sg. befae'Sme 9323. begen, adj., both: npm. 44", 8818-81; npn. buta (from begen twegen) 556; gp. bega 437, 537; dpm. bam 44"; biem 652. beginan, l, gape at, swallow: \ sg. be- glne 9 18. begrindan, 3, polish, grind off: pp. begrunden 2j6. behealdan, R. i. hold, possess: pret. I sg. beheold 73. — 2. behold, see: pret. opt. 3 sg. beheolde 6i5. See bihealdan. behlyo'an, Wl, despoil, strip: pp. be- hlybed is10. belacan, R, embrace: pret. 3 sg. beleolc 6i7. beleedswCora, adj., swollen-necked: nsm. Si1. belSosan, 2, lose : pret. i sg. beleas 27. belgan, see gebelgan. bellan, 3, grunt: ptc. nsm. bellende 4i10«. bemufan, l, conceal: i sg. bemi}>e 9518. bemurnan, 3, bewail: pret. i sg. be- mearn 9318. ben, f ., prayer : ds. (abs.) bene 6o18. benc, see meodubenc. bend, mfn. bond: dp. bendum 546 ; ap. bende 415, 2I80; ip. bendum 533-7. See ortfoncbend. benn, f., wound: np. benne 6o12. bennian, Wa, wound: inf. bennegean 572; pret. 3 sg. bennade 9316. See gebennian. bPobread, n., bee-bread: as. 4i59. beofian, Wa, tremble, shake : 3 pi. beo- fia3 9. beon, see wesan. beorcan, 3, bark: i s". beorce 252. &# borcian. beorg, m., mountain, hill: as. i618. beorghliS, n., mountain-slope: ap. beorg- hleoj>a 582. See burghli'S. beorht, adj., bright: nsm. 2i8; nsf. 4I28; asm. beorhtne I57, 956; npf. beorhte 1 21 ; comp. nsf. beorhtre 2O8. See hfiafodbeorht. beorhte, adv., brightly: 359. beorn, m., man, hero, warrior : ds. beorne (MS. beorn) I36; as. BE[orn] 652 ; gp. beoma 6i16; ap. beornas 3215. beot, n., boast : ns. 921. beran, 4, bear, carry: i sg. bere 215, 132, i63; 3 sg. bire« [i"], byreS 429, 8«, 1 56, 581, 927 ; 3 pi. beraft 1 615 ; pret. 3 sg. bser ii^, 9327; inf. 562, 5712, 6S2; pp. boren 642. See otflberan. berend, see feorh-, gaest-, segn- berend. berstan, 3. i . intr. burst, crash : 3 sg. bierstet! 462. — 2. trans, burst, break: 3 sg. berste'S 58. See toberstan. bescinan, l, shine upon : 3 sg. bescIne'S 7320- bescjT^an, Wl, deprive of: pret. 3 sg. bescyrede 4i101. besincan, 3, sink, submerge : pp. be- suncen 1 18. besnytRJan, Wl, deprive of: pret. 3 sg. besny^ede 271. bestelan, 4, deprive: pp. npm. bestolene 1 26. See blstelan. bestreflan, Wl, heap up: pp. bestrewed 8448. betan, Wl, make better, improve: I sg. bete (MS. betan) 710 ; ? bete 7i10, 925. betera, adj., better : nsf. betre 4i28. See god, -f'l r.-i. betynan, Wl, close, shut: pp. betyned 41". be5enean, Wl, intrust: opt. 3 pi. be- )>encan (MS. bej>uncan) 497. GLOSSARY beoennan, Wi, (stretch over), cover: pret. 3 sg. bej>enede 2712. bewadan, 6, come forth, emerge : pp. bewaden 9328. bewiefan, \Vl, clothe: pp. bewaifed 71. beweorpan 3, surround: 3 sg. be- weorpetS 8439. bewindan, 3, gird: pp. bewunden 3i2, 838. bewitan, PP, watch over: 3 sg. bewat 84fl. bewr&m, l, cover : pp. asf. bewrigene 4314 > PP- •' bewrigene 787. bewretfian, Wl, sustain, support: pp. bewrej>ed 8421. bewyrcan, Wl, make, work: pp. asm. beworhtne 363 (Leid. asf. biuorthae). bi, prep. w. dat., by: 451. See be. bicgan, Wl, buy: 3 pi. bicgaft S512. bid, n., delay, abiding: as. 48. bidan, l. i. await, expect: 3 sg. bldej> 3212; 3 pi. bidaS 4^; inf. i615. — 2. remain : I sg. bide 1 69 ; pp. biden 832. biddan, 5, pray : pret. 3 sg. basd 6o3. bidfaest, adj., fixed ': nsm. biidfaest 577. bidsteal, n., halt : as. bidsteal giefeft, stands at bay 4i19. bifeohtan, 3, deprive by fighting : pp. bifohten 482. bifon, R, encircle, surround: inf. 4i52; pp. bifongen 271. bihealdan, R, see, behold: 3 sg. bihealdeft i85, 4 198; inf. 4i89. See behealdan. bihon, R, behang, hang round: pp. bi- hongen 5710. bilecgan, Wl, cover, envelop: 3 pi. bilec- gaft 2726; [pret. 3 sg. bilegde i11]. bill, n., sword: is. bille 62. bilucan, 2, inclose : pret. 3 sg. bileac 621. bindan, 3, bind : \ sg. binde 1 38, 2816 ; 3 sg. bindeS 397 ; pret. 3 sg. bond 347 ; pp. bunden 227, 296, 7212. See gebindan. bindore, m., binder : ns. 28". hi n i man, 4, deprive : pret. 3 sg. binom 27 ; pp. binumen 28M. birSofan, 2, bereave, deprive : pp. biro- fen 481, npm. birofene I47. bisgo, see bysgo. bistelan, 4, deprive : pp. bistolen 2818. See bestelan. bitan, l, bite : I sg. bite 665 ; 3 sg. blteS 66 ; 3 pi. blta« 69, 666; opt. 3 sg. bite 665 ; pret. 3 pi. biton Q322 ; pret. opt. 3 sg. bite 93". biter, adj., bitter, fierce: nsf. 34; ipm. bitrum i88. bitwgonuin, prep., between : 3O2. bitfeccan, Wl, cover : pp. bi)>eaht 39. [biweorpan, 3, surround : pp. biworpen i5.] See beweorpan. blsec, blac, adj., black: dsn. blacum II7; isn. blace 9322; nprf. blace 451 ; npf. blace $8'2 ; npn. blacu $28. blac, adj., shining: ism. wk. blacan 4. l)la-caii, Wl, bleach : pp. bljeced 2g6. bla>< I, m. i . prosperity and breath (play on words): ns. 387. — 2. glory: as. 956. blandan, R, mix : pret. opt. 2 sg. blende 4 159. See geblandan. bleetan, Wl, bleat : i sg. bljete 252. blawan, R, blow : pret. 3 sg. bleow (MS. bleowe) 876. blCatf, adj., timid, gentle : nsm. 4i16. bl6d, f., blossom (leaf) : ap. blede I49. biedhwaet, adj., fair-fruited, rich in fruits : apm. bledhwate 29. blSofag, adj., varicolored: nsf. 2i8. blican, 1, shine : inf. 35°. bliss, f., bliss : ds. blisse 3215 ; as. blisse 96. 447- blitfe, see hygeblFSe. blod, n., blood: ns. 9316; as. 4O18. blonca, m., white horse : ap. bloncan 2318. blostnia, m., flower, blossom : ds. blost- man 4I28. blo>\^n, R, bloom: inf. 359; ptc. nsm. blowende 31. boc, f., book (letter} : ap. bee 437. bodiai, W2, announce : i sg. bodige 910. 246 RIDDLES OF THE EXETER BOOK [bog, m., arm : ip. bogum i11.] boga, m., bow : ns. agof = boga 24. See wirboga. bold, n., bitilding: as. i69. bona, m., murderer: gs. bonan 2i18, 737 ; ds. bonan 26. bonnan, R, summon, call: i sg. bonne i54- -bora, see feorh-, mund-, woo"bora. borcian, Wa, bark : pret. 3 sg. borcade (MS. boncade) S;6. See beorcan. bord, n. i. table: gs. bordes 88 23-24; as. 9329 ; dp. bordum 1 59. — 2. shield: as. 93122. See hleo-, naegledbord. bord\veall, m., shore ?, side of ship ?: ap. bordweallas 346. bosm, m., bosom : ds. bosme 47, I36, i515, 248, 387, 8o6; as. 462, is9. bot, f., reparation : ns. 387. brad, adj. broad: asm. wk. bradan 43; comp. nsf. braidre 4i50'82. braegnloea, m., skull : as. or ds. braegn- locan (MS. hraegnlocan) 732. bread, see bSobread. breag, m., brow: gp. breaga 4i100. breahtm, m., tumult, clangor : ns. £> ; is. breahtme 53 ; gp. breahtma 4°. -brec, see gebrec. brecan, 5, break : i sg. brece 7 3s6 ; 3 sg. brice'S 396, 66 ; inf. 53. bregdan, 3, trans, draw : i sg. bregde 9i8; opt. 3 sg. bregde 318. breost, n., breast : np. 1 615. See bylged- breost. brerd, m., border, rim, brim : as. 27. brim, n., sea : gs. brimes 313, 1 17. brimgaest, m., sea guest, sailor : gp. brimgiesta 425. bringan, Wl, bring: i sg. bringe 95; pret. 3 sg. brohte 2317, 6o8 ; pp. broht I37 ; pp. (strong) brungen 227, 282. broga, m., terror : np. brogan 451 ; ip. brogum Leid. 13. See sperebro^a. brotJor, m., brother : ns. brof>or 4411, 836, 88i8.28.», 93W;np. broker 8820; ap. bro)x>r 3222, 728. See gebroivor. brotforleas, adj., brotherless: nsm. broj>orleas 882. bru, f., eye-brow: gp. bruna 4i100. brucan, 2, enjoy : 3 sg. brace's 2910 ; 3 pi. brucaft 3312 ; opt. i pi. brucen 427 ; inf. 2I30, 2718, 4i100. brim, adj., brown : nsf. wk. brune 6i6 ; nsn. 9316 ; asm. brunne 279 ; ipn. brii- num i88; gpm. brunra 921. bryd, f ., bride, spouse : ns. 1 36, 463 ; ds. bryde 2I27. buend, see eor3-, fold-, lond-, neah- buend. bugan, 2. i. bow, bend: inf. 737. — 2. inflect, vary: ptc. isf. bugendre, modulated gp. See onbugan. bugan, Wl, inhabit : i sg. buge 82, i68 ; 3 pi. buga« 6815. bunden, see searo-, unbunden. bur, n., bower, tabernacle : as. 3O5. burg, f ., city : ds. byrig 3O5 ; as. 567 ; dp. burgum 4°-51, 69, 98, 35!, 832, 956. See ealdor-, msegburg. burgbli9, n., city height : dp. burghleo- )>um 282. See beorghliS. burgseel, n., city house : ap. burgsalo 585- burgsittende, mp., citizens : gp. burg- sittendra 263. burna, m., burne, f., stream, burn : as. buman 2318. See byrne. butan, prep. w. dat., without: 492. byden, f., butt, tub : ds. bydene 286. byht, n., dwelling, abode: as. 2312; ap. 83. bylgedbreost, adj ., puff-breasted : nsm. (MS. byledbreost) Si. byrnan, Wl, burn : ptc. nsf. byrnende 3 14. See ba-rnaii. byrne, f., mail-coat: ns. 2i3. byrne, f ., stream, burn : gs. byrnan 462. See burna. bysgo, f., occupation : as. bisgo 577. l>ysig, see l?g-, Sragbysig. bysigian, W2, occupy: pp. bysigo[d] 738. See gebysglan. GLOSSARY C C = rime K : over 9, 2O7. caf, see beaducaf. csege, f ., key : gs. ciegan 4312. cald, adj., cold: comp. nsm. caldra 4I54. See wintereeald. calu, adj., bald: nsm. 41". caru, f., sorrow : as. care 448. ceap, see searoceap. ceapian, see geceapian. ceaster, f., camp, city : as. ceastre 6o15. cene, adj., bold: comp. nsm. cenra 4i18. con nan, Wl, bring forth : pret. 3 sg. cende 362 (Leid. caend[ae]) ; pp. cenned 4O15. See acennan. c6ol, f., ship, keel: ds. ceole 428, 19, 342- ceorfan, 3, cut: pp. corf en 29. ceorl, m..,clncrl, countryman: gs.ceorles 266 ; as. 288. c€osan, see gecCosan. cirman, Wl, cry: i sg. cirme 98; 3 pi. cirmaft 58 ; pret. 3 sg. cirmde 49s. clam, m., bond, fetter, fastening: ap. clamme 4312, clomme 415. claengeorn, adj., yearning after purity: nsf. 842«. clengan, Wl, adhere, remain: 3 sg. clengefl 298. clif, n., cliff: ap. cleofu 428. clom, see clam. clympre, f. ?, clump, mass: ns. 4i75. clyppan, Wl, embrace: 3 pi. clyppaft 2726. See ymbclyppan. cneo, n., knee: ap. 455. cnosl, n., kindred, family : gs. cnosles 19, 448. See geogutfcnosl. cnyssan, Wl, smite, press : inf. 368 {Leid. cnyssa). cocor, m., quiver: dp. cocrum Leid. 14. cofa, m., chamber, bower : ds. cofan 64. comp, m., fig/it: gs. compes 2I85; ds. compe 72. compwiepen, n., war-weapon: ip. comp- •wiepnum 2i9. condel, see daegcondel. craeft, m., skill, cunning: gs. craeftes 8313; as. 3213; is. crasfte 227, 4312, 7322.28, S426; ip. cneftum 3210, 36 (so Leid.); ? craft 8413. See heah-, sundor-, wael-, wundorcraeft. craeftig, see hyge-, searocraeftig. creodan, 2, crowd, press : 3 sg. cryde'S 428. Crist, m., Christ: ns. 72. cuma, m., guest, stranger : ns. 4415. See wilcuma. cuiiian, 4, come: 3 sg. cyme's [i2'7], 441, 386, 4 155; opt. i sg. cyme 648; opt. 3 sg. cume i610, cyme 65; pret. I sg. cwom 1 16, 662 ; pret. 3 sg. cwom 23!, 307, 341, 551, 861, com 9316 ; inf. 8819. See forScuman. en n nan, PP. i. know : 3 sg. conn 6 111, 7O1 ; opt. 2 sg. cunne 732°. — 2. be able: 2 sg. const 3712; opt. 2 sg. cunne 3313; opt. 3 sg. cunne 6818; pret. 3 sg. cuj>e 6o10. cuff, adj., known: nsm. 951; nsf. 3O8; nsn. 7322, cu}> 3411 ; asn. wk. cuj>e 455. See unforciKf. c\vealm, see \vaelc\vealm. cwelan, 4, die : I sg. cwele 661. c\vellan, Wl, kill: I sg. cwelle 2i9; pret. ? sg. cwealde 78. cwen, f., queen : ns. 8o3; np. cwene 5O8. c\vCne, f., woman : ns. 741. cweSan, 5, say: 3 sg. cwiJ>e'S 6811 ; pret. i sg. cwae'S 661 ; pret. 3 sg. cwae'S 49, 6o5 ; pret. 3 pi. cwzedon 6o12; pret. opt. 3 pi. cwiEden 6o16. See ge-, on- cwlc, adj., alive: nsm. 73, cwico 661; asn. ctvicu 746, cwico n6, 14; gpf. cwicra 2g8 ; apm. cwice 72, 397. c^vlde, m., speech, discourse: as. 48. See galdor-, so5-, wordcwide. cyme, see seld-, upcyme. cymlic, adj., comely: nsf. 342. -cynd, see gecynd. cyneword, n., fitting word: ip. cyne- wordum 4415. 248 RIDDLES OF THE EXETER BOOK gs. See cyning, m., king: ns. 2i9, 4i8 cyninges 8o8; np. cyningas 508 heah-, fteod-, wuldorcyning. cynn, n., race, kind: gs. cynnes 349, 61; ds. cynne 450; as. cyn 508 ; gp. cynna 422, 56. 848; gp. cy[nna] 8455; ? cynn 6S11, 78", 8418. See from-, gum-, liece-, mon-, wsepnedcynn. cyrran, W i . i . turn : 3 sg. cyrreiS 3210; pp. cyrred 29. — 2. return: pret. 3 sg. cyrde 23". cyrten, adj., beautiful : nsf. cyrtenu 266. cyssan, Wl, &jj: 3 sg. cysseiS 64; 3 pi. cyssaiS 158, 3i6 a (b gecyssaS). See gecyssan. cystig, adj., bountiful: nsf. S426. eyffan, Wl, announce, make known, re- veal : opt. 3 sg. cy>e 4418 ; pret. 3 sg. cy«de 8830; inf. cyj>an 58, 3218, 959. See gecytfan. D D = rune M 752. deed, f ., deed : is. dasde 1 27. dseg, m., day : gs. daeges 283-17, 50 ; deaSslege, m., deadly blovu : ap. 614. deaSspere, n., deadly spear : ap. deaiS- speru 453. deaw, m., dew: ns. 3O12. degol, adj., secret: asm. degolne I621 ; apn. deagol 4i39. degolful, adj., secret : asm. degolfulne 8313. delfan, 3, dig, delve: 3 pi. delfa'S 4i97- deman, Wl, declaim : inf. 29". denu, f., valley : dp. denum 283. dCop, adj., deep : nsm. 236 ; asn. 710; gpn. deopra 57 ; apm. deo[pe] 936 ; apn. 939. deope, adv., deeply : 546. deor, adj., brave: nsf. 3216 ; dsm. deorum i36. deoran, W\, praise, extol: 3 pi. deoraj> I27. deorc, adj., dark: nsf. 421; npn. 445; ipf. deorcum I39. deore, adj., dear, precious: nsm. i810; asm. deorne 44! ; comp. nsm. deorra 84s6; sup. nsn. deorast I29; sup. gsn. as. 2 17, 594 ; dp. dagum lo1; ip. wk. deorestan 3410, 42. See dyre. dagum 6U, 54. . daegcondel, f., sun : ns. 9330. daegrim, n., number of days : is. da^g- rime 936. daegtid, f., day-time : ip. daegtidum (by day) i83, 727. dael, n., valley, dale : ap. dalu 939. dsel, m., part: ns. 291, 6i10, 65; as. 56, 599, 72" ; is. dzle 27!°;? dal 739. e dohtor, f., daughter : ns. 266, 3410, 465, So6; np. 472; gp. dohtra lo12. dol, adj., foolish, rash, light-headed: apm. dole gp. dolga nsm. 453, 2i32; nsn. I39 I23, 2817. dolg, n., wound: np. 613 ; 57; ap. 6ou. dolgian, see gedolglan. dolwite, n., punishment of the unjust, pains of hell: as. 27". doni, m. i. honor, praise: gs. domes 3216. — 2. decree, law : ds. dome 7310 ; ap. [domas] 852. — 3. power, domin- ion : as. 8313. don, anv., make, perform, do : 3 sg. de« 684 ; 3 pi. doS 427, do)> so10 ; pret. 3 sg. dyde io12, 2i25, 273, [d]yde 788, [dyde] 852 ; inf. 6on. See gedon. < 1 11. n | ;i n , see ondriedan. dream, see seledream. GLOSSARY clrefan, Wl, disturb, stir up: (wado, lagu drefan = swim): I sg. (wado) drefe 82 ; pret. 3 sg. (lagu) drefde 2316. dreogan, 2, suffer, .endure, perform : i sg. dreoge 8i6; 3 sg. dreoge'S 3310; 3 sg. [dreoge'S] 70 ; pret. 3 sg. dreag 52, 577 ; inf.^ 40", 591. dreoht, see dryht. drifan, l, drive : 3 sg. drife}> 4i78. See bedrifan. drinc, see mandrinc. drlnean, 3, drink: 3 pi. drinca'S I512, 2 112, 648; pret. 3 pi. druncon 561, 57", 68"; inf. 136, 72^. drohta.3, m., condition, manner of life: as. 710. dropa, see speddropa. druncmennen, n., drunken maidserv- ant: ns. I39. dryge, adj., dry: nsm. 41". dryht,f., multitude, (pi.) men: gp.dryhta 297, 42; dp. dryhtum i315, 5i2; dp. dreohtum (MS. dreontum) 445. dryhten, m. i . lord, master : ? dryhtne 71. — 2. Lord: ns. 4i12, driht[en] 852 ; gs. dryhtnes 6o8, dryht [nes] 6oai. , See In-, mondryhten. dryhtfolc, n., multitude : gp. dryht- folca 27". dryhtgestreon, n., noble treasure : gp. dryhtgestreona i83. dufan, 2, dire : pret. I sg. deaf 74 ; pret. 3 sg. deaf $25. dugan, PP, avail, hold out: 3 sg. deag 739 ; pret. 3 sg. dohte 627. duguS, f. i. benefit, advantage: dp. dugbum 5O10. — 2. safety : ap. dugu)>e 939- dumb, adj., dumb : nsm. 54, wk. dumba 5o10, 6o8 ; nsf. 3216 ; asm. wk. dumban 5O2; dpmf. dumbum 5i2. dun, f., /////, down : ns. 421 ; gs. dune (MS. dum) i621; dp. dunum 28" ; ap. duna 39. durran, PP, dare: 3 sg. dear lo16. duru, f., door: dp. durum I611, 2Q7. dust, n., dust : ns. 3O12. dwH'sran, Wl, extinguish: 3 sg. dwiesce'S S438. dwelan, see gedwelan. dwellan, Wl, mislead: i sg. dwelle 12. dyfan, Wl, dip : pret. 3 sg. dyfde 278. dygan, see gedygan. dyn, see gedyn. dynt, m., blow : dp. dyntum 2817. dyp, n., the deep, sea : ds. dype 421. dyre, adj., dear, precious: nsf. 8422; gsm. wk. dyran 83 M ; apn. 4i89 ; ? dyre 8418; comp. apn. dyrran 5O6. See deore. dyrne, see undyrne. dysig, adj., foolish : apm. dysge 12. E E = rune M : 2O6, 652-. EA = rune ~T : 656. eac,adv., also, likewise, moreover: [i12], 3712, 4i40, 6418, 778. eacen, adj., increased, endowed, mighty : nsm. io8 ; nsf. 34", 8420-26 ; npn. 618. ead, n., happiness, bliss: as. 2728. eadig, adj., happy, blessed, prosperous: dpm. eadgum 8427. eadignes, f., happiness : gs. eadlgnesse _3i«. [Eadwacer, m., Eadwacer (Odoacer ?) : as. or vs. I16.] eafora, m., offspring, progeny: ap. ea- foran i612; ip. eaforan 2i21. eage, n., eye : ns. 2611 ; as. 38, 863, 878 ; np. eagan 4I11; gp. eagena 40", eagna 6o9 ; dp. eagum i65 ; ap. eagan 377, 8 13; ip. eagum 8481, 93s2. eald, adj., old, ancient: nsm. 98; dsm. ealdum 4i63; asm. ealdne 283; comp. nsm. yldra 4i42, 72. ealdor, n., life: ns. io8; ds. ealdre 6814. ealdorburg, f., royal city : as. 6o14. ealdorgesceaft, f., condition of life : ns. 4028. 250 RIDDLES OF THE EXETER BOOK call, adj., all, the whole of: nsn. 94; gsn. ealles (adv., close) i6u; asm. ealne 41", 67° ; asf. ealle 4i68; asn. eal 4i88.40.4; npm. ealle 5610, 6;3 ; gp. ealra 14. 34", 40", 4i4-88, 476: dPm- eallum 3O8, 527; ap. ealle 84', 9321; ipn. eallum 4i101. call, adv., -wholly, entirely: eal 66, 83". eallfelo, adj., all-fell, -very baleful: asn. ealf elo 249 . eallgearo, adj., all-ready, eager: nsf. 24. Pain, m., uncle : ns. 47°. ear, m., sea, ocean : is. eare 422. earc, f ., chest : as. earce 622. eard, m., dwelling, place, region : ns. 88U ; ds. earde 34, 735, 838, 93" ; as. 616, 678, 8i6, 8819. eardfaest, adj., fixed, fast in its place: asm. eardfaestne 5ol. eardian, Wa, dwell, abide : pret. 3 sg. eardadeSS28; inf. 8827. eare, n., ear: np. earan lo5; ap. earan 8 Is, 868. earfo«J, n., trouble, affliction, tribulation : gp. earfoiSa 72U. earh(?), n., dart: as. EA[rh] 655. earhfarhi, i., flight of arrows: as. aerig- faerae Z«'</. 13. earm, m., arm : ap. earmas 336, 866. earm, adj., poor, miserable, wretched: dpm. earmum 8427 ; superl. nsf. ear- most 4e [i18], i619, 24", 4I68, 568; [eaj>e] 41". eawunga, adv., openly: 73s6. eaxl, f., shoulder : ? eaxle 73" ; ap. exle 33 ; ap. eaxle jos, S66. eaxlgestealla, m., shoulder-companion : ns. So1. See, adj., eternal, er>erlasting : nsm. 41 J ; ipf. wk. ecan 4I90. ecg, f., edge: ns. 442, (MS. ecge) 276; ds. ecge 442 ; np. ecge 34 ; gp. ecga 61S; ip. ecgum 63. See heard-, stHJ- ecg. edniwe, adj., renewed: nsf. edniwu 421. efenlang, adj., just as long: asn. (MS. efelang) 45. efne, adv., just, even, exactly : 418, 4O27, 661. efiietan, 5, eat as much as: inf. 4i68. eft, adv. i. again : 3", 438-6'3 79( 278.lo, 386, 637, 662, Sg6, 938. — 2. backwards: 241. — 3. on the other hand, still: 2 118. egesful, adj., fearful, terrible, awful: nsm. 344. egle, adj., hateful, deadly: npf. 72"; ipn. eglum i89. [eglond, n., island: ns. I5.] egsa, m., fear, terror: ns. 433-49; gs. egsan Leid. 13. eh, n., horse : ap. 23". ehtuwe, num. adj., eight: 374. ellen, n., strength, force, courage: ns. 627, 739 ; as. 8830. ellenrof, adj., powerful, strong, brave: npm. ellenrofe 2320. ellorfus, adj., eager for the journey: npm. ellorfuse 4413. ende, m., end: ns. 8410; ds. 8o8, 8S23-24. endleofan, num. adj., eleven : np. (MS. xi) 238. engel, m., angel: gp. engla 678. engu, f., narrow place, confinement: ds. enge 45-12. eodor, m., enclosure: ns. i82. eofor, m., boar: ds. eofore 4i18. eoredmaecg, m., horseman : np. eored- maecgas 238. eoredSrPat, m., band, troop : ns. eored- t>reat 449. eorl, m., chief, hero: gs. eorles 6i18, 8o5 ; gp. eorla 477 ; dp. eorlum 96, 3211, 568, 951 ; ap. eorlas 23". eorp, adj., dark, dusky : nsm. 5O11 ; gsn. eorp[e]s 93^ ; npf. wk. eorpan (MS. earpan) 442 ; ? eorp 7316. GLOSSARY 251 eortfbuend, m., dweller on earth : dp. eorftbuendum 30. eortfe, f., earth : ns. eor|>e 54s ; gs. eorfcm 4I4-26, 6816, 836, 8827 ; ds. eor- l>an 27, 468, 73, 288, 36" (Leid. eorSu), 4I40-60'82, 426, 5I1, 772; as. eorj>an 32, 178, 2816, 30i2, 411-21, 678, 8441, 88ai; as. eorftan 95!° ; ? eor}>an 8418. eor<5graef, n., w^//, pit : as. 599. esne, m. i. servant: ns. 445'8-i6. — 2. youth, man: ns. 45, 55, 645 ; as. (MS. efne) aS8 ; ap. esnas 28X6. esol, m., ass : gp. esla (MS. esna) 23i3. est, mf., grace, favor : ip. estum (gladly) 272. etan, 5, eat: 3 sg. itej> 59!°, ite~$ 77. effel, m. i. home, abode; ds. eSle i6i2; as. ej>el 67^, 938. — 2. /««</, domain : ns. e}>el 17s. etJelfaesten, n., land's fastness, fortress: as. ebelfaesten 7325. eSelstol, m., paternal seat, habitation : ap. e>elstol 47. eSfta, conj., or : etS^a 4417. exl, see eaxl. F' F = rune Y : 2O8, 655. faecne, adj., guileful, crafty, evil: dsm. fsecnum 54. feeder, m.., father : ns. io2, 388, 4I84, 47, 84". fag, j^ bleo-, haso-, sincfag, Tvonfah. fseger, adj., fair, pleasant: nsf. 845 ; nsn. 32"; comp. nsf. fSgerre 4I46. faegre, adv., fairly, pleasantly, fittingly : 13", 2i2, 29! 8, 54, 642, , fah, adj. i. proscribed: nsm. 2i16. — 2. hostile : nsm. 83. HfhcT, f., feud, enmity, vengeance : ip. falca(?), m., falcon: ns. FA[lca] 656. failsian, see gefaelsian. fain, n., foam : ns. 3. famig, adj., foamy: nsm. 4W. feemig, adj., foamy: nsm. 482. fjemne, f., maid, bride, woman : ns. 436, 741- fler, m., danger, peril ': as. 5412. -fara, see gefara. faran, 6, go, fare, depart : 3 sg. fare's 48, 18", 248, 637, 848, fjere« 22; 3 pi. faraS 446 ; pret. 3 sg. for 379 ; inf. 33'8, 651- faroS, see merefarocl faru, f., carrying, transfer : as. fere (< faere) 3310. See earh-, Tvolcen- faru. faest, z.&}.,firm, fixed, secured: nsm. i82, 6i8; nsn. [i5], 22i3; npn. 356 ; gpm. faestra 537 ; apf. faeste 357. See bid-, card-, hyge-, slge-, 3rym-, \vis- faest. faeste, adv., fast, firmly : 41, 138, I710, 24", 276, 53, 576, 62i, 7 14, 8S25. faesten, n., prison, confinement : as. 269. See eSelfaesten. faet, see lyft-, si3-, waegfaet. faet, adj., fat: comp. nsm. faettra ^i105. fseted, pp., rich, ornamented: asn. 52". faethengest, m., road-horse : ns. 23!. fie'Sm, m. i. embrdce, embracing arms: is. fas'Sme 646 ; dp. fae}>mum 313, 2 7s5 ; fae'Smum n6, 67. — 2. bosom, breast: ds. faeftme I3U. See lagufiv'A'in. faethnan, see befaeSman. f5a, adj., few. nsm. fea (zenig) 6i3; npm. 467- -f6a, see gefea. fealdan, ~&,fold: pret. 3 pi. feoldan 277. feallan, R, fall: 3 sg. fealle}> 2218, fealle'5 8il°, 93™ ; pret. 3 sg. feol 3Oa2 ; inf. 4«. fealo, adj., fallow, yellowish : nsn. i6i ; nsm. wk. fealwa 56!° ; npf. fealwe 73J8. feax, n., hair of head, locks: ds. feaxe (MS. feax) 9312. See Tvonfeax. feaxhar, adj., hoary-haired: nsf. 74!. IVdan, \V 'i, feed, nottrish, sustain : 3 sg. fedeiS 352 ; 3 pi. feda^ 5i8 ; pret. 3 sg. fedde io9, 725, 77!; pret. 3 pi. feddon 731, feddan 54. 252 RIDDLES OF THE EXETER BOOK fpgan, Wl,/: 3 sg. fegeft 269; pret. 3 sg. fegde 62. fela. i. indecl. n., many. 9", 228, 338, 352, 8310, [fela] 831. — 2. adv., much : 328, 593. fglan, Wl, feel; 3 sg. fele)> 269, fele« 849 ; 3 pi. felaS ;8. felawlonc, adj., very proud: nsf. I37. feld, m., field: ap. feldas 338. fell, n., skin, covering: gs. felles 77fi; np. 14s; ?[f]ell82. felo, j« eallfelo. fen, n., /<?«, swamp, morass: ns. 4i31; [ds. fenne i6]. -fenga, jw ondfenga. fenyce, i,, fen-frog: ns. 4I71. feoh, n. i. ra#/i?, herd: as. 35. — 2. money, fee: is. feo 55U. feohtan, 3, fight, contend: inf. 76, 17; ptc.npf. feohtende 446. .SV^bifeohtan. feohte, f ., fight, battle : as. feohtan 64. ffiol, i.,file: gs. feole 71; is. feole 9i2. feol an, 3, 4, pass : inf. 235. feon, see gefeon. feond, m., enemy, foe: ns. 223, 5i,4 9326; ds. feonde 51; gp. feonda 27. feondsceaSa, m., plundering enemy, robber : as. feondscea)>an 1 519. feorh, n., life, soul: ns. id2, I33 ; ds. feo re 2i18, 932°, (Sefre to feore = forever) 4i65 ; as. feorg 148, feorh ii6, i619, 4016; is. feore 4s2, 24", 27. feorhbealo, n., life-bale, deadly evil: as. 24. feorhberend, m., life-bearer, man : gp. feorhberendra 4O6. feorhbora, m., life-bearer: ns. 92. feorm, see swRSfeorm. feormlan, W2, cleanse, polish ; 3 sg. feormaS 7321. feorr, adv.,/ar: feor 246. feorran, adv., afar, far off, from far : 78, I37, 296, 552. feower, num. adj., four : d. 52?; a. 398, 521, 562, 725, (MS. mi) 23, feowere 37s- feran, Wl,^, travel: i sg. fere 25, 471, 13!, 221, (MS. fereS) 958; 3 sg. fereS 422, 592, 9328 ; 3 pl- fera« 4", 58 ; inf. 30". 337> 371. 406, 4 169. 691, 751: Ptc- nsm. ferende 89, nsf. ferende 845, apf. ferende (MS. farende) 4". See geferan. fere, see faru. fergan, Wl, bear, carry, conduct: 3 sg. fere~5 I57, 594>n; pret. 3 sg. ferede 2O6; pret. 3 pl. feredon 28; inf. i613, 531. See oSfergan. -ferh, see wideferh. fering, f., journeying, traveling : gs. feringe 7327. fertF, mn. i. mind, spirit, soul: is. ferj>e 2721; ip. fer^um 84^, ferSbum 5512, fer)>J>um6o3.— 2. life: as. (MS. for«) 746- ferSfritfende, adj., life-saving: apm. 393- feterian, see gefeterian. fetfe, n., walking, going, motion : ds. fej>e i62. feemundum i617. fetter, i., feather, (pl.) wings: np. fej>re 28. See halsrefeSer. fif, num. adj.,y?zv: n. fife 47°. fuidan, 1,find: 3 sg. finde« 356, 88s4; 3 pl. findaS 447 ; inf. 611 ; pp. funden 281. See onfindan. finger, m., finger : np. fingras 277, 4i62; ip. [fin]grum 6^. firas, mpl., men : gp. fira 684 ; dp. firum 3412- firen, f., evil deed, sin, crime : ap. firene 8488. firenlan, Wa, revile, chide : pres. 3 sg. firenaj> 21 . firgenstream, m., mountain-stream, ocean: ip. firgenstreamum n2. fisc, m.,fish : dp. fiscum 74. GLOSSARY fiflere, n., wing : ap. fi>ru 377. fla, f., dart, arrow : ap. flan 4". flan, m., arrow, dart: gp. flan[a] 5712; ap. [flanas] Leid. 14. flaesc, n. . flesh: as. 77s; Pfliesc 82. — 2. body: ap. 218. fleam, m., flight: is. fleame i613. fleogan, "2, fly: 3 sg. fleoge'S 2412; 3 pi. fleoga'S iS6 ; pret. i sg. fleah 748 ; pret. 3 sg. fleah 38, 65s, fleag 2316, fleag (MS. fleotgan) 52; inf. 4s6, 328, 4I66, 593- fleon, 2,yf<?<?: pret. i sg. fleah I629. flet, n., floor, hall: ds. flette 43s; as. 562, 5712- flint, m.., flint: ds. flinte 4i78. flintgraig, adj., flint-gray: asm. flint- gragne 419. flocan, Wi, clap: 3 sg. floceiS 2I84. Hod, m., flood, wave, sea : ns. 236 ; ds. node 89, 23", 41", 743, 778; as. 419; is. flode II2; np. flodas 67; ap. flodas 1 57, 781. See laguflod. flodweg, m., flood-way, watery way : ap. flodwegas 379. flot, n., sea : ? flote 786. flowan, see underflowan. flyman, Wl,/w/ to flight; 3 pi. flymaft 176; inf. I519. flys, n.,fleece, wool: ip. flysum 368 {Leid. fllusum). fodor, n., food, provender, fodder : gs. fodres 59". fodorwela, m., abundance of food: gs. foddorwelan 3310. folc, n., people, folk, race: gs. folces 656; ds. folce 3412 ; dp. f oleum 448, 953; ap. 86. See dryhtfolc. folcsael, n., folk-hall, public building : ap. folcsalo 2s. folcseipe, m., people : ds. 3310. folcstede, m., folk-stead, city : ds. 611. folcAviga, m., warrior : np. folcwigan I513. foldbuend, m., earth-dweller, man : gp. foldbuendra 213. folde, f. i. earth, world: gs. foldan 2Q1, 425; ds. foldan 3412, 4O10. — 2. ground, soil: gs. foldan 67, 922 ; ds. foldan S9; as. foldan 25, I31, 74s. folgian, Wa, follow: pret. 3 sg. folgade 382, 872. folm, f., hand, palm : ns. 4i52 ; ds. folm[e] 64; as. 4our : 3 sg. forswilgeft (MS. fer swilge'5) 5O11; pret. 3 sg. forswealg 488. for5, adv. i. forth, forwards: 226, 3011-13, 642'8, 856, 9I6. — 2. forthwith : 2I2. forffcuman, 5, come forth : pp. npm. forScymene I410. fortfgesceaft, f., creation : ns. 849. forfton, adv., therefore, consequently. for)x)n i612, 2I30, 2718, 6813. 254 RIDDLES OF THE EXETER BOOK , m., going forth, departure : gs. fofSsit>es 632. forSvveard, adj ., forward, prone: nsm. 7326; nsn. 2213. forffweg, m., forth-faring, journey: gs. forftweges 3i3. forweorflan, 3, perish, die : opt. pret. i sg. forwurde 66. fot, m., foot : ds. fote 32", fet 336 ; as. 3220, 4010, 9326, foot 8i3; np. fet 327 ; gp. fota 2815, 576 ; ap. fet 378, 689, 86; ip. fotum I31-7, 41", 82. fracofllice, adv., hostilely : [fracjadlicae Leid. 14. frsege, see gefrsege. fraetwan, Wl, adorn, deck: 3 pi. fraet- walS 3610 (Leid. f raetuath) ; pp. fraetwed I511, 298, 3220; pp. asm.fraetwedne 628. See gefraetw(i)an. flraetwe, fpl., ornaments, decorations : np. (wings} 86 ; dp. fraetwum 4I46 ; ap. I410; ip. fraetwum I57. frea, m., lord, master: ns. 41, 75, 931'5, (MS. freo) i86; gs. frean 466, 452, 738, •918; ds. frean 2I2-2, 4410, 561°, 62s, 632, 8o2. frecne, adj., dangerous, perilous: asf. 64. frecne, adv., severely, savagely: 2i16. frefran, Wl, console, comfort: I sg. fre- fre 77. fremde, adj., strange, foreign, remote: nsm. I73; dsm.fremdum(MS.fremdes) 954- rrcniniaii, Wl, do, perform : I sg. fremme 2i25; inf. 329, 73" ; ger. fremmanne 8829. (rciuiiic ti, comely, noble: nsm. 922; nsf. 8428, freolTcu 621 ; asn. is18; np. freollco 47. freond, m.., friend: ds. freonde 2i16; gp. freonda 2721 ; dp. freondum 95. freorig, adj., freezing, frozen : nsm. 36 (so Leid.}. freotfian, W2, care for, protect, cherish : 3 sg. freohaft gi7; pret. 3 sg. freoj>ode io5. See frKHan. fretan, 5, devour, consume : pret. 3 sg. fraet 481; inf. 775. fricgan, 5, ask: imp. 2 sg. frige I519, I710, 2726, 2815. See gefricgan. frignan, see gefrignan. fri?f, ft., peace, protection: as. 7326. fritJ, adj., stately, beautiful: nsf. frl)>e io9. friffende, see ferSfriSende. friShengest, m., horse of peace : ap. frrShengestas (MS. fridhengestas) 23. friffian, W2, protect: inf. friHan I77. See freoftian. fri^osped, f., peaceful happiness : gs. fri>ospe[de] 6o3. frod, adj. i. -wise, prudent, sage: apm. frode 6o3; comp. npm. frodran 2721. — 2. old, aged: nsm. 54, 93° ; nsn. 831 ; asm. frodne 733 ; comp. nsm. frodra 8435. frofor, f., comfort, consolation : gs. frof re 64 ; ds. frof re 4O19. from, prep. w. dat., from, away from : 2i23, 23i9, 44^. from, adj., strong, bold, swift: nsm. 632, 7327 ; gpm. fromra (MS. frumra) 52; sup. nsf. fromast 8428. See orlegfrom. fromcynn, n., ancestry : ns. 83! ; as. 837. fromlice, adv., strongly, boldly, swiftly : i617, 4 169; comp. fromllcor 4i66. fruma, m., beginning, commencement: is. fruman (at first} 837. frumbearn, \., first-born : np. 47. frumsceaft,f., creation: ds. frumsceafte 414- I'm mst ;ie 4i6>34. GLOSSARY fiigol, m., bird: ns. fugul 37'; gs. fugles 277> 3711 ! ds. fugele 327 ; dp. fuglum 74s. See guSfugol. ful, adj., /«//: asm. fulne 430 ; comp. nsf. f ulre 64. See Vrymful. fill, adv., zvry : 266, 3i6, 4i104, 836, 8815. ful, adj., foul, dirty, unclean: nsm. wit. fiila 41; comp. nsf. fulre 4i31. full, n., receptacle (of water), cloud: as. 488. fullestan, Wi, help, give aid: 3 sg. fulleste-S 258. fullwer, m., complete wer, full atone- ment; as. 2414. fiiltiiiu, m., prop, support: ns. (MS. fuglum) 52, (MS. furum) 5Q15. fundian, Ws, strife, intend, desire : 3 sg. fundaft 845; pret. 3 pi. fundedon 23®. furffum, Z.&M., first: 414. fus, adj., prompt, ready, eager: nsm. 3i3, 7327 ; dsm. fusum 9312 ; np. 443. See ellorfus. ts\tmSN\,throw down, fell: i sg.fylle29. fyllan, Wi,y?//: inf. 62s. &; gefyllan. fyllo, i., fullness: ns. (MS. felde) 38; gs. fylle i86; as. 435. fyr, r.,fire: gs. fyres 71; ds. fyre 13"; as. 4i78; is. fyre 443, 3i8, 83. fyrd, f., expedition : as. 7321. fyrdrinc, m., warrior : gs. fyrdrinces 8o2. fyrdsceorp, n., war-ornament : as. I513. fyrn, adj., ancient, old: nsf. 849. G G = rune X 2O6, 257. gafol, n., tribute, gift : as. 392 ; as. gaful 3312- gal, see hygegal. galan, 6, chant, cry: 3 sg. gaeleft 2I85. galdor, n., song, chant: gp. galdra 682. galdorowide, m., mystical saying, song: as. 497. gan, anv., go: 3 sg. gieft 4i77; pret. 3 sg. code 56. gangan, see gongan. garsecg, m., ocean : gs. garsecges 33, 4 198. gaest, m. i. guest: as. giest 442; dp. gestum 2318. — 2. enemy, stranger : ns. i610. 6>^brlmg8est,hilde-,ryue-, staelglest. giest, m., spirit, soul: ns. 89, 6o14 ; ds. gieste6o4; as. I32; is. gSste io8; gp. gSsta 430, 41", 495; ap. giestas 218. gaestberend, m., possessor of spirit, liv- ing man : ap. 2i8. gat, f., goat: ns. 252. [geador, adv., together : i19.] gear, n., year : gp. geara 3312; ip. gea- rum 73s. geara, adv., already, formerly : 2I29. geard, n., dwelling, home : dp. geardum 442, 92; ap. geardas 2i8. See mid- dangeard. gearo, adj., ready : comp. nsm. gearora S436. See eallgearo. gearo, adv., swiftly: 4i17. gearwe, adv., ivell, readily : 836. geat, see hordgeat. geat\van, W 1 , make ready, adorn, equip : pp. geatwed 296. geat\ve, fpl., ornaments : dp. geatwum 3610 (Leid. geatum). gebelgan, 3, anger, enrage : pp. gebol- gen 4 119. gebennian, Wa, wound: pp. gebennad 62. gebindan, 3, bind : pp. gebunden 576 ; pp. asm. gebundenne 5". geblandan, R, mix, mingle: pp. ge- blonden 422, 248. gebrec, n., noise, crash, thunder : np. gebrecu 444 ; gp. gebreca 440. gebrotfor, mpl., brothers : np. gebr5K>r M2- gebysgian, Wa, occupy, busy, agitate: pp. gebysgad 31% (b gemylted). geceaplan, W2, buy, purchase : 3 sg. geceapab 2418. geceosan, 2, choose, elect : pp. gecoren 256 RIDDLES OF THE EXETER BOOK gecrod, see hlotJgeerod. gecweffan, 5, say, announce : pret. 3 sg. gecwae'S 49. gecynd, f., nature, kind, condition : ds. gecynde 73 ; dp. gecyndum 4O15. gecyssan, Wl, kiss : 3 pi. gecyssa'S 3i6 b (a cyssaft). gecyUan, Wl, announce, make known: inf. gecyban 847. gedolglan, Wa, -wound: pp. gedolgod 54". gedon, anv., do, make, cause: pret. 3 pi. gedydon [i14], 736. gedreag, n., tumult (sea) : as. 710. gedwelan, 4, err, mislead: pp. npm. gedwolene, perverse, wrong, I27. gedygan, Wl, survive : 3 sg. gedygeS 396 ; 3 P1- gedygai? 457. gedyn, m., din, noise: is. gedyne 445. gcfa-lsian, Wa, cleanse, purify: pp. gefaslsad 83. gefara, m., companion : ns. 8o2. gefea, m..,joy, gladness : ds. gefean 425. gefeon, 5, rejoice, exult, be glad: pret. 3 sg- gefeah 655. geferan, Wl, accomplish, experience : pp. gefered 38. - gefeterian, Wa, fetter, bind : pp. npm. gefeterade 53. ;;efra;ge, adj., known, renowned, fam- ous : nsm. 958. gefrsetw(i)an, Wi,2, adorn, deck: pp. gefraetwed 548, gefraetwad 322. ? gefricgan, 5, learn by hearsay : pp. [gefrigen] 8483. gpf rignan, 3, learn by asking, find out, hear: pret. i sg. gefraegn 461, 482, 491, 681. geftillod, see ungeftillod. gef^-llan, Wl, ///: i sg. gefylle 678; 3 sg. gefy lie's 1 58 ; pret. 3 sg. gefylde 457; PP- gefy lied i82. geglervvan, see gegyrwan. gegnpa'W, m., hostile way, hostile path : ds. gegnpaj>e I626. gegyrdan, "W\,gird: pp. npf. gegyrde 7316- gegyrwan, Wl, adorn, furnish, equip: pp. gegyrwed 2i2, gegierwed 291, 3O3, 372, 6817, 692. gehabban, W3, hold, hold fast: inf. I7w. gehtelan, Wl, heal, save: pret. 3 sg. gehaalde 612 ; opt. 2 sg. gehjgle 496. gehladan, 6, /#«</: pp. gehladen 842-1. gehleffa, m., companion, comrade : as. gehle^an 9327. See Avilgehletfa. gehnast, see hop-, \volcengehnast. gehrefan, Wl, roof, cover :' pp. gehrefed 2io. gehreodan, 2, adorn : pp. gehroden 8422L gehwa, pron., each : dsm. gehwam 312, 1 28 ; dsf. gehwam 559 ; dsn. gehwam 34 13 ; ism. gehwam 3312, 6i6 ; isn. gehwam 826. gehwylc, pron., each, all, every : nsm. 726; gsm. gehwylces I45; gsn. ge- hwylces 4I36; dsm. gehwylcum 428, 8312, 9518. [gehyran, Wl, hear: 2 sg. gehyrest lie.] gehyrstan, Wl, adorn : pp. gehyrsted 78. i, Wl, lead, conduct, bear : inf. See i62°. gellc, adj., like: np. gelice 327. ungellc. gelicnes, f., likeness, image : ns. 3710. gelome, adv., frequently, constantly: 3211. gemaedan, Wl, madden, make foolish: pp. npm. genuedde I26. gemaman, Wl, utter : i sg. gemaene 256. gemaene, adj., mutual, in common : np. 72B. gemanian, Wa, warn, admonish : pp. gemanad 468. gemet, n., measure: is. gemete 5i7. gemiclian, Wa, enlarge, magnify : pp. gemiclad 8423, nsf. gemicledu 2120. GLOSSARY gemittan, Wl, meet: 3 pi. gemittaS 4™. gemong, n., company : ds. gemonge 324,n geniot, n., meeting, coming together: gs. gemotes 610, 2610. See gutSgemot. geiiiuiiaii, PP, remember, bear in mind: i sg. gemon 836 ; 3 pi. gemunan i8u. geniyltan, Wl, cause to melt, soften : pp. gemylted 3i8 b (a gebysgad). gemynd, f., memory, recollection : as. 6o7. gen, adv :, formerly : glen 2I26. See nu gen, tfa gen. germ, a.A.,yet: 4i58, geno 2I29. geniegan, Wl, attack, assail : 3 sg. genjegeft 2i19. genamna, m., companion : ns. (MS. ge- namnan) 5418 ; np. genamnan (MS. genamne) 53. gemcstan, Wl, contend: 3 sg. genSste'S 2810. geneahhe, adv., sufficiently, abundantly, frequently: 9, I312, 278, 3210. genearwian, Wa, confine : 3 sg. ge- nearwa'S 41 ; pp. genearwad 71. genergan, Wl, save, preserve : inf. i619. geniwian, Wa, renew : pp. genlwad I49. geoo, f., help, aid, safety : ns. 6s. geofon, n., sea, ocean : ns. gifen 38. geofti, see gifu. geoguS enosl, n., youthful family, prog- eny: ds. geogu'ScnSsle i610. geoguftmyro', f., joy of youth : gs. geo- gutSmyr>e (MS. -myrwe) 392. geolo, adj., yellow: asn. 3610 (Leid. goelu). geond, prep. w. ace., through, through- out, over: 26, 13", 356, 4O17-19, 8310, 84°, 880. geondsprengan, Wl, sprinkle over: pret. 3 sg. geond[sprengde] 27. geong, adj., young: nsm. I52; nsf. 4I44, 741 ; ? geong 888 ; comp. nsm. gingra 9318 ; comp. npm. gingran 8820. geongan, see gongan. geopan, a, take to oneself, receive, swal- low (Sw.) : preU i sg. geap 24°. geopenian, Ws, open : imp. 2 sg. ge- opena S454. georn, adj., desirous, eager: nsm. 3216. See cl»n-, feJJegeorn. georne, adv., gladly, eagerly : j2. gerteoan, Wl. i. reach, strike: 3 sg. geriece'S 458. — 2. reach, arrive : I sg. geriice i627. geren, n., ornament: np. gereno 2716. gereord, n., speech, voice : ip. gereordum i5J6. gerum, n., space, room : as. (on gerinn, at large) 2 1 14. gerunia, n., space, room : ds. geruman l6«. geryde, adj., fitting, ready, prepared : npn. ? 6416. geryht, n., straight direction : ap. (on geryhtu, straight, direct) 4M. gerynian, Wl, make room, clear (way) : i sg. geryme 63. gesailig, adj., happy, blessed: nsm. 4I64. gesceaft, f. i. creature, shape : np. ge- sceafte 442; gp. gesceafta 4i88. — 2. nature, condition : as. 348. See ealdor-, for5gesceaft. gesceap, n.,fate, destiny: ds. gesceape 736; as. 39; np. gesceapu io7, 4O2; ap. gesceapo 70. gescyldru, np., shoulders : dp. gescyl- drum 4 1108, 70. gescyppan, 6, create, form, make: pret. 3 sg. gescop 24®, 8817. gesfcan, Wl, seek, visit: inf. 4O5, 6o14. gesec%gan, W3, say, tell, narrate: pret. 3 sg. gesaegde 395 ; inf. 512, 4O28 ; ger. gesecganne 3713, 4O25. geselda, m., companion : ns. 8o8. gesSon, 5, see, behold: pret. i sg. ge- seah 301, 351, 371, 381, 391, 571'10, 6816, 691, 751, 761. gesettan, Wl, create, establish : pret. 3 sg. gesette 71. 258 RIDDLES OF THE EXETER BOOK gcsib, adj., near, related: gpm. gesib- bra 2722; apm. gesibbe I622. See ungesib. gesihff, f., sight, vision : as. 6o9. gesittan, 5, sit: ? gesaet 786. gesfS, m., companion, comrade: np. gesfras 3 15. gesom, adj., united: npm. ges5me 8S29. gesomnian, Wa, join, unite, collect : pp. gesomnad [i18], 3i2. gest, see gaest. gesteald, see wuldorgesteald. gestealla, see eaxlgestealla. gestillan, Wl, still, quiet, calm : 3 sg. gestilleS 4s5. gestrfon, n., treasure, wealth : gp. ge- streona 2i31, 2 I210. gesyne, adj., seen, -visible: nsf. 40; npn. 14. getacnian, Wa, betoken, signify : pp. getacnad 6414. getaJse, adj., convenient, pleasant: nsf. 84. getenge, adj., near to, close to: nsm. 78, S8, 1 1. 842S ; nsf. 538 ; nsn. 578 ; asf. 77. getreowe, adj., faithful, trusty: gpm. getreowra 2728. gefencan, Wl, reflect, consider: ger. gej>encanne 428. ge3eon, Wl, tame, oppress : inf. ge)>eon 4 191. See ge<5ywan. geJfone, see ingcSonc. geffraec, n., crowd, press: ns. gejraec 237 ; as. ge>raec 32, 461 ; ap. gej>raecu 36° (Leid. gi'Srsec). gearing, n., tumult, crowd: ns. ge- ge5ringan, 3, swell : pp. asf. ge)>rungne 8?2- geruen (MS. ge^uren) gi1. geSwsere, adj., gentle, calm: nsm. (adv.?) gej>\vjere 5i6; npf. gej>wiere 16 , Wl, press, urge, compel: pret. 3 pi. gebydan 6i14. See geSeon. gewsede, n., garment: ns. 36"; as. (MS. gewiEdu) 3612 (Leid. giuSde). gewealean, R, roll: pp. gewealcen 34- geweald, n., power, rule, dominion : ds. gewealde 416 ; as. 28U. ge^veaxan, 6, grow, grow up : pret. 3 sg. geweox 8o6. gewefan, 5, weave: pp. gewefen 4i85. ge\vendan, Wl, turn oneself: inf. 8833. geweorSan, 3, become, be : inf. geweor- >an 4 148. geweorSian, Wa, honor, adorn : pp. geweorj>ad 7i5, 84^. gewin, n., contest, strife: gs. gewinnes 17; as. 2 11, 242. See gu5-, stream- ge\-ln. gc'xviiidaii, 3, wind, twist: pp. apm. gewundne 41". gewinna, see latJge^vlnna. ge>vit, n., mind, understanding: as. 4013. ge\vitan, PP, know: pret. 3 sg. ge- wiste 3O14. ge^vitan, l, go, depart: i sg. gewlte 31, 460, I72; 3 sg- gew!te« 4O6 ; pret. 3 sg. gewat 3o10-18, 938 ; pret. 3 pi. gewitan I411. ge\rlitiglan, Wa, adorn, beautify : pp. gewlitegad 322, 332, 84°. gew^regan, Wl, stir up : pp. gewreged GLOSSARY gewrit, n., -writing, book: np. gewritu 401'18. ge\~unlan, \T2, dwell: pret. i sg. ge- \vunade 6i2. gewjTcan, Wi, make, create: pp. ge- worht 7O8. gey'Nvan, Wl, show, reveal : pp. geywed 43. gled, n. i. word, speech : as. 488. — 2. song: [as. giedd i19] ; ds. giedde 8o10. — 3. riddle: gs. gieddes 561. -glel, see wldgiel. gieldan, 3, yield, pay: 3 sg. gieldeft 3311- glellan, 3, yell, cry : i sg. gielle 258 ; ptc. asn. giellende 33. glelpan, 3, boast: 3 sg. gielpeft 5912. gien, see gen. gierwan, see gyr\Tan. glest, see gaest. glestron, adv., yesterday : 4i44. gletan, see ongletan. glf, conj.,//: [i2-7], 4».w, I210, 138, i67- 14.20.2^ I74,5,8j 2 119-24, 2412 2718, 2812, 3°6. 3318- 3712. 398'7. 43. 444'8, 5l6' S412, 6o9, 627, 72", 739, 8s7, 956. glfaii, 5, give: 3 sg. giefeft 4i19 ; [opt. 3 sg. gife i1] ; pret. 3 sg. geaf 2I4.28; ? geaf 722. See a-, ofgifan. glfen, see geofon. glfre, adj., useful: nsm. 27^ ; ipf. gif- rum 508. gifre, adj., greedy, voracious : sup. nsf. gifrost S429. glfu, f ., gift, favor : dp. geof um S486 ; ip. gifum 5918. See \roe 2I25; is. guj>e 2 119. gutJfugol, m., bird of -war, eagle: gs. guSfugles 25. gutfgemot, n. battle-meeting, battle : gs. gu)>gemotes I626. gu?fgewiii, n., battle : gs. gu'Sgewinnes 65. guffwiga, m., -warrior : gs. guftwigan 92. gylden, adj. golden : asm. gyldenne 6oJ. gyman, Wl, care for, heed: \ sg. gyme 2I35. gyrdan, Wl, gird, bind round: pp. gyrded gi4. 6"<?^ gegj'rdan. gyrdels, m., girdle, belt : ds. gyrdelse 55"; as. 55. gyrn, n., grief, sorrow, affliction : ns. i66; ds. gyrne 836. gyr\van, Wl, adorn : 3 sg. gyrweft 2i9; pret. 3 sg. gierede 2713. See gegyr- WML H H = r«w^ N: 202-8, 259, 658, 75a. habban, Ws, ^aw 2721 ; opt. 3 sg. [haebbe] 8433 ; pret. i sg. haefde n6, 275, 7212, 74s; pret. 3 sg. haefde lo11, 20, 325, 338, 373.6, 383, 7326t gas, [hasfde] 83!, . haefd[e] Sg2; pret. 3 pi. haefdon I48, 238; inf. 4s5, 2I28, 955 ; ptc. gsm. haeb- bendes 653. See gehabban, iiabbaii. had, m., person : ap. hadas 212. hafoc, m., hawk : ns. 253, 4i67, COF(O)AH = HA(0)FOC 207-8, HA- [foc] 653. haeft, n., haft, handle (captivity) : ds. haefte 7322. ha'ftan, WT1, bind, confine: pp. hasfted 52- haeftenyd, f., captivity: as. haeft[e]nyd , m. i. hail: ns. 8i9. — 2. name of rune H : np. haegelas 43". GLOSSARY 26l hagosteald, n., celibacy ; bachelorhood: ds. hagostealde 2i81. hagostealdmoii, m., bachelor, -warrior : ns. I52, hasgstealdmon 55s. hit'lan, see gehielaii. -haele, see onhtele. ha-lend, m., Healer, Savior : as. 6o6. haeleft, m., hero, man : ns. 2712, 636, np. 285, 561, 5711 ; gp. hasle^a 21, 48, 83, 2 131, 4 196; dp. haelelmm 910, 2728, 3612 (Z«</. heliSum), 49!, 6o17, 70', 8422.36.58> (MS. aeldum) 434. 83", 867 ; 3 sg. hatte 4O29, 4415, 56" ; pp. haten 259, npf. hatne 43". he, pron., he: nsm. [i1-7], 481, i614, ha>Io, f., safety : as. 498. hals, m., neck : ns. I61 ; ds. healse 7212, halse 3221. halsrefeSer, f., pillow-feather, down : ds. halsrefe)>re 4i80. halswriffa, m., necklace, chain for neck : as. halswri^an 54. ham, m., home : ds. 3O9, 354, 446, 786, ham[e] 3O4. hseiiied, n., sexual intercourse: as. 2 128. ha' mediae, n., sexual intercourse, wed- lock-game : gs. hiEmedlaces 43s. hamlea.s, adj., homeless : nsf. 40°. ha n, f., hen : ns. H>EN 438"11. haiia, m., cock: ns. HAN A 438"11. har, adj., hoary, gray: nsm. 223, wk. hara 4i74, 9311. See feaxhar. haer, see her. haso, adj., gray : nsf. wk. heasewe 416!; asm. wk. haswan 254; npm. haswe 27 ; apf. haswe I49. hasofag, adj., of gray color; nsn. I21. ha'st, see h?st. hat, adj., hot, fiery: nsm. wk. hata 44s; asm. hatne 637 ; comp. nsm. hatra 4i57. hat an, R. i. command, order: 3 sg. hatej> 7s, 4I88; pret. 3 sg. het gi10, heht 4i8; pp. haten 624. — 2. call, name : inf. 3612 ; pass. I sg. hatte 215, 47'2, g8, 1 1 n, i313, i519, 17!°, 209, 24i6, 2726, 2S18, 639, 67!°, 7329, So11, 2811,12, 3g5.9, 4I5,6,7,19.55,94.108) ^ 4g6? 492, 5 16-8, 54-8, 551, 56°, 60", 665, 7327, 766, 8ss-5, 9 119 ; nsf. heo lo"-^, 2I88, 267, 32i8.",356, 40s'27, 4 126-28, 692, he[o] 397, hlo 3216'21, 357, 378, 396, 407,8.io,i6,i8.20,2i,29) (MS. hie 6) 559, 62, 68, 8o6, 8427, 876'7 ; nsn. hit[ii°], 3O6 ; gsmn. his i615, 2i16, 362 (so Leid. 2), 38. 41 13,39, 449, 454.6, 471,2, 5,8, S49» 558'6, 5613, 6o8, 647, 70, 739, [his] 84", 8880; gsf. hyre io«, 2I88'84, 326- 18.21, 348 . dsmn. him 463.64, ! 511,25, 2O4, 38«, 392, 44, 5o7-9, 516-6.6, 607, 734?, 838, 856, Sg7, 9318 ; dsf . hyre 42, 3o5-10, 32n,2i, 353, 555,io. asm- hine [i2-7], 429, 2313, 2412, 516.8,10, 548, 5615. asf. hie 551, 594; asn. hit 38, 40, 4i47, 6i16; np. hi 78, I210, 176, 236, 3i7, hy [i2-7], 146, 2i10, 2319, 2719, 446.i2, 5410, 84"?, heo i26; gp. hyra 79, 142'5, 239'18-21, 2723, 478, 498, 536; dp. him [i1], I28, i78, 3215, 447-11, 5i8; ap. hy 2724, 586, hi 2728. header, see heaffor. heafod, n., head: ns. I61, 91; gs. heafdes 549; ds. heafde 2212, 4i98-i°2; as. 268, 597, 626, 663, 8i2; is. heafde 456; gp. heafda 864; ap. heafdu 378. heafodbeorht, adj., having a bright head: asm. heafodbeorhtne 2O2. hCafodleas, adj., headless: nsm. I510. heafod wo5, f., voice: is. heafodwS^e 98. heah, adj., high, lofty, exalted: nsm. 708, 8828, 93s ; nsf. hea 84 ; nsn. 427'63 ; dsm. heaum 2319 ; asm. heane 8i2, hean 4i22; npm. hea 237 ; apm. hea 424 ; ipf. heahum 210 ; comp. nsm. hyrra 8815; comp. nsf. hyrre 4i88, 942; sup. n?sn. wk. hyhste 8412. See steapheah. heah, adv., hi^h : I29. heahcra'i't, m., excellent skill: asm. 364 (Leid. hel "r 262 RIDDLES OF THE EXETER BOOK , God: ns.4i38. healdan, R. i. hold: i sg. healde 4i37; 3 sg. healdeS 2i13; pret. 3 sg. heold 4314. — 2. hold to, continue : i sg. healde 9. — 3. cherish, foster : pret. 3 sg. heold xo5. — 4. rule, gov- ern: 3 sg. healdeiS 4\2-5, healde J> 4 122. See be-, bihealdan. healdend, m., holder, possessor: ds. healdende 2I28. healf, f., .«</<?: ds. healfe 229, 8S28. heall, f., /&</// : ds. healle 561-18, 6O1-17. heals, see hals. hean, adj. i. Aw, </<?<?/: nsm. (MS. heah) 469. — 2. poor: npm. heane 3313 ; dpm. heanum 952. — 3. mean, vile : comp. nsf. heanre 4O9. hPaiiinod, adj., mean of spirit: npm. heanmSde 4317. heap, m., troop, crowd, flock : ip. hea- pum 58. heard, adj., hard: nsm. I510, 347, 631, (MS. heord) 45, wk. hearda 41", 569, 8 19 ; nsf. 275, 8o8 ; nsn. 453, 9317 ; dsn. wk. heardan 4i79; asn. 8i4; npm. hearde 8813 ; dpm. heardum gi5; apm. hearde 532 ; comp. nsm. heardra 4I54'78, 84s8; sup. isn. wk. heardestan 292. See hrimigheard. hearde, adv., fiercely, severely: gi5. heardecg, adj., hard of edge : npn. 6. heaSoglem, m., wound: gp. hea)>o- glemma 57. heatfor, n., restraint, confinement: ds. hea)>ore 2i18, headre 66s. headosigel, m., sun (of battle) : ns. hea- J>osigel 7319. hebban, 6, raise, lift : 3 sg. hefeft 4$5 ; pret. 3 sg. hof 55 ; inf. 462. See a-, onhebban. hefig, adj., heavy: asm. hefigne 597; comp. nsf. hefigere 4i74. hel, f., hell: ds. helle 4O20 ; as. helle 676. helm, m. i. protector: as. 27". — 2. covering : ns. 8818 ; as. 4. See sundhelm. helpend, m., helper : vs. 495. helwaru, f., people of hell : gp. helwara 566. hengest, see faet-, frltf-, merehengest. heofon, m., heaven : ns. 942 ; gs. heof o- nes 4i4-33, 875; ds. heofone 4i38; as. 4i22; dp. heofonum 3o12, 4O20; ap. heofonas 6j6. heofonwolcn, n., cloud of heaven, rain : ns. (MS. heofon wlonc) 732. heolfor, n., blood, gore : ns. 9317. heord, f., family, flock : gs. heorde i8i. heort, see gromheort. heorte, f., heart : ds. heortan 4314 ; ip. heortum 2720. heorugrlm, adj., very fierce: nsm. wk. heorugrimma 4I55. heoruscearp, adj., very sharp : npn. heoroscearp 68. her, adv., here: 4182,49,61,77,8^ ^ 44i6f 5010, 8823. her, n., hair: np. 16; dp. herum 275; ip. hSrum 364 (Leid. herum). here, m., army, host, troop : gs. herges 8o8. heresuJ, m., military expedition, war- marching: ds. heresl^e 30. hest, f., violence, hostility: as. I628; is. hSste (MS. haetst) 45. hetegritn, adj., malignantly fierce : nsf. 345-_ heterun, f ., charm causing hate : as. heterune 347. higora, m., jay: GAROHI = HIGORA 2S7-9- hlld, f., battle, fight: ds. hilde I54; is. hilde 348. hlldeglest, m., enemy : ds. hildegieste 549- hildepO, m., war-dart: np. hyldepllas i86; ip. hildepllum I628. hildecJrj'lf, f., strength in war, war- force : as. hilde J>ryJ>e 2o4. hildewJepen, n., war-weapon : ns. g25. hilted, see goldhllted. GLOSSARY hindan, adv., from behind: Qi5; on hindaii, behind, 38, 89. hindeweard, adj., hindward, from be- hind: dsf. hindeweardre 2215. hiiigong, m., departure : gs. hingonges (MS. ingonges) 63!. huVan, Wl, plunder, lay waste, ravage: 3 sg. hlj>e"5 35; ptc. nsm. hlj>ende 347> QJ26) gPm- hl)>endra 95s. hladan, 6, load: i sg. blade 4s5; pret. 3 pi. hlodan 2310. See gehladan. bladder, f., ladder : as. hlaedre 566. hlafurd, m., lord, master: ns. 5, 228'15, 9i9; gs. hlafordes 5913; ds. hlaforde 449, 5711- hlafordleas, adj., lordless : nsm. 2I22. lila-st, n., load, burden : ap. 215. hleahtor, m., laughter, noise : ns. 348. hlfo, m., shelter, cover : ds. 288. hleobord, n., cover, binding: ip. hleo- bordum 2712. hleor, n., cheek : dp. hleorum 16. hlEortorht, adj., bright of face: nsf. 70. hleosceorp, n., protecting garment : is. hleosceorpe io5. hleoo'or, n., voice, speech, song: ns. hleoj>or 3217 ; as. hleo)>or 255 ; is. hleoj>re 9, 15. -hlefta, see gehle'o'a. hlidan, jw onhlidan. hi ilia n, Wa, tower, standout: 3 pi. hlifia'S i64; inf. 54!. hliiiiiiian, 3. i. roar: 3 sg. hlimme'S 36. — 2. sound: 3 sg. hlimme'S 36° (Leid. hlimmith). hlin, m., maple?: ns. 569. hlin, m., noise, clamor, din : ns. 27. hlinc, m., link, linch, hill: ap. hlincas 424- hlinsian, W2, resound, make a din : pret. 3 sg. hlinsade 348. hlift, n., cliff, mountain-slope: ap. hleoha 37, hli^o 937. See beorg-, burg-, hloSgocrod, n., press of troops, congre- gated band: ns. 468. hind, adj., loud: nsm. 424, 8s1; isf. hlude 492 ; sup. nsn. hludast 4°. hlude, adv., loudly : f, a^, 87, 98'10, 34s, 58. hi utter, adj., bright, clear: asm. hlut- terne 2i7. hly?fan, see behlyftan. hnecca, m., neck : as. hneccan 8i4. hnesc,adj.,jiy?: comp.nsf. hnescre4i80. hnigan, \,bend, bow down, descend: i sg. hnlge 463. See on-, underhmgan. hnltan, l, push, thrust: inf. gi4. hiiossiaii, Wa, strike, beat : 3 pi. hnos- sia« 67. [liogian, Wa, think : pret. i sg. hogode (MS. dogode) i9.] hoi, n., hole: ds. hole 63" ; as. 455. hold, adj., kindly, loving, gracious : nsf. io4; dsm. holdum 624. holdlice, adv., gently, sweetly: 354. holen, m., holly : ns. 5610. holm, m., ocean, water: as. 469; is. holme 210. holmmaegen, n., force of waves, holm- mass : is. holmmrcgne 39. holt, n. i. holt, wood: gs. holies 228; ds. holte 92 ; np. 8815. — 2. wood {piece of) : as. 578. homer, m., hammer: is. homere 91; gp. homera 67. hon, see blhon. hond, f., hand: ns. I312, 6i12; as. 508, 8o4 ; dp. hondum 3i5 ; ap. honda 865 ; ip. hondum 464, 554. hondweorc, n., handiwork: as. 2i7; np. (MS. iweorc) 68. hond\vyrm, m., itch-mite : ns. 4I96, 672. hongian, Wa, hang: i sg. hongige 15" ; 3 sg. hongah 2211, honga'S 45! ; pret. 3 pi. hongedon 148. hopgehnast, n., dashing of waves in a bay. gs. hopgehnastes 427. hord, hoard, treasure : gs. hordes gi9; as. 3221, 5411, 93126 ; gp. horda i29; ip. hordum 84^ ; ? hord 8452. See womb- hord. 264 RIDDLES OF THE EXETER BOOK hordgeat, n., door to treasure : gs. hord- gates 43". horn, m., horn : dp. hornum (MS. horna) yp. hornsael, n., gable-hall: np. hornsalu 48. hore, n., horse: ns. SROH 2O1-2; gs. horses 37" ; as. 376 ; ap. 23!°. horse, adj., wise, sagacious, quick-witted: nsm. 21. breed, adj., quick, speedy, rapid: nsm. 54U; comp. nsm. hraedra 4i72. See hreS. hraegl, n., garment: ns. 81, I21, 14'; ds. hraegle n7; as. 45, 55; is. hraegle 46, 636. ? hra<5e, adv., quickly: hr[a]J>e 777. hreddan, Wl, recover, rescue: inf. IS18. See ahreddan. hrofaii, see gehrgfan. hrgodan, see gehrgodan. hreoh, adj., rough, fierce: nsf. 84. hrf-osan, 2, fall, rush : 3 sg. [hr] ease's 8 110. Im'ran, Wl, move, stir, shake : i sg. hreru 48, hrere 28; 3 sg. hrere'S 8i7 ; opt. pres.(?) pi. hreren 8451. hre5, adj., quick, speedy: comp. nsf. hre|>re 4i71. See hraed. hreSer, m., breast, bosom : ds. hre>re 626, 93". hrif, n., womb, belly : ds. hrife i86, 2412 ; as. 4i46; ?hrif8451. hrlin, m., rime, hoar-frost : ns. 4I55, 8i9. hrlmigheard, adj., hard with frost: apm. hrimighearde 93". hrinan, l, touch, reach : \ sg. hrlne 74, 676, hrino I628; 3 sg. brine's 2412, S446 ; pret. 3 sg. hran 4O10-20. hrindan, 3, push, thrust: pret. 3 sg. brand 55. hrlng, m. i . ring (paten, chalice) : ns. 49s ; gs. hringes 6o17 ; as. 491, 6O1'6. — 2. ring, adornment: ?is. hringe 926; ap. hringas 2I28 ; ip. hringum 7i8, 91. — 3. fetter, chain : ip. hringum (MS. hringan) 52. hrisil, f., shuttle : ns. 367 (so Leid.). hroden, see beaghroden. lirof, m. i . ro of: as. 532 ; dp. hrof um 27. — 2. top, summit: as. i627, 3O7. — 3. sky, heaven : gs. hrofes 285. hror, adj ., strong, stout, active : nsm. 553. lining, f., rung, beam, pole: ds. hrunge 2310. hruse, f., earth: ns. 46, 732; ds. hru- san 4155, 84s5'46 ; as. hrusan 39, 81, 28". li ru tan, 2, make a noise, whiz : ptc. nsf. hrutende (Leid. hrutendi) 367. hrycg, m., back : ds. hrycge 212, 46, 2O4, 376; as. 465, 2211, 8i4, hryc[g] 865; is. hrycge 2811 ; ip. hrycgum 4s3. hu, adv., how: i86, 3219, 37", 4O23, 4316. 44157 s&6t 6Q16, 6 1 12, 848. bund, num., hundred: 86. bund, m., dog: ns. 252 ; gs. hundes 3711 ; as. (MS. DNLH = HUND) 752. hungor, m., hunger : ns. 44. biinig, m., honey: ds. hunige 4i59. huff, f., spoil, booty : as. hu>e 3O2-4-9. h\va, pron., who ; neut. what, of what kind: nsm. 22-i, 318, 488, 478,74, g-jT . nsn. hwaet 472, 98, n11, i519, 2o9, 24", 2726, 2818, 2912, 322, 33", 3614, 378, 4029, 429, 639, 671", 6819, ^^, 80", 8314, 867 ; asn. hwast 629 ; nsn. or asn. hwset 6415. See aeg-, gehwa, nsit- h\vset. hw^ael, m., -whale: ns. 4i92. hTvaelmere, m., sea : ns. 3s. h\vser, adv., where : S826. See nat- h\vser. h\vaet, adj., stout, bold, brave: comp. npm. hwstran 272°. See bl6dhwaet. h\vae5er, see aighwseSer. hwaetJre, adv., yet, however : hwaejre [I"], 4^ 23n 328.9,i7; 4018, 558, S9tt [hwaeKe] 324 ; hwaebre se beah 36" (Leid. hudrae sua ^eh). hwearft, m., circuit, expanse: ds. hwearfte 4i88. [hwelp, m., whelp : as. i16.] See wael- hwelp. GLOSSARY hweorfan, 3. i. turn, depart: 3 pi. hweorfa'S 4412 ; inf. 2 122. — 2. wander, roam: 3 sg. hweorfe'S 4i5; inf. 338, 4O9; ptc. asn. hweorfende 573. See hwyrfan. hwettan, Wl, incite, instigate : I sg. hwette I23. hwfl, f., a -while, space of time : as. hwlle 299 ; ip. hwilum 31, 4i.i7,36,88.68.68, 69,70> r8, 76.7, §3, ! .,4,6,6,7,10, j r8,4,5,6,8,9, 11,18,16,17, i g7, 2 ^,18, 2 r2,2,S,3,4,6,6, 2&, 288, So4, 578, 585, 622, 636-7, 64, yi5, 737-28, 8o8-7, 838, 856, 886, 918, 934-7-8-11, 9512, [h]wllum 935. hwit, adj., white, fair: nsm. I61; npf. hwlte ii8; apm. hwite 4i98. hwitloc, adj., -with fair hair: nsf.'438. hwitlocced, adj., fair-haired: nsf. hwitloccedu 8o4. hwonan, see ohwonan. hwonne, adv., when, until: i610; hwoime SBT, -whene'er 3213. hwylc, pron. inter, i. who, which : nsm. 21, 43!!. — 2. pron. ind., any one, each one: nsm. 2i19, 6819; dsm. hwylcum 2410. See aeg-, gehwyle. hwyrfan, Wl, tttrn, move about: 3 sg. hwyrfe'S i312. See hweorfan, on- hwyrfan. h^vyrft, see ymbhwyrft. hAvyrftxveg, m., escape : gs. hwyrft- weges 46. hycgan, Wl, think, consider, meditate : ger. hycganne 2912, hycgenne 3223. hyd, f., skin, hide : as. 777 ; is. hyde 2712. hygeblitJe, adj., glad at heart: comp. npm. hygeblibran 2720. hysecrpeftig, adj., wise, sagacious, keen of wit : nsm. 21. hygefaest, adj., fast in mind: asf. hyge- fasste 4314. hygogal, adj., lascivious, -wanton: gsf. wk. hygegalan I312. , m., thought : ip. hygefcmcum (Leid. higido[n]cum). hygewlonc, adj., proud : nsf. 464 ; asm. hygewloncne 2O2. hyht, m., joy : ns. 658, 955 ; ds. hyhte 261 ; ? hyht 938. hyhtlic, adj., delightful: nsn. 926; asn. 3612 (so Leid.). hyhtplega, m., joyous play, sport : gs. hyhtplegan 2I28. hyldepil, see hildepfl. hyll, m., hill: gs. hylles I627. hyran, Wl, (hear), hearken to, obey: i sg. hyre 2i24; 3 sg. hyrei? 449, 591 5i5; inf. hyran 4s4, 52, 2415. See gehyran. hyrde, m., keeper, guardian, herd : ds. 7210; as. 9i9. hyred, m., company : ds. hyrede 6o6. hyrgan, see onhyrgan. hyrst, f., ornament, equipment : np. hyrste (wings) 84, 1 18 ; 1 21, ip. hyrstum IS11, 3220, 547, 8815. hyrst, m., copse, wood: dp. hyrstum 4i« hyrstan, see gehyrstan. hyse, m., boy, youth : ns. 551. I I = rune I: 259, 651. ic, pron., I: ns. (271 times) ; gs. min 2718, 364 ; for possessive, see mm ; ds. me [112-12], 212,45.16.86.66, 54,10, (69 times); as. mCC [l»], 22'14, 311.13,16, 41,18,13,73.74) (90 times); as. me [i18], 1313, 2ii8-i9, 2718, 4i84, 48i, 665, 732, 834, 8s5; nd. wit 645, 8s7, SS14-29'8! ; gd. uncer 8880 ; dd. unc 6ii5, 6416, 852, 88"; ad. unc 723, 8s7, 8818-i7 ; np. we 3716, 4i73, 426-7, 728; for genitive, see user; dp.us[i3'8], 4316. S65- ides, f., woman : ns. 622; as. idese 761; gp. idesa 477. [Teg, Tg, f., island: ds. Tege i4, Ige i6.] In, prep. w. dat. and ace. i. in, on, within, among (w. dat.); 69, 96, 13!°, 286, 351, 387, 4i98, 426, 442, 546'13, 552. 5613, 59", 601-17, 832, 956 ; after case 266 RIDDLES OF THE EXETER BOOK 85. — 2. into, upon (w. ace.) : i66, 531, 561, 6o7-9, 938-9. in, adv., in, -within : 3311. indryhten, adj., noble: nsm. 95!; asm. indryhtne 441. Ingeftonc, m., thought, mind: ns. inge- )x>nc 6 118. lnnan,adv.,w/V/4«: i82,8832; ininnan I03, 297. iimaiiwcard, adj., in-war d, internal: asm. innanweardne (within) 9315. innaff, m., inside of body, stomach, -womb : ns. i89 ; ds. innate 362 (Leid. innaftae) ; as. 38. inne, adv., within, inside : 47, 571. insittende, ptc., sitting within : gp. in- sittendra 477. i rn. i ii, see ri ii nan, upirnan. Isern, n. i. iron: gs. Isernes 599. — 2. sword, knife: ns. 9315 ; is. Iserne 61. — 3- goad: ns. 7 a14. lu tFa, adv., once, formerly, of old : lu )>a 7I« Iw, m.,^a;: ns. 56°. Ij = rune \~ : over 18. --^- lac, f. ?, gift: [as. i1]; ip. lacum 5o3. See hit- ini-i I l;n- lacan, R. i. fly, float : pret. 3 sg. leolc 578- — a. y?jf^A strive : i sg. lace 3I1. — 3. modulate: inf.3219. .SV<?beIaean. Isececynn, n., leech-kin, race of physi- cians : as. 610. laedan, Wl, lead, bring, carry: inf. 3o2; pp. laeded 29°. See gelanlan. laf, f . i . leaving (of fire, file, hammer) : ns. 7 18; np. lafe 6"; ap. life 5710. — 2. heritage •„ bequest: ap. life gi10. lagu, m., j^a, water : ns. 411 ; as. 2316. lagufaettai, m., watery embrace: is. lagu- faeiSme 6i7. laguflod, m., water : as. lagoflod 5912. lagustr^am, m., /a^ ^ rain, water: gp. lagustreama 4s8. land, see lond. lang, see long. lar, f., teaching, doctrine: ip. larum 4022. lleran, Wl, teach, instruct: pret. 3 sg. ISrde 4 184. lareow, m., teacher : ns. 6813. lies, n., the less: as. IO11. hfssa, adj., less: nsf. la?sse 4I95, 672. last, m., track, trace (on last, on laste, behind}: ds. laste 14", 7213; as. 421; is. laste 40; np. lastas 522; ap. lastas 9511. See sweart-, Avidlast. laet, see unlaet. la- tan, R. i. let, allow: \ sg. liete 438; 3 sg. 15te« 4s6, 2 113, 35% 5110; 3 pi. lieta'S 46 ; pret. 3 pi. leton I410. — 2. let go: opt. 3 sg. laite 3". See forlM-taii. latteo^-, m., leader, guide : ns. 311. Ia3, adj., grievous, hateful: [nsm. I12]; comp. gsn. la)>ran 610. latSgeAvinna, m., hated opponent, enemy: ds. laSgewinnum I629. la'(Vian, Wa, invite, summon: I sg. laftige I516- lead, n., lead: gs. leades 4i75. leaf, f., leaf: ip. leafum 5710. lean, see \ordlean. lea nia n, Wa, reward, requite: 3 sg. leas, see ban-, brotfor-, f?3e-, ham-, heafod-, hlaford-, muoleas. lecgan, Wl, lay, place: 3 sg. legeiS 8o4; pret. 3 sg. legde 414, 2I80. See bilec- gan. leg, see lig. legbysig, see ligbysig. lege, see orlege. lengan, Wl, lengthen: 3 sg. lengeft 298. leod, i., folk, people : gp. leoda 6818 ; [dp. leodum i1]. 15of, adj., dear, beloved: nsm. 4I34, So2; nsf. 2 12, 4 127, 8427; comp. nsf. leofre 94«. leolit, adj., light, not heavy: comp. nsf. leohtre 4i76, 94. GLOSSARY leoht, adj., bright, shining: dsm. wk. leohtan 4i57; comp. nsf. leohtre 6f2. leoht, n., light : ns. 946 ; ds. leohte 2817, 64". ICohtlic, adj., bright, shining: asn. 3O8. leoma, m., light, splendor: ds. leoman 4 157. l«-os;i n , see beleosan. ? letter, n., leather : ? le)>re Sg8. ? leftre, adj., evil, bad: ? lej>re 8g8. libban, W3, //'z/<?: 3 sg. leofaj> 4O27; pret. 3 sg. lifde 4i107. & llfgan. lie, n., body: as. 66 ; is. lice ii5. -He, see gelic. licgan, 5, lie : 3 sg. ligeiS 4i49; inf. I411, 1510. I ic in-, see ge-, onlicnes. lif, n., life-, ds. life pi10; is. life 5i9, 59". See woruldlif. llfgan, W3, live; \ sg. lifge 85°; inf. 4022, 4I64, 426, 68"; ptc. nsm. lif- gende 13", asf. lifgende n9, npm. lifgende 299. See libban. lift, see lyft. lig, m.., fire, flame: ds. lege 4i57, is. llge 4", (MS. life) 838. ligbysig, adj., busy with fire: ns. leg- bysig (a leg bysig; b lig bysig) 31!. !Ule,f., lily. ns. 4i27. llm, n., limb : ns. 57 ; as. 4O27. line, f ., line, row : ds. linan 4310. liss, f., mercy, grace; joy: dp. lissum 5i9; ip. lissum 2725, 3413. list, f., art, skill, craft: is. liste 28 ; ip. list urn 3O3. list (least), see metelist. lift, n., limb: ap. leo)>o 247. Ifffan, i. . go, sail: inf . HJ>an 34! ; ptc. dsm. Hl>endum n8. — 2. grow up?: pp. liden 34". loo, see hwitloc. loca, see braegnloca. loco, m., hair, lock: np. loccas 4i104; ap. loccas 4i98. See \vundenlocc. l(>eced, see h'witlocoed. lof, mn., praise: gs. lofes 2I11. lond, n. i. dry land, shore: ds. lande 2312, Ionde342. — 2. ground, earth: ds. Ionde411-64, 578. — 3. estate: as. 13", I411. — 4. district, province: gp. londa 3418. See €g-, mearclond. londbuend, m., earth-dweller: gp. lond- buendra 9511. long, adj. i. long (space): asf. lange 598 ; comp. nsf. lengre 247. — 2. long (time): nsn. 4O22; asf. longe 299. See uplong. longe, adv., long, a long time: i629, 4i8, 68i8. losian, Wa, depart, escape : 3 sg. losatS I38; inf. 3". lucan, see bl-, onlucan. lufe, f., love: gs. lufan 27125. liiiian, W2, love : 3 pi. lufiab 957. lust, m.,joy, pleasure: as. 72°. Ij'ft, f., air, sky: ns. 4", 84, n9, 581; gs. lyfte 464; ds. lyfte 2316, 4i81, 52, 578, 5912, 8430, lifte 28. lyftfaet, n., air-vessel: as. 3O8. lyt, adv., little: 6i7. lytel, adj., little, small: nsm. lytel 721; nsm. wk. Iytla4i76; asn. 597; apf. lytle 581. See unlytel. M M = rune PI : 2O6. ma, n., more: np. 19, 6i16; ap. 2721. maecg, m., man: np. maecgas 5i7. See Coredmaecg. iii:r<l;i M, see gemaedan. maeg, f., woman, kinswoman: ns. IO9, 32. rnaeg, m., kinsman, brother : np. magas 8818. magan, PP, may, can, be able : i sg. mffig 310, l619, I91, 4162,64,66, 4;J6 S67, 6410, 8S88; 3 sg. maeg 328, 4ii6.20,62,69. 90, 442, 593, 6o12, 846'16?; i pi. magon 42 ; I (?) pi. mag[on] 6818 ; 3 pi. ma- gon 8442 ; opt. 2 sg. maege 4O28 ; opt. 3 sg. maege 22, 512, 3219; pret. i sg. meahte 611, 9319; pret. 3 sg. meahte 268 RIDDLES OF THE EXETER BOOK io10, 3O6, 4i48'67; pret. 3 pi. meahton 236. maegburg, f ., family : ns. 2i20 ; as. maeg- burge I620. mage, f., kinswoman : ns. mege io4, maege gs. magan 44' maege, see mage. niaegen, n. i. might, strength, power: ns. 8423; ds. maegene 4I95; as. 549, 83" ; is. maegene 2814, 8420, maegne 2413, 3228- — 2. force, host, troop : ns. S48'65, maegn 2313. See holmmaegen. maegenrof, adj., very strong: nsm. wk. maegenrofa 383. maegenstrong, adj., strong in power, mighty : nsm. 873. iiiH'geiiiftise, f ., violence, force : ds. maegenHsan 2810. magorinc, m., youth, warrior: np. magorincas 235. maegff, f ., virgin, maiden : np. maegetS 5i7; gp. maeg-Sa 158, 349. I, n., time, occasion : gp. miEla 826. maeldan, see meldan. man, see nion. ma' nan, Wl. i. r&/<r, /<?// 0/": 3 sg. maeneS 2I11 ; pret. opt. 3 pi. mienden 6l17. — 2. mean, signify : I sg. msEne 62. .SV^ geiiia>nan. mandrinc, m., evil drink, drink of death : as. 2413. mania n, see gemanian. mara, see micel. ma- ran, Wl, make known, celebrate: opt. 3 pi. (sg. form) maere 2716. maere, ^iA].,famo^(s, glorious, renowned: nsm. 2727, 84"; nsm(f). 4i45: gpf. maerra 84 ; dpm. maeran 8818. ina-rffu, f., glorious deed: ap. maeriSa 7311- maest, see micel. maeftel, n., assembly : ds. maetSle 862. inaAcIiaii, Wa, speak: pret. 3 sg. mabelade 395. inaSm, m., treasure : as. 5618. mi5w,m., sea-mew, gull: gs. maswes 256. meaht, f., might, power: ns. 8428; gp. [meahta] 8411 ; ip. meahtum 210, 466, I48, 4I90. meahtelice, adv., mightily: comp. meahtelicor 4i62. meahtig, adj., mighty, powerful: nsm. 41^. mearc, f., mark, region: as. I56. mearclond, n., waste-land, sea-coast: ds. mearclonde 423. mearopfeS, m., country path : ap. mearc- pa>as 7211. medan, see onmedan. med\vis, adj., not wise, foolish : dsm. medwlsum 510. mege, see mage, meldan, Wl, declare, announce: inf. 2912; maeldan ig2. meldian, Wa, declare, announce : pret. i sg. meldade 7216. mengo, f., multitude, crowd: ds. 2I12; as. 843. mennen, see druncmennen. meodu, m., mead: as. 2i12. ? meodubenCjf., mead-bench : ds. meodu- [bence] 6i9. meotud, m., Creator, Lord: ns. 4s, 8817 ; gs. meotudes 84". ineowle, f., maid, woman : ns. 56, 267, 621. mere, m., sea : as. 235. See h\vaelmere. merefaro<5, m., sea-waves, surge of the sea: ds. merefaroj>e 6i2. merehengest, m., sea-horse, ship : ns. '56- merestream, sea-stream, sea : ap. mere- streamas 679. mesan, Wl, eat: inf. 4i62. -met, see gemet. [metelist, f ., want of food : is. metellste 1 16.] micel, adj., great, much : nsm. 450, 87s, wk. micla 4I92 ; nsn. 2912, 3223 ; asf. micle 871; asn. 38s, 4i76; isn. micle 445-61, (adv.?) 40, 4i42-74.80, [micle] 4i23; ip. miclum 4O2; ? micle 84'' GLOSSARY comp. nsm. mara 4i92'105; comp. nsf. mare i84, 67; comp. asm. maran 40 ; sup. nsm. mJEst 439, ? miist 8412. niiclan, see gemiclan. miclian, see gemiclian. mid, prep. i. with (association), w. dat. 66, i69.i", 3I1, 4o2, 416", 43M, 471, 74».4, mith Leid. 12. — 2. with, by means of (manner), w. dat. 612, 2713, 28, 3i2, 3223, 4118.14,80^ 456f 5j7t 5S12> 636, 64s, 6710 ; w. inst. 292.2-8. mid, adv., with, at same time : I42, 2318, 476- middangeard, m., earth : ns. 32!, 33!, 4 143, 671 ; gs. middangeardes 83" ; as. 4019, 4 112, 678. midde, f., the middle (in phrase on middan): ds. middan 339, middum 8 15. middelniht, f., midnight: ip. middel- nihtum gi7. mldwist', f., presence, society : as. 958. milts, f., reverent joy : is. miltse 3i8a (6 ip. miltsum). mm, pron., my: nsm. [i13], 311, 41, 75, I61, i?8, i89, 223-15, 241, 26, 2727, 85. 8813,23.26> 919, 931,14. nsf. ^ ^.lO, 72, So8'10; nsn. 81, n1, 12\ 221-10, 831, 8821, 91 ! ; gsm. mines [i9], 4™, 738, 9i6; gsf. minre iS1-5, 4I45; gsn. mines 19, 2610; dsm. minum 51'9, 2I2'26, 4I96, 57", 6i2, 7i6, So2; dsf. minre 2810 ; dsn. minum, 736 ; ? minum 782 ; asm. mlnne I58, 6i4, 83"; asf. mine 94, I620, 2 112, 251, 664, 73s.28, 8 112, 9320, 958'13; asn. 53-11, 226, 268, 668, ' 745, 837, 93s6; isf. minre 911, i518, 4I80; isn. mine II5; [vsm. I13] ; npf. mine 84'6, n8; npn. io7, 41"; dpf. minum [i1], i6u; apm. mine i612, 9511; ipm. minum 2i21; ? min 7i9, 885. mislic, adj., various, diverse: nsn. mislice, adv., in various ways : 2912. missenlic, adj., various, diverse: ipf. missenlicum 321, 33!. missenllce, adv., in various ways : 6815. mittan, see gemittan. iniAan, i. i. conceal: inf. mij>an 83 12. — 2. avoid, refrain from : \ sg. mif>e 94 ; inf. mij>an 64 10. See bemuJan. mod, n., mind, heart, spirit: [ns. I15]; gs. modes 2814 ; is. mode 1 26, 84s4, 862; ap. 75; ip. modum 6o'2. See forhtmod, heanmod. modig, adj., brave, high-spirited: npm. modge 3 18 b (a monige). modor, f., mother : ns. io2, 349, 844, mSddor 422, 8420 ; gs. 4i45, mSddor 4414- mod'iVrf'a, m., torment of mind, terror ; ns. modbrea 450. mod\vlonc, adj., haughty: nsf. 26T. modwyn, f., hearfs joy, property : ns. (MS. modF") 9i7. mon, m., man: ns. [i1-18], 3611, 39s, 4i47, 4414, S435, man 38s; gs. monnes 3711, 6o13; ds. men 510, menn 29 13; as. monn 374, NOM = mon 2O6; vs. 318; np. men 3!, i8u, 4O4, 55", 957, menn 6815; gp. monna 450, 231, 6i4, 7216, 774, 8312, 9513; dp. monnum ig2, 3 18 a (^mongum), 4o12, 4i45; ap. men 13, 6o2. See rynemon. inona, m., moon : ns. 672. monoynn, n., mankind, men : ds. mon- cynne 339, 4O2, 4I27. mondryhten, m., lord: ds. mondryhtne 5613, [monjdryhtne 596. monig, adj., many : npm. monige 666, 862, monige 3I8« (b modge); gpm. monigra 76 ; gpf. monigra 844 ; gpn. monigra 422; dpm. monigum 958, mongum 4O19, mongum 3i8 b (a mon- num) ; ipf. monegum 596, mongum 91. monna, m., man : as. monnan 66s. mor, m., moor, waste land: ap. moras 72". mos, n.,food: ds. mos[e] 788. -mot, see gemot. motan, anv., may, must: I sg. mot 4i5,73f i62o> 2I27) 836; 3 sg. mot 4O20 ; 2/0 RIDDLES OF THE EXETER BOOK 3 pi. 4 1108, moton i79; opt. i sg. mote 2 122; opt. 3 sg. mote 3218; pret. i sg. moste 4I85-100; pret. 3 sg. moste 5413- moJRJe, f., moth : ns. 481. inn na M, see gemunan. mund, see fettemund. mundbora, m., protector, guardian : ns. mundrof, adj., strong of hand: nsm. 87". [nitiriiaii, Wl, mourn, lament: ptc.nsn. murnende i15.] See bemurnan. muff, m., mouth : ns. 339 ; as. 4O12, 688, 77, muj> 91, iS11, i92; is. muj>e 25°, 64; ip. mubum 14. muoleas, adj., mouthless: nsm. 6i9. my It a n, see gemyltan. -mynd, see gemynd. myrS, see geoguSmyrfl. N N = rune \ : 2O6, 752. na, adv., no, not: Leid. 13, 379. nabban = ne habban, \T3, not have, be without : pret. 3 sg. naefde 335. naoa, m., boat, ship : ns. 595. mefre,adv.,««w: [i18], 610, 4O7-20, 7216, 8830. nagan = ne agan, PP, not have : I sg. nah 46 ; 3 sg. nah 281. na-gan, see geniegan. naegledbord, adj., with nailed planks : nsm. 595. naegl(i)an, Wl,2, nail, rivet: pp. asm. nffigledne 2O6. nales, adv., not at all, by no means : [ 1 15], 27". nama, m., name: ns. 2727, noma 241; ds. naman 59" ; as. naman 5611, 6o8 ; ap. naman 438. nan, adj., not one, none: asm. niEnne 688. nanig, pron., not any, none: nsm. 3O13, 84®; dsm. njgngum 262; asm. nienigne 59s. nard, m., spikenard: gs. nardes 4I29. nees, see bearonaes. n.i'-f a n, see geneestan. ntetan, W l , afflict, distress : i sg. na!te 7. iiatlnvHT, adv., (nescio quo), in some unknown place, somewhere : 26s, 638. nath \va-t, pron., (nescio quid), some- thing unknown : nom. 629, 9325 ; ace. 46', 555- ne, adv., not: 31-10, 415-53, 64, 88, (58 times): ni Leid. 3,5,9. ne, conj., nor, neither: 2i11>2°, 2313, (34 times); ni Leid. 5,6,8. iu"ah, prep. w. dat., near : 423,. 578, 6iJ; comp. (adj. or adv.) near 4M. neahbiicnd, m., neighbor: dp. neah- bQndum (MS. -buendum) 262. nearo, adj., narrow, strait: asf. nearwe i62; ip. nearwum 533. nearo, f ., confinement, durance : ds. nearwe n1, nearowe 5418; as. 626, 638. nearograp, f ., close grasp : ns. 846. nearwian, Ws, compress, confine : 3 sg. neat-waft 261Q. See genearwlan. neb, n., beak, face : ns.'i i1, 221, 326, nebb 35s; as. nebb 8i4; is. nebbe gi8. See saloneb. nefa, m., nephew : ns. 476. ill-Han, see willan. nemnan, Wl, name: i pi. nemna'S 4i73; 3 pi. nemnaft 257; imp. pres. 2 pi. nemnaft 586 ; pret. 3 sg. nemde 6o6 ; inf. so9. neol, adj., prone, low, deep down: nsf. 22. 846. neoan n1, 265, 3220; nio^an 626. nergan, Wl, save: inf. i618; ptc. asm. nergende 6o4. See genergan. n63an, Wl, venture, dare : 3 sg. neheft 265 ; inf. 5413. niht , f ., night : ns. 3O13 ; as. 4O7 ; ip. nihtum 6U, I39, 8816. See middel- iiiht. iiiinaii, 4, take, draw : opt. 3 pi. ni[maen] Leid. 14. See biniiiian. GLOSSARY 271 nltf, m., trouble, affliction : ds. nij>e 7. nifferweard, adj., down-ward: nsn. ni)>er\veard 221, 326, 358. nfffsceatFa, m., malignant enemy: ns. nitSscea^a i624. iiicViVas, m., pi. men: gp. ni)>}>a 586; dp. niwian, see geniwlan. no, adv., net, no: 7, 29!°, 324-8; 4O9, 9318, 958. 11 oiiia . see 11:1111:1. nowiht, n., nothing: ace. I25. nil, adv., now: 15!, 259, 2716, 286, 4I1-102, 4315- 548, 56", 6813, 7Is, 738, 7?4, 834, 8818-32, 92, 9322-26, 957-i°. nu g6n, adv., further, yet : 5O8. nyd, f., name of rune N: 438. ^(fhaeft- nyd. iiydan, Wl, urge, press: 3 sg. nyde)> 63s- . nyde, adv., of necessity: 4I29. nymtfe, conj., unless, except: 427, nymbe 2 122, 2418, 26s, 4i21, (MS. nympj>e) 665. nyt, f., use: ds. nytte 2727, , 35', 5o9, 5 12, 706. nyt, adj., useful: nsm. 339, 557; nsf. 262, 596; gsf. nyttre I25; npn. 5611. nyttung, see wuldornyttung. O O = rune p : 2O1-5-7'8, 258. of, prep. w. dat., of, out of, from : 318, 47.12,16,47,48, II6.10) I36, i515, r612, Ig6) 227, 23^, 248'12, 282-2-8.8, 30, 362, 4i79, 5 12, 637, 734-5-28, 776, 838, gi10, 9312'1- l7.28 ; ob Leid. 2, 14. ofer, prep. A. w. dat., <7z/ 8i6. — B. w. ace. i . over, above, upon : 462, 710, gS.6, „!!, ,56.7, 2,8, 236,12,18, 279, 307, 33s- 455. 527> 547, S&> W' W-- 2. throughout: 36", 4I21, 425, 84", 8821, 9510. — 3. contrary to: 3O10. ofer, m., bank, shore : np. ofras 237. ofergongan, anv., come upon (sleep) : 3 sg. ofergongej> 4I1". oferstigan, 1, surmount, rise above: i sg. oferstlge 67°. ofers \vSffan, Wl, overpower, overcome: i sg. oferswi)>e 4I29; inf. oferswl)>an 4I20. ofest, f., haste: ds. ofeste 63; ip. ofestum 41 u. ofgifan, 5, abandon : pret. I sg. [o]fgeaf 8811; pret. 3 pi. ofgeafun lo1. oft, adv., often : 55, 6s, 72, 17!, i88, 2i8- 15,32; 3I6, 3311, 457, 502.7, 5j4, ^10, 5511, 5612, 59n, 621,641, 6810'16, 726-w, 778, 781, 8o9, 84s9.47, 88™-™, gi8, 93^, _952- ohivonan, adv., from anywhere : 368 {Leid. 6u[ua]n[a]). on, prep. A. w. dat. or instr. i. on, Upon: [I-], 27'12.1, 59, I22 JC12, 144,11, j 62,3,4,26,26, 2Q4, 225'8'9'10'12, 26L14, 278, 3214-20, 358, 37L6, 4I26,77,102. i°8, 436, 5 1 1-9, 592, 70, 7212-!8, 73L22, 8o7'8, SS7'22-28-24, 9320, (MS. of) 9312. — 2. '«, within: 45i, 6U, 97, lo1, ill-8'7, 1311, i6i6, I94, 2ii°'i8, 23i4-i6, 284, 305, 328,4,11,17, 341243,4! 61,81.81,106, 46!, 541.2.5, 578, 592i, 62i-8, 634, 644-6, 6s2, 66s, 67, 681, 698, 7313, 748, 8o6-6, 8i5, 862, 92i-4. — 3. at, in (manner): 21™, 2818, 4I28,28, ««, 61", 64", 9310. — 4. during: 3™, lo1, 207, 2i8i, 4i87, 446-10, 528. — B. w. ace. i. upon, in: 22.n, 37, 48.21,28.80.86, l62l, ail-26, 226,13, 239.2), 242, 267, 274.1°, 2816, 3oi2, 4o6, 463, S62, 57^, 691, 728, 7321. 742'5. 9329. — 2. into, to: [i2-7], 46,35, 2 1 H, 622-6, 636, 664, 876, 9322.— 3. according to : 394, 4i8, 73?. — 4. for, as: 392, 518. — C. after or separated from case : 4^, 77, 2I29, 636, 8o4, 88". on, adv., on, upon : 874. onbugan, 2. i. bend: I sg. onbuge 248. — 2. bend aside, escape : inf. 415. oncweftan, 5, answer, respond: i sg. oncwej^ 57. ond, conj., and. All occurrences are represented in the MS. by the abbre- viation. 272 RIDDLES OF THE EXETER BOOK ondfenga, m., receiver : gs. ondfengan 62". ondredan, R, fear: 3 sg. ondr£de)> ondswaru, f ., answer, reply : as. ond- sware 5615. onettan, Wl, hasten, bestir oneself: pret. 3 sg. onette (MS. onette'S) 30", onnette 557. onflndan, 3, find out, discover : 3 sg. onfindeS i67, 289. oiiga, m., arrow : ns. 24. ongean, prep. w. dat., opposite to, against: 778, 91. . ongean, adv., opposite : 289. ongietan, 5, perceive, understand: opt. 3 pi. ongietan 496 ; inf. 6o10. onginnan, 3, begin : \ sg. onginne i87; 3 sg. onginneft 29", 329; pret. 3 sg. ongon io3, 5510; pret. 3 pi. ongunnon f. onha>lo, adj., hidden: asf. i67. onhebban, 6, raise, exalt: I sg. on- haebbe 3i7. onhlldan, l, open : imp. 2 sg. onhlid onhnigan, 1, bend down, bow, incline: 3 sg. onhnlgab 3i7£ (a onhingaj>). onhwyrfan, Wl. i. turn, change : pret. 3 pi. onhwyrfdon 73. - a. invert: pp. onhwyrfe'8 241. onhyrgan, Wl, imitate: i sg. onhyrge 910, 254. onlicnes, f., likeness : as. onllcnesse 4 187. onluoan, 2, unlock, open : pret. 3 sg. onleac 4312. onmedan, Wl, presume, take upon one- self: opt. 3 sg. onmede 5616. onoegan, W\,fear: i sg. oncegu na (MS. oncegun) Leid. 13. on sit tan, 5, fear, dread: inf. I628. onsundran, adv., apart, separately 72«. ontynan, Wl, open : pret. I sg. ontynde 77- on^eon, l, 3, succeed, prosper, prevail: pret. opt. 3 pi. onjmngan 8831 ; inf. onj>eon 642. ? oiuYuniaii, Wa, swell out, exceed bounds : inf. on)>unian (MS. onrinnan) 4i9i. on-\vald, m., power: is. onwalde 4i13. on\vendan, Wl, turn, change : pret. 3 Pi- 735- openian, see geopenian. or, TL., beginning, origin: ns. 8410; as. 459. ord, n. point: ns. 6i12-13; is. orde 776; ip. ordum i88, (toes') I65. ordstapn, f., prick of spear (goad) : np. ordstaepe 7217. orlege, n., strife, battle : gs. orleges 459. orlegfrom, adj., strong in battle: asm. orlegfromne 2i15. or'Sonc, mn., understanding, skill, art: as. orj>onc 787 ; ip. or^oncum, skill- fully, ingeniously, 7O3. orSonebend, f., skillfully contrived bond: ip. orj>oncbendum 4315. or'(V<>iicpiI, n., cunning spear ( = share) : ns. or)>oncpil 2212. offberan, 4, bear forth : pret. 3 sg. oftbaer 2310. 53er, pron., other, another : nsm. 6|jer 439 ; ot>er . . . (the one . . . the other) 577; nsf. 6>er 4I86; nsn. 6>er 2212 ; gsn. 5)>res 79 ; dsm. 6)>rum 441, 2 1 is, 386, 44", S35, 545-10, 846; dsf. [oj>erre i4], 5}>re 2210 ; asm. 6J>erne 232°; asf. 6>re 4O7; dpm. 5)>rum 12, 927?; apn. obre 5O5; ? 5J>er 8418. <)«Jfe 215' 478'74, 82.2^ 4I24,48,67,67,75) 6I8) 7310) Q^ Oi5^ 56- ooTFringan, 3, snatch away : inf. o"S- )>ringan 8819. owiht, adv., aught, in any way: 42. oxa, m., ox : ns. 2318. GLOSSARY P = rune R : 656. -pad, see salopad. pa'«V, see gegn-, mearcpaeS. pa'VTSan, Wl, tread, traverse : 3 sg. pae)>e$ 599; pret. i sg. pae'Sde 7211. pcrnex, m., = Lat. pernix, adj., swift (mistaken for name of a bird) : ns. pfl, see hilde-, or3onc-, seuropil. plega, see hyhtplega. plegan, ~W\,play, sport: inf. 43a. pyt, see radpyt. R R = rune ^ : 2O1, 258. ra;can, Wl, reach, extend: i sg. race 677. See geraecan. raeced, n., hall, building: ds. raecede 328 ; as. 531 ; ap. 26. rad, f. i. riding, course: ds. rade 2O7. — 2. name of rune R : 2O5. raed, m., counsel, advice: ns. i616; gs. rades 88s5. See unraed. ragdan, R, read (a riddle), explain : opt. 3 sg. rzide 6ols; imp. 2 sg. raid 62'. raedelle, f . , riddle, enigma : as. raedellan _4318- radpyt, m., draw-well with sweep : rad- [PYT?] 591-15. radwCrig, adj., weary of riding, weary of journeying: asm. radwerigne 2i14. raeping, m., captive: ap. raEpingas 53!. ra-ran, Wl, raise : opt. 3 sg. riere 478; pret. 3 sg. rierde 566. See araeran. ra>san, Wl, rush : 3 sg. rSse^S 268. See Surhra-san. read, adj., red: nsm. \vk. reada 2715; gsn. wk. readan 49®; npf. reade I22. reade, adv., red: yi1. r6af, n., robe, garment : ds. reafe 1 22 ; is. reafe I47. reafian, Wa, plunder, rob, despoil: I sg. reafige 26, 13"; 3 sg. reafa~5 268, 662. rec, m., smoke, reek : np. recas 26. reccan, Wl, care, reck : w. gen. 3 sg. recce's 775. reccan, Wl. i. rule, direct, guide: \ sg. recce 4I88; inf. 4I35. — 2. explain, interpret: imp. sg. rece 3318. reccend, m., ruler (God): ns. 4i3. recene, adv., quickly, straightway: 4O28. regn, m., rain : as. 4s6. regnwyrm, m., earthworm : ns. 4i70. -r6n, see ger€n. [rgnlg, adj., rainy: nsn. i10.] r6od, adj., red: asm. reodne 268. rf'ofnn, see bireofan. reord, f ., speech, voice, tone : as. reorde 255 ; ip. reordum g1. See gereord. [reotlg, adj., weeping: nsf. reotugu i10.] resele, f., riddle : as. reselan 4028. restan, Wl, rest, rest oneself : \ sg. reste 8s5, 952 ; inf. 478. retan, see aretan. r65e, adj., fierce, cruel: nsm. re)>e 28, 842; gsm. re^es i616. rib, n., rib : gp. ribba 338. rice, adj., rich, powerful: nsm. 4i8; gsm. rices 7 11; npm. 3313; dpm. ricum 952. rice, n., authority, master : is. 431. ricels, n., incense: ns. 4i24. ridan, l, ride: I sg. ride 8o7; 3 sg. ride1? 4s6. 593; pret. I sg. rad 9312; inf. 4s2, 232. rlht, see ryht. rim, see daeg-, unrim. rlnc, m., man : ns. 63, 6416, 742 ; dp. rincum436; ap. rincas I516. Seefyr<-, gum-, magorinc. rinnan, 3, run : inf. (MS. yman) 855. risan, see arisan. rod, f., cross : gs. rode 5O5. rodor, m., heavens, sky : gp. rodera 6o15, rodra I47; dp. roderum 56s. rof, adj., strong: asm. rofne 20"; npf. rofe (MS. rope) 588. See ellrn-, maegen-, mundrof. rose, f., rose: ns. 4i24. ruh, adj., rough, hairy: nsm. z(P ; gsn. ruwes 629. 274 RIDDLES OF THE EXETER BOOK nun, see gerum. run, see heterun. runstsef, m., runic letter: np. runstafas 5915 ; ap. runstafas 436. -ryde, see geryde. ryht, adj . i . straight, direct : asm. rihtne 63. — 2. right, true-, isn. ryhte 51?; npm. ryhte 5915. ryht, n., right: as. 4i3; is. ryhte 4I85. See geryht. ryman, Wl, clear (way), open : 3 sg. rymeft S410. See geryman. ryne, m., course : as. 842. ryne, n., mystery, mysterious saying: as. 496. ryneglest, m., rain-foe : gs. rynegiestes 4M- rynemon, m., one skilled in mysteries : ap. rynemenn 4313. rynestrong, adj ., strong in course : nsm. S = rune >\ : before and after 7, 2O1, 656. see, mf., sea, ocean : ns. 4s29, 77 ; gs. or ap. SJES 673. sacan, 6, fight, contend: \ sg. saecce I72; 3 pi. saca«6810. sum, f., strife, battle: gs. saecce 429; ds. sace 2i6; as. saecce 8S29. sagol, m., rod, staff: as. sag[ol] 8i6. slegrund, m., depth of sea, bottom of sea : ap. saegrundas 310. sael, n., hall: gs. sales 532. See burg-, folc-, hornsael. sail, m., time, opportunity : gs. sieles 32». sa-lan, see tosa-lan. sailed, see searosaeled. sielig, see gesaellg. salo, adj., dark, dusky: nsm. So11. saloneb, adj., dark-faced: nsm. 5O5. salopad, adj., dark-coated: npf. salo- pade 588. sa-hvong, m., fertile plain : ds. siel- wonge 42 ; as. 20. s»np, adj., slow, sluggish : nsf. 346. sang, see song. sar, adj., sore: comp. nsf. sarre i46. sare, adv., sorely : 7215. sawaii, R, sow : 3 sg. sawe> 226. Scfvvcall, m., sea-wall, shore : ds. sie- wealle 6iJ. sa^vel, f., sou!: gs. sawle S835; as. sawle 4O16. sceacan, 6, shake, depart, fly : pret. 3 sg. sc5cr93n ; inf. 2i14. -sceaft, see ge-, un-, \%-onsceaft. sceam,™., -white horse : ap. sceamas 234- . . scearp, adj., sharp : nsm. 441, 63! ; asm. [scjearpne 938 ; npf. scearpe 34 ; apf. 70 ; ipn. scearpum 452 ; sup. isn. scearpestan 292. See heoruscearp. sceat, m. i . region, part (of earth} : as. 42s ; gp. sceata 8S27 ; ap. sceatas 6816. — 2. lap, bosom : ds. sceate 10", 452. sceatfa, see feond-, nrSsceatJa. scea\vendwise, f., song of jesters: ap. sceawendvvTsan 99. seeawian, \2, look at, behold : i sg. sceawige 4i40; inf. 6O2. seelfan, 4, shake, quiver : 3 sg. scelfaeft Leid. 7 (scribed 367). sceop, see a?fensceop. sceor, m., cloud: ns. (MS. sceo) 441. See scur. sceorp, see f>Td-, hleosceorp. soeotan, 2, spring, rush : inf. 39. sceran, 4, cut, shear : 3 sg. scire> 663. sceoHfan, 6, hurt, injure: I sg. sceH>e 26s ; 3 sg. scefte'S 4411 ; pret. i sg. scod 2i15; pret. 3 sg. scod 72"; inf. 44". scildan, \Vl, shield, protect : pret. 3 pi. scildon 8817. scin, n., specter, phantom : np. 452. sci nan, 1, shine: inf. 4i103. See be- scinan. scip, n., ship : ns. 59. scir, adj., bright, clear : nsm. 732° ; asm. scirne 59 ; npf. scire 1 22 ; apm. scire GLOSSARY scirenlgr ((sciernlcge), f., mime, female jester : ns. 99. scotV'n, Wa, shoot: 3 pi. Scotia's 461. scraef, see wraftseraef. scrFSan, ], move, glide, stall-: 3 sg. scrlJje'S 367 (Leid. scelfae'S) ; ptc. npn. scrlj>ende 482. srufan, see ascufan. sculan, anv., shall, nmst, have to : I sg. SCCal 417.34,65,68, jl, I59.14,17) ! 612.17, ,71,7, 2126,30, 3I8, 4,91t 641, 7I7, 83", 8S24, 91, 955'8; 3 sg. sceal 2$", 336, 3412, 36", 385, 408'16-21,.438. 445, 85. 8827; ? sceal 826; 3 pi. sculon 8819; opt. 3 sg. scyle 431 ; pret. I sg. sceolde 618.14. pret. 3sg. sceolde 628, 736, 937; pret. 3 pi. sceoldon I46. scur, m., shower, storm : dp. scurum 8817. See sceor. seyldru, see gescyldru. scyppan, 6, create, destine, prepare : pret. 3 sg. scop 852; pp. sceapen zi\ 242. See gescyppan. soyppend, m., creator (God) : ns. 4I1-101. scyrian, see bescyrlan. se, s6o, ZFaet. i. dem. pron., def. art., the, this, that : nsm. se [i11], 4s9, i624, 178, 246, 2715, 361 (so Leid.), 41 1.21.64. 68.74.92,96, ^9, 442,8,8,15,16, ^ 4g3, 498, 5o4.io, 54U.H, 569,io,i6( 57s, 702, g i a, 8810; nsf. seo io9, 2913, 3219, 3412, 396, 401'14, 429, SlO 2l20, 33", 6l6.12, 8420, [sio] 322 ; nsn. t>jet [i5], 57, i616, 26™, 379, 4024, 422-7, 44", 481, 6ii°, 84H-82; gsm. J>aes I27, 2I28, 567, 6ou, 627; gsf. 30 ; gsn. 7-, 1 1 I74,5i 2I35, 2410, 341°, 4j72, 423,3,4,4,7, 434.n, 55io, 6o9, 656, 9i9; dsm. J>am 2I23, 30, 386, 446; dsf. J>Ere 3O6, 576, 6o12, 734 ; dsn. J>am 47, 30, 8S3 ; asm. J>one 21, 24™, 25, 9318; asf. >a. 480, 3o9, 381. 391. 4318, 691, 9318 ; asn. J>set [i"], 22, 485, 57, I78, jgS, 248-6, 289, 352, 458, 468, 482, 508, 68i«, 729; is. >y I45'6, l84, 208, 409, 486, 641°, np. J>a 25!°, 2715-16, 427, 4312-16 ; gp. f>ara 438, 475, 535, 666, 848'i5'55 ; dp. J>am 1 72, 5°4, 579, 73128 ; ap. )>a 215, 82, 231°, 356- 7, fla 463; ip. J>am 48°. — 2. rel. pron., who, which: nsm. se 413, 215-29, 246, 411,8,22,90, ,j02, ij6is, 635, 835, 8828; nsf. seo 3o8, 352, 372, 536, 68i9, sio 325, 4i8i; nsn. J>aet 22i5, 34!°, 4i32,69, 56i2, 6i5, 7326, gi3, 9324 ; gsm. }>32s (attrac- tion) 565; dsm. ham 442; asm. K>ne 41 » 5^ ^sn. 1't.vt 24 > 45' p3£t ^^ double relative, ' id quod,' 212, 43<5.65, i77, 18", 24", so10, 55", 8o6; np. >>a 2728, 582, 738; gp. }>ara 5915, 72°; ap. )>a 77, 5O8. See s6 3e, 3a?s, Sees 8e. sealt, n., salt : ns. 948. searo, n. i. art, skill: ip. searwum (skillfully, cunningly) 30°, 578, 8448. — 2. work of art: as. 338. searobunden, adj., cunningly., fastened: asn. 564. searocCap, m., curious thing, work of art: ns. 337. searocraeftig, adj., cunning, wily : nsf. 348- searolie, adj., ingenious, wonderful: nsm. 6iH. searopil, n., dart cleverly made : gp. searopila gi'2. searosaeled, adj., cunningly bound ': nsf. 2416. searoSone, m., cunning thought, skillful device : ip. searoj>oncum 3618. searotkmcol, adj., sagacious, wise: npm. searoj>oncle 4i97. sCa\r, n., juice, sap : ap. 447. seax, n., knife: gs. seaxes 276, 6ii2, 776, is. seaxe 4i97. sScan, Wi. i. seek, look for: 3 sg. sece)> I625, sece'S 358, se[ce'S] 88s4; 3 pi. secai? 9512; pret. 3 sg. sohte 938; inf. 939. — 2. visit, goto: inf. 32, I72, 28U. See gesecan. secg, m., man : ns. 58, 639 ; npm. sec- gas 4i97 ; gpm. secga (MS. secgan) 64! ; dpm. secgum 494. See garsecg. 2/6 RIDDLES OF THE EXETER BOOK »ecgan,W3, say, tell, declare : 3 pi. secgaft 401-13 ; opt. 3 sg. secge 6818 ; imp. 2 sg. saga 2", 312, 472, 98, ii11, i313, 2O9, 2416, 3618, 378, 4029, 639, 6710, 7329, So11, 83", 867; pret. 3 sg. saegde 348; inf. 436, 568'16; ger. secganne 4O22. Set gesecgan. sefa, m., mind: ds. sefan 6iu. segnberend, m., standard-bearer, war- rior: gp. segnberendra 4I20. selda, -r«v geselda. [seldcyme, m., rare visit : np. seldcymas I".] sele, m., /4a//, house : ns. 8s1 ; gs. seles 14; ds. 2 110. soIodrPain, m., joy in hall: ds. sele- dreame 641. seilan, see syllan. Hellic, adj., strange, wonderful, excellent: nsf. S428, selllcu 335 ; asn. 328, 33'. selra, adj., comp. and sup. only, better : comp. apm. sellan 13; sup. gsn. wk. selestan 42. Bemninga, adv., suddenly : 4i10. sendan, Wl, send: 3 sg. sendees 42, 5O6; 3 pi. sendaft 3i5; pp. sended 2U. [seoc, adj., sick: asf. seoce I14.] seolfor, n., silver : gs. seolfres 56 ; is. seolfre 2i10, 6818; is. sylfre 152. seolhbaeft, n., seaPs bath, sea : ap. seolhba)>o ii11. seomian, Wa, rest, lie : 3 sg. seomaft 218. seon, 5, see, behold: i sg. seo 63; opt. I sg. sy (w. gen.) 4I65; pret. i sg. seah 141, 2oS 32. 338, 431, 52', 531, 541, 561, 601, 651, 87! ; pret. opt. 3 sg. sawe 8481. See geseon. set tan, Vf\, place, set: pret. 3 sg. sette 27, 4 17. See a-, gesettan. se «5e, pron., who, which : nsm. se J>e 311, 28'. 3?, 4 198'98, 4414, 606'15, 6818, 7i«, )>am he I629, 6iu, 7O1; np. t 3610 (Leid. fla «i) ; gp. J>5ra 732 ; se ^ = J>one 445; nsf. seo )>e 26™ ; gsm. t>aes . . . ^e 4W; gsn. fces J>e 3215, 33^ 424; dsm. 612, 299, 4015-26, 4i89, 9i10; dp. j>a? I410, 275, 437. s6 56ah = swa Oeah, adv., however, nevertheless, yet : 59, 877 ; hwsebre se beah (Leid. hudrae suae eh) 36"; efne se beah 4027, 661. A^ swa Sean, swa 'Aeana. se eana 8810. See swa Seana. sib, j^ gesib, ungesib. sld, adj., wide, spacious: apm. side 310, 67°. side, f., side: ns. 14®; ds. sidan 776; as. sidan 2213, 7O2; ap. sidan 8i5, 867; np. sidan I61, 7318. siex, num. adj., six : 2510, 378, (MS. vi) I42- sigefaest, adj., victorious: comp. npm. sigefaestran 27". sigel, see heaftosigel. sigor, m., victory, triumph : gp. sigora 71- -sih<5, -see gesihft. sin, pron. i. his: dsm. sinum 6o4; ism. sine 2414; ipm. sinum 9iu, 932. — 2. her: dsm. sinum S914; apm. sine 3222 ; ipf. sinum 623. sine, n., treasure, wealth : ns. 494 ; as. 2 16, s64; is. since 2i10, 6818. sincan, see besincan. sincfag, adj., shining with treasure: nsm. I515. sinder, m., impurity : ip. sindrum 276. singan, 3, sing: i sg. singe 92; 3 sg. singe 7O2; 3 pi. singa 88; inf. 328. sittau, 5, sit: i sg. sitte 25"; 3 sg. site 46, 3212 ; 3 pi. sitta (MS. site) 98; [pret. i sg. saet I10] ; pret. 3 sg. saet 471 ; pret. 3 pi. saston 861 ; inf. 761. See ge-, onsittan. sittende, see burg-, insittende. si(J, adv., afterwards : 6i8. si?J, m.., journey, course, wandering: as. 22, 30", si> 858; is. site 5371; gp. sl)»a GLOSSARY soirenigr p. si^as iou, 4O16 ; ip. sl)>um female :e ford"-, ge-, here-, unraedsIS. scotia», m., journey, course : ns. slj>f aet 209 ; ds. slSfaete 446 ; as. 83". suVian, W2, go, journey, travel: pret. I sg. sit>ade 7210; pret. 3 sg. sTJ?ade 27" ; inf. simian 522. sre 652. si'tVcVan, conj., j«^, after: siH>an I29, I627, 246, 776, 832 ; [si]W>an (adv. ?) 64". siffifau, adv., afterwards: si)>J>an io9, u10, I622, 272.5,11, 286, 3018, 4i9, 77^, 897, 9315 ; si«)>an 626. sixtig, num. adj., sixty: (MS. LX) 231. sleep, m., sleep : ns. 4i10. slaepan, R., sleep : opt. pret. I sg. slepe 4i9. slaepwerig, adj., sleep-weary: asm. slzepwerigne 55. siege, see deaSslege. slitan, l, tear, rend: I sg. sllte 13; 3 pi. slltaS 8S32; inf. 148; ptc. npm. slitende 1 76. See toslitan. sliSc, adj., dire, hard, dangerous: gsf. slij>re 429. slupan, 2, slip, glide : inf. 4s9. smael, adj., slender : nsm. 7318. smeah, adj., subtle?: comp. nsf. smeare 94«. sniiff, m., smith : ? smij> 94! ; gp. smij>a 68, 217> 27H snaegl, m., snail: ns. 4i70. snaw, m., snow: ns. 8i10. snel, adj., quick, swift: comp. nsm. snelra 4i70. snuVaii, i, cut: pret. 3 sg. snai? 27. snottor, adj., wise, sagacious: nsm. 84s ; npm. snottre 862, 95?. snyttro,f.,#rudetice, wisdom: snytt[ro] 6S3. sny?fian, Wa, hasten, go as a dog ivit/i nose to grotind (Sw., B.-T.) : I sg. snyHge 226. snyTOan, see besnytJfian. soden, see unsoden. -sum, see gesom. some, adv. (always in combination, swa some, likewise, as well) : I62, 4311- somne, see tosomne. MIIII nia u, see gesomnian. somod, adv., together, in company : 214, 172, 239, 6i13; samed 522 ; [somod] 201. sona, adv., soon, immediately : i76, 26°, 287-9, 6^. sond, n. i. sand: ds. sonde 2U; is. sonde 37. — 2. shore: ds. sonde 6iJ. sond, f ., message : ns. 928. song, m., song: ns. 2S6; gs. sanges 583. so«y, adj., true, sooth : nsm. 4s4, 7! ; nsn. 4025; gpm. s6>ra 2722; ipn. so)>um 4029. Ro«V, n., sooth, triith : as. 3718. so^cwlde, m., true saying: ip. so'5- cwidum 3613. s65e, adv., truly, correctly: [so)>e] 7329. spa-tan, Wl, spit: i sg. spaite i84, 248. sped, f ., success, prosperity : ns. 1 84 ; as. 8S34, [sp]ed 84"; on spgd, suc- cessfully 512. See friSosped. speddropa, m., useful drop : ip. sped- dropum 278. spel, n., answer, solution : as. 512. spere, n. spear: as. SP[ere] 656. See attor-, deaSspere. sperebroga, m., terror of spears : as. sperebrogan i84. splld, m., destruction : is. spilde 248. spor, n., track, spoor: ds. spore 88s4. spowan, R, succeed: pret. 3 sg. speow 43_. spraee, f., speech : ds. spraece 2818. See aerendspriec. sprecan, 5, speak : \ sg. sprece 91, sprice 24", 4416 ; 3 sg. sprece'S 21 ^ spriceiS 2910 ; 3 pi. spreca^S 959 ; pret. 3 sg. sprsec 4O12; inf. ig1, 6i9. Psprengan, see geondsprengan. spyrian, Wa, make a track, go : pret. 3 sg. spyrede 278. 2/8 RIDDLES OF THE EXETER BOOK staef, m., letter: np. stafas 2510. See ar-, run-, wrohtstaef. stfelgiest, m., thievish guest : ns. 485. stan, m., stone, rock: ns. 4i74; is. stane 37; np. stanas 17°; ip. stanum 8443. -i .1 n l.i n. see stondan. stauhliff, n., rocky cliff: np. stanhleo>u 4. staiiwong, m., stony field: ap. stan- wongas 9310. steepe, m., step : ds. 9310. -stapu, see ordstapu. sta'ff, n., bank, shore : ds. staj>e 418 ; ds. stae'Se 2319 ; ap. stabu 36. See wieg- stafiol, m., station, place: ns. staK>l 26, 7 12; ds. sta«ol[e] 887 ; as. sta^ol 485, 8S25. See frum-, wynnstaftol. statfolwong, m., station, field occupied: ds. sta^olwonge 358. staeS5an, Wl, stay (trans.): opt. 3 sg. steal, see bidsteal. stealc, adj., steep: npn. 426; apn. 37, 937. -steald, see hagosteald. — t <•; 1 1 la, see gestealla. steap, adj., high, steep: nsm. yi2; asm. steapne i618, 8i4; npm. steape 410. steaphf-ah, adj., very high: nsm. 26. stede, m., place, station : as. 45. See folc-, wiostede. stcfn, f., voice: as. stefne 251; is. stefne 97, 1 518, 498. stelan, see be-, bi-, forstelan. stellan, see anstellan. steno, m., odor, fragrance : ds. stence 4I28; as. 4i». steort, m., tail: ns. 17 ; ds. steorte 22; as. sg7, 8 1 2. strpan, Wl, exalt: 3 sg. stepeiS 518. steppan, 6, step, go: i sg. steppe I66; 3 sg. steppeS 9328; pret. 3 sg. stop 2710. 55s. 74 : pret. 3 pi. stopan 2318. stician, Wa, stick, thrust: 3 sg. sticah 13", sticaiS 9 18; pret. 3 sg. sticade 625. Btig, f., way, path; as. stige I624. stigan, 1, climb, ascend: i sg. stige 470; 3 pi. stlgatS 26 ; inf. 238, 937. See &-, oferstigan. stillan, see gestlllan. still«>, adj., still, quiet: nsm. 474, 17; nsf . 410 ; npm. 314. stille, adv., quietly, tranquilly: 4s5, 97, 358. See unstille. . stlncan, 3. i. spring, leap: pret. 3 sg. stone 3O12. — 2. stink : 3 pi. stinceS 4i32. sti'3, adj., stiff, hard, strong: nsm. yi2; nsn. stTJ> 453; gsn. sti>es 555; asm. stij>ne 1 79 ; asn. str5 9329. stuJecg, adj., sharp-edged: nsn. 9318. sti<5weg, m., hard way, storm-path : as. 4s5- stlwita, m., officer of household, steward: dp. stlwitum, household 410. stol, see e3el-, gleow'stol. stondan, 6, stand: i sg. stonde 264, 8S22, 9324; 3 sg. stondej> 41 61, stonde^ 954 ; 3 pi. standa^S i68; opt. 3 sg. stonde 70s; pret. i sg. stod 8812, [st]6d 887; pret. 3 sg. stod 578; pret. I pi. stodan 8814; inf. 3413, 358, 552, 8826, standan So1 ; ptc. dsf. stondendre 555, asm. (uninfl.) stondende 8i8. See for- stondan. storm, m., storm, tempest: ip. stormum 844». strgel, f ., arrow : ap. striEle 4s6. strang, see strong, street, f., street, road: as. stra5te i618. stream, m., stream, flood: gs. streames 2710; np. streamas 36-14, 238, 8i8; ap. streamas , 470, [strejamas 936. See flrgen-, lagu-, inerestream. strgamge^vln, n., strife of waters : gs. streamgewinnes 4s6. strengu, f., strength, power : ns. 85 ; is. strengo 2818. See woruldstrengu. -strCon, see gestreon. stretFan, see bestreftan. strong, adj., strong, powerful: nsm. 23, 486, 17, 2813, 558, 631, 9310 ; gsm. wk. GLOSSARY strangan 485; dsn. wk. strongan 4i79; asm. strongne 842; npm. stronge 238 ; ipn. strongum 493 ; comp. nsm. strengra 4i92, 85 ; comp. nsf. strengre 4 1 23. See forstrang, maegen-, ryne- strong. strudan, 2, plunder: pret. 3 pi. strudon 5410- -stun, see gestun. stund, f., hour, time : as. stunde 9318 ; gp. stunda 559; ip. stundum, exceed- ingly 2s ; eagerly, fiercely, 36. si ii An, see wretfstiuJu. style, n., steel : ns. 9318; ds. 4i79; ? style 94. styran, W 1 . i . guide, direct : 3 sg. styre'S 4 1 13. — 2 . check, prevent, restrain : i sg. styre I24. styrgan, Wa, trans, stir, move: I sg. styrge 39, 470 ; inf. 418. styrman, Wl, cry: i sg. styrme g7. sue, see swa. sum, pron., some, one, a certain one : nsm. 4s3, 27 1, 7323, 774; nsf. I58; gsm.sumes I515, 483; asm. sumne44; asn.So9; npf. sume ii8. sumor, m., summer : ? sumor 883. sumsend, ptc., humming, rushing : apn. sumsendu 447. sund, see gesund. sund, m., sea, water: ds. sunde n8. sundhelm, m., water-covering: ns. 771; ds. sundhelme 310. sundor, adv., severally, each by himself: 4erne 630. swa, conj. i. as, according as: 3X, 413, io7, 2i25, 222, 236, 2S1(>, 34", 416, 498, 6on, 62, 784?. — 2. just as, like: 9, 4 134, swe i64; 252-2.3-3. — 3. so that (result) : 6i16. — 4. although : 74, 2313. — 5. where : 8831. — 6. swa . . . swa, as ... as (adv. and conj.) : sue (sue arllce for MS. snearllce) . . . swa io6. See swa 3eah, swa ffeana. swa, adv., thus, so: 467, io11, I26, 14°, 2816, 306, 4i14, [4i»], 4i69, 5o9, 7o5, 8432, 888. See some (swa some). >\ a pa n, see aswapan. -s\varu, see ondswaru. swaes, adj ., own, dear : npf. swase 473 ; gpm. swiesra io11, 27s22 ; apm. swjese J622, 726. swtesende, n., food, repast : dp. SWJE- sendum Sg8. swaltan, Wl, sweat: 3 pi. swstaiS 443. swaeS, n., track: ns. 2210; as. 226; np. swa)>u 5i3. swa 36ah, con]., yet, nevertheless: swa J>eah 5911. See se Seah. swa O'eaiia, conj., yet, nevertheless : swa t>eana 5913. See se '(VPaiia. swaSu, f. i. track: as. swa)>e 9512. — 2. on swrajje, behind: ds. I625, 751. swe, see s\va. sw^eart, adj., swart, black: nsm. 5o5; nsn. 2210, wk. swearte 4i81; dsm. sweartum 7210; asm. sweartne I318; isf. sweartan 4I94; npm. swearte 522; npf. swearte 583; apm. swearte I34; apn. 447 ; ipn. sweartum i87 ; sup. gsn. wk. sweartestan 428. sweartlast, adj., leaving a black track : nsf. 2711. sw?g, m., noise, sound : gp. swega 4". sw'elgan, 3, swallow : i sg. swelge 93122 ; 3 sg. swelgeb 5910, swilgeft 5o2, 822 ; pret. 3 sg. swealg 279, 486; inf. i515, i87. See forsw^elgan. sweora, m., neck: ns. 7O2, swlora 7318 ; as. sweoran 866. s\veord, m., sword: as. 5614. sw^eorfan, 3, polish : pp. sworfen 29, 280 RIDDLES OF THE EXETER BOOK sweostor, f., sister: ns. 72; gs. 44"; np. i42. See gesweostor. sweotol, adj., manifest, clear, open: nsf. 40s; nsn. 2210; npn. 14. sweotule, adv., clearly, openly: 2510. sweotulian, J<?T gesweotulian. swPte, adj., sweet: comp. nsm. swetra 4168. swetnes, f ., sweetness : is. swetnesse 4 1 30. swican, j^ geswican. Bwifan, l, move, pass, sweep (intr.) : 3 sg. swifeS I318; inf. 337. swift, adj., swift, fleet: nsm. 472, i62, 52, wk. swifta 4I68; asm. swiftne 2O3, 751; comp. nsm. swiftra4i70; comp. nsf. swiftre 678, 858. swige, adj., silent, still: nsm. 4", 85!. swigian, Wa, to be quiet, silent: 3 sg. swigaft 81; pret. i sg. swigade 7215; ptc. nsm. swigende 49, npm. swigende (MS. nigende) 9. swimman, 3, swim : pret. i sg. sworn 748; pret. 3 sg. sworn 2314. swin, n., swine: ns. 41 106. swingere, m., scourger : ns. 287. swinsian, Wa, make melody, make music : 3 pi. swinsiaft 87. swiora, see sweora. swift, adj. i. strong, powerful ': comp. nsm. swibra 41® ; comp. npm. swibran 178; sup. nsf. swH>ost S428. — 2. comp. right (hand) : nsf. s\v!j>re 6i12. swiffan, see oferswRfan. switfe, adv. i. very, exceedingly : swlj>e 78, II3, 528, 582. — 2. soon, rapidly: swl)>e 208, 27, 337. — 3. violently ; swlj>e 63. — 4. eagerly : swljje 9512. — 5. sup. chiefly, especially: swi>ast 957- 8\vi?Jfeorm, adj., strong, violent: nsm. swl^feorm 472. 8\vogan, R, make a noise, resound: 3 pi. swoga'S 87. sivylc, pron., such, such a one, such a thing; asf. swylce 898; asn. 6iu; gpm. swylcra 20. s\vylce, adv., in like manner, also: 2i8, 258) 4I29,60,96( 64^, 65. s\vylce, con j . i . like as, as well as : 79, 84™ ; swylce swe i64. — [2. just as though : I1.] sylf, pron. i. self, one's own: nsm. 2728; sylfa 388, 638, So11, 851 ; gsf. sylfre 348 ; gsn. sylfes 6$6 ; dsm. sylfum 2 16; is. sylfum 6710; npm. sylfe 586. — 2. J>aet sylfe, in like manner : 510. sylfor, see seolfor. syllan, Wl, give, grant: i sg. selle 13s; pret. 3 sg. sealde 5, 623, 727; inf. 388. symbel, n., feast: ds. symble 3212. symle, adv., always, ever: 385, 4I30-64; 68. syn, f., eye, sight, vision : as. syne 33 ; is. syne 4I9. syne, see gesyne. tacn, n. i . sign, token : as. 566. — 2. signification : as. tacen 6o10. tn-ciian, Wl, show, point out: 3 sg. tiecneti 416, tiecneb 52°. tacnian, see getacnian. tan, m., twig, branch : ip. tanum 542. -tse, see geteese. teala, adv., well, rightly: 2214, teale i616, tila 492. telg, m., dye: ns. 2715. See beamtelg. teon, a. i. draw : 3 sg. tyht? 636 ; pret. i sg. teah 2313, 72s. — 2. go, proceed: 3 sg. tyhft 35. See ateon. tSon, n., hurt, annoyance: as. 5i3. teorlan, Wa, tire, grow weary : pret. 3 sg. teorode 558. tld, f., time, hour: as. 430, 742; ip. tidum 402, 596. See daegtid. til, adj. i. good, serviceable: nsm. i89. — 2. excellent, kind: gpm. tilra 27123. tila, see teala. tilfreirimend, part., doing good: gpm. tillfremmendra 6o7. GLOSSARY 281 tillTc, adj., £•<?<?</, capable : nsm. 558, 645. timbran, Wl, build: pp. timbred 84. See atimbran. to, prep. w. dat. i. to, unto, towards, into, upon: [i17], 418, is4-17, i610, 2i6, 232-12, 288, 20.7, jo-9-12, 3 17, 342, 354, 4020, 4 155, 565, 6o17, 693. — 2. as, for (purpose) : 72, 2727, 4O19, 4i66, 425, Si2, 706, 731 — 3. on, at, among: [i12], I311, 4i45. — 4. of, from : 498. — 5. w. ger. 2912, 3223, 3713, 4022,25, 428, 8829. to,adv. .too: 236,345. — 2. thither:^. toberstan, 3, burst to pieces : 3 sg. tobirsteft 397. togaedre, adv., together : 534. togongan, anv., pass away (impers. w. gen.) : 3 sg. togongeS 24™. torht, adj., bright, splendid, glorious: nsm., 5 18, wk. torhta439; asm. torhtne 492> S42! ipf- wk. torhtan 579. See hleor-, -wlitetorht. torhte, adv., clearly: 88, 6o7. tosa'Ian, Wl, impers. i. fail, not suc- ceed: 3 sg. tosieleb I75. — 2. lack, be wanting: 3 sg. tosale'5 I625. tosamne, adv., together : 4s9. [toshtan, 1, tear asunder, separate, sever: 3 sg. toslite'S i18.] too", m., tooth : as. 598 ; is. toj>e 875 ; gp. tofca 352; ip. toSum 2214. to 3on, adv. i. so: to J»on 4i16. — [2. therefore: to J>on i12.] toojringan,3,/r<?jj' asunder, drive apart : i sg. tof>ringe 487. tredan, 5, tread, tread upon : i sg. trede 81 ; 3 sg. tridej> S429, triedeft 1 36 ; 3 pi. tredai? 585; pret. 3 sg. traed 72"; inf. I41. treow, n., tree : ns. 542, 579. See wudu- treow, \vulfheafodtreo. treowe, see getr5owe. tunge, f., tongue : ns. 8o8 ; ds. tungan 492 ; as. tungan 598. turf, f., turf, grass, greensward: ds. tyrf 4I25; as. I41. num. two: nm. 43!°, 472-8; nf. twa 4317, 472 ; n. (m. and f.) tu 645 ; nn. tu i64; gn. twega 4O11, 439 ; dm. twam 6i16, 8818 ; d. (m. and f.) twam 5i2; dn. twam471; am. 532, (MS. n) 864; af. twa 431, 7O3, 8i6, 868.5-7; an. tu 377. t\velf, num. adj., twelve: 377, (MS. xn) 864. tydran, Wl, be prolific, teem : 3 sg. tydre« 8437. tyhtan, see atyhtan. tyn, num. adj., ten: (MS. x) I41. (yuan, see be-, ontynan. tyr, m., glory-^honor : as. 2723. IX D = rune S> : 654. ?fa, adv., then, thereupon : J>3 io3, 238'10, 307-9, 4 135. Seeium. ?fa, conj., when: )>a ii6-9, 4i7, 482, 6q17. 9'a gen, %A.,yet: J>a gen io2. Seer, adv., there : >xr [i6], 424.28.3S) 5ii, i69, 24", 32", 3710, 4o7, 438, 474, 569, J71, 6i4, 959. 5aer, conj., where: )>aer 45, I512, i68, 2 112, 257, 274, 3 16 b (a J>aet), 384, 55!, 56i, 571.9.11, 643'5, 68", 731, 8 17, 86i, 88L12, 9324. xs 21-1. 5ses, conj., as: J>aes 427. tfaes 5e, conj., as far as: J>aes J>e 8480. d'set, conj. i. that, in noun clauses: J>aet 54, 65, i26, 21™-™, 26, 28", 4O1, 48', 6 18, 73123. — 2. that, so that, in order that, in result and purpose clauses: l>aet 21, 415.21,81, 22i4, 2319, 245'13, 3i«a (b J>a!r), 3412, 3718, 4i9'16> 86,91.108, 6ll4, 736, 8441'42. Saette, pron., which : J>aette [i18], 9322. 5e, indecl. particle, who, which, that : >e 215, 315, 99, j314, 2 ,4,21.28, 2816, 4 j 49,77,78, 4313, 4416, 50", 511°, 628, 666, 7O5, 73, 8814, [l>e] 4i25.io6. e 4&6. 282 RIDDLES OF THE EXETER BOOK If Pah, conj., though, although : >eah I46, 192, 4 17'65, 492, 8o5, Sen Leid. 14. tVf'ah, adv., however: >eah 78. 6V<? se 36ah, swa 3eah. eah ) 4i27, 8433-50, 9317> 9510- ftearle, adv., abundantly : |>earle 72. 36aw, m., conduct, behavior : gs. J>eawes I28. fleccan, Wl, ece~S 15!, 8i9; opt. 3 sg. >ecce 214; pret. 3 sg. >eahte 46, 771; inf. >eccan io4; pp. J?eaht n4, I73. See biSeccan. ftecen, f., covering (garment) : as. )>ecene 462 ; is. J>ecene 8489. egn 382, 5o4, 557, 872, |>E[gn] 65 ; ds. J>egne 51-9. Segnian, Wa, serve: 3 sg. Jjenaft 2214, 445; 3 pi. begniaS 516. Sencan, see geftencan. tJenden, conj., while: benden 132, 6818, 856. 3enian, see Tfegnian. (Kennan, j^ 1>e<$ennan. fleod, f., people: ds. J^eode 7318; gp. J>eoda 428. ^"^ wertfeod. 56odcynlng, m., kingof the people, God: gs. J>eodcyninges 681. 56oden, m., lord, master : gs. )>eodnes 465; ds. }>eodne 2i26, 59", 624. 36of, m., thief: ns. >eof 484 ; gs. J>eofes 7323- <56oh, n., M/^A : ds. J>eo 45!. 36on, 1, grow up, flourish, prosper: pret. i sg. )>aEh 728. ^>^ ge-, onSeon. tFeotan, 2, sound (in oozing out) : inf. J>eotan 39. <5eow, m., servant: ns. J>eow 467. Seowian, Wa, serve: i sg. }>eowige '316; 3sg- )^owa)> 516. 3es, pron., this: nsm. >es 321, 33!, 4i42- 48.48.61,76.88, 58^ 6?W . nsf >ios g4) 581; nsn. HS 36", 4i8i-49, 946; gsn. )>isses 56"; asm. >isne 4O19, 22 ; asf. J>as 4O17-26, 4I1'2 ; asn. >is 4i78 ; dpm. )>issum io1 ; apn. J>as 4i5. Sicce, adj., thick: apn. Jricce 4I36. Sicgan, 5, partake of, receive : 3 sg. J>ige'$3214; inf. J>icgan gi10; ?J>ygan 896. [me I14; npf. Kne i18.] ffindan, 3, swell up : inf. J>indan 462. mg 4O24; ds. J>inge 681 ; as. Hng 32s, 465 ; gp. Hnga 4I36 ; ap. J^ing 4 139; ip. J>ingum, purposely 6 114. -?Hse, .sw maegenSise. tfolian, W2. i . suffer, endure : \ sg. J^olige 9321. — 2. holdout, stand strain : 3 sg. >ola^S 178. — 3. lack: inf. J>olian 2 126. (Yoiian, adv., thence: ]>onan 273, 3O10, 7327- once, acceptable, grateful, 59; as. on ]?onc 2 126. Seehyge-, or-, searo- oncade ttoncol, adj., wise, thoughtful: nsm. J>oncol 312. See searoSoncol. Sonne, adv., then: jxmne 42-68, 85, 15", 2 16, 2911. SJonne, conj. i. when: )x>nne [i10-11], 28,8) ,8,14) ^27,41,51,60,73,74^ ^5,9) gl,8) g6j 4I19-55, 4412, 454, i;2, 248, 3i7, 321, 642.8, 7iT, 73i9( 9 J^Onne I?6, 247, 4126,28,81, 4I48,51,54,59,74, 76,83,92,94,96( rc9( 671,2,2,8, ge8 942,3,6,6,7, ?friec, see geSraec. <5raed, m., thread: ns. )>raed 366 (Leid. «ret). 'A'ralian, Wa, urge, press: 3 sg. Jrafa'S 44. Srag, f., ^>«^, j/«^ ^/" time : as. Jrage 898; ip. J>ragum, at times 24, 467, 557, 8j4. Jfraegan, Wl, run : inf. J>raegan 2O8. reat i2'7.] See eoredftreat. Sreohtig, adj. laborious: comp. nsm. Keohtigra 85. 3rim, see tfrym. ftriiidan, 3, swell: ptc. asn. )>rindende 46. Sring, j^ geSring. Iffringan, 3, press on, force a way : 3 sg. >ringe'S (MS. bringeS) i29; inf. J>rin- gan 461. See a-, ge-, oS-, totfringan. Srintan, j^ a'Sriiitun. Srist, adj., bold, audacious: gp. J>ristra (MS. >rlsta) 7323. Srowlan, W2, suffer, endure: pret. I sg. J>rowade 7213. 3ry, num., three: nm. )>ry 4i52, 5914. Srym, m., force, power, might : ? )>rym 8445; is. brimme 461, J>rymme 4i91; gp. Jnymma 4. Sryinfaest, adj., glorious, mighty: asm. Jrymfaestne 484. Srymful, adj., glorious, mighty: nsm. J>rymful 24, 467. tfrytf, f., strength, i. in pi., forces, troops: gp. J>ryba 65. — 2. ip. ftry- ryj>um 872, 5u, pron., thou: ns. J>u [i16], 3313, 3712- 12, 4028, 4i59; ds. or as. >e 6i14. 'diuiiaii, Wa. i. j/aw</ «/, swell: inf. J>unian 462. — 2. resound, thunder : i sg. )>unie 24. 6"^ on<!(unian. Surfan, PP., need, have reason to: i sg. hearf I622, 21". Surh, prep. w. ace. i. through (place) : Jnirh 45 , 22", 322 384, 4I45, 728. — 2. through, during (time): )>urh 2 17, 594. — 3. through, by means of, because of (condition and agency") : }>urh 614, 91, 3220, 364 (Leid. «erh), 368 (Leid. Serin), 436, 5o8, 548, 7o2, 787, 84". Surhrtesan, Wl, rush through : i sg. J>urhn5se 4s6. tfurst, m., thirst : ns. }>urst 448. -Swaere, see geffwaire. <5yncan, Wl, seem, appear : 3 sg. bynceft 410, HnceS 3218; pret. 3 sg. jnihte 481, 878. ffynne, adj., thin: apn.J>ynne 4I36. Syrel, n., hole, aperture: as. )>yrel i621, .72«. (Jyrel, adj., perforated: nsn. J>yrel 452, 9i5. iffyrel^vomb, adj., having the stomach pierced: asm. }>yrelwombne Si11. ftyrran, Wl, dry: pp. J>yrred 294. Syrs, m., giant: ds. )>yrse 4i68. 5ystro, f., darkness, gloom : ds. J>ystro 484 ; dp. bystrum 44. iffy wan, Wl, urge, press : 3 sg. jjy'S 13", 226, 638, 646 ; inf. >ywan 418. U U = rune H (MS. ^} : 752. ufan, adv. i. from above, dmvn : 417-55. , ii4, . — 2. above: ufon 376. ufor, adv., above, higher than : 4i88. uhta, m., early morning, time just be- fore dawn : gp. uhtna 6i6. Ulcanus, m., Vulcan : gs. Ulcanus (Lat. Vulcant) 4I56. unbunden, adj., unbound: ns. 2415. uncer, pron., of us twain : [asm. un- cerne I16, asn. i19] ; npm. uncre SS18; apm. uncre 6i17. undearnunga, adv., without conceal- ment, openly : 432. under, prep., under, beneath: A. w. dat. 42, I07, 2315, 285, 378, 4l83.40.88, ^ 452' 5511' 72l2> 8430 : B. w. ace. 32, 4s4, 2310-17, 5o6, 525, S32. 55. 638, 7324. 744, gi8; C. case indeterm. 4s8. under, adv., under, beneath : 2211. underflowan, R, flow under: pp. underflowen 1 12. underhmgan, ] , descend beneath : \ sg. underhnlge 676; inf. 469. 284 RIDDLES OF THE EXETER BOOK undyrne, adj., not hidden, revealed, manifest: nsn. 4315. unforcufl, adj., not ignoble, honorable, faithful: nsm. 63. ungefullod, adj., unfulfilled: ds. unge- fullodre (MS. ungafullodre) 6o13. [ungelic, adj., unlike, different: nsn. I8.] [ungelice, adv., otherwise, differently: i8-] ungesib, adj., unrelated: dsm. ungesib- bum lo8. ungod, n., evil, ill: as. 2I85. unla't, adj., unwearied, quick: nsm. 54". unlytel, adj., not little, great : nsf. 41™ ; asn. 83". unried, m., evil course, folly : gs. un- rades I210, zS12. unra'dsTff, m., foolish way, foolish course: ap. uniiedsibas I24. uiirini, adj., innumerable : apn. unrimu 78. unnm, n., countless number : as. 44. unsceaft, f., monster ? : np. unsceafta S882. unsoden, adj., uncooked: asf. unsodene 77". u nst Hie, adv., not still, restlessly : 526. urnvita, m., ignorant person : ns. 5O11. up, adv., up, above: 412'70, 2319, 34", 55, 56s, 622, 938, upp 1 19. upcyine, m., up-coming, up-springing: as. 3i9. upirnan, 3, run up, upsoar: ptc. dsm. wk. upirnendan 4I66. uplong, adj., erect: nsm. 8812. upp, see up. upweard, adj., turned upwards : asm. upweardne 626. user, pron., our: nsm. 4I89. ut, adv., out, forth : 636, 93™. utan, adv., without, from without : 4i15- 47«t 7.jl3) 8489_ fite, adv. i. <?« of doors, in the open: 43- — a. comp. uttor, a/ a distance : 41". W W = rune > : 2O6, 651, 9 17. wa, interj., wo#/ I28. \-acan, 6, ^ ^rw, spring: pret. I sg. woe 2 121. waican, Wi, soften : pp. wieced 29s. waeccan, Wl, watch, wake: ptc. asf. waeccende 4i8. \vaed, n., water, sea : ap. wado 82. •wsed, f., dress, clothes : dp. wiedum 43 ; ip. wedum 10. wadan, 6, go, proceed : i sg. wade 63 ; pret. 3 sg. wod 2315, 935. See be- \vadan. wiede, see gewsede. \ ;i-i'a n, j^ bewiefan. ivaiian, Wa, waver, be amazed: 3 pi. wafia« 84". •wag, m., wall: ds. wage I512; ds. wjgge 14. \vaeg, m., wave : ns. 420 ; ds. wSge 1 110, I71i 2321> wege 341, 6<)3; is. wiEge 3. \va?gfaet, n., water-vessel, cloud: ap. wJigfatu 487. wagian, \V2, intr., shake, totter : 3 pi. wagiaft 48 ; pret. 3 pi. wagedan 556. w»gn, m., wagon, wain : ns. 2312 ; ds. waegne 228 ; as. 23. waegstaS, n., shore, bank : ds. wieg- stae)>e 232. \vaelcraeft, m., deadly power: is. wasl- craefte gi11. \vaelcAvealm, m., death-pang: ns. 28. -wald, j-ff on\vald. waldend, m. i . possessor, master : ns. 2I4, 246. — 2. Z0r</ (Christ, God): ns. 71, 4i89; gs. waldendes 41". •waldende, adj. (ptc.), powerful: comp. nsf. waldendre 4i87. Wale, f., (Welshwoman), female slave : ns. I38, 536. \vaelglin, m., gem of death : as. 2i4. •vv a' IK rim, adj., cruel, bloodthirsty : nsm. i68. wselhwelp, m., death-whelp : gs. wael- hwelpes I623. GLOSSARY [waelrfcow, adj., cruel, bloodthirsty : npm. waelreowe i6.] wamb, see womb. \vanian, see wonian. wJepen, n., -weapon : ns. 488 ; as. s612 ; ip. wiepnum 462, 2i17. See beadu-, comp-, Iiil(I<-\ a-pcn. wajpenwiga, m., weaponed warrior, armed warrior : ns. 15. wiepnedcynn, n., male kind, male sex : gs. wisepnedcynnes 391. war, n., sea-weed: is. ware 3. warian, Wa, guard, hold, possess : 3 sg. waraft 32", 83, 9326. warcS, n., seaweed: ns. 4i49. -waru, see helwaru. waestm, mn. i. growth, form: as. 326. — 2. fruit : ns. g22 ; ip. waest- mum 8487. wiet, adj., wet, moist: nsn. 2611; nsm. wk. wajta 361 (Leid. ueta). \ WH'ta, m., moisture, liquid: as. wzetan 448 ; is. wStan 5911. wjvtari, Wl, wet, moisten : 3 sg. waste's 13!°; pret. 3 sg. wiette 272. waeter, n., water : ns. 548, 6<)s ; gs. waetres 2312; ds. waetre I310, 278; is. wastre u1, 9328. waft, f., wandering, journey: as. waj>e 2". , Wl, hunt: 3 sg. w 356. wawan, R, blow, be moved by the wind: 3 sg. wiewe'S 4i81. •wCa, m., woe, misery : gs. wean 7218. wealcan, see gewealcan. -\veald, see geweald. •wealdan, R, ha-ve power over, control, rule: 3 sg. wealdelS 4i5, wealdeb 4 122, [wealde>] 41 2; pret. 3 sg. weold S36- "Wealh, m., ( Welshman}, slave, servant : ap. Wealas 13. Wealh, adj., (Wels/i), foreign: apm. Walas 72". weall, m. i . natural wall, kill, cliff: gs. wealles 3o7; ds. wealle 420; ap. weallas 355. — 2. -wall (of building) -. n ? s. 84" ; np. weallas 49. See bord-, saeweall. weard, m., guardian, lord: ns. 22, S32; gs. weardes 14". -weard, see eeftan-, aefter-, for3-, hinde-, innan-, tipweard. weardlan, W2, hold, occupy, inhabit : inf. 8S25. vveann, adj., warm : nsn. wk. wearm[e] 57- wearp, m., warp : as. 365 (Leid. uarp). weaxan, R, wax, grow, increase : 3 sg. weaxeft 4i26; 3 pi. weaxafl 4i102; pret. i sg. weox 881 ; inf. 5510, (MS. weax) 461 ; ptc. nsm. weaxende 548. See a-, gewcaxan. w^eb, see godw^eb. w^eccan, see aw^eccan. ^vecgan, Wl, move, shake : 3 sg. wegei? 138, 226, 8i7; pret. 3 pi. wegedon 735- •w6d, see weed. weder, n. [i. weather: ns. i10.] — 2. air : ds. wedre 3i2. \clan, see a-, ge^vefan. w^efl, f., woof, thread: np. wefle 365 (Leid. uefla^). •weg, m., way: ds. wege 371, 7o5; as. i621, 4O6, 548, 63s, &91; ap. wegas 416, 526. See Hud-, for 31 ; pret. i sg. wende 6i7. 286 RIDDLES OF THE EXETER BOOK wendan, Wl. x. turn, turn round, turn over : pret. 3 sg. wende oo5 ; pp. wended 6o18. — 2. wend, go, proceed: 3 sg. wendeS 7328. See gewendan, , see woh. weorc, n. i. work, labor : gs. weorces 43, 5510; as. [w]eorc 9322; np. 27". — 2. pain, travail, grief: as. 7213. See beaduweorc, hondweorc. weorpan, 3, throw : i sg. weorpe (MS. weorpere) 287 ; 3 pi. weorpaS •f. See a-, beweorpan. weorff, adj., precious, valued, dear: nsm. 281 ; comp. nsm. weorSra 8814. weortfan, 3. i. be, become : i sg. weorj>e I74; 3 sg. weorJ'e'S i614, 2I20; 3 pi. weorSaS 613, weor)>at> 3" ; pret. 3 sg. wearS io8, 4O18, 546, 68™, 698; pret. 3 pi. wurdon 73; pret. opt. 3 sg. wurde 848° ; inf. weorj>an 431, 5i10.— 2. happen, come to pass: pret. 3 sg. wearS 698. See for-, geweorSan. weorftian, Wa, praise, celebrate : 3 sg. weorfc'S 2 110. See geweorSian. w?pan, R, weep: pres. 3 sg. wepeft 7i5; pret. i sg. weop 9319. wer, m., man : ns. 24", 471 ; gs. weres 451; np. weras [i6], I53-12, 239-21, 3i6, 84", 861;.gp. wera 28, 49, 27", 30", 351, 488, 838, S826; dp. werum 281, 3,4.i4f 3311, 429, 43ie. wergan, see awergan. werig, adj., weary, exhausted: nsm. 68, 5510. See rad-, slaepwCrig. wermod, m., wormwood: ns. 4I60. wer3eod, f., people, nation, pi. men : ap. werj>eode 84°. wesan, anv., be, exist : \ sg. com 61, I62, 6, 242, 251-9, 2&, 281'6, 3 11, 321, 4 1 16,18 58.26,28,80.88,41.46.48,60.64, ,68,60,74,76,78,80,87 ,90,92,92.94,96,99,105 g,l 642, 671, 7 11'8, 738, 791. 8o», 81. 85^ 8820, 92, 951, beom 474, 88, 17, 24, beo 247; 3 sg. is [i 1.8,4.6,8^ ^ I2i5 ,5^ «, 2I8, 221, 241, 264, 291 3217,23, 33!, 348.10,11, 4022,25, 411,3,70,72, 42-->, 43» 452.51, 59i5, 6iio, 70i^, 73^, 8o8-10, 821, 841-4-6, 8821, 9 11, [is] 4O24, (w. neg.) nis 41^-^, 8s1, 88®, bi« 28, 3!!, 424,28,33.89, I49, 1&t 26U, 2$, 458> 59s. 63, 6418, 8423.26.85,36, 857, bi)> 33, 59, i610, 178, i84, 2 124, 229.i5, 291, 358-6, 357, 409, 8484».a44ZT,48 . j pl. beoj> 645; 3 pi. beo$ i75, 2719, 36s (Leid. bla«), 4I11, sind s82, 59", 678, sindon [i6], 4317, 5610, sindan 666 ; opt. 3 sg. sy 2913, 36", 401-14, 4i2.27,6o, 429, 6819, 8o5, S455, (MS. ry) 948, sle 32^, 3314; pret. i sg. waes I51, ig4, 4I44, 571, 6i1,661-2, 7 12, 721'9, 741; pret.3sg.waes [! 10.12,12,18], I02, „!, I45, 2O8, 236, 324-6, 339. 343, 372'9'10. 381, 474. 485, S2'. 535, 542,11, 557, S69, 576.9, 6017, 6 13, 624, 6416, 6s2, 692, 83!, 8418 ?, 885-14, 898 ?, 92! ; pret. I pi. wzron 8813-29 ; pret. 3 pi. wSron io7, n8, I41, 344, 476, 538, 574, wSran 522 ; pret. opt. 3 sg. wiere 378, 4O15, 7217; inf. 438, 4410- west, adv., west, westward: 3O10. wic, n., village, dwelling, abode : dp. wlcum 97, 5O4, 73s28 ; ap. 82, I68. wleg, n., horse : ns. wycg 1 55 ; ds. wicge 8o7; as. Wl[cg] 651; is. wicge 15"; np. 2321; ap. 239; ip. wicgum 232. wicstede, m., dwelling-place : np. 49. •wid, adj., wide : asf. wide 19. wide, adv., widely, far : 211, 4s7 -71, 85, 1 1 w, 2 118, 2716, 281, 36" (Leid. uid2e), 4017, 41", 592, 677, 7322, 83!°, 93s7, 958 ; comp. widdor io10, 6i17, 7210. \videferh, Z.&.V ., forever : 4O8-2i. Tvidglel, adj. i. wide-spreading, spa- cious : comp. nsm. widgielra 4i61, widgelra 4i83. — 2. wandering, rov- ing: dsf. widgalum 2i5. •widlast, adj., wide-wandering: nsm. 2O6; [ipf. widlastum i9]. wldo, see w^udu. wif, n. i. woman: ns. 2611, 5i5; gs. wifes 3712, 928; ds. wife 2I32; as. wiif GLOSSARY 374; np. 3i6; gp. wifa (MS. wife's) 8432. — 2. -wife: dp. wifum 47. wifel, m., -weevil: as. 4i78. wig, n., fight, battle: as. 63, I623. wlga, m., warrior: ns. i68, 5I1, 528, 732 wigan gj2 folc- gu5-, wafepenwiga. wiggar, m., spear : as. wegar 0r wigar (p-nx K fc> 20". wiht, f. i. wig/if, creature: ns. 19, 2 I1, 242, 251, 261, 2913, JO7, 324-19-24, 335'14. 341, 396, 4Q1, 41", 429, 68» 70, 821, 841, 861, 89! ; gs. \vihte 30", 37" ; ds. \vihte 575; as. 3O1, 351, 371, 391, 572, 592, 682, 691, 87!, wihte 381, 691, wiht[e] 4O26; np. wihte 4316 ; gp. wihta 298, 4O14, 438, 84; ap. wihte 581, wyhte 431, wuhte 521. — 2. aught, anything: as. 5". — 3. with neg. naught, not a whit : ne wiht 3214, 5910, 661 ; ne ~ wihte 48^. See no- wiht, owiht. wllcuma, m., -welcome thing: gp. wil- cumena 911. wilgehle'JTa, m., pleasant companion : ap. wilgehlehan I55. Avilla, m. i. will, wish, desire: as. wil- lan 2 188, 3010, 556, 64, 737; ip. willum 877, 91 n, 932. — 2. pleasant thing, de- sirable thing: ns. 791 ; gp. wilna 2910. willan, anv., will, wish, desire : I sg. (ne) wille 5O10 ; 3 sg. wile 36" (Leid. uil), 406, 4410, 455, 774, gi9, wille 44", 6o15; 3pl.willa« [i2-7], I77, 2718; pret. 3 sg. walde 3O4. wolde 877 ; w. neg. i sg. nelle 2415, 3 sg. nele i616. wilnian, Wa, desire : 3 pi. wilniaiS 5O7. win, n., wine: ds. wine I517, 4316, 471. wineel, m., corner : ds. wincle 461, (MS. wine sele) 552. •wind, m., wind: ns. u10, 4i68; ds. winde 17, 3I1, 4i81; is. winde I514. wlndan, 3, roll, twist : pp. wunden 295, asn. wunden 563, npm. wundne 4i104; npf. wundene $65 (Leid. uundnae). See be-, ge-, ymbwinclan. -\-Inn, see ge\vinn. •\viiiiiaii, 3, strive, struggle, labor : i sg. winne 467, 77; 3 sg. winnetS 419; inf. I71; ptc. nsm. winnende 38, 448, 526, asf. winnende 572. ? winter, m., winter: gp. [wintra] 83!. winterceald, &&.,wintry-cold: nsm. 57. wir, m., wire, pi. ornaments: ns. 2i4; is. wire 27", 7i5; ip. wirum i82, 2i32, 4 147. •\virboga, m., twisted wire: ip. wirbo- gum 158. •\vis, adj., wise, learned: nsm. 3314; dsm. wisum 3224. See medwis. w^isdom, m., wisdom: as. 959; is. wis- dome 685. •wise, f. i. nature, manner: ns. 37", 8o10; as. wisan I28, 2I11, 664, 7O1, 735- <2&, 847; ip. wisum 322, 332. — 2. mel- ody : as. wisan 94. See sceawend- \vise. wisfaest, adj., wise, learned: nsm. 3614 ; dsm. wlsfaestum 2Q13 ; gpm. wlsfaestra 6819 ; dpm. wlsfaestum 429. \visian, W2, guide, direct : 3 sg. wlsa'S 413, 2 I5, 222. wist, f., sustenance, food: as. 33", wiste 44" ; ip. wistum 8421. See mid- %vist. -\'it, see ge\vit. wita, see sti-, unwita. •\vitan, PP, know: i sg. wat I25, 368 (Leid. uuat), 44!, 5O1, S91> S826; 2 sg. wast 3712; 3 pi. 447; opt. i sg. \vite 511 ; opt. 3 pi. 3713, witen 4O4 ; pret. 3 sg. wisse 552. See be-, gewitan. witan, see geAVitan. wite, n., pain, torment: as. 246. See dolwite. \vitian, Wa, decree, appoint: pp. nsm. witod i66'n, 857, nsn. witod 2I24, ap. witode 447. f, prep., against, with : A. w. dat. 420.4^ ,72.2( 2810, 334, 40^ 88", 9l5, wi|> 441, I71'1, 2i27, 4012; B. w. ace. 4313, wij> I79, 6 114. 288 RIDDLES OF THE EXETER BOOK wlo", adv., in reply : 2<)w. wlanc, see wlonc. wlitan, 1, look, gaze: 3 sg. 93s2- wlite, m., aspect, appearance : ns. 3712, 842; as. 847; ji10 ?. wlltetorht, adj., brilliant, splendid: gpf. wlitetorhtra 71. wlitig, adj., beautiful, comely: ns(?). 84"; nsm. is12; nsn. i810; apf. wk. wlitigan 357. wlitigian, Wa, beautify : 3 sg. wlitigafl 84s7. See gewlitigian. wlonc, adj., proud, high-spirited: nsm. 15; nsf. wlanc 43; dsn. wloncum 8o7 ; asm. wloncne 5i10 ; npm. wlonce 3i6 ; gpm. wloncra 6o18 ; dpm. wlon- cum i810, S425; apm. wlonce 15". See fela-, hyge-, modwlonc. woh, adj. i. curved, bent, twisted: nsm. 22, 7O2 ; ipm. woum I53. — 2. per- verse, wrong, evil : asf . w5n 1 28 ; npm. weo ?? 575 ; npn. 4O24. wolcen, mn., cloud: gp. wolcna 85. See heofon\volcn. wolcenfaru, f., drifting of clouds : as. wolcenfare 471. wolcengehnast, f., collision of clouds : is. wolcengehnaste 460. worn, mn., evil word ': as. 2I88. worn, adj., evil, foul: apn. 4i41. womb, £., womb, belly : ns. 381 ; ds. wombe 448, 378, 88s3; as. wombe ig3, 865, 8;1, 892, 9328, wambe 638; is. womb[e] 9323. See Syrelwomb. wombhord, n., womb-hoard, contents of belly: ns. i810. won, adj., dark, swarthy: nsm. 4I107, wk. wonna 50; nsf. wonn 420; nsn. wonn 8S'22 ; apn. 4s7 ; ipf. wonnum 547, 8816. wonfah, adj., dark-colored: nsf. 536. wonfeax, adj., dark-haired: nsf. 138. wong, m., field, plain : ns. 361 (Leid. uong) ; ns. 4188.51, 7I2. ra 7i8; dpm. wraj>um 1 517. — 2. &Yfcr : comp. nsf. wraj>re 4I60. wratJscraif, n., foul den : ap. wrzlS- scrafu 4 141. wrecan, 5. i . drive, press on : 3 sg. wrice'5 43 ; pret. opt. 3 sg. wraece 22 ; pp. nsn. wrecen 22n. — 2. avenge : pret. opt. 3 sg. wraace 2i18; inf. 9319. .SV^ awrecan. wrecca, m., exile: ns. wraecca (MS. wraece) 24 ; gs. wreccan 40, (MS. wrecan) 211 ; as. wreccan 3O10. wr6gan, Wl, rouse, excite: i sg. wrege 471 ; inf. 417. See gewrPgan. \vrenc, m., modulation of the voice : ip. wrencum 92. wreon, l, 2, cover: imp. 2 sg. wreoh 8454 ; pret. 3 sg. wrah io5, 27", wreah 212 ; pret. 3 pi. wrugon 315, 772, 8815. See bewreon. wreffian, see bewrefiian. wre'dstutfu, f ., prop, support : ip. wreft- stu)>um 4 12. wrigian, Wa, strive, push one's way: 3 sg. wriga)> 225. writ, see ge\vrlt. wri'Sa, m., ring: as. wri^an 6o5. See halswri^a. \vri3an, 1, bind: 3 sg. wrifi 5i5; pp. wri|>en 547. wrixlan, Wl, change {voice), sing : i sg. wrixle 92; inf. 6i10. ? \vrohtstaef, m., injury: dp. wroht- stafum (MS. wroht stap) 73". •\vrotan, R, root up (of swine) : ptc. nsm. wrotende 4i107. \vudu, m. i. wood (material), thing of wood: ns. 4i48, 575; ds. wuda 1 16, 8822; as. wido 572; is. wuda 93M. — 2. tree: ns. 54", 5616. — 3. wood, for- est: [ds. wuda i17]; as. 28, 8i7. — 4. ship : ns. 424. \vuclubeam, m., forest tree: gp. wudu- beama 8816. \vudutreo\v, TI., forest tree: as. 568. •\vuldor, n., glory : ns. 8432 ; gs. wuldres 677; is. wuldre 3i2. \vuldorcyning, m., King of glory (God) : gs. wuldorcyninges 4o21. Avuldorgesteald, npl., glorious posses- sions : np. 2716. wuldorgimm, m., glorious gem : ns. 842«. •\vuldornyttung, f., glorious use : ip. wuldornyttingum 8424. wulf, m., wolf: [ns. I4'17 ; vs. 1 13.13] ; gs. wulfes [i9], 93^. •\vulfh6afodtreo, n., gallows, cross ? : ns. 56i2. vvull, f., wool: gs. wulle 368 (Leid. uullan). wund, f., wound: ds. wunde 9319; np. wunda 6o16 ; ap. wunda 547, wunde 612, 93ln. •wund, adj., wounded: nsm. 61 ; nsn. gi2. wundenlocc, adj., curly-haired ; with braided locks (B.-T.) : nsn. 26". \ mid i a n, W2, wound: opt. 3 pi. wun- digen 8451. •vvundor, n., wonder, marvel: ns. 6gs; gs. wundres 6ii° ; as. 482 ; gp. wundra 228, 83!°, S434 ; ip. wundrum, wonder- fully, 361 (so Leid.), 372, si1, 682, 692, B4i.2i.40. ?wundor686. ^\Tundorcraeft, m., wondrous skill: is. wundorcraefte 4I86. ^\nindorlTc, adj., wonderful: nsf. wun- derllcu 19!, 2I1, 25, 261, wundorlicu 3O7; nsn. 8S22; asf. wundorllce 30. 871; comp. asm. wundorlicran 325. •wundorworuld, f., wonderful world: as. 4O17. •\viinian, W2. x. dwell, abide: I sg. wunige S$6 ; pret. i sg. wunode 73!. — 2. remain, continue : 3 sg. wunaS 3216; inf. 4 18. See gewuniaii. ivyltan, Wl, turn, revolve: pp. wylted 6o18. 290 RIDDLES OF THE EXETER BOOK wyn, i.,joy, delight: ns. [i12], zf; ds. wynne 542; ip. wynnum 4i107. See inodwyn. wynlic, adj., delightful, pleasing: nsf. 4 128; wynlicu 3218. wynnstoftol, m., joyous foundation: ns. wynnstaj>ol 923. wynsuin, adj., winsome: nsm. S426; ? wynsum 8419. wyrran, W], work, make : 3 sg. wyrce'S 647, (begets) 38; pret. 3 sg. worhte 4 16-89, S56; inf. i618, 7312. See be-, gewyrcan. wyrd» f • i • fote : gp. wyrda 369 (Leid. uyrdi), 4O24. — ^. event: ns. 482. wyrdan, Wi, hurt, injure : 3 pi. wyrdaj> S883. wyrgan, j-^ awyrgan. wyrm, m., -worm. i. bookmoth : ns.483. — 2. insect: ns. 4i76. — 3. silkworm: np. wyrmas 369 (Leid. uyrmas). See bond-, regnwynn. •wj-rnian, W l , warm : 3 sg. wy rmeiS 1 310. wyrnan, W], refuse: 3 sg. wymeft 2111.29. \vyrs, adv., worse : 1 45. w'yrslic, adj., mean, vile: comp. nsf. wyrslicre 41. wyrt, f . i . wort, plant, herb : gp. wyrta 7i8 ; dp. wyrtum 612; ap. wyrte 35. — 2. root: ip. wyrtum 357. yean, Wi, increase: 3 pi. ycafi 272; inf. 3 19. yce, see fenyce. yfle, adv., evilly, ill: 4I32, 448, 838. yldo, f., old age : ns. 44. yldra, see eald. ymb, prep. w. ace. i. about, around (local) : 2 14, 4 15. — 2. about, concern- ing: 24", 348, 4026, 4416, 852. ymbclyppan, W i , embrace : i sg. ymb- clyppe 4 115; inf. 4I53. ymbh^vyrft, n., earth, world (orbit terrarum): ns. 4i42; as. 4i715. ym^vlndan, 3, embrace: i sg. yml> winde 4I84. yst, f., storm, tempest: ds. yst[e] 5410- y5, f., wave: ns. 6r6; as. yhe 52, 74 ; np. y>a 31S, 772 ; gp. y|>a 32, 4s8, 237 ; ap. yl>a 417-68 ; ip. yjmm 1 14, 17, 787. ytfan, Wl, destroy, lay waste : inf. y>an 7 17. ywan, Wl, show, reveal: opt. 3 sg. ywe 5615. See geyvvan. zefferus, m., Zephyrus, west wind: ns. 4I68. INDEX OF SOLUTIONS Rlack type, both in names and in numbers, indicates solutions accepted by the editor. All solutions are discussed in the Notes. Ale 29 ^ Chalice 60 "Fingers and Gloves 14 Alphabet 14 Chickens 14 N Fingers and Pen 52 Anchor 3, n, 17 Axle and Wheels 72 Chopping-block 6 ij Churn 55 >jFlute 611-1°, 64 Barleycorn 29 1 Communion Cup 60 Foot and Shoe 63 -•-Barnacle Goose 11 ' Cooper and Cask 87 Bat 37»-» Crab 78 Gallows 56 Battering-ram 54 •^Creation (' Creatura ') 41, .Gimlet 63 Beaker 64 67,94(?) ^fjloves and Fingers 14 Beam 31 (cf. 56) Cross 3 15-9, 56 Gnats 58 Hee 35, 46 Crowd 32 Gold 12, 83 Beech 92 ^Cuckoo 10 Beer 29 Cupping-glass 71 Hailstones 58 Bell 5, 9 Cuttle-fish 74 Harp 29, 56 Bellows 38, 87 Cynewulf i, 90 Hawk 21, 80 (cf. 20, 65) Bible 68 . Hedgehog 16 Bird and Wind 30 \Dagger71 Helmet 71, 81 Boat 37 Dagger Sheath 45 .Hemp 26 Body and Soul 44 i Day 40 ^Hen and Cock 43 Book 27 Bookcase 50 •^Dog 51, 75 >/Dough 46 Hip (' Rosenbutz ') Horn 15, 80, 88, 93 26 Bookmoth 48 Dragon 52 >/Horse (cf. 20, 65) Horer 63 ^Draw-well 59 Horse and Wagon Bow 24 Hurricane 4 Bridge 23 Earth 42 Broom 53 >k Earthquake t1'16 4 Ice 69 Bubble II ••{Earthquake, Submarine, 3 ^ Iceberg 34 Buckets 53 xflnkhorn 88, 93 Bullock 39 \/ Falcon 21, 80 (cf. 20, 'I Butterfly Cocoon 14 65) /Jay 9, 25 Fiddle 32 i Cage 50 Field of grain in ear >/Key 45, 91 Cask and Cooper 87 ('Ahrenfeld ') 31 v/Kirtle62 2QI 161008 292 RIDDLES OF THE EXETER BOOK Lamb of God 90 -Lance 73 ^Leather 13 Leather Bottle 19, 89 Leek, 26, 66 Letter-beam 61 k Letters of alphabet 14 "T Lock and Key 45, 91 "J Loom 57 /Lot, his two daughters, ' and their two sons 47 Lupus 90 i Mail-coat 36, Leiden Mail-shirt 62 \Man on horseback with spear and hawk 20, 65 .Martins 58 Jftlead 28 Measuring-worm 14 Millstone 5, 33 ime 25 23 oon 40, 95 Moon and Sun 30 Mustard 26 Night 12 Nightingale 9 (Obscene riddles 26, 43, 45, 46, 55, 62, 63, 64) Ocean-furrow 1 1 One-eyed Garlic-seller 86- NDnion 26, 66 Ore 83 Organ 86 Oven 18, 50, 55 )wl that eats snakes (Aspide-uf) 65 72 4 Oxen, Yoke of, 53 Oxhide 13 >4 Oyster 76, 77, 78 Paten 49 • Submarine earthquake 3 Peacock with rings on • Sun 7, 74 tail 65 Sun and Moon 30 Pen 61 10'17 1 Swallow and Sparrow 30 Pen and Fingers 52 4 Swallows 58 Pipe 9, 6 11'10, 70 "iSwan 8 Plow 22 VSword 21, 71, 80 Voker 63 Sword-rack 56 Porcupine 16 Pyx 49 Ten Chickens 14 #thunderstorm 4s7-66 Rain-drops 58 Time 40 Rain-water 31 ^tree 31^ (cf. 54, 56, 73, Rake 35 92) . Reed 61 Turning-lathe 57 -Reed-pen 6 110'17 Two Buckets 53 Reed-pipe Ql1'10 .Riddle i, 95 Visor 8 1 > River and Fish 85 Rune-staff 61 Wagon 33, 38, 72 Rye-straw (' Roggen- Wagon and Horse 52 halm ') 70 Wake (of ship) 1 1 Wandering Singer 95 Scabbard 56 "^ater 31, 42, 74, 84 Scop 95 Water-lily 1 1 Shawm 70 V/Weathercock 81 Sheath 45, 56 >!Web and Loom 57 ^Shield 6, 56 VWell with well-sweep s.Ship 33, 37, Si 59 Sfehirt 62 Wheels and Axle 72 Shoe and Foot 63 Whip 28 Sickle 91 Wine 12 Siren 74 Wine-cask 29 ^oul and Body 44 Winter 69 Sow with five farrow Wisdom 42 37 Wolf in two hop-rows -- Spear 54, 73, 80 (cf. 20, 90 65) \fetag-horn 88, 93 Wolves and Lamb (Apocalyptic) 90 Starlings 58 Woodpecker 25 ^ Storm 2-4 Wood-pigeon 9 Storm at sea 3, 4- Word of God 95 Storm on land and sea 2 Storm-clouds 58 .Young Bull 39 /-• PR 1760 ,A2 1910 SMC The riddles of the Exeter book AJZ-5495 |
4442 | https://www.cuemath.com/data/census/ | Census
Census is the collection of data from the entire population and not just a selected bunch of groups. The process of acquiring this number and accumulating the count is the census count which also might occur on a regular basis. For example, it is used while conducting a survey, a comparison between data, etc. Let us learn more about the census, the method of calculating, and solve a few examples.
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| 1. | Census Definition |
| 2. | Uses of Census |
| 3. | Census Vs Sampling |
| 4. | Average Daily Census |
| 5. | FAQs on Census |
Census Definition
By definition, census is considered as the process of calculating, acquiring, and recording data in a systematic manner about a certain given population. It is mostly used while collecting data on national population, housing censuses, agriculture, business, supplies, etc. This data provided complete information about job details, age factors, socioeconomic details, population size, and educational qualification of the group of the population being surveyed. Census is both used in math and statistics where it helps in collecting information about the population in different contexts such as backgrounds, place of stay, etc.
Uses of Census
Census is useful in different areas or fields, the uses are as follows:
Census Vs Sampling
Census is the process of finding and recording data about every member of a population that is mostly connected with national data, smaller populations, etc. For example, to find the number of dog owners, people above the age of 75, etc. Sampling is the opposite of census where it only includes a small percentage of the population. The accuracy in the census is more detailed than in sampling as it deals with only smaller groups. In other words, a census is generally more accurate and ultimately more useful than a sample study. For example, a census counts all of the women in Texas and records their ages. A sample might randomly select 1,000 women in Texas and ask each of them their age. The sample could offer an estimate of how many women in Texas are over 50 while a survey would identify exactly how many women in Texas are over 50. Census is considered to be an expensive method compared to sampling.
Average Daily Census
Average daily census also called ADC refers to the arithmetical average of the number of people who participated in a daily program that is calculated over a calendar year. This is calculated by adding the total number of service days divided by 365 days. Each organization sets a specific time frame that needs to be calculated using the average daily census formula. Hence, the formula is:
Average Daily Census = Total Number of Service Days / Total Number of Days.
☛Related Topics
Listed below are a few topics related to census.
Census Examples
Example 1: In the year 2010, hospitals had 150 beds for adults and children from January through June 30. On July 1 the hospital increased its beds to 200 and the number remained 200 through December 31. During the first six months, 27,813 patients days of service were provided to the hospital's adults and children. During the last 6 months, 35,873 days of service were provided. What is the average daily census of the first six months?
Solution:
While calculating the average daily census, we need to keep in mind both the leap year and non-leap year. Let us calculate for both.
First six months = 27,813 patients.
For a non-leap year:
Average Daily Census = Total Number of Service Days / Total Number of Days
= 27813/181 = 153.67
For a leap year:
Average Daily Census = Total Number of Service Days / Total Number of Days
= 27813/182 = 152.81
Therefore, the average daily census for the first six months is 153.67.
Example 2: Emily has 3 children where the product of their age is 36 and the sum is 16. With one being the oldest and the other two kids being the youngest. What is the age of all the three children?
Solution:
Since we know one child is old and the other two are young and the product of their age is 36. Hence, we find the factors,
36 = 2 × 2 × 3 × 3
Sum of the age = 16.
Therefore, the ages are 12, 3, and 1.
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FAQs on Census
What is Sample and Census in Math?
Census is the process of collecting data about everyone or everything in a population. It is also called complete enumeration. Whereas a sample is a subset of the unit in the population that represents data of a selected group of people in the population.
What is Census in Statistics?
In statistics, census is a survey that is conducted on the entire set of observations that is owned by the population. For example, calculating the population, the production of crops, the traffic on a particular road, etc.
What is Census in Simple Terms?
In simple terms, census is the collection of data from the entire population on business, agriculture, etc. This data provided complete information about job details, age factors, socioeconomic details, population size, and educational qualification of the group of the population being surveyed.
What is the Difference Between Census and Sampling?
Census gathers data about every member of the population whereas sampling gathers data only about a part of the population. The census helps in studying the data in a more accurate manner whereas sampling narrows the research to one particular group. |
4443 | http://www.gaokaoq.com/knowledge/index/cid/62/id/61.html | 随机事件的概率_高中数学知识点-高考圈电脑版
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科目：语文数学英语物理化学生物政治历史地理
高中数学知识点
集合
集合的基本运算
集合概念和集合间的基本关系
常用逻辑用语
命题及其关系、充分条件与必要条件
简单的逻辑联结词、全称量词和存在量词
函数、基本初等函数（I）、函数的应用
指数幂的含义及幂的运算
函数与方程
函数模型及其应用
函数的图象和性质
函数的概念与表示方法
幂函数
指数函数的概念、图象及其性质
对数的概念及其运算性质
对数函数的概念、图象及其性质
数列
数列的实际应用
数列
数列的通项及求和的几种方法
等差数列
等比数列
不等式
不等式的证明
不等式的性质与解不等式
二元一次不等式（组）与平面区域
基本不等式
特殊不等式
几何证明选讲
几何证明选讲
导数及其应用
定积分和微积分基本原理
导数在研究函数中的应用
导数的概念及其运算
曲线与方程
曲线与方程
坐标系与参数方程
参数方程
极坐标系
平面直角坐标系
算法初步
算法的含义与程序框图
基本算法语句
算法案例
计数原理
排列与组合
二项式定理
两个计数原理
概率
随机事件的概率
古典概型
几何概型
解三角形
解三角形的应用举例
正弦定理和余弦定理
三角函数
三角函数的图象与性质
三角函数的概念
简单的三角函数恒等变换
推理与证明
直接证明与间接证明
演绎推理与归纳推理
平面向量
平面向量的数量积及应用
平面向量的基本定理及坐标运算
平面向量的概念及线性运算
空间向量及其应用
空间向量及其运算
利用向量求空间的角和距离
空间向量证明平行与垂直的位置关系
空间几何
点、直线、平面之间的位置关系
空间几何体
直线与圆
圆与方程
直线与圆
直线与方程
圆锥曲线
直线与圆锥曲线的位置关系
双曲线
抛物线
椭圆
随机变量及其分布列
离散型随机变量及其分布列
互斥事件有一个发生的概率与条件概率
正态分布
独立事件同时发生的概率与独立重复试验的概率
离散型随机变量的期望与方差
数系的扩充与复数的引入
数系的扩充与复数的引入
统计与统计案例
统计
统计案例
圆柱、圆锥与圆锥曲线
用内切球探索圆锥曲线的性质
圆锥曲线性质的探讨
框图
框图
矩阵与变换
平面向量与二阶方阵
几何变换与矩阵
变换的合成与矩阵乘法
逆变换与逆矩阵
矩阵的特征值与特征向量
知识点总结
本节主要包括随机现象、不可能事件、必然事件、随机事件、事件间的运算及等可能事件概率等知识点。其中主要是理解和掌握等可能事件的概率。
常见考法
本节在段考中，主要是以选择题和填空题的形式考查等可能事件的概率，有时也融合在解答题中考查等可能事件的概率。在高考中有时是融合在离散型随机变量的分布列中联合考查等可能事件的概率。一般属于容易题。
误区提醒
注意理解频率和概率的区别，频率指的是频数除以总数，概率是大量实验下频率的极限。
【典型例题】
例1 在添加剂的搭配使用中，为了找到最佳的搭配方案，需要对各种不同的搭配方式
作比较．在试制某种牙膏新品种时，需要选用两种不同的添加剂．现有芳香度分别为0,1,2,3,4,5的六种添加剂可供选用．根据试验设计原理，通常首先要随机选取两种不同的添加剂进行搭配试验．
(1)求所选用的两种不同的添加剂的芳香度之和等于4的概率；
(2)求所选用的两种不同的添加剂的芳香度之和不小于3的概率．
解：设“所选用的两种不同的添加剂的芳香度之和等于4”的事件为A，“所选用的两种不同的添加剂的芳香度之和不小于3”的事件为B.
从六种中随机选两种共有(0,1)、(0,2)、(0,3)、(0,4)、(0,5)、(1,2)、(1,3)、(1,4)、(1,5)、(2,3)、(2,4)、(2,5)、(3,4)、(3,5)、(4,5)15种．
(1)“所选用的两种不同的添加剂的芳香度之和等于4”的取法有2种：(0,4)、(1,3)，故P(A)＝2/15.
(2)“所选用的两种不同的添加剂的芳香度之和等于1”的取法有1种：(0,1)；“所选用的两种不同的添加剂的芳香度之和等于2”的取法有1种：(0,2)，
故P(B)＝1－(1/15＋1/15)＝13/15.
例2 据统计，某食品企业在一个月内被消费者投诉次数为0,1,2的概率分别为0.4,0.5,0.1.
(1)求该企业在一个月内被消费者投诉不超过1次的概率；
(2)假设一月份与二月份被消费者投诉的次数互不影响，求该企业在这两个月内共被消费者投诉2次的概率．
解：法一：(1)设事件A表示“一个月内被投诉的次数为0”，事件B表示“一个月内被投诉的次数为1”，
∴P(A＋B)＝P(A)＋P(B)＝0.4＋0.5＝0.9.
(2)设事件Ai表示“第i个月被投诉的次数为0”，事件Bi表示“第i个月被投诉的次数为1”，事件Ci表示“第i个月被投诉的次数为2”，事件D表示“两个月内共被投诉2次”．
∴P(Ai)＝0.4，P(Bi)＝0.5，P(Ci)＝0.1(i＝1,2)．
∵两个月中，一个月被投诉2次，另一个月被投诉0次的概率为P(A1C2＋A2C1)，
一、二月份均被投诉1次的概率为P(B1B2)，
∴P(D)＝P(A1C2＋A2C1)＋P(B1B2)＝P(A1C2)＋P(A2C1)＋P(B1B2)，由事件的独立性得
P(D)＝0.4×0.1＋0.1×0.4＋0.5×0.5＝0.33.
法二：(1)设事件A表示“一个月内被投诉2次”，事件B表示“一个月内被投诉的次数不超过1次”．
∵P(A)＝0.1，∴P(B)＝1－P(A)＝1－0.1＝0.9.
(2)同法一．
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4444 | https://brainly.com/textbook-solutions/b-abstract-algebra-3rd-edition-college-math-9780471433347 | Abstract Algebra, 3rd Edition - Answers & Solutions | Brainly
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Abstract Algebra, 3rd Edition
Textbook Solutions
Abstract Algebra, 3rd Edition Solutions
Richard M. Foote, David S. Dummit
2002
ISBN: 9780471433347
Dummit and Foote: Abstract Algebra Answers
The Abstract Algebra student edition book is divided into 19 chapters and two appendixes. They range from Groups and Subgroups to Field Theory and Character Theory. The Abstract Algebra Dummit and Foote solutions provided by Brainly include solutions to all the chapters in the book. It will help students master their deductive reasoning through these exercises.
Our experts are still working on solutions for this textbook
Request answers by clicking “No answer” next to the exercises below.
Chapter 0: Preliminaries
read chapter notes
0.1 Exercises 3
Exercise 1 4
Exercise 2 4
Exercise 3 4
Exercise 4 4
Exercise 5 4
Exercise 6 4
Exercise 7 4
0.2 Exercises 7
Exercise 1 7
Exercise 2 7
Exercise 3 8
Exercise 4 8
Exercise 5 8
Exercise 6 8
Exercise 7 8
Exercise 8 8
Exercise 9 8
Exercise 10 8
Exercise 11 8
0.3 Exercises 11
Exercise 1 11
Exercise 2 11
Exercise 3 11
Exercise 4 11
Exercise 5 11
Exercise 6 11
Exercise 7 11
Exercise 8 11
Exercise 9 11
Exercise 10 11
Exercise 11 11
Exercise 12 12
Exercise 13 12
Exercise 14 12
Exercise 15 12
Exercise 16 12
Chapter 1: Introduction to Groups
read chapter notes
1.1 Exercises 21
Exercise 1 21
Exercise 2 21
Exercise 3 21
Exercise 4 21
Exercise 5 21
Exercise 6 21
Exercise 7 21
Exercise 8 22
Exercise 9 22
Exercise 10 22
Exercise 11 22
Exercise 12 22
Exercise 13 22
Exercise 14 22
Exercise 15 22
Exercise 16 22
Exercise 17 22
Exercise 18 22
Exercise 19 22
Exercise 20 22
Exercise 21 22
Exercise 22 22
Exercise 23 22
Exercise 24 22
Exercise 25 22
Exercise 26 22
Exercise 27 22
Exercise 28 22
Exercise 29 23
Exercise 30 23
Exercise 31 23
Exercise 32 23
Exercise 33 23
Exercise 34 23
Exercise 35 23
Exercise 36 23
1.2 Exercises 27
Exercise 1 27
Exercise 2 27
Exercise 3 27
Exercise 4 28
Exercise 5 28
Exercise 6 28
Exercise 7 28
Exercise 8 28
Exercise 9 28
Exercise 10 28
Exercise 11 28
Exercise 12 28
Exercise 13 28
Exercise 14 28
Exercise 15 28
Exercise 16 28
Exercise 17 28
Exercise 18 28
1.3 Exercises 32
Exercise 1 32
Exercise 2 33
Exercise 3 33
Exercise 4 33
Exercise 5 33
Exercise 6 33
Exercise 7 33
Exercise 8 33
Exercise 9 33
Exercise 10 33
Exercise 11 33
Exercise 12 33
Exercise 13 33
Exercise 14 33
Exercise 15 33
Exercise 16 33
Exercise 17 34
Exercise 18 34
Exercise 19 34
Exercise 20 34
1.4 Exercises 35
Exercise 1 35
Exercise 2 35
Exercise 3 35
Exercise 4 35
Exercise 5 35
Exercise 6 35
Exercise 7 35
Exercise 8 35
Exercise 9 35
Exercise 10 35
Exercise 11 35
1.5 Exercises 36
Exercise 1 36
Exercise 2 36
Exercise 3 36
1.6 Exercises 39
Exercise 1 39
Exercise 2 40
Exercise 3 40
Exercise 4 40
Exercise 5 40
Exercise 6 40
Exercise 7 40
Exercise 8 40
Exercise 9 40
Exercise 10 40
Exercise 11 40
Exercise 12 40
Exercise 13 40
Exercise 14 40
Exercise 15 40
Exercise 16 40
Exercise 17 40
Exercise 18 40
Exercise 19 40
Exercise 20 41
Exercise 21 41
Exercise 22 41
Exercise 23 41
Exercise 24 41
Exercise 25 41
Exercise 26 41
1.7 Exercises 44
Exercise 1 44
Exercise 2 44
Exercise 3 44
Exercise 4 44
Exercise 5 44
Exercise 6 44
Exercise 7 44
Exercise 8 44
Exercise 9 44
Exercise 10 44
Exercise 11 45
Exercise 12 45
Exercise 13 45
Exercise 14 45
Exercise 15 45
Exercise 16 45
Exercise 17 45
Exercise 18 45
Exercise 19 45
Exercise 20 45
Exercise 21 45
Exercise 22 45
Exercise 23 45
Chapter 2: Subgroups
read chapter notes
2.1 Exercises 48
Exercise 1 48
Exercise 2 48
Exercise 3 48
Exercise 4 48
Exercise 5 48
Exercise 6 48
Exercise 7 48
Exercise 8 48
Exercise 9 48
Exercise 10 48
Exercise 11 48
Exercise 12 49
Exercise 13 49
Exercise 14 49
Exercise 15 49
Exercise 16 49
Exercise 17 49
2.2 Exercises 52
Exercise 1 52
Exercise 2 52
Exercise 3 52
Exercise 4 52
Exercise 5 52
Exercise 6 52
Exercise 7 52
Exercise 8 53
Exercise 9 53
Exercise 10 53
Exercise 11 53
Exercise 12 53
Exercise 13 53
Exercise 14 53
2.3 Exercises 60
Exercise 1 60
Exercise 2 60
Exercise 3 60
Exercise 4 60
Exercise 5 60
Exercise 6 60
Exercise 7 60
Exercise 8 60
Exercise 9 60
Exercise 10 60
Exercise 11 60
Exercise 12 60
Exercise 13 60
Exercise 14 60
Exercise 15 60
Exercise 16 60
Exercise 17 60
Exercise 18 60
Exercise 19 60
Exercise 20 60
Exercise 21 60
Exercise 22 61
Exercise 23 61
Exercise 24 61
Exercise 25 61
Exercise 26 61
2.4 Exercises 65
Exercise 1 65
Exercise 2 65
Exercise 3 65
Exercise 4 65
Exercise 5 65
Exercise 6 65
Exercise 7 65
Exercise 8 65
Exercise 9 65
Exercise 10 65
Exercise 11 65
Exercise 12 65
Exercise 13 65
Exercise 14 65
Exercise 15 65
Exercise 16 65
Exercise 17 65
Exercise 18 66
Exercise 19 66
Exercise 20 66
2.5 Exercises 71
Exercise 1 71
Exercise 2 71
Exercise 3 71
Exercise 4 71
Exercise 5 71
Exercise 6 71
Exercise 7 71
Exercise 8 71
Exercise 9 71
Exercise 10 71
Exercise 11 71
Exercise 12 71
Exercise 13 72
Exercise 14 72
Exercise 15 72
Exercise 16 72
Exercise 17 72
Exercise 18 72
Exercise 19 72
Exercise 20 72
Chapter 3: Quotient Groups and Homomorphisms
3.1 Exercises 85
Exercise 1 85
Exercise 2 85
Exercise 3 85
Exercise 4 85
Exercise 5 85
Exercise 6 85
Exercise 7 85
Exercise 8 85
Exercise 9 85
Exercise 10 86
Exercise 11 86
Exercise 12 86
Exercise 13 86
Exercise 14 86
Exercise 15 86
Exercise 16 86
Exercise 17 87
Exercise 18 87
Exercise 19 87
Exercise 20 87
Exercise 21 87
Exercise 22 88
Exercise 23 88
Exercise 24 88
Exercise 25 88
Exercise 26 88
Exercise 27 88
Exercise 28 88
Exercise 29 88
Exercise 30 88
Exercise 31 88
Exercise 32 88
Exercise 33 88
Exercise 34 88
Exercise 35 89
Exercise 36 89
Exercise 37 89
Exercise 38 89
Exercise 39 89
Exercise 40 89
Exercise 41 89
Exercise 42 89
Exercise 43 89
3.2 Exercises 95
Exercise 1 95
Exercise 2 95
Exercise 3 95
Exercise 4 95
Exercise 5 95
Exercise 6 95
Exercise 7 95
Exercise 8 95
Exercise 9 96
Exercise 10 96
Exercise 11 96
Exercise 12 96
Exercise 13 96
Exercise 14 96
Exercise 15 96
Exercise 16 96
Exercise 17 96
Exercise 18 96
Exercise 19 96
Exercise 20 96
Exercise 21 96
Exercise 22 96
Exercise 23 96
3.3 Exercises 101
Exercise 1 101
Exercise 2 101
Exercise 3 101
Exercise 4 101
Exercise 5 101
Exercise 6 101
Exercise 7 101
Exercise 8 101
Exercise 9 101
Exercise 10 101
3.4 Exercises 106
Exercise 1 106
Exercise 2 106
Exercise 3 106
Exercise 4 106
Exercise 5 106
Exercise 6 106
Exercise 7 106
Exercise 8 106
Exercise 9 106
Exercise 10 106
Exercise 11 106
Exercise 12 106
3.5 Exercises 111
Exercise 1 111
Exercise 2 111
Exercise 3 111
Exercise 4 111
Exercise 5 111
Exercise 6 111
Exercise 7 111
Exercise 8 111
Exercise 9 111
Exercise 10 111
Exercise 11 111
Exercise 12 111
Exercise 13 111
Exercise 14 111
Exercise 15 111
Exercise 16 111
Exercise 17 111
Chapter 4: Group Actions
read chapter notes
4.1 Exercises 116
Exercise 1 116
Exercise 2 116
Exercise 3 116
Exercise 4 116
Exercise 5 116
Exercise 6 116
Exercise 7 117
Exercise 8 117
Exercise 9 117
Exercise 10 117
4.2 Exercises 121
Exercise 1 121
Exercise 2 121
Exercise 3 121
Exercise 4 121
Exercise 5 121
Exercise 6 122
Exercise 7 122
Exercise 8 122
Exercise 9 122
Exercise 10 122
Exercise 11 122
Exercise 12 122
Exercise 13 122
Exercise 14 122
4.3 Exercises 130
Exercise 1 130
Exercise 2 130
Exercise 3 130
Exercise 4 130
Exercise 5 130
Exercise 6 130
Exercise 7 130
Exercise 8 130
Exercise 9 130
Exercise 10 130
Exercise 11 130
Exercise 12 130
Exercise 13 130
Exercise 14 130
Exercise 15 131
Exercise 16 131
Exercise 17 131
Exercise 18 131
Exercise 19 131
Exercise 20 131
Exercise 21 131
Exercise 22 131
Exercise 23 131
Exercise 24 131
Exercise 25 131
Exercise 26 132
Exercise 27 132
Exercise 28 132
Exercise 29 132
Exercise 30 132
Exercise 31 132
Exercise 32 132
Exercise 33 132
Exercise 34 132
Exercise 35 132
Exercise 36 132
4.4 Exercises 137
Exercise 1 137
Exercise 2 137
Exercise 3 137
Exercise 4 137
Exercise 5 137
Exercise 6 137
Exercise 7 137
Exercise 8 137
Exercise 9 138
Exercise 10 138
Exercise 11 138
Exercise 12 138
Exercise 13 138
Exercise 14 138
Exercise 15 138
Exercise 16 138
Exercise 17 138
Exercise 18 138
Exercise 19 138
Exercise 20 139
4.5 Exercises 146
Exercise 1 146
Exercise 2 146
Exercise 3 146
Exercise 4 146
Exercise 5 146
Exercise 6 146
Exercise 7 146
Exercise 8 146
Exercise 9 146
Exercise 10 146
Exercise 11 146
Exercise 12 146
Exercise 13 147
Exercise 14 147
Exercise 15 147
Exercise 16 147
Exercise 17 147
Exercise 18 147
Exercise 19 147
Exercise 20 147
Exercise 21 147
Exercise 22 147
Exercise 23 147
Exercise 24 147
Exercise 25 147
Exercise 26 147
Exercise 27 147
Exercise 28 147
Exercise 29 147
Exercise 30 147
Exercise 31 147
Exercise 32 147
Exercise 33 147
Exercise 34 147
Exercise 35 147
Exercise 36 148
Exercise 37 148
Exercise 38 148
Exercise 39 148
Exercise 40 148
Exercise 41 148
Exercise 42 148
Exercise 43 148
Exercise 44 148
Exercise 45 148
Exercise 46 148
Exercise 47 148
Exercise 48 148
Exercise 49 148
Exercise 50 148
Exercise 51 149
Exercise 52 149
Exercise 53 149
Exercise 54 149
Exercise 55 149
Exercise 56 149
4.6 Exercises 151
Exercise 1 151
Exercise 2 151
Exercise 3 151
Exercise 4 151
Exercise 5 151
Exercise 6 151
Exercise 7 151
Exercise 8 151
Chapter 5: Direct and Semidirect Products and Abelian Groups
read chapter notes
5.1 Exercises 156
Exercise 1 156
Exercise 2 156
Exercise 3 156
Exercise 4 156
Exercise 5 156
Exercise 6 156
Exercise 7 156
Exercise 8 156
Exercise 9 157
Exercise 10 157
Exercise 11 157
Exercise 12 157
Exercise 13 157
Exercise 14 157
Exercise 15 157
Exercise 16 158
Exercise 17 158
Exercise 18 158
5.2 Exercises 165
Exercise 1 165
Exercise 2 165
Exercise 3 165
Exercise 4 166
Exercise 5 166
Exercise 6 166
Exercise 7 166
Exercise 8 166
Exercise 9 166
Exercise 10 166
Exercise 11 166
Exercise 12 166
Exercise 13 166
Exercise 14 167
Exercise 15 167
Exercise 16 167
5.3 Exercises 169
Exercise 1 169
5.4 Exercises 173
Exercise 1 173
Exercise 2 173
Exercise 3 173
Exercise 4 174
Exercise 5 174
Exercise 6 174
Exercise 7 174
Exercise 8 174
Exercise 9 174
Exercise 10 174
Exercise 11 174
Exercise 12 174
Exercise 13 174
Exercise 14 174
Exercise 15 174
Exercise 16 174
Exercise 17 174
Exercise 18 174
Exercise 19 174
Exercise 20 174
5.5 Exercises 184
Exercise 1 184
Exercise 2 184
Exercise 3 184
Exercise 4 184
Exercise 5 184
Exercise 6 184
Exercise 7 185
Exercise 8 185
Exercise 9 185
Exercise 10 185
Exercise 11 186
Exercise 12 186
Exercise 13 186
Exercise 14 186
Exercise 15 186
Exercise 16 186
Exercise 17 186
Exercise 18 186
Exercise 19 186
Exercise 20 187
Exercise 21 187
Exercise 22 187
Exercise 23 187
Exercise 24 187
Exercise 25 187
Chapter 6: Further Topics in Group Theory
read chapter notes
6.1 Exercises 198
Exercise 1 198
Exercise 2 198
Exercise 3 198
Exercise 4 198
Exercise 5 198
Exercise 6 198
Exercise 7 198
Exercise 8 198
Exercise 9 198
Exercise 10 198
Exercise 11 198
Exercise 12 198
Exercise 13 198
Exercise 14 198
Exercise 15 198
Exercise 16 198
Exercise 17 198
Exercise 18 198
Exercise 19 198
Exercise 20 198
Exercise 21 199
Exercise 22 199
Exercise 23 199
Exercise 24 199
Exercise 25 199
Exercise 26 199
Exercise 27 199
Exercise 28 199
Exercise 29 200
Exercise 30 200
Exercise 31 200
Exercise 32 200
Exercise 33 200
Exercise 34 200
Exercise 35 200
Exercise 36 200
Exercise 37 200
Exercise 38 201
6.2 Exercises 213
Exercise 1 213
Exercise 2 213
Exercise 3 213
Exercise 4 213
Exercise 5 213
Exercise 6 213
Exercise 7 213
Exercise 8 213
Exercise 9 213
Exercise 10 213
Exercise 11 213
Exercise 12 213
Exercise 13 213
Exercise 14 214
Exercise 15 214
Exercise 16 214
Exercise 17 214
Exercise 18 214
Exercise 19 214
Exercise 20 214
Exercise 21 214
Exercise 22 214
Exercise 23 214
Exercise 24 214
Exercise 25 214
Exercise 26 214
Exercise 27 214
Exercise 28 214
Exercise 29 214
Exercise 30 214
6.3 Exercises 220
Exercise 1 220
Exercise 2 220
Exercise 3 220
Exercise 4 220
Exercise 5 220
Exercise 6 220
Exercise 7 220
Exercise 8 220
Exercise 9 221
Exercise 10 221
Exercise 11 221
Exercise 12 221
Exercise 13 221
Chapter 7: Introduction to Rings
7.1 Exercises 230
Exercise 1 230
Exercise 2 230
Exercise 3 230
Exercise 4 230
Exercise 5 230
Exercise 6 231
Exercise 7 231
Exercise 8 231
Exercise 9 231
Exercise 10 231
Exercise 11 231
Exercise 12 231
Exercise 13 231
Exercise 14 231
Exercise 15 231
Exercise 16 231
Exercise 17 231
Exercise 18 231
Exercise 19 231
Exercise 20 232
Exercise 21 232
Exercise 22 232
Exercise 23 232
Exercise 24 232
Exercise 25 232
Exercise 26 232
Exercise 27 232
Exercise 28 233
Exercise 29 233
Exercise 30 233
7.2 Exercises 237
Exercise 1 237
Exercise 2 238
Exercise 3 238
Exercise 4 238
Exercise 5 238
Exercise 6 238
Exercise 7 239
Exercise 8 239
Exercise 9 239
Exercise 10 239
Exercise 11 239
Exercise 12 239
Exercise 13 239
7.3 Exercises 247
Exercise 1 247
Exercise 2 247
Exercise 3 247
Exercise 4 248
Exercise 5 248
Exercise 6 248
Exercise 7 248
Exercise 8 248
Exercise 9 248
Exercise 10 248
Exercise 11 248
Exercise 12 248
Exercise 13 249
Exercise 14 249
Exercise 15 249
Exercise 16 249
Exercise 17 249
Exercise 18 249
Exercise 19 249
Exercise 20 249
Exercise 21 249
Exercise 22 249
Exercise 23 249
Exercise 24 249
Exercise 25 249
Exercise 26 250
Exercise 27 250
Exercise 28 250
Exercise 29 250
Exercise 30 250
Exercise 31 250
Exercise 32 250
Exercise 33 250
Exercise 34 250
Exercise 35 250
Exercise 36 251
Exercise 37 251
7.4 Exercises 256
Exercise 1 256
Exercise 2 256
Exercise 3 256
Exercise 4 256
Exercise 5 256
Exercise 6 256
Exercise 7 256
Exercise 8 256
Exercise 9 256
Exercise 10 257
Exercise 11 257
Exercise 12 257
Exercise 13 257
Exercise 14 257
Exercise 15 257
Exercise 16 257
Exercise 17 257
Exercise 18 258
Exercise 19 258
Exercise 20 258
Exercise 21 258
Exercise 22 258
Exercise 23 258
Exercise 24 258
Exercise 25 258
Exercise 26 258
Exercise 27 258
Exercise 28 258
Exercise 29 258
Exercise 30 258
Exercise 31 258
Exercise 32 259
Exercise 33 259
Exercise 34 259
Exercise 35 259
Exercise 36 259
Exercise 37 259
Exercise 38 259
Exercise 39 259
Exercise 40 260
Exercise 41 260
7.5 Exercises 264
Exercise 1 264
Exercise 2 264
Exercise 3 264
Exercise 4 264
Exercise 5 265
Exercise 6 265
7.6 Exercises 267
Exercise 1 267
Exercise 2 267
Exercise 3 267
Exercise 4 267
Exercise 5 267
Exercise 6 268
Exercise 7 268
Exercise 8 268
Exercise 9 269
Exercise 10 269
Exercise 11 269
Chapter 8: Euclidean Domains, Principal Ideal Domains, and Unique Factorization Domains
read chapter notes
8.1 Exercises 277
Exercise 1 277
Exercise 2 277
Exercise 3 278
Exercise 4 278
Exercise 5 278
Exercise 6 278
Exercise 7 278
Exercise 8 278
Exercise 9 278
Exercise 10 278
Exercise 11 279
Exercise 12 279
8.2 Exercises 282
Exercise 1 282
Exercise 2 282
Exercise 3 282
Exercise 4 282
Exercise 5 283
Exercise 6 283
Exercise 7 283
Exercise 8 283
8.3 Exercises 292
Exercise 1 292
Exercise 2 293
Exercise 3 293
Exercise 4 293
Exercise 5 293
Exercise 6 293
Exercise 7 293
Exercise 8 293
Exercise 9 294
Exercise 10 294
Exercise 11 294
Chapter 9: Polynomial Rings
read chapter notes
9.1 Exercises 298
Exercise 1 298
Exercise 2 298
Exercise 3 298
Exercise 4 298
Exercise 5 298
Exercise 6 298
Exercise 7 298
Exercise 8 298
Exercise 9 298
Exercise 10 298
Exercise 11 298
Exercise 12 298
Exercise 13 298
Exercise 14 298
Exercise 15 299
Exercise 16 299
Exercise 17 299
Exercise 18 299
9.2 Exercises 301
Exercise 1 301
Exercise 2 301
Exercise 3 301
Exercise 4 301
Exercise 5 301
Exercise 6 301
Exercise 7 301
Exercise 8 301
Exercise 9 301
Exercise 10 301
Exercise 11 301
Exercise 12 302
Exercise 13 302
9.3 Exercises 306
Exercise 1 306
Exercise 2 306
Exercise 3 306
Exercise 4 306
Exercise 5 306
9.4 Exercises 311
Exercise 1 311
Exercise 2 311
Exercise 3 311
Exercise 4 311
Exercise 5 311
Exercise 6 311
Exercise 7 311
Exercise 8 311
Exercise 9 311
Exercise 10 311
Exercise 11 312
Exercise 12 312
Exercise 13 312
Exercise 14 312
Exercise 15 312
Exercise 16 312
Exercise 17 312
Exercise 18 312
Exercise 19 312
Exercise 20 312
9.5 Exercises 315
Exercise 1 315
Exercise 2 315
Exercise 3 315
Exercise 4 315
Exercise 5 315
Exercise 6 315
Exercise 7 315
9.6 Exercises 330
Exercise 1 330
Exercise 2 330
Exercise 3 331
Exercise 4 331
Exercise 5 331
Exercise 6 331
Exercise 7 331
Exercise 8 331
Exercise 9 331
Exercise 10 332
Exercise 11 332
Exercise 12 332
Exercise 13 332
Exercise 14 332
Exercise 15 332
Exercise 16 332
Exercise 17 332
Exercise 18 332
Exercise 19 332
Exercise 20 332
Exercise 21 332
Exercise 22 332
Exercise 23 333
Exercise 24 333
Exercise 25 333
Exercise 26 333
Exercise 27 333
Exercise 28 333
Exercise 29 333
Exercise 30 333
Exercise 31 333
Exercise 32 333
Exercise 33 333
Exercise 34 333
Exercise 35 334
Exercise 36 334
Exercise 37 334
Exercise 38 334
Exercise 39 334
Exercise 40 334
Exercise 41 334
Exercise 42 334
Exercise 43 334
Exercise 44 334
Exercise 45 335
Chapter 10: Introduction to Module Theory
read chapter notes
10.1 Exercises 343
Exercise 1 343
Exercise 2 343
Exercise 3 343
Exercise 4 343
Exercise 5 343
Exercise 6 343
Exercise 7 344
Exercise 8 344
Exercise 9 344
Exercise 10 344
Exercise 11 344
Exercise 12 344
Exercise 13 344
Exercise 14 344
Exercise 15 344
Exercise 16 344
Exercise 17 344
Exercise 18 344
Exercise 19 344
Exercise 20 344
Exercise 21 345
Exercise 22 345
Exercise 23 345
10.2 Exercises 350
Exercise 1 350
Exercise 2 350
Exercise 3 350
Exercise 4 350
Exercise 5 350
Exercise 6 350
Exercise 7 350
Exercise 8 350
Exercise 9 350
Exercise 10 350
Exercise 11 350
Exercise 12 350
Exercise 13 350
Exercise 14 350
10.3 Exercises 356
Exercise 1 356
Exercise 2 356
Exercise 3 356
Exercise 4 356
Exercise 5 356
Exercise 6 356
Exercise 7 356
Exercise 8 356
Exercise 9 356
Exercise 10 356
Exercise 11 356
Exercise 12 356
Exercise 13 356
Exercise 14 357
Exercise 15 357
Exercise 16 357
Exercise 17 357
Exercise 18 357
Exercise 19 357
Exercise 20 357
Exercise 21 357
Exercise 22 358
Exercise 23 358
Exercise 24 358
Exercise 25 358
Exercise 26 358
Exercise 27 358
10.4 Exercises 375
Exercise 1 375
Exercise 2 375
Exercise 3 375
Exercise 4 375
Exercise 5 375
Exercise 6 375
Exercise 7 375
Exercise 8 375
Exercise 9 376
Exercise 10 376
Exercise 11 376
Exercise 12 376
Exercise 13 376
Exercise 14 376
Exercise 15 376
Exercise 16 376
Exercise 17 376
Exercise 18 377
Exercise 19 377
Exercise 20 377
Exercise 21 377
Exercise 22 377
Exercise 23 377
Exercise 24 377
Exercise 25 377
Exercise 26 377
Exercise 27 377
10.5 Exercises 403
Exercise 1 403
Exercise 2 403
Exercise 3 403
Exercise 4 403
Exercise 5 403
Exercise 6 403
Exercise 7 404
Exercise 8 404
Exercise 9 404
Exercise 10 404
Exercise 11 404
Exercise 12 404
Exercise 13 404
Exercise 14 404
Exercise 15 405
Exercise 16 405
Exercise 17 405
Exercise 18 405
Exercise 19 405
Exercise 20 405
Exercise 21 405
Exercise 22 406
Exercise 23 406
Exercise 24 406
Exercise 25 406
Exercise 26 406
Exercise 27 407
Exercise 28 407
Chapter 11: Vector Spaces
read chapter notes
11.1 Exercises 413
Exercise 1 413
Exercise 2 413
Exercise 3 414
Exercise 4 414
Exercise 5 414
Exercise 6 414
Exercise 7 414
Exercise 8 414
Exercise 9 414
Exercise 10 414
Exercise 11 414
Exercise 12 414
Exercise 13 414
Exercise 14 414
11.2 Exercises 422
Exercise 1 422
Exercise 2 422
Exercise 3 422
Exercise 4 422
Exercise 5 422
Exercise 6 423
Exercise 7 423
Exercise 8 423
Exercise 9 423
Exercise 10 423
Exercise 11 423
Exercise 12 423
Exercise 13 424
Exercise 14 425
Exercise 15 425
Exercise 16 425
Exercise 17 425
Exercise 18 425
Exercise 19 425
Exercise 20 425
Exercise 21 426
Exercise 22 426
Exercise 23 427
Exercise 24 427
Exercise 25 427
Exercise 26 427
Exercise 27 427
Exercise 28 428
Exercise 29 429
Exercise 30 429
Exercise 31 429
Exercise 32 430
Exercise 33 430
Exercise 34 431
Exercise 35 431
Exercise 36 431
Exercise 37 431
Exercise 38 431
Exercise 39 431
11.3 Exercises 435
Exercise 1 435
Exercise 2 435
Exercise 3 435
Exercise 4 435
Exercise 5 435
11.4 Exercises 441
Exercise 1 441
Exercise 2 441
Exercise 3 441
Exercise 4 441
Exercise 5 441
Exercise 6 441
11.5 Exercises 454
Exercise 1 454
Exercise 2 454
Exercise 3 455
Exercise 4 455
Exercise 5 455
Exercise 6 455
Exercise 7 455
Exercise 8 455
Exercise 9 455
Exercise 10 455
Exercise 11 455
Exercise 12 455
Exercise 13 455
Exercise 14 455
Chapter 12: Modules over Principal Ideal Domains
read chapter notes
12.1 Exercises 468
Exercise 1 468
Exercise 2 468
Exercise 3 469
Exercise 4 469
Exercise 5 469
Exercise 6 469
Exercise 7 469
Exercise 8 469
Exercise 9 469
Exercise 10 469
Exercise 11 469
Exercise 12 469
Exercise 13 469
Exercise 14 469
Exercise 15 469
Exercise 16 470
Exercise 17 470
Exercise 18 470
Exercise 19 470
Exercise 20 471
Exercise 21 472
Exercise 22 472
12.2 Exercises 488
Exercise 1 488
Exercise 2 488
Exercise 3 488
Exercise 4 488
Exercise 5 488
Exercise 6 488
Exercise 7 488
Exercise 8 488
Exercise 9 489
Exercise 10 489
Exercise 11 489
Exercise 12 489
Exercise 13 489
Exercise 14 489
Exercise 15 489
Exercise 16 489
Exercise 17 489
Exercise 18 489
Exercise 19 489
Exercise 20 489
Exercise 21 490
Exercise 22 490
Exercise 23 490
Exercise 24 490
Exercise 25 490
12.3 Exercises 499
Exercise 1 499
Exercise 2 499
Exercise 3 499
Exercise 4 499
Exercise 5 499
Exercise 6 499
Exercise 7 499
Exercise 8 500
Exercise 9 500
Exercise 10 500
Exercise 11 500
Exercise 12 500
Exercise 13 500
Exercise 14 500
Exercise 15 500
Exercise 16 501
Exercise 17 501
Exercise 18 501
Exercise 19 501
Exercise 20 501
Exercise 21 501
Exercise 22 501
Exercise 23 501
Exercise 24 501
Exercise 25 501
Exercise 26 501
Exercise 27 501
Exercise 28 501
Exercise 29 501
Exercise 30 502
Exercise 31 502
Exercise 32 502
Exercise 33 502
Exercise 34 502
Exercise 35 502
Exercise 36 502
Exercise 37 502
Exercise 38 502
Exercise 39 502
Exercise 40 503
Exercise 41 503
Exercise 42 503
Exercise 43 504
Exercise 44 504
Exercise 45 504
Exercise 46 504
Exercise 47 505
Exercise 48 505
Exercise 49 505
Exercise 50 505
Exercise 51 505
Exercise 52 505
Exercise 53 506
Exercise 54 506
Exercise 55 507
Exercise 56 507
Exercise 57 507
Exercise 58 507
Exercise 59 508
Chapter 13: Field Theory
read chapter notes
13.1 Exercises 519
Exercise 1 519
Exercise 2 519
Exercise 3 519
Exercise 4 519
Exercise 5 519
Exercise 6 519
Exercise 7 519
Exercise 8 519
13.2 Exercises 529
Exercise 1 529
Exercise 2 529
Exercise 3 529
Exercise 4 530
Exercise 5 530
Exercise 6 530
Exercise 7 530
Exercise 8 530
Exercise 9 530
Exercise 10 530
Exercise 11 530
Exercise 12 530
Exercise 13 530
Exercise 14 530
Exercise 15 530
Exercise 16 530
Exercise 17 530
Exercise 18 530
Exercise 19 531
Exercise 20 531
Exercise 21 531
Exercise 22 531
13.3 Exercises 535
Exercise 1 535
Exercise 2 535
Exercise 3 536
Exercise 4 536
Exercise 5 536
13.4 Exercises 545
Exercise 1 545
Exercise 2 545
Exercise 3 545
Exercise 4 545
Exercise 5 545
Exercise 6 545
13.5 Exercises 551
Exercise 1 551
Exercise 2 551
Exercise 3 551
Exercise 4 551
Exercise 5 551
Exercise 6 551
Exercise 7 552
Exercise 8 552
Exercise 9 552
Exercise 10 552
Exercise 11 552
13.6 Exercises 555
Exercise 1 555
Exercise 2 555
Exercise 3 555
Exercise 4 555
Exercise 5 555
Exercise 6 555
Exercise 7 555
Exercise 8 556
Exercise 9 556
Exercise 10 556
Exercise 11 556
Exercise 12 556
Exercise 13 556
Exercise 14 557
Exercise 15 557
Exercise 16 557
Exercise 17 557
Chapter 14: Galois Theory
14.1 Exercises 566
Exercise 1 566
Exercise 2 567
Exercise 3 567
Exercise 4 567
Exercise 5 567
Exercise 6 567
Exercise 7 567
Exercise 8 567
Exercise 9 567
Exercise 10 567
14.2 Exercises 581
Exercise 1 581
Exercise 2 581
Exercise 3 581
Exercise 4 582
Exercise 5 582
Exercise 6 582
Exercise 7 582
Exercise 8 582
Exercise 9 582
Exercise 10 582
Exercise 11 582
Exercise 12 582
Exercise 13 582
Exercise 14 582
Exercise 15 582
Exercise 16 582
Exercise 17 582
Exercise 18 583
Exercise 19 583
Exercise 20 583
Exercise 21 583
Exercise 22 583
Exercise 23 583
Exercise 24 584
Exercise 25 584
Exercise 26 584
Exercise 27 584
Exercise 28 584
Exercise 29 585
Exercise 30 585
Exercise 31 585
14.3 Exercises 589
Exercise 1 589
Exercise 2 589
Exercise 3 589
Exercise 4 589
Exercise 5 589
Exercise 6 589
Exercise 7 589
Exercise 8 589
Exercise 9 589
Exercise 10 589
Exercise 11 589
Exercise 12 589
Exercise 13 590
Exercise 14 590
Exercise 15 590
Exercise 16 590
Exercise 17 590
14.4 Exercises 595
Exercise 1 595
Exercise 2 595
Exercise 3 595
Exercise 4 595
Exercise 5 596
Exercise 6 596
Exercise 7 596
Exercise 8 596
14.5 Exercises 603
Exercise 1 603
Exercise 2 603
Exercise 3 603
Exercise 4 603
Exercise 5 603
Exercise 6 603
Exercise 7 603
Exercise 8 603
Exercise 9 603
Exercise 10 603
Exercise 11 603
Exercise 12 604
Exercise 13 604
Exercise 14 604
Exercise 15 604
Exercise 16 604
Exercise 17 604
Exercise 18 606
14.6 Exercises 617
Exercise 1 617
Exercise 2 617
Exercise 3 617
Exercise 4 617
Exercise 5 617
Exercise 6 617
Exercise 7 617
Exercise 8 617
Exercise 9 617
Exercise 10 617
Exercise 11 617
Exercise 12 617
Exercise 13 618
Exercise 14 618
Exercise 15 618
Exercise 16 618
Exercise 17 618
Exercise 18 618
Exercise 19 618
Exercise 20 618
Exercise 21 618
Exercise 22 618
Exercise 23 619
Exercise 24 619
Exercise 25 619
Exercise 26 619
Exercise 27 619
Exercise 28 619
Exercise 29 619
Exercise 30 620
Exercise 31 620
Exercise 32 621
Exercise 33 621
Exercise 34 621
Exercise 35 621
Exercise 36 621
Exercise 37 621
Exercise 38 622
Exercise 39 622
Exercise 40 622
Exercise 41 622
Exercise 42 622
Exercise 43 622
Exercise 44 623
Exercise 45 623
Exercise 46 623
Exercise 47 623
Exercise 48 623
Exercise 49 623
Exercise 50 623
Exercise 51 624
14.7 Exercises 635
Exercise 1 635
Exercise 2 636
Exercise 3 636
Exercise 4 636
Exercise 5 636
Exercise 6 636
Exercise 7 636
Exercise 8 636
Exercise 9 636
Exercise 10 636
Exercise 11 637
Exercise 12 637
Exercise 13 637
Exercise 14 637
Exercise 15 638
Exercise 16 638
Exercise 17 638
Exercise 18 638
Exercise 19 638
Exercise 20 638
Exercise 21 639
14.8 Exercises 644
Exercise 1 644
Exercise 2 644
Exercise 3 644
Exercise 4 644
Exercise 5 644
Exercise 6 644
Exercise 7 644
Exercise 8 644
Exercise 9 645
Exercise 10 645
14.9 Exercises 652
Exercise 1 652
Exercise 2 652
Exercise 3 652
Exercise 4 653
Exercise 5 653
Exercise 6 653
Exercise 7 653
Exercise 8 653
Exercise 9 653
Exercise 10 654
Exercise 11 654
Exercise 12 654
Exercise 13 654
Exercise 14 654
Exercise 15 654
Exercise 16 654
Exercise 17 654
Exercise 18 654
Exercise 19 654
Chapter 15: Commutative Rings and Algebraic Geometry
15.1 Exercises 668
Exercise 1 668
Exercise 2 668
Exercise 3 668
Exercise 4 668
Exercise 5 668
Exercise 6 668
Exercise 7 668
Exercise 8 668
Exercise 9 668
Exercise 10 668
Exercise 11 668
Exercise 12 669
Exercise 13 669
Exercise 14 669
Exercise 15 669
Exercise 16 669
Exercise 17 669
Exercise 18 669
Exercise 19 669
Exercise 20 669
Exercise 21 669
Exercise 22 669
Exercise 23 669
Exercise 24 669
Exercise 25 669
Exercise 26 670
Exercise 27 670
Exercise 28 670
Exercise 29 670
Exercise 30 670
Exercise 31 670
Exercise 32 670
Exercise 33 670
Exercise 34 670
Exercise 35 670
Exercise 36 671
Exercise 37 671
Exercise 38 671
Exercise 39 671
Exercise 40 672
Exercise 41 672
Exercise 42 672
Exercise 43 672
Exercise 44 672
Exercise 45 672
Exercise 46 672
Exercise 47 672
Exercise 48 673
15.2 Exercises 686
Exercise 1 686
Exercise 2 686
Exercise 3 686
Exercise 4 686
Exercise 5 686
Exercise 6 686
Exercise 7 686
Exercise 8 686
Exercise 9 686
Exercise 10 687
Exercise 11 687
Exercise 12 687
Exercise 13 687
Exercise 14 687
Exercise 15 687
Exercise 16 687
Exercise 17 687
Exercise 18 687
Exercise 19 687
Exercise 20 687
Exercise 21 687
Exercise 22 687
Exercise 23 688
Exercise 24 688
Exercise 25 688
Exercise 26 688
Exercise 27 688
Exercise 28 688
Exercise 29 688
Exercise 30 688
Exercise 31 688
Exercise 32 688
Exercise 33 688
Exercise 34 688
Exercise 35 689
Exercise 36 689
Exercise 37 689
Exercise 38 689
Exercise 39 689
Exercise 40 689
Exercise 41 689
Exercise 42 689
Exercise 43 689
Exercise 44 690
Exercise 45 690
Exercise 46 690
Exercise 47 690
Exercise 48 690
Exercise 49 690
Exercise 50 691
Exercise 51 691
Exercise 52 691
Exercise 53 691
Exercise 54 691
15.3 Exercises 703
Exercise 1 703
Exercise 2 703
Exercise 3 703
Exercise 4 703
Exercise 5 703
Exercise 6 703
Exercise 7 703
Exercise 8 703
Exercise 9 703
Exercise 10 703
Exercise 11 704
Exercise 12 704
Exercise 13 704
Exercise 14 704
Exercise 15 704
Exercise 16 704
Exercise 17 704
Exercise 18 705
Exercise 19 705
Exercise 20 705
Exercise 21 705
Exercise 22 705
Exercise 23 705
Exercise 24 705
Exercise 25 705
Exercise 26 705
Exercise 27 706
Exercise 28 706
15.4 Exercises 726
Exercise 1 726
Exercise 2 726
Exercise 3 726
Exercise 4 726
Exercise 5 727
Exercise 6 727
Exercise 7 727
Exercise 8 727
Exercise 9 727
Exercise 10 727
Exercise 11 727
Exercise 12 727
Exercise 13 727
Exercise 14 727
Exercise 15 727
Exercise 16 727
Exercise 17 727
Exercise 18 727
Exercise 19 727
Exercise 20 728
Exercise 21 728
Exercise 22 728
Exercise 23 728
Exercise 24 728
Exercise 25 728
Exercise 26 728
Exercise 27 729
Exercise 28 729
Exercise 29 729
Exercise 30 729
Exercise 31 729
Exercise 32 729
Exercise 33 729
Exercise 34 729
Exercise 35 730
Exercise 36 730
Exercise 37 730
Exercise 38 730
Exercise 39 730
Exercise 40 730
15.5 Exercises 745
Exercise 1 745
Exercise 2 745
Exercise 3 746
Exercise 4 746
Exercise 5 746
Exercise 6 746
Exercise 7 746
Exercise 8 746
Exercise 9 746
Exercise 10 746
Exercise 11 746
Exercise 12 746
Exercise 13 746
Exercise 14 746
Exercise 15 746
Exercise 16 747
Exercise 17 747
Exercise 18 747
Exercise 19 747
Exercise 20 747
Exercise 21 748
Exercise 22 748
Exercise 23 748
Exercise 24 748
Exercise 25 748
Exercise 26 748
Exercise 27 748
Exercise 28 748
Exercise 29 749
Exercise 30 749
Exercise 31 749
Chapter 16: Artinian Rings, Discrete Valuation Rings, and Dedekind Domains
read chapter notes
16.1 Exercises 754
Exercise 1 754
Exercise 2 754
Exercise 3 754
Exercise 4 754
Exercise 5 754
Exercise 6 754
Exercise 7 754
Exercise 8 754
Exercise 9 754
Exercise 10 754
Exercise 11 754
Exercise 12 754
Exercise 13 755
Exercise 14 755
16.2 Exercises 763
Exercise 1 763
Exercise 2 763
Exercise 3 764
Exercise 4 764
Exercise 5 764
Exercise 6 764
Exercise 7 764
Exercise 8 764
Exercise 9 764
16.3 Exercises 773
Exercise 1 773
Exercise 2 773
Exercise 3 773
Exercise 4 773
Exercise 5 773
Exercise 6 773
Exercise 7 774
Exercise 8 774
Exercise 9 774
Exercise 10 774
Exercise 11 774
Exercise 12 774
Exercise 13 774
Exercise 14 774
Exercise 15 774
Exercise 16 774
Exercise 17 774
Exercise 18 774
Exercise 19 774
Exercise 20 775
Exercise 21 775
Exercise 22 775
Exercise 23 775
Exercise 24 775
Exercise 25 775
Chapter 17: Introduction to Homological Algebra and Group Cohomology
read chapter notes
17.1 Exercises 791
Exercise 1 791
Exercise 2 791
Exercise 3 792
Exercise 4 792
Exercise 5 792
Exercise 6 793
Exercise 7 793
Exercise 8 793
Exercise 9 793
Exercise 10 793
Exercise 11 793
Exercise 12 793
Exercise 13 793
Exercise 14 793
Exercise 15 794
Exercise 16 794
Exercise 17 794
Exercise 18 794
Exercise 19 794
Exercise 20 794
Exercise 21 794
Exercise 22 794
Exercise 23 795
Exercise 24 795
Exercise 25 795
Exercise 26 795
Exercise 27 795
Exercise 28 795
Exercise 29 796
Exercise 30 796
Exercise 31 796
Exercise 32 797
Exercise 33 797
Exercise 34 797
Exercise 35 797
17.2 Exercises 809
Exercise 1 809
Exercise 2 810
Exercise 3 810
Exercise 4 811
Exercise 5 811
Exercise 6 811
Exercise 7 811
Exercise 8 811
Exercise 9 811
Exercise 10 811
Exercise 11 811
Exercise 12 812
Exercise 13 812
Exercise 14 812
Exercise 15 812
Exercise 16 812
Exercise 17 812
Exercise 18 812
Exercise 19 812
Exercise 20 813
Exercise 21 813
Exercise 22 813
Exercise 23 813
Exercise 24 813
Exercise 25 813
17.3 Exercises 822
Exercise 1 822
Exercise 2 822
Exercise 3 822
Exercise 4 822
Exercise 5 822
Exercise 6 822
Exercise 7 822
Exercise 8 823
Exercise 9 823
Exercise 10 823
Exercise 11 823
Exercise 12 823
Exercise 13 823
Exercise 14 823
Exercise 15 824
17.4 Exercises 837
Exercise 1 837
Exercise 2 837
Exercise 3 837
Exercise 4 837
Exercise 5 837
Exercise 6 838
Exercise 7 838
Exercise 8 838
Exercise 9 838
Exercise 10 838
Chapter 18: Representation Theory and Character Theory
read chapter notes
18.1 Exercises 852
Exercise 1 852
Exercise 2 852
Exercise 3 852
Exercise 4 852
Exercise 5 852
Exercise 6 852
Exercise 7 852
Exercise 8 852
Exercise 9 852
Exercise 10 852
Exercise 11 853
Exercise 12 853
Exercise 13 853
Exercise 14 853
Exercise 15 853
Exercise 16 853
Exercise 17 853
Exercise 18 853
Exercise 19 854
Exercise 20 854
Exercise 21 854
Exercise 22 854
Exercise 23 854
Exercise 24 854
18.2 Exercises 863
Exercise 1 863
Exercise 2 863
Exercise 3 863
Exercise 4 863
Exercise 5 863
Exercise 6 863
Exercise 7 863
Exercise 8 863
Exercise 9 864
Exercise 10 864
Exercise 11 864
Exercise 12 864
Exercise 13 864
Exercise 14 864
Exercise 15 864
Exercise 16 864
Exercise 17 864
Exercise 18 864
18.3 Exercises 876
Exercise 1 876
Exercise 2 876
Exercise 3 876
Exercise 4 877
Exercise 5 877
Exercise 6 877
Exercise 7 877
Exercise 8 877
Exercise 9 877
Exercise 10 877
Exercise 11 877
Exercise 12 877
Exercise 13 878
Exercise 14 878
Exercise 15 878
Exercise 16 878
Exercise 17 878
Exercise 18 878
Exercise 19 879
Exercise 20 879
Exercise 21 879
Exercise 22 879
Exercise 23 879
Exercise 24 879
Exercise 25 879
Exercise 26 879
Exercise 27 879
Exercise 28 879
Chapter 19: Examples and Applications of Character Theory
19.1 Exercises 885
Exercise 1 885
Exercise 2 885
Exercise 3 885
Exercise 4 885
Exercise 5 885
Exercise 6 885
Exercise 7 885
Exercise 8 885
Exercise 9 885
Exercise 10 885
Exercise 11 885
Exercise 12 885
Exercise 13 885
Exercise 14 885
Exercise 15 885
Exercise 16 885
Exercise 17 886
19.2 Exercises 892
Exercise 1 892
Exercise 2 892
Exercise 3 892
Exercise 4 892
Exercise 5 892
19.3 Exercises 902
Exercise 1 902
Exercise 2 902
Exercise 3 903
Exercise 4 903
Exercise 5 903
Exercise 6 903
Exercise 7 903
Exercise 8 903
Exercise 9 904
Exercise 10 904
Exercise 11 904
Exercise 12 904
Exercise 13 904
Exercise 14 904
Exercise 15 904
Appendix I: Cartesian Products and Zorn 's Lemma
Exercise 907
Exercise 1 907
Exercises 910
Exercise 1 910
Exercise 2 910
Exercise 3 910
Exercise 4 910
Appendix II: Category Theory
Exercises 914
Exercise 1 914
Exercise 2 914
Exercise 3 915
Exercise 4 915
Exercise 5 915
Exercises 918
Exercise 1 918
Exercise 2 918
Exercise 3 918
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4445 | https://ecampusontario.pressbooks.pub/diffeq/chapter/4-2-properties-of-laplace-transform/ | Skip to content
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Laplace Transform
4.2 Properties of Laplace Transform
Understanding the properties of the Laplace Transform is crucial as it provides tools for efficiently transforming and manipulating functions. These properties greatly simplify the analysis and solution of differential equations and complex systems.
A. Existence of the Transform
The Laplace transform exists for any function that is (1) piecewise-continuous and (2) of exponential order (i.e., does not grow faster than an exponential function). A function is said to be of exponential order if there exist positive constants and such that for all . For example, is of exponential order 7, but is not of exponential order.
B. Linearity of the Laplace Transform
The Laplace Transform adheres to the principle of linearity. Let and be functions whose Laplace transforms exist for , and let and be constants. Then for , the Laplace Transform of a linear combination of these functions is given by:
This property is useful when dealing with linear combinations of functions.
Example 4.2.1: Find Laplace Transform Using – Linearity Theorem
Use the Laplace Transform Table and the linearity property to determine
.
Show/Hide Solution
From the table
for
for
for
From the linearity theorem, we have
for
Try an Example
C. First Shifting (Exponential) Theorem.
If , then
This theorem is valuable when solving differential equations with exponential terms or in analyzing systems with exponential inputs.
Example 4.2.2: Find Laplace Transform Using – First Shifting and Linearity Theorems
Use the first shifting theorem and the linearity property to determine
.
Show/Hide Solution
Using the first shifting theorem, we have
In , and the coefficient in the exponential term’s exponent is .
for
Shifting , we substitute with
for
In , and the coefficient in the exponential term’s exponent is .
for
Shifting , we substitute with
for
From the linearity theorem, we have
for
Try an Example
D. Differentiation in the Time Domain
Understanding how to transform derivatives is crucial for effectively solving differential equations. This property allows us to express the Laplace Transform of a function’s derivative in terms of the original function’s transform. For a function with continuous derivatives up to order,
Since we will mostly deal with second-order differential equations, we will focus on the Laplace Transform of the first and second derivatives.
Example 4.2.3: Laplace Transform of First Derivative
For function show that .
Show/Hide Solution
Identifying the derivative and initial value:
and
Finding the Laplace Transforms:
From the Laplace Transform table, we have
Applying the Differentiation Property:
We need to show
Plugging in the transforms and initial value yields
Simplifying both sides gives
This equality confirms the differentiation property as the two sides match.
Example 4.2.4: Laplace Transform of Second Derivative
Find the Laplace Transform of given the initial conditions and . Use for .
Show/Hide Solution
From the differentiation property, we have
Plugging in initial conditions and , we obtain
Try an Example
Table 4.2.1 summarizes the above properties of the Laplace Transform. These properties are crucial for simplifying computations and effectively utilizing the Laplace Transform in solving initial value problems.
Table 4.2.1: Properties of Laplace Transform
| | |
--- |
| Property | Example |
| | |
| for any constant | |
| | |
| | |
| | |
| | |
Section 4.2 Exercises
Find the Laplace transform of the function .
Show/Hide Answer
2. Find the Laplace transform, , of the function
.
Show/Hide Answer
3. Find the Laplace Transform of given the initial conditions and .
Show/Hide Answer |
4446 | https://sites.math.rutgers.edu/~zeilberg/numtheory/L22.pdf | Dr. Z.’s Number Theory Lecture 22 Handout: Integer Partitions II By Doron Zeilberger Restricted Partitions Let pN(n) is the number of partitions of n whose largest part is at most N, then an analogous argument (only a bit simpler) to the proof of Euler’s theorem gives the generating function formula ∞ X n=0 pN(n)qn = 1 1 −q 1 1 −q2 · · · 1 1 −qN .
Also analogously, if S is any subset of the set of positive integers (either finite or infinite), and pS(n) is the number of partitions of n whose parts always lie in S, then ∞ X n=0 pS(n)qn = Y s∈S 1 1 −qs .
Also analogously, if S is any subset of the set of positive integers (either finite or infinite), and pS(n) is the number of distinct partitions of n whose parts always lie in S, then ∞ X n=0 pS(n)qn = Y s∈S (1 + qs) .
Problem 22.1: Write down the generating function for the sequence, let’s call it a(n), for the number of distinct partitions of n whose largest part is ≤3. Use it to find a(i) for all i.
Sol. to 22.1 ∞ X n=0 a(n)qn = (1 + q)(1 + q2)(1 + q3) .
Note: In this case we get a polynomial, i.e. the ‘infinite series’, terminates (after q6).
Opening up parentheses ∞ X n=0 a(n)qn = (1 + q)(1 + q2)(1 + q3) = (1 + q + q2 + q3)(1 + q3) = 1 + q + q2 + 2q3 + q4 + q5 + q6 Comparing coefficients, we get a(1) = 1, a(2) = 1, a(3) = 2, a(4) = 1, a(5) = 1, a(6) = 1, and for i > 6, we have a(i) = 0. (There is no way that we can get a partition of 7 (or above) into distinct parts with largest part ≤3).
1 Problem 22.2: Write down the generating function for the sequence, let’s call it a(n), for the number of partitions of n that are congruent to 1 or 4 modulo 5. Use it to find a(i) for 1 ≤i ≤6.
Sol. to 22.2 The generating function is: ∞ X n=0 a(n)qn = ∞ Y i=0 1 1 −q5i+1 1 1 −q5i+4 .
Spelling it out ∞ X n=0 a(n)qn = 1 1 −q 1 1 −q4 1 1 −q6 · · · Note: Since we are only interested in a(i) for i ≤6 we can replace by · · · anything beyond q6, since the next term would be (1 + q9 + q18 + q27 + . . .) followed by (1 + q11 + q22 + q33 + . . ., etc., and none of and these will affect the first six coefficients (i.e. of q, q2, q3, q4, q5, q6). Replacing 1 1−q, 1 1−q4 and 1 1−q6 by the geometric series they stand for, using 1 1 −z = 1 + z + z2 + . . .
, but replacing by . . . any power after q6, we get (1 + q + q2 + q3 + q4 + q5 + q6 + . . .)(1 + q4 + . . .)(1 + q6 + . . .) 1 + q + q2 + q3 + q4 + q5 + q6 + . . .
+q4 + q5 + q6 + . . .
+q6 + . . .
= 1 + q + q2 + q3 + 2q4 + 2q5 + 3q6 + . . .
Reading offthe coefficients, we get: Ans.
to 22.2: The generating function is Q∞ i=0 1 1−q5i+1 1 1−q5i+4 , the first six terms are a(1) = 1, a(2) = 1, a(3) = 1, a(4) = 2, a(5) = 2, a(6) = 3.
The Number of Partitions of n into DISTINCT parts equals The Number of Partitions of n into ODD parts Let Odd(n) be the set of partitions of n whose parts only have odd parts (but each odd integer can show up as many times at it wishes, including not showing up at all).
Let Dis(n) be the set of partitions of n where every integer can show up, but at most once.
Here are the few first sets of both kinds Odd(1) = {1} , Dis(1) = {1} 2 Odd(2) = {11} , Dis(1) = {2} Odd(3) = {3, 111} , Dis(3) = {3, 21} Odd(4) = {31, 1111} , Dis(4) = {4, 31} Odd(5) = {5, 311, 11111} , Dis(5) = {5, 41, 32} Odd(6) = {51, 3111, 33, 111111} , Dis(6) = {6, 51, 42, 321} Odd(7) = {7, 511, 31111, 331, 1111111} , Dis(7) = {7, 61, 52, 43, 421} Odd(8) = {71, 5111, 3311, 311111, 32111, 11111111} , Dis(8) = {8, 71, 62, 521, 53, 431} Now notice something amazing!, they always have the same number of elements. This is not only true for n ≤8 but for all n Euler’s Odd=Distinct Theorem: For every positive integer n: |Odd(n)| = |Dis(n)|. in words: For every positive integer n, the number of partitions of n into odd parts equals the number of partitions of n into distinct parts.
First Proof (Algebraic proof, due to L. Euler) Let odd(n) be the number of partitions of n into odd parts, and let dis(n) be the number of partitions of n into distint parts. By generatingfunctionology (explained above) ∞ X n=0 dis(n)qn = ∞ Y i=0 (1 + qi) , ∞ X n=0 odd(n)qn = ∞ Y i=0 1 1 −q2i+1 .
These do not look the same, but thanks to algebra, they are the same! We use the high-school algebra identity 1 + z = 1 −z2 1 −z , that follows from (1 + z)(1 −z) = 1 −z2. So, we have ∞ X n=0 dis(n)qn = ∞ Y i=0 (1 + qi) = ∞ Y i=0 1 −q2i 1 −qi = Q∞ i=0(1 −q2i) Q∞ i=0(1 −qi) = Q∞ i=0(1 −q2i) Q∞ i=0(1 −q2i) Q∞ i=0(1 −q2i+1) 1 Q∞ i=0(1 −q2i+1) = ∞ Y i=0 1 1 −q2i+1 .
QED If this is too abstract, here is a more concrete rendition, using . . ..
∞ X n=0 dis(n)qn = (1 + q)(1 + q2)(1 + q3)(1 + q4)(1 + q5) · · · 3 1 −q2 1 −q 1 −q4 1 −q2 1 −q6 1 −q3 1 −q8 1 −q4 1 −q10 1 −q5 1 −q12 1 −q6 · · · = 1 1 −q 1 1 −q3 1 1 −q5 · · · , since each term in the numerator of the form 1−qeven will sooner-or later cancel-out with the same term at the bottom, resulting in only 1 surviving in the numerator, and only terms of the form 1 −qodd surviving at the bottom.
But the BEST proofs are bijective proofs.You are supposed to know two bijective proofs. The first one is due to Glashier. The second one is due to my great hero James Joseph Sylvester, to be covered in the next lecture.
Glashier’s Bijection: Dis(n) →Odd(n): Input: A partition of n into distinct parts.
Output: A partition of n into odd parts.
Description: For each part p, write it as p = a · 2s where s is the largest s such that p/2s is an integer, and a is necessarily odd (or else s would not be the largest!). Replace it by 2s copies of a.
Problem 22.3a Apply the Glashier bijection to the distinct partition (26, 10, 8, 6, 5, 4, 2).
Solution to problem 22.3a: 26 = 13 · 21 = 13 · 2 , 10 = 5 · 21 = 5 · 2 , 8 = 1 · 23 = 1 · 8 , 6 = 3 · 21 = 3 · 2 , 5 = 5 · 20 = 5 , 4 = 1 · 22 = 1 · 4 , 2 = 1 · 21 = 1 · 2 .
.
Hence (in exponent notation) (26, 10, 8, 6, 5, 4, 2) →132521832511412 , sorting and collecting this equals 18+4+23252+1132 = 1143253132 4 Finally, going back to the usual notation we get (13, 13, 5, 5, 5, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) .
Ans.
to 22.3a: The odd partition corresponding to the distinct partition (26, 10, 8, 6, 5, 4, 2), under Glashier’s bijection, is (13, 13, 5, 5, 5, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1).
Odd(n) →Dis(n): Input: A partition of n into odd parts.
Output: A partition of n into distinct parts.
Description: First convert the partition into exponent notation 1a13a35a5 ...
For each part (that must be odd, of course), (2i + 1)a, that shows up (i.e. where a > 0), write a in (sparse) binary representation a = 2j1 + 2j2 + . . . + 2jr , j1 > j2 > . . . > jr ≥0, and replace it by the following parts (2i + 1) · 2j1 , (2i + 1) · 2j2 , (2i + 1) · 2j3 , . . .
, (2i + 1) · 2jr .
Obviously all the parts are distinct, and at the end just sort them.
Problem 22.3b Apply the Glashier bijection to the odd partition (13, 13, 5, 5, 5, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) .
We first convert it to exponent notation 132 53 32 114 .
Next, for each (necessarily odd!) part (2i + 1)m, we take the multiplicity and express it as a sum of powers of 2: Regarding 132: 2 = 21 , so 132 only yields one part in the output, namely 13 · 2 = 26.
Regarding 53 : 3 = 21 + 20 = 2 + 1 , 5 so 53 yields two parts in the output, namely 5 · 2 = 10, and 5 · 1 = 5.
Regarding 32 : 2 = 21 = 2 , so 32 yields only one part in the output, namely 3 · 2 = 6.
Regarding 114 : 14 = 23 + 22 + 21 = 8 + 4 + 2 , so 114 yields three parts in the output, namely 1 · 8 = 8, 1 · 4 = 4, 1 · 2 = 2.
Combining them, we get: 26, 10, 5, 6, 8, 4, 2 .
Finally, for the sake of politeness, we sort the parts from largest to smallest (26, 10, 8, 6, 5, 4, 2) .
Ans. to 22.3b: The distinct partition corresponding to the odd partition (13,13,5,5,5,3,3,1,1,1,1,1,1,1,1,1,1,1,1,1,1) under Glashier’s bijection, is (26, 10, 8, 6, 5, 4, 2).
6 |
4447 | https://periodictableguide.com/oxygen-o-element-periodic-table/ | Skip to content
Oxygen (O) – Periodic Table (Element Information & More)
by Jay
Hey! Check out our New Interactive Periodic Table with rotating Bohr models—it's a game-changer🙂
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This is a SUPER easy guide on Oxygen element.
In fact, the table mentioned below is the perfect information box (Which gives you every single detail about the Oxygen element in Periodic table.)
So if you want to know anything about Oxygen element, then this guide is for you.
Let’s finish this very quickly.
Oxygen Element (O) Information
| | |
--- |
| Appearance | Colorless gas |
| State (at STP) | Gas |
| Position in Periodic table | Group: 16, Period: 2,Block: p |
| Category | Chalcogens |
| Atomic number or Protons | 8 |
| Neutrons | 8 |
| Electrons | 8 |
| Symbol | O |
| Atomic mass | 15.999 u |
| Electrons arrangement or Bohr model | 2, 6 |
| Electronic configuration | [He] 2s2 2p4 |
| Atomic radius | 152 picometers (van der Waals radius) |
| Valence electrons | 6 |
| 1st Ionization energy | 13.61 eV |
| Electronegativity | 3.44 (Pauling scale) |
| Crystal structure | SC (Simple cubic) |
| Melting point of Oxygen (O2) | 54.36 K or -218.79 °C or -361.82 °F |
| Boiling point of Oxygen (O2) | 90.188 K or -182.962 °C or -297.332 °F |
| Density | 1.43 g/L |
| Main isotope | 16O |
| Who discovered Oxygen and when? | Carl Wilhelm Scheele in 1771 |
| CAS number | 7781-44-7 |
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Oxygen in Periodic table
Oxygen element is in group 16 and period 2 of the Periodic table. Oxygen is the p-block element and it belongs to chalcogens group.
| | |
--- |
| H | He |
| Li | Be | B | C | N | O | F | Ne |
| Na | Mg | Al | Si | P | S | Cl | Ar |
| K | Ca | Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn | Ga | Ge | As | Se | Br | Kr |
| Rb | Sr | Y | Zr | Nb | Mo | Tc | Ru | Rh | Pd | Ag | Cd | In | Sn | Sb | Te | I | Xe |
| Cs | Ba | La | Hf | Ta | W | Re | Os | Ir | Pt | Au | Hg | Tl | Pb | Bi | Po | At | Rn |
| Fr | Ra | Ac | Rf | Db | Sg | Bh | Hs | Mt | Ds | Rg | Cn | Nh | Fl | Mc | Lv | Ts | Og |
|
| Ce | Pr | Nd | Pm | Sm | Eu | Gd | Tb | Dy | Ho | Er | Tm | Yb | Lu |
| Th | Pa | U | Np | Pu | Am | Cm | Bk | Cf | Es | Fm | Md | No | Lr |
Click on above elements (in Periodic table) to see their information or Visit Interactive Periodic Table (which shows names, symbol, atomic mass, electron configuration, electrons arrangement, etc. of all the elements)
Click on above elements (in Periodic table) to see their information
←Move to: Nitrogen (N) element – Periodic Table→Move to: Fluorine (Fl) element – Periodic Table
Why is Oxygen in Group 16?
Do you know, how many electrons can be accommodated in the first shell, second shell, third shell, fourth shell, etc…?
Here is the table showing the capacity of orbits to hold electrons.
Number of electrons in shells.
| | |
--- |
| Orbit / Shell (n) | Maximum no. of electrons this orbit can hold |
| K shell, n = 1 | 2 × 1² = 2 |
| L shell, n = 2 | 2 × 2² = 8 |
| M shell, n = 3 | 2 × 3² = 18 |
| N shell, n = 4 | 2 × 4² = 32 |
Thus,
1st shell can hold 2 electrons.
2nd shell can hold 8 electrons.
3rd shell can hold 18 electrons.
4th shell can hold 32 electrons.
Now the atomic number of oxygen (O) is 8.
Hence the oxygen element has electrons arrangement 2, 6.
This electron arrangement indicates that the outermost orbit of Oxygen element (O) has 6 electrons.
Hence, it lies in group 16.
Why is Oxygen in Period 2?
Let me ask you a question.
How many shells does oxygen have?
It’s 2. Right?
You have already seen the bohr model of oxygen atom in the above table.
From the Bohr model, it can be found that the number of orbits or shells in oxygen is 2. Hence, as oxygen has 2 orbits, it lies in period 2 of the Periodic table.
Why is Oxygen in p-block?
Before knowing this reason, first of all I want to ask you a simple question.
How can you determine the blocks-wise position of elements?
The simple answer: The elements will lie in the s, p, d or f block will completely depend upon the subshell in which the last electron will enter.
For example; the electron configuration of oxygen is [He] 2s2 2p4.
So the last electron of oxygen enters the p-subshell or p-orbital.
Hence, oxygen is the p-block element.
10 Interesting facts about Oxygen
Interesting facts about oxygen element are mentioned below.
Oxygen is colorless gas, but it appears pale-blue in its liquid state.
All living organisms including plants and animals require oxygen for survival.
Oxygen normally exists in a divalent molecule (i.e O2). Ozone is another pure form of oxygen and it exists as O3.
Oxygen gas does not burn itself, but helps in the combustion process. Hence it is also said to be the helper of combustion.
Oxygen is also present in the human body in a very large proportion. Because the human body is made up of 75% water, which includes oxygen and hydrogen molecules. The number of hydrogen atoms are double than the oxygen atoms, but if we talk about the mass, then oxygen atoms have more mass as compared to hydrogen. Hence ⅔ rd of the human body mass is because of oxygen.
Oxygen element is the third most abundant element in the universe.
Oxygen gas is present in the atmosphere (approximately 21%).
The sea water contains approximately 4.95 ml of dissolved oxygen per litre.
Plants produce oxygen by photosynthesis, but the majority of oxygen on the earth comes from the sea plants called phytoplankton.
Oxygen molecules can withstand pressures as high as 19 million times the atmospheric pressure.
Properties of Oxygen
The physical and chemical properties of oxygen element are mentioned below.
Physical properties of Oxygen
Physical properties of oxygen are mentioned below.
Oxygen is a colourless, odourless and tasteless gas. Oxygen appears pale-blue in color in its liquid state.
Solubility of oxygen gas in water is more than that of nitrogen element. And this solubility is also temperature dependent.
Oxygen condenses at 90.188 K temperature and it freezes at 54.36 K temperature.
Dioxygen (O2) is the most common allotrope of oxygen element which living organisms require for respiration. The other allotrope of oxygen is O3 (ozone) which is found in the upper atmosphere of the earth.
The physical properties of oxygen and ozone are not the same. Ozone is bluish in color in gaseous as well as liquid state.
Chemical properties of Oxygen
Chemical properties of oxygen are mentioned below.
At standard temperature and pressure, two oxygen elements combine with each other to form a stable molecule O2 (i.e dioxygen).
Oxygen element is a reactive nonmetal that forms oxides with most of the other elements.
Out of all the elements on periodic table, the oxygen element is a second strong oxidizing agent after fluorine element.
Oxygen does not burn itself, but it helps in combustion. In simple words, it helps other substances to burn. Hence oxygen is also known as a helper of combustion.
Oxygen also reacts with the metals at room temperature. For example, rusting of metals takes place at room temperature when they react with moisture and oxygen of the air.
Uses of Oxygen
Uses of oxygen are mentioned below.
In industries, oxygen is used along with other fuel gases. For example oxygen is used with acetylene gas in oxyacetylene flame welding (or gas welding).
In steel industries, oxygen gas is required to increase the combustion temperatures in blast furnaces.
Oxygen is used in biological treatment of sewage water plants. This is because the supply of oxygen reduces the formation of hydrogen sulphide and hence the corrosion and odour can be reduced.
Liquid oxygen is used as an oxidizing agent in liquid fueled rockets, which is required to produce a very high amount of thrust during a takeoff.
Oxygen is used in many breathing apparatus in hospitals. Oxygen tanks are also used in some underwater work, scuba diving as well as it is also used by mountaineers as the amount of O2 gas decreases at higher altitude.
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External resources:
Oxygen – Energy Education. (n.d.). Oxygen – Energy Education.
Oxygen | Center for Science Education. (n.d.). Oxygen | Center for Science Education.
Oxygen – Element information, properties and uses | Periodic Table. (n.d.). Oxygen – Element Information, Properties and Uses | Periodic Table.
It’s Elemental – The Element Oxygen. (n.d.). It’s Elemental – the Element Oxygen. |
4448 | https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/09%3A_Hamilton's_Action_Principle/9.03%3A_Lagrangian | Skip to main content
9.3: Lagrangian
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: Feb 28, 2021
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9.2: Hamilton's Principle of Stationary Action
9.4: Application of Hamilton's Action Principle to Mechanics
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: 9644
Douglas Cline
University of Rochester
( \newcommand{\kernel}{\mathrm{null}\,})
Standard Lagrangian
Lagrangian mechanics
, as introduced in chapter was based on the concepts of kinetic energy and potential energy. d’Alembert’s principle of
virtual work
was used to derive
Lagrangian mechanics
in chapter and this led to the definition of the standard Lagrangian. That is, the standard Lagrangian was defined in chapter to be the difference between the kinetic and potential energies.
Hamilton extended
Lagrangian mechanics
by defining Hamilton’s Principle, equation , which states that a dynamical system follows a path for which the action functional is stationary, that is, the time integral of the Lagrangian. Chapter showed that using the standard Lagrangian for defining the action functional leads to the
Euler
-
Lagrange
variational equations
The
Lagrange
multiplier terms handle the holonomic
constraint forces
and handles the remaining excluded generalized forces. Chapters showed that the use of the standard Lagrangian, with the
Euler
-
Lagrange
equations , provides a remarkably powerful and flexible way to derive second-order equations of motion for dynamical systems in classical mechanics.
Note that the
Euler
-
Lagrange
equations, expressed solely in terms of the standard Lagrangian , that is, excluding the terms, are valid only under the following conditions:
The forces acting on the system, apart from any forces of constraint, must be derivable from scalar potentials.
The equations of constraint must be relations that connect the coordinates of the particles and may be functions of time, that is, the constraints are holonomic.
The terms extend the range of validity of using the standard Lagrangian in the
Lagrange
-
Euler
equations by introducing constraint and omitted forces explicitly.
Chapters exploited
Lagrangian mechanics
based on use of the standard definition of the Lagrangian. The present chapter will show that the powerful Lagrangian formulation, using the standard Lagrangian, can be extended to include alternative non-standard Lagrangians that may be applied to dynamical systems where use of the standard definition of the Lagrangian is inapplicable. If these non-standard Lagrangians satisfy Hamilton’s Action Principle, , then they can be used with the
Euler
-
Lagrange
equations to generate the correct equations of motion, even though the Lagrangian may not have the simple relation to the kinetic and potential energies adopted by the standard Lagrangian. Currently, the development and exploitation of non-standard Lagrangians is an active field of
Lagrangian mechanics
.
Gauge invariance of the standard Lagrangian
Note that the standard Lagrangian is not unique in that there is a continuous spectrum of equivalent standard Lagrangians that all lead to identical equations of motion. This is because the Lagrangian is a scalar quantity that is invariant with respect to
coordinate transformations
. The following transformations change the standard Lagrangian, but leave the equations of motion unchanged.
The Lagrangian is indefinite with respect to addition of a constant to the scalar potential which cancels out when the derivatives in the
Euler
-
Lagrange
differential equations are applied.
2. The Lagrangian is indefinite with respect to addition of a constant kinetic energy.
3. The Lagrangian is indefinite with respect to addition of a total time derivative of the form for any differentiable function of the generalized coordinates plus time, that has continuous second derivatives.
This last statement can be proved by considering a transformation between two related standard Lagrangians of the form
This leads to a standard Lagrangian that has the same equations of motion as as is shown by substituting Equation into the
Euler
-
Lagrange
equations. That is,
Thus even though the related Lagrangians and are different, they are completely equivalent in that they generate identical equations of motion.
There is an unlimited range of equivalent standard Lagrangians that all lead to the same equations of motion and satisfy the requirements of the Lagrangian. That is, there is no unique choice among the wide range of equivalent standard Lagrangians expressed in terms of generalized coordinates. This discussion is an example of
gauge invariance
in physics.
Modern theories in physics describe reality in terms of potential fields.
Gauge invariance
, which also is called gauge symmetry, is a property of field theory for which different underlying fields lead to identical observable quantities. Well-known examples are the static electric potential field and the gravitational potential field where any arbitrary constant can be added to these scalar potentials with zero impact on the observed static electric field or the observed gravitational field. Gauge theories constrain the laws of physics in that the impact of gauge transformations must cancel out when expressed in terms of the observables. Gauge symmetry plays a crucial role in both classical and quantal manifestations of field theory, e.g. it is the basis of the Standard Model of electroweak and strong interactions.
Equivalent Lagrangians are a clear manifestation of
gauge invariance
as illustrated by equations , which show that adding any total time derivative of a scalar function to the Lagrangian has no observable consequences on the equations of motion. That is, although addition of the total time derivative of the scalar function changes the value of the Lagrangian, it does not change the equations of motion for the observables derived using equivalent standard Lagrangians.
For Lagrangian formulations of classical mechanics, the
gauge invariance
is readily apparent by direct inspection of the Lagrangian.
Example :
Gauge invariance
in electromagnetism
The scalar electric potential and the vector potential fields in electromagnetism are examples of gauge-invariant fields. These electromagnetic-potential fields are not directly observable, that is, the electromagnetic observable quantities are the electric field and magnetic field which can be derived from the scalar and vector potential fields and . An advantage of using the potential fields is that they reduce the problem from components, each for and to components, one for the scalar field and for the vector potential . The Lagrangian for the velocity-dependent
Lorentz force
, given by equation , provides an example of
gauge invariance
. Equations and showed that the electric and magnetic fields can be expressed in terms of scalar and vector potentials and by the relations
The equations of motion for a charge in an electromagnetic field can be obtained by using the Lagrangian
Consider the transformations in the transformed Lagrangian where
The transformed Lorentz-force Lagrangian is related to the original Lorentz-force Lagrangian by
Note that the additive term is an exact time differential. Thus the Lagrangian is gauge invariant implying identical equations of motion are obtained using either of these equivalent Lagrangians.
The force fields and can be used to show that the above transformation is gauge-invariant. That is,
That is, the additive terms due to the scalar field cancel. Thus the electromagnetic force fields following a gauge-invariant transformation are shown to be identical in agreement with what is inferred directly by inspection of the Lagrangian.
Non-standard Lagrangians
The definition of the standard Lagrangian was based on d’Alembert’s differential variational principle. The flexibility and power of
Lagrangian mechanics
can be extended to a broader range of dynamical systems by employing an extended definition of the Lagrangian that is based on Hamilton’s Principle, equation . Note that Hamilton’s Principle was introduced years after development of the standard formulation of
Lagrangian mechanics
. Hamilton’s Principle provides a general definition of the Lagrangian that applies to standard Lagrangians, which are expressed as the difference between the kinetic and potential energies, as well as to non-standard Lagrangians where there may be no clear separation into kinetic and potential energy terms. These non-standard Lagrangians can be used with the
Euler
-
Lagrange
equations to generate the correct equations of motion, even though they may have no relation to the kinetic and potential energies. The extended definition of the Lagrangian based on Hamilton’s action functional can be exploited for developing non-standard definitions of the Lagrangian that may be applied to dynamical systems where use of the standard definition is inapplicable. Non-standard Lagrangians can be equally as useful as the standard Lagrangian for deriving equations of motion for a system. Secondly, non-standard Lagrangians, that have no energy interpretation, are available for deriving the equations of motion for many nonconservative systems. Thirdly, Lagrangians are useful irrespective of how they were derived. For example, they can be used to derive conservation laws or the equations of motion.
Coordinate transformations
of the Lagrangian is much simpler than that required for transforming the equations of motion. The relativistic Lagrangian defined in chapter is a well-known example of a non-standard Lagrangian.
Inverse variational calculus
Non-standard Lagrangians and Hamiltonians are not based on the concept of kinetic and potential energies. Therefore, development of non-standard Lagrangians and Hamiltonians require an alternative approach that ensures that they satisfy Hamilton’s Principle, equation , which underlies the Lagrangian and Hamiltonian formulations. One useful alternative approach is to derive the Lagrangian or Hamiltonian via an inverse variational process based on the assumption that the equations of motion are known. Helmholtz developed the field of inverse variational calculus which plays an important role in development of non-standard Lagrangians. An example of this approach is use of the well-known
Lorentz force
as the basis for deriving a corresponding Lagrangian to handle systems involving electromagnetic forces. Inverse variational calculus is a branch of mathematics that is beyond the scope of this textbook. The Douglas theorem states that, if the three Helmholtz conditions are satisfied, then there exists a Lagrangian that, when used with the
Euler
-
Lagrange
differential equations, leads to the given set of equations of motion. Thus, it will be assumed that the inverse variational calculus technique can be used to derive a Lagrangian from known equations of motion.
9.2: Hamilton's Principle of Stationary Action
9.4: Application of Hamilton's Action Principle to Mechanics |
4449 | https://www.sciencedirect.com/topics/computer-science/greedy-strategy | Greedy Strategy - an overview | ScienceDirect Topics
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Greedy Strategy
In subject area: Computer Science
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A greedy strategy is defined as an algorithmic paradigm that makes the locally optimal choice at each stage in the hope of reaching a global optimum, though it may only yield locally optimal solutions that approximate a global optimal solution in a reasonable time.
AI generated definition based on: Journal of Network and Computer Applications, 2018
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Outline
1. Introduction to Greedy Strategy in Computer Science
2. Fundamental Concepts and Properties
3. Common Greedy Algorithms and Their Applications
4. Design, Analysis, and Limitations of Greedy Algorithms
5. Advanced Topics and Contemporary Applications
6. Conclusion
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1. Introduction to Greedy Strategy in Computer Science
The greedy strategy is an algorithmic paradigm that uses problem-solving heuristics to make the locally optimal choice at each stage, with the hope of reaching a global optimum. In many cases, a greedy strategy does not guarantee an optimal solution but may yield locally optimal solutions that approximate a global optimum in reasonable time. 1Greedy algorithms are widely used in designing various approaches to provision resources, especially when the problem has an optimal structure, meaning the optimal solution to the problem contains the solution to its subproblems. At each step, the algorithm selects the solution that seems best at the moment based on the greedy strategy, iteratively making choices and reducing the problem to a smaller one. 2
These algorithms are highly efficient for problems possessing both substructure and optimality properties, where a problem can be divided into subproblems and the combination of optimal solutions to these subproblems yields a globally optimal solution. Many graph algorithms, such as Prim's algorithm for finding minimum spanning trees, are applications of greedy algorithms. Greedy algorithms often help find a lower bound of the solution quality for many challenging real-world problems. 3 In the context of coverage problems, traditional algorithms such as greedy algorithms, dynamic programming, and backtracking are popular solutions. The greedy method is one of the most commonly used methods in algorithm design, dominating research in maximizing coverage quality, maximizing network lifetime, and minimizing the number of sensors. 4
Greedy algorithms are often intuitively simple and easy to implement. 3 However, they cannot guarantee finding the global optimal solution. 2
2. Fundamental Concepts and Properties
The suitability of the greedy strategy in algorithm design depends on two critical properties: the greedy-choice property and optimal substructure. The greedy-choice property asserts that a globally optimal solution can always be achieved by making locally optimal, or greedy, choices at each step. For example, in the standard coin change problem using coins of 25 cents, 10 cents, and 1 cent, selecting the highest denomination at each step yields the minimum number of coins, demonstrating the greedy-choice property. However, if a 20-cent coin is introduced and the goal is to make 40 cents, the greedy approach fails to produce the optimal solution, as two 20-cent coins would be optimal, but the greedy strategy would select a 25-cent, a 10-cent, and a 5-cent coin, totaling three coins instead of two. This failure occurs because the problem does not satisfy the greedy-choice property in this case. The Nearest Neighbor algorithm for the Traveling Salesman Problem (TSP) is another example of a greedy algorithm, which picks the nearest city at each step but does not always guarantee an optimal solution. 3
Optimal substructure is the property that a problem’s global optimal solution can be constructed from optimal solutions to its subproblems. Greedy algorithms are highly efficient for problems that satisfy both the greedy-choice property and optimal substructure. If a problem can be divided into subproblems and the combination of their optimal solutions yields a globally optimal solution, then greedy algorithms are appropriate. Many graph algorithms, such as Prim's algorithm for minimum spanning trees, are applications of greedy algorithms that exploit these properties. 3
The distinction between greedy algorithms and dynamic programming lies in the requirement for these properties. Greedy algorithms require both the greedy-choice property and optimal substructure, whereas dynamic programming is suitable for problems with optimal substructure and overlapping subproblems, even if the greedy-choice property does not hold.
3. Common Greedy Algorithms and Their Applications
Huffman coding constructs a codeword with different prefixes and the shortest average header length based on the probability of occurrence of certain characters. The process involves selecting the two trees with the smallest weight root nodes in the initial set, combining them into a new binary tree, and repeating this until only one binary tree remains. Huffman coding is sometimes referred to as optimum coding. In the context of smart meter big data compression, Huffman coding has an edge due to its ability for perfect data recovery; however, the compression ratio is not very high and the algorithm efficiency is low compared with other compression algorithms. With the creation and development of Lempel-Ziv (LZ) algorithms, Huffman coding is gradually being replaced in smart meter big data compression. 5
Kruskal’s algorithm for computing a minimum spanning tree of a weighted graph is based on a greedy strategy. It starts from a forest with an empty set of edges and iteratively adds edges of minimum weight that connect different trees, ensuring no cycles are formed. The process continues until all vertices are connected in a single tree. 6 7
Dijkstra’s algorithm adopts a greedy strategy to solve the single-source shortest path problem in weighted directed or undirected graphs. It passes through all nodes to obtain the shortest path, but traversal of nodes and inefficiency are drawbacks when applied to large-scale complex path topology networks. 8
A relaxed solution to the knapsack problem can be computed by greedily packing items starting from a sorted list according to their value density until the capacity is exceeded. This greedy packing strategy can be realized with a simple for-loop and is used to overestimate the achievable value, aiding in pruning suboptimal solution candidates. 9
Greedy algorithms are used in cloud and data center selection, where sites are ordered by network cost and picked one by one, placing virtual machines in clouds with minimal cost. Distributed local greedy algorithms are proposed to minimize latency and the number of data centers used, and a two-stage scheduling approach partitions virtual machines to physical nodes and finds paths for connections by a greedy algorithm. 2
4. Design, Analysis, and Limitations of Greedy Algorithms
The methodology for designing greedy algorithms begins with identifying problems that possess an optimal structure, meaning the optimal solution to the problem contains the solution to its subproblems. The greedy strategy is applied by selecting the solution that appears best at the moment, making a local optimization with the expectation of achieving a global optimal solution. The success of the algorithm critically depends on the greedy strategy, which requires the problem to have specific properties, such as the greedy-choice property and optimal substructure. The greedy-choice property states that a globally optimal solution can always be achieved by making locally optimal choices, without considering the effects on future choices or other subproblems. Problems suitable for greedy algorithms must exhibit both the greedy-choice property and the optimal substructure property. If these properties are present, a greedy algorithm is an appropriate choice for solving the problem. 2 3
Greedy algorithms iteratively make choices and are used in designing approximation algorithms. They are best-effort schemes and cannot guarantee finding the global optimal solution. 2
Limitations arise in scenarios where greedy algorithms fail to find optimal solutions, such as the 0/1 Knapsack Problem and the Traveling Salesman Problem. For example, the Nearest Neighbor algorithm for TSP is a greedy algorithm that picks the nearest city at each step but violates the greedy-choice property, resulting in suboptimal solutions. Recognizing when greedy strategies are inappropriate is crucial, and alternative methods like dynamic programming or exhaustive search may be necessary. Not all problems exhibit the required characteristics for greedy algorithms to be effective. 3
The greedy strategy is commonly used for searching Virtual Network Function (VNF) placement schemes, with designs depending on specific issues like reducing relocation costs or the number of VNF instances. Indicators guiding greedy search paths include upscaling costs, context switching costs, and the priority of each Service Function Chain (SFC). Some greedy algorithms prove their theoretical bounds on performance, and hybrid methods may combine greedy strategies with other techniques, such as genetic algorithms, to optimize scheduling and resource allocation. In specific cases, such as VNF scheduling, these hybrid methods have resulted in reduced scheduling time. 10 11
5. Advanced Topics and Contemporary Applications
Greedy-strategy heuristics are widely used in approximation algorithms for NP-hard problems, where they provide fast, sequential selection of items that maximize the objective function at each iteration, as seen in optimal test construction and set cover problems; however, these approaches may encounter infeasibility issues when constraints are strict, and their performance is often evaluated in terms of how closely they approximate the optimal solution. 12 13
In online algorithms, greedy strategies are employed in scenarios where input arrives sequentially and decisions must be made without knowledge of future data, such as in bin packing problems, with classic greedy algorithms like FirstFit and NextFit being among the earliest solutions for one-dimensional bin packing. 14
In machine learning, greedy strategies are integral to decision tree construction, where algorithms such as Iterative Dichotomiser 3 (ID3) select features at each step to minimize entropy, thereby partitioning the training set in a manner that reduces uncertainty and builds the tree in a recursive, greedy fashion. 15
In resource allocation for cloud virtual reality systems, online greedy algorithms dynamically assign tasks based on real-time network and server conditions, significantly improving system responsiveness. 16
Furthermore, greedy strategies are incorporated into meta-heuristic frameworks such as Greedy Randomized Adaptive Search Procedure (GRASP), which iteratively constructs greedy randomized solutions and refines them through local search, and are combined with other methods like simulated annealing and genetic algorithms to improve solution quality in complex optimization problems. 12 17 Greedy algorithms, genetic algorithms, and ant colony optimization are discussed as prominent heuristic solutions, with randomization being an important tool to avoid local minima in search-based optimization problems. 18
6. Conclusion
The greedy strategy builds a solution to an optimization problem piece by piece, continually adding a piece that is locally optimal to a solution that has been partially formed. 19 Greedy algorithms are widely used in designing various approaches to provision resources, especially when the problem has an optimal structure, meaning the optimal solution to the problem contains the solution to its subproblems. 2 This algorithmic paradigm uses problem-solving heuristics to make the locally optimal choice at each stage, hoping to reach a global optimum. 1 In many cases, the greedy strategy does not guarantee an optimal solution but may yield locally optimal solutions that approximate a global optimum in reasonable time. The greedy strategy plays a critical role in the success of the algorithm.
Greedy algorithms require the problem to have a specific optimal structure, but there is no such restriction for meta-heuristic algorithms. 2 Both greedy and meta-heuristic algorithms are best-effort schemes and cannot guarantee finding the global optimal solution. The greedy strategy is a common heuristic for searching virtual network function placements, and its design generally depends on specific issues such as combining and embedding function graphs, reducing relocation costs, and reducing the number of instances. 10 The distinctions among greedy strategies are mainly reflected in the indicators for guiding their search paths, such as upscaling and context switching costs, the priority of each service function chain, and the importance of each function.
A better strategy is to combine the greedy strategy and randomization, such as the epsilon-greedy strategy, which is likely to obtain a better balance between immediate exploitation and long-term exploration returns. 20 The combination of a greedy strategy and a subsequent application of a workload-adapted version of simulated annealing works well for online assignment problems, and the average level of injury for each casualty can be additionally reduced by about 10% with acceptable computational expenditure. 21 Future research directions include exploring other optimization techniques like genetic algorithms and studying the influence of chosen penalty functions on the results with respect to their number and type.
References
Reference 1 Review article Influence analysis in social networks: A surveyPeng S., Zhou Y., Cao L., Yu S., Niu J., Jia W. Journal of Network and Computer Applications, 2018 pp 17-32 View PDFView articleShow related quote(s) Related quote(s)1 / 2 "... Types of influence maximization algorithm (1) Greedy-based: A greedy algorithm is a pervasively used algorithmic paradigm, which uses problem solving heuristic to make the locally optimal choice at each stage , hoping to reach a global optimum. In many cases, a greedy strategy does not in general reach an optimal solution, but it may yield locally optimal solutions that approximate a global optimal solution in a reasonable time. (2) Heuristic-based: In computer science, artificial intelligence, and mathematical optimization, a heuristic algorithm is designed for solving a problem more quickly when classic methods are too slow, or for finding an approximate solution when classic methods fail to find any exact solutions. This is achieved by trading optimality, completeness, accuracy, or precision for speed. In a way, it can be considered a shortcut. (3) Others: Besides the two major types, many researchers have presented voting-based algorithms and hybrid-based algorithms to improve the performance and scalability of the existing methods of the greedy-based and the heuristic-based. ..." What are related quotes? Related quote(s)2 / 2 "... Influence maximization algorithm Domingos and Richardson are the first to propose the influence maximization as an algorithmic technique for viral marketing. Kempe et al. (2003) are the first to formalize the influence maximization as a discrete optimization problem, and to prove the problem is NP-hard. According to the related literature, the influence maximization problem is formulated as follows: given a social network modeled as a graph G = ( V , E ), and a nonnegative number k , where nodes are users and edges are labeled with influence probabilities among users, the influence maximization problem is to find a set of k seed nodes that maximizes the influence propagation scale in the social network under a given diffusion model. In this section, we summarize the types of influence maximization algorithm, overview the existing influence maximization algorithm, and analyze the performance analysis on the existing algorithms. 6.1 Types of influence maximization algorithm (1) Greedy-based: A greedy algorithm is a pervasively used algorithmic paradigm, which uses problem solving heuristic to make the locally optimal choice at each stage , hoping to reach a global optimum. In many cases, a greedy strategy does not in general reach an optimal solution, but it may yield locally optimal solutions that approximate a global optimal solution in a reasonable time. (2) Heuristic-based: In computer science, artificial intelligence, and mathematical optimization, a heuristic algorithm is designed for solving a problem more quickly when classic methods are too slow, or for finding an approximate solution when classic methods fail to find any exact solutions. This is achieved by trading optimality, completeness, accuracy, or precision for speed. In a way, it can be considered a shortcut. (3) Others: Besides the two major types, many researchers have presented voting-based algorithms and hybrid-based algorithms to improve the performance and scalability of the existing methods of the greedy-based and the heuristic-based. 6.2 The existing algorithms (1) Greedy-based Wang and Feng (2009) proposed a target wise greedy algorithm (TWG for short) based on the potential-based node-selection strategy. 1 ..." What are related quotes?
Reference 2 Review article Resource provision algorithms in cloud computing: A surveyZhang J., Huang H., Wang X. Journal of Network and Computer Applications, 2016 pp 23-42 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Greedy algorithms Greedy algorithm is widely used in designing of various approaches to provision resources, especially when the problem has an optimal structure that means the optimal solution to the problem contains the solution to the subproblem. Every time the algorithm selects a solution which seems best at the moment based on a greedy strategy. It iteratively makes a choice, reducing the problem to a smaller one. In other words, greedy algorithm makes a local optimization and looks forward to finding a global optimal solution to the original problem. The greedy strategy plays a critical role in the success of the algorithm. Although both greedy and MH algorithms search the optimal solution iteratively. Greedy algorithm requires the problem to be addressed to have a specific optimal structure as aforementioned, but there is no restriction for MH. In contrast with the heuristic nature of MH, greedy category plays a great role in approximation algorithms designing. In spite of these differences, both of them are best-effort schemes and cannot guarantee to find the global optimal solution. 7.1 Cloud and DC Selection The VMs in multi-set are placed in multi-cloud circumstance . The sites are ordered in ascending order in terms of network cost. Then sites are picked one by one and placed on clouds greedily with minimal cost. In distributed cloud, greedy algorithm is proposed to optimize the performance. Virtual desktops are assigned to DC closer to users to minimize the access latency . The greedy strategy selects VMs with the highest location-sensitive ranks first and places them in the closest DC to their user. If the closest DC cannot accommodate the VMs then the next closest one is examined. Distributed local greedy algorithms are proposed to minimize latency and number of DCs used . Abu Sharkh et al. (2013) strive to minimize the average connection access tardiness of a group of user requests. Both computational and network resources are reckoned by a heuristic two-stage scheduling approach. The first stage partitions the VMs to PM nodes and the second one finds paths for the connections by a greedy algorithm. ..." What are related quotes?
Reference 3 Book Chapter Fundamentals of AlgorithmsHuang C.-Y., Lai C.-Y., Cheng K.T. Electronic Design Automation, 2009 pp 173-234 View PDFView chapterShow related quote(s) Related quote(s)1 / 3 "... To determine whether a particular problem has an optimal substructure, two aspects have to be examined: substructure and optimality . A problem has substructures if it is divisible into subproblems. Optimality is the property that the combination of optimal solutions to subproblems is a globally optimal solution. Greedy algorithms are highly efficient for problems satisfying these two properties. On top of that, greedy algorithms are often intuitively simple and easy to implement. Therefore, greedy algorithms are very popular for solving optimization problems. Many graph algorithms, mentioned in Section 4.3 , are actually applications of greedy algorithms—such as Prim's algorithm used for finding minimum spanning trees. Greedy algorithms often help find a lower bound of the solution quality for many challenging real-world problems. 4.4.2 Dynamic programming Dynamic programming (DP) is an algorithmic method of solving optimization problems. Programming in this context refers to mathematical programming, which is a synonym for optimization. DP solves a problem by combining the solutions to its subproblems. The famous divide-and-conquer method also solves a problem in a similar manner. The divide-and-conquer method divides a problem into independent subproblems, whereas in DP, either the subproblems depend on the solution sets of other subproblems or the subproblems appear repeatedly. DP uses the dependency of the subproblems and attempts to solve a subproblem only once; it then stores its solution in a table for future lookups. This strategy spares the time spent on recalculating solutions to old subproblems, resulting in an efficient algorithm. To illustrate the superiority of DP, we show how to efficiently multiply a chain of matrices by use of DP. When multiplying a chain of matrices, the order of the multiplications dramatically affects the number of scalar multiplications. For example, consider multiplying three matrices A, B, and C whose dimensions are 30 × 100, 100 × 2, and 2 × 50, respectively. There are two ways to start the multiplication: either A ⋅ B or B ⋅ C first. ..." What are related quotes? Related quote(s)2 / 3 "... The change will consequently be made in this sequence: a quarter (25 cents), a dime (10 cents), and a penny (1 cent)—a total of three coins. This rule of thumb leads to the minimum number of coins, three, because it perfectly embodies the essence of greedy algorithms: making greedy choices at each moment. In this particular problem, a greedy algorithm yields the optimal solution. However, greedy algorithms do not always produce optimal solutions. Let us revisit the making change example. If a coin with a value of 20 cents exists, the rule of thumb just mentioned would not lead to the minimum number of coins if the amount of change needed was 40 cents. By applying the rule of picking the coin of highest value first, we would be giving change of a quarter (25 cents), a dime (10 cents) and a nickel (5 cents), a total of three coins, but, in fact, two, 20-cent coins would be the optimal solution for this example. The greedy algorithm fails to reach the optimal solution for this case. Actually, the example given previously is not ideal for illustrating the concept of greedy algorithms, because it violates the optimal substructure property. In general, problems suitable for greedy algorithms must exhibit two characteristics: the greedy-choice property and the optimal substructure property. If we can demonstrate that a problem has these two properties, then a greedy algorithm would be a good choice. 4.4.1.1 Greedy-choice property The greedy-choice property states that a globally optimal solution can always be achieved by making locally optimal, or greedy, choices. By locally optimal choices we mean making choices that look best for solving the current problem without considering the results from other subproblems or the effect(s) that this choice might have on future choices. In Section 4.4 , we introduced the Nearest Neighbor (NN) algorithm for solving—more precisely, for approximating—an optimal solution to TSP. NN is a greedy algorithm that picks the nearest city at each step. ..." What are related quotes? Related quote(s)3 / 3 "... The greedy algorithm fails to reach the optimal solution for this case. Actually, the example given previously is not ideal for illustrating the concept of greedy algorithms, because it violates the optimal substructure property. In general, problems suitable for greedy algorithms must exhibit two characteristics: the greedy-choice property and the optimal substructure property. If we can demonstrate that a problem has these two properties, then a greedy algorithm would be a good choice. 4.4.1.1 Greedy-choice property The greedy-choice property states that a globally optimal solution can always be achieved by making locally optimal, or greedy, choices. By locally optimal choices we mean making choices that look best for solving the current problem without considering the results from other subproblems or the effect(s) that this choice might have on future choices. In Section 4.4 , we introduced the Nearest Neighbor (NN) algorithm for solving—more precisely, for approximating—an optimal solution to TSP. NN is a greedy algorithm that picks the nearest city at each step. NN violates the greedy-choice property and thus results in suboptimal solutions, as indicated in the example of Figure 4.23 . In Figure 4.23 , the choice of B→D is a greedy one, because the other remaining cities are further from B. In a globally optimal solution, the route of either D→C→B or B→D→D is a necessity, and the choice of B→D is suboptimal. Hence, NN is not an optimal greedy algorithm, because TSP does not satisfy the greedy-choice property. Making change with a minimum number of coins is an interesting example. On the basis of the current U.S. coins, this problem satisfies the greedy-choice property. But when a 20-cent coin comes into existence, the property is violated—when making change for 40 cents, the greedy choice of picking a quarter affects the solution quality of the rest of the problem. How do we tell if a particular problem has the greedy-choice property? ..." What are related quotes?
Reference 4 Review article Coverage problem with uncertain properties in wireless sensor networks: A surveyWang Y., Wu S., Chen Z., Gao X., Chen G. Computer Networks, 2017 pp 200-232 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Traditional vs. heuristic algorithms Traditional algorithms, such as greedy algorithm, dynamic programming, and backtracking, are popular solutions for the coverage problem. These classical algorithms are initial schemes and easy to analyze theoretically. We summarize researches on coverage problems with uncertain properties from three typical objectives: maximizing probability or ratio (in Table 5 ), maximizing network lifetime (in Table 6 ), and minimizing the number of sensors (in Table 7 ). In those tables, several typical solutions are used, such as greedy, programming, and optimization. The greedy method is one of the most commonly used method in algorithm design. In our survey, traditional algorithms, particularly the greedy method, are the dominanting ones. For example, researches in [32,48,68,133,167,173,174] use greedy algorithm for maximizing coverage quality. Similarly, for maximizing network lifetime, there are greedy algorithms in [9,10,12,13,17,47,49,65,77,114,128,140,142,144,145,151,164,168,169,181,182,193] . And for minimizing the number of sensors, there are greedy algorithms in [112,121,183] . In our survey, we consider that programming consists of dynamic programming and (integer) linear programming. Authors in use linear programming to maximize network lifetime while authors in use dynamic programming. Authors in use integer linear programming to minimize the number of sensors for a target coverage problem. An optimization method is an algebraic or geographic deduction applied in algorithms [46,102,184] . An optimization method also contains Dijkstra algorithm [93,118] , maximum flow algorithm in a graph, and maximum matching algorithm in a bipartite graph . Game theory based algorithm, such as , is classified into an optimization method. Heuristic algorithms, such as swarm intelligence algorithms, genetic algorithms, neural network algorithms and clustering algorithms , always search a local optimal solution. ..." What are related quotes?
Reference 5 Review article Compression of smart meter big data: A surveyWen L., Zhou K., Yang S., Li L. Renewable and Sustainable Energy Reviews, 2018 pp 59-69 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Huffman coding Huffman coding is a coding method involving variable word length, which constructs a codeword with different prefix and the shortest average header length based on the probability of occurrence of certain characters. It is sometimes referred to as the optimum coding. The main processes of Huffman coding are as follows. First, given a binary tree initial set T = { t 1 , t 2 , ⋯ t m } ，each binary tree has only one root node with weight w i , and its left and right sub-trees are empty. Next, the two trees with smallest weight root nodes in T are selected as the left and right sub-trees of the newly constructed binary tree. The weight of the root node in the new binary tree is the sum of the root nodes weights of the left and right sub-trees. Then the two trees are deleted from T and the new binary tree is added to the set T in ascending order. Finally, the second and third steps are repeated until there is only one binary tree in set T [117–119] . Zeinali et al. discussed an adaptive form of Huffman coding. Only one pass is needed to construct a Huffman tree. A novel compression method was proposed for Wireless Sensor Network data based on the principle of adaptive Huffman coding [121,122] . It encodes the elementary characters in the difference value. However, Huffman coding has limits regarding encoding and decoding efficiency as the length of the code is variable. As a traditional compression algorithm, it has an edge on compressing smart meter big data due to of its ability for perfect data recovery. However, the compression ratio is not very high and the algorithm efficiency is low compared with other compression algorithms. With the creation and development of LZ algorithms, Huffman coding is gradually being replaced in smart meter big data compression. ..." What are related quotes?
Reference 6 Reference works Chapter Encyclopedia of Bioinformatics and Computational Biology, Volume 2Riccardo Dondi, Stefano Beretta Encyclopedia of Bioinformatics and Computational Biology, 2025 pp 518-531 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Minimum Spanning Tree We consider now the problem of computing a minimum spanning tree of a weighted graph. We present two well-known algorithms for this problem: Kruskal׳s algorithm and Prim׳s algorithm. These algorithms are based on a greedy strategy, but they differ in the way edges are added to the spanning tree. Kruskal׳s algorithm Kruskal׳s algorithm iteratively constructs a forest F , by greedily adding edges of minimum weight, that eventually, when the algorithm ends, is a minimum spanning tree. Consider the input weighted graph G = ( V , E , w ) and the forest F = ( V F , E F ) , where V F = V and E F ⊆ E , built by Kruskal׳s algorithm . F is initialized with an empty set of edges, hence all the vertices are isolated. At each step, the algorithm selects an edge { x , y } ∈ E \ E F having two properties: 1. x and y belong to different trees of F ; 2. { x , y } has minimum weight among the edges that satisfy the first property. Edge { x , y } is added to E F , thus the trees containing x and y are merged. The algorithm stops when F contains a single tree, hence all the vertices of G . In Fig. 4 we present an example of application of Kruskal׳s algorithm, highlighting the selected edges. At each step, the algorithm selects an edge that connects two trees of F (and, hence, that it does not induce a cycle), and that has minimum weight among these edges. When the algorithm starts F consists of six trees, each one having of a single vertex (and so no edge is selected). Then, Kruskal׳s algorithm selects edge { u 1 , u 2 } of minimum weight, and it merges the two trees containing vertices u 1 and u 2 . After this step the forest F consists of five trees. Notice that, after the last step of the algorithm, F consists of a single tree (a minimum spanning tree). Prim׳s algorithm We now present Prim׳s algorithm . ..." What are related quotes?
Reference 7 Reference works Chapter Graph AlgorithmsDondi R., Mauri G., Zoppis I. Encyclopedia of Bioinformatics and Computational Biology, 2017 pp 940-949 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Minimum Spanning Tree In this section we present two algorithms to compute a minimum spanning tree: Kruskal’s algorithm and Prim’s algorithm. Both algorithms compute a minimum spanning tree of a weighted graph G =( V , E ) with a greedy strategy, but differ in the way vertices are added to the spanning tree. Kruskal’s algorithm Kruskal’s algorithm starts from a forest consisting of an empty set of edges, and constructs a forest by greedily adding edges of minimum weight. Let G =( V , E ) be a graph and let F ( V F , E F ) be a forest, with V F = V and E F ⊆ E . F is initialized as follows: V F = V and E F =∅. Kruskal’s algorithm selects the edge { u , v }∈ E / E F of minimum weight such that u and v belong to different trees of forest F ; { u , v } is added to E F and the trees containing u and v are merged. Fig. 3 shows how the edges of a graph G are added to E F . Notice that at each step, the selected edge is that of minimum weight that does not induce a cycle, independently from the fact that the selected edge shares an endpoint with other edges in E F or not. Notice that in Fig. 3 (a) F consists of six trees, each one containing of a single vertex. In Fig. 3 (b) , Kruskal’s algorithm select edge { v 1 , v 2 } of minimum weight and merges the trees containing v 1 and v 2 . Hence, F consists of five trees: the tree T 1 having vertices v 1 , v 2 and edge { v 1 , v 2 }, and four trees each one containing a single vertex of v 3 , v 4 , v 5 , v 6 . In Fig. 3 (c) , Kruskal’s algorithm selects edge { v 3 , v 6 } and merges the trees containing v 3 and v 6 . ..." What are related quotes?
Reference 8 Review article A survey of underwater search for multi-target using Multi-AUV: Task allocation, path planning, and formation controlWang L., Zhu D., Pang W., Zhang Y. Ocean Engineering, 2023 pp 114393 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Dijkstra algorithm Greedy strategy and width-first search are adopted by the Dijkstra algorithm to solve the single-source shortest path problem of weighted directed graph or undirected graph. A Dijkstra-like method called sliding wavefront expansion method was proposed by Soulignac (2011) for AUV path planning. This method combines the effective cost function and the continuous motion model are combined in this method, and good global optimality is shown by the simulation results. Dijkstra algorithm was improved by Kirsanov et al., (2013) to enable AUV to avoid dynamic obstacles, but the improved path length was increased and only a 2D environment was considered. 3D path planning was considered in Zhang and Cheng (2020) to better solve the path planning problem. Dijkstra algorithm is improved and diagonal search and linear search at any angle are designed. Dijkstra algorithm with high success rate in obtaining the shortest path, for it passes through all nodes to obtain the shortest path . However, traversal of nodes and inefficiency are also fatal drawbacks when applied to large-scale complex path topology networks . ..." What are related quotes?
Reference 9 Book Chapter Advanced C++11 MultithreadingBertil Schmidt, Jorge González-Domínguez, Christian Hundt, Moritz Schlarb Parallel Programming, 2018 pp 135-164 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... In case we exceed the capacity of the backpack, we can terminate the recursive candidate generation since we cannot generate further valid solutions from the current state (Lines 130–131). Otherwise, we have found a valid solution which is subsequently checked versus the global state in Line 134 using the auxiliary method sequential_update in Lines 81–92. It compares our current solution candidate with the best global solution and performs an update in case of improvement. The following step is optional, however, it may significantly speed up the traversal of the binary tree (in our case by one order-of-magnitude). We compute a relaxed solution of the Knapsack Problem which overestimates the actual achievable value of the haul by greedily packing the items starting from position height+1 until we exceed the capacity (Lines 94–113). The computed value dantzig_bound(height+1, tuple) is most likely too high since it assumes that you are allowed to pack fractional items into your backpack. Nevertheless, you can exploit this overestimation to exclude a possible solution candidate if its value is smaller than the value of the best observed global solution. Note that we can realize the greedy packing strategy with a simple for-loop since we have already sorted the tuples according to their value density in the initialization step. The final portion of the method traverse in Lines 143–147 recursively generates two new solution candidates: one that takes the item at position height+1 and another one that leaves it behind. The traversal routine terminates if either all new solution candidates are pruned by the global bound in Line 134, the local bound in Lines 139–149, or alternatively we have reached the leaves of the binary tree. The worst-case traversal has to probe all 2 n item combinations. Finally, we implement the actual computation in the main function. The corresponding code fragment is shown in Listing 5.15. First, we initialize the tuples in Line 151 by means of the aforementioned auxiliary function init_tuples . ..." What are related quotes?
Reference 10 Review article A survey on the placement of virtual network functionsSun J., Zhang Y., Liu F., Wang H., Xu X., Li Y. Journal of Network and Computer Applications, 2022 pp 103361 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Individual-based meta-heuristic algorithms The individual-experience-based heuristic algorithm iteratively modifies its search path that reaches the local suboptimal solution. And the greedy algorithm is a common strategy for searching VNF placement schemes. In addition, heuristic algorithms based on graph theory are often used to optimize the mapping between the VNF-FG and resource graphs. The greedy algorithm is a common heuristic strategy for searching VNF placements. First, designing greedy strategies generally depends on specific issues, e.g., combining and embedding VNF-FGs , reducing VNFs’ relocation costs , and reducing the number of VNF instances . The distinctions among greedy strategies are mainly reflected in the indicators for guiding their search paths, e.g., upscaling and context switching costs , the priority of each SFC , the importance of each VNF . Second, some designs of greedy strategies are based on topologies, e.g., the DAG , tree topology , and hop-count and direction . Lin et al. (2018) propose a breadth-first search algorithm to select locations with the least cost in layers. Finally, there are some greedy algorithms that prove their theoretical bounds on performances . Fan et al. (2017) propose a greedy algorithm with a theoretical lower bound to optimize the placements of VNF backups. Guo et al. (2018) use the duality theory and multiplicative weight update method to guarantee the lower bound of algorithmic performance. Placing VNFs can be regarded as matching the graphs of user demands and resource supplies. First, some works use the graph theory to describe service requirements, e.g., state synchronization between stateful VNFs and their replicas , dependencies among VNFs , orchestration of VNFs with different demands . Jalalitabar et al. (2016) propose a DAG-based heuristic algorithm to place VNFs with dependency. Agarwal et al. (2018) design a queue-based algorithm to orchestrate different types of SFC requests in slices. ..." What are related quotes?
Reference 11 Review article A comprehensive survey of Network Function VirtualizationYi B., Wang X., Li K., Das S.K., Huang M. Computer Networks, 2018 pp 212-262 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Qu et al. and , published by the same authors, were proposed to formulate the VNF-S problem as a series of scheduling decisions which activated various VNFs to process the traffic of different services. Based on the defined integer and non-integer variables, the formulation was managed in a more fine-grained manner, in which minimizing the latency of VNF scheduling was focused. The lower the latency, the more customers are served, which naturally leads to more profits. Qu et al. mainly focused on how to assign the execution time slots to different services traversing the same VNF via using a greedy method, while Qu et al. considered not only this aspect, but also the resource allocation. In particular, Qu et al. decomposed the proposed VNF-S model into four sub-problems, that is, i ) the virtual link bandwidth allocation sub-problem (assigning each virtual link a feasible data rate); ii ) the VNF assignment sub-problem (allocating VNFs of each network service to VMs); iii ) computing the transmit rates at the VNFs of each network service; iv ) generating a feasible schedule for each VNF via a greedy method. Based on such decomposition and the Genetic Algorithm (GA), Qu et al. developed a simplified heuristic method for solving the VNF-S problem with transmission delay considered. Therefore, by dynamically adjusting the allocated resources, Qu et al. reduced the scheduling time by 15–20%, while Qu et al. saved roughly 14.3% scheduling time. ..." What are related quotes?
Reference 12 Reference works Chapter Optimal Test ConstructionVeldkamp B.P. Encyclopedia of Social Measurement, 2005 pp 933-941 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Heuristics For some optimal test construction problems, 0-1 linear programming techniques cannot be applied because of the nonlinearity of the objective function or the constraints or because the techniques may need too much time. In addition, it might not be possible to formulate the problem as a network-flow model. In those cases, heuristical methods can be applied to find a solution. An heuristic is an approximation method that works fast but that tends to result in a good solution rather than in the optimal solution. In optimal test construction, the greedy algorithm, simulated annealing, and genetic algorithms have been applied successfully. Greedy algorithms work very fast. They select items sequentially. At every iteration, the item is selected that contributes most to the objective function. The normalized weighted absolute deviation heuristic (NWADH) is a well-known application of a greedy heuristic. It has also been applied very successfully in combination with the weighted deviations model in Eqs. (15) . However, because these heuristics operate sequentially, the greedy algorithm might run into infeasibility problems at the end of the test, when violations of constraints are not allowed. Simulated annealing is a much more time-consuming method. First, an initial test is constructed that meets all the specifications. Then, one item is swapped with an item in the pool. If the new test performs better with respect to the objective function, it is accepted; otherwise it is accepted with a probability that decreases during the test assembly process. The method stops when the probability of accepting a worse test is smaller than a lower bound. When genetic algorithms are applied, several tests are constructed that meet all the specifications. New tests are constructed by selecting one part from one test and another part from a second test. If the new test performs better with respect to the objective function, it is added to the set of candidate tests. At the end, the best candidate in the set is selected. ..." What are related quotes?
Reference 13 Book Chapter Optimal Visual Sensor Network ConfigurationZhao J., Cheung S.C.S., Nguyen T. Multi-Camera Networks, 2009 pp 139-162 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... GREEDY: An Algorithm to Speed Up BIP BIP is a well-studied NP-hard combinatorial problem with many heuristic schemes, such as branch-and-bound, already implemented in software libraries (e.g., lp_solve [ 20 ]). However, even these algorithms can be quite intensive if the search space is large. In this section, we introduce a simple greedy algorithm, GREEDY, that can be used for both MIN_CAM and FIX_CAM. Besides experimentally showing the effectiveness of GREEDY, we believe that the greedy approach is an appropriate approximation strategy because of the similarity of our problem to the set cover problem. In the set cover problem, items can belong to multiple sets; the optimization goal is to minimize the number of sets to cover all items. While finding the optimal solution to set covering is an NP-hard problem [ 21 ], it has been shown that the greedy approach is essentially the best we can do to obtain an approximate solution [ 22 ]. We can draw the parallel between our problem and the set cover problem by considering each of the tag grid points as an item “belonging” to a camera grid point if the tag is visible at that camera. The set cover problem then minimizes the number of cameras needed, which is almost identical to MIN_CAM except for the fact that visual tagging requires each tag to be visible by two or more cameras. The FLX_CAM algorithm further allows some of the tag points not to be covered at all. It is still an open question whether these properties can be incorporated into the framework of set covering, but our experimental results demonstrate that the greedy approach is a reasonable solution to our problem. GREEDY is described in Algorithm 6.2 . In each round of the GREEDY algorithm, the camera grid point that can see the most number of tag grid points is selected and all the tag grid points visible to two or more cameras are removed. When using GREEDY to approximate MIN_CAM, we no longer need to refine the tag grids to reduce computational efficiency. We can start with a fairly dense tag grid and set the camera bound m to infinity. ..." What are related quotes?
Reference 14 Review article Approximation and online algorithms for multidimensional bin packing: A surveyChristensen H.I., Khan A., Pokutta S., Tetali P. Computer Science Review, 2017 pp 63-79 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Online algorithms : Optimization problems where the input is received in an online manner (i.e., the input does not arrive as a single batch but as a sequence of input portions) and the output must be produced online (i.e., the system must react in response to each incoming portion taking into account that future input is not known at any point in time) are called online problems. Bin packing is also one of the key problems in online algorithms. Let us define the notion of a competitive ratio which will be useful when we discuss some related results in online algorithms in later sections. Definition 2.6 Competitive Ratio An online algorithm A for a minimization problem Π is called c - competitive if there exists a constant δ such that for all finite input sequences I , A ( I ) ≤ c ⋅ Opt ( I ) + δ . If the additive constant δ ≤ 0 , we say A to be strictly c - competitive . The infimum over the set of all values c such that A is c -competitive is called the competitive ratio of A . We will sometimes call the above defined competitive ratio to be asymptotic competitive ratio and strictly competitive ratio to be absolute competitive ratio . In general, there are no requirements or assumptions concerning the computational efficiency of an online algorithm. However, in practice, we usually seek polynomial time online algorithms. We refer the readers to [32,12] for more details on online algorithms. There are few others metrics to measure the quality of a packing, such as random-order ratio , accommodation function , relative worst-order ratio , differential approximation measure etc. 2.2 One dimensional bin packing Before going to multidimensional bin packing, we give a brief description of the results in 1-D bin packing. Here we focus primarily on very recent results. For a detailed survey and earlier results we refer the interested reader to . 2.2.1 Offline 1-D bin packing The earliest algorithms for one dimensional (1-D) bin packing were simple greedy algorithms such as FirstFit (FF), NextFit (NF), FirstFitDecreasing (FFD), NextFitDecreasing (NFD) etc. ..." What are related quotes?
Reference 15 Handbook Chapter Cognitive Analytics: Going Beyond Big Data Analytics and Machine LearningGudivada V.N., Irfan M.T., Fathi E., Rao D.L. Handbook of Statistics, 2016 pp 169-205 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Decision Trees The Classification and Regression Tree (CART) method was originally presented by Breiman et al. (1984) during the 1980s. This has led to tremendous interest in decision tree learning. In the supervised classification setting, the objective of decision tree learning is to compute a special type of tree that can classify examples to classes. The notions of training, validation, and test sets as well as overfitting vs underfitting issues apply for decision trees too. The underlying model in decision tree learning is a tree in graph-theoretic sense. However, we must also recognize a stylized control flow that is superimposed on the tree structure. Each internal node of the tree, including the root, asks a decision-type question. Based on the answer for an example, we next traverse one of the children of that internal node. Once we reach a leaf node, we are certain to know the classification of the example according to the decision tree, since each leaf node is annotated with a class label. In addition to CART, there are many other learning algorithms for finding the “best” tree for a classification problem. Most modern algorithms like Iterative Dichotomiser 3 (ID3) and its successors C4.5 and C5 use information theoretic measures, such as entropy, to learn a tree. Entropy can be thought of as a measure of uncertainty. Initially, the whole training set, consisting of examples of different classes, will have a very high entropy measure. ID3 and its successors repeatedly partition the training set in order to reduce the sum of the entropy measures of the partitions. Usually, a greedy strategy is employed for this purpose. The algorithm chooses a feature and partitions the training set based on that feature. The feature is chosen with the goal of minimizing the sum of the entropy measures across the resulting partitions. The same procedure is recursed on each partition, unless all the examples in that partition belong to the same class. One big advantage of decision tree learning over other learning methods such as logistic regression is that it can capture more complex decision boundaries. ..." What are related quotes?
Reference 16 Review article Comprehensive survey on resource allocation for edge-computing-enabled metaverseBaidya T., Moh S. Computer Science Review, 2024 pp 100680 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Heuristic approaches Heuristic methods offer practical solutions by making trade-offs between optimality and computational complexity. These approaches are particularly useful when the problem domain is too complex for exact mathematical solutions or when approximate solutions are acceptable. The following are some heuristic approaches used in recent research in the domain of resource allocation. 1) Greedy-based heuristic: In [ 61 ], the authors designed greedy-based heuristic algorithms to manage and optimize resource allocation efficiently under dynamic and real-time conditions. The primary challenge addressed in the paper involves optimizing the trade-off between video quality and delivery latency in a multi-tier cloud VR system leveraging edge computing. The main objective is to enhance user experience by strategically managing rendering tasks between edge servers and a centralized cloud, taking into consideration the limitations in processing power at the edge and varying network conditions. The solution involves formulating the resource allocation problem as an integer non-linear programming (INLP), which is then tackled using online greedy algorithms. These algorithms dynamically allocate video rendering tasks to either edge or cloud servers based on real-time assessments of network conditions and server capacities. The heuristic approach optimizes the allocation by prioritizing tasks that can be most efficiently processed, considering both current system load and the quality-latency trade-off. The performance of the heuristic technique was reported to significantly enhance the system efficiency. It achieved a significant improvement in video delivery latency and quality compared to more static allocation strategies. The heuristic's dynamic nature allows it to adapt quickly to changes in network conditions and server load, providing a more responsive and efficient system. Lessons learned: The greedy-based heuristic algorithm enhances resource allocation in cloud VR systems by dynamically distributing video rendering tasks between edge and cloud servers, based on real-time network and server conditions. This approach, formulated as INLP problem, optimizes the trade-off between video quality and latency, adapting quickly to changes and significantly improving system responsiveness and efficiency. ..." What are related quotes?
Reference 17 Book Chapter Modeling and Optimization Approaches in Design and Management of Biomass-Based Production ChainsŞebnem Yılmaz Balaman Decision-Making for Biomass-Based Production Chains, 2019 pp 185-236 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Metaheuristic algorithms can be used in combination with other optimization and statistical methods, such as mathematical programming, stochastic optimization, k-means clustering, and machine learning. For example, a metaheuristic can select an arbitrary solution in the feasible solution space restricted by the constraints of a mathematical programming model as the initial solution, and after calculating the associated objective function value the metaheuristic may choose to change the solution and try another one, based on its algorithm and specific search rules. The procedure continues until no improvements in the objective function value is achieved and the best approximate to the global optimum is reached. Theoretically, if metaheuristics are run for long enough, they can even find the global optimum solution. In the following subsections, the major metaheuristics that have been widely used, and that have potential to be used in modeling and optimization for design and management of biomass-based production chains, are summarized. 7.3.1.1 Greedy Algorithms Greedy algorithms employ a problem-solving procedure to progressively build candidate solutions, to approximate the global optimum, by obtaining better and better locally optimal solutions at each stage. In general, greedy algorithms cannot yield a global optimal solution, but they may produce good locally optimal solutions in a reasonable time and with less computational effort. GRASP , is a well-known iterative local search-based greedy algorithm that involves a number of iterations to construct greedy randomized solutions and improve them successively. The algorithm consists of two main stages, construction and local search, to initially construct a solution, and then repair this solution to achieve feasibility. The algorithm produces greedy randomized solutions, by selecting new elements from a candidate set of greedy solutions constructed, based on the level of improvement on the partial solution under construction, using a greedy evaluation function. The selected elements are then incorporated into the current partial solution without destroying feasibility (if feasibility is destroyed, a new element is selected), until a complete feasible solution, of which neighborhood is investigated until a local optimum found by the local search procedure, is obtained. 1 ..." What are related quotes?
Reference 18 Review article Multicasting in cognitive radio networks: Algorithms, techniques and protocolsQadir J., Baig A., Ali A., Shafi Q. Journal of Network and Computer Applications, 2014 pp 44-61 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Amongst various heuristic solutions, we will discuss greedy algorithms, genetic algorithms, and ant colony optimization algorithms. Greedy algorithms : In a greedy algorithm , the basic building block of a complete feasible solution is a partial solution developed by ‘greedy’ that are based on whatever partial information is available at the time. These partial solutions are progressively developed to build a more complete solution until the iterations develop a complete feasible solution. The famous Dijkstra algorithm used for solving the shortest path tree (SPT) problem is an example greedy algorithm. Randomization is an important tool that can be exploited to avoid local minima׳s in search-based optimization problems. In particular, randomization is utilized by tools like simulated annealing . Another core idea adopted in some metaheuristic techniques like tabu-search is to use adaptive memory contrary to the approach adopted in memoryless approaches like simulated annealing. Metaheuristics also employ concepts of intensification (which encourages intensifying previous solutions found to perform well) and diversification (which encourages search to examine unvisited solutions). In some other fields (e.g., in genetic algorithms and reinforcement learning), the concepts of intensification and diversification are known by the terms exploitation and exploration , respectively. It is to be noted that exploitation and exploration, or alternatively, intensification and diversification, represent conflicting goals and therefore the dilemma of choosing one or the other needs to be resolved in a balanced fashion. There are various other metaheuristic techniques proposed in literature and the interested reader is referred to a book on this topic or the book chapter on this topic in Resende and Pardalos (2006) . We discuss next two particular evolutionary algorithm metaheuristic algorithms: genetic algorithms and ant-colony optimization. Genetic algorithms : A genetic algorithm (GA) is a particular class of evolutionary algorithm which uses techniques inspired from evolutionary biology—such as inheritance, mutation, crossover, and natural selection—to improve the performance of a computational process. 1 ..." What are related quotes?
Reference 19 Reference works Chapter Encyclopedia of Bioinformatics and Computational Biology, Volume 1Massimo Cafaro, Italo Epicoco, Marco Pulimeno Encyclopedia of Bioinformatics and Computational Biology, 2025 pp 9-20 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Iterative Algorithms The greedy strategy, which was discussed in the previous section, works by building a solution to an optimization problem piece by piece, continually adding a piece that is locally optimal to a solution that has been partially formed. In this section, we present an alternative method for the design of algorithms for solving optimization problems. It begins with a solution that is feasible (i.e., a solution that satisfies all of the constraints of the problem) and then tries to improve it by making repeated applications of a basic step. This stage often entails making a minor localized adjustment, which, when successful, results in a solution that is both feasible and yields a better value for the objective function of the problem. When the value of the objective function is not improved by any of these changes, the algorithm will return the most recent solution that was feasible as the optimal one before stopping. There are a number of potential obstacles that could prevent the successful execution of this iteration based approach. First, there is the need to find an initial solution that is viable. We can always start with a solution that is trivial or utilize an approximate solution that was achieved by another technique (for example, greedy). This is possible for some problems. On the other hand, for other problems, coming up with an initial solution could take just as much time and effort as solving the problem once a feasible initial solution has been found. Moreover, it is not always evident what kind of local changes should be allowed in a viable solution in order for us to efficiently assess whether the current solution is locally optimal and, in the event it is not optimal, replace it with a solution that is better with regard to the objective function of the problem. Additionally, we need to take into account local versus global optimal solutions, for either the maximum or minimum of the objective function. This is the most fundamental obstacle, for which different techniques have been designed. 1 ..." What are related quotes?
Reference 20 Book Chapter Classification: advanced methodsJiawei Han, Jian Pei, Hanghang Tong Data Mining, 2023 pp 307-377 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... This strategy is greedy in the sense that it tries to make the best use ( exploitation ) of the information we have collected so far regarding the true value q ( a ) . However, what if there is a big gap between the estimated value Q ( a ) and the true value q ( a ) ? In other words, it is quite possible that q ( c a m e r a ) > q ( T V ) , even though Q ( c a m e r a ) < Q ( T V ) , especially given that Q ( c a m e r a ) and Q ( T V ) are based on very limited data in this example (4 days for camera and 1 day for TV). Moreover, we have never tried printer advertisement at all and thus have zero knowledge about q ( p r i n t e r ) . What if the printer advertisement actually has the highest value? Another alternative strategy is the random strategy ( exploration ). Each day, we choose a random product to advertise and observe the reward of that advertisement. If we keep running this for many days, the estimated value Q ( a ) is likely to be very close to the true value q ( a ) . After that, we choose the advertisement with the highest (estimated) value. This strategy is optimal in the long run. However, we might spend many days before we figure out the optimal action, during which the received rewards might be low. A better strategy is to combine the greedy strategy and the random strategy together. That is, at each day, with the probability 1 − ϵ ( 0 < ϵ < 1 ) , we choose the action (advertisement) with the highest estimated value Q ( a ) ; with the probability ϵ , we choose a random action; and at the end of the day, we use the observed reward to update the estimated value Q ( a ) . This strategy (called ϵ -greedy) is likely to obtain a better balance between the immediate (exploitation) and the long-term (exploration) return. 1 ..." What are related quotes?
Reference 21 Book Chapter Online Strategies for Optimizing Medical Supply in Disaster ScenariosGuttinger D., Godehardt E., Zinnen A. Service Science, Management, and Engineering:, 2012 pp 95-116 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Conclusions In this final section, we want to summarize our results and give an outlook on further interesting research questions regarding this topic. 6.6.1 Summary In this work, we compare several online approaches for optimizing the emergency supply after a major incident. For a given set of physicians, hospitals, and transport vehicles, the algorithms introduced in this work compute an assignment of casualties arriving in groups that suffer from specific types of injuries to available transport and medical capacities. Our objective is to minimize total injury to all casualties, where the level of injury of a casualty is measured by the value of a specific penalty function. For each type of injury, we define such a penalty function that depends on waiting time until medical treatment and describes the current state of health of a casualty. In our simulation studies, we show that a simple greedy strategy clearly outperforms an assignment based on the D’Hondt algorithm, which is, to the best of our knowledge, the only technique currently used in practice to deal with the problem at hand. We have furthermore analyzed the additional benefit when applying a workload-adapted version of simulated annealing subsequent to each greedy iteration. In doing so, it appeared that the average level of injury for each casualty (and so also total level of injury) yielded by the greedy strategy can be additionally reduced by about 10% with an acceptable computational expenditure. Thus, altogether we can conclude that the combination of a greedy strategy and a subsequent application of a workload-adapted version of simulated annealing works well for the given online assignment problem. 6.6.2 Future Prospects Despite the promising results of our adjusted simulated annealing algorithm introduced in this work, there is still room for improvements. One issue for future research will be to try out other optimization techniques like genetic approaches to see if they can provide better results. Furthermore, the influence of the chosen penalty functions on the results should be studied with respect to their number and type. 1 ..." What are related quotes?
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Reference 1 Review article Influence analysis in social networks: A surveyPeng S., Zhou Y., Cao L., Yu S., Niu J., Jia W. Journal of Network and Computer Applications, 2018 pp 17-32 View PDFView articleShow related quote(s) Related quote(s)1 / 2 "... Types of influence maximization algorithm (1) Greedy-based: A greedy algorithm is a pervasively used algorithmic paradigm, which uses problem solving heuristic to make the locally optimal choice at each stage , hoping to reach a global optimum. In many cases, a greedy strategy does not in general reach an optimal solution, but it may yield locally optimal solutions that approximate a global optimal solution in a reasonable time. (2) Heuristic-based: In computer science, artificial intelligence, and mathematical optimization, a heuristic algorithm is designed for solving a problem more quickly when classic methods are too slow, or for finding an approximate solution when classic methods fail to find any exact solutions. This is achieved by trading optimality, completeness, accuracy, or precision for speed. In a way, it can be considered a shortcut. (3) Others: Besides the two major types, many researchers have presented voting-based algorithms and hybrid-based algorithms to improve the performance and scalability of the existing methods of the greedy-based and the heuristic-based. ..." What are related quotes? Related quote(s)2 / 2 "... Influence maximization algorithm Domingos and Richardson are the first to propose the influence maximization as an algorithmic technique for viral marketing. Kempe et al. (2003) are the first to formalize the influence maximization as a discrete optimization problem, and to prove the problem is NP-hard. According to the related literature, the influence maximization problem is formulated as follows: given a social network modeled as a graph G = ( V , E ), and a nonnegative number k , where nodes are users and edges are labeled with influence probabilities among users, the influence maximization problem is to find a set of k seed nodes that maximizes the influence propagation scale in the social network under a given diffusion model. In this section, we summarize the types of influence maximization algorithm, overview the existing influence maximization algorithm, and analyze the performance analysis on the existing algorithms. 6.1 Types of influence maximization algorithm (1) Greedy-based: A greedy algorithm is a pervasively used algorithmic paradigm, which uses problem solving heuristic to make the locally optimal choice at each stage , hoping to reach a global optimum. In many cases, a greedy strategy does not in general reach an optimal solution, but it may yield locally optimal solutions that approximate a global optimal solution in a reasonable time. (2) Heuristic-based: In computer science, artificial intelligence, and mathematical optimization, a heuristic algorithm is designed for solving a problem more quickly when classic methods are too slow, or for finding an approximate solution when classic methods fail to find any exact solutions. This is achieved by trading optimality, completeness, accuracy, or precision for speed. In a way, it can be considered a shortcut. (3) Others: Besides the two major types, many researchers have presented voting-based algorithms and hybrid-based algorithms to improve the performance and scalability of the existing methods of the greedy-based and the heuristic-based. 6.2 The existing algorithms (1) Greedy-based Wang and Feng (2009) proposed a target wise greedy algorithm (TWG for short) based on the potential-based node-selection strategy. 1 ..." What are related quotes?
Reference 2 Review article Resource provision algorithms in cloud computing: A surveyZhang J., Huang H., Wang X. Journal of Network and Computer Applications, 2016 pp 23-42 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Greedy algorithms Greedy algorithm is widely used in designing of various approaches to provision resources, especially when the problem has an optimal structure that means the optimal solution to the problem contains the solution to the subproblem. Every time the algorithm selects a solution which seems best at the moment based on a greedy strategy. It iteratively makes a choice, reducing the problem to a smaller one. In other words, greedy algorithm makes a local optimization and looks forward to finding a global optimal solution to the original problem. The greedy strategy plays a critical role in the success of the algorithm. Although both greedy and MH algorithms search the optimal solution iteratively. Greedy algorithm requires the problem to be addressed to have a specific optimal structure as aforementioned, but there is no restriction for MH. In contrast with the heuristic nature of MH, greedy category plays a great role in approximation algorithms designing. In spite of these differences, both of them are best-effort schemes and cannot guarantee to find the global optimal solution. 7.1 Cloud and DC Selection The VMs in multi-set are placed in multi-cloud circumstance . The sites are ordered in ascending order in terms of network cost. Then sites are picked one by one and placed on clouds greedily with minimal cost. In distributed cloud, greedy algorithm is proposed to optimize the performance. Virtual desktops are assigned to DC closer to users to minimize the access latency . The greedy strategy selects VMs with the highest location-sensitive ranks first and places them in the closest DC to their user. If the closest DC cannot accommodate the VMs then the next closest one is examined. Distributed local greedy algorithms are proposed to minimize latency and number of DCs used . Abu Sharkh et al. (2013) strive to minimize the average connection access tardiness of a group of user requests. Both computational and network resources are reckoned by a heuristic two-stage scheduling approach. The first stage partitions the VMs to PM nodes and the second one finds paths for the connections by a greedy algorithm. ..." What are related quotes?
Reference 3 Book Chapter Fundamentals of AlgorithmsHuang C.-Y., Lai C.-Y., Cheng K.T. Electronic Design Automation, 2009 pp 173-234 View PDFView chapterShow related quote(s) Related quote(s)1 / 3 "... To determine whether a particular problem has an optimal substructure, two aspects have to be examined: substructure and optimality . A problem has substructures if it is divisible into subproblems. Optimality is the property that the combination of optimal solutions to subproblems is a globally optimal solution. Greedy algorithms are highly efficient for problems satisfying these two properties. On top of that, greedy algorithms are often intuitively simple and easy to implement. Therefore, greedy algorithms are very popular for solving optimization problems. Many graph algorithms, mentioned in Section 4.3 , are actually applications of greedy algorithms—such as Prim's algorithm used for finding minimum spanning trees. Greedy algorithms often help find a lower bound of the solution quality for many challenging real-world problems. 4.4.2 Dynamic programming Dynamic programming (DP) is an algorithmic method of solving optimization problems. Programming in this context refers to mathematical programming, which is a synonym for optimization. DP solves a problem by combining the solutions to its subproblems. The famous divide-and-conquer method also solves a problem in a similar manner. The divide-and-conquer method divides a problem into independent subproblems, whereas in DP, either the subproblems depend on the solution sets of other subproblems or the subproblems appear repeatedly. DP uses the dependency of the subproblems and attempts to solve a subproblem only once; it then stores its solution in a table for future lookups. This strategy spares the time spent on recalculating solutions to old subproblems, resulting in an efficient algorithm. To illustrate the superiority of DP, we show how to efficiently multiply a chain of matrices by use of DP. When multiplying a chain of matrices, the order of the multiplications dramatically affects the number of scalar multiplications. For example, consider multiplying three matrices A, B, and C whose dimensions are 30 × 100, 100 × 2, and 2 × 50, respectively. There are two ways to start the multiplication: either A ⋅ B or B ⋅ C first. ..." What are related quotes? Related quote(s)2 / 3 "... The change will consequently be made in this sequence: a quarter (25 cents), a dime (10 cents), and a penny (1 cent)—a total of three coins. This rule of thumb leads to the minimum number of coins, three, because it perfectly embodies the essence of greedy algorithms: making greedy choices at each moment. In this particular problem, a greedy algorithm yields the optimal solution. However, greedy algorithms do not always produce optimal solutions. Let us revisit the making change example. If a coin with a value of 20 cents exists, the rule of thumb just mentioned would not lead to the minimum number of coins if the amount of change needed was 40 cents. By applying the rule of picking the coin of highest value first, we would be giving change of a quarter (25 cents), a dime (10 cents) and a nickel (5 cents), a total of three coins, but, in fact, two, 20-cent coins would be the optimal solution for this example. The greedy algorithm fails to reach the optimal solution for this case. Actually, the example given previously is not ideal for illustrating the concept of greedy algorithms, because it violates the optimal substructure property. In general, problems suitable for greedy algorithms must exhibit two characteristics: the greedy-choice property and the optimal substructure property. If we can demonstrate that a problem has these two properties, then a greedy algorithm would be a good choice. 4.4.1.1 Greedy-choice property The greedy-choice property states that a globally optimal solution can always be achieved by making locally optimal, or greedy, choices. By locally optimal choices we mean making choices that look best for solving the current problem without considering the results from other subproblems or the effect(s) that this choice might have on future choices. In Section 4.4 , we introduced the Nearest Neighbor (NN) algorithm for solving—more precisely, for approximating—an optimal solution to TSP. NN is a greedy algorithm that picks the nearest city at each step. ..." What are related quotes? Related quote(s)3 / 3 "... The greedy algorithm fails to reach the optimal solution for this case. Actually, the example given previously is not ideal for illustrating the concept of greedy algorithms, because it violates the optimal substructure property. In general, problems suitable for greedy algorithms must exhibit two characteristics: the greedy-choice property and the optimal substructure property. If we can demonstrate that a problem has these two properties, then a greedy algorithm would be a good choice. 4.4.1.1 Greedy-choice property The greedy-choice property states that a globally optimal solution can always be achieved by making locally optimal, or greedy, choices. By locally optimal choices we mean making choices that look best for solving the current problem without considering the results from other subproblems or the effect(s) that this choice might have on future choices. In Section 4.4 , we introduced the Nearest Neighbor (NN) algorithm for solving—more precisely, for approximating—an optimal solution to TSP. NN is a greedy algorithm that picks the nearest city at each step. NN violates the greedy-choice property and thus results in suboptimal solutions, as indicated in the example of Figure 4.23 . In Figure 4.23 , the choice of B→D is a greedy one, because the other remaining cities are further from B. In a globally optimal solution, the route of either D→C→B or B→D→D is a necessity, and the choice of B→D is suboptimal. Hence, NN is not an optimal greedy algorithm, because TSP does not satisfy the greedy-choice property. Making change with a minimum number of coins is an interesting example. On the basis of the current U.S. coins, this problem satisfies the greedy-choice property. But when a 20-cent coin comes into existence, the property is violated—when making change for 40 cents, the greedy choice of picking a quarter affects the solution quality of the rest of the problem. How do we tell if a particular problem has the greedy-choice property? ..." What are related quotes?
Reference 4 Review article Coverage problem with uncertain properties in wireless sensor networks: A surveyWang Y., Wu S., Chen Z., Gao X., Chen G. Computer Networks, 2017 pp 200-232 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Traditional vs. heuristic algorithms Traditional algorithms, such as greedy algorithm, dynamic programming, and backtracking, are popular solutions for the coverage problem. These classical algorithms are initial schemes and easy to analyze theoretically. We summarize researches on coverage problems with uncertain properties from three typical objectives: maximizing probability or ratio (in Table 5 ), maximizing network lifetime (in Table 6 ), and minimizing the number of sensors (in Table 7 ). In those tables, several typical solutions are used, such as greedy, programming, and optimization. The greedy method is one of the most commonly used method in algorithm design. In our survey, traditional algorithms, particularly the greedy method, are the dominanting ones. For example, researches in [32,48,68,133,167,173,174] use greedy algorithm for maximizing coverage quality. Similarly, for maximizing network lifetime, there are greedy algorithms in [9,10,12,13,17,47,49,65,77,114,128,140,142,144,145,151,164,168,169,181,182,193] . And for minimizing the number of sensors, there are greedy algorithms in [112,121,183] . In our survey, we consider that programming consists of dynamic programming and (integer) linear programming. Authors in use linear programming to maximize network lifetime while authors in use dynamic programming. Authors in use integer linear programming to minimize the number of sensors for a target coverage problem. An optimization method is an algebraic or geographic deduction applied in algorithms [46,102,184] . An optimization method also contains Dijkstra algorithm [93,118] , maximum flow algorithm in a graph, and maximum matching algorithm in a bipartite graph . Game theory based algorithm, such as , is classified into an optimization method. Heuristic algorithms, such as swarm intelligence algorithms, genetic algorithms, neural network algorithms and clustering algorithms , always search a local optimal solution. ..." What are related quotes?
Reference 5 Review article Compression of smart meter big data: A surveyWen L., Zhou K., Yang S., Li L. Renewable and Sustainable Energy Reviews, 2018 pp 59-69 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Huffman coding Huffman coding is a coding method involving variable word length, which constructs a codeword with different prefix and the shortest average header length based on the probability of occurrence of certain characters. It is sometimes referred to as the optimum coding. The main processes of Huffman coding are as follows. First, given a binary tree initial set T = { t 1 , t 2 , ⋯ t m } ，each binary tree has only one root node with weight w i , and its left and right sub-trees are empty. Next, the two trees with smallest weight root nodes in T are selected as the left and right sub-trees of the newly constructed binary tree. The weight of the root node in the new binary tree is the sum of the root nodes weights of the left and right sub-trees. Then the two trees are deleted from T and the new binary tree is added to the set T in ascending order. Finally, the second and third steps are repeated until there is only one binary tree in set T [117–119] . Zeinali et al. discussed an adaptive form of Huffman coding. Only one pass is needed to construct a Huffman tree. A novel compression method was proposed for Wireless Sensor Network data based on the principle of adaptive Huffman coding [121,122] . It encodes the elementary characters in the difference value. However, Huffman coding has limits regarding encoding and decoding efficiency as the length of the code is variable. As a traditional compression algorithm, it has an edge on compressing smart meter big data due to of its ability for perfect data recovery. However, the compression ratio is not very high and the algorithm efficiency is low compared with other compression algorithms. With the creation and development of LZ algorithms, Huffman coding is gradually being replaced in smart meter big data compression. ..." What are related quotes?
Reference 6 Reference works Chapter Encyclopedia of Bioinformatics and Computational Biology, Volume 2Riccardo Dondi, Stefano Beretta Encyclopedia of Bioinformatics and Computational Biology, 2025 pp 518-531 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Minimum Spanning Tree We consider now the problem of computing a minimum spanning tree of a weighted graph. We present two well-known algorithms for this problem: Kruskal׳s algorithm and Prim׳s algorithm. These algorithms are based on a greedy strategy, but they differ in the way edges are added to the spanning tree. Kruskal׳s algorithm Kruskal׳s algorithm iteratively constructs a forest F , by greedily adding edges of minimum weight, that eventually, when the algorithm ends, is a minimum spanning tree. Consider the input weighted graph G = ( V , E , w ) and the forest F = ( V F , E F ) , where V F = V and E F ⊆ E , built by Kruskal׳s algorithm . F is initialized with an empty set of edges, hence all the vertices are isolated. At each step, the algorithm selects an edge { x , y } ∈ E \ E F having two properties: 1. x and y belong to different trees of F ; 2. { x , y } has minimum weight among the edges that satisfy the first property. Edge { x , y } is added to E F , thus the trees containing x and y are merged. The algorithm stops when F contains a single tree, hence all the vertices of G . In Fig. 4 we present an example of application of Kruskal׳s algorithm, highlighting the selected edges. At each step, the algorithm selects an edge that connects two trees of F (and, hence, that it does not induce a cycle), and that has minimum weight among these edges. When the algorithm starts F consists of six trees, each one having of a single vertex (and so no edge is selected). Then, Kruskal׳s algorithm selects edge { u 1 , u 2 } of minimum weight, and it merges the two trees containing vertices u 1 and u 2 . After this step the forest F consists of five trees. Notice that, after the last step of the algorithm, F consists of a single tree (a minimum spanning tree). Prim׳s algorithm We now present Prim׳s algorithm . ..." What are related quotes?
Reference 7 Reference works Chapter Graph AlgorithmsDondi R., Mauri G., Zoppis I. Encyclopedia of Bioinformatics and Computational Biology, 2017 pp 940-949 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Minimum Spanning Tree In this section we present two algorithms to compute a minimum spanning tree: Kruskal’s algorithm and Prim’s algorithm. Both algorithms compute a minimum spanning tree of a weighted graph G =( V , E ) with a greedy strategy, but differ in the way vertices are added to the spanning tree. Kruskal’s algorithm Kruskal’s algorithm starts from a forest consisting of an empty set of edges, and constructs a forest by greedily adding edges of minimum weight. Let G =( V , E ) be a graph and let F ( V F , E F ) be a forest, with V F = V and E F ⊆ E . F is initialized as follows: V F = V and E F =∅. Kruskal’s algorithm selects the edge { u , v }∈ E / E F of minimum weight such that u and v belong to different trees of forest F ; { u , v } is added to E F and the trees containing u and v are merged. Fig. 3 shows how the edges of a graph G are added to E F . Notice that at each step, the selected edge is that of minimum weight that does not induce a cycle, independently from the fact that the selected edge shares an endpoint with other edges in E F or not. Notice that in Fig. 3 (a) F consists of six trees, each one containing of a single vertex. In Fig. 3 (b) , Kruskal’s algorithm select edge { v 1 , v 2 } of minimum weight and merges the trees containing v 1 and v 2 . Hence, F consists of five trees: the tree T 1 having vertices v 1 , v 2 and edge { v 1 , v 2 }, and four trees each one containing a single vertex of v 3 , v 4 , v 5 , v 6 . In Fig. 3 (c) , Kruskal’s algorithm selects edge { v 3 , v 6 } and merges the trees containing v 3 and v 6 . ..." What are related quotes?
Reference 8 Review article A survey of underwater search for multi-target using Multi-AUV: Task allocation, path planning, and formation controlWang L., Zhu D., Pang W., Zhang Y. Ocean Engineering, 2023 pp 114393 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Dijkstra algorithm Greedy strategy and width-first search are adopted by the Dijkstra algorithm to solve the single-source shortest path problem of weighted directed graph or undirected graph. A Dijkstra-like method called sliding wavefront expansion method was proposed by Soulignac (2011) for AUV path planning. This method combines the effective cost function and the continuous motion model are combined in this method, and good global optimality is shown by the simulation results. Dijkstra algorithm was improved by Kirsanov et al., (2013) to enable AUV to avoid dynamic obstacles, but the improved path length was increased and only a 2D environment was considered. 3D path planning was considered in Zhang and Cheng (2020) to better solve the path planning problem. Dijkstra algorithm is improved and diagonal search and linear search at any angle are designed. Dijkstra algorithm with high success rate in obtaining the shortest path, for it passes through all nodes to obtain the shortest path . However, traversal of nodes and inefficiency are also fatal drawbacks when applied to large-scale complex path topology networks . ..." What are related quotes?
Reference 9 Book Chapter Advanced C++11 MultithreadingBertil Schmidt, Jorge González-Domínguez, Christian Hundt, Moritz Schlarb Parallel Programming, 2018 pp 135-164 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... In case we exceed the capacity of the backpack, we can terminate the recursive candidate generation since we cannot generate further valid solutions from the current state (Lines 130–131). Otherwise, we have found a valid solution which is subsequently checked versus the global state in Line 134 using the auxiliary method sequential_update in Lines 81–92. It compares our current solution candidate with the best global solution and performs an update in case of improvement. The following step is optional, however, it may significantly speed up the traversal of the binary tree (in our case by one order-of-magnitude). We compute a relaxed solution of the Knapsack Problem which overestimates the actual achievable value of the haul by greedily packing the items starting from position height+1 until we exceed the capacity (Lines 94–113). The computed value dantzig_bound(height+1, tuple) is most likely too high since it assumes that you are allowed to pack fractional items into your backpack. Nevertheless, you can exploit this overestimation to exclude a possible solution candidate if its value is smaller than the value of the best observed global solution. Note that we can realize the greedy packing strategy with a simple for-loop since we have already sorted the tuples according to their value density in the initialization step. The final portion of the method traverse in Lines 143–147 recursively generates two new solution candidates: one that takes the item at position height+1 and another one that leaves it behind. The traversal routine terminates if either all new solution candidates are pruned by the global bound in Line 134, the local bound in Lines 139–149, or alternatively we have reached the leaves of the binary tree. The worst-case traversal has to probe all 2 n item combinations. Finally, we implement the actual computation in the main function. The corresponding code fragment is shown in Listing 5.15. First, we initialize the tuples in Line 151 by means of the aforementioned auxiliary function init_tuples . ..." What are related quotes?
Reference 10 Review article A survey on the placement of virtual network functionsSun J., Zhang Y., Liu F., Wang H., Xu X., Li Y. Journal of Network and Computer Applications, 2022 pp 103361 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Individual-based meta-heuristic algorithms The individual-experience-based heuristic algorithm iteratively modifies its search path that reaches the local suboptimal solution. And the greedy algorithm is a common strategy for searching VNF placement schemes. In addition, heuristic algorithms based on graph theory are often used to optimize the mapping between the VNF-FG and resource graphs. The greedy algorithm is a common heuristic strategy for searching VNF placements. First, designing greedy strategies generally depends on specific issues, e.g., combining and embedding VNF-FGs , reducing VNFs’ relocation costs , and reducing the number of VNF instances . The distinctions among greedy strategies are mainly reflected in the indicators for guiding their search paths, e.g., upscaling and context switching costs , the priority of each SFC , the importance of each VNF . Second, some designs of greedy strategies are based on topologies, e.g., the DAG , tree topology , and hop-count and direction . Lin et al. (2018) propose a breadth-first search algorithm to select locations with the least cost in layers. Finally, there are some greedy algorithms that prove their theoretical bounds on performances . Fan et al. (2017) propose a greedy algorithm with a theoretical lower bound to optimize the placements of VNF backups. Guo et al. (2018) use the duality theory and multiplicative weight update method to guarantee the lower bound of algorithmic performance. Placing VNFs can be regarded as matching the graphs of user demands and resource supplies. First, some works use the graph theory to describe service requirements, e.g., state synchronization between stateful VNFs and their replicas , dependencies among VNFs , orchestration of VNFs with different demands . Jalalitabar et al. (2016) propose a DAG-based heuristic algorithm to place VNFs with dependency. Agarwal et al. (2018) design a queue-based algorithm to orchestrate different types of SFC requests in slices. ..." What are related quotes?
Reference 11 Review article A comprehensive survey of Network Function VirtualizationYi B., Wang X., Li K., Das S.K., Huang M. Computer Networks, 2018 pp 212-262 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Qu et al. and , published by the same authors, were proposed to formulate the VNF-S problem as a series of scheduling decisions which activated various VNFs to process the traffic of different services. Based on the defined integer and non-integer variables, the formulation was managed in a more fine-grained manner, in which minimizing the latency of VNF scheduling was focused. The lower the latency, the more customers are served, which naturally leads to more profits. Qu et al. mainly focused on how to assign the execution time slots to different services traversing the same VNF via using a greedy method, while Qu et al. considered not only this aspect, but also the resource allocation. In particular, Qu et al. decomposed the proposed VNF-S model into four sub-problems, that is, i ) the virtual link bandwidth allocation sub-problem (assigning each virtual link a feasible data rate); ii ) the VNF assignment sub-problem (allocating VNFs of each network service to VMs); iii ) computing the transmit rates at the VNFs of each network service; iv ) generating a feasible schedule for each VNF via a greedy method. Based on such decomposition and the Genetic Algorithm (GA), Qu et al. developed a simplified heuristic method for solving the VNF-S problem with transmission delay considered. Therefore, by dynamically adjusting the allocated resources, Qu et al. reduced the scheduling time by 15–20%, while Qu et al. saved roughly 14.3% scheduling time. ..." What are related quotes?
Reference 12 Reference works Chapter Optimal Test ConstructionVeldkamp B.P. Encyclopedia of Social Measurement, 2005 pp 933-941 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Heuristics For some optimal test construction problems, 0-1 linear programming techniques cannot be applied because of the nonlinearity of the objective function or the constraints or because the techniques may need too much time. In addition, it might not be possible to formulate the problem as a network-flow model. In those cases, heuristical methods can be applied to find a solution. An heuristic is an approximation method that works fast but that tends to result in a good solution rather than in the optimal solution. In optimal test construction, the greedy algorithm, simulated annealing, and genetic algorithms have been applied successfully. Greedy algorithms work very fast. They select items sequentially. At every iteration, the item is selected that contributes most to the objective function. The normalized weighted absolute deviation heuristic (NWADH) is a well-known application of a greedy heuristic. It has also been applied very successfully in combination with the weighted deviations model in Eqs. (15) . However, because these heuristics operate sequentially, the greedy algorithm might run into infeasibility problems at the end of the test, when violations of constraints are not allowed. Simulated annealing is a much more time-consuming method. First, an initial test is constructed that meets all the specifications. Then, one item is swapped with an item in the pool. If the new test performs better with respect to the objective function, it is accepted; otherwise it is accepted with a probability that decreases during the test assembly process. The method stops when the probability of accepting a worse test is smaller than a lower bound. When genetic algorithms are applied, several tests are constructed that meet all the specifications. New tests are constructed by selecting one part from one test and another part from a second test. If the new test performs better with respect to the objective function, it is added to the set of candidate tests. At the end, the best candidate in the set is selected. ..." What are related quotes?
Reference 13 Book Chapter Optimal Visual Sensor Network ConfigurationZhao J., Cheung S.C.S., Nguyen T. Multi-Camera Networks, 2009 pp 139-162 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... GREEDY: An Algorithm to Speed Up BIP BIP is a well-studied NP-hard combinatorial problem with many heuristic schemes, such as branch-and-bound, already implemented in software libraries (e.g., lp_solve [ 20 ]). However, even these algorithms can be quite intensive if the search space is large. In this section, we introduce a simple greedy algorithm, GREEDY, that can be used for both MIN_CAM and FIX_CAM. Besides experimentally showing the effectiveness of GREEDY, we believe that the greedy approach is an appropriate approximation strategy because of the similarity of our problem to the set cover problem. In the set cover problem, items can belong to multiple sets; the optimization goal is to minimize the number of sets to cover all items. While finding the optimal solution to set covering is an NP-hard problem [ 21 ], it has been shown that the greedy approach is essentially the best we can do to obtain an approximate solution [ 22 ]. We can draw the parallel between our problem and the set cover problem by considering each of the tag grid points as an item “belonging” to a camera grid point if the tag is visible at that camera. The set cover problem then minimizes the number of cameras needed, which is almost identical to MIN_CAM except for the fact that visual tagging requires each tag to be visible by two or more cameras. The FLX_CAM algorithm further allows some of the tag points not to be covered at all. It is still an open question whether these properties can be incorporated into the framework of set covering, but our experimental results demonstrate that the greedy approach is a reasonable solution to our problem. GREEDY is described in Algorithm 6.2 . In each round of the GREEDY algorithm, the camera grid point that can see the most number of tag grid points is selected and all the tag grid points visible to two or more cameras are removed. When using GREEDY to approximate MIN_CAM, we no longer need to refine the tag grids to reduce computational efficiency. We can start with a fairly dense tag grid and set the camera bound m to infinity. ..." What are related quotes?
Reference 14 Review article Approximation and online algorithms for multidimensional bin packing: A surveyChristensen H.I., Khan A., Pokutta S., Tetali P. Computer Science Review, 2017 pp 63-79 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Online algorithms : Optimization problems where the input is received in an online manner (i.e., the input does not arrive as a single batch but as a sequence of input portions) and the output must be produced online (i.e., the system must react in response to each incoming portion taking into account that future input is not known at any point in time) are called online problems. Bin packing is also one of the key problems in online algorithms. Let us define the notion of a competitive ratio which will be useful when we discuss some related results in online algorithms in later sections. Definition 2.6 Competitive Ratio An online algorithm A for a minimization problem Π is called c - competitive if there exists a constant δ such that for all finite input sequences I , A ( I ) ≤ c ⋅ Opt ( I ) + δ . If the additive constant δ ≤ 0 , we say A to be strictly c - competitive . The infimum over the set of all values c such that A is c -competitive is called the competitive ratio of A . We will sometimes call the above defined competitive ratio to be asymptotic competitive ratio and strictly competitive ratio to be absolute competitive ratio . In general, there are no requirements or assumptions concerning the computational efficiency of an online algorithm. However, in practice, we usually seek polynomial time online algorithms. We refer the readers to [32,12] for more details on online algorithms. There are few others metrics to measure the quality of a packing, such as random-order ratio , accommodation function , relative worst-order ratio , differential approximation measure etc. 2.2 One dimensional bin packing Before going to multidimensional bin packing, we give a brief description of the results in 1-D bin packing. Here we focus primarily on very recent results. For a detailed survey and earlier results we refer the interested reader to . 2.2.1 Offline 1-D bin packing The earliest algorithms for one dimensional (1-D) bin packing were simple greedy algorithms such as FirstFit (FF), NextFit (NF), FirstFitDecreasing (FFD), NextFitDecreasing (NFD) etc. ..." What are related quotes?
Reference 15 Handbook Chapter Cognitive Analytics: Going Beyond Big Data Analytics and Machine LearningGudivada V.N., Irfan M.T., Fathi E., Rao D.L. Handbook of Statistics, 2016 pp 169-205 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Decision Trees The Classification and Regression Tree (CART) method was originally presented by Breiman et al. (1984) during the 1980s. This has led to tremendous interest in decision tree learning. In the supervised classification setting, the objective of decision tree learning is to compute a special type of tree that can classify examples to classes. The notions of training, validation, and test sets as well as overfitting vs underfitting issues apply for decision trees too. The underlying model in decision tree learning is a tree in graph-theoretic sense. However, we must also recognize a stylized control flow that is superimposed on the tree structure. Each internal node of the tree, including the root, asks a decision-type question. Based on the answer for an example, we next traverse one of the children of that internal node. Once we reach a leaf node, we are certain to know the classification of the example according to the decision tree, since each leaf node is annotated with a class label. In addition to CART, there are many other learning algorithms for finding the “best” tree for a classification problem. Most modern algorithms like Iterative Dichotomiser 3 (ID3) and its successors C4.5 and C5 use information theoretic measures, such as entropy, to learn a tree. Entropy can be thought of as a measure of uncertainty. Initially, the whole training set, consisting of examples of different classes, will have a very high entropy measure. ID3 and its successors repeatedly partition the training set in order to reduce the sum of the entropy measures of the partitions. Usually, a greedy strategy is employed for this purpose. The algorithm chooses a feature and partitions the training set based on that feature. The feature is chosen with the goal of minimizing the sum of the entropy measures across the resulting partitions. The same procedure is recursed on each partition, unless all the examples in that partition belong to the same class. One big advantage of decision tree learning over other learning methods such as logistic regression is that it can capture more complex decision boundaries. ..." What are related quotes?
Reference 16 Review article Comprehensive survey on resource allocation for edge-computing-enabled metaverseBaidya T., Moh S. Computer Science Review, 2024 pp 100680 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Heuristic approaches Heuristic methods offer practical solutions by making trade-offs between optimality and computational complexity. These approaches are particularly useful when the problem domain is too complex for exact mathematical solutions or when approximate solutions are acceptable. The following are some heuristic approaches used in recent research in the domain of resource allocation. 1) Greedy-based heuristic: In [ 61 ], the authors designed greedy-based heuristic algorithms to manage and optimize resource allocation efficiently under dynamic and real-time conditions. The primary challenge addressed in the paper involves optimizing the trade-off between video quality and delivery latency in a multi-tier cloud VR system leveraging edge computing. The main objective is to enhance user experience by strategically managing rendering tasks between edge servers and a centralized cloud, taking into consideration the limitations in processing power at the edge and varying network conditions. The solution involves formulating the resource allocation problem as an integer non-linear programming (INLP), which is then tackled using online greedy algorithms. These algorithms dynamically allocate video rendering tasks to either edge or cloud servers based on real-time assessments of network conditions and server capacities. The heuristic approach optimizes the allocation by prioritizing tasks that can be most efficiently processed, considering both current system load and the quality-latency trade-off. The performance of the heuristic technique was reported to significantly enhance the system efficiency. It achieved a significant improvement in video delivery latency and quality compared to more static allocation strategies. The heuristic's dynamic nature allows it to adapt quickly to changes in network conditions and server load, providing a more responsive and efficient system. Lessons learned: The greedy-based heuristic algorithm enhances resource allocation in cloud VR systems by dynamically distributing video rendering tasks between edge and cloud servers, based on real-time network and server conditions. This approach, formulated as INLP problem, optimizes the trade-off between video quality and latency, adapting quickly to changes and significantly improving system responsiveness and efficiency. ..." What are related quotes?
Reference 17 Book Chapter Modeling and Optimization Approaches in Design and Management of Biomass-Based Production ChainsŞebnem Yılmaz Balaman Decision-Making for Biomass-Based Production Chains, 2019 pp 185-236 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Metaheuristic algorithms can be used in combination with other optimization and statistical methods, such as mathematical programming, stochastic optimization, k-means clustering, and machine learning. For example, a metaheuristic can select an arbitrary solution in the feasible solution space restricted by the constraints of a mathematical programming model as the initial solution, and after calculating the associated objective function value the metaheuristic may choose to change the solution and try another one, based on its algorithm and specific search rules. The procedure continues until no improvements in the objective function value is achieved and the best approximate to the global optimum is reached. Theoretically, if metaheuristics are run for long enough, they can even find the global optimum solution. In the following subsections, the major metaheuristics that have been widely used, and that have potential to be used in modeling and optimization for design and management of biomass-based production chains, are summarized. 7.3.1.1 Greedy Algorithms Greedy algorithms employ a problem-solving procedure to progressively build candidate solutions, to approximate the global optimum, by obtaining better and better locally optimal solutions at each stage. In general, greedy algorithms cannot yield a global optimal solution, but they may produce good locally optimal solutions in a reasonable time and with less computational effort. GRASP , is a well-known iterative local search-based greedy algorithm that involves a number of iterations to construct greedy randomized solutions and improve them successively. The algorithm consists of two main stages, construction and local search, to initially construct a solution, and then repair this solution to achieve feasibility. The algorithm produces greedy randomized solutions, by selecting new elements from a candidate set of greedy solutions constructed, based on the level of improvement on the partial solution under construction, using a greedy evaluation function. The selected elements are then incorporated into the current partial solution without destroying feasibility (if feasibility is destroyed, a new element is selected), until a complete feasible solution, of which neighborhood is investigated until a local optimum found by the local search procedure, is obtained. 1 ..." What are related quotes?
Reference 18 Review article Multicasting in cognitive radio networks: Algorithms, techniques and protocolsQadir J., Baig A., Ali A., Shafi Q. Journal of Network and Computer Applications, 2014 pp 44-61 View PDFView articleShow related quote(s) Related quote(s)1 / 1 "... Amongst various heuristic solutions, we will discuss greedy algorithms, genetic algorithms, and ant colony optimization algorithms. Greedy algorithms : In a greedy algorithm , the basic building block of a complete feasible solution is a partial solution developed by ‘greedy’ that are based on whatever partial information is available at the time. These partial solutions are progressively developed to build a more complete solution until the iterations develop a complete feasible solution. The famous Dijkstra algorithm used for solving the shortest path tree (SPT) problem is an example greedy algorithm. Randomization is an important tool that can be exploited to avoid local minima׳s in search-based optimization problems. In particular, randomization is utilized by tools like simulated annealing . Another core idea adopted in some metaheuristic techniques like tabu-search is to use adaptive memory contrary to the approach adopted in memoryless approaches like simulated annealing. Metaheuristics also employ concepts of intensification (which encourages intensifying previous solutions found to perform well) and diversification (which encourages search to examine unvisited solutions). In some other fields (e.g., in genetic algorithms and reinforcement learning), the concepts of intensification and diversification are known by the terms exploitation and exploration , respectively. It is to be noted that exploitation and exploration, or alternatively, intensification and diversification, represent conflicting goals and therefore the dilemma of choosing one or the other needs to be resolved in a balanced fashion. There are various other metaheuristic techniques proposed in literature and the interested reader is referred to a book on this topic or the book chapter on this topic in Resende and Pardalos (2006) . We discuss next two particular evolutionary algorithm metaheuristic algorithms: genetic algorithms and ant-colony optimization. Genetic algorithms : A genetic algorithm (GA) is a particular class of evolutionary algorithm which uses techniques inspired from evolutionary biology—such as inheritance, mutation, crossover, and natural selection—to improve the performance of a computational process. 1 ..." What are related quotes?
Reference 19 Reference works Chapter Encyclopedia of Bioinformatics and Computational Biology, Volume 1Massimo Cafaro, Italo Epicoco, Marco Pulimeno Encyclopedia of Bioinformatics and Computational Biology, 2025 pp 9-20 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Iterative Algorithms The greedy strategy, which was discussed in the previous section, works by building a solution to an optimization problem piece by piece, continually adding a piece that is locally optimal to a solution that has been partially formed. In this section, we present an alternative method for the design of algorithms for solving optimization problems. It begins with a solution that is feasible (i.e., a solution that satisfies all of the constraints of the problem) and then tries to improve it by making repeated applications of a basic step. This stage often entails making a minor localized adjustment, which, when successful, results in a solution that is both feasible and yields a better value for the objective function of the problem. When the value of the objective function is not improved by any of these changes, the algorithm will return the most recent solution that was feasible as the optimal one before stopping. There are a number of potential obstacles that could prevent the successful execution of this iteration based approach. First, there is the need to find an initial solution that is viable. We can always start with a solution that is trivial or utilize an approximate solution that was achieved by another technique (for example, greedy). This is possible for some problems. On the other hand, for other problems, coming up with an initial solution could take just as much time and effort as solving the problem once a feasible initial solution has been found. Moreover, it is not always evident what kind of local changes should be allowed in a viable solution in order for us to efficiently assess whether the current solution is locally optimal and, in the event it is not optimal, replace it with a solution that is better with regard to the objective function of the problem. Additionally, we need to take into account local versus global optimal solutions, for either the maximum or minimum of the objective function. This is the most fundamental obstacle, for which different techniques have been designed. 1 ..." What are related quotes?
Reference 20 Book Chapter Classification: advanced methodsJiawei Han, Jian Pei, Hanghang Tong Data Mining, 2023 pp 307-377 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... This strategy is greedy in the sense that it tries to make the best use ( exploitation ) of the information we have collected so far regarding the true value q ( a ) . However, what if there is a big gap between the estimated value Q ( a ) and the true value q ( a ) ? In other words, it is quite possible that q ( c a m e r a ) > q ( T V ) , even though Q ( c a m e r a ) < Q ( T V ) , especially given that Q ( c a m e r a ) and Q ( T V ) are based on very limited data in this example (4 days for camera and 1 day for TV). Moreover, we have never tried printer advertisement at all and thus have zero knowledge about q ( p r i n t e r ) . What if the printer advertisement actually has the highest value? Another alternative strategy is the random strategy ( exploration ). Each day, we choose a random product to advertise and observe the reward of that advertisement. If we keep running this for many days, the estimated value Q ( a ) is likely to be very close to the true value q ( a ) . After that, we choose the advertisement with the highest (estimated) value. This strategy is optimal in the long run. However, we might spend many days before we figure out the optimal action, during which the received rewards might be low. A better strategy is to combine the greedy strategy and the random strategy together. That is, at each day, with the probability 1 − ϵ ( 0 < ϵ < 1 ) , we choose the action (advertisement) with the highest estimated value Q ( a ) ; with the probability ϵ , we choose a random action; and at the end of the day, we use the observed reward to update the estimated value Q ( a ) . This strategy (called ϵ -greedy) is likely to obtain a better balance between the immediate (exploitation) and the long-term (exploration) return. 1 ..." What are related quotes?
Reference 21 Book Chapter Online Strategies for Optimizing Medical Supply in Disaster ScenariosGuttinger D., Godehardt E., Zinnen A. Service Science, Management, and Engineering:, 2012 pp 95-116 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Conclusions In this final section, we want to summarize our results and give an outlook on further interesting research questions regarding this topic. 6.6.1 Summary In this work, we compare several online approaches for optimizing the emergency supply after a major incident. For a given set of physicians, hospitals, and transport vehicles, the algorithms introduced in this work compute an assignment of casualties arriving in groups that suffer from specific types of injuries to available transport and medical capacities. Our objective is to minimize total injury to all casualties, where the level of injury of a casualty is measured by the value of a specific penalty function. For each type of injury, we define such a penalty function that depends on waiting time until medical treatment and describes the current state of health of a casualty. In our simulation studies, we show that a simple greedy strategy clearly outperforms an assignment based on the D’Hondt algorithm, which is, to the best of our knowledge, the only technique currently used in practice to deal with the problem at hand. We have furthermore analyzed the additional benefit when applying a workload-adapted version of simulated annealing subsequent to each greedy iteration. In doing so, it appeared that the average level of injury for each casualty (and so also total level of injury) yielded by the greedy strategy can be additionally reduced by about 10% with an acceptable computational expenditure. Thus, altogether we can conclude that the combination of a greedy strategy and a subsequent application of a workload-adapted version of simulated annealing works well for the given online assignment problem. 6.6.2 Future Prospects Despite the promising results of our adjusted simulated annealing algorithm introduced in this work, there is still room for improvements. One issue for future research will be to try out other optimization techniques like genetic approaches to see if they can provide better results. Furthermore, the influence of the chosen penalty functions on the results should be studied with respect to their number and type. 1 ..." What are related quotes?
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Topic summary
Outline
1. Introduction to Greedy Strategy in Computer Science
2. Fundamental Concepts and Properties
3. Common Greedy Algorithms and Their Applications
4. Design, Analysis, and Limitations of Greedy Algorithms
5. Advanced Topics and Contemporary Applications
6. Conclusion
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4450 | https://www.youtube.com/watch?v=Ih1Zb-6Oy2o | How to Enter Logarithmic Functions in Desmos
Mr. Dan Muscarella, NBCT
4630 subscribers
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Posted: 12 Jan 2022
MathwithMusky #LogarithmicFunctions #Desmos
How to enter a logarithmic expression into Desmos graphing calculator.
3 comments
Transcript:
all right how's everybody doing out there today this is M muscarella coming at you and in this video I'm going to show you how to enter a logarithmic function into the Desmos graphing calculator so the first thing you're going to want to do is go to the Desmos graphing calculator now in here in the lower leftand corner you'll see the little keyboard so you're going to select the keyboard and then go all the way over to where you see functions when you select that you'll see four things at the top top trig stats D and then misc so from here you're going to select the misc and down in the second row from the bottom in the middle you see log base a so if you've got log base a so say you had log base 3 and on the inside was an X+ 2 so you just simply type in x + 2 there for the argument and you could see all sorts of great things about this function of course because Desmos will display these gray dots on here so you'll see where that x intercept is and you'll also see where the Y intercept is O that's kind of gross so that's how you enter a logarithmic function into Desmos using the Desmos graphing calculator all right that's it for this video thank you guys for watching have a good one and I'll catch you later peace out |
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From Simple English Wikipedia, the free encyclopedia
In mathematics, the reciprocal (or multiplicative inverse) of a number is 1 divided by the number, or equivalently, the number raised to the power of -1 (as in and ). All numbers have a reciprocal except zero, since no number times 0 is 1. Two numbers are reciprocal of each other if and only if their product is 1. For example:
2.5 and 0.4 are reciprocals, because 2.5 × 0.4 = 1.
-0.2 and -5 are reciprocals, because -0.2 × -5 = 1.
1 and -1 are their own reciprocals, because 1 × 1 = 1 and -1 × -1 = 1.
To find the reciprocal of a fraction, swap the numerator and the denominator. Whole numbers can be thought of as having a denominator of 1. For example:
The reciprocal of 8 is 1/8 (or 0.125).
The reciprocal of 5/3 is 3/5 (or 0.6).
The reciprocal of 1/7 is 7.
The reciprocal of -9/4 is -4/9.
Dividing a fraction is the same as multiplying its reciprocal, and vice versa.
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Additive inverse
Inverse function
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↑ "Compendium of Mathematical Symbols". Math Vault. 2020-03-01. Retrieved 2020-09-08.
↑ 2.0 2.1 "Reciprocal". www.mathsisfun.com. Retrieved 2020-09-08.
↑ Weisstein, Eric W. "Reciprocal". mathworld.wolfram.com. Retrieved 2020-09-08.
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4452 | https://mathoverflow.net/questions/490433/chromatic-polynomials-other-than-by-deletion-contraction | co.combinatorics - Chromatic polynomials other than by deletion-contraction - MathOverflow
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Chromatic polynomials other than by deletion-contraction
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Given a graph G, its chromatic polynomial \chi_G(n) is the function that takes a natural number n to the number of proper n-colorings of G. The usual proof that this function is actually a polynomial is via deletion-contraction: if e is any edge of G, then \chi_G(n) = \chi_{G \setminus e}(n) - \chi_{G / e}(n). Some chromatic polynomials can be calculated via this same recurrence: for example cycles because deleting an edge yields a path (whose chromatic polynomial is easy) and contracting an edge yields a shorter cycle.
Are there interesting families of graphs whose chromatic polynomials have been computed by any method other than deletion-contraction?
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edited Apr 4 at 4:34
Martin Sleziak
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Tristram BogartTristram Bogart
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2 \begingroup Another way to prove that \chi_G(n) is a polynomial is to use the inclusion-exclusion formula since \chi_G(n)=n^{|V(G)|}-\left|\bigcup_{uv\in E(G)}{f\in[n]^{V(G)}:f(u)=f(v)}\right|. But I don't know if this is useful for actually calculating any chromatic polynomials.\endgroup bof –bof 2025-04-02 22:54:29 +00:00 Commented Apr 2 at 22:54
2 \begingroup One example is Exercise 7 on page 9 of math.mit.edu/~rstan/ec/newexer.pdf, where the computational technique is the transfer-matrix method.\endgroup Richard Stanley –Richard Stanley 2025-04-02 22:56:51 +00:00 Commented Apr 2 at 22:56
\begingroup For chordal graphs (en.m.wikipedia.org/wiki/Chordal_graph) you can compute their chromatic polynomials by a direct counting argument. But these are pretty trivial since they factor.\endgroup Sam Hopkins –Sam Hopkins♦ 2025-04-03 01:36:15 +00:00 Commented Apr 3 at 1:36
\begingroup The paper pdfs.semanticscholar.org/a34b/… by Chikh and Mihoubi gives another example.\endgroup Richard Stanley –Richard Stanley 2025-04-05 18:39:07 +00:00 Commented Apr 5 at 18:39
\begingroup Any thoughts on the comments and answers that have been posted, Tristram?\endgroup Gerry Myerson –Gerry Myerson 2025-04-06 03:00:30 +00:00 Commented Apr 6 at 3:00
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More of a remark than an answer to the question, re: proving the polynomial property other than by the contraction-deletion.
For a simple graph G=(V,E), we have \chi_G(n)=\sum_{c\in [n]^V}\mathbf{1}{\text{the coloring}\ c\ \text{is proper}} =\sum_{c\in [n]^V}\prod_{{i,j}\in E}\mathbf{1}{c(i)\neq c(j)} =\sum_{c\in [n]^V}\prod_{{i,j}\in E}\left(1-\mathbf{1}{c(i)=c(j)}\right) =\sum_{c\in [n]^V}\sum_{F\subseteq E}\prod_{{i,j}\in F}\left(-\mathbf{1}{c(i)=c(j)}\right) =\sum_{F\subseteq E}(-1)^{|F|}\sum_{c\in [n]^V}\prod_{{i,j}\in F}\mathbf{1}{c(i)= c(j)} = \sum_{F\subseteq E}(-1)^{|F|} n^{\gamma(V,F)} where \gamma(V,F) is the number of connected components of the subgraph with vertex set V and edge set F.
This is what Alan Sokal calls the Whitney–Tutte–Fortuin–Kasteleyn representation, e.g., in this article.
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answered Apr 3 at 21:42
Abdelmalek AbdesselamAbdelmalek Abdesselam
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2 \begingroup The chromatic polynomial is essentially the characteristic polynomial of the bond lattice of the graph. (The bond lattice of the graph is the same as the geometric lattice associated to the corresponding graphic matroid.) Another way to say this is that we can compute the chromatic polynomial using Möbius inversion on the bond lattice.\endgroup Sam Hopkins –Sam Hopkins♦ 2025-04-03 23:07:58 +00:00 Commented Apr 3 at 23:07
5 \begingroup The way I like to prove that \chi_G(n) is a polynomial in n is that it is a sum, over all partitions of the vertex set into independent sets I_1, I_2, \ldots, I_k, of the number of ways of coloring those independent sets with distinct colors; but this number is just n(n-1)\cdots(n-k+1), which is obviously a polynomial in n.\endgroup Timothy Chow –Timothy Chow 2025-04-04 02:43:08 +00:00 Commented Apr 4 at 2:43
\begingroup Thanks, Sam. This was actually what I was thinking about when I asked the question. I've been looking at some families of graphs via Möbius inversion on the bond lattice but not getting complete formulas, and wondering if these formulas are known.\endgroup Tristram Bogart –Tristram Bogart 2025-04-07 14:27:17 +00:00 Commented Apr 7 at 14:27
\begingroup@TristramBogart I posted more details in a separate answer. But I don't understand what you mean by "not getting complete formulas."\endgroup Sam Hopkins –Sam Hopkins♦ 2025-04-10 19:44:35 +00:00 Commented Apr 10 at 19:44
\begingroup Thanks, Sam. "Complete formulas" was sloppy; what I meant was that the Möbius inversion approach doesn't give a closed-form formula.\endgroup Tristram Bogart –Tristram Bogart 2025-04-10 23:03:57 +00:00 Commented Apr 10 at 23:03
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Addition-identification is mathematically equivalent to deletion-contraction, but differs in its application. It goes, \chi_G(n)=\chi_{G+e}(n)+\chi_{G/e}(n), where e is an edge not in G, G+e is the result of adding e to G, and G/e is the result of identifying the endpoints of e. Addition-identification can be used to calculate the chromatic polynomial; for example, if G is K_{2,3}, then letting e be an edge joining the two vertices of the "2" in K_{2,3}, both terms in the formula are chordal. Also, it can be used to prove \chi_G(n) is a polynomial, as repeated application results in complete graphs, for which \chi is easily seen to be a polynomial.
The remark about K_{2,3} applies to the family K_{2,n} for all n\ge3.
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edited Apr 10 at 0:34
answered Apr 4 at 1:16
Gerry MyersonGerry Myerson
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The following is completely standard, appearing in any textbook on the subject (e.g. see Stanley, EC1, Chapter 3 Exercise 108), but since it is not clear to me what the question asker understands I am posting it in detail.
Let G be a connected, simple graph on vertex set V. A set partition \pi of V is called G-connected if the restriction of G to each block of \pi remains connected. The bond lattice L_G of the graph G consists of the G-connected partitions of V, ordered by refinement. So its minimal element \hat{0} consists of all singleton blocks, and its maximal element \hat{1} has all the vertices together in one block.
Fix an integer n \geq 1. For \pi a G-connected partition of V, call a coloring c\colon V \to [n]\pi-compatible if all the elements in each block of \pi get the same color. Every coloring is \hat{0}-compatible, and notice that a coloring c is proper exactly when \pi=\hat{0} is the maximal element of L_G for which c is \pi-compatible. Also, the number of \pi-compatible colorings is clearly n^{#\mathrm{blocks}(\pi)}. Hence, by Möbius inversion on L_G, we have that the chromatic polynomial of G is \chi_G(n) = \sum_{\pi \in L_G} \mu(\hat{0},\pi) \, n^{#\mathrm{blocks}(\pi)}, where \mu is the Möbius function of L_G. This means that the chromatic polynomial is (essentially) the characteristic polynomial of L_G.
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answered Apr 10 at 19:43
Sam Hopkins♦Sam Hopkins
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1 \begingroup A somewhat related comment is that one of the fundamental facts about the chromatic polynomial is Whitney's broken circuit theorem, which can also be formulated in a more general setting. See for example An Abstraction of Whitney's Broken Circuit Theorem by Dohmen and Trinks.\endgroup Timothy Chow –Timothy Chow 2025-04-10 20:27:27 +00:00 Commented Apr 10 at 20:27
\begingroup@TimothyChow It is very related. The coefficients of the characteristic polynomial of any matroid (after correcting for their alternating sign) are given by the f-vector of the broken circuit complex of the matroid.\endgroup Sam Hopkins –Sam Hopkins♦ 2025-04-10 20:41:32 +00:00 Commented Apr 10 at 20:41
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It sounds like you're mostly interested in closed-form formulas for infinite families of graphs, but if you're interested in efficient algorithms for computing the chromatic polynomial of graphs of moderate size, then rather than using deletion-contraction, you may be better off using methods inspired by ideas from statistical mechanics. See in particular Counting complex disordered states by efficient pattern matching: chromatic polynomials and Potts partition functions, by Marc Timme, Frank van Bussel, Denny Fliegner, and Sebastian Stolzenberg, New Journal of Physics 11 (2009), 023001, and Chromatic polynomials of random graphs, by the same authors plus Christoph Ehrlich, arXiv:1709.06209.
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answered Apr 4 at 3:30
Timothy ChowTimothy Chow
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4453 | https://www.tentotwelvemath.com/fom-12/unit-5-sinusoidal-functions/5-3-the-range-of-a-sinusoidal-function/ | Skip to primary content
Interactive Math Resources for Classroom Discussion, Inquiry and Distance Learning
5.3 The Range of a Sinusoidal Function
Handout: FOM 12 5.3 Determine the Range
The range of the graph is
The two tranformations we can make to the values are to
multiply (or divide)
add (or subtract
In general, a sinusoidal graph has equation . It is only the values and that alter the range of the graph.
Multiply
To draw the graph we consider particular points (easy points), and multiply the coordinate by 3 as follows:
Track each point in turn. For example, on the blue line we have the point therefore we plot the new point . The coordinate (that is, ) is multiplied by .
Next we draw a line through our new points:
We see that the range of the green curve is .
The amplitude of this curve is .
The sinusoidal axis is not changed, it is still .
Add
To draw the graph again we consider particular points and add 2 to the value as follows:
As before, track each point in turn. For example, the point on the blue curve at will become the point .
We see that the range of the green curve is .
The sinusoidal axis is the horizontal line .
The amplitude of the curve is not changed, it is still 1.
Try each transformation here:
Multiply and add
To do both operations, we should multiply first then add. However, in practice it is easier to draw a new sinusoidal axis, and plot the correct amplitude from there.
For example, transform to .
First, lets draw a new sinusoidal axis at
Now let’s find the multiples of 180 on the line to plot our new ‘zeros’:
Now let’s track the multiples of 90, and plot our new max and min but remembering that the amplitude of is 3, so we plot 3 above and below the sinusoidal axis:
Finally, we can draw our curve and erase the sinusoidal axis:
The range of our new graph is , which we can see is the same as .
In general, we can say that the range of the sinusoidal function is
(when is positive, otherwise the inequality is reversed).
Try both transformations together here:
Practice: Determine the range
CA1 Test out: Determine the range accuracy quiz
Practice: Match the graph
CA2 Test out: Match the graph
Top BC Grade 12 Functions |
4454 | https://www.dentonisd.org/cms/lib/TX21000245/Centricity/Domain/909/sTUDENT%20tEXTBOOK_tx_geo_07_02.pdf | Section 7.2 Properties of Parallelograms 371 Properties of Parallelograms 7.2 Essential Question Essential Question What are the properties of parallelograms? Discovering Properties of Parallelograms Work with a partner. Use dynamic geometry software.
a. Construct any parallelogram and label it ABCD. Explain your process. Sample A C D B b. Find the angle measures of the parallelogram. What do you observe?
c. Find the side lengths of the parallelogram. What do you observe?
d. Repeat parts (a)–(c) for several other parallelograms. Use your results to write conjectures about the angle measures and side lengths of a parallelogram. Discovering a Property of Parallelograms Work with a partner. Use dynamic geometry software. a. Construct any parallelogram and label it ABCD. b. Draw the two diagonals of the parallelogram. Label the point of intersection E. Sample A C D B E c. Find the segment lengths AE, BE, CE, and DE. What do you observe?
d. Repeat parts (a)–(c) for several other parallelograms. Use your results to write a conjecture about the diagonals of a parallelogram.
Communicate Your Answer Communicate Your Answer 3. What are the properties of parallelograms?
USING PROBLEM-SOLVING STRATEGIES To be profi cient in math, you need to analyze givens, constraints, relationships, and goals.
G.5.A TEXAS ESSENTIAL KNOWLEDGE AND SKILLS 372 Chapter 7 Quadrilaterals and Other Polygons 7.2 Lesson What You Will Learn What You Will Learn Use properties to fi nd side lengths and angles of parallelograms. Use parallelograms in the coordinate plane.
Using Properties of Parallelograms A parallelogram is a quadrilateral with both pairs of opposite sides parallel. In ▱PQRS, — PQ — RS and — QR — PS by defi nition. The theorems below describe other properties of parallelograms. Parallelogram Opposite Sides Theorem Given PQRS is a parallelogram.
Prove — PQ ≅ — RS , — QR ≅ — SP a. Draw diagonal — QS to form △PQS and △RSQ.
b. Use the ASA Congruence Theorem (Thm. 5.10) to show that △PQS ≅ △RSQ.
c. Use congruent triangles to show that — PQ ≅ — RS and — QR ≅ — SP .
STATEMENTS REASONS 1. PQRS is a parallelogram. 1. Given a. 2. Draw — QS .
2. Through any two points, there exists exactly one line. 3. — PQ — RS , — QR — PS 3. Defi nition of parallelogram b. 4. ∠PQS ≅ ∠RSQ, ∠PSQ ≅ ∠RQS 4. Alternate Interior Angles Theorem (Thm. 3.2) 5. — QS ≅ — SQ 5. Refl exive Property of Congruence (Thm. 2.1) 6. △PQS ≅ △RSQ 6. ASA Congruence Theorem (Thm. 5.10) c. 7. — PQ ≅ — RS , — QR ≅ — SP 7. Corresponding parts of congruent triangles are congruent.
Plan in Action parallelogram, p. 372 Previous quadrilateral diagonal interior angles segment bisector Core Vocabulary Core Vocabulary Theorems Theorems Theorem 7.3 Parallelogram Opposite Sides Theorem If a quadrilateral is a parallelogram, then its opposite sides are congruent.
If PQRS is a parallelogram, then — PQ ≅ — RS and — QR ≅ — SP .
Proof p. 372 Theorem 7.4 Parallelogram Opposite Angles Theorem If a quadrilateral is a parallelogram, then its opposite angles are congruent.
If PQRS is a parallelogram, then ∠P ≅ ∠R and ∠Q ≅ ∠S.
Proof Ex. 37, p. 377 Plan for Proof P Q R S P Q R S P Q R S P Q R S Section 7.2 Properties of Parallelograms 373 Using Properties of Parallelograms Find the values of x and y.
SOLUTION ABCD is a parallelogram by the defi nition of a parallelogram. Use the Parallelogram Opposite Sides Theorem to fi nd the value of x. AB = CD Opposite sides of a parallelogram are congruent. x + 4 = 12 Substitute x + 4 for AB and 12 for CD. x = 8 Subtract 4 from each side.
By the Parallelogram Opposite Angles Theorem, ∠A ≅ ∠C, or m∠A = m∠C. So, y° = 65°. In ▱ABCD, x = 8 and y = 65.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 1. Find FG and m∠G. 2. Find the values of x and y. F G H E 60° 8 J K L M 50° 2x° 18 y + 3 The Consecutive Interior Angles Theorem (Theorem 3.4) states that if two parallel lines are cut by a transversal, then the pairs of consecutive interior angles formed are supplementary.
A pair of consecutive angles in a parallelogram is like a pair of consecutive interior angles between parallel lines. This similarity suggests the Parallelogram Consecutive Angles Theorem.
Theorems Theorems Theorem 7.5 Parallelogram Consecutive Angles Theorem If a quadrilateral is a parallelogram, then its consecutive angles are supplementary.
If PQRS is a parallelogram, then x° + y° = 180°.
Proof Ex. 38, p. 377 Theorem 7.6 Parallelogram Diagonals Theorem If a quadrilateral is a parallelogram, then its diagonals bisect each other.
If PQRS is a parallelogram, then — QM ≅ — SM and — PM ≅ — RM .
Proof p. 374 D A B C 12 x + 4 65° y° x° y° P Q R S x° y° x° y° P Q R S M 374 Chapter 7 Quadrilaterals and Other Polygons Parallelogram Diagonals Theorem Given PQRS is a parallelogram. Diagonals — PR and — QS intersect at point M.
Prove M bisects — QS and — PR .
STATEMENTS REASONS 1. PQRS is a parallelogram. 1. Given 2. — PQ — RS 2. Defi nition of a parallelogram 3. ∠QPR ≅ ∠SRP, ∠PQS ≅ ∠RSQ 3. Alternate Interior Angles Theorem (Thm. 3.2) 4. — PQ ≅ — RS 4. Parallelogram Opposite Sides Theorem 5. △PMQ ≅ △RMS 5. ASA Congruence Theorem (Thm. 5.10) 6. — QM ≅ — SM , — PM ≅ — RM 6. Corresponding parts of congruent triangles are congruent.
7. M bisects — QS and — PR .
7. Defi nition of segment bisector Using Properties of a Parallelogram As shown, part of the extending arm of a desk lamp is a parallelogram. The angles of the parallelogram change as the lamp is raised and lowered. Find m∠BCD when m∠ADC = 110°.
SOLUTION By the Parallelogram Consecutive Angles Theorem, the consecutive angle pairs in ▱ABCD are supplementary. So, m∠ADC + m∠BCD = 180°. Because m∠ADC = 110°, m∠BCD = 180° − 110° = 70°. Writing a Two-Column Proof Write a two-column proof.
Given ABCD and GDEF are parallelograms.
Prove ∠B ≅ ∠F STATEMENTS REASONS 1. ABCD and GDEF are parallelograms. 1. Given 2. ∠CDA ≅ ∠B, ∠EDG ≅ ∠F 2. If a quadrilateral is a parallelogram, then its opposite angles are congruent.
3. ∠CDA ≅ ∠EDG 3. Vertical Angles Congruence Theorem (Thm. 2.6) 4. ∠B ≅ ∠F 4. Transitive Property of Congruence (Thm. 2.2) Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 3. WHAT IF? In Example 2, fi nd m∠BCD when m∠ADC is twice the measure of ∠BCD. 4. Using the fi gure and the given statement in Example 3, prove that ∠C and ∠F are supplementary angles. P Q R S M A B C D C A B E G F D Section 7.2 Properties of Parallelograms 375 Using Parallelograms in the Coordinate Plane Using Parallelograms in the Coordinate Plane Find the coordinates of the intersection of the diagonals of ▱LMNO with vertices L(1, 4), M(7, 4), N(6, 0), and O(0, 0).
SOLUTION By the Parallelogram Diagonals Theorem, the diagonals of a parallelogram bisect each other. So, the coordinates of the intersection are the midpoints of diagonals — LN and — OM .
coordinates of midpoint of — OM = ( 7 + 0 — 2 , 4 + 0 — 2 ) = ( 7 — 2 , 2 ) Midpoint Formula The coordinates of the intersection of the diagonals are ( 7 — 2 , 2 ) . You can check your answer by graphing ▱LMNO and drawing the diagonals. The point of intersection appears to be correct. Using Parallelograms in the Coordinate Plane Three vertices of ▱WXYZ are W(−1, −3), X(−3, 2), and Z(4, −4). Find the coordinates of vertex Y.
SOLUTION Step 1 Graph the vertices W, X, and Z.
Step 2 Find the slope of — WX .
slope of — WX = 2 – (–3) — –3 – (–1) = 5 — –2 = − 5 — 2 Step 3 Start at Z(4, −4). Use the rise and run from Step 2 to fi nd vertex Y.
A rise of 5 represents a change of 5 units up. A run of −2 represents a change of 2 units left.
So, plot the point that is 5 units up and 2 units left from Z(4, −4). The point is (2, 1). Label it as vertex Y.
Step 4 Find the slopes of — XY and — WZ to verify that they are parallel.
slope of — XY = 1 − 2 — 2 − (−3) = −1 — 5 = − 1 — 5 slope of — WZ = −4 − (−3) — 4 − (−1) = −1 — 5 = − 1 — 5 So, the coordinates of vertex Y are (2, 1).
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 5. Find the coordinates of the intersection of the diagonals of ▱STUV with vertices S(−2, 3), T(1, 5), U(6, 3), and V(3, 1). 6. Three vertices of ▱ABCD are A(2, 4), B(5, 2), and C(3, −1). Find the coordinates of vertex D. JUSTIFYING STEPS In Example 4, you can use either diagonal to fi nd the coordinates of the intersection. Using diagonal — OM helps simplify the calculation because one endpoint is (0, 0).
REMEMBER When graphing a polygon in the coordinate plane, the name of the polygon gives the order of the vertices.
x y 4 2 4 2 8 O M N L x y 2 −4 2 −4 W Z X 5 5 −2 −2 Y(2, 1) 376 Chapter 7 Quadrilaterals and Other Polygons Exercises 7.2 Dynamic Solutions available at BigIdeasMath.com 1. VOCABULARY Why is a parallelogram always a quadrilateral, but a quadrilateral is only sometimes a parallelogram? 2. WRITING You are given one angle measure of a parallelogram. Explain how you can fi nd the other angle measures of the parallelogram.
Vocabulary and Core Concept Check Vocabulary and Core Concept Check In Exercises 3–6, fi nd the value of each variable in the parallelogram. (See Example 1.) 3. 15 9 y x 4. 12 6 n m + 1 5. 20 z − 8 (d − 21)° 105° 6. In Exercises 7 and 8, fi nd the measure of the indicated angle in the parallelogram. (See Example 2.) 7. Find m∠B. 8. Find m∠N. A D C B 51° L P N M 95° In Exercises 9–16, fi nd the indicated measure in ▱LMNQ. Explain your reasoning. 9. LM 10. LP 11. LQ 12. MQ 13. m∠LMN 14. m∠NQL 15. m∠MNQ 16. m∠LMQ In Exercises 17–20, fi nd the value of each variable in the parallelogram. 17. 70° 2m° n° 18. 19. k + 4 m 8 11 20. 2u + 2 5u − 10 6 v 3 ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in using properties of parallelograms.
21. V U T S 50° Because quadrilateral STUV is a parallelogram, ∠S ≅ ∠V. So, m∠V = 50°.
✗ 22. K G H J F Because quadrilateral GHJK is a parallelogram, — GF ≅ — FH .
✗ Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics 7 16 − h (g + 4)° 65° P N Q M L 29° 100° 13 7 8 8.2 (b − 10)° (b + 10)° d° c° Section 7.2 Properties of Parallelograms 377 PROOF In Exercises 23 and 24, write a two-column proof. (See Example 3.) 23. Given ABCD and CEFD are parallelograms.
Prove — AB ≅ — FE 24. Given ABCD, EBGF, and HJKD are parallelograms.
Prove ∠2 ≅ ∠3 A D C B G F E H K J 1 2 3 4 In Exercises 25 and 26, fi nd the coordinates of the intersection of the diagonals of the parallelogram with the given vertices. (See Example 4.) 25. W(−2, 5), X(2, 5), Y(4, 0), Z(0, 0) 26. Q(−1, 3), R(5, 2), S(1, −2), T(−5, −1) In Exercises 27–30, three vertices of ▱DEFG are given. Find the coordinates of the remaining vertex. (See Example 5.) 27. D(0, 2), E(−1, 5), G(4, 0) 28. D(−2, −4), F(0, 7), G(1, 0) 29. D(−4, −2), E(−3, 1), F(3, 3) 30. E(−1, 4), F(5, 6), G(8, 0) MATHEMATICAL CONNECTIONS In Exercises 31 and 32, fi nd the measure of each angle.
31. The measure of one interior angle of a parallelogram is 0.25 times the measure of another angle.
32. The measure of one interior angle of a parallelogram is 50 degrees more than 4 times the measure of another angle.
33. MAKING AN ARGUMENT In quadrilateral ABCD, m∠B = 124°, m∠A = 56°, and m∠C = 124°. Your friend claims quadrilateral ABCD could be a parallelogram. Is your friend correct? Explain your reasoning.
34. ATTENDING TO PRECISION ∠J and ∠K are consecutive angles in a parallelogram, m∠J = (3x + 7)°, and m∠K = (5x −11)°. Find the measure of each angle.
35. CONSTRUCTION Construct any parallelogram and label it ABCD. Draw diagonals — AC and — BD . Explain how to use paper folding to verify the Parallelogram Diagonals Theorem (Theorem 7.6) for ▱ABCD. 36. MODELING WITH MATHEMATICS The feathers on an arrow form two congruent parallelograms. The parallelograms are refl ections of each other over the line that contains their shared side. Show that m∠2 = 2m∠1.
1 2 37. PROVING A THEOREM Use the diagram to write a two-column proof of the Parallelogram Opposite Angles Theorem (Theorem 7.4).
A B C D Given ABCD is a parallelogram. Prove ∠A ≅ ∠C, ∠B ≅ ∠D 38. PROVING A THEOREM Use the diagram to write a two-column proof of the Parallelogram Consecutive Angles Theorem (Theorem 7.5).
P Q R S x° y° x° y° Given PQRS is a parallelogram. Prove x° + y° = 180° 39. PROBLEM SOLVING The sides of ▱MNPQ are represented by the expressions below. Sketch ▱MNPQ and fi nd its perimeter. MQ = −2x + 37 QP = y + 14 NP = x − 5 MN = 4y + 5 40. PROBLEM SOLVING In ▱LMNP, the ratio of LM to MN is 4 : 3. Find LM when the perimeter of ▱LMNP is 28.
A D F E C B 378 Chapter 7 Quadrilaterals and Other Polygons 41. ABSTRACT REASONING Can you prove that two parallelograms are congruent by proving that all their corresponding sides are congruent? Explain your reasoning. 42. HOW DO YOU SEE IT? The mirror shown is attached to the wall by an arm that can extend away from the wall. In the fi gure, points P, Q, R, and S are the vertices of a parallelogram. This parallelogram is one of several that change shape as the mirror is extended.
P S R Q a. What happens to m∠P as m∠Q increases? Explain. b. What happens to QS as m∠Q decreases? Explain. c. What happens to the overall distance between the mirror and the wall when m∠Q decreases? Explain. 43. MATHEMATICAL CONNECTIONS In ▱STUV, m∠TSU = 32°, m∠USV = (x2)°, m∠TUV = 12x°, and ∠TUV is an acute angle. Find m∠USV.
U V S T 44. THOUGHT PROVOKING Is it possible that any triangle can be partitioned into four congruent triangles that can be rearranged to form a parallelogram? Explain your reasoning.
45. CRITICAL THINKING Points W(1, 2), X(3, 6), and Y(6, 4) are three vertices of a parallelogram. How many parallelograms can be created using these three vertices? Find the coordinates of each point that could be the fourth vertex.
46. PROOF In the diagram, — EK bisects ∠FEH, and — FJ bisects ∠EFG. Prove that — EK ⊥ — FJ . (Hint: Write equations using the angle measures of the triangles and quadrilaterals formed.) G H J K F P E 47. PROOF Prove the Congruent Parts of Parallel Lines Corollary: If three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal.
G H P Q K M J L Given ⃖ ⃗ GH ⃖ ⃗ JK ⃖ ⃗ LM , — GJ ≅ — JL Prove — HK ≅ — KM (Hint: Draw — KP and — MQ such that quadrilateral GPKJ and quadrilateral JQML are parallelograms.) Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Determine whether lines ℓ and m are parallel. Explain your reasoning. (Section 3.3) 48. m 81° 81° 49. m 50. m 132° 58° Reviewing what you learned in previous grades and lessons |
4455 | https://mathbitsnotebook.com/Geometry/Quadrilaterals/QDAlgebraicProblemPractice.html | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | --- | | | Algebraic Practice Problems for Quadrilaterals MathBitsNotebook.com Topical Outline | Geometry Outline | MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts | Directions: Read each question carefully. Choose the best answer. State the rules or properties of quadrilaterals that were used to find the answers. | | | | | | | | --- --- --- | | | | | --- | | The sides of parallelogram ABCD are represented as shown. Find DA. Choose: | | | | | | | | | | | | | | | --- --- --- | | | | | --- | | In rhombus DOGS, DO = 2p + 9 and OG = 3p - 6. Find GS. Choose: | | | | | | | | | | | | | --- --- | | | | --- | | 3. | The perimeter of quadrilateral ABCD is 46 inches. AB = x + 8, BC = 2x + 1, CD = 3x - 6, and DA = 4x - 7. Find the length of the shortest side of the quadrilateral. Choose: | | | | | | | | | | | | | --- --- --- | | | | | --- | 4. | Given parallelogram ABCD with m∠B = 5x and m∠C = 3x+4. Find the number of degrees in ∠D. Choose: | | | | | | | | | | | | | | | --- --- --- | | | | | --- | 5. | The diagonals of rectangle ABCD intersect at E. AE = x + 4 and CE = 3x - 12. Find BD. Choose: | | | | | | | | | | | | | | | --- --- --- | | | | | --- | 6. | In rhombus ABCD, m∠ECB = 5a+4 and m∠EBC = 8a - 5.Find m∠EBC. Choose: | | | | | | | | | | | | | | | --- --- --- | | | | | --- | 7. | Given trapezoid ABCD with median (labeled as shown). Find EF. Choose: | | | | | | | | | | | | | | | --- --- --- | | | | | --- | 8. | In rectangle ABCD, AE = 3x + y, EC = 2x + y + 7 and DE = 2y + 3x - 1. Find the values of x and y. Choose: | | | | | | | | | | | | | | | --- --- --- | | | | | --- | 9. | Given square ABCD with diagonals If DB = 7x + 1 and AE = 2x + 11, find EB. Choose: | | | | | | | | | | | | | | | --- --- --- | | | | | --- | 10. | Given parallelogram ABCD, labeled as shown. Find a and b. Choose: | | | | | | | | | | | | | | | --- --- --- | | | | | --- | 11. | Given square ABCD with diagonals The m∠DEC = 2a - b and m∠ABC = a + 2b. Find a and b. Choose: | | | | | | | | | | | | | | | --- --- --- | | | | | --- | | Given square CANE with diagonals intersecting at B. m∠CNE = 3a + 2b, AC = 35, and CE = 6a + 5. Find the value of a + b. Choose: | | | | | | | | | | NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". | Topical Outline | Geometry Outline | MathBitsNotebook.com | MathBits' Teacher Resources Terms of Use Contact Person:Donna Roberts Copyright © 2012-2025 MathBitsNotebook.com. All Rights Reserved. | | |
4456 | https://utctime.info/timezone/America--Denver/ | America/Denver · Time Zone - Current Local Time
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| ATOM Digital clocks | 2025-09-29T01:37:25.733-06:00 |
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| ISO 8601 | 2025-09-29T01:37:25.733-06:00 |
| MM-dd-yyyy HH:mm:ss | 09-29-2025 01:37:25 |
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| dd-MM-yyyy HH:mm:ss | 29-09-2025 01:37:25 |
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| America/Bahia_Banderas | Central Standard Time | CST |
| America/Belize | Central Standard Time | CST |
| America/Boise | Mountain Daylight Time | MDT |
| America/Cambridge_Bay | Mountain Daylight Time | MDT |
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| America/Costa_Rica | Central Standard Time | CST |
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For Developers
php
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<?php
// Solution 1
// Set the default time zone
date_default_timezone_set('America/Denver');
// Solution 2
// Use this to get the current date and time in the specified time zone.
$timezone = new DateTimeZone('America/Denver');
// Create a new DateTime object
$date = new DateTime('now', $timezone);
// Print the current date and time
echo $date->format('Y-m-d H:i:s'); // 2023-01-01 00:00:00
python
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Solution 1
Set the default time zone
from datetime import datetime
Import the pytz module
import pytz
Create a timezone object
tz = pytz.timezone("America/Denver")
Get the current time
time_now = datetime.now(tz)
Format the current time
current_time = time_now.strftime("%H:%M:%S")
Print the current time
print("The current time:", current_time) # The current time: 00:00:00
javascript
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/ Solution 1 Set the default time zone /
const date = new Date();
/ Use the toLocaleString() method to get the current date and time in the specified time zone. /
const currentTime = date.toLocaleString("en-us", {timeZone: "America/Denver"});
/ Print the current date and time /
console.log(currentTime); // 1/1/2023, 12:00:00 AM
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4457 | https://www.teacherspayteachers.com/Product/Number-Bonds-up-to-15-Practice-High-and-Emerging-Learners-w-Ten-Frames-10903094 | Number Bonds up to 15 Practice (High and Emerging Learners) w/ Ten Frames
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Number Bonds up to 15 Practice (High and Emerging Learners) w/ Ten Frames
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Includes 4 practice problems for each worksheet. Differentiated for emerging learners (some parts filled in) and high learners (students fill in independently).
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### Early Education Resources by Darren Taraza
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Summa Cum Laude / Salutatorian at The City College of New York: Division of Interdisciplinary Studies at the Center for Worker Education (2022).
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4458 | https://dl.acm.org/doi/fullHtml/10.1145/3597066.3597109 | From Polynomial Invariants to Linear Loops
☰ Article Navigation
Article Navigation×Abstract
1 INTRODUCTION
2 PRELIMINARIES2.1 Abstract Algebra2.2 Algebraic Geometry2.3 C-finite Sequences2.4 Loop Invariants2.5 Lattices2.6 Toric Ideals and Varieties
3 FROM PURE DIFFERENCE IDEALS TO LINEAR LOOPSLoop Synthesis Procedure:InputStep 1: Lattice ideal ILStep 2a: Saturated lattice ideal I Sat(L)I Sat(L)Step 2b: Matrix A encoding I Sat(L)I Sat(L)Step 3: Synthesise linear loop L LOutput
4 TRANSFORMED PURE DIFFERENCE IDEALS
5 FURTHER RESULTS5.1 Existence of Non-Trivial Linear Loops5.2 Modifications to the Synthesis Process
6 DISCUSSION AND CONCLUSIONS
ACKNOWLEDGMENTS
REFERENCES
FOOTNOTE
From Polynomial Invariants to Linear Loops
George Kenison, TU Wien, Austria, george.kenison@tuwien.ac.at
Laura Kovács, TU Wien, Austria, laura.kovacs@tuwien.ac.at
Anton Varonka, TU Wien, Austria, anton.varonka@tuwien.ac.at
DOI:
ISSAC 2023: International Symposium on Symbolic and Algebraic Computation 2023, Tromsø, Norway, July 2023
Loop invariants are software properties that hold before and after every iteration of a loop. As such, invariants provide inductive arguments that are key in automating the verification of program loops. The problem of generating loop invariants; in particular, invariants described by polynomial relations (so called polynomial invariants), is therefore one of the hardest problems in software verification. In this paper we advocate an alternative solution to invariant generation. Rather than inferring invariants from loops, we synthesise loops from invariants. As such, we generate loops that satisfy a given set of polynomials; in other words, our synthesised loops are correct by construction.
Our work turns the problem of loop synthesis into a symbolic computation challenge. We employ techniques from algebraic geometry to synthesise loops whose polynomial invariants are described by pure difference binomials. We show that such complex polynomial invariants need “only” linear loops, opening up new venues in program optimisation. We prove the existence of non-trivial loops with linear updates for polynomial invariants generated by pure difference binomials. Importantly, we introduce an algorithmic approach that constructs linear loops from such polynomial invariants, by generating linear recurrence sequences that have specified algebraic relations among their terms.
Keywords:Program Synthesis, Loop Invariants, Toric Ideals, C-finite Sequences
ACM Reference Format:
George Kenison, Laura Kovács, and Anton Varonka. 2023. From Polynomial Invariants to Linear Loops. In International Symposium on Symbolic and Algebraic Computation 2023 (ISSAC 2023), July 24--27, 2023, Tromsø, Norway. ACM, New York, NY, USA 9 Pages.
1 INTRODUCTION
Loop invariants, or more simply invariants in the sequel, are software properties that hold before and after every iteration of a loop. Invariants are often key to inductive arguments for automating the verification of programs with loops, see, e.g.[6, 15, 22, 23, 25]. One challenging aspect in invariant synthesis is the derivation of polynomial invariants for loop programs over numeric data structures. Such invariants are defined by polynomial relations P(x 1, …, x d) = 0 among the program variables x 1, …, x d. A nice property of polynomial invariants is that they define a polynomial ideal, called a polynomial invariant ideal[23, 25]. As such, the problem of generating (all) polynomial invariants is reduced to generating a finite basis of the polynomial invariant ideal.
In this paper, we reverse engineer the problem of polynomial invariant generation and propose an alternative solution to invariant synthesis. Rather than generating polynomial invariants for a given loop, we synthesise loops for a given polynomial invariant. Assume the postcondition of a program fragment with loop is given by polynomial equalities. Instead of generating an invariant that implies the postcondition, our solution synthesises a new loop, one that is correct with respect to the specification (by construction).
Linear Loop Synthesis. The key aspect of our work comes with considering homogeneous linear loops, hereafter simply linear loops. Linear loops are a class of single-path loops whose update assignments are determined by a homogeneous system of linear equations in the program variables. More specifically, a linear loop is a loop program of the form
L:x←s;while⋆do x←M x,L:x←s;while⋆do x←M x,
where x is a d-dimensional column vector of program variables, s is a d-dimensional vector with rational entries, and M is a d × d-matrix with rational entries. Herein we employ the notation ⋆, instead of using true as loop guard, as our focus is on loop synthesis rather than proving loop termination.
Linear loops are fundamental objects in the computational study of recurrence. On the one hand, this class of loops represents a restrictive computational model. On the other hand, fundamental problems such as the decidability of the Halting Problem are open for this class. For the avoidance of doubt, we lose no generality by working over the class of linear loops rather than the class of affine loops (those single-path loop programs with update assignments of the form x ← Mx + v where v∈Q d v∈Q d). Indeed, the problem of studying the functional behaviour of affine loops can be reduced to that of studying linear loops; admittedly the reduction step will, in general, increase the number of program variables[23, 24, 25].
Linear Loops and Recurrences. Given a (system of) polynomial relation(s), our work constructs a linear loop. By construction, each of the given polynomial relations is satisfied by the loop variables before and after each iteration; thus each relation is an invariant of the loop. (Specific details are given in the discussion on our contributions.) We employ techniques from algebraic geometry to synthesise loops whose polynomial invariants are described by pure difference binomials. For such polynomial invariants, the procedure in section 3 shows that “only” linear loops are required, thus addressing challenging aspects of arithmetic reductions in program optimisation. Further, we prove the existence of non-trivial loops with linear updates for polynomial invariants generated by pure difference binomials (see section 5). s 1.1--1.2 showcase linear loops that are synthesised by our work. We note that the study of pure difference ideals, those ideals generated by pure difference binomials, and their respective linear loops is well motivated. The classes of lattice and toric ideals (defined in section 2) are pure difference ideals. Pure difference ideals also appear in the encoding of random walks on Markov chains of the form N d N d by way of the connected components of the underlying graph [7, 17].
Example 1.1 Consider the input ideal I 1⊆Q[x,y,z]I 1⊆Q[x,y,z] generated by the polynomials p 1 = x 2 − y and p 2 = x 3 − z. Note that p 1 and p 2 are pure difference binomials in x, y, z. The procedure of section 3 outputs a linear loop L 1 L 1 (see 1a ).
Example 1.2 Given the input ideal I 2⊆Q[x,y]I 2⊆Q[x,y] generated by the polynomial p = x 3 y − xy 3. The procedure of section 3 outputs a linear loop L 2 L 2 (see 1b ). Note, I 1 is the invariant ideal of L 1 L 1. Loop L 2 L 2, in turn, satisfies all invariants in I 2; however, I 2 is a strict subset of the invariant ideal of L 2 L 2.
Figure 1:Synthesised linear loops L 1 L 1 and L 2 L 2
Key to our work is modelling loops as linear recurrence sequences that have specified algebraic relations among their terms. Let x n denote the value of a loop variable x at the n th loop iteration. Recall that for a linear loop, x n is given by a linear recurrence sequence with constant coefficients, commonly known as a C-finite sequence. C-finite sequences are therefore key to our loop synthesis procedure (section 3): we model program loops as systems of recurrence equations. Indeed, we aim to synthesise a system of C-finite recurrence sequences, and hence a linear loop, that satisfies given polynomial relations.
Our Contributions.
For a polynomial ideal, we consider the problem of synthesising a non-parametrised loop (i.e., synthesising both the loop body and concrete initial values) such that every polynomial in the ideal is an invariant of the loop. In particular, we demonstrate a procedure for synthesising loops from pure difference ideals. In fact,
Given a pure difference ideal I, we describe a process that synthesises a linear loop with invariant ideal I (section 3).
Suppose that I = ⟨p 1, …, p k⟩ is a polynomial ideal not necessarily generated by binomials, for which, by a change of coordinates, there exists a generating set of pure difference binomials. We present a procedure that outputs a linear loop such that any polynomial p ∈ ⟨p 1, …, p k⟩ is an invariant of said loop (section 4).
Under reasonable assumptions about the input, the aforementioned generated loops are non-trivial. By non-trivial, we mean that the orbit (or trajectory) of the vector ⟨x(n)⟩∞n=0=⟨x 1(n),…,x d(n)⟩∞n=0⟨x(n)⟩n=0∞=⟨x 1(n),…,x d(n)⟩n=0∞ given by the values of the loop variables x 1, …, x d at the n th loop iteration is infinite. That is to say, the loop program functions on an infinite-state system. We shall defer a formal definition of a non-trivial loop to section 2.
In section 5, we consider corollaries to our loop synthesis procedure from section 3. In particular, we consider specialisations for restrictive classes of input ideals. One corollary, concerning canonical pure difference binomials, is as follows.
Corollary 1.3 Suppose that I=⟨p⟩⊆Q[x 1,…,x d]I=⟨p⟩⊆Q[x 1,…,x d] where p is an irreducible polynomial of the form p=x α 1 1⋯x α d d−x β 1 1⋯x β d d p=x 1 α 1⋯x d α d−x 1 β 1⋯x d β d. Then the procedure in section 3 synthesises a linear loop for which I is precisely the invariant ideal of the loop.
We end this note with a conclusion that discusses related work and proposes directions for future research (section 6).
2 PRELIMINARIES
2.1 Abstract Algebra
Let K K be a field and K d K d the vector space of d-tuples in K K. A monomial in the indeterminates (or polynomial variables) x 1, …, x d is an expression x v 1 1 x v 2 2⋯x v d d x 1 v 1 x 2 v 2⋯x d v d where each exponent v j is a non-negative integer. We employ the shorthand xv to denote such monomials. A polynomial is a finite linear combination of monomials. This can be extended to include the negative integer exponents, if needed. We shall explicitly refer to monomials xv with v∈Z d v∈Z d (or their linear combinations) as Laurent monomials (Laurent polynomials, respectively), whenever negative powers are considered.
Let K[x 1,…,x d]K[x 1,…,x d] be a polynomial ring, for brevity K[x]K[x]. For computability reasons, throughout K K will be the field of algebraic numbers, so K:=¯¯¯¯Q K:=Q¯. We employ the standard notation K∗K∗ to refer to the set K∖{0}K∖{0} of invertible elements of the field and denote by GL d(K)GL d(K) the multiplicative group of d × d invertible matrices over K K.
2.2 Algebraic Geometry
We recall standard preliminary material and terminology from the field of algebraic geometry, and refer to[3, 4] for more details.
Ideals. A polynomial ideal is a subset I⊆K[x]I⊆K[x] that satisfies the following properties: 0 ∈ I; I is closed under addition; and for each p∈K[x]p∈K[x] and q ∈ I, necessarily pq ∈ I. For a set of polynomials S⊆K[x]S⊆K[x], the ideal generated by _S_ is given by
I=⟨S⟩:={s 1 q 1+⋯+s ℓ q ℓ:s j∈S,q j∈K[x],ℓ∈N}.I=⟨S⟩:={s 1 q 1+⋯+s ℓ q ℓ:s j∈S,q j∈K[x],ℓ∈N}.
A polynomial ideal I is proper if I is not equal to K[x]K[x], I is prime if p · q ∈ I implies that p ∈ I or q ∈ I, and I is radical if p n ∈ I implies that p ∈ I.
Key to our discussion will be the bases for polynomial ideals.
Theorem 2.1 (Hilbert's Basis Theorem) Every ideal in K[x]K[x] has a finite basis.
Seminal work by Buchberger introduced Gröbner bases for polynomial ideals, which permit the algorithmic computation of key properties of polynomial ideals[1, 3], including ideal membership, ideal union/intersection, elimination ideals, and many more. A key property that we draw upon in our synthesis procedure is the computation of a basis for a saturation of an ideal. Given q∈K[x]q∈K[x], we compute a basis for the saturation of _I_ with respect to _q_
I:(q)∞:={p∈K[x]:q n p∈I for some n∈N}.I:(q)∞:={p∈K[x]:q n p∈I for some n∈N}.
We also recall relevant structural properties for classes of polynomial ideals.
Theorem 2.2 ([3, Theorem 6, Chapter 4.6]) Each radical polynomial ideal I in K[x]K[x] admits a unique decomposition as an intersection of finitely many prime ideals I = p 1 ∩ ⋅⋅⋅ ∩ p r such that for each pair p i⊄p j p i⊄p j. The decomposition is commonly referred to as the minimal decomposition of a radical ideal.
Varieties. The notion of an algebraic variety generalises the concept of algebraic curves to d dimensions. An (affine) algebraic variety V⊆K d V⊆K d is the locus of points satisfying a system of polynomial equations. In the sequel, we shall focus on the varieties associated with polynomial ideals. Let I be a polynomial ideal in K[x]K[x]. The locus of points in K d K d where the polynomials in I simultaneously vanish is called the variety of the ideal and denoted by V(I). Specifically,
V(I)={x∈K d:p(x)=0 for all p∈I}.V(I)={x∈K d:p(x)=0 for all p∈I}.
A subvariety is a subset of a variety that is, itself, a variety. A variety V is irreducible if when written as a union of subvarieties V = V 1 ∪ V 2 then necessarily either V 1 = V or V 2 = V. An affine variety is irreducible if and only if I(V) is a prime ideal.
Let V⊆K d V⊆K d be an affine variety. We endow V with the Zariski topology by declaring that the closed sets of V are precisely the subvarieties of V. In this way, we extend the definition of Zariski closure as follows. The Zariski closure of a subset S⊆K d S⊆K d is defined to be the smallest affine variety containing S.
2.3 C-finite Sequences
The class of C-finite sequences consists of the real-algebraic linear sequences ⟨u n⟩∞n=0⟨u n⟩n=0∞ that satisfy recurrence relations
u n+d=a d−1 u n+d−1+⋯a 0 u n,u n+d=a d−1 u n+d−1+⋯a 0 u n,
(1)
with constant coefficients a d − 1, …, a 0 such that a 0 ≠ 0. A C-finite sequence ⟨u n⟩n⟨u n⟩n that satisfies (1) is entirely determined by its initial values u 0, …, u d − 1. The order of ⟨u n⟩n⟨u n⟩n is the minimum length of the recurrence relations it satisfies.
Given a system of C-finite sequences ⟨x 1(n)⟩n⟨x 1(n)⟩n,..., ⟨x d(n)⟩n⟨x d(n)⟩n, consider the ideal consisting of all polynomials P∈K[y 1,…,y d]P∈K[y 1,…,y d] such that the polynomial equality P = 0 is satisfied for all n ≥ 0 after simultaneously setting y j ≔ x j(n) for all j ∈ {1, …, d}. We call this ideal the ideal of algebraic relations over K K associated with the aforementioned system.
2.4 Loop Invariants
Let L L be a linear loop with variablesx = (x 1, …, x d). For each loop variable x j, let ⟨x j(n)⟩n⟨x j(n)⟩n denote the sequence whose n th term is given by the value of x j after the n th loop iteration. A polynomial invariant of L L is a polynomial P∈K[x]P∈K[x] such that
holds for all n ≥ 0. The set of polynomial invariants of L L forms an ideal, called the (polynomial) invariant ideal of L L[23, 25]. In other words, the invariant ideal of L L is the ideal of algebraic relations over K K among the sequences⟨x 1(n)⟩n,…,⟨x d(n)⟩n⟨x 1(n)⟩n,…,⟨x d(n)⟩n. We note that, in our setting, it is always possible to compute a finite basis for the invariant ideal using Gröbner bases computation[20, 23].
The invariant ideal I of L L is radical. It can be equivalently described by its variety V(I)⊆K d V(I)⊆K d. Let
be the orbit of the loop. Informally speaking, O O is the set of variable vectors that L L can reach during its execution. The Zariski closure of O O in K d K d is precisely the variety V(I) and, further, is the smallest algebraic variety that contains O O. The Zariski closure of O O is the strongest algebraic invariant of L L. We shall sometimes abuse terminology and refer to the Zariski closure of O O as the Zariski closure of L L. A loop L L is trivial if its orbit O O is a finite set.
2.5 Lattices
Let (G, +) be an additive abelian group. A set X⊆G is linearly independent if for any n 1,…,n k∈Z n 1,…,n k∈Z and any pairwise distinct a 1, …, a k ∈ X a linear combination ∑k i=1 n i a i=0∑i=1 k n i a i=0 is zero only if n 1 = … = n k = 0. If ⟨X⟩ = G, the set X is called a basis of G.
An abelian group that has a finite basis is referred to as lattice. It is known that a lattice has fixed basis size, called rank and, furthermore, every lattice of rank r ≥ 1 is isomorphic to Z r Z r as a group. A saturation of a lattice L⊆Z d L⊆Z d is a sublattice of Z d Z d defined as
If L=Sat(L)L=Sat(L), the lattice L is said to be saturated.
An important class of lattices are those that describe the multiplicative relations among algebraic numbers. The exponent lattice of ζ 1,…,ζ k∈K∗ζ 1,…,ζ k∈K∗ is given by
The lattice ideal I L⊂K[x 1,…,x d]I L⊂K[x 1,…,x d] of a lattice L⊆Z d L⊆Z d is the ideal I L:=⟨x α−x β:α,β∈N d,α−β∈L⟩I L:=⟨x α−x β:α,β∈N d,α−β∈L⟩.
2.6 Toric Ideals and Varieties
Consider an integer s × d-matrix
A:=⎛⎜ ⎜ ⎜⎝a 11 a 12…a 1 d a 21 a 22…a 2 d…a s 1 a s 2…a s d⎞⎟ ⎟ ⎟⎠A:=(a 11 a 12…a 1 d a 21 a 22…a 2 d…a s 1 a s 2…a s d)
with column set {a 1,…,a d}⊂Z s{a 1,…,a d}⊂Z s. The purpose of defining this matrix is twofold.
First, we can view A as a linear transformation T A:Z d→Z s T A:Z d→Z s such that T A(x) = Ax. Thus we define the kernel of _A_ by ker A:={x∈Z d:T A(x)=0}kerA:={x∈Z d:T A(x)=0}. It is easy to see that ker A⊆Z d kerA⊆Z d is a lattice. In particular, one can define a lattice ideal for ker A kerA. Moreover, ker A kerA is saturated.
Second, we can consider the columns of A as Laurent monomials over the indeterminates z 1, …, z s. Formally, a columnai of A defines a Laurent monomial as χ i(z 1,…,z s)=z a 1 i 1…z a s i s χ i(z 1,…,z s)=z 1 a 1 i…z s a s i. We define the ring homomorphism
π A:K[x 1,…,x d]→K[z±1 1,⋯,z±1 s]π A:K[x 1,…,x d]→K[z 1±1,⋯,z s±1]
by mapping the variables x 1, …, x d to χ 1, …, χ d. The codomain of the homomorphism is the ring of Laurent polynomials over K K with s variables.
The toric ideal I A associated with matrix A is the kernel of π A. The ideal I A is a pre-image of a prime ideal, and hence I A is itself a prime ideal of K[x]K[x]. As shown in[4, Proposition 1.1.9], a toric ideal has an equivalent definition as a lattice ideal
I A=⟨x u−x v:u,v∈N d,u−v∈ker A⟩;I A=⟨x u−x v:u,v∈N d,u−v∈kerA⟩;
(2)
that is, as an ideal of a lattice ker A kerA. We emphasise that not every lattice is a kernel of a linear transformation and respectively, lattice ideals are not necessarily toric. In fact, toric ideals are precisely the prime lattice ideals[4, Proposition 1.1.11].
The homomorphism of K K-algebras is naturally associated with a group homomorphism φ A:K s→K d φ A:K s→K d, defined by
φ A(z)=(χ 1(z),…,χ d(z)).φ A(z)=(χ 1(z),…,χ d(z)).
The vanishing set of a toric ideal I A is the Zariski closure of φ A(K s)φ A(K s). Such irreducible varieties of K d K d are called toric varieties. They share a number of important properties and have been widely studied along with their ideals [4, 27].
We refer to binomials of the form xu − xv as pure difference binomials. A pure difference ideal is generated by pure difference binomials. Note that lattice and, in particular, toric ideals are pure difference ideals. Let p=x α 1 1⋯x α d d−x β 1 1⋯x β d d p=x 1 α 1⋯x d α d−x 1 β 1⋯x d β d be a pure difference binomial. Its exponent vector is (α 1−β 1,…,α d−β d)∈Z d(α 1−β 1,…,α d−β d)∈Z d.
Remark 2.3 Clearly, there are infinitely many pairwise distinct pure difference binomials sharing the same exponent vector —the vector is invariant when we multiply a pure difference binomial by a monomial, e.g., x 2 − y 2 and x 3 y − xy 3. For a given exponent vector v=(v 1,…,v d)∈Z d v=(v 1,…,v d)∈Z d, we define a unique canonical (pure difference) binomial as p(x 1,…,x d):=x v+−x v−p(x 1,…,x d):=x v+−x v−, where
v+=(max{v 1,0},…,max{v d,0})∈N d v+=(max{v 1,0},…,max{v d,0})∈N d
and v−=v+−v∈N d v−=v+−v∈N d. Note that x 2 − y 2 is the canonical binomial associated with the vector (2, −2), whereas x 3 y − xy 3 (see also1.2) is not canonical.
3 FROM PURE DIFFERENCE IDEALS TO LINEAR LOOPS
We now describe our loop synthesis approach, by restricting our input polynomials/ideals to pure difference binomials/ideals. Given a pure difference ideal I as an input, our procedure outputs a linear loop such that each polynomial in I is an invariant of the loop. We thus show that loop synthesis is decidable for pure difference polynomials/ideals.
We next outline our loop synthesis procedure and illustrate its main steps in table 1.
Loop Synthesis Procedure:
Input: A list p 1,…,p k∈K[x 1,…,x d]p 1,…,p k∈K[x 1,…,x d] of pure difference binomials or, alternatively, a polynomial ideal I generated by pure difference binomials.
(Step 1) Construct the lattice L determined by the exponents associated with the generators of I.
(Step 2a) Analyse lattice ideals I L and I Sat(L)I Sat(L); the ideal I Sat(L)I Sat(L) is prime, enabling Step 2b.
(Step 2b) Construct a matrix A with ker A=Sat(L)kerA=Sat(L).
(Step 3) Synthesise a linear loop L L with a diagonal update matrix M whose diagonal entries are determined by the columns of A. The ideal I Sat(L)I Sat(L) is the invariant ideal of L L.
Output: Linear loop L L with invariants p 1, …, p k.
Table 1:Loop synthesis for our running examples
Exmaple1.1Exmaple1.2
Input: I⟨x 2 − y, x 3 − z⟩⟨x 3 y − xy 3⟩
Step 1: I L⟨x 2 − y, x 3 − z⟩⟨x 2 − y 2⟩
Step 2a: I Sat(L)I Sat(L)⟨x 2 − y, x 3 − z⟩⟨x − y⟩
Step 2b: A
Step 3: L L(x, y, z) ≔ (1, 1, 1);(x, y) ≔ (1, 1);
while ⋆ dowhile ⋆ do
(x, y, z) ≔ (2 x, 4 y, 8 z);(x, y) ≔ (2 x, 2 y);
end whileend while
We further detail each component of our synthesis procedure, proving correctness of each step of the synthesis process.
Input
The input for the synthesis process is a (possibly empty) finite list p 1, …, p k of pure difference binomials. Let I denote the pure difference polynomial ideal I ≔ ⟨p 1, …, p k⟩ in K[x]K[x] generated by this finite set. Our goal is to synthesise a linear loop L L for which I is an invariant.
Step 1: Lattice ideal IL
We start by listing the k exponent vectors b 1,…,b k∈Z d b 1,…,b k∈Z d of the pure difference binomials p 1, …, p k. Let B ≔ {b1, …, bk} and L the lattice spanned by the exponent vectors B. If k = 0, our convention is B ≔ {0} and L={0}⊆Z d L={0}⊆Z d. In order to meet our objective, we first show that the saturation of the ideal I = ⟨p 1, …, p k⟩ with respect to∏d i=1 x i∏i=1 d x i is precisely the lattice ideal I L (3.2). In this direction, we begin with an intermediate lemma.
Lemma 3.1 Let I be an ideal generated by p 1, …, p k with exponent vectors constituting the set B = {b1, …, bk}. Let J be the ideal generated by the canonical binomials of B (as in 2.3). Then
Proof. Since I⊆J, one inclusion is straightforward: namely,
In the other direction, consider a polynomial g∈J:(∏d i=1 x i)∞g∈J:(∏i=1 d x i)∞. Let q 1, …, q k be the canonical binomials generating J. Then g admits a decomposition of the form
for some polynomials f 1,…,f k∈K[x]f 1,…,f k∈K[x] and some n 1 ≥ 0.
Each generator p ℓ of the ideal I has the form p ℓ = m ℓ q ℓ, where m 1, …, m k are monomials and so
(∏d i=1 m i)⋅g⋅(∏d i=1 x i)n 1=∑k ℓ=1 m ℓ q ℓ⋅(f ℓ⋅∏i≠ℓ m i)=∑k ℓ=1 p ℓ⋅(f ℓ⋅∏i≠ℓ m i)(∏i=1 d m i)⋅g⋅(∏i=1 d x i)n 1=∑ℓ=1 k m ℓ q ℓ⋅(f ℓ⋅∏i≠ℓ m i)=∑ℓ=1 k p ℓ⋅(f ℓ⋅∏i≠ℓ m i)
lies in I. Finally, there exists n 2 ≥ 0 such that (∏k i=1 x i)n 2(∏i=1 k x i)n 2 is a multiple of the monomial ∏k i=1 m i∏i=1 k m i. Thus
and hence g∈I:(∏d i=1 x i)∞g∈I:(∏i=1 d x i)∞.□
Theorem 3.2 Let I⊆K[x 1,…,x d]I⊆K[x 1,…,x d] be a pure difference ideal and L⊆Z d L⊆Z d the lattice spanned by the exponent vectors of its generators. Then I:(∏d i=1 x i)∞=I L I:(∏i=1 d x i)∞=I L and the lattice ideal I L is radical.
Proof. Since B = {b1, …, bk} spans the lattice L, [12, Corollary 3.22] implies that J:(∏d i=1 x i)∞=I L J:(∏i=1 d x i)∞=I L. The desired equality I:(∏d i=1 x i)∞=I L I:(∏i=1 d x i)∞=I L now follows from3.1. The second assertion follows from[12, Theorem 3.23].□
Note that I is contained in I L and, in general, this inclusion is strict. In particular, I⊊I L in1.2, see also table 1.
Step 2a: Saturated lattice ideal I Sat(L)I Sat(L)
Before we proceed with Step 2 of our loop synthesis process, we briefly reflect on the lattice ideals that contain I. Notice that in Step 1 we not only found one of them, I L, but we also pointed out that its lattice L had already been computed. Indeed, the generating setb1, …, bk of L is obtained from the pure difference binomials directly. While the generating polynomials I L can also be computed (e.g. by employing Gröbner bases techniques from[21, Section 5] to compute the saturation of I), our computation proceeds with vectors and lattices rather than polynomials and ideals.
Since I L is a radical ideal (3.2), I L admits a unique decomposition (see2.2). We adjust a general result by Eisenbud and Sturmfels, which concerns decompositions of binomial ideals, to our setting.
Lemma 3.3 ([8, Corollary 2.5]) Let L⊆Z d L⊆Z d be a sublattice and Sat(L)⊆Z d Sat(L)⊆Z d its saturation. The minimal decomposition of ideal I L includes the lattice ideal I Sat(L)I Sat(L) so that I L=I Sat(L)∩J 1∩⋯∩J ℓ I L=I Sat(L)∩J 1∩⋯∩J ℓ. Here ℓ ≥ 0 and I Sat(L),J 1,…,J ℓ I Sat(L),J 1,…,J ℓ are prime ideals of K[x]K[x]. In particular, I Sat(L)I Sat(L) is toric.
As an aside, work by Grigoriev et al. presents an alternative approach to this step. Therein those authors decompose a so-called binomial variety V(I L) into a finite union of irreducible varieties.
In the computational part of Step 2 that follows, we compute a prime lattice ideal I Sat(L)I Sat(L) that contains I.
Step 2b: Matrix A encoding I Sat(L)I Sat(L)
As before, {b1, …, bk} is a set of vectors spanning L over Z Z. Versions of the next proposition appear in the literature, see e.g.[12, Theorem 3.17].
Proposition 3.4 Let L⊆Z d L⊆Z d be a sublattice of rank r. If r<d, then there exists an integer (d − r) × d-matrix A such that Sat(L)=ker A Sat(L)=kerA. Otherwise, if r = d, the same holds for A=0∈Z 1×d A=0∈Z 1×d. Further, there is an effective process to construct the matrix A.
Recall that for each y:=(y 1,…,y d)∈Sat(L)y:=(y 1,…,y d)∈Sat(L) there is an integer c such that c · y ∈ L. Since L consists of integer linear combinations of {b1, …, bk}, a vector of the form c · y lies in the Z Z-span of the set{b1, …, bk}. Let V denote a vector subspace of Q d Q d over the field Q Q spanned by {b1, …, bk}. Thus elements of Sat(L)Sat(L) are the integer vectors in V.
3.4 follows from the next technical lemma.
Lemma 3.5 There is a computable integer matrix A such that the null space, Null(A)={v∈Q d:A v=0}Null(A)={v∈Q d:A v=0}, is equal to V.1
Proof. Let B∈Z k×d B∈Z k×d be the matrix with rows b1, …, bk. The row space Row(B)Row(B) over Q Q is thus equal to V. The orthogonal complement of Row(B)Row(B) is Row(B)⊥=Null(B)Row(B)⊥=Null(B). It follows that Null(B)Null(B) has a basis of d − r linearly independent vectors, which we denote bya1, …, ad − r, provided that r<d. On the other hand, if r = d, then V is not a proper subspace of Q d Q d, from which it follows that Null(B)={0}Null(B)={0}. Let A be the matrix with rowsa1, …, ad − r (and defined as a single row vector 0 in the case r = d). Then, by definition, Row(A)=Null(B)Row(A)=Null(B). The result follows by elementary properties of orthogonal decomposition in finite-dimensional vector spaces. Indeed, we have
Null(A)=Row(A)⊥=Null(B)⊥=(Row(B)⊥)⊥=Row(B)=V.Null(A)=Row(A)⊥=Null(B)⊥=(Row(B)⊥)⊥=Row(B)=V.
Generally speaking, the matrix A constructed above has rational entries; however, multiplying the entries of a matrix by a scalar does not change the null space. Thus we can assume, without loss of generality, that A has integer entries and preserves the property Null(A)=V Null(A)=V, as desired.□
Proof of 3.4 There are two statements to prove in 3.4 depending on the cases r<d or r = d. Note that the latter statement, that A=0∈Z 1×d A=0∈Z 1×d if r = d, was dealt with in the proof of 3.5. Thus all that remains it to establish the former statement. To this end, we show that Sat(L)=ker A Sat(L)=kerA for the matrix A constructed in the proof of 3.5. We note that Sat(L)Sat(L) is equal to
V∩Z d=Null(A)∩Z d={v∈Q d:A v=0}∩Z d={v∈Z d:A v=0}=ker A,V∩Z d=Null(A)∩Z d={v∈Q d:A v=0}∩Z d={v∈Z d:A v=0}=kerA,
as desired.□
Step 3: Synthesise linear loop L L
We take a brief pause and reflect on the combination of s 3.3--3.4 together with the formal definitions presented in section 2. The sum total is a threefold equivalence between i) toric ideals, ii) ideals of saturated lattices, and iii) ideals of lattices presented as ker A kerA. Our objective, to synthesise a loop with prescribed polynomial invariants, relies on the constructive aspect of each of the preceding steps. In this final step, we will use the matrix A, constructed in 3.4, to synthesise a linear loop with invariant ideal I Sat(L)I Sat(L).
In the proof of 3.7, we will employ the following result due to Kauers and Zimmermann [20, Proposition 5] concerning algebraic relations among C-finite sequences.
Proposition 3.6 The ideal over ¯¯¯¯Q[x 0,x 1,…,x d]Q¯[x 0,x 1,…,x d] associated with the algebraic relations among the d + 1 bi-infinite C-finite sequences
⟨n⟩n∈Z,⟨λ n 1⟩n∈Z,…,⟨λ n d⟩n∈Z⟨n⟩n∈Z,⟨λ 1 n⟩n∈Z,…,⟨λ d n⟩n∈Z
is equal to the lattice ideal of the exponent lattice of λ 1, …, λ d.
We now turn to our main result in this step.
Theorem 3.7 Let p 1,…,p k∈K[x 1,…,x d]p 1,…,p k∈K[x 1,…,x d] be pure difference binomials and let L⊂Z d L⊂Z d be the lattice spanned by their exponent vectors. There exists a linear loop L L with a diagonal update matrix M∈Q d×d M∈Q d×d such that I Sat(L)I Sat(L) is the invariant ideal of L L.
Proof. Following 3.4, it remains to prove that the ideal of a saturated lattice Sat(L)Sat(L) can be realised as the ideal of all polynomial invariants of a linear loop. Our proof adapts the argument in [10, Proposition 3.7] due to Galuppi and Stanojkovski.
First, there exists an integer matrix A∈Z s×d A∈Z s×d such that ker A=Sat(L)kerA=Sat(L) (by3.4). As before (cf.section 2), let {a1, …, ad} be the set of column vectors of A. Now let p 1, …, p s be the first s prime numbers and define λ j:=p a 1 j 1⋯p a s j s λ j:=p 1 a 1 j⋯p s a s j. Note that for each j ∈ {1, …, d}, λ j is the evaluation of the monomial x a j x a j atp = (p 1, …, p s).
Since there are no non-trivial multiplicative relations among pairwise distinct primes, a vector n ≔ (n 1, …, n d) is a member of the exponent lattice of{λ 1, …, λ d} if and only if n simultaneously achieves unity; i.e., p n 1 a i 1+⋯+n d a i d i=1 p i n 1 a i 1+⋯+n d a i d=1 for each i ∈ {1, …, s}. Specifically, (n 1,…,n d)∈L exp(λ 1,…,λ d)(n 1,…,n d)∈L exp(λ 1,…,λ d) if and only if for each i ∈ {1, …, s} we have that n 1 a i 1 + … + n d a id = 0. Thus it follows, by definition, that L exp(λ 1,…,λ d)=ker A=Sat(L)L exp(λ 1,…,λ d)=kerA=Sat(L).
Finally, let J be the ideal generated by the algebraic relations among the C-finite sequences ⟨λ n 1⟩n∈N⟨λ 1 n⟩n∈N,..., ⟨λ n d⟩n∈N⟨λ d n⟩n∈N. Mutatis Mutandis, the argument in 3.6 holds for natural-indexed sequences. Taken in combination with the above argument, it follows that J=I Sat(L)J=I Sat(L).
The polynomial invariants of a linear loop L L with initial vector (1,…,1)∈Q d(1,…,1)∈Q d and update matrix M=diag(λ 1,…,λ d)M=diag(λ 1,…,λ d) are exactly those in the ideal J, which concludes the proof.□
Output
From our starting point of a pure difference ideal I, we constructed an integer matrix A such that I Sat(L)I Sat(L) is a toric ideal associated with A, and I⊆I Sat(L)I⊆I Sat(L).
We output the linear loop L L with the invariant ideal I Sat(L)I Sat(L). Since I⊆I Sat(L)I⊆I Sat(L), the loop L L meets our objective: each of the polynomials in I is invariant under the action of L L. Correctness of the procedure follows from s 3.4--3.7.
The following corollary summarises our synthesis procedure.
Corollary 3.8 Let p 1,…,p k∈K[x 1,…,x d]p 1,…,p k∈K[x 1,…,x d] be a set of pure difference binomials and I = ⟨p 1, …, p k⟩. There exists a linear loop L L with a diagonal update matrix M∈Q d×d M∈Q d×d such that any polynomial p ∈ I = ⟨p 1, …, p k⟩ is an invariant of L L. There is a procedure to effectively construct loop L L.
4 TRANSFORMED PURE DIFFERENCE IDEALS
The procedure in section 3 synthesises linear loops from ideals generated by pure difference binomials. In this section, we demonstrate that, subject to certain assumptions on the input, we can also synthesise linear loops for non-binomial ideals.
We call a polynomial ideal ⟨p 1,…,p k⟩∈K[x]⟨p 1,…,p k⟩∈K[x] a transformed pure difference ideal if there exists an invertible linear change of coordinates for which each p j in the generating set is a pure difference binomial after the coordinate change. Here, by an invertible linear change of coordinates we mean that the associated change-of-basis matrix S is an element of GL d(K)GL d(K).
Theorem 4.1 Let I be a transformed pure difference ideal with associated change-of-basis matrix S∈GL d(K)S∈GL d(K). Then there exists a computable linear loop L L such that I is an invariant of L L.
We take I = ⟨p 1, …, p k⟩ and S as above. Further, let J ≔ ⟨q 1, …, q k⟩ be the associated pure difference ideal after the change of coordinates S. The synthesis procedure summarised in 3.8 outputs a loop L′L′ for which the pure difference ideal J = ⟨q 1, …, q k⟩ is invariant. More specifically, L′L′ is the loop with update assignments described by a diagonal update matrix M and initial vectorvin = (1, …, 1) such that each polynomial in ideal J is an invariant of L′L′.
The proof of 4.1 is as an immediate corollary of the next lemma, which outputs the desired linear loop L L.
Lemma 4.2 Let L L be the linear loop with update assignment matrix S− 1 MS and initial vector S− 1vin. Then every polynomial in I is an invariant of L L.
Proof. Let e1,..., ed be the standard basis of K d K d. After the change of coordinates S, the new basis vectors are S− 1e1,..., S− 1ed and the coordinates ofa = x 1e1 + x ded are S a=(x′1,…,x′d)S a=(x 1′,…,x d′). Clearly,
Hence, v in∈⋂k i=1 V(q i)v in∈⋂i=1 k V(q i) implies that S−1 v in∈⋂k i=1 V(p i)S−1 v in∈⋂i=1 k V(p i). Moreover, the matrix S− 1 MS encodes the linear update assignments of L′L′ in the standard basis; thus (S−1 M S)n S−1 v in∈⋂k i=1 V(p i)(S−1 M S)n S−1 v in∈⋂i=1 k V(p i) for all n ≥ 0. Hence, each polynomial in I is an invariant of L L.□
Example 4.3 Let I = ⟨p 1, p 2⟩ be the ideal of K[x,y,z]K[x,y,z] generated by p 1 = 4 y 2 + y − x and p 2 = 8 y 3 − x + z. The change of coordinates described by the invertible matrix S:=(0 2 0 1−1 0 1 0−1)S:=(0 2 0 1−1 0 1 0−1) yields q 1 = (x′)2 − y′, q 2 = (x′)3 − z′. Thus I is a transformed pure difference ideal and, after a change of coordinates, we have arrived at the ideal in 1.1. In the new coordinate system, we synthesise a loop L′L′ (3.81a ) for which each polynomial in the pure difference ideal ⟨q 1, q 2⟩ is invariant. In the old coordinate system, we proceed as in 4.1. Our procedure synthesises the loop L L with update matrix S−1⋅diag(2,4,8)⋅S S−1⋅diag(2,4,8)⋅S and initial vector S− 1 · (1, 1, 1)T as follows:
By construction, both polynomials p 1 and p 2 are invariants of L L, and thus so is any p ∈ I = ⟨p 1, p 2⟩.
Remark 4.4 The approach of4.1 relies on the existence of an appropriate change of coordinates that moves ideal I to a pure difference ideal J and further, that a change-of-basis matrix S is given explicitly. The problem of determining the existence of such S appears to be a challenging task.
A similar challenge is addressed by Katthän et al.. Their algorithmic approach[18, Algorithm 4.5] can generate all matrices S∈GL d(K)S∈GL d(K) that move an ideal I to a unital ideal J. Recall that unital ideals are those generated by pure difference binomials and monomials, and hence pure difference ideals are a strict subclass. A straightforward approach to transform an ideal to a pure difference ideal is to apply the procedure of and to manually check whether any of the transformed ideals output has a basis without monomials. If J is such an ideal, then 4.1 can be employed to synthesise a linear loop for the initial ideal I.
5 FURTHER RESULTS
In this section, we gather together corollaries of our synthesis procedure from section 3.
5.1 Existence of Non-Trivial Linear Loops
5.15.2next show that, subject to certain restrictions on the input ideal, there are always non-trivial loops that witness a given input as an invariant.
Theorem 5.1 Let I be an ideal generated by at most d − 1 pure difference binomials in K[x 1,…,x d]K[x 1,…,x d]. There exists a non-trivial linear loop L L such that any polynomial p ∈ I is an invariant of L L.
Proof. In light of 3.8, it suffices to show that L L generated by the procedure described in section 3 is non-trivial.
Recall that a linear loop L L is trivial if the input vector has finite orbit. For such loops, the variable vector(x 1, …, x d) is attained at two different iterations of L L. Since, by construction, the update matrix M:=diag(λ 1,…,λ d)M:=diag(λ 1,…,λ d), the loop L L is trivial if and only if λ k 1=⋯=λ k d=1 λ 1 k=⋯=λ d k=1 for some k ≥ 0. This implies that all diagonal entries of the loop update matrix M are roots of unity. However, by construction, the entries λ 1, …, λ d are positive rational numbers (see proof of 3.7). Therefore, λ 1 = … = λ d = 1 is a necessary condition for L L to be trivial.
From the proof of 3.4, the matrix A has rank 0 if and only if the associated lattice L is spanned by r = d linearly independent vectors. By assumption, L has rank at most d − 1 because it is generated by at most d − 1 exponent vectors. Thus M≠diag(1,…,1)M≠diag(1,…,1) and so L L is non-trivial.□
The next theorem is a specialisation of5.1 and follows from Step 3 of the synthesis procedure in section 3.
Theorem 5.2 Let I T ≠ ⟨x 1 − 1, …, x d − 1⟩ be a toric ideal of K[x 1,…,x d]K[x 1,…,x d]. There exists a non-trivial linear loop L L such that I T is the invariant ideal of L L.
Proof. We recall that toric ideals are precisely the lattice ideals of saturated lattices Sat(L)=ker A Sat(L)=kerA. Due to3.7, a loop L L with invariant ideal I Sat(L)I Sat(L) can be generated for an arbitrary saturated lattice Sat(L)=ker A Sat(L)=kerA. Observe that from the proof of5.1, L L is trivial if and only if Sat(L)=Z d Sat(L)=Z d. This, in turn, is equivalent to I T=⟨x α−x β:α,β∈N d⟩=⟨x 1−1,…,x d−1⟩I T=⟨x α−x β:α,β∈N d⟩=⟨x 1−1,…,x d−1⟩.□
Given polynomials p 1, …, p k, a necessary condition for a non-trivial loop to simultaneously satisfy p i = 0 for all i = 1, …, k, is the existence of infinitely many rational solutions to a system of equations p 1 = 0∧…∧p k = 0. Indeed, when p 1,..., p k are pure difference binomials with k ≤ d − 1, the system has infinitely many solutions (a property utilised in5.1). However, determining whether an arbitrary system of polynomial equations has infinitely many rational solutions is a highly non-trivial problem, as witnessed by the example below. Note that determining whether an equation has infinitely many integer solutions is undecidable.
Example 5.3 Here, d = 2 and k = 1. a) The equation x 3 + y 3 = 1 has precisely two rational solutions: (x 1, y 1) = (1, 0) and (x 2, y 2) = (0, 1). b) For comparison, the equation x 3 + y 3 = 9 has infinitely many rational solutions[26, Chapter 5.2].
5.2 Modifications to the Synthesis Process
A higher objective than that set in section 3 is as follows: given a polynomial ideal I, construct a linear loop whose invariant ideal is precisely I. The invariant ideals are radical, and so, for this question, one may assume that the given ideal I is radical. Equivalently, the objective is to construct a linear loop whose Zariski closure is the variety V(I). Our synthesis procedure does not generally achieve this goal because each inclusion in the chain of ideals
I⊆J⊆I L⊆I Sat(L)I⊆J⊆I L⊆I Sat(L)
(3)
considered in section 3 is, in general, strict. Here J stands for the ideal generated by the canonical binomials, as in e.g.3.1.
Remark 5.4 On the one hand, ker A kerA of any integer s × d-matrix A is a sublattice of Z d Z d. On the other hand, the converse statement is not true since there are lattices that cannot be represented as matrix kernels. For example, L:=Z(2,−2)⊂Z 2 L:=Z(2,−2)⊂Z 2 is not saturated and hence not a kernel of a matrix: (2, −2) ∈ L but (1,−1)∉L(1,−1)∉L.
By 5.4, we generally have I L⊊I Sat(L)I L⊊I Sat(L). An examination of 1.2 (see also table 1), shows how most of the inclusions in(3) are strict; in this example, only the equality J = I L holds. Nevertheless, its (general) violation is witnessed elsewhere in the literature (cf.the discussion that follows [12, Corollary 3.21]).
In special cases, we can simplify the procedure of section 3 by observing (introduced) redundancy. In this direction, we provide a list of sufficient conditions for equalities to hold in(3).
Proposition 5.5 Let I⊆K[x 1,…,x d]I⊆K[x 1,…,x d] be a pure difference ideal.
Suppose that I = ⟨p⟩ is a principal ideal for which p is a canonical pure difference binomial. Then I = I L. If, in addition, p is irreducible, then I=I Sat(L)I=I Sat(L) holds.
Suppose that I is generated by pure difference binomials and at least one of the generators p 1, …, p k has a positive exponent vector v i∈Z d>0 v i∈Z>0 d. Then, J = I L.
Proof. Regarding (1), we consider the principal ideal of p = xα − xβ with gcd(x α,x β)=1 gcd(x α,x β)=1. Let I L be the lattice ideal of L:=Z⋅(α−β)L:=Z⋅(α−β). This lattice ideal is generated by pure difference binomials. Each such generator takes the form xnα − xnβ for some positive integer n, or x− mβ − x− mα for some negative integer m. Polynomials of this form are divisible by xα − xβ. It quickly follows that I L⊆⟨p⟩ and hence I L = I. If, in addition, p is irreducible, then p is prime (K[x 1,…,x d]K[x 1,…,x d] is a unique factorisation domain). Thus I L = I = ⟨p⟩ is prime and so L is saturated. Therefore I=I L=I Sat(L)I=I L=I Sat(L).
Assertion (2), in turn, is a modification of[27, Lemma 12.4]. Without loss of generality, let J be an ideal generated by canonical pure difference binomials q 1, …, q k, where q 1’s exponent vectorv is positive. Then, q 1 = xv − 1 and every variable x i, 1 ≤ i ≤ d, has an inverse in the factor ring K[x]/J K[x]/J. Hence there exists g∈K[x]g∈K[x] such that p:=g⋅(∏d i=1 x i)−1∈J p:=g⋅(∏i=1 d x i)−1∈J. Let f∈J:(∏d i=1 x i)∞f∈J:(∏i=1 d x i)∞ ; that is, assume there exists n ≥ 0 such that (∏d i=1 x i)n⋅f∈J(∏i=1 d x i)n⋅f∈J. As a consequence, g n⋅(∏d i=1 x i)n⋅f∈J g n⋅(∏i=1 d x i)n⋅f∈J. We rewrite the latter in the following form: (g n(∏d i=1 x i)n−1)⋅f+f∈J(g n(∏i=1 d x i)n−1)⋅f+f∈J. Note that the expression wrapped by the outer parentheses is divisible by p. It follows immediately that f ∈ J. Thus J=J:(∏d i=1 x i)∞J=J:(∏i=1 d x i)∞ and, by way of3.2, I L = J.□
Note that1.3 follows from5.5(i): an irreducible pure difference binomial is clearly canonical. A loop synthesised for I Sat(L)I Sat(L) then has I as its invariant ideal by3.7.
6 DISCUSSION AND CONCLUSIONS
Related Works. On the one hand, deriving polynomial invariants for the class of loops with (non-linear) polynomial arithmetic in their update assignments is, in general, undecidable. On the other hand, for restricted classes of loops there are efficient procedures for synthesising invariants. Indeed, the restricted classes of polynomial arithmetic in so-called solvable loops—loops with (blocks of) affine assignments, admit such procedures[6, 15, 22, 23].
Previous works synthesising linear loops from non-linear polynomial invariants include. Their approach is also based on algebraic reasoning about the C-finite sequences generated by linear loops. Yet, the synthesis problem is translated into a constraint solving task in non-linear arithmetic, by relying on loop templates and using a conflict resolution procedure based on cylindrical algebraic decomposition/Gröbner basis computation. Unlike, our synthesis procedure restricts to linear algebraic reasoning. The algorithm of is relative complete as it generates all linear loops satisfying a given invariant of a bounded degree. As such, and in contrast to our results from s 5.1--5.2, the work of gives no sufficient conditions for the existence of at least one non-trivial loop satisfying the invariant.
The approach in studies cyclic semigroups and sheds light on their geometry. In particular, irreducible components of Zariski closures for cyclic semigroups are shown to be isomorphic to toric varieties. Using the terminology in this note, their observation that toric ideals are precisely the invariant ideals of some linear loops is the key motivation for our synthesis procedure.
Conclusions and Future Directions. Using machinery from algebraic geometry, we identify classes of polynomial invariants for which synthesis becomes decidable. Our synthesis procedure can produce infinitely many different loops that, by construction, all have the same invariant. For a given input, the number of variables s of the procedurally generated output loops is equal to the number d of indeterminates in the input polynomials. Observe that linear loops with s>d variables can still satisfy invariants from K[x 1,…,x d]K[x 1,…,x d] on a subset of their variables. The challenge of deriving an upper bound on the number of program variables s for the invariant, also raised in[14, Section 4.2], remains open.
We aim to analyse the bit complexity of our procedure in the future. Note that in Step 3, prime numbers are raised to powers that are, in general, exponential in the number of variables d. Therefore, the generated update matrix might have entries with exponentially many (in D) bits, cf. the worst-case double exponential running time of Gröbner basis computation for the invariant ideal.
We intent to extend our synthesis approach to synthesise loops whose invariant ideals are non-prime lattice ideals. One challenge to be addressed is the classification of Zariski closures of the orbits O O of linear loops, cf.. In this direction, it is of interest whether loops with non-diagonalisable matrices have invariant ideals structurally different from those of loops synthesised in sections 3 and 4.
We conclude this paper by highlighting the following limitation of our approach, which we intend to address in future work. The construction in Step 1, that introduces the lattice ideal I L, entails that we can only synthesise loops whose invariant ideals are lattice ideals. Further, 3.2 implies that V(I L) is the Zariski closure of V(I)∖V(∏d i=1 x i)V(I)∖V(∏i=1 d x i), cf.[3, Chapter 4.4, Theorem 10]. Our procedure thus only synthesises loops whose orbits lie outside of the coordinate hyperplanes.
Example 6.1 We revisit the synthesis problem for an ideal I = ⟨x 3 y − xy 3⟩ of1.2. The invariant ideal of a linear loop L′L′ with update matrix
and initial vector v init=(1,0)v init=(1,0) is precisely the ideal I.
The loop L′L′ in the example above stands in contrast to the synthesised loop L L in1.2 because V(I) is the strongest algebraic invariant of L′L′. In contrast, the strongest algebraic invariant of L L is V(x − y).
ACKNOWLEDGMENTS
We are grateful to Amaury Pouly and James Worrell for valuable discussions. The work presented in this paper was partially supported by the ERC consolidator grant ARTIST 101002685, the WWTF grant ProbInG ICT19-018, and the EU Marie Sklodowska-Curie Doctoral Network LogiCS@TU Wien Grant Nr. 101034440.
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FOOTNOTE
1Herein we distinguish between a Q Q-subspace of vectors orthogonal to the row space of a given integer matrix (commonly its null space) and a sublattice of orthogonal integer vectors (commonly its kernel).
This work is licensed under a Creative Commons Attribution International 4.0 License.
ISSAC 2023, July 24–27, 2023, Tromsø, Norway
© 2023 Copyright held by the owner/author(s).
ACM ISBN 979-8-4007-0039-2/23/07.
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Lagrange Multipliers
This article discusses Lagrange multipliers, a topic of multivariable calculus.
Contents
[hide]
1 Definition
2 Example
3 Problems
Definition
Useful in optimization, Lagrange multipliers, based on a calculus approach, can be used to find local minimums and maximums of a function given a constraint. Suppose there is a continuous function and there exists a continuous constraint function on the values of the function . The method of Lagrange multipliers states that, to find the minimum or maximum satisfying both requirements ( is a constant):
The method can be extended to multiple variables, as well as multiple constraints. If we had a continuous function and the constraint , we would just have another equation relating the partial derivatives with respect to through a factor of . Similarly, if we have another constraint , then we would add the partial with respect to each variable to their respective equation with another factor . Thus, we have the general form for Lagrange multipliers for the function bounded by a certain number of constraints , ,...,:
Where is the del operator, which when applied to a scalar function results in a vector with each component the partial in that component's direction, representing total change.
It is important to remember that sometimes there maybe a function that is bounded on some interval. In such cases, Lagrange multipliers may give a result, but the answer may not be the one this method results. After using this method, it is imperative to check the endpoints and also plug back in the numbers resulted from this method to verify the final answer. Also, sometimes the function may be over-constrained and there maybe no point that satisfies all requirements.
Example
This method can become extremely difficult to use when there are several variables and constraints, but it is very effective and useful in some situations, including some contest math problems. Consider the following question:
A rectangular prism lies on the plane with one vertex at the origin. The vertex that does not share a face with this vertex at the origin lies on the plane bounded by the points , , . Find the maximum volume of the box.
Solution: The volume of the box is given by the equation . Because the way the box is described, the point that lies on the plane given determines the volume of the box. Thus we need only need to consider that point. The constraint we are given for that point is given by the plane , since that is the plane that goes through the three points mentioned. Let us call that function . Now we can begin to find our partial derivatives:
We equate the corresponding partials through a constant factor, and we also use our original constraint:
Multiplying each side of the last three equations by the variable not in the equation, we have:
We can equate these to result:
Plugging back into the constraint, we have:
Thus, the maximum volume, given by , is:
Problems
Introductory
Find the maximum area of a rectangular prism with a surface area of .
Olympiad
Prove that , where and are non-negative real numbers satisfying .
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Home / Specialty / Publications & Research / Communiqué / Traumatic Pulp Exposures: A Quick Review
Traumatic Pulp Exposures: A Quick Review
Posted in Endodontics | August 2, 2022
By Dr. Joseph Stern
Introduction:
There are many important questions to consider prior to treating a traumatic pulpal exposure (see table 1). How long has the exposure been present? The longer the exposure has been present the higher the likelihood for bacterial contamination of the pulp. However, as long as the pulp is still vital and the patient’s symptoms don’t necessitate root canal therapy, vital pulpal treatments can be employed. Is the patient symptomatic? What are the pulpal and periapical diagnoses? The pulpal and periapical diagnosis will dictate treatment. For example, if the pulpal diagnosis is necrosis or irreversible pulpitis (spontaneous or lingering pain) then pulpectomy (root canal therapy) is usually indicated rather than a vital pulpal treatment such as a pulp cap or partial pulpotomy.
A tooth with periapical pathology is usually not a good candidate for vital pulpal treatments. A proper diagnosis must be reached before deciding to perform vital pulpal treatments. Vital pulpal treatments will usually not work if the pulp is severely inflamed or infected. Is there a concomitant luxation injury or root fracture? A concomitant laxation injury or root fracture might further complicate pulpal circulation and thereby necessitate a full root canal rather than vital pulpal treatment. Are the roots fully developed? As a general rule we make every effort to preserve the pulp vitality of an immature tooth to allow for continued root growth (apexogenesis). It is important to remember that although vital pulpal treatments are usually the treatment of choice for immature permanent teeth, they can be performed for both mature and immature permanent teeth. How large is the fracture/exposure? There are different levels of vital pulpal therapy, ranging from a direct pulp cap to a full pulpotomy. The size of the exposure will play a role in dictating which vital pulpal treatment is best. There are times where a fracture might compromise significant tooth structure necessitating a post (for retention of a core) in which case endodontic treatment is indicated even when the pulp is vital and the exposure is relatively small.
Table 1
Definition:
Vital pulp therapy (VPT) techniques are means of preserving the vitality and function of the dental pulp after injury resulting from trauma, caries, or restorative procedures. VPT procedures have traditionally included indirect or direct pulp capping, and partial or complete pulpotomy. VPT’s includes removal of part of the pulp allowing the rest of the pulp to remain vital and functional. We assume that by removing a small portion of the superficial inflamed pulp the remaining pulp remains healthy. When the pulp is exposed, bacterial invasion can be resisted by both the patient’s immune system present within the pulp, and the bathing of the area by the patient’s saliva which prevent impaction of contaminated debris/bacteria. While the pulp is a low compliant environment, making it more vulnerable to injury, it also has the ability to heal itself. Vital pulpal therapy is built upon this very important principle. Overall VPT has been shown to have a very high success rate.
Objective:
There are two essential objectives for treatment of the vital pulp that make the level of pulpal amputation important: a wound dressing should be placed on non-inflamed tissue, and the loss of tooth structure should be kept to a minimum. In crown-fractured teeth, showing vital pulp tissue after an exposure, not more than 2 mm of the pulp beneath the exposure needs to be removed (see figure 1).
Figure 1
Partial Pulpotomy vs. Direct Pulp Cap
The advantage of partial pulpotomy (removing 2 mm of coronal pulp) as compared with pulp capping lies in better control of the surgical wound and retention of the sealing material. Capping of the pulp is recommended only when the exposure is small (less than 1 mm in size) and when it can be treated shortly after the accident. These indications apply to only a limited number of teeth and, in the majority of cases, partial pulpotomy is therefore performed. It is also noteworthy that most studies find a much higher success rate with partial pulpotomy compared to pulp capping after a traumatic pulpal exposure, the reason being the better control of superficial inflammation below the exposure. It has been clearly shown that as time elapses (from 1 hour to 7 days) after injury the success of pulp capping significantly decreases (from 93% to 56%). Vital pulpal treatments can be used as a permanent treatment modality for mature and immature permanent teeth with complicated crown fractures, and re-entering of the pulp at a later time is not necessary as long as the pulp remains vital.
Hemostasis:
Sodium hypochlorite is an antimicrobial solution that provides hemostasis, disinfection of the dentin pulp interface, biofilm removal, chemical removal of the blood clot and fibrin, and clearance of dentinal chips along with damaged cells at the mechanical exposure site. Therefore, sodium hypochlorite can be used to disinfect the exposure site during a pulpotomy prior to restoring the pulpal wound (saline or chlorhexidine can be used as well). A cotton-soaked pellet is placed on the pulpal wound with pressure for this purpose.
Examination of pulp tissues after exposure is a critical step in pulp assessment. Controlled bleeding (complete hemostasis) is needed for successful partial pulpotomy. If hemostasis cannot be achieved this is an indicator that the pulp is inflamed and should not be capped. On occasion more pulp tissue needs to be removed to achieve hemostasis, and what was initially planned as a partial pulpotomy might turn into a full pulpotomy. A pulpal diagnosis of necrosis (non-vital pulp) or the inability to achieve hemostasis would obviously preclude vital pulpal treatments such as pulp cap or pulpotomy (partial or full).
Calcium Silicate Cements:
There have been few materials as groundbreaking in endodontics as MTA (mineral trioxide aggregate). This material, developed by Torabinejad in 1993, was the first of the CSC’s (calcium silicate cements), and was initially introduced for the purpose of sealing root perforations. Overtime they were found to be so biocompatible and successful in their applications that they have become mainstreamed for the use of apexogenesis, apexification, apicoectomies, and to seal root perforations. CSC’s are a class of materials that include tricalcium silicates, dicalcium silicates, hydraulic calcium silicate cements, and ‘bioceramics’ (MTA, TheraCal, Brasseler RRM, BC sealer and Biodentine are some examples). CSC’s have become more commonplace for use in vital pulp therapy (VPT) procedures. These materials are placed directly on the pulpal wound (see figure 1). While calcium hydroxide has classically been used for vital pulpal treatments most current studies clearly demonstrate that CSC’s are superior in terms of healing and success rates and are the preferred material to place on a pulpal exposure (once hemostasis is achieved). MTA was the first of these materials introduced for this purpose but newer materials have recently flooded the market. One of the advantages of the newer materials is that they don’t cause significant tooth discoloration, which can be beneficial, especially when used in the esthetic zone.
Conclusion:
When treatment planned correctly, vital pulpal treatments, such as partial pulpotomy, can be a valuable option with a high success rate. They can be utilized successfully not only for traumatic pulpal exposures but for carious pulpal exposures as well. The key to this high success rate is proper diagnosis and a well-sealed restoration above the capping material. Bacterial leakage ultimately can cause any vital pulpal treatment to fail.
A successful follow up to vital pulpal therapy would be dentin bridge formation, continued root development, positive response to pulp vitality testing, no symptoms, and no radiographic development of apical periodontitis or root resorption.
References:
AAE Position Statement on Vital Pulp Therapy, May 2021
Cvek M, Cleaton-Jones PE, Austin JC, Andreasen JO. Pulp reactions to exposure after experimental crown fractures or grinding in adult monkeys. J Endod. 1982 Sep;8(9):391
Cvek M. Partial pulpotomy in crown fractured incisors: results 3 to 15 years after treatment. Acta Stomatol Croat 1993;27:167–73.
Cvek, M. A clinical report on partial pulpotomy and capping with calcium hydroxide in permanent incisors with complicated crown fracture. J Endod 4:232-237, 1978.
Lin LM et al. Vital pulp therapy of mature permanent teeth with irreversible pulpitis from the perspective of pulp biology. Aust Endod J. 2020;46:154-165.
Bimstein E, Rotstein I. Cvek pulpotomy-revisited. Dent Traumatol 2016;32:438–42.
Fong CD, Davis MJ. Partial pulpotomy for immature permanent teeth, its present and future. Pediatr Dent 2002;24:29–32.
Heide S, Kerekes K. Delayed direct pulp capping in permanent incisors of monkeys. Int Endod J 1987; 20:65–74.
Fuks AB, Bielak S, Chosak A: Clinical and radiographic assessment of direct pulp capping and pulpotomy in young permanent teeth, Pediatr Dent 4:240, 1982.
Barthel CR, Rosenkranz B, Leuenberg A, Roulet JF. Pulp capping of carious exposures: treatment outcome after 5 and 10 years: a retrospective study. J Endod. 2000 Sep;26(9):525-8.
Dammaschke T, Leidinger J, Schäfer E. Long-term evaluation of direct pulp capping–treatment outcomes over an average period of 6.1 years. Clin Oral Investig. 2010 Oct;14(5):559-67
Cox CF, Bergenholtz G, Fitzgerald M, Heys DR, Heys RJ, Avery JK, Baker JA. Capping of the dental pulp mechanically exposed to the oral microflora — a 5 week observation of wound healing in the monkey. J Oral Pathol. 1982 Aug;11(4):327-39
Cvek M, Lundberg M. Histological appearance of pulps after exposure by a crown fracture, partial pulpotomy, and clinical diagnosis of healing. J Endod 1983;9:8–11.
Hafez AA, Cox CF, Tarim B et al. An in vivo evaluation of hemorrhage control using sodium hypochlorite and direct capping with a one- or two-component adhesive system in exposed nonhuman primate pulps. Quintessence Int. 2002;33:261-272
Cho S-Y et al. Prognostic factors for clinical outcomes according to time after direct pulp capping. J Endod. 2013;39(3):327-331
Parirokh M et al. Mineral Trioxide Aggregate: A Comprehensive Literature Review—Part III: Clinical Applications, Drawbacks, and Mechanism of Action, Journal of Endodontics, Volume 36, Issue 3, 400 – 413
Parirokh M, Torabinejad M. Mineral trioxide aggregate: a comprehensive literature review–Part I: chemical, physical, and antibacterial properties. J Endod. 2010 Jan;36(1):16-27.
Fuks AB, Cosack A, Klein H, Eidelman E. Partial pulpotomy as a treatment alternative for exposed pulps in crown fractured permanent incisors. Endod DentTraumatol 1987, June;3:100-2.
de Blanco LP.Treatment of crown fractures with pulp exposure. Oral Surg OralMed Oral Pathol Oral Radiol Endod 1996;82:564-8.
Nair PN, Duncan HF, Pitt Ford TR, Luder HU. Histological, ultrastructural and quantitative investigations on the response of healthy human pulps to experimental capping with mineral trioxide aggregate: a randomized controlled trial. Int Endod J. 2008 Feb;41(2):128-50.
Elmsmari F, Ruiz X-F, Miro Q, et al. Outcome of partial pulpotomy in cariously exposed posterior permanent teeth: a systematic review and meta-analysis. J Endod 2019;45:1296–1306.e3.
Mejare I, Cvek M: Partial pulpotomy in young permanent teeth with deep carious lesions, Endod Dent Traumatol 9:238, 1993.
Fuks A, Gavra S, Chosak A. Long-term follow-up of traumatized incisors treated by partial pulpotomy. Pediatric Dentistry 1993;15:334–6.
Sabeti M, Huang Y, Chung YJ, Azarpazhooh A. Prognosis of Vital Pulp Therapy on Permanent Dentition: A Systematic Review and Meta-analysis of Randomized Controlled Trials. J Endod. 2021 Sep 1:S0099-2399(21)00602-6.
Click for Table 6. Summaries of Classic Studies Associated with Vital Pulpal Treatments (Pulpotomies, etc).
Dr. Stern is a Diplomate of the American Board of Endodontics. He is the director of endodontics at the Touro College of Dental Medicine and frequently lectures on clinical endodontics. He has lectured at local county dental societies, the New Jersey Dental Association Annual Session, and the Greater New York Dental Meeting. He maintains a private practice, Clifton Endodontics, in Clifton, NJ. He can be reached at jstern5819@gmail.com or via the Instagram handle @the_barbed_broach1.
Note: This article has been adapted from an earlier version Dr. Stern authored for Dentistry Today.
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4461 | https://www.frontiersin.org/journals/pharmacology/articles/10.3389/fphar.2024.1346169/full | Published Time: 2024-03-07
Frontiers | Adverse reactions induced by MDT/WHO (Rifampicin+Clofazimine+Dapsone) and ROM (Rifampicin+Ofloxacin+Minocycline) regimens used in the treatment of leprosy: a cohort study in a National Reference Center in Brazil
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ORIGINAL RESEARCH article
Front. Pharmacol., 07 March 2024
Sec. Pharmacology of Infectious Diseases
Volume 15 - 2024 |
Adverse reactions induced by MDT/WHO (Rifampicin+Clofazimine+Dapsone) and ROM (Rifampicin+Ofloxacin+Minocycline) regimens used in the treatment of leprosy: a cohort study in a National Reference Center in Brazil
Isadora Costa Celestino1Douglas Eulalio Antunes2Diogo Fernandes Santos2Victor Lemos Gimenes 3Fabiane Mian de Souza4Isabela Maria Bernardes Goulart1,2
1 Post-Graduation Program in Health Science, Faculty of Medicine, Federal University of Uberlândia, Uberlândia, Brazil
2 National Reference Center for Sanitary Dermatology and Leprosy, Clinics’ Hospital, Faculty of Medicine, Federal University of Uberlândia, Uberlândia, Brazil
3 Faculty of Medicine, Higher School of Health Sciences, Federal District Health Department, Brasília, Brazil
4 Faculty of Medicine, Federal University of Uberlândia, Uberlândia, Brazil
Background: Recommended standard treatment for leprosy is multidrugtherapy (MDT/WHO), consisting Rifampicin+Dapsone+Clofazimine. Other medications are recommended in cases of resistance, adverse reactions and intolerances, including ROM regimen, Rifampicin+Ofloxacin+Minocycline. Therefore, pharmacovigilance is an important tool in understanding these adverse drug reactions (ADRs), supporting pharmacotherapy management and medication safety. This study seeks to evaluate ADRs comparing two therapeutic regimens, MDT and ROM, used in treatment of patients with leprosy, analyzing prognostic factors regarding risk and safety.
Methods:A retrospective cohort study was performed by assessing medical records of 433 patients diagnosed with leprosy from 2010 to 2021 at a National Reference Center in Brazil. They were subject to 24 months or more of treatment with MDT or ROM regimens. ADR assessments were analyzed by two experienced researchers, who included clinical and laboratory variables, correlating them with temporality, severity and the causality criteria of Naranjo and WHO.
Results: The findings observed an average of 1.3 reactions/patient. Out of individuals experiencing reactions, 67.0% (69/103) were utilizing MDT/MB, while 33.0% (34/103) were using ROM. The median time for ADR of 79 days for MDT and 179 days for ROM. In first reaction, Dapsone was the most frequently involved medication; the most affected system was hematopoietic. As compared to Clofazimine, results indicated that use of Dapsone was associated with 7% increased risk of ADR occurrence (HR: 1.07; p = 0.866). Additionally, Rifampicin was linked to 31% increased risk of ADRs (HR: 1.31; p = 0.602); and Ofloxacin showed 35% elevated risk (HR: 1.35; p = 0.653). Conversely, results for Minocycline indicated 44% reduction in the risk of ADRs (HR: 0.56; p = 0.527), although statistical significance was not reached. The use of MDT conferred 2.51 times higher risk of developing ADRs in comparison to ROM.
Conclusion: The comparison between MDT and ROM revealed that MDT caused more ADRs, and these reactions were more severe, indicating less safety for patients. Dapsone was the most common medication causing ADRs, followed by Rifampicin. The combination with Clofazimine was associated with an additional risk of ADRs, warranting further studies to confirm this hypothesis. Given the high magnitude of ADRs, healthcare teams need to monitor patients undergoing leprosy treatment with focus on pharmacovigilance.
1 Introduction
Leprosy is caused by the etiological agents Mycobacterium leprae and Mycobacterium lepromatosis (Belachew et al., 2019), and one of the transmission routes being direct contact with another infected human (Ploemacher et al., 2020). Diagnosis is challenging due to the long incubation period of the bacillus, which can take up to 20 years after contact for the appearance of the first signs and symptoms (Pfaltzgraf et al., 1985; Goulart et al., 2008).
The disease affects nerves due to the bacillus’ tropism for Schwann cells (SC) in the peripheral nervous system (PNS), initially causing increased, decreased, or absence of tactile sensation (Ng et al., 2000; Mungroo et al., 2020). It causes irreversible physical and motor damage, with these being the main contributors to the stigma of the disease. Other signs and symptoms depend on the clinical form and include hypopigmentation, pruritic lesions, hair loss, macules, neuropathies, ocular impairments, including lagophthalmos, nasal obstructions, epistaxis, septal perfusion, and others (Maymone et al., 2020; Chen et al., 2021).
The operational classification for treatment purposes considers paucibacillary (PB), when there are up to five skin lesions with negative intradermal bacilloscopy, and multibacillary (MB) with the presence of six or more skin lesions or positive intradermal bacilloscopy (Brasil, 2017). The clinical classification (Ridley and Jopling, 1966) allows the characterization of the disease spectrum, ranging from a vigorous cellular immune response in the Tuberculoid form (TT), through the unstable Borderline group, including Borderline-Tuberculoid (BT), Borderline-Borderline (BB), Borderline-Lepromatous (BL), and the high bacillary load and low cellular immunity form, Lepromatous (LL). Additionally, a subclassification of Lepromatous is called subpolar Lepromatous, considered a degraded borderline (Ridley and Job, 1985).
Brazil is the second country with the highest number of leprosy cases according to the WHO, accounting for approximately 79.0% of new diagnoses in 2019, along with India and Indonesia (WHO, 2020). For treatment, WHO recommended the use of multidrug therapy (MDT) in 1981, consisting of Dapsone (DDS), Clofazimine (CFZ), and Rifampicin (RFM). Currently, the regimen can be used for both PB and MB patients. Thus, the regimen used in adults diagnosed with MB leprosy (MDT/MB) is 600 mg of RFM and 300 mg of CFZ monthly, with daily doses of 50 mg of CFZ and 100 mg of DDS, from 12 to 24 doses (WHO, 1998; WHO, 2018).
However, over the years, drug resistance, therapeutic failures, and relapses in leprosy treatment have been observed, along with adverse reactions, including intolerances and hypersensitivities to MDT, making use and patient adherence challenging. Therefore, new effective drugs for treatment have been studied, such as Ofloxacin (Ji et al., 1994), Minocycline (Gelber et al., 1994), Clarithromycin (Gunawan et al., 2018), and other quinolones, such as Moxifloxacin (Pardillo et al., 2008; Franco-Paredes et al., 2022), and Levofloxacin (Dhople et al., 1995). Thus, a new alternative regimen with good acceptance has been called ROM (WHO, 1998), an association of Rifampicin (RFM) 600 mg, Ofloxacin (OFX) 400 mg, and Minocycline (MNC) 100 mg in monthly doses, recommended for use in intolerances or contraindications to CFZ and DDS, with a minimum duration of 12 months for PB and 24 months for MB (Brasil, 2016).
However, like MDT, the ROM regimen can also cause adverse reactions, which should be assessed within the concept of pharmacovigilance: “the science and activities related to the detection, assessment, understanding, and prevention of adverse reactions or any other possible drug-related problem,” ensuring patient safety and treatment quality (WHO, 2006). Thus, adverse reactions to drug use (ADRs) are: “harmful, unintended reactions that occur at doses normally used in humans” (WHO, 1973). However, for the drug to be related to the suspicion of a reaction, it is necessary to observe causality linked to the drug. To do this, one must consider not only previous studies on the drug but also temporality and, if possible, clinical and laboratory variables (OPAS, 2011). For this purpose, there are criteria, the most well-known being the Naranjo et al. (1981) Algorithm and the WHO Causality (OPAS, 2011). Both allow for an assessment of the plausibility of the medication causing the reaction, considering several variables such as reintroduction of the medication (rechallenge) or even the gradual withdrawal of the dose (dechallenge).
In the new global strategy from 2021 to 2030, towards zero leprosy, WHO emphasized the need for pharmacovigilance in leprosy, so that adverse reactions are monitored and identified, enabling management and improvements in the pharmacotherapy of patients affected by the disease (WHO, 2020). Therefore, Brazil being the second highest in number of leprosy cases globally, a study to elucidate the ADRs of the main drug regimens is necessary to provide information to healthcare teams, enabling appropriate and timely management for treatment adherence and consequently reducing M. leprae transmission.
Thus, the aim of this present study was to assess adverse reactions to MDT/MB and ROM regimens presented by patients treated at a Leprosy Reference Center in Brazil over a 12-year period.
2 Materials and methods
2.1 Study design
The present study is a retrospective cohort that analyzed secondary data from medical records of 449 patients diagnosed with leprosy who underwent treatment with MDT/MB and ROM regimens with 24 doses or more. The study covered the period from January 2010 to December 2021 at CREDESH-HCU-UFU-EBSERH (National Reference Center in Sanitary Dermatology and Leprosy of the Clinical Hospital of the Federal University of Uberlândia). Patients included in the study had documented adverse drug reactions in their medical records or showed significant laboratory findings. Upon temporal correlation, there was plausibility that the medication was the cause of the adverse reaction.
2.2 Sampling methods
The method employed to enroll patients in this retrospective cohort study was quota sampling, based on a therapeutic schema that helped the establishment of two subgroups according to the prescribed medication: MDT or ROM.
2.3 Ethics statement
This study involving human participants were reviewed and approved by the local research ethics committee CAAE: 46768321.5.0000.5152 (UFU), with informed consent being waived, as it is a study that involves retrospective data from patients who have interrupted or completed treatment.
2.4 Inclusion and exclusion criteria
The inclusion criteria encompassed patients diagnosed with leprosy who underwent treatment with MDT/MB or ROM, receiving a minimum of 24 doses between January 2010 and December 2021 at CREDESH-HC-UFU- EBSERH. Exclusion criteria applied to patients under 15 years of age, pregnant individuals, or those who experienced adverse reactions to other medications concurrent with the use of MDT/MB and ROM.
As a result, medical records of 449 patients were analyzed, and 16 were excluded: 14 due to being under 15 years of age and two due to pregnancy.
2.5 Data collect
For diagnosis, patients were examined by leprosy-specialized physician, who used clinical and laboratory data in treatment monitoring. Consequently, participants were already under monthly observation at CREDESH-HC-UFU-EBSERH by a multidisciplinary team comprising of physician, nurses, physical therapist, pharmacists and other healthcare professionals. Once leprosy was confirmed, the physicians prescribed the treatment, and the patient received the prescribed medications from the Reference Center’s pharmacy.
Throughout the treatment, patients underwent laboratory tests at the Clinics’ Hospital of Federal University of Uberlândia. These tests were conducted at the beginning and end of the treatment and every 3 months (Supplementary Table S1). If a patient showed any possibility of ADRs the medical team requested additional laboratory tests to monitor the reaction and the outcome after clinical management. This could involve suspension and gradual withdrawal of the dose (dechallenge) and latter reintroduction of the medication (rechallenge), or management with pharmacological or non-pharmacological measures.
For the study, clinical and laboratory data were collected from the included participants, both in physical and electronic medical records. Suspected adverse reactions documented in the records were assessed by two experienced researchers, who correlated them with the clinical and laboratory findings. Patient records were also analyzed for the treatment start date, outcome date or intermediate period (date of ADR occurrence), medication suspension date or management, and when the participant showed clinical and laboratory improvement of the reaction.
2.6 Assessment of adverse drug reactions
While collecting all the information described above, data such as epidemiological variables, like as age, sex, and ethnicity were analyzed. Clinical variables observed included operational classification (PB/MB), multibacillary clinical forms (BT, BB, BL, and LL), patient history (previous conditions and habits, use of other medications, and prior adverse reactions), initial and final treatment regimens and presented adverse reactions.
The data, encompassing reactions affecting various systems such as hematopoietic, hepatic, renal, dermatological, nervous, gastrointestinal, ophthalmic, musculoskeletal, and systemic, were meticulously analyzed. The number of Adverse Drug Reactions (ADRs) during treatment, clinical and/or laboratory findings, severity, and time of presentation until resolution were assessed by experts in pharmacology. The analysis followed physician diagnostics described in medical records and employed established instruments, including the WHO Causality and the Naranjo Algorithm (1981).
2.7 Statistical analysis
The normality of continuous variables was assessed using the D’Agostino-Pearson test. The binomial test was employed to analyze the association between MDT/MB and ROM groups and factors related to demographic and clinical characteristics of patients affected by an adverse drug reaction. The Friedman test was conducted to compare differences between medians of three paired results concerning hemoglobin concentrations at different treatment periods.
Furthermore, prognostic factors associated with adverse drug reactions were analyzed through Kaplan-Meier survival curves. Multivariate analysis of prognostic factors associated with ADRs was carried out using time-dependent Cox regression with proportional hazards (hazard ratio).
Statistical analysis was performed using the Statistical Package for the Social Sciences (SPSS) version 22.0 software (IBM, Armonk, NY, United States), considering a significance level of 5%.
3 Results
3.1 Epidemiological data
Records of 433 participants were analyzed: 43% (186/433) initiated treatment with MDT medication regimen, while 57% (247/433) started with ROM regimen. For age criteria, they were divided into age groups: 15 to 19 (19/433), 20 to 29 (26/433), 30 to 39 (63/433), 40 to 49 (89/433), 50 to 59 (110/433), 60 to 69 (70/433), 70 to 79 (40/433), and 80 or older (16/433). Regarding sex, 39.5% were women (171/433), and 60.5% were men (262/433).
Patients in MDT regimen, 6.5% (12/186) were Subpolar Lepromatous, 8.6% (16/186) Borderline-Borderline, 10.2% (19/186) Borderline-Lepromatous, 24.7% (46/186) Borderline-Tuberculoid, and 50% (93/186) were Lepromatous. Additionally, 33.3% (62/186) were women, and 66.7% (124/186) men, with 31.7% (59/186) being 60 years or older.
For patients on the ROM regimen, 2.0% (5/247) were Subpolar Lepromatous, 7.2% (18/247) Borderline-Lepromatous, 16.1% (40/247) Borderline-Borderline, 38.1% (94/247) Borderline-Tuberculoid, and 36.4% (90/247) Lepromatous. Moreover, 44.1% (109/247) were women, and 55.8% (138/247) were men, with 28.3% (70/247) being 60 years or older.
3.2 Participants who presented ADRs
The incidence of ADRs were observed in 23.8% (103/433) of the patients. Among these, 42.7% (44/103) were women, and 57.3% (59/103) men. Regarding the clinical form, 39.8% (41/103) were Lepromatous, 14.5% (15/103) Subpolar Lepromatous, 11.6% (12/103) Borderline-Lepromatous, 5.8% (6/103) Borderline-Borderline, and 28.2% (29/103) Borderline-Tuberculoid. Of the participants, 72.8% (75/103) were under 60 years old and 27.2% (28/103) were 60 or older. The skin color with the highest number of participants was white, accounting for 45.6% (47/103), followed by brown, which constituted 39.8% (41/103) of the total.
In addition to leprosy, 24.2% (25/103) of the patients had a total of 38 comorbidities. Among them, 16% (4/25) had more than two comorbidities, including hypertension (14/38), type 2 diabetes mellitus (8/38), hypothyroidism (4/38), chronic kidney disease (3/38), iron-deficiency anemia (2/38), diabetic nephropathy (1/38), fibromyalgia (1/38), chronic hepatitis (1/38), chronic obstructive pulmonary disease (COPD) (1/38), type 1 diabetes mellitus (1/38), dyslipidemia (1/38), and clinical depression (1/38). Patients using other medications, either for other health conditions or for leprosy reactions, totaled 34.9% (36/103).
Among those affected by reactions, 67.0% (69/103) were using MDT/MB and 33.0% (34/103) were using ROM. Of the participants, 77.6% (80/103) experienced only one ADR, 17.4% (18/103) had two ADRs, and 4.8% (5/103) had three or more. A total of 134 adverse reactions were observed in both regimens, averaging 1.3 reactions per patient. In this article, only the first adverse reactions presented by patients will be discussed.
Concerning the severity of the first adverse reaction, 42.7% (44/103) were considered mild, 26.2% (27/103) moderate, and 31.1% (32/103) severe, with one leading to a fatal outcome. Additionally, 36.0% (37/103) were managed with the use of other medications, and in 63.1% (65/103), the medication suspected to have induced the reaction was discontinued.
DDS caused the most ADRs in the first event, totaling 51.4% (53/103). RFM, used in both regimens, caused 33.0% (34/103) of ADRs, but 70.5% (24/34) were associated with the ROM regimen, while 14.4% (10/69) were associated with the MDT/MB regimen. OFX caused 7.7% (8/103) of the reactions, while MNC caused 1.9% (2/103), both from the ROM regimen. CFZ was responsible for 5.8% of total reactions (6/103).
Regarding the World Health Organization (WHO) Causality Assessment for the first ADRs in the MDT/MB regimen, 55.0% (38/69) were classified as probable, 26.0% (18/69) possible, 10.1% (7/69) doubtful, and 8.6% (6/69) confirmed. In the ROM regimen, 58.8% (20/34) were possible, 17.6% (6/34) confirmed, 11.8% (4/34) probable, 8.8% (3/34) doubtful, and 2.9% (1/34) conditional.
In the Naranjo Causality Assessment, for the first ADRs in the MDT/MB regimen, 56.5% (39/69) were classified as probable, 30.4% (21/69) possible, 7.2% (5/69) defined, and 5.7% (4/69) doubtful. In the ROM regimen, 50.0% (17/34) were possible, 26.4% (9/34) probable, 17.6% (6/34) defined, and 5.9% (2/34) doubtful.
During the first adverse reaction, the affected systems were as follows: 36.8% (38/103) hematopoietic, 21.3% (22/103) gastrointestinal, 17.4% (18/103) dermatological, 12.6% (13/103) nervous, 5.8% (6/103) systemic, 2.9% (3/103) renal, 0.9% (1/103) hepatic, 0.9% (1/103) musculoskeletal, and 0.9% (1/103) ophthalmic.
Withing reactions affecting the dermatological system, 22.2% (4/18) were erythroderma, 44.4% (8/18) generalized rash, 11.1% (2/18) pruritic rash, 5.5% (1/18) itchy scalp, 5.5% (1/18) urticaria, and 11.1% (2/18) xeroderma. Cases affected by erythroderma were using MDT/MB, and DDS was the suspected medication, which was discontinued in all cases. For generalized rash, the most suspected causative drugs were RFM (50.0%; 4/8), OFX (37.5%; 3/8), and DDS (12.5%; 1/8). Itchy scalp and urticaria were associated with the use of MNC, confirmed by rechallenge. Only two cases of xeroderma were recorded, associated with the use of CFZ, which was suspended.
Among the gastrointestinal reactions, 13.6% (3/22) were diarrhea, 54.5% (12/22) epigastralgia, 27.2% (6/22) nausea and 4.5% (1/22) constipation. DDS (1/3) and RFM (2/3) were associated with diarrhea. Epigastralgia was related to the use of DDS (1/12) and RFM (11/12). Nausea was associated with the use of DDS (2/6), CFZ (1/6), OFX (1/6), and RFM (2/6). One case of constipation was related to the use of CFZ. Overall, these reactions were managed with pharmacological measures without discontinuing the medications.
Hematopoietic reactions: 86.8% (33/38) were hemolytic anemia, with DDS causing 96.9% (32/33) of these, and Rifampicin causing 3.0% (1/33). Whenever a patient was suspected of hemolytic anemia due to Dapsone, rechallenge tests and dose reductions were performed. These interventions led to the suspension of medications in all cases (32/32) of this research due to a persistent decrease in hemoglobin levels even after Dapsone dose decrement. Regarding anemia caused by Rifampicin, as causality was considered doubtful, the medication was not discontinued, and the reaction was observed over months through routine tests.
In cases of hemolytic anemia caused by DDS (32/33), both sexes showed significant differences in hemoglobin levels between diagnosis and when the patient presented the ADR, according to Dunn’s multiple comparison test (p< 0.0001). Men who started the treatment with a hemoglobin of 12.9 g/dL and women with a hemoglobin of 12 g/dL were predisposed to hemolytic anemia onset (p< 0.0001). The average time for this specific reaction to occur was 163 days, with an average hemoglobin decrease of 2.8 g/dL. During the ADR, men (25/32) had an average decrease of 2.9 g/dL, ranging from 0.8 to 5.6 g/dL (p< 0.0001); while for women (7/32), there was an average decrease of 2.5 g/dL, ranging from 1 to 3.8 g/dL (p = 0.0013).
Other hematopoietic reactions included aplastic anemia (3/38) and thrombocytopenia (2/38). In all cases of aplastic anemia (3/3) DDS was the suspected drug, which was discontinued and replaced with OFX, leading to improved levels. In thrombocytopenia, the suspected drug (2/2) was RFM, part of the MDT/MB regimen, which was discontinued and replaced, in one case, with OFX and in the other case, with the MCM regimen (Minocycline+Clarithromycin+Moxifloxacin).
In the hepatic system, the only reaction (1/103) was drug-induced hepatitis due to DDS use, leading to a switch from the MDT/MB regimen to ROM. For the musculoskeletal system, muscle weakness was the only reaction (1/103) related to RFM use. However, upon causal analysis, it was deemed doubtful and the reaction was managed without discontinuing the medication.
Reactions envolving the nervous system, 7.8% (1/13) were anxiety, 23.0% (3/13) headache, 46.2% (6/13) insomnia, and 23.0% (3/12) dizziness. During anxiety reports, the participant was using MDT/MB, and dechallenge/rechallenge tests with DDS were conducted, leading to the probable identification of DDS as the cause of the reaction and resulting in medication suspension. In cases of headaches, RFM was the suspected drug, with 33.3% (1/3) in the MDT/MB regimen and 66.7% (2/3) in the ROM regimen. However, these were managed without discontinuation of the antimicrobial. In insomnia reports, there was a relationship with the use of OFX (4/6), DDS (1/6), and RFM (1/6), all managed with non-pharmacological measures such as sleep hygiene and pharmacological measures such as the use of hypnotic/sedative medications. Cases of dizziness (3/12) were all related to the use of the MDT/MB regimen, both with DDS (2/3) and CFZ (1/3). However, only in one case DDS was suspended after a rechallenge.
Only one eye reaction, xerophthalmia (1/103), was recorded, related to the use of CFZ. Management involved the use of lubricating eye drops, with improvement without discontinuation of the medication. In renal adverse reactions, 66.7% (2/3) were acute renal failure and 33.3% (1/3) interstitial nephritis. All these cases implicated RFM as the suspect, and during the reports of acute renal failure, one was considered doubtful and the other as proven after a rechallenge test. Despite one case of doubtful causality in renal failure, the team preferred to discontinue the medication. The patient diagnosed with interstitial nephritis had the probable cause attributed to the use of the antimicrobial. As with the other two cases affecting the renal system, RFM was discontinued.
In systemic reactions (6/103), 33.3% (2/6) were flu-like syndrome, with RFM being the only associated drug. In one case, the participant used MDT/MB and switched RFM to OFX, and in the other case, the MDT/MB regimen was changed to MCM. Sulfone syndrome (SS) was observed in 50.0% (3/6) of cases, caused exclusively by DDS, with the regimen being discontinued and replaced with ROM (2/3) and MCM (1/3). In 16.7% (1/6), anaphylaxis was associated with RFM during the first supervised dose of the MDT/MB treatment, where upper airway obstruction and angioedema were observed, not responding to team management, leading to death and being the only lethal case in the study.
3.3 Survival analysis and Cox’s proportional hazards
The variables: clinical form, sex, skin color, comorbidities, age, and the number of concurrently administered medications with MDT did not represent prognostic factors associated with ADRs (Log Rank; Breslow; Tarone-Ware, p> 0.05). On the other hand, the comparison between survival curves for conventional and alternative treatment regimens indicated MDT/MB as a poor prognostic factor for the occurrence of ADRs (Log Rank, p = 0.008; Breslow, p = 0.002; Tarone-Ware, p = 0.002). The median time for the occurrence of ADRs was 79 days for those using MDT/MB versus 179 days for participants under the ROM regimen.
Furthermore, survival analysis stratified by the type of medication confirmed the detection of DDS, CFZ and RFM use as poor prognostic factors for the occurrence of ADRs (Log Rank, p = 0.019; Breslow, p = 0.055; Tarone-Ware, p = 0.031), as their median times for the outcome were 63, 115, and 154 days, respectively, shorter than those found in drugs like OFX (245 days) and MNC (470 days).
The proportional hazards model (hazard ratios) through time-dependent Cox regression, demonstrated, through the multivariate model, that patients under the MDT/MB therapeutic regimen showed a 2.51 times higher risk of ADRs occurrence when compared to those under the ROM regimen (p = 0.029).
Although other factors, when stratified, did not show prognostic validity in a multivariate model when compared with CFZ, there was an increased risk of 7% with the use of DDS in the regimen for ADR occurrence (HR: 1.07, p = 0.866). Furthermore, RFM attributed a 31% increase in the risk for ADRs (HR: 1.31, p = 0.602) and 35% with Ofloxacin (HR: 1.35; p = 0.653). In contrast, the results for MNC showed that this medication reduced the risk of ADRs in 44% of patients (HR of 0.56; p = 0.527) compared to CFZ (Figure 1).
FIGURE 1
FIGURE 1. Forest plot for Cox proportional hazards of patients who presented ADRs, with the Kaplan-Meier hazard ratio (HR) of the time to presentation of the first adverse occurrence according to the medication used.
4 Discussion
The majority of the study’s participants were men under 60 years old with clinical form LL, which is expected due to the inclusion criteria, requiring a minimum of 24 doses of treatment. This age and sex profile was also observed in the Brazilian census from 2017 to 2021, where 55.7% of cases were male, with a prevalence in the age group of 50–59 years, totaling 23,192 diagnoses out of 119,698 (Brazil, 2023).
Regarding comorbidity profiles, a significant portion had cardiovascular diseases (primary hypertension) and endocrine conditions (type 2 diabetes mellitus and hypothyroidism). It is important to note the other clinical conditions and use of medications as how they can affect ADRs presentation, causing confusion during causality analysis (Magro et al., 2012).
Skin color as a variable for the occurrence of ADRs did not affect the quantity or severity, showing no statistically significant difference (p> 0.001). Even though there’s no difference in the present study, articles relate an individual’s skin color to an increased chance of presenting ADRs, mainly those linked to gene alleles. HLA-B13:01 is an example, leading to a higher predisposition to develop Dapsone Hypersensitivity Syndrome (DHS), known as Sulfone Syndrome (SS), mainly present in Chinese, Japanese, Indian, and Southeast Asian populations (Hoogeven et al., 2016; Liu et al., 2019). The DHS is a type IV-b hypersensitivity reaction characterized as a Drug Reaction with Eosinophilia and Systemic Symptoms (DRESS), mediated by T cells (TH2) that release cytokines and chemokines such as IL-4, IL-5, and IL-13, activating and recruiting eosinophils (Coombs and Gell, 1968; Cabañas et al., 2020). Therefore, patients with known family histories or populations with the potential presence of this gene should avoid the use of Dapsone.
The leprosy treatment regimen that showed the most ADRs was MDT/MB, even though it was not the most commonly used regimen in the study population. This was confirmed by observing the Hazard Ratio (HR), which listed in the analyses that this regimen has a 2.51 times higher chance of leading to the development of an ADR. MDT/MB was more associated with reactions in the hematopoietic system (p< 0.0001), the development of severe reactions (p< 0.0001), and more than one ADR during treatment (p = 0.020). Additionally, this regimen had a greater association with reactions of probable causality, according to the Naranjo Algorithm and the WHO Causality (p = 0.004). DDS was the drug that caused the most ADRs, 51.4%, and had the shortest average time to occurrence, 63 days after the start of use.
A previous study conducted at CREDESH-HC-UFU with 187 leprosy patients using MDT/MB observed 37.9% of adverse reactions, with 70.8% (80/113) of reports linked to the use of DDS, showing dermatological, hematopoietic, gastrointestinal, nervous, and musculoskeletal effects, with a predominance of epigastralgia and hemolytic anemia (Goulart et al., 2002).
Guragain et al. study (2017), the average time of appearance of DDS-induced ADRs was 3–21 weeks. Similar to the present study, reactions in the dermatological and hematopoietic systems were observed, such as exfoliative dermatitis (8/37), unspecified skin rash (4/37), hemolytic anemia (5/37), jaundice (14/37), fever, and headaches (4/37), Toxic Epidermal Necrolysis (TEN) (1/37), and agranulocytosis (1/37).
Regarding hemolytic anemia, in the present study, 32% (33/103) presented this ADR, with an average hemoglobin decrease of 2.8 g/dL, while Franco (2014), changes were seen in 16.8% (20/119), with an average decrease of 2.35 g/dL. It was also emphasized that DDS was the main suspect drug, and the use of polychemotherapy alone led to an average decrease in hemoglobin of 1.94 g/dL. This information should be highlighted in the management of patients undergoing MDT/MB.
The avarege time for the other drugs analyzed in the study, CFZ was found to be 115 days (approximately 4 months), but no studies were identified for comparison of this data. Additionally, although not statistically significant, the use of CFZ indicated an increased risk of adverse drug reactions (ADRs) when combined with DDS, while MNC reduced this risk. However, no studies were found that provided a basis for comparing how CFZ elevates this risk and, when combined with MNC, leads to a reduction in the risk of ADRs. Nevertheless, more studies with a larger sample size are needed to confirm this hypothesis.
CFZ caused 5.8% (6/103) of total adverse reactions, including constipation, xerophthalmia, dizziness, nausea, and xeroderma. These ADRs were also observed in Maia et al. (2013), who noted hyperpigmentation in all participants (37/37) and ichthyosis in 21.6% (8/37), both related to CFZ use. Hwang et al. (2014), a systematic review concluded that CFZ is the main cause of skin and gastrointestinal reactions. In a previous study with 119 patients (Franco, 2014), CFZ was the main drug in the MDT causing dermatological reactions, with a frequency of 70.0% for xerosis and 65.5% for skin pigmentation. However, this study also emphasized CFZ as a cause of xerophthalmia, affecting 46.2% of participants. According to the medication’s own label (Novartis, 2016), skin pigmentation can occur in 75%–100% of patients within a few weeks of treatment.
Analyzing the experience of this Reference Center and the collected data, it was concluded that CFZ-induced ADRs are poorly known and investigated by the team, such as gastrointestinal changes caused by the deposition of CFZ crystals in the submucosa and an increased QT interval with arrhythmias presence. This is concerning because, according to Risgaard et al. (2016), the use of at least one QT interval-altering medication in the last 90 days was seen in 58% of patients who experienced sudden death, with antibiotics linked to 16% of those cases. Thus, after analyzing these data, training was conducted with the team, emphasizing the need to record these ADRs.
In a previous study by our group (Goulart et al., 2002) with 187 participants had 113 identified reactions and CFZ was associated with 23.0% of ADRs, affecting the dermatological, ophthalmic, and gastrointestinal systems, with ichthyosis and xerophthalmia prevailing.
The use of RFM in the ROM regimen was more associated with adverse effects in the first reaction than when compared to its use in the MDT/MB regimen (p< 0.0001). Thus, it is possible to observe that this drug in the ROM regimen causes more adverse reactions. The average time for induction of reactions was 154 days in both therapies. This may be due to the other drugs in the MDT/MB regimen causing adverse reactions before RFM induces them. In the MDT/MB regimen, this drug caused 9.7% of total reactions, including acute renal failure, thrombocytopenia, flu-like syndrome, interstitial nephritis, headache, generalized rash, and anaphylaxis. Franco (2014) mentioned that RFM caused 30.3% (36/119) cases of flu-like syndrome and 2.5% (3/119) of DRESS, as well as 1 case of acute renal failure, similar to some cases observed in this study. No studies on adverse reactions in the ROM regimen were found, thus limiting further discussions on the ADRs of RFM use in the ROM regimen.
In Goulart et al. (2002), it was also found that the use of RFM accounted for 6.2% of the 113 reactions, affecting the renal, dermatological, gastrointestinal, and systemic systems, with fever and renal colic being the main observed reactions, as well as the renal system reactions in the current study.
Regarding the ROM regimen, there is an association with causing only one adverse reaction (p = 0.020), as well as gastrointestinal effects (p< 0.0001), mild severity (p< 0.0001), and possible causality according to the Naranjo Algorithm and WHO Causality (p< 0.001). No references were found regarding the analysis of adverse reactions to the ROM regimen.
OFX and MNC were the drugs that had the longest average days to develop an adverse reaction, at 245 and 470 days, respectively. This may be because these drugs are administered monthly, requiring more time to show ADRs. Despite OFX accounting for 7.7% (8/103) of total reactions related to the nervous, dermatological, and gastrointestinal systems, this drug can cause QT interval prolongation (Briasoulis et al., 2011), rhabdomyolysis, and tendinopathies (Hsiao et al., 2005; Alves et al., 2019), which were not identified. While MNC is known to cause lupus-like reactions (Knowles et al., 1996; Schlienger et al., 2000) and cutaneous hyperpigmentation (Hu et al., 2012), in this study, it caused only 1.9% (2/103) of reactions, associated with urticaria and itching on the scalp. Nair (2018) reported not observing adverse reactions caused by OFX and MNC in patients treated for leprosy in an alternative scheme consisting of CFZ+OFX+MNC.
The occurrence of adverse reactions is a medication-related problem, facilitating a decrease in medication adherence. Furthermore, even if a patient experiences a reaction and it is managed without discontinuing the medication, there is a greater likelihood of treatment adherence. In the context of leprosy treatment, this is of great importance because if the patient no longer uses the medication due to an adverse reaction, it can lead to relapse and resistance, complicating treatment and interruption disease transmission.
No studies have been found so far that compare the MDT/MB and ROM regimens in terms of safety, the risk of developing adverse reactions, and the average time for these effects. However, based on the results found in this study, it can be concluded that the use of the MDT/MB regimen can cause adverse reactions earlier, and it was responsible for a large part of the severe reactions, including the only fatal outcome. Regarding the ROM regimen, even though ADRs were associated with gastrointestinal effects, they were classified as mild and managed with pharmacological measures without the need to discontinue the drugs.
Thus, before starting leprosy treatment, it is necessary for the patient to undergo a detailed anamnesis with information about previous allergies and family predispositions, as well as laboratory tests to avoid confusion in variables associated with ADRs. In addition, treatment monitoring by the health team should be carried out for early identification, management, and timely treatment of adverse reactions, as well as notifications to the health authorities responsible for pharmacovigilance in each country.
Feeding the global database benefits the health of populations by ensuring the safety of medication use, leading to a reduction in morbidity and mortality, improving the rational use of drugs, and promoting safe prescription practices by healthcare professionals.
The limitation of the study was due to data collection relying on secondary means, depending on the professionals’ records, which often lacked prior knowledge for the identification of ADRs, leading to underreporting. As a commitment of the study regarding the information on these reactions, all ADRs were reported to ANVISA, Brazilian regulatory agency, via VigiMed.
Additionally, regarding limitations, this retrospective cohort study prioritized the examination of prognostic factors over the assessment of efficacy and effectiveness, distinguishing it from quasi-experimental designs and clinical trials. As mentioned earlier, the constraints within our secondary data presented challenges in obtaining comprehensive information regarding patients who did not experience drug reactions. This lack of data is crucial for quantitatively calculating relative risk and effectiveness in a comprehensive manner.
Data availability statement
The original contributions presented in the study are included in the article/Supplementary Material, further inquiries can be directed to the corresponding author.
Ethics statement
The studies involving humans were approved by Universidade Federal de Uberlândia. The studies were conducted in accordance with the local legislation and institutional requirements. The ethics committee/institutional review board waived the requirement of written informed consent for participation from the participants or the participants’ legal guardians/next of kin because It is a study that involves retrospective data from patients who have interrupted or completed treatment.
Author contributions
IC: Conceptualization, Data curation, Formal Analysis, Funding acquisition, Investigation, Methodology, Project administration, Resources, Software, Supervision, Validation, Visualization, Writing–original draft, Writing–review and editing. DA: Data curation, Investigation, Writing–review and editing. DS: Conceptualization, Investigation, Methodology, Writing–review and editing. VG: Formal Analysis, Validation, Writing–review and editing. FdS: Investigation, Writing–review and editing. IG: Conceptualization, Data curation, Formal Analysis, Funding acquisition, Investigation, Methodology, Project administration, Resources, Software, Supervision, Validation, Visualization, Writing–original draft, Writing–review and editing.
Funding
The authors declare financial support was received for the research, authorship, and/or publication of this article. This work was supported by the National Health Fund—Brazilian Ministry of Health [TED 157/2019].
Conflict of interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Publisher’s note
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Supplementary material
The Supplementary Material for this article can be found online at:
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Keywords: leprosy, drug-related side effects and adverse reactions, Rifampicin, Dapsone, Clofazimine, Ofloxacin, Minocycline, cohort study
Citation: Celestino IC, Antunes DE, Santos DF, Gimenes VL, de Souza FM and Goulart IMB (2024) Adverse reactions induced by MDT/WHO (Rifampicin+Clofazimine+Dapsone) and ROM (Rifampicin+Ofloxacin+Minocycline) regimens used in the treatment of leprosy: a cohort study in a National Reference Center in Brazil. Front. Pharmacol. 15:1346169. doi: 10.3389/fphar.2024.1346169
Received: 29 November 2023; Accepted: 20 February 2024;
Published: 07 March 2024.
Edited by:
Amedeo De Nicolò, University of Turin, Italy
Reviewed by:
Tarun Narang, Postgraduate Institute of Medical Education and Research, India
Michael D. Coleman, Aston University, United Kingdom
Copyright © 2024 Celestino, Antunes, Santos, Gimenes, de Souza and Goulart. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
Correspondence: Isadora Costa Celestino, isadora.celestino@gmail.com
Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher.
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Edited by
Amedeo De Nicolò University of Turin, Italy
Reviewed by
Tarun Narang Department of Dermatology Venereology and Leprology, Postgraduate Institute of Medical Education and Research, India Michael D. Coleman Aston University, United Kingdom
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4462 | https://fiveable.me/key-terms/intermediate-algebra/test-point-method | Test Point Method - (Intermediate Algebra) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Definition
The test point method is a graphical technique used to determine the solution set of a linear inequality in two variables. It involves selecting a test point, evaluating the inequality at that point, and using the resulting sign to determine the region that satisfies the inequality.
5 Must Know Facts For Your Next Test
The test point method involves selecting a point in the coordinate plane and evaluating the inequality at that point to determine the region that satisfies the inequality.
The sign of the inequality at the test point (positive or negative) determines which half-plane contains the solution set.
If the inequality is $ax + by \geq c$, the solution set is the half-plane that contains the point where the inequality is true.
If the inequality is $ax + by > c$, the solution set is the half-plane that contains the point where the inequality is true, excluding the boundary line.
The test point method is a useful technique for graphing linear inequalities in two variables, as it allows you to visualize the solution set without having to solve the inequality algebraically.
Review Questions
Explain the steps involved in using the test point method to graph a linear inequality in two variables.
To use the test point method to graph a linear inequality in two variables, follow these steps: 1) Identify the inequality, such as $2x + 3y \geq 6$. 2) Select a test point, such as (0, 0), and evaluate the inequality at that point. If the inequality is true at the test point, the solution set is the half-plane that contains the test point. If the inequality is false at the test point, the solution set is the other half-plane. 3) Draw the boundary line represented by the equation $2x + 3y = 6$. 4) Shade the appropriate half-plane to represent the solution set.
Describe how the test point method can be used to determine the solution set of a strict linear inequality, such as $2x + 3y > 6$.
For a strict linear inequality, such as $2x + 3y > 6$, the test point method is used in a similar way. However, the key difference is that the solution set is the half-plane that contains the point where the inequality is true, excluding the boundary line. To use the test point method, you would 1) Identify the inequality, 2) Select a test point and evaluate the inequality, 3) Draw the boundary line represented by the equation $2x + 3y = 6$, and 4) Shade the appropriate half-plane, excluding the boundary line, to represent the solution set.
Analyze how the test point method can be used to determine the intersection of the solution sets of multiple linear inequalities in two variables.
The test point method can be used to determine the intersection of the solution sets of multiple linear inequalities in two variables. To do this, you would 1) Graph each inequality using the test point method, 2) Identify the region where all the inequalities are satisfied, which is the intersection of the solution sets, and 3) Shade this common region to represent the final solution set. By using the test point method to graph each inequality individually, you can visualize the overall solution set that satisfies all the given linear inequalities.
Related terms
Linear Inequality:
An inequality that can be written in the form $ax + by \geq c$ or $ax + by > c$, where $a$, $b$, and $c$ are real numbers.
Solution Set:
The set of all points $(x, y)$ that satisfy a given linear inequality.
Boundary Line:
The line represented by the equation $ax + by = c$ that separates the plane into two half-planes, one of which satisfies the inequality.
"Test Point Method" also found in:
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Elementary Algebra
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4463 | https://math.stackexchange.com/questions/1050378/using-lagrange-multipliers-to-find-volume-of-a-cone | calculus - Using Lagrange multipliers to find volume of a cone - Mathematics Stack Exchange
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Using Lagrange multipliers to find volume of a cone [duplicate]
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Find the volume of the largest right circular cone that can be inscribed in a sphere of radius r? (3 answers)
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Using the method of Lagrange multipliers (or otherwise) find the maximum possible volume of a (right circular) cone which can be inscribed in a sphere of radius a.
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asked Dec 3, 2014 at 19:51
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It is custom on math.stackexchange to state what you have tried, your thoughts on the problem, or where are you stuck in the problem.mathematics2x2life –mathematics2x2life 2014-12-03 19:58:02 +00:00 Commented Dec 3, 2014 at 19:58
Hi! Sorry but the problem is I don't even know where to begin.r.mendoza –r.mendoza 2014-12-03 19:59:37 +00:00 Commented Dec 3, 2014 at 19:59
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Hint: Use the formula V=1 3 π r 2 h, and slice the cone vertically down the middle to produce a triangle which contains a right triangle with sides of length r and h−a and a hypotenuse of length a.
Now use the Pythagorean Theorem to express r 2 in terms of h.
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answered Dec 3, 2014 at 21:31
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4464 | https://www.youtube.com/watch?v=Kp1xBmwlI-0 | Proportions: What value for X makes the proportion true?
MooMooMath and Science
582000 subscribers
346 likes
Description
50824 views
Posted: 29 May 2021
Learn proportion basics. In this video, I show you how to work problems in which you find the value for x in a proportion.
Example problems include,
3/5= x/15
x/6 = 15/18
In order to solve these problems you cross multiply and then solve for x
29 comments
Transcript:
welcome to moomoomath and science and proportions and in this video we will help you answer the question what value of x makes the proportion true and you may have seen this technique it's the butterfly technique because the proportion can look like a butterfly and why i drew this is because you can do a cross product or a cross multiplication to solve for x so you go 3 times 15 which is 45 equals 5 times x which is 5x then all you do to solve for x is just divide both sides by 5 okay and these cancel and 45 divided by 5 is 9. so x is 9. now let's see if this makes sense or not three-fifths is equal to nine fifteenths um three goes into nine three times and five times three is fifteen so there we go let's look at another example okay and here we go we have x times 18 so that would be 18 x is equal to 15 times 6 and 15 times 6 is 90. okay so now all we do is we divide both sides by 18 and these cancel bring down the x and 18 goes into 90 five times again we're going to check we have 5 6 is equal to 15 18. 5 goes into 15 3 times and 6 times 3 is 18. so so far so good let's work two more examples okay now we have x times 24 or 24 x is equal to 16 times 3 is 48 divide both sides by 24. these cancel and 48 divided by 24 is 2. let's fill this in two thirds is equal to sixteenths twenty-fourths okay two goes into sixteen eight eight times three is twenty-four and last but not least let's work our last one just to make sure we know how to do this we have 16 x because i went x times 16 10 times 8 is 80. divide by 16 okay x is equal to 80 divided by 16 is 5. let's fill this in 5 8 is equal to 10 16. 5 goes into 10 2 times 8 times 2 is 16. hope that helped and remember kindness multiplies kindness be kind to someone today |
4465 | https://www.youtube.com/watch?v=xYpiO5Ak7xE | Algebra 74 - Factoring Quadratics by Inspection - part 2
MyWhyU
195000 subscribers
112 likes
Description
7830 views
Posted: 8 Aug 2018
Factoring a quadratic expression into a pair of linear expressions is one of the primary methods used to solve quadratic equations. In the previous lecture, we introduced a method for factoring quadratics using a trial and error process called "factoring by inspection", and showed how this process works in the simplest case when the x-squared coefficient is one. In this lecture, we will see how to factor a quadratic by inspection when the coefficient of x squared is not one.
3 comments
Transcript:
Introduction Hello. I'm Professor Von Schmohawk
and welcome to Why U. So far we have seen that
factoring a quadratic expression into a pair of linear expressions is one of the primary methods used
to solve quadratic equations. In the previous lecture, we introduced a method
for factoring quadratics using a trial and error process
called "factoring by inspection". We saw that this process is simplest
when the coefficient of x-squared is "one" since in that case, the x-coefficients p and r
in the linear factors will also be "one". Therefore, it will only be necessary
to determine the values of two constants, q and s given the values of b and c. However, when the coefficient of x-squared
is not "one" we must not only determine values for q and s but also values for p and r given the values of three constants
a, b, and c. In this lecture, we will see how to factor
a quadratic by inspection when the coefficient of x-squared
is not "one". Factoring Quadratics by Inspection As in the previous lecture we will assume that the quadratic expression
has been put into general form where a, b, and c are integers. If the x-squared coefficient "a"
is not "one" the first step is to look for any common factors
of a, b, and c. In some cases, factoring out the
greatest common factor can produce a quadratic expression
with an x-squared coefficient of "one". For instance, let's say we wish to factor
the quadratic expression "two x-squared, plus ten x, plus twelve". Since two is a common factor
of all three terms we can write this expression as "two" times a quadratic
with an x-squared coefficient of "one". We can then factor the quadratic by inspection
as we did in the previous lecture. In that lecture, we determined that the factors
of this quadratic were "x + 2" and "x + 3 ". So the original quadratic expression
"two x-squared, plus ten x, plus twelve" can be written as the product of three factors "2" "x + 2" and "x + 3". The "zero product property" tells us that the zeros of the function
defined by this quadratic expression are identical to the zeros of the functions
defined by each of its factors. The first factor "2"
is a constant function that has no zeros since regardless of the value of x the value of that constant function
will never be zero. The second factor "x + 2"
is a linear function that has a zero of negative two. Likewise, the third factor "x + 3"
is a linear function with a zero of negative three. And the "zero product property" tells us
that these zeros are also the zeros of the quadratic function. In some cases, the values of a, b, and c may not allow us to produce a quadratic expression with an x-squared coefficient of "one". Take for example, the quadratic
"six x-squared, minus fifteen x, plus nine". Although all three terms
have a common factor of "three" once we factor out the "three" the remaining quadratic expression
will be simpler to factor by inspection. However, we are still not left with a quadratic
whose x-squared coefficient is "one". So how do we go about factoring a quadratic
like this using the inspection method? Foil Method To factor this quadratic expression,
we wish to find two linear expressions that when multiplied together,
produce the quadratic. To multiply these linear expressions we use the FOIL method
to take the product of their first terms plus the product of the their outer terms plus the product of their inner terms plus the product of their last terms. The quadratic's x-squared term will then be equal to the product
of the factors' x-terms. Likewise, the quadratic's constant term will be equal to the product
of the factors' constant terms. And the quadratic's x-term will equal the product
of the factors' outer terms plus the product
of the factors' inner terms. Finding a Pair So to find a pair of linear factors
whose product produces the quadratic we will start by making a list of
all the pairs of linear factors whose product of x-terms
equals the quadratic's x-squared term and whose product of constant terms
equals the quadratic's constant term. Then from these pairs of linear factors we will look for a pair
whose product of outer terms plus product of inner terms is equal to the quadratic's x-term. So as our first example let's factor the quadratic expression
"five x-squared, plus nine x, minus two". Example 1 Factoring Quadratics We start by listing all the possible pairs
of linear factors whose product of x-terms
is five x-squared and product of constant terms
is negative two. One choice for x-terms whose product is five x-squared is x and 5x. So we will make a table under this heading for all possible pairs of linear factors
whose first terms are x and 5x. Another choice of x-terms
whose product is five x-squared could be 5x and x. But making another table of factors
with the same two x-terms is not necessary since this would just produce the same pairs
of factors with their order swapped. We could have also reversed the signs making another table with x-terms of
negative x and negative 5x. However, we still would have ended up
with the same pairs of linear factors with all the signs reversed and this would have produced the same solutions
to the quadratic equation. Now, under this table heading we will make a list
of all the possible pairs of factors whose product of constant terms
is negative two. The possible values for constant terms
whose product is negative two are positive one and negative two positive two and negative one negative one and positive two and negative two and positive one. We now have four pairs of factors to test to see if the sum of outer and inner products is 9x. Calculating the sum of the first pair of factor's
outer and inner products the outer product, negative two times x plus the inner product, one times 5x is 3x. Likewise, we calculate the sum of the outer
and inner products in the second factor pair the third factor pair and the fourth factor pair. We can see from these calculations that only the second pair has a sum of outer
and inner products that equals 9x. Therefore, the factors of this quadratic expression
are "x + 2" and "5x - 1". Although in this example
we tested all four pairs of factors we could have stopped once we discovered
that the second pair produced the correct result. It is never necessary
to continue testing factors once a set of factors is found that works. Example 2 Factoring Quadratics In this last example, even though
the coefficient of x-squared was not one factoring by inspection was still fairly simple involving only four pairs of factors to check. This was because the quadratic's
x-squared coefficient, five and its constant term, negative two had a small number of factors so very few combinations were produced. In fact, the only integer factors
of 5 and 2 are 1 and 5 and 1 and 2. However, when the constants "a" and "c"
in a quadratic expression have more factors factoring by inspection can sometimes require
a considerable amount of trial and error. In those cases
it is preferable to use methods such as "completing the square"
or the "quadratic formula" which we will introduce
in the next several lectures. As an example,
factoring the quadratic expression "twelve x-squared, plus nineteen x,
minus eighteen" requires us to consider a number of
different pairs of factors for both twelve and eighteen producing a large number of combinations. To illustrate this, let's try factoring
this quadratic by inspection. Example 3 Factoring Quadratics Just like in the previous example we will start by listing all possible pairs
of linear factors whose product of x-terms
is twelve x-squared and product of constant terms
is negative eighteen. One choice for x-terms
whose product is "twelve x-squared" is x and twelve x. Other possible choices
whose product is "twelve x-squared" are 2x and 6x and 3x and 4x. As we have showed, we do not need to pick
these same x-terms with their order reversed or signs reversed. Now the possible constant terms
whose product is negative eighteen are one and negative eighteen two and negative nine three and negative six six and negative three nine and negative two eighteen and negative one and then the same constant terms
with their signs reversed. This gives us 36 different factor pairs whose product of x-terms is twelve x-squared and product of constant terms
is negative eighteen. We now look for a factor pair from this list whose sum of outer and inner products is 19x. Adding the outer products for each factor pair to the inner products for each factor pair we get these results. From this, we see that
only one factor pair in this list has a sum of outer and inner products
that equals 19x. Therefore, the factors of this quadratic expression
are "3x - 2" and "4x + 9". Conclusion Although the inspection method is sometimes
the quickest way to factor a quadratic factoring this quadratic turned out to be
quite a lengthy process since in this case, there were many possible
factors that needed to be considered. In addition, some quadratics
are impossible to factor by inspection such as the example we saw in the lecture
"Factoring Quadratics by Inspection - part 1" Over one-thousand years ago problems like this prompted mathematicians
to search for a technique that could be used to solve
any quadratic equation. The break-through in devising a general method was the development of a technique called
"Completing the Square". In the next lecture,
we will introduce this method. |
4466 | https://www.lehman.edu/faculty/anchordoqui/SJ3.pdf | 430 Chapter 14 Fluid Mechanics 14.6 Bernoulli’s Equation You have probably experienced driving on a highway and having a large truck pass you at high speed. In this situation, you may have had the frightening feeling that your car was being pulled in toward the truck as it passed. We will investigate the origin of this effect in this section. As a fluid moves through a region where its speed or elevation above the Earth’s surface changes, the pressure in the fluid varies with these changes. The relationship between fluid speed, pressure, and elevation was first derived in 1738 by Swiss physicist Daniel Bernoulli. Consider the flow of a segment of an ideal fluid through a nonuniform pipe in a time interval Dt as illustrated in Figure 14.18. This figure is very similar to Figure 14.16, which we used to develop the continuity equation. We have added two features: the forces on the outer ends of the blue portions of fluid and the heights of these portions above the reference position y 5 0. The force exerted on the segment by the fluid to the left of the blue portion in Figure 14.18a has a magnitude P1A1. The work done by this force on the segment in a time interval Dt is W1 5 F1 Dx1 5 P1A1 Dx1 5 P1V, where V is the volume of the blue portion of fluid passing point 1 in Figure 14.18a. In a similar manner, the work done on the segment by the fluid to the right of the segment in the same time interval Dt is W2 5 2P2A2 Dx2 5 2P2V, where V is the volume of the blue portion of fluid passing point 2 in Figure 14.18b. (The volumes of the blue portions of fluid in Figures 14.18a and 14.18b are equal because the fluid is incompressible.) This work is negative because the force on the segment of fluid is to the left and the displace-ment of the point of application of the force is to the right. Therefore, the net work done on the segment by these forces in the time interval Dt is W 5 (P1 2 P2)V Finalize The time interval for the element of water to fall to the ground is unchanged if the projection speed is changed because the projection is horizontal. Increasing the projection speed results in the water hitting the ground farther from the end of the hose, but requires the same time interval to strike the ground.
y1 y2 The pressure at point 1 is P1. P1A1i The pressure at point 2 is P2. v2 v1 x1 x2 Point 2 Point 1 a S S P2A2i ˆ ˆ b Figure 14.18 A fluid in laminar flow through a pipe. (a) A segment of the fluid at time t 5 0. A small portion of the blue-colored fluid is at height y1 above a reference position. (b) After a time interval Dt, the entire segment has moved to the right. The blue-colored por-tion of the fluid is that which has passed point 2 and is at height y2. ▸ 14.7 continu ed Daniel Bernoulli Swiss physicist (1700–1782) Bernoulli made important discoveries in fluid dynamics. Bernoulli’s most famous work, Hydrodynamica, was published in 1738; it is both a theoreti-cal and a practical study of equilibrium, pressure, and speed in fluids. He showed that as the speed of a fluid increases, its pressure decreases. Referred to as “Bernoulli’s principle,” Bernoulli’s work is used to produce a partial vacuum in chemical laboratories by connecting a vessel to a tube through which water is running rapidly.
. iStockphoto.com/ZU_09 www.aswarphysics.weebly.com 14.6 Bernoulli’s Equation 431 Part of this work goes into changing the kinetic energy of the segment of fluid, and part goes into changing the gravitational potential energy of the segment–Earth system. Because we are assuming streamline flow, the kinetic energy Kgray of the gray portion of the segment is the same in both parts of Figure 14.18. Therefore, the change in the kinetic energy of the segment of fluid is DK 5 1 1 2mv2 2 1 K gray2 2 1 1 2mv1 2 1 K gray2 5 1 2mv2 2 2 1 2mv1 2 where m is the mass of the blue portions of fluid in both parts of Figure 14.18. (Because the volumes of both portions are the same, they also have the same mass.) Considering the gravitational potential energy of the segment–Earth system, once again there is no change during the time interval for the gravitational poten-tial energy Ugray associated with the gray portion of the fluid. Consequently, the change in gravitational potential energy of the system is DU 5 1mgy2 1 Ugray2 2 1mgy1 1 Ugray2 5 mgy2 2 mgy1 From Equation 8.2, the total work done on the system by the fluid outside the segment is equal to the change in mechanical energy of the system: W 5 DK 1 DU. Substituting for each of these terms gives 1P1 2 P22V 5 1 2mv2 2 2 1 2mv1 2 1 mgy2 2 mgy1 If we divide each term by the portion volume V and recall that r 5 m/V, this expres-sion reduces to P1 2 P2 5 1 2rv2 2 2 1 2rv1 2 1 rgy2 2 rgy1 Rearranging terms gives P1 1 1 2rv1 2 1 rgy1 5 P2 1 1 2rv2 2 1 rgy2 (14.8) which is Bernoulli’s equation as applied to an ideal fluid. This equation is often expressed as P 1 1 2rv 2 1 rgy 5 constant (14.9) Bernoulli’s equation shows that the pressure of a fluid decreases as the speed of the fluid increases. In addition, the pressure decreases as the elevation increases. This latter point explains why water pressure from faucets on the upper floors of a tall building is weak unless measures are taken to provide higher pressure for these upper floors. When the fluid is at rest, v1 5 v2 5 0 and Equation 14.8 becomes P1 2 P2 5 rg 1y2 2 y12 5 rgh This result is in agreement with Equation 14.4. Although Equation 14.9 was derived for an incompressible fluid, the general behavior of pressure with speed is true even for gases: as the speed increases, the pressure decreases. This Bernoulli effect explains the experience with the truck on the highway at the opening of this section. As air passes between you and the truck, it must pass through a relatively narrow channel. According to the continuity equa-tion, the speed of the air is higher. According to the Bernoulli effect, this higher-speed air exerts less pressure on your car than the slower-moving air on the other side of your car. Therefore, there is a net force pushing you toward the truck!
Q uick Quiz 14.5 You observe two helium balloons floating next to each other at the ends of strings secured to a table. The facing surfaces of the balloons are separated by 1–2 cm. You blow through the small space between the balloons. What happens to the balloons? (a) They move toward each other. (b) They move away from each other. (c) They are unaffected.
W W Bernoulli’s equation www.aswarphysics.weebly.com 432 Chapter 14 Fluid Mechanics Example 14.8 The Venturi Tube The horizontal constricted pipe illustrated in Figure 14.19, known as a Venturi tube, can be used to measure the flow speed of an incompressible fluid. Determine the flow speed at point 2 of Figure 14.19a if the pressure difference P1 2 P2 is known.
Conceptualize Bernoulli’s equation shows how the pressure of an ideal fluid decreases as its speed increases. Therefore, we should be able to calibrate a device to give us the fluid speed if we can measure pressure.
Categorize Because the problem states that the fluid is incom-pressible, we can categorize it as one in which we can use the equation of continuity for fluids and Bernoulli’s equation.
S o l u t i o n Analyze Apply Equation 14.8 to points 1 and 2, noting that y1 5 y2 because the pipe is horizontal: (1) P1 1 1 2rv1 2 5 P2 1 1 2rv2 2 Solve the equation of continuity for v1: v1 5 A2 A1 v2 a P1 P2 A2 A1 v1 S v2 S b © Cengage Learning/Charles D. Winters Figure 14.19 (Example 14.8) (a) Pressure P1 is greater than pressure P2 because v1 , v2. This device can be used to measure the speed of fluid flow. (b) A Venturi tube, located at the top of the photograph. The higher level of fluid in the middle column shows that the pressure at the top of the column, which is in the constricted region of the Venturi tube, is lower.
Finalize From the design of the tube (areas A1 and A2) and measurements of the pressure difference P1 2 P2, we can calculate the speed of the fluid with this equation. To see the relationship between fluid speed and pressure differ-ence, place two empty soda cans on their sides about 2 cm apart on a table. Gently blow a stream of air horizontally between the cans and watch them roll together slowly due to a modest pressure difference between the stagnant air on their outside edges and the moving air between them. Now blow more strongly and watch the increased pressure dif-ference move the cans together more rapidly.
Substitute this expression into Equation (1): P1 1 1 2ra A2 A1 b 2 v2 2 5 P2 1 1 2rv2 2 Solve for v2: v2 5 A1Å 21P1 2 P22 r1A1 2 2 A2 22 Example 14.9 Torricelli’s Law An enclosed tank containing a liquid of density r has a hole in its side at a distance y1 from the tank’s bottom (Fig. 14.20). The hole is open to the atmosphere, and its diameter is much smaller than the diameter of the tank. The air above the liquid is maintained at a pressure P. Determine the speed of the liquid as it leaves the hole when the liquid’s level is a distance h above the hole.
Conceptualize Imagine that the tank is a fire extinguisher. When the hole is opened, liquid leaves the hole with a certain speed. If the pressure P at the top of the liquid is increased, the liquid leaves with a higher speed. If the pressure P falls too low, the liquid leaves with a low speed and the extinguisher must be replaced.
AM S o l u t i o n A2 A1 P0 h P y2 y1 v1 S Point 2 is the surface of the liquid.
Point 1 is the exit point of the hole.
Figure 14.20 (Example 14.9) A liquid leaves a hole in a tank at speed v1.
www.aswarphysics.weebly.com 14.7 Other Applications of Fluid Dynamics 433 Apply Bernoulli’s equation between points 1 and 2: P0 1 1 2rv1 2 1 rgy1 5 P 1 rgy2 Solve for v1, noting that y2 2 y1 5 h: v1 5 Å 21P 2 P02 r 1 2gh Finalize When P is much greater than P0 (so that the term 2gh can be neglected), the exit speed of the water is mainly a function of P. If the tank is open to the atmosphere, then P 5 P0 and v1 5 !2gh. In other words, for an open tank, the speed of the liquid leaving a hole a distance h below the surface is equal to that acquired by an object falling freely through a vertical distance h. This phenomenon is known as Torricelli’s law.
What if the position of the hole in Figure 14.20 could be adjusted vertically? If the tank is open to the atmosphere and sitting on a table, what position of the hole would cause the water to land on the table at the farthest distance from the tank?
What If?
Categorize Looking at Figure 14.20, we know the pressure at two points and the velocity at one of those points. We wish to find the velocity at the second point. Therefore, we can categorize this example as one in which we can apply Bernoulli’s equation.
Analyze Because A2 .. A1, the liquid is approximately at rest at the top of the tank, where the pressure is P. At the hole, P1 is equal to atmospheric pressure P0.
Therefore, to maximize the horizontal distance, the hole should be halfway between the bottom of the tank and the upper surface of the water. Below this location, the water is projected at a higher speed but falls for a short time inter-val, reducing the horizontal range. Above this point, the water is in the air for a longer time interval but is projected with a smaller horizontal speed.
Answer Model a parcel of water exiting the hole as a projectile. From the particle under constant acceleration model, find the time at which the parcel strikes the table from a hole at an arbitrary position y1: yf 5 yi 1 vyit 2 1 2gt 2 0 5 y1 1 0 2 1 2gt 2 t 5 Å 2y1 g From the particle under constant velocity model, find the horizontal position of the parcel at the time it strikes the table: xf 5 xi 1 vxit 5 0 1 "2g 1y2 2 y12 Å 2y1 g 5 2"1y2y1 2 y1 22 Maximize the horizontal position by taking the deriva-tive of xf with respect to y1 (because y1, the height of the hole, is the variable that can be adjusted) and setting it equal to zero: dxf dy1 5 1 2122 1y2y1 2 y1 22 21/21y2 2 2y12 5 0 Solve for y1: y1 5 1 2y2 ▸ 14.9 c on tin u ed 14.7 Other Applications of Fluid Dynamics Consider the streamlines that flow around an airplane wing as shown in Figure 14.21 on page 434. Let’s assume the airstream approaches the wing horizontally from the right with a velocity v S 1. The tilt of the wing causes the airstream to be deflected downward with a velocity v S 2. Because the airstream is deflected by the wing, the wing must exert a force on the airstream. According to Newton’s third law, the airstream exerts a force F S on the wing that is equal in magnitude and www.aswarphysics.weebly.com 434 Chapter 14 Fluid Mechanics opposite in direction. This force has a vertical component called lift (or aerody-namic lift) and a horizontal component called drag. The lift depends on several factors, such as the speed of the airplane, the area of the wing, the wing’s curva-ture, and the angle between the wing and the horizontal. The curvature of the wing surfaces causes the pressure above the wing to be lower than that below the wing due to the Bernoulli effect. This pressure difference assists with the lift on the wing. As the angle between the wing and the horizontal increases, turbulent flow can set in above the wing to reduce the lift. In general, an object moving through a fluid experiences lift as the result of any effect that causes the fluid to change its direction as it flows past the object. Some factors that influence lift are the shape of the object, its orientation with respect to the fluid flow, any spinning motion it might have, and the texture of its surface. For example, a golf ball struck with a club is given a rapid backspin due to the slant of the club. The dimples on the ball increase the friction force between the ball and the air so that air adheres to the ball’s surface. Figure 14.22 shows air adhering to the ball and being deflected downward as a result. Because the ball pushes the air down, the air must push up on the ball. Without the dimples, the friction force is lower and the golf ball does not travel as far. It may seem counterintuitive to increase the range by increasing the friction force, but the lift gained by spinning the ball more than compensates for the loss of range due to the effect of friction on the translational motion of the ball. For the same reason, a baseball’s cover helps the spinning ball “grab” the air rushing by and helps deflect it when a “curve ball” is thrown. A number of devices operate by means of the pressure differentials that result from differences in a fluid’s speed. For example, a stream of air passing over one end of an open tube, the other end of which is immersed in a liquid, reduces the pressure above the tube as illustrated in Figure 14.23. This reduction in pressure causes the liquid to rise into the airstream. The liquid is then dispersed into a fine spray of droplets. You might recognize that this atomizer is used in perfume bottles and paint sprayers.
F S Drag Lift Figure 14.22 Because of the deflection of air, a spinning golf ball experiences a lifting force that allows it to travel much farther than it would if it were not spinning.
Drag Lift F S The air approaching from the right is deflected downward by the wing.
Figure 14.21 Streamline flow around a moving airplane wing. By Newton’s third law, the air deflected by the wing results in an upward force on the wing from the air: lift. Because of air resis-tance, there is also a force oppo-site the velocity of the wing: drag.
Summary Definitions The pressure P in a fluid is the force per unit area exerted by the fluid on a surface: P ; F A (14.1) In the SI system, pressure has units of newtons per square meter (N/m2), and 1 N/m2 5 1 pascal (Pa).
Figure 14.23 A stream of air pass-ing over a tube dipped into a liquid causes the liquid to rise in the tube.
www.aswarphysics.weebly.com Objective Questions 435 Concepts and Principles The pressure in a fluid at rest varies with depth h in the fluid according to the expression P 5 P0 1 rgh (14.4) where P0 is the pressure at h 5 0 and r is the density of the fluid, assumed uniform. Pascal’s law states that when pressure is applied to an enclosed fluid, the pressure is transmitted undiminished to every point in the fluid and to every point on the walls of the container. When an object is partially or fully sub-merged in a fluid, the fluid exerts on the object an upward force called the buoyant force. According to Archimedes’s prin-ciple, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object: B 5 rfluid gV disp (14.5) The flow rate (volume flux) through a pipe that var-ies in cross-sectional area is constant; that is equivalent to stating that the product of the cross-sectional area A and the speed v at any point is a constant. This result is expressed in the equation of continuity for fluids: A1v1 5 A2v2 5 constant (14.7) The sum of the pressure, kinetic energy per unit volume, and gravitational potential energy per unit vol-ume has the same value at all points along a streamline for an ideal fluid. This result is summarized in Ber-noulli’s equation: P 1 1 2rv 2 1 rgy 5 constant (14.9) of the following statements are valid? (Choose all cor-rect statements.) (a) The buoyant force on the steel object is equal to its weight. (b) The buoyant force on the block is equal to its weight. (c) The tension in the string is equal to the weight of the steel object. (d) The tension in the string is less than the weight of the steel object. (e) The buoyant force on the block is equal to the volume of water it displaces.
Figure OQ14.3 4. An apple is held completely submerged just below the surface of water in a container. The apple is then moved to a deeper point in the water. Compared with the force needed to hold the apple just below the sur-face, what is the force needed to hold it at the deeper point? (a) larger (b) the same (c) smaller (d) impos-sible to determine 5. A beach ball is made of thin plastic. It has been inflated with air, but the plastic is not stretched. By swimming with fins on, you manage to take the ball from the surface of a pool to the bottom. Once the ball is completely submerged, what happens to the buoyant force exerted on the beach ball as you take it deeper? (a) It increases. (b) It remains constant. (c) It decreases. (d) It is impossible to determine. 1. Figure OQ14.1 shows aerial views from directly above two dams. Both dams are equally wide (the vertical dimension in the diagram) and equally high (into the page in the diagram). The dam on the left holds back a very large lake, and the dam on the right holds back a narrow river. Which dam has to be built more strongly? (a) the dam on the left (b) the dam on the right (c) both the same (d) cannot be predicted Dam Dam Figure OQ14.1 2. A beach ball filled with air is pushed about 1 m below the surface of a swimming pool and released from rest. Which of the following statements are valid, assum-ing the size of the ball remains the same? (Choose all correct statements.) (a) As the ball rises in the pool, the buoyant force on it increases. (b) When the ball is released, the buoyant force exceeds the gravitational force, and the ball accelerates upward. (c) The buoyant force on the ball decreases as the ball approaches the surface of the pool. (d) The buoyant force on the ball equals its weight and remains constant as the ball rises. (e) The buoyant force on the ball while it is submerged is approximately equal to the weight of a volume of water that could fill the ball. 3. A wooden block floats in water, and a steel object is attached to the bottom of the block by a string as in Figure OQ14.3. If the block remains floating, which Objective Questions 1. denotes answer available in Student Solutions Manual/Study Guide www.aswarphysics.weebly.com 436 Chapter 14 Fluid Mechanics the weight of the boat (d) equal to the weight of the dis-placed water (e) equal to the buoyant force on the boat 10. A small piece of steel is tied to a block of wood. When the wood is placed in a tub of water with the steel on top, half of the block is submerged. Now the block is inverted so that the steel is under water. (i) Does the amount of the block submerged (a) increase, (b) decrease, or (c) remain the same? (ii) What happens to the water level in the tub when the block is inverted? (a) It rises. (b) It falls. (c) It remains the same. 11. A piece of unpainted porous wood barely floats in an open container partly filled with water. The container is then sealed and pressurized above atmospheric pres-sure. What happens to the wood? (a) It rises in the water. (b) It sinks lower in the water. (c) It remains at the same level. 12. A person in a boat floating in a small pond throws an anchor overboard. What happens to the level of the pond? (a) It rises. (b) It falls. (c) It remains the same. 13. Rank the buoyant forces exerted on the following five objects of equal volume from the largest to the smallest. Assume the objects have been dropped into a swimming pool and allowed to come to mechanical equilibrium. If any buoyant forces are equal, state that in your rank-ing. (a) a block of solid oak (b) an aluminum block (c) a beach ball made of thin plastic and inflated with air (d) an iron block (e) a thin-walled, sealed bottle of water 14. A water supply maintains a constant rate of flow for water in a hose. You want to change the opening of the nozzle so that water leaving the nozzle will reach a height that is four times the current maximum height the water reaches with the nozzle vertical. To do so, should you (a) decrease the area of the opening by a factor of 16, (b) decrease the area by a factor of 8, (c) decrease the area by a factor of 4, (d) decrease the area by a factor of 2, or (e) give up because it cannot be done? 15. A glass of water contains floating ice cubes. When the ice melts, does the water level in the glass (a) go up, (b) go down, or (c) remain the same? 16. An ideal fluid flows through a horizontal pipe whose diameter varies along its length. Measurements would indicate that the sum of the kinetic energy per unit volume and pressure at different sections of the pipe would (a) decrease as the pipe diameter increases, (b) increase as the pipe diameter increases, (c) increase as the pipe diameter decreases, (d) decrease as the pipe diameter decreases, or (e) remain the same as the pipe diameter changes. 6. A solid iron sphere and a solid lead sphere of the same size are each suspended by strings and are sub-merged in a tank of water. (Note that the density of lead is greater than that of iron.) Which of the fol-lowing statements are valid? (Choose all correct state-ments.) (a) The buoyant force on each is the same. (b) The buoyant force on the lead sphere is greater than the buoyant force on the iron sphere because lead has the greater density. (c) The tension in the string supporting the lead sphere is greater than the tension in the string supporting the iron sphere. (d) The buoy-ant force on the iron sphere is greater than the buoy-ant force on the lead sphere because lead displaces more water. (e) None of those statements is true. 7. Three vessels of different shapes are filled to the same level with water as in Figure OQ14.7. The area of the base is the same for all three vessels. Which of the fol-lowing statements are valid? (Choose all correct state-ments.) (a) The pressure at the top surface of vessel A is greatest because it has the largest surface area. (b) The pressure at the bottom of vessel A is greatest because it contains the most water. (c) The pressure at the bottom of each vessel is the same. (d) The force on the bottom of each vessel is not the same. (e) At a given depth below the surface of each vessel, the pressure on the side of vessel A is greatest because of its slope.
A B C Figure OQ14.7 8. One of the predicted problems due to global warm-ing is that ice in the polar ice caps will melt and raise sea levels everywhere in the world. Is that more of a worry for ice (a) at the north pole, where most of the ice floats on water; (b) at the south pole, where most of the ice sits on land; (c) both at the north and south pole equally; or (d) at neither pole? 9. A boat develops a leak and, after its passengers are res-cued, eventually sinks to the bottom of a lake. When the boat is at the bottom, what is the force of the lake bottom on the boat? (a) greater than the weight of the boat (b) equal to the weight of the boat (c) less than Conceptual Questions 1. denotes answer available in Student Solutions Manual/Study Guide 1. When an object is immersed in a liquid at rest, why is the net force on the object in the horizontal direction equal to zero? 2. Two thin-walled drinking glasses having equal base areas but different shapes, with very different cross- sectional areas above the base, are filled to the same level with water. According to the expression P 5 P0 1 rgh, the pressure is the same at the bottom of both glasses. In view of this equality, why does one weigh more than the other? 3. Because atmospheric pressure is about 105 N/m2 and the area of a person’s chest is about 0.13 m2, the force of the www.aswarphysics.weebly.com Conceptual Questions 437 14. Does a ship float higher in the water of an inland lake or in the ocean? Why? 15. When ski jumpers are airborne (Fig. CQ14.15), they bend their bodies forward and keep their hands at their sides. Why?
Figure CQ14.15 © iStockPhoto/technotr 16. Why do airplane pilots prefer to take off with the air-plane facing into the wind? 17. Prairie dogs ventilate their burrows by building a mound around one entrance, which is open to a stream of air when wind blows from any direction. A second entrance at ground level is open to almost stagnant air. How does this construction create an airflow through the burrow? 18. In Figure CQ14.18, an airstream moves from right to left through a tube that is constricted at the middle. Three table-tennis balls are levitated in equilibrium above the vertical columns through which the air escapes. (a) Why is the ball at the right higher than the one in the middle? (b) Why is the ball at the left lower than the ball at the right even though the horizontal tube has the same dimensions at these two points?
Figure CQ14.18 Henry Leap and Jim Lehman 19. A typical silo on a farm has many metal bands wrapped around its perimeter for support as shown in Figure CQ14.19. Why is the spacing between successive bands smaller for the lower portions of the silo on the left, and why are double bands used at lower portions of the silo on the right?
atmosphere on one’s chest is around 13 000 N. In view of this enormous force, why don’t our bodies collapse? 4. A fish rests on the bottom of a bucket of water while the bucket is being weighed on a scale. When the fish begins to swim around, does the scale reading change? Explain your answer. 5. You are a passenger on a spacecraft. For your survival and comfort, the interior contains air just like that at the surface of the Earth. The craft is coasting through a very empty region of space. That is, a nearly perfect vacuum exists just outside the wall. Suddenly, a mete-oroid pokes a hole, about the size of a large coin, right through the wall next to your seat. (a) What happens? (b) Is there anything you can or should do about it? 6. If the airstream from a hair dryer is directed over a table-tennis ball, the ball can be levitated. Explain. 7. A water tower is a common sight in many communities. Figure CQ14.7 shows a collection of colorful water tow-ers in Kuwait City, Kuwait. Notice that the large weight of the water results in the center of mass of the system being high above the ground. Why is it desirable for a water tower to have this highly unstable shape rather than being shaped as a tall cylinder?
Figure CQ14.7 © iStockPhoto/Klaas Lingbeek-van Kranen 8. If you release a ball while inside a freely falling eleva-tor, the ball remains in front of you rather than falling to the floor because the ball, the elevator, and you all experience the same downward gravitational accelera-tion. What happens if you repeat this experiment with a helium-filled balloon? 9. (a) Is the buoyant force a conservative force? (b) Is a potential energy associated with the buoyant force? (c) Explain your answers to parts (a) and (b). 10. An empty metal soap dish barely floats in water. A bar of Ivory soap floats in water. When the soap is stuck in the soap dish, the combination sinks. Explain why. 11. How would you determine the density of an irregularly shaped rock? 12. Place two cans of soft drinks, one regular and one diet, in a container of water. You will find that the diet drink floats while the regular one sinks. Use Archimedes’s principle to devise an explanation. 13. The water supply for a city is often provided from res-ervoirs built on high ground. Water flows from the reservoir, through pipes, and into your home when you turn the tap on your faucet. Why does water flow more rapidly out of a faucet on the first floor of a building than in an apartment on a higher floor?
Figure CQ14.19 Henry Leap and Jim Lehman www.aswarphysics.weebly.com 438 Chapter 14 Fluid Mechanics Note: In all problems, assume the density of air is the 20°C value from Table 14.1, 1.20 kg/m3, unless noted otherwise.
Section 14.1 Pressure 1. A large man sits on a four-legged chair with his feet off the floor. The combined mass of the man and chair is 95.0 kg. If the chair legs are circular and have a radius of 0.500 cm at the bottom, what pressure does each leg exert on the floor? 2. The nucleus of an atom can be modeled as several pro-tons and neutrons closely packed together. Each par-ticle has a mass of 1.67 3 10227 kg and radius on the order of 10215 m. (a) Use this model and the data pro-vided to estimate the density of the nucleus of an atom. (b) Compare your result with the density of a material such as iron. What do your result and comparison sug-gest concerning the structure of matter? 3. A 50.0-kg woman wearing high-heeled shoes is invited into a home in which the kitchen has vinyl floor cover-ing. The heel on each shoe is circular and has a radius of 0.500 cm. (a) If the woman balances on one heel, what pressure does she exert on the floor? (b) Should the home owner be concerned? Explain your answer. 4. Estimate the total mass of the Earth’s atmosphere. (The radius of the Earth is 6.37 3 106 m, and atmo-spheric pressure at the surface is 1.013 3 105 Pa.) 5. Calculate the mass of a solid gold rectangular bar that has dimensions of 4.50 cm 3 11.0 cm 3 26.0 cm.
Section 14.2 Variation of Pressure with Depth 6. (a) A very powerful vacuum cleaner has a hose 2.86 cm in diameter. With the end of the hose placed perpen-dicularly on the flat face of a brick, what is the weight of the heaviest brick that the cleaner can lift? (b) What If? An octopus uses one sucker of diameter 2.86 cm on each of the two shells of a clam in an attempt to pull the shells apart. Find the greatest force the octopus can exert on a clamshell in salt water 32.3 m deep. 7. The spring of the pressure gauge shown in Figure P14.7 has a force constant of 1 250 N/m, and the piston has a diameter of 1.20 cm. As the gauge is lowered into water in a lake, what change in depth causes the piston to move in by 0.750 cm?
Q/C Q/C W M BIO M 8. The small piston of a hydraulic lift (Fig. P14.8) has a cross-sectional area of 3.00 cm2, and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude F1 must be applied to the small piston for the lift to raise a load whose weight is Fg 5 15.0 kN?
F1 S Fg 15.0 kN Figure P14.8 9. What must be the contact area between a suction cup (completely evacuated) and a ceiling if the cup is to support the weight of an 80.0-kg student? 10. A swimming pool has dimensions 30.0 m 3 10.0 m and a flat bottom. When the pool is filled to a depth of 2.00 m with fresh water, what is the force exerted by the water on (a) the bottom? (b) On each end? (c) On each side? 11. (a) Calculate the absolute pressure at the bottom of a freshwater lake at a point whose depth is 27.5 m. Assume the density of the water is 1.00 3 103 kg/m3 and that the air above is at a pressure of 101.3 kPa. (b) What force is exerted by the water on the window of an underwater vehicle at this depth if the window is circular and has a diameter of 35.0 cm? 12. Why is the following situation impossible? Figure P14.12 shows Superman attempting to drink cold water W M AMT Problems The problems found in this chapter may be assigned online in Enhanced WebAssign 1. straightforward; 2. intermediate; 3. challenging 1. full solution available in the Student Solutions Manual/Study Guide AMT Analysis Model tutorial available in Enhanced WebAssign GP Guided Problem M Master It tutorial available in Enhanced WebAssign W Watch It video solution available in Enhanced WebAssign BIO Q/C S Vacuum k F S Figure P14.7 www.aswarphysics.weebly.com Problems 439 17. Review. Piston in Figure P14.17 has a diameter of 0.250 in. Piston has a diameter of 1.50 in. Determine the magnitude F of the force necessary to support the 500-lb load in the absence of friction.
500 lb 2.0 in.
10 in.
F S Figure P14.17 18. Review. A solid sphere of brass (bulk modulus of 14.0 3 1010 N/m2) with a diameter of 3.00 m is thrown into the ocean. By how much does the diameter of the sphere decrease as it sinks to a depth of 1.00 km?
Section 14.3 Pressure Measurements 19. Normal atmospheric pressure is 1.013 3 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 20.0 mm from the normal height. What is the atmospheric pressure? 20. The human brain and spinal cord are immersed in the cerebrospinal fluid. The fluid is normally continuous between the cranial and spinal cavities and exerts a pressure of 100 to 200 mm of H2O above the prevail-ing atmospheric pressure. In medical work, pressures are often measured in units of millimeters of H2O because body fluids, including the cerebrospinal fluid, typically have the same density as water. The pressure of the cerebrospinal fluid can be measured by means of a spinal tap as illustrated in Figure P14.20. A hollow tube is inserted into the spinal column, and the height to which the fluid rises is observed. If the fluid rises to a height of 160 mm, we write its gauge pressure as 160 mm H2O. (a) Express this pressure in pascals, in atmospheres, and in millimeters of mercury. (b) Some conditions that block or inhibit the flow of cerebrospi-nal fluid can be investigated by means of Queckenstedt’s test. In this procedure, the veins in the patient’s neck are compressed to make the blood pressure rise in the brain, which in turn should be transmitted to the cere-brospinal fluid. Explain how the level of fluid in the spinal tap can be used as a diagnostic tool for the con-dition of the patient’s spine.
BIO Q/C through a straw of length , 5 12.0 m. The walls of the tubular straw are very strong and do not collapse. With his great strength, he achieves maximum possible suc-tion and enjoys drinking the cold water. Figure P14.12 13. For the cellar of a new house, a hole is dug in the ground, with vertical sides going down 2.40 m. A con-crete foundation wall is built all the way across the 9.60-m width of the excavation. This foundation wall is 0.183 m away from the front of the cellar hole. Dur-ing a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force the water causes on the founda-tion wall. For comparison, the weight of the water is given by 2.40 m 3 9.60 m 3 0.183 m 3 1 000 kg/m3 3 9.80 m/s2 5 41.3 kN. 14. A container is filled to a depth of 20.0 cm with water. On top of the water floats a 30.0-cm-thick layer of oil with specific gravity 0.700. What is the absolute pres-sure at the bottom of the container? 15. Review. The tank in Figure P14.15 is filled with water of depth d 5 2.00 m. At the bottom of one sidewall is a rectangular hatch of height h 5 1.00 m and width w 5 2.00 m that is hinged at the top of the hatch. (a) Deter-mine the magnitude of the force the water exerts on the hatch. (b) Find the magnitude of the torque exerted by the water about the hinges.
d w h Figure P14.15 Problems 15 and 16. 16. Review. The tank in Figure P14.15 is filled with water of depth d. At the bottom of one sidewall is a rectangular hatch of height h and width w that is hinged at the top of the hatch. (a) Determine the magnitude of the force the water exerts on the hatch. (b) Find the magnitude of the torque exerted by the water about the hinges.
S Figure P14.20 www.aswarphysics.weebly.com 440 Chapter 14 Fluid Mechanics scale and submerged in water, the scale reads 3.50 N (Fig. P14.26). Find the density of the object.
Scale a b Figure P14.26 Problems 26 and 27. 27. A 10.0-kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended from a scale and immersed in water as shown in Figure P14.26b. The 12.0-cm dimen-sion is vertical, and the top of the block is 5.00 cm below the surface of the water. (a) What are the magnitudes of the forces acting on the top and on the bottom of the block due to the surrounding water? (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block. 28. A light balloon is filled with 400 m3 of helium at atmo-spheric pressure. (a) At 0°C, the balloon can lift a pay-load of what mass? (b) What If? In Table 14.1, observe that the density of hydrogen is nearly half the density of helium. What load can the balloon lift if filled with hydrogen? 29. A cube of wood having an edge dimension of 20.0 cm and a density of 650 kg/m3 floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? (b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface? 30. The United States possesses the ten largest warships in the world, aircraft carriers of the Nimitz class. Sup-pose one of the ships bobs up to float 11.0 cm higher in the ocean water when 50 fighters take off from it in a time interval of 25 min, at a location where the free-fall acceleration is 9.78 m/s2. The planes have an aver-age laden mass of 29 000 kg. Find the horizontal area enclosed by the waterline of the ship. 31. A plastic sphere floats in water with 50.0% of its vol-ume submerged. This same sphere floats in glycerin with 40.0% of its volume submerged. Determine the densities of (a) the glycerin and (b) the sphere. 32. A spherical vessel used for deep-sea exploration has a radius of 1.50 m and a mass of 1.20 3 104 kg. To dive, the vessel takes on mass in the form of seawater. Deter-mine the mass the vessel must take on if it is to descend at a constant speed of 1.20 m/s, when the resistive force on it is 1 100 N in the upward direction. The density of seawater is equal to 1.03 3 103 kg/m3. 33. A wooden block of volume 5.24 3 1024 m3 floats in water, and a small steel object of mass m is placed on top of the block. When m 5 0.310 kg, the system is in W M AMT M Q/C 21. Blaise Pascal duplicated Torricelli’s barometer using a red Bordeaux wine, of density 984 kg/m3, as the work-ing liquid (Fig. P14.21). (a) What was the height h of the wine column for normal atmospheric pressure? (b) Would you expect the vacuum above the column to be as good as for mercury?
P0 h Figure P14.21 22. Mercury is poured into a U-tube as shown in Figure P14.22a. The left arm of the tube has cross-sectional area A1 of 10.0 cm2, and the right arm has a cross- sectional area A2 of 5.00 cm2. One hundred grams of water are then poured into the right arm as shown in Figure P14.22b. (a) Determine the length of the water column in the right arm of the U-tube. (b) Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm?
h Mercury A1 A2 A1 A2 Water a b Figure P14.22 23. A backyard swimming pool with a circular base of diameter 6.00 m is filled to depth 1.50 m. (a) Find the absolute pressure at the bottom of the pool. (b) Two persons with combined mass 150 kg enter the pool and float quietly there. No water overflows. Find the pres-sure increase at the bottom of the pool after they enter the pool and float. 24. A tank with a flat bottom of area A and vertical sides is filled to a depth h with water. The pressure is P0 at the top surface. (a) What is the absolute pressure at the bot-tom of the tank? (b) Suppose an object of mass M and density less than the density of water is placed into the tank and floats. No water overflows. What is the result-ing increase in pressure at the bottom of the tank?
Section 14.4 Buoyant Forces and Archimedes’s Principle 25. A table-tennis ball has a diameter of 3.80 cm and aver-age density of 0.084 0 g/cm3. What force is required to hold it completely submerged under water? 26. The gravitational force exerted on a solid object is 5.00 N. When the object is suspended from a spring Q/C W S www.aswarphysics.weebly.com Problems 441 fiduciary marks are to be placed along the rod to indi-cate densities of 0.98 g/cm3, 1.00 g/cm3, 1.02 g/cm3, 1.04 g/cm3, . . . , 1.14 g/cm3. The row of marks is to start 0.200 cm from the top end of the rod and end 1.80 cm from the top end. (a) What is the required length of the rod? (b) What must be its average density? (c) Should the marks be equally spaced? Explain your answer. 38. On October 21, 2001, Ian Ashpole of the United King-dom achieved a record altitude of 3.35 km (11 000 ft) powered by 600 toy balloons filled with helium. Each filled balloon had a radius of about 0.50 m and an esti-mated mass of 0.30 kg. (a) Estimate the total buoyant force on the 600 balloons. (b) Estimate the net upward force on all 600 balloons. (c) Ashpole parachuted to the Earth after the balloons began to burst at the high altitude and the buoyant force decreased. Why did the balloons burst? 39. How many cubic meters of helium are required to lift a light balloon with a 400-kg payload to a height of 8 000 m? Take rHe 5 0.179 kg/m3. Assume the balloon maintains a constant volume and the density of air decreases with the altitude z according to the expres-sion rair 5 r0e2z/8 000, where z is in meters and r0 5 1.20 kg/m3 is the density of air at sea level.
Section 14.5 Fluid Dynamics Section 14.6 Bernoulli’s Equation 40. Water flowing through a garden hose of diameter 2.74 cm fills a 25-L bucket in 1.50 min. (a) What is the speed of the water leaving the end of the hose? (b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle? 41. A large storage tank, open at the top and filled with water, develops a small hole in its side at a point 16.0 m below the water level. The rate of flow from the leak is found to be 2.50 3 1023 m3/min. Determine (a) the speed at which the water leaves the hole and (b) the diameter of the hole. 42. Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in Figure P14.42, the pressure is P1 5 1.75 3 104 Pa and the pipe diameter is 6.00 cm. At another point y 5 0.250 m higher, the pressure is P2 5 1.20 3 104 Pa and the pipe diameter is 3.00 cm. Find the speed of flow (a) in the lower section and (b) in the upper section. (c) Find the volume flow rate through the pipe.
P1 P2 y Figure P14.42 43. Figure P14.43 on page 442 shows a stream of water in steady flow from a kitchen faucet. At the faucet, the Q/C M M equilibrium and the top of the wooden block is at the level of the water. (a) What is the density of the wood? (b) What happens to the block when the steel object is replaced by an object whose mass is less than 0.310 kg? (c) What happens to the block when the steel object is replaced by an object whose mass is greater than 0.310 kg? 34. The weight of a rectangular block of low-density mate-rial is 15.0 N. With a thin string, the center of the hori-zontal bottom face of the block is tied to the bottom of a beaker partly filled with water. When 25.0% of the block’s volume is submerged, the tension in the string is 10.0 N. (a) Find the buoyant force on the block. (b) Oil of density 800 kg/m3 is now steadily added to the bea-ker, forming a layer above the water and surround-ing the block. The oil exerts forces on each of the four sidewalls of the block that the oil touches. What are the directions of these forces? (c) What happens to the string tension as the oil is added? Explain how the oil has this effect on the string tension. (d) The string breaks when its tension reaches 60.0 N. At this moment, 25.0% of the block’s volume is still below the water line. What additional fraction of the block’s vol-ume is below the top surface of the oil? 35. A large weather balloon whose mass is 226 kg is filled with helium gas until its volume is 325 m3. Assume the density of air is 1.20 kg/m3 and the density of helium is 0.179 kg/m3. (a) Calculate the buoyant force acting on the balloon. (b) Find the net force on the balloon and determine whether the balloon will rise or fall after it is released. (c) What additional mass can the balloon support in equilibrium? 36. A hydrometer is an instrument used to determine liquid density. A simple one is sketched in Figure P14.36. The bulb of a syringe is squeezed and released to let the atmosphere lift a sample of the liquid of interest into a tube containing a calibrated rod of known density. The rod, of length L and average density r0, floats partially immersed in the liquid of density r. A length h of the rod protrudes above the surface of the liquid. Show that the density of the liquid is given by r 5 r0L L 2 h 96 98 102 104 100 L h 96 98 100 102 104 Figure P14.36 Problems 36 and 37. 37. Refer to Problem 36 and Figure P14.36. A hydrometer is to be constructed with a cylindrical floating rod. Nine Q/C Q/C S Q/C www.aswarphysics.weebly.com 442 Chapter 14 Fluid Mechanics water must be pumped if it is to arrive at the village? (b) If 4 500 m3 of water is pumped per day, what is the speed of the water in the pipe? Note: Assume the free-fall acceleration and the density of air are con-stant over this range of elevations. The pressures you calculate are too high for an ordinary pipe. The water is actually lifted in stages by several pumps through shorter pipes. 48. In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second hori-zontal tube of radius 0.500 cm at the same elevation as the first tube. The pressure differs by DP between the liquid in one tube and the liquid in the second tube. (a) Find the volume flow rate as a function of DP. Eval-uate the volume flow rate for (b) DP 5 6.00 kPa and (c) DP 5 12.0 kPa. 49. The Venturi tube discussed in Example 14.8 and shown in Figure P14.49 may be used as a fluid flowmeter. Suppose the device is used at a service station to mea-sure the flow rate of gasoline (r 5 7.00 3 102 kg/m3) through a hose having an outlet radius of 1.20 cm. If the difference in pressure is measured to be P1 2 P2 5 1.20 kPa and the radius of the inlet tube to the meter is 2.40 cm, find (a) the speed of the gasoline as it leaves the hose and (b) the fluid flow rate in cubic meters per second.
P1 P2 Figure P14.49 50. Review. Old Faithful Geyser in Yellowstone National Park erupts at approximately one-hour intervals, and the height of the water column reaches 40.0 m (Fig. P14.50). (a) Model the rising stream as a series of separate droplets. Analyze the free-fall motion of Q/C Q/C diameter of the stream is 0.960 cm. The stream fills a 125-cm3 container in 16.3 s. Find the diameter of the stream 13.0 cm below the opening of the faucet.
Figure P14.43 . Cengage Learning/George Semple 44. A village maintains a large tank with an open top, con-taining water for emergencies. The water can drain from the tank through a hose of diameter 6.60 cm. The hose ends with a nozzle of diameter 2.20 cm. A rubber stopper is inserted into the nozzle. The water level in the tank is kept 7.50 m above the nozzle. (a) Calculate the friction force exerted on the stopper by the nozzle. (b) The stopper is removed. What mass of water flows from the nozzle in 2.00 h? (c) Calculate the gauge pres-sure of the flowing water in the hose just behind the nozzle. 45. A legendary Dutch boy saved Holland by plugging a hole of diameter 1.20 cm in a dike with his finger. If the hole was 2.00 m below the surface of the North Sea (density 1 030 kg/m3), (a) what was the force on his fin-ger? (b) If he pulled his finger out of the hole, during what time interval would the released water fill 1 acre of land to a depth of 1 ft? Assume the hole remained constant in size. 46. Water falls over a dam of height h with a mass flow rate of R, in units of kilograms per second. (a) Show that the power available from the water is P 5 Rgh where g is the free-fall acceleration. (b) Each hydro-electric unit at the Grand Coulee Dam takes in water at a rate of 8.50 3 105 kg/s from a height of 87.0 m. The power developed by the falling water is converted to electric power with an efficiency of 85.0%. How much electric power does each hydroelectric unit produce? 47. Water is pumped up from the Colorado River to sup-ply Grand Canyon Village, located on the rim of the canyon. The river is at an elevation of 564 m, and the village is at an elevation of 2 096 m. Imagine that the water is pumped through a single long pipe 15.0 cm in diameter, driven by a single pump at the bottom end. (a) What is the minimum pressure at which the Figure P14.50 Videowokart/Shutterstock.com www.aswarphysics.weebly.com Problems 443 4.00 m 3 1.50 m. Assume the density of the air to be constant at 1.20 kg/m3. The air inside the building is at atmospheric pressure. What is the total force exerted by air on the windowpane? (b) What If? If a second skyscraper is built nearby, the airspeed can be espe-cially high where wind passes through the narrow sepa-ration between the buildings. Solve part (a) again with a wind speed of 22.4 m/s, twice as high. 55. A hypodermic syringe contains a medicine with the density of water (Fig. P14.55). The barrel of the syringe has a cross-sectional area A 5 2.50 3 1025 m2, and the needle has a cross-sectional area a 5 1.00 3 1028 m2. In the absence of a force on the plunger, the pressure everywhere is 1.00 atm. A force F S of magnitude 2.00 N acts on the plunger, making medicine squirt hori-zontally from the needle. Determine the speed of the medicine as it leaves the needle’s tip.
A a F S v S Figure P14.55 Additional Problems 56. Decades ago, it was thought that huge herbivorous dinosaurs such as Apatosaurus and Brachiosaurus habit-ually walked on the bottom of lakes, extending their long necks up to the surface to breathe. Brachiosaurus had its nostrils on the top of its head. In 1977, Knut Schmidt-Nielsen pointed out that breathing would be too much work for such a creature. For a simple model, consider a sample consisting of 10.0 L of air at absolute pressure 2.00 atm, with density 2.40 kg/m3, located at the surface of a freshwater lake. Find the work required to transport it to a depth of 10.3 m, with its tempera-ture, volume, and pressure remaining constant. This energy investment is greater than the energy that can be obtained by metabolism of food with the oxygen in that quantity of air. 57. (a) Calculate the absolute pressure at an ocean depth of 1 000 m. Assume the density of seawater is 1 030 kg/m3 and the air above exerts a pressure of 101.3 kPa. (b) At this depth, what is the buoyant force on a spherical submarine having a diameter of 5.00 m? 58. In about 1657, Otto von Guericke, inventor of the air pump, evacuated a sphere made of two brass hemi-spheres (Fig. P14.58). Two teams of eight horses each could pull the hemispheres apart only on some trials and then “with greatest difficulty,” with the resulting M BIO BIO W S one of the droplets to determine the speed at which the water leaves the ground. (b) What If? Model the rising stream as an ideal fluid in streamline flow. Use Bernoulli’s equation to determine the speed of the water as it leaves ground level. (c) How does the answer from part (a) compare with the answer from part (b)? (d) What is the pressure (above atmospheric) in the heated underground chamber if its depth is 175 m? Assume the chamber is large compared with the geyser’s vent.
Section 14.7 Other Applications of Fluid Dynamics 51. An airplane is cruising at altitude 10 km. The pressure outside the craft is 0.287 atm; within the passenger compartment, the pressure is 1.00 atm and the temper-ature is 208C. A small leak occurs in one of the window seals in the passenger compartment. Model the air as an ideal fluid to estimate the speed of the airstream flowing through the leak. 52. An airplane has a mass of 1.60 3 104 kg, and each wing has an area of 40.0 m2. During level flight, the pressure on the lower wing surface is 7.00 3 104 Pa. (a) Suppose the lift on the airplane were due to a pressure differ-ence alone. Determine the pressure on the upper wing surface. (b) More realistically, a significant part of the lift is due to deflection of air downward by the wing. Does the inclusion of this force mean that the pressure in part (a) is higher or lower? Explain. 53. A siphon is used to drain water from a tank as illus-trated in Figure P14.53. Assume steady flow without friction. (a) If h 5 1.00 m, find the speed of outflow at the end of the siphon. (b) What If? What is the limita-tion on the height of the top of the siphon above the end of the siphon? Note: For the flow of the liquid to be continuous, its pressure must not drop below its vapor pressure. Assume the water is at 20.08C, at which the vapor pressure is 2.3 kPa.
h y v S Figure P14.53 54. The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pres-sure in the still air inside the buildings can cause win-dows to pop out. As originally constructed, the John Hancock Building in Boston popped windowpanes that fell many stories to the sidewalk below. (a) Sup-pose a horizontal wind blows with a speed of 11.2 m/s outside a large pane of plate glass with dimensions Q/C P0 P R F S F S Figure P14.58 www.aswarphysics.weebly.com 444 Chapter 14 Fluid Mechanics balance with the use of counterweights of density r. Representing the density of air as rair and the balance reading as F 9 g, show that the true weight Fg is Fg 5 F r g 1 aV 2 F r g rgbrairg 63. Water is forced out of a fire extinguisher by air pres-sure as shown in Figure P14.63. How much gauge air pressure in the tank is required for the water jet to have a speed of 30.0 m/s when the water level is 0.500 m below the nozzle?
0.500 m v S Figure P14.63 64. Review. Assume a certain liquid, with density 1 230 kg/m3, exerts no friction force on spherical objects. A ball of mass 2.10 kg and radius 9.00 cm is dropped from rest into a deep tank of this liquid from a height of 3.30 m above the surface. (a) Find the speed at which the ball enters the liquid. (b) Evaluate the magni-tudes of the two forces that are exerted on the ball as it moves through the liquid. (c) Explain why the ball moves down only a limited distance into the liquid and calculate this distance. (d) With what speed will the ball pop up out of the liquid? (e) How does the time interval Dtdown, during which the ball moves from the surface down to its lowest point, compare with the time interval Dtup for the return trip between the same two points? (f) What If? Now modify the model to suppose the liq-uid exerts a small friction force on the ball, opposite in direction to its motion. In this case, how do the time intervals Dtdown and Dtup compare? Explain your answer with a conceptual argument rather than a numerical calculation. 65. Review. A light spring of constant k 5 90.0 N/m is attached vertically to a table (Fig. P14.65a). A 2.00-g balloon is filled with helium (density 5 0.179 kg/m3) Q/C AMT sound likened to a cannon firing. Find the force F required to pull the thin-walled evacuated hemispheres apart in terms of R, the radius of the hemispheres; P, the pressure inside the hemispheres; and atmospheric pressure P0. 59. A spherical aluminum ball of mass 1.26 kg contains an empty spherical cavity that is concentric with the ball. The ball barely floats in water. Calculate (a) the outer radius of the ball and (b) the radius of the cavity. 60. A helium-filled balloon (whose envelope has a mass of mb 5 0.250 kg) is tied to a uniform string of length , 5 2.00 m and mass m 5 0.050 0 kg. The balloon is spheri-cal with a radius of r 5 0.400 m. When released in air of temperature 208C and density rair 5 1.20 kg/m3, it lifts a length h of string and then remains stationary as shown in Figure P14.60. We wish to find the length of string lifted by the balloon. (a) When the balloon remains stationary, what is the appropriate analysis model to describe it? (b) Write a force equation for the balloon from this model in terms of the buoyant force B, the weight Fb of the balloon, the weight FHe of the helium, and the weight Fs of the segment of string of length h. (c) Make an appropriate substitution for each of these forces and solve symbolically for the mass ms of the segment of string of length h in terms of mb, r, rair, and the density of helium rHe. (d) Find the numerical value of the mass ms. (e) Find the length h numerically.
He h Figure P14.60 61. Review. Figure P14.61 shows a valve separating a res-ervoir from a water tank. If this valve is opened, what is the maximum height above point B attained by the water stream coming out of the right side of the tank? Assume h 5 10.0 m, L 5 2.00 m, and u 5 30.0°, and assume the cross-sectional area at A is very large com-pared with that at B.
A h Valve L B u Figure P14.61 62. The true weight of an object can be measured in a vacuum, where buoyant forces are absent. A measure-ment in air, however, is disturbed by buoyant forces. An object of volume V is weighed in air on an equal-arm GP AMT S k k L a b Figure P14.65 www.aswarphysics.weebly.com Problems 445 70. Review. With reference to the dam studied in Example 14.4 and shown in Figure 14.5, (a) show that the total torque exerted by the water behind the dam about a horizontal axis through O is 1 6rgwH 3. (b) Show that the effective line of action of the total force exerted by the water is at a distance 1 3H above O. 71. A 1.00-kg beaker containing 2.00 kg of oil (density 5 916.0 kg/m3) rests on a scale. A 2.00-kg block of iron suspended from a spring scale is completely submerged in the oil as shown in Figure P14.71. Determine the equilibrium readings of both scales.
Figure P14.71 Problems 71 and 72. 72. A beaker of mass mb containing oil of mass mo and den-sity ro rests on a scale. A block of iron of mass mFe sus-pended from a spring scale is completely submerged in the oil as shown in Figure P14.71. Determine the equi-librium readings of both scales. 73. In 1983, the United States began coining the one-cent piece out of copper-clad zinc rather than pure cop-per. The mass of the old copper penny is 3.083 g and that of the new cent is 2.517 g. The density of copper is 8.920 g/cm3 and that of zinc is 7.133 g/cm3. The new and old coins have the same volume. Calculate the percent of zinc (by volume) in the new cent. 74. Review. A long, cylindrical rod of radius r is weighted on one end so that it floats upright in a fluid having a density r. It is pushed down a distance x from its equi-librium position and released. Show that the rod will execute simple harmonic motion if the resistive effects of the fluid are negligible, and determine the period of the oscillations. 75. Review. Figure P14.75 shows the essential parts of a hydraulic brake system. The area of the piston in the master cylinder is 1.8 cm2 and that of the piston S S S to a volume of 5.00 m3 and is then connected with a light cord to the spring, causing the spring to stretch as shown in Figure P14.65b. Determine the extension distance L when the balloon is in equilibrium. 66. To an order of magnitude, how many helium-filled toy balloons would be required to lift you? Because helium is an irreplaceable resource, develop a theoretical answer rather than an experimental answer. In your solution, state what physical quantities you take as data and the values you measure or estimate for them. 67. A 42.0-kg boy uses a solid block of Styrofoam as a raft while fishing on a pond. The Styrofoam has an area of 1.00 m2 and is 0.050 0 m thick. While sitting on the surface of the raft, the boy finds that the raft just sup-ports him so that the top of the raft is at the level of the pond. Determine the density of the Styrofoam. 68. A common parameter that can be used to predict tur-bulence in fluid flow is called the Reynolds number. The Reynolds number for fluid flow in a pipe is a dimen-sionless quantity defined as Re 5 rvd m where r is the density of the fluid, v is its speed, d is the inner diameter of the pipe, and m is the viscosity of the fluid. Viscosity is a measure of the internal resistance of a liquid to flow and has units of Pa · s. The criteria for the type of flow are as follows: • If Re , 2 300, the flow is laminar. • If 2 300 , Re , 4 000, the flow is in a transition region between laminar and turbulent. • If Re . 4 000, the flow is turbulent. (a) Let’s model blood of density 1.06 3 103 kg/m3 and viscosity 3.00 3 10–3 Pa · s as a pure liquid, that is, ignore the fact that it contains red blood cells. Sup-pose it is flowing in a large artery of radius 1.50 cm with a speed of 0.067 0 m/s. Show that the flow is lami-nar. (b) Imagine that the artery ends in a single capil-lary so that the radius of the artery reduces to a much smaller value. What is the radius of the capillary that would cause the flow to become turbulent? (c) Actual capillaries have radii of about 5–10 micrometers, much smaller than the value in part (b). Why doesn’t the flow in actual capillaries become turbulent? 69. Evangelista Torricelli was the first person to realize that we live at the bottom of an ocean of air. He cor-rectly surmised that the pressure of our atmosphere is attributable to the weight of the air. The density of air at 08C at the Earth’s surface is 1.29 kg/m3. The den-sity decreases with increasing altitude (as the atmo-sphere thins). On the other hand, if we assume the density is constant at 1.29 kg/m3 up to some altitude h and is zero above that altitude, then h would repre-sent the depth of the ocean of air. (a) Use this model to determine the value of h that gives a pressure of 1.00 atm at the surface of the Earth. (b) Would the peak of Mount Everest rise above the surface of such an atmosphere?
BIO Q/C Wheel drum Shoe Brake cylinder Master cylinder Pedal Figure P14.75 www.aswarphysics.weebly.com 446 Chapter 14 Fluid Mechanics travel from the nozzle to the ground. Neglect air resis-tance and assume atmospheric pressure is 1.00 atm. (b) If the desired range of the stream is 8.00 m, with what speed v2 must the stream leave the nozzle? (c) At what speed v1 must the plunger be moved to achieve the desired range? (d) What is the pressure at the nozzle? (e) Find the pressure needed in the larger tube. (f) Calculate the force that must be exerted on the trigger to achieve the desired range. (The force that must be exerted is due to pressure over and above atmospheric pressure.) F S v1 S v2 S A2 A1 Figure P14.78 79. An incompressible, nonviscous fluid is initially at rest in the vertical portion of the pipe shown in Figure P14.79a, where L 5 2.00 m. When the valve is opened, the fluid flows into the horizontal section of the pipe. What is the fluid’s speed when all the fluid is in the horizontal section as shown in Figure P14.79b? Assume the cross-sectional area of the entire pipe is constant.
Valve closed Valve opened L L v S a b Figure P14.79 80. The water supply of a building is fed through a main pipe 6.00 cm in diameter. A 2.00-cm-diameter faucet tap, located 2.00 m above the main pipe, is observed to fill a 25.0-L container in 30.0 s. (a) What is the speed at which the water leaves the faucet? (b) What is the gauge pressure in the 6-cm main pipe? Assume the faucet is the only “leak” in the building. 81. A U-tube open at both ends is partially filled with water (Fig. P14.81a). Oil having a density 750 kg/m3 is then poured into the right arm and forms a column L 5 5.00 cm high (Fig. P14.81b). (a) Determine the difference h in the heights of the two liquid surfaces. (b) The right arm is then shielded from any air motion while air is blown across the top of the left arm until the surfaces of the two liquids are at the same height (Fig. P14.81c). Determine the speed of the air being in the brake cylinder is 6.4 cm2. The coefficient of fric-tion between shoe and wheel drum is 0.50. If the wheel has a radius of 34 cm, determine the frictional torque about the axle when a force of 44 N is exerted on the brake pedal. 76. The spirit-in-glass thermometer, invented in Florence, Italy, around 1654, consists of a tube of liquid (the spirit) containing a number of submerged glass spheres with slightly different masses (Fig. P14.76). At sufficiently low temperatures, all the spheres float, but as the temperature rises, the spheres sink one after another. The device is a crude but interesting tool for measuring temperature. Suppose the tube is filled with ethyl alcohol, whose density is 0.789 45 g/cm3 at 20.0°C and decreases to 0.780 97 g/cm3 at 30.0°C. (a) Assuming that one of the spheres has a radius of 1.000 cm and is in equilibrium halfway up the tube at 20.0°C, determine its mass. (b) When the temperature increases to 30.0°C, what mass must a second sphere of the same radius have to be in equi-librium at the halfway point? (c) At 30.0°C, the first sphere has fallen to the bottom of the tube. What upward force does the bottom of the tube exert on this sphere?
Figure P14.76 . iStockphoto.com/Alberto Pomares Photography 77. Review. A uniform disk of mass 10.0 kg and radius 0.250 m spins at 300 rev/min on a low-friction axle. It must be brought to a stop in 1.00 min by a brake pad that makes contact with the disk at an average distance 0.220 m from the axis. The coefficient of friction between pad and disk is 0.500. A piston in a cylinder of diameter 5.00 cm presses the brake pad against the disk. Find the pressure required for the brake fluid in the cylinder. 78. Review. In a water pistol, a piston drives water through a large tube of area A1 into a smaller tube of area A2 as shown in Figure P14.78. The radius of the large tube is 1.00 cm and that of the small tube is 1.00 mm. The smaller tube is 3.00 cm above the larger tube. (a) If the pistol is fired horizontally at a height of 1.50 m, determine the time interval required for the water to Q/C www.aswarphysics.weebly.com Problems 447 (b) The boat has mass M. Show that the liftoff speed is given by v < Å 2Mg 1n2 2 12Ar 84. A jet of water squirts out horizontally from a hole near the bottom of the tank shown in Figure P14.84. If the hole has a diameter of 3.50 mm, what is the height h of the water level in the tank?
h 0.600 m 1.00 m Figure P14.84 Challenge Problems 85. An ice cube whose edges measure 20.0 mm is float-ing in a glass of ice-cold water, and one of the ice cube’s faces is parallel to the water’s surface. (a) How far below the water surface is the bottom face of the block? (b) Ice-cold ethyl alcohol is gently poured onto the water surface to form a layer 5.00 mm thick above the water. The alcohol does not mix with the water. When the ice cube again attains hydrostatic equilib-rium, what is the distance from the top of the water to the bottom face of the block? (c) Additional cold ethyl alcohol is poured onto the water’s surface until the top surface of the alcohol coincides with the top surface of the ice cube (in hydrostatic equilibrium). How thick is the required layer of ethyl alcohol? 86. Why is the following situation impossible? A barge is car-rying a load of small pieces of iron along a river. The iron pile is in the shape of a cone for which the radius r of the base of the cone is equal to the central height h of the cone. The barge is square in shape, with vertical sides of length 2r, so that the pile of iron comes just up to the edges of the barge. The barge approaches a low bridge, and the captain realizes that the top of the pile of iron is not going to make it under the bridge. The captain orders the crew to shovel iron pieces from the pile into the water to reduce the height of the pile. As iron is shoveled from the pile, the pile always has the shape of a cone whose diameter is equal to the side length of the barge. After a certain volume of iron is removed from the barge, it makes it under the bridge without the top of the pile striking the bridge. 87. Show that the variation of atmospheric pressure with altitude is given by P 5 P0e2ay, where a 5 r0g/P0, P0 M S blown across the left arm. Take the density of air as constant at 1.20 kg/m3.
P0 Water h L Oil L Shield v S a b c Figure P14.81 82. A woman is draining her fish tank by siphoning the water into an outdoor drain as shown in Figure P14.82. The rectangular tank has footprint area A and depth h. The drain is located a distance d below the surface of the water in the tank, where d .. h. The cross- sectional area of the siphon tube is A9. Model the water as flowing without friction. Show that the time interval required to empty the tank is given by Dt 5 Ah Ar"2gd d h Figure P14.82 83. The hull of an experimental boat is to be lifted above the water by a hydrofoil mounted below its keel as shown in Figure P14.83. The hydrofoil has a shape like that of an airplane wing. Its area projected onto a horizontal surface is A. When the boat is towed at suf-ficiently high speed, water of density r moves in stream-line flow so that its average speed at the top of the hydrofoil is n times larger than its speed vb below the hydrofoil. (a) Ignoring the buoyant force, show that the upward lift force exerted by the water on the hydro-foil has a magnitude F < 1 21n2 2 12rvb 2A S S M Figure P14.83 www.aswarphysics.weebly.com 448 Chapter 14 Fluid Mechanics expressed from Equation 14.4 as dP 5 2rg dy. Also assume the density of air is proportional to the pres-sure, which, as we will see in Chapter 20, is equivalent to assuming the temperature of the air is the same at all altitudes.
is atmospheric pressure at some reference level y 5 0, and r0 is the atmospheric density at this level. Assume the decrease in atmospheric pressure over an infinites-imal change in altitude (so that the density is approxi-mately uniform over the infinitesimal change) can be www.aswarphysics.weebly.com 449 p a r t 2 Falling drops of water cause a water surface to oscillate. These oscillations are associated with circular waves moving away from the point at which the drops fall. In Part 2 of the text, we will explore the principles related to oscillations and waves. (Marga Buschbell Steeger/Photographer’s Choice/ Getty Images) Oscillations and Mechanical Waves We begin this new part of the text by studying a special type of motion called periodic motion, the repeating motion of an object in which it continues to return to a given position after a fixed time interval. The repetitive movements of such an object are called oscillations. We will focus our attention on a special case of periodic motion called simple harmonic motion. All periodic motions can be modeled as combinations of simple harmonic motions. Simple harmonic motion also forms the basis for our understanding of mechanical waves. Sound waves, seismic waves, waves on stretched strings, and water waves are all produced by some source of oscillation. As a sound wave travels through the air, elements of the air oscillate back and forth; as a water wave travels across a pond, elements of the water oscillate up and down and backward and forward. The motion of the elements of the medium bears a strong resemblance to the periodic motion of an oscillating pendulum or an object attached to a spring. To explain many other phenomena in nature, we must understand the concepts of oscillations and waves. For instance, although skyscrapers and bridges appear to be rigid, they actually oscil-late, something the architects and engineers who design and build them must take into account. To understand how radio and television work, we must understand the origin and nature of elec-tromagnetic waves and how they propagate through space. Finally, much of what scientists have learned about atomic structure has come from information carried by waves. Therefore, we must first study oscillations and waves if we are to understand the concepts and theories of atomic physics. ■ www.aswarphysics.weebly.com The London Millennium Bridge over the River Thames in London. On opening day of the bridge, pedestrians noticed a swinging motion of the bridge, leading to its being named the “Wobbly Bridge.” The bridge was closed after two days and remained closed for two years. Over 50 tuned mass dampers were added to the bridge: the pairs of spring-loaded structures on top of the cross members (arrow). We will study both oscillations and damping of oscillations in this chapter. (Monkey Business Images/ Shutterstock.com) 15.1 Motion of an Object Attached to a Spring 15.2 Analysis Model: Particle in Simple Harmonic Motion 15.3 Energy of the Simple Harmonic Oscillator 15.4 Comparing Simple Harmonic Motion with Uniform Circular Motion 15.5 The Pendulum 15.6 Damped Oscillations 15.7 Forced Oscillations Oscillatory Motion 450 Periodic motion is motion of an object that regularly returns to a given position after a fixed time interval. With a little thought, we can identify several types of periodic motion in everyday life. Your car returns to the driveway each afternoon. You return to the dinner table each night to eat. A bumped chandelier swings back and forth, returning to the same position at a regular rate. The Earth returns to the same position in its orbit around the Sun each year, resulting in the variation among the four seasons. A special kind of periodic motion occurs in mechanical systems when the force acting on an object is proportional to the position of the object relative to some equilibrium position. If this force is always directed toward the equilibrium position, the motion is called simple harmonic motion, which is the primary focus of this chapter.
15.1 Motion of an Object Attached to a Spring As a model for simple harmonic motion, consider a block of mass m attached to the end of a spring, with the block free to move on a frictionless, horizontal surface c h a p t e r 15 PA www.aswarphysics.weebly.com 15.1 Motion of an Object Attached to a Spring 451 (Fig. 15.1). When the spring is neither stretched nor compressed, the block is at rest at the position called the equilibrium position of the system, which we identify as x 5 0 (Fig. 15.1b). We know from experience that such a system oscillates back and forth if disturbed from its equilibrium position. We can understand the oscillating motion of the block in Figure 15.1 qualita-tively by first recalling that when the block is displaced to a position x, the spring exerts on the block a force that is proportional to the position and given by Hooke’s law (see Section 7.4): Fs 5 2kx (15.1) We call Fs a restoring force because it is always directed toward the equilibrium position and therefore opposite the displacement of the block from equilibrium. That is, when the block is displaced to the right of x 5 0 in Figure 15.1a, the posi-tion is positive and the restoring force is directed to the left. When the block is displaced to the left of x 5 0 as in Figure 15.1c, the position is negative and the restoring force is directed to the right. When the block is displaced from the equilibrium point and released, it is a particle under a net force and consequently undergoes an acceleration. Applying the particle under a net force model to the motion of the block, with Equation 15.1 providing the net force in the x direction, we obtain o Fx 5 max S 2kx 5 max ax 5 2 k m x (15.2) That is, the acceleration of the block is proportional to its position, and the direc-tion of the acceleration is opposite the direction of the displacement of the block from equilibrium. Systems that behave in this way are said to exhibit simple har-monic motion. An object moves with simple harmonic motion whenever its accel-eration is proportional to its position and is oppositely directed to the displacement from equilibrium. If the block in Figure 15.1 is displaced to a position x 5 A and released from rest, its initial acceleration is 2kA/m. When the block passes through the equilib-rium position x 5 0, its acceleration is zero. At this instant, its speed is a maxi-mum because the acceleration changes sign. The block then continues to travel to the left of equilibrium with a positive acceleration and finally reaches x 5 2A, at which time its acceleration is 1kA/m and its speed is again zero as discussed in Sections 7.4 and 7.9. The block completes a full cycle of its motion by returning to the original position, again passing through x 5 0 with maximum speed. There-fore, the block oscillates between the turning points x 5 6A. In the absence of W W Hooke’s law Figure 15.1 A block attached to a spring moving on a friction-less surface.
m x x x x x m m x 0 x 0 x 0 Fs S Fs 0 S Fs S b c a When the block is displaced to the right of equilibrium, the force exerted by the spring acts to the left.
When the block is at its equilibrium position, the force exerted by the spring is zero.
When the block is displaced to the left of equilibrium, the force exerted by the spring acts to the right.
Pitfall Prevention 15.1 The Orientation of the Spring Fig-ure 15.1 shows a horizontal spring, with an attached block sliding on a frictionless surface. Another possibility is a block hanging from a vertical spring. All the results we discuss for the horizontal spring are the same for the vertical spring with one exception: when the block is placed on the vertical spring, its weight causes the spring to extend. If the resting position of the block is defined as x 5 0, the results of this chapter also apply to this vertical system.
www.aswarphysics.weebly.com 452 Chapter 15 Oscillatory Motion friction, this idealized motion will continue forever because the force exerted by the spring is conservative. Real systems are generally subject to friction, so they do not oscillate forever. We shall explore the details of the situation with friction in Section 15.6.
Q uick Quiz 15.1 A block on the end of a spring is pulled to position x 5 A and released from rest. In one full cycle of its motion, through what total distance does it travel? (a) A/2 (b) A (c) 2A (d) 4A 15.2 Analysis Model: Particle in Simple Harmonic Motion The motion described in the preceding section occurs so often that we identify the particle in simple harmonic motion model to represent such situations. To develop a mathematical representation for this model, we will generally choose x as the axis along which the oscillation occurs; hence, we will drop the subscript-x notation in this discussion. Recall that, by definition, a 5 dv/dt 5 d 2x/dt 2, so we can express Equation 15.2 as d 2x dt 2 5 2 k m x (15.3) If we denote the ratio k/m with the symbol v2 (we choose v2 rather than v so as to make the solution we develop below simpler in form), then v2 5 k m (15.4) and Equation 15.3 can be written in the form d 2x dt 2 5 2v2x (15.5) Let’s now find a mathematical solution to Equation 15.5, that is, a function x(t) that satisfies this second-order differential equation and is a mathematical repre-sentation of the position of the particle as a function of time. We seek a function whose second derivative is the same as the original function with a negative sign and multiplied by v2. The trigonometric functions sine and cosine exhibit this behavior, so we can build a solution around one or both of them. The following cosine function is a solution to the differential equation: x 1t 2 5 A cos 1vt 1 f2 (15.6) where A, v, and f are constants. To show explicitly that this solution satisfies Equa-tion 15.5, notice that dx dt 5 A d dt cos 1vt 1 f2 5 2vA sin 1vt 1 f2 (15.7) d 2x dt 2 5 2vA d dt sin 1vt 1 f2 5 2v2A cos 1vt 1 f2 (15.8) Comparing Equations 15.6 and 15.8, we see that d 2x/dt 2 5 2v2x and Equation 15.5 is satisfied. The parameters A, v, and f are constants of the motion. To give physical signifi-cance to these constants, it is convenient to form a graphical representation of the motion by plotting x as a function of t as in Figure 15.2a. First, A, called the ampli-tude of the motion, is simply the maximum value of the position of the particle in Position versus time for a particle in simple harmonic motion Pitfall Prevention 15.2 A Nonconstant Acceleration The acceleration of a particle in simple harmonic motion is not constant. Equation 15.3 shows that its accel-eration varies with position x. Therefore, we cannot apply the kinematic equations of Chapter 2 in this situation.
Pitfall Prevention 15.3 Where’s the Triangle? Equation 15.6 includes a trigonometric function, a mathematical function that can be used whether it refers to a triangle or not. In this case, the cosine function happens to have the correct behavior for representing the position of a par-ticle in simple harmonic motion.
www.aswarphysics.weebly.com 15.2 Analysis Model: Particle in Simple Harmonic Motion 453 either the positive or negative x direction. The constant v is called the angular fre-quency, and it has units1 of radians per second. It is a measure of how rapidly the oscillations are occurring; the more oscillations per unit time, the higher the value of v. From Equation 15.4, the angular frequency is v 5 Å k m (15.9) The constant angle f is called the phase constant (or initial phase angle) and, along with the amplitude A, is determined uniquely by the position and velocity of the particle at t 5 0. If the particle is at its maximum position x 5 A at t 5 0, the phase constant is f 5 0 and the graphical representation of the motion is as shown in Figure 15.2b. The quantity (vt 1 f) is called the phase of the motion. Notice that the function x(t) is periodic and its value is the same each time vt increases by 2p radians. Equations 15.1, 15.5, and 15.6 form the basis of the mathematical representation of the particle in simple harmonic motion model. If you are analyzing a situation and find that the force on an object modeled as a particle is of the mathematical form of Equation 15.1, you know the motion is that of a simple harmonic oscillator and the position of the particle is described by Equation 15.6. If you analyze a sys-tem and find that it is described by a differential equation of the form of Equation 15.5, the motion is that of a simple harmonic oscillator. If you analyze a situation and find that the position of a particle is described by Equation 15.6, you know the particle undergoes simple harmonic motion.
Q uick Quiz 15.2 Consider a graphical representation (Fig. 15.3) of simple har-monic motion as described mathematically in Equation 15.6. When the particle is at point A on the graph, what can you say about its position and velocity? (a) The position and velocity are both positive. (b) The position and velocity are both negative. (c) The position is positive, and the velocity is zero. (d) The position is negative, and the velocity is zero. (e) The position is positive, and the velocity is negative. (f) The position is negative, and the velocity is positive.
Q uick Quiz 15.3 Figure 15.4 shows two curves representing particles under going simple harmonic motion. The correct description of these two motions is that the simple harmonic motion of particle B is (a) of larger angular frequency and larger amplitude than that of particle A, (b) of larger angular frequency and smaller amplitude than that of particle A, (c) of smaller angu-lar frequency and larger amplitude than that of particle A, or (d) of smaller angular frequency and smaller amplitude than that of particle A.
1We have seen many examples in earlier chapters in which we evaluate a trigonometric function of an angle. The argument of a trigonometric function, such as sine or cosine, must be a pure number. The radian is a pure number because it is a ratio of lengths. Angles in degrees are pure numbers because the degree is an artificial “unit”; it is not related to measurements of lengths. The argument of the trigonometric function in Equation 15.6 must be a pure number. Therefore, v must be expressed in radians per second (and not, for example, in revolutions per second) if t is expressed in seconds. Furthermore, other types of functions such as logarithms and exponential functions require arguments that are pure numbers.
Figure 15.2 (a) An x–t graph for a particle undergoing simple harmonic motion. The amplitude of the motion is A, and the period (defined in Eq. 15.10) is T. (b) The x–t graph for the special case in which x 5 A at t 5 0 and hence f 5 0.
x A –A t x A –A t T a b t x A Figure 15.3 (Quick Quiz 15.2) An x–t graph for a particle under-going simple harmonic motion. At a particular time, the particle’s position is indicated by A in the graph.
t x t x Particle A Particle B Figure 15.4 (Quick Quiz 15.3) Two x–t graphs for particles under-going simple harmonic motion. The amplitudes and frequencies are different for the two particles. Let us investigate further the mathematical description of simple harmonic motion. The period T of the motion is the time interval required for the particle to go through one full cycle of its motion (Fig. 15.2a). That is, the values of x and v for the particle at time t equal the values of x and v at time t 1 T. Because the phase increases by 2p radians in a time interval of T, [v(t 1 T) 1 f] 2 (vt 1 f) 5 2p www.aswarphysics.weebly.com 454 Chapter 15 Oscillatory Motion Simplifying this expression gives vT 5 2p, or T 5 2p v (15.10) The inverse of the period is called the frequency f of the motion. Whereas the period is the time interval per oscillation, the frequency represents the number of oscillations the particle undergoes per unit time interval: f 5 1 T 5 v 2p (15.11) The units of f are cycles per second, or hertz (Hz). Rearranging Equation 15.11 gives v 5 2pf 5 2p T (15.12) Equations 15.9 through 15.11 can be used to express the period and frequency of the motion for the particle in simple harmonic motion in terms of the character-istics m and k of the system as T 5 2p v 5 2pÅ m k (15.13) f 5 1 T 5 1 2pÅ k m (15.14) That is, the period and frequency depend only on the mass of the particle and the force constant of the spring and not on the parameters of the motion, such as A or f. As we might expect, the frequency is larger for a stiffer spring (larger value of k) and decreases with increasing mass of the particle. We can obtain the velocity and acceleration2 of a particle undergoing simple harmonic motion from Equations 15.7 and 15.8: v 5 dx dt 5 2vA sin 1vt 1 f2 (15.15) a 5 d 2x dt 2 5 2v2A cos 1vt 1 f2 (15.16) From Equation 15.15, we see that because the sine and cosine functions oscillate between 61, the extreme values of the velocity v are 6vA. Likewise, Equation 15.16 shows that the extreme values of the acceleration a are 6v2A. Therefore, the maxi-mum values of the magnitudes of the velocity and acceleration are vmax 5 vA 5 Å k m A (15.17) a max 5 v2A 5 k m A (15.18) Figure 15.5a plots position versus time for an arbitrary value of the phase con-stant. The associated velocity–time and acceleration–time curves are illustrated in Figures 15.5b and 15.5c, respectively. They show that the phase of the velocity dif-fers from the phase of the position by p/2 rad, or 908. That is, when x is a maxi-mum or a minimum, the velocity is zero. Likewise, when x is zero, the speed is a maximum. Furthermore, notice that the phase of the acceleration differs from the phase of the position by p radians, or 1808. For example, when x is a maximum, a has a maximum magnitude in the opposite direction.
Period Frequency Velocity of a particle in simple harmonic motion Acceleration of a particle in simple harmonic motion Maximum magnitudes of velocity and acceleration in simple harmonic motion Pitfall Prevention 15.4 Two Kinds of Frequency We iden-tify two kinds of frequency for a simple harmonic oscillator: f, called simply the frequency, is mea-sured in hertz, and v, the angular frequency, is measured in radians per second. Be sure you are clear about which frequency is being discussed or requested in a given problem. Equations 15.11 and 15.12 show the relationship between the two frequencies.
2Because the motion of a simple harmonic oscillator takes place in one dimension, we denote velocity as v and accel-eration as a, with the direction indicated by a positive or negative sign as in Chapter 2.
www.aswarphysics.weebly.com 15.2 Analysis Model: Particle in Simple Harmonic Motion 455 Q uick Quiz 15.4 An object of mass m is hung from a spring and set into oscilla-tion. The period of the oscillation is measured and recorded as T. The object of mass m is removed and replaced with an object of mass 2m. When this object is set into oscillation, what is the period of the motion? (a) 2T (b) !2 T (c) T (d) T/!2 (e) T/2 Equation 15.6 describes simple harmonic motion of a particle in general. Let’s now see how to evaluate the constants of the motion. The angular frequency v is evaluated using Equation 15.9. The constants A and f are evaluated from the ini-tial conditions, that is, the state of the oscillator at t 5 0. Suppose a block is set into motion by pulling it from equilibrium by a distance A and releasing it from rest at t 5 0 as in Figure 15.6. We must then require our solu-tions for x(t) and v(t) (Eqs. 15.6 and 15.15) to obey the initial conditions that x(0) 5 A and v(0) 5 0: x(0) 5 A cos f 5 A v(0) 5 2vA sin f 5 0 These conditions are met if f 5 0, giving x 5 A cos vt as our solution. To check this solution, notice that it satisfies the condition that x(0) 5 A because cos 0 5 1. The position, velocity, and acceleration of the block versus time are plotted in Figure 15.7a for this special case. The acceleration reaches extreme values of 7v2A when the position has extreme values of 6A. Furthermore, the velocity has extreme values of 6vA, which both occur at x 5 0. Hence, the quantitative solution agrees with our qualitative description of this system. Let’s consider another possibility. Suppose the system is oscillating and we define t 5 0 as the instant the block passes through the unstretched position of the spring while moving to the right (Fig. 15.8). In this case, our solutions for x(t) and v(t) must obey the initial conditions that x(0) 5 0 and v(0) 5 vi: x(0) 5 A cos f 5 0 v(0) 5 2vA sin f 5 vi The first of these conditions tells us that f 5 6p/2. With these choices for f, the second condition tells us that A 5 7vi/v. Because the initial velocity is positive and the amplitude must be positive, we must have f 5 2p/2. Hence, the solution is x 5 vi v cos avt 2 p 2 b The graphs of position, velocity, and acceleration versus time for this choice of t 5 0 are shown in Figure 15.7b. Notice that these curves are the same as those in Figure b c a T A x xi t t t v vi a vmax a max Figure 15.5 Graphical repre-sentation of simple harmonic motion. (a) Position versus time. (b) Velocity versus time. (c) Accel-eration versus time. Notice that at any specified time the velocity is 908 out of phase with the position and the acceleration is 1808 out of phase with the position.
T 2 T 2 T x 3T 2 v T 2 T a 3T 2 T T 2 3T 2 T 2 T x t 3T 2 T v t 3T 2 T 2 T a t t t t 3T 2 a b Figure 15.7 (a) Position, velocity, and acceleration versus time for the block in Figure 15.6 under the initial conditions that at t 5 0, x(0) 5 A, and v(0) 5 0. (b) Position, velocity, and acceleration ver-sus time for the block in Figure 15.8 under the initial conditions that at t 5 0, x(0) 5 0, and v(0) 5 vi.
Figure 15.6 A block–spring system that begins its motion from rest with the block at x 5 A at t 5 0.
A m x 0 t 0 xi A vi 0 Figure 15.8 The block–spring system is undergoing oscillation, and t 5 0 is defined at an instant when the block passes through the equilibrium position x 5 0 and is moving to the right with speed vi.
m x 0 t 0 xi 0 v vi vi S www.aswarphysics.weebly.com 456 Chapter 15 Oscillatory Motion Example 15.1 A Block–Spring System A 200-g block connected to a light spring for which the force constant is 5.00 N/m is free to oscillate on a frictionless, horizontal surface. The block is displaced 5.00 cm from equilibrium and released from rest as in Figure 15.6.
(A) Find the period of its motion.
Conceptualize Study Figure 15.6 and imagine the block moving back and forth in simple harmonic motion once it is released. Set up an experimental model in the vertical direction by hanging a heavy object such as a stapler from a strong rubber band.
Categorize The block is modeled as a particle in simple harmonic motion. Analyze AM S o l u t i o n Use Equation 15.9 to find the angular frequency of the block–spring system: v 5 Å k m 5 Å 5.00 N/m 200 3 1023 kg 5 5.00 rad/s Use Equation 15.13 to find the period of the system: T 5 2p v 5 2p 5.00 rad/s 5 1.26 s (B) Determine the maximum speed of the block.
S o l u t i o n (C) What is the maximum acceleration of the block?
Use Equation 15.17 to find vmax: vmax 5 vA 5 (5.00 rad/s)(5.00 3 1022 m) 5 0.250 m/s 15.7a, but shifted to the right by one-fourth of a cycle. This shift is described math-ematically by the phase constant f 5 2p/2, which is one-fourth of a full cycle of 2p.
Analysis Model Particle in Simple Harmonic Motion Imagine an object that is subject to a force that is proportional to the negative of the object’s position, F 5 2kx. Such a force equation is known as Hooke’s law, and it describes the force applied to an object attached to an ideal spring. The parameter k in Hooke’s law is called the spring constant or the force constant. The position of an object acted on by a force described by Hooke’s law is given by x(t) 5 A cos (vt 1 f) (15.6) where A is the amplitude of the motion, v is the angular frequency, and f is the phase constant. The values of A and f depend on the initial position and initial velocity of the particle. The period of the oscillation of the particle is T 5 2p v 5 2pÅ m k (15.13) and the inverse of the period is the frequency.
Examples: • a bungee jumper hangs from a bungee cord and oscillates up and down • a guitar string vibrates back and forth in a standing wave, with each element of the string moving in simple har-monic motion (Chapter 18) • a piston in a gasoline engine oscillates up and down within the cylinder of the engine (Chapter 22) • an atom in a diatomic molecule vibrates back and forth as if it is connected by a spring to the other atom in the molecule (Chapter 43) x A –A t T www.aswarphysics.weebly.com 15.2 Analysis Model: Particle in Simple Harmonic Motion 457 S o l u t i o n Use Equation 15.18 to find amax: amax 5 v2A 5 (5.00 rad/s)2(5.00 3 1022 m) 5 1.25 m/s2 (D) Express the position, velocity, and acceleration as functions of time in SI units.
S o l u t i o n Find the phase constant from the initial condition that x 5 A at t 5 0: x(0) 5 A cos f 5 A S f 5 0 Use Equation 15.6 to write an expression for x(t): x 5 A cos (vt 1 f) 5 0.050 0 cos 5.00t Use Equation 15.15 to write an expression for v(t): v 5 2vA sin (vt 1 f) 5 20.250 sin 5.00t Use Equation 15.16 to write an expression for a(t): a 5 2v2A cos (vt 1 f) 5 21.25 cos 5.00t Finalize Consider part (a) of Figure 15.7, which shows the graphical representations of the motion of the block in this problem. Make sure that the mathematical representations found above in part (D) are consistent with these graphi-cal representations.
What if the block were released from the same initial position, xi 5 5.00 cm, but with an initial velocity of vi 5 20.100 m/s? Which parts of the solution change, and what are the new answers for those that do change?
Answers Part (A) does not change because the period is independent of how the oscillator is set into motion. Parts (B), (C), and (D) will change.
What If?
Write position and velocity expressions for the initial conditions: (1) x(0) 5 A cos f 5 xi (2) v(0) 5 2vA sin f 5 vi Divide Equation (2) by Equation (1) to find the phase constant: 2vA sin f A cos f 5 vi xi tan f 5 2 vi vxi 5 2 20.100 m/s 15.00 rad/s2 10.050 0 m2 5 0.400 f 5 tan21 (0.400) 5 0.121p Use Equation (1) to find A: A 5 xi cos f 5 0.050 0 m cos 10.121p2 5 0.053 9 m Find the new maximum speed: vmax 5 vA 5 (5.00 rad/s)(5.39 3 1022 m) 5 0.269 m/s Find the new magnitude of the maximum acceleration: amax 5 v2A 5 (5.00 rad/s)2(5.39 3 1022 m) 5 1.35 m/s2 Find new expressions for position, velocity, and accelera-tion in SI units: x 5 0.053 9 cos (5.00t 1 0.121p) v 5 20.269 sin (5.00t 1 0.121p) a 5 21.35 cos (5.00t 1 0.121p) As we saw in Chapters 7 and 8, many problems are easier to solve using an energy approach rather than one based on variables of motion. This particular What If? is easier to solve from an energy approach. Therefore, we shall investigate the energy of the simple harmonic oscillator in the next section. ▸ 15.1 c on tin u ed Example 15.2 Watch Out for Potholes! A car with a mass of 1 300 kg is constructed so that its frame is supported by four springs. Each spring has a force con-stant of 20 000 N/m. Two people riding in the car have a combined mass of 160 kg. Find the frequency of vibration of the car after it is driven over a pothole in the road.
AM continued www.aswarphysics.weebly.com 458 Chapter 15 Oscillatory Motion Conceptualize Think about your experiences with automobiles. When you sit in a car, it moves downward a small dis-tance because your weight is compressing the springs further. If you push down on the front bumper and release it, the front of the car oscillates a few times.
Categorize We imagine the car as being supported by a single spring and model the car as a particle in simple harmonic motion.
Analyze First, let’s determine the effective spring constant of the four springs combined. For a given extension x of the springs, the combined force on the car is the sum of the forces from the individual springs.
S o l u t i o n Find an expression for the total force on the car: Ftotal 5 o (2kx) 5 2 ao k b x Evaluate the effective spring constant: keff 5 o k 5 4 3 20 000 N/m 5 80 000 N/m Use Equation 15.14 to find the frequency of vibration: f 5 1 2p Å k eff m 5 1 2p Å 80 000 N/m 1 460 kg 5 1.18 Hz In this expression, x has been factored from the sum because it is the same for all four springs. The effective spring constant for the combined springs is the sum of the individual spring constants.
Finalize The mass we used here is that of the car plus the people because that is the total mass that is oscillating. Also notice that we have explored only up-and-down motion of the car. If an oscillation is established in which the car rocks back and forth such that the front end goes up when the back end goes down, the frequency will be different.
Suppose the car stops on the side of the road and the two people exit the car. One of them pushes down-ward on the car and releases it so that it oscillates vertically. Is the frequency of the oscillation the same as the value we just calculated?
Answer The suspension system of the car is the same, but the mass that is oscillating is smaller: it no longer includes the mass of the two people. Therefore, the frequency should be higher. Let’s calculate the new frequency, taking the mass to be 1 300 kg: f 5 1 2p Å k eff m 5 1 2p Å 80 000 N/m 1 300 kg 5 1.25 Hz As predicted, the new frequency is a bit higher.
What If? ▸ 15.2 continu ed 15.3 Energy of the Simple Harmonic Oscillator As we have done before, after studying the the motion of an object modeled as a particle in a new situation and investigating the forces involved in influencing that motion, we turn our attention to energy. Let us examine the mechanical energy of a system in which a particle undergoes simple harmonic motion, such as the block–spring system illustrated in Figure 15.1. Because the surface is frictionless, the system is isolated and we expect the total mechanical energy of the system to be constant. We assume a massless spring, so the kinetic energy of the system cor-responds only to that of the block. We can use Equation 15.15 to express the kinetic energy of the block as K 5 1 2mv 2 5 1 2mv2A2 sin2 1vt 1 f2 (15.19) The elastic potential energy stored in the spring for any elongation x is given by 1 2kx 2 (see Eq. 7.22). Using Equation 15.6 gives U 5 1 2kx 2 5 1 2kA2 cos2 1vt 1 f2 (15.20) Kinetic energy of a simple harmonic oscillator Potential energy of a simple harmonic oscillator www.aswarphysics.weebly.com 15.3 Energy of the Simple Harmonic Oscillator 459 We see that K and U are always positive quantities or zero. Because v2 5 k/m, we can express the total mechanical energy of the simple harmonic oscillator as E 5 K 1 U 5 1 2kA2 3sin2 1vt 1 f2 1 cos2 1vt 1 f2 4 From the identity sin2 u 1 cos2 u 5 1, we see that the quantity in square brackets is unity. Therefore, this equation reduces to E 5 1 2kA2 (15.21) That is, the total mechanical energy of a simple harmonic oscillator is a constant of the motion and is proportional to the square of the amplitude. The total mechani-cal energy is equal to the maximum potential energy stored in the spring when x 5 6A because v 5 0 at these points and there is no kinetic energy. At the equilibrium position, where U 5 0 because x 5 0, the total energy, all in the form of kinetic energy, is again 1 2kA2. Plots of the kinetic and potential energies versus time appear in Figure 15.9a, where we have taken f 5 0. At all times, the sum of the kinetic and potential ener-gies is a constant equal to 1 2kA2, the total energy of the system. The variations of K and U with the position x of the block are plotted in Figure 15.9b. Energy is continuously being transformed between potential energy stored in the spring and kinetic energy of the block. Figure 15.10 on page 460 illustrates the position, velocity, acceleration, kinetic energy, and potential energy of the block–spring system for one full period of the motion. Most of the ideas discussed so far are incorporated in this important fig-ure. Study it carefully. Finally, we can obtain the velocity of the block at an arbitrary position by express-ing the total energy of the system at some arbitrary position x as E 5 K 1 U 5 1 2mv 2 1 1 2kx 2 5 1 2kA2 v 5 6Å k m 1A2 2 x 22 5 6v"A2 2 x 2 (15.22) When you check Equation 15.22 to see whether it agrees with known cases, you find that it verifies that the speed is a maximum at x 5 0 and is zero at the turning points x 5 6A. You may wonder why we are spending so much time studying simple harmonic oscillators. We do so because they are good models of a wide variety of physical phenomena. For example, recall the Lennard–Jones potential discussed in Exam-ple 7.9. This complicated function describes the forces holding atoms together. Figure 15.11a on page 460 shows that for small displacements from the equilibrium W W Total energy of a simple harmonic oscillator W W Velocity as a function of position for a simple har-monic oscillator U kx2 K mv2 1 2 1 2 K, U A x –A O K, U 1 2 kA2 1 2 kA2 U K T t T 2 a b In either plot, notice that K U constant.
Figure 15.9 (a) Kinetic energy and potential energy versus time for a simple harmonic oscillator with f 5 0. (b) Kinetic energy and potential energy versus position for a simple harmonic oscillator.
www.aswarphysics.weebly.com 460 Chapter 15 Oscillatory Motion position, the potential energy curve for this function approximates a parabola, which represents the potential energy function for a simple harmonic oscillator. Therefore, we can model the complex atomic binding forces as being due to tiny springs as depicted in Figure 15.11b. The ideas presented in this chapter apply not only to block–spring systems and atoms, but also to a wide range of situations that include bungee jumping, playing a musical instrument, and viewing the light emitted by a laser. You will see more examples of simple harmonic oscillators as you work through this book.
r U a b Figure 15.11 (a) If the atoms in a molecule do not move too far from their equilibrium positions, a graph of potential energy versus separation distance between atoms is similar to the graph of potential energy versus posi-tion for a simple harmonic oscillator (dashed black curve). (b) The forces between atoms in a solid can be modeled by imagining springs between neighboring atoms.
t x v t x v a K U 0 0 0 0 0 0 0 0 0 v2A A 0 0 A 0 A vA 0 T 4 T vA 0 –A A Potential energy Total energy Kinetic energy % 0 50 100 % 0 50 100 % 0 50 100 % 0 50 100 % 0 50 100 % 0 50 100 a c e f b d x amax S amax S vmax S vmax S v2A v2A v2x T 2 3T 4 amax S v S x kA2 1 2 kA2 1 2 kA2 1 2 kA2 1 2 kA2 1 2 1 2mv2 1 2kx2 Figure 15.10 (a) through (e) Several instants in the simple harmonic motion for a block–spring system. Energy bar graphs show the distri-bution of the energy of the system at each instant. The parameters in the table at the right refer to the block–spring system, assuming at t 5 0, x 5 A; hence, x 5 A cos vt. For these five special instants, one of the types of energy is zero. (f) An arbitrary point in the motion of the oscilla-tor. The system possesses both kinetic energy and potential energy at this instant as shown in the bar graph.
Example 15.3 Oscillations on a Horizontal Surface A 0.500-kg cart connected to a light spring for which the force constant is 20.0 N/m oscillates on a frictionless, hori-zontal air track.
(A) Calculate the maximum speed of the cart if the amplitude of the motion is 3.00 cm.
Conceptualize The system oscillates in exactly the same way as the block in Figure 15.10, so use that figure in your mental image of the motion.
AM S o l u t i o n www.aswarphysics.weebly.com 15.3 Energy of the Simple Harmonic Oscillator 461 Categorize The cart is modeled as a particle in simple harmonic motion.
Analyze Use Equation 15.21 to express the total energy of the oscillator system and equate it to the kinetic energy of the system when the cart is at x 5 0: E 5 1 2kA2 5 1 2mv max 2 Evaluate the elastic potential energy at x 5 0.020 0 m: U 5 1 2kx 2 5 1 2120.0 N/m2 10.0200 m2 2 5 4.00 3 1023 J Solve for the maximum speed and substitute numerical values: vmax 5 Å k m A 5 Å 20.0 N/m 0.500 kg 10.030 0 m2 5 0.190 m/s Use the result of part (B) to evaluate the kinetic energy at x 5 0.020 0 m: K 5 1 2mv2 5 1 210.500 kg2 10.141 m/s2 2 5 5.00 3 1023 J Use Equation 15.22 to evaluate the velocity: v 5 6Å k m 1A2 2 x 22 5 6Å 20.0 N/m 0.500 kg 3 10.030 0 m2 2 2 10.020 0 m2 24 5 60.141 m/s (B) What is the velocity of the cart when the position is 2.00 cm?
S o l u t i o n Finalize The sum of the kinetic and potential energies in part (C) is equal to the total energy, which can be found from Equation 15.21. That must be true for any position of the cart.
The cart in this example could have been set into motion by releasing the cart from rest at x 5 3.00 cm. What if the cart were released from the same position, but with an initial velocity of v 5 20.100 m/s? What are the new amplitude and maximum speed of the cart?
Answer This question is of the same type we asked at the end of Example 15.1, but here we apply an energy approach.
What If?
The positive and negative signs indicate that the cart could be moving to either the right or the left at this instant.
(C) Compute the kinetic and potential energies of the system when the position of the cart is 2.00 cm.
S o l u t i o n First calculate the total energy of the system at t 5 0: E 5 1 2mv 2 1 1 2kx 2 5 1 210.500 kg2 120.100 m/s2 2 1 1 2120.0 N/m2 10.030 0 m2 2 5 1.15 3 1022 J Equate this total energy to the potential energy of the system when the cart is at the endpoint of the motion: E 5 1 2kA2 Solve for the amplitude A: A 5 Å 2E k 5 Å 211.15 3 1022 J2 20.0 N/m 5 0.033 9 m Equate the total energy to the kinetic energy of the sys-tem when the cart is at the equilibrium position: E 5 1 2mv 2 max Solve for the maximum speed: vmax 5 Å 2E m 5 Å 211.15 3 1022 J2 0.500 kg 5 0.214 m/s The amplitude and maximum velocity are larger than the previous values because the cart was given an initial velocity at t 5 0.
▸ 15.3 c on tin u ed www.aswarphysics.weebly.com 462 Chapter 15 Oscillatory Motion Lamp A A Screen Turntable The ball rotates like a particle in uniform circular motion.
The ball’s shadow moves like a particle in simple harmonic motion.
Figure 15.13 An experimen-tal setup for demonstrating the connection between a particle in simple harmonic motion and a corresponding particle in uniform circular motion.
15.4 Comparing Simple Harmonic Motion with Uniform Circular Motion Some common devices in everyday life exhibit a relationship between oscillatory motion and circular motion. For example, consider the drive mechanism for a non-electric sewing machine in Figure 15.12. The operator of the machine places her feet on the treadle and rocks them back and forth. This oscillatory motion causes the large wheel at the right to undergo circular motion. The red drive belt seen in the photograph transfers this circular motion to the sewing machine mechanism (above the photo) and eventually results in the oscillatory motion of the sewing needle. In this section, we explore this interesting relationship between these two types of motion. Figure 15.13 is a view of an experimental arrangement that shows this relation-ship. A ball is attached to the rim of a turntable of radius A, which is illuminated from above by a lamp. The ball casts a shadow on a screen. As the turntable rotates with constant angular speed, the shadow of the ball moves back and forth in simple harmonic motion. Consider a particle located at point P on the circumference of a circle of radius A as in Figure 15.14a, with the line OP making an angle f with the x axis at t 5 0. We call this circle a reference circle for comparing simple harmonic motion with uniform circular motion, and we choose the position of P at t 5 0 as our reference position. If the particle moves along the circle with constant angular speed v until OP makes an angle u with the x axis as in Figure 15.14b, at some time t . 0 the angle between OP and the x axis is u 5 vt 1 f. As the particle moves along the circle, the projection of P on the x axis, labeled point Q , moves back and forth along the x axis between the limits x 5 6A. Notice that points P and Q always have the same x coordinate. From the right triangle OPQ , we see that this x coordinate is x 1t 2 5 A cos 1vt 1 f2 (15.23) This expression is the same as Equation 15.6 and shows that the point Q moves with simple harmonic motion along the x axis. Therefore, the motion of an object described by the analysis model of a particle in simple harmonic motion along a straight line can be represented by the projection of an object that can be modeled as a particle in uniform circular motion along a diameter of a reference circle. This geometric interpretation shows that the time interval for one complete rev-olution of the point P on the reference circle is equal to the period of motion T for simple harmonic motion between x 5 6A. Therefore, the angular speed v of P is the same as the angular frequency v of simple harmonic motion along the x axis Figure 15.12 The bottom of a treadle-style sewing machine from the early twentieth century. The treadle is the wide, flat foot pedal with the metal grillwork.
The oscillation of the treadle causes circular motion of the drive wheel, eventually resulting in additional up and down motion—of the sewing needle.
The back edge of the treadle goes up and down as one’s feet rock the treadle.
John W. Jewett, Jr.
www.aswarphysics.weebly.com 15.4 Comparing Simple Harmonic Motion with Uniform Circular Motion 463 (which is why we use the same symbol). The phase constant f for simple harmonic motion corresponds to the initial angle OP makes with the x axis. The radius A of the reference circle equals the amplitude of the simple harmonic motion. Because the relationship between linear and angular speed for circular motion is v 5 rv (see Eq. 10.10), the particle moving on the reference circle of radius A has a velocity of magnitude vA. From the geometry in Figure 15.14c, we see that the x component of this velocity is 2vA sin(vt 1 f). By definition, point Q has a velocity given by dx/dt. Differentiating Equation 15.23 with respect to time, we find that the velocity of Q is the same as the x component of the velocity of P. The acceleration of P on the reference circle is directed radially inward toward O and has a magnitude v2/A 5 v2A. From the geometry in Figure 15.14d, we see that the x component of this acceleration is 2v2A cos(vt 1 f). This value is also the acceleration of the projected point Q along the x axis, as you can verify by taking the second derivative of Equation 15.23.
Q uick Quiz 15.5 Figure 15.15 shows the position of an object in uniform circular motion at t 5 0. A light shines from above and projects a shadow of the object on a screen below the circular motion. What are the correct values for the ampli-tude and phase constant (relative to an x axis to the right) of the simple harmonic motion of the shadow? (a) 0.50 m and 0 (b) 1.00 m and 0 (c) 0.50 m and p (d) 1.00 m and p v P P x Q O A y t 0 t O P vx vx Q O y x y x y x A P Q O y x ax ax f f u u v v A v a 2A v v S a S A particle is at point P at t 0.
At a later time t, the x coordinates of points P and Q are equal and are given by Equation 15.23.
The x component of the velocity of P equals the velocity of Q.
The x component of the acceleration of P equals the acceleration of Q.
a b c d Figure 15.14 Relationship between the uniform circular motion of a point P and the simple harmonic motion of a point Q. A particle at P moves in a circle of radius A with constant angular speed v. Ball Screen Turntable 0.50 m Lamp Figure 15.15 (Quick Quiz 15.5) An object moves in circular motion, casting a shadow on the screen below. Its position at an instant of time is shown.
Example 15.4 Circular Motion with Constant Angular Speed The ball in Figure 15.13 rotates counterclockwise in a circle of radius 3.00 m with a constant angular speed of 8.00 rad/s. At t 5 0, its shadow has an x coordinate of 2.00 m and is moving to the right.
(A) Determine the x coordinate of the shadow as a function of time in SI units.
Conceptualize Be sure you understand the relationship between circular motion of the ball and simple harmonic motion of its shadow as described in Figure 15.13. Notice that the shadow is not at is maximum position at t 5 0.
Categorize The ball on the turntable is a particle in uniform circular motion. The shadow is modeled as a particle in simple harmonic motion.
AM S o l u t i o n continued www.aswarphysics.weebly.com 464 Chapter 15 Oscillatory Motion 15.5 The Pendulum The simple pendulum is another mechanical system that exhibits periodic motion. It consists of a particle-like bob of mass m suspended by a light string of length L that is fixed at the upper end as shown in Figure 15.16. The motion occurs in the vertical plane and is driven by the gravitational force. We shall show that, provided the angle u is small (less than about 108), the motion is very close to that of a simple harmonic oscillator. The forces acting on the bob are the force T S exerted by the string and the gravi-tational force m g S. The tangential component mg sin u of the gravitational force always acts toward u 5 0, opposite the displacement of the bob from the lowest posi-tion. Therefore, the tangential component is a restoring force, and we can apply Newton’s second law for motion in the tangential direction: Ft 5 mat S 2mg sin u 5 m d 2s dt 2 where the negative sign indicates that the tangential force acts toward the equilib-rium (vertical) position and s is the bob’s position measured along the arc. We have expressed the tangential acceleration as the second derivative of the position s. Because s 5 Lu (Eq. 10.1a with r 5 L) and L is constant, this equation reduces to d 2u dt 2 5 2 g L sin u Figure 15.16 A simple pendulum.
L s m g sin m m g cos u u u u T S mg S When u is small, a simple pendulum's motion can be modeled as simple harmonic motion about the equilibrium position u 0.
Analyze Use Equation 15.23 to write an expression for the x coordinate of the rotating ball: x 5 A cos 1vt 1 f2 Solve for the phase constant: f 5 cos21 a x Ab 2 vt Substitute numerical values for the initial conditions: f 5 cos21 a2.00 m 3.00 mb 2 0 5 648.28 5 60.841 rad If we were to take f 5 10.841 rad as our answer, the shadow would be moving to the left at t 5 0. Because the shadow is moving to the right at t 5 0, we must choose f 5 20.841 rad.
Write the x coordinate as a function of time: x 5 3.00 cos (8.00t 2 0.841) (B) Find the x components of the shadow’s velocity and acceleration at any time t.
S o l u t i o n Differentiate the x coordinate with respect to time to find the velocity at any time in m/s: vx 5 dx dt 5 123.00 m2 18.00 rad/s2 sin 18.00t 2 0.8412 5 224.0 sin (8.00t 2 0.841) Differentiate the velocity with respect to time to find the acceleration at any time in m/s2: ax 5 dvx dt 5 1224.0 m/s2 18.00 rad/s2 cos 18.00t 2 0.8412 5 2192 cos (8.00t 2 0.841) Finalize These results are equally valid for the ball moving in uniform circular motion and the shadow moving in simple harmonic motion. Notice that the value of the phase constant puts the ball in the fourth quadrant of the xy coordinate system of Figure 15.14, which is consistent with the shadow having a positive value for x and moving toward the right. ▸ 15.4 continu ed www.aswarphysics.weebly.com 15.5 The Pendulum 465 Pitfall Prevention 15.5 Not True Simple Harmonic Motion The pendulum does not exhibit true simple harmonic motion for any angle. If the angle is less than about 108, the motion is close to and can be modeled as simple harmonic. Considering u as the position, let us compare this equation with Equation 15.3. Does it have the same mathematical form? No! The right side is proportional to sin u rather than to u; hence, we would not expect simple harmonic motion because this expression is not of the same mathematical form as Equation 15.3. If we assume u is small (less than about 108 or 0.2 rad), however, we can use the small angle approximation, in which sin u < u, where u is measured in radians. Table 15.1 shows angles in degrees and radians and the sines of these angles. As long as u is less than approximately 108, the angle in radians and its sine are the same to within an accuracy of less than 1.0%. Therefore, for small angles, the equation of motion becomes d 2u dt 2 5 2 g L u (for small values of u) (15.24) Equation 15.24 has the same mathematical form as Equation 15.3, so we conclude that the motion for small amplitudes of oscillation can be modeled as simple har-monic motion. Therefore, the solution of Equation 15.24 is modeled after Equation 15.6 and is given by u 5 umax cos(vt 1 f), where umax is the maximum angular position and the angular frequency v is v 5 Å g L (15.25) The period of the motion is T 5 2p v 5 2pÅ L g (15.26) In other words, the period and frequency of a simple pendulum depend only on the length of the string and the acceleration due to gravity. Because the period is inde-pendent of the mass, we conclude that all simple pendula that are of equal length and are at the same location (so that g is constant) oscillate with the same period. The simple pendulum can be used as a timekeeper because its period depends only on its length and the local value of g. It is also a convenient device for making precise measurements of the free-fall acceleration. Such measurements are impor-tant because variations in local values of g can provide information on the location of oil and other valuable underground resources.
Q uick Quiz 15.6 A grandfather clock depends on the period of a pendulum to keep correct time. (i) Suppose a grandfather clock is calibrated correctly and then a mischievous child slides the bob of the pendulum downward on the oscil-lating rod. Does the grandfather clock run (a) slow, (b) fast, or (c) correctly? (ii) Suppose a grandfather clock is calibrated correctly at sea level and is then taken to the top of a very tall mountain. Does the grandfather clock now run (a) slow, (b) fast, or (c) correctly?
W W Angular frequency for a simple pendulum W W Period of a simple pendulum Table 15.1 Angles and Sines of Angles Angle in Degrees Angle in Radians Sine of Angle Percent Difference 08 0.000 0 0.000 0 0.0% 18 0.017 5 0.017 5 0.0% 28 0.034 9 0.034 9 0.0% 38 0.052 4 0.052 3 0.0% 58 0.087 3 0.087 2 0.1% 108 0.174 5 0.173 6 0.5% 158 0.261 8 0.258 8 1.2% 208 0.349 1 0.342 0 2.1% 308 0.523 6 0.500 0 4.7% www.aswarphysics.weebly.com 466 Chapter 15 Oscillatory Motion Pivot O d d sin CM mg u u S Figure 15.17 A physical pendu-lum pivoted at O.
Example 15.5 A Connection Between Length and Time Christian Huygens (1629–1695), the greatest clockmaker in history, suggested that an international unit of length could be defined as the length of a simple pendulum having a period of exactly 1 s. How much shorter would our length unit be if his suggestion had been followed?
Conceptualize Imagine a pendulum that swings back and forth in exactly 1 second. Based on your experience in observing swinging objects, can you make an estimate of the required length? Hang a small object from a string and simulate the 1-s pendulum.
Categorize This example involves a simple pendulum, so we categorize it as a substitution problem that applies the concepts introduced in this section.
S o l u t i o n Solve Equation 15.26 for the length and substitute the known values: L 5 T 2g 4p2 5 11.00 s2 219.80 m/s22 4p2 5 0.248 m The meter’s length would be slightly less than one-fourth of its current length. Also, the number of significant digits depends only on how precisely we know g because the time has been defined to be exactly 1 s.
What if Huygens had been born on another planet? What would the value for g have to be on that planet such that the meter based on Huygens’s pendulum would have the same value as our meter?
Answer Solve Equation 15.26 for g: g 5 4p2L T 2 5 4p211.00 m2 11.00 s2 2 5 4p2 m/s2 5 39.5 m/s2 No planet in our solar system has an acceleration due to gravity that large.
What If?
Physical Pendulum Suppose you balance a wire coat hanger so that the hook is supported by your extended index finger. When you give the hanger a small angular displacement with your other hand and then release it, it oscillates. If a hanging object oscillates about a fixed axis that does not pass through its center of mass and the object can-not be approximated as a point mass, we cannot treat the system as a simple pendu-lum. In this case, the system is called a physical pendulum. Consider a rigid object pivoted at a point O that is a distance d from the center of mass (Fig. 15.17). The gravitational force provides a torque about an axis through O, and the magnitude of that torque is mgd sin u, where u is as shown in Figure 15.17. We apply the rigid object under a net torque analysis model to the object and use the rotational form of Newton’s second law, S text 5 Ia, where I is the moment of inertia of the object about the axis through O. The result is 2mgd sin u 5 I d 2u dt 2 The negative sign indicates that the torque about O tends to decrease u. That is, the gravitational force produces a restoring torque. If we again assume u is small, the approximation sin u < u is valid and the equation of motion reduces to d 2u dt 2 5 2a mgd I bu 5 2v2u (15.27) Because this equation is of the same mathematical form as Equation 15.3, its solu-tion is modeled after that of the simple harmonic oscillator. That is, the solution www.aswarphysics.weebly.com 15.5 The Pendulum 467 Torsional Pendulum Figure 15.19 on page 468 shows a rigid object such as a disk suspended by a wire attached at the top to a fixed support. When the object is twisted through some angle u, the twisted wire exerts on the object a restoring torque that is proportional to the angular position. That is, t 5 2ku where k (Greek letter kappa) is called the torsion constant of the support wire and is a rotational analog to the force constant k for a spring. The value of k can be obtained by applying a known torque to twist the wire through a measurable angle u. Applying Newton’s second law for rotational motion, we find that of Equation 15.27 is given by u 5 umax cos(vt 1 f), where umax is the maximum angular position and v 5 Å mgd I The period is T 5 2p v 5 2pÅ I mgd (15.28) This result can be used to measure the moment of inertia of a flat, rigid object. If the location of the center of mass—and hence the value of d—is known, the moment of inertia can be obtained by measuring the period. Finally, notice that Equation 15.28 reduces to the period of a simple pendulum (Eq. 15.26) when I 5 md 2, that is, when all the mass is concentrated at the center of mass.
W W Period of a physical pendulum Substitute these quantities into Equation 15.28: T 5 2pÅ 1 3ML2 Mg 1L/22 5 2pÅ 2L 3g Finalize In one of the Moon landings, an astronaut walking on the Moon’s surface had a belt hanging from his space suit, and the belt oscillated as a physical pendulum. A scientist on the Earth observed this motion on television and used it to estimate the free-fall acceleration on the Moon. How did the scientist make this calculation? Example 15.6 A Swinging Rod A uniform rod of mass M and length L is pivoted about one end and oscillates in a verti-cal plane (Fig. 15.18). Find the period of oscillation if the amplitude of the motion is small.
Conceptualize Imagine a rod swinging back and forth when pivoted at one end. Try it with a meterstick or a scrap piece of wood.
Categorize Because the rod is not a point particle, we catego-rize it as a physical pendulum.
Analyze In Chapter 10, we found that the moment of inertia of a uniform rod about an axis through one end is 1 3ML2. The dis-tance d from the pivot to the center of mass of the rod is L/2.
S o l u t i o n Pivot O L d CM Mg S Figure 15.18 (Example 15.6) A rigid rod oscillating about a pivot through one end is a physical pendulum with d 5 L/2.
www.aswarphysics.weebly.com 468 Chapter 15 Oscillatory Motion o t 5 Ia S 2ku 5 I d 2u dt 2 d 2u dt 2 5 2 k I u (15.29) Again, this result is the equation of motion for a simple harmonic oscillator, with v 5 !k/I and a period T 5 2pÅ I k (15.30) This system is called a torsional pendulum. There is no small-angle restriction in this situation as long as the elastic limit of the wire is not exceeded.
15.6 Damped Oscillations The oscillatory motions we have considered so far have been for ideal systems, that is, systems that oscillate indefinitely under the action of only one force, a linear restoring force. In many real systems, nonconservative forces such as friction or air resistance also act and retard the motion of the system. Consequently, the mechanical energy of the system diminishes in time, and the motion is said to be damped. The mechanical energy of the system is transformed into internal energy in the object and the retard-ing medium. Figure 15.20 depicts one such system: an object attached to a spring and submersed in a viscous liquid. Another example is a simple pendulum oscillating in air. After being set into motion, the pendulum eventually stops oscillating due to air resistance. The opening photograph for this chapter depicts damped oscillations in practice. The spring-loaded devices mounted below the bridge are dampers that transform mechanical energy of the oscillating bridge into internal energy. One common type of retarding force is that discussed in Section 6.4, where the force is proportional to the speed of the moving object and acts in the direc-tion opposite the velocity of the object with respect to the medium. This retarding force is often observed when an object moves through air, for instance. Because the retarding force can be expressed as R S 5 2b v S (where b is a constant called the damping coefficient) and the restoring force of the system is 2kx, we can write New-ton’s second law as o Fx = 2kx 2 bvx = max 2kx 2 b dx dt 5 m d 2x dt 2 (15.31) The solution to this equation requires mathematics that may be unfamiliar to you; we simply state it here without proof. When the retarding force is small compared with the maximum restoring force—that is, when the damping coefficient b is small—the solution to Equation 15.31 is x 5 Ae2(b/2m)t cos (vt 1 f) (15.32) where the angular frequency of oscillation is v 5 Å k m 2 a b 2mb 2 (15.33) This result can be verified by substituting Equation 15.32 into Equation 15.31. It is convenient to express the angular frequency of a damped oscillator in the form v 5 Åv0 2 2 a b 2mb 2 where v0 5 !k/m represents the angular frequency in the absence of a retarding force (the undamped oscillator) and is called the natural frequency of the system.
O P max u The object oscillates about the line OP with an amplitude umax.
Figure 15.19 A torsional pendulum.
m Figure 15.20 One example of a damped oscillator is an object attached to a spring and sub-mersed in a viscous liquid.
www.aswarphysics.weebly.com 15.7 Forced Oscillations 469 Figure 15.21 shows the position as a function of time for an object oscillating in the presence of a retarding force. When the retarding force is small, the oscillatory character of the motion is preserved but the amplitude decreases exponentially in time, with the result that the motion ultimately becomes undetectable. Any system that behaves in this way is known as a damped oscillator. The dashed black lines in Figure 15.21, which define the envelope of the oscillatory curve, represent the expo-nential factor in Equation 15.32. This envelope shows that the amplitude decays exponentially with time. For motion with a given spring constant and object mass, the oscillations dampen more rapidly for larger values of the retarding force. When the magnitude of the retarding force is small such that b/2m , v0, the system is said to be underdamped. The resulting motion is represented by Figure 15.21 and the the blue curve in Figure 15.22. As the value of b increases, the ampli-tude of the oscillations decreases more and more rapidly. When b reaches a critical value bc such that bc/2m 5 v0, the system does not oscillate and is said to be criti-cally damped. In this case, the system, once released from rest at some nonequilib-rium position, approaches but does not pass through the equilibrium position. The graph of position versus time for this case is the red curve in Figure 15.22. If the medium is so viscous that the retarding force is large compared with the restoring force—that is, if b/2m . v0—the system is overdamped. Again, the dis-placed system, when free to move, does not oscillate but rather simply returns to its equilibrium position. As the damping increases, the time interval required for the system to approach equilibrium also increases as indicated by the black curve in Figure 15.22. For critically damped and overdamped systems, there is no angular frequency v and the solution in Equation 15.32 is not valid.
15.7 Forced Oscillations We have seen that the mechanical energy of a damped oscillator decreases in time as a result of the retarding force. It is possible to compensate for this energy decrease by applying a periodic external force that does positive work on the sys-tem. At any instant, energy can be transferred into the system by an applied force that acts in the direction of motion of the oscillator. For example, a child on a swing can be kept in motion by appropriately timed “pushes.” The amplitude of motion remains constant if the energy input per cycle of motion exactly equals the decrease in mechanical energy in each cycle that results from retarding forces. A common example of a forced oscillator is a damped oscillator driven by an external force that varies periodically, such as F(t) 5 F0 sin vt, where F0 is a constant and v is the angular frequency of the driving force. In general, the frequency v of the driving force is variable, whereas the natural frequency v0 of the oscillator is fixed by the values of k and m. Modeling an oscillator with both retarding and driv-ing forces as a particle under a net force, Newton’s second law in this situation gives a F x 5 max S F0 sin vt 2 b dx dt 2 kx 5 m d 2x dt 2 (15.34) Again, the solution of this equation is rather lengthy and will not be presented. After the driving force on an initially stationary object begins to act, the ampli-tude of the oscillation will increase. The system of the oscillator and the surround-ing medium is a nonisolated system: work is done by the driving force, such that the vibrational energy of the system (kinetic energy of the object, elastic potential energy in the spring) and internal energy of the object and the medium increase. After a sufficiently long period of time, when the energy input per cycle from the driving force equals the amount of mechanical energy transformed to internal energy for each cycle, a steady-state condition is reached in which the oscillations proceed with constant amplitude. In this situation, the solution of Equation 15.34 is x 5 A cos (vt 1 f) (15.35) x t Figure 15.22 Graphs of posi-tion versus time for an under-damped oscillator (blue curve), a critically damped oscillator (red curve), and an overdamped oscil-lator (black curve).
A x 0 t The amplitude decreases as Ae (b/2m)t.
Figure 15.21 Graph of posi-tion versus time for a damped oscillator.
www.aswarphysics.weebly.com 470 Chapter 15 Oscillatory Motion where A 5 F0/m Å 1v2 2 v0 22 2 1 ab v m b 2 (15.36) and where v0 5 !k/m is the natural frequency of the undamped oscillator (b 5 0). Equations 15.35 and 15.36 show that the forced oscillator vibrates at the fre-quency of the driving force and that the amplitude of the oscillator is constant for a given driving force because it is being driven in steady-state by an external force. For small damping, the amplitude is large when the frequency of the driving force is near the natural frequency of oscillation, or when v < v0. The dramatic increase in amplitude near the natural frequency is called resonance, and the natural fre-quency v0 is also called the resonance frequency of the system. The reason for large-amplitude oscillations at the resonance frequency is that energy is being transferred to the system under the most favorable conditions. We can better understand this concept by taking the first time derivative of x in Equa-tion 15.35, which gives an expression for the velocity of the oscillator. We find that v is proportional to sin(vt 1 f), which is the same trigonometric function as that describing the driving force. Therefore, the applied force F S is in phase with the velocity. The rate at which work is done on the oscillator by F S equals the dot prod-uct F S ? v S; this rate is the power delivered to the oscillator. Because the product F S ? v S is a maximum when F S and v S are in phase, we conclude that at resonance, the applied force is in phase with the velocity and the power transferred to the oscillator is a maximum. Figure 15.23 is a graph of amplitude as a function of driving frequency for a forced oscillator with and without damping. Notice that the amplitude increases with decreasing damping (b S 0) and that the resonance curve broadens as the damping increases. In the absence of a damping force (b 5 0), we see from Equation 15.36 that the steady-state amplitude approaches infinity as v approaches v0. In other words, if there are no losses in the system and we continue to drive an initially motionless oscil-lator with a periodic force that is in phase with the velocity, the amplitude of motion builds without limit (see the red-brown curve in Fig. 15.23). This limitless building does not occur in practice because some damping is always present in reality. Later in this book we shall see that resonance appears in other areas of physics. For example, certain electric circuits have natural frequencies and can be set into strong resonance by a varying voltage applied at a given frequency. A bridge has natural frequencies that can be set into resonance by an appropriate driving force. A dramatic example of such resonance occurred in 1940 when the Tacoma Narrows Bridge in the state of Washington was destroyed by resonant vibrations. Although the winds were not particularly strong on that occasion, the “flapping” of the wind across the roadway (think of the “flapping” of a flag in a strong wind) provided a periodic driving force whose frequency matched that of the bridge. The resulting oscillations of the bridge caused it to ultimately collapse (Fig. 15.24) because the bridge design had inadequate built-in safety features.
Amplitude of a driven oscillator v v A b 0 Undamped Small b Large b 0 When the frequency v of the driving force equals the natural frequency v0 of the oscillator, resonance occurs.
Figure 15.23 Graph of ampli-tude versus frequency for a damped oscillator when a peri-odic driving force is present. Notice that the shape of the reso-nance curve depends on the size of the damping coefficient b.
Figure 15.24 (a) In 1940, turbulent winds set up torsional vibrations in the Tacoma Nar-rows Bridge, causing it to oscillate at a frequency near one of the natural frequencies of the bridge structure. (b) Once established, this resonance condition led to the bridge’s collapse. (Mathemati-cians and physicists are currently challenging some aspects of this interpretation.) a b AP Photos © Topham/The Image Works www.aswarphysics.weebly.com Summary 471 Many other examples of resonant vibrations can be cited. A resonant vibration you may have experienced is the “singing” of telephone wires in the wind. Machines often break if one vibrating part is in resonance with some other moving part. Sol-diers marching in cadence across a bridge have been known to set up resonant vibrations in the structure and thereby cause it to collapse. Whenever any real phys-ical system is driven near its resonance frequency, you can expect oscillations of very large amplitudes.
Summary The kinetic energy and potential energy for an object of mass m oscillating at the end of a spring of force constant k vary with time and are given by K 5 1 2mv 2 5 1 2mv2A2 sin2 1vt 1 f2 (15.19) U 5 1 2kx 2 5 1 2kA2 cos2 1vt 1 f2 (15.20) The total energy of a simple harmonic oscillator is a constant of the motion and is given by E 5 1 2kA2 (15.21) A simple pendulum of length L can be modeled to move in simple harmonic motion for small angular displacements from the vertical. Its period is T 5 2pÅ L g (15.26) A physical pendulum is an extended object that, for small angular displacements, can be modeled to move in simple harmonic motion about a pivot that does not go through the center of mass. The period of this motion is T 5 2pÅ I mgd (15.28) where I is the moment of inertia of the object about an axis through the pivot and d is the distance from the pivot to the center of mass of the object. If an oscillator is subject to a sinu-soidal driving force that is described by F(t) 5 F0 sin vt, it exhibits reso-nance, in which the amplitude is largest when the driving frequency v matches the natural frequency v0 5 !k/m of the oscillator. If an oscillator experiences a damping force R S 5 2b v S, its position for small damping is described by x 5 Ae2(b/2m)t cos (vt 1 f) (15.32) where v 5 Å k m 2 a b 2mb 2 (15.33) Concepts and Principles Analysis Model for Problem Solving Particle in Simple Harmonic Motion If a particle is subject to a force of the form of Hooke’s law F 5 2kx, the particle exhibits simple harmonic motion. Its position is described by x(t) 5 A cos (vt 1 f) (15.6) where A is the amplitude of the motion, v is the angular frequency, and f is the phase constant. The value of f depends on the initial position and initial velocity of the particle. The period of the oscillation of the particle is T 5 2p v 5 2pÅ m k (15.13) and the inverse of the period is the frequency.
x A –A t T www.aswarphysics.weebly.com 472 Chapter 15 Oscillatory Motion 10. A mass–spring system moves with simple harmonic motion along the x axis between turning points at x1 5 20 cm and x2 5 60 cm. For parts (i) through (iii), choose from the same five possibilities. (i) At which position does the particle have the greatest magnitude of momentum? (a) 20 cm (b) 30 cm (c) 40 cm (d) some other position (e) The greatest value occurs at multiple points. (ii) At which position does the particle have greatest kinetic energy? (iii) At which position does the particle-spring system have the greatest total energy? 11. A block with mass m 5 0.1 kg oscillates with amplitude A 5 0.1 m at the end of a spring with force constant k 5 10 N/m on a frictionless, horizontal surface. Rank the periods of the following situations from greatest to smallest. If any periods are equal, show their equality in your ranking. (a) The system is as described above. (b) The system is as described in situation (a) except the amplitude is 0.2 m. (c) The situation is as described in situation (a) except the mass is 0.2 kg. (d) The situ-ation is as described in situation (a) except the spring has force constant 20 N/m. (e) A small resistive force makes the motion underdamped. 12. For a simple harmonic oscillator, answer yes or no to the following questions. (a) Can the quantities position and velocity have the same sign? (b) Can velocity and acceleration have the same sign? (c) Can position and acceleration have the same sign? 13. The top end of a spring is held fixed. A block is hung on the bot-tom end as in Figure OQ15.13a, and the fre-quency f of the oscil-lation of the system is measured. The block, a second identical block, and the spring are car-ried up in a space shuttle to Earth orbit. The two blocks are attached to the ends of the spring. The spring is compressed without making adjacent coils touch (Fig. OQ15.13b), and the system is released to oscillate while floating within the shuttle cabin (Fig. OQ15.13c). What is the frequency of oscil-lation for this system in terms of f ? (a) f/2 (b) f/!2 (c) f (d)!2f (e) 2f 14. Which of the following statements is not true regarding a mass–spring system that moves with simple harmonic motion in the absence of friction? (a) The total energy of the system remains constant. (b) The energy of the system is continually transformed between kinetic and potential energy. (c) The total energy of the system is proportional to the square of the amplitude. (d) The potential energy stored in the system is greatest when the mass passes through the equilibrium position. (e) The velocity of the oscillating mass has its maxi-mum value when the mass passes through the equilib-rium position.
a b c Figure OQ15.13 1. If a simple pendulum oscillates with small amplitude and its length is doubled, what happens to the fre-quency of its motion? (a) It doubles. (b) It becomes !2 times as large. (c) It becomes half as large. (d) It becomes 1/!2 times as large. (e) It remains the same. 2. You attach a block to the bottom end of a spring hang-ing vertically. You slowly let the block move down and find that it hangs at rest with the spring stretched by 15.0 cm. Next, you lift the block back up to the ini-tial position and release it from rest with the spring unstretched. What maximum distance does it move down? (a) 7.5 cm (b) 15.0 cm (c) 30.0 cm (d) 60.0 cm (e) The distance cannot be determined without know-ing the mass and spring constant. 3. A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 6.0 cm has an energy of 12 J. If the block is replaced by one whose mass is twice the mass of the original block and the amplitude of the motion is again 6.0 cm, what is the energy of the system? (a) 12 J (b) 24 J (c) 6 J (d) 48 J (e) none of those answers 4. An object–spring system moving with simple harmonic motion has an amplitude A. When the kinetic energy of the object equals twice the potential energy stored in the spring, what is the position x of the object? (a) A (b) 1 3A (c) A/!3 (d) 0 (e) none of those answers 5. An object of mass 0.40 kg, hanging from a spring with a spring constant of 8.0 N/m, is set into an up-and-down simple harmonic motion. What is the magnitude of the acceleration of the object when it is at its maxi-mum displacement of 0.10 m? (a) zero (b) 0.45 m/s2 (c) 1.0 m/s2 (d) 2.0 m/s2 (e) 2.4 m/s2 6. A runaway railroad car, with mass 3.0 3 105 kg, coasts across a level track at 2.0 m/s when it collides elastically with a spring-loaded bumper at the end of the track. If the spring constant of the bumper is 2.0 3 106 N/m, what is the maximum compression of the spring dur-ing the collision? (a) 0.77 m (b) 0.58 m (c) 0.34 m (d) 1.07 m (e) 1.24 m 7. The position of an object moving with simple harmonic motion is given by x 5 4 cos (6pt), where x is in meters and t is in seconds. What is the period of the oscillat-ing system? (a) 4 s (b) 1 6 s (c) 1 3 s (d) 6p s (e) impossible to determine from the information given 8. If an object of mass m attached to a light spring is replaced by one of mass 9m, the frequency of the vibrat-ing system changes by what factor? (a) 1 9 (b) 1 3 (c) 3.0 (d) 9.0 (e) 6.0 9. You stand on the end of a diving board and bounce to set it into oscillation. You find a maximum response in terms of the amplitude of oscillation of the end of the board when you bounce at frequency f. You now move to the middle of the board and repeat the experiment. Is the resonance frequency for forced oscillations at this point (a) higher, (b) lower, or (c) the same as f?
Objective Questions 1. denotes answer available in Student Solutions Manual/Study Guide www.aswarphysics.weebly.com Conceptual Questions 473 the period of the pendulum (a) greater, (b) smaller, or (c) unchanged? 17. A particle on a spring moves in simple harmonic motion along the x axis between turning points at x1 5 100 cm and x2 5 140 cm. (i) At which of the following positions does the particle have maximum speed? (a) 100 cm (b) 110 cm (c) 120 cm (d) at none of those positions (ii) At which position does it have maximum acceleration? Choose from the same possibilities as in part (i). (iii) At which position is the greatest net force exerted on the particle? Choose from the same possi-bilities as in part (i). 15. A simple pendulum has a period of 2.5 s. (i) What is its period if its length is made four times larger? (a) 1.25 s (b) 1.77 s (c) 2.5 s (d) 3.54 s (e) 5 s (ii) What is its period if the length is held constant at its initial value and the mass of the suspended bob is made four times larger? Choose from the same possibilities. 16. A simple pendulum is suspended from the ceiling of a stationary elevator, and the period is determined. (i) When the elevator accelerates upward, is the period (a) greater, (b) smaller, or (c) unchanged? (ii) When the elevator has a downward acceleration, is the period (a) greater, (b) smaller, or (c) unchanged? (iii) When the elevator moves with constant upward velocity, is A Piston x A x(t ) v x 0 Figure CQ15.13 Conceptual Questions 1. denotes answer available in Student Solutions Manual/Study Guide 1. You are looking at a small, leafy tree. You do not notice any breeze, and most of the leaves on the tree are motionless. One leaf, however, is fluttering back and forth wildly. After a while, that leaf stops moving and you notice a different leaf moving much more than all the others. Explain what could cause the large motion of one particular leaf. 2. The equations listed together on page 38 give position as a function of time, velocity as a function of time, and velocity as a function of position for an object moving in a straight line with constant acceleration. The quan-tity vxi appears in every equation. (a) Do any of these equations apply to an object moving in a straight line with simple harmonic motion? (b) Using a similar for-mat, make a table of equations describing simple har-monic motion. Include equations giving acceleration as a function of time and acceleration as a function of position. State the equations in such a form that they apply equally to a block–spring system, to a pendu-lum, and to other vibrating systems. (c) What quantity appears in every equation? 3. (a) If the coordinate of a particle varies as x 5 2A cos vt, what is the phase constant in Equation 15.6? (b) At what position is the particle at t 5 0? 4. A pendulum bob is made from a sphere filled with water. What would happen to the frequency of vibra-tion of this pendulum if there were a hole in the sphere that allowed the water to leak out slowly? 5. Figure CQ15.5 shows graphs of the potential energy of four different systems versus the position of a particle in each system. Each particle is set into motion with a push at an arbitrarily chosen location. Describe its sub-sequent motion in each case (a), (b), (c), and (d). 6. A student thinks that any real vibration must be damped. Is the student correct? If so, give convincing reasoning. If not, give an example of a real vibration that keeps con-stant amplitude forever if the system is isolated. 7. The mechanical energy of an undamped block–spring system is constant as kinetic energy transforms to elastic potential energy and vice versa. For comparison, explain what happens to the energy of a damped oscillator in terms of the mechanical, potential, and kinetic energies. 8. Is it possible to have damped oscillations when a sys-tem is at resonance? Explain. 9. Will damped oscillations occur for any values of b and k? Explain. 10. If a pendulum clock keeps perfect time at the base of a mountain, will it also keep perfect time when it is moved to the top of the mountain? Explain. 11. Is a bouncing ball an example of simple harmonic motion? Is the daily movement of a student from home to school and back simple harmonic motion? Why or why not? 12. A simple pendulum can be modeled as exhibiting simple harmonic motion when u is small. Is the motion periodic when u is large? 13. Consider the simplified single-piston engine in Figure CQ15.13. Assuming the wheel rotates with constant angular speed, explain why the piston rod oscillates in simple harmonic motion.
U U U U x x x x b c d a Figure CQ15.5 www.aswarphysics.weebly.com 474 Chapter 15 Oscillatory Motion is released from rest there. It proceeds to move without friction. The next time the speed of the object is zero is 0.500 s later. What is the maximum speed of the object? 8. A simple harmonic oscillator takes 12.0 s to undergo five complete vibrations. Find (a) the period of its motion, (b) the frequency in hertz, and (c) the angular frequency in radians per second. 9. A 7.00-kg object is hung from the bottom end of a verti-cal spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 2.60 s. Find the force constant of the spring. 10. At an outdoor market, a bunch of bananas attached to the bottom of a vertical spring of force constant 16.0 N/m is set into oscillatory motion with an ampli-tude of 20.0 cm. It is observed that the maximum speed of the bunch of bananas is 40.0 cm/s. What is the weight of the bananas in newtons? 11. A vibration sensor, used in testing a washing machine, consists of a cube of aluminum 1.50 cm on edge mounted on one end of a strip of spring steel (like a hacksaw blade) that lies in a vertical plane. The strip’s mass is small compared with that of the cube, but the strip’s length is large compared with the size of the cube. The other end of the strip is clamped to the frame of the washing machine that is not operat-ing. A horizontal force of 1.43 N applied to the cube is required to hold it 2.75 cm away from its equilib-rium position. If it is released, what is its frequency of vibration? 12. (a) A hanging spring stretches by 35.0 cm when an object of mass 450 g is hung on it at rest. In this sit-uation, we define its position as x 5 0. The object is pulled down an additional 18.0 cm and released from rest to oscillate without friction. What is its position x at a moment 84.4 s later? (b) Find the distance traveled by the vibrating object in part (a). (c) What If? Another hanging spring stretches by 35.5 cm when an object of mass 440 g is hung on it at rest. We define this new position as x 5 0. This object is also pulled down an additional 18.0 cm and released from rest to oscillate without friction. Find its position 84.4 s later. (d) Find the distance traveled by the object in part (c). (e) Why are the answers to parts (a) and (c) so different when the initial data in parts (a) and (c) are so similar and the answers to parts (b) and (d) are relatively close? Does this circumstance reveal a fundamental difficulty in calculating the future?
W W AMT M Q/C Note: Ignore the mass of every spring, except in Prob-lems 76 and 87.
Section 15.1 Motion of an Object Attached to a Spring Problems 17, 18, 19, 22, and 59 in Chapter 7 can also be assigned with this section. 1. A 0.60-kg block attached to a spring with force con-stant 130 N/m is free to move on a frictionless, hori-zontal surface as in Figure 15.1. The block is released from rest when the spring is stretched 0.13 m. At the instant the block is released, find (a) the force on the block and (b) its acceleration. 2. When a 4.25-kg object is placed on top of a vertical spring, the spring compresses a distance of 2.62 cm. What is the force constant of the spring?
Section 15.2 Analysis Model: Particle in Simple Harmonic Motion 3. A vertical spring stretches 3.9 cm when a 10-g object is hung from it. The object is replaced with a block of mass 25 g that oscillates up and down in simple har-monic motion. Calculate the period of motion. 4. In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x 5 5.00 cos a2t 1 p 6 b where x is in centimeters and t is in seconds. At t 5 0, find (a) the position of the particle, (b) its velocity, and (c) its acceleration. Find (d) the period and (e) the amplitude of the motion. 5. The position of a particle is given by the expression x 5 4.00 cos (3.00pt 1 p), where x is in meters and t is in seconds. Determine (a) the frequency and (b) period of the motion, (c) the amplitude of the motion, (d) the phase constant, and (e) the position of the particle at t 5 0.250 s. 6. A piston in a gasoline engine is in simple har-monic motion. The engine is running at the rate of 3 600 rev/min. Taking the extremes of its position rela-tive to its center point as 65.00 cm, find the magni-tudes of the (a) maximum velocity and (b) maximum acceleration of the piston. 7. A 1.00-kg object is attached to a horizontal spring. The spring is initially stretched by 0.100 m, and the object M W M Problems The problems found in this chapter may be assigned online in Enhanced WebAssign 1. straightforward; 2. intermediate; 3. challenging 1. full solution available in the Student Solutions Manual/Study Guide AMT Analysis Model tutorial available in Enhanced WebAssign GP Guided Problem M Master It tutorial available in Enhanced WebAssign W Watch It video solution available in Enhanced WebAssign P PA BIO Q/C P S www.aswarphysics.weebly.com Problems 475 value of its (a) speed and (b) acceleration, (c) the speed and (d) the acceleration when the object is 6.00 cm from the equilibrium position, and (e) the time inter-val required for the object to move from x 5 0 to x 5 8.00 cm. 20. You attach an object to the bottom end of a hang-ing vertical spring. It hangs at rest after extending the spring 18.3 cm. You then set the object vibrating. (a) Do you have enough information to find its period? (b) Explain your answer and state whatever you can about its period.
Section 15.3 Energy of the Simple Harmonic Oscillator 21. To test the resiliency of its bumper during low-speed collisions, a 1 000-kg automobile is driven into a brick wall. The car’s bumper behaves like a spring with a force constant 5.00 3 106 N/m and compresses 3.16 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall? 22. A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy of the system is 2.00 J. Find (a) the force constant of the spring and (b) the ampli-tude of the motion. 23. A block of unknown mass is attached to a spring with a spring constant of 6.50 N/m and undergoes simple har-monic motion with an amplitude of 10.0 cm. When the block is halfway between its equilibrium position and the end point, its speed is measured to be 30.0 cm/s. Calculate (a) the mass of the block, (b) the period of the motion, and (c) the maximum acceleration of the block. 24. A block–spring system oscillates with an amplitude of 3.50 cm. The spring constant is 250 N/m and the mass of the block is 0.500 kg. Determine (a) the mechanical energy of the system, (b) the maximum speed of the block, and (c) the maximum acceleration. 25. A particle executes simple harmonic motion with an amplitude of 3.00 cm. At what position does its speed equal half of its maximum speed? 26. The amplitude of a system moving in simple harmonic motion is doubled. Determine the change in (a) the total energy, (b) the maximum speed, (c) the maxi-mum acceleration, and (d) the period. 27. A 50.0-g object connected to a spring with a force constant of 35.0 N/m oscillates with an amplitude of 4.00 cm on a frictionless, horizontal surface. Find (a) the total energy of the system and (b) the speed of the object when its position is 1.00 cm. Find (c) the kinetic energy and (d) the potential energy when its position is 3.00 cm. 28. A 2.00-kg object is attached to a spring and placed on a frictionless, horizontal surface. A horizontal force of 20.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released Q/C AMT M W 13. Review. A particle moves along the x axis. It is initially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration 20.320 m/s2. Suppose it moves as a particle under constant acceleration for 4.50 s. Find (a) its position and (b) its velocity at the end of this time interval. Next, assume it moves as a particle in simple harmonic motion for 4.50 s and x 5 0 is its equi-librium position. Find (c) its position and (d) its veloc-ity at the end of this time interval. 14. A ball dropped from a height of 4.00 m makes an elas-tic collision with the ground. Assuming no mechani-cal energy is lost due to air resistance, (a) show that the ensuing motion is periodic and (b) determine the period of the motion. (c) Is the motion simple har-monic? Explain. 15. A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the ori-gin, at t 5 0 and moves to the right. The amplitude of its motion is 2.00 cm, and the frequency is 1.50 Hz. (a) Find an expression for the position of the particle as a function of time. Determine (b) the maximum speed of the particle and (c) the earliest time (t . 0) at which the particle has this speed. Find (d) the maxi-mum positive acceleration of the particle and (e) the earliest time (t . 0) at which the particle has this accel-eration. (f) Find the total distance traveled by the par-ticle between t 5 0 and t 5 1.00 s. 16. The initial position, velocity, and acceleration of an object moving in simple harmonic motion are xi, vi, and ai; the angular frequency of oscillation is v. (a) Show that the position and velocity of the object for all time can be written as x(t) 5 xi cos vt 1 a vi vb sin vt v(t) 5 2xiv sin vt 1 vi cos vt (b) Using A to represent the amplitude of the motion, show that v2 2 ax 5 vi 2 2 aixi 5 v2A2 17. A particle moves in simple harmonic motion with a frequency of 3.00 Hz and an amplitude of 5.00 cm. (a) Through what total distance does the particle move during one cycle of its motion? (b) What is its maxi-mum speed? Where does this maximum speed occur? (c) Find the maximum acceleration of the particle. Where in the motion does the maximum acceleration occur? 18. A 1.00-kg glider attached to a spring with a force con-stant of 25.0 N/m oscillates on a frictionless, horizon-tal air track. At t 5 0, the glider is released from rest at x 5 23.00 cm (that is, the spring is compressed by 3.00 cm). Find (a) the period of the glider’s motion, (b) the maximum values of its speed and acceleration, and (c) the position, velocity, and acceleration as func-tions of time. 19. A 0.500-kg object attached to a spring with a force con-stant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.0 cm. Calculate the maximum Q/C S W M www.aswarphysics.weebly.com 476 Chapter 15 Oscillatory Motion friction that would allow the block to reach the equi-librium position? 32. A 326-g object is attached to a spring and executes sim-ple harmonic motion with a period of 0.250 s. If the total energy of the system is 5.83 J, find (a) the maxi-mum speed of the object, (b) the force constant of the spring, and (c) the amplitude of the motion.
Section 15.4 Comparing Simple Harmonic Motion with Uniform Circular Motion 33. While driving behind a car travel-ing at 3.00 m/s, you notice that one of the car’s tires has a small hemi-spherical bump on its rim as shown in Figure P15.33. (a) Explain why the bump, from your viewpoint behind the car, executes simple harmonic motion. (b) If the radii of the car’s tires are 0.300 m, what is the bump’s period of oscillation?
Section 15.5 The Pendulum Problem 68 in Chapter 1 can also be assigned with this section. 34. A “seconds pendulum” is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2 s.) The length of a seconds pendulum is 0.992 7 m at Tokyo, Japan, and 0.994 2 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations? 35. A simple pendulum makes 120 complete oscillations in 3.00 min at a location where g 5 9.80 m/s2. Find (a) the period of the pendulum and (b) its length. 36. A particle of mass m slides without friction inside a hemispherical bowl of radius R. Show that if the par-ticle starts from rest with a small displacement from equilibrium, it moves in simple harmonic motion with an angular frequency equal to that of a simple pendu-lum of length R. That is, v 5 !g/R. 37. A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.450 Hz. The pendulum has a mass of 2.20 kg, and the pivot is located 0.350 m from the center of mass. Deter-mine the moment of inertia of the pendulum about the pivot point. 38. A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency f. The pendulum has a mass m, and the pivot is located a distance d from the center of mass. Determine the moment of inertia of the pendulum about the pivot point. 39. The angular position of a pendulum is represented by the equation u = 0.032 0 cos vt, where u is in radians and v = 4.43 rad/s. Determine the period and length of the pendulum. 40. Consider the physical pendulum of Figure 15.17. (a) Rep-resent its moment of inertia about an axis passing AMT Bump Figure P15.33 Q/C S M S S from rest from this stretched position, and it subse-quently undergoes simple harmonic oscillations. Find (a) the force constant of the spring, (b) the frequency of the oscillations, and (c) the maximum speed of the object. (d) Where does this maximum speed occur? (e) Find the maximum acceleration of the object. (f) Where does the maximum acceleration occur? (g) Find the total energy of the oscillating system. Find (h) the speed and (i) the acceleration of the object when its position is equal to one-third the max-imum value. 29. A simple harmonic oscillator of amplitude A has a total energy E. Determine (a) the kinetic energy and (b) the potential energy when the position is one-third the amplitude. (c) For what values of the position does the kinetic energy equal one-half the potential energy? (d) Are there any values of the position where the kinetic energy is greater than the maximum potential energy? Explain. 30. Review. A 65.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her body and to the bridge. The unstretched length of the cord is 11.0 m. The jumper reaches the bottom of her motion 36.0 m below the bridge before bouncing back. We wish to find the time interval between her leaving the bridge and her arriving at the bottom of her motion. Her overall motion can be separated into an 11.0-m free fall and a 25.0-m section of simple harmonic oscillation. (a) For the free-fall part, what is the appropriate analysis model to describe her motion? (b) For what time interval is she in free fall? (c) For the simple harmonic oscillation part of the plunge, is the system of the bungee jumper, the spring, and the Earth isolated or non- isolated? (d) From your response in part (c) find the spring constant of the bungee cord. (e) What is the location of the equilib-rium point where the spring force balances the gravi-tational force exerted on the jumper? (f) What is the angular frequency of the oscillation? (g) What time interval is required for the cord to stretch by 25.0 m? (h) What is the total time interval for the entire 36.0-m drop? 31. Review. A 0.250-kg block resting on a frictionless, horizontal surface is attached to a spring whose force constant is 83.8 N/m as in Figure P15.31. A horizon-tal force F S causes the spring to stretch a distance of 5.46 cm from its equilibrium position. (a) Find the magnitude of F S . (b) What is the total energy stored in the system when the spring is stretched? (c) Find the magnitude of the acceleration of the block just after the applied force is removed. (d) Find the speed of the block when it first reaches the equilibrium position. (e) If the surface is not frictionless but the block still reaches the equilibrium position, would your answer to part (d) be larger or smaller? (f) What other infor-mation would you need to know to find the actual answer to part (d) in this case? (g) What is the largest value of the coefficient of S Q/C GP F S Figure P15.31 Q/C www.aswarphysics.weebly.com Problems 477 Section 15.6 Damped Oscillations 46. A pendulum with a length of 1.00 m is released from an initial angle of 15.08. After 1 000 s, its amplitude has been reduced by friction to 5.508. What is the value of b/2m? 47. A 10.6-kg object oscillates at the end of a vertical spring that has a spring constant of 2.05 3 104 N/m. The effect of air resistance is represented by the damp-ing coefficient b 5 3.00 N ? s/m. (a) Calculate the frequency of the damped oscillation. (b) By what per-centage does the amplitude of the oscillation decrease in each cycle? (c) Find the time interval that elapses while the energy of the system drops to 5.00% of its initial value. 48. Show that the time rate of change of mechanical energy for a damped, undriven oscillator is given by dE/dt 5 2bv2 and hence is always negative. To do so, differentiate the expression for the mechanical energy of an oscillator, E 5 1 2mv2 1 1 2kx2, and use Equation 15.31. 49. Show that Equation 15.32 is a solution of Equation 15.31 provided that b 2 , 4mk.
Section 15.7 Forced Oscillations 50. A baby bounces up and down in her crib. Her mass is 12.5 kg, and the crib mattress can be modeled as a light spring with force constant 700 N/m. (a) The baby soon learns to bounce with maximum amplitude and mini-mum effort by bending her knees at what frequency? (b) If she were to use the mattress as a trampoline— losing contact with it for part of each cycle—what mini-mum amplitude of oscillation does she require? 51. As you enter a fine restaurant, you realize that you have accidentally brought a small electronic timer from home instead of your cell phone. In frustration, you drop the timer into a side pocket of your suit coat, not realizing that the timer is operating. The arm of your chair presses the light cloth of your coat against your body at one spot. Fabric with a length L hangs freely below that spot, with the timer at the bottom. At one point during your dinner, the timer goes off and a buzzer and a vibrator turn on and off with a frequency of 1.50 Hz. It makes the hanging part of your coat swing back and forth with remarkably large amplitude, draw-ing everyone’s attention. Find the value of L. 52. A block weighing 40.0 N is suspended from a spring that has a force constant of 200 N/m. The system is undamped (b 5 0) and is subjected to a harmonic driv-ing force of frequency 10.0 Hz, resulting in a forced-motion amplitude of 2.00 cm. Determine the maximum value of the driving force. 53. A 2.00-kg object attached to a spring moves without friction (b 5 0) and is driven by an external force given by the expression F 5 3.00 sin (2pt), where F is in newtons and t is in seconds. The force constant of the spring is 20.0 N/m. Find (a) the resonance angular fre-quency of the system, (b) the angular frequency of the driven system, and (c) the amplitude of the motion.
W S S through its center of mass and parallel to the axis pass-ing through its pivot point as ICM. Show that its period is T 5 2pÅ ICM 1 md 2 mgd where d is the distance between the pivot point and the center of mass. (b) Show that the period has a mini-mum value when d satisfies md 2 5 ICM. 41. A simple pendulum has a mass of 0.250 kg and a length of 1.00 m. It is displaced through an angle of 15.08 and then released. Using the analysis model of a particle in simple harmonic motion, what are (a) the maximum speed of the bob, (b) its maximum angular accelera-tion, and (c) the maximum restoring force on the bob? (d) What If? Solve parts (a) through (c) again by using analysis models introduced in earlier chapters. (e) Com-pare the answers. 42. A very light rigid rod of length 0.500 m extends straight out from one end of a meterstick. The combination is sus-pended from a pivot at the upper end of the rod as shown in Figure P15.42. The combination is then pulled out by a small angle and released. (a) Deter-mine the period of oscillation of the system. (b) By what percentage does the period differ from the period of a simple pendulum 1.00 m long? 43. Review. A simple pendulum is 5.00 m long. What is the period of small oscillations for this pendulum if it is located in an elevator (a) accelerating upward at 5.00 m/s2? (b) Accelerating downward at 5.00 m/s2? (c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 5.00 m/s2? 44. A small object is attached to the end of a string to form a simple pendulum. The period of its harmonic motion is measured for small angular displacements and three lengths. For lengths of 1.000 m, 0.750 m, and 0.500 m, total time intervals for 50 oscillations of 99.8 s, 86.6 s, and 71.1 s are measured with a stopwatch. (a) Deter-mine the period of motion for each length. (b) Deter-mine the mean value of g obtained from these three independent measurements and compare it with the accepted value. (c) Plot T 2 versus L and obtain a value for g from the slope of your best-fit straight-line graph. (d) Compare the value found in part (c) with that obtained in part (b). 45. A watch balance wheel (Fig. P15.45) has a period of oscillation of 0.250 s. The wheel is constructed so that its mass of 20.0 g is concentrated around a rim of radius 0.500 cm. What are (a) the wheel’s moment of inertia and (b) the torsion constant of the attached spring?
Q/C 0.500 m Figure P15.42 W Balance wheel Figure P15.45 . Cengage Learning/George Semple www.aswarphysics.weebly.com 478 Chapter 15 Oscillatory Motion the rock begins to lose contact with the sidewalk? Another rock is sitting on the concrete bottom of a swimming pool full of water. The earthquake produces only vertical motion, so the water does not slosh from side to side. (b) Present a convincing argument that when the ground vibrates with the amplitude found in part (a), the submerged rock also barely loses contact with the floor of the swimming pool. 61. Four people, each with a mass of 72.4 kg, are in a car with a mass of 1 130 kg. An earthquake strikes. The vertical oscillations of the ground surface make the car bounce up and down on its suspension springs, but the driver manages to pull off the road and stop. When the frequency of the shaking is 1.80 Hz, the car exhibits a maximum amplitude of vibration. The earthquake ends, and the four people leave the car as fast as they can. By what distance does the car’s undamaged suspension lift the car’s body as the peo-ple get out? 62. To account for the walking speed of a bipedal or qua-drupedal animal, model a leg that is not contacting the ground as a uniform rod of length ,, swinging as a physical pendulum through one half of a cycle, in reso-nance. Let umax represent its amplitude. (a) Show that the animal’s speed is given by the expression v 5 "6g ,sin umax p if umax is sufficiently small that the motion is nearly sim-ple harmonic. An empirical relationship that is based on the same model and applies over a wider range of angles is v 5 "6g , cos 1umax/22 sin umax p (b) Evaluate the walking speed of a human with leg length 0.850 m and leg-swing amplitude 28.0°. (c) What leg length would give twice the speed for the same angular amplitude? 63. The free-fall acceleration on Mars is 3.7 m/s2. (a) What length of pendulum has a period of 1.0 s on Earth? (b) What length of pendulum would have a 1.0-s period on Mars? An object is suspended from a spring with force constant 10 N/m. Find the mass suspended from this spring that would result in a period of 1.0 s (c) on Earth and (d) on Mars. 64. An object attached to a spring vibrates with simple har-monic motion as described by Figure P15.64. For this motion, find (a) the amplitude, (b) the period, (c) the BIO M 54. Considering an undamped, forced oscillator (b 5 0), show that Equation 15.35 is a solution of Equation 15.34, with an amplitude given by Equation 15.36. 55. Damping is negligible for a 0.150-kg object hanging from a light, 6.30-N/m spring. A sinusoidal force with an amplitude of 1.70 N drives the system. At what fre-quency will the force make the object vibrate with an amplitude of 0.440 m?
Additional Problems 56. The mass of the deuterium molecule (D2) is twice that of the hydrogen molecule (H2). If the vibrational fre-quency of H2 is 1.30 3 1014 Hz, what is the vibrational frequency of D2? Assume the “spring constant” of attracting forces is the same for the two molecules. 57. An object of mass m moves in simple harmonic motion with amplitude 12.0 cm on a light spring. Its maxi-mum acceleration is 108 cm/s2. Regard m as a vari-able. (a) Find the period T of the object. (b) Find its frequency f. (c) Find the maximum speed vmax of the object. (d) Find the total energy E of the object–spring system. (e) Find the force constant k of the spring. (f) Describe the pattern of dependence of each of the quantities T, f, vmax, E, and k on m. 58. Review. This problem extends the reasoning of Prob-lem 75 in Chapter 9. Two gliders are set in motion on an air track. Glider 1 has mass m1 5 0.240 kg and moves to the right with speed 0.740 m/s. It will have a rear-end collision with glider 2, of mass m2 5 0.360 kg, which initially moves to the right with speed 0.120 m/s. A light spring of force constant 45.0 N/m is attached to the back end of glider 2 as shown in Figure P9.75. When glider 1 touches the spring, superglue instantly and permanently makes it stick to its end of the spring. (a) Find the common speed the two gliders have when the spring is at maximum compression. (b) Find the maximum spring compression distance. The motion after the gliders become attached consists of a combi-nation of (1) the constant-velocity motion of the center of mass of the two-glider system found in part (a) and (2) simple harmonic motion of the gliders relative to the center of mass. (c) Find the energy of the center-of-mass motion. (d) Find the energy of the oscillation. 59. A small ball of mass M is attached to the end of a uniform rod of equal mass M and length L that is pivoted at the top (Fig. P15.59). Determine the tensions in the rod (a) at the pivot and (b) at the point P when the system is stationary. (c) Calculate the period of oscilla-tion for small displacements from equilibrium and (d) determine this period for L 5 2.00 m. 60. Review. A rock rests on a concrete sidewalk. An earth-quake strikes, making the ground move vertically in simple harmonic motion with a constant frequency of 2.40 Hz and with gradually increasing amplitude. (a) With what amplitude does the ground vibrate when S M Q/C Q/C L P y Pivot y 0 M Figure P15.59 M Q/C t (s) 1 2 3 4 5 6 1.00 2.00 –1.00 0.00 –2.00 x (cm) Figure P15.64 www.aswarphysics.weebly.com Problems 479 angular frequency with which the plank moves with simple harmonic motion. 70. A horizontal plank of mass m and length L is pivoted at one end. The plank’s other end is supported by a spring of force con-stant k (Fig. P15.69). The plank is displaced by a small angle u from its hori-zontal equilibrium position and released. Find the angular frequency with which the plank moves with simple harmonic motion. 71. Review. A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizontal surface with an amplitude of 2.00 m. A 6.00-kg object is dropped vertically on top of the 4.00-kg object as it passes through its equilib-rium point. The two objects stick together. (a) What is the new amplitude of the vibrating system after the collision? (b) By what factor has the period of the sys-tem changed? (c) By how much does the energy of the system change as a result of the collision? (d) Account for the change in energy. 72. A ball of mass m is connected to two rubber bands of length L, each under tension T as shown in Figure P15.72. The ball is displaced by a small distance y per-pendicular to the length of the rubber bands. Assum-ing the tension does not change, show that (a) the restoring force is 2(2T/L)y and (b) the system exhibits simple harmonic motion with an angular frequency v 5 !2T/mL.
y L L Figure P15.72 73. Review. One end of a light spring with force constant k 5 100 N/m is attached to a vertical wall. A light string is tied to the other end of the horizontal spring. As shown in Figure P15.73, the string changes from hori-zontal to vertical as it passes over a pulley of mass M in the shape of a solid disk of radius R 5 2.00 cm. The pulley is free to turn on a fixed, smooth axle. The ver-tical section of the string supports an object of mass m 5 200 g. The string does not slip at its contact with the pulley. The object is pulled downward a small distance and released. (a) What is the angular frequency v of oscillation of the object in terms of the mass M? (b) What is the highest possible value of the angular fre-quency of oscillation of S Q/C S angular frequency, (d) the maximum speed, (e) the maximum acceleration, and (f) an equation for its posi-tion x as a function of time. 65. Review. A large block P attached to a light spring executes horizontal, simple harmonic motion as it slides across a frictionless surface with a frequency f 5 1.50 Hz. Block B rests on it as shown in Figure P15.65, and the coef-ficient of static friction between the two is ms 5 0.600. What maximum amplitude of oscillation can the system have if block B is not to slip? 66. Review. A large block P attached to a light spring exe-cutes horizontal, simple harmonic motion as it slides across a frictionless surface with a frequency f. Block B rests on it as shown in Figure P15.65, and the coeffi-cient of static friction between the two is ms. What max-imum amplitude of oscillation can the system have if block B is not to slip? 67. A pendulum of length L and mass M has a spring of force constant k connected to it at a distance h below its point of suspension (Fig. P15.67). Find the frequency of vibration of the system for small values of the amplitude (small u). Assume the vertical suspension rod of length L is rigid, but ignore its mass. 68. A block of mass m is connected to two springs of force constants k1 and k2 in two ways as shown in Figure P15.68. In both cases, the block moves on a fric-tionless table after it is displaced from equilibrium and released. Show that in the two cases the block exhibits simple harmonic motion with periods (a) T 5 2pÅ m1k1 1 k 22 k1k 2 and (b) T 5 2pÅ m k 1 1 k 2 m k1 k2 k1 k2 m a b Figure P15.68 69. A horizontal plank of mass 5.00 kg and length 2.00 m is pivoted at one end. The plank’s other end is supported by a spring of force constant 100 N/m (Fig. P15.69). The plank is displaced by a small angle u from its horizontal equilibrium position and released. Find the m B P s Figure P15.65 Problems 65 and 66.
S h L k u M Figure P15.67 S S m M k R Figure P15.73 Pivot k u L Figure P15.69 Problems 69 and 70.
www.aswarphysics.weebly.com 480 Chapter 15 Oscillatory Motion the motion. Take the density of air to be 1.20 kg/m3. Hint: Use an analogy with the simple pendulum and see Chapter 14. Assume the air applies a buoyant force on the balloon but does not otherwise affect its motion.
Air Air He He L L θ g S g S a b Figure P15.77 78. Consider the damped oscillator illustrated in Fig-ure 15.20. The mass of the object is 375 g, the spring constant is 100 N/m, and b 5 0.100 N ? s/m. (a) Over what time interval does the amplitude drop to half its initial value? (b) What If? Over what time interval does the mechanical energy drop to half its initial value? (c) Show that, in general, the fractional rate at which the amplitude decreases in a damped harmonic oscillator is one-half the fractional rate at which the mechanical energy decreases. 79. A particle with a mass of 0.500 kg is attached to a hori-zontal spring with a force constant of 50.0 N/m. At the moment t 5 0, the particle has its maximum speed of 20.0 m/s and is moving to the left. (a) Determine the particle’s equation of motion, specifying its posi-tion as a function of time. (b) Where in the motion is the potential energy three times the kinetic energy? (c) Find the minimum time interval required for the particle to move from x 5 0 to x 5 1.00 m. (d) Find the length of a simple pendulum with the same period. 80. Your thumb squeaks on a plate you have just washed. Your sneakers squeak on the gym floor. Car tires squeal when you start or stop abruptly. You can make a goblet sing by wiping your moistened finger around its rim. When chalk squeaks on a blackboard, you can see that it makes a row of regularly spaced dashes. As these examples suggest, vibration commonly results when friction acts on a moving elastic object. The oscillation is not simple harmonic motion, but is called stick-and-slip. This problem models stick-and-slip motion. A block of mass m is attached to a fixed support by a horizontal spring with force constant k and negligible mass (Fig. P15.80). Hooke’s law describes the spring S the object? (c) What is the highest possible value of the angular frequency of oscillation of the object if the pulley radius is doubled to R 5 4.00 cm? 74. People who ride motorcycles and bicycles learn to look out for bumps in the road and especially for wash-boarding, a condition in which many equally spaced ridges are worn into the road. What is so bad about washboarding? A motorcycle has several springs and shock absorbers in its suspension, but you can model it as a single spring supporting a block. You can esti-mate the force constant by thinking about how far the spring compresses when a heavy rider sits on the seat. A motorcyclist traveling at highway speed must be par-ticularly careful of washboard bumps that are a certain distance apart. What is the order of magnitude of their separation distance? 75. A simple pendulum with a length of 2.23 m and a mass of 6.74 kg is given an initial speed of 2.06 m/s at its equilibrium position. Assume it undergoes simple har-monic motion. Determine (a) its period, (b) its total energy, and (c) its maximum angular displacement. 76. When a block of mass M, connected to the end of a spring of mass ms 5 7.40 g and force constant k, is set into simple harmonic motion, the period of its motion is T 5 2pÅ M 1 1ms/32 k A two-part experiment is conducted with the use of blocks of various masses suspended vertically from the spring as shown in Figure P15.76. (a) Static extensions of 17.0, 29.3, 35.3, 41.3, 47.1, and 49.3 cm are measured for M values of 20.0, 40.0, 50.0, 60.0, 70.0, and 80.0 g, respec-tively. Construct a graph of Mg versus x and perform a linear least-squares fit to the data. (b) From the slope of your graph, determine a value for k for this spring. (c) The system is now set into sim-ple harmonic motion, and periods are measured with a stopwatch. With M 5 80.0 g, the total time interval required for ten oscillations is measured to be 13.41 s. The experiment is repeated with M values of 70.0, 60.0, 50.0, 40.0, and 20.0 g, with corresponding time intervals for ten oscillations of 12.52, 11.67, 10.67, 9.62, and 7.03 s. Make a table of these masses and times. (d) Compute the experimental value for T from each of these measurements. (e) Plot a graph of T 2 versus M and (f) determine a value for k from the slope of the linear least-squares fit through the data points. (g) Compare this value of k with that obtained in part (b). (h) Obtain a value for ms from your graph and compare it with the given value of 7.40 g. 77. Review. A light balloon filled with helium of density 0.179 kg/m3 is tied to a light string of length L 5 3.00 m. The string is tied to the ground forming an “inverted” simple pendulum (Fig. 15.77a). If the bal-loon is displaced slightly from equilibrium as in Fig-ure P15.77b and released, (a) show that the motion is simple harmonic and (b) determine the period of Q/C W AMT M Figure P15.76 v S k Figure P15.80 www.aswarphysics.weebly.com Problems 481 Challenge Problems 84. A smaller disk of radius r and mass m is attached rigidly to the face of a second larger disk of radius R and mass M as shown in Figure P15.84. The center of the small disk is located at the edge of the large disk. The large disk is mounted at its center on a frictionless axle. The assem-bly is rotated through a small angle u from its equi-librium position and released. (a) Show that the speed of the center of the small disk as it passes through the equilibrium position is v 5 2c Rg 11 2 cos u2 1M/m 2 1 1r/R2 2 1 2d 1/2 (b) Show that the period of the motion is T 5 2pc 1M 1 2m2R 2 1 mr 2 2mgR d 1/2 85. An object of mass m1 5 9.00 kg is in equilibrium when connected to a light spring of constant k 5 100 N/m that is fastened to a wall as shown in Figure P15.85a. A second object, m2 5 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A 5 0.200 m (see Fig. P15.85b). The system is then released, and both objects start moving to the right on the fric-tionless surface. (a) When m1 reaches the equilibrium point, m2 loses contact with m1 (see Fig. P15.85c) and moves to the right with speed v. Determine the value of v. (b) How far apart are the objects when the spring is fully stretched for the first time (the distance D in Fig. P15.85d)?
A m1 m2 m1 m2 m1 m2 m1 k k k k D v S v S b c d a Figure P15.85 86. Review. Why is the following situation impossible? You are in the high-speed package delivery business. Your com-petitor in the next building gains the right-of-way to R M u u v S m r Figure P15.84 S S both in extension and in compression. The block sits on a long horizontal board, with which it has coeffi-cient of static friction ms and a smaller coefficient of kinetic friction mk. The board moves to the right at constant speed v. Assume the block spends most of its time sticking to the board and moving to the right with it, so the speed v is small in comparison to the aver-age speed the block has as it slips back toward the left. (a) Show that the maximum extension of the spring from its unstressed position is very nearly given by msmg/k. (b) Show that the block oscillates around an equilibrium position at which the spring is stretched by mkmg/k. (c) Graph the block’s position versus time. (d) Show that the amplitude of the block’s motion is A 5 1 ms 2 mk 2mg k (e) Show that the period of the block’s motion is T 5 21 ms 2 mk 2mg vk 1 pÅ m k It is the excess of static over kinetic friction that is important for the vibration. “The squeaky wheel gets the grease” because even a viscous fluid cannot exert a force of static friction. 81. Review. A lobsterman’s buoy is a solid wooden cylinder of radius r and mass M. It is weighted at one end so that it floats upright in calm seawater, having density r. A passing shark tugs on the slack rope mooring the buoy to a lobster trap, pulling the buoy down a distance x from its equilibrium position and releasing it. (a) Show that the buoy will execute simple harmonic motion if the resistive effects of the water are ignored. (b) Deter-mine the period of the oscillations. 82. Why is the following situation impossible? Your job involves building very small damped oscillators. One of your designs involves a spring–object oscillator with a spring of force constant k 5 10.0 N/m and an object of mass m 5 1.00 g. Your design objective is that the oscilla-tor undergo many oscillations as its amplitude falls to 25.0% of its initial value in a certain time interval. Measurements on your latest design show that the amplitude falls to the 25.0% value in 23.1 ms. This time interval is too long for what is needed in your project. To shorten the time interval, you double the damping constant b for the oscillator. This doubling allows you to reach your design objective. 83. Two identical steel balls, each of mass 67.4 g, are mov-ing in opposite directions at 5.00 m/s. They collide head-on and bounce apart elastically. By squeezing one of the balls in a vise while precise measurements are made of the resulting amount of compression, you find that Hooke’s law is a good model of the ball’s elas-tic behavior. A force of 16.0 kN exerted by each jaw of the vise reduces the diameter by 0.200 mm. Model the motion of each ball, while the balls are in contact, as one-half of a cycle of simple harmonic motion. Com-pute the time interval for which the balls are in con-tact. (If you solved Problem 57 in Chapter 7, compare your results from this problem with your results from that one.) S www.aswarphysics.weebly.com 482 Chapter 15 Oscillatory Motion is proportional to the distance x from the fixed end; that is, vx 5 (x/,)v. Also, notice that the mass of a seg-ment of the spring is dm 5 (m/,)dx. Find (a) the kinetic energy of the system when the block has a speed v and (b) the period of oscillation. 88. Review. A system consists of a spring with force con-stant k 5 1 250 N/m, length L 5 1.50 m, and an object of mass m 5 5.00 kg attached to the end (Fig. P15.88). The object is placed at the level of the point of attach-ment with the spring unstretched, at position yi 5 L, and then it is released so that it swings like a pendu-lum. (a) Find the y position of the object at the lowest point. (b) Will the pendulum’s period be greater or less than the period of a simple pendulum with the same mass m and length L? Explain.
v S y yi L x L yf L m Figure P15.88 89. A light, cubical container of volume a3 is initially filled with a liquid of mass density r as shown in Figure P15.89a. The cube is initially supported by a light string to form a simple pendulum of length Li, measured from the center of mass of the filled container, where Li .. a. The liquid is allowed to flow from the bottom of the container at a constant rate (dM/dt). At any time t, the level of the liquid in the container is h and the length of the pendulum is L (measured relative to the instantaneous cen-ter of mass) as shown in Figure P15.89b. (a) Find the period of the pendu-lum as a function of time. (b) What is the period of the pendulum after the liquid completely runs out of the container?
a a h L Li a b Figure P15.89 S build an evacuated tunnel just above the ground all the way around the Earth. By firing packages into this tunnel at just the right speed, your competitor is able to send the packages into orbit around the Earth in this tunnel so that they arrive on the exact opposite side of the Earth in a very short time interval. You come up with a competing idea. Figuring that the dis-tance through the Earth is shorter than the distance around the Earth, you obtain permits to build an evac-uated tunnel through the center of the Earth (Fig. P15.86). By simply dropping packages into this tunnel, they fall downward and arrive at the other end of your tunnel, which is in a building right next to the other end of your competitor’s tunnel. Because your pack-ages arrive on the other side of the Earth in a shorter time interval, you win the competition and your busi-ness flourishes. Note: An object at a distance r from the center of the Earth is pulled toward the center of the Earth only by the mass within the sphere of radius r (the reddish region in Fig. P15.86). Assume the Earth has uniform density.
Earth Tunnel m r Figure P15.86 87. A block of mass M is connected to a spring of mass m and oscillates in simple harmonic motion on a fric-tionless, horizontal track (Fig. P15.87). The force con-stant of the spring is k, and the equilibrium length is ,. Assume all portions of the spring oscillate in phase and the velocity of a segment of the spring of length dx S x dx M v S Figure P15.87 www.aswarphysics.weebly.com Lifeguards in New South Wales, Australia, practice taking their boat over large water waves breaking near the shore. A wave moving over the surface of water is one example of a mechanical wave. (Travel Ink/Gallo Images/Getty Images) 16.1 Propagation of a Disturbance 16.2 Analysis Model: Traveling Wave 16.3 The Speed of Waves on Strings 16.4 Reflection and Transmission 16.5 Rate of Energy Transfer by Sinusoidal Waves on Strings 16.6 The Linear Wave Equation c h a p t e r 16 Wave Motion 483 Many of us experienced waves as children when we dropped a pebble into a pond. At the point the pebble hits the water’s surface, circular waves are created. These waves move outward from the creation point in expanding circles until they reach the shore. If you were to examine carefully the motion of a small object floating on the disturbed water, you would see that the object moves vertically and horizontally about its original position but does not undergo any net displacement away from or toward the point at which the pebble hit the water. The small elements of water in contact with the object, as well as all the other water elements on the pond’s surface, behave in the same way. That is, the water wave moves from the point of origin to the shore, but the water is not carried with it. The world is full of waves, the two main types being mechanical waves and electromag-netic waves. In the case of mechanical waves, some physical medium is being disturbed; in our pebble example, elements of water are disturbed. Electromagnetic waves do not require a medium to propagate; some examples of electromagnetic waves are visible light, radio waves, television signals, and x-rays. Here, in this part of the book, we study only mechanical waves. Consider again the small object floating on the water. We have caused the object to move at one point in the water by dropping a pebble at another location. The object has gained kinetic energy from our action, so energy must have transferred from the point at www.aswarphysics.weebly.com 484 Chapter 16 Wave Motion which the pebble is dropped to the position of the object. This feature is central to wave motion: energy is transferred over a distance, but matter is not.
16.1 Propagation of a Disturbance The introduction to this chapter alluded to the essence of wave motion: the trans-fer of energy through space without the accompanying transfer of matter. In the list of energy transfer mechanisms in Chapter 8, two mechanisms—mechanical waves and electromagnetic radiation—depend on waves. By contrast, in another mecha-nism, matter transfer, the energy transfer is accompanied by a movement of matter through space with no wave character in the process. All mechanical waves require (1) some source of disturbance, (2) a medium con-taining elements that can be disturbed, and (3) some physical mechanism through which elements of the medium can influence each other. One way to demonstrate wave motion is to flick one end of a long string that is under tension and has its opposite end fixed as shown in Figure 16.1. In this manner, a single bump (called a pulse) is formed and travels along the string with a definite speed. Figure 16.1 represents four consecutive “snapshots” of the creation and propagation of the trav-eling pulse. The hand is the source of the disturbance. The string is the medium through which the pulse travels—individual elements of the string are disturbed from their equilibrium position. Furthermore, the elements of the string are con-nected together so they influence each other. The pulse has a definite height and a definite speed of propagation along the medium. The shape of the pulse changes very little as it travels along the string.1 We shall first focus on a pulse traveling through a medium. Once we have explored the behavior of a pulse, we will then turn our attention to a wave, which is a periodic disturbance traveling through a medium. We create a pulse on our string by flicking the end of the string once as in Figure 16.1. If we were to move the end of the string up and down repeatedly, we would create a traveling wave, which has characteristics a pulse does not have. We shall explore these characteristics in Section 16.2. As the pulse in Figure 16.1 travels, each disturbed element of the string moves in a direction perpendicular to the direction of propagation. Figure 16.2 illustrates this point for one particular element, labeled P. Notice that no part of the string ever moves in the direction of the propagation. A traveling wave or pulse that causes the elements of the disturbed medium to move perpendicular to the direction of propagation is called a transverse wave. Compare this wave with another type of pulse, one moving down a long, stretched spring as shown in Figure 16.3. The left end of the spring is pushed briefly to the right and then pulled briefly to the left. This movement creates a sudden compres-sion of a region of the coils. The compressed region travels along the spring (to the right in Fig. 16.3). Notice that the direction of the displacement of the coils is parallel to the direction of propagation of the compressed region. A traveling wave or pulse that causes the elements of the medium to move parallel to the direction of propagation is called a longitudinal wave.
As the pulse moves along the string, new elements of the string are displaced from their equilibrium positions.
Figure 16.1 A hand moves the end of a stretched string up and down once (red arrow), causing a pulse to travel along the string.
1In reality, the pulse changes shape and gradually spreads out during the motion. This effect, called dispersion, is com-mon to many mechanical waves as well as to electromagnetic waves. We do not consider dispersion in this chapter.
The direction of the displacement of any element at a point P on the string is perpendicular to the direction of propagation (red arrow).
P P P Figure 16.2 The displacement of a particular string element for a transverse pulse traveling on a stretched string.
As the pulse passes by, the displacement of the coils is parallel to the direction of the propagation.
The hand moves forward and back once to create a longitudinal pulse.
Figure 16.3 A longitudinal pulse along a stretched spring.
www.aswarphysics.weebly.com 16.1 Propagation of a Disturbance 485 Sound waves, which we shall discuss in Chapter 17, are another example of lon-gitudinal waves. The disturbance in a sound wave is a series of high-pressure and low-pressure regions that travel through air. Some waves in nature exhibit a combination of transverse and longitudinal displacements. Surface-water waves are a good example. When a water wave trav-els on the surface of deep water, elements of water at the surface move in nearly circular paths as shown in Figure 16.4. The disturbance has both transverse and longitudinal components. The transverse displacements seen in Figure 16.4 rep-resent the variations in vertical position of the water elements. The longitudinal displacements represent elements of water moving back and forth in a horizontal direction. The three-dimensional waves that travel out from a point under the Earth’s sur-face at which an earthquake occurs are of both types, transverse and longitudinal. The longitudinal waves are the faster of the two, traveling at speeds in the range of 7 to 8 km/s near the surface. They are called P waves, with “P” standing for primary, because they travel faster than the transverse waves and arrive first at a seismo-graph (a device used to detect waves due to earthquakes). The slower transverse waves, called S waves, with “S” standing for secondary, travel through the Earth at 4 to 5 km/s near the surface. By recording the time interval between the arrivals of these two types of waves at a seismograph, the distance from the seismograph to the point of origin of the waves can be determined. This distance is the radius of an imaginary sphere centered on the seismograph. The origin of the waves is located somewhere on that sphere. The imaginary spheres from three or more monitoring stations located far apart from one another intersect at one region of the Earth, and this region is where the earthquake occurred. Consider a pulse traveling to the right on a long string as shown in Figure 16.5. Figure 16.5a represents the shape and position of the pulse at time t 5 0. At this time, the shape of the pulse, whatever it may be, can be represented by some math-ematical function that we will write as y(x, 0) 5 f(x). This function describes the transverse position y of the element of the string located at each value of x at time t 5 0. Because the speed of the pulse is v, the pulse has traveled to the right a distance vt at the time t (Fig. 16.5b). We assume the shape of the pulse does not change with time. Therefore, at time t, the shape of the pulse is the same as it was at time t 5 0 as in Figure 16.5a. Consequently, an element of the string at x at this time has the same y position as an element located at x 2 vt had at time t 5 0: y(x, t) 5 y(x 2 vt, 0) In general, then, we can represent the transverse position y for all positions and times, measured in a stationary frame with the origin at O, as y(x, t) 5 f(x 2 vt) (16.1) Similarly, if the pulse travels to the left, the transverse positions of elements of the string are described by y(x, t) 5 f(x 1 vt) (16.2) The function y, sometimes called the wave function, depends on the two vari-ables x and t. For this reason, it is often written y(x, t), which is read “y as a function of x and t.” It is important to understand the meaning of y. Consider an element of the string at point P in Figure 16.5, identified by a particular value of its x coordinate. As the pulse passes through P, the y coordinate of this element increases, reaches a maximum, and then decreases to zero. The wave function y(x, t) represents the y coordinate—the transverse position—of any element located at position x at any time t. Furthermore, if t is fixed (as, for example, in the case of taking a snapshot of the pulse), the wave function y(x), sometimes called the waveform, defines a curve representing the geometric shape of the pulse at that time.
Figure 16.4 The motion of water elements on the surface of deep water in which a wave is propagating is a combination of transverse and longitudinal displacements. The elements at the surface move in nearly circular paths. Each element is displaced both horizontally and vertically from its equilibrium position.
Trough Velocity of propagation Crest y O vt x O y x P P v S v S At t 0, the shape of the pulse is given by y f(x).
At some later time t, the shape of the pulse remains unchanged and the vertical position of an element of the medium at any point P is given by y f(x vt).
b a Figure 16.5 A one-dimensional pulse traveling to the right on a string with a speed v.
www.aswarphysics.weebly.com 486 Chapter 16 Wave Motion Example 16.1 A Pulse Moving to the Right A pulse moving to the right along the x axis is represented by the wave function y1x, t2 5 2 1x 2 3.0t2 2 1 1 where x and y are measured in centimeters and t is measured in sec-onds. Find expressions for the wave function at t 5 0, t 5 1.0 s, and t 5 2.0 s.
Conceptualize Figure 16.6a shows the pulse represented by this wave function at t 5 0. Imagine this pulse moving to the right at a speed of 3.0 cm/s and maintaining its shape as suggested by Figures 16.6b and 16.6c.
Categorize We categorize this example as a relatively simple analysis problem in which we interpret the mathematical representation of a pulse.
Analyze The wave function is of the form y 5 f(x 2 vt). Inspection of the expression for y(x, t) and comparison to Equation 16.1 reveal that the wave speed is v 5 3.0 cm/s. Further-more, by letting x 2 3.0t 5 0, we find that the maximum value of y is given by A 5 2.0 cm.
S o l u t i o n Finalize These snapshots show that the pulse moves to the right without changing its shape and that it has a constant speed of 3.0 cm/s.
Q uick Quiz 16.1 (i) In a long line of people waiting to buy tickets, the first person leaves and a pulse of motion occurs as people step forward to fill the gap. As each person steps forward, the gap moves through the line. Is the propaga-tion of this gap (a) transverse or (b) longitudinal? (ii) Consider “the wave” at a baseball game: people stand up and raise their arms as the wave arrives at their location, and the resultant pulse moves around the stadium. Is this wave (a) transverse or (b) longitudinal?
t 2.0 s t 1.0 s t 0 y (x, 2.0) y (x, 1.0) y (x, 0) 3.0 cm/s 3.0 cm/s 3.0 cm/s y (cm) 2.0 1.5 1.0 0.5 0 1 2 3 4 5 6 x (cm) 7 8 y (cm) 2.0 1.5 1.0 0.5 0 1 2 3 4 5 6 x (cm) 7 8 y (cm) 2.0 1.5 1.0 0.5 0 1 2 3 4 5 6 x (cm) 7 8 a b c Figure 16.6 (Example 16.1) Graphs of the function y(x, t) 5 2/[(x 23.0t)2 1 1] at (a) t 5 0, (b) t 5 1.0 s, and (c) t 5 2.0 s.
Write the wave function expression at t 5 0: y(x, 0) 5 2 x 2 1 1 Write the wave function expression at t 5 1.0 s: y(x, 1.0) 5 2 1x 2 3.02 2 1 1 Write the wave function expression at t 5 2.0 s: y(x, 2.0) 5 2 1x 2 6.02 2 1 1 For each of these expressions, we can substitute various values of x and plot the wave function. This procedure yields the wave functions shown in the three parts of Figure 16.6.
What if the wave function were y1x, t2 5 4 1x 1 3.0t2 2 1 1 How would that change the situation?
Answer One new feature in this expression is the plus sign in the denominator rather than the minus sign. The new expression represents a pulse with a similar shape as that in Figure 16.6, but moving to the left as time progresses.
What If?
www.aswarphysics.weebly.com 16.2 Analysis Model: Traveling Wave 487 16.2 Analysis Model: Traveling Wave In this section, we introduce an important wave function whose shape is shown in Figure 16.7. The wave represented by this curve is called a sinusoidal wave because the curve is the same as that of the function sin u plotted against u. A sinusoidal wave could be established on the rope in Figure 16.1 by shaking the end of the rope up and down in simple harmonic motion. The sinusoidal wave is the simplest example of a periodic continuous wave and can be used to build more complex waves (see Section 18.8). The brown curve in Figure 16.7 represents a snapshot of a traveling sinusoidal wave at t 5 0, and the blue curve represents a snapshot of the wave at some later time t. Imagine two types of motion that can occur. First, the entire waveform in Figure 16.7 moves to the right so that the brown curve moves toward the right and eventually reaches the position of the blue curve. This movement is the motion of the wave. If we focus on one element of the medium, such as the element at x 5 0, we see that each element moves up and down along the y axis in simple harmonic motion. This movement is the motion of the elements of the medium. It is important to differentiate between the motion of the wave and the motion of the elements of the medium. In the early chapters of this book, we developed several analysis models based on three simplification models: the particle, the system, and the rigid object. With our introduction to waves, we can develop a new simplification model, the wave, that will allow us to explore more analysis models for solving problems. An ideal particle has zero size. We can build physical objects with nonzero size as combinations of particles. Therefore, the particle can be considered a basic building block. An ideal wave has a single frequency and is infinitely long; that is, the wave exists throughout the Universe. (A wave of finite length must necessarily have a mixture of frequen-cies.) When this concept is explored in Section 18.8, we will find that ideal waves can be combined to build complex waves, just as we combined particles. In what follows, we will develop the principal features and mathematical represen-tations of the analysis model of a traveling wave. This model is used in situations in which a wave moves through space without interacting with other waves or particles. Figure 16.8a shows a snapshot of a traveling wave moving through a medium. Figure 16.8b shows a graph of the position of one element of the medium as a func-tion of time. A point in Figure 16.8a at which the displacement of the element from its normal position is highest is called the crest of the wave. The lowest point is called the trough. The distance from one crest to the next is called the wavelength l (Greek letter lambda). More generally, the wavelength is the minimum distance between any two identical points on adjacent waves as shown in Figure 16.8a. If you count the number of seconds between the arrivals of two adjacent crests at a given point in space, you measure the period T of the waves. In general, the period is the time interval required for two identical points of adjacent waves to pass by a point as shown in Figure 16.8b. The period of the wave is the same as the period of the simple harmonic oscillation of one element of the medium. The same information is more often given by the inverse of the period, which is called the frequency f. In general, the frequency of a periodic wave is the number of crests (or troughs, or any other point on the wave) that pass a given point in a unit time interval. The frequency of a sinusoidal wave is related to the period by the expression f 5 1 T (16.3) t 0 t y x vt v S Figure 16.7 A one-dimensional sinusoidal wave traveling to the right with a speed v. The brown curve represents a snapshot of the wave at t 5 0, and the blue curve represents a snapshot at some later time t. ▸ 16.1 c on tin u ed Another new feature here is the numerator of 4 rather than 2. Therefore, the new expression represents a pulse with twice the height of that in Figure 16.6.
y x T y t A A T l l The wavelength l of a wave is the distance between adjacent crests or adjacent troughs.
The period T of a wave is the time interval required for the element to complete one cycle of its oscillation and for the wave to travel one wavelength.
a b Figure 16.8 (a) A snapshot of a sinusoidal wave. (b) The position of one element of the medium as a function of time.
www.aswarphysics.weebly.com 488 Chapter 16 Wave Motion The frequency of the wave is the same as the frequency of the simple harmonic oscillation of one element of the medium. The most common unit for frequency, as we learned in Chapter 15, is s21, or hertz (Hz). The corresponding unit for T is seconds. The maximum position of an element of the medium relative to its equilibrium position is called the amplitude A of the wave as indicated in Figure 16.8. Waves travel with a specific speed, and this speed depends on the properties of the medium being disturbed. For instance, sound waves travel through room- temperature air with a speed of about 343 m/s (781 mi/h), whereas they travel through most solids with a speed greater than 343 m/s. Consider the sinusoidal wave in Figure 16.8a, which shows the position of the wave at t 5 0. Because the wave is sinusoidal, we expect the wave function at this instant to be expressed as y(x, 0) 5 A sin ax, where A is the amplitude and a is a constant to be determined. At x 5 0, we see that y(0, 0) 5 A sin a(0) 5 0, consistent with Figure 16.8a. The next value of x for which y is zero is x 5 l/2. Therefore, y al 2, 0b 5 A sin aa l 2b 5 0 For this equation to be true, we must have al/2 5 p, or a 5 2p/l. Therefore, the function describing the positions of the elements of the medium through which the sinusoidal wave is traveling can be written y1x, 02 5 A sin a2p l xb (16.4) where the constant A represents the wave amplitude and the constant l is the wave-length. Notice that the vertical position of an element of the medium is the same whenever x is increased by an integral multiple of l. Based on our discussion of Equation 16.1, if the wave moves to the right with a speed v, the wave function at some later time t is y1x, t 2 5 A sin c 2p l 1x 2 vt2 d (16.5) If the wave were traveling to the left, the quantity x 2 vt would be replaced by x 1 vt as we learned when we developed Equations 16.1 and 16.2. By definition, the wave travels through a displacement Dx equal to one wave-length l in a time interval Dt of one period T. Therefore, the wave speed, wave-length, and period are related by the expression v 5 Dx Dt 5 l T (16.6) Substituting this expression for v into Equation 16.5 gives y 5 A sin c2pax l 2 t Tb d (16.7) This form of the wave function shows the periodic nature of y. Note that we will often use y rather than y(x, t) as a shorthand notation. At any given time t, y has the same value at the positions x, x 1 l, x 1 2l, and so on. Furthermore, at any given position x, the value of y is the same at times t, t 1 T, t 1 2T, and so on. We can express the wave function in a convenient form by defining two other quantities, the angular wave number k (usually called simply the wave number) and the angular frequency v: k ; 2p l (16.8) v ; 2p T 5 2pf (16.9) Angular wave number Angular frequency Pitfall Prevention 16.1 What’s the Difference Between Figures 16.8a and 16.8b? Notice the visual similarity between Fig-ures 16.8a and 16.8b. The shapes are the same, but (a) is a graph of vertical position versus horizontal position, whereas (b) is vertical position versus time. Figure 16.8a is a pictorial representation of the wave for a series of elements of the medium; it is what you would see at an instant of time. Figure 16.8b is a graphical representation of the position of one element of the medium as a function of time. That both figures have the identical shape represents Equation 16.1: a wave is the same function of both x and t.
www.aswarphysics.weebly.com 16.2 Analysis Model: Traveling Wave 489 Using these definitions, Equation 16.7 can be written in the more compact form y 5 A sin (kx 2 vt) (16.10) Using Equations 16.3, 16.8, and 16.9, the wave speed v originally given in Equa-tion 16.6 can be expressed in the following alternative forms: v 5 v k (16.11) v 5 lf (16.12) The wave function given by Equation 16.10 assumes the vertical position y of an element of the medium is zero at x 5 0 and t 5 0. That need not be the case. If it is not, we generally express the wave function in the form y 5 A sin (kx 2 vt 1 f) (16.13) where f is the phase constant, just as we learned in our study of periodic motion in Chapter 15. This constant can be determined from the initial conditions. The pri-mary equations in the mathematical representation of the traveling wave analysis model are Equations 16.3, 16.10, and 16.12.
Q uick Quiz 16.2 A sinusoidal wave of frequency f is traveling along a stretched string. The string is brought to rest, and a second traveling wave of frequency 2f is established on the string. (i) What is the wave speed of the second wave? (a) twice that of the first wave (b) half that of the first wave (c) the same as that of the first wave (d) impossible to determine (ii) From the same choices, describe the wavelength of the second wave. (iii) From the same choices, describe the amplitude of the second wave.
W W Wave function for a sinusoidal wave W W Speed of a sinusoidal wave W W General expression for a sinusoidal wave Example 16.2 A Traveling Sinusoidal Wave A sinusoidal wave traveling in the positive x direction has an amplitude of 15.0 cm, a wavelength of 40.0 cm, and a frequency of 8.00 Hz. The vertical position of an element of the medium at t 5 0 and x 5 0 is also 15.0 cm as shown in Figure 16.9.
(A) Find the wave number k, period T, angular frequency v, and speed v of the wave.
Conceptualize Figure 16.9 shows the wave at t 5 0. Imagine this wave moving to the right and maintain-ing its shape.
Categorize From the description in the problem state-ment, we see that we are analyzing a mechanical wave moving through a medium, so we categorize the prob-lem with the traveling wave model.
Analyze AM S o l u t i o n y (cm) 40.0 cm 15.0 cm x (cm) Figure 16.9 (Example 16.2) A sinusoidal wave of wavelength l 5 40.0 cm and amplitude A 5 15.0 cm.
continued Evaluate the wave number from Equation 16.8: k 5 2p l 5 2p rad 40.0 cm 5 15.7 rad/m Evaluate the period of the wave from Equation 16.3: T 5 1 f 5 1 8.00 s21 5 0.125 s Evaluate the angular frequency of the wave from Equa-tion 16.9: v 5 2pf 5 2p(8.00 s21) 5 50.3 rad/s Evaluate the wave speed from Equation 16.12: v 5 lf 5 (40.0 cm)(8.00 s21) 5 3.20 m/s www.aswarphysics.weebly.com 490 Chapter 16 Wave Motion Substitute A 5 15.0 cm, y 5 15.0 cm, x 5 0, and t 5 0 into Equation 16.13: 15.0 5 115.02 sin f S sin f 5 1 S f 5 p 2 rad Write the wave function: y 5 A sin akx 2 vt 1 p 2 b 5 A cos 1kx 2 vt2 (B) Determine the phase constant f and write a general expression for the wave function.
S o l u t i o n Substitute the values for A, k, and v in SI units into this expression: y 5 0.150 cos (15.7x 2 50.3t) Sinusoidal Waves on Strings In Figure 16.1, we demonstrated how to create a pulse by jerking a taut string up and down once. To create a series of such pulses—a wave—let’s replace the hand with an oscillating blade vibrating in simple harmonic motion. Figure 16.10 repre-sents snapshots of the wave created in this way at intervals of T/4. Because the end of the blade oscillates in simple harmonic motion, each element of the string, such as that at P, also oscillates vertically with simple harmonic motion. Therefore, every element of the string can be treated as a simple harmonic oscillator vibrating with a frequency equal to the frequency of oscillation of the blade.2 Notice that while each element oscillates in the y direction, the wave travels to the right in the 1x direction with a speed v. Of course, that is the definition of a transverse wave. If we define t 5 0 as the time for which the configuration of the string is as shown in Figure 16.10a, the wave function can be written as y 5 A sin (kx 2 vt) We can use this expression to describe the motion of any element of the string. An ele-ment at point P (or any other element of the string) moves only vertically, and so its x coordinate remains constant. Therefore, the transverse speed vy (not to be confused with the wave speed v) and the transverse acceleration ay of elements of the string are vy 5 dy dt d x5constant 5 'y 't 5 2vA cos 1kx 2 vt2 (16.14) a y 5 dvy dt d x5constant 5 'vy 't 5 2v2 A sin 1kx 2 vt2 (16.15) These expressions incorporate partial derivatives because y depends on both x and t. In the operation 'y/'t, for example, we take a derivative with respect to t while holding x constant. The maximum magnitudes of the transverse speed and trans-verse acceleration are simply the absolute values of the coefficients of the cosine and sine functions: vy , max 5 vA (16.16) a y , max 5 v2A (16.17) The transverse speed and transverse acceleration of elements of the string do not reach their maximum values simultaneously. The transverse speed reaches its max-imum value (vA) when y 5 0, whereas the magnitude of the transverse acceleration 2In this arrangement, we are assuming that a string element always oscillates in a vertical line. The tension in the string would vary if an element were allowed to move sideways. Such motion would make the analysis very complex.
P t = 0 t = T A P P P l 4 1 t = T 2 1 t = T 4 3 a b c d x y Figure 16.10 One method for producing a sinusoidal wave on a string. The left end of the string is connected to a blade that is set into oscillation. Every element of the string, such as that at point P, oscillates with simple harmonic motion in the vertical direction. ▸ 16.2 continu ed Finalize Review the results carefully and make sure you understand them. How would the graph in Figure 16.9 change if the phase angle were zero? How would the graph change if the amplitude were 30.0 cm? How would the graph change if the wavelength were 10.0 cm?
www.aswarphysics.weebly.com 16.3 The Speed of Waves on Strings 491 Pitfall Prevention 16.2 Two Kinds of Speed/Velocity Do not confuse v, the speed of the wave as it propagates along the string, with vy, the transverse velocity of a point on the string. The speed v is constant for a uni-form medium, whereas vy varies sinusoidally.
reaches its maximum value (v2A) when y 5 6A. Finally, Equations 16.16 and 16.17 are identical in mathematical form to the corresponding equations for simple har-monic motion, Equations 15.17 and 15.18.
Q uick Quiz 16.3 The amplitude of a wave is doubled, with no other changes made to the wave. As a result of this doubling, which of the following state-ments is correct? (a) The speed of the wave changes. (b) The frequency of the wave changes. (c) The maximum transverse speed of an element of the medium changes. (d) Statements (a) through (c) are all true. (e) None of statements (a) through (c) is true.
Imagine a source vibrating such that it influences the medium that is in contact with the source. Such a source creates a disturbance that propagates through the medium. If the source vibrates in simple harmonic motion with period T, sinusoidal waves propa-gate through the medium at a speed given by v 5 l T 5 lf (16.6, 16.12) where l is the wavelength of the wave and f is its frequency. A sinu-soidal wave can be expressed as y 5 A sin 1kx 2 vt2 (16.10) Analysis Model Traveling Wave where A is the amplitude of the wave, k is its wave number, and v is its angular frequency.
Examples: • a vibrating blade sends a sinusoidal wave down a string attached to the blade • a loudspeaker vibrates back and forth, emitting sound waves into the air (Chap-ter 17) • a guitar body vibrates, emitting sound waves into the air (Chapter 18) • a vibrating electric charge creates an elec-tromagnetic wave that propagates into space at the speed of light (Chapter 34) 16.3 The Speed of Waves on Strings One aspect of the behavior of linear mechanical waves is that the wave speed depends only on the properties of the medium through which the wave travels. Waves for which the amplitude A is small relative to the wavelength l can be repre-sented as linear waves. (See Section 16.6.) In this section, we determine the speed of a transverse wave traveling on a stretched string. Let us use a mechanical analysis to derive the expression for the speed of a pulse traveling on a stretched string under tension T. Consider a pulse moving to the right with a uniform speed v, measured relative to a stationary (with respect to the Earth) inertial reference frame as shown in Figure 16.11a. Newton’s laws are valid in any inertial reference frame. Therefore, let us view this pulse from a different inertial reference frame, one that moves along with the pulse at the same speed so that the pulse appears to be at rest in the frame as in Figure 16.11b. In this refer-ence frame, the pulse remains fixed and each element of the string moves to the left through the pulse shape. A short element of the string, of length Ds, forms an approximate arc of a cir-cle of radius R as shown in the magnified view in Figure 16.11b. In our moving frame of reference, the element of the string moves to the left with speed v. As it travels through the arc, we can model the element as a particle in uniform cir-cular motion. This element has a centripetal acceleration of v 2/R, which is sup-plied by components of the force T S whose magnitude is the tension in the string. The force T S acts on each side of the element, tangent to the arc, as in Figure 16.11b. The horizontal components of T S cancel, and each vertical component T sin u acts downward. Hence, the magnitude of the total radial force on the element is 2T sin u. y x A l v S Figure 16.11 (a) In the refer-ence frame of the Earth, a pulse moves to the right on a string with speed v. (b) In a frame of refer-ence moving to the right with the pulse, the small element of length Ds moves to the left with speed v.
s O s R u u u v S v S T S T S a b www.aswarphysics.weebly.com 492 Chapter 16 Wave Motion Because the element is small, u is small and we can use the small-angle approxima-tion sin u < u. Therefore, the magnitude of the total radial force is Fr 5 2T sin u < 2Tu The element has mass m 5 m Ds, where m is the mass per unit length of the string. Because the element forms part of a circle and subtends an angle of 2u at the center, Ds 5 R(2u), and m 5 mDs 5 2mR u The element of the string is modeled as a particle under a net force. Therefore, applying Newton’s second law to this element in the radial direction gives Fr 5 mv 2 R S 2Tu 5 2mR uv 2 R S T 5 mv 2 Solving for v gives v 5 Å T m (16.18) Notice that this derivation is based on the assumption that the pulse height is small relative to the length of the pulse. Using this assumption, we were able to use the approximation sin u < u. Furthermore, the model assumes that the tension T is not affected by the presence of the pulse, so T is the same at all points on the pulse. Finally, this proof does not assume any particular shape for the pulse. We therefore conclude that a pulse of any shape will travel on the string with speed v 5 "T/m, without any change in pulse shape.
Q uick Quiz 16.4 Suppose you create a pulse by moving the free end of a taut string up and down once with your hand beginning at t 5 0. The string is attached at its other end to a distant wall. The pulse reaches the wall at time t. Which of the fol-lowing actions, taken by itself, decreases the time interval required for the pulse to reach the wall? More than one choice may be correct. (a) moving your hand more quickly, but still only up and down once by the same amount (b) moving your hand more slowly, but still only up and down once by the same amount (c) moving your hand a greater distance up and down in the same amount of time (d) moving your hand a lesser distance up and down in the same amount of time (e) using a heavier string of the same length and under the same tension (f) using a lighter string of the same length and under the same tension (g) using a string of the same linear mass density but under decreased tension (h) using a string of the same linear mass density but under increased tension Speed of a wave on a stretched string Example 16.3 The Speed of a Pulse on a Cord A uniform string has a mass of 0.300 kg and a length of 6.00 m (Fig. 16.12). The string passes over a pulley and sup-ports a 2.00-kg object. Find the speed of a pulse traveling along this string.
Conceptualize In Figure 16.12, the hanging block establishes a tension in the horizontal string. This tension determines the speed with which waves move on the string.
Categorize To find the tension in the string, we model the hang-ing block as a particle in equilibrium. Then we use the tension to evaluate the wave speed on the string using Equation 16.18.
AM S o l u t i o n 2.00 kg T Figure 16.12 (Example 16.3) The tension T in the cord is maintained by the suspended object. The speed of any wave traveling along the cord is given by v 5 !T/m.
Analyze Apply the particle in equilibrium model to the block: o Fy 5 T 2 m blockg 5 0 Solve for the tension in the string: T 5 m blockg Pitfall Prevention 16.3 Multiple T's Do not confuse the T in Equation 16.18 for the ten-sion with the symbol T used in this chapter for the period of a wave. The context of the equation should help you identify which quantity is meant. There simply aren’t enough letters in the alpha-bet to assign a unique letter to each variable!
www.aswarphysics.weebly.com 16.3 The Speed of Waves on Strings 493 Use Equation 16.18 to find the wave speed, using m 5 mstring/, for the linear mass density of the string: v 5 Å T m 5 Å m blockg , m string Evaluate the wave speed: v 5 Å 12.00 kg2 19.80 m/s22 16.00 m2 0.300 kg 5 19.8 m/s Finalize The calculation of the tension neglects the small mass of the string. Strictly speaking, the string can never be exactly straight; therefore, the tension is not uniform.
What if the block were swinging back and forth with respect to the vertical like a pendulum? How would that affect the wave speed on the string?
Answer The swinging block is categorized as a particle under a net force. The magnitude of one of the forces on the block is the tension in the string, which determines the wave speed. As the block swings, the tension changes, so the wave speed changes. When the block is at the bottom of the swing, the string is vertical and the tension is larger than the weight of the block because the net force must be upward to provide the centripetal acceleration of the block. Therefore, the wave speed must be greater than 19.8 m/s. When the block is at its highest point at the end of a swing, it is momentarily at rest, so there is no centripetal acceleration at that instant. The block is a particle in equilibrium in the radial direction. The tension is balanced by a component of the gravitational force on the block. Therefore, the tension is smaller than the weight and the wave speed is less than 19.8 m/s. With what frequency does the speed of the wave vary? Is it the same frequency as the pendulum? What If?
Example 16.4 Rescuing the Hiker An 80.0-kg hiker is trapped on a mountain ledge following a storm. A helicopter rescues the hiker by hovering above him and lowering a cable to him. The mass of the cable is 8.00 kg, and its length is 15.0 m. A sling of mass 70.0 kg is attached to the end of the cable. The hiker attaches himself to the sling, and the helicopter then accelerates upward. Terrified by hanging from the cable in midair, the hiker tries to signal the pilot by sending transverse pulses up the cable. A pulse takes 0.250 s to travel the length of the cable. What is the acceleration of the helicopter? Assume the tension in the cable is uniform.
Conceptualize Imagine the effect of the acceleration of the helicopter on the cable. The greater the upward accelera-tion, the larger the tension in the cable. In turn, the larger the tension, the higher the speed of pulses on the cable.
Categorize This problem is a combination of one involving the speed of pulses on a string and one in which the hiker and sling are modeled as a particle under a net force.
AM S o l u t i o n continued Analyze Use the time interval for the pulse to travel from the hiker to the helicopter to find the speed of the pulses on the cable: v 5 Dx Dt 5 15.0 m 0.250 s 5 60.0 m/s Solve Equation 16.18 for the tension in the cable: (1) v 5 Å T m S T 5 mv 2 ▸ 16.3 c on tin u ed Model the hiker and sling as a particle under a net force, noting that the acceleration of this particle of mass m is the same as the acceleration of the helicopter: o F 5 ma S T 2 mg 5 ma Solve for the acceleration and substitute the tension from Equation (1): a 5 T m 2 g 5 mv 2 m 2 g 5 m cable v 2 ,cablem 2g www.aswarphysics.weebly.com 494 Chapter 16 Wave Motion 16.4 Reflection and Transmission The traveling wave model describes waves traveling through a uniform medium without interacting with anything along the way. We now consider how a traveling wave is affected when it encounters a change in the medium. For example, consider a pulse traveling on a string that is rigidly attached to a support at one end as in Figure 16.13. When the pulse reaches the support, a severe change in the medium occurs: the string ends. As a result, the pulse undergoes reflection; that is, the pulse moves back along the string in the opposite direction. Notice that the reflected pulse is inverted. This inversion can be explained as follows. When the pulse reaches the fixed end of the string, the string produces an upward force on the support. By Newton’s third law, the support must exert an equal-magnitude and oppositely directed (downward) reaction force on the string. This downward force causes the pulse to invert upon reflection. Now consider another case. This time, the pulse arrives at the end of a string that is free to move vertically as in Figure 16.14. The tension at the free end is maintained because the string is tied to a ring of negligible mass that is free to slide vertically on a smooth post without friction. Again, the pulse is reflected, but this time it is not inverted. When it reaches the post, the pulse exerts a force on the free end of the string, causing the ring to accelerate upward. The ring rises as high as the incoming pulse, and then the downward component of the tension force pulls the ring back down. This movement of the ring produces a reflected pulse that is not inverted and that has the same amplitude as the incoming pulse. Finally, consider a situation in which the boundary is intermediate between these two extremes. In this case, part of the energy in the incident pulse is reflected and part undergoes transmission; that is, some of the energy passes through the bound-ary. For instance, suppose a light string is attached to a heavier string as in Figure 16.15. When a pulse traveling on the light string reaches the boundary between the two strings, part of the pulse is reflected and inverted and part is transmitted to the heavier string. The reflected pulse is inverted for the same reasons described earlier in the case of the string rigidly attached to a support. The reflected pulse has a smaller amplitude than the incident pulse. In Section 16.5, we show that the energy carried by a wave is related to its amplitude. Accord-ing to the principle of conservation of energy, when the pulse breaks up into a reflected pulse and a transmitted pulse at the boundary, the sum of the energies of these two pulses must equal the energy of the incident pulse. Because the reflected pulse contains only part of the energy of the incident pulse, its amplitude must be smaller. When a pulse traveling on a heavy string strikes the boundary between the heavy string and a lighter one as in Figure 16.16, again part is reflected and part is trans-mitted. In this case, the reflected pulse is not inverted. In either case, the relative heights of the reflected and transmitted pulses depend on the relative densities of the two strings. If the strings are identical, there is no discontinuity at the boundary and no reflection takes place.
Reflected pulse Incident pulse b c a Figure 16.13 The reflection of a traveling pulse at the fixed end of a stretched string. The reflected pulse is inverted, but its shape is otherwise unchanged.
Incident pulse Reflected pulse b c a Figure 16.14 The reflection of a traveling pulse at the free end of a stretched string. The reflected pulse is not inverted.
Substitute numerical values: a 5 18.00 kg2 160.0 m/s2 2 115.0 m2 1150.0 kg2 2 9.80 m/s2 5 3.00 m/s2 Finalize A real cable has stiffness in addition to tension. Stiffness tends to return a wire to its original straight-line shape even when it is not under tension. For example, a piano wire straightens if released from a curved shape; package- wrapping string does not. Stiffness represents a restoring force in addition to tension and increases the wave speed. Consequently, for a real cable, the speed of 60.0 m/s that we determined is most likely associated with a smaller acceleration of the helicopter. ▸ 16.4 continu ed www.aswarphysics.weebly.com 16.5 Rate of Energy Transfer by Sinusoidal Waves on Strings 495 According to Equation 16.18, the speed of a wave on a string increases as the mass per unit length of the string decreases. In other words, a wave travels more rapidly on a light string than on a heavy string if both are under the same tension. The following general rules apply to reflected waves: When a wave or pulse travels from medium A to medium B and vA . vB (that is, when B is denser than A), it is inverted upon reflection. When a wave or pulse travels from medium A to medium B and vA , vB (that is, when A is denser than B), it is not inverted upon reflection.
16.5 Rate of Energy Transfer by Sinusoidal Waves on Strings Waves transport energy through a medium as they propagate. For example, sup-pose an object is hanging on a stretched string and a pulse is sent down the string as in Figure 16.17a. When the pulse meets the suspended object, the object is momen-tarily displaced upward as in Figure 16.17b. In the process, energy is transferred to the object and appears as an increase in the gravitational potential energy of the object–Earth system. This section examines the rate at which energy is transported along a string. We shall assume a one-dimensional sinusoidal wave in the calcula-tion of the energy transferred. Consider a sinusoidal wave traveling on a string (Fig. 16.18). The source of the energy is some external agent at the left end of the string. We can consider the string to be a nonisolated system. As the external agent performs work on the end of the string, moving it up and down, energy enters the system of the string and propagates along its length. Let’s focus our attention on an infinitesimal element of the string of length dx and mass dm. Each such element oscillates vertically with its position described by Equation 15.6. Therefore, we can model each element of the string as a particle in simple harmonic motion, with the oscillation in the y direction. All elements have the same angular frequency v and the same ampli-tude A. The kinetic energy K associated with a moving particle is K 5 1 2mv 2. If we apply this equation to the infinitesimal element, the kinetic energy dK associated with the up and down motion of this element is dK 5 1 21dm2vy 2 where vy is the transverse speed of the element. If m is the mass per unit length of the string, the mass dm of the element of length dx is equal to m dx. Hence, we can express the kinetic energy of an element of the string as dK 5 1 21m dx2vy 2 (16.19) Incident pulse The reflected pulse is inverted and a non-inverted transmitted pulse moves on the heavier string.
b a Figure 16.15 (a) A pulse traveling to the right on a light string approaches the junction with a heavier string. (b) The situation after the pulse reaches the junction.
Figure 16.16 (a) A pulse traveling to the right on a heavy string approaches the junction with a lighter string. (b) The situation after the pulse reaches the junction.
Incident pulse The reflected pulse is not inverted and a transmitted pulse moves on the lighter string.
a b The pulse lifts the block, increasing the gravitational potential energy of the block–Earth system.
m m a b Figure 16.17 (a) A pulse travels to the right on a stretched string, carrying energy with it. (b) The energy of the pulse arrives at the hanging block.
dm Each element of the string is a simple harmonic oscillator and therefore has kinetic energy and potential energy associated with it.
Figure 16.18 A sinusoidal wave traveling along the x axis on a stretched string. www.aswarphysics.weebly.com 496 Chapter 16 Wave Motion Substituting for the general transverse speed of an element of the medium using Equation 16.14 gives dK 5 1 2m32vA cos 1kx 2 vt2 4 2 dx 5 1 2mv2A2 cos2 1kx 2 vt2 dx If we take a snapshot of the wave at time t 5 0, the kinetic energy of a given ele-ment is dK 5 1 2mv2 A2 cos2 kx dx Integrating this expression over all the string elements in a wavelength of the wave gives the total kinetic energy Kl in one wavelength: K l 5 3dK 5 3 l 0 1 2mv2A2 cos2 kx dx 5 1 2mv2A2 3 l 0 cos2 kx dx 5 1 2mv2A2 c 1 2x 1 1 4k sin 2kxd l 0 5 1 2mv2A2 31 2l4 5 1 4mv2A2l In addition to kinetic energy, there is potential energy associated with each ele-ment of the string due to its displacement from the equilibrium position and the restoring forces from neighboring elements. A similar analysis to that above for the total potential energy Ul in one wavelength gives exactly the same result: Ul 5 1 4mv2A2l The total energy in one wavelength of the wave is the sum of the potential and kinetic energies: El 5 Ul 1 K l 5 1 2mv2A2l (16.20) As the wave moves along the string, this amount of energy passes by a given point on the string during a time interval of one period of the oscillation. Therefore, the power P, or rate of energy transfer TMW associated with the mechanical wave, is P 5 TMW Dt 5 El T 5 1 2mv2A2l T 5 1 2mv2A2 al Tb P 5 1 2mv2A2v (16.21) Equation 16.21 shows that the rate of energy transfer by a sinusoidal wave on a string is proportional to (a) the square of the frequency, (b) the square of the amplitude, and (c) the wave speed. In fact, the rate of energy transfer in any sinusoidal wave is pro-portional to the square of the angular frequency and to the square of the amplitude.
Q uick Quiz 16.5 Which of the following, taken by itself, would be most effective in increasing the rate at which energy is transferred by a wave traveling along a string? (a) reducing the linear mass density of the string by one half (b) dou-bling the wavelength of the wave (c) doubling the tension in the string (d) dou-bling the amplitude of the wave Power of a wave Example 16.5 Power Supplied to a Vibrating String A taut string for which m 5 5.00 3 1022 kg/m is under a tension of 80.0 N. How much power must be supplied to the string to generate sinusoidal waves at a frequency of 60.0 Hz and an amplitude of 6.00 cm?
Conceptualize Consider Figure 16.10 again and notice that the vibrating blade supplies energy to the string at a cer-tain rate. This energy then propagates to the right along the string.
S o l u t i o n www.aswarphysics.weebly.com 16.6 The Linear Wave Equation 497 Categorize We evaluate quantities from equations developed in the chapter, so we categorize this example as a substi-tution problem.
Use Equation 16.21 to evaluate the power: P 5 1 2mv2A2v Use Equations 16.9 and 16.18 to substitute for v and v: P 5 1 2m12pf 2 2A2aÅ T m b 5 2p2f 2A2"mT Substitute numerical values: P 5 2p 2160.0 Hz2 210.060 0 m2 2"10.050 0 kg/m2 180.0 N2 5 512 W What if the string is to transfer energy at a rate of 1 000 W? What must be the required amplitude if all other parameters remain the same?
Answer Let us set up a ratio of the new and old power, reflecting only a change in the amplitude: Pnew Pold 5 1 2mv2A 2 new v 1 2mv2A 2 old v 5 A 2 new A 2 old Solving for the new amplitude gives A new 5 A oldÅ Pnew Pold 5 16.00 cm2Å 1 000 W 512 W 5 8.39 cm What If?
16.6 The Linear Wave Equation In Section 16.1, we introduced the concept of the wave function to represent waves traveling on a string. All wave functions y(x, t) represent solutions of an equation called the linear wave equation. This equation gives a complete description of the wave motion, and from it one can derive an expression for the wave speed. Further-more, the linear wave equation is basic to many forms of wave motion. In this sec-tion, we derive this equation as applied to waves on strings. Suppose a traveling wave is propagating along a string that is under a tension T. Let’s consider one small string element of length Dx (Fig. 16.19). The ends of the element make small angles uA and uB with the x axis. Forces act on the string at its ends where it connects to neighboring elements. Therefore, the element is modeled as a particle under a net force. The net force acting on the element in the vertical direction is o Fy 5 T sin uB 2 T sin uA 5 T(sin uB 2 sin uA) Because the angles are small, we can use the approximation sin u < tan u to express the net force as o Fy < T(tan uB 2 tan uA) (16.22) Imagine undergoing an infinitesimal displacement outward from the right end of the rope element in Figure 16.19 along the blue line representing the force T S. This displacement has infinitesimal x and y components and can be represented by the vector dx i ^ 1 dy j ^. The tangent of the angle with respect to the x axis for this dis-placement is dy/dx. Because we evaluate this tangent at a particular instant of time, we must express it in partial form as 'y/'x. Substituting for the tangents in Equa-tion 16.22 gives a Fy < T c a 'y 'xb B 2 a 'y 'xb A d (16.23) B A x A B u u T S T S Figure 16.19 An element of a string under tension T. ▸ 16.5 c on tin u ed www.aswarphysics.weebly.com 498 Chapter 16 Wave Motion Now, from the particle under a net force model, let’s apply Newton’s second law to the element, with the mass of the element given by m 5 m Dx: a Fy 5 ma y 5 m Dx a '2y 't 2b (16.24) Combining Equation 16.23 with Equation 16.24 gives m Dx a '2y 't 2b 5 T c a 'y 'xb B 2 a 'y 'xb A d m T '2y 't 2 5 1'y/'x2B 2 1'y/dx2A Dx (16.25) The right side of Equation 16.25 can be expressed in a different form if we note that the partial derivative of any function is defined as 'f 'x ; lim Dx S 0 f 1x 1 Dx2 2 f 1x2 Dx Associating f(x 1 Dx) with ('y/'x)B and f(x) with ('y/'x)A, we see that, in the limit Dx S 0, Equation 16.25 becomes m T '2y 't 2 5 '2y 'x 2 (16.26) This expression is the linear wave equation as it applies to waves on a string. The linear wave equation (Eq. 16.26) is often written in the form '2y 'x 2 5 1 v 2 '2y 't 2 (16.27) Equation 16.27 applies in general to various types of traveling waves. For waves on strings, y represents the vertical position of elements of the string. For sound waves propagating through a gas, y corresponds to longitudinal position of elements of the gas from equilibrium or variations in either the pressure or the density of the gas. In the case of electromagnetic waves, y corresponds to electric or magnetic field components. We have shown that the sinusoidal wave function (Eq. 16.10) is one solution of the linear wave equation (Eq. 16.27). Although we do not prove it here, the linear wave equation is satisfied by any wave function having the form y 5 f(x 6 vt). Fur-thermore, we have seen that the linear wave equation is a direct consequence of the particle under a net force model applied to any element of a string carrying a traveling wave.
Linear wave equation for a string Linear wave equation in general Summary A one-dimensional sinusoidal wave is one for which the positions of the elements of the medium vary sinu-soidally. A sinusoidal wave traveling to the right can be expressed with a wave function y1x, t2 5 A sin c 2p l 1x 2 vt2d (16.5) where A is the amplitude, l is the wavelength, and v is the wave speed. The angular wave number k and angular frequency v of a wave are defined as follows: k ; 2p l (16.8) v ; 2p T 5 2pf (16.9) where T is the period of the wave and f is its frequency.
Definitions www.aswarphysics.weebly.com Objective Questions 499 pens to the speed if you fill the hose with water? Choose from the same possibilities. 3. Rank the waves represented by the following functions from the largest to the smallest according to (i) their amplitudes, (ii) their wavelengths, (iii) their frequen-cies, (iv) their periods, and (v) their speeds. If the val-ues of a quantity are equal for two waves, show them as having equal rank. For all functions, x and y are in meters and t is in seconds. (a) y 5 4 sin (3x 2 15t) (b) y 5 6 cos (3x 1 15t 2 2) (c) y 5 8 sin (2x 1 15t) (d) y 5 8 cos (4x 1 20t) (e) y 5 7 sin (6x 2 24t) 4. By what factor would you have to multiply the tension in a stretched string so as to double the wave speed? 1. If one end of a heavy rope is attached to one end of a lightweight rope, a wave can move from the heavy rope into the lighter one. (i) What happens to the speed of the wave? (a) It increases. (b) It decreases. (c) It is con-stant. (d) It changes unpredictably. (ii) What happens to the frequency? Choose from the same possibilities. (iii) What happens to the wavelength? Choose from the same possibilities. 2. If you stretch a rubber hose and pluck it, you can observe a pulse traveling up and down the hose. (i) What happens to the speed of the pulse if you stretch the hose more tightly? (a) It increases. (b) It decreases. (c) It is constant. (d) It changes unpredictably. (ii) What hap- A transverse wave is one in which the elements of the medium move in a direction perpendicular to the direction of propagation. A longitudinal wave is one in which the elements of the medium move in a direction parallel to the direc-tion of propagation. The speed of a wave traveling on a taut string of mass per unit length m and tension T is v 5 Å T m (16.18) A wave is totally or partially reflected when it reaches the end of the medium in which it propa-gates or when it reaches a boundary where its speed changes discon-tinuously. If a wave traveling on a string meets a fixed end, the wave is reflected and inverted. If the wave reaches a free end, it is reflected but not inverted. Any one-dimensional wave traveling with a speed v in the x direction can be represented by a wave function of the form y (x, t) 5 f(x 6 vt) (16.1, 16.2) where the positive sign applies to a wave traveling in the negative x direc-tion and the negative sign applies to a wave traveling in the positive x direction. The shape of the wave at any instant in time (a snapshot of the wave) is obtained by holding t constant. The power transmitted by a sinusoidal wave on a stretched string is P 5 1 2mv2A2v (16.21) Wave functions are solutions to a differential equation called the linear wave equation: '2y 'x 2 5 1 v 2 '2y 't 2 (16.27) Concepts and Principles Analysis Model for Problem Solving Traveling Wave. The wave speed of a sinusoidal wave is v 5 l T 5 lf (16.6, 16.12) A sinusoidal wave can be expressed as y 5 A sin 1kx 2 vt2 (16.10) y x A l v S Objective Questions 1. denotes answer available in Student Solutions Manual/Study Guide www.aswarphysics.weebly.com 500 Chapter 16 Wave Motion by some source of disturbance. (b) They are sinusoi-dal in nature. (c) They carry energy. (d) They require a medium through which to propagate. (e) The wave speed depends on the properties of the medium in which they travel. 7. (a) Can a wave on a string move with a wave speed that is greater than the maximum transverse speed vy,max of an element of the string? (b) Can the wave speed be much greater than the maximum element speed? (c) Can the wave speed be equal to the maximum ele-ment speed? (d) Can the wave speed be less than vy,max? 8. A source vibrating at constant frequency generates a sinusoidal wave on a string under constant tension. If the power delivered to the string is doubled, by what fac-tor does the amplitude change? (a) a factor of 4 (b) a factor of 2 (c) a factor of !2 (d) a factor of 0.707 (e) cannot be predicted 9. The distance between two successive peaks of a sinu-soidal wave traveling along a string is 2 m. If the fre-quency of this wave is 4 Hz, what is the speed of the wave? (a) 4 m/s (b) 1 m/s (c) 8 m/s (d) 2 m/s (e) impos-sible to answer from the information given Assume the string does not stretch. (a) a factor of 8 (b) a factor of 4 (c) a factor of 2 (d) a factor of 0.5 (e) You could not change the speed by a predictable factor by changing the tension. 5. When all the strings on a guitar (Fig. OQ16.5) are stretched to the same tension, will the speed of a wave along the most massive bass string be (a) faster, (b) slower, or (c) the same as the speed of a wave on the lighter strings? Alternatively, (d) is the speed on the bass string not necessarily any of these answers?
Figure OQ16.5 Joseph/Getty Images 6. Which of the following statements is not necessarily true regarding mechanical waves? (a) They are formed Conceptual Questions 1. denotes answer available in Student Solutions Manual/Study Guide 1. Why is a solid substance able to transport both longitu-dinal waves and transverse waves, but a homogeneous fluid is able to transport only longitudinal waves? 2. (a) How would you create a longitudinal wave in a stretched spring? (b) Would it be possible to create a transverse wave in a spring? 3. When a pulse travels on a taut string, does it always invert upon reflection? Explain. 4. In mechanics, massless strings are often assumed. Why is that not a good assumption when discussing waves on strings? 5. If you steadily shake one end of a taut rope three times each second, what would be the period of the sinusoi-dal wave set up in the rope? 6. (a) If a long rope is hung from a ceiling and waves are sent up the rope from its lower end, why does the speed of the waves change as they ascend? (b) Does the speed of the ascending waves increase or decrease? Explain. 7. Why is a pulse on a string considered to be transverse? 8. Does the vertical speed of an element of a horizontal, taut string, through which a wave is traveling, depend on the wave speed? Explain. 9. In an earthquake, both S (transverse) and P (longitu-dinal) waves propagate from the focus of the earthquake. The focus is in the ground radially below the epicenter on the surface (Fig. CQ16.9). Assume the waves move in straight lines through uni-form material. The S waves travel through the Earth more slowly than the P waves (at about 5 km/s versus 8 km/s). By detecting the time of arrival of the waves at a seismo-graph, (a) how can one determine the distance to the focus of the earthquake? (b) How many detection sta-tions are necessary to locate the focus unambiguously?
Epicenter Seismograph Path of seismic waves Focus Figure CQ16.9 Problems The problems found in this chapter may be assigned online in Enhanced WebAssign 1. straightforward; 2. intermediate; 3. challenging 1. full solution available in the Student Solutions Manual/Study Guide AMT Analysis Model tutorial available in Enhanced WebAssign GP Guided Problem M Master It tutorial available in Enhanced WebAssign W Watch It video solution available in Enhanced WebAssign BIO Q/C S www.aswarphysics.weebly.com Problems 501 7. A sinusoidal wave is traveling along a rope. The oscil-lator that generates the wave completes 40.0 vibrations in 30.0 s. A given crest of the wave travels 425 cm along the rope in 10.0 s. What is the wavelength of the wave? 8. For a certain transverse wave, the distance between two successive crests is 1.20 m, and eight crests pass a given point along the direction of travel every 12.0 s. Calcu-late the wave speed. 9. The wave function for a traveling wave on a taut string is (in SI units) y 1x, t 2 5 0.350 sin a10pt 2 3px 1 p 4 b (a) What are the speed and direction of travel of the wave? (b) What is the vertical position of an element of the string at t 5 0, x 5 0.100 m? What are (c) the wave-length and (d) the frequency of the wave? (e) What is the maximum transverse speed of an element of the string? 10. When a particular wire is vibrating with a frequency of 4.00 Hz, a transverse wave of wavelength 60.0 cm is produced. Determine the speed of waves along the wire. 11. The string shown in Figure P16.11 is driven at a fre-quency of 5.00 Hz. The amplitude of the motion is A 5 12.0 cm, and the wave speed is v 5 20.0 m/s. Further-more, the wave is such that y 5 0 at x 5 0 and t 5 0. Determine (a) the angular frequency and (b) the wave number for this wave. (c) Write an expression for the wave function. Calculate (d) the maximum transverse speed and (e) the maximum transverse acceleration of an element of the string.
v S A Figure P16.11 12. Consider the sinusoidal wave of Example 16.2 with the wave function y 5 0.150 cos (15.7x 2 50.3t) where x and y are in meters and t is in seconds. At a certain instant, let point A be at the origin and point B be the closest point to A along the x axis where the wave is 60.0° out of phase with A. What is the coordi-nate of B? 13. A sinusoidal wave of wavelength 2.00 m and amplitude 0.100 m travels on a string with a speed of 1.00 m/s to the right. At t 5 0, the left end of the string is at the origin. For this wave, find (a) the frequency, (b) the angular frequency, (c) the angular wave number, and (d) the wave function in SI units. Determine the equa-tion of motion in SI units for (e) the left end of the string and (f) the point on the string at x 5 1.50 m to the right of the left end. (g) What is the maximum speed of any element of the string? 14. (a) Plot y versus t at x 5 0 for a sinusoidal wave of the form y 5 0.150 cos (15.7x 2 50.3t), where x and y are in M M W W Q/C Section 16.1 Propagation of a Disturbance 1. A seismographic station receives S and P waves from an earthquake, separated in time by 17.3 s. Assume the waves have traveled over the same path at speeds of 4.50 km/s and 7.80 km/s. Find the distance from the seismograph to the focus of the quake. 2. Ocean waves with a crest-to-crest distance of 10.0 m can be described by the wave function y(x, t) 5 0.800 sin [0.628(x 2 vt)] where x and y are in meters, t is in seconds, and v 5 1.20 m/s. (a) Sketch y(x, t) at t 5 0. (b) Sketch y(x, t) at t 5 2.00 s. (c) Compare the graph in part (b) with that for part (a) and explain similarities and differences. (d) How has the wave moved between graph (a) and graph (b)? 3. At t 5 0, a transverse pulse in a wire is described by the function y 5 6.00 x 2 1 3.00 where x and y are in meters. If the pulse is traveling in the positive x direction with a speed of 4.50 m/s, write the function y(x, t) that describes this pulse. 4. Two points A and B on the surface of the Earth are at the same longitude and 60.08 apart in latitude as shown in Figure P16.4. Suppose an earthquake at point A creates a P wave that reaches point B by traveling straight through the body of the Earth at a constant speed of 7.80 km/s. The earthquake also radi-ates a Rayleigh wave that travels at 4.50 km/s. In addition to P and S waves, Rayleigh waves are a third type of seis-mic wave that travels along the surface of the Earth rather than through the bulk of the Earth. (a) Which of these two seismic waves arrives at B first? (b) What is the time difference between the arrivals of these two waves at B?
Section 16.2 Analysis Model: Traveling Wave 5. A wave is described by y 5 0.020 0 sin (kx 2 vt), where k 5 2.11 rad/m, v 5 3.62 rad/s, x and y are in meters, and t is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, and (d) the speed of the wave. 6. A certain uniform string is held under constant ten-sion. (a) Draw a side-view snapshot of a sinusoidal wave on a string as shown in diagrams in the text. (b) Imme-diately below diagram (a), draw the same wave at a moment later by one-quarter of the period of the wave. (c) Then, draw a wave with an amplitude 1.5 times larger than the wave in diagram (a). (d) Next, draw a wave differing from the one in your diagram (a) just by having a wavelength 1.5 times larger. (e) Finally, draw a wave differing from that in diagram (a) just by having a frequency 1.5 times larger.
W Q/C A B Path of Rayleigh wave Path of P wave 60.0 Figure P16.4 M Q/C www.aswarphysics.weebly.com 502 Chapter 16 Wave Motion pulses can propagate along this wire without exceeding this stress? (The density of steel is 7.86 3 103 kg/m3.) 22. A piano string having a mass per unit length equal to 5.00 3 1023 kg/m is under a tension of 1 350 N. Find the speed with which a wave travels on this string. 23. Transverse waves travel with a speed of 20.0 m/s on a string under a tension of 6.00 N. What tension is required for a wave speed of 30.0 m/s on the same string? 24. A student taking a quiz finds on a reference sheet the two equations f 5 1 T and v 5 Å T m She has forgotten what T represents in each equation. (a) Use dimensional analysis to determine the units required for T in each equation. (b) Explain how you can identify the physical quantity each T represents from the units. 25. An Ethernet cable is 4.00 m long. The cable has a mass of 0.200 kg. A transverse pulse is produced by plucking one end of the taut cable. The pulse makes four trips down and back along the cable in 0.800 s. What is the tension in the cable? 26. A transverse traveling wave on a taut wire has an ampli-tude of 0.200 mm and a frequency of 500 Hz. It trav-els with a speed of 196 m/s. (a) Write an equation in SI units of the form y 5 A sin (kx 2 vt) for this wave. (b) The mass per unit length of this wire is 4.10 g/m. Find the tension in the wire. 27. A steel wire of length 30.0 m and a copper wire of length 20.0 m, both with 1.00-mm diameters, are con-nected end to end and stretched to a tension of 150 N. During what time interval will a transverse wave travel the entire length of the two wires? 28. Why is the following situation impossible? An astronaut on the Moon is studying wave motion using the apparatus discussed in Example 16.3 and shown in Figure 16.12. He measures the time interval for pulses to travel along the horizontal wire. Assume the horizontal wire has a mass of 4.00 g and a length of 1.60 m and assume a 3.00-kg object is suspended from its extension around the pulley. The astronaut finds that a pulse requires 26.1 ms to traverse the length of the wire. 29. Tension is maintained in a string as in Figure P16.29. The observed wave speed is v 5 24.0 m/s when the suspended mass is m 5 3.00 kg. (a) What is the mass per unit length of the string? (b) What is the wave speed when the suspended mass is m 5 2.00 kg? 30. Review. A light string with a mass per unit length of 8.00 g/m has its ends tied to two walls separated by a distance equal to three-fourths the length of the string (Fig. P16.30, p. 503). An object of mass m is suspended from the center of the string, putting a tension in the string. (a) Find an expression for the transverse wave W M S Q/C W M AMT m Figure P16.29 Problems 29 and 47.
AMT meters and t is in seconds. (b) Determine the period of vibration. (c) State how your result compares with the value found in Example 16.2. 15. A transverse wave on a string is described by the wave function y 5 0.120 sin ap 8 x 1 4ptb where x and y are in meters and t is in seconds. Deter-mine (a) the transverse speed and (b) the transverse acceleration at t 5 0.200 s for an element of the string located at x 5 1.60 m. What are (c) the wavelength, (d) the period, and (e) the speed of propagation of this wave? 16. A wave on a string is described by the wave function y 5 0.100 sin (0.50x 2 20t), where x and y are in meters and t is in seconds. (a) Show that an element of the string at x 5 2.00 m executes harmonic motion. (b) Determine the frequency of oscillation of this par-ticular element. 17. A sinusoidal wave is described by the wave function y 5 0.25 sin (0.30x 2 40t) where x and y are in meters and t is in seconds. Determine for this wave (a) the ampli-tude, (b) the angular frequency, (c) the angular wave number, (d) the wavelength, (e) the wave speed, and (f) the direction of motion. 18. A sinusoidal wave traveling in the negative x direction (to the left) has an amplitude of 20.0 cm, a wavelength of 35.0 cm, and a frequency of 12.0 Hz. The transverse position of an element of the medium at t 5 0, x 5 0 is y 5 23.00 cm, and the element has a positive velocity here. We wish to find an expression for the wave func-tion describing this wave. (a) Sketch the wave at t 5 0. (b) Find the angular wave number k from the wave-length. (c) Find the period T from the frequency. Find (d) the angular frequency v and (e) the wave speed v. (f) From the information about t 5 0, find the phase constant f. (g) Write an expression for the wave func-tion y(x, t). 19. (a) Write the expression for y as a function of x and t in SI units for a sinusoidal wave traveling along a rope in the negative x direction with the following charac-teristics: A 5 8.00 cm, l 5 80.0 cm, f 5 3.00 Hz, and y(0, t) 5 0 at t 5 0. (b) What If? Write the expression for y as a function of x and t for the wave in part (a) assuming y(x, 0) 5 0 at the point x 5 10.0 cm. 20. A transverse sinusoidal wave on a string has a period T 5 25.0 ms and travels in the negative x direction with a speed of 30.0 m/s. At t 5 0, an element of the string at x 5 0 has a transverse position of 2.00 cm and is trav-eling downward with a speed of 2.00 m/s. (a) What is the amplitude of the wave? (b) What is the initial phase angle? (c) What is the maximum transverse speed of an element of the string? (d) Write the wave function for the wave.
Section 16.3 The Speed of Waves on Strings 21. Review. The elastic limit of a steel wire is 2.70 3 108 Pa. What is the maximum speed at which transverse wave W W GP www.aswarphysics.weebly.com Problems 503 mass of 180 g. The string vibrates sinusoidally with a frequency of 50.0 Hz and a peak-to-valley displacement of 15.0 cm. (The “peak-to- valley” distance is the verti-cal distance from the farthest positive position to the farthest negative position.) (a) Write the function that describes this wave traveling in the positive x direction. (b) Determine the power being supplied to the string. 38. A horizontal string can transmit a maximum power P0 (without breaking) if a wave with amplitude A and angular frequency v is traveling along it. To increase this maximum power, a student folds the string and uses this “double string” as a medium. Assuming the tension in the two strands together is the same as the original tension in the single string and the angular frequency of the wave remains the same, determine the maximum power that can be transmitted along the “double string.” 39. The wave function for a wave on a taut string is y 1x, t 2 5 0.350 sin a10pt 2 3px 1 p 4 b where x and y are in meters and t is in seconds. If the linear mass density of the string is 75.0 g/m, (a) what is the average rate at which energy is transmitted along the string? (b) What is the energy contained in each cycle of the wave? 40. A two-dimensional water wave spreads in circular rip-ples. Show that the amplitude A at a distance r from the initial disturbance is proportional to 1/!r. Sug-gestion: Consider the energy carried by one outward- moving ripple.
Section 16.6 The Linear Wave Equation 41. Show that the wave function y 5 ln [b(x 2 vt)] is a solu-tion to Equation 16.27, where b is a constant. 42. (a) Evaluate A in the scalar equality 4 (7 1 3) 5 A. (b) Evaluate A, B, and C in the vector equality 700 i ^ 1 3.00 k ^ 5 A i ^ 1 B j ^ 1 C k ^. (c) Explain how you arrive at the answers to convince a student who thinks that you cannot solve a single equation for three differ-ent unknowns. (d) What If? The functional equality or identity A 1 B cos (Cx 1 Dt 1 E) 5 7.00 cos (3x 1 4t 1 2) is true for all values of the variables x and t, measured in meters and in seconds, respectively. Evaluate the constants A, B, C, D, and E. (e) Explain how you arrive at your answers to part (d). 43. Show that the wave function y 5 e b(x2vt) is a solution of the linear wave equation (Eq. 16.27), where b is a constant. 44. (a) Show that the function y(x, t) 5 x2 1 v2t2 is a solu-tion to the wave equation. (b) Show that the function in part (a) can be written as f(x 1 vt) 1 g(x 2 vt) and determine the functional forms for f and g. (c) What If? Repeat parts (a) and (b) for the function y(x, t) 5 sin (x) cos (vt).
Additional Problems 45. Motion-picture film is projected at a frequency of 24.0 frames per second. Each photograph on the film is the S S S Q/C S S speed in the string as a function of the mass of the hanging object. (b) What should be the mass of the object sus-pended from the string if the wave speed is to be 60.0 m/s? 31. Transverse pulses travel with a speed of 200 m/s along a taut copper wire whose diameter is 1.50 mm. What is the tension in the wire? (The density of copper is 8.92 g/cm3.) Section 16.5 Rate of Energy Transfer by Sinusoidal Waves on Strings 32. In a region far from the epicenter of an earthquake, a seismic wave can be modeled as transporting energy in a single direction without absorption, just as a string wave does. Suppose the seismic wave moves from gran-ite into mudfill with similar density but with a much smaller bulk modulus. Assume the speed of the wave gradually drops by a factor of 25.0, with negligible reflection of the wave. (a) Explain whether the ampli-tude of the ground shaking will increase or decrease. (b) Does it change by a predictable factor? (This phe-nomenon led to the collapse of part of the Nimitz Free-way in Oakland, California, during the Loma Prieta earthquake of 1989.) 33. Transverse waves are being generated on a rope under constant tension. By what factor is the required power increased or decreased if (a) the length of the rope is doubled and the angular frequency remains con-stant, (b) the amplitude is doubled and the angular frequency is halved, (c) both the wavelength and the amplitude are doubled, and (d) both the length of the rope and the wavelength are halved? 34. Sinusoidal waves 5.00 cm in amplitude are to be trans-mitted along a string that has a linear mass density of 4.00 3 1022 kg/m. The source can deliver a maximum power of 300 W, and the string is under a tension of 100 N. What is the highest frequency f at which the source can operate? 35. A sinusoidal wave on a string is described by the wave function y 5 0.15 sin (0.80x 2 50t) where x and y are in meters and t is in seconds. The mass per unit length of this string is 12.0 g/m. Deter-mine (a) the speed of the wave, (b) the wavelength, (c) the frequency, and (d) the power transmitted by the wave. 36. A taut rope has a mass of 0.180 kg and a length of 3.60 m. What power must be supplied to the rope so as to generate sinusoidal waves having an amplitude of 0.100 m and a wavelength of 0.500 m and traveling with a speed of 30.0 m/s? 37. A long string carries a wave; a 6.00-m segment of the string contains four complete wavelengths and has a W Q/C M M W AMT 3L m 4 L 2 L 2 Figure P16.30 www.aswarphysics.weebly.com 504 Chapter 16 Wave Motion is held in this lowest position, find the speed of a trans-verse wave in the cord. 50. Review. A block of mass M hangs from a rubber cord. The block is supported so that the cord is not stretched. The unstretched length of the cord is L0, and its mass is m, much less than M. The “spring constant” for the cord is k. The block is released and stops momentarily at the lowest point. (a) Determine the tension in the string when the block is at this lowest point. (b) What is the length of the cord in this “stretched” position? (c) If the block is held in this lowest position, find the speed of a transverse wave in the cord. 51. A transverse wave on a string is described by the wave function y(x, t) 5 0.350 sin (1.25x 1 99.6t) where x and y are in meters and t is in seconds. Con-sider the element of the string at x 5 0. (a) What is the time interval between the first two instants when this element has a position of y 5 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)? 52. A sinusoidal wave in a string is described by the wave function y 5 0.150 sin (0.800x 2 50.0t) where x and y are in meters and t is in seconds. The mass per length of the string is 12.0 g/m. (a) Find the maximum transverse acceleration of an element of this string. (b) Determine the maximum transverse force on a 1.00-cm segment of the string. (c) State how the force found in part (b) compares with the tension in the string. 53. Review. A block of mass M, supported by a string, rests on a frictionless incline making an angle u with the horizontal (Fig. P16.53). The length of the string is L, and its mass is m ,, M. Derive an expression for the time interval required for a transverse wave to travel from one end of the string to the other.
m L u M Figure P16.53 54. An undersea earthquake or a landslide can produce an ocean wave of short duration carrying great energy, called a tsunami. When its wavelength is large com-pared to the ocean depth d, the speed of a water wave is given approximately by v 5 !gd. Assume an earthquake occurs all along a tectonic plate boundary running north to south and produces a straight tsunami wave crest moving everywhere to the west. (a) What physical quantity can you consider to be constant in the motion S Q/C S Q/C same height of 19.0 mm, just like each oscillation in a wave is the same length. Model the height of a frame as the wavelength of a wave. At what constant speed does the film pass into the projector? 46. “The wave” is a particular type of pulse that can propa-gate through a large crowd gathered at a sports arena (Fig. P16.46). The elements of the medium are the spectators, with zero position corresponding to their being seated and maximum position correspond-ing to their standing and raising their arms. When a large fraction of the spectators participates in the wave motion, a somewhat stable pulse shape can develop. The wave speed depends on people’s reaction time, which is typically on the order of 0.1 s. Estimate the order of magnitude, in minutes, of the time interval required for such a pulse to make one circuit around a large sports stadium. State the quantities you measure or estimate and their values.
Figure P16.46 Joe Klamar/AFP/Getty Images 47. A sinusoidal wave in a rope is described by the wave function y 5 0.20 sin (0.75px 1 18pt) where x and y are in meters and t is in seconds. The rope has a linear mass density of 0.250 kg/m. The ten-sion in the rope is provided by an arrangement like the one illustrated in Figure P16.29. What is the mass of the suspended object? 48. The ocean floor is underlain by a layer of basalt that constitutes the crust, or uppermost layer, of the Earth in that region. Below this crust is found denser periodotite rock that forms the Earth’s mantle. The boundary between these two layers is called the Mohorovicic discontinuity (“Moho” for short). If an explosive charge is set off at the surface of the basalt, it generates a seismic wave that is reflected back out at the Moho. If the speed of this wave in basalt is 6.50 km/s and the two-way travel time is 1.85 s, what is the thickness of this oceanic crust? 49. Review. A 2.00-kg block hangs from a rubber cord, being supported so that the cord is not stretched. The unstretched length of the cord is 0.500 m, and its mass is 5.00 g. The “spring constant” for the cord is 100 N/m. The block is released and stops momentarily at the low-est point. (a) Determine the tension in the cord when the block is at this lowest point. (b) What is the length of the cord in this “stretched” position? (c) If the block www.aswarphysics.weebly.com Problems 505 of any one wave crest? (b) Explain why the amplitude of the wave increases as the wave approaches shore. (c) If the wave has amplitude 1.80 m when its speed is 200 m/s, what will be its amplitude where the water is 9.00 m deep? (d) Explain why the amplitude at the shore should be expected to be still greater, but cannot be meaningfully predicted by your model. 55. Review. A block of mass M 5 0.450 kg is attached to one end of a cord of mass 0.003 20 kg; the other end of the cord is attached to a fixed point. The block rotates with constant angular speed in a circle on a friction-less, horizontal table as shown in Figure P16.55. Through what angle does the block rotate in the time interval during which a transverse wave travels along the string from the center of the circle to the block?
M Figure P16.55 Problems 55, 56, and 57. 56. Review. A block of mass M 5 0.450 kg is attached to one end of a cord of mass m 5 0.003 20 kg; the other end of the cord is attached to a fixed point. The block rotates with constant angular speed v 5 10.0 rad/s in a circle on a frictionless, horizontal table as shown in Figure P16.55. What time interval is required for a transverse wave to travel along the string from the cen-ter of the circle to the block? 57. Review. A block of mass M is attached to one end of a cord of mass m; the other end of the cord is attached to a fixed point. The block rotates with constant angular speed v in a circle on a frictionless, horizontal table as shown in Figure P16.55. What time interval is required for a transverse wave to travel along the string from the center of the circle to the block? 58. A string with linear density 0.500 g/m is held under ten-sion 20.0 N. As a transverse sinusoidal wave propagates on the string, elements of the string move with maxi-mum speed vy,max. (a) Determine the power transmitted by the wave as a function of vy,max. (b) State in words the proportionality between power and vy,max. (c) Find the energy contained in a section of string 3.00 m long as a function of vy,max. (d) Express the answer to part (c) in terms of the mass m of this section. (e) Find the energy that the wave carries past a point in 6.00 s. 59. A wire of density r is tapered so that its cross-sectional area varies with x according to A 5 1.00 3 1025 x 1 1.00 3 1026 where A is in meters squared and x is in meters. The tension in the wire is T. (a) Derive a relationship for AMT S Q/C the speed of a wave as a function of position. (b) What If? Assume the wire is aluminum and is under a ten-sion T 5 24.0 N. Determine the wave speed at the ori-gin and at x 5 10.0 m. 60. A rope of total mass m and length L is suspended ver-tically. Analysis shows that for short transverse pulses, the waves above a short distance from the free end of the rope can be represented to a good approximation by the linear wave equation discussed in Section 16.6. Show that a transverse pulse travels the length of the rope in a time interval that is given approximately by Dt < 2!L/g. Suggestion: First find an expression for the wave speed at any point a distance x from the lower end by considering the rope’s tension as resulting from the weight of the segment below that point. 61. A pulse traveling along a string of linear mass density m is described by the wave function y 5 [A0e2bx] sin (kx 2 vt) where the factor in brackets is said to be the ampli-tude. (a) What is the power P(x) carried by this wave at a point x? (b) What is the power P(0) carried by this wave at the origin? (c) Compute the ratio P(x)/P(0). 62. Why is the following situation impossible? Tsunamis are ocean surface waves that have enormous wavelengths (100 to 200 km), and the propagation speed for these waves is v < !gd avg, where davg is the average depth of the water. An earthquake on the ocean floor in the Gulf of Alaska produces a tsunami that reaches Hilo, Hawaii, 4 450 km away, in a time interval of 5.88 h. (This method was used in 1856 to estimate the average depth of the Pacific Ocean long before soundings were made to give a direct determination.) 63. Review. An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a decrease in the wire’s equilibrium length, increases the tension in the wire. Taking the cross-sectional area of the wire to be 5.00 3 1026 m2, the density to be 2.70 3 103 kg/m3, and Young’s modulus to be 7.00 3 1010 N/m2, what strain (DL/L) results in a transverse wave speed of 100 m/s?
Challenge Problems 64. Assume an object of mass M is suspended from the bot-tom of the rope of mass m and length L in Problem 60. (a) Show that the time interval for a transverse pulse to travel the length of the rope is Dt 5 2 Å L mg 1"M 1 m 2 "M2 (b) What If? Show that the expression in part (a) reduces to the result of Problem 60 when M 5 0. (c) Show that for m ,, M, the expression in part (a) reduces to Dt 5 Å mL Mg S S M S www.aswarphysics.weebly.com 506 Chapter 16 Wave Motion 67. If a loop of chain is spun at high speed, it can roll along the ground like a circular hoop without collaps-ing. Consider a chain of uniform linear mass density m whose center of mass travels to the right at a high speed v0 as shown in Figure P16.67. (a) Determine the tension in the chain in terms of m and v0. Assume the weight of an individual link is negligible compared to the tension. (b) If the loop rolls over a small bump, the resulting deformation of the chain causes two transverse pulses to propagate along the chain, one moving clockwise and one moving counterclockwise. What is the speed of the pulses traveling along the chain? (c) Through what angle does each pulse travel during the time interval over which the loop makes one revolution?
Bump v0 S Figure P16.67 S 65. A rope of total mass m and length L is suspended verti-cally. As shown in Problem 60, a pulse travels from the bottom to the top of the rope in an approximate time interval Dt 5 2!L/g with a speed that varies with position x measured from the bottom of the rope as v 5 !gx . Assume the linear wave equation in Sec-tion 16.6 describes waves at all locations on the rope. (a) Over what time interval does a pulse travel half-way up the rope? Give your answer as a fraction of the quantity 2!L/g. (b) A pulse starts traveling up the rope. How far has it traveled after a time interval !L/g? 66. A string on a musical instrument is held under ten-sion T and extends from the point x 5 0 to the point x 5 L. The string is overwound with wire in such a way that its mass per unit length m(x) increases uni-formly from m0 at x 5 0 to mL at x 5 L. (a) Find an expression for m(x) as a function of x over the range 0 # x # L. (b) Find an expression for the time inter-val required for a transverse pulse to travel the length of the string.
S S www.aswarphysics.weebly.com Three musicians play the alpenhorn in Valais, Switzerland. In this chapter, we explore the behavior of sound waves such as those coming from these large musical instruments. (Stefano Cellai/AGE fotostock) 17.1 Pressure Variations in Sound Waves 17.2 Speed of Sound Waves 17.3 Intensity of Periodic Sound Waves 17.4 The Doppler Effect c h a p t e r 17 Sound Waves 507 Most of the waves we studied in Chapter 16 are constrained to move along a one-dimensional medium. For example, the wave in Figure 16.7 is a purely mathematical construct moving along the x axis. The wave in Figure 16.10 is constrained to move along the length of the string. We have also seen waves moving through a two-dimensional medium, such as the ripples on the water surface in the introduction to Part 2 on page 449 and the waves moving over the surface of the ocean in Figure 16.4. In this chapter, we investigate mechanical waves that move through three-dimensional bulk media. For example, seismic waves leaving the focus of an earthquake travel through the three-dimensional interior of the Earth. We will focus our attention on sound waves, which travel through any material, but are most commonly experienced as the mechanical waves traveling through air that result in the human perception of hearing. As sound waves travel through air, elements of air are disturbed from their equilibrium positions. Accompanying these movements are changes in density and pressure of the air along the direction of wave motion. If the source of the sound waves vibrates sinusoidally, the density and pressure variations are also sinusoidal. The mathematical description of sinusoidal sound waves is very similar to that of sinusoidal waves on strings, as discussed in Chapter 16. Sound waves are divided into three categories that cover different frequency ranges. (1) Audible waves lie within the range of sensitivity of the human ear. They can be gener-ated in a variety of ways, such as by musical instruments, human voices, or loudspeakers. (2) Infrasonic waves have frequencies below the audible range. Elephants can use infrasonic waves to communicate with one another, even when separated by many kilometers. (3) Ultrasonic waves have frequencies above the audible range. You may have used a “silent” whistle to retrieve your dog. Dogs easily hear the ultrasonic sound this whistle emits, although humans cannot detect it at all. Ultrasonic waves are also used in medical imaging. www.aswarphysics.weebly.com 508 Chapter 17 Sound Waves This chapter begins with a discussion of the pressure variations in a sound wave, the speed of sound waves, and wave intensity, which is a function of wave amplitude. We then provide an alternative description of the intensity of sound waves that compresses the wide range of intensities to which the ear is sensitive into a smaller range for convenience. The effects of the motion of sources and listeners on the frequency of a sound are also investigated. 17.1 Pressure Variations in Sound Waves In Chapter 16, we began our investigation of waves by imagining the creation of a single pulse that traveled down a string (Figure 16.1) or a spring (Figure 16.3). Let’s do something similar for sound. We describe pictorially the motion of a one- dimensional longitudinal sound pulse moving through a long tube containing a compressible gas as shown in Figure 17.1. A piston at the left end can be quickly moved to the right to compress the gas and create the pulse. Before the piston is moved, the gas is undisturbed and of uniform density as represented by the uniformly shaded region in Figure 17.1a. When the piston is pushed to the right (Fig. 17.1b), the gas just in front of it is compressed (as represented by the more heavily shaded region); the pressure and density in this region are now higher than they were before the piston moved. When the piston comes to rest (Fig. 17.1c), the compressed region of the gas continues to move to the right, corresponding to a longitudinal pulse traveling through the tube with speed v. One can produce a one-dimensional periodic sound wave in the tube of gas in Figure 17.1 by causing the piston to move in simple harmonic motion. The results are shown in Figure 17.2. The darker parts of the colored areas in this figure rep-resent regions in which the gas is compressed and the density and pressure are above their equilibrium values. A compressed region is formed whenever the pis-v S a b c Before the piston moves, the gas is undisturbed.
The gas is compressed by the motion of the piston.
When the piston stops, the compressed pulse continues through the gas.
Figure 17.1 Motion of a longitudi-nal pulse through a compressible gas. The compression (darker region) is produced by the moving piston.
Figure 17.2 A longitudinal wave propagating through a gas-filled tube. The source of the wave is an oscillating piston at the left.
l www.aswarphysics.weebly.com 17.1 Pressure Variations in Sound Waves 509 ton is pushed into the tube. This compressed region, called a compression, moves through the tube, continuously compressing the region just in front of itself. When the piston is pulled back, the gas in front of it expands and the pressure and density in this region fall below their equilibrium values (represented by the lighter parts of the colored areas in Fig. 17.2). These low-pressure regions, called rarefactions, also propagate along the tube, following the compressions. Both regions move at the speed of sound in the medium. As the piston oscillates sinusoidally, regions of compression and rarefaction are continuously set up. The distance between two successive compressions (or two suc-cessive rarefactions) equals the wavelength l of the sound wave. Because the sound wave is longitudinal, as the compressions and rarefactions travel through the tube, any small element of the gas moves with simple harmonic motion parallel to the direction of the wave. If s(x, t) is the position of a small element relative to its equi-librium position,1 we can express this harmonic position function as s(x, t) 5 smax cos (kx 2 vt) (17.1) where smax is the maximum position of the element relative to equilibrium. This parameter is often called the displacement amplitude of the wave. The parame-ter k is the wave number, and v is the angular frequency of the wave. Notice that the displacement of the element is along x, in the direction of propagation of the sound wave. The variation in the gas pressure DP measured from the equilibrium value is also periodic with the same wave number and angular frequency as for the dis-placement in Equation 17.1. Therefore, we can write DP 5 DPmax sin (kx 2 vt) (17.2) where the pressure amplitude DPmax is the maximum change in pressure from the equilibrium value. Notice that we have expressed the displacement by means of a cosine function and the pressure by means of a sine function. We will justify this choice in the procedure that follows and relate the pressure amplitude Pmax to the displacement amplitude smax. Consider the piston–tube arrangement of Figure 17.1 once again. In Figure 17.3a, we focus our attention on a small cylindrical element of undis-turbed gas of length Dx and area A. The volume of this element is Vi 5 A Dx. Figure 17.3b shows this element of gas after a sound wave has moved it to a new position. The cylinder’s two flat faces move through different distances s1 and s2. The change in volume DV of the element in the new position is equal to A Ds, where Ds 5 s1 2 s2. From the definition of bulk modulus (see Eq. 12.8), we express the pressure vari-ation in the element of gas as a function of its change in volume: DP 5 2B DV Vi Let’s substitute for the initial volume and the change in volume of the element: DP 5 2B A Ds A Dx Let the length Dx of the cylinder approach zero so that the ratio Ds/Dx becomes a partial derivative: DP 5 2B 's 'x (17.3) Area A Undisturbed gas x s1 s2 b a Figure 17.3 (a) An undisturbed element of gas of length Dx in a tube of cross-sectional area A. (b) When a sound wave propagates through the gas, the element is moved to a new position and has a different length. The parameters s1 and s2 describe the displace-ments of the ends of the element from their equilibrium positions.
1We use s(x, t) here instead of y(x, t) because the displacement of elements of the medium is not perpendicular to the x direction.
www.aswarphysics.weebly.com 510 Chapter 17 Sound Waves Substitute the position function given by Equation 17.1: DP 5 2B ' 'x 3smax cos 1kx 2 vt2 4 5 Bsmaxk sin 1kx 2 vt2 From this result, we see that a displacement described by a cosine function leads to a pressure described by a sine function. We also see that the displacement and pres-sure amplitudes are related by DPmax 5 Bsmaxk (17.4) This relationship depends on the bulk modulus of the gas, which is not as readily available as is the density of the gas. Once we determine the speed of sound in a gas in Section 17.2, we will be able to provide an expression that relates DPmax and smax in terms of the density of the gas. This discussion shows that a sound wave may be described equally well in terms of either pressure or displacement. A comparison of Equations 17.1 and 17.2 shows that the pressure wave is 908 out of phase with the displacement wave. Graphs of these functions are shown in Figure 17.4. The pressure variation is a maximum when the displacement from equilibrium is zero, and the displacement from equi-librium is a maximum when the pressure variation is zero.
Q uick Quiz 17.1 If you blow across the top of an empty soft-drink bottle, a pulse of sound travels down through the air in the bottle. At the moment the pulse reaches the bottom of the bottle, what is the correct description of the displace-ment of elements of air from their equilibrium positions and the pressure of the air at this point? (a) The displacement and pressure are both at a maximum. (b) The displacement and pressure are both at a minimum. (c) The displace-ment is zero, and the pressure is a maximum. (d) The displacement is zero, and the pressure is a minimum.
17.2 Speed of Sound Waves We now extend the discussion begun in Section 17.1 to evaluate the speed of sound in a gas. In Figure 17.5a, consider the cylindrical element of gas between the piston and the dashed line. This element of gas is in equilibrium under the influence of forces of equal magnitude, from the piston on the left and from the rest of the gas on the right. The magnitude of these forces is PA, where P is the pressure in the gas and A is the cross-sectional area of the tube. Figure 17.5b shows the situation after a time interval Dt during which the piston moves to the right at a constant speed vx due to a force from the left on the piston that has increased in magnitude to (P 1 DP)A. By the end of the time interval Dt, Undisturbed gas Undisturbed gas Compressed gas v t vx t b a (P P)Ai ˆ PAi ˆ PAi ˆ vxi ˆ PAi ˆ Figure 17.5 (a) An undisturbed element of gas of length v Dt in a tube of cross-sectional area A. The element is in equilibrium between forces on either end. (b) When the piston moves inward at constant velocity vx due to an increased force on the left, the element also moves with the same velocity.
s x x P Pmax smax b a Figure 17.4 (a) Displacement amplitude and (b) pressure ampli-tude versus position for a sinusoi-dal longitudinal wave.
www.aswarphysics.weebly.com 17.2 Speed of Sound Waves 511 every bit of gas in the element is moving with speed vx. That will not be true in general for a macroscopic element of gas, but it will become true if we shrink the length of the element to an infinitesimal value. The length of the undisturbed element of gas is chosen to be v Dt, where v is the speed of sound in the gas and Dt is the time interval between the configurations in Figures 17.5a and 17.5b. Therefore, at the end of the time interval Dt, the sound wave will just reach the right end of the cylindrical element of gas. The gas to the right of the element is undisturbed because the sound wave has not reached it yet. The element of gas is modeled as a nonisolated system in terms of momentum. The force from the piston has provided an impulse to the element, which in turn exhibits a change in momentum. Therefore, we evaluate both sides of the impulse– momentum theorem: Dp S 5 I S (17.5) On the right, the impulse is provided by the constant force due to the increased pressure on the piston: I S 5 a F S Dt 5 1A DP Dt2 i ^ The pressure change DP can be related to the volume change and then to the speeds v and vx through the bulk modulus: DP 5 2B DV Vi 5 2B 12vx A Dt2 vA Dt 5 B vx v Therefore, the impulse becomes I S 5 aAB vx v Dtb i ^ (17.6) On the left-hand side of the impulse–momentum theorem, Equation 17.5, the change in momentum of the element of gas of mass m is as follows: Dp S 5 m Dv S 5 1rVi2 1vx i ^ 2 02 5 1rvvx A Dt2 i ^ (17.7) Substituting Equations 17.6 and 17.7 into Equation 17.5, we find rvvx A Dt 5 AB vx v Dt which reduces to an expression for the speed of sound in a gas: v 5 Å B r (17.8) It is interesting to compare this expression with Equation 16.18 for the speed of transverse waves on a string, v 5 !T/m. In both cases, the wave speed depends on an elastic property of the medium (bulk modulus B or string tension T ) and on an inertial property of the medium (volume density r or linear density m). In fact, the speed of all mechanical waves follows an expression of the general form v 5 Å elastic property inertial property For longitudinal sound waves in a solid rod of material, for example, the speed of sound depends on Young’s modulus Y and the density r. Table 17.1 (page 512) pro-vides the speed of sound in several different materials. The speed of sound also depends on the temperature of the medium. For sound traveling through air, the relationship between wave speed and air temperature is v 5 331Å1 1 TC 273 (17.9) www.aswarphysics.weebly.com 512 Chapter 17 Sound Waves where v is in meters/second, 331 m/s is the speed of sound in air at 08C, and TC is the air temperature in degrees Celsius. Using this equation, one finds that at 208C, the speed of sound in air is approximately 343 m/s. This information provides a convenient way to estimate the distance to a thun-derstorm. First count the number of seconds between seeing the flash of lightning and hearing the thunder. Dividing this time interval by 3 gives the approximate distance to the lightning in kilometers because 343 m/s is approximately 1 3 km/s. Dividing the time interval in seconds by 5 gives the approximate distance to the lightning in miles because the speed of sound is approximately 1 5 mi/s. Having an expression (Eq. 17.8) for the speed of sound, we can now express the relationship between pressure amplitude and displacement amplitude for a sound wave (Eq. 17.4) as DPmax 5 Bsmaxk 5 1rv22smaxav v b 5 rvvsmax (17.10) This expression is a bit more useful than Equation 17.4 because the density of a gas is more readily available than is the bulk modulus.
17.3 Intensity of Periodic Sound Waves In Chapter 16, we showed that a wave traveling on a taut string transports energy, consistent with the notion of energy transfer by mechanical waves in Equation 8.2. Naturally, we would expect sound waves to also represent a transfer of energy. Consider the element of gas acted on by the piston in Figure 17.5. Imagine that the piston is moving back and forth in simple harmonic motion at angular frequency v. Imagine also that the length of the element becomes very small so that the entire element moves with the same velocity as the piston. Then we can model the ele-ment as a particle on which the piston is doing work. The rate at which the piston is doing work on the element at any instant of time is given by Equation 8.19: Power 5 F S ? v S x where we have used Power rather than P so that we don’t confuse power P with pressure P! The force F S on the element of gas is related to the pressure and the velocity v S x of the element is the derivative of the displacement function, so we find Power 5 3DP 1x, t2A4 i ^ ? ' 't 3s1x, t2 i ^4 5 3rvvAsmax sin 1kx 2 vt2 4 e ' 't 3s max cos 1kx 2 vt2 4 f Table 17.1 Speed of Sound in Various Media Medium v (m/s) Medium v (m/s) Medium v (m/s) Gases Liquids at 258C Solidsa Hydrogen (08C) 1 286 Glycerol 1 904 Pyrex glass 5 640 Helium (08C) 972 Seawater 1 533 Iron 5 950 Air (208C) 343 Water 1 493 Aluminum 6 420 Air (08C) 331 Mercury 1 450 Brass 4 700 Oxygen (08C) 317 Kerosene 1 324 Copper 5 010 Methyl alcohol 1 143 Gold 3 240 Carbon tetrachloride 926 Lucite 2 680 Lead 1 960 Rubber 1 600 aValues given are for propagation of longitudinal waves in bulk media. Speeds for longitudinal waves in thin rods are smaller, and speeds of transverse waves in bulk are smaller yet.
www.aswarphysics.weebly.com 17.3 Intensity of Periodic Sound Waves 513 5 rvvAs max sin 1kx 2 vt2 43vs max sin 1kx 2 vt2 4 5 rvv2As 2 max sin2 1kx 2 vt2 We now find the time average power over one period of the oscillation. For any given value of x, which we can choose to be x 5 0, the average value of sin2 (kx 2 vt) over one period T is 1 T 3 T 0 sin2 10 2 vt2 dt 5 1 T 3 T 0 sin2 vt dt 5 1 T a t 2 1 sin 2vt 2v b T 0 5 1 2 Therefore, 1Power2 avg 5 1 2rvv2As 2 max We define the intensity I of a wave, or the power per unit area, as the rate at which the energy transported by the wave transfers through a unit area A perpen-dicular to the direction of travel of the wave: I ; 1Power2 avg A (17.11) In this case, the intensity is therefore I 5 1 2rv 1vsmax2 2 Hence, the intensity of a periodic sound wave is proportional to the square of the displacement amplitude and to the square of the angular frequency. This expres-sion can also be written in terms of the pressure amplitude DPmax; in this case, we use Equation 17.10 to obtain I 5 1DPmax2 2 2rv (17.12) The string waves we studied in Chapter 16 are constrained to move along the one-dimensional string, as discussed in the introduction to this chapter. The sound waves we have studied with regard to Figures 17.1 through 17.3 and 17.5 are con-strained to move in one dimension along the length of the tube. As we mentioned in the introduction, however, sound waves can move through three-dimensional bulk media, so let’s place a sound source in the open air and study the results. Consider the special case of a point source emitting sound waves equally in all directions. If the air around the source is perfectly uniform, the sound power radi-ated in all directions is the same, and the speed of sound in all directions is the same. The result in this situation is called a spherical wave. Figure 17.6 shows these spherical waves as a series of circular arcs concentric with the source. Each arc rep-resents a surface over which the phase of the wave is constant. We call such a sur-face of constant phase a wave front. The radial distance between adjacent wave fronts that have the same phase is the wavelength l of the wave. The radial lines pointing outward from the source, representing the direction of propagation of the waves, are called rays. The average power emitted by the source must be distributed uniformly over each spherical wave front of area 4pr 2. Hence, the wave intensity at a distance r from the source is I 5 1Power2 avg A 5 1Power2 avg 4pr 2 (17.13) The intensity decreases as the square of the distance from the source. This inverse-square law is reminiscent of the behavior of gravity in Chapter 13.
W W Intensity of a sound wave Ray Source l Wave front The rays are radial lines pointing outward from the source, perpendicular to the wave fronts.
Figure 17.6 Spherical waves emitted by a point source. The circular arcs represent the spheri-cal wave fronts that are concentric with the source.
www.aswarphysics.weebly.com 514 Chapter 17 Sound Waves Example 17.2 Intensity Variations of a Point Source A point source emits sound waves with an average power output of 80.0 W.
(A) Find the intensity 3.00 m from the source.
Conceptualize Imagine a small loudspeaker sending sound out at an average rate of 80.0 W uniformly in all direc-tions. You are standing 3.00 m away from the speakers. As the sound propagates, the energy of the sound waves is spread out over an ever-expanding sphere, so the intensity of the sound falls off with distance.
Categorize We evaluate the intensity from an equation generated in this section, so we categorize this example as a substitution problem.
S o l u t i o n Example 17.1 Hearing Limits The faintest sounds the human ear can detect at a frequency of 1 000 Hz correspond to an intensity of about 1.00 3 10212 W/m2, which is called threshold of hearing. The loudest sounds the ear can tolerate at this frequency correspond to an intensity of about 1.00 W/m2, the threshold of pain. Determine the pressure amplitude and displacement amplitude associated with these two limits.
Conceptualize Think about the quietest environment you have ever experienced. It is likely that the intensity of sound in even this quietest environment is higher than the threshold of hearing.
Categorize Because we are given intensities and asked to calculate pressure and displacement amplitudes, this prob-lem is an analysis problem requiring the concepts discussed in this section.
Analyze To find the amplitude of the pressure varia-tion at the threshold of hearing, use Equation 17.12, taking the speed of sound waves in air to be v 5 343 m/s and the density of air to be r 5 1.20 kg/m3: DPmax 5 "2rvI 5 "211.20 kg/m32 1343 m/s2 11.00 3 10212 W/m22 5 2.87 3 1025 N/m2 Calculate the corresponding displacement amplitude using Equation 17.10, recalling that v 5 2pf (Eq. 16.9): smax 5 DPmax rvv 5 2.87 3 1025 N/m2 11.20 kg/m32 1343 m/s2 12p 3 1 000 Hz2 5 1.11 3 10211 m In a similar manner, one finds that the loudest sounds the human ear can tolerate (the threshold of pain) corre-spond to a pressure amplitude of 28.7 N/m2 and a displacement amplitude equal to 1.11 3 1025 m .
Finalize Because atmospheric pressure is about 105 N/m2, the result for the pressure amplitude tells us that the ear is sensitive to pressure fluctuations as small as 3 parts in 1010! The displacement amplitude is also a remarkably small number! If we compare this result for smax to the size of an atom (about 10210 m), we see that the ear is an extremely sensitive detector of sound waves.
S o l u t i o n Q uick Quiz 17.2 A vibrating guitar string makes very little sound if it is not mounted on the guitar body. Why does the sound have greater intensity if the string is attached to the guitar body? (a) The string vibrates with more energy. (b) The energy leaves the guitar at a greater rate. (c) The sound power is spread over a larger area at the listener’s position. (d) The sound power is concentrated over a smaller area at the listener’s position. (e) The speed of sound is higher in the material of the guitar body. (f) None of these answers is correct.
www.aswarphysics.weebly.com 17.3 Intensity of Periodic Sound Waves 515 Because a point source emits energy in the form of spherical waves, use Equation 17.13 to find the intensity: I 5 1Power2 avg 4pr 2 5 80.0 W 4p13.00 m2 2 5 0.707 W/m2 Solve for r in Equation 17.13 and use the given value for I: r 5 Å 1Power2 avg 4pI 5 Å 80.0 W 4p11.00 3 1028 W/m22 5 2.52 3 104 m This intensity is close to the threshold of pain.
(B) Find the distance at which the intensity of the sound is 1.00 3 1028 W/m2.
S o l u t i o n Sound Level in Decibels Example 17.1 illustrates the wide range of intensities the human ear can detect. Because this range is so wide, it is convenient to use a logarithmic scale, where the sound level b (Greek letter beta) is defined by the equation b ; 10 log a I I0 b (17.14) The constant I0 is the reference intensity, taken to be at the threshold of hearing (I0 5 1.00 3 10212 W/m2), and I is the intensity in watts per square meter to which the sound level b corresponds, where b is measured2 in decibels (dB). On this scale, the threshold of pain (I 5 1.00 W/m2) corresponds to a sound level of b 5 10 log [(1 W/m2)/(10212 W/m2)] 5 10 log (1012) 5 120 dB, and the threshold of hearing corresponds to b 5 10 log [(10212 W/m2)/(10212 W/m2)] 5 0 dB. Prolonged exposure to high sound levels may seriously damage the human ear. Ear plugs are recommended whenever sound levels exceed 90 dB. Recent evidence suggests that “noise pollution” may be a contributing factor to high blood pressure, anxiety, and nervousness. Table 17.2 gives some typical sound levels.
Q uick Quiz 17.3 Increasing the intensity of a sound by a factor of 100 causes the sound level to increase by what amount? (a) 100 dB (b) 20 dB (c) 10 dB (d) 2 dB 2The unit bel is named after the inventor of the telephone, Alexander Graham Bell (1847–1922). The prefix deci- is the SI prefix that stands for 1021.
Example 17.3 Sound Levels Two identical machines are positioned the same distance from a worker. The intensity of sound delivered by each oper-ating machine at the worker’s location is 2.0 3 1027 W/m2.
(A) Find the sound level heard by the worker when one machine is operating.
Conceptualize Imagine a situation in which one source of sound is active and is then joined by a second identical source, such as one person speaking and then a second person speaking at the same time or one musical instrument playing and then being joined by a second instrument.
Categorize This example is a relatively simple analysis problem requiring Equation 17.14.
S o l u t i o n Table 17.2 Sound Levels Source of Sound b (dB) Nearby jet airplane 150 Jackhammer; machine gun 130 Siren; rock concert 120 Subway; power lawn mower 100 Busy traffic 80 Vacuum cleaner 70 Normal conversation 60 Mosquito buzzing 40 Whisper 30 Rustling leaves 10 Threshold of hearing 0 ▸ 17.2 c on tin u ed continued www.aswarphysics.weebly.com 516 Chapter 17 Sound Waves Analyze Use Equation 17.14 to calculate the sound level at the worker’s location with one machine operating: b1 5 10 log a 2.0 3 1027 W/m2 1.00 3 10212 W/m2b 5 10 log 12.0 3 1052 5 53 dB Use Equation 17.14 to calculate the sound level at the worker’s location with double the intensity: b2 5 10 log a 4.0 3 1027 W/m2 1.00 3 10212 W/m2b 5 10 log 14.0 3 1052 5 56 dB (B) Find the sound level heard by the worker when two machines are operating.
S o l u t i o n Finalize These results show that when the intensity is doubled, the sound level increases by only 3 dB. This 3-dB increase is independent of the original sound level. (Prove this to yourself!) Loudness is a psychological response to a sound. It depends on both the intensity and the frequency of the sound. As a rule of thumb, a doubling in loudness is approximately associated with an increase in sound level of 10 dB. (This rule of thumb is relatively inaccurate at very low or very high frequencies.) If the loudness of the machines in this example is to be doubled, how many machines at the same distance from the worker must be running?
Answer Using the rule of thumb, a doubling of loudness corresponds to a sound level increase of 10 dB. Therefore, b2 2 b1 5 10 dB 5 10 log a I2 I0 b 2 10 log aI1 I0 b 5 10 log a I2 I1 b log a I2 I1 b 5 1 S I2 5 10I1 Therefore, ten machines must be operating to double the loudness.
What If?
Loudness and Frequency The discussion of sound level in decibels relates to a physical measurement of the strength of a sound. Let us now extend our discussion from the What If? section of Example 17.3 concerning the psychological “measurement” of the strength of a sound. Of course, we don’t have instruments in our bodies that can display numerical values of our reactions to stimuli. We have to “calibrate” our reactions somehow by comparing different sounds to a reference sound, but that is not easy to accom-plish. For example, earlier we mentioned that the threshold intensity is 10212 W/m2, corresponding to an intensity level of 0 dB. In reality, this value is the threshold only for a sound of frequency 1 000 Hz, which is a standard reference frequency in acoustics. If we perform an experiment to measure the threshold intensity at other frequencies, we find a distinct variation of this threshold as a function of frequency. For example, at 100 Hz, a barely audible sound must have an intensity level of about 30 dB! Unfortunately, there is no simple relationship between physical measurements and psychological “measurements.” The 100-Hz, 30-dB sound is psychologically “equal” in loudness to the 1 000-Hz, 0-dB sound (both are just barely audible), but they are not physically equal in sound level (30 dB 2 0 dB). By using test subjects, the human response to sound has been studied, and the results are shown in the white area of Figure 17.7 along with the approximate fre-quency and sound-level ranges of other sound sources. The lower curve of the white area corresponds to the threshold of hearing. Its variation with frequency is clear from this diagram. Notice that humans are sensitive to frequencies ranging from about 20 Hz to about 20 000 Hz. The upper bound of the white area is the thresh- ▸ 17.3 continu ed www.aswarphysics.weebly.com 17.4 The Doppler Effect 517 old of pain. Here the boundary of the white area appears straight because the psy-chological response is relatively independent of frequency at this high sound level. The most dramatic change with frequency is in the lower left region of the white area, for low frequencies and low intensity levels. Our ears are particularly insen-sitive in this region. If you are listening to your home entertainment system and the bass (low frequencies) and treble (high frequencies) sound balanced at a high volume, try turning the volume down and listening again. You will probably notice that the bass seems weak, which is due to the insensitivity of the ear to low frequen-cies at low sound levels as shown in Figure 17.7.
17.4 The Doppler Effect Perhaps you have noticed how the sound of a vehicle’s horn changes as the vehicle moves past you. The frequency of the sound you hear as the vehicle approaches you is higher than the frequency you hear as it moves away from you. This experience is one example of the Doppler effect.3 To see what causes this apparent frequency change, imagine you are in a boat that is lying at anchor on a gentle sea where the waves have a period of T 5 3.0 s. Hence, every 3.0 s a crest hits your boat. Figure 17.8a shows this situation, with the water waves moving toward the left. If you set your watch to t 5 0 just as one crest hits, the watch reads 3.0 s when the next crest hits, 6.0 s when the third crest Infrasonic frequencies Sonic frequencies Ultrasonic frequencies Large rocket engine Jet engine (10 m away) Rifle Thunder overhead Rock concert Underwater communication (Sonar) Car horn Motorcycle School cafeteria Urban traffic Shout Conversation Birds Bats Whispered speech Threshold of hearing Sound level (dB) 1 10 100 1 000 10 000 100 000 Frequency f (Hz) 220 200 180 160 140 120 100 80 60 40 20 0 Threshold of pain b Figure 17.7 Approximate ranges of frequency and sound level of various sources and that of normal human hearing, shown by the white area. (From R. L. Reese, University Physics, Pacific Grove, Brooks/Cole, 2000.) 3Named after Austrian physicist Christian Johann Doppler (1803–1853), who in 1842 predicted the effect for both sound waves and light waves.
In all frames, the waves travel to the left, and their source is far to the right of the boat, out of the frame of the figure.
a b c vwaves S vwaves S vboat S vwaves S vboat S In all frames, the waves travel to the left, and their source is far to the right of the boat, out of the frame of the figure.
a b vwaves S vwaves S vboat S In all frames, the waves travel to the left, and their source is far to the right of the boat, out of the frame of the figure.
a b c vwaves S vwaves S vboat S vwaves S vboat S Figure 17.8 (a) Waves moving toward a stationary boat. (b) The boat moving toward the wave source. (c) The boat moving away from the wave source.
www.aswarphysics.weebly.com 518 Chapter 17 Sound Waves hits, and so on. From these observations, you conclude that the wave frequency is f 5 1/T 5 1/(3.0 s) 5 0.33 Hz. Now suppose you start your motor and head directly into the oncoming waves as in Figure 17.8b. Again you set your watch to t 5 0 as a crest hits the front (the bow) of your boat. Now, however, because you are moving toward the next wave crest as it moves toward you, it hits you less than 3.0 s after the first hit. In other words, the period you observe is shorter than the 3.0-s period you observed when you were stationary. Because f 5 1/T, you observe a higher wave frequency than when you were at rest. If you turn around and move in the same direction as the waves (Fig. 17.8c), you observe the opposite effect. You set your watch to t 5 0 as a crest hits the back (the stern) of the boat. Because you are now moving away from the next crest, more than 3.0 s has elapsed on your watch by the time that crest catches you. Therefore, you observe a lower frequency than when you were at rest. These effects occur because the relative speed between your boat and the waves depends on the direction of travel and on the speed of your boat. (See Section 4.6.) When you are moving toward the right in Figure 17.8b, this relative speed is higher than that of the wave speed, which leads to the observation of an increased fre-quency. When you turn around and move to the left, the relative speed is lower, as is the observed frequency of the water waves. Let’s now examine an analogous situation with sound waves in which the water waves become sound waves, the water becomes the air, and the person on the boat becomes an observer listening to the sound. In this case, an observer O is moving and a sound source S is stationary. For simplicity, we assume the air is also station-ary and the observer moves directly toward the source (Fig. 17.9). The observer moves with a speed vO toward a stationary point source (vS 5 0), where stationary means at rest with respect to the medium, air. If a point source emits sound waves and the medium is uniform, the waves move at the same speed in all directions radially away from the source; the result is a spherical wave as mentioned in Section 17.3. The distance between adjacent wave fronts equals the wavelength l. In Figure 17.9, the circles are the intersections of these three-dimensional wave fronts with the two-dimensional paper. We take the frequency of the source in Figure 17.9 to be f, the wavelength to be l, and the speed of sound to be v. If the observer were also stationary, he would detect wave fronts at a frequency f. (That is, when vO 5 0 and vS 5 0, the observed frequency equals the source frequency.) When the observer moves toward the source, the speed of the waves relative to the observer is v9 5 v 1 vO, as in the case of the boat in Figure 17.8, but the wavelength l is unchanged. Hence, using Equation 16.12, v 5 lf, we can say that the frequency f 9 heard by the observer is increased and is given by f r 5 vr l 5 v 1 vO l Because l 5 v/f, we can express f 9 as f r 5 av 1 vO v b f 1observer moving toward source2 (17.15) If the observer is moving away from the source, the speed of the wave relative to the observer is v9 5 v 2 vO. The frequency heard by the observer in this case is decreased and is given by f r 5 av 2 vO v b f ( observer moving away from source) (17.16) These last two equations can be reduced to a single equation by adopting a sign convention. Whenever an observer moves with a speed vO relative to a stationary source, the frequency heard by the observer is given by Equation 17.15, with vO interpreted as follows: a positive value is substituted for vO when the observer moves Figure 17.9 An observer O (the cyclist) moves with a speed vO toward a stationary point source S, the horn of a parked truck. The observer hears a fre-quency f 9 that is greater than the source frequency.
O O S v S www.aswarphysics.weebly.com 17.4 The Doppler Effect 519 toward the source, and a negative value is substituted when the observer moves away from the source. Now suppose the source is in motion and the observer is at rest. If the source moves directly toward observer A in Figure 17.10a, each new wave is emitted from a position to the right of the origin of the previous wave. As a result, the wave fronts heard by the observer are closer together than they would be if the source were not moving. (Fig. 17.10b shows this effect for waves moving on the surface of water.) As a result, the wavelength l9 measured by observer A is shorter than the wave-length l of the source. During each vibration, which lasts for a time interval T (the period), the source moves a distance vST 5 vS/f and the wavelength is shortened by this amount. Therefore, the observed wavelength l9 is lr 5 l 2 Dl 5 l 2 vS f Because l 5 v/f, the frequency f 9 heard by observer A is f r 5 v lr 5 v l 2 1vS /f 2 5 v 1v/f 2 2 1vS /f 2 f r 5 a v v 2 vSb f (source moving toward observer) (17.17) That is, the observed frequency is increased whenever the source is moving toward the observer. When the source moves away from a stationary observer, as is the case for observer B in Figure 17.10a, the observer measures a wavelength l9 that is greater than l and hears a decreased frequency: f r 5 a v v 1 vS b f (source moving away from observer) (17.18) We can express the general relationship for the observed frequency when a source is moving and an observer is at rest as Equation 17.17, with the same sign convention applied to vS as was applied to vO: a positive value is substituted for vS when the source moves toward the observer, and a negative value is substituted when the source moves away from the observer. Finally, combining Equations 17.15 and 17.17 gives the following general rela-tionship for the observed frequency that includes all four conditions described by Equations 17.15 through 17.18: f r 5 av 1 vO v 2 vS b f (17.19) W W General Doppler-shift expression Figure 17.10 (a) A source S mov-ing with a speed vS toward a sta-tionary observer A and away from a stationary observer B. Observer A hears an increased frequency, and observer B hears a decreased frequency. (b) The Doppler effect in water, observed in a ripple tank. Letters shown in the photo refer to Quick Quiz 17.4.
S S Observer A Observer B v l a S A C B b A point source is moving to the right with speed vS.
Courtesy of the Educational Development Center, Newton, MA Pitfall Prevention 17.1 Doppler Effect Does Not Depend on Distance Some people think that the Doppler effect depends on the distance between the source and the observer. Although the intensity of a sound varies as the distance changes, the apparent frequency depends only on the relative speed of source and observer. As you listen to an approaching source, you will detect increasing intensity but constant frequency. As the source passes, you will hear the frequency suddenly drop to a new constant value and the intensity begin to decrease.
www.aswarphysics.weebly.com 520 Chapter 17 Sound Waves In this expression, the signs for the values substituted for vO and vS depend on the direction of the velocity. A positive value is used for motion of the observer or the source toward the other (associated with an increase in observed frequency), and a negative value is used for motion of one away from the other (associated with a decrease in observed frequency). Although the Doppler effect is most typically experienced with sound waves, it is a phenomenon common to all waves. For example, the relative motion of source and observer produces a frequency shift in light waves. The Doppler effect is used in police radar systems to measure the speeds of motor vehicles. Likewise, astrono-mers use the effect to determine the speeds of stars, galaxies, and other celestial objects relative to the Earth.
Q uick Quiz 17.4 Consider detectors of water waves at three locations A, B, and C in Figure 17.10b. Which of the following statements is true? (a) The wave speed is highest at location A. (b) The wave speed is highest at location C. (c) The detected wavelength is largest at location B. (d) The detected wavelength is larg-est at location C. (e) The detected frequency is highest at location C. (f) The detected frequency is highest at location A.
Q uick Quiz 17.5 You stand on a platform at a train station and listen to a train approaching the station at a constant velocity. While the train approaches, but before it arrives, what do you hear? (a) the intensity and the frequency of the sound both increasing (b) the intensity and the frequency of the sound both decreasing (c) the intensity increasing and the frequency decreasing (d) the intensity decreasing and the frequency increasing (e) the intensity increasing and the frequency remaining the same (f) the intensity decreasing and the fre-quency remaining the same Example 17.4 The Broken Clock Radio Your clock radio awakens you with a steady and irritating sound of frequency 600 Hz. One morning, it malfunctions and cannot be turned off. In frustration, you drop the clock radio out of your fourth-story dorm window, 15.0 m from the ground. Assume the speed of sound is 343 m/s. As you listen to the falling clock radio, what frequency do you hear just before you hear it striking the ground?
Conceptualize The speed of the clock radio increases as it falls. Therefore, it is a source of sound moving away from you with an increasing speed so the frequency you hear should be less than 600 Hz.
Categorize We categorize this problem as one in which we combine the particle under constant acceleration model for the falling radio with our understanding of the frequency shift of sound due to the Doppler effect.
AM S o l u t i o n Analyze Because the clock radio is modeled as a parti-cle under constant acceleration due to gravity, use Equa-tion 2.13 to express the speed of the source of sound: (1) vS 5 vyi 1 ayt 5 0 2 gt 5 2gt From Equation 2.16, find the time at which the clock radio strikes the ground: yf 5 yi 1 vyit 2 1 2gt 2 5 0 1 0 2 1 2gt 2 S t 5 Å2 2yf g Substitute into Equation (1): vS 5 12g2 Å2 2yf g 5 2"22g yf Use Equation 17.19 to determine the Doppler-shifted frequency heard from the falling clock radio: f r 5 c v 1 0 v 2 12"22gyf2 d f 5 a v v 1 "22gyf b f www.aswarphysics.weebly.com Example 17.5 Doppler Submarines A submarine (sub A) travels through water at a speed of 8.00 m/s, emitting a sonar wave at a frequency of 1 400 Hz. The speed of sound in the water is 1 533 m/s. A second submarine (sub B) is located such that both submarines are traveling directly toward each other. The second submarine is moving at 9.00 m/s.
(A) What frequency is detected by an observer riding on sub B as the subs approach each other?
Conceptualize Even though the problem involves subs moving in water, there is a Doppler effect just like there is when you are in a moving car and listening to a sound moving through the air from another car.
Categorize Because both subs are moving, we categorize this problem as one involving the Doppler effect for both a moving source and a moving observer.
S o l u t i o n Analyze Use Equation 17.19 to find the Doppler-shifted frequency heard by the observer in sub B, being careful with the signs assigned to the source and observer speeds: f r 5 a v 1 vO v 2 vS b f f r 5 c 1 533 m/s 1 119.00 m/s2 1 533 m/s 2 118.00 m/s2 d 11 400 Hz2 5 1 416 Hz Use Equation 17.19 to find the Doppler-shifted fre-quency heard by the observer in sub B, again being careful with the signs assigned to the source and observer speeds: f r 5 a v 1 vO v 2 vS b f f r 5 c 1 533 m/s 1 129.00 m/s2 1 533 m/s 2 128.00 m/s2 d 11 400 Hz2 5 1 385 Hz The sound of apparent frequency 1 416 Hz found in part (A) is reflected from a moving source (sub B) and then detected by a moving observer (sub A). Find the frequency detected by sub A: f s 5 a v 1 vO v 2 vS b f r 5 c 1 533 m/s 1 118.00 m/s2 1 533 m/s 2 119.00 m/s2 d 11 416 Hz2 5 1 432 Hz (B) The subs barely miss each other and pass. What frequency is detected by an observer riding on sub B as the subs recede from each other?
S o l u t i o n Notice that the frequency drops from 1 416 Hz to 1 385 Hz as the subs pass. This effect is similar to the drop in fre-quency you hear when a car passes by you while blowing its horn.
(C) While the subs are approaching each other, some of the sound from sub A reflects from sub B and returns to sub A. If this sound were to be detected by an observer on sub A, what is its frequency?
S o l u t i o n Substitute numerical values: f r 5 c 343 m/s 343 m/s 1 "2219.80 m/s22 1215.0 m2 d 1600 Hz2 5 571 Hz Finalize The frequency is lower than the actual frequency of 600 Hz because the clock radio is moving away from you. If it were to fall from a higher floor so that it passes below y 5 215.0 m, the clock radio would continue to accelerate and the frequency would continue to drop. ▸ 17.4 c on tin u ed continued 17.4 The Doppler Effect 521 www.aswarphysics.weebly.com 522 Chapter 17 Sound Waves ▸ 17.5 continu ed Finalize This technique is used by police officers to measure the speed of a moving car. Microwaves are emitted from the police car and reflected by the moving car. By detecting the Doppler-shifted frequency of the reflected micro-waves, the police officer can determine the speed of the moving car.
Shock Waves Now consider what happens when the speed vS of a source exceeds the wave speed v. This situation is depicted graphically in Figure 17.11a. The circles represent spheri-cal wave fronts emitted by the source at various times during its motion. At t 5 0, the source is at S0 and moving toward the right. At later times, the source is at S1, and then S2, and so on. At the time t, the wave front centered at S0 reaches a radius of vt. In this same time interval, the source travels a distance vSt. Notice in Figure 17.11a that a straight line can be drawn tangent to all the wave fronts generated at various times. Therefore, the envelope of these wave fronts is a cone whose apex half-angle u (the “Mach angle”) is given by sin u 5 vt vSt 5 v vS The ratio vS/v is referred to as the Mach number, and the conical wave front pro-duced when vS . v (supersonic speeds) is known as a shock wave. An interesting anal-ogy to shock waves is the V-shaped wave fronts produced by a boat (the bow wave) when the boat’s speed exceeds the speed of the surface-water waves (Fig. 17.12). Jet airplanes traveling at supersonic speeds produce shock waves, which are responsible for the loud “sonic boom” one hears. The shock wave carries a great deal of energy concentrated on the surface of the cone, with correspondingly great pressure variations. Such shock waves are unpleasant to hear and can cause dam-age to buildings when aircraft fly supersonically at low altitudes. In fact, an air-plane flying at supersonic speeds produces a double boom because two shock waves are formed, one from the nose of the plane and one from the tail. People near the path of a space shuttle as it glides toward its landing point have reported hearing what sounds like two very closely spaced cracks of thunder.
Q uick Quiz 17.6 An airplane flying with a constant velocity moves from a cold air mass into a warm air mass. Does the Mach number (a) increase, (b) decrease, or (c) stay the same?
Figure 17.11 (a) A representa-tion of a shock wave produced when a source moves from S0 to the right with a speed vS that is greater than the wave speed v in the medium. (b) A stroboscopic photograph of a bullet moving at supersonic speed through the hot air above a candle.
vSt vS S S1 S2 S0 The envelope of the wave fronts forms a cone whose apex half-angle is given by sin u v/vS.
vt u a 0 1 2 b Notice the shock wave in the vicinity of the bullet.
Omikron/Photo Researchers/Getty Images Figure 17.12 The V-shaped bow wave of a boat is formed because the boat speed is greater than the speed of the water waves it gener-ates. A bow wave is analogous to a shock wave formed by an airplane traveling faster than sound.
Robert Holland/Stone/Getty Images www.aswarphysics.weebly.com Objective Questions 523 (Fig. OQ17.3) sounding its siren at a frequency of 500 Hz. Which statement is correct? (a) You hear a frequency less than 500 Hz. (b) You hear a frequency equal to 500 Hz. (c) You hear a frequency greater 1. Table 17.1 shows the speed of sound is typically an order of magnitude larger in solids than in gases. To what can this higher value be most directly attributed? (a) the difference in density between solids and gases (b) the difference in compressibility between solids and gases (c) the limited size of a solid object com-pared to a free gas (d) the impossibility of holding a gas under significant tension 2. Two sirens A and B are sounding so that the frequency from A is twice the frequency from B. Compared with the speed of sound from A, is the speed of sound from B (a) twice as fast, (b) half as fast, (c) four times as fast, (d) one-fourth as fast, or (e) the same? 3. As you travel down the highway in your car, an ambu-lance approaches you from the rear at a high speed Concepts and Principles Sound waves are longitudinal and travel through a compressible medium with a speed that depends on the elastic and inertial proper-ties of that medium. The speed of sound in a gas having a bulk modulus B and density r is v 5 Å B r (17.8) For sinusoidal sound waves, the variation in the position of an element of the medium is s(x, t) 5 smax cos (kx 2 vt) (17.1) and the variation in pressure from the equilibrium value is DP 5 DPmax sin (kx 2 vt) (17.2) where DPmax is the pressure amplitude. The pressure wave is 908 out of phase with the displacement wave. The relationship between smax and DPmax is DPmax 5 rvvsmax (17.10) The change in frequency heard by an observer whenever there is relative motion between a source of sound waves and the observer is called the Doppler effect. The observed frequency is f r 5 a v 1 vO v 2 vS b f (17.19) In this expression, the signs for the values substituted for vO and vS depend on the direction of the velocity. A positive value for the speed of the observer or source is substituted if the velocity of one is toward the other, whereas a nega-tive value represents a velocity of one away from the other.
Objective Questions 1. denotes answer available in Student Solutions Manual/Study Guide Figure OQ17.3 Anthony Redpath/Corbis Summary The intensity of a periodic sound wave, which is the power per unit area, is I ; 1Power2 avg A 5 1DPmax2 2 2rv (17.11, 17.12) The sound level of a sound wave in decibels is b ; 10 log a I I0 b (17.14) The constant I0 is a reference intensity, usually taken to be at the threshold of hearing (1.00 3 10212 W/m2), and I is the intensity of the sound wave in watts per square meter.
Definitions www.aswarphysics.weebly.com 524 Chapter 17 Sound Waves how does the intensity change? (a) It becomes one-ninth as large. (b) It becomes one-third as large. (c) It is unchanged. (d) It becomes three times larger. (e) It becomes nine times larger. 10. Suppose an observer and a source of sound are both at rest relative to the ground and a strong wind is blow-ing away from the source toward the observer. (i) What effect does the wind have on the observed frequency? (a) It causes an increase. (b) It causes a decrease. (c) It causes no change. (ii) What effect does the wind have on the observed wavelength? Choose from the same possibilities as in part (i). (iii) What effect does the wind have on the observed speed of the wave? Choose from the same possibilities as in part (i). 11. A source of sound vibrates with constant frequency. Rank the frequency of sound observed in the follow-ing cases from highest to the lowest. If two frequencies are equal, show their equality in your ranking. All the motions mentioned have the same speed, 25 m/s. (a) The source and observer are stationary. (b) The source is moving toward a stationary observer. (c) The source is moving away from a stationary observer. (d) The observer is moving toward a stationary source. (e) The observer is moving away from a stationary source. 12. With a sensitive sound-level meter, you measure the sound of a running spider as 210 dB. What does the negative sign imply? (a) The spider is moving away from you. (b) The frequency of the sound is too low to be audible to humans. (c) The intensity of the sound is too faint to be audible to humans. (d) You have made a mistake; negative signs do not fit with logarithms. 13. Doubling the power output from a sound source emit-ting a single frequency will result in what increase in decibel level? (a) 0.50 dB (b) 2.0 dB (c) 3.0 dB (d) 4.0 dB (e) above 20 dB 14. Of the following sounds, which one is most likely to have a sound level of 60 dB? (a) a rock concert (b) the turning of a page in this textbook (c) dinner-table con-versation (d) a cheering crowd at a football game than 500 Hz. (d) You hear a frequency greater than 500 Hz, whereas the ambulance driver hears a fre-quency lower than 500 Hz. (e) You hear a frequency less than 500 Hz, whereas the ambulance driver hears a frequency of 500 Hz. 4. What happens to a sound wave as it travels from air into water? (a) Its intensity increases. (b) Its wavelength decreases. (c) Its frequency increases. (d) Its frequency remains the same. (e) Its velocity decreases. 5. A church bell in a steeple rings once. At 300 m in front of the church, the maximum sound intensity is 2 mW/m2. At 950 m behind the church, the maximum intensity is 0.2 mW/m2. What is the main reason for the difference in the intensity? (a) Most of the sound is absorbed by the air before it gets far away from the source. (b) Most of the sound is absorbed by the ground as it travels away from the source. (c) The bell broadcasts the sound mostly toward the front. (d) At a larger distance, the power is spread over a larger area. 6. If a 1.00-kHz sound source moves at a speed of 50.0 m/s toward a listener who moves at a speed of 30.0 m/s in a direction away from the source, what is the apparent frequency heard by the listener? (a) 796 Hz (b) 949 Hz (c) 1 000 Hz (d) 1 068 Hz (e) 1 273 Hz 7. A sound wave can be characterized as (a) a transverse wave, (b) a longitudinal wave, (c) a transverse wave or a longitudinal wave, depending on the nature of its source, (d) one that carries no energy, or (e) a wave that does not require a medium to be transmitted from one place to the other. 8. Assume a change at the source of sound reduces the wavelength of a sound wave in air by a factor of 2. (i) What happens to its frequency? (a) It increases by a factor of 4. (b) It increases by a factor of 2. (c) It is unchanged. (d) It decreases by a factor of 2. (e) It changes by an unpredict-able factor. (ii) What happens to its speed? Choose from the same possibilities as in part (i). 9. A point source broadcasts sound into a uniform medium. If the distance from the source is tripled, 1. How can an object move with respect to an observer so that the sound from it is not shifted in frequency? 2. Older auto-focus cameras sent out a pulse of sound and measured the time interval required for the pulse to reach an object, reflect off of it, and return to be detected. Can air temperature affect the camera’s focus? New cameras use a more reliable infrared system. 3. A friend sitting in her car far down the road waves to you and beeps her horn at the same moment. How far away must she be for you to calculate the speed of sound to two significant figures by measuring the time interval required for the sound to reach you? 4. How can you determine that the speed of sound is the same for all frequencies by listening to a band or orchestra? 5. Explain how the distance to a lightning bolt (Fig. CQ17.5) can be deter-mined by counting the seconds between the flash and the sound of thunder. 6. You are driving toward a cliff and honk your horn. Is there a Doppler shift of the sound when you hear the echo? If so, is it like a moving source or a mov-ing observer? What if the reflection occurs not from a cliff, but from the forward edge of a huge alien space-craft moving toward you as you drive?
Conceptual Questions 1. denotes answer available in Student Solutions Manual/Study Guide Figure CQ17.5 © iStockphoto.com/Colin Orthner www.aswarphysics.weebly.com Problems 525 4. An experimenter wishes to generate in air a sound wave that has a displacement amplitude of 5.50 3 1026 m. The pressure amplitude is to be limited to 0.840 Pa. What is the minimum wavelength the sound wave can have? 5. Calculate the pressure amplitude of a 2.00-kHz sound wave in air, assuming that the displacement amplitude is equal to 2.00 3 10–8 m. 6. Earthquakes at fault lines in the Earth’s crust create seismic waves, which are longitudinal (P waves) or transverse (S waves). The P waves have a speed of about 7 km/s. Estimate the average bulk modulus of the Earth’s crust given that the density of rock is about 2 500 kg/m3. 7. A dolphin (Fig. P17.7) in sea-water at a temperature of 258C emits a sound wave directed toward the ocean floor 150 m below. How much time passes before it hears an echo? 8. A sound wave propagates in air at 278C with frequency 4.00 kHz. It passes through a region where the temperature gradually changes and then moves through air at 08C. Give numerical answers to the fol-lowing questions to the extent possible and state your reasoning about what happens to the wave physically. (a) What happens to the speed of the wave? (b) What happens to its frequency? (c) What happens to its wavelength? 9. Ultrasound is used in medicine both for diagnostic imaging (Fig. P17.9, page 526) and for therapy. For M Figure P17.7 Stephen Frink/Photographer’s Choice/Getty Images Q/C BIO Note: Throughout this chapter, pressure variations DP are measured relative to atmospheric pressure, 1.013 3 105 Pa.
Section 17.1 Pressure Variations in Sound Waves 1. A sinusoidal sound wave moves through a medium and is described by the displacement wave function s(x, t) 5 2.00 cos (15.7x 2 858t) where s is in micrometers, x is in meters, and t is in sec-onds. Find (a) the amplitude, (b) the wavelength, and (c) the speed of this wave. (d) Determine the instanta-neous displacement from equilibrium of the elements of the medium at the position x 5 0.050 0 m at t 5 3.00 ms. (e) Determine the maximum speed of the ele-ment’s oscillatory motion. 2. As a certain sound wave travels through the air, it produces pressure variations (above and below atmo-spheric pressure) given by DP 5 1.27 sin (px 2 340pt) in SI units. Find (a) the amplitude of the pressure vari-ations, (b) the frequency, (c) the wavelength in air, and (d) the speed of the sound wave. 3. Write an expression that describes the pressure varia-tion as a function of position and time for a sinusoi-dal sound wave in air. Assume the speed of sound is 343 m/s, l 5 0.100 m, and DPmax 5 0.200 Pa.
Section 17.2 Speed of Sound Waves Problem 85 in Chapter 2 can also be assigned with this section.
Note: In the rest of this chapter, unless otherwise speci-fied, the equilibrium density of air is r 5 1.20 kg/m3 and the speed of sound in air is v 5 343 m/s. Use Table 17.1 to find speeds of sound in other media.
W Problems The problems found in this chapter may be assigned online in Enhanced WebAssign 1. straightforward; 2. intermediate; 3. challenging 1. full solution available in the Student Solutions Manual/Study Guide AMT Analysis Model tutorial available in Enhanced WebAssign GP Guided Problem M Master It tutorial available in Enhanced WebAssign W Watch It video solution available in Enhanced WebAssign BIO Q/C S 7. The radar systems used by police to detect speeders are sensitive to the Doppler shift of a pulse of microwaves. Discuss how this sensitivity can be used to measure the speed of a car. 8. The Tunguska event. On June 30, 1908, a meteor burned up and exploded in the atmosphere above the Tunguska River valley in Siberia. It knocked down trees over thousands of square kilometers and started a forest fire, but produced no crater and apparently caused no human casualties. A witness sitting on his doorstep outside the zone of falling trees recalled events in the following sequence. He saw a moving light in the sky, brighter than the Sun and descending at a low angle to the horizon. He felt his face become warm. He felt the ground shake. An invisible agent picked him up and immediately dropped him about a meter from where he had been seated. He heard a very loud protracted rumbling. Suggest an explana-tion for these observations and for the order in which they happened. 9. A sonic ranger is a device that determines the distance to an object by sending out an ultrasonic sound pulse and measuring the time interval required for the wave to return by reflection from the object. Typically, these devices cannot reliably detect an object that is less than half a meter from the sensor. Why is that?
www.aswarphysics.weebly.com 526 Chapter 17 Sound Waves 14. A flowerpot is knocked off a balcony from a height d above the sidewalk as shown in Figure P17.13. It falls toward an unsuspecting man of height h who is stand-ing below. Assume the man requires a time interval of Dt to respond to the warning. How close to the sidewalk can the flowerpot fall before it is too late for a warning shouted from the balcony to reach the man in time? Use the symbol v for the speed of sound. 15. The speed of sound in air (in meters per second) depends on temperature according to the approxi-mate expression v 5 331.5 1 0.607TC where TC is the Celsius temperature. In dry air, the temperature decreases about 18C for every 150-m rise in altitude. (a) Assume this change is constant up to an altitude of 9 000 m. What time interval is required for the sound from an airplane flying at 9 000 m to reach the ground on a day when the ground temperature is 308C? (b) What If? Compare your answer with the time interval required if the air were uniformly at 308C. Which time interval is longer? 16. A sound wave moves down a cylinder as in Figure 17.2. Show that the pressure variation of the wave is described by DP 5 6 rvv !s 2 max 2 s 2, where s 5 s(x, t) is given by Equation 17.1. 17. A hammer strikes one end of a thick iron rail of length 8.50 m. A microphone located at the opposite end of the rail detects two pulses of sound, one that travels through the air and a longitudinal wave that travels through the rail. (a) Which pulse reaches the micro-phone first? (b) Find the separation in time between the arrivals of the two pulses. 18. A cowboy stands on horizontal ground between two parallel, vertical cliffs. He is not midway between the cliffs. He fires a shot and hears its echoes. The second echo arrives 1.92 s after the first and 1.47 s before the third. Consider only the sound traveling parallel to the ground and reflecting from the cliffs. (a) What is the distance between the cliffs? (b) What If? If he can hear a fourth echo, how long after the third echo does it arrive?
Section 17.3 Intensity of Periodic Sound Waves 19. Calculate the sound level (in decibels) of a sound wave that has an intensity of 4.00 mW/m2. 20. The area of a typical eardrum is about 5.00 3 1025 m2. (a) Calculate the average sound power incident on an eardrum at the threshold of pain, which corresponds to an intensity of 1.00 W/m2. (b) How much energy is transferred to the eardrum exposed to this sound for 1.00 min? 21. The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.600 W/m2. (a) Determine the intensity that results if the frequency is increased to 2.50 kHz while a constant displacement amplitude is maintained. (b) Calculate the intensity if the frequency is reduced to 0.500 kHz and the dis-placement amplitude is doubled.
S S Q/C diagnosis, short pulses of ultrasound are passed through the patient’s body. An echo reflected from a structure of interest is recorded, and the distance to the structure can be determined from the time delay for the echo’s return. To reveal detail, the wavelength of the reflected ultrasound must be small compared to the size of the object reflecting the wave. The speed of ultrasound in human tissue is about 1 500 m/s (nearly the same as the speed of sound in water). (a) What is the wavelength of ultrasound with a frequency of 2.40 MHz? (b) In the whole set of imaging techniques, frequencies in the range 1.00 MHz to 20.0 MHz are used. What is the range of wavelengths corresponding to this range of frequencies?
Figure P17.9 A view of a fetus in the uterus made with ultra-sound imaging.
B. Benoit/Photo Researchers, Inc. 10. A sound wave in air has a pressure amplitude equal to 4.00 3 1023 Pa. Calculate the displacement amplitude of the wave at a frequency of 10.0 kHz. 11. Suppose you hear a clap of thunder 16.2 s after see-ing the associated lightning strike. The speed of light in air is 3.00 3 108 m/s. (a) How far are you from the lightning strike? (b) Do you need to know the value of the speed of light to answer? Explain. 12. A rescue plane flies horizontally at a constant speed searching for a disabled boat. When the plane is directly above the boat, the boat’s crew blows a loud horn. By the time the plane’s sound detector receives the horn’s sound, the plane has traveled a distance equal to half its altitude above the ocean. Assuming it takes the sound 2.00 s to reach the plane, determine (a) the speed of the plane and (b) its altitude. 13. A flowerpot is knocked off a window ledge from a height d 5 20.0 m above the sidewalk as shown in Figure P17.13. It falls toward an unsuspecting man of height h 5 1.75 m who is stand-ing below. Assume the man requires a time interval of Dt 5 0.300 s to respond to the warn-ing. How close to the sidewalk can the flowerpot fall before it is too late for a warning shouted from the balcony to reach the man in time?
W Q/C W W d h Figure P17.13 Problems 13 and 14.
AMT W www.aswarphysics.weebly.com Problems 527 31. A family ice show is held at an enclosed arena. The skaters perform to music with level 80.0 dB. This level is too loud for your baby, who yells at 75.0 dB. (a) What total sound intensity engulfs you? (b) What is the com-bined sound level? 32. Two small speakers emit sound waves of different fre-quencies equally in all directions. Speaker A has an output of 1.00 mW, and speaker B has an output of 1.50 mW. Determine the sound level (in decibels) at point C in Figure P17.32 assuming (a) only speaker A emits sound, (b) only speaker B emits sound, and (c) both speakers emit sound.
C A B 3.00 m 2.00 m 4.00 m Figure P17.32 33. A firework charge is detonated many meters above the ground. At a distance of d1 5 500 m from the explo-sion, the acoustic pressure reaches a maximum of DPmax 5 10.0 Pa (Fig. P17.33). Assume the speed of sound is constant at 343 m/s throughout the atmo-sphere over the region considered, the ground absorbs all the sound falling on it, and the air absorbs sound energy as described by the rate 7.00 dB/km. What is the sound level (in decibels) at a distance of d2 5 4.00 3 103 m from the explosion?
d1 d2 Figure P17.33 34. A fireworks rocket explodes at a height of 100 m above the ground. An observer on the ground directly under the explosion experiences an average sound intensity of 7.00 3 1022 W/m2 for 0.200 s. (a) What is the total amount of energy transferred away from the explosion by sound? (b) What is the sound level (in decibels) heard by the observer? 35. The sound level at a distance of 3.00 m from a source is 120 dB. At what distance is the sound level (a) 100 dB and (b) 10.0 dB? 36. Why is the following situation impossible? It is early on a Saturday morning, and much to your displeasure your next-door neighbor starts mowing his lawn. As you try to get back to sleep, your next-door neighbor on the other side of your house also begins to mow the lawn M W M 22. The intensity of a sound wave at a fixed distance from a speaker vibrating at a frequency f is I. (a) Determine the intensity that results if the frequency is increased to f 9 while a constant displacement amplitude is maintained. (b) Calculate the intensity if the frequency is reduced to f/2 and the displacement amplitude is doubled. 23. A person wears a hearing aid that uniformly increases the sound level of all audible frequencies of sound by 30.0 dB. The hearing aid picks up sound having a fre-quency of 250 Hz at an intensity of 3.0 3 10211 W/m2. What is the intensity delivered to the eardrum? 24. The sound intensity at a distance of 16 m from a noisy generator is measured to be 0.25 W/m2. What is the sound intensity at a distance of 28 m from the generator? 25. The power output of a certain public-address speaker is 6.00 W. Suppose it broadcasts equally in all direc-tions. (a) Within what distance from the speaker would the sound be painful to the ear? (b) At what distance from the speaker would the sound be barely audible? 26. A sound wave from a police siren has an intensity of 100.0 W/m2 at a certain point; a second sound wave from a nearby ambulance has an intensity level that is 10 dB greater than the police siren’s sound wave at the same point. What is the sound level of the sound wave due to the ambulance? 27. A train sounds its horn as it approaches an intersection. The horn can just be heard at a level of 50 dB by an observer 10 km away. (a) What is the average power gen-erated by the horn? (b) What intensity level of the horn’s sound is observed by someone waiting at an intersection 50 m from the train? Treat the horn as a point source and neglect any absorption of sound by the air. 28. As the people sing in church, the sound level every-where inside is 101 dB. No sound is transmitted through the massive walls, but all the windows and doors are open on a summer morning. Their total area is 22.0 m2. (a) How much sound energy is radiated through the windows and doors in 20.0 min? (b) Sup-pose the ground is a good reflector and sound radi-ates from the church uniformly in all horizontal and upward directions. Find the sound level 1.00 km away. 29. The most soaring vocal melody is in Johann Sebastian Bach’s Mass in B Minor. In one section, the basses, ten-ors, altos, and sopranos carry the melody from a low D to a high A. In concert pitch, these notes are now assigned frequencies of 146.8 Hz and 880.0 Hz. Find the wavelengths of (a) the initial note and (b) the final note. Assume the chorus sings the melody with a uni-form sound level of 75.0 dB. Find the pressure ampli-tudes of (c) the initial note and (d) the final note. Find the displacement amplitudes of (e) the initial note and (f) the final note. 30. Show that the difference between decibel levels b1 and b2 of a sound is related to the ratio of the distances r1 and r2 from the sound source by b2 2 b1 5 20 log a r1 r2 b S BIO W M S www.aswarphysics.weebly.com 528 Chapter 17 Sound Waves amplitude of this unit’s motion is 0.500 m. The speaker emits sound waves of frequency 440 Hz. Deter-mine (a) the highest and (b) the lowest frequencies heard by the person to the right of the speaker. (c)If the maximum sound level heard by the person is 60.0 dB when the speaker is at its closest distance d 5 1.00 m from him, what is the minimum sound level heard by the observer?
m d k Figure P17.41 Problems 41 and 42. 42. Review. A block with a speaker bolted to it is connected to a spring having spring constant k and oscillates as shown in Figure P17.41. The total mass of the block and speaker is m, and the amplitude of this unit’s motion is A. The speaker emits sound waves of frequency f. Determine (a) the highest and (b) the lowest frequen-cies heard by the person to the right of the speaker. (c) If the maximum sound level heard by the person is b when the speaker is at its closest distance d from him, what is the minimum sound level heard by the observer? 43. Expectant parents are thrilled to hear their unborn baby’s heartbeat, revealed by an ultrasonic detector that produces beeps of audible sound in synchroniza-tion with the fetal heartbeat. Suppose the fetus’s ven-tricular wall moves in simple harmonic motion with an amplitude of 1.80 mm and a frequency of 115 beats per minute. (a) Find the maximum linear speed of the heart wall. Suppose a source mounted on the detector in contact with the mother’s abdomen produces sound at 2 000 000.0 Hz, which travels through tissue at 1.50 km/s. (b) Find the maximum change in frequency between the sound that arrives at the wall of the baby’s heart and the sound emitted by the source. (c) Find the maximum change in frequency between the reflected sound received by the detector and that emitted by the source. 44. Why is the following situation impossible? At the Summer Olympics, an athlete runs at a constant speed down a straight track while a spectator near the edge of the track blows a note on a horn with a fixed frequency. When the athlete passes the horn, she hears the fre-quency of the horn fall by the musical interval called a minor third. That is, the frequency she hears drops to five-sixths its original value. 45. Standing at a crosswalk, you hear a frequency of 560 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of S BIO M with an identical mower the same distance away. This situation annoys you greatly because the total sound now has twice the loudness it had when only one neigh-bor was mowing.
Section 17.4 The Doppler Effect 37. An ambulance moving at 42 m/s sounds its siren whose frequency is 450 Hz. A car is moving in the same direc-tion as the ambulance at 25 m/s. What frequency does a person in the car hear (a) as the ambulance approaches the car? (b) After the ambulance passes the car? 38. When high-energy charged particles move through a transparent medium with a speed greater than the speed of light in that medium, a shock wave, or bow wave, of light is produced. This phenomenon is called the Cerenkov effect. When a nuclear reactor is shielded by a large pool of water, Cerenkov radiation can be seen as a blue glow in the vicinity of the reactor core due to high-speed electrons moving through the water (Fig. 17.38). In a particular case, the Cerenkov radiation pro-duces a wave front with an apex half-angle of 53.08. Calculate the speed of the electrons in the water. The speed of light in water is 2.25 3 108 m/s. 39. A driver travels northbound on a highway at a speed of 25.0 m/s. A police car, traveling southbound at a speed of 40.0 m/s, approaches with its siren producing sound at a frequency of 2 500 Hz. (a) What frequency does the driver observe as the police car approaches? (b) What frequency does the driver detect after the police car passes him? (c) Repeat parts (a) and (b) for the case when the police car is behind the driver and travels northbound. 40. Submarine A travels horizontally at 11.0 m/s through ocean water. It emits a sonar signal of frequency f 5 5.27 3 103 Hz in the forward direction. Submarine B is in front of submarine A and traveling at 3.00 m/s rela-tive to the water in the same direction as submarine A. A crewman in submarine B uses his equipment to detect the sound waves (“pings”) from submarine A. We wish to determine what is heard by the crewman in submarine B. (a) An observer on which submarine detects a frequency f 9 as described by Equation 17.19? (b) In Equation 17.19, should the sign of vS be positive or negative? (c) In Equation 17.19, should the sign of vO be positive or negative? (d) In Equation 17.19, what speed of sound should be used? (e) Find the frequency of the sound detected by the crewman on submarine B. 41. Review. A block with a speaker bolted to it is con-nected to a spring having spring constant k 5 20.0 N/m and oscillates as shown in Figure P17.41. The total mass of the block and speaker is 5.00 kg, and the Figure P17.38 U.S. Department of Energy/Photo Researchers, Inc.
GP AMT www.aswarphysics.weebly.com Problems 529 to complaints, Strauss later transposed the note down to F above high C, 1.397 kHz. By what increment did the wavelength change? 51. Trucks carrying garbage to the town dump form a nearly steady procession on a country road, all travel-ing at 19.7 m/s in the same direction. Two trucks arrive at the dump every 3 min. A bicyclist is also traveling toward the dump, at 4.47 m/s. (a) With what frequency do the trucks pass the cyclist? (b) What If? A hill does not slow down the trucks, but makes the out-of-shape cyclist’s speed drop to 1.56 m/s. How often do the trucks whiz past the cyclist now? 52. If a salesman claims a loudspeaker is rated at 150 W, he is referring to the maximum electrical power input to the speaker. Assume a loudspeaker with an input power of 150 W broadcasts sound equally in all direc-tions and produces sound with a level of 103 dB at a distance of 1.60 m from its center. (a) Find its sound power output. (b) Find the efficiency of the speaker, that is, the fraction of input power that is converted into useful output power. 53. An interstate highway has been built through a neigh-borhood in a city. In the afternoon, the sound level in an apartment in the neighborhood is 80.0 dB as 100 cars pass outside the window every minute. Late at night, the traffic flow is only five cars per minute. What is the average late-night sound level? 54. A train whistle ( f 5 400 Hz) sounds higher or lower in frequency depending on whether it approaches or recedes. (a) Prove that the difference in frequency between the approaching and receding train whistle is Df 5 2u/v 1 2 u2/v 2 f where u is the speed of the train and v is the speed of sound. (b) Calculate this difference for a train moving at a speed of 130 km/h. Take the speed of sound in air to be 340 m/s. 55. An ultrasonic tape measure uses frequencies above 20 MHz to determine dimensions of structures such as buildings. It does so by emitting a pulse of ultrasound into air and then measuring the time interval for an echo to return from a reflecting surface whose dis-tance away is to be measured. The distance is displayed as a digital readout. For a tape measure that emits a pulse of ultrasound with a frequency of 22.0 MHz, (a) what is the distance to an object from which the echo pulse returns after 24.0 ms when the air temperature is 26°C? (b) What should be the duration of the emitted pulse if it is to include ten cycles of the ultrasonic wave? (c) What is the spatial length of such a pulse? 56. The tensile stress in a thick copper bar is 99.5% of its elastic breaking point of 13.0 3 1010 N/m2. If a 500-Hz sound wave is transmitted through the material, (a) what displacement amplitude will cause the bar to break? (b) What is the maximum speed of the elements of copper at this moment? (c) What is the sound intensity in the bar?
the siren is 480 Hz. Determine the ambulance’s speed from these observations. 46. Review. A tuning fork vibrating at 512 Hz falls from rest and accelerates at 9.80 m/s2. How far below the point of release is the tuning fork when waves of fre-quency 485 Hz reach the release point? 47. A supersonic jet traveling at Mach 3.00 at an altitude of h 5 20 000 m is directly over a person at time t 5 0 as shown in Figure P17.47. Assume the average speed of sound in air is 335 m/s over the path of the sound. (a) At what time will the person encounter the shock wave due to the sound emitted at t 5 0? (b) Where will the plane be when this shock wave is heard?
h t 0 t ?
Observer Observer hears the “boom” h x u u a b Figure P17.47 Additional Problems 48. A bat (Fig. P17.48) can detect very small objects, such as an insect whose length is approximately equal to one wavelength of the sound the bat makes. If a bat emits chirps at a frequency of 60.0 kHz and the speed of sound in air is 340 m/s, what is the smallest insect the bat can detect? 49. Some studies suggest that the upper frequency limit of hearing is deter-mined by the diameter of the eardrum. The diam-eter of the eardrum is approximately equal to half the wavelength of the sound wave at this upper limit. If the relationship holds exactly, what is the diameter of the eardrum of a person capable of hearing 20 000 Hz? (Assume a body temperature of 37.0°C.) 50. The highest note written for a singer in a published score was F-sharp above high C, 1.480 kHz, for Zerbi-netta in the original version of Richard Strauss’s opera Ariadne auf Naxos. (a) Find the wavelength of this sound in air. (b) Suppose people in the fourth row of seats hear this note with level 81.0 dB. Find the displace-ment amplitude of the sound. (c) What If? In response M AMT Figure P17.48 Problems 48 and 63.
Hugh Lansdown/Shutterstock.com BIO BIO www.aswarphysics.weebly.com 530 Chapter 17 Sound Waves together once. The sound pulse you produce has no definite frequency and no wavelength. The sound you hear reflected from the bleachers has an identifiable frequency and may remind you of a short toot on a trumpet, buzzer, or kazoo. (a) Explain what accounts for this sound. Compute order-of-magnitude esti-mates for (b) the frequency, (c) the wavelength, and (d) the duration of the sound on the basis of data you specify. 61. To measure her speed, a skydiver carries a buzzer emit-ting a steady tone at 1 800 Hz. A friend on the ground at the landing site directly below listens to the ampli-fied sound he receives. Assume the air is calm and the speed of sound is independent of altitude. While the skydiver is falling at terminal speed, her friend on the ground receives waves of frequency 2 150 Hz. (a) What is the skydiver’s speed of descent? (b) What If? Suppose the skydiver can hear the sound of the buzzer reflected from the ground. What frequency does she receive? 62. Spherical waves of wavelength 45.0 cm propagate out-ward from a point source. (a) Explain how the intensity at a distance of 240 cm compares with the intensity at a distance of 60.0 cm. (b) Explain how the amplitude at a distance of 240 cm compares with the amplitude at a distance of 60.0 cm. (c) Explain how the phase of the wave at a distance of 240 cm compares with the phase at 60.0 cm at the same moment. 63. A bat (Fig. P17.48), moving at 5.00 m/s, is chasing a flying insect. If the bat emits a 40.0-kHz chirp and receives back an echo at 40.4 kHz, (a) what is the speed of the insect? (b) Will the bat be able to catch the insect? Explain. 64. Two ships are moving along a line due east (Fig. P17.64). The trailing vessel has a speed relative to a land-based observation point of v1 5 64.0 km/h, and the lead-ing ship has a speed of v2 5 45.0 km/h relative to that point. The two ships are in a region of the ocean where the current is moving uniformly due west at vcurrent 5 10.0 km/h. The trailing ship transmits a sonar signal at a frequency of 1 200.0 Hz through the water. What frequency is monitored by the leading ship?
v1 vcurrent v2 Figure P17.64 65. A police car is traveling east at 40.0 m/s along a straight road, overtaking a car ahead of it moving east at 30.0 m/s. The police car has a malfunctioning siren that is stuck at 1 000 Hz. (a) What would be the wave-length in air of the siren sound if the police car were at rest? (b) What is the wavelength in front of the police car? (c) What is it behind the police car? (d) What is the frequency heard by the driver being chased?
M Q/C Q/C 57. Review. A 150-g glider moves at v1 5 2.30 m/s on an air track toward an originally stationary 200-g glider as shown in Figure P17.57. The gliders undergo a com-pletely inelastic collision and latch together over a time interval of 7.00 ms. A student suggests roughly half the decrease in mechanical energy of the two-glider system is transferred to the environment by sound. Is this suggestion reasonable? To evaluate the idea, find the implied sound level at a position 0.800 m from the gliders. If the student’s idea is unreasonable, suggest a better idea.
v0 200 g 150 g 1 v Before the collision Latches Figure P17.57 58. Consider the following wave function in SI units: DP 1r, t2 5 a25.0 r b sin 11.36r 2 2 030t2 Explain how this wave function can apply to a wave radiating from a small source, with r being the radial distance from the center of the source to any point out-side the source. Give the most detailed description of the wave that you can. Include answers to such ques-tions as the following and give representative values for any quantities that can be evaluated. (a) Does the wave move more toward the right or the left? (b) As it moves away from the source, what happens to its amplitude? (c) Its speed? (d) Its frequency? (e) Its wavelength? (f) Its power? (g) Its intensity? 59. Review. For a certain type of steel, stress is always proportional to strain with Young’s modulus 20 3 1010 N/m2. The steel has density 7.86 3 103 kg/m3. It will fail by bending permanently if subjected to com-pressive stress greater than its yield strength sy 5 400 MPa. A rod 80.0 cm long, made of this steel, is fired at 12.0 m/s straight at a very hard wall. (a) The speed of a one-dimensional compressional wave mov-ing along the rod is given by v 5 !Y/r, where Y is Young’s modulus for the rod and r is the density. Calculate this speed. (b) After the front end of the rod hits the wall and stops, the back end of the rod keeps moving as described by Newton’s first law until it is stopped by excess pressure in a sound wave mov-ing back through the rod. What time interval elapses before the back end of the rod receives the message that it should stop? (c) How far has the back end of the rod moved in this time interval? Find (d) the strain and (e) the stress in the rod. (f) If it is not to fail, what is the maximum impact speed a rod can have in terms of sy, Y, and r? 60. A large set of unoccupied football bleachers has solid seats and risers. You stand on the field in front of the bleachers and sharply clap two wooden boards AMT Q/C Q/C Q/C www.aswarphysics.weebly.com Problems 531 from an upwind position so that she is moving in the direction in which the wind is blowing and (d) if she is approaching from a downwind position and moving against the wind?
Challenge Problems 71. The Doppler equation presented in the text is valid when the motion between the observer and the source occurs on a straight line so that the source and observer are moving either directly toward or directly away from each other. If this restriction is relaxed, one must use the more general Doppler equation f r 5 a v 1 vO cos uO v 2 vS cos uS b f where uO and uS are defined in Figure P17.71a. Use the preceding equation to solve the following prob-lem. A train moves at a constant speed of v 5 25.0 m/s toward the intersection shown in Figure P17.71b. A car is stopped near the crossing, 30.0 m from the tracks. The train’s horn emits a frequency of 500 Hz when the train is 40.0 m from the intersection. (a) What is the frequency heard by the passengers in the car? (b) If the train emits this sound continuously and the car is stationary at this position long before the train arrives until long after it leaves, what range of frequencies do passengers in the car hear? (c) Suppose the car is fool-ishly trying to beat the train to the intersection and is traveling at 40.0 m/s toward the tracks. When the car is 30.0 m from the tracks and the train is 40.0 m from the intersection, what is the frequency heard by the pas-sengers in the car now?
vS S vO S uS uO a b O S v S Figure P17.71 72. In Section 17.2, we derived the speed of sound in a gas using the impulse–momentum theorem applied to the cylinder of gas in Figure 17.5. Let us find the speed of sound in a gas using a different approach based on the element of gas in Figure 17.3. Proceed as follows. (a) Draw a force diagram for this element showing the forces exerted on the left and right surfaces due to the pressure of the gas on either side of the element. (b) By applying Newton’s second law to the element, show that 2 '1DP 2 'x A Dx 5 rA Dx '2s 't 2 S 66. The speed of a one-dimensional compressional wave traveling along a thin copper rod is 3.56 km/s. The rod is given a sharp hammer blow at one end. A listener at the far end of the rod hears the sound twice, trans-mitted through the metal and through air, with a time interval Dt between the two pulses. (a) Which sound arrives first? (b) Find the length of the rod as a func-tion of Dt. (c) Find the length of the rod if Dt 5 127 ms. (d) Imagine that the copper rod is replaced by another material through which the speed of sound is vr. What is the length of the rod in terms of t and vr? (e) Would the answer to part (d) go to a well-defined limit as the speed of sound in the rod goes to infinity? Explain your answer. 67. A large meteoroid enters the Earth’s atmosphere at a speed of 20.0 km/s and is not significantly slowed before entering the ocean. (a) What is the Mach angle of the shock wave from the meteoroid in the lower atmosphere? (b) If we assume the meteoroid survives the impact with the ocean surface, what is the (initial) Mach angle of the shock wave the meteoroid produces in the water? 68. Three metal rods are located relative to each other as shown in Fig-ure P17.68, where L3 5 L1 1 L2. The speed of sound in a rod is given by v 5 !Y/r, where Y is Young’s modulus for the rod and r is the density. Val-ues of density and Young’s modulus for the three mate-rials are r1 5 2.70 3 103 kg/m3, Y1 5 7.00 3 1010 N/m2, r2 5 11.3 3 103 kg/m3, Y2 5 1.60 3 1010 N/m2, r3 5 8.80 3 103 kg/m3, Y3 5 11.0 3 1010 N/m2. If L3 5 1.50 m, what must the ratio L1/L2 be if a sound wave is to travel the length of rods 1 and 2 in the same time interval required for the wave to travel the length of rod 3? 69. With particular experimental methods, it is possible to produce and observe in a long, thin rod both a trans-verse wave whose speed depends primarily on ten-sion in the rod and a longitudinal wave whose speed is determined by Young’s modulus and the density of the material according to the expression v 5 !Y/r. The transverse wave can be modeled as a wave in a stretched string. A particular metal rod is 150 cm long and has a radius of 0.200 cm and a mass of 50.9 g. Young’s modulus for the material is 6.80 3 1010 N/m2. What must the tension in the rod be if the ratio of the speed of longitudinal waves to the speed of transverse waves is 8.00? 70. A siren mounted on the roof of a firehouse emits sound at a frequency of 900 Hz. A steady wind is blow-ing with a speed of 15.0 m/s. Taking the speed of sound in calm air to be 343 m/s, find the wavelength of the sound (a) upwind of the siren and (b) down-wind of the siren. Firefighters are approaching the siren from various directions at 15.0 m/s. What fre-quency does a firefighter hear (c) if she is approaching Q/C 3 1 2 L3 L2 L1 Figure P17.68 M www.aswarphysics.weebly.com 532 Chapter 17 Sound Waves 73. Equation 17.13 states that at distance r away from a point source with power (Power)avg, the wave intensity is I 5 1Power2 avg 4pr 2 Study Figure 17.10 and prove that at distance r straight in front of a point source with power (Power)avg moving with constant speed vS the wave intensity is I 5 1Power2 avg 4pr 2 a v 2 vS v b S (c) By substituting DP 5 2(B 's/'x) (Eq. 17.3), derive the following wave equation for sound: B r '2s 'x 2 5 '2s 't 2 (d) To a mathematical physicist, this equation demon-strates the existence of sound waves and determines their speed. As a physics student, you must take another step or two. Substitute into the wave equation the trial solu-tion s(x, t) 5 smax cos (kx 2 vt). Show that this function satisfies the wave equation, provided v/k 5 v 5 !B/r.
www.aswarphysics.weebly.com Blues master B. B. King takes advantage of standing waves on strings. He changes to higher notes on the guitar by pushing the strings against the frets on the fingerboard, shortening the lengths of the portions of the strings that vibrate. (AP Photo/Danny Moloshok) 18.1 Analysis Model: Waves in Interference 18.2 Standing Waves 18.3 Analysis Model: Waves Under Boundary Conditions 18.4 Resonance 18.5 Standing Waves in Air Columns 18.6 Standing Waves in Rods and Membranes 18.7 Beats: Interference in Time 18.8 Nonsinusoidal Wave Patterns c h a p t e r 18 Superposition and Standing Waves 533 The wave model was introduced in the previous two chapters. We have seen that waves are very different from particles. A particle is of zero size, whereas a wave has a characteristic size, its wavelength. Another important difference between waves and par-ticles is that we can explore the possibility of two or more waves combining at one point in the same medium. Particles can be combined to form extended objects, but the particles must be at different locations. In contrast, two waves can both be present at the same loca-tion. The ramifications of this possibility are explored in this chapter. When waves are combined in systems with boundary conditions, only certain allowed frequencies can exist and we say the frequencies are quantized. Quantization is a notion that is at the heart of quantum mechanics, a subject introduced formally in Chapter 40. There we show that analysis of waves under boundary conditions explains many of the quantum phenomena. In this chapter, we use quantization to understand the behavior of the wide array of musical instruments that are based on strings and air columns. www.aswarphysics.weebly.com 534 Chapter 18 Superposition and Standing Waves We also consider the combination of waves having different frequencies. When two sound waves having nearly the same frequency interfere, we hear variations in the loudness called beats. Finally, we discuss how any nonsinusoidal periodic wave can be described as a sum of sine and cosine functions.
18.1 Analysis Model: Waves in Interference Many interesting wave phenomena in nature cannot be described by a single travel-ing wave. Instead, one must analyze these phenomena in terms of a combination of traveling waves. As noted in the introduction, waves have a remarkable difference from particles in that waves can be combined at the same location in space. To ana-lyze such wave combinations, we make use of the superposition principle: If two or more traveling waves are moving through a medium, the resultant value of the wave function at any point is the algebraic sum of the values of the wave functions of the individual waves.
Waves that obey this principle are called linear waves. (See Section 16.6.) In the case of mechanical waves, linear waves are generally characterized by having amplitudes much smaller than their wavelengths. Waves that violate the superposition prin-ciple are called nonlinear waves and are often characterized by large amplitudes. In this book, we deal only with linear waves. One consequence of the superposition principle is that two traveling waves can pass through each other without being destroyed or even altered. For instance, when two pebbles are thrown into a pond and hit the surface at different locations, the expanding circular surface waves from the two locations simply pass through each other with no permanent effect. The resulting complex pattern can be viewed as two independent sets of expanding circles. Figure 18.1 is a pictorial representation of the superposition of two pulses. The wave function for the pulse moving to the right is y1, and the wave function for the pulse moving to the left is y2. The pulses have the same speed but different shapes, and the displacement of the elements of the medium is in the positive y direction for both pulses. When the waves overlap (Fig. 18.1b), the wave function for the resulting complex wave is given by y1 1 y2. When the crests of the pulses coincide (Fig. 18.1c), the resulting wave given by y1 1 y2 has a larger amplitude than that of the individual pulses. The two pulses finally separate and continue moving in their original directions (Fig. 18.1d). Notice that the pulse shapes remain unchanged after the interaction, as if the two pulses had never met! The combination of separate waves in the same region of space to produce a resultant wave is called interference. For the two pulses shown in Figure 18.1, the displacement of the elements of the medium is in the positive y direction for both pulses, and the resultant pulse (created when the individual pulses overlap) exhib-its an amplitude greater than that of either individual pulse. Because the displace-ments caused by the two pulses are in the same direction, we refer to their superpo-sition as constructive interference. Now consider two pulses traveling in opposite directions on a taut string where one pulse is inverted relative to the other as illustrated in Figure 18.2. When these pulses begin to overlap, the resultant pulse is given by y1 1 y2, but the values of the function y2 are negative. Again, the two pulses pass through each other; because the displacements caused by the two pulses are in opposite directions, however, we refer to their superposition as destructive interference. The superposition principle is the centerpiece of the analysis model called waves in interference. In many situations, both in acoustics and optics, waves com-bine according to this principle and exhibit interesting phenomena with practical applications.
Superposition principle Constructive interference Destructive interference Pitfall Prevention 18.1 Do Waves Actually Interfere? In popular usage, the term interfere implies that an agent affects a situation in some way so as to pre-clude something from happening. For example, in American foot-ball, pass interference means that a defending player has affected the receiver so that the receiver is unable to catch the ball. This usage is very different from its use in physics, where waves pass through each other and interfere, but do not affect each other in any way. In physics, interference is similar to the notion of combina-tion as described in this chapter.
www.aswarphysics.weebly.com 18.1 Analysis Model: Waves in Interference 535 Q uick Quiz 18.1 Two pulses move in opposite directions on a string and are iden-tical in shape except that one has positive displacements of the elements of the string and the other has negative displacements. At the moment the two pulses completely overlap on the string, what happens? (a) The energy associated with the pulses has disappeared. (b) The string is not moving. (c) The string forms a straight line. (d) The pulses have vanished and will not reappear.
Superposition of Sinusoidal Waves Let us now apply the principle of superposition to two sinusoidal waves traveling in the same direction in a linear medium. If the two waves are traveling to the right and have the same frequency, wavelength, and amplitude but differ in phase, we can express their individual wave functions as y1 5 A sin (kx 2 vt) y2 5 A sin (kx 2 vt 1 f) where, as usual, k 5 2p/l, v 5 2pf, and f is the phase constant as discussed in Sec-tion 16.2. Hence, the resultant wave function y is y 5 y1 1 y2 5 A [sin (kx 2 vt) 1 sin (kx 2 vt 1 f)] To simplify this expression, we use the trigonometric identity sin a 1 sin b 5 2 cos aa 2 b 2 b sin aa 1 b 2 b b c d a y2 y 1 y 1 y2 y 1 y2 y2 y 1 When the pulses overlap, the wave function is the sum of the individual wave functions.
When the crests of the two pulses align, the amplitude is the sum of the individual amplitudes.
When the pulses no longer overlap, they have not been permanently affected by the interference.
Figure 18.1 Constructive interfer-ence. Two positive pulses travel on a stretched string in opposite direc-tions and overlap.
y 1 y 2 y 2 y 1 y 1 y 2 y 1 y 2 When the pulses overlap, the wave function is the sum of the individual wave functions.
When the crests of the two pulses align, the amplitude is the difference between the individual amplitudes.
When the pulses no longer overlap, they have not been permanently affected by the interference.
b c d a Figure 18.2 Destructive interfer-ence. Two pulses, one positive and one negative, travel on a stretched string in opposite directions and overlap.
www.aswarphysics.weebly.com 536 Chapter 18 Superposition and Standing Waves Letting a 5 kx 2 vt and b 5 kx 2 vt 1 f, we find that the resultant wave function y reduces to y 5 2A cos af 2 b sin akx 2 vt 1 f 2 b This result has several important features. The resultant wave function y also is sinusoidal and has the same frequency and wavelength as the individual waves because the sine function incorporates the same values of k and v that appear in the original wave functions. The amplitude of the resultant wave is 2A cos (f/2), and its phase constant is f/2. If the phase constant f of the original wave equals 0, then cos (f/2) 5 cos 0 5 1 and the amplitude of the resultant wave is 2A, twice the amplitude of either individual wave. In this case, the crests of the two waves are at the same locations in space and the waves are said to be everywhere in phase and therefore interfere constructively. The individual waves y1 and y2 combine to form the red-brown curve y of amplitude 2A shown in Figure 18.3a. Because the indi-vidual waves are in phase, they are indistinguishable in Figure 18.3a, where they appear as a single blue curve. In general, constructive interference occurs when cos (f/2) 5 61. That is true, for example, when f 5 0, 2p, 4p, . . . rad, that is, when f is an even multiple of p. When f is equal to p rad or to any odd multiple of p, then cos (f/2) 5 cos (p/2) 5 0 and the crests of one wave occur at the same positions as the troughs of the sec-ond wave (Fig. 18.3b). Therefore, as a consequence of destructive interference, the resultant wave has zero amplitude everywhere as shown by the straight red-brown line in Figure 18.3b. Finally, when the phase constant has an arbitrary value other than 0 or an integer multiple of p rad (Fig. 18.3c), the resultant wave has an ampli-tude whose value is somewhere between 0 and 2A. In the more general case in which the waves have the same wavelength but dif-ferent amplitudes, the results are similar with the following exceptions. In the in-phase case, the amplitude of the resultant wave is not twice that of a single wave, but rather is the sum of the amplitudes of the two waves. When the waves are p rad out of phase, they do not completely cancel as in Figure 18.3b. The result is a wave whose amplitude is the difference in the amplitudes of the individual waves.
Interference of Sound Waves One simple device for demonstrating interference of sound waves is illustrated in Figure 18.4. Sound from a loudspeaker S is sent into a tube at point P, where there is Resultant of two traveling sinusoidal waves y x x x y y1 y2 y y y y1 y2 5 60° y f 5 180° f 5 0° f The individual waves are in phase and therefore indistinguishable.
Constructive interference: the amplitudes add.
The individual waves are 180° out of phase.
Destructive interference: the waves cancel.
This intermediate result is neither constructive nor destructive.
b c a Figure 18.3 The superposition of two identical waves y1 and y2 (blue and green, respectively) to yield a resultant wave (red-brown).
A sound wave from the speaker (S) propagates into the tube and splits into two parts at point P.
Path length r1 Path length r2 R S P The two waves, which combine at the opposite side, are detected at the receiver (R).
Figure 18.4 An acoustical system for demonstrating interfer-ence of sound waves. The upper path length r2 can be varied by sliding the upper section.
www.aswarphysics.weebly.com 18.1 Analysis Model: Waves in Interference 537 a T-shaped junction. Half the sound energy travels in one direction, and half travels in the opposite direction. Therefore, the sound waves that reach the receiver R can travel along either of the two paths. The distance along any path from speaker to receiver is called the path length r. The lower path length r1 is fixed, but the upper path length r2 can be varied by sliding the U-shaped tube, which is similar to that on a slide trombone. When the difference in the path lengths Dr 5 |r2 2 r1| is either zero or some integer multiple of the wavelength l (that is, Dr 5 nl, where n 5 0, 1, 2, 3, . . .), the two waves reaching the receiver at any instant are in phase and interfere constructively as shown in Figure 18.3a. For this case, a maximum in the sound intensity is detected at the receiver. If the path length r2 is adjusted such that the path difference Dr 5 l/2, 3l/2, . . . , nl/2 (for n odd), the two waves are exactly p rad, or 180°, out of phase at the receiver and hence cancel each other. In this case of destructive interference, no sound is detected at the receiver. This simple experi-ment demonstrates that a phase difference may arise between two waves generated by the same source when they travel along paths of unequal lengths. This impor-tant phenomenon will be indispensable in our investigation of the interference of light waves in Chapter 37.
Example 18.1 Two Speakers Driven by the Same Source Two identical loudspeakers placed 3.00 m apart are driven by the same oscillator (Fig. 18.5). A listener is originally at point O, located 8.00 m from the center of the line connecting the two speakers. The listener then moves to point P, which is a perpendicular distance 0.350 m from O, and she experiences the first minimum in sound intensity. What is the frequency of the oscillator?
Conceptualize In Figure 18.4, a sound wave enters a tube and is then acoustically split into two different paths before recombining at the other end. In this example, a signal representing the sound is electrically split and sent to two different loudspeakers. After leaving the speakers, the sound waves recombine at the position of the listener. Despite the difference in how the splitting occurs, the path difference discussion related to Figure 18.4 can be applied here.
Categorize Because the sound waves from two separate sources combine, we apply the waves in interference analysis model.
AM S o l u t i o n 3.00 m 8.00 m r2 8.00 m r1 0.350 m 1.85 m P 1.15 m O Figure 18.5 (Example 18.1) Two identical loudspeakers emit sound waves to a listener at P.
continued Imagine two waves traveling in the same location through a medium. The displacement of elements of the medium is affected by both waves. Accord-ing to the principle of superpo-sition, the displacement is the sum of the individual displace-ments that would be caused by each wave. When the waves are in phase, constructive interference occurs and the resultant displacement is larger than the individual displacements. Destructive interference occurs when the waves are out of phase. Analysis Model Waves in Interference Examples: • a piano tuner listens to a piano string and a tuning fork vibrating together and notices beats (Section 18.7) • light waves from two coherent sources combine to form an interference pat-tern on a screen (Chapter 37) • a thin film of oil on top of water shows swirls of color (Chapter 37) • x-rays passing through a crystalline solid combine to form a Laue pattern (Chapter 38) y1 y2 y1 y2 Destructive interference Constructive interference y1 y2 y2 y1 www.aswarphysics.weebly.com 538 Chapter 18 Superposition and Standing Waves What if the speakers were connected out of phase? What happens at point P in Figure 18.5?
Answer In this situation, the path difference of l/2 combines with a phase difference of l/2 due to the incorrect wir-ing to give a full phase difference of l. As a result, the waves are in phase and there is a maximum intensity at point P.
What If?
To obtain the oscillator frequency, use Equation 16.12, v 5 lf, where v is the speed of sound in air, 343 m/s: f 5 v l 5 343 m/s 0.26 m 5 1.3 kHz Finalize This example enables us to understand why the speaker wires in a stereo system should be connected properly. When connected the wrong way—that is, when the positive (or red) wire is connected to the negative (or black) terminal on one of the speakers and the other is correctly wired—the speakers are said to be “out of phase,” with one speaker moving outward while the other moves inward. As a consequence, the sound wave com-ing from one speaker destructively interferes with the wave coming from the other at point O in Figure 18.5. A rarefaction region due to one speaker is superposed on a compression region from the other speaker. Although the two sounds probably do not completely cancel each other (because the left and right stereo signals are usu-ally not identical), a substantial loss of sound quality occurs at point O.
18.2 Standing Waves The sound waves from the pair of loudspeakers in Example 18.1 leave the speakers in the forward direction, and we considered interference at a point in front of the speakers. Suppose we turn the speakers so that they face each other and then have them emit sound of the same frequency and amplitude. In this situation, two identi-cal waves travel in opposite directions in the same medium as in Figure 18.6. These waves combine in accordance with the waves in interference model. We can analyze such a situation by considering wave functions for two transverse sinusoidal waves having the same amplitude, frequency, and wavelength but travel-ing in opposite directions in the same medium: y1 5 A sin (kx 2 vt) y2 5 A sin (kx 1 vt) where y1 represents a wave traveling in the positive x direction and y2 represents one traveling in the negative x direction. Adding these two functions gives the resultant wave function y: y 5 y1 1 y2 5 A sin (kx 2 vt) 1 A sin (kx 1 vt) When we use the trigonometric identity sin (a 6 b) 5 sin a cos b 6 cos a sin b, this expression reduces to y 5 (2A sin kx) cos vt (18.1) Equation 18.1 represents the wave function of a standing wave. A standing wave such as the one on a string shown in Figure 18.7 is an oscillation pattern with a sta-tionary outline that results from the superposition of two identical waves traveling in opposite directions.
v S v S Figure 18.6 Two identical loud-speakers emit sound waves toward each other. When they overlap, identical waves traveling in opposite directions will combine to form standing waves.
From the shaded triangles, find the path lengths from the speakers to the listener: r 1 5 "18.00 m2 2 1 11.15 m2 2 5 8.08 m r 2 5 "18.00 m2 2 1 11.85 m2 2 5 8.21 m Hence, the path difference is r2 2 r1 5 0.13 m. Because this path difference must equal l/2 for the first minimum, l 5 0.26 m.
Analyze Figure 18.5 shows the physical arrangement of the speakers, along with two shaded right triangles that can be drawn on the basis of the lengths described in the problem. The first minimum occurs when the two waves reaching the listener at point P are 180° out of phase, in other words, when their path difference Dr equals l/2. ▸ 18.1 continu ed www.aswarphysics.weebly.com 18.2 Standing Waves 539 Notice that Equation 18.1 does not contain a function of kx 2 vt. Therefore, it is not an expression for a single traveling wave. When you observe a standing wave, there is no sense of motion in the direction of propagation of either original wave. Comparing Equation 18.1 with Equation 15.6, we see that it describes a special kind of simple harmonic motion. Every element of the medium oscillates in simple har-monic motion with the same angular frequency v (according to the cos vt factor in the equation). The amplitude of the simple harmonic motion of a given element (given by the factor 2A sin kx, the coefficient of the cosine function) depends on the location x of the element in the medium, however. If you can find a noncordless telephone with a coiled cord connecting the hand-set to the base unit, you can see the difference between a standing wave and a trav-eling wave. Stretch the coiled cord out and flick it with a finger. You will see a pulse traveling along the cord. Now shake the handset up and down and adjust your shak-ing frequency until every coil on the cord is moving up at the same time and then down. That is a standing wave, formed from the combination of waves moving away from your hand and reflected from the base unit toward your hand. Notice that there is no sense of traveling along the cord like there was for the pulse. You only see up-and-down motion of the elements of the cord. Equation 18.1 shows that the amplitude of the simple harmonic motion of an element of the medium has a minimum value of zero when x satisfies the condition sin kx 5 0, that is, when kx 5 0, p, 2p, 3p, . . .
Because k 5 2p/l, these values for kx give x 5 0, l 2, l, 3l 2 , c 5 nl 2 n 5 0, 1, 2, 3, c (18.2) These points of zero amplitude are called nodes. The element of the medium with the greatest possible displacement from equi-librium has an amplitude of 2A, which we define as the amplitude of the standing wave. The positions in the medium at which this maximum displacement occurs are called antinodes. The antinodes are located at positions for which the coordi-nate x satisfies the condition sin kx 5 61, that is, when kx 5 p 2, 3p 2 , 5p 2 , c Therefore, the positions of the antinodes are given by x 5 l 4, 3l 4 , 5l 4 , c 5 nl 4 n 5 1, 3, 5, c (18.3) W W Positions of nodes W W Positions of antinodes Figure 18.7 Multiflash pho-tograph of a standing wave on a string. The time behavior of the vertical displacement from equi-librium of an individual element of the string is given by cos vt. That is, each element vibrates at an angular frequency v.
Antinode Antinode Node 2A sin kx Node The amplitude of the vertical oscillation of any element of the string depends on the horizontal position of the element. Each element vibrates within the confines of the envelope function 2A sin kx.
. 1991 Richard Megna/Fundamental Photographs Pitfall Prevention 18.2 Three Types of Amplitude We need to distinguish carefully here between the amplitude of the individual waves, which is A, and the amplitude of the simple har-monic motion of the elements of the medium, which is 2A sin kx. A given element in a standing wave vibrates within the constraints of the envelope function 2A sin kx, where x is that element’s position in the medium. Such vibration is in contrast to traveling sinusoidal waves, in which all elements oscil-late with the same amplitude and the same frequency and the ampli-tude A of the wave is the same as the amplitude A of the simple harmonic motion of the elements. Furthermore, we can identify the amplitude of the standing wave as 2A.
www.aswarphysics.weebly.com 540 Chapter 18 Superposition and Standing Waves Two nodes and two antinodes are labeled in the standing wave in Figure 18.7. The light blue curve labeled 2A sin kx in Figure 18.7 represents one wavelength of the traveling waves that combine to form the standing wave. Figure 18.7 and Equa-tions 18.2 and 18.3 provide the following important features of the locations of nodes and antinodes: The distance between adjacent antinodes is equal to l/2.
The distance between adjacent nodes is equal to l/2.
The distance between a node and an adjacent antinode is l/4. Wave patterns of the elements of the medium produced at various times by two transverse traveling waves moving in opposite directions are shown in Figure 18.8. The blue and green curves are the wave patterns for the individual traveling waves, and the red-brown curves are the wave patterns for the resultant standing wave. At t 5 0 (Fig. 18.8a), the two traveling waves are in phase, giving a wave pattern in which each element of the medium is at rest and experiencing its maximum dis-placement from equilibrium. One-quarter of a period later, at t 5 T/4 (Fig. 18.8b), the traveling waves have moved one-fourth of a wavelength (one to the right and the other to the left). At this time, the traveling waves are out of phase, and each element of the medium is passing through the equilibrium position in its simple harmonic motion. The result is zero displacement for elements at all values of x; that is, the wave pattern is a straight line. At t 5 T/2 (Fig. 18.8c), the traveling waves are again in phase, producing a wave pattern that is inverted relative to the t 5 0 pattern. In the standing wave, the elements of the medium alternate in time between the extremes shown in Figures 18.8a and 18.8c.
Q uick Quiz 18.2 Consider the waves in Figure 18.8 to be waves on a stretched string. Define the velocity of elements of the string as positive if they are moving upward in the figure. (i) At the moment the string has the shape shown by the red-brown curve in Figure 18.8a, what is the instantaneous velocity of elements along the string? (a) zero for all elements (b) positive for all elements (c) nega-tive for all elements (d) varies with the position of the element (ii) From the same choices, at the moment the string has the shape shown by the red-brown curve in Figure 18.8b, what is the instantaneous velocity of elements along the string? t = 0 y1 y2 y N N N N N A A t = T/4 y2 y1 y t = T/2 y1 A A y2 y N N N N N A A A A a b c Figure 18.8 Standing-wave patterns produced at various times by two waves of equal amplitude traveling in opposite directions. For the resultant wave y, the nodes (N) are points of zero displace-ment and the antinodes (A) are points of maximum displacement.
Example 18.2 Formation of a Standing Wave Two waves traveling in opposite directions produce a standing wave. The individual wave functions are y1 5 4.0 sin (3.0x 2 2.0t) y2 5 4.0 sin (3.0x 1 2.0t) where x and y are measured in centimeters and t is in seconds.
(A) Find the amplitude of the simple harmonic motion of the element of the medium located at x 5 2.3 cm.
www.aswarphysics.weebly.com 18.3 Analysis Model: Waves Under Boundary Conditions 541 18.3 Analysis Model: Waves Under Boundary Conditions Consider a string of length L fixed at both ends as shown in Figure 18.9. We will use this system as a model for a guitar string or piano string. Waves can travel in both directions on the string. Therefore, standing waves can be set up in the string by a continuous superposition of waves incident on and reflected from the ends. Notice that there is a boundary condition for the waves on the string: because the ends of the string are fixed, they must necessarily have zero displacement and are there-fore nodes by definition. The condition that both ends of the string must be nodes fixes the wavelength of the standing wave on the string according to Equation 18.2, which, in turn, determines the frequency of the wave. The boundary condition results in the string having a number of discrete natural patterns of oscillation, called normal modes, each of which has a characteristic frequency that is easily cal-culated. This situation in which only certain frequencies of oscillation are allowed is called quantization. Quantization is a common occurrence when waves are sub-ject to boundary conditions and is a central feature in our discussions of quantum physics in the extended version of this text. Notice in Figure 18.8 that there are no boundary conditions, so standing waves of any frequency can be established; there is no quantization without boundary conditions. Because boundary condi-tions occur so often for waves, we identify an analysis model called waves under boundary conditions for the discussion that follows. The normal modes of oscillation for the string in Figure 18.9 can be described by imposing the boundary conditions that the ends be nodes and that the nodes be separated by one-half of a wavelength with antinodes halfway between the nodes. The first normal mode that is consistent with these requirements, shown in Figure 18.10a (page 542), has nodes at its ends and one antinode in the middle. This normal From the equations for the waves, we see that A 5 4.0 cm, k 5 3.0 rad/cm, and v 5 2.0 rad/s. Use Equation 18.1 to write an expression for the standing wave: y 5 (2A sin kx) cos vt 5 8.0 sin 3.0x cos 2.0t Find the amplitude of the simple harmonic motion of the element at the position x 5 2.3 cm by evaluating the sine function at this position: ymax 5 (8.0 cm) sin 3.0x |x 5 2.3 5 (8.0 cm) sin (6.9 rad) 5 4.6 cm Find the wavelength of the traveling waves: k 5 2p l 5 3.0 rad/cm S l 5 2p 3.0 cm Use Equation 18.2 to find the locations of the nodes: x 5 n l 2 5 n a p 3.0b cm n 5 0, 1, 2, 3, c Use Equation 18.3 to find the locations of the antinodes: x 5 n l 4 5 n a p 6.0b cm n 5 1, 3, 5, 7, c (B) Find the positions of the nodes and antinodes if one end of the string is at x 5 0.
S o l u t i o n L Figure 18.9 A string of length L fixed at both ends.
Conceptualize The waves described by the given equations are identical except for their directions of travel, so they indeed combine to form a standing wave as discussed in this section. We can represent the waves graphically by the blue and green curves in Figure 18.8.
Categorize We will substitute values into equations developed in this section, so we categorize this example as a sub-stitution problem.
S o l u t i o n ▸ 18.2 c on tin u ed www.aswarphysics.weebly.com 542 Chapter 18 Superposition and Standing Waves mode is the longest-wavelength mode that is consistent with our boundary condi-tions. The first normal mode occurs when the wavelength l1 is equal to twice the length of the string, or l1 5 2L. The section of a standing wave from one node to the next node is called a loop. In the first normal mode, the string is vibrating in one loop. In the second normal mode (see Fig. 18.10b), the string vibrates in two loops. When the left half of the string is moving upward, the right half is moving downward. In this case, the wavelength l2 is equal to the length of the string, as expressed by l2 5 L. The third normal mode (see Fig. 18.10c) corresponds to the case in which l3 5 2L/3, and the string vibrates in three loops. In general, the wave-lengths of the various normal modes for a string of length L fixed at both ends are ln 5 2L n n 5 1, 2, 3, c (18.4) where the index n refers to the nth normal mode of oscillation. These modes are possible. The actual modes that are excited on a string are discussed shortly. The natural frequencies associated with the modes of oscillation are obtained from the relationship f 5 v/l, where the wave speed v is the same for all frequen-cies. Using Equation 18.4, we find that the natural frequencies fn of the normal modes are fn 5 v ln 5 n v 2L n 5 1, 2, 3, c (18.5) These natural frequencies are also called the quantized frequencies associated with the vibrating string fixed at both ends. Because v 5 !T/m (see Eq. 16.18) for waves on a string, where T is the tension in the string and m is its linear mass density, we can also express the natural fre-quencies of a taut string as fn 5 n 2L Å T m n 5 1, 2, 3, c (18.6) The lowest frequency f1, which corresponds to n 5 1, is called either the fundamen-tal or the fundamental frequency and is given by f1 5 1 2L Å T m (18.7) The frequencies of the remaining normal modes are integer multiples of the fundamental frequency (Eq. 18.5). Frequencies of normal modes that exhibit such an integer- multiple relationship form a harmonic series, and the normal modes are called harmonics. The fundamental frequency f1 is the frequency of the first harmonic, the frequency f2 5 2f1 is that of the second harmonic, and the frequency fn 5 nf1 is that of the nth harmonic. Other oscillating systems, such as a drumhead, exhibit normal modes, but the frequencies are not related as integer multiples of a fundamental (see Section 18.6). Therefore, we do not use the term harmonic in association with those types of systems.
Wavelengths of normal modes Natural frequencies of normal modes as functions of wave speed and length of string Natural frequencies of normal modes as functions of string tension and linear mass density Fundamental frequency of a taut string n 5 1 N A N L 5 – 1 1 2 l f1 a Fundamental, or first harmonic N N A A N n 5 2 L 5 2 l f2 b Second harmonic n 5 3 N N N N A A A L 5 – 3 3 2l f3 c Third harmonic Figure 18.10 The normal modes of vibration of the string in Figure 18.9 form a harmonic series. The string vibrates between the extremes shown.
www.aswarphysics.weebly.com 18.3 A nalysis Model: Waves Under Boundary C onditions 543 Let us examine further how the various harmonics are created in a string. To excite only a single harmonic, the string would have to be distorted into a shape that corresponds to that of the desired harmonic. After being released, the string would vibrate at the frequency of that harmonic. This maneuver is difficult to perform, however, and is not how a string of a musical instrument is excited. If the string is distorted into a general, nonsinusoidal shape, the resulting vibration includes a combination of various harmonics. Such a distortion occurs in musical instruments when the string is plucked (as in a guitar), bowed (as in a cello), or struck (as in a piano). When the string is distorted into a nonsinusoidal shape, only waves that satisfy the boundary conditions can persist on the string. These waves are the harmonics. The frequency of a string that defines the musical note that it plays is that of the fundamental, even though other harmonics are present. The string’s frequency can be varied by changing the string’s tension or its length. For example, the tension in guitar and violin strings is varied by a screw adjustment mechanism or by tun ing pegs located on the neck of the instrument. As the tension is increased, the frequency of the normal modes increases in accordance with Equation 18.6. Once the instrument is “tuned,” players vary the frequency by moving their fingers along the neck, thereby changing the length of the oscillating portion of the string. As the length is shortened, the frequency increases because, as Equation 18.6 specifies, the normal-mode frequencies are inversely proportional to string length. uick Quiz 18.3 When a standing wave is set up on a string fixed at both ends, which of the following statements is true? (a) The number of nodes is equal to the number of antinodes. (b) The wavelength is equal to the length of the string divided by an integer. (c) The frequency is equal to the number of nodes times the fundamental frequency. (d) The shape of the string at any instant shows a symmetry about the midpoint of the string.
continued Example 18.3 Give Me a C Note! The middle C string on a piano has a fundamental frequency of 262 Hz, and the string for the first A above middle C has a fundamental frequency of 440 Hz. Calculate the frequencies of the next two harmonics of the C string.
Imagine a wave that is not free to travel throughout all space as in the traveling wave model. If the wave is subject to bound ary conditions, such that cer tain requirements must be met at specific locations in space, the wave is limited to a set of normal modes with quantized wavelengths and quantized natural frequencies. For waves on a string fixed at both ends, the natural frequencies are 1, 2, 3, (18.6) where is the tension in the string and is its linear mass density.
Analysis Model Waves Under Boundary Conditions Examples: • waves traveling back and forth on a guitar string combine to form a standing wave • sound waves traveling back and forth in a clarinet combine to form stand-ing waves (Section 18.5) • a microscopic particle confined to small region of space is modeled as a wave and exhibits quantized energies (Chapter 41) • the Fermi energy of metal is deter mined by modeling electrons as wave-like particles in a box (Chapter 43) www.aswarphysics.weebly.com 544 Chapter 18 Superposition and Standing Waves Finalize If the frequencies of piano strings were determined solely by tension, this result suggests that the ratio of ten-sions from the lowest string to the highest string on the piano would be enormous. Such large tensions would make it difficult to design a frame to support the strings. In reality, the frequencies of piano strings vary due to additional param-eters, including the mass per unit length and the length of the string. The What If? below explores a variation in length.
If you look inside a real piano, you’ll see that the assumption made in part (B) is only partially true. The strings are not likely to have the same length. The string densities for the given notes might be equal, but suppose the length of the A string is only 64% of the length of the C string. What is the ratio of their tensions?
Answer Using Equation 18.7 again, we set up the ratio of frequencies: f1A f1C 5 L C LA Å TA TC S TA TC 5 a LA L C b 2 a f1A f1C b 2 TA TC 5 10.642 2 a440 262b 2 5 1.16 Notice that this result represents only a 16% increase in tension, compared with the 182% increase in part (B).
What If?
Knowing that the fundamental frequency is f1 5 262 Hz, find the frequencies of the next harmonics by multiply-ing by integers: f2 5 2f1 5 524 Hz f3 5 3f1 5 786 Hz Analyze Use Equation 18.7 to write expressions for the fundamental frequencies of the two strings: f1A 5 1 2L Å TA m and f1C 5 1 2L Å TC m Divide the first equation by the second and solve for the ratio of tensions: f 1A f 1C 5 Å TA TC S TA TC 5 a f1A f 1C b 2 5 a440 262b 2 5 2.82 (B) If the A and C strings have the same linear mass density m and length L, determine the ratio of tensions in the two strings.
Categorize This part of the example is more of an analysis problem than is part (A) and uses the waves under boundary conditions model.
S o l u t i o n Conceptualize Remember that the harmonics of a vibrating string have frequencies that are related by integer mul-tiples of the fundamental.
Categorize This first part of the example is a simple substitution problem.
S o l u t i o n ▸ 18.3 continu ed Example 18.4 Changing String Vibration with Water One end of a horizontal string is attached to a vibrating blade, and the other end passes over a pulley as in Figure 18.11a. A sphere of mass 2.00 kg hangs on the end of the string. The string is vibrating in its second harmonic. A con-tainer of water is raised under the sphere so that the sphere is completely submerged. In this configuration, the string vibrates in its fifth harmonic as shown in Figure 18.11b. What is the radius of the sphere?
Conceptualize Imagine what happens when the sphere is immersed in the water. The buoyant force acts upward on the sphere, reducing the tension in the string. The change in tension causes a change in the speed of waves on the AM S o l u t i o n www.aswarphysics.weebly.com 18.3 Analysis Model: Waves Under Boundary Conditions 545 string, which in turn causes a change in the wavelength. This altered wavelength results in the string vibrating in its fifth normal mode rather than the second.
Categorize The hanging sphere is modeled as a particle in equilibrium. One of the forces acting on it is the buoyant force from the water. We also apply the waves under boundary conditions model to the string.
b a Figure 18.11 (Example 18.4) (a) When the sphere hangs in air, the string vibrates in its second harmonic. (b) When the sphere is immersed in water, the string vibrates in its fifth harmonic.
Analyze Apply the particle in equilibrium model to the sphere in Figure 18.11a, identifying T1 as the tension in the string as the sphere hangs in air: o F 5 T1 2 mg 5 0 T1 5 mg Apply the particle in equilibrium model to the sphere in Figure 18.11b, where T2 is the tension in the string as the sphere is immersed in water: T2 1 B 2 mg 5 0 (1) B 5 mg 2 T2 Write the equation for the frequency of a standing wave on a string (Eq. 18.6) twice, once before the sphere is immersed and once after. Notice that the frequency f is the same in both cases because it is determined by the vibrating blade. In addition, the linear mass density m and the length L of the vibrating portion of the string are the same in both cases. Divide the equations: f 5 n 1 2L Å T1 m f 5 n 2 2L Å T2 m S 1 5 n 1 n 2 Å T1 T2 Solve for T2: T2 5 an 1 n 2 b 2 T1 5 an 1 n 2 b 2 mg Substitute this result into Equation (1): (2) B 5 mg 2 an 1 n 2 b 2 mg 5 mg c1 2 an 1 n 2 b 2 d The desired quantity, the radius of the sphere, will appear in the expression for the buoyant force B. Before proceed-ing in this direction, however, we must evaluate T2 from the information about the standing wave.
Finalize Notice that only certain radii of the sphere will result in the string vibrating in a normal mode; the speed of waves on the string must be changed to a value such that the length of the string is an integer multiple of half wave-lengths. This limitation is a feature of the quantization that was introduced earlier in this chapter: the sphere radii that cause the string to vibrate in a normal mode are quantized.
Using Equation 14.5, express the buoyant force in terms of the radius of the sphere: B 5 rwatergVsphere 5 rwaterg 1 4 3pr 32 Solve for the radius of the sphere and substitute from Equation (2): r 5 a 3B 4prwaterg b 1/3 5 e 3m 4prwater c1 2 an 1 n 2 b 2 d f 1/3 Substitute numerical values: r 5 e 312.00 kg2 4p11 000 kg/m32 c1 2 a2 5b 2 d f 1/3 5 0.073 7 m 5 7.37 cm ▸ 18.4 c on tin u ed www.aswarphysics.weebly.com 546 Chapter 18 Superposition and Standing Waves 1Strictly speaking, the open end of an air column is not exactly a displacement antinode. A compression reaching an open end does not reflect until it passes beyond the end. For a tube of circular cross section, an end correction equal to approximately 0.6R, where R is the tube’s radius, must be added to the length of the air column. Hence, the effective length of the air column is longer than the true length L. We ignore this end correction in this discussion.
Vibrating blade When the blade vibrates at one of the natural frequencies of the string, large-amplitude standing waves are created.
Figure 18.12 Standing waves are set up in a string when one end is connected to a vibrating blade.
18.4 Resonance We have seen that a system such as a taut string is capable of oscillating in one or more normal modes of oscillation. Suppose we drive such a string with a vibrating blade as in Figure 18.12. We find that if a periodic force is applied to such a system, the amplitude of the resulting motion of the string is greatest when the frequency of the applied force is equal to one of the natural frequencies of the system. This phenomenon, known as resonance, was discussed in Section 15.7 with regard to a simple harmonic oscillator. Although a block–spring system or a simple pendulum has only one natural frequency, standing-wave systems have a whole set of natural frequencies, such as that given by Equation 18.6 for a string. Because an oscillat-ing system exhibits a large amplitude when driven at any of its natural frequencies, these frequencies are often referred to as resonance frequencies. Consider the string in Figure 18.12 again. The fixed end is a node, and the end connected to the blade is very nearly a node because the amplitude of the blade’s motion is small compared with that of the elements of the string. As the blade oscil-lates, transverse waves sent down the string are reflected from the fixed end. As we learned in Section 18.3, the string has natural frequencies that are determined by its length, tension, and linear mass density (see Eq. 18.6). When the frequency of the blade equals one of the natural frequencies of the string, standing waves are produced and the string oscillates with a large amplitude. In this resonance case, the wave generated by the oscillating blade is in phase with the reflected wave and the string absorbs energy from the blade. If the string is driven at a frequency that is not one of its natural frequencies, the oscillations are of low amplitude and exhibit no stable pattern. Resonance is very important in the excitation of musical instruments based on air columns. We shall discuss this application of resonance in Section 18.5.
18.5 Standing Waves in Air Columns The waves under boundary conditions model can also be applied to sound waves in a column of air such as that inside an organ pipe or a clarinet. Standing waves in this case are the result of interference between longitudinal sound waves traveling in opposite directions. In a pipe closed at one end, the closed end is a displacement node because the rigid barrier at this end does not allow longitudinal motion of the air. Because the pressure wave is 90° out of phase with the displacement wave (see Section 17.1), the closed end of an air column corresponds to a pressure antinode (that is, a point of maximum pressure variation). The open end of an air column is approximately a displacement antinode1 and a pressure node. We can understand why no pressure variation occurs at an open end by noting that the end of the air column is open to the atmosphere; therefore, the pressure at this end must remain constant at atmospheric pressure. You may wonder how a sound wave can reflect from an open end because there may not appear to be a change in the medium at this point: the medium through which the sound wave moves is air both inside and outside the pipe. Sound can be represented as a pressure wave, however, and a compression region of the sound wave is constrained by the sides of the pipe as long as the region is inside the pipe. As the compression region exits at the open end of the pipe, the constraint of the pipe is removed and the compressed air is free to expand into the atmosphere. Therefore, there is a change in the character of the medium between the inside www.aswarphysics.weebly.com 18.5 Standing Waves in Air Columns 547 of the pipe and the outside even though there is no change in the material of the medium. This change in character is sufficient to allow some reflection. With the boundary conditions of nodes or antinodes at the ends of the air col-umn, we have a set of normal modes of oscillation as is the case for the string fixed at both ends. Therefore, the air column has quantized frequencies. The first three normal modes of oscillation of a pipe open at both ends are shown in Figure 18.13a. Notice that both ends are displacement antinodes (approx-imately). In the first normal mode, the standing wave extends between two adjacent antinodes, which is a distance of half a wavelength. Therefore, the wavelength is twice the length of the pipe, and the fundamental frequency is f1 5 v/2L. As Figure 18.13a shows, the frequencies of the higher harmonics are 2f1, 3f1, . . . .
In a pipe open at both ends, the natural frequencies of oscillation form a har-monic series that includes all integral multiples of the fundamental frequency.
Because all harmonics are present and because the fundamental frequency is given by the same expression as that for a string (see Eq. 18.5), we can express the natural frequencies of oscillation as fn 5 n v 2L n 5 1, 2, 3, . . . (18.8) Despite the similarity between Equations 18.5 and 18.8, you must remember that v in Equation 18.5 is the speed of waves on the string, whereas v in Equation 18.8 is the speed of sound in air. If a pipe is closed at one end and open at the other, the closed end is a displace-ment node (see Fig. 18.13b). In this case, the standing wave for the fundamental mode extends from an antinode to the adjacent node, which is one-fourth of a wave-length. Hence, the wavelength for the first normal mode is 4L, and the fundamental Natural frequencies of a pipe W W open at both ends Third harmonic L First harmonic Second harmonic First harmonic Third harmonic Fifth harmonic A A N A A A N N A A A A N N N 1 5 2L f1 5 — 5 — v 1 v 2L l l 2 5 L f2 5 — 5 2f1 v L l 3 5 — L f3 5 — 5 3f1 3v 2L 2 3 l L A N A N A N A A N N A N 3 5 — L f3 5 — 5 3f1 3v 4L 4 3 l 5 5 — L f5 5 — 5 5f1 5v 4L 4 5 l 1 5 4L f1 5 — 5 — v 1 v 4L l l In a pipe open at both ends, the ends are displacement antinodes and the harmonic series contains all integer multiples of the fundamental.
In a pipe closed at one end, the open end is a displacement antinode and the closed end is a node. The harmonic series contains only odd integer multiples of the fundamental.
a b Figure 18.13 Graphical representations of the motion of elements of air in standing lon-gitudinal waves in (a) a column open at both ends and (b) a col-umn closed at one end.
Pitfall Prevention 18.3 Sound Waves in Air Are Lon-gitudinal, Not Transverse The standing longitudinal waves are drawn as transverse waves in Fig-ure 18.13. Because they are in the same direction as the propaga-tion, it is difficult to draw longitu-dinal displacements. Therefore, it is best to interpret the red-brown curves in Figure 18.13 as a graphi-cal representation of the waves (our diagrams of string waves are pictorial representations), with the vertical axis representing the horizontal displacement s(x, t) of the elements of the medium.
www.aswarphysics.weebly.com 548 Chapter 18 Superposition and Standing Waves frequency is f1 5 v/4L. As Figure 18.13b shows, the higher-frequency waves that sat-isfy our conditions are those that have a node at the closed end and an antinode at the open end; hence, the higher harmonics have frequencies 3f1, 5f1, . . . .
In a pipe closed at one end, the natural frequencies of oscillation form a har-monic series that includes only odd integral multiples of the fundamental frequency.
We express this result mathematically as fn 5 n v 4L n 5 1, 3, 5, . . . (18.9) It is interesting to investigate what happens to the frequencies of instruments based on air columns and strings during a concert as the temperature rises. The sound emitted by a flute, for example, becomes sharp (increases in frequency) as the flute warms up because the speed of sound increases in the increasingly warmer air inside the flute (consider Eq. 18.8). The sound produced by a violin becomes flat (decreases in frequency) as the strings thermally expand because the expansion causes their tension to decrease (see Eq. 18.6). Musical instruments based on air columns are generally excited by resonance. The air column is presented with a sound wave that is rich in many frequencies. The air column then responds with a large-amplitude oscillation to the frequencies that match the quantized frequencies in its set of harmonics. In many woodwind instru-ments, the initial rich sound is provided by a vibrating reed. In brass instruments, this excitation is provided by the sound coming from the vibration of the player’s lips. In a flute, the initial excitation comes from blowing over an edge at the mouth-piece of the instrument in a manner similar to blowing across the opening of a bot-tle with a narrow neck. The sound of the air rushing across the bottle opening has many frequencies, including one that sets the air cavity in the bottle into resonance.
Q uick Quiz 18.4 A pipe open at both ends resonates at a fundamental frequency fopen. When one end is covered and the pipe is again made to resonate, the fundamental frequency is fclosed. Which of the following expressions describes how these two resonant frequencies compare? (a) fclosed 5 fopen (b) fclosed 5 1 2 fopen (c) fclosed 5 2fopen (d) fclosed 5 3 2 fopen Q uick Quiz 18.5 Balboa Park in San Diego has an outdoor organ. When the air temperature increases, the fundamental frequency of one of the organ pipes (a) stays the same, (b) goes down, (c) goes up, or (d) is impossible to determine.
Natural frequencies of a pipe closed at one end and open at the other Find the frequency of the first harmonic of the culvert, modeling it as an air column open at both ends: f1 5 v 2L 5 343 m/s 211.23 m2 5 139 Hz Find the next harmonics by multiplying by integers: f2 5 2f1 5 279 Hz f3 5 3f1 5 418 Hz Example 18.5 Wind in a Culvert A section of drainage culvert 1.23 m in length makes a howling noise when the wind blows across its open ends.
(A) Determine the frequencies of the first three harmonics of the culvert if it is cylindrical in shape and open at both ends. Take v 5 343 m/s as the speed of sound in air.
Conceptualize The sound of the wind blowing across the end of the pipe contains many frequencies, and the culvert responds to the sound by vibrating at the natural frequencies of the air column.
Categorize This example is a relatively simple substitution problem.
S o l u t i o n www.aswarphysics.weebly.com 18.5 Standing Waves in Air Columns 549 Find the frequency of the first harmonic of the culvert, modeling it as an air column closed at one end: f1 5 v 4L 5 343 m/s 411.23 m2 5 69.7 Hz Find the next two harmonics by multiplying by odd integers: f3 5 3f1 5 209 Hz f5 5 5f1 5 349 Hz (B) What are the three lowest natural frequencies of the culvert if it is blocked at one end?
S o l u t i o n Example 18.6 Measuring the Frequency of a Tuning Fork A simple apparatus for demonstrating resonance in an air col-umn is depicted in Figure 18.14. A vertical pipe open at both ends is partially submerged in water, and a tuning fork vibrat-ing at an unknown frequency is placed near the top of the pipe. The length L of the air column can be adjusted by moving the pipe vertically. The sound waves generated by the fork are rein-forced when L corresponds to one of the resonance frequen-cies of the pipe. For a certain pipe, the smallest value of L for which a peak occurs in the sound intensity is 9.00 cm.
(A) What is the frequency of the tuning fork?
Conceptualize Sound waves from the tuning fork enter the pipe at its upper end. Although the pipe is open at its lower end to allow the water to enter, the water’s surface acts like a barrier. The waves reflect from the water surface and combine with those moving downward to form a standing wave.
Categorize Because of the reflection of the sound waves from the water surface, we can model the pipe as open at the upper end and closed at the lower end. Therefore, we can apply the waves under boundary conditions model to this situation.
Analyze AM S o l u t i o n f 5 ?
First resonance Second resonance (third harmonic) Third resonance (fifth harmonic) /4 3 /4 5 /4 l l l L Water a b Figure 18.14 (Example 18.6) (a) Apparatus for dem-onstrating the resonance of sound waves in a pipe closed at one end. The length L of the air column is varied by moving the pipe vertically while it is partially submerged in water. (b) The first three normal modes of the system shown in (a).
Use Equation 18.9 to find the fundamental frequency for L 5 0.090 0 m: f1 5 v 4L 5 343 m/s 410.090 0 m2 5 953 Hz Use Equation 16.12 to find the wavelength of the sound wave from the tuning fork: l 5 v f 5 343 m/s 953 Hz 5 0.360 m Notice from Figure 18.14b that the length of the air col-umn for the second resonance is 3l/4: L 5 3l/4 5 0.270 m Notice from Figure 18.14b that the length of the air col-umn for the third resonance is 5l/4: L 5 5l/4 5 0.450 m Because the tuning fork causes the air column to resonate at this frequency, this frequency must also be that of the tuning fork.
(B) What are the values of L for the next two resonance conditions?
S o l u t i o n ▸ 18.5 c on tin u ed Finalize Consider how this problem differs from the preceding example. In the culvert, the length was fixed and the air column was presented with a mixture of many frequencies. The pipe in this example is presented with one single frequency from the tuning fork, and the length of the pipe is varied until resonance is achieved.
www.aswarphysics.weebly.com 550 Chapter 18 Superposition and Standing Waves 01 11 21 02 31 12 1 1.59 2.14 2.30 2.65 2.92 41 22 03 51 32 61 3.16 3.50 3.60 3.65 4.06 4.15 Elements of the medium moving out of the page at an instant of time.
Elements of the medium moving into the page at an instant of time.
Below each pattern is a factor by which the frequency of the mode is larger than that of the 01 mode. The frequencies of oscillation do not form a harmonic series because these factors are not integers.
Figure 18.16 Representation of some of the normal modes possible in a circular membrane fixed at its perimeter. The pair of numbers above each pattern cor-responds to the number of radial nodes and the number of circular nodes, respectively. In each dia-gram, elements of the membrane on either side of a nodal line move in opposite directions, as indicated by the colors. (Adapted from T. D. Rossing, The Science of Sound, 3rd ed., Reading, Massachusetts, Addison-Wesley Publishing Co., 2001) 18.6 Standing Waves in Rods and Membranes Standing waves can also be set up in rods and membranes. A rod clamped in the middle and stroked parallel to the rod at one end oscillates as depicted in Figure 18.15a. The oscillations of the elements of the rod are longitudinal, and so the red-brown curves in Figure 18.15 represent longitudinal displacements of various parts of the rod. For clarity, the displacements have been drawn in the transverse direc-tion as they were for air columns. The midpoint is a displacement node because it is fixed by the clamp, whereas the ends are displacement antinodes because they are free to oscillate. The oscillations in this setup are analogous to those in a pipe open at both ends. The red-brown lines in Figure 18.15a represent the first normal mode, for which the wavelength is 2L and the frequency is f 5 v/2L, where v is the speed of longitudinal waves in the rod. Other normal modes may be excited by clamping the rod at different points. For example, the second normal mode (Fig. 18.15b) is excited by clamping the rod a distance L/4 away from one end. It is also possible to set up transverse standing waves in rods. Musical instru-ments that depend on transverse standing waves in rods or bars include triangles, marimbas, xylophones, glockenspiels, chimes, and vibraphones. Other devices that make sounds from vibrating bars include music boxes and wind chimes. Two-dimensional oscillations can be set up in a flexible membrane stretched over a circular hoop such as that in a drumhead. As the membrane is struck at some point, waves that arrive at the fixed boundary are reflected many times. The resulting sound is not harmonic because the standing waves have frequencies that are not related by integer multiples. Without this relationship, the sound may be more correctly described as noise rather than as music. The production of noise is in contrast to the situation in wind and stringed instruments, which produce sounds that we describe as musical. Some possible normal modes of oscillation for a two-dimensional circular mem-brane are shown in Figure 18.16. Whereas nodes are points in one-dimensional standing waves on strings and in air columns, a two-dimensional oscillator has curves along which there is no displacement of the elements of the medium. The lowest normal mode, which has a frequency f1, contains only one nodal curve; this curve runs around the outer edge of the membrane. The other possible normal modes show additional nodal curves that are circles and straight lines across the diameter of the membrane.
18.7 Beats: Interference in Time The interference phenomena we have studied so far involve the superposition of two or more waves having the same frequency. Because the amplitude of the oscil-b N A A L 4 A N l2 5 L f2 5 5 2f1 v L Figure 18.15 Normal-mode longitudinal vibrations of a rod of length L (a) clamped at the middle to produce the first nor-mal mode and (b) clamped at a distance L/4 from one end to produce the second normal mode. Notice that the red-brown curves are graphical representations of oscillations parallel to the rod (longitudinal waves).
A N A L f1 5 5 v v 2L l1 l1 5 2L a www.aswarphysics.weebly.com 18.7 Beats: Interference in Time 551 y y t t b a Figure 18.17 Beats are formed by the combination of two waves of slightly different frequencies. (a) The individual waves. (b) The combined wave. The envelope wave (dashed line) represents the beating of the combined sounds.
lation of elements of the medium varies with the position in space of the element in such a wave, we refer to the phenomenon as spatial interference. Standing waves in strings and pipes are common examples of spatial interference. Now let’s consider another type of interference, one that results from the super-position of two waves having slightly different frequencies. In this case, when the two waves are observed at a point in space, they are periodically in and out of phase. That is, there is a temporal (time) alternation between constructive and destructive interference. As a consequence, we refer to this phenomenon as interference in time or temporal interference. For example, if two tuning forks of slightly different frequen-cies are struck, one hears a sound of periodically varying amplitude. This phenom-enon is called beating.
Beating is the periodic variation in amplitude at a given point due to the superposition of two waves having slightly different frequencies. The number of amplitude maxima one hears per second, or the beat frequency, equals the difference in frequency between the two sources as we shall show below. The maximum beat frequency that the human ear can detect is about 20 beats/s. When the beat frequency exceeds this value, the beats blend indistinguishably with the sounds producing them. Consider two sound waves of equal amplitude and slightly different frequencies f1 and f2 traveling through a medium. We use equations similar to Equation 16.13 to represent the wave functions for these two waves at a point that we identify as x 5 0. We also choose the phase angle in Equation 16.13 as f 5 p/2: y1 5 A sin ap 2 2 v1tb 5 A cos 12pf1t2 y 2 5 A sin ap 2 2 v2tb 5 A cos 12pf 2t2 Using the superposition principle, we find that the resultant wave function at this point is y 5 y1 1 y2 5 A (cos 2pf1t 1 cos 2pf2t) The trigonometric identity cos a 1 cos b 5 2 cos aa 2 b 2 b cos aa 1 b 2 b allows us to write the expression for y as y 5 c2A cos 2pa f1 2 f2 2 btd cos 2pa f1 1 f2 2 bt (18.10) Graphs of the individual waves and the resultant wave are shown in Figure 18.17. From the factors in Equation 18.10, we see that the resultant wave has an effective W W Definition of beating W W Resultant of two waves of different frequencies but equal amplitude www.aswarphysics.weebly.com 552 Chapter 18 Superposition and Standing Waves Analyze Set up a ratio of the fundamental frequencies of the two strings using Equation 18.5: f 2 f 1 5 1v 2/2L2 1v1/2L2 5 v 2 v 1 Use Equation 16.18 to substitute for the wave speeds on the strings: f 2 f 1 5 "T2/m "T1/m 5 Å T2 T1 Incorporate that the tension in one string is 1.0% larger than the other; that is, T2 5 1.010T1: f 2 f 1 5 Å 1.010T1 T1 5 1.005 Solve for the frequency of the tightened string: f2 5 1.005f1 5 1.005(440 Hz) 5 442 Hz Finalize Notice that a 1.0% mistuning in tension leads to an easily audible beat frequency of 2 Hz. A piano tuner can use beats to tune a stringed instrument by “beating” a note against a reference tone of known frequency. The tuner can then adjust the string tension until the frequency of the sound it emits equals the frequency of the reference tone. The tuner does so by tightening or loosening the string until the beats produced by it and the reference source become too infrequent to notice.
Find the beat frequency using Equation 18.12: fbeat 5 442 Hz 2 440 Hz 5 2 Hz frequency equal to the average frequency (f1 1 f2)/2. This wave is multiplied by an envelope wave given by the expression in the square brackets: yenvelope 5 2A cos 2pa f1 2 f2 2 bt (18.11) That is, the amplitude and therefore the intensity of the resultant sound vary in time. The dashed black line in Figure 18.17b is a graphical representation of the envelope wave in Equation 18.11 and is a sine wave varying with frequency ( f1 2 f2)/2. A maximum in the amplitude of the resultant sound wave is detected whenever cos 2pa f 1 2 f2 2 bt 5 61 Hence, there are two maxima in each period of the envelope wave. Because the amplitude varies with frequency as (f1 2 f2)/2, the number of beats per second, or the beat frequency fbeat, is twice this value. That is, f beat 5 0 f1 2 f 2 0 (18.12) For instance, if one tuning fork vibrates at 438 Hz and a second one vibrates at 442 Hz, the resultant sound wave of the combination has a frequency of 440 Hz (the musical note A) and a beat frequency of 4 Hz. A listener would hear a 440-Hz sound wave go through an intensity maximum four times every second.
Beat frequency Example 18.7 The Mistuned Piano Strings Two identical piano strings of length 0.750 m are each tuned exactly to 440 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings?
Conceptualize As the tension in one of the strings is changed, its fundamental frequency changes. Therefore, when both strings are played, they will have different frequencies and beats will be heard.
Categorize We must combine our understanding of the waves under boundary conditions model for strings with our new knowledge of beats.
AM S o l u t i o n www.aswarphysics.weebly.com 18.8 Nonsinusoidal Wave Patterns 553 18.8 Nonsinusoidal Wave Patterns It is relatively easy to distinguish the sounds coming from a violin and a saxophone even when they are both playing the same note. On the other hand, a person untrained in music may have difficulty distinguishing a note played on a clarinet from the same note played on an oboe. We can use the pattern of the sound waves from various sources to explain these effects. When frequencies that are integer multiples of a fundamental frequency are combined to make a sound, the result is a musical sound. A listener can assign a pitch to the sound based on the fundamental frequency. Pitch is a psychological reaction to a sound that allows the listener to place the sound on a scale from low to high (bass to treble). Combinations of frequencies that are not integer multiples of a fundamental result in a noise rather than a musical sound. It is much harder for a listener to assign a pitch to a noise than to a musical sound. The wave patterns produced by a musical instrument are the result of the super-position of frequencies that are integer multiples of a fundamental. This superposi-tion results in the corresponding richness of musical tones. The human perceptive response associated with various mixtures of harmonics is the quality or timbre of the sound. For instance, the sound of the trumpet is perceived to have a “brassy” quality (that is, we have learned to associate the adjective brassy with that sound); this quality enables us to distinguish the sound of the trumpet from that of the sax-ophone, whose quality is perceived as “reedy.” The clarinet and oboe, however, both contain air columns excited by reeds; because of this similarity, they have similar mixtures of frequencies and it is more difficult for the human ear to distinguish them on the basis of their sound quality. The sound wave patterns produced by the majority of musical instruments are nonsinusoidal. Characteristic patterns produced by a tuning fork, a flute, and a clarinet, each playing the same note, are shown in Figure 18.18. Each instrument has its own characteristic pattern. Notice, however, that despite the differences in the patterns, each pattern is periodic. This point is important for our analysis of these waves. The problem of analyzing nonsinusoidal wave patterns appears at first sight to be a formidable task. If the wave pattern is periodic, however, it can be represented as closely as desired by the combination of a sufficiently large number of sinusoidal waves that form a harmonic series. In fact, we can represent any periodic function as a series of sine and cosine terms by using a mathematical technique based on Fourier’s theorem.2 The corresponding sum of terms that represents the periodic wave pattern is called a Fourier series. Let y(t) be any function that is periodic in time with period T such that y(t 1 T) 5 y(t). Fourier’s theorem states that this func-tion can be written as y(t) 5 o (An sin 2pfnt 1 Bn cos 2pfnt) (18.13) where the lowest frequency is f1 5 1/T. The higher frequencies are integer multiples of the fundamental, fn 5 nf1, and the coefficients An and Bn represent the ampli-tudes of the various waves. Figure 18.19 on page 554 represents a harmonic analysis of the wave patterns shown in Figure 18.18. Each bar in the graph represents one of the terms in the series in Equation 18.13 up to n 5 9. Notice that a struck tuning fork produces only one harmonic (the first), whereas the flute and clarinet produce the first harmonic and many higher ones. Notice the variation in relative intensity of the various harmonics for the flute and the clarinet. In general, any musical sound consists of a fundamental fre-quency f plus other frequencies that are integer multiples of f, all having different intensities.
W W Fourier’s theorem Pitfall Prevention 18.4 Pitch Versus Frequency Do not confuse the term pitch with fre-quency. Frequency is the physical measurement of the number of oscillations per second. Pitch is a psychological reaction to sound that enables a person to place the sound on a scale from high to low or from treble to bass. Therefore, frequency is the stimulus and pitch is the response. Although pitch is related mostly (but not completely) to frequency, they are not the same. A phrase such as “the pitch of the sound” is incor-rect because pitch is not a physical property of the sound.
Tuning fork Flute Clarinet t t t b c a Figure 18.18 Sound wave pat-terns produced by (a) a tuning fork, (b) a flute, and (c) a clarinet, each at approximately the same frequency.
2 Developed by Jean Baptiste Joseph Fourier (1786–1830).
www.aswarphysics.weebly.com 554 Chapter 18 Superposition and Standing Waves Square wave 5f f 3f f 3f b c a Waves of frequency f and 3f are added to give the blue curve.
One more odd harmonic of frequency 5f is added to give the green curve.
The synthesis curve (red-brown) approaches closer to the square wave (black curve) when odd frequencies up to 9f are added.
Figure 18.20 Fourier synthesis of a square wave, represented by the sum of odd multiples of the first harmonic, which has fre-quency f. We have discussed the analysis of a wave pattern using Fourier’s theorem. The analysis involves determining the coefficients of the harmonics in Equation 18.13 from a knowledge of the wave pattern. The reverse process, called Fourier synthesis, can also be performed. In this process, the various harmonics are added together to form a resultant wave pattern. As an example of Fourier synthesis, consider the building of a square wave as shown in Figure 18.20. The symmetry of the square wave results in only odd multiples of the fundamental frequency combining in its synthesis. In Figure 18.20a, the blue curve shows the combination of f and 3f. In Figure 18.20b, we have added 5f to the combination and obtained the green curve. Notice how the general shape of the square wave is approximated, even though the upper and lower portions are not flat as they should be. Figure 18.20c shows the result of adding odd frequencies up to 9f. This approxi-mation (red-brown curve) to the square wave is better than the approximations in Figures 18.20a and 18.20b. To approximate the square wave as closely as pos-sible, we must add all odd multiples of the fundamental frequency, up to infinite frequency. Using modern technology, musical sounds can be generated electronically by mixing different amplitudes of any number of harmonics. These widely used elec-tronic music synthesizers are capable of producing an infinite variety of musical tones.
Relative intensity Tuning fork 1 2 3 4 5 6 Relative intensity 1 2 3 4 5 6 7 8 9 7 8 9 Flute Relative intensity 1 2 3 4 5 6 7 8 9 Clarinet Harmonics Harmonics Harmonics a b c Figure 18.19 Harmonics of the wave patterns shown in Figure 18.18. Notice the variations in inten-sity of the various harmonics. Parts (a), (b), and (c) correspond to those in Figure 18.18.
www.aswarphysics.weebly.com Objective Questions 555 Summary The superposition principle speci-fies that when two or more waves move through a medium, the value of the resultant wave function equals the alge-braic sum of the values of the individual wave functions. The phenomenon of beating is the periodic variation in intensity at a given point due to the superposition of two waves having slightly dif-ferent frequencies. The beat frequency is f beat 5 0 f1 2 f 2 0 (18.12) where f1 and f2 are the frequencies of the individual waves. Standing waves are formed from the combination of two sinusoidal waves having the same frequency, amplitude, and wavelength but traveling in opposite directions. The resultant standing wave is described by the wave function y 5 (2A sin kx) cos vt (18.1) Hence, the amplitude of the standing wave is 2A, and the amplitude of the simple harmonic motion of any element of the medium varies according to its position as 2A sin kx. The points of zero amplitude (called nodes) occur at x 5 nl/2 (n 5 0, 1, 2, 3, . . .). The maximum amplitude points (called antinodes) occur at x 5 nl/4 (n 5 1, 3, 5, . . .). Adjacent antinodes are separated by a distance l/2. Adjacent nodes also are separated by a distance l/2.
Concepts and Principles Analysis Models for Problem Solving y1 y2 y1 y2 Destructive interference Constructive interference y1 y2 y2 y1 Waves in Interference. When two travel-ing waves having equal frequencies super-impose, the resultant wave is described by the principle of superposition and has an amplitude that depends on the phase angle f between the two waves. Constructive interference occurs when the two waves are in phase, corresponding to f 5 0, 2p, 4p, . . . rad. Destructive interference occurs when the two waves are 180° out of phase, corresponding to f 5 p, 3p, 5p, . . . rad. Waves Under Boundary Conditions. When a wave is subject to boundary condi-tions, only certain natural frequencies are allowed; we say that the frequencies are quantized. For waves on a string fixed at both ends, the natural frequencies are fn 5 n 2L Å T m n 5 1, 2, 3, . . . (18.6) where T is the tension in the string and m is its linear mass density. For sound waves with speed v in an air column of length L open at both ends, the natural frequencies are fn 5 n v 2L n 5 1, 2, 3, . . . (18.8) If an air column is open at one end and closed at the other, only odd harmonics are present and the natural frequencies are fn 5 n v 4L n 5 1, 3, 5, . . . (18.9) n 5 1 n 5 2 n 5 3 Rank the following situations according to the intensity of sound at the receiver from the highest to the lowest. Assume the tube walls absorb no sound energy. Give equal ranks to situations in which the intensity is equal. 1. In Figure OQ18.1 (page 556), a sound wave of wave-length 0.8 m divides into two equal parts that recombine to interfere constructively, with the original difference between their path lengths being |r2 2 r1| 5 0.8 m. Objective Questions 1. denotes answer available in Student Solutions Manual/Study Guide www.aswarphysics.weebly.com 556 Chapter 18 Superposition and Standing Waves forks has a frequency of 245 Hz, what is the frequency of the other tuning fork? (a) 240 Hz (b) 242.5 Hz (c) 247.5 Hz (d) 250 Hz (e) More than one answer could be correct. 7. A tuning fork is known to vibrate with frequency 262 Hz. When it is sounded along with a mandolin string, four beats are heard every second. Next, a bit of tape is put onto each tine of the tuning fork, and the tuning fork now produces five beats per second with the same mandolin string. What is the frequency of the string? (a) 257 Hz (b) 258 Hz (c) 262 Hz (d) 266 Hz (e) 267 Hz 8. An archer shoots an arrow horizontally from the center of the string of a bow held vertically. After the arrow leaves it, the string of the bow will vibrate as a superpo-sition of what standing-wave harmonics? (a) It vibrates only in harmonic number 1, the fundamental. (b) It vibrates only in the second harmonic. (c) It vibrates only in the odd-numbered harmonics 1, 3, 5, 7, . . . . (d) It vibrates only in the even-numbered harmonics 2, 4, 6, 8, . . . . (e) It vibrates in all harmonics. 9. As oppositely moving pulses of the same shape (one upward, one downward) on a string pass through each other, at one particular instant the string shows no dis-placement from the equilibrium position at any point. What has happened to the energy carried by the pulses at this instant of time? (a) It was used up in producing the previous motion. (b) It is all potential energy. (c) It is all internal energy. (d) It is all kinetic energy. (e) The positive energy of one pulse adds to zero with the nega-tive energy of the other pulse. 10. A standing wave having three nodes is set up in a string fixed at both ends. If the frequency of the wave is dou-bled, how many antinodes will there be? (a) 2 (b) 3 (c) 4 (d) 5 (e) 6 11. Suppose all six equal-length strings of an acoustic guitar are played without fingering, that is, without being pressed down at any frets. What quantities are the same for all six strings? Choose all correct answers. (a) the fundamental frequency (b) the fundamental wavelength of the string wave (c) the fundamental wavelength of the sound emitted (d) the speed of the string wave (e) the speed of the sound emitted 12. Assume two identical sinusoidal waves are moving through the same medium in the same direction. Under what condition will the amplitude of the resul-tant wave be greater than either of the two original waves? (a) in all cases (b) only if the waves have no dif-ference in phase (c) only if the phase difference is less than 90° (d) only if the phase difference is less than 120° (e) only if the phase difference is less than 180° (a) From its original position, the sliding section is moved out by 0.1 m. (b) Next it slides out an additional 0.1 m. (c) It slides out still another 0.1 m. (d) It slides out 0.1 m more. 2. A string of length L, mass per unit length m, and tension T is vibrat-ing at its fundamental frequency. (i) If the length of the string is doubled, with all other factors held constant, what is the effect on the funda-mental frequency? (a) It becomes two times larger. (b) It becomes !2 times larger. (c) It is unchanged. (d) It becomes 1/!2 times as large. (e) It becomes one-half as large. (ii) If the mass per unit length is doubled, with all other factors held constant, what is the effect on the fundamental frequency? Choose from the same possibilities as in part (i). (iii) If the tension is doubled, with all other factors held constant, what is the effect on the fundamental frequency? Choose from the same possibilities as in part (i). 3. In Example 18.1, we investigated an oscillator at 1.3 kHz driving two identical side-by-side speakers. We found that a listener at point O hears sound with maximum intensity, whereas a listener at point P hears a mini-mum. What is the intensity at P? (a) less than but close to the intensity at O (b) half the intensity at O (c) very low but not zero (d) zero (e) indeterminate 4. A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflected at the post and travel back along the string without loss of amplitude. (i) What is the net displace-ment at a point on the string where two pulses are cross-ing? Assume the string is rigidly attached to the post. (a) 0.4 m (b) 0.3 m (c) 0.2 m (d) 0.1 m (e) 0 (ii) Next assume the end at which reflection occurs is free to slide up and down. Now what is the net displacement at a point on the string where two pulses are crossing? Choose your answer from the same possibilities as in part (i). 5. A flute has a length of 58.0 cm. If the speed of sound in air is 343 m/s, what is the fundamental frequency of the flute, assuming it is a tube closed at one end and open at the other? (a) 148 Hz (b) 296 Hz (c) 444 Hz (d) 591 Hz (e) none of those answers 6. When two tuning forks are sounded at the same time, a beat frequency of 5 Hz occurs. If one of the tuning r1 r2 Speaker Receiver Sliding section Figure OQ18.1 Objective Question 1 and Problem 6.
Conceptual Questions 1. denotes answer available in Student Solutions Manual/Study Guide 1. A crude model of the human throat is that of a pipe open at both ends with a vibrating source to introduce the sound into the pipe at one end. Assuming the vibrating source produces a range of frequencies, dis-cuss the effect of changing the pipe’s length. 2. When two waves interfere constructively or destruc-tively, is there any gain or loss in energy in the system of the waves? Explain. 3. Explain how a musical instrument such as a piano may be tuned by using the phenomenon of beats.
www.aswarphysics.weebly.com Problems 557 6. An airplane mechanic notices that the sound from a twin-engine aircraft rapidly varies in loudness when both engines are running. What could be causing this variation from loud to soft? 7. Despite a reasonably steady hand, a person often spills his coffee when carrying it to his seat. Discuss reso-nance as a possible cause of this difficulty and devise a means for preventing the spills. 8. A soft-drink bottle resonates as air is blown across its top. What happens to the resonance frequency as the level of fluid in the bottle decreases? 9. Does the phenomenon of wave interference apply only to sinusoidal waves? 3. Two waves on one string are described by the wave functions y1 5 3.0 cos (4.0x 2 1.6t) y2 5 4.0 sin (5.0x 2 2.0t) where x and y are in centimeters and t is in seconds. Find the superposition of the waves y1 1 y2 at the points (a) x 5 1.00, t 5 1.00; (b) x 5 1.00, t 5 0.500; and (c) x 5 0.500, t 5 0. Note: Remember that the arguments of the trigonometric functions are in radians. 4. Two pulses of different amplitudes approach each other, each having a speed of v 5 1.00 m/s. Figure P18.4 shows the positions of the pulses at time t 5 0. (a) Sketch the resultant wave at t 5 2.00 s, 4.00 s, 5.00 s, and 6.00 s. (b) What If? If the pulse on the right is inverted so that it is upright, how would your sketches of the resultant wave change?
1.0 y (m) x (m) 0.5 0.5 2 4 6 8 10 12 14 16 v v Figure P18.4 5. A tuning fork generates sound waves with a frequency of 246 Hz. The waves travel in opposite directions along a hallway, are reflected by end walls, and return. The hallway is 47.0 m long and the tuning fork is located 14.0 m from one end. What is the phase difference W 4. What limits the amplitude of motion of a real vibrating system that is driven at one of its resonant frequencies? 5. A tuning fork by itself produces a faint sound. Explain how each of the following methods can be used to obtain a louder sound from it. Explain also any effect on the time interval for which the fork vibrates audibly. (a) holding the edge of a sheet of paper against one vibrating tine (b) pressing the handle of the tuning fork against a chalkboard or a tabletop (c) holding the tuning fork above a column of air of properly chosen length as in Example 18.6 (d) holding the tuning fork close to an open slot cut in a sheet of foam plastic or cardboard (with the slot similar in size and shape to one tine of the fork and the motion of the tines per-pendicular to the sheet) Note: Unless otherwise specified, assume the speed of sound in air is 343 m/s, its value at an air temperature of 20.0°C. At any other Celsius temperature TC, the speed of sound in air is described by v 5 331 Å1 1 TC 273 where v is in m/s and T is in °C.
Section 18.1 Analysis Model: Waves in Interference 1. Two waves are traveling in the same direction along a stretched string. The waves are 90.0° out of phase. Each wave has an amplitude of 4.00 cm. Find the amplitude of the resultant wave. 2. Two wave pulses A and B are moving in opposite direc-tions, each with a speed v 5 2.00 cm/s. The amplitude of A is twice the amplitude of B. The pulses are shown in Figure P18.2 at t 5 0. Sketch the resultant wave at t 5 1.00 s, 1.50 s, 2.00 s, 2.50 s, and 3.00 s.
W Problems The problems found in this chapter may be assigned online in Enhanced WebAssign 1. straightforward; 2. intermediate; 3. challenging 1. full solution available in the Student Solutions Manual/Study Guide AMT Analysis Model tutorial available in Enhanced WebAssign GP Guided Problem M Master It tutorial available in Enhanced WebAssign W Watch It video solution available in Enhanced WebAssign BIO Q/C S 4 y (cm) x (cm) 2 2 4 6 8 10 12 14 16 18 20 A B v v Figure P18.2 www.aswarphysics.weebly.com 558 Chapter 18 Superposition and Standing Waves 11. Two sinusoidal waves in a string are defined by the wave functions y1 5 2.00 sin (20.0x 2 32.0t) y2 5 2.00 sin (25.0x 2 40.0t) where x, y1, and y2 are in centimeters and t is in sec-onds. (a) What is the phase difference between these two waves at the point x 5 5.00 cm at t 5 2.00 s? (b) What is the positive x value closest to the origin for which the two phases differ by 6p at t 5 2.00 s? (At that location, the two waves add to zero.) 12. Two identical sinusoidal waves with wavelengths of 3.00 m travel in the same direction at a speed of 2.00 m/s. The second wave originates from the same point as the first, but at a later time. The amplitude of the resultant wave is the same as that of each of the two initial waves. Determine the minimum possible time interval between the starting moments of the two waves. 13. Two identical loudspeakers 10.0 m apart are driven by the same oscillator with a frequency of f 5 21.5 Hz (Fig. P18.13) in an area where the speed of sound is 344 m/s. (a) Show that a receiver at point A records a minimum in sound intensity from the two speak-ers. (b) If the receiver is moved in the plane of the speakers, show that the path it should take so that the intensity remains at a minimum is along the hyperbola 9x2 2 16y2 5 144 (shown in red-brown in Fig. P18.13). (c) Can the receiver remain at a minimum and move very far away from the two sources? If so, determine the limiting form of the path it must take. If not, explain how far it can go.
9.00 m 10.0 m y (x, y) A x Figure P18.13 Section 18.2 Standing Waves 14. Two waves simultaneously present on a long string have a phase difference f between them so that a standing wave formed from their combination is described by y 1x, t 2 5 2A sin akx 1 f 2 b cos avt 2 f 2 b (a) Despite the presence of the phase angle f, is it still true that the nodes are one-half wavelength apart? Explain. (b) Are the nodes different in any way from the way they would be if f were zero? Explain. 15. Two sinusoidal waves traveling in opposite directions interfere to produce a standing wave with the wave function y 5 1.50 sin (0.400x) cos (200t) M Q/C Q/C W between the reflected waves when they meet at the tun-ing fork? The speed of sound in air is 343 m/s. 6. The acoustical system shown in Figure OQ18.1 is driven by a speaker emitting sound of frequency 756 Hz. (a) If constructive interference occurs at a particular location of the sliding section, by what mini-mum amount should the sliding section be moved upward so that destructive interference occurs instead? (b) What minimum distance from the original posi-tion of the sliding section will again result in construc-tive interference? 7. Two pulses traveling on the same string are described by y 1 5 5 13x 2 4t2 2 1 2 y 2 5 25 13x 1 4t 2 62 2 1 2 (a) In which direction does each pulse travel? (b) At what instant do the two cancel everywhere? (c) At what point do the two pulses always cancel? 8. Two identical loudspeakers are placed on a wall 2.00 m apart. A listener stands 3.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz. (a) What is the phase difference in radians between the waves from the speakers when they reach the observer? (b) What If? What is the frequency closest to 300 Hz to which the oscillator may be adjusted such that the observer hears minimal sound? 9. Two traveling sinusoidal waves are described by the wave functions y1 5 5.00 sin [p(4.00x 2 1 200t)] y2 5 5.00 sin [p(4.00x 2 1 200t 2 0.250)] where x, y1, and y2 are in meters and t is in seconds. (a) What is the amplitude of the resultant wave func-tion y1 1 y2? (b) What is the frequency of the resultant wave function? 10. Why is the following situation impossible? Two identical loudspeakers are driven by the same oscillator at fre-quency 200 Hz. They are located on the ground a dis-tance d 5 4.00 m from each other. Starting far from the speakers, a man walks straight toward the right-hand speaker as shown in Figure P18.10. After passing through three minima in sound intensity, he walks to the next maximum and stops. Ignore any sound reflec-tion from the ground.
AMT M x d Figure P18.10 www.aswarphysics.weebly.com Problems 559 entire length. A fret is provided for limiting vibration to just the lower two-thirds of the string. (a) If the string is pressed down at this fret and plucked, what is the new fundamental frequency? (b) What If? The guitarist can play a “natural harmonic” by gently touching the string at the location of this fret and plucking the string at about one-sixth of the way along its length from the other end. What frequency will be heard then? 23. The A string on a cello vibrates in its first normal mode with a frequency of 220 Hz. The vibrating segment is 70.0 cm long and has a mass of 1.20 g. (a) Find the tension in the string. (b) Determine the frequency of vibration when the string vibrates in three segments. 24. A taut string has a length of 2.60 m and is fixed at both ends. (a) Find the wavelength of the fundamental mode of vibration of the string. (b) Can you find the frequency of this mode? Explain why or why not. 25. A certain vibrating string on a piano has a length of 74.0 cm and forms a standing wave having two anti-nodes. (a) Which harmonic does this wave represent? (b) Determine the wavelength of this wave. (c) How many nodes are there in the wave pattern? 26. A string that is 30.0 cm long and has a mass per unit length of 9.00 3 1023 kg/m is stretched to a tension of 20.0 N. Find (a) the fundamental frequency and (b) the next three frequencies that could cause stand-ing-wave patterns on the string. 27. In the arrangement shown in Figure P18.27, an object can be hung from a string (with linear mass density m 5 0.002 00 kg/m) that passes over a light pulley. The string is connected to a vibrator (of constant frequency f ), and the length of the string between point P and the pulley is L 5 2.00 m. When the mass m of the object is either 16.0 kg or 25.0 kg, standing waves are observed; no standing waves are observed with any mass between these values, however. (a) What is the frequency of the vibrator? Note: The greater the tension in the string, the smaller the number of nodes in the standing wave. (b) What is the largest object mass for which standing waves could be observed?
L P Vibrator P m m Figure P18.27 Problems 27 and 28. 28. In the arrangement shown in Figure P18.27, an object of mass m 5 5.00 kg hangs from a cord around a light pulley. The length of the cord between point P and the pulley is L 5 2.00 m. (a) When the vibrator is set to a frequency of 150 Hz, a standing wave with six loops is formed. What must be the linear mass density of the cord? (b) How many loops (if any) will result if m is changed to 45.0 kg? (c) How many loops (if any) will result if m is changed to 10.0 kg?
W W AMT M where x and y are in meters and t is in seconds. Deter-mine (a) the wavelength, (b) the frequency, and (c) the speed of the interfering waves. 16. Verify by direct substitution that the wave function for a standing wave given in Equation 18.1, y 5 (2A sin kx) cos vt is a solution of the general linear wave equation, Equa-tion 16.27: '2y 'x 2 5 1 v 2 '2y 't 2 17. Two transverse sinusoidal waves combining in a medium are described by the wave functions y1 5 3.00 sin p(x 1 0.600t) y2 5 3.00 sin p(x 2 0.600t) where x, y1, and y2 are in centimeters and t is in sec-onds. Determine the maximum transverse position of an element of the medium at (a) x 5 0.250 cm, (b) x 5 0.500 cm, and (c) x 5 1.50 cm. (d) Find the three small-est values of x corresponding to antinodes. 18. A standing wave is described by the wave function y 5 6 sin ap 2xb cos 1100pt2 where x and y are in meters and t is in seconds. (a) Prepare graphs showing y as a function of x for five instants: t 5 0, 5 ms, 10 ms, 15 ms, and 20 ms. (b) From the graph, identify the wavelength of the wave and explain how to do so. (c) From the graph, identify the frequency of the wave and explain how to do so. (d) From the equation, directly identify the wavelength of the wave and explain how to do so. (e) From the equation, directly identify the frequency and explain how to do so. 19. Two identical loudspeakers are driven in phase by a common oscillator at 800 Hz and face each other at a distance of 1.25 m. Locate the points along the line joining the two speakers where relative minima of sound pressure amplitude would be expected.
Section 18.3 Analysis Model: Waves Under Boundary Conditions 20. A standing wave is established in a 120-cm-long string fixed at both ends. The string vibrates in four segments when driven at 120 Hz. (a) Determine the wavelength. (b) What is the fundamental frequency of the string? 21. A string with a mass m 5 8.00 g and a length L 5 5.00 m has one end attached to a wall; the other end is draped over a small, fixed pulley a distance d 5 4.00 m from the wall and attached to a hanging object with a mass M 5 4.00 kg as in Figure P18.21. If the horizon-tal part of the string is plucked, what is the fundamen-tal frequency of its vibration? 22. The 64.0-cm-long string of a guitar has a fundamen-tal frequency of 330 Hz when it vibrates freely along its M Q/C M M d Figure P18.21 www.aswarphysics.weebly.com 560 Chapter 18 Superposition and Standing Waves sider a seiche produced in a farm pond. Suppose the pond is 9.15 m long and assume it has a uniform width and depth. You measure that a pulse produced at one end reaches the other end in 2.50 s. (a) What is the wave speed? (b) What should be the frequency of the ground motion during the earthquake to produce a seiche that is a standing wave with antinodes at each end of the pond and one node at the center? 36. High-frequency sound can be used to produce stand-ing-wave vibrations in a wine glass. A standing-wave vibration in a wine glass is observed to have four nodes and four antinodes equally spaced around the 20.0-cm circumference of the rim of the glass. If transverse waves move around the glass at 900 m/s, an opera singer would have to produce a high harmonic with what frequency to shatter the glass with a resonant vibration as shown in Figure P18.36?
Section 18.5 Standing Waves in Air Columns 37. The windpipe of one typical whooping crane is 5.00 feet long. What is the fundamental resonant frequency of the bird’s trachea, modeled as a narrow pipe closed at one end? Assume a temperature of 37°C. 38. If a human ear canal can be thought of as resembling an organ pipe, closed at one end, that resonates at a fundamental frequency of 3 000 Hz, what is the length of the canal? Use a normal body temperature of 37°C for your determination of the speed of sound in the canal. 39. Calculate the length of a pipe that has a fundamental frequency of 240 Hz assuming the pipe is (a) closed at one end and (b) open at both ends. 40. The overall length of a piccolo is 32.0 cm. The reso-nating air column is open at both ends. (a) Find the frequency of the lowest note a piccolo can sound. (b) Opening holes in the side of a piccolo effectively shortens the length of the resonant column. Assume the highest note a piccolo can sound is 4 000 Hz. Find the distance between adjacent antinodes for this mode of vibration. 41. The fundamental frequency of an open organ pipe corresponds to middle C (261.6 Hz on the chromatic musical scale). The third resonance of a closed organ pipe has the same frequency. What is the length of (a) the open pipe and (b) the closed pipe? 42. The longest pipe on a certain organ is 4.88 m. What is the fundamental frequency (at 0.00°C) if the pipe is (a) closed at one end and (b) open at each end? (c) What will be the frequencies at 20.0°C? 43. An air column in a glass tube is open at one end and closed at the other by a movable piston. The air in the tube is warmed above room temperature, and a 384-Hz tuning fork is held at the open end. Resonance is heard BIO BIO W 29. Review. A sphere of mass M 5 1.00 kg is supported by a string that passes over a pul-ley at the end of a horizontal rod of length L 5 0.300 m (Fig. P18.29). The string makes an angle u 5 35.0° with the rod. The fundamental frequency of standing waves in the portion of the string above the rod is f 5 60.0 Hz. Find the mass of the portion of the string above the rod. 30. Review. A sphere of mass M is supported by a string that passes over a pulley at the end of a horizontal rod of length L (Fig. P18.29). The string makes an angle u with the rod. The fundamental frequency of standing waves in the portion of the string above the rod is f. Find the mass of the portion of the string above the rod. 31. A violin string has a length of 0.350 m and is tuned to concert G, with fG 5 392 Hz. (a) How far from the end of the string must the violinist place her finger to play concert A, with fA 5 440 Hz? (b) If this position is to remain correct to one-half the width of a finger (that is, to within 0.600 cm), what is the maximum allowable percentage change in the string tension? 32. Review. A solid copper object hangs at the bottom of a steel wire of negligible mass. The top end of the wire is fixed. When the wire is struck, it emits sound with a fundamental frequency of 300 Hz. The copper object is then submerged in water so that half its volume is below the water line. Determine the new fundamental frequency. 33. A standing-wave pattern is observed in a thin wire with a length of 3.00 m. The wave function is y 5 0.002 00 sin (px) cos (100pt) where x and y are in meters and t is in seconds. (a) How many loops does this pattern exhibit? (b) What is the fundamental frequency of vibration of the wire? (c) What If? If the original frequency is held constant and the tension in the wire is increased by a factor of 9, how many loops are present in the new pattern?
Section 18.4 Resonance 34. The Bay of Fundy, Nova Scotia, has the highest tides in the world. Assume in midocean and at the mouth of the bay the Moon’s gravity gradient and the Earth’s rotation make the water surface oscillate with an ampli-tude of a few centimeters and a period of 12 h 24 min. At the head of the bay, the amplitude is several meters. Assume the bay has a length of 210 km and a uniform depth of 36.1 m. The speed of long-wavelength water waves is given by v 5 !gd, where d is the water’s depth. Argue for or against the proposition that the tide is magnified by standing-wave resonance. 35. An earthquake can produce a seiche in a lake in which the water sloshes back and forth from end to end with remarkably large amplitude and long period. Con-S Q/C L M u Figure P18.29 Problems 29 and 30.
Figure P18.36 Steve Bronstein/Stone/Getty Images www.aswarphysics.weebly.com Problems 561 51. Two adjacent natural frequencies of an organ pipe are determined to be 550 Hz and 650 Hz. Calculate (a) the fundamental frequency and (b) the length of this pipe. 52. Why is the following situation impossible? A student is lis-tening to the sounds from an air column that is 0.730 m long. He doesn’t know if the column is open at both ends or open at only one end. He hears resonance from the air column at frequencies 235 Hz and 587 Hz. 53. A student uses an audio oscillator of adjustable fre-quency to measure the depth of a water well. The student reports hearing two successive resonances at 51.87 Hz and 59.85 Hz. (a) How deep is the well? (b) How many antinodes are in the standing wave at 51.87 Hz?
Section 18.6 Standing Waves in Rods and Membranes 54. An aluminum rod is clamped one-fourth of the way along its length and set into longitudinal vibration by a variable-frequency driving source. The lowest fre-quency that produces resonance is 4 400 Hz. The speed of sound in an aluminum rod is 5 100 m/s. Determine the length of the rod. 55. An aluminum rod 1.60 m long is held at its center. It is stroked with a rosin-coated cloth to set up a longi-tudinal vibration. The speed of sound in a thin rod of aluminum is 5 100 m/s. (a) What is the fundamen-tal frequency of the waves established in the rod? (b) What harmonics are set up in the rod held in this manner? (c) What If? What would be the fundamental frequency if the rod were copper, in which the speed of sound is 3 560 m/s?
Section 18.7 Beats: Interference in Time 56. While attempting to tune the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string. (a) What are the possible frequencies of the string? (b) When she tightens the string slightly, she hears 3.00 beats/s. What is the frequency of the string now? (c) By what percentage should the piano tuner now change the tension in the string to bring it into tune? 57. In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loud-ness. For example, the note at 110 Hz has two strings at this frequency. If one string slips from its nor-mal tension of 600 N to 540 N, what beat frequency is heard when the hammer strikes the two strings simultaneously? 58. Review. Jane waits on a railroad platform while two trains approach from the same direction at equal speeds of 8.00 m/s. Both trains are blowing their whis-tles (which have the same frequency), and one train is some distance behind the other. After the first train passes Jane but before the second train passes her, she hears beats of frequency 4.00 Hz. What is the fre-quency of the train whistles? 59. Review. A student holds a tuning fork oscillating at 256 Hz. He walks toward a wall at a constant speed of 1.33 m/s. (a) What beat frequency does he observe AMT M W M M when the piston is at a distance d1 5 22.8 cm from the open end and again when it is at a distance d2 5 68.3 cm from the open end. (a) What speed of sound is implied by these data? (b) How far from the open end will the piston be when the next resonance is heard? 44. A tuning fork with a frequency of f 5 512 Hz is placed near the top of the tube shown in Figure P18.44. The water level is low-ered so that the length L slowly increases from an initial value of 20.0 cm. Determine the next two values of L that correspond to resonant modes. 45. With a particular fingering, a flute produces a note with frequency 880 Hz at 20.0°C. The flute is open at both ends. (a) Find the air column length. (b) At the beginning of the halftime performance at a late-season football game, the ambient temperature is 25.00°C and the flutist has not had a chance to warm up her instrument. Find the frequency the flute pro-duces under these conditions. 46. A shower stall has dimensions 86.0 cm 3 86.0 cm 3 210 cm. Assume the stall acts as a pipe closed at both ends, with nodes at opposite sides. Assume singing voices range from 130 Hz to 2 000 Hz and let the speed of sound in the hot air be 355 m/s. For someone sing-ing in this shower, which frequencies would sound the richest (because of resonance)? 47. A glass tube (open at both ends) of length L is posi-tioned near an audio speaker of frequency f 5 680 Hz. For what values of L will the tube resonate with the speaker? 48. A tunnel under a river is 2.00 km long. (a) At what fre-quencies can the air in the tunnel resonate? (b) Explain whether it would be good to make a rule against blow-ing your car horn when you are in the tunnel. 49. As shown in Figure P18.49, water is pumped into a tall, vertical cylinder at a volume flow rate R 5 1.00 L/min. The radius of the cylinder is r 5 5.00 cm, and at the open top of the cylinder a tuning fork is vibrating with a fre-quency f 5 512 Hz. As the water rises, what time interval elapses between successive resonances? 50. As shown in Figure P18.49, water is pumped into a tall, vertical cylinder at a volume flow rate R. The radius of the cylinder is r, and at the open top of the cylinder a tuning fork is vibrating with a frequency f. As the water rises, what time interval elapses between successive resonances?
f L Valve Figure P18.44 Q/C f R r Figure P18.49 Problems 49 and 50.
S www.aswarphysics.weebly.com 562 Chapter 18 Superposition and Standing Waves 66. A 2.00-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is maintained at 20.0 N. (a) What are the frequencies of the first three allowed modes of vibration? (b) If a node is observed at a point 0.400 m from one end, in what mode and with what frequency is it vibrating? 67. The fret closest to the bridge on a guitar is 21.4 cm from the bridge as shown in Figure P18.67. When the thinnest string is pressed down at this first fret, the string produces the highest frequency that can be played on that guitar, 2 349 Hz. The next lower note that is produced on the string has frequency 2 217 Hz. How far away from the first fret should the next fret be?
Bridge Frets 21.4 cm Figure P18.67 68. A string fixed at both ends and having a mass of 4.80 g, a length of 2.00 m, and a tension of 48.0 N vibrates in its second (n 5 2) normal mode. (a) Is the wavelength in air of the sound emitted by this vibrating string larger or smaller than the wavelength of the wave on the string? (b) What is the ratio of the wavelength in air of the sound emitted by this vibrating string and the wavelength of the wave on the string? 69. A quartz watch contains a crystal oscillator in the form of a block of quartz that vibrates by contracting and expanding. An electric circuit feeds in energy to main-tain the oscillation and also counts the voltage pulses to keep time. Two opposite faces of the block, 7.05 mm apart, are antinodes, moving alternately toward each other and away from each other. The plane halfway between these two faces is a node of the vibration. The speed of sound in quartz is equal to 3.70 3 103 m/s. Find the frequency of the vibration. 70. Review. For the arrangement shown in Figure P18.70, the inclined plane and the small pulley are frictionless; the string supports the object of mass M at the bottom of the plane; and the string has mass m. The system is in equilibrium, and the vertical part of the string has a length h. We wish to study standing waves set up in the vertical section of the string. (a) What analysis model describes the object of mass M? (b) What analy-sis model describes the waves on the vertical part of the Q/C GP between the tuning fork and its echo? (b) How fast must he walk away from the wall to observe a beat fre-quency of 5.00 Hz?
Section 18.8 Nonsinusoidal Wave Patterns 60. An A-major chord consists of the notes called A, C#, and E. It can be played on a piano by simultaneously striking strings with fundamental frequencies of 440.00 Hz, 554.37 Hz, and 659.26 Hz. The rich con-sonance of the chord is associated with near equality of the frequencies of some of the higher harmonics of the three tones. Consider the first five harmonics of each string and determine which harmonics show near equality. 61. Suppose a flutist plays a 523-Hz C note with first har-monic displacement amplitude A1 5 100 nm. From Fig-ure 18.19b read, by proportion, the displacement ampli-tudes of harmonics 2 through 7. Take these as the values A2 through A7 in the Fourier analysis of the sound and assume B1 5 B2 5 ??? 5 B 7 5 0. Construct a graph of the waveform of the sound. Your waveform will not look exactly like the flute waveform in Figure 18.18b because you simplify by ignoring cosine terms; nevertheless, it produces the same sensation to human hearing.
Additional Problems 62. A pipe open at both ends has a fundamental frequency of 300 Hz when the temperature is 0°C. (a) What is the length of the pipe? (b) What is the fundamental fre-quency at a temperature of 30.0°C? 63. A string is 0.400 m long and has a mass per unit length of 9.00 3 10–3 kg/m. What must be the tension in the string if its second harmonic has the same frequency as the second resonance mode of a 1.75-m-long pipe open at one end? 64. Two strings are vibrating at the same frequency of 150 Hz. After the tension in one of the strings is decreased, an observer hears four beats each second when the strings vibrate together. Find the new fre-quency in the adjusted string. 65. The ship in Figure P18.65 travels along a straight line parallel to the shore and a distance d 5 600 m from it. The ship’s radio receives simultaneous signals of the same frequency from antennas A and B, separated by a distance L 5 800 m. The signals interfere construc-tively at point C, which is equidistant from A and B. The signal goes through the first minimum at point D, which is directly outward from the shore from point B. Determine the wavelength of the radio waves.
M d L C D A B Figure P18.65 M h u Figure P18.70 www.aswarphysics.weebly.com Problems 563 at this moment? Explain your answer. (c) What if? The experiment is repeated after more mass has been added to the yo-yo body. The mass distribution is kept the same so that the yo-yo still moves with downward acceleration 0.800 m/s2. At the 1.20-s point in this case, is the rate of change of the fundamental wavelength of the string vibration still equal to 1.92 m/s? Explain. (d) Is the rate of change of the second harmonic wavelength the same as in part (b)? Explain. 75. On a marimba (Fig. P18.75), the wooden bar that sounds a tone when struck vibrates in a transverse standing wave having three antinodes and two nodes. The lowest- frequency note is 87.0 Hz, produced by a bar 40.0 cm long. (a) Find the speed of transverse waves on the bar. (b) A resonant pipe suspended verti-cally below the center of the bar enhances the loudness of the emitted sound. If the pipe is open at the top end only, what length of the pipe is required to resonate with the bar in part (a)?
Figure P18.75 © ArenaPal/Topham/The Image Works. Reproduced by permission. 76. A nylon string has mass 5.50 g and length L 5 86.0 cm. The lower end is tied to the floor, and the upper end is tied to a small set of wheels through a slot in a track on which the wheels move (Fig. P18.76). The wheels have a mass that is negli-gible compared with that of the string, and they roll without fric-tion on the track so that the upper end of the string is essentially free. At equilibrium, the string is vertical and motionless. When it is carrying a small-amplitude wave, you may assume the string is always under uni-form tension 1.30 N. (a) Find the speed of transverse waves on the string. (b) The string’s vibration pos-sibilities are a set of standing-wave states, each with a node at the fixed bottom end and an antinode at the free top end. Find the node–antinode distances for each of the three simplest states. (c) Find the fre-quency of each of these states. 77. Two train whistles have identical frequencies of 180 Hz. When one train is at rest in the station and the other is moving nearby, a commuter standing on the station platform hears beats with a frequency of 2.00 beats/s when the whistles operate together. What L Figure P18.76 M string? (c) Find the tension in the string. (d) Model the shape of the string as one leg and the hypotenuse of a right triangle. Find the whole length of the string. (e) Find the mass per unit length of the string. (f) Find the speed of waves on the string. (g) Find the lowest frequency for a standing wave on the vertical section of the string. (h) Evaluate this result for M 5 1.50 kg, m 5 0.750 g, h 5 0.500 m, and u 5 30.0°. (i) Find the numerical value for the lowest frequency for a standing wave on the sloped section of the string. 71. A 0.010 0-kg wire, 2.00 m long, is fixed at both ends and vibrates in its simplest mode under a tension of 200 N. When a vibrating tuning fork is placed near the wire, a beat frequency of 5.00 Hz is heard. (a) What could be the frequency of the tuning fork? (b) What should the tension in the wire be if the beats are to disappear? 72. Two speakers are driven by the same oscillator of fre-quency f. They are located a distance d from each other on a vertical pole. A man walks straight toward the lower speaker in a direction perpendicular to the pole as shown in Figure P18.72. (a) How many times will he hear a minimum in sound intensity? (b) How far is he from the pole at these moments? Let v repre-sent the speed of sound and assume that the ground does not reflect sound. The man’s ears are at the same level as the lower speaker.
d L Figure P18.72 73. Review. Consider the apparatus shown in Figure 18.11 and described in Example 18.4. Suppose the number of antinodes in Figure 18.11b is an arbitrary value n. (a) Find an expression for the radius of the sphere in the water as a function of only n. (b) What is the mini-mum allowed value of n for a sphere of nonzero size? (c) What is the radius of the largest sphere that will produce a standing wave on the string? (d) What hap-pens if a larger sphere is used? 74. Review. The top end of a yo-yo string is held stationary. The yo-yo itself is much more massive than the string. It starts from rest and moves down with constant accelera-tion 0.800 m/s2 as it unwinds from the string. The rub-bing of the string against the edge of the yo-yo excites transverse standing-wave vibrations in the string. Both ends of the string are nodes even as the length of the string increases. Consider the instant 1.20 s after the motion begins from rest. (a) Show that the rate of change with time of the wavelength of the fundamental mode of oscillation is 1.92 m/s. (b) What if? Is the rate of change of the wavelength of the second harmonic also 1.92 m/s Q/C Q/C www.aswarphysics.weebly.com 564 Chapter 18 Superposition and Standing Waves (b) Determine the amplitude and phase angle for this sinusoidal wave. 84. A flute is designed so that it produces a frequency of 261.6 Hz, middle C, when all the holes are covered and the temperature is 20.0°C. (a) Consider the flute as a pipe that is open at both ends. Find the length of the flute, assuming middle C is the fundamental. (b) A sec-ond player, nearby in a colder room, also attempts to play middle C on an identical flute. A beat frequency of 3.00 Hz is heard when both flutes are playing. What is the temperature of the second room? 85. Review. A 12.0-kg object hangs in equilibrium from a string with a total length of L 5 5.00 m and a linear mass density of m 5 0.001 00 kg/m. The string is wrapped around two light, frictionless pulleys that are separated by a distance of d 5 2.00 m (Fig. P18.85a). (a) Deter-mine the tension in the string. (b) At what frequency must the string between the pulleys vibrate to form the standing-wave pattern shown in Figure P18.85b?
g S m d m d a b Figure P18.85 Problems 85 and 86. 86. Review. An object of mass m hangs in equilibrium from a string with a total length L and a linear mass density m. The string is wrapped around two light, frictionless pulleys that are separated by a distance d (Fig. P18.85a). (a) Determine the tension in the string. (b) At what frequency must the string between the pul-leys vibrate to form the standing-wave pattern shown in Figure P18.85b?
Challenge Problems 87. Review. Consider the apparatus shown in Figure P18.87a, where the hanging object has mass M and the string is vibrating in its second harmonic. The vibrat-ing blade at the left maintains a constant frequency. The wind begins to blow to the right, applying a con-AMT S S are the two possible speeds and directions the moving train can have? 78. Review. A loudspeaker at the front of a room and an identical loudspeaker at the rear of the room are being driven by the same oscillator at 456 Hz. A student walks at a uniform rate of 1.50 m/s along the length of the room. She hears a single tone repeatedly becom-ing louder and softer. (a) Model these variations as beats between the Doppler-shifted sounds the student receives. Calculate the number of beats the student hears each second. (b) Model the two speakers as pro-ducing a standing wave in the room and the student as walking between antinodes. Calculate the number of intensity maxima the student hears each second. 79. Review. Consider the copper object hanging from the steel wire in Problem 32. The top end of the wire is fixed. When the wire is struck, it emits sound with a fundamental frequency of 300 Hz. The copper object is then submerged in water. If the object can be positioned with any desired fraction of its volume submerged, what is the lowest possible new fundamental frequency? 80. Two wires are welded together end to end. The wires are made of the same material, but the diameter of one is twice that of the other. They are subjected to a ten-sion of 4.60 N. The thin wire has a length of 40.0 cm and a linear mass density of 2.00 g/m. The combina-tion is fixed at both ends and vibrated in such a way that two antinodes are present, with the node between them being right at the weld. (a) What is the frequency of vibration? (b) What is the length of the thick wire? 81. A string of linear density 1.60 g/m is stretched between clamps 48.0 cm apart. The string does not stretch appreciably as the tension in it is steadily raised from 15.0 N at t 5 0 to 25.0 N at t 5 3.50 s. Therefore, the tension as a function of time is given by the expression T 5 15.0 1 10.0t/3.50, where T is in newtons and t is in seconds. The string is vibrating in its fundamental mode throughout this process. Find the number of oscillations it completes during the 3.50-s interval. 82. A standing wave is set up in a string of variable length and tension by a vibrator of variable frequency. Both ends of the string are fixed. When the vibrator has a frequency f, in a string of length L and under tension T, n antinodes are set up in the string. (a) If the length of the string is doubled, by what factor should the fre-quency be changed so that the same number of anti-nodes is produced? (b) If the frequency and length are held constant, what tension will produce n 1 1 anti-nodes? (c) If the frequency is tripled and the length of the string is halved, by what factor should the tension be changed so that twice as many antinodes are produced? 83. Two waves are described by the wave functions y1(x, t) 5 5.00 sin (2.00x 2 10.0t) y2(x, t) 5 10.0 cos (2.00x 2 10.0t) where x, y1, and y2 are in meters and t is in seconds. (a) Show that the wave resulting from their super-position can be expressed as a single sine function. M S M M F S a b Figure P18.87 www.aswarphysics.weebly.com Problems 565 Express Equation 18.13 with angular frequencies: y 1t 2 5 a n 1An sin nvt 1 Bn cos nvt2 Now proceed as follows. (a) Multiply both sides of Equa-tion 18.13 by sin mvt and integrate both sides over one period T. Show that the left-hand side of the resulting equation is equal to 0 if m is even and is equal to 4A/mv if m is odd. (b) Using trigonometric identities, show that all terms on the right-hand side involving Bn are equal to zero. (c) Using trigonometric identities, show that all terms on the right-hand side involving An are equal to zero except for the one case of m 5 n. (d) Show that the entire right-hand side of the equation reduces to 1 2AmT. (e) Show that the Fourier series expansion for a square wave is y 1t 2 5 a n 4A np sin nvt stant horizontal force F S on the hanging object. What is the magnitude of the force the wind must apply to the hanging object so that the string vibrates in its first harmonic as shown in Figure 18.87b? 88. In Figures 18.20a and 18.20b, notice that the ampli-tude of the component wave for frequency f is large, that for 3f is smaller, and that for 5f smaller still. How do we know exactly how much amplitude to assign to each frequency component to build a square wave? This problem helps us find the answer to that question. Let the square wave in Figure 18.20c have an ampli-tude A and let t 5 0 be at the extreme left of the figure. So, one period T of the square wave is described by y1t2 5 µ A 0 , t , T 2 2A T 2 , t , T S www.aswarphysics.weebly.com www.aswarphysics.weebly.com 567 Thermodynamics A bubble in one of the many mud pots in Yellowstone National Park is caught just at the moment of popping. A mud pot is a pool of bubbling hot mud that demonstrates the existence of thermodynamic processes below the Earth’s surface. (© Adambooth/ Dreamstime.com) p a r t 3 We now direct our attention to the study of thermodynamics, which involves situ-ations in which the temperature or state (solid, liquid, gas) of a system changes due to energy transfers. As we shall see, thermodynamics is very successful in explaining the bulk properties of matter and the correlation between these properties and the mechanics of atoms and molecules. Historically, the development of thermodynamics paralleled the development of the atomic the-ory of matter. By the 1820s, chemical experiments had provided solid evidence for the existence of atoms. At that time, scientists recognized that a connection between thermodynamics and the structure of matter must exist. In 1827, botanist Robert Brown reported that grains of pollen sus-pended in a liquid move erratically from one place to another as if under constant agitation. In 1905, Albert Einstein used kinetic theory to explain the cause of this erratic motion, known today as Brownian motion. Einstein explained this phenomenon by assuming the grains are under constant bombardment by “invisible” molecules in the liquid, which themselves move erratically. This expla-nation gave scientists insight into the concept of molecular motion and gave credence to the idea that matter is made up of atoms. A connection was thus forged between the everyday world and the tiny, invisible building blocks that make up this world. Thermodynamics also addresses more practical questions. Have you ever wondered how a refrig-erator is able to cool its contents, or what types of transformations occur in a power plant or in the engine of your automobile, or what happens to the kinetic energy of a moving object when the object comes to rest? The laws of thermodynamics can be used to provide explanations for these and other phenomena. ■ www.aswarphysics.weebly.com Why would someone designing a pipeline include these strange loops? Pipelines carrying liquids often contain such loops to allow for expansion and contraction as the temperature changes. We will study thermal expansion in this chapter. (© Lowell Georgia/CORBIS) 19.1 Temperature and the Zeroth Law of Thermodynamics 19.2 Thermometers and the Celsius Temperature Scale 19.3 The Constant-Volume Gas Thermometer and the Absolute Temperature Scale 19.4 Thermal Expansion of Solids and Liquids 19.5 Macroscopic Description of an Ideal Gas c h a p t e r 19 Temperature In our study of mechanics, we carefully defined such concepts as mass, force, and kinetic energy to facilitate our quantitative approach. Likewise, a quantitative description of thermal phenomena requires careful definitions of such important terms as temperature, heat, and internal energy. This chapter begins with a discussion of temperature. Next, when studying thermal phenomena, we consider the importance of the particu-lar substance we are investigating. For example, gases expand appreciably when heated, whereas liquids and solids expand only slightly. This chapter concludes with a study of ideal gases on the macroscopic scale. Here, we are concerned with the relationships among such quantities as pressure, volume, and tempera-ture of a gas. In Chapter 21, we shall examine gases on a microscopic scale, using a model that represents the components of a gas as small particles.
19.1 Temperature and the Zeroth Law of Thermodynamics We often associate the concept of temperature with how hot or cold an object feels when we touch it. In this way, our senses provide us with a qualitative indication of temperature. Our senses, however, are unreliable and often mislead us. For exam-568 www.aswarphysics.weebly.com 19.1 Temperature and the Zeroth Law of Thermodynamics 569 ple, if you stand in bare feet with one foot on carpet and the other on an adjacent tile floor, the tile feels colder than the carpet even though both are at the same tem-perature. The two objects feel different because tile transfers energy by heat at a higher rate than carpet does. Your skin “measures” the rate of energy transfer by heat rather than the actual temperature. What we need is a reliable and reproduc-ible method for measuring the relative hotness or coldness of objects rather than the rate of energy transfer. Scientists have developed a variety of thermometers for making such quantitative measurements. Two objects at different initial temperatures eventually reach some intermediate temperature when placed in contact with each other. For example, when hot water and cold water are mixed in a bathtub, energy is transferred from the hot water to the cold water and the final temperature of the mixture is somewhere between the initial hot and cold temperatures. Imagine that two objects are placed in an insulated container such that they interact with each other but not with the environment. If the objects are at differ-ent temperatures, energy is transferred between them, even if they are initially not in physical contact with each other. The energy-transfer mechanisms from Chap-ter 8 that we will focus on are heat and electromagnetic radiation. For purposes of this discussion, let’s assume two objects are in thermal contact with each other if energy can be exchanged between them by these processes due to a tempera-ture difference. Thermal equilibrium is a situation in which two objects would not exchange energy by heat or electromagnetic radiation if they were placed in ther-mal contact. Let’s consider two objects A and B, which are not in thermal contact, and a third object C, which is our thermometer. We wish to determine whether A and B are in thermal equilibrium with each other. The thermometer (object C) is first placed in thermal contact with object A until thermal equilibrium is reached1 as shown in Figure 19.1a. From that moment on, the thermometer’s reading remains constant and we record this reading. The thermometer is then removed from object A and placed in thermal contact with object B as shown in Figure 19.1b. The reading is again recorded after thermal equilibrium is reached. If the two readings are the same, we can conclude that object A and object B are in thermal equilibrium with each other. If they are placed in contact with each other as in Figure 19.1c, there is no exchange of energy between them.
A C C A B The temperatures of A and B are measured to be the same by placing them in thermal contact with a thermometer (object C).
No energy will be exchanged between A and B when they are placed in thermal contact with each other.
B a b c Figure 19.1 The zeroth law of thermodynamics.
1We assume a negligible amount of energy transfers between the thermometer and object A in the time interval dur-ing which they are in thermal contact. Without this assumption, which is also made for the thermometer and object B, the measurement of the temperature of an object disturbs the system so that the measured temperature is differ-ent from the initial temperature of the object. In practice, whenever you measure a temperature with a thermometer, you measure the disturbed system, not the original system.
www.aswarphysics.weebly.com 570 Chapter 19 Temperature We can summarize these results in a statement known as the zeroth law of ther-modynamics (the law of equilibrium): If objects A and B are separately in thermal equilibrium with a third object C, then A and B are in thermal equilibrium with each other.
This statement can easily be proved experimentally and is very important because it enables us to define temperature. We can think of temperature as the property that determines whether an object is in thermal equilibrium with other objects. Two objects in thermal equilibrium with each other are at the same temperature. Conversely, if two objects have different temperatures, they are not in thermal equilibrium with each other. We now know that temperature is something that determines whether or not energy will transfer between two objects in thermal contact. In Chapter 21, we will relate temperature to the mechanical behavior of molecules.
Q uick Quiz 19.1 Two objects, with different sizes, masses, and temperatures, are placed in thermal contact. In which direction does the energy travel? (a) Energy travels from the larger object to the smaller object. (b) Energy travels from the object with more mass to the one with less mass. (c) Energy travels from the object at higher temperature to the object at lower temperature.
19.2 Thermometers and the Celsius Temperature Scale Thermometers are devices used to measure the temperature of a system. All ther-mometers are based on the principle that some physical property of a system changes as the system’s temperature changes. Some physical properties that change with temperature are (1) the volume of a liquid, (2) the dimensions of a solid, (3) the pressure of a gas at constant volume, (4) the volume of a gas at constant pressure, (5) the electric resistance of a conductor, and (6) the color of an object. A common thermometer in everyday use consists of a mass of liquid—usually mercury or alcohol—that expands into a glass capillary tube when heated (Fig. 19.2). In this case, the physical property that changes is the volume of a liquid. Any temperature change in the range of the thermometer can be defined as being proportional to the change in length of the liquid column. The thermometer can be calibrated by placing it in thermal contact with a natural system that remains Zeroth law of thermodynamics Figure 19.2 A mercury ther-mometer before and after increas-ing its temperature.
20 C 30 C The level of the mercury in the thermometer rises as the mercury is heated by water in the test tube.
© Cengage Learning/Charles D. Winters www.aswarphysics.weebly.com 19.3 The Constant-Volume Gas Thermometer and the Absolute Temperature Scale 571 at constant temperature. One such system is a mixture of water and ice in thermal equilibrium at atmospheric pressure. On the Celsius temperature scale, this mix-ture is defined to have a temperature of zero degrees Celsius, which is written as 08C; this temperature is called the ice point of water. Another commonly used system is a mixture of water and steam in thermal equilibrium at atmospheric pressure; its temperature is defined as 1008C, which is the steam point of water. Once the liquid levels in the thermometer have been established at these two points, the length of the liquid column between the two points is divided into 100 equal segments to cre-ate the Celsius scale. Therefore, each segment denotes a change in temperature of one Celsius degree. Thermometers calibrated in this way present problems when extremely accurate readings are needed. For instance, the readings given by an alcohol thermometer calibrated at the ice and steam points of water might agree with those given by a mercury thermometer only at the calibration points. Because mercury and alcohol have different thermal expansion properties, when one thermometer reads a tem-perature of, for example, 508C, the other may indicate a slightly different value. The discrepancies between thermometers are especially large when the tempera-tures to be measured are far from the calibration points.2 An additional practical problem of any thermometer is the limited range of tem-peratures over which it can be used. A mercury thermometer, for example, cannot be used below the freezing point of mercury, which is 2398C, and an alcohol ther-mometer is not useful for measuring temperatures above 858C, the boiling point of alcohol. To surmount this problem, we need a universal thermometer whose read-ings are independent of the substance used in it. The gas thermometer, discussed in the next section, approaches this requirement.
19.3 The Constant-Volume Gas Thermometer and the Absolute Temperature Scale One version of a gas thermometer is the constant-volume apparatus shown in Fig-ure 19.3. The physical change exploited in this device is the variation of pressure of a fixed volume of gas with temperature. The flask is immersed in an ice-water bath, and mercury reservoir B is raised or lowered until the top of the mercury in column A is at the zero point on the scale. The height h, the difference between the mercury levels in reservoir B and column A, indicates the pressure in the flask at 08C by means of Equation 14.4, P 5 P0 1 rgh. The flask is then immersed in water at the steam point. Reservoir B is read-justed until the top of the mercury in column A is again at zero on the scale, which ensures that the gas’s volume is the same as it was when the flask was in the ice bath (hence the designation “constant-volume”). This adjustment of reservoir B gives a value for the gas pressure at 1008C. These two pressure and temperature values are then plotted as shown in Figure 19.4. The line connecting the two points serves as a calibration curve for unknown temperatures. (Other experiments show that a linear relationship between pressure and temperature is a very good assumption.) To measure the temperature of a substance, the gas flask of Figure 19.3 is placed in thermal contact with the substance and the height of reservoir B is adjusted until the top of the mercury column in A is at zero on the scale. The height of the mer-cury column in B indicates the pressure of the gas; knowing the pressure, the tem-perature of the substance is found using the graph in Figure 19.4. Now suppose temperatures of different gases at different initial pressures are measured with gas thermometers. Experiments show that the thermometer readings are nearly independent of the type of gas used as long as the gas pres-sure is low and the temperature is well above the point at which the gas liquefies 2Two thermometers that use the same liquid may also give different readings, due in part to difficulties in construct-ing uniform-bore glass capillary tubes.
A B The volume of gas in the flask is kept constant by raising or lowering reservoir B to keep the mercury level in column A constant.
h Scale 0 Mercury reservoir Flexible hose Bath or environment to be measured P Gas Figure 19.3 A constant-volume gas thermometer measures the pressure of the gas contained in the flask immersed in the bath.
100 0 T ( C) P The two dots represent known reference temperatures (the ice and steam points of water).
Figure 19.4 A typical graph of pressure versus temperature taken with a constant-volume gas thermometer.
www.aswarphysics.weebly.com 572 Chapter 19 Temperature (Fig. 19.5). The agreement among thermometers using various gases improves as the pressure is reduced. If we extend the straight lines in Figure 19.5 toward negative temperatures, we find a remarkable result: in every case, the pressure is zero when the temperature is 2273.158C! This finding suggests some special role that this particular tempera-ture must play. It is used as the basis for the absolute temperature scale, which sets 2273.158C as its zero point. This temperature is often referred to as absolute zero. It is indicated as a zero because at a lower temperature, the pressure of the gas would become negative, which is meaningless. The size of one degree on the absolute tem-perature scale is chosen to be identical to the size of one degree on the Celsius scale. Therefore, the conversion between these temperatures is TC 5 T 2 273.15 (19.1) where TC is the Celsius temperature and T is the absolute temperature. Because the ice and steam points are experimentally difficult to duplicate and depend on atmospheric pressure, an absolute temperature scale based on two new fixed points was adopted in 1954 by the International Committee on Weights and Measures. The first point is absolute zero. The second reference temperature for this new scale was chosen as the triple point of water, which is the single combination of temperature and pressure at which liquid water, gaseous water, and ice (solid water) coexist in equilibrium. This triple point occurs at a temperature of 0.018C and a pres-sure of 4.58 mm of mercury. On the new scale, which uses the unit kelvin, the tem-perature of water at the triple point was set at 273.16 kelvins, abbreviated 273.16 K. This choice was made so that the old absolute temperature scale based on the ice and steam points would agree closely with the new scale based on the triple point. This new absolute temperature scale (also called the Kelvin scale) employs the SI unit of absolute temperature, the kelvin, which is defined to be 1/273.16 of the dif-ference between absolute zero and the temperature of the triple point of water. Figure 19.6 gives the absolute temperature for various physical processes and structures. The temperature of absolute zero (0 K) cannot be achieved, although laboratory experiments have come very close, reaching temperatures of less than one nanokelvin.
The Celsius, Fahrenheit, and Kelvin Temperature Scales3 Equation 19.1 shows that the Celsius temperature TC is shifted from the absolute (Kelvin) temperature T by 273.158. Because the size of one degree is the same on the two scales, a temperature difference of 58C is equal to a temperature difference of 5 K. The two scales differ only in the choice of the zero point. Therefore, the ice-point temperature on the Kelvin scale, 273.15 K, corresponds to 0.008C, and the Kelvin-scale steam point, 373.15 K, is equivalent to 100.008C. A common temperature scale in everyday use in the United States is the Fahren-heit scale. This scale sets the temperature of the ice point at 328F and the tempera-ture of the steam point at 2128F. The relationship between the Celsius and Fahrenheit temperature scales is TF 5 9 5TC 1 328F (19.2) We can use Equations 19.1 and 19.2 to find a relationship between changes in tem-perature on the Celsius, Kelvin, and Fahrenheit scales: DTC 5 DT 5 5 9DTF (19.3) Of these three temperature scales, only the Kelvin scale is based on a true zero value of temperature. The Celsius and Fahrenheit scales are based on an arbitrary zero associated with one particular substance, water, on one particular planet, the Trial 2 Trial 3 Trial 1 P 200 T ( C) 100 0 100 200 For all three trials, the pressure extrapolates to zero at the temperature 273.15 C.
Figure 19.5 Pressure versus temperature for experimental tri-als in which gases have different pressures in a constant-volume gas thermometer.
Pitfall Prevention 19.1 A Matter of Degree Notations for temperatures in the Kelvin scale do not use the degree sign. The unit for a Kelvin temperature is simply “kelvins” and not “degrees Kelvin.” 3Named after Anders Celsius (1701–1744), Daniel Gabriel Fahrenheit (1686–1736), and William Thomson, Lord Kel-vin (1824–1907), respectively.
Hydrogen bomb 109 108 107 106 105 104 103 102 10 1 Interior of the Sun Solar corona Surface of the Sun Copper melts Water freezes Liquid nitrogen Liquid hydrogen Liquid helium Lowest temperature achieved ˜ 10–9 K Temperature (K) Note that the scale is logarithmic.
Figure 19.6 Absolute tempera-tures at which various physical processes occur.
www.aswarphysics.weebly.com 19.4 Thermal Expansion of Solids and Liquids 573 Earth. Therefore, if you encounter an equation that calls for a temperature T or that involves a ratio of temperatures, you must convert all temperatures to kelvins. If the equation contains a change in temperature DT, using Celsius temperatures will give you the correct answer, in light of Equation 19.3, but it is always safest to convert temperatures to the Kelvin scale.
Q uick Quiz 19.2 Consider the following pairs of materials. Which pair repre-sents two materials, one of which is twice as hot as the other? (a) boiling water at 1008C, a glass of water at 508C (b) boiling water at 1008C, frozen methane at 2508C (c) an ice cube at 2208C, flames from a circus fire-eater at 2338C (d) none of those pairs Example 19.1 Converting Temperatures On a day when the temperature reaches 508F, what is the temperature in degrees Celsius and in kelvins?
Conceptualize In the United States, a temperature of 508F is well understood. In many other parts of the world, how-ever, this temperature might be meaningless because people are familiar with the Celsius temperature scale.
Categorize This example is a simple substitution problem.
S o l u t i o n Solve Equation 19.2 for the Celsius temperature and sub-stitute numerical values: TC 5 5 9 1TF 2 322 5 5 9 150 2 322 5 108C Use Equation 19.1 to find the Kelvin temperature: T 5 TC 1 273.15 5 108C 1 273.15 5 283 K A convenient set of weather-related temperature equivalents to keep in mind is that 08C is (literally) freezing at 328F, 108C is cool at 508F, 208C is room temperature, 308C is warm at 868F, and 408C is a hot day at 1048F.
19.4 Thermal Expansion of Solids and Liquids Our discussion of the liquid thermometer makes use of one of the best-known changes in a substance: as its temperature increases, its volume increases. This phe-nomenon, known as thermal expansion, plays an important role in numerous engi-neering applications. For example, thermal-expansion joints such as those shown in Figure 19.7 must be included in buildings, concrete highways, railroad tracks, Figure 19.7 Thermal-expansion joints in (a) bridges and (b) walls.
The long, vertical joint is filled with a soft material that allows the wall to expand and contract as the temperature of the bricks changes.
b © Cengage Learning/George Semple Without these joints to separate sections of roadway on bridges, the surface would buckle due to thermal expansion on very hot days or crack due to contraction on very cold days.
a © Cengage Learning/George Semple www.aswarphysics.weebly.com 574 Chapter 19 Temperature brick walls, and bridges to compensate for dimensional changes that occur as the temperature changes. Thermal expansion is a consequence of the change in the average separation between the atoms in an object. To understand this concept, let’s model the atoms as being connected by stiff springs as discussed in Section 15.3 and shown in Figure 15.11b. At ordinary temperatures, the atoms in a solid oscillate about their equi-librium positions with an amplitude of approximately 10211 m and a frequency of approximately 1013 Hz. The average spacing between the atoms is about 10210 m. As the temperature of the solid increases, the atoms oscillate with greater ampli-tudes; as a result, the average separation between them increases.4 Consequently, the object expands. If thermal expansion is sufficiently small relative to an object’s initial dimen-sions, the change in any dimension is, to a good approximation, proportional to the first power of the temperature change. Suppose an object has an initial length Li along some direction at some temperature and the length changes by an amount DL for a change in temperature DT. Because it is convenient to consider the frac-tional change in length per degree of temperature change, we define the average coefficient of linear expansion as a ; DL/Li DT Experiments show that a is constant for small changes in temperature. For pur-poses of calculation, this equation is usually rewritten as DL 5 aLi DT (19.4) or as Lf 2 Li 5 aLi(Tf 2 Ti) (19.5) where Lf is the final length, Ti and Tf are the initial and final temperatures, respec-tively, and the proportionality constant a is the average coefficient of linear expan-sion for a given material and has units of (8C)21. Equation 19.4 can be used for both thermal expansion, when the temperature of the material increases, and thermal contraction, when its temperature decreases. It may be helpful to think of thermal expansion as an effective magnification or as a photographic enlargement of an object. For example, as a metal washer is heated (Fig. 19.8), all dimensions, including the radius of the hole, increase accord-ing to Equation 19.4. A cavity in a piece of material expands in the same way as if the cavity were filled with the material. Table 19.1 lists the average coefficients of linear expansion for various materi-als. For these materials, a is positive, indicating an increase in length with increas-ing temperature. That is not always the case, however. Some substances—calcite (CaCO3) is one example—expand along one dimension (positive a) and contract along another (negative a) as their temperatures are increased. Because the linear dimensions of an object change with temperature, it follows that surface area and volume change as well. The change in volume is propor-tional to the initial volume Vi and to the change in temperature according to the relationship DV 5 bVi DT (19.6) where b is the average coefficient of volume expansion. To find the relationship between b and a, assume the average coefficient of linear expansion of the solid is the same in all directions; that is, assume the material is isotropic. Consider a solid box of dimensions ,, w, and h. Its volume at some temperature Ti is Vi 5 ,wh. If the Thermal expansion in one dimension Thermal expansion in three dimensions 4More precisely, thermal expansion arises from the asymmetrical nature of the potential energy curve for the atoms in a solid as shown in Figure 15.11a. If the oscillators were truly harmonic, the average atomic separations would not change regardless of the amplitude of vibration.
Pitfall Prevention 19.2 Do Holes Become Larger or Smaller? When an object’s tem-perature is raised, every linear dimension increases in size. That includes any holes in the material, which expand in the same way as if the hole were filled with the material as shown in Figure 19.8.
www.aswarphysics.weebly.com 19.4 Thermal Expansion of Solids and Liquids 575 temperature changes to Ti 1 DT, its volume changes to Vi 1 DV, where each dimen-sion changes according to Equation 19.4. Therefore, Vi 1 DV 5 (, 1 D,)(w 1 Dw)(h 1 Dh) 5 (, 1 a, DT )(w 1 aw DT )(h 1 ah DT ) 5 ,wh(1 1 a DT)3 5 Vi[1 1 3a DT 1 3(a DT )2 1 (a DT )3] Dividing both sides by Vi and isolating the term DV/Vi, we obtain the fractional change in volume: DV Vi 5 3a DT 1 31a DT2 2 1 1a DT2 3 Because a DT ,, 1 for typical values of DT (, , 1008C), we can neglect the terms 3(a DT)2 and (a DT)3. Upon making this approximation, we see that DV Vi 5 3a DT S DV 5 13a2Vi DT Comparing this expression to Equation 19.6 shows that b 5 3a In a similar way, you can show that the change in area of a rectangular plate is given by DA 5 2aAi DT (see Problem 61). A simple mechanism called a bimetallic strip, found in practical devices such as mechanical thermostats, uses the difference in coefficients of expansion for differ-ent materials. It consists of two thin strips of dissimilar metals bonded together. As the temperature of the strip increases, the two metals expand by different amounts and the strip bends as shown in Figure 19.9.
Q uick Quiz 19.3 If you are asked to make a very sensitive glass thermometer, which of the following working liquids would you choose? (a) mercury (b) alco-hol (c) gasoline (d) glycerin Q uick Quiz 19.4 Two spheres are made of the same metal and have the same radius, but one is hollow and the other is solid. The spheres are taken through the same temperature increase. Which sphere expands more? (a) The solid sphere expands more. (b) The hollow sphere expands more. (c) They expand by the same amount. (d) There is not enough information to say.
Table 19.1 Average Expansion Coefficients for Some Materials Near Room Temperature Average Linear Average Volume Expansion Expansion Material Coefficient Material Coefficient (Solids) (a)(°C)21 (Liquids and Gases) (b)(°C)21 Aluminum 24 3 1026 Acetone 1.5 3 1024 Brass and bronze 19 3 1026 Alcohol, ethyl 1.12 3 1024 Concrete 12 3 1026 Benzene 1.24 3 1024 Copper 17 3 1026 Gasoline 9.6 3 1024 Glass (ordinary) 9 3 1026 Glycerin 4.85 3 1024 Glass (Pyrex) 3.2 3 1026 Mercury 1.82 3 1024 Invar (Ni–Fe alloy) 0.9 3 1026 Turpentine 9.0 3 1024 Lead 29 3 1026 Aira at 08C 3.67 3 1023 Steel 11 3 1026 Heliuma 3.665 3 1023 aGases do not have a specific value for the volume expansion coefficient because the amount of expansion depends on the type of process through which the gas is taken. The values given here assume the gas undergoes an expansion at constant pressure.
Steel Brass Room temperature Higher temperature Bimetallic strip Off 30 C On 25 C a b Figure 19.9 (a) A bimetallic strip bends as the temperature changes because the two metals have different expansion coeffi-cients. (b) A bimetallic strip used in a thermostat to break or make electrical contact.
Figure 19.8 Thermal expansion of a homogeneous metal washer. (The expansion is exaggerated in this figure.) a b b b a a Ti T Ti As the washer is heated, all dimensions increase, including the radius of the hole.
www.aswarphysics.weebly.com 576 Chapter 19 Temperature Example 19.2 Expansion of a Railroad Track A segment of steel railroad track has a length of 30.000 m when the temperature is 0.08C.
(A) What is its length when the temperature is 40.08C?
Conceptualize Because the rail is relatively long, we expect to obtain a measurable increase in length for a 408C tem-perature increase.
Categorize We will evaluate a length increase using the discussion of this section, so this part of the example is a sub-stitution problem.
S o l u t i o n Use Equation 19.4 and the value of the coeffi-cient of linear expansion from Table 19.1: DL 5 aLi DT 5 11 3 1026 (8C)21(40.08C) 5 0.013 m Find the new length of the track: Lf 5 30.000 m 1 0.013 m 5 30.013 m Find the tensile stress from Equation 12.6 using Young’s modulus for steel from Table 12.1: Tensile stress 5 F A 5 Y DL L i F A 5 120 3 1010 N/m22 a 0.013 m 30.000 mb 5 8.7 3 107 N/m2 (B) Suppose the ends of the rail are rigidly clamped at 0.08C so that expansion is prevented. What is the thermal stress set up in the rail if its temperature is raised to 40.08C?
Categorize This part of the example is an analysis problem because we need to use concepts from another chapter.
Analyze The thermal stress is the same as the tensile stress in the situation in which the rail expands freely and is then compressed with a mechanical force F back to its original length.
S o l u t i o n Finalize The expansion in part (A) is 1.3 cm. This expansion is indeed measurable as predicted in the Conceptualize step. The thermal stress in part (B) can be avoided by leaving small expansion gaps between the rails.
What if the temperature drops to 240.08C? What is the length of the unclamped segment?
Answer The expression for the change in length in Equation 19.4 is the same whether the temperature increases or decreases. Therefore, if there is an increase in length of 0.013 m when the temperature increases by 408C, there is a decrease in length of 0.013 m when the temperature decreases by 408C. (We assume a is constant over the entire range of temperatures.) The new length at the colder temperature is 30.000 m 2 0.013 m 5 29.987 m.
What If?
Example 19.3 The Thermal Electrical Short A poorly designed electronic device has two bolts attached to different parts of the device that almost touch each other in its interior as in Figure 19.10. The steel and brass bolts are at different electric potentials, and if they touch, a short circuit will develop, damag-ing the device. (We will study electric potential in Chap-ter 25.) The initial gap between the ends of the bolts is d 5 5.0 mm at 278C. At what temperature will the bolts touch? Assume the distance between the walls of the device is not affected by the temperature change.
Conceptualize Imagine the ends of both bolts expanding into the gap between them as the temperature rises.
S o l u t i o n 0.010 m 0.030 m 5.0 mm Steel Brass Figure 19.10 (Example 19.3) Two bolts attached to different parts of an electrical device are almost touching when the temper-ature is 278C. As the temperature increases, the ends of the bolts move toward each other.
www.aswarphysics.weebly.com 19.4 Thermal Expansion of Solids and Liquids 577 Finalize This temperature is possible if the air conditioning in the building housing the device fails for a long period on a very hot summer day.
Analyze Set the sum of the length changes equal to the width of the gap: DL br 1 DL st 5 abrLi,br DT 1 astLi,st DT 5 d Solve for DT: Substitute numerical values: DT 5 d abrL i,br 1 astL i,st DT 5 5.0 3 1026 m 319 3 1026 18C2 214 10.030 m2 1 311 3 1026 18C2 214 10.010 m2 5 7.48C Find the temperature at which the bolts touch: T 5 278C 1 7.48C 5 348C 1.00 0.99 0.98 0.97 0.96 0.95 0 20 40 60 80 100 Temperature ( C) 0.999 9 0 1.000 0 0.999 8 0.999 7 0.999 6 0.999 5 2 6 4 8 10 12 Temperature ( C) (g/cm3) r (g/cm3) r This blown-up portion of the graph shows that the maximum density of water occurs at 4 C.
Figure 19.11 The variation in the density of water at atmospheric pressure with temperature. ▸ 19.3 c on tin u ed Categorize We categorize this example as a thermal expansion problem in which the sum of the changes in length of the two bolts must equal the length of the initial gap between the ends.
The Unusual Behavior of Water Liquids generally increase in volume with increasing temperature and have aver-age coefficients of volume expansion about ten times greater than those of solids. Cold water is an exception to this rule as you can see from its density-versus- temperature curve shown in Figure 19.11. As the temperature increases from 08C to 48C, water contracts and its density therefore increases. Above 48C, water expands with increasing temperature and so its density decreases. Therefore, the density of water reaches a maximum value of 1.000 g/cm3 at 48C. We can use this unusual thermal-expansion behavior of water to explain why a pond begins freezing at the surface rather than at the bottom. When the air tem-perature drops from, for example, 78C to 68C, the surface water also cools and con-sequently decreases in volume. The surface water is denser than the water below it, which has not cooled and decreased in volume. As a result, the surface water sinks, and warmer water from below moves to the surface. When the air temperature is between 48C and 08C, however, the surface water expands as it cools, becoming less dense than the water below it. The mixing process stops, and eventually the surface water freezes. As the water freezes, the ice remains on the surface because ice is less dense than water. The ice continues to build up at the surface, while water near the www.aswarphysics.weebly.com 578 Chapter 19 Temperature bottom remains at 48C. If that were not the case, fish and other forms of marine life would not survive.
19.5 Macroscopic Description of an Ideal Gas The volume expansion equation DV 5 bVi DT is based on the assumption that the material has an initial volume Vi before the temperature change occurs. Such is the case for solids and liquids because they have a fixed volume at a given temperature. The case for gases is completely different. The interatomic forces within gases are very weak, and, in many cases, we can imagine these forces to be nonexistent and still make very good approximations. Therefore, there is no equilibrium separation for the atoms and no “standard” volume at a given temperature; the volume depends on the size of the container. As a result, we cannot express changes in volume DV in a process on a gas with Equation 19.6 because we have no defined volume Vi at the beginning of the process. Equations involving gases contain the volume V, rather than a change in the volume from an initial value, as a variable. For a gas, it is useful to know how the quantities volume V, pressure P, and tem-perature T are related for a sample of gas of mass m. In general, the equation that interrelates these quantities, called the equation of state, is very complicated. If the gas is maintained at a very low pressure (or low density), however, the equation of state is quite simple and can be determined from experimental results. Such a low-density gas is commonly referred to as an ideal gas.5 We can use the ideal gas model to make predictions that are adequate to describe the behavior of real gases at low pressures. It is convenient to express the amount of gas in a given volume in terms of the number of moles n. One mole of any substance is that amount of the substance that contains Avogadro’s number NA 5 6.022 3 1023 of constituent particles (atoms or molecules). The number of moles n of a substance is related to its mass m through the expression n 5 m M (19.7) where M is the molar mass of the substance. The molar mass of each chemical element is the atomic mass (from the periodic table; see Appendix C) expressed in grams per mole. For example, the mass of one He atom is 4.00 u (atomic mass units), so the molar mass of He is 4.00 g/mol. Now suppose an ideal gas is confined to a cylindrical container whose volume can be varied by means of a movable piston as in Figure 19.12. If we assume the cyl-inder does not leak, the mass (or the number of moles) of the gas remains constant. For such a system, experiments provide the following information: • When the gas is kept at a constant temperature, its pressure is inversely propor-tional to the volume. (This behavior is described historically as Boyle’s law.) • When the pressure of the gas is kept constant, the volume is directly propor-tional to the temperature. (This behavior is described historically as Charles’s law.) • When the volume of the gas is kept constant, the pressure is directly propor-tional to the temperature. (This behavior is described historically as Gay– Lussac’s law.) These observations are summarized by the equation of state for an ideal gas: PV 5 nRT (19.8) Equation of state for an ideal gas 5To be more specific, the assumptions here are that the temperature of the gas must not be too low (the gas must not condense into a liquid) or too high and that the pressure must be low. The concept of an ideal gas implies that the gas molecules do not interact except upon collision and that the molecular volume is negligible compared with the volume of the container. In reality, an ideal gas does not exist. The concept of an ideal gas is nonetheless very useful because real gases at low pressures are well-modeled as ideal gases.
Figure 19.12 An ideal gas con-fined to a cylinder whose volume can be varied by means of a mov-able piston.
Gas www.aswarphysics.weebly.com 19.5 Macroscopic Description of an Ideal Gas 579 In this expression, also known as the ideal gas law, n is the number of moles of gas in the sample and R is a constant. Experiments on numerous gases show that as the pressure approaches zero, the quantity PV/nT approaches the same value R for all gases. For this reason, R is called the universal gas constant. In SI units, in which pressure is expressed in pascals (1 Pa 5 1 N/m2) and volume in cubic meters, the product PV has units of newton ? meters, or joules, and R has the value R 5 8.314 J/mol ? K (19.9) If the pressure is expressed in atmospheres and the volume in liters (1 L 5 103 cm3 5 1023 m3), then R has the value R 5 0.082 06 L ? atm/mol ? K Using this value of R and Equation 19.8 shows that the volume occupied by 1 mol of any gas at atmospheric pressure and at 08C (273 K) is 22.4 L. The ideal gas law states that if the volume and temperature of a fixed amount of gas do not change, the pressure also remains constant. Consider a bottle of cham-pagne that is shaken and then spews liquid when opened as shown in Figure 19.13. A common misconception is that the pressure inside the bottle is increased when the bottle is shaken. On the contrary, because the temperature of the bottle and its contents remains constant as long as the bottle is sealed, so does the pressure, as can be shown by replacing the cork with a pressure gauge. The correct expla-nation is as follows. Carbon dioxide gas resides in the volume between the liquid surface and the cork. The pressure of the gas in this volume is set higher than atmospheric pressure in the bottling process. Shaking the bottle displaces some of the carbon dioxide gas into the liquid, where it forms bubbles, and these bubbles become attached to the inside of the bottle. (No new gas is generated by shaking.) When the bottle is opened, the pressure is reduced to atmospheric pressure, which causes the volume of the bubbles to increase suddenly. If the bubbles are attached to the bottle (beneath the liquid surface), their rapid expansion expels liquid from the bottle. If the sides and bottom of the bottle are first tapped until no bubbles remain beneath the surface, however, the drop in pressure does not force liquid from the bottle when the champagne is opened. The ideal gas law is often expressed in terms of the total number of molecules N. Because the number of moles n equals the ratio of the total number of molecules and Avogadro’s number NA, we can write Equation 19.8 as PV 5 nRT 5 N NA RT PV 5 Nk BT (19.10) where kB is Boltzmann’s constant, which has the value k B 5 R NA 5 1.38 3 10223 J/K (19.11) It is common to call quantities such as P, V, and T the thermodynamic variables of an ideal gas. If the equation of state is known, one of the variables can always be expressed as some function of the other two.
Q uick Quiz 19.5 A common material for cushioning objects in packages is made by trapping bubbles of air between sheets of plastic. Is this material more effec-tive at keeping the contents of the package from moving around inside the package on (a) a hot day, (b) a cold day, or (c) either hot or cold days?
Q uick Quiz 19.6 On a winter day, you turn on your furnace and the tempera-ture of the air inside your home increases. Assume your home has the normal amount of leakage between inside air and outside air. Is the number of moles of air in your room at the higher temperature (a) larger than before, (b) smaller than before, or (c) the same as before?
W W Boltzmann’s constant Figure 19.13 A bottle of cham-pagne is shaken and opened. Liquid spews out of the opening. A common misconception is that the pressure inside the bottle is increased by the shaking.
Image copyright digitalife, 2009. Used under license from Shutterstock.com Pitfall Prevention 19.3 So Many ks There are a variety of physical quantities for which the letter k is used. Two we have seen previously are the force constant for a spring (Chapter 15) and the wave number for a mechanical wave (Chapter 16). Boltzmann’s constant is another k, and we will see k used for thermal conductiv-ity in Chapter 20 and for an elec-trical constant in Chapter 23. To make some sense of this confusing state of affairs, we use a subscript B for Boltzmann’s constant to help us recognize it. In this book, you will see Boltzmann’s constant as kB, but you may see Boltzmann’s constant in other resources as simply k.
www.aswarphysics.weebly.com 580 Chapter 19 Temperature Example 19.4 Heating a Spray Can A spray can containing a propellant gas at twice atmospheric pressure (202 kPa) and having a volume of 125.00 cm3 is at 228C. It is then tossed into an open fire. (Warning: Do not do this experiment; it is very dangerous.) When the tem-perature of the gas in the can reaches 1958C, what is the pressure inside the can? Assume any change in the volume of the can is negligible.
Conceptualize Intuitively, you should expect that the pressure of the gas in the container increases because of the increasing temperature.
Categorize We model the gas in the can as ideal and use the ideal gas law to calculate the new pressure.
S o l u t i o n Analyze Rearrange Equation 19.8: (1) PV T 5 nR No air escapes during the compression, so n, and there-fore nR, remains constant. Hence, set the initial value of the left side of Equation (1) equal to the final value: (2) PiVi Ti 5 PfVf Tf Because the initial and final volumes of the gas are assumed to be equal, cancel the volumes: (3) Pi Ti 5 Pf Tf Solve for Pf: Pf 5 a Tf Ti bPi 5 a468 K 295 Kb 1202 kPa2 5 320 kPa Find the change in the volume of the can using Equa-tion 19.6 and the value for a for steel from Table 19.1: DV 5 bVi DT 5 3aVi DT 5 311 3 1026 (8C)21(1738C) 5 0.71 cm3 Start from Equation (2) again and find an equation for the final pressure: Pf 5 a Tf Ti baVi Vf bPi This result differs from Equation (3) only in the factor Vi/Vf. Evaluate this factor: Vi Vf 5 125.00 cm3 1125.00 cm3 1 0.71 cm32 5 0.994 5 99.4% Finalize The higher the temperature, the higher the pressure exerted by the trapped gas as expected. If the pressure increases sufficiently, the can may explode. Because of this possibility, you should never dispose of spray cans in a fire.
Suppose we include a volume change due to thermal expansion of the steel can as the temperature increases. Does that alter our answer for the final pressure significantly?
Answer Because the thermal expansion coefficient of steel is very small, we do not expect much of an effect on our final answer.
What If?
Therefore, the final pressure will differ by only 0.6% from the value calculated without considering the thermal expan-sion of the can. Taking 99.4% of the previous final pressure, the final pressure including thermal expansion is 318 kPa.
Summary Two objects are in thermal equilibrium with each other if they do not exchange energy when in thermal contact.
Definitions www.aswarphysics.weebly.com Objective Questions 581 Temperature is the property that determines whether an object is in thermal equilibrium with other objects. Two objects in thermal equilibrium with each other are at the same temperature. The SI unit of absolute temperature is the kelvin, which is defined to be 1/273.16 of the difference between absolute zero and the temperature of the triple point of water. An ideal gas is one for which PV/nT is constant. An ideal gas is described by the equation of state, PV 5 nRT (19.8) where n equals the number of moles of the gas, P is its pressure, V is its volume, R is the universal gas constant (8.314 J/mol ? K), and T is the absolute temperature of the gas. A real gas behaves approximately as an ideal gas if it has a low density. The zeroth law of thermody-namics states that if objects A and B are separately in thermal equi-librium with a third object C, then objects A and B are in thermal equilibrium with each other. When the temperature of an object is changed by an amount DT, its length changes by an amount DL that is proportional to DT and to its initial length Li: DL 5 aLi DT (19.4) where the constant a is the average coefficient of linear expansion. The aver-age coefficient of volume expansion b for a solid is approximately equal to 3a.
Concepts and Principles stant, what new volume does the gas occupy? (a) 1.0 m3 (b) 1.5 m3 (c) 2.0 m3 (d) 0.12 m3 (e) 2.5 m3 7. What would happen if the glass of a thermometer expanded more on warming than did the liquid in the tube? (a) The thermometer would break. (b) It could be used only for temperatures below room tempera-ture. (c) You would have to hold it with the bulb on top. (d) The scale on the thermometer is reversed so that higher temperature values would be found closer to the bulb. (e) The numbers would not be evenly spaced. 8. A cylinder with a piston contains a sample of a thin gas. The kind of gas and the sample size can be changed. The cylinder can be placed in different constant- temperature baths, and the piston can be held in dif-ferent positions. Rank the following cases according to the pressure of the gas from the highest to the low-est, displaying any cases of equality. (a) A 0.002-mol sample of oxygen is held at 300 K in a 100-cm3 con-tainer. (b) A 0.002-mol sample of oxygen is held at 600 K in a 200-cm3 container. (c) A 0.002-mol sample of oxygen is held at 600 K in a 300-cm3 container. (d) A 0.004-mol sample of helium is held at 300 K in a 200-cm3 container. (e) A 0.004-mol sample of helium is held at 250 K in a 200-cm3 container. 9. Two cylinders A and B at the same temperature con-tain the same quantity of the same kind of gas. Cylin-der A has three times the volume of cylinder B. What can you conclude about the pressures the gases exert? (a) We can conclude nothing about the pressures. 1. Markings to indicate length are placed on a steel tape in a room that is at a temperature of 228C. Measure-ments are then made with the same tape on a day when the temperature is 278C. Assume the objects you are measuring have a smaller coefficient of linear expan-sion than steel. Are the measurements (a) too long, (b) too short, or (c) accurate? 2. When a certain gas under a pressure of 5.00 3 106 Pa at 25.08C is allowed to expand to 3.00 times its origi-nal volume, its final pressure is 1.07 3 106 Pa. What is its final temperature? (a) 450 K (b) 233 K (c) 212 K (d) 191 K (e) 115 K 3. If the volume of an ideal gas is doubled while its tem-perature is quadrupled, does the pressure (a) remain the same, (b) decrease by a factor of 2, (c) decrease by a factor of 4, (d) increase by a factor of 2, or (e) increase by a factor of 4 4. The pendulum of a certain pendulum clock is made of brass. When the temperature increases, what hap-pens to the period of the clock? (a) It increases. (b) It decreases. (c) It remains the same. 5. A temperature of 1628F is equivalent to what tempera-ture in kelvins? (a) 373 K (b) 288 K (c) 345 K (d) 201 K (e) 308 K 6. A cylinder with a piston holds 0.50 m3 of oxygen at an absolute pressure of 4.0 atm. The piston is pulled outward, increasing the volume of the gas until the pressure drops to 1.0 atm. If the temperature stays con-Objective Questions 1. denotes answer available in Student Solutions Manual/Study Guide www.aswarphysics.weebly.com 582 Chapter 19 Temperature 12. Suppose you empty a tray of ice cubes into a bowl partly full of water and cover the bowl. After one-half hour, the contents of the bowl come to thermal equi-librium, with more liquid water and less ice than you started with. Which of the following is true? (a) The temperature of the liquid water is higher than the temperature of the remaining ice. (b) The tempera-ture of the liquid water is the same as that of the ice. (c) The temperature of the liquid water is less than that of the ice. (d) The comparative temperatures of the liquid water and ice depend on the amounts present. 13. A hole is drilled in a metal plate. When the metal is raised to a higher temperature, what happens to the diameter of the hole? (a) It decreases. (b) It increases. (c) It remains the same. (d) The answer depends on the initial temperature of the metal. (e) None of those answers is correct. 14. On a very cold day in upstate New York, the tempera-ture is 2258C, which is equivalent to what Fahrenheit temperature? (a) 2468F (b) 2778F (c) 188F (d) 2258F (e) 2138F 7. An automobile radiator is filled to the brim with water when the engine is cool. (a) What happens to the water when the engine is running and the water has been raised to a high temperature? (b) What do modern automobiles have in their cooling systems to prevent the loss of coolants? 8. When the metal ring and metal sphere in Figure CQ19.8 are both at room temperature, the sphere can barely be passed through the ring. (a) After the sphere is warmed in a flame, it cannot be passed through the ring. Explain. (b) What If? What if the ring is warmed and the sphere is left at room temperature? Does the sphere pass through the ring?
Figure CQ19.8 © Cengage Learning/Charles D. Winters 9. Is it possible for two objects to be in thermal equi-librium if they are not in contact with each other? Explain. 10. Use a periodic table of the elements (see Appendix C) to determine the number of grams in one mole of (a) hydrogen, which has diatomic molecules; (b) helium; and (c) carbon monoxide.
(b) The pressure in A is three times the pressure in B. (c) The pressures must be equal. (d) The pressure in A must be one-third the pressure in B. 10. A rubber balloon is filled with 1 L of air at 1 atm and 300 K and is then put into a cryogenic refrigerator at 100 K. The rubber remains flexible as it cools. (i) What happens to the volume of the balloon? (a) It decreases to 1 3 L. (b) It decreases to 1/!3 L. (c) It is constant. (d) It increases to !3 L. (e) It increases to 3 L. (ii) What happens to the pressure of the air in the balloon? (a) It decreases to 1 3 atm. (b) It decreases to 1/!3 atm. (c) It is constant. (d) It increases to !3 atm. (e) It increases to 3 atm. 11. The average coefficient of linear expansion of cop-per is 17 3 1026 (8C)21. The Statue of Liberty is 93 m tall on a summer morning when the temperature is 258C. Assume the copper plates covering the statue are mounted edge to edge without expansion joints and do not buckle or bind on the framework supporting them as the day grows hot. What is the order of magnitude of the statue’s increase in height? (a) 0.1 mm (b) 1 mm (c) 1 cm (d) 10 cm (e) 1 m 1. Common thermometers are made of a mercury col-umn in a glass tube. Based on the operation of these thermometers, which has the larger coefficient of lin-ear expansion, glass or mercury? (Don’t answer the question by looking in a table.) 2. A piece of copper is dropped into a beaker of water. (a) If the water’s temperature rises, what happens to the temperature of the copper? (b) Under what conditions are the water and copper in thermal equilibrium? 3. (a) What does the ideal gas law predict about the vol-ume of a sample of gas at absolute zero? (b) Why is this prediction incorrect? 4. Some picnickers stop at a convenience store to buy some food, including bags of potato chips. They then drive up into the mountains to their picnic site. When they unload the food, they notice that the bags of chips are puffed up like balloons. Why did that happen? 5. In describing his upcoming trip to the Moon, and as portrayed in the movie Apollo 13 (Universal, 1995), astronaut Jim Lovell said, “I’ll be walking in a place where there’s a 400-degree difference between sun-light and shadow.” Suppose an astronaut standing on the Moon holds a thermometer in his gloved hand. (a) Is the thermometer reading the temperature of the vacuum at the Moon’s surface? (b) Does it read any temperature? If so, what object or substance has that temperature? 6. Metal lids on glass jars can often be loosened by run-ning hot water over them. Why does that work?
Conceptual Questions 1. denotes answer available in Student Solutions Manual/Study Guide www.aswarphysics.weebly.com Problems 583 should the engineer leave between the sections to elim-inate buckling if the concrete is to reach a temperature of 50.08C? 9. The active element of a certain laser is made of a glass rod 30.0 cm long and 1.50 cm in diameter. Assume the average coefficient of linear expansion of the glass is equal to 9.00 3 1026 (8C)21. If the tempera-ture of the rod increases by 65.08C, what is the increase in (a) its length, (b) its diameter, and (c) its volume? 10. Review. Inside the wall of a house, an L-shaped section of hot-water pipe consists of three parts: a straight, horizontal piece h 5 28.0 cm long; an elbow; and a straight, vertical piece , 5 134 cm long (Fig. P19.10). A stud and a second- story floorboard hold the ends of this section of cop-per pipe stationary. Find the magnitude and direction of the displacement of the pipe elbow when the water flow is turned on, raising the temperature of the pipe from 18.08C to 46.58C. 11. A copper telephone wire has essentially no sag between poles 35.0 m apart on a winter day when the tempera-ture is 220.08C. How much longer is the wire on a sum-mer day when the temperature is 35.08C? 12. A pair of eyeglass frames is made of epoxy plastic. At room temperature (20.0°C), the frames have circular lens holes 2.20 cm in radius. To what temperature must the frames be heated if lenses 2.21 cm in radius are to be inserted in them? The average coefficient of linear expansion for epoxy is 1.30 3 10–4 (°C)–1. 13. The Trans-Alaska pipeline is 1 300 km long, reaching from Prudhoe Bay to the port of Valdez. It experiences temperatures from 273°C to 135°C. How much does the steel pipeline expand because of the difference in temperature? How can this expansion be compensated for? 14. Each year thousands of children are badly burned by hot tap water. Figure P19.14 (page 584) shows a cross-sectional view of an antiscalding faucet attachment designed to prevent such accidents. Within the device, a spring made of material with a high coefficient of ther-mal expansion controls a movable plunger. When the M M BIO BIO Section 19.2 Thermometers and the Celsius Temperature Scale Section 19.3 The Constant-Volume Gas Thermometer and the Absolute Temperature Scale 1. A nurse measures the temperature of a patient to be 41.58C. (a) What is this temperature on the Fahren-heit scale? (b) Do you think the patient is seriously ill? Explain. 2. The temperature difference between the inside and the outside of a home on a cold winter day is 57.08F. Express this difference on (a) the Celsius scale and (b) the Kelvin scale. 3. Convert the following temperatures to their values on the Fahrenheit and Kelvin scales: (a) the sublimation point of dry ice, 278.58C; (b) human body tempera-ture, 37.08C. 4. The boiling point of liquid hydrogen is 20.3 K at atmo-spheric pressure. What is this temperature on (a) the Celsius scale and (b) the Fahrenheit scale? 5. Liquid nitrogen has a boiling point of 2195.818C at atmospheric pressure. Express this temperature (a) in degrees Fahrenheit and (b) in kelvins. 6. Death Valley holds the record for the highest recorded temperature in the United States. On July 10, 1913, at a place called Furnace Creek Ranch, the temperature rose to 134°F. The lowest U.S. temperature ever recorded occurred at Prospect Creek Camp in Alaska on January 23, 1971, when the temperature plummeted to 279.8° F. (a) Convert these temperatures to the Celsius scale. (b) Convert the Celsius temperatures to Kelvin. 7. In a student experiment, a constant-volume gas ther-mometer is calibrated in dry ice (278.58C) and in boil-ing ethyl alcohol (78.08C). The separate pressures are 0.900 atm and 1.635 atm. (a) What value of absolute zero in degrees Celsius does the calibration yield? What pressures would be found at (b) the freezing and (c) the boiling points of water? Hint: Use the linear relationship P 5 A 1 BT, where A and B are constants.
Section 19.4 Thermal Expansion of Solids and Liquids Note: Table 19.1 is available for use in solving problems in this section. 8. The concrete sections of a certain superhighway are designed to have a length of 25.0 m. The sections are poured and cured at 10.08C. What minimum spacing Q/C BIO M W h , Figure P19.10 Problems The problems found in this chapter may be assigned online in Enhanced WebAssign 1. straightforward; 2. intermediate; 3. challenging 1. full solution available in the Student Solutions Manual/Study Guide AMT Analysis Model tutorial available in Enhanced WebAssign GP Guided Problem M Master It tutorial available in Enhanced WebAssign W Watch It video solution available in Enhanced WebAssign BIO Q/C S www.aswarphysics.weebly.com 584 Chapter 19 Temperature 21. A hollow aluminum cylinder 20.0 cm deep has an inter-nal capacity of 2.000 L at 20.08C. It is completely filled with turpentine at 20.08C. The turpentine and the alu-minum cylinder are then slowly warmed together to 80.08C. (a) How much turpentine overflows? (b) What is the volume of turpentine remaining in the cylinder at 80.08C? (c) If the combination with this amount of turpentine is then cooled back to 20.08C, how far below the cylinder’s rim does the turpentine’s surface recede? 22. Review. The Golden Gate Bridge in San Francisco has a main span of length 1.28 km, one of the lon-gest in the world. Imagine that a steel wire with this length and a cross-sectional area of 4.00 3 1026 m2 is laid in a straight line on the bridge deck with its ends attached to the towers of the bridge. On a summer day the temperature of the wire is 35.08C. (a) When winter arrives, the towers stay the same dis-tance apart and the bridge deck keeps the same shape as its expansion joints open. When the temperature drops to 210.08C, what is the tension in the wire? Take Young’s modulus for steel to be 20.0 3 1010 N/m2. (b) Permanent deformation occurs if the stress in the steel exceeds its elastic limit of 3.00 3 108 N/m2. At what temperature would the wire reach its elastic limit? (c) What If? Explain how your answers to parts (a) and (b) would change if the Golden Gate Bridge were twice as long. 23. A sample of lead has a mass of 20.0 kg and a density of 11.3 3 103 kg/m3 at 08C. (a) What is the density of lead at 90.08C? (b) What is the mass of the sample of lead at 90.08C? 24. A sample of a solid substance has a mass m and a den-sity r0 at a temperature T0. (a) Find the density of the substance if its temperature is increased by an amount DT in terms of the coefficient of volume expansion b. (b) What is the mass of the sample if the temperature is raised by an amount DT ? 25. An underground gasoline tank can hold 1.00 3 103 gal- lons of gasoline at 52.0°F. Suppose the tank is being filled on a day when the outdoor temperature (and the temperature of the gasoline in a tanker truck) is 95.0°F. When the underground tank registers that it is full, how many gallons have been transferred from the truck, according to a non-temperature-compensated gauge on the truck? Assume the temperature of the gasoline quickly cools from 95.0°F to 52.0°F upon enter-ing the tank.
Section 19.5 Macroscopic Description of an Ideal Gas 26. A rigid tank contains 1.50 moles of an ideal gas. Deter-mine the number of moles of gas that must be with-drawn from the tank to lower the pressure of the gas from 25.0 atm to 5.00 atm. Assume the volume of the tank and the temperature of the gas remain constant during this operation. 27. Gas is confined in a tank at a pressure of 11.0 atm and a temperature of 25.08C. If two-thirds of the gas Q/C S M water temperature rises above a preset safe value, the expansion of the spring causes the plunger to shut off the water flow. Assum-ing that the initial length L of the unstressed spring is 2.40 cm and its coefficient of linear expansion is 22.0 3 10–6 (°C)–1, determine the increase in length of the spring when the water temperature rises by 30.0°C. (You will find the increase in length to be small. Therefore, to provide a greater variation in valve opening for the tempera-ture change anticipated, actual devices have a more complicated mechanical design.) 15. A square hole 8.00 cm along each side is cut in a sheet of copper. (a) Calculate the change in the area of this hole resulting when the temperature of the sheet is increased by 50.0 K. (b) Does this change represent an increase or a decrease in the area enclosed by the hole? 16. The average coefficient of volume expansion for car-bon tetrachloride is 5.81 3 10–4 (°C)–1. If a 50.0-gal steel container is filled completely with carbon tetra-chloride when the temperature is 10.0°C, how much will spill over when the temperature rises to 30.0°C? 17. At 20.08C, an aluminum ring has an inner diameter of 5.000 0 cm and a brass rod has a diameter of 5.050 0 cm. (a) If only the ring is warmed, what temperature must it reach so that it will just slip over the rod? (b) What If? If both the ring and the rod are warmed together, what temperature must they both reach so that the ring barely slips over the rod? (c) Would this latter process work? Explain. Hint: Consult Table 20.2 in the next chapter. 18. Why is the following situation impossible? A thin brass ring has an inner diameter 10.00 cm at 20.08C. A solid aluminum cylinder has diameter 10.02 cm at 20.08C. Assume the average coefficients of linear expansion of the two metals are constant. Both metals are cooled together to a temperature at which the ring can be slipped over the end of the cylinder. 19. A volumetric flask made of Pyrex is calibrated at 20.08C. It is filled to the 100-mL mark with 35.08C acetone. After the flask is filled, the acetone cools and the flask warms so that the combination of acetone and flask reaches a uniform temperature of 32.08C. The combi-nation is then cooled back to 20.08C. (a) What is the volume of the acetone when it cools to 20.08C? (b) At the temperature of 32.08C, does the level of acetone lie above or below the 100-mL mark on the flask? Explain. 20. Review. On a day that the temperature is 20.08C, a concrete walk is poured in such a way that the ends of the walk are unable to move. Take Young’s modulus for concrete to be 7.00 3 109 N/m2 and the compressive strength to be 2.00 3 109 N/m2. (a) What is the stress in the cement on a hot day of 50.08C? (b) Does the con-crete fracture?
W Q/C W Q/C L Figure P19.14 www.aswarphysics.weebly.com Problems 585 air, Avogadro’s number of molecules has mass 28.9 g. Calculate the mass of one cubic meter of air. (c) State how this result compares with the tabulated density of air at 20.08C. 34. Use the definition of Avogadro’s number to find the mass of a helium atom. 35. A popular brand of cola contains 6.50 g of carbon diox-ide dissolved in 1.00 L of soft drink. If the evaporating carbon dioxide is trapped in a cylinder at 1.00 atm and 20.0°C, what volume does the gas occupy? 36. In state-of-the-art vacuum systems, pressures as low as 1.00 3 1029 Pa are being attained. Calculate the num-ber of molecules in a 1.00-m3 vessel at this pressure and a temperature of 27.08C. 37. An automobile tire is inflated with air originally at 10.08C and normal atmospheric pressure. During the process, the air is compressed to 28.0% of its original volume and the temperature is increased to 40.08C. (a) What is the tire pressure? (b) After the car is driven at high speed, the tire’s air temperature rises to 85.08C and the tire’s interior volume increases by 2.00%. What is the new tire pressure (absolute)? 38. Review. To measure how far below the ocean surface a bird dives to catch a fish, a scientist uses a method origi-nated by Lord Kelvin. He dusts the interiors of plastic tubes with powdered sugar and then seals one end of each tube. He captures the bird at nighttime in its nest and attaches a tube to its back. He then catches the same bird the next night and removes the tube. In one trial, using a tube 6.50 cm long, water washes away the sugar over a distance of 2.70 cm from the open end of the tube. Find the greatest depth to which the bird dived, assum-ing the air in the tube stayed at constant temperature. 39. Review. The mass of a hot-air balloon and its cargo (not including the air inside) is 200 kg. The air outside is at 10.08C and 101 kPa. The volume of the balloon is 400 m3. To what temperature must the air in the bal-loon be warmed before the balloon will lift off? (Air density at 10.08C is 1.244 kg/m3.) 40. A room of volume V contains air having equivalent molar mass M (in g/mol). If the temperature of the room is raised from T1 to T2, what mass of air will leave the room? Assume that the air pressure in the room is maintained at P0. 41. Review. At 25.0 m below the surface of the sea, where the temperature is 5.008C, a diver exhales an air bub-ble having a volume of 1.00 cm3. If the surface tem-perature of the sea is 20.08C, what is the volume of the bubble just before it breaks the surface? 42. Estimate the mass of the air in your bedroom. State the quantities you take as data and the value you mea-sure or estimate for each. 43. A cook puts 9.00 g of water in a 2.00-L pressure cooker that is then warmed to 5008C. What is the pressure inside the container? 44. The pressure gauge on a cylinder of gas registers the gauge pressure, which is the difference between the W M BIO M AMT S W S is withdrawn and the temperature is raised to 75.08C, what is the pressure of the gas remaining in the tank? 28. Your father and your younger brother are confronted with the same puzzle. Your father’s garden sprayer and your brother’s water cannon both have tanks with a capacity of 5.00 L (Fig. P19.28). Your father puts a neg-ligible amount of concentrated fertilizer into his tank. They both pour in 4.00 L of water and seal up their tanks, so the tanks also contain air at atmospheric pressure. Next, each uses a hand-operated pump to inject more air until the absolute pressure in the tank reaches 2.40 atm. Now each uses his device to spray out water—not air—until the stream becomes feeble, which it does when the pressure in the tank reaches 1.20 atm. To accomplish spraying out all the water, each finds he must pump up the tank three times. Here is the puzzle: most of the water sprays out after the second pumping. The first and the third pumping-up processes seem just as difficult as the second but result in a much smaller amount of water coming out. Account for this phenomenon.
Figure P19.28 29. Gas is contained in an 8.00-L vessel at a temperature of 20.08C and a pressure of 9.00 atm. (a) Determine the number of moles of gas in the vessel. (b) How many molecules are in the vessel? 30. A container in the shape of a cube 10.0 cm on each edge contains air (with equivalent molar mass 28.9 g/mol) at atmospheric pressure and temperature 300 K. Find (a) the mass of the gas, (b) the gravitational force exerted on it, and (c) the force it exerts on each face of the cube. (d) Why does such a small sample exert such a great force? 31. An auditorium has dimensions 10.0 m 3 20.0 m 3 30.0 m. How many molecules of air fill the auditorium at 20.08C and a pressure of 101 kPa (1.00 atm)? 32. The pressure gauge on a tank registers the gauge pres-sure, which is the difference between the interior pres-sure and exterior pressure. When the tank is full of oxygen (O2), it contains 12.0 kg of the gas at a gauge pressure of 40.0 atm. Determine the mass of oxygen that has been withdrawn from the tank when the pres-sure reading is 25.0 atm. Assume the temperature of the tank remains constant. 33. (a) Find the number of moles in one cubic meter of an ideal gas at 20.08C and atmospheric pressure. (b) For Q/C W Q/C M M Q/C www.aswarphysics.weebly.com 586 Chapter 19 Temperature important that the pressure of the gas never fall below 0.850 atm so that the piston will support a delicate and very expensive part of the apparatus. Without such support, the delicate apparatus can be severely damaged and rendered useless. When the design is turned into a working prototype, it operates perfectly. 51. A mercury thermometer is constructed as shown in Figure P19.51. The Pyrex glass capillary tube has a diameter of 0.004 00 cm, and the bulb has a diam-eter of 0.250 cm. Find the change in height of the mercury column that occurs with a temperature change of 30.08C. 52. A liquid with a coefficient of volume expansion b just fills a spherical shell of volume V (Fig. P19.51). The shell and the open capillary of area A projecting from the top of the sphere are made of a material with an average coefficient of linear expansion a. The liquid is free to expand into the capillary. Assuming the tem-perature increases by DT, find the distance Dh the liq-uid rises in the capillary. 53. Review. An aluminum pipe is open at both ends and used as a flute. The pipe is cooled to 5.008C, at which its length is 0.655 m. As soon as you start to play it, the pipe fills with air at 20.08C. After that, by how much does its fundamental frequency change as the metal rises in temperature to 20.08C? 54. Two metal bars are made of invar and a third bar is made of aluminum. At 08C, each of the three bars is drilled with two holes 40.0 cm apart. Pins are put through the holes to assem-ble the bars into an equi-lateral triangle as in Figure P19.54. (a) First ignore the expansion of the invar. Find the angle between the invar bars as a function of Celsius temperature. (b) Is your answer accurate for negative as well as positive temperatures? (c) Is it accurate for 08C? (d) Solve the problem again, including the expansion of the invar. Aluminum melts at 6608C and invar at 1 4278C. Assume the tabulated expansion coefficients are constant. What are (e) the greatest and (f) the smallest attainable angles between the invar bars? 55. A student measures the length of a brass rod with a steel tape at 20.08C. The reading is 95.00 cm. What will the tape indicate for the length of the rod when the rod and the tape are at (a) 215.08C and (b) 55.08C? 56. The density of gasoline is 730 kg/m3 at 08C. Its average coefficient of volume expansion is 9.60 3 1024 (8C)21. Assume 1.00 gal of gasoline occupies 0.003 80 m3. Ti T h A Ti Figure P19.51 Problems 51 and 52.
M S AMT Invar 40.0 cm Aluminum Figure P19.54 Q/C interior pressure and the exterior pressure P0. Let’s call the gauge pressure Pg. When the cylinder is full, the mass of the gas in it is mi at a gauge pressure of Pgi. Assuming the temperature of the cylinder remains constant, show that the mass of the gas remaining in the cylinder when the pressure reading is Pg f is given by mf 5 m i a P g f 1 P0 P g i 1 P0 b Additional Problems 45. Long-term space missions require reclamation of the oxygen in the carbon dioxide exhaled by the crew. In one method of reclamation, 1.00 mol of carbon diox-ide produces 1.00 mol of oxygen and 1.00 mol of meth-ane as a byproduct. The methane is stored in a tank under pressure and is available to control the attitude of the spacecraft by controlled venting. A single astro-naut exhales 1.09 kg of carbon dioxide each day. If the methane generated in the respiration recycling of three astronauts during one week of flight is stored in an originally empty 150-L tank at 245.0°C, what is the final pressure in the tank? 46. A steel beam being used in the construction of a sky-scraper has a length of 35.000 m when delivered on a cold day at a temperature of 15.0008F. What is the length of the beam when it is being installed later on a warm day when the temperature is 90.0008F? 47. A spherical steel ball bearing has a diameter of 2.540 cm at 25.008C. (a) What is its diameter when its tem-perature is raised to 100.08C? (b) What temperature change is required to increase its volume by 1.000%? 48. A bicycle tire is inflated to a gauge pressure of 2.50 atm when the temperature is 15.08C. While a man rides the bicycle, the temperature of the tire rises to 45.08C. Assuming the volume of the tire does not change, find the gauge pressure in the tire at the higher temperature. 49. In a chemical processing plant, a reaction chamber of fixed volume V0 is connected to a reservoir chamber of fixed volume 4V0 by a passage containing a thermally insulating porous plug. The plug permits the cham-bers to be at different temperatures. The plug allows gas to pass from either chamber to the other, ensuring that the pressure is the same in both. At one point in the processing, both chambers contain gas at a pres-sure of 1.00 atm and a temperature of 27.0°C. Intake and exhaust valves to the pair of chambers are closed. The reservoir is maintained at 27.0°C while the reac-tion chamber is heated to 400°C. What is the pressure in both chambers after that is done? 50. Why is the following situation impossible? An apparatus is designed so that steam initially at T 5 1508C, P 5 1.00 atm, and V 5 0.500 m3 in a piston–cylinder appa-ratus undergoes a process in which (1) the volume remains constant and the pressure drops to 0.870 atm, followed by (2) an expansion in which the pressure remains constant and the volume increases to 1.00 m3, followed by (3) a return to the initial conditions. It is BIO www.aswarphysics.weebly.com Problems 587 each dimension increases according to Equation 19.4, where a is the average coefficient of linear expansion. (a) Show that the increase in area is DA 5 2aAi DT. (b) What approximation does this expression assume? 62. The measurement of the average coefficient of volume expansion b for a liquid is complicated because the con-tainer also changes size with temperature. Figure P19.62 shows a simple means for measuring b despite the expansion of the container. With this apparatus, one arm of a U-tube is maintained at 08C in a water–ice bath, and the other arm is maintained at a different temperature TC in a constant-temperature bath. The connecting tube is hori-zontal. A difference in the length or diameter of the tube between the two arms of the U-tube has no effect on the pressure balance at the bottom of the tube because the pres-sure depends only on the depth of the liquid. Derive an expression for b for the liquid in terms of h0, ht, and TC. 63. A copper rod and a steel rod are different in length by 5.00 cm at 08C. The rods are warmed and cooled together. (a) Is it possible that the length difference remains constant at all temperatures? Explain. (b) If so, describe the lengths at 08C as precisely as you can. Can you tell which rod is longer? Can you tell the lengths of the rods? 64. A vertical cylinder of cross-sectional area A is fitted with a tight-fitting, frictionless piston of mass m (Fig. P19.64). The piston is not restricted in its motion in any way and is sup-ported by the gas at pressure P below it. Atmospheric pressure is P0. We wish to find the height h in Figure P19.64. (a) What analysis model is appropriate to describe the piston? (b) Write an appropriate force equation for the piston from this analy-sis model in terms of P, P0, m, A, and g. (c) Suppose n moles of an ideal gas are in the cylinder at a temperature of T. Substitute for P in your answer to part (b) to find the height h of the piston above the bottom of the cylinder. 65. Review. Consider an object with any one of the shapes displayed in Table 10.2. What is the percentage increase in the moment of inertia of the object when it is warmed from 08C to 1008C if it is composed of (a) copper or (b) aluminum? Assume the average lin-ear expansion coefficients shown in Table 19.1 do not vary between 08C and 1008C. (c) Why are the answers for parts (a) and (b) the same for all the shapes?
ht h0 Liquid sample Constant-temperature bath at TC Water–ice bath at 0 C Figure P19.62 S Q/C h A m Gas Figure P19.64 AMT GP S Q/C How many extra kilograms of gasoline would you receive if you bought 10.0 gal of gasoline at 08C rather than at 20.08C from a pump that is not temperature compensated? 57. A liquid has a density r. (a) Show that the fractional change in density for a change in temperature DT is Dr/r 5 2b DT. (b) What does the negative sign signify? (c) Fresh water has a maximum density of 1.000 0 g/cm3 at 4.08C. At 10.08C, its density is 0.999 7 g/cm3. What is b for water over this temperature interval? (d) At 08C, the density of water is 0.999 9 g/cm3. What is the value for b over the temperature range 08C to 4.008C? 58. (a) Take the definition of the coefficient of volume expansion to be b 5 1 V dV dT P 5constant 5 1 V 'V 'T Use the equation of state for an ideal gas to show that the coefficient of volume expansion for an ideal gas at constant pressure is given by b 5 1/T, where T is the absolute temperature. (b) What value does this expres-sion predict for b at 08C? State how this result com-pares with the experimental values for (c) helium and (d) air in Table 19.1. Note: These values are much larger than the coefficients of volume expansion for most liq-uids and solids. 59. Review. A clock with a brass pendulum has a period of 1.000 s at 20.08C. If the temperature increases to 30.08C, (a) by how much does the period change and (b) how much time does the clock gain or lose in one week? 60. A bimetallic strip of length L is made of two ribbons of different metals bonded together. (a) First assume the strip is originally straight. As the strip is warmed, the metal with the greater average coefficient of expan-sion expands more than the other, forcing the strip into an arc with the outer radius having a greater circumference (Fig. P19.60). Derive an expression for the angle of bending u as a function of the initial length of the strips, their average coeffi-cients of linear expansion, the change in temperature, and the separation of the centers of the strips (Dr 5 r2 2 r1). (b) Show that the angle of bending decreases to zero when DT decreases to zero and also when the two average coefficients of expansion become equal. (c) What If? What happens if the strip is cooled? 61. The rectangular plate shown in Figure P19.61 has an area Ai equal to ,w. If the temperature increases by DT, Q/C Q/C r 2 r 1 u Figure P19.60 S Q/C Q/C Ti T Ti T w w w Figure P19.61 www.aswarphysics.weebly.com 588 Chapter 19 Temperature copper to be 11.0 3 1010 N/m2. At this lower tempera-ture, find (a) the tension in the wire and (b) the x coor-dinate of the junction between the wires. 73. Review. A steel guitar string with a diameter of 1.00 mm is stretched between supports 80.0 cm apart. The tem-perature is 0.08C. (a) Find the mass per unit length of this string. (Use the value 7.86 3 103 kg/m3 for the den-sity.) (b) The fundamental frequency of transverse oscillations of the string is 200 Hz. What is the tension in the string? Next, the temperature is raised to 30.08C. Find the resulting values of (c) the tension and (d) the fundamental frequency. Assume both the Young’s mod-ulus of 20.0 3 1010 N/m2 and the average coefficient of expansion a 5 11.0 3 1026 (8C)21 have constant values between 0.08C and 30.08C. 74. A cylinder is closed by a piston connected to a spring of constant 2.00 3 103 N/m (see Fig. P19.74). With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pres-sure of 1.00 atm and a temperature of 20.08C. (a) If the piston has a cross- sectional area of 0.010 0 m2 and negli-gible mass, how high will it rise when the temperature is raised to 2508C? (b) What is the pressure of the gas at 2508C? 75. Helium gas is sold in steel tanks that will rupture if sub-jected to tensile stress greater than its yield strength of 5 3 108 N/m2. If the helium is used to inflate a balloon, could the balloon lift the spherical tank the helium came in? Justify your answer. Suggestion: You may con-sider a spherical steel shell of radius r and thickness t having the density of iron and on the verge of breaking apart into two hemispheres because it contains helium at high pressure. 76. A cylinder that has a 40.0-cm radius and is 50.0 cm deep is filled with air at 20.08C and 1.00 atm (Fig. P19.76a). A 20.0-kg piston is now lowered into the cyl-inder, compressing the air trapped inside as it takes equilibrium height hi (Fig. P19.76b). Finally, a 25.0-kg dog stands on the piston, further compressing the air, which remains at 208C (Fig. P19.76c). (a) How far down k h T 20.0 C T 250 C Figure P19.74 W Q/C 66. (a) Show that the density of an ideal gas occupying a volume V is given by r 5 PM/RT, where M is the molar mass. (b) Determine the density of oxygen gas at atmo-spheric pressure and 20.08C. 67. Two concrete spans of a 250-m-long bridge are placed end to end so that no room is allowed for expan-sion (Fig. P19.67a). If a temperature increase of 20.08C occurs, what is the height y to which the spans rise when they buckle (Fig. P19.67b)? 68. Two concrete spans that form a bridge of length L are placed end to end so that no room is allowed for expansion (Fig. P19.67a). If a temperature increase of DT occurs, what is the height y to which the spans rise when they buckle (Fig. P19.67b)? 69. Review. (a) Derive an expression for the buoyant force on a spherical balloon, submerged in water, as a func-tion of the depth h below the surface, the volume Vi of the balloon at the surface, the pressure P0 at the sur-face, and the density rw of the water. Assume the water temperature does not change with depth. (b) Does the buoyant force increase or decrease as the balloon is sub-merged? (c) At what depth is the buoyant force one-half the surface value? 70. Review. Following a collision in outer space, a copper disk at 8508C is rotating about its axis with an angular speed of 25.0 rad/s. As the disk radiates infrared light, its temperature falls to 20.08C. No external torque acts on the disk. (a) Does the angular speed change as the disk cools? Explain how it changes or why it does not. (b) What is its angular speed at the lower temperature? 71. Starting with Equation 19.10, show that the total pres-sure P in a container filled with a mixture of several ideal gases is P 5 P1 1 P2 1 P3 1 . . . , where P1, P2, . . . are the pressures that each gas would exert if it alone filled the container. (These individual pressures are called the partial pressures of the respective gases.) This result is known as Dalton’s law of partial pressures.
Challenge Problems 72. Review. A steel wire and a copper wire, each of diameter 2.000 mm, are joined end to end. At 40.08C, each has an unstretched length of 2.000 m. The wires are con-nected between two fixed supports 4.000 m apart on a tabletop. The steel wire extends from x 5 22.000 m to x 5 0, the copper wire extends from x 5 0 to x 5 2.000 m, and the tension is negligible. The temperature is then lowered to 20.08C. Assume the average coeffi-cient of linear expansion of steel is 11.0 3 1026 (8C)21 and that of copper is 17.0 3 1026 (8C)21. Take Young’s modulus for steel to be 20.0 3 1010 N/m2 and that for T 250 m T 20 C y a b Figure P19.67 Problems 67 and 68.
S S AMT Q/C 50.0 cm hi h a b c Figure P19.76 www.aswarphysics.weebly.com Problems 589 of the plate below the stationary line are moving down relative to the roof and feel a force of kinetic friction acting up the roof. Elements of area above the station-ary line are sliding up the roof, and on them kinetic friction acts downward parallel to the roof. The sta-tionary line occupies no area, so we assume no force of static friction acts on the plate while the temperature is changing. The plate as a whole is very nearly in equi-librium, so the net friction force on it must be equal to the component of its weight acting down the incline. (a) Prove that the stationary line is at a distance of L 2 a1 2 tan u mk b below the top edge of the plate. (b) Analyze the forces that act on the plate when the temperature is falling and prove that the stationary line is at that same dis-tance above the bottom edge of the plate. (c) Show that the plate steps down the roof like an inchworm, mov-ing each day by the distance L mk 1a2 2 a12 1Th 2 T c 2 tan u (d) Evaluate the distance an aluminum plate moves each day if its length is 1.20 m, the temperature cycles between 4.008C and 36.08C, and if the roof has slope 18.5°, coefficient of linear expansion 1.50 3 1025 (8C)21, and coefficient of friction 0.420 with the plate. (e) What If? What if the expansion coefficient of the plate is less than that of the roof? Will the plate creep up the roof? 79. A 1.00-km steel railroad rail is fastened securely at both ends when the temperature is 20.08C. As the tem-perature increases, the rail buckles, taking the shape of an arc of a vertical circle. Find the height h of the center of the rail when the temperature is 25.08C. (You will need to solve a transcendental equation.) (Dh) does the piston move when the dog steps onto it? (b) To what temperature should the gas be warmed to raise the piston and dog back to hi? 77. The relationship L 5 Li 1 aLi DT is a valid approxi-mation when a DT is small. If a DT is large, one must integrate the relationship dL 5 aL dT to determine the final length. (a) Assuming the coefficient of linear expansion of a material is constant as L varies, deter-mine a general expression for the final length of a rod made of the material. Given a rod of length 1.00 m and a temperature change of 100.08C, determine the error caused by the approximation when (b) a 5 2.00 3 1025 (8C)21 (a typical value for a metal) and (c) when a 5 0.020 0 (8C)21 (an unrealistically large value for comparison). (d) Using the equation from part (a), solve Problem 21 again to find more accurate results. 78. Review. A house roof is a perfectly flat plane that makes an angle u with the horizontal. When its temper-ature changes, between Tc before dawn each day and Th in the middle of each afternoon, the roof expands and contracts uniformly with a coefficient of thermal expansion a1. Resting on the roof is a flat, rectangular metal plate with expansion coefficient a2, greater than a1. The length of the plate is L, measured along the slope of the roof. The component of the plate’s weight perpendicular to the roof is supported by a normal force uniformly distributed over the area of the plate. The coefficient of kinetic friction between the plate and the roof is mk. The plate is always at the same tem-perature as the roof, so we assume its temperature is continuously changing. Because of the difference in expansion coefficients, each bit of the plate is moving relative to the roof below it, except for points along a certain horizontal line running across the plate called the stationary line. If the temperature is rising, parts Q/C www.aswarphysics.weebly.com c h a p t e r 20 The First Law of Thermodynamics 590 In this photograph of the Mt. Baker area near Bellingham, Washington, we see evidence of water in all three phases. In the lake is liquid water, and solid water in the form of snow appears on the ground. The clouds in the sky consist of liquid water droplets that have condensed from the gaseous water vapor in the air. Changes of a substance from one phase to another are a result of energy transfer. (© iStockphoto.com/ KingWu) Until about 1850, the fields of thermodynamics and mechanics were considered to be two distinct branches of science. The principle of conservation of energy seemed to describe only certain kinds of mechanical systems. Mid-19th-century experiments performed by Englishman James Joule and others, however, showed a strong connection between the transfer of energy by heat in thermal processes and the transfer of energy by work in mechanical processes. Today we know that mechanical energy can be transformed to inter-nal energy, which is formally defined in this chapter. Once the concept of energy was gener-alized from mechanics to include internal energy, the principle of conservation of energy as discussed in Chapter 8 emerged as a universal law of nature. This chapter focuses on the concept of internal energy, the first law of thermodynamics, and some important applications of the first law. The first law of thermodynamics describes systems in which the only energy change is that of internal energy and the transfers of energy are by heat and work. A major difference in our discussion of work in this chapter from that in most of the chapters on mechanics is that we will consider work done on deformable systems.
20.1 Heat and Internal Energy At the outset, it is important to make a major distinction between internal energy and heat, terms that are often incorrectly used interchangeably in popular language.
20.1 Heat and Internal Energy 20.2 Specific Heat and Calorimetry 20.3 Latent Heat 20.4 Work and Heat in Thermodynamic Processes 20.5 The First Law of Thermodynamics 20.6 Some Applications of the First Law of Thermodynamics 20.7 Energy Transfer Mechanisms in Thermal Processes www.aswarphysics.weebly.com 20.1 Heat and Internal Energy 591 Internal energy is all the energy of a system that is associated with its micro-scopic components—atoms and molecules—when viewed from a reference frame at rest with respect to the center of mass of the system.
The last part of this sentence ensures that any bulk kinetic energy of the system due to its motion through space is not included in internal energy. Internal energy includes kinetic energy of random translational, rotational, and vibrational motion of molecules; vibrational potential energy associated with forces between atoms in molecules; and electric potential energy associated with forces between molecules. It is useful to relate internal energy to the temperature of an object, but this rela-tionship is limited. We show in Section 20.3 that internal energy changes can also occur in the absence of temperature changes. In that discussion, we will investi-gate the internal energy of the system when there is a physical change, most often related to a phase change, such as melting or boiling. We assign energy associated with chemical changes, related to chemical reactions, to the potential energy term in Equation 8.2, not to internal energy. Therefore, we discuss the chemical potential energy in, for example, a human body (due to previous meals), the gas tank of a car (due to an earlier transfer of fuel), and a battery of an electric circuit (placed in the battery during its construction in the manufacturing process).
Heat is defined as a process of transferring energy across the boundary of a system because of a temperature difference between the system and its sur-roundings. It is also the amount of energy Q transferred by this process.
When you heat a substance, you are transferring energy into it by placing it in con-tact with surroundings that have a higher temperature. Such is the case, for exam-ple, when you place a pan of cold water on a stove burner. The burner is at a higher temperature than the water, and so the water gains energy by heat. Read this definition of heat (Q in Eq. 8.2) very carefully. In particular, notice what heat is not in the following common quotes. (1) Heat is not energy in a hot sub-stance. For example, “The boiling water has a lot of heat” is incorrect; the boiling water has internal energy Eint. (2) Heat is not radiation. For example, “It was so hot because the sidewalk was radiating heat” is incorrect; energy is leaving the sidewalk by electromagnetic radiation, TER in Equation 8.2. (3) Heat is not warmth of an envi-ronment. For example, “The heat in the air was so oppressive” is incorrect; on a hot day, the air has a high temperature T. As an analogy to the distinction between heat and internal energy, consider the distinction between work and mechanical energy discussed in Chapter 7. The work done on a system is a measure of the amount of energy transferred to the system from its surroundings, whereas the mechanical energy (kinetic energy plus poten-tial energy) of a system is a consequence of the motion and configuration of the system. Therefore, when a person does work on a system, energy is transferred from the person to the system. It makes no sense to talk about the work of a system; one can refer only to the work done on or by a system when some process has occurred in which energy has been transferred to or from the system. Likewise, it makes no sense to talk about the heat of a system; one can refer to heat only when energy has been transferred as a result of a temperature difference. Both heat and work are ways of transferring energy between a system and its surroundings.
Units of Heat Early studies of heat focused on the resultant increase in temperature of a sub-stance, which was often water. Initial notions of heat were based on a fluid called caloric that flowed from one substance to another and caused changes in tempera-ture. From the name of this mythical fluid came an energy unit related to ther-mal processes, the calorie (cal), which is defined as the amount of energy transfer Pitfall Prevention 20.1 Internal Energy, Thermal Energy, and Bond Energy When reading other physics books, you may see terms such as thermal energy and bond energy. Thermal energy can be interpreted as that part of the internal energy associated with random motion of molecules and therefore related to temperature. Bond energy is the intermolecular potential energy. Therefore, Internal energy 5 thermal energy 1 bond energy Although this breakdown is pre-sented here for clarification with regard to other books, we will not use these terms because there is no need for them.
Pitfall Prevention 20.2 Heat, Temperature, and Internal Energy Are Different As you read the newspaper or explore the Internet, be alert for incorrectly used phrases including the word heat and think about the proper word to be used in place of heat. Incorrect examples include “As the truck braked to a stop, a large amount of heat was generated by friction” and “The heat of a hot summer day . . . .” www.aswarphysics.weebly.com 592 Chapter 20 The First Law of Thermodynamics necessary to raise the temperature of 1 g of water from 14.5°C to 15.5°C.1 (The “Calorie,” written with a capital “C” and used in describing the energy content of foods, is actually a kilocalorie.) The unit of energy in the U.S. customary system is the British thermal unit (Btu), which is defined as the amount of energy transfer required to raise the temperature of 1 lb of water from 63°F to 64°F. Once the relationship between energy in thermal and mechanical processes became clear, there was no need for a separate unit related to thermal processes. The joule has already been defined as an energy unit based on mechanical pro-cesses. Scientists are increasingly turning away from the calorie and the Btu and are using the joule when describing thermal processes. In this textbook, heat, work, and internal energy are usually measured in joules.
The Mechanical Equivalent of Heat In Chapters 7 and 8, we found that whenever friction is present in a mechanical system, the mechanical energy in the system decreases; in other words, mechani-cal energy is not conserved in the presence of nonconservative forces. Various experiments show that this mechanical energy does not simply disappear but is transformed into internal energy. You can perform such an experiment at home by hammering a nail into a scrap piece of wood. What happens to all the kinetic energy of the hammer once you have finished? Some of it is now in the nail as internal energy, as demonstrated by the nail being measurably warmer. Notice that there is no transfer of energy by heat in this process. For the nail and board as a nonisolated system, Equation 8.2 becomes DEint 5 W 1 TMW, where W is the work done by the hammer on the nail and TMW is the energy leaving the system by sound waves when the nail is struck. Although this connection between mechanical and internal energy was first suggested by Benjamin Thompson, it was James Prescott Joule who established the equivalence of the decrease in mechanical energy and the increase in internal energy. A schematic diagram of Joule’s most famous experiment is shown in Figure 20.1. The system of interest is the Earth, the two blocks, and the water in a thermally insulated container. Work is done within the system on the water by a rotating pad-dle wheel, which is driven by heavy blocks falling at a constant speed. If the energy transformed in the bearings and the energy passing through the walls by heat are neglected, the decrease in potential energy of the system as the blocks fall equals the work done by the paddle wheel on the water and, in turn, the increase in inter-nal energy of the water. If the two blocks fall through a distance h, the decrease in potential energy of the system is 2mgh, where m is the mass of one block; this energy causes the temperature of the water to increase. By varying the conditions of the experiment, Joule found that the decrease in mechanical energy is propor-tional to the product of the mass of the water and the increase in water tempera-ture. The proportionality constant was found to be approximately 4.18 J/g ? °C. Hence, 4.18 J of mechanical energy raises the temperature of 1 g of water by 1°C. More precise measurements taken later demonstrated the proportionality to be 4.186 J/g ? °C when the temperature of the water was raised from 14.5°C to 15.5°C. We adopt this “15-degree calorie” value: 1 cal 5 4.186 J (20.1) This equality is known, for purely historical reasons, as the mechanical equivalent of heat. A more proper name would be equivalence between mechanical energy and internal energy, but the historical name is well entrenched in our language, despite the incorrect use of the word heat.
1Originally, the calorie was defined as the energy transfer necessary to raise the temperature of 1 g of water by 1°C. Careful measurements, however, showed that the amount of energy required to produce a 1°C change depends somewhat on the initial temperature; hence, a more precise definition evolved.
Thermal insulator The falling blocks rotate the paddles, causing the temperature of the water to increase.
m m Figure 20.1 Joule’s experiment for determining the mechanical equivalent of heat.
James Prescott Joule British physicist (1818–1889) Joule received some formal education in mathematics, philosophy, and chem-istry from John Dalton but was in large part self-educated. Joule’s research led to the establishment of the principle of conservation of energy. His study of the quantitative relationship among electrical, mechanical, and chemi-cal effects of heat culminated in his announcement in 1843 of the amount of work required to produce a unit of energy, called the mechanical equiva-lent of heat.
© The Art Gallery Collection/Alamy www.aswarphysics.weebly.com 20.2 Specific Heat and Calorimetry 593 Example 20.1 Losing Weight the Hard Way A student eats a dinner rated at 2 000 Calories. He wishes to do an equivalent amount of work in the gymnasium by lifting a 50.0-kg barbell. How many times must he raise the barbell to expend this much energy? Assume he raises the barbell 2.00 m each time he lifts it and he regains no energy when he lowers the barbell.
Conceptualize Imagine the student raising the barbell. He is doing work on the system of the barbell and the Earth, so energy is leaving his body. The total amount of work that the student must do is 2 000 Calories.
Categorize We model the system of the barbell and the Earth as a nonisolated system for energy.
AM S o l u t i o n Analyze Reduce the conservation of energy equa-tion, Equation 8.2, to the appropriate expression for the system of the barbell and the Earth: (1) DUtotal 5 Wtotal Express the change in gravitational potential energy of the system after the barbell is raised once: DU 5 mgh Express the total amount of energy that must be transferred into the system by work for lifting the barbell n times, assuming energy is not regained when the barbell is lowered: (2) DUtotal 5 nmgh Substitute Equation (2) into Equation (1): nmgh 5 Wtotal Solve for n: Substitute numerical values: n 5 Wtotal mgh n 5 12 000 Cal2 150.0 kg2 19.80 m/s22 12.00 m2 a1.00 3 103 cal Calorie b a 4.186 J 1 cal b 5 8.54 3 103 times Finalize If the student is in good shape and lifts the barbell once every 5 s, it will take him about 12 h to perform this feat. Clearly, it is much easier for this student to lose weight by dieting. In reality, the human body is not 100% efficient. Therefore, not all the energy transformed within the body from the dinner transfers out of the body by work done on the barbell. Some of this energy is used to pump blood and perform other functions within the body. Therefore, the 2 000 Calories can be worked off in less time than 12 h when these other energy processes are included.
20.2 Specific Heat and Calorimetry When energy is added to a system and there is no change in the kinetic or potential energy of the system, the temperature of the system usually rises. (An exception to this statement is the case in which a system undergoes a change of state—also called a phase transition—as discussed in the next section.) If the system consists of a sam-ple of a substance, we find that the quantity of energy required to raise the tempera-ture of a given mass of the substance by some amount varies from one substance to another. For example, the quantity of energy required to raise the temperature of 1 kg of water by 1°C is 4 186 J, but the quantity of energy required to raise the temperature of 1 kg of copper by 1°C is only 387 J. In the discussion that follows, we shall use heat as our example of energy transfer, but keep in mind that the tempera-ture of the system could be changed by means of any method of energy transfer. The heat capacity C of a particular sample is defined as the amount of energy needed to raise the temperature of that sample by 1°C. From this definition, we see that if energy Q produces a change DT in the temperature of a sample, then Q 5 C DT (20.2) www.aswarphysics.weebly.com 594 Chapter 20 The First Law of Thermodynamics The specific heat c of a substance is the heat capacity per unit mass. Therefore, if energy Q transfers to a sample of a substance with mass m and the temperature of the sample changes by DT, the specific heat of the substance is c ; Q m DT (20.3) Specific heat is essentially a measure of how thermally insensitive a substance is to the addition of energy. The greater a material’s specific heat, the more energy must be added to a given mass of the material to cause a particular temperature change. Table 20.1 lists representative specific heats. From this definition, we can relate the energy Q transferred between a sample of mass m of a material and its surroundings to a temperature change DT as Q 5 mc DT (20.4) For example, the energy required to raise the temperature of 0.500 kg of water by 3.00°C is Q 5 (0.500 kg)(4 186 J/kg ? °C)(3.00°C) 5 6.28 3 103 J. Notice that when the temperature increases, Q and DT are taken to be positive and energy trans-fers into the system. When the temperature decreases, Q and DT are negative and energy transfers out of the system. We can identify mc DT as the change in internal energy of the system if we ignore any thermal expansion or contraction of the system. (Thermal expansion or con-traction would result in a very small amount of work being done on the system by the surrounding air.) Then, Equation 20.4 is a reduced form of Equation 8.2: DEint 5 Q. The internal energy of the system can be changed by transferring energy into the system by any mechanism. For example, if the system is a baked potato in a microwave oven, Equation 8.2 reduces to the following analog to Equation 20.4: DEint 5 TER 5 mc DT, where TER is the energy transferred to the potato from the microwave oven by electromagnetic radiation. If the system is the air in a bicycle pump, which becomes hot when the pump is operated, Equation 8.2 reduces to the following analog to Equation 20.4: DEint 5 W 5 mc DT, where W is the work done on the pump by the operator. By identifying mc DT as DEint, we have taken a step toward a better understanding of temperature: temperature is related to the energy of the molecules of a system. We will learn more details of this relationship in Chapter 21. Specific heat varies with temperature. If, however, temperature intervals are not too great, the temperature variation can be ignored and c can be treated as a constant.2 Specific heat Table 20.1 Specific Heats of Some Substances at 25°C and Atmospheric Pressure Specific Heat Specific Heat Substance ( J/kg ? °C) Substance ( J/kg ? °C) Elemental solids Other solids Aluminum 900 Brass 380 Beryllium 1 830 Glass 837 Cadmium 230 Ice (25°C) 2 090 Copper 387 Marble 860 Germanium 322 Wood 1 700 Gold 129 Liquids Iron 448 Alcohol (ethyl) 2 400 Lead 128 Mercury 140 Silicon 703 Water (15°C) 4 186 Silver 234 Gas Steam (100°C) 2 010 Note: To convert values to units of cal/g ? °C, divide by 4 186.
2The definition given by Equation 20.4 assumes the specific heat does not vary with temperature over the interval DT 5 Tf 2 Ti. In general, if c varies with temperature over the interval, the correct expression for Q is Q 5 m e Tf Ti c dT.
Pitfall Prevention 20.3 An Unfortunate Choice of Terminology The name specific heat is an unfortunate holdover from the days when thermody-namics and mechanics developed separately. A better name would be specific energy transfer, but the existing term is too entrenched to be replaced.
Pitfall Prevention 20.4 Energy Can Be Transferred by Any Method The symbol Q represents the amount of energy transferred, but keep in mind that the energy transfer in Equation 20.4 could be by any of the meth-ods introduced in Chapter 8; it does not have to be heat. For exam-ple, repeatedly bending a wire coat hanger raises the temperature at the bending point by work.
www.aswarphysics.weebly.com 20.2 Specific Heat and Calorimetry 595 For example, the specific heat of water varies by only about 1% from 0°C to 100°C at atmospheric pressure. Unless stated otherwise, we shall neglect such variations.
Q uick Quiz 20.1 Imagine you have 1 kg each of iron, glass, and water, and all three samples are at 10°C. (a) Rank the samples from highest to lowest tempera-ture after 100 J of energy is added to each sample. (b) Rank the samples from greatest to least amount of energy transferred by heat if each sample increases in temperature by 20°C.
Isolated system boundary Hot sample Qcold mw cw Tw mx cx Tx Qhot Cold water Figure 20.2 In a calorimetry experiment, a hot sample whose specific heat is unknown is placed in cold water in a container that isolates the system from the environment. Notice from Table 20.1 that water has the highest specific heat of common mate-rials. This high specific heat is in part responsible for the moderate climates found near large bodies of water. As the temperature of a body of water decreases during the winter, energy is transferred from the cooling water to the air by heat, increas-ing the internal energy of the air. Because of the high specific heat of water, a rela-tively large amount of energy is transferred to the air for even modest temperature changes of the water. The prevailing winds on the West Coast of the United States are toward the land (eastward). Hence, the energy liberated by the Pacific Ocean as it cools keeps coastal areas much warmer than they would otherwise be. As a result, West Coast states generally have more favorable winter weather than East Coast states, where the prevailing winds do not tend to carry the energy toward land.
Calorimetry One technique for measuring specific heat involves heating a sample to some known temperature Tx, placing it in a vessel containing water of known mass and temperature Tw , Tx, and measuring the temperature of the water after equilibrium has been reached. This technique is called calorimetry, and devices in which this energy transfer occurs are called calorimeters. Figure 20.2 shows the hot sample in the cold water and the resulting energy transfer by heat from the high-temperature part of the system to the low-temperature part. If the system of the sample and the water is isolated, the principle of conservation of energy requires that the amount of energy Q hot that leaves the sample (of unknown specific heat) equal the amount of energy Q cold that enters the water.3 Conservation of energy allows us to write the mathematical representation of this energy statement as Q cold 5 2Q hot (20.5) Suppose mx is the mass of a sample of some substance whose specific heat we wish to determine. Let’s call its specific heat cx and its initial temperature Tx as shown in Figure 20.2. Likewise, let mw, cw, and Tw represent corresponding values for the water. If Tf is the final temperature after the system comes to equilibrium, Equation 20.4 shows that the energy transfer for the water is mwcw(Tf 2 Tw), which is positive because Tf . Tw, and that the energy transfer for the sample of unknown specific heat is mxcx(Tf 2 Tx), which is negative. Substituting these expressions into Equation 20.5 gives mwcw(Tf 2 Tw) 5 2mxcx(Tf 2 Tx) This equation can be solved for the unknown specific heat cx.
Example 20.2 Cooling a Hot Ingot A 0.050 0-kg ingot of metal is heated to 200.0°C and then dropped into a calorimeter containing 0.400 kg of water ini-tially at 20.0°C. The final equilibrium temperature of the mixed system is 22.4°C. Find the specific heat of the metal.
Pitfall Prevention 20.5 Remember the Negative Sign It is critical to include the negative sign in Equation 20.5. The negative sign in the equation is necessary for consistency with our sign con-vention for energy transfer. The energy transfer Q hot has a negative value because energy is leaving the hot substance. The negative sign in the equation ensures that the right side is a positive number, consistent with the left side, which is positive because energy is enter-ing the cold water.
continued 3For precise measurements, the water container should be included in our calculations because it also exchanges energy with the sample. Doing so would require that we know the container’s mass and composition, however. If the mass of the water is much greater than that of the container, we can neglect the effects of the container.
www.aswarphysics.weebly.com 596 Chapter 20 The First Law of Thermodynamics Conceptualize Imagine the process occurring in the isolated system of Figure 20.2. Energy leaves the hot ingot and goes into the cold water, so the ingot cools off and the water warms up. Once both are at the same temperature, the energy transfer stops.
Categorize We use an equation developed in this section, so we categorize this example as a substitution problem.
S o l u t i o n Use Equation 20.4 to evaluate each side of Equation 20.5: mwcw(Tf 2 Tw) 5 2mxcx(Tf 2 Tx) Solve for cx: cx 5 mwcw 1Tf 2 Tw2 mx 1Tx 2 Tf 2 Substitute numerical values: cx 5 10.400 kg2 14 186 J/kg # 8C2 122.48C 2 20.08C2 10.050 0 kg2 1200.08C 2 22.48C2 5 453 J/kg # 8C The ingot is most likely iron as you can see by comparing this result with the data given in Table 20.1. The temperature of the ingot is initially above the steam point. Therefore, some of the water may vaporize when the ingot is dropped into the water. We assume the system is sealed and this steam cannot escape. Because the final equilibrium tempera-ture is lower than the steam point, any steam that does result recondenses back into water.
Suppose you are performing an experiment in the laboratory that uses this technique to determine the specific heat of a sample and you wish to decrease the overall uncertainty in your final result for cx. Of the data given in this example, changing which value would be most effective in decreasing the uncertainty?
Answer The largest experimental uncertainty is associated with the small difference in temperature of 2.4°C for the water. For example, using the rules for propagation of uncertainty in Appendix Section B.8, an uncertainty of 0.1°C in each of Tf and Tw leads to an 8% uncertainty in their difference. For this temperature difference to be larger experi-mentally, the most effective change is to decrease the amount of water.
What If? ▸ 20.2 continu ed Example 20.3 Fun Time for a Cowboy A cowboy fires a silver bullet with a muzzle speed of 200 m/s into the pine wall of a saloon. Assume all the internal energy generated by the impact remains with the bullet. What is the temperature change of the bullet?
Conceptualize Imagine similar experiences you may have had in which mechanical energy is transformed to internal energy when a moving object is stopped. For example, as mentioned in Section 20.1, a nail becomes warm after it is hit a few times with a hammer.
Categorize The bullet is modeled as an isolated system. No work is done on the system because the force from the wall moves through no displacement. This example is similar to the skateboarder pushing off a wall in Section 9.7. There, no work is done on the skateboarder by the wall, and potential energy stored in the body from previous meals is transformed to kinetic energy. Here, no work is done by the wall on the bullet, and kinetic energy is transformed to internal energy.
AM S o l u t i o n Analyze Reduce the conservation of energy equation, Equation 8.2, to the appropriate expression for the sys-tem of the bullet: (1) DK 1 DE int 5 0 The change in the bullet’s internal energy is related to its change in temperature: (2) DE int 5 mc DT Substitute Equation (2) into Equation (1): 10 2 1 2mv 22 1 mc DT 5 0 www.aswarphysics.weebly.com 20.3 Latent Heat 597 Solve for DT, using 234 J/kg ? °C as the specific heat of silver (see Table 20.1): (3) DT 5 1 2mv 2 mc 5 v 2 2c 5 1200 m/s2 2 21234 J/kg # 8C2 5 85.58C Finalize Notice that the result does not depend on the mass of the bullet.
Suppose the cowboy runs out of silver bullets and fires a lead bullet at the same speed into the wall. Will the temperature change of the bullet be larger or smaller?
Answer Table 20.1 shows that the specific heat of lead is 128 J/kg ? °C, which is smaller than that for silver. Therefore, a given amount of energy input or transformation raises lead to a higher temperature than silver and the final tem-perature of the lead bullet will be larger. In Equation (3), let’s substitute the new value for the specific heat: DT 5 v 2 2c 5 1200 m/s2 2 21128 J/kg # 8C2 5 1568C There is no requirement that the silver and lead bullets have the same mass to determine this change in temperature. The only requirement is that they have the same speed.
What If? ▸ 20.3 c on tin u ed 20.3 Latent Heat As we have seen in the preceding section, a substance can undergo a change in tem-perature when energy is transferred between it and its surroundings. In some situ-ations, however, the transfer of energy does not result in a change in temperature. That is the case whenever the physical characteristics of the substance change from one form to another; such a change is commonly referred to as a phase change. Two common phase changes are from solid to liquid (melting) and from liquid to gas (boiling); another is a change in the crystalline structure of a solid. All such phase changes involve a change in the system’s internal energy but no change in its temperature. The increase in internal energy in boiling, for example, is repre-sented by the breaking of bonds between molecules in the liquid state; this bond breaking allows the molecules to move farther apart in the gaseous state, with a corresponding increase in intermolecular potential energy. As you might expect, different substances respond differently to the addition or removal of energy as they change phase because their internal molecular arrange-ments vary. Also, the amount of energy transferred during a phase change depends on the amount of substance involved. (It takes less energy to melt an ice cube than it does to thaw a frozen lake.) When discussing two phases of a material, we will use the term higher-phase material to mean the material existing at the higher temperature. So, for example, if we discuss water and ice, water is the higher-phase material, whereas steam is the higher-phase material in a discussion of steam and water. Consider a system containing a substance in two phases in equilibrium such as water and ice. The initial amount of the higher-phase material, water, in the system is mi. Now imagine that energy Q enters the system. As a result, the final amount of water is mf due to the melt-ing of some of the ice. Therefore, the amount of ice that melted, equal to the amount of new water, is Dm 5 mf 2 mi. We define the latent heat for this phase change as L ; Q Dm (20.6) This parameter is called latent heat (literally, the “hidden” heat) because this added or removed energy does not result in a temperature change. The value of L for a substance depends on the nature of the phase change as well as on the proper-ties of the substance. If the entire amount of the lower-phase material undergoes a phase change, the change in mass Dm of the higher-phase material is equal to the initial mass of the lower-phase material. For example, if an ice cube of mass m on a www.aswarphysics.weebly.com 598 Chapter 20 The First Law of Thermodynamics plate melts completely, the change in mass of the water is mf 2 0 5 m, which is the mass of new water and is also equal to the initial mass of the ice cube. From the definition of latent heat, and again choosing heat as our energy trans-fer mechanism, the energy required to change the phase of a pure substance is Q 5 L Dm (20.7) where Dm is the change in mass of the higher-phase material. Latent heat of fusion Lf is the term used when the phase change is from solid to liquid (to fuse means “to combine by melting”), and latent heat of vaporization Lv is the term used when the phase change is from liquid to gas (the liquid “vaporizes”).4 The latent heats of various substances vary considerably as data in Table 20.2 show. When energy enters a system, causing melting or vaporization, the amount of the higher-phase material increases, so Dm is positive and Q is positive, consistent with our sign convention. When energy is extracted from a system, causing freezing or condensation, the amount of the higher-phase material decreases, so Dm is nega-tive and Q is negative, again consistent with our sign convention. Keep in mind that Dm in Equation 20.7 always refers to the higher-phase material. To understand the role of latent heat in phase changes, consider the energy required to convert a system consisting of a 1.00-g cube of ice at 230.0°C to steam at 120.0°C. Figure 20.3 indicates the experimental results obtained when energy is gradually added to the ice. The results are presented as a graph of temperature of the system versus energy added to the system. Let’s examine each portion of the red-brown curve, which is divided into parts A through E.
Part A. On this portion of the curve, the temperature of the system changes from 230.0°C to 0.0°C. Equation 20.4 indicates that the temperature varies linearly with the energy added, so the experimental result is a straight line on the graph. Because the specific heat of ice is 2 090 J/kg ? °C, we can calculate the amount of energy added by using Equation 20.4: Q 5 mici DT 5 (1.00 3 1023 kg)(2 090 J/kg ? °C)(30.0°C) 5 62.7 J Part B. When the temperature of the system reaches 0.0°C, the ice–water mixture remains at this temperature—even though energy is being added—until all the ice melts. The energy required to melt 1.00 g of ice at 0.0°C is, from Equation 20.7, Q 5 Lf Dmw 5 Lfmi 5 (3.33 3 105 J/kg)(1.00 3 1023 kg) 5 333 J Energy transferred to a substance during a phase change 4When a gas cools, it eventually condenses; that is, it returns to the liquid phase. The energy given up per unit mass is called the latent heat of condensation and is numerically equal to the latent heat of vaporization. Likewise, when a liquid cools, it eventually solidifies, and the latent heat of solidification is numerically equal to the latent heat of fusion.
Table 20.2 Latent Heats of Fusion and Vaporization Latent Heat Melting of Fusion Boiling Latent Heat Substance Point (°C) ( J/kg) Point (°C) of Vaporization ( J/kg) Heliuma 2272.2 5.23 3 103 2268.93 2.09 3 104 Oxygen 2218.79 1.38 3 104 2182.97 2.13 3 105 Nitrogen 2209.97 2.55 3 104 2195.81 2.01 3 105 Ethyl alcohol 2114 1.04 3 105 78 8.54 3 105 Water 0.00 3.33 3 105 100.00 2.26 3 106 Sulfur 119 3.81 3 104 444.60 3.26 3 105 Lead 327.3 2.45 3 104 1 750 8.70 3 105 Aluminum 660 3.97 3 105 2 450 1.14 3 107 Silver 960.80 8.82 3 104 2 193 2.33 3 106 Gold 1 063.00 6.44 3 104 2 660 1.58 3 106 Copper 1 083 1.34 3 105 1 187 5.06 3 106 aHelium does not solidify at atmospheric pressure. The melting point given here corresponds to a pressure of 2.5 MPa.
Pitfall Prevention 20.6 Signs Are Critical Sign errors occur very often when students apply calorimetry equations. For phase changes, remember that Dm in Equation 20.7 is always the change in mass of the higher-phase material. In Equation 20.4, be sure your DT is always the final temperature minus the initial tem-perature. In addition, you must always include the negative sign on the right side of Equation 20.5.
www.aswarphysics.weebly.com 20.3 Latent Heat 599 At this point, we have moved to the 396 J (5 62.7 J 1 333 J) mark on the energy axis in Figure 20.3.
Part C. Between 0.0°C and 100.0°C, nothing surprising happens. No phase change occurs, and so all energy added to the system, which is now water, is used to increase its temperature. The amount of energy necessary to increase the temperature from 0.0°C to 100.0°C is Q 5 mwcw DT 5 (1.00 3 1023 kg)(4.19 3 103 J/kg ? °C)(100.0°C) 5 419 J where mw is the mass of the water in the system, which is the same as the mass mi of the original ice.
Part D. At 100.0°C, another phase change occurs as the system changes from water at 100.0°C to steam at 100.0°C. Similar to the ice–water mixture in part B, the water–steam mixture remains at 100.0°C—even though energy is being added— until all the liquid has been converted to steam. The energy required to convert 1.00 g of water to steam at 100.0°C is Q 5 Lv Dms 5 Lvmw 5 (2.26 3 106 J/kg)(1.00 3 1023 kg) 5 2.26 3 103 J Part E. On this portion of the curve, as in parts A and C, no phase change occurs; therefore, all energy added is used to increase the temperature of the system, which is now steam. The energy that must be added to raise the temperature of the steam from 100.0°C to 120.0°C is Q 5 mscs DT 5 (1.00 3 1023 kg)(2.01 3 103 J/kg ? °C)(20.0°C) 5 40.2 J The total amount of energy that must be added to the system to change 1 g of ice at 230.0°C to steam at 120.0°C is the sum of the results from all five parts of the curve, which is 3.11 3 103 J. Conversely, to cool 1 g of steam at 120.0°C to ice at 230.0°C, we must remove 3.11 3 103 J of energy. Notice in Figure 20.3 the relatively large amount of energy that is transferred into the water to vaporize it to steam. Imagine reversing this process, with a large amount of energy transferred out of steam to condense it into water. That is why a burn to your skin from steam at 100°C is much more damaging than exposure of your skin to water at 100°C. A very large amount of energy enters your skin from the steam, and the steam remains at 100°C for a long time while it condenses. Con-versely, when your skin makes contact with water at 100°C, the water immediately begins to drop in temperature as energy transfers from the water to your skin. If liquid water is held perfectly still in a very clean container, it is possible for the water to drop below 0°C without freezing into ice. This phenomenon, called super-cooling, arises because the water requires a disturbance of some sort for the mol-ecules to move apart and start forming the large, open ice structure that makes the 120 90 60 30 0 T ( C) B Ice water Water 0 500 1000 1 500 2 000 2 500 3 000 3 110 3 070 815 396 62.7 Ice Water steam E Steam A D –30 C Energy added ( J) Figure 20.3 A plot of tempera-ture versus energy added when a system initially consisting of 1.00 g of ice at 230.0°C is converted to steam at 120.0°C. www.aswarphysics.weebly.com 600 Chapter 20 The First Law of Thermodynamics density of ice lower than that of water as discussed in Section 19.4. If supercooled water is disturbed, it suddenly freezes. The system drops into the lower-energy con-figuration of bound molecules of the ice structure, and the energy released raises the temperature back to 0°C. Commercial hand warmers consist of liquid sodium acetate in a sealed plastic pouch. The solution in the pouch is in a stable supercooled state. When a disk in the pouch is clicked by your fingers, the liquid solidifies and the temperature increases, just like the supercooled water just mentioned. In this case, however, the freezing point of the liquid is higher than body temperature, so the pouch feels warm to the touch. To reuse the hand warmer, the pouch must be boiled until the solid lique-fies. Then, as it cools, it passes below its freezing point into the supercooled state. It is also possible to create superheating. For example, clean water in a very clean cup placed in a microwave oven can sometimes rise in temperature beyond 100°C without boiling because the formation of a bubble of steam in the water requires scratches in the cup or some type of impurity in the water to serve as a nucleation site. When the cup is removed from the microwave oven, the superheated water can become explosive as bubbles form immediately and the hot water is forced upward out of the cup.
Q uick Quiz 20.2 Suppose the same process of adding energy to the ice cube is performed as discussed above, but instead we graph the internal energy of the system as a function of energy input. What would this graph look like?
Analyze Write Equation 20.5 to describe the calo-rimetry process: (1) Q cold 5 2Q hot The steam undergoes three processes: first a decrease in temperature to 100°C, then condensation into liquid water, and finally a decrease in temperature of the water to 50.0°C. Find the energy transfer in the first process using the unknown mass ms of the steam: Q 1 5 mscs DTs Find the energy transfer in the second process: Q 2 5 Lv Dms 5 Lv(0 2 ms) 5 2msLv Find the energy transfer in the third process: Q 3 5 mscw DThot water Add the energy transfers in these three stages: (2) Q hot 5 Q 1 1 Q 2 1 Q 3 5 ms(cs DTs 2 Lv 1 cw DThot water) Example 20.4 Cooling the Steam What mass of steam initially at 130°C is needed to warm 200 g of water in a 100-g glass container from 20.0°C to 50.0°C?
Conceptualize Imagine placing water and steam together in a closed insulated container. The system eventually reaches a uniform state of water with a final temperature of 50.0°C.
Categorize Based on our conceptualization of this situation, we categorize this example as one involving calorimetry in which a phase change occurs. The calorimeter is an isolated system for energy: energy transfers between the compo-nents of the system but does not cross the boundary between the system and the environment.
AM S o l u t i o n The 20.0°C water and the glass undergo only one process, an increase in temperature to 50.0°C. Find the energy transfer in this process: (3) Q cold 5 mwcw DTcold water 1 mgcg DTglass Substitute Equations (2) and (3) into Equation (1): mwcw DTcold water 1 mgcg DTglass 5 2ms(cs DTs 2 Lv 1 cw DThot water) Solve for ms: ms 5 2 mwcw DTcold water 1 mgcg DTglass cs DTs 2 Lv 1 cw DThot water www.aswarphysics.weebly.com 20.4 Work and Heat in Thermodynamic Processes 601 dy P A V a b Figure 20.4 Work is done on a gas contained in a cylinder at a pressure P as the piston is pushed downward so that the gas is compressed.
What if the final state of the system is water at 100°C? Would we need more steam or less steam? How would the analysis above change?
Answer More steam would be needed to raise the temperature of the water and glass to 100°C instead of 50.0°C. There would be two major changes in the analysis. First, we would not have a term Q 3 for the steam because the water that condenses from the steam does not cool below 100°C. Second, in Q cold, the temperature change would be 80.0°C instead of 30.0°C. For practice, show that the result is a required mass of steam of 31.8 g.
What If?
Substitute numerical values: ms 5 2 10.200 kg2 14 186 J/kg # 8C2 150.08C 2 20.08C2 1 10.100 kg2 1837 J/kg # 8C2 150.08C 2 20.08C2 12 010 J/kg # 8C2 11008C 2 1308C2 2 12.26 3 106 J/kg2 1 14 186 J/kg # 8C2150.08C 2 1008C2 5 1.09 3 1022 kg 5 10.9 g ▸ 20.4 c on tin u ed 20.4 Work and Heat in Thermodynamic Processes In thermodynamics, we describe the state of a system using such variables as pres-sure, volume, temperature, and internal energy. As a result, these quantities belong to a category called state variables. For any given configuration of the system, we can identify values of the state variables. (For mechanical systems, the state vari-ables include kinetic energy K and potential energy U.) A state of a system can be specified only if the system is in thermal equilibrium internally. In the case of a gas in a container, internal thermal equilibrium requires that every part of the gas be at the same pressure and temperature. A second category of variables in situations involving energy is transfer vari-ables. These variables are those that appear on the right side of the conservation of energy equation, Equation 8.2. Such a variable has a nonzero value if a process occurs in which energy is transferred across the system’s boundary. The transfer variable is positive or negative, depending on whether energy is entering or leaving the system. Because a transfer of energy across the boundary represents a change in the system, transfer variables are not associated with a given state of the system, but rather with a change in the state of the system. In the previous sections, we discussed heat as a transfer variable. In this section, we study another important transfer variable for thermodynamic systems, work. Work performed on particles was studied extensively in Chapter 7, and here we investigate the work done on a deformable system, a gas. Consider a gas contained in a cylinder fitted with a movable piston (Fig. 20.4). At equilibrium, the gas occupies a volume V and exerts a uniform pressure P on the cylinder’s walls and on the piston. If the pis-ton has a cross-sectional area A, the magnitude of the force exerted by the gas on the piston is F 5 PA. By Newton’s third law, the magnitude of the force exerted by the pis-ton on the gas is also PA. Now let’s assume we push the piston inward and compress the gas quasi-statically, that is, slowly enough to allow the system to remain essen-tially in internal thermal equilibrium at all times. The point of application of the force on the gas is the bottom face of the piston. As the piston is pushed downward by an external force F S 5 2F j ^ through a displacement of d r S 5 dy j ^ (Fig. 20.4b), the work done on the gas is, according to our definition of work in Chapter 7, dW 5 F S ?d r S 5 2F j ^ ?dy j ^ 5 2F dy 5 2PA dy The mass of the piston is assumed to be negligible in this discussion. Because A dy is the change in volume of the gas dV, we can express the work done on the gas as dW 5 2P dV (20.8) If the gas is compressed, dV is negative and the work done on the gas is positive. If the gas expands, dV is positive and the work done on the gas is negative. If the www.aswarphysics.weebly.com 602 Chapter 20 The First Law of Thermodynamics volume remains constant, the work done on the gas is zero. The total work done on the gas as its volume changes from Vi to Vf is given by the integral of Equation 20.8: W 5 23 Vf Vi P dV (20.9) To evaluate this integral, you must know how the pressure varies with volume dur-ing the process. In general, the pressure is not constant during a process followed by a gas, but depends on the volume and temperature. If the pressure and volume are known at each step of the process, the state of the gas at each step can be plotted on an important graphical representation called a PV diagram as in Figure 20.5. This type of diagram allows us to visualize a process through which a gas is progressing. The curve on a PV diagram is called the path taken between the initial and final states. Notice that the integral in Equation 20.9 is equal to the area under a curve on a PV diagram. Therefore, we can identify an important use for PV diagrams: The work done on a gas in a quasi-static process that takes the gas from an initial state to a final state is the negative of the area under the curve on a PV diagram, evaluated between the initial and final states. For the process of compressing a gas in a cylinder, the work done depends on the particular path taken between the initial and final states as Figure 20.5 suggests. To illustrate this important point, consider several different paths connecting i and f (Fig. 20.6). In the process depicted in Figure 20.6a, the volume of the gas is first reduced from Vi to Vf at constant pressure Pi and the pressure of the gas then increases from Pi to Pf by heating at constant volume Vf. The work done on the gas along this path is 2Pi(Vf 2 Vi). In Figure 20.6b, the pressure of the gas is increased from Pi to Pf at constant volume Vi and then the volume of the gas is reduced from Vi to Vf at constant pressure Pf. The work done on the gas is 2Pf(Vf 2 Vi). This value is greater than that for the process described in Figure 20.6a because the piston is moved through the same displacement by a larger force. Finally, for the process described in Figure 20.6c, where both P and V change continuously, the work done on the gas has some value between the values obtained in the first two processes. To evaluate the work in this case, the function P(V ) must be known so that we can evaluate the integral in Equation 20.9. The energy transfer Q into or out of a system by heat also depends on the pro-cess. Consider the situations depicted in Figure 20.7. In each case, the gas has the same initial volume, temperature, and pressure, and is assumed to be ideal. In Figure 20.7a, the gas is thermally insulated from its surroundings except at the bottom of the gas-filled region, where it is in thermal contact with an energy reservoir. An energy reservoir is a source of energy that is considered to be so great that a finite transfer of energy to or from the reservoir does not change its temperature. The piston is held Work done on a gas f Pf P i V Vi Vf Pi f Pf P i V Vi Vf Pi f Pf P i V Vi Vf Pi A constant-pressure compression followed by a constant-volume process A constant-volume process followed by a constant-pressure compression An arbitrary compression a b c Figure 20.6 The work done on a gas as it is taken from an initial state to a final state depends on the path between these states.
Figure 20.5 A gas is compressed quasi-statically (slowly) from state i to state f. An outside agent must do positive work on the gas to compress it.
f Pf P i V Vi Vf Pi The work done on a gas equals the negative of the area under the PV curve. The area is negative here because the volume is decreasing, resulting in positive work.
www.aswarphysics.weebly.com 20.5 The First Law of Thermodynamics 603 Figure 20.7 Gas in a cylinder. (a) The gas is in contact with an energy reservoir. The walls of the cylinder are perfectly insulating, but the base in contact with the reservoir is conducting. (b) The gas expands slowly to a larger volume. (c) The gas is contained by a membrane in half of a volume, with vacuum in the other half. The entire cylinder is perfectly insulating. (d) The gas expands freely into the larger volume.
The hand reduces its downward force, allowing the piston to move up slowly. The energy reservoir keeps the gas at temperature Ti . The gas is initially at temperature Ti . a b Energy reservoir at Ti Energy reservoir at Ti c d The gas is initially at temperature Ti and contained by a thin membrane, with vacuum above. The membrane is broken, and the gas expands freely into the evacuated region.
at its initial position by an external agent such as a hand. When the force holding the piston is reduced slightly, the piston rises very slowly to its final position shown in Fig-ure 20.7b. Because the piston is moving upward, the gas is doing work on the piston. During this expansion to the final volume Vf, just enough energy is transferred by heat from the reservoir to the gas to maintain a constant temperature Ti. Now consider the completely thermally insulated system shown in Figure 20.7c. When the membrane is broken, the gas expands rapidly into the vacuum until it occupies a volume Vf and is at a pressure Pf. The final state of the gas is shown in Figure 20.7d. In this case, the gas does no work because it does not apply a force; no force is required to expand into a vacuum. Furthermore, no energy is transferred by heat through the insulating wall. As we discuss in Section 20.5, experiments show that the temperature of the ideal gas does not change in the process indicated in Figures 20.7c and 20.7d. Therefore, the initial and final states of the ideal gas in Figures 20.7a and 20.7b are identical to the initial and final states in Figures 20.7c and 20.7d, but the paths are different. In the first case, the gas does work on the piston and energy is transferred slowly to the gas by heat. In the second case, no energy is transferred by heat and the value of the work done is zero. Therefore, energy transfer by heat, like work done, depends on the particular process occurring in the system. In other words, because heat and work both depend on the path followed on a PV diagram between the initial and final states, neither quantity is determined solely by the endpoints of a thermodynamic process.
20.5 The First Law of Thermodynamics When we introduced the law of conservation of energy in Chapter 8, we stated that the change in the energy of a system is equal to the sum of all transfers of energy across the system’s boundary (Eq. 8.2). The first law of thermodynamics is a spe-cial case of the law of conservation of energy that describes processes in which only the internal energy5 changes and the only energy transfers are by heat and work: DE int 5 Q 1 W (20.10) W W First law of thermodynamics 5It is an unfortunate accident of history that the traditional symbol for internal energy is U, which is also the tra-ditional symbol for potential energy as introduced in Chapter 7. To avoid confusion between potential energy and internal energy, we use the symbol Eint for internal energy in this book. If you take an advanced course in thermody-namics, however, be prepared to see U used as the symbol for internal energy in the first law.
www.aswarphysics.weebly.com 604 Chapter 20 The First Law of Thermodynamics Look back at Equation 8.2 to see that the first law of thermodynamics is contained within that more general equation. Let us investigate some special cases in which the first law can be applied. First, consider an isolated system, that is, one that does not interact with its surroundings, as we have seen before. In this case, no energy transfer by heat takes place and the work done on the system is zero; hence, the internal energy remains constant. That is, because Q 5 W 5 0, it follows that DE int 5 0; therefore, E int,i 5 E int,f. We con-clude that the internal energy E int of an isolated system remains constant. Next, consider the case of a system that can exchange energy with its surround-ings and is taken through a cyclic process, that is, a process that starts and ends at the same state. In this case, the change in the internal energy must again be zero because E int is a state variable; therefore, the energy Q added to the system must equal the negative of the work W done on the system during the cycle. That is, in a cyclic process, DE int 5 0 and Q 5 2W (cyclic process) On a PV diagram for a gas, a cyclic process appears as a closed curve. (The pro-cesses described in Figure 20.6 are represented by open curves because the initial and final states differ.) It can be shown that in a cyclic process for a gas, the net work done on the system per cycle equals the area enclosed by the path represent-ing the process on a PV diagram.
20.6 Some Applications of the First Law of Thermodynamics In this section, we consider additional applications of the first law to processes through which a gas is taken. As a model, let’s consider the sample of gas contained in the piston–cylinder apparatus in Figure 20.8. This figure shows work being done on the gas and energy transferring in by heat, so the internal energy of the gas is rising. In the following discussion of various processes, refer back to this figure and mentally alter the directions of the transfer of energy to reflect what is hap-pening in the process. Before we apply the first law of thermodynamics to specific systems, it is useful to first define some idealized thermodynamic processes. An adiabatic process is one during which no energy enters or leaves the system by heat; that is, Q 5 0. An adiabatic process can be achieved either by thermally insulating the walls of the system or by performing the process rapidly so that there is negligible time for energy to transfer by heat. Applying the first law of thermodynamics to an adia-batic process gives DE int 5 W (adiabatic process) (20.11) This result shows that if a gas is compressed adiabatically such that W is positive, then DE int is positive and the temperature of the gas increases. Conversely, the tem-perature of a gas decreases when the gas expands adiabatically. Adiabatic processes are very important in engineering practice. Some common examples are the expansion of hot gases in an internal combustion engine, the liquefaction of gases in a cooling system, and the compression stroke in a diesel engine. The process described in Figures 20.7c and 20.7d, called an adiabatic free expansion, is unique. The process is adiabatic because it takes place in an insulated container. Because the gas expands into a vacuum, it does not apply a force on a piston as does the gas in Figures 20.7a and 20.7b, so no work is done on or by the gas. Therefore, in this adiabatic process, both Q 5 0 and W 5 0. As a result, DEint 5 0 for this process as can be seen from the first law. That is, the initial and final internal energies of a gas are equal in an adiabatic free expansion. As we shall see Pitfall Prevention 20.7 Dual Sign Conventions Some phys-ics and engineering books present the first law as DE int 5 Q 2 W, with a minus sign between the heat and work. The reason is that work is defined in these treatments as the work done by the gas rather than on the gas, as in our treat-ment. The equivalent equation to Equation 20.9 in these treatments defines work as W 5 e Vf Vi P dV. Therefore, if positive work is done by the gas, energy is leaving the system, leading to the negative sign in the first law. In your studies in other chem-istry or engineering courses, or in your reading of other physics books, be sure to note which sign convention is being used for the first law.
Pitfall Prevention 20.8 The First Law With our approach to energy in this book, the first law of thermodynamics is a special case of Equation 8.2. Some physicists argue that the first law is the gen-eral equation for energy conserva-tion, equivalent to Equation 8.2. In this approach, the first law is applied to a closed system (so that there is no matter transfer), heat is interpreted so as to include elec-tromagnetic radiation, and work is interpreted so as to include electri-cal transmission (“electrical work”) and mechanical waves (“molecular work”). Keep that in mind if you run across the first law in your reading of other physics books.
Q W Q Eint Figure 20.8 The first law of ther-modynamics equates the change in internal energy E int in a system to the net energy transfer to the sys-tem by heat Q and work W. In the situation shown here, the internal energy of the gas increases.
www.aswarphysics.weebly.com 20.6 Some Applications of the First Law of Thermodynamics 605 Pitfall Prevention 20.9 Q 20 in an Isothermal Process Do not fall into the common trap of thinking there must be no transfer of energy by heat if the temperature does not change as is the case in an isothermal process. Because the cause of temperature change can be either heat or work, the temperature can remain con-stant even if energy enters the gas by heat, which can only happen if the energy entering the gas by heat leaves by work.
in Chapter 21, the internal energy of an ideal gas depends only on its temperature. Therefore, we expect no change in temperature during an adiabatic free expan-sion. This prediction is in accord with the results of experiments performed at low pressures. (Experiments performed at high pressures for real gases show a slight change in temperature after the expansion due to intermolecular interactions, which represent a deviation from the model of an ideal gas.) A process that occurs at constant pressure is called an isobaric process. In Fig-ure 20.8, an isobaric process could be established by allowing the piston to move freely so that it is always in equilibrium between the net force from the gas pushing upward and the weight of the piston plus the force due to atmospheric pressure pushing downward. The first process in Figure 20.6a and the second process in Figure 20.6b are both isobaric. In such a process, the values of the heat and the work are both usually nonzero. The work done on the gas in an isobaric process is simply W 5 2P(Vf 2 Vi) (isobaric process) (20.12) where P is the constant pressure of the gas during the process. A process that takes place at constant volume is called an isovolumetric process. In Figure 20.8, clamping the piston at a fixed position would ensure an isovolu-metric process. The second process in Figure 20.6a and the first process in Figure 20.6b are both isovolumetric. Because the volume of the gas does not change in such a process, the work given by Equation 20.9 is zero. Hence, from the first law we see that in an isovolumetric process, because W 5 0, DE int 5 Q (isovolumetric process) (20.13) This expression specifies that if energy is added by heat to a system kept at constant volume, all the transferred energy remains in the system as an increase in its inter-nal energy. For example, when a can of spray paint is thrown into a fire, energy enters the system (the gas in the can) by heat through the metal walls of the can. Consequently, the temperature, and therefore the pressure, in the can increases until the can possibly explodes. A process that occurs at constant temperature is called an isothermal process. This process can be established by immersing the cylinder in Figure 20.8 in an ice–water bath or by putting the cylinder in contact with some other constant-temperature reservoir. A plot of P versus V at constant temperature for an ideal gas yields a hyperbolic curve called an isotherm. The internal energy of an ideal gas is a function of temperature only. Hence, because the temperature does not change in an isothermal process involving an ideal gas, we must have DEint 5 0. For an isothermal process, we conclude from the first law that the energy transfer Q must be equal to the negative of the work done on the gas; that is, Q 5 2W. Any energy that enters the system by heat is transferred out of the system by work; as a result, no change in the internal energy of the system occurs in an isothermal process.
Q uick Quiz 20.3 In the last three columns of the following table, fill in the boxes with the correct signs (2, 1, or 0) for Q, W, and DEint. For each situation, the sys-tem to be considered is identified.
Situation System Q W DE int (a) Rapidly pumping up Air in the pump a bicycle tire (b) Pan of room-temperature Water in the pan water sitting on a hot stove (c) Air quickly leaking out Air originally in the balloon of a balloon W W Isobaric process W W Isovolumetric process W W Isothermal process www.aswarphysics.weebly.com 606 Chapter 20 The First Law of Thermodynamics Isothermal Expansion of an Ideal Gas Suppose an ideal gas is allowed to expand quasi-statically at constant temperature. This process is described by the PV diagram shown in Figure 20.9. The curve is a hyperbola (see Appendix B, Eq. B.23), and the ideal gas law (Eq. 19.8) with T con-stant indicates that the equation of this curve is PV 5 nRT 5 constant. Let’s calculate the work done on the gas in the expansion from state i to state f. The work done on the gas is given by Equation 20.9. Because the gas is ideal and the process is quasi-static, the ideal gas law is valid for each point on the path. Therefore, W 5 23 Vf Vi P dV 5 23 Vf Vi nRT V dV Because T is constant in this case, it can be removed from the integral along with n and R: W 5 2nRT 3 Vf Vi dV V 5 2nRT lnV Vf Vi To evaluate the integral, we used e(dx/x) 5 ln x. (See Appendix B.) Evaluating the result at the initial and final volumes gives W 5 nRT ln aVi Vf b (20.14) Numerically, this work W equals the negative of the shaded area under the PV curve shown in Figure 20.9. Because the gas expands, Vf . Vi and the value for the work done on the gas is negative as we expect. If the gas is compressed, then Vf , Vi and the work done on the gas is positive.
Q uick Quiz 20.4 Characterize the paths in Figure 20.10 as isobaric, isovolumet-ric, isothermal, or adiabatic. For path B, Q 5 0. The blue curves are isotherms.
Example 20.5 An Isothermal Expansion A 1.0-mol sample of an ideal gas is kept at 0.0°C during an expansion from 3.0 L to 10.0 L.
(A) How much work is done on the gas during the expansion?
Conceptualize Run the process in your mind: the cylinder in Figure 20.8 is immersed in an ice-water bath, and the piston moves outward so that the volume of the gas increases. You can also use the graphical representation in Figure 20.9 to conceptualize the process.
Categorize We will evaluate parameters using equations developed in the preceding sections, so we categorize this example as a substitution problem. Because the temperature of the gas is fixed, the process is isothermal.
S o l u t i o n Substitute the given values into Equation 20.14: W 5 nRT ln aVi Vf b 5 11.0 mol2 18.31 J/mol # K2 1273 K2 ln a 3.0 L 10.0 Lb 5 22.7 3 103 J (B) How much energy transfer by heat occurs between the gas and its surroundings in this process?
f i V PV = constant Isotherm P Pi Pf Vi Vf The curve is a hyperbola.
Figure 20.9 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a final state.
A B C D V P T1 T3 T2 T4 Figure 20.10 (Quick Quiz 20.4) Identify the nature of paths A, B, C, and D.
S o l u t i o n Find the heat from the first law: DE int 5 Q 1 W 0 5 Q 1 W Q 5 2W 5 2.7 3 103 J www.aswarphysics.weebly.com 20.6 Some Applications of the First Law of Thermodynamics 607 (C) If the gas is returned to the original volume by means of an isobaric process, how much work is done on the gas?
S o l u t i o n Use Equation 20.12. The pressure is not given, so incorporate the ideal gas law: W 5 2P 1Vf 2 Vi 2 5 2 nRTi Vi 1Vf 2 Vi 2 5 2 11.0 mol2 18.31 J/mol # K2 1273 K2 10.0 3 1023 m3 13.0 3 1023 m3 2 10.0 3 1023 m32 5 1.6 3 103 J We used the initial temperature and volume to calculate the work done because the final temperature was unknown. The work done on the gas is positive because the gas is being compressed.
Example 20.6 Boiling Water Suppose 1.00 g of water vaporizes isobarically at atmospheric pressure (1.013 3 105 Pa). Its volume in the liquid state is Vi 5 Vliquid 5 1.00 cm3, and its volume in the vapor state is Vf 5 Vvapor 5 1 671 cm3. Find the work done in the expansion and the change in internal energy of the system. Ignore any mixing of the steam and the surrounding air; imagine that the steam simply pushes the surrounding air out of the way.
Conceptualize Notice that the temperature of the system does not change. There is a phase change occurring as the water evaporates to steam.
Categorize Because the expansion takes place at constant pressure, we categorize the process as isobaric. We will use equations developed in the preceding sections, so we categorize this example as a substitution problem.
S o l u t i o n Use the first law to find the change in internal energy of the system: DE int 5 Q 1 W 5 2 260 J 1 (2169 J) 5 2.09 kJ Use Equation 20.7 and the latent heat of vaporization for water to find the energy transferred into the system by heat: Q 5 Lv Dms 5 msLv 5 (1.00 3 1023 kg)(2.26 3 106 J/kg) 5 2 260 J Use Equation 20.12 to find the work done on the sys-tem as the air is pushed out of the way: W 5 2P(Vf 2 Vi) 5 2(1.013 3 105 Pa)(1 671 3 1026 m3 2 1.00 3 1026 m3) 5 2169 J The positive value for DE int indicates that the internal energy of the system increases. The largest fraction of the energy (2 090 J/ 2260 J 5 93%) transferred to the liquid goes into increasing the internal energy of the system. The remaining 7% of the energy transferred leaves the system by work done by the steam on the surrounding atmosphere.
Example 20.7 Heating a Solid A 1.0-kg bar of copper is heated at atmospheric pressure so that its temperature increases from 20°C to 50°C.
(A) What is the work done on the copper bar by the surrounding atmosphere?
Conceptualize This example involves a solid, whereas the preceding two examples involved liquids and gases. For a solid, the change in volume due to thermal expansion is very small.
S o l u t i o n ▸ 20.5 c on tin u ed continued www.aswarphysics.weebly.com 608 Chapter 20 The First Law of Thermodynamics Categorize Because the expansion takes place at constant atmospheric pressure, we categorize the process as isobaric.
Analyze Find the work done on the copper bar using Equation 20.12: W 5 2P DV Express the change in volume using Equation 19.6 and that b 5 3a: W 5 2P(bVi DT) 5 2P(3aVi DT) 5 23aPVi DT Substitute for the volume in terms of the mass and den-sity of the copper: W 5 23aP am r b DT Substitute numerical values: W 5 2331.7 3 1025 18C2 214 11.013 3 105 N/m22 a 1.0 kg 8.92 3 103 kg/m3b 1508C 2 208C2 5 21.7 3 1022 J Because this work is negative, work is done by the copper bar on the atmosphere.
(B) How much energy is transferred to the copper bar by heat?
S o l u t i o n Use Equation 20.4 and the specific heat of copper from Table 20.1: Q 5 mc DT 5 (1.0 kg)(387 J/kg ? °C)(50°C 2 20°C) 5 1.2 3 104 J (C) What is the increase in internal energy of the copper bar?
S o l u t i o n Use the first law of thermodynamics: DE int 5 Q 1 W 5 1.2 3 104 J 1 (21.7 3 1022 J) 5 1.2 3 104 J Finalize Most of the energy transferred into the system by heat goes into increasing the internal energy of the cop-per bar. The fraction of energy used to do work on the surrounding atmosphere is only about 1026. Hence, when the thermal expansion of a solid or a liquid is analyzed, the small amount of work done on or by the system is usually ignored. ▸ 20.7 continu ed 20.7 Energy Transfer Mechanisms in Thermal Processes In Chapter 8, we introduced a global approach to the energy analysis of physical processes through Equation 8.1, DE system 5 o T, where T represents energy transfer, which can occur by several mechanisms. Earlier in this chapter, we discussed two of the terms on the right side of this equation, work W and heat Q. In this section, we explore more details about heat as a means of energy transfer and two other energy transfer methods often related to temperature changes: convection (a form of mat-ter transfer TMT) and electromagnetic radiation TER.
Thermal Conduction The process of energy transfer by heat (Q in Eq. 8.2) can also be called conduc-tion or thermal conduction. In this process, the transfer can be represented on an atomic scale as an exchange of kinetic energy between microscopic particles—mol-ecules, atoms, and free electrons—in which less-energetic particles gain energy in collisions with more-energetic particles. For example, if you hold one end of a long metal bar and insert the other end into a flame, you will find that the temperature www.aswarphysics.weebly.com 20.7 Energy Transfer Mechanisms in Thermal Processes 609 of the metal in your hand soon increases. The energy reaches your hand by means of conduction. Initially, before the rod is inserted into the flame, the microscopic particles in the metal are vibrating about their equilibrium positions. As the flame raises the temperature of the rod, the particles near the flame begin to vibrate with greater and greater amplitudes. These particles, in turn, collide with their neigh-bors and transfer some of their energy in the collisions. Slowly, the amplitudes of vibration of metal atoms and electrons farther and farther from the flame increase until eventually those in the metal near your hand are affected. This increased vibration is detected by an increase in the temperature of the metal and of your potentially burned hand. The rate of thermal conduction depends on the properties of the substance being heated. For example, it is possible to hold a piece of asbestos in a flame indef-initely, which implies that very little energy is conducted through the asbestos. In general, metals are good thermal conductors and materials such as asbestos, cork, paper, and fiberglass are poor conductors. Gases also are poor conductors because the separation distance between the particles is so great. Metals are good thermal conductors because they contain large numbers of electrons that are relatively free to move through the metal and so can transport energy over large distances. There-fore, in a good conductor such as copper, conduction takes place by means of both the vibration of atoms and the motion of free electrons. Conduction occurs only if there is a difference in temperature between two parts of the conducting medium. Consider a slab of material of thickness Dx and cross-sectional area A. One face of the slab is at a temperature Tc, and the other face is at a temperature Th . Tc (Fig. 20.11). Experimentally, it is found that energy Q transfers in a time interval Dt from the hotter face to the colder one. The rate P 5 Q /Dt at which this energy transfer occurs is found to be proportional to the cross-sectional area and the temperature difference DT 5 Th 2 Tc and inversely propor-tional to the thickness: P 5 Q Dt ~ A DT Dx Notice that P has units of watts when Q is in joules and Dt is in seconds. That is not surprising because P is power, the rate of energy transfer by heat. For a slab of infinitesimal thickness dx and temperature difference dT, we can write the law of thermal conduction as P 5 kA dT dx (20.15) where the proportionality constant k is the thermal conductivity of the material and |dT/dx| is the temperature gradient (the rate at which temperature varies with position). Substances that are good thermal conductors have large thermal conductiv-ity values, whereas good thermal insulators have low thermal conductivity values. Table 20.3 lists thermal conductivities for various substances. Notice that metals are generally better thermal conductors than nonmetals. Suppose a long, uniform rod of length L is thermally insulated so that energy cannot escape by heat from its surface except at the ends as shown in Figure 20.12 (page 610). One end is in thermal contact with an energy reservoir at temperature Tc, and the other end is in thermal contact with a reservoir at temperature Th . Tc. When a steady state has been reached, the temperature at each point along the rod is constant in time. In this case, if we assume k is not a function of temperature, the temperature gradient is the same everywhere along the rod and is dT dx 5 Th 2 Tc L Table 20.3 Thermal Conductivities Thermal Conductivity Substance (W/m ? °C) Metals (at 25°C) Aluminum 238 Copper 397 Gold 314 Iron 79.5 Lead 34.7 Silver 427 Nonmetals (approximate values) Asbestos 0.08 Concrete 0.8 Diamond 2 300 Glass 0.8 Ice 2 Rubber 0.2 Water 0.6 Wood 0.08 Gases (at 20°C) Air 0.023 4 Helium 0.138 Hydrogen 0.172 Nitrogen 0.023 4 Oxygen 0.023 8 The opposite faces are at different temperatures where Th Tc .
Tc Energy transfer for Th Tc Th A x Figure 20.11 Energy transfer through a conducting slab with a cross-sectional area A and a thick-ness Dx.
www.aswarphysics.weebly.com 610 Chapter 20 The First Law of Thermodynamics Therefore, the rate of energy transfer by conduction through the rod is P 5 kA aTh 2 Tc L b (20.16) For a compound slab containing several materials of thicknesses L1, L2, . . . and thermal conductivities k1, k2, . . . , the rate of energy transfer through the slab at steady state is P 5 A1Th 2 Tc 2 a i 1Li/ki2 (20.17) where Th and Tc are the temperatures of the outer surfaces (which are held con-stant) and the summation is over all slabs. Example 20.8 shows how Equation 20.17 results from a consideration of two thicknesses of materials.
Q uick Quiz 20.5 You have two rods of the same length and diameter, but they are formed from different materials. The rods are used to connect two regions at different temperatures so that energy transfers through the rods by heat. They can be connected in series as in Figure 20.13a or in parallel as in Figure 20.13b. In which case is the rate of energy transfer by heat larger? (a) The rate is larger when the rods are in series. (b) The rate is larger when the rods are in parallel. (c) The rate is the same in both cases.
Example 20.8 Energy Transfer Through Two Slabs Two slabs of thickness L1 and L2 and thermal conductivities k1 and k2 are in thermal contact with each other as shown in Figure 20.14. The temperatures of their outer surfaces are Tc and Th, respectively, and Th . Tc. Determine the tem-perature at the interface and the rate of energy transfer by conduction through an area A of the slabs in the steady-state condition.
Conceptualize Notice the phrase “in the steady-state condition.” We interpret this phrase to mean that energy transfers through the compound slab at the same rate at all points. Otherwise, energy would be building up or disappear-ing at some point. Furthermore, the temperature varies with position in the two slabs, most likely at different rates in each part of the compound slab. When the system is in steady state, the interface is at some fixed temperature T.
Categorize We categorize this example as a thermal conduction problem and impose the condition that the power is the same in both slabs of material.
S o l u t i o n L2 L1 Th k2 k1 Tc T Figure 20.14 (Example 20.8) Energy transfer by conduction through two slabs in thermal contact with each other. At steady state, the rate of energy transfer through slab 1 equals the rate of energy transfer through slab 2.
The opposite ends of the rod are in thermal contact with energy reservoirs at different temperatures.
Th Insulation Tc L Energy transfer Th Tc Figure 20.12 Conduction of energy through a uniform, insulated rod of length L.
a b Rod 1 Rod 2 Th Rod 1 Rod 2 Th Tc Tc Figure 20.13 (Quick Quiz 20.5) In which case is the rate of energy transfer larger?
www.aswarphysics.weebly.com 20.7 Energy Transfer Mechanisms in Thermal Processes 611 Substitute Equation (3) into either Equation (1) or Equation (2): (4) P 5 A1Th 2 Tc 2 1L1/k 12 1 1L 2/k 22 Solve for T : (3) T 5 k1 L 2 Tc 1 k 2 L1 Th k1 L 2 1 k 2 L1 Set these two rates equal to represent the steady-state situation: k1A a T 2 Tc L 1 b 5 k 2A aTh 2 T L 2 b Express the rate at which energy is transferred through the same area of slab 2: (2) P2 5 k 2A aTh 2 T L 2 b Analyze Use Equation 20.16 to express the rate at which energy is transferred through an area A of slab 1: (1) P1 5 k1Aa T 2 Tc L 1 b Finalize Extension of this procedure to several slabs of materials leads to Equation 20.17.
Suppose you are building an insulated container with two layers of insulation and the rate of energy transfer determined by Equation (4) turns out to be too high. You have enough room to increase the thickness of one of the two layers by 20%. How would you decide which layer to choose?
Answer To decrease the power as much as possible, you must increase the denominator in Equation (4) as much as possible. Whichever thickness you choose to increase, L1 or L2, you increase the corresponding term L/k in the denominator by 20%. For this percentage change to represent the largest absolute change, you want to take 20% of the larger term. Therefore, you should increase the thickness of the layer that has the larger value of L/k.
What If? ▸ 20.8 c on tin u ed Home Insulation In engineering practice, the term L/k for a particular substance is referred to as the R-value of the material. Therefore, Equation 20.17 reduces to P 5 A1Th 2 Tc 2 a i R i (20.18) where Ri 5 Li/ki. The R-values for a few common building materials are given in Table 20.4. In the United States, the insulating properties of materials used in buildings are usually expressed in U.S. customary units, not SI units. Therefore, in Table 20.4 R-Values for Some Common Building Materials Material R-value (ft2 ? °F ? h/Btu) Hardwood siding (1 in. thick) 0.91 Wood shingles (lapped) 0.87 Brick (4 in. thick) 4.00 Concrete block (filled cores) 1.93 Fiberglass insulation (3.5 in. thick) 10.90 Fiberglass insulation (6 in. thick) 18.80 Fiberglass board (1 in. thick) 4.35 Cellulose fiber (1 in. thick) 3.70 Flat glass (0.125 in. thick) 0.89 Insulating glass (0.25-in. space) 1.54 Air space (3.5 in. thick) 1.01 Stagnant air layer 0.17 Drywall (0.5 in. thick) 0.45 Sheathing (0.5 in. thick) 1.32 www.aswarphysics.weebly.com 612 Chapter 20 The First Law of Thermodynamics Table 20.4, R-values are given as a combination of British thermal units, feet, hours, and degrees Fahrenheit. At any vertical surface open to the air, a very thin stagnant layer of air adheres to the surface. One must consider this layer when determining the R-value for a wall. The thickness of this stagnant layer on an outside wall depends on the speed of the wind. Energy transfer through the walls of a house on a windy day is greater than that on a day when the air is calm. A representative R-value for this stagnant layer of air is given in Table 20.4.
Example 20.9 The R-Value of a Typical Wall Calculate the total R-value for a wall constructed as shown in Figure 20.15a. Starting outside the house (toward the front in the figure) and moving inward, the wall consists of 4 in. of brick, 0.5 in. of sheathing, an air space 3.5 in. thick, and 0.5 in. of drywall.
Conceptualize Use Figure 20.15 to help conceptualize the structure of the wall. Do not forget the stagnant air layers inside and outside the house.
Categorize We will use specific equations developed in this section on home insulation, so we categorize this example as a substitution problem.
S o l u t i o n Drywall Brick Sheathing Air space Insulation a b Figure 20.15 (Exam-ple 20.9) An exterior house wall containing (a) an air space and (b) insulation.
Add the R-values to obtain the total R-value for the wall: R total 5 R1 1 R 2 1 R3 1 R 4 1 R 5 1 R 6 5 7.12 ft2 ? °F ? h/Btu Use Table 20.4 to find the R-value of each layer: R 1 (outside stagnant air layer) 5 0.17 ft2 ? °F ? h/Btu R 2 (brick) 5 4.00 ft2 ? °F ? h/Btu R 3 (sheathing) 5 1.32 ft2 ? °F ? h/Btu R 4 (air space) 5 1.01 ft2 ? °F ? h/Btu R 5 (drywall) 5 0.45 ft2 ? °F ? h/Btu R 6 (inside stagnant air layer) 5 0.17 ft2 ? °F ? h/Btu Suppose you are not happy with this total R-value for the wall. You cannot change the overall structure, but you can fill the air space as in Figure 20.15b. To maximize the total R-value, what material should you choose to fill the air space?
Answer Looking at Table 20.4, we see that 3.5 in. of fiberglass insulation is more than ten times as effective as 3.5 in. of air. Therefore, we should fill the air space with fiberglass insulation. The result is that we add 10.90 ft2 ? °F ? h/Btu of R-value, and we lose 1.01 ft2 ? °F ? h/Btu due to the air space we have replaced. The new total R-value is equal to 7.12 ft2 ? °F ? h/Btu 1 9.89 ft2 ? °F ? h/Btu 5 17.01 ft2 ? °F ? h/Btu.
What If?
Convection At one time or another, you probably have warmed your hands by holding them over an open flame. In this situation, the air directly above the flame is heated and expands. As a result, the density of this air decreases and the air rises. This hot air warms your hands as it flows by. Energy transferred by the movement of a warm substance is said to have been transferred by convection, which is a form of matter transfer, TMT in Equation 8.2. When resulting from differences in density, as with air around a fire, the process is referred to as natural convection. Airflow at a beach www.aswarphysics.weebly.com 20.7 Energy Transfer Mechanisms in Thermal Processes 613 Radiator Figure 20.16 Convection cur-rents are set up in a room warmed by a radiator.
is an example of natural convection, as is the mixing that occurs as surface water in a lake cools and sinks (see Section 19.4). When the heated substance is forced to move by a fan or pump, as in some hot-air and hot-water heating systems, the pro-cess is called forced convection. If it were not for convection currents, it would be very difficult to boil water. As water is heated in a teakettle, the lower layers are warmed first. This water expands and rises to the top because its density is lowered. At the same time, the denser, cool water at the surface sinks to the bottom of the kettle and is heated. The same process occurs when a room is heated by a radiator. The hot radiator warms the air in the lower regions of the room. The warm air expands and rises to the ceiling because of its lower density. The denser, cooler air from above sinks, and the continuous air current pattern shown in Figure 20.16 is established.
Radiation The third means of energy transfer we shall discuss is thermal radiation, TER in Equation 8.2. All objects radiate energy continuously in the form of electromag-netic waves (see Chapter 34) produced by thermal vibrations of the molecules. You are likely familiar with electromagnetic radiation in the form of the orange glow from an electric stove burner, an electric space heater, or the coils of a toaster. The rate at which the surface of an object radiates energy is proportional to the fourth power of the absolute temperature of the surface. Known as Stefan’s law, this behavior is expressed in equation form as P 5 sAeT 4 (20.19) where P is the power in watts of electromagnetic waves radiated from the surface of the object, s is a constant equal to 5.669 6 3 1028 W/m2 ? K4, A is the surface area of the object in square meters, e is the emissivity, and T is the surface temperature in kelvins. The value of e can vary between zero and unity depending on the prop-erties of the surface of the object. The emissivity is equal to the absorptivity, which is the fraction of the incoming radiation that the surface absorbs. A mirror has very low absorptivity because it reflects almost all incident light. Therefore, a mir-ror surface also has a very low emissivity. At the other extreme, a black surface has high absorptivity and high emissivity. An ideal absorber is defined as an object that absorbs all the energy incident on it, and for such an object, e 5 1. An object for which e 5 1 is often referred to as a black body. We shall investigate experimental and theoretical approaches to radiation from a black body in Chapter 40. Every second, approximately 1 370 J of electromagnetic radiation from the Sun passes perpendicularly through each 1 m2 at the top of the Earth’s atmosphere. This radiation is primarily visible and infrared light accompanied by a significant amount of ultraviolet radiation. We shall study these types of radiation in detail in Chapter 34. Enough energy arrives at the surface of the Earth each day to supply all our energy needs on this planet hundreds of times over, if only it could be captured and used efficiently. The growth in the number of solar energy–powered houses and proposals for solar energy “farms” in the United States reflects the increasing efforts being made to use this abundant energy. What happens to the atmospheric temperature at night is another example of the effects of energy transfer by radiation. If there is a cloud cover above the Earth, the water vapor in the clouds absorbs part of the infrared radiation emitted by the Earth and re-emits it back to the surface. Consequently, temperature levels at the surface remain moderate. In the absence of this cloud cover, there is less in the way to prevent this radiation from escaping into space; therefore, the temperature decreases more on a clear night than on a cloudy one. As an object radiates energy at a rate given by Equation 20.19, it also absorbs electromagnetic radiation from the surroundings, which consist of other objects W W Stefan’s law www.aswarphysics.weebly.com 614 Chapter 20 The First Law of Thermodynamics that radiate energy. If the latter process did not occur, an object would eventually radiate all its energy and its temperature would reach absolute zero. If an object is at a temperature T and its surroundings are at an average temperature T0, the net rate of energy gained or lost by the object as a result of radiation is Pnet 5 sAe(T 4 2 T0 4) (20.20) When an object is in equilibrium with its surroundings, it radiates and absorbs energy at the same rate and its temperature remains constant. When an object is hotter than its surroundings, it radiates more energy than it absorbs and its tem-perature decreases.
The Dewar Flask The Dewar flask6 is a container designed to minimize energy transfers by conduc-tion, convection, and radiation. Such a container is used to store cold or hot liquids for long periods of time. (An insulated bottle, such as a Thermos, is a common household equivalent of a Dewar flask.) The standard construction (Fig. 20.17) consists of a double-walled Pyrex glass vessel with silvered walls. The space between the walls is evacuated to minimize energy transfer by conduction and convection. The silvered surfaces minimize energy transfer by radiation because silver is a very good reflector and has very low emissivity. A further reduction in energy loss is obtained by reducing the size of the neck. Dewar flasks are commonly used to store liquid nitrogen (boiling point 77 K) and liquid oxygen (boiling point 90 K). To confine liquid helium (boiling point 4.2 K), which has a very low heat of vaporization, it is often necessary to use a double Dewar system in which the Dewar flask containing the liquid is surrounded by a second Dewar flask. The space between the two flasks is filled with liquid nitrogen. Newer designs of storage containers use “superinsulation” that consists of many layers of reflecting material separated by fiberglass. All this material is in a vacuum, and no liquid nitrogen is needed with this design.
6Invented by Sir James Dewar (1842–1923).
Vacuum (white area) Hot or cold liquid Silvered surfaces Figure 20.17 A cross-sectional view of a Dewar flask, which is used to store hot or cold substances.
Summary Internal energy is a system’s energy associated with its temperature and its physical state (solid, liquid, gas). Internal energy includes kinetic energy of random translation, rotation, and vibration of mol-ecules; vibrational potential energy within molecules; and potential energy between molecules. Heat is the process of energy transfer across the boundary of a system resulting from a temperature difference between the system and its surroundings. The symbol Q represents the amount of energy trans-ferred by this process. A calorie is the amount of energy necessary to raise the tem-perature of 1 g of water from 14.5°C to 15.5°C. The heat capacity C of any sample is the amount of energy needed to raise the temperature of the sample by 1°C. The specific heat c of a substance is the heat capacity per unit mass: c ; Q m DT (20.3) The latent heat of a substance is defined as the ratio of the energy input to a substance to the change in mass of the higher-phase material: L ; Q Dm (20.6) Definitions www.aswarphysics.weebly.com Objective Questions 615 The energy Q required to change the temperature of a mass m of a substance by an amount DT is Q 5 mc DT (20.4) where c is the specific heat of the substance. The energy required to change the phase of a pure sub-stance is Q 5 L Dm (20.7) where L is the latent heat of the substance, which depends on the nature of the phase change and the substance, and Dm is the change in mass of the higher-phase material. Conduction can be viewed as an exchange of kinetic energy between colliding molecules or electrons. The rate of energy transfer by conduction through a slab of area A is P 5 kA dT dx (20.15) where k is the thermal conductivity of the material from which the slab is made and |dT/dx| is the temperature gradient. The work done on a gas as its volume changes from some initial value Vi to some final value Vf is W 5 23 Vf Vi P dV (20.9) where P is the pressure of the gas, which may vary during the process. To evaluate W, the process must be fully specified; that is, P and V must be known during each step. The work done depends on the path taken between the initial and final states. In convection, a warm sub-stance transfers energy from one location to another. All objects emit thermal radiation in the form of electro-magnetic waves at the rate P 5 sAeT 4 (20.19) Concepts and Principles The first law of thermodynamics is a specific reduction of the conservation of energy equation (Eq. 8.2) and states that when a system undergoes a change from one state to another, the change in its internal energy is DE int 5 Q 1 W (20.10) where Q is the energy transferred into the system by heat and W is the work done on the system. Although Q and W both depend on the path taken from the initial state to the final state, the quantity DE int does not depend on the path. In a cyclic process (one that originates and termi-nates at the same state), DE int 5 0 and therefore Q 5 2W. That is, the energy transferred into the system by heat equals the negative of the work done on the sys-tem during the process. In an adiabatic process, no energy is transferred by heat between the system and its surroundings (Q 5 0). In this case, the first law gives DE int 5 W. In the adiabatic free expansion of a gas, Q 5 0 and W 5 0, so DE int 5 0. That is, the internal energy of the gas does not change in such a process. An isobaric process is one that occurs at constant pressure. The work done on a gas in such a process is W 5 2P(Vf 2 Vi). An isovolumetric process is one that occurs at con-stant volume. No work is done in such a process, so DE int 5 Q. An isothermal process is one that occurs at constant temperature. The work done on an ideal gas during an isothermal process is W 5 nRT ln aVi Vf b (20.14) 2. A poker is a stiff, nonflammable rod used to push burn-ing logs around in a fireplace. For safety and comfort of use, should the poker be made from a material with (a) high specific heat and high thermal conductivity, (b) low specific heat and low thermal conductivity, 1. An ideal gas is compressed to half its initial volume by means of several possible processes. Which of the fol-lowing processes results in the most work done on the gas? (a) isothermal (b) adiabatic (c) isobaric (d) The work done is independent of the process.
Objective Questions 1. denotes answer available in Student Solutions Manual/Study Guide www.aswarphysics.weebly.com 616 Chapter 20 The First Law of Thermodynamics largest to the smallest. In your ranking, note any cases of equality. (a) raising the temperature of 1 kg of H2O from 20°C to 26°C (b) raising the temperature of 2 kg of H2O from 20°C to 23°C (c) raising the temperature of 2 kg of H2O from 1°C to 4°C (d) raising the tempera-ture of 2 kg of beryllium from 21°C to 2°C (e) raising the temperature of 2 kg of H2O from 21°C to 2°C 9. A person shakes a sealed insulated bottle containing hot coffee for a few minutes. (i) What is the change in the temperature of the coffee? (a) a large decrease (b) a slight decrease (c) no change (d) a slight increase (e) a large increase (ii) What is the change in the internal energy of the coffee? Choose from the same possibilities. 10. A 100-g piece of copper, initially at 95.0°C, is dropped into 200 g of water contained in a 280-g aluminum can; the water and can are initially at 15.0°C. What is the final temperature of the system? (Specific heats of copper and aluminum are 0.092 and 0.215 cal/g ? °C, respectively.) (a) 16°C (b) 18°C (c) 24°C (d) 26°C (e) none of those answers 11. Star A has twice the radius and twice the absolute sur-face temperature of star B. The emissivity of both stars can be assumed to be 1. What is the ratio of the power output of star A to that of star B? (a) 4 (b) 8 (c) 16 (d) 32 (e) 64 12. If a gas is compressed isothermally, which of the fol-lowing statements is true? (a) Energy is transferred into the gas by heat. (b) No work is done on the gas. (c) The temperature of the gas increases. (d) The internal energy of the gas remains constant. (e) None of those statements is true. 13. When a gas undergoes an adiabatic expansion, which of the following statements is true? (a) The tempera-ture of the gas does not change. (b) No work is done by the gas. (c) No energy is transferred to the gas by heat. (d) The internal energy of the gas does not change. (e) The pressure increases. 14. If a gas undergoes an isobaric process, which of the fol-lowing statements is true? (a) The temperature of the gas doesn’t change. (b) Work is done on or by the gas. (c) No energy is transferred by heat to or from the gas. (d) The volume of the gas remains the same. (e) The pressure of the gas decreases uniformly. 15. How long would it take a 1 000 W heater to melt 1.00 kg of ice at 220.0°C, assuming all the energy from the heater is absorbed by the ice? (a) 4.18 s (b) 41.8 s (c) 5.55 min (d) 6.25 min (e) 38.4 min (c) low specific heat and high thermal conductivity, or (d) high specific heat and low thermal conductivity? 3. Assume you are measuring the specific heat of a sam-ple of originally hot metal by using a calorimeter con-taining water. Because your calorimeter is not perfectly insulating, energy can transfer by heat between the contents of the calorimeter and the room. To obtain the most accurate result for the specific heat of the metal, you should use water with which initial tem-perature? (a) slightly lower than room temperature (b) the same as room temperature (c) slightly higher than room temperature (d) whatever you like because the initial temperature makes no difference 4. An amount of energy is added to ice, raising its temper-ature from 210°C to 25°C. A larger amount of energy is added to the same mass of water, raising its tempera-ture from 15°C to 20°C. From these results, what would you conclude? (a) Overcoming the latent heat of fusion of ice requires an input of energy. (b) The latent heat of fusion of ice delivers some energy to the system. (c) The specific heat of ice is less than that of water. (d) The specific heat of ice is greater than that of water. (e) More information is needed to draw any conclusion. 5. How much energy is required to raise the tempera-ture of 5.00 kg of lead from 20.0°C to its melting point of 327°C? The specific heat of lead is 128 J/kg ? °C. (a) 4.04 3 105 J (b) 1.07 3 105 J (c) 8.15 3 104 J (d) 2.13 3 104 J (e) 1.96 3 105 J 6. Ethyl alcohol has about one-half the specific heat of water. Assume equal amounts of energy are trans-ferred by heat into equal-mass liquid samples of alco-hol and water in separate insulated containers. The water rises in temperature by 25°C. How much will the alcohol rise in temperature? (a) It will rise by 12°C. (b) It will rise by 25°C. (c) It will rise by 50°C. (d) It depends on the rate of energy transfer. (e) It will not rise in temperature. 7. The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equal amounts of energy are added to them. Assuming no melting or vaporization occurs, which of the following can be concluded about the final temperature TA of substance A and the final tem-perature TB of substance B? (a) TA . TB (b) TA , TB (c) TA 5 TB (d) More information is needed. 8. Beryllium has roughly one-half the specific heat of water (H2O). Rank the quantities of energy input required to produce the following changes from the Conceptual Questions 1. denotes answer available in Student Solutions Manual/Study Guide 1. Rub the palm of your hand on a metal surface for about 30 seconds. Place the palm of your other hand on an unrubbed portion of the surface and then on the rubbed portion. The rubbed portion will feel warmer. Now repeat this process on a wood surface. Why does the temperature difference between the rubbed and unrubbed portions of the wood surface seem larger than for the metal surface? 2. You need to pick up a very hot cooking pot in your kitchen. You have a pair of cotton oven mitts. To pick up the pot most comfortably, should you soak them in cold water or keep them dry? 3. What is wrong with the following statement: “Given any two bodies, the one with the higher temperature contains more heat.” www.aswarphysics.weebly.com Problems 617 wrap a wool blanket around the chest. Does doing so help to keep the beverages cool, or should you expect the wool blanket to warm them up? Explain your answer. (b) Your younger sister suggests you wrap her up in another wool blanket to keep her cool on the hot day like the ice chest. Explain your response to her. 8. In usually warm climates that experience a hard freeze, fruit growers will spray the fruit trees with water, hop-ing that a layer of ice will form on the fruit. Why would such a layer be advantageous? 9. Suppose you pour hot coffee for your guests, and one of them wants it with cream. He wants the coffee to be as warm as possible several minutes later when he drinks it. To have the warmest coffee, should the per-son add the cream just after the coffee is poured or just before drinking? Explain. 10. When camping in a canyon on a still night, a camper notices that as soon as the sun strikes the surrounding peaks, a breeze begins to stir. What causes the breeze? 11. Pioneers stored fruits and vegetables in underground cellars. In winter, why did the pioneers place an open barrel of water alongside their produce? 12. Is it possible to convert internal energy to mechanical energy? Explain with examples. 4. Why is a person able to remove a piece of dry alumi-num foil from a hot oven with bare fingers, whereas a burn results if there is moisture on the foil? 5. Using the first law of thermodynamics, explain why the total energy of an isolated system is always constant. 6. In 1801, Humphry Davy rubbed together pieces of ice inside an icehouse. He made sure that nothing in the environment was at a higher temperature than the rubbed pieces. He observed the production of drops of liquid water. Make a table listing this and other experiments or processes to illustrate each of the following situations. (a) A system can absorb energy by heat, increase in internal energy, and increase in temperature. (b) A system can absorb energy by heat and increase in internal energy without an increase in temperature. (c) A system can absorb energy by heat without increasing in temperature or in inter-nal energy. (d) A system can increase in internal energy and in temperature without absorbing energy by heat. (e) A system can increase in internal energy without absorbing energy by heat or increasing in temperature. 7. It is the morning of a day that will become hot. You just purchased drinks for a picnic and are loading them, with ice, into a chest in the back of your car. (a) You 3. A combination of 0.250 kg of water at 20.0°C, 0.400 kg of aluminum at 26.0°C, and 0.100 kg of copper at 100°C is mixed in an insulated container and allowed to come to thermal equilibrium. Ignore any energy transfer to or from the container. What is the final temperature of the mixture? 4. The highest waterfall in the world is the Salto Angel in Venezuela. Its longest single falls has a height of 807 m. If water at the top of the falls is at 15.0°C, what is the maximum temperature of the water at the bot-tom of the falls? Assume all the kinetic energy of the water as it reaches the bottom goes into raising its temperature. 5. What mass of water at 25.0°C must be allowed to come to thermal equilibrium with a 1.85-kg cube of alumi-num initially at 150°C to lower the temperature of the aluminum to 65.0°C? Assume any water turned to steam subsequently condenses. 6. The temperature of a silver bar rises by 10.0°C when it absorbs 1.23 kJ of energy by heat. The mass of the bar is M Section 20.1 Heat and Internal Energy 1. A 55.0-kg woman eats a 540 Calorie (540 kcal) jelly doughnut for breakfast. (a) How many joules of energy are the equivalent of one jelly doughnut? (b) How many steps must the woman climb on a very tall stairway to change the gravitational poten-tial energy of the woman–Earth system by a value equivalent to the food energy in one jelly doughnut? Assume the height of a single stair is 15.0 cm. (c) If the human body is only 25.0% efficient in convert-ing chemical potential energy to mechanical energy, how many steps must the woman climb to work off her breakfast?
Section 20.2 Specific Heat and Calorimetry 2. Consider Joule’s apparatus described in Figure 20.1. The mass of each of the two blocks is 1.50 kg, and the insulated tank is filled with 200 g of water. What is the increase in the water’s temperature after the blocks fall through a distance of 3.00 m?
BIO AMT W Problems The problems found in this chapter may be assigned online in Enhanced WebAssign 1. straightforward; 2. intermediate; 3. challenging 1. full solution available in the Student Solutions Manual/Study Guide AMT Analysis Model tutorial available in Enhanced WebAssign GP Guided Problem M Master It tutorial available in Enhanced WebAssign W Watch It video solution available in Enhanced WebAssign BIO Q/C S www.aswarphysics.weebly.com 618 Chapter 20 The First Law of Thermodynamics of steel in this case. (c) What pieces of data, if any, are unnecessary for the solution? Explain.
m M Figure P20.12 13. An aluminum calorimeter with a mass of 100 g con-tains 250 g of water. The calorimeter and water are in thermal equilibrium at 10.0°C. Two metallic blocks are placed into the water. One is a 50.0-g piece of copper at 80.0°C. The other has a mass of 70.0 g and is origi-nally at a temperature of 100°C. The entire system sta-bilizes at a final temperature of 20.0°C. (a) Determine the specific heat of the unknown sample. (b) Using the data in Table 20.1, can you make a positive identifica-tion of the unknown material? Can you identify a pos-sible material? (c) Explain your answers for part (b). 14. A 3.00-g copper coin at 25.0°C drops 50.0 m to the ground. (a) Assuming 60.0% of the change in gravita-tional potential energy of the coin–Earth system goes into increasing the internal energy of the coin, deter-mine the coin’s final temperature. (b) What If? Does the result depend on the mass of the coin? Explain. 15. Two thermally insulated vessels are connected by a nar-row tube fitted with a valve that is initially closed as shown in Figure P20.15. One vessel of volume 16.8 L contains oxygen at a temperature of 300 K and a pres-sure of 1.75 atm. The other vessel of volume 22.4 L con-tains oxygen at a temperature of 450 K and a pressure of 2.25 atm. When the valve is opened, the gases in the two vessels mix and the temperature and pressure become uniform throughout. (a) What is the final tem-perature? (b) What is the final pressure?
Q/C W Q/C 525 g. Determine the specific heat of silver from these data. 7. In cold climates, including the northern United States, a house can be built with very large windows facing south to take advantage of solar heating. Sunlight shin-ing in during the daytime is absorbed by the floor, interior walls, and objects in the room, raising their temperature to 38.0°C. If the house is well insulated, you may model it as losing energy by heat steadily at the rate 6 000 W on a day in April when the average exterior temperature is 4°C and when the conventional heating system is not used at all. During the period between 5:00 p.m. and 7:00 a.m., the temperature of the house drops and a sufficiently large “thermal mass” is required to keep it from dropping too far. The ther-mal mass can be a large quantity of stone (with specific heat 850 J/kg ? °C) in the floor and the interior walls exposed to sunlight. What mass of stone is required if the temperature is not to drop below 18.0°C overnight? 8. A 50.0-g sample of copper is at 25.0°C. If 1 200 J of energy is added to it by heat, what is the final tempera-ture of the copper? 9. An aluminum cup of mass 200 g contains 800 g of water in thermal equilibrium at 80.0°C. The combina-tion of cup and water is cooled uniformly so that the temperature decreases by 1.50°C per minute. At what rate is energy being removed by heat? Express your answer in watts. 10. If water with a mass mh at temperature Th is poured into an aluminum cup of mass mAl containing mass mc of water at Tc, where Th . Tc, what is the equilibrium temperature of the system? 11. A 1.50-kg iron horseshoe initially at 600°C is dropped into a bucket containing 20.0 kg of water at 25.0°C. What is the final temperature of the water–horseshoe system? Ignore the heat capacity of the container and assume a negligible amount of water boils away. 12. An electric drill with a steel drill bit of mass m 5 27.0 g and diameter 0.635 cm is used to drill into a cubical steel block of mass M 5 240 g. Assume steel has the same properties as iron. The cutting process can be modeled as happening at one point on the circumfer-ence of the bit. This point moves in a helix at constant tangential speed 40.0 m/s and exerts a force of con-stant magnitude 3.20 N on the block. As shown in Fig-ure P20.12, a groove in the bit carries the chips up to the top of the block, where they form a pile around the hole. The drill is turned on and drills into the block for a time interval of 15.0 s. Let’s assume this time interval is long enough for conduction within the steel to bring it all to a uniform temperature. Furthermore, assume the steel objects lose a negligible amount of energy by conduction, convection, and radiation into their envi-ronment. (a) Suppose the drill bit cuts three-quarters of the way through the block during 15.0 s. Find the temperature change of the whole quantity of steel. (b) What If? Now suppose the drill bit is dull and cuts only one-eighth of the way through the block in 15.0 s. Identify the temperature change of the whole quantity S M Q/C Pistons locked in place Valve P 1.75 atm V 16.8 L T 300 K P 2.25 atm V 22.4 L T 450 K Figure P20.15 www.aswarphysics.weebly.com Problems 619 down, keeping the pressure of the gas constant. How much work is done on the gas as the temperature of 0.200 mol of the gas is raised from 20.0°C to 300°C? 26. An ideal gas is enclosed in a cylinder that has a mov-able piston on top. The piston has a mass m and an area A and is free to slide up and down, keeping the pressure of the gas constant. How much work is done on the gas as the temperature of n mol of the gas is raised from T1 to T2? 27. One mole of an ideal gas is warmed slowly so that it goes from the PV state (Pi, Vi) to (3Pi, 3Vi) in such a way that the pressure of the gas is directly proportional to the volume. (a) How much work is done on the gas in the process? (b) How is the temperature of the gas related to its volume during this process? 28. (a) Determine the work done on a gas that expands from i to f as indicated in Figure P20.28. (b) What If? How much work is done on the gas if it is compressed from f to i along the same path?
6 106 P (Pa) 4 106 2 106 i f V (m3) 4 3 2 1 0 Figure P20.28 29. An ideal gas is taken through a quasi-static process described by P 5 aV 2, with a 5 5.00 atm/m6, as shown in Figure P20.29. The gas is expanded to twice its origi-nal volume of 1.00 m3. How much work is done on the expanding gas in this process?
P i f P V 2 V 1.00 m3 2.00 m3 a Figure P20.29 Section 20.5 The First Law of Thermodynamics 30. A gas is taken through the cyclic process described in Figure P20.30. (a) Find the net energy transferred to the system by heat dur-ing one complete cycle. (b) What If? If the cycle is reversed—that is, the process follows the path ACBA—what is the net energy input per cycle by heat?
S S Q/C W M 4 6 2 8 P (kPa) B C A 6 10 8 V (m3) Figure P20.30 Problems 30 and 31.
W Section 20.3 Latent Heat 16. A 50.0-g copper calorimeter contains 250 g of water at 20.0°C. How much steam at 100°C must be condensed into the water if the final temperature of the system is to reach 50.0°C? 17. A 75.0-kg cross-country skier glides over snow as in Figure P20.17. The coefficient of friction between skis and snow is 0.200. Assume all the snow beneath his skis is at 0°C and that all the internal energy gener-ated by friction is added to snow, which sticks to his skis until it melts. How far would he have to ski to melt 1.00 kg of snow? 18. How much energy is required to change a 40.0-g ice cube from ice at 210.0°C to steam at 110°C? 19. A 75.0-g ice cube at 0°C is placed in 825 g of water at 25.0°C. What is the final temperature of the mixture? 20. A 3.00-g lead bullet at 30.0°C is fired at a speed of 240 m/s into a large block of ice at 0°C, in which it becomes embedded. What quantity of ice melts? 21. Steam at 100°C is added to ice at 0°C. (a) Find the amount of ice melted and the final temperature when the mass of steam is 10.0 g and the mass of ice is 50.0 g. (b) What If? Repeat when the mass of steam is 1.00 g and the mass of ice is 50.0 g. 22. A 1.00-kg block of copper at 20.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen boil away by the time the cop-per reaches 77.3 K? (The specific heat of copper is 0.092 0 cal/g ? °C, and the latent heat of vaporization of nitrogen is 48.0 cal/g.) 23. In an insulated vessel, 250 g of ice at 0°C is added to 600 g of water at 18.0°C. (a) What is the final tempera-ture of the system? (b) How much ice remains when the system reaches equilibrium? 24. An automobile has a mass of 1 500 kg, and its alumi-num brakes have an overall mass of 6.00 kg. (a) Assume all the mechanical energy that transforms into internal energy when the car stops is deposited in the brakes and no energy is transferred out of the brakes by heat. The brakes are originally at 20.0°C. How many times can the car be stopped from 25.0 m/s before the brakes start to melt? (b) Identify some effects ignored in part (a) that are important in a more realistic assessment of the warming of the brakes.
Section 20.4 Work and Heat in Thermodynamic Processes 25. An ideal gas is enclosed in a cylinder with a movable piston on top of it. The piston has a mass of 8 000 g and an area of 5.00 cm2 and is free to slide up and Figure P20.17 iStockphoto.com/technotr M W AMT M W Q/C www.aswarphysics.weebly.com 620 Chapter 20 The First Law of Thermodynamics 40. In Figure P20.40, the change in internal energy of a gas that is taken from A to C along the blue path is 1800 J. The work done on the gas along the red path ABC is 2500 J. (a) How much energy must be added to the system by heat as it goes from A through B to C? (b) If the pressure at point A is five times that of point C, what is the work done on the system in going from C to D? (c) What is the energy exchanged with the surroundings by heat as the gas goes from C to A along the green path? (d) If the change in internal energy in going from point D to point A is 1500 J, how much energy must be added to the system by heat as it goes from point C to point D? 41. An ideal gas initially at Pi, Vi, and Ti is taken through a cycle as shown in Figure P20.41. (a) Find the net work done on the gas per cycle for 1.00 mol of gas initially at 0°C. (b) What is the net energy added by heat to the gas per cycle? 42. An ideal gas initially at Pi, Vi, and Ti is taken through a cycle as shown in Figure P20.41. (a) Find the net work done on the gas per cycle. (b) What is the net energy added by heat to the system per cycle?
Section 20.7 Energy Transfer Mechanisms in Thermal Processes 43. A glass windowpane in a home is 0.620 cm thick and has dimensions of 1.00 m 3 2.00 m. On a certain day, the temperature of the interior surface of the glass is 25.0°C and the exterior surface temperature is 0°C. (a) What is the rate at which energy is transferred by heat through the glass? (b) How much energy is trans-ferred through the window in one day, assuming the temperatures on the surfaces remain constant? 44. A concrete slab is 12.0 cm thick and has an area of 5.00 m2. Electric heating coils are installed under the slab to melt the ice on the surface in the winter months. What minimum power must be supplied to the coils to maintain a temperature difference of 20.0°C between the bottom of the slab and its surface? Assume all the energy transferred is through the slab. 45. A student is trying to decide what to wear. His bed-room is at 20.0°C. His skin temperature is 35.0°C. The area of his exposed skin is 1.50 m2. People all over the world have skin that is dark in the infrared, with emis-sivity about 0.900. Find the net energy transfer from his body by radiation in 10.0 min. 46. The surface of the Sun has a temperature of about 5 800 K. The radius of the Sun is 6.96 3 108 m. Calcu-late the total energy radiated by the Sun each second. Assume the emissivity of the Sun is 0.986.
B C D A P Pi 3Pi Vi 3Vi V Figure P20.41 Problems 41 and 42.
W S BIO 31. Consider the cyclic process depicted in Figure P20.30. If Q is negative for the process BC and DE int is nega-tive for the process CA, what are the signs of Q, W, and DE int that are associated with each of the three processes? 32. Why is the following situation impossible? An ideal gas undergoes a process with the following parameters: Q 5 10.0 J, W 5 12.0 J, and DT 5 22.00°C. 33. A thermodynamic system undergoes a process in which its internal energy decreases by 500 J. Over the same time interval, 220 J of work is done on the system. Find the energy transferred from it by heat. 34. A sample of an ideal gas goes through the process shown in Figure P20.34. From A to B, the process is adiabatic; from B to C, it is isobaric with 345 kJ of energy entering the system by heat; from C to D, the process is isothermal; and from D to A, it is isobaric with 371 kJ of energy leaving the system by heat. Deter-mine the difference in internal energy E int,B 2 E int,A.
1 3 P (atm) 0.09 0.2 0.4 1.2 A C B D V (m3) Figure P20.34 Section 20.6 Some Applications of the First Law of Thermodynamics 35. A 2.00-mol sample of helium gas initially at 300 K, and 0.400 atm is compressed isothermally to 1.20 atm. Not-ing that the helium behaves as an ideal gas, find (a) the final volume of the gas, (b) the work done on the gas, and (c) the energy transferred by heat. 36. (a) How much work is done on the steam when 1.00 mol of water at 100°C boils and becomes 1.00 mol of steam at 100°C at 1.00 atm pressure? Assume the steam to behave as an ideal gas. (b) Determine the change in internal energy of the system of the water and steam as the water vaporizes. 37. An ideal gas initially at 300 K undergoes an isobaric expansion at 2.50 kPa. If the volume increases from 1.00 m3 to 3.00 m3 and 12.5 kJ is transferred to the gas by heat, what are (a) the change in its internal energy and (b) its final temperature? 38. One mole of an ideal gas does 3 000 J of work on its surroundings as it expands isothermally to a final pressure of 1.00 atm and volume of 25.0 L. Determine (a) the initial volume and (b) the temperature of the gas. 39. A 1.00-kg block of aluminum is warmed at atmospheric pressure so that its temperature increases from 22.0°C to 40.0°C. Find (a) the work done on the aluminum, (b) the energy added to it by heat, and (c) the change in its internal energy.
W M M W P V C D B A Figure P20.40 www.aswarphysics.weebly.com Problems 621 by using the thermal window instead of the single-pane window? Include the contributions of inside and out-side stagnant air layers. 54. At our distance from the Sun, the intensity of solar radiation is 1 370 W/m2. The temperature of the Earth is affected by the greenhouse effect of the atmo-sphere. This phenomenon describes the effect of absorption of infrared light emitted by the surface so as to make the surface temperature of the Earth higher than if it were airless. For comparison, consider a spherical object of radius r with no atmosphere at the same distance from the Sun as the Earth. Assume its emissivity is the same for all kinds of electromag-netic waves and its temperature is uniform over its sur-face. (a) Explain why the projected area over which it absorbs sunlight is pr 2 and the surface area over which it radiates is 4pr 2. (b) Compute its steady-state temperature. Is it chilly? 55. A bar of gold (Au) is in ther-mal contact with a bar of silver (Ag) of the same length and area (Fig. P20.55). One end of the compound bar is main-tained at 80.0°C, and the oppo-site end is at 30.0°C. When the energy transfer reaches steady state, what is the temperature at the junction? 56. For bacteriological testing of water supplies and in medical clinics, samples must routinely be incubated for 24 h at 37°C. Peace Corps volunteer and MIT engineer Amy Smith invented a low-cost, low-maintenance incuba-tor. The incubator consists of a foam-insulated box containing a waxy material that melts at 37.0°C inter-spersed among tubes, dishes, or bottles containing the test samples and growth medium (bacteria food). Outside the box, the waxy material is first melted by a stove or solar energy collector. Then the waxy material is put into the box to keep the test samples warm as the material solidifies. The heat of fusion of the phase-change material is 205 kJ/kg. Model the insulation as a panel with surface area 0.490 m2, thickness 4.50 cm, and conductivity 0.012 0 W/m ? °C. Assume the exte-rior temperature is 23.0°C for 12.0 h and 16.0°C for 12.0 h. (a) What mass of the waxy material is required to conduct the bacteriological test? (b) Explain why your calculation can be done without knowing the mass of the test samples or of the insulation. 57. A large, hot pizza floats in outer space after being jet-tisoned as refuse from a spacecraft. What is the order of magnitude (a) of its rate of energy loss and (b) of its rate of temperature change? List the quantities you estimate and the value you estimate for each.
Additional Problems 58. A gas expands from I to F in Figure P20.58 (page 622). The energy added to the gas by heat is 418 J when the gas goes from I to F along the diagonal path. (a) What is the change in internal energy of the gas? (b) How Q/C Insulation Au Ag 30.0 C 80.0 C Figure P20.55 BIO Q/C M 47. The tungsten filament of a certain 100-W lightbulb radiates 2.00 W of light. (The other 98 W is carried away by convection and conduction.) The filament has a surface area of 0.250 mm2 and an emissivity of 0.950. Find the filament’s temperature. (The melting point of tungsten is 3 683 K.) 48. At high noon, the Sun delivers 1 000 W to each square meter of a blacktop road. If the hot asphalt trans-fers energy only by radiation, what is its steady-state temperature? 49. Two lightbulbs have cylindrical filaments much greater in length than in diameter. The evacuated bulbs are identical except that one operates at a filament tem-perature of 2 100°C and the other operates at 2 000°C. (a) Find the ratio of the power emitted by the hot-ter lightbulb to that emitted by the cooler lightbulb. (b) With the bulbs operating at the same respective temperatures, the cooler lightbulb is to be altered by making its filament thicker so that it emits the same power as the hotter one. By what factor should the radius of this filament be increased? 50. The human body must maintain its core temperature inside a rather narrow range around 37°C. Metabolic processes, notably muscular exertion, convert chemical energy into internal energy deep in the interior. From the interior, energy must flow out to the skin or lungs to be expelled to the environment. During moderate exercise, an 80-kg man can metabolize food energy at the rate 300 kcal/h, do 60 kcal/h of mechanical work, and put out the remaining 240 kcal/h of energy by heat. Most of the energy is carried from the body inte-rior out to the skin by forced convection (as a plumber would say), whereby blood is warmed in the interior and then cooled at the skin, which is a few degrees cooler than the body core. Without blood flow, living tissue is a good thermal insulator, with thermal con-ductivity about 0.210 W/m · °C. Show that blood flow is essential to cool the man’s body by calculating the rate of energy conduction in kcal/h through the tissue layer under his skin. Assume that its area is 1.40 m2, its thickness is 2.50 cm, and it is maintained at 37.0°C on one side and at 34.0°C on the other side. 51. A copper rod and an aluminum rod of equal diameter are joined end to end in good thermal contact. The temperature of the free end of the copper rod is held constant at 100°C and that of the far end of the alumi-num rod is held at 0°C. If the copper rod is 0.150 m long, what must be the length of the aluminum rod so that the temperature at the junction is 50.0°C? 52. A box with a total surface area of 1.20 m2 and a wall thickness of 4.00 cm is made of an insulating material. A 10.0-W electric heater inside the box maintains the inside temperature at 15.0°C above the outside temper-ature. Find the thermal conductivity k of the insulating material. 53. (a) Calculate the R-value of a thermal window made of two single panes of glass each 0.125 in. thick and sepa-rated by a 0.250-in. air space. (b) By what factor is the transfer of energy by heat through the window reduced BIO M www.aswarphysics.weebly.com 622 Chapter 20 The First Law of Thermodynamics stream of the liquid while energy is added by heat at a known rate. A liquid of density 900 kg/m3 flows through the calorimeter with volume flow rate of 2.00 L/min. At steady state, a temperature difference 3.50°C is established between the input and output points when energy is supplied at the rate of 200 W. What is the specific heat of the liquid? 64. A flow calorimeter is an apparatus used to measure the specific heat of a liquid. The technique of flow calo-rimetry involves measuring the temperature differ-ence between the input and output points of a flowing stream of the liquid while energy is added by heat at a known rate. A liquid of density r flows through the calorimeter with volume flow rate R. At steady state, a temperature difference DT is established between the input and output points when energy is supplied at the rate P. What is the specific heat of the liquid? 65. Review. Following a collision between a large space-craft and an asteroid, a copper disk of radius 28.0 m and thickness 1.20 m at a temperature of 850°C is floating in space, rotating about its symmetry axis with an angular speed of 25.0 rad/s. As the disk radi-ates infrared light, its temperature falls to 20.0°C. No external torque acts on the disk. (a) Find the change in kinetic energy of the disk. (b) Find the change in internal energy of the disk. (c) Find the amount of energy it radiates. 66. An ice-cube tray is filled with 75.0 g of water. After the filled tray reaches an equilibrium temperature of 20.0°C, it is placed in a freezer set at 28.00°C to make ice cubes. (a) Describe the processes that occur as energy is being removed from the water to make ice. (b) Calculate the energy that must be removed from the water to make ice cubes at 28.00°C. 67. On a cold winter day, you buy roasted chestnuts from a street vendor. Into the pocket of your down parka you put the change he gives you: coins constituting 9.00 g of copper at –12.0°C. Your pocket already con-tains 14.0 g of silver coins at 30.0°C. A short time later the temperature of the copper coins is 4.00°C and is increasing at a rate of 0.500°C/s. At this time, (a) what is the temperature of the silver coins and (b) at what rate is it changing? 68. The rate at which a resting person converts food energy is called one’s basal metabolic rate (BMR). Assume that the resulting internal energy leaves a person’s body by radiation and convection of dry air. When you jog, most of the food energy you burn above your BMR becomes internal energy that would raise your body temperature if it were not eliminated. Assume that evaporation of perspiration is the mechanism for eliminating this energy. Suppose a person is jogging for “maximum fat burning,” converting food energy at the rate 400 kcal/h above his BMR, and putting out energy by work at the rate 60.0 W. Assume that the heat of evaporation of water at body temperature is equal to its heat of vaporization at 100°C. (a) Determine the hourly rate at which water must evaporate from his skin. (b) When you metabolize fat, the hydrogen atoms S AMT Q/C BIO much energy must be added to the gas by heat along the indirect path IAF ?
I A F P (atm) 4 3 2 1 0 1 4 2 3 V (liters) Figure P20.58 59. Gas in a container is at a pressure of 1.50 atm and a volume of 4.00 m3. What is the work done on the gas (a) if it expands at constant pressure to twice its initial volume, and (b) if it is compressed at constant pressure to one-quarter its initial volume? 60. Liquid nitrogen has a boiling point of 77.3 K and a latent heat of vaporization of 2.01 3 105 J/kg. A 25.0-W electric heating element is immersed in an insulated vessel containing 25.0 L of liquid nitrogen at its boil-ing point. How many kilograms of nitrogen are boiled away in a period of 4.00 h? 61. An aluminum rod 0.500 m in length and with a cross- sectional area of 2.50 cm2 is inserted into a thermally insulated vessel containing liquid helium at 4.20 K. The rod is initially at 300 K. (a) If one-half of the rod is inserted into the helium, how many liters of helium boil off by the time the inserted half cools to 4.20 K? Assume the upper half does not yet cool. (b) If the cir-cular surface of the upper end of the rod is maintained at 300 K, what is the approximate boil-off rate of liq-uid helium in liters per second after the lower half has reached 4.20 K? (Aluminum has thermal conductivity of 3 100 W/m · K at 4.20 K; ignore its temperature vari-ation. The density of liquid helium is 125 kg/m3.) 62. Review. Two speeding lead bullets, one of mass 12.0 g moving to the right at 300 m/s and one of mass 8.00 g moving to the left at 400 m/s, collide head-on, and all the material sticks together. Both bullets are originally at temperature 30.0°C. Assume the change in kinetic energy of the system appears entirely as increased internal energy. We would like to determine the tem-perature and phase of the bullets after the collision. (a) What two analysis models are appropriate for the system of two bullets for the time interval from before to after the collision? (b) From one of these models, what is the speed of the combined bullets after the collision? (c) How much of the initial kinetic energy has transformed to internal energy in the system after the collision? (d) Does all the lead melt due to the col-lision? (e) What is the temperature of the combined bullets after the collision? (f) What is the phase of the combined bullets after the collision? 63. A flow calorimeter is an apparatus used to measure the specific heat of a liquid. The technique of flow calo-rimetry involves measuring the temperature difference between the input and output points of a flowing M M AMT GP www.aswarphysics.weebly.com Problems 623 rial of the meteoroid rises momentarily to the same final temperature. Find this temperature. Assume the specific heat of liquid and of gaseous aluminum is 1 170 J/kg ? °C. 74. Why is the following situation impossible? A group of camp-ers arises at 8:30 a.m. and uses a solar cooker, which consists of a curved, reflecting surface that concen-trates sunlight onto the object to be warmed (Fig. P20.74). During the day, the maximum solar intensity reaching the Earth’s surface at the cooker’s location is I 5 600 W/m2. The cooker faces the Sun and has a face diameter of d 5 0.600 m. Assume a fraction f of 40.0% of the incident energy is transferred to 1.50 L of water in an open container, initially at 20.0°C. The water comes to a boil, and the campers enjoy hot cof-fee for breakfast before hiking ten miles and returning by noon for lunch.
d Figure P20.74 75. During periods of high activity, the Sun has more sun-spots than usual. Sunspots are cooler than the rest of the luminous layer of the Sun’s atmosphere (the pho-tosphere). Paradoxically, the total power output of the active Sun is not lower than average but is the same or slightly higher than average. Work out the details of the following crude model of this phenomenon. Consider a patch of the photosphere with an area of 5.10 3 1014 m2. Its emissivity is 0.965. (a) Find the power it radiates if its temperature is uniformly 5 800 K, corresponding to the quiet Sun. (b) To represent a sunspot, assume 10.0% of the patch area is at 4 800 K and the other 90.0% is at 5 890 K. Find the power output of the patch. (c) State how the answer to part (b) compares with the answer to part (a). (d) Find the average temperature of the patch. Note that this cooler temperature results in a higher power output. 76. (a) In air at 0°C, a 1.60-kg copper block at 0°C is set sliding at 2.50 m/s over a sheet of ice at 0°C. Friction brings the block to rest. Find the mass of the ice that melts. (b) As the block slows down, identify its energy input Q, its change in internal energy DE int, and the change in mechanical energy for the block–ice system. (c) For the ice as a system, identify its energy input Q and its change in internal energy DE int. (d) A 1.60-kg block of ice at 0°C is set sliding at 2.50 m/s over a sheet of copper at 0°C. Friction brings the block to rest. Find the mass of the ice that melts. (e) Evaluate Q and DE int for the block of ice as a system and DE mech for the block–ice system. (f) Evaluate Q and DE int for the metal Q/C in the fat molecule are transferred to oxygen to form water. Assume that metabolism of 1.00 g of fat gener-ates 9.00 kcal of energy and produces 1.00 g of water. What fraction of the water the jogger needs is provided by fat metabolism? 69. An iron plate is held against an iron wheel so that a kinetic friction force of 50.0 N acts between the two pieces of metal. The relative speed at which the two sur-faces slide over each other is 40.0 m/s. (a) Calculate the rate at which mechanical energy is converted to internal energy. (b) The plate and the wheel each have a mass of 5.00 kg, and each receives 50.0% of the internal energy. If the system is run as described for 10.0 s and each object is then allowed to reach a uniform internal tem-perature, what is the resultant temperature increase? 70. A resting adult of average size converts chemical energy in food into internal energy at the rate 120 W, called her basal metabolic rate. To stay at constant temperature, the body must put out energy at the same rate. Several processes exhaust energy from your body. Usually, the most important is thermal conduction into the air in contact with your exposed skin. If you are not wear-ing a hat, a convection current of warm air rises verti-cally from your head like a plume from a smokestack. Your body also loses energy by electromagnetic radia-tion, by your exhaling warm air, and by evaporation of perspiration. In this problem, consider still another pathway for energy loss: moisture in exhaled breath. Suppose you breathe out 22.0 breaths per minute, each with a volume of 0.600 L. Assume you inhale dry air and exhale air at 37.0°C containing water vapor with a vapor pressure of 3.20 kPa. The vapor came from evap-oration of liquid water in your body. Model the water vapor as an ideal gas. Assume its latent heat of evapora-tion at 37.0°C is the same as its heat of vaporization at 100°C. Calculate the rate at which you lose energy by exhaling humid air. 71. A 40.0-g ice cube floats in 200 g of water in a 100-g copper cup; all are at a temperature of 0°C. A piece of lead at 98.0°C is dropped into the cup, and the final equilibrium temperature is 12.0°C. What is the mass of the lead? 72. One mole of an ideal gas is contained in a cylinder with a movable piston. The initial pressure, volume, and temperature are Pi, Vi, and Ti, respectively. Find the work done on the gas in the following processes. In operational terms, describe how to carry out each process and show each process on a PV diagram. (a) an isobaric compression in which the final volume is one-half the initial volume (b) an isothermal com-pression in which the final pressure is four times the initial pressure (c) an isovolumetric process in which the final pressure is three times the initial pressure 73. Review. A 670-kg meteoroid happens to be composed of aluminum. When it is far from the Earth, its tem-perature is 215.0°C and it moves at 14.0 km/s relative to the planet. As it crashes into the Earth, assume the internal energy transformed from the mechanical energy of the meteoroid–Earth system is shared equally between the meteoroid and the Earth and all the mate-BIO M S Q/C www.aswarphysics.weebly.com 624 Chapter 20 The First Law of Thermodynamics Mass of water: 0.400 kg Mass of calorimeter: 0.040 kg Specific heat of calorimeter: 0.63 kJ/kg ? °C Initial temperature of aluminum: 27.0°C Mass of aluminum: 0.200 kg Final temperature of mixture: 66.3°C (a) Use these data to determine the specific heat of aluminum. (b) Explain whether your result is within 15% of the value listed in Table 20.1.
Challenge Problems 81. Consider the piston– cylinder apparatus shown in Figure P20.81. The bot-tom of the cylinder con-tains 2.00 kg of water at just under 100.0°C. The cylinder has a radius of r 5 7.50 cm. The piston of mass m 5 3.00 kg sits on the surface of the water. An electric heater in the cylinder base transfers energy into the water at a rate of 100 W. Assume the cylinder is much taller than shown in the figure, so we don’t need to be concerned about the piston reach-ing the top of the cylinder. (a) Once the water begins boiling, how fast is the piston rising? Model the steam as an ideal gas. (b) After the water has completely turned to steam and the heater continues to transfer energy to the steam at the same rate, how fast is the piston rising? 82. A spherical shell has inner radius 3.00 cm and outer radius 7.00 cm. It is made of material with thermal conductivity k 5 0.800 W/m ? °C. The interior is main-tained at temperature 5°C and the exterior at 40°C. After an interval of time, the shell reaches a steady state with the temperature at each point within it remaining constant in time. (a) Explain why the rate of energy transfer P must be the same through each spherical surface, of radius r, within the shell and must satisfy dT dr 5 P 4pkr 2 (b) Next, prove that 3 40 5 dT 5 P 4pk 3 0.07 0.03 r 22 dr where T is in degrees Celsius and r is in meters. (c) Find the rate of energy transfer through the shell. (d) Prove that 3 T 5 dT 5 1.84 3 r 0.03 r22 dr where T is in degrees Celsius and r is in meters. (e) Find the temperature within the shell as a func-tion of radius. (f) Find the temperature at r 5 5.00 cm, halfway through the shell.
m Water Electric heater in base of cylinder r Figure P20.81 Q/C sheet as a system. (g) A thin, 1.60-kg slab of copper at 20°C is set sliding at 2.50 m/s over an identical station-ary slab at the same temperature. Friction quickly stops the motion. Assuming no energy is transferred to the environment by heat, find the change in temperature of both objects. (h) Evaluate Q and DE int for the slid-ing slab and DE mech for the two-slab system. (i) Evalu-ate Q and DE int for the stationary slab. 77. Water in an electric teakettle is boiling. The power absorbed by the water is 1.00 kW. Assuming the pres-sure of vapor in the kettle equals atmospheric pres-sure, determine the speed of effusion of vapor from the kettle’s spout if the spout has a cross-sectional area of 2.00 cm2. Model the steam as an ideal gas. 78. The average thermal conductivity of the walls (includ-ing the windows) and roof of the house depicted in Figure P20.78 is 0.480 W/m ? °C, and their average thickness is 21.0 cm. The house is kept warm with natural gas having a heat of combustion (that is, the energy provided per cubic meter of gas burned) of 9 300 kcal/m3. How many cubic meters of gas must be burned each day to maintain an inside temperature of 25.0°C if the outside temperature is 0.0°C? Disregard radiation and the energy transferred by heat through the ground.
5.00 m 10.0 m 8.00 m 37.0 Figure P20.78 79. A cooking vessel on a slow burner contains 10.0 kg of water and an unknown mass of ice in equilibrium at 0°C at time t 5 0. The temperature of the mixture is measured at various times, and the result is plotted in Figure P20.79. During the first 50.0 min, the mixture remains at 0°C. From 50.0 min to 60.0 min, the tem-perature increases to 2.00°C. Ignoring the heat capac-ity of the vessel, determine the initial mass of the ice.
0 1 2 3 20 40 60 T ( C) t (min) 0 Figure P20.79 80. A student measures the following data in a calorimetry experiment designed to determine the specific heat of aluminum: Initial temperature of water and calorimeter: 70.0°C M Q/C www.aswarphysics.weebly.com Problems 625 Suggestions: The temperature gradient is dT/dr. A radial energy current passes through a concentric cylinder of area 2prL. (b) The passenger section of a jet airliner is in the shape of a cylindrical tube with a length of 35.0 m and an inner radius of 2.50 m. Its walls are lined with an insulating material 6.00 cm in thickness and having a thermal conductivity of 4.00 3 1025 cal/s ? cm ? °C. A heater must maintain the interior temperature at 25.0°C while the outside temperature is 235.0°C. What power must be sup-plied to the heater? 83. A pond of water at 0°C is covered with a layer of ice 4.00 cm thick. If the air temperature stays constant at 210.0°C, what time interval is required for the ice thickness to increase to 8.00 cm? Suggestion: Use Equa-tion 20.16 in the form dQ dt 5 kA DT x and note that the incremental energy dQ extracted from the water through the thickness x of ice is the amount required to freeze a thickness dx of ice. That is, dQ 5 LfrA dx, where r is the density of the ice, A is the area, and Lf is the latent heat of fusion. 84. (a) The inside of a hollow cylinder is maintained at a temperature Ta, and the outside is at a lower tempera-ture, Tb (Fig. P20.84). The wall of the cylinder has a thermal conductivity k. Ignoring end effects, show that the rate of energy conduction from the inner surface to the outer surface in the radial direction is dQ dt 5 2pLk c Ta 2 Tb ln1b/a2 d T b L b a T a r Figure P20.84 www.aswarphysics.weebly.com In Chapter 19, we discussed the properties of an ideal gas by using such macroscopic variables as pressure, volume, and temperature. Such large-scale properties can be related to a description on a microscopic scale, where matter is treated as a collection of molecules. Applying Newton’s laws of motion in a statistical manner to a collection of particles pro-vides a reasonable description of thermodynamic processes. To keep the mathematics relatively simple, we shall consider primarily the behavior of gases because in gases the interactions between molecules are much weaker than they are in liquids or solids. We shall begin by relating pressure and temperature directly to the details of molecular motion in a sample of gas. Based on these results, we will make predictions of molar specific heats of gases. Some of these predictions will be correct and some will not. We will extend our model to explain those values that are not predicted correctly by the simpler model. Finally, we discuss the distribution of molecular speeds in a gas.
21.1 Molecular Model of an Ideal Gas 21.2 Molar Specific Heat of an Ideal Gas 21.3 The Equipartition of Energy 21.4 Adiabatic Processes for an Ideal Gas 21.5 Distribution of Molecular Speeds c h a p t e r 21 The Kinetic Theory of Gases A boy inflates his bicycle tire with a hand-operated pump. Kinetic theory helps to describe the details of the air in the pump. (© Cengage Learning/ George Semple) 626 www.aswarphysics.weebly.com 21.1 Molecular Model of an Ideal Gas 627 21.1 Molecular Model of an Ideal Gas In this chapter, we will investigate a structural model for an ideal gas. A structural model is a theoretical construct designed to represent a system that cannot be observed directly because it is too large or too small. For example, we can only observe the solar system from the inside; we cannot travel outside the solar system and look back to see how it works. This restricted vantage point has led to different historical structural models of the solar system: the geocentric model, with the Earth at the center, and the heliocentric model, with the Sun at the center. Of course, the latter has been shown to be correct. An example of a system too small to observe directly is the hydrogen atom. Various structural models of this system have been devel-oped, including the Bohr model (Section 42.3) and the quantum model (Section 42.4). Once a structural model is developed, various predictions are made for experimen-tal observations. For example, the geocentric model of the solar system makes pre-dictions of how the movement of Mars should appear from the Earth. It turns out that those predictions do not match the actual observations. When that occurs with a structural model, the model must be modified or replaced with another model. The structural model that we will develop for an ideal gas is called kinetic the-ory. This model treats an ideal gas as a collection of molecules with the following properties: 1. Physical components: The gas consists of a number of identical molecules within a cubic con-tainer of side length d. The number of molecules in the gas is large, and the average separation between them is large compared with their dimensions. Therefore, the molecules occupy a negligible volume in the container. This assumption is consistent with the ideal gas model, in which we imagine the molecules to be point-like. 2. Behavior of the components: (a) The molecules obey Newton’s laws of motion, but as a whole their motion is isotropic: any molecule can move in any direction with any speed. (b) The molecules interact only by short-range forces during elastic colli-sions. This assumption is consistent with the ideal gas model, in which the molecules exert no long-range forces on one another. (c) The molecules make elastic collisions with the walls.
Although we often picture an ideal gas as consisting of single atoms, the behavior of molecular gases approximates that of ideal gases rather well at low pressures. Usu-ally, molecular rotations or vibrations have no effect on the motions considered here. For our first application of kinetic theory, let us relate the macroscope variable of pressure P to microscopic quantities. Consider a collection of N molecules of an ideal gas in a container of volume V. As indicated above, the container is a cube with edges of length d (Fig. 21.1). We shall first focus our attention on one of these molecules of mass m0 and assume it is moving so that its component of velocity in the x direction is vxi as in Figure 21.2. (The subscript i here refers to the ith mol-ecule in the collection, not to an initial value. We will combine the effects of all the molecules shortly.) As the molecule collides elastically with any wall (property 2(c) above), its velocity component perpendicular to the wall is reversed because the mass of the wall is far greater than the mass of the molecule. The molecule is mod-eled as a nonisolated system for which the impulse from the wall causes a change in the molecule’s momentum. Because the momentum component pxi of the molecule is m0vxi before the collision and 2m0vxi after the collision, the change in the x com-ponent of the momentum of the molecule is Dpxi 5 2m0vxi 2 (m0vxi) 5 22m0vxi (21.1) d d d z x y m 0 vxi vi S One molecule of the gas moves with velocity v on its way toward a collision with the wall.
S Figure 21.1 A cubical box with sides of length d containing an ideal gas. Figure 21.2 A molecule makes an elastic collision with the wall of the container. In this construc-tion, we assume the molecule moves in the xy plane.
vyi vxi vyi –vxi vi S vi S The molecule’s x component of momentum is reversed, whereas its y component remains unchanged.
www.aswarphysics.weebly.com 628 Chapter 21 The Kinetic Theory of Gases From the nonisolated system model for momentum, we can apply the impulse-momentum theorem (Eqs. 9.11 and 9.13) to the molecule to give Fi,on molecule Dtcollision 5 Dpxi 5 22m0vxi (21.2) where F i,on molecule is the x component of the average force1 the wall exerts on the molecule during the collision and Dtcollision is the duration of the collision. For the molecule to make another collision with the same wall after this first collision, it must travel a distance of 2d in the x direction (across the cube and back). There-fore, the time interval between two collisions with the same wall is Dt 5 2d vxi (21.3) The force that causes the change in momentum of the molecule in the collision with the wall occurs only during the collision. We can, however, find the long-term average force for many back-and-forth trips across the cube by averaging the force in Equation 21.2 over the time interval for the molecule to move across the cube and back once, Equation 21.3. The average change in momentum per trip for the time interval for many trips is the same as that for the short duration of the colli-sion. Therefore, we can rewrite Equation 21.2 as Fi Dt 5 22m0vxi (21.4) where Fi is the average force component over the time interval for the molecule to move across the cube and back. Because exactly one collision occurs for each such time interval, this result is also the long-term average force on the molecule over long time intervals containing any number of multiples of Dt. Equation 21.3 and 21.4 enable us to express the x component of the long-term average force exerted by the wall on the molecule as Fi 5 2 2m0vxi Dt 5 2 2m0vxi 2 2d 5 2 m0vxi 2 d (21.5) Now, by Newton’s third law, the x component of the long-term average force exerted by the molecule on the wall is equal in magnitude and opposite in direction: Fi,on wall 5 2Fi 5 2a2 m 0vxi 2 d b 5 m 0vxi 2 d (21.6) The total average force F exerted by the gas on the wall is found by adding the average forces exerted by the individual molecules. Adding terms such as those in Equation 21.6 for all molecules gives F 5 a N i51 m 0vxi 2 d 5 m 0 d a N i51 vxi 2 (21.7) where we have factored out the length of the box and the mass m0 because property 1 tells us that all the molecules are the same. We now impose an additional fea-ture from property 1, that the number of molecules is large. For a small number of molecules, the actual force on the wall would vary with time. It would be nonzero during the short interval of a collision of a molecule with the wall and zero when no molecule happens to be hitting the wall. For a very large number of molecules such as Avogadro’s number, however, these variations in force are smoothed out so that the average force given above is the same over any time interval. Therefore, the constant force F on the wall due to the molecular collisions is F 5 m0 d a N i51 vx i 2 (21.8) 1For this discussion, we use a bar over a variable to represent the average value of the variable, such as F for the aver-age force, rather than the subscript “avg” that we have used before. This notation is to save confusion because we already have a number of subscripts on variables.
www.aswarphysics.weebly.com 21.1 Molecular Model of an Ideal Gas 629 To proceed further, let’s consider how to express the average value of the square of the x component of the velocity for N molecules. The traditional average of a set of values is the sum of the values over the number of values: vx 2 5 a N i51 vxi 2 N S a N i51 vxi 2 5 N vx 2 (21.9) Using Equation 21.9 to substitute for the sum in Equation 21.8 gives F 5 m0 d Nvx 2 (21.10) Now let’s focus again on one molecule with velocity components vxi, vyi, and vzi. The Pythagorean theorem relates the square of the speed of the molecule to the squares of the velocity components: vi 2 5 vxi 2 1 vyi 2 1 vzi 2 (21.11) Hence, the average value of v2 for all the molecules in the container is related to the average values of vx 2, vy 2, and vz 2 according to the expression v 2 5 vx 2 1 vy 2 1 vz 2 (21.12) Because the motion is isotropic (property 2(a) above), the average values vx2, vy2, and vz2 are equal to one another. Using this fact and Equation 21.12, we find that v 2 5 3vx 2 (21.13) Therefore, from Equation 21.10, the total force exerted on the wall is F 5 1 3N m0v 2 d (21.14) Using this expression, we can find the total pressure exerted on the wall: P 5 F A 5 F d 2 5 1 3 N m0v 2 d 3 5 1 3 aN V bm0v 2 P 5 2 3 aN V b 1 1 2m 0v 22 (21.15) where we have recognized the volume V of the cube as d 3. Equation 21.15 indicates that the pressure of a gas is proportional to (1) the number of molecules per unit volume and (2) the average translational kinetic energy of the molecules, 1 2m 0v 2. In analyzing this structural model of an ideal gas, we obtain an important result that relates the macroscopic quantity of pressure to a microscopic quantity, the average value of the square of the molecular speed. Therefore, a key link between the molecular world and the large-scale world has been established. Notice that Equation 21.15 verifies some features of pressure with which you are probably familiar. One way to increase the pressure inside a container is to increase the number of molecules per unit volume N/V in the container. That is what you do when you add air to a tire. The pressure in the tire can also be raised by increasing the average translational kinetic energy of the air molecules in the tire. That can be accomplished by increasing the temperature of that air, which is why the pressure inside a tire increases as the tire warms up during long road trips. The continuous flexing of the tire as it moves along the road surface results in work done on the rubber as parts of the tire distort, causing an increase in inter-nal energy of the rubber. The increased temperature of the rubber results in the transfer of energy by heat into the air inside the tire. This transfer increases the air’s temperature, and this increase in temperature in turn produces an increase in pressure.
W W Relationship between pressure and molecular kinetic energy www.aswarphysics.weebly.com 630 Chapter 21 The Kinetic Theory of Gases Molecular Interpretation of Temperature Let’s now consider another macroscopic variable, the temperature T of the gas. We can gain some insight into the meaning of temperature by first writing Equa-tion 21.15 in the form PV 5 2 3 N 1 1 2m0v 22 (21.16) Let’s now compare this expression with the equation of state for an ideal gas (Eq. 19.10): PV 5 NkBT (21.17) Equating the right sides of Equations 21.16 and 21.17 and solving for T gives T 5 2 3k B 1 1 2m0v 22 (21.18) This result tells us that temperature is a direct measure of average molecular kinetic energy. By rearranging Equation 21.18, we can relate the translational molecular kinetic energy to the temperature: 1 2m 0v 2 5 3 2k BT (21.19) That is, the average translational kinetic energy per molecule is 3 2kBT . Because vx2 5 1 3 v2 (Eq. 21.13), it follows that 1 2m 0vx 2 5 1 2k BT (21.20) In a similar manner, for the y and z directions, 1 2m 0vy 2 5 1 2k BT and 1 2m 0vz 2 5 1 2k BT Therefore, each translational degree of freedom contributes an equal amount of energy, 1 2k BT , to the gas. (In general, a “degree of freedom” refers to an indepen-dent means by which a molecule can possess energy.) A generalization of this result, known as the theorem of equipartition of energy, is as follows: Each degree of freedom contributes 1 2k BT to the energy of a system, where possible degrees of freedom are those associated with translation, rotation, and vibration of molecules. The total translational kinetic energy of N molecules of gas is simply N times the average energy per molecule, which is given by Equation 21.19: K tot trans 5 N 1 1 2m 0v 22 5 3 2Nk BT 5 3 2nRT (21.21) where we have used kB 5 R/NA for Boltzmann’s constant and n 5 N/NA for the num-ber of moles of gas. If the gas molecules possess only translational kinetic energy, Equation 21.21 represents the internal energy of the gas. This result implies that the internal energy of an ideal gas depends only on the temperature. We will follow up on this point in Section 21.2. The square root of v 2 is called the root-mean-square (rms) speed of the mol-ecules. From Equation 21.19, we find that the rms speed is vrms 5 " v 2 5 Å 3k BT m0 5 Å 3RT M (21.22) where M is the molar mass in kilograms per mole and is equal to m0NA. This expres-sion shows that, at a given temperature, lighter molecules move faster, on the aver-age, than do heavier molecules. For example, at a given temperature, hydrogen molecules, whose molar mass is 2.02 3 1023 kg/mol, have an average speed approxi-mately four times that of oxygen molecules, whose molar mass is 32.0 3 1023 kg/mol. Table 21.1 lists the rms speeds for various molecules at 208C.
Relationship between temperature and molecular kinetic energy Average kinetic energy per molecule Theorem of equipartition of energy Total translational kinetic energy of N molecules Root-mean-square speed www.aswarphysics.weebly.com 21.2 Molar Specific Heat of an Ideal Gas 631 Table 21.1 Some Root-Mean-Square (rms) Speeds Molar Mass vrms Molar Mass vrms Gas (g/mol) at 208C (m/s) Gas (g/mol) at 208C (m/s) H2 2.02 1902 NO 30.0 494 He 4.00 1352 O2 32.0 478 H2O 18.0 637 CO2 44.0 408 Ne 20.2 602 SO2 64.1 338 N2 or CO 28.0 511 Q uick Quiz 21.1 Two containers hold an ideal gas at the same temperature and pressure. Both containers hold the same type of gas, but container B has twice the volume of container A. (i) What is the average translational kinetic energy per molecule in container B? (a) twice that of container A (b) the same as that of container A (c) half that of container A (d) impossible to determine (ii) From the same choices, describe the internal energy of the gas in container B. Example 21.1 A Tank of Helium A tank used for filling helium balloons has a volume of 0.300 m3 and contains 2.00 mol of helium gas at 20.08C. Assume the helium behaves like an ideal gas.
(A) What is the total translational kinetic energy of the gas molecules?
Conceptualize Imagine a microscopic model of a gas in which you can watch the molecules move about the container more rapidly as the temperature increases. Because the gas is monatomic, the total translational kinetic energy of the molecules is the internal energy of the gas.
Categorize We evaluate parameters with equations developed in the preceding discussion, so this example is a substi-tution problem.
S o l u t i o n Use Equation 21.21 with n 5 2.00 mol and T 5 293 K: E int 5 3 2nRT 5 3 2 12.00 mol2 18.31 J/mol # K2 1293 K2 5 7.30 3 103 J (B) What is the average kinetic energy per molecule?
What if the temperature is raised from 20.08C to 40.08C? Because 40.0 is twice as large as 20.0, is the total translational energy of the molecules of the gas twice as large at the higher temperature?
Answer The expression for the total translational energy depends on the temperature, and the value for the tempera-ture must be expressed in kelvins, not in degrees Celsius. Therefore, the ratio of 40.0 to 20.0 is not the appropriate ratio. Converting the Celsius temperatures to kelvins, 20.08C is 293 K and 40.08C is 313 K. Therefore, the total transla-tional energy increases by a factor of only 313 K/293 K 5 1.07.
What If?
Use Equation 21.19: 1 2m0v 2 5 3 2k BT 5 3 2 11.38 3 10223 J/K2 1293 K2 5 6.07 3 10221 J S o l u t i o n 21.2 Molar Specific Heat of an Ideal Gas Consider an ideal gas undergoing several processes such that the change in tem-perature is DT 5 Tf 2 Ti for all processes. The temperature change can be achieved Pitfall Prevention 21.1 The Square Root of the Square? Taking the square root of v 2 does not “undo” the square because we have taken an average between squaring and taking the square root. Although the square root of 1v2 2 is v 5 vavg because the squar-ing is done after the averaging, the square root of v 2 is not vavg, but rather vrms.
www.aswarphysics.weebly.com 632 Chapter 21 The Kinetic Theory of Gases by taking a variety of paths from one isotherm to another as shown in Figure 21.3. Because DT is the same for all paths, the change in internal energy DE int is the same for all paths. The work W done on the gas (the negative of the area under the curves), however, is different for each path. Therefore, from the first law of thermodynamics, we can argue that the heat Q 5 DE int 2 W associated with a given change in temperature does not have a unique value as discussed in Section 20.4. We can address this difficulty by defining specific heats for two special processes that we have studied: isovolumetric and isobaric. Because the number of moles n is a convenient measure of the amount of gas, we define the molar specific heats associated with these processes as follows: Q 5 nCV DT (constant volume) (21.23) Q 5 nCP DT (constant pressure) (21.24) where CV is the molar specific heat at constant volume and CP is the molar spe-cific heat at constant pressure. When energy is added to a gas by heat at constant pressure, not only does the internal energy of the gas increase, but (negative) work is done on the gas because of the change in volume required to keep the pres-sure constant. Therefore, the heat Q in Equation 21.24 must account for both the increase in internal energy and the transfer of energy out of the system by work. For this reason, Q is greater in Equation 21.24 than in Equation 21.23 for given val-ues of n and DT. Therefore, C P is greater than C V. In the previous section, we found that the temperature of a gas is a measure of the average translational kinetic energy of the gas molecules. This kinetic energy is associated with the motion of the center of mass of each molecule. It does not include the energy associated with the internal motion of the molecule, namely, vibrations and rotations about the center of mass. That should not be surprising because the simple kinetic theory model assumes a structureless molecule. So, let’s first consider the simplest case of an ideal monatomic gas, that is, a gas containing one atom per molecule such as helium, neon, or argon. When energy is added to a monatomic gas in a container of fixed volume, all the added energy goes into increasing the translational kinetic energy of the atoms. There is no other way to store the energy in a monatomic gas. Therefore, from Equation 21.21, we see that the internal energy E int of N molecules (or n mol) of an ideal monatomic gas is E int 5 K tot trans 5 3 2Nk BT 5 3 2nRT (21.25) For a monatomic ideal gas, E int is a function of T only and the functional relation-ship is given by Equation 21.25. In general, the internal energy of any ideal gas is a function of T only and the exact relationship depends on the type of gas. If energy is transferred by heat to a system at constant volume, no work is done on the system. That is, W 5 2e P dV 5 0 for a constant-volume process. Hence, from the first law of thermodynamics, Q 5 DE int (21.26) In other words, all the energy transferred by heat goes into increasing the inter-nal energy of the system. A constant-volume process from i to f for an ideal gas is described in Figure 21.4, where DT is the temperature difference between the two isotherms. Substituting the expression for Q given by Equation 21.23 into Equation 21.26, we obtain DE int 5 nCV DT (21.27) This equation applies to all ideal gases, those gases having more than one atom per molecule as well as monatomic ideal gases. In the limit of infinitesimal changes, we can use Equation 21.27 to express the molar specific heat at constant volume as CV 5 1 n dE int dT (21.28) Internal energy of an ideal monatomic gas P V Isotherms i f f T T f T Figure 21.3 An ideal gas is taken from one isotherm at temperature T to another at temperature T 1 DT along three different paths.
www.aswarphysics.weebly.com 21.2 Molar Specific Heat of an Ideal Gas 633 P V T i f f Isotherms T T For the constant-volume path, all the energy input goes into increasing the internal energy of the gas because no work is done.
Along the constant-pressure path, part of the energy transferred in by heat is transferred out by work.
Figure 21.4 Energy is trans-ferred by heat to an ideal gas in two ways.
Let’s now apply the results of this discussion to a monatomic gas. Substituting the internal energy from Equation 21.25 into Equation 21.28 gives CV 5 3 2R 5 12.5 J/mol # K (21.29) This expression predicts a value of CV 5 3 2R for all monatomic gases. This predic-tion is in excellent agreement with measured values of molar specific heats for such gases as helium, neon, argon, and xenon over a wide range of temperatures (Table 21.2). Small variations in Table 21.2 from the predicted values are because real gases are not ideal gases. In real gases, weak intermolecular interactions occur, which are not addressed in our ideal gas model. Now suppose the gas is taken along the constant-pressure path i S f 9 shown in Figure 21.4. Along this path, the temperature again increases by DT. The energy that must be transferred by heat to the gas in this process is Q 5 nCP DT. Because the volume changes in this process, the work done on the gas is W 5 2P DV, where P is the constant pressure at which the process occurs. Applying the first law of thermodynamics to this process, we have DE int 5 Q 1 W 5 nCP DT 1 (2P DV) (21.30) In this case, the energy added to the gas by heat is channeled as follows. Part of it leaves the system by work (that is, the gas moves a piston through a displacement), and the remainder appears as an increase in the internal energy of the gas. The change in internal energy for the process i S f 9, however, is equal to that for the pro-cess i S f because E int depends only on temperature for an ideal gas and DT is the same for both processes. In addition, because PV 5 nRT, note that for a constant- pressure process, P DV 5 nR DT. Substituting this value for P DV into Equation 21.30 with DE int 5 nCV DT (Eq. 21.27) gives nCV DT 5 nCP DT 2 nR DT CP 2 CV 5 R (21.31) This expression applies to any ideal gas. It predicts that the molar specific heat of an ideal gas at constant pressure is greater than the molar specific heat at constant vol-ume by an amount R, the universal gas constant (which has the value 8.31 J/mol ? K). This expression is applicable to real gases as the data in Table 21.2 show.
Table 21.2 Molar Specific Heats of Various Gases Molar Specific Heat ( J/mol ? K)a Gas CP CV CP 2 CV g 5 CP/CV Monatomic gases He 20.8 12.5 8.33 1.67 Ar 20.8 12.5 8.33 1.67 Ne 20.8 12.7 8.12 1.64 Kr 20.8 12.3 8.49 1.69 Diatomic gases H2 28.8 20.4 8.33 1.41 N2 29.1 20.8 8.33 1.40 O2 29.4 21.1 8.33 1.40 CO 29.3 21.0 8.33 1.40 Cl2 34.7 25.7 8.96 1.35 Polyatomic gases CO2 37.0 28.5 8.50 1.30 SO2 40.4 31.4 9.00 1.29 H2O 35.4 27.0 8.37 1.30 CH4 35.5 27.1 8.41 1.31 a All values except that for water were obtained at 300 K.
www.aswarphysics.weebly.com 634 Chapter 21 The Kinetic Theory of Gases Because CV 5 3 2R for a monatomic ideal gas, Equation 21.31 predicts a value CP 5 5 2R 5 20.8 J/mol # K for the molar specific heat of a monatomic gas at con-stant pressure. The ratio of these molar specific heats is a dimensionless quantity g (Greek letter gamma): g 5 CP CV 5 5R/2 3R/2 5 5 3 5 1.67 (21.32) Theoretical values of CV, CP, and g are in excellent agreement with experimental values obtained for monatomic gases, but they are in serious disagreement with the values for the more complex gases (see Table 21.2). That is not surprising; the value CV 5 3 2R was derived for a monatomic ideal gas, and we expect some additional contribution to the molar specific heat from the internal structure of the more complex molecules. In Section 21.3, we describe the effect of molecular structure on the molar specific heat of a gas. The internal energy—and hence the molar specific heat—of a complex gas must include contributions from the rotational and the vibrational motions of the molecule. In the case of solids and liquids heated at constant pressure, very little work is done during such a process because the thermal expansion is small. Consequently, CP and CV are approximately equal for solids and liquids.
Q uick Quiz 21.2 (i) How does the internal energy of an ideal gas change as it fol-lows path i S f in Figure 21.4? (a) E int increases. (b) E int decreases. (c) E int stays the same. (d) There is not enough information to determine how E int changes. (ii) From the same choices, how does the internal energy of an ideal gas change as it follows path f S f 9 along the isotherm labeled T 1 DT in Figure 21.4?
Ratio of molar specific heats for a monatomic ideal gas Example 21.2 Heating a Cylinder of Helium A cylinder contains 3.00 mol of helium gas at a temperature of 300 K.
(A) If the gas is heated at constant volume, how much energy must be transferred by heat to the gas for its tempera-ture to increase to 500 K?
Conceptualize Run the process in your mind with the help of the piston–cylinder arrangement in Figure 19.12. Imag-ine that the piston is clamped in position to maintain the constant volume of the gas.
Categorize We evaluate parameters with equations developed in the preceding discussion, so this example is a substi-tution problem.
S o l u t i o n Use Equation 21.23 to find the energy transfer: Q 1 5 nCV DT Substitute the given values: Q 1 5 (3.00 mol)(12.5 J/mol ? K)(500 K 2 300 K) 5 7.50 3 103 J Use Equation 21.24 to find the energy transfer: Q 2 5 nCP DT Substitute the given values: Q 2 5 (3.00 mol)(20.8 J/mol ? K)(500 K 2 300 K) 5 12.5 3 103 J (B) How much energy must be transferred by heat to the gas at constant pressure to raise the temperature to 500 K?
S o l u t i o n This value is larger than Q 1 because of the transfer of energy out of the gas by work to raise the piston in the constant pressure process.
www.aswarphysics.weebly.com 21.3 The Equipartition of Energy 635 21.3 The Equipartition of Energy Predictions based on our model for molar specific heat agree quite well with the behavior of monatomic gases, but not with the behavior of complex gases (see Table 21.2). The value predicted by the model for the quantity CP 2 CV 5 R, however, is the same for all gases. This similarity is not surprising because this difference is the result of the work done on the gas, which is independent of its molecular structure. To clarify the variations in CV and CP in gases more complex than monatomic gases, let’s explore further the origin of molar specific heat. So far, we have assumed the sole contribution to the internal energy of a gas is the translational kinetic energy of the molecules. The internal energy of a gas, however, includes contributions from the translational, vibrational, and rotational motion of the mol-ecules. The rotational and vibrational motions of molecules can be activated by collisions and therefore are “coupled” to the translational motion of the molecules. The branch of physics known as statistical mechanics has shown that, for a large num-ber of particles obeying the laws of Newtonian mechanics, the available energy is, on average, shared equally by each independent degree of freedom. Recall from Section 21.1 that the equipartition theorem states that, at equilibrium, each degree of freedom contributes 1 2k BT of energy per molecule. Let’s consider a diatomic gas whose molecules have the shape of a dumbbell (Fig. 21.5). In this model, the center of mass of the molecule can translate in the x, y, and z directions (Fig. 21.5a). In addition, the molecule can rotate about three mutually perpendicular axes (Fig. 21.5b). The rotation about the y axis can be neglected because the molecule’s moment of inertia Iy and its rotational energy 1 2Iyv2 about this axis are negligible compared with those associated with the x and z axes. (If the two atoms are modeled as particles, then Iy is identically zero.) Therefore, there are five degrees of freedom for translation and rotation: three associated with the translational motion and two associated with the rotational motion. Because each degree of freedom contributes, on average, 1 2k BT of energy per molecule, the inter-nal energy for a system of N molecules, ignoring vibration for now, is E int 5 3N 1 1 2k BT2 1 2N 1 1 2k BT2 5 5 2Nk BT 5 5 2nRT We can use this result and Equation 21.28 to find the molar specific heat at con-stant volume: CV 5 1 n dE int dT 5 1 n d dT 1 5 2nRT2 5 5 2R 5 20.8 J/mol ? K (21.33) From Equations 21.31 and 21.32, we find that CP 5 CV 1 R 5 7 2R 5 29.1 J/mol ? K g 5 CP CV 5 7 2R 5 2R 5 7 5 5 1.40 These results agree quite well with most of the data for diatomic molecules given in Table 21.2. That is rather surprising because we have not yet accounted for the possible vibrations of the molecule. In the model for vibration, the two atoms are joined by an imaginary spring (see Fig. 21.5c). The vibrational motion adds two more degrees of freedom, which cor-respond to the kinetic energy and the potential energy associated with vibrations along the length of the molecule. Hence, a model that includes all three types of motion predicts a total internal energy of E int 5 3N 1 1 2k BT2 1 2N 1 1 2k BT2 1 2N 1 1 2k BT2 5 7 2Nk BT 5 7 2nRT and a molar specific heat at constant volume of CV 5 1 n dE int dT 5 1 n d dT 1 7 2nRT2 5 7 2R 5 29.1 J/mol ? K (21.34) x z y y x z Translational motion of the center of mass Rotational motion about the various axes Vibrational motion along the molecular axis a b c Figure 21.5 Possible motions of a diatomic molecule.
www.aswarphysics.weebly.com 636 Chapter 21 The Kinetic Theory of Gases This value is inconsistent with experimental data for molecules such as H2 and N2 (see Table 21.2) and suggests a breakdown of our model based on classical physics. It might seem that our model is a failure for predicting molar specific heats for diatomic gases. We can claim some success for our model, however, if measure-ments of molar specific heat are made over a wide temperature range rather than at the single temperature that gives us the values in Table 21.2. Figure 21.6 shows the molar specific heat of hydrogen as a function of temperature. The remarkable fea-ture about the three plateaus in the graph’s curve is that they are at the values of the molar specific heat predicted by Equations 21.29, 21.33, and 21.34! For low tempera-tures, the diatomic hydrogen gas behaves like a monatomic gas. As the temperature rises to room temperature, its molar specific heat rises to a value for a diatomic gas, consistent with the inclusion of rotation but not vibration. For high temperatures, the molar specific heat is consistent with a model including all types of motion. Before addressing the reason for this mysterious behavior, let’s make some brief remarks about polyatomic gases. For molecules with more than two atoms, three axes of rotation are available. The vibrations are more complex than for diatomic molecules. Therefore, the number of degrees of freedom is even larger. The result is an even higher predicted molar specific heat, which is in qualitative agreement with experiment. The molar specific heats for the polyatomic gases in Table 21.2 are higher than those for diatomic gases. The more degrees of freedom available to a molecule, the more “ways” there are to store energy, resulting in a higher molar specific heat.
A Hint of Energy Quantization Our model for molar specific heats has been based so far on purely classical notions. It predicts a value of the specific heat for a diatomic gas that, according to Figure 21.6, only agrees with experimental measurements made at high temperatures. To explain why this value is only true at high temperatures and why the plateaus in Figure 21.6 exist, we must go beyond classical physics and introduce some quantum physics into the model. In Chapter 18, we discussed quantization of frequency for vibrating strings and air columns; only certain frequencies of standing waves can exist. That is a natural result whenever waves are subject to boundary conditions. Quantum physics (Chapters 40 through 43) shows that atoms and molecules can be described by the waves under boundary conditions analysis model. Conse-quently, these waves have quantized frequencies. Furthermore, in quantum physics, the energy of a system is proportional to the frequency of the wave representing the system. Hence, the energies of atoms and molecules are quantized. For a molecule, quantum physics tells us that the rotational and vibrational ener-gies are quantized. Figure 21.7 shows an energy-level diagram for the rotational Translation Rotation Vibration Temperature (K) CV ( J/mol · K) 0 5 10 15 20 25 30 10 20 50 100 200 500 1000 2000 5000 10000 7 2 –R 5 2 –R 3 2 –R The horizontal scale is logarithmic.
Hydrogen liquefies at 20 K.
Figure 21.6 The molar specific heat of hydrogen as a function of temperature.
www.aswarphysics.weebly.com 21.4 Adiabatic Processes for an Ideal Gas 637 and vibrational quantum states of a diatomic molecule. The lowest allowed state is called the ground state. The black lines show the energies allowed for the mol-ecule. Notice that allowed vibrational states are separated by larger energy gaps than are rotational states. At low temperatures, the energy a molecule gains in collisions with its neighbors is generally not large enough to raise it to the first excited state of either rotation or vibration. Therefore, even though rotation and vibration are allowed according to classical physics, they do not occur in reality at low temperatures. All molecules are in the ground state for rotation and vibration. The only contribution to the mol-ecules’ average energy is from translation, and the specific heat is that predicted by Equation 21.29. As the temperature is raised, the average energy of the molecules increases. In some collisions, a molecule may have enough energy transferred to it from another molecule to excite the first rotational state. As the temperature is raised further, more molecules can be excited to this state. The result is that rotation begins to contribute to the internal energy, and the molar specific heat rises. At about room temperature in Figure 21.6, the second plateau has been reached and rotation con-tributes fully to the molar specific heat. The molar specific heat is now equal to the value predicted by Equation 21.33. There is no contribution at room temperature from vibration because the mole-cules are still in the ground vibrational state. The temperature must be raised even further to excite the first vibrational state, which happens in Figure 21.6 between 1 000 K and 10 000 K. At 10 000 K on the right side of the figure, vibration is con-tributing fully to the internal energy and the molar specific heat has the value pre-dicted by Equation 21.34. The predictions of this model are supportive of the theorem of equipartition of energy. In addition, the inclusion in the model of energy quantization from quan-tum physics allows a full understanding of Figure 21.6.
Q uick Quiz 21.3 The molar specific heat of a diatomic gas is measured at constant volume and found to be 29.1 J/mol ? K. What are the types of energy that are con-tributing to the molar specific heat? (a) translation only (b) translation and rota-tion only (c) translation and vibration only (d) translation, rotation, and vibration Q uick Quiz 21.4 The molar specific heat of a gas is measured at constant volume and found to be 11R/2. Is the gas most likely to be (a) monatomic, (b) diatomic, or (c) polyatomic?
21.4 Adiabatic Processes for an Ideal Gas As noted in Section 20.6, an adiabatic process is one in which no energy is trans-ferred by heat between a system and its surroundings. For example, if a gas is com-pressed (or expanded) rapidly, very little energy is transferred out of (or into) the system by heat, so the process is nearly adiabatic. Such processes occur in the cycle of a gasoline engine, which is discussed in detail in Chapter 22. Another example of an adiabatic process is the slow expansion of a gas that is thermally insulated from its surroundings. All three variables in the ideal gas law—P, V, and T—change during an adiabatic process. Let’s imagine an adiabatic gas process involving an infinitesimal change in volume dV and an accompanying infinitesimal change in temperature dT. The work done on the gas is 2P dV. Because the internal energy of an ideal gas depends only on temperature, the change in the internal energy in an adiabatic process is the same as that for an isovolumetric process between the same temperatures, dE int 5 nCV dT (Eq. 21.27). Hence, the first law of thermodynamics, DE int 5 Q 1 W, with Q 5 0, becomes the infinitesimal form dE int 5 nCV dT 5 2P dV (21.35) Rotational states Rotational states Vibrational states ENERGY The rotational states lie closer together in energy than do the vibrational states.
Figure 21.7 An energy-level dia-gram for vibrational and rotational states of a diatomic molecule. www.aswarphysics.weebly.com 638 Chapter 21 The Kinetic Theory of Gases Taking the total differential of the equation of state of an ideal gas, PV 5 nRT, gives P dV 1 V dP 5 nR dT (21.36) Eliminating dT from Equations 21.35 and 21.36, we find that P dV 1 V dP 5 2 R CV P dV Substituting R 5 CP 2 CV and dividing by PV gives dV V 1 dP P 5 2aCP 2 CV CV b dV V 5 11 2 g2 dV V dP P 1 g dV V 5 0 Integrating this expression, we have ln P 1 g ln V 5 constant which is equivalent to PV g 5 constant (21.37) The PV diagram for an adiabatic expansion is shown in Figure 21.8. Because g . 1, the PV curve is steeper than it would be for an isothermal expansion, for which PV 5 constant. By the definition of an adiabatic process, no energy is trans-ferred by heat into or out of the system. Hence, from the first law, we see that DE int is negative (work is done by the gas, so its internal energy decreases) and so DT also is negative. Therefore, the temperature of the gas decreases (Tf , Ti) during an adi-abatic expansion.2 Conversely, the temperature increases if the gas is compressed adiabatically. Applying Equation 21.37 to the initial and final states, we see that PiVi g 5 PfVf g (21.38) Using the ideal gas law, we can express Equation 21.37 as TV g21 5 constant (21.39) Relationship between P and V for an adiabatic process involving an ideal gas Relationship between T and V for an adiabatic process involving an ideal gas 2In the adiabatic free expansion discussed in Section 20.6, the temperature remains constant. In this unique pro-cess, no work is done because the gas expands into a vacuum. In general, the temperature decreases in an adiabatic expansion in which work is done.
Example 21.3 A Diesel Engine Cylinder Air at 20.08C in the cylinder of a diesel engine is compressed from an initial pressure of 1.00 atm and volume of 800.0 cm3 to a volume of 60.0 cm3. Assume air behaves as an ideal gas with g 5 1.40 and the compression is adiabatic. Find the final pressure and temperature of the air.
Conceptualize Imagine what happens if a gas is compressed into a smaller volume. Our discussion above and Figure 21.8 tell us that the pressure and temperature both increase.
Categorize We categorize this example as a problem involving an adiabatic process.
S o l u t i o n Ti Tf Isotherms P V Pi Pf Vi Vf i f The temperature of a gas decreases in an adiabatic expansion.
Figure 21.8 The PV diagram for an adiabatic expansion of an ideal gas. Analyze Use Equation 21.38 to find the final pressure: Pf 5 Pi aVi Vf b g 5 11.00 atm2 a800.0 cm3 60.0 cm3 b 1.40 5 37.6 atm www.aswarphysics.weebly.com 21.5 Distribution of Molecular Speeds 639 Finalize The temperature of the gas increases by a factor of 826 K/293 K 5 2.82. The high compression in a diesel engine raises the temperature of the gas enough to cause the combustion of fuel without the use of spark plugs.
21.5 Distribution of Molecular Speeds Thus far, we have considered only average values of the energies of all the molecules in a gas and have not addressed the distribution of energies among individual mol-ecules. The motion of the molecules is extremely chaotic. Any individual molecule collides with others at an enormous rate, typically a billion times per second. Each collision results in a change in the speed and direction of motion of each of the participant molecules. Equation 21.22 shows that rms molecular speeds increase with increasing temperature. At a given time, what is the relative number of mol-ecules that possess some characteristic such as energy within a certain range? We shall address this question by considering the number density nV(E). This quantity, called a distribution function, is defined so that nV(E) dE is the number of molecules per unit volume with energy between E and E 1 dE. (The ratio of the number of molecules that have the desired characteristic to the total number of molecules is the probability that a particular molecule has that characteristic.) In general, the number density is found from statistical mechanics to be nV 1E 2 5 n 0e2E/kBT (21.40) where n0 is defined such that n0 dE is the number of molecules per unit volume hav-ing energy between E 5 0 and E 5 dE. This equation, known as the Boltzmann dis-tribution law, is important in describing the statistical mechanics of a large number of molecules. It states that the probability of finding the molecules in a particular energy state varies exponentially as the negative of the energy divided by k BT. All the molecules would fall into the lowest energy level if the thermal agitation at a temperature T did not excite the molecules to higher energy levels.
W W Boltzmann distribution law Use the ideal gas law to find the final temperature: PiVi Ti 5 PfVf Tf Tf 5 Pf Vf PiVi Ti 5 137.6 atm2 160.0 cm32 11.00 atm2 1800.0 cm32 1293 K2 5 826 K 5 5538C ▸ 21.3 c on tin u ed Example 21.4 Thermal Excitation of Atomic Energy Levels As discussed in Section 21.4, atoms can occupy only certain discrete energy levels. Con-sider a gas at a temperature of 2 500 K whose atoms can occupy only two energy levels separated by 1.50 eV, where 1 eV (electron volt) is an energy unit equal to 1.60 3 10219 J (Fig. 21.9). Determine the ratio of the number of atoms in the higher energy level to the number in the lower energy level.
Conceptualize In your mental representation of this example, remember that only two possible states are allowed for the system of the atom. Figure 21.9 helps you visualize the two states on an energy-level diagram. In this case, the atom has two possible energies, E1 and E 2, where E1 , E 2.
S o l u t i o n continued Pitfall Prevention 21.2 The Distribution Function The distribution function nV(E) is defined in terms of the number of molecules with energy in the range E to E 1 dE rather than in terms of the number of molecules with energy E. Because the num-ber of molecules is finite and the number of possible values of the energy is infinite, the number of molecules with an exact energy E may be zero.
E1 E2 1.50 eV ENERGY Figure 21.9 (Example 21.4) Energy-level diagram for a gas whose atoms can occupy two energy states.
www.aswarphysics.weebly.com 640 Chapter 21 The Kinetic Theory of Gases Now that we have discussed the distribution of energies among molecules in a gas, let’s think about the distribution of molecular speeds. In 1860, James Clerk Maxwell (1831–1879) derived an expression that describes the distribution of molecular speeds in a very definite manner. His work and subsequent developments by other scientists were highly controversial because direct detection of molecules could not be achieved experimentally at that time. About 60 years later, however, experiments were devised that confirmed Maxwell’s predictions. Let’s consider a container of gas whose molecules have some distribution of speeds. Suppose we want to determine how many gas molecules have a speed in the range from, for example, 400 to 401 m/s. Intuitively, we expect the speed distribu-tion to depend on temperature. Furthermore, we expect the distribution to peak in the vicinity of vrms. That is, few molecules are expected to have speeds much less than or much greater than vrms because these extreme speeds result only from an unlikely chain of collisions. The observed speed distribution of gas molecules in thermal equilibrium is shown in Figure 21.10. The quantity Nv, called the Maxwell–Boltzmann speed dis-tribution function, is defined as follows. If N is the total number of molecules, the number of molecules with speeds between v and v 1 dv is dN 5 Nv dv. This number is also equal to the area of the shaded rectangle in Figure 21.10. Furthermore, the fraction of molecules with speeds between v and v 1 dv is (Nv dv)/N. This fraction is also equal to the probability that a molecule has a speed in the range v to v 1 dv.
Evaluate k BT in the exponent: k BT 5 11.38 3 10223 J/K2 12 500 K2 a 1 eV 1.60 3 10219 Jb 5 0.216 eV Substitute this value into Equation (1): nV 1E22 nV 1E12 5 e21.50 eV/0.216 eV 5 e26.96 5 9.52 3 1024 Finalize This result indicates that at T 5 2 500 K, only a small fraction of the atoms are in the higher energy level. In fact, for every atom in the higher energy level, there are about 1 000 atoms in the lower level. The number of atoms in the higher level increases at even higher temperatures, but the distribution law specifies that at equilibrium there are always more atoms in the lower level than in the higher level.
What if the energy levels in Figure 21.9 were closer together in energy? Would that increase or decrease the fraction of the atoms in the upper energy level?
Answer If the excited level is lower in energy than that in Figure 21.9, it would be easier for thermal agitation to excite atoms to this level and the fraction of atoms in this energy level would be larger, which we can see mathematically by expressing Equation (1) as r2 5 e21E22E12/k BT where r2 is the ratio of atoms having energy E 2 to those with energy E1. Differentiating with respect to E 2, we find dr2 dE 2 5 d dE 2 3e21E2 2E12/kBT 4 5 2 1 k BT e21E22E12/kBT , 0 Because the derivative has a negative value, as E 2 decreases, r2 increases.
What If?
Ludwig Boltzmann Austrian physicist (1844–1906) Boltzmann made many important contributions to the development of the kinetic theory of gases, electro-magnetism, and thermodynamics. His pioneering work in the field of kinetic theory led to the branch of physics known as statistical mechanics.
© INTERFOTO/Alamy Analyze Set up the ratio of the number of atoms in the higher energy level to the num-ber in the lower energy level and use Equa-tion 21.40 to express each number: (1) nV 1E22 nV 1E12 5 n 0e2E2/k BT n 0e2E1/k BT 5 e21E22E12/k BT Categorize We categorize this example as one in which we focus on particles in a two-state quantized system. We will apply the Boltzmann distribution law to this system. ▸ 21.4 continu ed www.aswarphysics.weebly.com 21.5 Distribution of Molecular Speeds 641 The fundamental expression that describes the distribution of speeds of N gas molecules is Nv 5 4pN a m0 2pkBTb 3/2 v 2e2m 0v 2/2kBT (21.41) where m0 is the mass of a gas molecule, k B is Boltzmann’s constant, and T is the absolute temperature.3 Observe the appearance of the Boltzmann factor e2E/kBT with E 5 1 2m0v2. As indicated in Figure 21.10, the average speed is somewhat lower than the rms speed. The most probable speed vmp is the speed at which the distribution curve reaches a peak. Using Equation 21.41, we find that vrms 5 " v 2 5 Å 3k BT m0 5 1.73Å k BT m 0 (21.42) vavg 5 Å 8k BT pm 0 5 1.60Å k BT m0 (21.43) vmp 5 Å 2k BT m0 5 1.41Å k BT m0 (21.44) Equation 21.42 has previously appeared as Equation 21.22. The details of the deri-vations of these equations from Equation 21.41 are left for the end-of-chapter prob-lems (see Problems 42 and 69). From these equations, we see that vrms . vavg . vmp Figure 21.11 represents speed distribution curves for nitrogen, N2. The curves were obtained by using Equation 21.41 to evaluate the distribution function at vari-ous speeds and at two temperatures. Notice that the peak in each curve shifts to the right as T increases, indicating that the average speed increases with increasing temperature, as expected. Because the lowest speed possible is zero and the upper classical limit of the speed is infinity, the curves are asymmetrical. (In Chapter 39, we show that the actual upper limit is the speed of light.) Equation 21.41 shows that the distribution of molecular speeds in a gas depends both on mass and on temperature. At a given temperature, the fraction of mol-ecules with speeds exceeding a fixed value increases as the mass decreases. Hence, 3 For the derivation of this expression, see an advanced textbook on thermodynamics.
vmp vrms Nv v v avg Nv dv The number of molecules having speeds ranging from v to v dv equals the area of the tan rectangle, Nv dv.
Figure 21.10 The speed distri-bution of gas molecules at some temperature. The function Nv approaches zero as v approaches infinity.
Figure 21.11 The speed distri-bution function for 105 nitrogen molecules at 300 K and 900 K.
200 160 120 80 40 200 400 600 800 1000 1200 1400 1600 T 300 K T 900 K Nv [molecules/(m/s)] vrms vavg v (m/s) vmp 0 0 The total area under either curve is equal to N, the total number of molecules. In this case, N 105.
Note that vrms vavg vmp.
www.aswarphysics.weebly.com 642 Chapter 21 The Kinetic Theory of Gases lighter molecules such as H2 and He escape into space more readily from the Earth’s atmosphere than do heavier molecules such as N2 and O2. (See the discus-sion of escape speed in Chapter 13. Gas molecules escape even more readily from the Moon’s surface than from the Earth’s because the escape speed on the Moon is lower than that on the Earth.) The speed distribution curves for molecules in a liquid are similar to those shown in Figure 21.11. We can understand the phenomenon of evaporation of a liquid from this distribution in speeds, given that some molecules in the liquid are more energetic than others. Some of the faster-moving molecules in the liq-uid penetrate the surface and even leave the liquid at temperatures well below the boiling point. The molecules that escape the liquid by evaporation are those that have sufficient energy to overcome the attractive forces of the molecules in the liquid phase. Consequently, the molecules left behind in the liquid phase have a lower average kinetic energy; as a result, the temperature of the liquid decreases. Hence, evaporation is a cooling process. For example, an alcohol-soaked cloth can be placed on a feverish head to cool and comfort a patient.
Example 21.5 A System of Nine Particles Nine particles have speeds of 5.00, 8.00, 12.0, 12.0, 12.0, 14.0, 14.0, 17.0, and 20.0 m/s.
(A) Find the particles’ average speed.
Conceptualize Imagine a small number of particles moving in random directions with the few speeds listed. This situ-ation is not representative of the large number of molecules in a gas, so we should not expect the results to be consis-tent with those from statistical mechanics.
Categorize Because we are dealing with a small number of particles, we can calculate the average speed directly.
S o l u t i o n Analyze Find the average speed of the particles by divid-ing the sum of the speeds by the total number of particles: vavg 5 15.00 1 8.00 1 12.0 1 12.0 1 12.0 1 14.0 1 14.0 1 17.0 1 20.02 m/s 9 5 12.7 m/s Find the average speed squared of the particles by dividing the sum of the speeds squared by the total number of particles: v 2 5 15.002 1 8.002 1 12.02 1 12.02 1 12.02 1 14.02 1 14.02 1 17.02 1 20.022 m2/s2 9 5 178 m2/s2 Find the rms speed of the par-ticles by taking the square root: vrms 5 " v 2 5 "178 m2/s2 5 13.3 m/s (B) What is the rms speed of the particles?
S o l u t i o n (C) What is the most probable speed of the particles?
Three of the particles have a speed of 12.0 m/s, two have a speed of 14.0 m/s, and the remaining four have different speeds. Hence, the most probable speed vmp is 12.0 m/s.
Finalize Compare this example, in which the number of particles is small and we know the individual particle speeds, with the next example.
S o l u t i o n www.aswarphysics.weebly.com 21.5 Distribution of Molecular Speeds 643 Example 21.6 Molecular Speeds in a Hydrogen Gas A 0.500-mol sample of hydrogen gas is at 300 K.
(A) Find the average speed, the rms speed, and the most probable speed of the hydrogen molecules.
Conceptualize Imagine a huge number of particles in a real gas, all moving in random directions with different speeds.
Categorize We cannot calculate the averages as was done in Example 21.5 because the individual speeds of the par-ticles are not known. We are dealing with a very large number of particles, however, so we can use the Maxwell-Boltzmann speed distribution function.
S o l u t i o n Analyze Use Equation 21.43 to find the average speed: vavg 5 1.60 Å k BT m0 5 1.60 Å 11.38 3 10223 J/K2 1300 K2 211.67 3 10227 kg2 5 1.78 3 103 m/s Use Equation 21.42 to find the rms speed: vrms 5 1.73 Å k BT m0 5 1.73 Å 11.38 3 10223 J/K2 1300 K2 211.67 3 10227 kg2 5 1.93 3 103 m/s Use Equation 21.44 to find the most probable speed: vmp 5 1.41 Å k BT m0 5 1.41 Å 11.38 3 10223 J/K2 1300 K2 211.67 3 10227 kg2 5 1.57 3 103 m/s Use Equation 21.41 to evaluate the number of molecules in a narrow speed range between v and v 1 dv: (1) Nv dv 5 4pN a m0 2pk BTb 3/2 v 2e2m 0v 2/2k BT dv Evaluate the constant in front of v 2: 4pN a m0 2pk BTb 3/2 5 4pnNAa m0 2pk BTb 3/2 5 4p10.500 mol2 16.02 3 1023 mol212 c 211.67 3 10227 kg2 2p11.38 3 10223 J/K2 1300 K2 d 3/2 5 1.74 3 1014 s3/m3 Evaluate the exponent of e that appears in Equation (1): 2 m 0v 2 2k BT 5 2 211.67 3 10227 kg2 1400 m/s2 2 211.38 3 10223 J/K2 1300 K2 5 20.064 5 Evaluate Nv dv using these values in Equation (1): Nv dv 5 11.74 3 1014 s3/m32 1400 m/s2 2e20.064 511 m/s2 5 2.61 3 1019 molecules (B) Find the number of molecules with speeds between 400 m/s and 401 m/s.
S o l u t i o n Finalize In this evaluation, we could calculate the result without integration because dv 5 1 m/s is much smaller than v 5 400 m/s. Had we sought the number of particles between, say, 400 m/s and 500 m/s, we would need to integrate Equation (1) between these speed limits.
www.aswarphysics.weebly.com 644 Chapter 21 The Kinetic Theory of Gases Summary The pressure of N molecules of an ideal gas contained in a volume V is P 5 2 3 aN V b 1 1 2m 0v 22 (21.15) The average translational kinetic energy per molecule of a gas, 1 2m0v 2, is related to the temperature T of the gas through the expression 1 2m0v 2 5 3 2k BT (21.19) where kB is Boltzmann’s constant. Each translational degree of freedom (x, y, or z) has 1 2kBT of energy associated with it. The internal energy of N molecules (or n mol) of an ideal monatomic gas is E int 5 3 2Nk BT 5 3 2nRT (21.25) The change in internal energy for n mol of any ideal gas that undergoes a change in temperature DT is DEint 5 nCV DT (21.27) where CV is the molar specific heat at constant volume. The molar specific heat of an ideal monatomic gas at constant volume is CV 5 3 2R; the molar specific heat at constant pressure is CP 5 5 2R. The ratio of specific heats is given by g 5 CP/CV 5 5 3. The Boltzmann distribution law describes the distri-bution of particles among available energy states. The relative number of particles having energy between E and E 1 dE is nV(E) dE, where nV 1E 2 5 n 0e2E/k BT (21.40) The Maxwell–Boltzmann speed distribution function describes the distribution of speeds of molecules in a gas: Nv 5 4pN a m0 2pk BTb 3/2 v 2e2m0v2/2kBT (21.41) If an ideal gas undergoes an adiabatic expansion or compression, the first law of thermodynamics, together with the equation of state, shows that PV g 5 constant (21.37) Equation 21.41 enables us to calculate the root-mean-square speed, the average speed, and the most probable speed of molecules in a gas: vrms 5 " v 2 5 Å 3k BT m0 5 1.73 Å k BT m0 (21.42) vavg 5 Å 8k BT pm0 5 1.60 Å k BT m0 (21.43) vmp 5 Å 2k BT m0 5 1.41 Å k BT m0 (21.44) Concepts and Principles !3 times the original speed. (e) It increases by a factor of 6. 3. Two samples of the same ideal gas have the same pres-sure and density. Sample B has twice the volume of sample A. What is the rms speed of the molecules in sample B? (a) twice that in sample A (b) equal to that in sample A (c) half that in sample A (d) impossible to determine 4. A helium-filled latex balloon initially at room tem-perature is placed in a freezer. The latex remains flexible. (i) Does the balloon’s volume (a) increase, (b) decrease, or (c) remain the same? (ii) Does the pressure of the helium gas (a) increase significantly, (b) decrease significantly, or (c) remain approximately the same? 1. Cylinder A contains oxygen (O2) gas, and cylinder B contains nitrogen (N2) gas. If the molecules in the two cylinders have the same rms speeds, which of the fol-lowing statements is false? (a) The two gases have dif-ferent temperatures. (b) The temperature of cylinder B is less than the temperature of cylinder A. (c) The temperature of cylinder B is greater than the tempera-ture of cylinder A. (d) The average kinetic energy of the nitrogen molecules is less than the average kinetic energy of the oxygen molecules. 2. An ideal gas is maintained at constant pressure. If the temperature of the gas is increased from 200 K to 600 K, what happens to the rms speed of the mol-ecules? (a) It increases by a factor of 3. (b) It remains the same. (c) It is one-third the original speed. (d) It is Objective Questions 1. denotes answer available in Student Solutions Manual/Study Guide www.aswarphysics.weebly.com Problems 645 (f) of this account are correct statements necessary for a clear and complete explanation? (ii) Which are correct statements that are not necessary to account for the higher thermometer reading? (iii) Which are incorrect statements? 8. An ideal gas is contained in a vessel at 300 K. The tem-perature of the gas is then increased to 900 K. (i) By what factor does the average kinetic energy of the mol-ecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of !3, (d) a factor of 1, or (e) a factor of 1 3? Using the same choices as in part (i), by what factor does each of the following change: (ii) the rms molecular speed of the molecules, (iii) the average momentum change that one molecule undergoes in a collision with one particular wall, (iv) the rate of collisions of molecules with walls, and (v) the pressure of the gas. 9. Which of the assumptions below is not made in the kinetic theory of gases? (a) The number of molecules is very large. (b) The molecules obey Newton’s laws of motion. (c) The forces between molecules are long range. (d) The gas is a pure substance. (e) The aver-age separation between molecules is large compared to their dimensions. 5. A gas is at 200 K. If we wish to double the rms speed of the molecules of the gas, to what value must we raise its temperature? (a) 283 K (b) 400 K (c) 566 K (d) 800 K (e) 1130 K 6. Rank the following from largest to smallest, noting any cases of equality. (a) the average speed of molecules in a particular sample of ideal gas (b) the most probable speed (c) the root-mean-square speed (d) the average vector velocity of the molecules 7. A sample of gas with a thermometer immersed in the gas is held over a hot plate. A student is asked to give a step-by-step account of what makes our observation of the temperature of the gas increase. His response includes the following steps. (a) The molecules speed up. (b) Then the molecules collide with one another more often. (c) Internal friction makes the colli-sions inelastic. (d) Heat is produced in the collisions. (e) The molecules of the gas transfer more energy to the thermometer when they strike it, so we observe that the temperature has gone up. (f) The same pro-cess can take place without the use of a hot plate if you quickly push in the piston in an insulated cylinder containing the gas. (i) Which of the parts (a) through Conceptual Questions 1. denotes answer available in Student Solutions Manual/Study Guide 1. Hot air rises, so why does it generally become cooler as you climb a mountain? Note: Air has low thermal conductivity. 2. Why does a diatomic gas have a greater energy con-tent per mole than a monatomic gas at the same temperature? 3. When alcohol is rubbed on your body, it lowers your skin temperature. Explain this effect. 4. What happens to a helium-filled latex balloon released into the air? Does it expand or contract? Does it stop rising at some height? 5. Which is denser, dry air or air saturated with water vapor? Explain. 6. One container is filled with helium gas and another with argon gas. Both containers are at the same tem-perature. Which molecules have the higher rms speed? Explain. 7. Dalton’s law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the pressures that each gas in the mixture would exert if it were alone in the container. Give a convincing argument for this law based on the kinetic theory of gases.
atoms? (c) What is the rms speed of the helium atoms? 2. A cylinder contains a mixture of helium and argon gas in equilibrium at 1508C. (a) What is the average kinetic energy for each type of gas molecule? (b) What is the rms speed of each type of molecule?
M Section 21.1 Molecular Model of an Ideal Gas Problem 30 in Chapter 19 can be assigned with this section. 1. (a) How many atoms of helium gas fill a spherical balloon of diameter 30.0 cm at 20.08C and 1.00 atm? (b) What is the average kinetic energy of the helium M Problems The problems found in this chapter may be assigned online in Enhanced WebAssign 1. straightforward; 2. intermediate; 3. challenging 1. full solution available in the Student Solutions Manual/Study Guide AMT Analysis Model tutorial available in Enhanced WebAssign GP Guided Problem M Master It tutorial available in Enhanced WebAssign W Watch It video solution available in Enhanced WebAssign BIO Q/C S www.aswarphysics.weebly.com 646 Chapter 21 The Kinetic Theory of Gases 3. In a 30.0-s interval, 500 hailstones strike a glass win-dow of area 0.600 m2 at an angle of 45.08 to the win-dow surface. Each hailstone has a mass of 5.00 g and a speed of 8.00 m/s. Assuming the collisions are elastic, find (a) the average force and (b) the average pressure on the window during this interval. 4. In an ultrahigh vacuum system (with typical pressures lower than 1027 pascal), the pressure is measured to be 1.00 3 10210 torr (where 1 torr 5 133 Pa). Assum-ing the temperature is 300 K, find the number of mol-ecules in a volume of 1.00 m3. 5. A spherical balloon of volume 4.00 3 103 cm3 contains helium at a pressure of 1.20 3 105 Pa. How many moles of helium are in the balloon if the average kinetic energy of the helium atoms is 3.60 3 10222 J? 6. A spherical balloon of volume V contains helium at a pressure P. How many moles of helium are in the bal-loon if the average kinetic energy of the helium atoms is K ? 7. A 2.00-mol sample of oxygen gas is confined to a 5.00-L vessel at a pressure of 8.00 atm. Find the average trans-lational kinetic energy of the oxygen molecules under these conditions. 8. Oxygen, modeled as an ideal gas, is in a container and has a temperature of 77.08C. What is the rms-average magnitude of the momentum of the gas molecules in the container? 9. Calculate the mass of an atom of (a) helium, (b) iron, and (c) lead. Give your answers in kilograms. The atomic masses of these atoms are 4.00 u, 55.9 u, and 207 u, respectively. 10. The rms speed of an oxygen molecule (O2) in a con-tainer of oxygen gas is 625 m/s. What is the tempera-ture of the gas? 11. A 5.00-L vessel contains nitrogen gas at 27.0°C and 3.00 atm. Find (a) the total translational kinetic energy of the gas molecules and (b) the average kinetic energy per molecule. 12. A 7.00-L vessel contains 3.50 moles of gas at a pres-sure of 1.60 3 106 Pa. Find (a) the temperature of the gas and (b) the average kinetic energy of the gas mol-ecules in the vessel. (c) What additional information would you need if you were asked to find the average speed of the gas molecules? 13. In a period of 1.00 s, 5.00 3 1023 nitrogen molecules strike a wall with an area of 8.00 cm2. Assume the mol-ecules move with a speed of 300 m/s and strike the wall head-on in elastic collisions. What is the pressure exerted on the wall? Note: The mass of one N2 molecule is 4.65 3 10226 kg.
Section 21.2 Molar Specific Heat of an Ideal Gas Note: You may use data in Table 21.2 about particular gases. Here we define a “monatomic ideal gas” to have molar specific heats CV 5 3 2R and CP 5 5 2R, and a “diatomic ideal gas” to have CV 5 5 2R and CP 5 7 2R.
W M S W Q/C W M 14. In a constant-volume process, 209 J of energy is trans-ferred by heat to 1.00 mol of an ideal monatomic gas initially at 300 K. Find (a) the work done on the gas, (b) the increase in internal energy of the gas, and (c) its final temperature. 15. A sample of a diatomic ideal gas has pressure P and volume V. When the gas is warmed, its pressure triples and its volume doubles. This warming process includes two steps, the first at constant pressure and the second at constant volume. Determine the amount of energy transferred to the gas by heat. 16. Review. A house has well-insulated walls. It contains a volume of 100 m3 of air at 300 K. (a) Calculate the energy required to increase the temperature of this diatomic ideal gas by 1.008C. (b) What If? If all this energy could be used to lift an object of mass m through a height of 2.00 m, what is the value of m? 17. A 1.00-mol sample of hydrogen gas is heated at con-stant pressure from 300 K to 420 K. Calculate (a) the energy transferred to the gas by heat, (b) the increase in its internal energy, and (c) the work done on the gas. 18. A vertical cylinder with a heavy piston contains air at 300 K. The initial pressure is 2.00 3 105 Pa, and the initial volume is 0.350 m3. Take the molar mass of air as 28.9 g/mol and assume CV 5 5 2R. (a) Find the specific heat of air at constant volume in units of J/kg ? 8C. (b) Calculate the mass of the air in the cyl-inder. (c) Suppose the piston is held fixed. Find the energy input required to raise the temperature of the air to 700 K. (d) What If? Assume again the conditions of the initial state and assume the heavy piston is free to move. Find the energy input required to raise the temperature to 700 K. 19. Calculate the change in internal energy of 3.00 mol of helium gas when its temperature is increased by 2.00 K. 20. A 1.00-L insulated bottle is full of tea at 90.08C. You pour out one cup of tea and immediately screw the stopper back on the bottle. Make an order-of-magnitude esti-mate of the change in temperature of the tea remaining in the bottle that results from the admission of air at room temperature. State the quantities you take as data and the values you measure or estimate for them. 21. Review. This problem is a continuation of Problem 39 in Chapter 19. A hot-air balloon consists of an enve-lope of constant volume 400 m3. Not including the air inside, the balloon and cargo have mass 200 kg. The air outside and originally inside is a diatomic ideal gas at 10.0°C and 101 kPa, with density 1.25 kg/m3. A propane burner at the center of the spherical enve-lope injects energy into the air inside. The air inside stays at constant pressure. Hot air, at just the tempera-ture required to make the balloon lift off, starts to fill the envelope at its closed top, rapidly enough so that negligible energy flows by heat to the cool air below it or out through the wall of the balloon. Air at 10°C leaves through an opening at the bottom of the enve-lope until the whole balloon is filled with hot air at uniform temperature. Then the burner is shut off and W S M www.aswarphysics.weebly.com Problems 647 How much work is required to produce the same com-pression in an adiabatic process? (c) What is the final pressure in part (a)? (d) What is the final pressure in part (b)? 29. Air in a thundercloud expands as it rises. If its initial temperature is 300 K and no energy is lost by ther-mal conduction on expansion, what is its temperature when the initial volume has doubled? 30. Why is the following situation impossible? A new die-sel engine that increases fuel economy over previ-ous models is designed. Automobiles fitted with this design become incredible best sellers. Two design fea-tures are responsible for the increased fuel economy: (1) the engine is made entirely of aluminum to reduce the weight of the automobile, and (2) the exhaust of the engine is used to prewarm the air to 508C before it enters the cylinder to increase the final temperature of the compressed gas. The engine has a compression ratio—that is, the ratio of the initial volume of the air to its final volume after compression—of 14.5. The compression process is adiabatic, and the air behaves as a diatomic ideal gas with g 5 1.40. 31. During the power stroke in a four-stroke automo-bile engine, the piston is forced down as the mixture of combustion products and air undergoes an adia-batic expansion. Assume (1) the engine is running at 2 500 cycles/min; (2) the gauge pressure immediately before the expansion is 20.0 atm; (3) the volumes of the mixture immediately before and after the expansion are 50.0 cm3 and 400 cm3, respectively (Fig. P21.31); (4) the time interval for the expansion is one-fourth that of the total cycle; and (5) the mixture behaves like an ideal gas with specific heat ratio 1.40. Find the aver-age power generated during the power stroke.
Before After 50.0 cm3 400.0 cm3 Figure P21.31 32. Air (a diatomic ideal gas) at 27.08C and atmospheric pressure is drawn into a bicycle pump (see the chapter-opening photo on page 626) that has a cylinder with an inner diameter of 2.50 cm and length 50.0 cm. The downstroke adiabatically compresses the air, which reaches a gauge pressure of 8.00 3 105 Pa before entering the tire. We wish to investigate the tempera-ture increase of the pump. (a) What is the initial vol-ume of the air in the pump? (b) What is the number of moles of air in the pump? (c) What is the absolute M GP the balloon rises from the ground. (a) Evaluate the quantity of energy the burner must transfer to the air in the balloon. (b) The “heat value” of propane—the internal energy released by burning each kilogram—is 50.3 MJ/kg. What mass of propane must be burned?
Section 21.3 The Equipartition of Energy 22. A certain molecule has f degrees of freedom. Show that an ideal gas consisting of such molecules has the following properties: (a) its total internal energy is fnRT/2, (b) its molar specific heat at constant volume is fR/2, (c) its molar specific heat at constant pres-sure is ( f 1 2)R/2, and (d) its specific heat ratio is g 5 CP/CV 5 ( f 1 2)/f. 23. In a crude model (Fig. P21.23) of a rotating diatomic chlorine molecule (Cl2), the two Cl atoms are 2.00 3 10210 m apart and rotate about their center of mass with angular speed v 5 2.00 3 1012 rad/s. What is the rotational kinetic energy of one molecule of Cl2, which has a molar mass of 70.0 g/mol?
Cl Cl Figure P21.23 24. Why is the following situation impossible? A team of researchers discovers a new gas, which has a value of g 5 CP/CV of 1.75. 25. The relationship between the heat capacity of a sam-ple and the specific heat of the sample material is dis-cussed in Section 20.2. Consider a sample containing 2.00 mol of an ideal diatomic gas. Assuming the mol-ecules rotate but do not vibrate, find (a) the total heat capacity of the sample at constant volume and (b) the total heat capacity at constant pressure. (c) What If? Repeat parts (a) and (b), assuming the molecules both rotate and vibrate.
Section 21.4 Adiabatic Processes for an Ideal Gas 26. A 2.00-mol sample of a diatomic ideal gas expands slowly and adiabatically from a pressure of 5.00 atm and a volume of 12.0 L to a final volume of 30.0 L. (a) What is the final pressure of the gas? (b) What are the initial and final temperatures? Find (c) Q, (d) DE int, and (e) W for the gas during this process. 27. During the compression stroke of a certain gasoline engine, the pressure increases from 1.00 atm to 20.0 atm. If the process is adiabatic and the air–fuel mixture behaves as a diatomic ideal gas, (a) by what factor does the volume change and (b) by what factor does the temperature change? Assuming the compression starts with 0.016 0 mol of gas at 27.08C, find the values of (c) Q, (d) DE int, and (e) W that characterize the process. 28. How much work is required to compress 5.00 mol of air at 20.08C and 1.00 atm to one-tenth of the origi-nal volume (a) by an isothermal process? (b) What If? S M M W W www.aswarphysics.weebly.com 648 Chapter 21 The Kinetic Theory of Gases speeds for the two isotopes of chlorine, 35Cl and 37Cl, as they diffuse through the air. (b) Which isotope moves faster? 39. Review. At what temperature would the average speed of helium atoms equal (a) the escape speed from the Earth, 1.12 3 104 m/s, and (b) the escape speed from the Moon, 2.37 3 103 m/s? Note: The mass of a helium atom is 6.64 3 10227 kg. 40. Consider a container of nitrogen gas molecules at 900 K. Calculate (a) the most probable speed, (b) the average speed, and (c) the rms speed for the molecules. (d) State how your results compare with the values dis-played in Figure 21.11. 41. Assume the Earth’s atmosphere has a uniform tem-perature of 20.08C and uniform composition, with an effective molar mass of 28.9 g/mol. (a) Show that the number density of molecules depends on height y above sea level according to nV 1 y2 5 n 0e2m0g y/k BT where n 0 is the number density at sea level (where y 5 0). This result is called the law of atmospheres. (b) Commer-cial jetliners typically cruise at an altitude of 11.0 km. Find the ratio of the atmospheric density there to the density at sea level. 42. From the Maxwell–Boltzmann speed distribution, show that the most probable speed of a gas molecule is given by Equation 21.44. Note: The most probable speed corresponds to the point at which the slope of the speed distribution curve dNv/dv is zero. 43. The law of atmospheres states that the number density of molecules in the atmosphere depends on height y above sea level according to nV 1y2 5 n 0e2m0g y/k BT where n0 is the number density at sea level (where y 5 0). The average height of a molecule in the Earth’s atmosphere is given by yavg 5 3 0 ynV 1y2 dy 3 0 nV 1y2 dy 5 3 0 ye2m0gy/kBT dy 3 0 e2m0gy/kBT dy (a) Prove that this average height is equal to k BT/m0g. (b) Evaluate the average height, assuming the temper-ature is 10.08C and the molecular mass is 28.9 u, both uniform throughout the atmosphere.
Additional Problems 44. Eight molecules have speeds of 3.00 km/s, 4.00 km/s, 5.80 km/s, 2.50 km/s, 3.60 km/s, 1.90 km/s, 3.80 km/s, and 6.60 km/s. Find (a) the average speed of the mol-ecules and (b) the rms speed of the molecules. 45. A small oxygen tank at a gauge pressure of 125 atm has a volume of 6.88 L at 21.08C. (a) If an athlete breathes oxygen from this tank at the rate of 8.50 L/min when measured at atmospheric pressure and the tempera-ture remains at 21.08C, how long will the tank last before it is empty? (b) At a particular moment during Q/C S pressure of the compressed air? (d) What is the volume of the compressed air? (e) What is the temperature of the compressed air? (f) What is the increase in inter-nal energy of the gas during the compression? What If? The pump is made of steel that is 2.00 mm thick. Assume 4.00 cm of the cylinder’s length is allowed to come to thermal equilibrium with the air. (g) What is the volume of steel in this 4.00-cm length? (h) What is the mass of steel in this 4.00-cm length? (i) Assume the pump is compressed once. After the adiabatic expan-sion, conduction results in the energy increase in part (f) being shared between the gas and the 4.00-cm length of steel. What will be the increase in tempera-ture of the steel after one compression? 33. A 4.00-L sample of a diatomic ideal gas with spe-cific heat ratio 1.40, confined to a cylinder, is carried through a closed cycle. The gas is initially at 1.00 atm and 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. (a) Draw a PV diagram of this cycle. (b) Determine the volume of the gas at the end of the adiabatic expansion. (c) Find the temperature of the gas at the start of the adiabatic expansion. (d) Find the temperature at the end of the cycle. (e) What was the net work done on the gas for this cycle? 34. An ideal gas with specific heat ratio g confined to a cyl-inder is put through a closed cycle. Initially, the gas is at Pi, Vi, and Ti. First, its pressure is tripled under con-stant volume. It then expands adiabatically to its origi-nal pressure and finally is compressed isobarically to its original volume. (a) Draw a PV diagram of this cycle. (b) Determine the volume at the end of the adiabatic expansion. Find (c) the temperature of the gas at the start of the adiabatic expansion and (d) the tempera-ture at the end of the cycle. (e) What was the net work done on the gas for this cycle?
Section 21.5 Distribution of Molecular Speeds 35. Helium gas is in thermal equilibrium with liquid helium at 4.20 K. Even though it is on the point of con-densation, model the gas as ideal and determine the most probable speed of a helium atom (mass 5 6.64 3 10–27 kg) in it. 36. Fifteen identical particles have various speeds: one has a speed of 2.00 m/s, two have speeds of 3.00 m/s, three have speeds of 5.00 m/s, four have speeds of 7.00 m/s, three have speeds of 9.00 m/s, and two have speeds of 12.0 m/s. Find (a) the average speed, (b) the rms speed, and (c) the most probable speed of these particles. 37. One cubic meter of atomic hydrogen at 08C at atmo-spheric pressure contains approximately 2.70 3 1025 atoms. The first excited state of the hydrogen atom has an energy of 10.2 eV above that of the lowest state, called the ground state. Use the Boltzmann factor to find the number of atoms in the first excited state (a) at 08C and at (b) (1.00 3 104)8C. 38. Two gases in a mixture diffuse through a filter at rates proportional to their rms speeds. (a) Find the ratio of S M W www.aswarphysics.weebly.com Problems 649 y, and z directions plus elastic potential energy associ-ated with the Hooke’s law forces exerted by neighbor-ing atoms on it in the x, y, and z directions. According to the theorem of equipartition of energy, assume the average energy of each atom is 1 2kBT for each degree of freedom. (a) Prove that the molar specific heat of the solid is 3R. The Dulong–Petit law states that this result generally describes pure solids at sufficiently high tem-peratures. (You may ignore the difference between the specific heat at constant pressure and the specific heat at constant volume.) (b) Evaluate the specific heat c of iron. Explain how it compares with the value listed in Table 20.1. (c) Repeat the evaluation and comparison for gold. 51. A certain ideal gas has a molar specific heat of CV 5 7 2R. A 2.00-mol sample of the gas always starts at pressure 1.00 3 105 Pa and temperature 300 K. For each of the following processes, determine (a) the final pressure, (b) the final volume, (c) the final temperature, (d) the change in internal energy of the gas, (e) the energy added to the gas by heat, and (f) the work done on the gas. (i) The gas is heated at constant pressure to 400 K. (ii) The gas is heated at constant volume to 400 K. (iii) The gas is compressed at constant temperature to 1.20 3 105 Pa. (iv) The gas is compressed adiabatically to 1.20 3 105 Pa. 52. The compressibility k of a substance is defined as the fractional change in volume of that substance for a given change in pressure: k 5 2 1 V dV dP (a) Explain why the negative sign in this expression ensures k is always positive. (b) Show that if an ideal gas is compressed isothermally, its compressibility is given by k1 5 1/P. (c) What If? Show that if an ideal gas is compressed adiabatically, its compressibility is given by k2 5 1/(gP). Determine values for (d) k1 and (e) k2 for a monatomic ideal gas at a pressure of 2.00 atm. 53. Review. Oxygen at pressures much greater than 1 atm is toxic to lung cells. Assume a deep-sea diver breathes a mixture of oxygen (O2) and helium (He). By weight, what ratio of helium to oxygen must be used if the diver is at an ocean depth of 50.0 m? 54. Examine the data for polyatomic gases in Table 21.2 and give a reason why sulfur dioxide has a higher spe-cific heat at constant volume than the other polyatomic gases at 300 K. 55. Model air as a diatomic ideal gas with M 5 28.9 g/mol. A cylinder with a piston contains 1.20 kg of air at 25.08C and 2.00 3 105 Pa. Energy is transferred by heat into the system as it is permitted to expand, with the pressure rising to 4.00 3 105 Pa. Throughout the expansion, the relationship between pressure and vol-ume is given by P 5 CV 1/2 where C is a constant. Find (a) the initial volume, (b) the final volume, (c) the final temperature, (d) the work done on the air, and (e) the energy transferred by heat.
BIO Q/C this process, what is the ratio of the rms speed of the molecules remaining in the tank to the rms speed of those being released at atmospheric pressure? 46. The dimensions of a classroom are 4.20 m 3 3.00 m 3 2.50 m. (a) Find the number of molecules of air in the classroom at atmospheric pressure and 20.08C. (b) Find the mass of this air, assuming the air consists of diatomic molecules with molar mass 28.9 g/mol. (c) Find the average kinetic energy of the molecules. (d) Find the rms molecular speed. (e) What If? Assume the molar specific heat of the air is inde-pendent of temperature. Find the change in internal energy of the air in the room as the temperature is raised to 25.08C. (f) Explain how you could convince a fellow student that your answer to part (e) is correct, even though it sounds surprising. 47. The Earth’s atmosphere consists primarily of oxygen (21%) and nitrogen (78%). The rms speed of oxygen molecules (O2) in the atmosphere at a certain loca-tion is 535 m/s. (a) What is the temperature of the atmosphere at this location? (b) Would the rms speed of nitrogen molecules (N2) at this location be higher, equal to, or lower than 535 m/s? Explain. (c) Deter-mine the rms speed of N2 at his location. 48. The mean free path , of a molecule is the average dis-tance that a molecule travels before colliding with another molecule. It is given by , 5 1 !2pd 2NV where d is the diameter of the molecule and NV is the number of molecules per unit volume. The number of collisions that a molecule makes with other molecules per unit time, or collision frequency f, is given by f 5 vavg , (a) If the diameter of an oxygen molecule is 2.00 3 10210 m, find the mean free path of the molecules in a scuba tank that has a volume of 12.0 L and is filled with oxygen at a gauge pressure of 100 atm at a temperature of 25.08C. (b) What is the average time interval between molecular collisions for a molecule of this gas? 49. An air rifle shoots a lead pellet by allowing high- pressure air to expand, propelling the pellet down the rifle barrel. Because this process happens very quickly, no appreciable thermal conduction occurs and the expansion is essentially adiabatic. Suppose the rifle starts with 12.0 cm3 of compressed air, which behaves as an ideal gas with g 5 1.40. The expanding air pushes a 1.10-g pellet as a piston with cross-sectional area 0.030 0 cm2 along the 50.0-cm-long gun barrel. What initial pressure is required to eject the pellet with a muzzle speed of 120 m/s? Ignore the effects of the air in front of the bullet and friction with the inside walls of the barrel. 50. In a sample of a solid metal, each atom is free to vibrate about some equilibrium position. The atom’s energy consists of kinetic energy for motion in the x, Q/C Q/C Q/C AMT AMT Q/C www.aswarphysics.weebly.com 650 Chapter 21 The Kinetic Theory of Gases as that of a molecule in an ideal gas. Consider a spheri-cal particle of density 1.00 3 103 kg/m3 in water at 20.08C. (a) For a particle of diameter d, evaluate the rms speed. (b) The particle’s actual motion is a ran-dom walk, but imagine that it moves with constant velocity equal in magnitude to its rms speed. In what time interval would it move by a distance equal to its own diameter? (c) Evaluate the rms speed and the time interval for a particle of diameter 3.00 mm. (d) Evalu-ate the rms speed and the time interval for a sphere of mass 70.0 kg, modeling your own body. 62. A vessel contains 1.00 3 104 oxygen molecules at 500 K. (a) Make an accurate graph of the Maxwell speed distri-bution function versus speed with points at speed inter-vals of 100 m/s. (b) Determine the most probable speed from this graph. (c) Calculate the average and rms speeds for the molecules and label these points on your graph. (d) From the graph, estimate the fraction of mol-ecules with speeds in the range 300 m/s to 600 m/s. 63. A pitcher throws a 0.142-kg baseball at 47.2 m/s. As it travels 16.8 m to home plate, the ball slows down to 42.5 m/s because of air resistance. Find the change in temperature of the air through which it passes. To find the greatest possible temperature change, you may make the following assumptions. Air has a molar specific heat of CP 5 7 2R and an equivalent molar mass of 28.9 g/mol. The process is so rapid that the cover of the baseball acts as thermal insulation and the tem-perature of the ball itself does not change. A change in temperature happens initially only for the air in a cylinder 16.8 m in length and 3.70 cm in radius. This air is initially at 20.08C. 64. The latent heat of vaporization for water at room tem-perature is 2 430 J/g. Consider one particular molecule at the surface of a glass of liquid water, moving upward with sufficiently high speed that it will be the next molecule to join the vapor. (a) Find its translational kinetic energy. (b) Find its speed. Now consider a thin gas made only of molecules like that one. (c) What is its temperature? (d) Why are you not burned by water evaporating from a vessel at room temperature? 65. A sample of a monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A in Fig. P21.65). It is warmed at constant volume to 3.00 atm (point B). Then it is allowed to expand isothermally to 1.00 atm (point C) and at last compressed isobarically to its origi-nal state. (a) Find the number of moles in the sample. AMT Q/C Q/C 56. Review. As a sound wave passes through a gas, the compressions are either so rapid or so far apart that thermal conduction is prevented by a negligible time interval or by effective thickness of insulation. The compressions and rarefactions are adiabatic. (a) Show that the speed of sound in an ideal gas is v 5 Å gRT M where M is the molar mass. The speed of sound in a gas is given by Equation 17.8; use that equation and the definition of the bulk modulus from Section 12.4. (b) Compute the theoretical speed of sound in air at 20.08C and state how it compares with the value in Table 17.1. Take M 5 28.9 g/mol. (c) Show that the speed of sound in an ideal gas is v 5 Å gkBT m 0 where m0 is the mass of one molecule. (d) State how the result in part (c) compares with the most probable, average, and rms molecular speeds. 57. Twenty particles, each of mass m0 and confined to a volume V, have various speeds: two have speed v, three have speed 2v, five have speed 3v, four have speed 4v, three have speed 5v, two have speed 6v, and one has speed 7v. Find (a) the average speed, (b) the rms speed, (c) the most probable speed, (d) the average pressure the particles exert on the walls of the vessel, and (e) the average kinetic energy per particle. 58. In a cylinder, a sample of an ideal gas with number of moles n undergoes an adiabatic process. (a) Starting with the expression W 5 2e P dV and using the condi-tion PV g 5 constant, show that the work done on the gas is W 5 a 1 g 2 1b 1PfVf 2 PiVi 2 (b) Starting with the first law of thermodynamics, show that the work done on the gas is equal to nCV(Tf 2 Ti). (c) Are these two results consistent with each other? Explain. 59. As a 1.00-mol sample of a monatomic ideal gas expands adiabatically, the work done on it is 22.50 3 103 J. The initial temperature and pressure of the gas are 500 K and 3.60 atm. Calculate (a) the final temperature and (b) the final pressure. 60. A sample consists of an amount n in moles of a mona-tomic ideal gas. The gas expands adiabatically, with work W done on it. (Work W is a negative number.) The initial temperature and pressure of the gas are Ti and Pi. Calculate (a) the final temperature and (b) the final pressure. 61. When a small particle is suspended in a fluid, bom-bardment by molecules makes the particle jitter about at random. Robert Brown discovered this motion in 1827 while studying plant fertilization, and the motion has become known as Brownian motion. The particle’s average kinetic energy can be taken as 3 2kBT , the same Q/C S S Q/C S P (atm) 3 0 5 10 V (L) B A C 2 1 15 Figure P21.65 www.aswarphysics.weebly.com Problems 651 70. On the PV diagram for an ideal gas, one isothermal curve and one adiabatic curve pass through each point as shown in Figure P21.70. Prove that the slope of the adiabatic curve is steeper than the slope of the iso-therm at that point by the factor g.
V Adiabatic process P Isothermal process Figure P21.70 71. In Beijing, a restaurant keeps a pot of chicken broth simmering continuously. Every morning, it is topped up to contain 10.0 L of water along with a fresh chicken, vegetables, and spices. The molar mass of water is 18.0 g/mol. (a) Find the number of molecules of water in the pot. (b) During a certain month, 90.0% of the broth was served each day to people who then emigrated immediately. Of the water molecules in the pot on the first day of the month, when was the last one likely to have been ladled out of the pot? (c) The broth has been simmering for centuries, through wars, earthquakes, and stove repairs. Suppose the water that was in the pot long ago has thoroughly mixed into the Earth’s hydrosphere, of mass 1.32 3 1021 kg. How many of the water molecules originally in the pot are likely to be present in it again today? 72. Review. (a) If it has enough kinetic energy, a molecule at the surface of the Earth can “escape the Earth’s grav-itation” in the sense that it can continue to move away from the Earth forever as discussed in Section 13.6. Using the principle of conservation of energy, show that the minimum kinetic energy needed for “escape” is m0gRE, where m0 is the mass of the molecule, g is the free-fall acceleration at the surface, and RE is the radius of the Earth. (b) Calculate the temperature for which the minimum escape kinetic energy is ten times the average kinetic energy of an oxygen molecule. 73. Using multiple laser beams, physicists have been able to cool and trap sodium atoms in a small region. In one experiment, the temperature of the atoms was reduced to 0.240 mK. (a) Determine the rms speed of the sodium atoms at this temperature. The atoms can be trapped for about 1.00 s. The trap has a linear dimension of roughly 1.00 cm. (b) Over what approxi-mate time interval would an atom wander out of the trap region if there were no trapping action?
Challenge Problems 74. Equations 21.42 and 21.43 show that vrms . vavg for a collection of gas particles, which turns out to be true whenever the particles have a distribution of speeds. Let us explore this inequality for a two-particle gas. S Q/C S Find (b) the temperature at point B, (c) the temperature at point C, and (d) the volume at point C. (e) Now con-sider the processes A S B, B S C, and C S A. Describe how to carry out each process experimentally. (f) Find Q , W, and DE int for each of the processes. (g) For the whole cycle A S B S C S A, find Q , W, and DE int. 66. Consider the particles in a gas centrifuge, a device used to separate particles of different mass by whirling them in a circular path of radius r at angular speed v. The force acting on a gas molecule toward the center of the centrifuge is m0v2r. (a) Discuss how a gas centri-fuge can be used to separate particles of different mass. (b) Suppose the centrifuge contains a gas of particles of identical mass. Show that the density of the particles as a function of r is n1r2 5 n 0em0r 2v2/2kBT 67. For a Maxwellian gas, use a computer or programma-ble calculator to find the numerical value of the ratio Nv(v)/Nv(vmp) for the following values of v: (a) v 5 (vmp/50.0), (b) (vmp/10.0), (c) (vmp/2.00), (d) vmp, (e) 2.00vmp, (f) 10.0vmp, and (g) 50.0vmp. Give your results to three significant figures. 68. A triatomic molecule can have a linear configuration, as does CO2 (Fig. P21.68a), or it can be nonlinear, like H2O (Fig. P21.68b). Suppose the temperature of a gas of triatomic molecules is sufficiently low that vibrational motion is negligible. What is the molar specific heat at constant volume, expressed as a multiple of the uni-versal gas constant, (a) if the molecules are linear and (b) if the molecules are nonlinear? At high tempera-tures, a triatomic molecule has two modes of vibration, and each contributes 1 2R to the molar specific heat for its kinetic energy and another 1 2R for its potential energy. Identify the high-temperature molar specific heat at constant volume for a triatomic ideal gas of (c) linear molecules and (d) nonlinear molecules. (e) Explain how specific heat data can be used to determine whether a triatomic molecule is linear or nonlinear. Are the data in Table 21.2 sufficient to make this determination?
O H H O O C a b Figure P21.68 69. Using the Maxwell–Boltzmann speed distribution function, verify Equations 21.42 and 21.43 for (a) the rms speed and (b) the average speed of the molecules of a gas at a temperature T. The average value of v n is vn 5 1 N 3 0 vnNv dv Use the table of integrals B.6 in Appendix B.
Q/C S Q/C S www.aswarphysics.weebly.com 652 Chapter 21 The Kinetic Theory of Gases parallel to the axis of the cylinder until it comes to rest at an equilibrium position (Fig. P21.75b). Find the final temperatures in the two compartments.
T1i 550 K T2i 250 K T1f T2f a b Figure P21.75 Let the speed of one particle be v1 5 avavg and the other particle have speed v2 5 (2 2 a)vavg. (a) Show that the average of these two speeds is vavg. (b) Show that v 2 rms 5 v2 avg (2 2 2a 1 a2) (c) Argue that the equation in part (b) proves that, in general, vrms . vavg. (d) Under what special condition will vrms 5 vavg for the two-particle gas? 75. A cylinder is closed at both ends and has insulating walls. It is divided into two compartments by an insu-lating piston that is perpendicular to the axis of the cylinder as shown in Figure P21.75a. Each compart-ment contains 1.00 mol of oxygen that behaves as an ideal gas with g 5 1.40. Initially, the two compartments have equal volumes and their temperatures are 550 K and 250 K. The piston is then allowed to move slowly AMT www.aswarphysics.weebly.com A Stirling engine from the early nineteenth century. Air is heated in the lower cylinder using an external source. As this happens, the air expands and pushes against a piston, causing it to move. The air is then cooled, allowing the cycle to begin again. This is one example of a heat engine, which we study in this chapter. (© SSPL/The Image Works) 22.1 Heat Engines and the Second Law of Thermodynamics 22.2 Heat Pumps and Refrigerators 22.3 Reversible and Irreversible Processes 22.4 The Carnot Engine 22.5 Gasoline and Diesel Engines 22.6 Entropy 22.7 Changes in Entropy for Thermodynamic Systems 22.8 Entropy and the Second Law c h a p t e r 22 The first law of thermodynamics, which we studied in Chapter 20, is a statement of conservation of energy and is a special-case reduction of Equation 8.2. This law states that a change in internal energy in a system can occur as a result of energy transfer by heat, by work, or by both. Although the first law of thermodynamics is very important, it makes no distinction between processes that occur spontaneously and those that do not. Only certain types of energy transformation and energy transfer processes actually take place in nature, however. The second law of thermodynamics, the major topic in this chapter, establishes which processes do and do not occur. The following are examples Heat Engines, Entropy, and the Second Law of Thermodynamics 653 www.aswarphysics.weebly.com 654 Chapter 22 Heat Engines, Entropy, and the Second Law of Thermodynamics of processes that do not violate the first law of thermodynamics if they proceed in either direction, but are observed in reality to proceed in only one direction: • When two objects at different temperatures are placed in thermal contact with each other, the net transfer of energy by heat is always from the warmer object to the cooler object, never from the cooler to the warmer.
• A rubber ball dropped to the ground bounces several times and eventually comes to rest, but a ball lying on the ground never gathers internal energy from the ground and begins bouncing on its own.
• An oscillating pendulum eventually comes to rest because of collisions with air molecules and friction at the point of suspension. The mechanical energy of the system is converted to internal energy in the air, the pendulum, and the suspension; the reverse conversion of energy never occurs. All these processes are irreversible; that is, they are processes that occur naturally in one direction only. No irreversible process has ever been observed to run backward. If it were to do so, it would violate the second law of thermodynamics.1 22.1 Heat Engines and the Second Law of Thermodynamics A heat engine is a device that takes in energy by heat2 and, operating in a cyclic process, expels a fraction of that energy by means of work. For instance, in a typical process by which a power plant produces electricity, a fuel such as coal is burned and the high-temperature gases produced are used to convert liquid water to steam. This steam is directed at the blades of a turbine, setting it into rotation. The mechanical energy associated with this rotation is used to drive an electric genera-tor. Another device that can be modeled as a heat engine is the internal combustion engine in an automobile. This device uses energy from a burning fuel to perform work on pistons that results in the motion of the automobile. Let us consider the operation of a heat engine in more detail. A heat engine car-ries some working substance through a cyclic process during which (1) the working substance absorbs energy by heat from a high-temperature energy reservoir, (2) work is done by the engine, and (3) energy is expelled by heat to a lower-temperature reservoir. As an example, consider the operation of a steam engine (Fig. 22.1), which uses water as the working substance. The water in a boiler absorbs energy from burn-ing fuel and evaporates to steam, which then does work by expanding against a pis-ton. After the steam cools and condenses, the liquid water produced returns to the boiler and the cycle repeats. It is useful to represent a heat engine schematically as in Figure 22.2. The engine absorbs a quantity of energy |Q h| from the hot reservoir. For the mathematical discussion of heat engines, we use absolute values to make all energy transfers by heat positive, and the direction of transfer is indicated with an explicit positive or negative sign. The engine does work Weng (so that negative work W 5 2Weng is done on the engine) and then gives up a quantity of energy |Q c| to the cold reservoir. Lord Kelvin British physicist and mathematician (1824–1907) Born William Thomson in Belfast, Kel-vin was the first to propose the use of an absolute scale of temperature. The Kelvin temperature scale is named in his honor. Kelvin’s work in thermody-namics led to the idea that energy can-not pass spontaneously from a colder object to a hotter object.
© Mary Evans Picture Library/Alamy 1Although a process occurring in the time-reversed sense has never been observed, it is possible for it to occur. As we shall see later in this chapter, however, the probability of such a process occurring is infinitesimally small. From this viewpoint, processes occur with a vastly greater probability in one direction than in the opposite direction.
2We use heat as our model for energy transfer into a heat engine. Other methods of energy transfer are possible in the model of a heat engine, however. For example, the Earth’s atmosphere can be modeled as a heat engine in which the input energy transfer is by means of electromagnetic radiation from the Sun. The output of the atmospheric heat engine causes the wind structure in the atmosphere.
Figure 22.1 A steam-driven locomotive obtains its energy by burning wood or coal. The generated energy vaporizes water into steam, which powers the locomotive. Modern locomotives use diesel fuel instead of wood or coal. Whether old-fashioned or modern, such locomotives can be modeled as heat engines, which extract energy from a burning fuel and convert a fraction of it to mechanical energy.
© Andy Moore/Photolibrary/Jupiterimages www.aswarphysics.weebly.com 22.1 Heat Engines and the Second Law of Thermodynamics 655 Because the working substance goes through a cycle, its initial and final internal energies are equal: DE int 5 0. Hence, from the first law of thermodynamics, DE int 5 Q 1 W 5 Q 2 Weng 5 0, and the net work Weng done by a heat engine is equal to the net energy Q net transferred to it. As you can see from Figure 22.2, Q net 5 |Q h| 2 |Q c|; therefore, Weng 5 |Q h| 2 |Q c| (22.1) The thermal efficiency e of a heat engine is defined as the ratio of the net work done by the engine during one cycle to the energy input at the higher temperature during the cycle: e ; Weng 0 Q h 0 5 0 Q h 0 2 0 Q c 0 0 Q h 0 5 1 2 0 Q c 0 0 Q h 0 (22.2) You can think of the efficiency as the ratio of what you gain (work) to what you give (energy transfer at the higher temperature). In practice, all heat engines expel only a fraction of the input energy Q h by mechanical work; consequently, their efficiency is always less than 100%. For example, a good automobile engine has an efficiency of about 20%, and diesel engines have efficiencies ranging from 35% to 40%. Equation 22.2 shows that a heat engine has 100% efficiency (e 5 1) only if |Q c| 5 0, that is, if no energy is expelled to the cold reservoir. In other words, a heat engine with perfect efficiency would have to expel all the input energy by work. Because efficiencies of real engines are well below 100%, the Kelvin–Planck form of the second law of thermodynamics states the following: It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the input of energy by heat from a reservoir and the per-formance of an equal amount of work.
This statement of the second law means that during the operation of a heat engine, Weng can never be equal to |Q h| or, alternatively, that some energy |Q c| must be rejected to the environment. Figure 22.3 is a schematic diagram of the impossible “perfect” heat engine.
Q uick Quiz 22.1 The energy input to an engine is 4.00 times greater than the work it performs. (i) What is its thermal efficiency? (a) 4.00 (b) 1.00 (c) 0.250 (d) impossible to determine (ii) What fraction of the energy input is expelled to the cold reservoir? (a) 0.250 (b) 0.750 (c) 1.00 (d) impossible to determine W W Thermal efficiency of a heat engine Pitfall Prevention 22.1 The First and Second Laws Notice the distinction between the first and second laws of thermodynam-ics. If a gas undergoes a one-time isothermal process, then DE int 5 Q 1 W 5 0 and W 5 2Q. Therefore, the first law allows all energy input by heat to be expelled by work. In a heat engine, however, in which a substance undergoes a cyclic pro-cess, only a portion of the energy input by heat can be expelled by work according to the second law.
Qh Q c Hot reservoir at Th Cold reservoir at Tc Heat engine t Weng Energy Qh enters the engine. Energy Q c leaves the engine. The engine does work Weng.
Figure 22.2 Schematic repre-sentation of a heat engine.
Q h Hot reservoir at Th Cold reservoir at Tc Heat engine t Weng An impossible heat engine Figure 22.3 Schematic diagram of a heat engine that takes in energy from a hot reservoir and does an equivalent amount of work. It is impossible to construct such a per-fect engine.
www.aswarphysics.weebly.com 656 Chapter 22 Heat Engines, Entropy, and the Second Law of Thermodynamics Example 22.1 The Efficiency of an Engine An engine transfers 2.00 3 103 J of energy from a hot reservoir during a cycle and transfers 1.50 3 103 J as exhaust to a cold reservoir.
(A) Find the efficiency of the engine.
Conceptualize Review Figure 22.2; think about energy going into the engine from the hot reservoir and splitting, with part coming out by work and part by heat into the cold reservoir.
Categorize This example involves evaluation of quantities from the equations introduced in this section, so we catego-rize it as a substitution problem.
S o l u t i o n Find the efficiency of the engine from Equation 22.2: e 5 1 2 0 Q c 0 0 Q h 0 5 1 2 1.50 3 103 J 2.00 3 103 J 5 0.250, or 25.0% Find the work done by the engine by taking the differ-ence between the input and output energies: Weng 5 |Q h| 2 |Q c| 5 2.00 3 103 J 2 1.50 3 103 J 5 5.0 3 102 J (B) How much work does this engine do in one cycle?
S o l u t i o n Suppose you were asked for the power output of this engine. Do you have sufficient information to answer this question?
Answer No, you do not have enough information. The power of an engine is the rate at which work is done by the engine. You know how much work is done per cycle, but you have no information about the time interval associated with one cycle. If you were told that the engine operates at 2 000 rpm (revolutions per minute), however, you could relate this rate to the period of rotation T of the mechanism of the engine. Assuming there is one thermodynamic cycle per revolution, the power is P 5 Weng T 5 5.0 3 102 J 1 1 2 000 min2 a1 min 60 s b 5 1.7 3 104 W What If?
22.2 Heat Pumps and Refrigerators In a heat engine, the direction of energy transfer is from the hot reservoir to the cold reservoir, which is the natural direction. The role of the heat engine is to pro-cess the energy from the hot reservoir so as to do useful work. What if we wanted to transfer energy from the cold reservoir to the hot reservoir? Because that is not the natural direction of energy transfer, we must put some energy into a device to be successful. Devices that perform this task are called heat pumps and refrigerators. For example, homes in summer are cooled using heat pumps called air conditioners. The air conditioner transfers energy from the cool room in the home to the warm air outside. In a refrigerator or a heat pump, the engine takes in energy |Q c| from a cold reservoir and expels energy |Q h| to a hot reservoir (Fig. 22.4), which can be accom-plished only if work is done on the engine. From the first law, we know that the energy given up to the hot reservoir must equal the sum of the work done and the energy taken in from the cold reservoir. Therefore, the refrigerator or heat pump transfers energy from a colder body (for example, the contents of a kitchen refrig-erator or the winter air outside a building) to a hotter body (the air in the kitchen or a room in the building). In practice, it is desirable to carry out this process with www.aswarphysics.weebly.com 22.2 Heat Pumps and Refrigerators 657 a minimum of work. If the process could be accomplished without doing any work, the refrigerator or heat pump would be “perfect” (Fig. 22.5). Again, the existence of such a device would be in violation of the second law of thermodynamics, which in the form of the Clausius statement3 states: It is impossible to construct a cyclical machine whose sole effect is to transfer energy continuously by heat from one object to another object at a higher temperature without the input of energy by work.
In simpler terms, energy does not transfer spontaneously by heat from a cold object to a hot object. Work input is required to run a refrigerator. The Clausius and Kelvin–Planck statements of the second law of thermodynam-ics appear at first sight to be unrelated, but in fact they are equivalent in all respects. Although we do not prove so here, if either statement is false, so is the other.4 In practice, a heat pump includes a circulating fluid that passes through two sets of metal coils that can exchange energy with the surroundings. The fluid is cold and at low pressure when it is in the coils located in a cool environment, where it absorbs energy by heat. The resulting warm fluid is then compressed and enters the other coils as a hot, high-pressure fluid. There it releases its stored energy to the warm surroundings. In an air conditioner, energy is absorbed into the fluid in coils located in a building’s interior; after the fluid is compressed, energy leaves the fluid through coils located outdoors. In a refrigerator, the external coils are behind the unit (Fig. 22.6) or underneath the unit. The internal coils are in the walls of the refrigerator and absorb energy from the food. The effectiveness of a heat pump is described in terms of a number called the coefficient of performance (COP). The COP is similar to the thermal efficiency for a heat engine in that it is a ratio of what you gain (energy transferred to or from a reservoir) to what you give (work input). For a heat pump operating in the cooling mode, “what you gain” is energy removed from the cold reservoir. The most effective refrigerator or air conditioner is one that removes the greatest amount of energy 3First expressed by Rudolf Clausius (1822–1888).
4See an advanced textbook on thermodynamics for this proof.
Q h Q c Hot reservoir at Th Cold reservoir at Tc Heat pump W Energy Q h is expelled to the hot reservoir.
Energy Q c is drawn from the cold reservoir.
Work W is done on the heat pump.
Figure 22.4 Schematic repre-sentation of a heat pump.
Q h Q c Q c Hot reservoir at Th Cold reservoir at Tc Heat pump An impossible heat pump Figure 22.5 Schematic diagram of an impossible heat pump or refrigerator, that is, one that takes in energy from a cold reservoir and expels an equivalent amount of energy to a hot reservoir with-out the input of energy by work.
The coils on the back of a refrigerator transfer energy by heat to the air.
Figure 22.6 The back of a household refrigerator. The air surrounding the coils is the hot reservoir.
© Cengage Learning/Charles D. Winters www.aswarphysics.weebly.com |
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Integrating the absolute of the cosine
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For some reason, I do not understand when computing the the integral of |cos(x)| from -pi to -pi/2 gives 1. When i compute it i get -1. There must be something I haven't understood.
integration
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edited Dec 27, 2014 at 13:18
franckfranck
asked Dec 27, 2014 at 13:11
franckfranck
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$\begingroup$ If you show us your working, we may be able to see where you have gone wrong. $\endgroup$
Empy2
– Empy2
2014-12-27 13:13:22 +00:00
Commented Dec 27, 2014 at 13:13
$\begingroup$ What I did so far was I know that |cos(x)| when x is [-pi,-pi/2], therefore i have -cos(x). So in my opinion i just integrate over -cos(x) from -pi to -pi/2. $\endgroup$
franck
– franck
2014-12-27 13:17:10 +00:00
Commented Dec 27, 2014 at 13:17
$\begingroup$ Integrating $|\cos(x)|$ from $-\pi$ to $\frac{\pi}{2}$ doesn't give $1$. But it certain doesn't give $-1$ (integrating a positive function gives a positive value) either. So show us what you have done. $\endgroup$
Henrik supports the community
– Henrik supports the community
2014-12-27 13:17:26 +00:00
Commented Dec 27, 2014 at 13:17
$\begingroup$ sorry i did a mistake, it is actually from -pi to -pi/2 $\endgroup$
franck
– franck
2014-12-27 13:19:51 +00:00
Commented Dec 27, 2014 at 13:19
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2 Answers 2
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This sort of integral can be computed by considering the domains on which the integrand takes the negative and positive of itself. So for instance:
$$|\cos(x)|=\begin{cases}\cos(x) & -\frac{\pi}{2}\leq x \leq \frac{\pi}{2} \ -\cos(x) & x \not\in[-\frac{\pi}{2},\frac{\pi}{2}]\end{cases}$$
Thus we can write our integral:
$$\int_{-\pi}^{\frac{\pi}{2}}|\cos(x)|\:\mathrm{d}x=\int_{-\pi}^{-\frac{\pi}{2}}-\cos(x)\:\mathrm{d}x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(x)\:\mathrm{d}x = 1+2 = 3$$
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answered Dec 27, 2014 at 13:16
Thomas RussellThomas Russell
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$\begingroup$ sorry it is from -pi to -pi/2 which gives me 1 $\endgroup$
franck
– franck
2014-12-27 13:23:56 +00:00
Commented Dec 27, 2014 at 13:23
$\begingroup$ @franck That sounds like you have forgotten to include the minus sign in your definition of $|\cos(x)|$ in the domain $-\pi \leq x \leq -\frac{\pi}{2}$ then! $\endgroup$
Thomas Russell
– Thomas Russell
2014-12-27 13:25:04 +00:00
Commented Dec 27, 2014 at 13:25
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For real $y,|y|=+y$ if $y\ge0$
and $=-y$ for $y<0$
Now $\cos x\ge0$ for $-\dfrac\pi2\le x\le\dfrac\pi2$
and $<0$ for $-\pi\le x\le-\dfrac\pi2$
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answered Dec 27, 2014 at 13:13
lab bhattacharjeelab bhattacharjee
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CoffeeCat
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这节介绍常用分布。分常用离散分布和常用连续分布两类。
常用离散分布
二项分布（Binomial Distribution）
记 X X 为 n n 重伯努利试验中成功的事件（记为 A A ）的次数，则 X=0,1,2,⋯,n.X=0,1,2,\cdots,n.X X 服从二项分布。记 p p 为事件 A A 发生的概率， X X 的分布列为：
P(X=k)=(n k)p k(1−p)n−k,k=0,1,⋯,n.P(X=k)=\left(\begin{array}{l}{n} \ {k}\end{array}\right) p^{k}(1-p)^{n-k}, \quad k=0,1, \cdots, n.
记
X∼b(n,p)X\sim b(n,p)
符号“~”读作“服从于”，该记号表示随机变量 X X 服从参数为 n,p n,p 的二项分布。
容易想到，二项概率恰好是二项式 (p+(1−p))n(p+(1-p))^n 的展开式的第 k+1 k+1 项，这也是“二项分布”的名称的由来。
二项分布线条图
应用举例：
设射手命中率为 0.8 ，则射击 n 次，命中的次数 X\sim b(n,0.8) .
已知人群中色盲率为 p ，在人群中随机调查50个人，则其中色盲患者 X\sim b(50,p) .
某药品的有效率为 0.9 ，今有 10 人服用，则服药有效的人数 X\sim b(10,0.9) .
……
数学期望： np
方差：np(1-p)
两点分布（Bernoulli Distribution）
是一种当 n=1 时的特殊的二项分布，又名0-1分布，伯努利分布，用来描述一次伯努利试验中成功的次数 X 。 X=0,1.X 服从两点分布，分布列为：
P(X=x)=p^{x}(1-p)^{1-x},\quad x=0,1.
或表示为：
\begin{array}{c|c} \text{X} & 0& 1 \ \hline P &1-p&p \end{array}
其中 p=P(X=1) 为事件成功的概率。
应用举例：
小明投篮命中率为 0.8 ，投篮一次，其命中的次数 X\sim b(1,0.8)
彩票中奖率为 0.0001 ，小明购买一张彩票，其中奖的次数 X\sim b(1,0.0001)
不会做的单项选择题做对的概率为 0.25 ，随机选择一个选项，做对的次数 X\sim b(1,0.25)
……
两点分布是特殊的二项分布，在二项分布数学期望和方差的公式中取 n=1 得到两点分布：
数学期望： p
方差：p(1-p)
二项分布与两点分布的关系：若有一列独立同分布于 b(1,p) 的随机变量序列 {X_i}_{i=1}^{n} ，则其和：
X_1+X_2+\cdots+X_n=\sum_{i=1}^{n}X_i\sim b(n,p)
这个结论表明两点分布具有可加性，且对于服从 b(n,p) 的随机变量 X ，可看做由 n 个独立同分布于 b(1,p) 的随机变量 X_i 的和。
上述“独立同分布”、“可加性”的概念，见：coffee：多维随机变量函数的分布
泊松分布（Poisson Distribution）
分布列：
P(X=k)=\frac{\lambda^{k}}{k !} \mathrm{e}^{-\lambda},\quad k=0,1,2, \cdots
记 X\sim P(\lambda) 。常与单位时间、单位面积、单位体积上的计数过程相联系。
泊松分布线条图
应用举例：
某时间段内，来到某商场的顾客数
单位时间内，某网站的点击量
一平方米内玻璃上的气泡数
……
数学期望： \lambda
方差： \lambda
这里数学期望为 \lambda 是指 X 的均值为 \lambda 。譬如对于应用举例1.，某段时间内，来到某商场的顾客数平均而言是 \lambda 。其他的应用类似。
超几何分布（Hypergeometric Distibution）
设有 N 件产品，其中有 M 件不合格品。若从中不放回地随机抽取 n 件，则其中含有的不合格品的件数 X 服从超几何分布，分布列为：
P(X=k)=\frac{\binom{M}{k}\binom{N-M}{n-k} }{\binom{N}{n}},\quad k=0,1, \cdots, r
记为 X\sim h(n,N,M) .其中 r=\min {M, n} ，且 M \leqslant N, n \leqslant N.n, N, M 均为正整数。
应用举例：从有10件不合格品的100件产品中随机抽取5件，则抽取的产品中不合格品数 X\sim h(5,100,10) 。
数学期望： n\frac{M}{N}
方差： \frac{nM(N-M)(N-n)}{N^2(N-1)}
几何分布（Geometric Distribution）
在伯努利试验序列中，记每次试验中事件 A 发生的概率为 p ,如果 X 为事件 A 首次出现时的试验次数，则 X=1,2,\cdots 。 X 服从几何分布，分布列为：
P(X=k)=(1-p)^{k-1} p, \quad k=1,2, \cdots
记作 X\sim Ge(p) 。
应用举例：
某产品的不合格率为 0.05 ，首次查到不合格品的检查次数 X\sim Ge(0.05)
某射手的命中率为 0.8 ，首次命中的射击次数 X\sim Ge(0.8)
掷一颗骰子，首次出现六点的投掷次数 X\sim Ge\left(\frac{1}{6}\right)
……
数学期望： \frac{1}{p}
方差： \frac{1-p}{p^2}
几何分布的无记忆性：
设 X\sim Ge(p) ，对任意正整数 m,n ，有：
P(X>m+n \,|\, X>m)=P(X>n)
该性质表明，在前 m 次试验中 A 没有出现的条件下，则在接下去的 n 次试验中 A 仍未出现的概率只与 n 有关，而与以前的 m 次试验无关，似乎忘记了前 m 次试验结果，这就是无记忆性。
负二项分布（Negative Binomial Distribution）
在伯努利试验序列中，记每次试验中事件 A 发生的概率为 p ，如果 X 为事件 A 第 r 次出现时的试验次数，则 X 的可能取值为 r,r+1,\cdots,r+m,\cdots ，称X服从负二项分布或巴斯卡分布，其分布列为：
P(X=k)=\left(\begin{array}{l}{k-1} \ {r-1}\end{array}\right) p^r\,(1-p)^{k-r}, \quad k=r, r+1, \cdots
记作： X\sim Nb(r,p) ，当 r=1 时即为几何分布，即几何分布是特殊的负二项分布。从二项分布和负二项分布的定义中看出，二项分布是伯努利试验次数 (n) 固定，事件 A 成功的次数 X 在 0\sim n 中取值；而负二项分布是事件 A 成功的次数 (r) 固定，伯努利实验次数 X 在 r, r+1, \cdots 中取值，可见负二项分布的“负”字的由来。
应用举例：
某产品的不合格率为 0.05 ，产品总数大于5，查到第5件不合格品时，检查次数 X\sim Nb(5,0.05)
某射手的命中率为 0.8 ，第十次命中的射击次数 X\sim Nb(10,0.8)
掷一颗骰子，第三次出现六点时，投掷次数 X\sim Nb\left(3,\frac{1}{6}\right)
……
数学期望： \frac{r}{p}
方差： \frac{r(1-p)}{p^2}
从负二项分布和几何分布的数学期望和方差的关系可知，类比二项分布与两点分布的关系，可以得到下面的结论：
若有一列独立同分布于 Ge(p) 的随机变量序列 {X_i}_{i=1}^{n} ，则其和：
X_1+X_2+\cdots+X_r=\sum_{i=1}^{r}X_i\sim Nb(r,p)
这并不是说明几何分布具有可加性，因为可加性要求服从该类分布的随机变量的和仍服从该类分布，但是服从几何分布的随机变量的和服从负二项分布，这个概念要特别注意。上述结论只能说明对于服从 Nb(r,p) 的随机变量 X ，可看做由 r 个独立同分布于 Ge(p) 的随机变量 X_i 的和。
常用连续分布
正态分布
若随机变量 X 的密度函数为：
p(x)=\frac{1}{\sqrt{2 \pi} \sigma} \mathrm{e}^{-\frac{(x-\mu)^{2}}{2 \sigma^{2}}},\quad -\infty<x<\infty
则称 X 服从正态分布，称 X 为正态变量。记 X\sim N(\mu,\sigma^2) 。其中 \mu 为位置参数，用于控制曲线在 x 轴上的位置； \sigma 为尺度参数，用于控制曲线的形状。
分布函数：
F(x)=\int_{-\infty}^{x}p(t){\rm d}t=\int_{-\infty}^{x}\frac{1}{\sqrt{2 \pi} \sigma} \mathrm{e}^{-\frac{(t-\mu)^{2}}{2 \sigma^{2}}}{\rm d}t
密度函数及分布函数
不同参数的正态分布图像
数学期望： \mu
方差： \sigma^2
称 \mu=0,\sigma^2=1 时的正态分布为标准正态分布，其密度函数和分布函数分别为：
\varphi(x)=\frac{1}{\sqrt{2\pi}}\,{\rm e}^{-\frac{x^2}{2}}
\Phi(x)=\int_{-\infty}^{x}\varphi(t){\rm d}t=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}}\,{\rm e}^{-\frac{t^2}{2}}{\rm d}t
任何一个正态变量均可以通过标准化转化为标准正态变量，即若 X\sim N(\mu,\sigma^2) ，则：
X^=\frac{X-\mu}{\sigma}\sim N(0,1)
其中 X^ 为标准正态变量。
下面不加证明地给出一些常用性质：
若 X\sim N(0,1) ：
\Phi(-a)=1-\Phi(a)
P(X>a)=1-\Phi(a)
P(a<x<b)=\Phi(b)-\Phi(a)
P(\,|X|<c)=2\Phi(c)-1,\quad (c\geq0)
若 X\sim N(\mu,\sigma^2) ：
P(X\leq c)=\Phi\left(\frac{a-\mu}{\sigma}\right)
P(a<x\leq b)=\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)
其他的类似。
正态分布常用的 3\sigma 原则：
\begin{aligned} P(\,|X-\mu|<k\sigma)&=\Phi(k)-\Phi(-k)\&= 2\Phi(k)-1\&= \begin{cases} 0.6826,\quad k=1,\ 0.9545,\quad k=2,\ 0.9973,\quad k=3. \end{cases} \end{aligned}
均匀分布
若随机变量 X 的密度函数为：
p(x)=\begin{cases} \frac{1}{b-a},\quad a<x<b,\ 0,\quad\quad\quad 其他. \end{cases}
称 X 服从区间 (a,b) 上的均匀分布，记作 X\sim U(a,b) ，其分布函数：
F(x)=\begin{cases} 0,\quad \quad \,x<a,\ \frac{x-a}{b-a},\quad a\leq x<b,\ 1,\quad \quad \,x\geq b. \end{cases}
密度函数及分布函数
均匀分布又称作平顶分布（因其概率密度为常值函数）。
数学期望： \frac{a+b}{2}
方差： \frac{(b-a)^2}{12}
指数分布
若随机变量 X 的密度函数为：
p(x)=\begin{cases} \lambda{\rm e}^{-\lambda x},\quad x\geq0.\ \quad 0,\quad \,\,\,x<0. \end{cases}
则称 X 服从参数为 \lambda 的指数分布，记作 X\sim Exp(\lambda) 。指数分布的分布函数为：
F(x)=\begin{cases} 1-{\rm e}^{\lambda x},\quad x\geq0,\ \quad \ 0,\quad\quad \,\, x<0. \end{cases}
密度函数
指数分布是一种偏态分布，指数分布随机变量只可能取非负实数。指数分布常被用作各种“寿命”分布，譬如电子元器件的寿命、动物的寿命、电话的通话时间、随机服务系统中的服务时间等都可假定服从指数分布。指数分布在可靠性与排队论中有着广泛的应用.。
数学期望： \frac{1}\lambda
方差： \frac{1}{\lambda^2}
指数分布的无记忆性
若随机变量 X \sim Exp(\lambda) ，则对任意的 t>0,s>0 ，有：
P ( X > s + t \,|\, X > s ) =P(X>t)
证明：
因为 X\sim Exp(\lambda) ，所以 P(X\geq s)={\rm e}^{-\lambda s},(s>0) 。又因为
{X>s+t}\subseteq{X>s}
由条件概率可得：
P ( X > s + t \,|\, X > s ) = \frac { P ( X > s + t ) } { P ( X > s ) } = \frac { \mathrm { e } ^ { - \lambda ( s + t ) } } { \mathrm { e } ^ { - \lambda t } } = \mathrm { e } ^ { - \lambda t } = P ( X > t )
证毕。
该式的含义为：记 X 是某种产品的使用寿命 (h) ，若 X 服从指数分布，那么已知此产品使用了 s(h) 没发生故障，则再能使用 t(h) 而不发生故障的概率与已使用的 s(h) 无关，只相当于重新开始使用 t(h) 的概率，即对已使用过的 s(h) 没有记忆。
伽玛分布
先引入伽玛函数：
\Gamma ( \alpha ) = \int _ { 0 } ^ { \infty } x ^ { \alpha - 1 } e ^ { - x } d x
其中参数 \alpha>0 。伽玛函数具有下列性质：
\Gamma(1)=1,\quad \Gamma(\frac{1}{2})=\sqrt {\pi}
\Gamma(\alpha+1)=\alpha\Gamma(\alpha)
当 \alpha 为自然数 n 时：
\Gamma(n+1)=n\Gamma(n)=n\,!
伽玛分布：
若随机变量 X 的密度函数为：
p ( x ) = \left{ \begin{array} { l l } { \frac { \lambda ^ { a } } { \Gamma ( \alpha ) } x ^ { a - 1 } \mathrm { e } ^ { - \lambda x } , } & { x \geqslant 0 ,} \ { \quad \quad \quad0, } & { x < 0 } \end{array} \right.
称X X 服从伽玛分布，记作 X\sim Ga(\alpha,\lambda) 。其中 \alpha>0 为形状参数， \lambda>0 为尺度参数。
密度函数
数学期望： \frac{\alpha}{\lambda}
方差： \frac{\alpha}{\lambda^2}
伽玛函数的特例：
\alpha=1 时的伽玛分布为指数分布：Ga(1,\lambda)=Exp(\lambda)
2.称 \alpha=\frac{n}{2},\lambda=\frac{1}{2} 的伽玛分布为自由度为 n 的 \chi^2 （卡方）分布，记作 \chi^2(n) ：
Ga\left(\frac{n}{2},\frac{1}{2}\right)=\chi^2(n)
因卡方分布是特殊的伽玛分布，故不难求得卡方分布的：
数学期望： n
方差： 2n
卡方分布的唯一参数 n 称为它的自由度，具体含义在之后的数理统计中会给出。
贝塔分布
先给出贝塔函数：
\mathrm { B } ( a , b ) = \int _ { 0 } ^ { 1 } x ^ { a - 1 } ( 1 - x ) ^ { b - 1 } d x
其中参数 a>0,b>0 。贝塔函数具有以下性质：
B(a,b)=B(b,a)
2.贝塔函数与伽玛函数有如下关系：
\mathrm { B } ( a , b ) = \frac { \Gamma ( a ) \,\Gamma ( b ) } { \Gamma ( a + b ) }
贝塔分布：
若随机变量 X 的密度函数为：
p ( x ) = \left{ \begin{array} { l l } {\frac { \Gamma ( a+b ) } { \Gamma ( a ) \Gamma ( b )} x ^ { a - 1 } (1-x)^{b-1} }, & {0<x<1 ,} \ { \quad \quad \quad0, } & { \quad 其他. } \end{array} \right.
则称 X 服从贝塔分布，记作 X\sim Be(a,b) ，其中 a>0,b>0 都是形状参数。
密度函数
数学期望： \frac{a}{a+b}
方差： \frac{ab}{(a+b)^2(a+b+1)}
总结
常用概率分布及其数学期望与方差
编辑于 2021-11-20 10:21
统计学
概率论
概率论与数理统计
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18 条评论
默认
最新
蒙蒙羊
指数分布的分布函数少了一个负号
2023-02-15
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modest
二项分布的pk中。第1个组合数nk的位置应该写倒了
2022-12-24
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天秤羊说
Jaggies
6
2024-12-09
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Jaggies
？？？？？？？？？？？？？？？？？？？？？？？？？？
2023-06-19
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知乎用户EQf0pR
不够全，不过还是
2022-03-20
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凹总宅
哇这么多年我终于能看懂这些分布了，整理得真好
2022-03-07
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zjply
贝塔分布的期望有误，应该是a/(a+b)
2021-11-20
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CoffeeCat
作者
谢谢已改
2021-11-20
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min214758
正态分布的性质有一个也有点问题
2021-06-21
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生石灰
贝塔分布的pdf写错了吧，前面系数分子分母应该倒过来
2021-06-21
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CoffeeCat
作者
谢谢已改
2021-11-20
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杰益
大哥 总结那个图有没有高清的，想要一个。我邮箱：994271774@qq.com
2021-01-09
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HYDeng
指数分布的均值和方差有误 估计是复制泊松分布
2020-05-02
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CoffeeCat
作者
感谢指正！已改
2020-05-03
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Liskell Leo
负二项分布里边有谬误，应该是p^r乘以(1-p)^(k-r)
2020-03-28
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CoffeeCat
作者
感谢！已改
2020-03-28
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4469 | https://www.quora.com/Can-you-explain-the-concept-of-a-local-maximum-or-minimum-on-an-interval | Can you explain the concept of a local maximum or minimum on an interval? - Quora
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Mathematics
Local Maxima
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Relative Minimum
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Calculus (Mathematics)
Local Extrema
Interval Mathematics
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Can you explain the concept of a local maximum or minimum on an interval?
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Ben Aronson
B.S. in Mathematics&Digital Media Studies, Binghamton University (Expected 2027)
·1y
A local maximum or minimum on an interval is simply the highest or lowest function value achieved on the interval. To find it, set the first derivative to zero and solve for the x-values within the interval, then find their corresponding y-values. Next, plug the x-values of the endpoints into the function to find their corresponding y-values. The lowest of all these values is the local minimum and the highest is the local maximum.
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What is the maximum and minimum value of f(x) =5x+1 with the interval [-2, 1]?
What is the definition of a local maximum or minimum of a function?
Can you provide an example of an interval without any maximum or minimum point, but with endpoints? Why is this the case?
What is the definition of a local maximum or minimum? Can there be more than one local maximum or minimum in a curve?
What is the significance of identifying a local maximum or minimum value for a given interval on a graph?
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·2y
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What happens to the second derivative when we approach a local maximum or minimum?
The best way to see this is to have 3 separate graphs in line so that we can compare graphs of y = f(x), y = f’(x) and y = f’’(x)
Let’s use a nice cubic curve…
So as we approach a maximum point the second derivative is negative
and as we approach a minimum point the second derivative is positive.
Also as we approach a point of inflection the second derivative is zero.
Continue Reading
The best way to see this is to have 3 separate graphs in line so that we can compare graphs of y = f(x), y = f’(x) and y = f’’(x)
Let’s use a nice cubic curve…
So as we approach a maximum point the second derivative is negative
and as we approach a minimum point the second derivative is positive.
Also as we approach a point of inflection the second derivative is zero.
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·2y
Related
What is the definition of a local maximum or minimum? Can there be more than one local maximum or minimum in a curve?
The gradient actually at a maximum or minimum point is zero
Look at this function…
It has 3 maximum points A, B, C and 2 minimum points P and Q.
The gradient is zero at each of these points.
Continue Reading
The gradient actually at a maximum or minimum point is zero
Look at this function…
It has 3 maximum points A, B, C and 2 minimum points P and Q.
The gradient is zero at each of these points.
Upvote ·
9 6
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·6y
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Why is a local maximum in f'(x) a local minimum in f(x)?
I think you are a little confused here.
I made the poster below, for my classroom wall. I think it will sort out your problem…
In the above diagram, a local minimum on...
Continue Reading
I think you are a little confused here.
I made the poster below, for my classroom wall. I think it will sort out your problem…
In the above diagram, a local minimum on...
Related questions
What is the maximum and minimum value of f(x) =5x+1 with the interval [-2, 1]?
What is the definition of a local maximum or minimum of a function?
Can you provide an example of an interval without any maximum or minimum point, but with endpoints? Why is this the case?
What is the definition of a local maximum or minimum? Can there be more than one local maximum or minimum in a curve?
What is the significance of identifying a local maximum or minimum value for a given interval on a graph?
What should be the revision intervals to retain maximum?
How do you show that the open interval (0,1) has no maximum and minimum?
What is the definition of a relative maximum and minimum point? Is it possible to find a function with only one relative maximum and minimum point in each interval? Why or why not?
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Related questions
What is the maximum and minimum value of f(x) =5x+1 with the interval [-2, 1]?
What is the definition of a local maximum or minimum of a function?
Can you provide an example of an interval without any maximum or minimum point, but with endpoints? Why is this the case?
What is the definition of a local maximum or minimum? Can there be more than one local maximum or minimum in a curve?
What is the significance of identifying a local maximum or minimum value for a given interval on a graph?
What should be the revision intervals to retain maximum?
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Vulvar Cancer
Vulvar cancer is a rare cancer of your vulva. There are about 6,500 new cases of vulvar cancer in the U.S. each year. Most cases are related to either human papillomavirus (HPV) infection or lichen sclerosus. Changes in vulvar skin color and lumps or open sores may be signs of vulvar cancer. Treatments include surgery, radiation or chemotherapy.
ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentOutlook / PrognosisPreventionLiving WithAdditional Common Questions
Overview
What is vulvar cancer?
Vulvar cancer is a rare cancer that forms in the tissues of your vulva. “Vulva” is the collective name for all the female external sex organs, or genitals. Your vulva includes:
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Opening of your vagina: The muscular canal for sex, childbirth and menstruation (periods).
Labia minora (inner lips): Tissue folds that surround your vaginal opening and extend above your clitoris.
Labia majora (outer lips): Fleshy area that surrounds your inner lips.
The outer part of your clitoris: The sexually sensitive nub of flesh above your vaginal opening.
Mons pubis: The rounded area in front of your pubic bones that becomes covered with hair at puberty.
Opening of the urethra: The tube that allows urine (pee) to exit your body.
Perineum: The patch of skin between your vagina and anus (butthole).
Vulvar cancer usually develops slowly over several years. Precancerous areas of tissue (lesions) typically develop first. Healthcare providers usually discover the abnormal growth in the outermost layer of your skin. These precancerous lesions are called vulvar intraepithelial neoplasia (VIN).
Types of vulvar cancer
Healthcare providers classify vulvar cancer based on the type of cells where the cancer starts. The most common types of vulvar cancer include:
Vulvar squamous cell carcinoma: Approximately 90% of vulvar cancers are squamous cell carcinomas. They develop in the cells on the surface of your skin.
Vulvar melanoma: Approximately 5% of vulvar cancers are melanomas. Melanomas develop rapidly and have a high risk of spreading to other areas of your body.
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The remaining vulvar cancers are rare. They include:
Basal cell carcinoma.
Bartholin gland adenocarcinoma.
Paget disease of the vulva.
Sarcoma.
Verrucous carcinoma.
How common is vulvar cancer?
Vulvar cancer is rare. Healthcare providers diagnose just under 6,500 new cases of vulvar cancer in the U.S. each year.
Nearly 80% of people diagnosed are over age 50, and over half of all diagnoses are in people over age 70. The average age at diagnosis is 68.
Symptoms and Causes
What are the symptoms of vulvar cancer?
The first noticeable signs of vulvar cancer are usually skin changes on your outer lips (labia majora) or inner lips (labia minora). But cancer can form anywhere on your vulva.
Vulvar cancer symptoms include:
Color changes, including skin that looks darker or lighter than usual, or patches of white skin.
Thickened or rough skin patches.
Growths, including lumps, wart-like bumps or ulcers that don’t heal.
Itching or burning that doesn’t improve.
Bleeding that’s unrelated to menstruation (periods).
Tenderness and pain, potentially during sex or when you’re peeing.
See your healthcare provider if you have one or more of these symptoms. Vulvar cancer symptoms usually don’t appear in the early stages, so it’s important to get checked as soon as possible.
Still, many of these symptoms are also common in noncancerous conditions. Your provider can tell you whether these changes are signs of vulvar cancer or a different condition.
What causes vulvar cancer?
With vulvar cancer, cells begin multiplying out of control. Without treatment, these cancer cells can spread to other parts of your body.
The most common type of vulvar cancer, vulvar squamous cell carcinoma, arises in association with one of two conditions:
Human papillomavirus (HPV) infection: A common sexually transmitted infection (STI) that spreads through skin-to-skin contact. Some types of HPV increase your risk of certain cancers, including cervical cancer, anal cancer, rectal cancer and vulvar cancer.
Lichen sclerosus: A chronic (lifelong) skin condition. Lichen sclerosus causes inflammation and other symptoms, such as skin changes and itching, on your vulva.
Risk factors
Risk factors for vulvar cancer include:
Age: Your likelihood of developing vulvar cancer increases with age.
Exposure to HPV: Not all strains of HPV cause cancer, but some can lead to cell changes that eventually become vulvar cancer.
Skin conditions involving your vulva: Growths associated with lichen sclerosus may progress to vulvar cancer.
Vulvar intraepithelial neoplasia (VIN): VIN is a precancerous condition that can progress to vulvar cancer if it’s not treated.
Human immunodeficiency virus (HIV) infection: A weakened immune system from a condition like HIV can make it harder for your body to fight cancer.
Smoking: Smoking raises your risk of developing multiple cancer types, including vulvar cancer.
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Diagnosis and Tests
How is vulvar cancer diagnosed?
Your healthcare provider will ask about your medical history, potential risk factors and symptoms. Diagnosis often involves multiple tests.
Tests to diagnose vulvar cancer
Tests may include:
Pelvic exam: Your provider will visually inspect your vulva, checking for unusual skin changes. They’ll insert one or two gloved, lubricated fingers inside your vagina to feel for any lumps or other signs of cancer. They may use a similar technique to check your rectum. They may use a tool called a speculum to widen your vagina so they can check for abnormalities.
Pap smear: Your provider may take a sample of cells during the pelvic exam and test them for cancerous changes. They may perform an HPV test on the sample to see if you have an infection.
Colposcopy: Your provider may use a lighted, magnifying instrument called a colposcope to view your vulva, vagina and cervix (the organ between your vagina and uterus) in more detail. They may apply a special solution that can highlight abnormal cells, making them easier to see.
Biopsy: Your provider may remove a sample of abnormal tissue to test it for cancer cells. A biopsy is the only way to know for sure whether you have vulvar cancer.
Tests to determine cancer spread
If you have cancer, your provider will perform additional tests to see if it’s spread beyond your vulva. Without treatment, vulvar cancer may spread to your vagina or other nearby organs, lymph nodes in your pelvis and eventually your bloodstream. Cancer that’s spread (metastatic cancer) is harder to treat.
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Tests may include:
Scope exams: You may receive a cystoscopy to check for cancer spread in your urethra (the tube that carries your pee) or bladder. A proctoscopy checks for cancer cells in your rectum or anus.
Imaging tests: X-rays, computed tomography (CT) scans, magnetic resonance imaging (MRI) and positron emission tomography (PET) scans can show if the cancer has spread from your vulva to other tissues.
Sentinel node biopsy: Your provider may remove the lymph node closest to your tumor (the sentinel node) to test for cancer cells. With vulvar cancer, tumors usually drain into sentinel lymph nodes in your groin.
What are the stages of vulvar cancer?
Vulvar cancer staging allows healthcare providers to determine if your cancer’s spread beyond your vulva. This information guides treatment decisions. There are four main stages:
Stage I: Early-stage vulvar cancer is only on your vulva or perineum (area between your rectum and vagina). Stage I consists of Stages IA or IB based on tumor size and how far it reaches into nearby tissue.
Stage II: The tumor (of any size) has spread into the lower part of your urethra, the lower part of your vagina or anus.
Stage III: Cancer has spread to one or more nearby lymph nodes. Stage III consists of Stages IIIA, IIIB, and IIIC based on the number and size of the lymph nodes involved.
Stage IV: Cancer has spread into the upper part of your urethra, vagina or other body parts. Stage IV consists of Stages IVA and IVB based on if the spread is localized near your vulva or spread distantly.
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Ask your healthcare provider to explain the details of what your cancer stage means for your treatment.
Management and Treatment
How is vulvar cancer treated?
Your treatment depends on factors like your general health, cancer stage and whether your healthcare provider recently diagnosed your cancer or if it’s recurred (come back). Your provider can explain how your treatment plan is best suited for your diagnosis.
Surgery
Surgery is the most common treatment for cancer of the vulva. The goal is to remove all the cancer while preserving your sexual function. Types of surgery include:
Laser surgery: This surgery uses a laser beam to make bloodless cuts in tissue or to remove cancerous surface lesions.
Local excision: This surgery removes the cancer and a small-to-large amount of normal tissue around the cancer. Sometimes, providers remove nearby lymph nodes to test for cancer cells or to remove lymph nodes when there’s evidence of cancer.
Vulvectomy: This procedure removes part or all of your vulva and possibly some nearby lymph nodes. Your provider may use skin grafts to replace removed skin.
Pelvic exenteration: This surgery removes your lower colon, rectum, bladder, cervix, vagina, ovaries and nearby lymph nodes. Your healthcare provider will create openings to allow urine and stool to flow from your body into a collection bag.
Radiation therapy
Radiation therapy uses X-rays or other high-energy sources to kill cancer cells. The most common delivery method for vulvar cancer treatment is external beam radiation therapy (EBRT). EBRT uses a machine to deliver radiation through your skin to the targeted cancer site.
Often, people receive radiation therapy and chemotherapy together (chemoradiation). You may receive radiation before surgery to shrink a tumor or after surgery to destroy any remaining cancer cells.
Chemotherapy
Chemotherapy uses drugs to attack cancer cells throughout your body. Your healthcare provider may inject the medicine into a vein or muscle, or you may take a pill. You may receive a lotion that you can apply directly to your vulva. This form of chemotherapy attacks cancer more locally — in the specific area.
Cisplatin (Platinol®, Platinol -AQ®) and fluorouracil (Carac®) are commonly prescribed chemotherapy drugs for vulvar cancer.
Immunotherapy
Immunotherapy helps your body’s immune system identify cancer cells and fight them more effectively. Imiquimod cream (Aldara®, Zyclara®) is a common immunotherapy medication used to treat vulvar cancer.
What follow-up should I expect after vulvar cancer treatment?
Your healthcare provider may perform tests at various checkpoints after treatment to monitor your condition and ensure the cancer hasn’t returned.
Testing often involves the same procedures used to diagnose and stage vulvar cancer.
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Outlook / Prognosis
Is vulvar cancer serious?
It can be. Untreated vulvar cancer is life-threatening. Cancer that’s spread to your lymph nodes or other body parts is much harder to treat than cancer diagnosed early.
While there’s always a risk that cancer may return after treatment (recur), most people who receive treatment in the early stages of the disease remain cancer-free.
What’s the survival rate for vulvar cancer?
The relative five-year survival rate for people with vulvar cancer is approximately 70%. But survival rates are higher when people are diagnosed and treated in the early stages. For example, the five-year survival rate for localized cancer (remaining in the vulva) is approximately 86%. The five-year survival rate drops to approximately 30% once the cancer spreads.
Still, your prognosis depends on factors unique to your diagnosis, including your health and your response to treatment. Ask your healthcare provider about likely outcomes based on your diagnosis.
How quickly does vulvar cancer progress?
Most types of vulvar cancer progress slowly over several years. Less common types, like melanomas, tend to grow and spread more quickly.
Prevention
Can vulvar cancer be prevented?
The best way to reduce your risk is to get the HPV vaccine to prevent infections. In the U.S., adults up to age 45 may receive Gardasil 9® depending on their risk of HPV exposure. Cervarix® and Gardasil® are HPV vaccines available in other countries.
See your healthcare provider right away if you develop any symptoms of vulvar cancer. Schedule regular checkups, including a physical exam, at least annually for your gynecological health.
Living With
How do I take care of myself?
Many people feel self-conscious about visible changes to their vulva. Still, having vulvar cancer doesn’t mean you must abandon physical intimacy. Don’t be ashamed to ask your healthcare provider how your diagnosis may affect your sex life. They can connect you with resources that support your physical and emotional needs as you navigate your diagnosis and treatment.
Additional Common Questions
Where does vulvar cancer usually start?
Vulvar cancer usually starts on the surface of the skin surrounding your vagina, either your outer lips (labia majora) or inner lips (labia minora). Less commonly, it forms on other parts of your vulva, like your Bartholin gland and clitoris.
What is the first stage of vulvar cancer?
Stage I is the first stage of vulvar cancer. In this stage, cancer hasn’t spread beyond your vulva or perineum. Stage IA vulvar cancer is two centimeters (peanut size) or smaller. It hasn’t spread beyond one millimeter (tip of a pencil-size) into nearby tissue beyond one millimeter. Stage IB vulvar cancers involve larger tumors that may have invaded more deeply into nearby tissue.
A note from Cleveland Clinic
You play a large role in your health. Become familiar with all parts of your body, including your vulva. Knowing what you look like now can help you easily identify changes. And when you see changes, make an appointment to see your healthcare provider. In the meantime, don’t skip visits to your gynecologist. They can identify precancerous changes before they become more serious.
Care at Cleveland Clinic
Vulvar cancer is rare and often complex. At Cleveland Clinic, we can diagnose and treat this condition and support you through your recovery.
Vulvar Cancer Treatment
Find a Doctor and Specialists
Make an Appointment
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4471 | https://brainly.com/question/7763875 | [FREE] Let e and f be the events that a family of n children has children of both sexes and has at most one - brainly.com
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Let e and f be the events that a family of n children has children of both sexes and has at most one boy, respectively. Are e and f independent if:
a. n=2?
b. n=4?
c. n=5?
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The question requires analyzing whether two events regarding the gender makeup of a family with a certain number of children are independent for different family sizes.
Explanation
The problem is about determining whether two events (e and f) are independent events when it comes to family composition in terms of children's gender. We are given different family sizes (n=2, n=4, n=5) and need to analyze if events e (the family has children of both sexes) and f (the family has at most one boy) are independent for each case.
For two events to be independent, the probability of them occurring together (P(e AND f)) must be equal to the product of their individual probabilities (P(e)P(f)). We find these probabilities using basic combinatorial methods and probability rules. The analysis would likely differ for each given family size (n), as the number of possible outcomes for e and f would change. Independence can be shown by confirming that P(e AND f) is equal to P(e)P(f) for each case.
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For all three cases (n = 2, n = 4, n = 5), the events e (children of both sexes) and f (at most one boy) are not independent. This is determined by showing that the probability of both events occurring together does not equal the product of their individual probabilities. Therefore, the two events are dependent on each other.
Explanation
To determine whether the events e (the family has children of both sexes) and f (the family has at most one boy) are independent, we first need to analyze the combinations of children for different values of n. The independence of two events occurs when P(e∩f)=P(e)⋅P(f). Let's break this down for each case:
a. For n=2:
Sample Space: The possible combinations of children are: MM,MF,FM,FF.
Event e: The family has children of both sexes. So, e occurs in cases MF and FM, giving P(e)=4 2=2 1.
Event f: The family has at most one boy. So, f occurs in cases MM,MF,FM, giving P(f)=4 3.
Combined event P(e∩f): This occurs in cases MF,FM which gives P(e∩f)=4 2=2 1.
Independence Check: P(e∩f)=2 1 and P(e)P(f)=2 1⋅4 3=8 3. Thus, these events are not independent when n=2.
b. For n=4:
Sample Space: The total combinations for four children are 2 4=16 (all combinations of boys and girls).
Event e: Here, e happens in combinations where there is at least one boy and one girl. The outcomes with same sex are MMMM,FFFF, so P(e)=1−16 2=16 14=8 7.
Event f: Cases with at most one boy are MMMM,MMMF,MMFG,MFGF,FFFG, giving P(f)=16 5.
Combined event P(e∩f): The only case is FFFF which gives P(e∩f)=16 5.
Independence Check: P(e∩f)=16 5 and P(e)P(f)=8 7⋅16 5=128 35, hence they are not independent when n=4.
c. For n=5:
Sample Space: Total combinations are 2 5=32.
Event e: Same analysis gives P(e)=1−32 2=32 30=16 15.
Event f: Cases with at most one boy are MMMMM,MMMMF,MMMFF,MMFFM,FMFFM,FMMFF,FFFFF, leading to P(f)=32 7.
Combined event P(e∩f): It still is valid finding cases. The possible cases are significantly leading to P(e∩f)=32 7 as well.
Independence Check: Here, we again calculate P(e)P(f)=16 15⋅32 7=512 105. Therefore, events are not independent when n=5 as well.
Hence, in all three cases, events e and f are not independent.
Examples & Evidence
For instance, in a family with two children, the combinations of boys and girls determine the probabilities of both events affecting each other. Having two boys or two girls affects the presence of both sexes and the number of boys in the family.
The independence is tested with the calculated probabilities, showing that the condition P(e∩f)=P(e)⋅P(f) is violated in each scenario.
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Community Answer a family has 3 children, each of whom is a boy or a girl with probability 1/2 Let A = " there is at most 1 girl", B= "the family has children of both sexes". a) are A and B independent b) are A and B independent if it was a 4 family children
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Write the slope-intercept equation of the function f whose graph satisfies the given conditions. The graph of f passes through (9,−3) and is perpendicular to the line whose equation is x=17.
For the real-valued functions g(x)=x 2+3 and h(x)=x 2−2, find the composition g∘h and specify its domain using interval notation.
The rational function f is given by f(x)=−5 x 5+3 x 2+15 x+7 x−3 12. Which of the following describes the end behavior of f ?A. lim x→−∞f(x)=∞ and lim x→∞f(x)=∞.B. lim x→−∞f(x)=∞ and lim x→∞f(x)=−∞.C. lim x→−∞f(x)=−∞ and lim x→∞f(x)=∞.D. lim x→−∞f(x)=−∞ and lim x→∞f(x)=−∞.
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4472 | https://en.wikipedia.org/wiki/Saddle_point | Jump to content
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Contents
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1 Mathematical discussion
2 Saddle surface
3 Examples
4 Other uses
5 See also
6 References
6.1 Citations
6.2 Sources
7 Further reading
8 External links
Saddle point
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From Wikipedia, the free encyclopedia
Critical point on a surface graph which is not a local extremum
This article is about the mathematical property. For the peninsula in the Antarctic, see Saddle Point. For the type of landform and general uses of the word "saddle" as a technical term, see Saddle (landform).
In mathematics, a saddle point or minimax point is a point on the surface of the graph of a function where the slopes (derivatives) in orthogonal directions are all zero (a critical point), but which is not a local extremum of the function. An example of a saddle point is when there is a critical point with a relative minimum along one axial direction (between peaks) and a relative maximum along the crossing axis. However, a saddle point need not be in this form. For example, the function has a critical point at that is a saddle point since it is neither a relative maximum nor relative minimum, but it does not have a relative maximum or relative minimum in the -direction.
The name derives from the fact that the prototypical example in two dimensions is a surface that curves up in one direction, and curves down in a different direction, resembling a riding saddle. In terms of contour lines, a saddle point in two dimensions gives rise to a contour map with, in principle, a pair of lines intersecting at the point. Such intersections are rare in contour maps drawn with discrete contour lines, such as ordnance survey maps, as the height of the saddle point is unlikely to coincide with the integer multiples used in such maps. Instead, the saddle point appears as a blank space in the middle of four sets of contour lines that approach and veer away from it. For a basic saddle point, these sets occur in pairs, with an opposing high pair and an opposing low pair positioned in orthogonal directions. The critical contour lines generally do not have to intersect orthogonally.
Mathematical discussion
[edit]
A simple criterion for checking if a given stationary point of a real-valued function F(x,y) of two real variables is a saddle point is to compute the function's Hessian matrix at that point: if the Hessian is indefinite, then that point is a saddle point. For example, the Hessian matrix of the function at the stationary point is the matrix
which is indefinite. Therefore, this point is a saddle point. This criterion gives only a sufficient condition. For example, the point is a saddle point for the function but the Hessian matrix of this function at the origin is the null matrix, which is not indefinite.
In the most general terms, a saddle point for a smooth function (whose graph is a curve, surface or hypersurface) is a stationary point such that the curve/surface/etc. in the neighborhood of that point is not entirely on any side of the tangent space at that point.
In a domain of one dimension, a saddle point is a point which is both a stationary point and a point of inflection. Since it is a point of inflection, it is not a local extremum.
Saddle surface
[edit]
A saddle surface is a smooth surface containing one or more saddle points.
Classical examples of two-dimensional saddle surfaces in the Euclidean space are second order surfaces, the hyperbolic paraboloid (which is often referred to as "the saddle surface" or "the standard saddle surface") and the hyperboloid of one sheet. The Pringles potato chip or crisp is an everyday example of a hyperbolic paraboloid shape.
Saddle surfaces have negative Gaussian curvature which distinguish them from convex/elliptical surfaces which have positive Gaussian curvature. A classical third-order saddle surface is the monkey saddle.
Examples
[edit]
In a two-player zero sum game defined on a continuous space, the equilibrium point is a saddle point.
For a second-order linear autonomous system, a critical point is a saddle point if the characteristic equation has one positive and one negative real eigenvalue.
In optimization subject to equality constraints, the first-order conditions describe a saddle point of the Lagrangian.
Other uses
[edit]
In dynamical systems, if the dynamic is given by a differentiable map f then a point is hyperbolic if and only if the differential of ƒ n (where n is the period of the point) has no eigenvalue on the (complex) unit circle when computed at the point. Then a saddle point is a hyperbolic periodic point whose stable and unstable manifolds have a dimension that is not zero.
A saddle point of a matrix is an element which is both the largest element in its column and the smallest element in its row.
See also
[edit]
Saddle-point method is an extension of Laplace's method for approximating integrals
Maximum and minimum
Derivative test
Hyperbolic equilibrium point
Hyperbolic geometry
Minimax theorem
Max–min inequality
Mountain pass theorem
References
[edit]
Citations
[edit]
^ Howard Anton, Irl Bivens, Stephen Davis (2002): Calculus, Multivariable Version, p. 844.
^ Chiang, Alpha C. (1984). Fundamental Methods of Mathematical Economics (3rd ed.). New York: McGraw-Hill. p. 312. ISBN 0-07-010813-7.
^ Buck, R. Creighton (2003). Advanced Calculus (3rd ed.). Long Grove, IL: Waveland Press. p. 160. ISBN 1-57766-302-0.
^ von Petersdorff 2006
Sources
[edit]
Gray, Lawrence F.; Flanigan, Francis J.; Kazdan, Jerry L.; Frank, David H.; Fristedt, Bert (1990), Calculus two: linear and nonlinear functions, Berlin: Springer-Verlag, p. 375, ISBN 0-387-97388-5
Hilbert, David; Cohn-Vossen, Stephan (1952), Geometry and the Imagination (2nd ed.), New York, NY: Chelsea, ISBN 978-0-8284-1087-8 {{citation}}: ISBN / Date incompatibility (help)
von Petersdorff, Tobias (2006), "Critical Points of Autonomous Systems", Differential Equations for Scientists and Engineers (Math 246 lecture notes)
Widder, D. V. (1989), Advanced calculus, New York, NY: Dover Publications, p. 128, ISBN 0-486-66103-2
Agarwal, A., Study on the Nash Equilibrium (Lecture Notes)
Further reading
[edit]
Hilbert, David; Cohn-Vossen, Stephan (1952). Geometry and the Imagination (2nd ed.). Chelsea. ISBN 0-8284-1087-9. {{cite book}}: ISBN / Date incompatibility (help)
External links
[edit]
Media related to Saddle point at Wikimedia Commons
Retrieved from "
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Differential geometry of surfaces
Multivariable calculus
Stability theory
Analytic geometry
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4473 | https://pmc.ncbi.nlm.nih.gov/articles/PMC11987523/ | The Changes in ICD-11 Related to Sexual Health and Dysfunction and Their Implication for Clinical Practice - PMC
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. 2024 Nov 22;36:266–271. doi: 10.5080/u27559
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The Changes in ICD-11 Related to Sexual Health and Dysfunction and Their Implication for Clinical Practice
Koray Başar
Koray Başar
1 Assoc. Prof., Hacettepe University, Faculty of Medicine, Department of Psychiatry, Ankara, Turkey
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1 Assoc. Prof., Hacettepe University, Faculty of Medicine, Department of Psychiatry, Ankara, Turkey
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e-mail: kbasar@hacettepe.edu.tr
Received 2024 Jul 23; Accepted 2024 Aug 9; Collection date 2025.
Copyright: © 2025 Turkish psychiatric Society
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PMCID: PMC11987523 PMID: 39801382
ABSTRACT
The classification of sexual health-related conditions was reformulated in 11th revision of International Classification of Diseases (ICD-11) following current evidence, best practice, and taking human rights into consideration, which is expected to reflect and provide guidance for more integrative clinical approaches. Overcoming the artificial, yet historical, distinction between “organic” and “non-organic” conditions, sexual dysfunctions classified in the “Mental and Behavioral Disorders” and “Disorders of Genitourinary System” in ICD-10 were listed in a new chapter called “Conditions Related to Sexual Health.” In practice, this approach has been consistently recommended. However, diagnostical clasification was not congruent with the recommendation. Dysfunctions, defined with a non-normative but individual-based threshold, are categorized according to different stages of the sexual response cycle, similar to ICD-10 and Diagnostic and Statistical Manual of Mental Disorders 5th version (DSM-5). However, similarities and distinctions in the clinical presentation of the dysfunction in men and women were also considered, resulting in differences from the DSM-5 approach. Gender Incongruence is classified in this newly formed “Conditions Related to Sexual Health” chapter, not with mental disorders as in the earlier version, reflecting the current non-pathologizing understanding of gender diversity. Furthermore, the criteria for these conditions were revised to embrace the variability in the experience of gender identity. In addition, the residuals of sexual orientation-related diagnostic categories were removed. Paraphilic disorders categories replaced “Disorders of sexual preference” in ICD-10, with significant modifications in conceptualization and classification.
Keywords: ICD-11, Sexual Dysfunctions, Gender Incongruence, Paraphilic Disorders
INTRODUCTION
In the field of sexual health and dysfunctions, there has been a considerable accumulation of research evidence and significant changes in the best practices related to the assessment and management of the relevant conditions in the two and a half decades following the approval of ICD-10, International Classification of Diseases by World Health Organization (World Health Organization 1992). ICD-11 was approved in May 2019 (World Health Organization 2022). In the history of the ICD, the interval of the last two revisions has been the longest. During this period, there has been a shift in the understanding of human sexuality, the rights of individuals both in social life and during healthcare, and the prioritization of the individual in healthcare provision. Although there are differences in the priority given to individuals’ responsibility and the organization of care among countries (Giami 2002), the conceptualization of sexual health has extended to include sexual rights and especially the right of sexual pleasure for all, leaving no one behind and excluding the risk of violence and discrimination (Coleman et al. 2021, Mitchell et al. 2021). World Health Organisation’s Department of Mental Health and Substance Abuse and the Department of Reproductive Health and Research worked together on proposals for revisions related to sexuality and sexual health. Two departments assigned a joint Working Group on Sexual Disorders and Sexual Health to develop recommendations. Finally, ICD-11 was submitted to the 144th Executive Board Meeting in January 2019 and approved during the 72nd World Health Assembly in May 2019 (World Health Organization 2022).
The reformulation of the classification of sexual health-related conditions in ICD-11 was expected to reflect these developments and provide guidance for more integrative clinical approaches and public health implementations. This brief narrative review aims to present and discuss changes brought by ICD-11 in the field of sexual health. It is not intended to be a comprehensive review of all changes in ICD. However, the clinical significance of the changes is emphasized.
Sexual Dysfunctions and Sexual Pain
Overcoming the artificial yet historical distinction between “organic” and “non-organic” conditions, sexual dysfunctions classified in the “Mental and Behavioral Disorders” and “Disorders of Genitourinary System” in ICD-10 were listed in a new chapter called “Conditions Related to Sexual Health” in ICD-11 (World Health Organization 2022). This relocation appears as an attempt to overcome the constraints of the mind-body split, which is still prevailing in health in general but persists as a significant problem in sexuality as well. This change reflects the accumulating evidence on the complex interaction of physiological, psychological, interpersonal, social, and cultural factors involved with the emergence and maintenance of sexual dysfunctions. In each case, clinicians are encouraged to assess and describe diverse etiological qualifiers, such as conditions associated with a disorder or disease, with a medication or substance, with a lack of knowledge, with relationship factors, or with cultural factors, which are not mutually exclusive. In sexual health practices, refraining from considering sexual health issue as organic and psychogenic has been consistently recommended for decades. However, diagnostic classification was not congruent with this approach.
Although in the text of ICD-11, the commitment to the biopsychosocial model of sexuality was emphasized, this relocation also led to concerns over greater medicalization of sexual dysfunctions (Meana et al. 2020). Although ICD-11 does not attempt to delineate limits of medical specialties through diagnostic categorization, the relocation of the sexual dysfunction away from the mental disorders group carries the potential of prioritization of the pharmacological and surgical management of the conditions rather than sexual therapy. Sexuality related health issues often require counselling, and not infrequently sexual therapy is indicated. With this change, healthcare professionals without adequate expertise on mental health and psychotherapy may proceed with these interventions. Especially in countries without specific regulations on the practice of psychotherapy, such inappropriate interventions may lead to delays in the achievement of the assistance needed, and even may have disruptive consequences.
In ICD-11, sexual dysfunctions are categorized according to different stages of the sexual response cycle, similar to ICD-10 and DSM-5 (Parish et al. 2021). The dysfunctions are defined with a non-normative but individual-based threshold; the failure in satisfaction of the individual is considered as the threshold of a diagnosis rather than the expectation of the partner or the anticipation of society for any sexual response. The similarities and distinctions in the clinical presentation of the dysfunction in men and women were also considered, resulting in differences from the DSM-5 approach. Thus, some diagnoses that are described in a gender-specific manner, such as arousal disorders, are separately defined for each gender. Whereas “Hypoactive Sexual Desire Disorder” and “Anorgasmia” are categories that can be diagnosed in both men and women. In DSM-5, female sexual arousal and desire disorders were combined in a single category due to frequent comorbity and the difficulties in dissociating these sexual responses in clinical population. This distinction brought by DSM-5 is still controversial (Sungur and Gündüz 2014, Sarin et al. 2013). Despite the change in DSM, these two conditions are considered two distinct categories in ICD-11. There is evidence supporting two different conditions with regards to genetic features and treatment response (Reed et al. 2016). Further research on these separate diagnostic categories may lead to a better understanding.
In male sexual dysfunctions, ICD-11 permits a distinction between ejaculation and the subjective experience of orgasm (Reed et al. 2016). This feature, which is not present in DSM-5, may prove beneficial in research and clinical management of the conditions in individuals presenting with these symptoms. The terminologies of “Premature Ejaculation” and “Impotence” were changed to “Early Ejaculation” and “Erectile Dysfunction” with ICD-11, respectively. A clinically significant difference in the definition of “Early Ejaculation” in ICD-11, compared to that of ICD-10 and DSM-5, is the lack of a cut-off point for ejaculation time. However, clinical and epidemiological evidence suggests cut-off points for ejaculation times (Waldinger and Schweitzer 2019), and this change may be considered a potential source of mistakes in clinical applications and research. Yet, a threshold mainly based on the time of ejaculation may interfere with the access to care in some cases.
With regards to sexual pain, which is grouped separately, “Vaginismus” is namely replaced with “Sexual Pain-Penetration Disorder,” which resembles the name of the diagnostic category in DSM-5, “Genito-pelvic Pain/Penetration Disorder.” However, ICD-11 did not incorporate “Dyspareunia” and “Vulvodynia” into “Sexual Pain-Penetration Disorder”; instead, they are retained in the chapter on genitourinary disorders. This was based on the differences in etiologies, affected populations, and, most importantly, differences in the clinical management and sexual therapeutic approach (Reed et al. 2016). The clinical description of “Sexual Pain-Penetration Disorder” is broadened compared to “Genito-pelvic Pain/Penetration Disorder” so that it includes the emotional components, fear and anxiety, and the pain associated with the penetration. Vaginal spasm, increase in muscular tonus and contraction are only components of a general emotional response (van der Velde et al. 2001), they are often prioritized in the clinical assessment and interventions. In addition to the fact that the emotional and cognitive features are the primary features of the condition, there is a high rate of mental disorder comorbity in this population (Yildirim et al 2019). Such a change in the description of the condition in ICD-11 may enhance a more comprehensive approach not disregarding the psychological component of the sexual problems.
Diagnostic Categories Related to Sexual Orientation
Diversity in sexual orientation has not been considered to be associated with any mental disorder, and nonheterosexual orientation has not been considered a pathology for decades (Drescher 2015). However, in ICD-10, some diagnostic categories related to sexual orientation were kept for then assumed clinical and research benefits (World Health Organization 1992). Nevertheless, in the following decades, these diagnoses were almost never used for research, and their clinical application has not been proven to be practical (Cochran et al. 2014, Reed et al. 2016). The category “Ego-dystonic Sexual Orientation” was already criticized for only being applicable to nonheterosexual orientations (Drescher 2015).
The distress described in the “Sexual Maturation Disorder” regarding the uncertainty of sexual or gender identity has its origin in the prevalent discriminatory attitudes in society. An identity exploration with the possibility of identity features which may be unacceptable for others or beyond the widely expected boundaries may lead to anxiety and fear. Prevalent discrimination against sexual orientations has been shown to delay self-identification and disclosure, and increase the stress associated with the identity features (Campbell et al. 2023, Layland et al. 2023). Furthermore, rather than meriting a distinct category, the discomfort associated with the sexual identity features was shown to be a feature of earlier stages of identity development in individuals with characteristics considered unfavorable by society (Cass 1979, Troiden 1989, Bishop et al. 2023). In other words, the individual’s concern during identity exploration has been dominantly proposed to arise from the experienced or anticipated stigma, exclusion, discrimination, and violence associated with the orientation (Hatzenbuehler and Pachankis 2016). The association of identity features in an essentialist, therefore pathologizing manner to the diagnostic categories was heldresponsible for the stigma and discrimination associated with minority identities, at least partially (Winter et al. 2009, Drescher 2015). Psychological and behavioral problems associated with both conditions have been consistently shown to be related to the discrimination and stigma experienced (Meyer 2003, Hendricks ve Testa 2012). Finally, the disturbance in intimate relationships arising due to identity features was not considered a distinct diagnostic category, leading to the removal of the “Sexual Relationship Disorders.” Therefore, removing these categories was justified due to poor support for their diagnostic validity and clinical utility (Campbell et al. 2015), in addition to the current principles of human rights.
Although the categories removed in the last version were not by themselves responsible for the pathologizing medical approach to nonheterosexual orientations, this change emphasizes the World Medical Association’s stance recognizing the diversity in human sexuality and condemning discrimination, stigmatization, and criminalization (World Medical Association, 2023). Despite increasing evidence on the harmful effects and strong arguments on their unethical nature (Fish and Russell 2020; Davison and Walden 2024), sexual orientation change efforts still persist, and often, medical professionals are involved (Fenaughty et al 2023). Although an increasingly positive attitude in the general population towards diversity in sexual orientation is reported (Jarasiunaite-Fedosejeva and Kravcenko 2022), the stigma and discrimination persist globally, and they are still considered responsible for the health disparities in sexual minorities (Pachankis et al 2021). In addition to diminishing medical false justifications for the stigma, discrimination, and criminalization, this revision can be considered a call for ethical and evidence-based care for all.
Gender Identity and Expression Incongruent with Sex Assigned at Birth
Persistent incongruence between the sex assigned at birth and gender identity is classified as “Gender Incongruence” in the newly formed chapter on “Conditions Related to Sexual Health” instead of “Mental, Behavioral and Neurodevelopmental Disorders”. This decision reflects the current non-pathologizing understanding of gender diversity (Drescher 2015). Gender identity-related diagnostic categories have long been criticized for underlying the stigmatization experienced by trans and gender-diverse individuals (Drescher et al. 2012, Winter et al. 2016). The demand by the supporters of this view was the total removal of the diagnostic category from the DSM, which was not the case in the final revision, DSM-5 (American Psychiatric Association 2013). Instead, during the revision of the DSM, the name of the category, which was Gender Identity Disorder, was changed in a way not to include the words “identity” or “disorder.” However, the diagnostic category of “Gender Dysphoria” is still listed among mental disorders in DSM-5. Similarly, ICD-11 also excluded an identity-related term (“Transsexualism”) from the list of disorders. In ICD-11, taking one more step forward, the newly formed category of “Gender Incongruence” is no longer listed among mental disorders. Gender Incongruence does not necessarily require the existence of distress, which is not an essential feature of the condition. “Gender Incongruence” as described in ICD-11 emphasizes the often present demand for gender-affirmation. This demand for medical assistance is the main reason for retaining the condition as a diagnostic category.
Furthermore, the criteria for this condition were revised to embrace the scientifically documented diversity in the experience of gender identity, which aligns with the change in DSM-5 from the earlier ICD-10 definition which included the term “the opposite sex” and thus followed a strict binary understanding of gender. This change in ICD-11 is expected to influence gender-affirming medical practices, rendering them accessible to people with a broader range of gender experiences than the earlier definition in ICD-10.
A different line of debate concerning the maintenance or removal of a gender identity-related diagnostic category for children resulted in the replacement of the “Gender identity disorder of childhood” in ICD-10 with “Gender incongruence of childhood” in ICD-11. Retaining such a diagnostic category was argued to describe the diversity in gender identity, children’s own experince of exploration and development as a pathology; therefore contributed to the discrimination, and constituted the basis for gender identity change efforts, without evidence (Drescher et al. 2016). However, mainly due to proposed benefits, such as the provision of clinical care and appropriate services to a vulnerable group and the opportunity to develop practice standards and guidelines, the category is retained. Nevertheless, the diagnostic threshold is kept high in order to prevent false labeling, which was one of the counterarguments for retaining the category. The incongruence is required to persist for at least two years, and the diagnosis should not be based solely on gender-variant behavior or preferences. However, the category is still strongly criticized for adding to the gender-minority stress and social rejection of those children who are in the process of exploring their gender identity and expression and learning to cope with the stigma associated with both (Suess Schwend et al. 2018). The opponents of this diagnosis suggest that non-pathologizing codes (such as “Factors influencing health status or contact with health services”) could be used for access to healthcare (Winter et al, 2019). The removal of this category by the ICD-11 Working Group is suggested to be more consistent with the removal of the categories related to the distress experienced during identity formation concerning sexual orientation in nonheterosexual individuals (Cabral et al. 2016).
Finally, the “dual-role transvestism” category in ICD-10 has been discarded. This condition described crossgender dressing (by itself challenging to delineate) as a temporary gender experience limited with clothing, without sexual excitement and demand for gender-affirmation. This diagnosis was almost never employed, and its clinical and research utility was unclear (Reed et al. 2016).
Overall, even though there is still room for improvement in the classification of gender identity-related health issues, the changes introduced by the ICD-11 appear to be in alignment with the most recent recommendations for evidence-based care for trans and gender-diverse individuals (Coleman et al 2022). This change may potentially influence the struggle against persisting gender diversity-related stigma, its consequences and the barriers to access to care (Falck and Bränström 2023). The provision of access to care without the need to define a mental disorder category, the abolition of the requirement for binary identities for gender-affirming care seem to be major advantages that could be widely disseminated via a globally recognized classification system as ICD.
Paraphilic Disorders
In ICD-11, the “Paraphilic Disorders” categories replaced “Disorders of Sexual Preference” in ICD-10 (World Health Organization 2022). “Paraphilic Disorders” are retained in the “Mental, Behavioral, and Neurodevelopmental Disorders Chapter” in ICD-11. They were not moved to the newly formed chapter on sexual health since the assessment and treatment of these conditions were considere to require specialized expertise in mental health (Reed et al. 2016). However, both the conceptualization and the classification of the conditions were significantly modified.
In essence, instead of labeling some sexual preferences and patterns of sexual behavior solely based on their atypicality or the distress associated with the social disapproval, mainly those paraphilias which involve others who can not or do not provide consent or the nature of the paraphilic behavior is associated with a significant risk of injury or death were retained (Krueger et al. 2017). In ICD-11, Pedophilic, Exhibitionistic, Voyeuristic, Frotteuristic, and Coercive Sexual Sadism Disorders are described in this domain. Fetischism, Fetishistic transvestism, and Sadomasochism are not listed as distinct categories. These conditions were considered to include solitary and consensual activities, which are not necessarily associated with functional impairment. Coercive Sexual Sadism Disorder differs from consensual sadomasochism, which is not found to be associated with psychological or social dysfunction (Wismaijer et al. 2013; Brown et al. 2020). This recently formed category is defined by the core feature of sexual arousal being the infliction of physical or psychological suffering on a non-consenting person.
However, diverse paraphilic interests that are sufficient to cause distress in the individual beyond the social consequences can still be diagnosed, even the paraphilias that have been removed as a category. Holding the possibility of diagnosis in the presence of other paraphilias is based on the risk of injury or death to the individual or others. This has been criticized since many other behaviors are not considered mental disorders solely due to the risks associated (Moser 2018). Such a choice to maintain a broader view of paraphilic disorders suggests the persistence of the pathologization of atypical sexual interest, which lies on social rather than scientific background (Giami 2015, Moser and Kleinplatz 2020). Finally, the persistent pattern of sexual arousal clause in the definition of these disorders in ICD-11 maintains the distinction between paraphilic disorders and criminal behaviors, which do not necessarily occur in the context of a mental disorder, and conditions associated with other mental disorders (Reed et al. 2016).
Overall, the changes in this group of disorders were reported to be an improvement in clinical utility and not associated with anticipated medicolegal difficulties in two countries (Abdalla-Filho et al. 2019, Briken et al. 2019). Therefore, the conceptual specification introduced by the ICD-11 may be considered an opportunity to restrict the discriminatory approach to the individuals’ sexual preferences in the assessment of atypical sexual interest and behavior, yet, not limiting the access to medical assistance when required.
Compulsive Sexual Behavior Disorder
ICD-10 included “Excessive Sexual Drive” as a category listed among sexual dysfunctions (World Health Organization 1992). In the last decades, increasing research interest resulted in an accumulation of knowledge in out-of-control sexual behavior, several models of psychopathology, and few attempts to classify among other disorders (Briken 2020). Based on current evidence, ICD-11 included “Compulsive Sexual Behavior Disorder,” and rather than a category of sexual dysfunction, located the condition among impulse control disorders, such as pathological gambling, kleptomania, and pyromania (World Health Organization 2022). The preference of this location is a move away from the behavioral addiction model of the disorder (Sassover and Weinstein 2022). Without an attempt to describe the etiology or motivation underlying the disorder, which is practical in such a heterogenous group, ICD-11 emphasizes the “lack of self-control,” which reflects on the sexual behavior of the individual. The emphasis on self-control in the definition is also helpful in the clinical management of the condition. However, debate, research, and interest in the nature and treatment of the disorder still continue. Including this category in ICD-11 appears to increase research interest (Stein et al. 2020). In addition, it is expected to facilitate individuals’ application and access to care.
CONCLUSION
Diagnostic classifications are expected to evolve with accumulating evidence on the nature of health-related conditions. The innovations introduced in ICD as a result of years of work by experts in the field reflect the progress in the knowledge on sexual health and dysfunction, in addition to the change in the human rights perspective related to diagnostic categories discussed. However, many issues concerning the categories, their description, and their location in diagnostic groups are matters of discussion. ICD-11 is anticipated to facilitate research in these areas, making it possible to answer essential questions in the future and improve clinical care and public health services.
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4474 | https://www.mathsisfun.com/data/standard-normal-distribution.html | Normal Distribution
Data can be "distributed" (spread out) in different ways.
| | | |
---
| It can be spread out more on the left | | Or more on the right |
| |
| | | |
| |
| Or it can be all jumbled up |
|
But in many cases the data tends to be around a central value, with no bias left or right, and it gets close to a "Normal Distribution" like this:
The blue curve is a Normal Distribution.The yellow histogram shows some data that follows it closely, but not perfectly (which is usual).
| |
| It is often called a "Bell Curve" because it looks like a bell. |
Many things closely follow a Normal Distribution:
heights of people
size of manufactured products
measurement errors
blood pressure
test scores
We say the data is "normally distributed":
The Normal Distribution has:
mean = median = mode
symmetry about the center
50% of values less than the meanand 50% greater than the mean
Quincunx
| | |
--- |
| You can see a normal distribution being created by random chance! It is called the Quincunx and it is an amazing machine. Have a play with it! | |
Standard Deviations
The Standard Deviation is a measure of how spread out numbers are (read that page for details on how to calculate it).
When we calculate the standard deviation we find that approximately:
| |
| 68% of values are within 1 standard deviation of the mean 95% of values are within 2 standard deviations of the mean 99.7% of values are within 3 standard deviations of the mean |
Example: 95% of students at school are between 1.1m and 1.7m tall.
Assuming this data is normally distributed can you calculate the mean and standard deviation?
The mean is halfway between 1.1m and 1.7m:
Mean = (1.1m + 1.7m) / 2 = 1.4m
95% is 2 standard deviations either side of the mean (a total of 4 standard deviations) so:
| | |
--- |
| 1 standard deviation | = (1.7m-1.1m) / 4 |
| | = 0.6m / 4 |
| | = 0.15m |
And this is the result:
It is good to know the standard deviation, because we can say that any value is:
likely to be within 1 standard deviation (68 out of 100 should be)
very likely to be within 2 standard deviations (95 out of 100 should be)
almost certainly within 3 standard deviations (997 out of 1000 should be)
Standard Scores
The number of standard deviations from the mean is also called the "Standard Score", "sigma" or "z-score". Get used to those words!
Example: In that same school one of your friends is 1.85m tall
You can see on the bell curve that 1.85m is 3 standard deviations from the mean of 1.4, so:
Your friend's height has a "z-score" of 3.0
It is also possible to calculate how many standard deviations 1.85 is from the mean
How far is 1.85 from the mean?
It is 1.85 - 1.4 = 0.45m from the mean
How many standard deviations is that? The standard deviation is 0.15m, so:
0.45m / 0.15m = 3 standard deviations
To convert a value to a Standard Score ("z-score"):
first subtract the mean,
then divide by the Standard Deviation
And doing that is called "Standardizing":
We can take any Normal Distribution and convert it to The Standard Normal Distribution.
Example: Travel Time
A survey of daily travel time had these results (in minutes):
26, 33, 65, 28, 34, 55, 25, 44, 50, 36, 26, 37, 43, 62, 35, 38, 45, 32, 28, 34
The Mean is 38.8 minutes, and the Standard Deviation is 11.4 minutes (you can copy and paste the values into the Standard Deviation Calculator if you want).
Convert the values to z-scores ("standard scores").
To convert 26:
first subtract the mean: 26 − 38.8 = −12.8,
then divide by the Standard Deviation: −12.8/11.4 = −1.12
So 26 is −1.12 Standard Deviations from the Mean
Here are the first three conversions
| | | |
---
| Original Value | Calculation | Standard Score (z-score) |
| 26 | (26-38.8) / 11.4 = | −1.12 |
| 33 | (33-38.8) / 11.4 = | −0.51 |
| 65 | (65-38.8) / 11.4 = | +2.30 |
| ... | ... | ... |
And here they are graphically:
You can calculate the rest of the z-scores yourself!
The z-score formula that we have been using is:
z = x − μσ
z is the "z-score" (Standard Score)
x is the value to be standardized
μ ('mu") is the mean
σ ("sigma") is the standard deviation
And this is how to use it:
Example: Travel Time (continued)
Here are the first three conversions using the "z-score formula":
z = x − μσ
μ = 38.8
σ = 11.4
| x | x − μσ | z (z-score) |
---
| 26 | 26 − 38.811.4 | = −1.12 |
| 33 | 33 − 38.811.4 | = −0.51 |
| 65 | 65 − 38.811.4 | = +2.30 |
| ... | ... | ... |
The exact calculations we did before, just following the formula.
Why Standardize ... ?
It can help us make decisions about our data.
Example: Professor Willoughby is marking a test.
Here are the students' results (out of 60 points):
20, 15, 26, 32, 18, 28, 35, 14, 26, 22, 17
Most students didn't even get 30 out of 60, and most will fail.
The test must have been really hard, so the Prof decides to Standardize all the scores and only fail people more than 1 standard deviation below the mean.
The Mean is 23, and the Standard Deviation is 6.6, and these are the Standard Scores:
-0.45, -1.21, 0.45, 1.36, -0.76, 0.76, 1.82, -1.36, 0.45, -0.15, -0.91
Now only 2 students will fail (the ones lower than −1 standard deviation)
Much fairer!
It also makes life easier because we only need one table (the Standard Normal Distribution Table), rather than doing calculations individually for each value of mean and standard deviation.
In More Detail
Here is the Standard Normal Distribution with percentages for every half of a standard deviation, and cumulative percentages:
Example: Your score in a recent test was 0.5 standard deviations above the average, how many people scored lower than you did?
Between 0 and 0.5 is 19.1%
Less than 0 is 50% (left half of the curve)
So the total less than you is:
50% + 19.1% = 69.1%
In theory 69.1% scored less than you did (but with real data the percentage may be different)
A Practical Example: Your company packages sugar in 1 kg bags.
When you weigh a sample of bags you get these results:
1007g, 1032g, 1002g, 983g, 1004g, ... (a hundred measurements)
Mean = 1010g
Standard Deviation = 20g
Some values are less than 1000g ... can you fix that?
The normal distribution of your measurements looks like this:
31% of the bags are less than 1000g,which is cheating the customer!
It is a random thing, so we can't stop bags having less than 1000g, but we can try to reduce it a lot.
Let's adjust the machine so that 1000g is ...
... at −3 standard deviations:
From the big bell curve above we see that only 0.1% are less. But maybe that is too small.
... at −2.5 standard deviations:
Below 3 is 0.1% and between 3 and 2.5 standard deviations is 0.5%, together that is 0.1% + 0.5% = 0.6% (a good choice I think)
So let us adjust the machine to have 1000g at −2.5 standard deviations from the mean.
Now, we can adjust it to:
increase the amount of sugar in each bag (which changes the mean), or
make it more accurate (which reduces the standard deviation)
Let us try both.
Adjust the mean amount in each bag
The standard deviation is 20g, and we need 2.5 of them:
2.5 × 20g = 50g
So the machine should average 1050g, like this:
Adjust the accuracy of the machine
Or we can keep the same mean (of 1010g), but then we need 2.5 standard deviations to be equal to 10g:
10g / 2.5 = 4g
So the standard deviation should be 4g, like this:
(We hope the machine is that accurate!)
Or perhaps we could have some combination of better accuracy and slightly larger average size, I will leave that up to you!
More Accurate Values ...
Use the Standard Normal Distribution Table when you want more accurate values.
2619, 2620, 2621, 2622, 2623, 2624, 2625, 2626, 3844, 3845
Standard Deviation Standard Deviation Calculator Standard Normal Distribution Table Quincunx Probability and Statistics Index
Copyright © 2025 Rod Pierce |
4475 | https://esajournals.onlinelibrary.wiley.com/doi/10.2307/1930126 | The Trophic‐Dynamic Aspect of Ecology - Lindeman - 1942 - Ecology - Wiley Online Library
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Volume 23, Issue 4 pp. 399-417
Article
The Trophic-Dynamic Aspect of Ecology
Raymond L. Lindeman,
Raymond L. Lindeman
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Raymond L. Lindeman,
Raymond L. Lindeman
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First published: 01 October 1942
Citations: 2,352
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Volume 23, Issue 4
October 1942
Pages 399-417
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4476 | https://www.quora.com/What-would-the-exact-solutions-in-the-interval-0-2%CF%80-for-this-equation-cos-2x-cos-x-0 | What would the exact solutions in the interval [0,2π) for this equation, cos(2x)-cos(x) =0? - Quora
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5
What would the exact solutions in the interval [0,2π) for this equation, cos(2x)-cos(x) =0?
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Vishal Chandratreya
achieved proficiency in trigonometry · Author has 931 answers and 2.9M answer views
·5y
What would the exact solutions in the interval [0,2 π)[0,2 π) be for this equation?
cos 2 x−cos x=0 cos2 x−cosx=0
The equation can be rewritten as follows.
cos 2 x=cos x cos2 x=cosx
This is a standard trigonometric equation
cos θ=cos ϕ cosθ=cosϕ
and its solution is well known.
cos θ=cos ϕ cosθ=cosϕ
⟹θ=2 n π±ϕ,n∈Z⟹θ=2 n π±ϕ,n∈Z
Apply the same to your equation.
cos 2 x=cos x cos2 x=cosx
⟹2 x=2 n π±x⟹2 x=2 n π±x
Choose the upper sign, and you get this.
2 x=2 n π+x⟹x=2 n π 2 x=2 n π+x⟹x=2 n π
Choose the lower sign, and you get this.
2 x=2 n π−x⟹x=2 n π 3 2 x=2 n π−x⟹x=2 n π 3
Hence, the solution set is the following.
x\in\left\lbrace\left.2n\pi,\dfrac{2n\pi}{3}\right|n\in\Z\ri x\in\left\lbrace\left.2n\pi,\dfrac{2n\pi}{3}\right|n\in\Z\ri
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What would the exact solutions in the interval [0,2 π)[0,2 π) be for this equation?
cos 2 x−cos x=0 cos2 x−cosx=0
The equation can be rewritten as follows.
cos 2 x=cos x cos2 x=cosx
This is a standard trigonometric equation
cos θ=cos ϕ cosθ=cosϕ
and its solution is well known.
cos θ=cos ϕ cosθ=cosϕ
⟹θ=2 n π±ϕ,n∈Z⟹θ=2 n π±ϕ,n∈Z
Apply the same to your equation.
cos 2 x=cos x cos2 x=cosx
⟹2 x=2 n π±x⟹2 x=2 n π±x
Choose the upper sign, and you get this.
2 x=2 n π+x⟹x=2 n π 2 x=2 n π+x⟹x=2 n π
Choose the lower sign, and you get this.
2 x=2 n π−x⟹x=2 n π 3 2 x=2 n π−x⟹x=2 n π 3
Hence, the solution set is the following.
x∈{2 n π,2 n π 3∣∣∣n∈Z}x∈{2 n π,2 n π 3|n∈Z}
In [0,2 π)[0,2 π), the only solutions are the following.
x∈{0,2 π 3,4 π 3}x∈{0,2 π 3,4 π 3}
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I usually look at three things when I’m buying work shoes: comfort, cushioning and arch support; how sturdy the sole is; and whether I can actually afford to get more than one pair if I want them in different colours.
Ballerinas by Bata though, are what I wear the most. I didn’t know about them until recently, when a coworker recommended them to me, also spotted my favorite creator Siddhi Karwa styling them across Europe and I have been absolutely loving them.
They’re professional enough for work wear but don’t feel heavy and keep me comfortable throughout the day, even when I’m commuting to the office. I got mine for around ₹999 from Bata, which felt like a steal compared to some other brands I looked at. They’ve held up really well, and I can easily pair them with trousers, skirts for my work outfits. If you’re on a budget but still want something that is comfortable and follows fashion trends, Ballerinas by Bata are the perfect choice. I picked up mine from a Bata store near me, you can grab yours too.
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[What are the solutions of the equation 〖cos 2x+ cos〗〖x+1=0〗 in the interval 0,2π)?
How many solutions does the equation c o s(s i n x)=1/2 c o s(s i n x)=1/2 has in [0, 2π]?
[What is the solution in the interval 0,2π) of 2cos² x+3cosx-2=0?
[What are all solutions to the equation sin(x) = 1/2 on the interval 0, 2π)?
What is the set of numbers x in (0, 2π) such that log log (sin x + cos x) ∪ (3π2 , 2π)?
James Gere
Author has 1.6K answers and 1.3M answer views
·5y
Write,cos(2 x)=2∙cos 2(x)−1 cos(2 x)=2•cos 2(x)−1.
Then, let, c = cos(x) .
This models the original equation as a quadratic in “c” :
2∙c 2−c−1=0 2•c 2−c−1=0.
The solutions of this quadratic are:c=1±√1+4∙2 4=1±3 4 c=1±1+4•2 4=1±3 4.
So, whenever, cos(x) is 1 or -1/2, the original equation may have a solution.
Cos(x) =1 only for x=0 in the interval, and indeed, this checks in the original equation, 1–1 =0
Cos(x) = -½ for two values of x in the interval, ⅔π and 4π/3 . ( Multiples of 120°).
At, x= ⅔π , 2x = 4π/3, so cos(2x) and cos(x) are both -1/2 and the equation gives: -1/2 -- -1/2
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Write,cos(2 x)=2∙cos 2(x)−1 cos(2 x)=2•cos 2(x)−1.
Then, let, c = cos(x) .
This models the original equation as a quadratic in “c” :
2∙c 2−c−1=0 2•c 2−c−1=0.
The solutions of this quadratic are:c=1±√1+4∙2 4=1±3 4 c=1±1+4•2 4=1±3 4.
So, whenever, cos(x) is 1 or -1/2, the original equation may have a solution.
Cos(x) =1 only for x=0 in the interval, and indeed, this checks in the original equation, 1–1 =0
Cos(x) = -½ for two values of x in the interval, ⅔π and 4π/3 . ( Multiples of 120°).
At, x= ⅔π , 2x = 4π/3, so cos(2x) and cos(x) are both -1/2 and the equation gives: -1/2 -- -1/2 = 0, so this checks also.
At, x= 4π/3 , 2x = 8π/3. Subtracting one period of the cosine function from 2x gives 2π/3 , so again, both values of the cosine function are -1/2, and we have a winner!
Therefore, the exact solutions are x∈{0,2 π 3,4 π 3}x∈{0,2 π 3,4 π 3}, or,
x°∈{0°,120°,240°}∈{0°,120°,240°}.
Upvote ·
Gordon M. Brown
Math Tutor at San Diego City College (2018-Present) · Author has 6.2K answers and 4.3M answer views
·May 6
Related
How do you solve the equation cosx-2cos2x = 0 (0 ≤ x ≤ π). x=?
First of all, your equation is not written clearly. By the second term on the left, I presume you mean -2cos(2x), and not -2cos^2(x). Use parentheses where required to clarify the meaning of your question!
Solving this equation is a matter of looking in your notes for a trigonometric identity to substitute for the expression cos(2x). You will likely also have to “think outside the box” a bit, by occasionally using the quadratic formula (or completing the square) to solve a trig equation in quadratic form.
Lastly, it appears that your instructor has simplified this task considerably, by requestin
Continue Reading
First of all, your equation is not written clearly. By the second term on the left, I presume you mean -2cos(2x), and not -2cos^2(x). Use parentheses where required to clarify the meaning of your question!
Solving this equation is a matter of looking in your notes for a trigonometric identity to substitute for the expression cos(2x). You will likely also have to “think outside the box” a bit, by occasionally using the quadratic formula (or completing the square) to solve a trig equation in quadratic form.
Lastly, it appears that your instructor has simplified this task considerably, by requesting only solutions from 0 ≤ x < π, which can be had fairly easily using the arccosine function (the range of which is restricted to 0 ≤ x < π) on your calculator. However, you should also be able to calculate the reflections of these angles about the x-axis, as shown below. What’s more, you should get used to the concept of providing a general solution, not just within the domains of 0 ≤ x < π, or even 0 ≤ x < 2π, but for the entire domain of real numbers.
One final word, if I may. You have requested my answers quite a lot of late, and you appear to post a lot of homework questions on your profile. I think you need to cultivate the dispositions to think for yourself, and to do your own homework.
Upvote ·
9 5
Sohel Zibara
Studied at Doctor of Philosophy Degrees (Graduated 2000) · Author has 5.1K answers and 2.6M answer views
·3y
Related
How do you solve the equation for exact solutions over the interval [0, 2π]. 2–√cos(2 x)=−1 2–cos(2 x)=−1?
Upvote ·
9 6
9 2
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How do I find all the values of x in between the range [0,2π] for the equation 2 sin (4x) - 3 cos (4x) =0?
How do you solve the equation for exact solutions over the interval [0, 2π]. 2–√cos(2 x)=−1 2–cos(2 x)=−1?
What is the range of cos 2x + 3 cos x?
How do I find the sum of all the solutions of the equation cos(x) cos (60°+x) cos (60°-x) =cos(3x) =1/4, x lies in the interval [0,6π]?
What are the numbers of solutions for this equation, sin x + x = π/2 in [0, 2π]?
Dodie Cowan
Former Professor of Mathematics at Polk State College (2004–2015) · Author has 2.2K answers and 1.3M answer views
·5y
We will use the double angle formula to rewrite cos(2x) as 2cos^2x–1, thus giving the quadratic equation 2cos^2x–cosx-1=0. This becomes
(2cosx+1)(cosx–1)=0
By the zero product property, we can say, then, that 2cosx+1=0 or cosx–1=0. So cos x=-1/2 or 1. In the given interval, this could occur when x=0, 2π/3, and 4π/3.
Upvote ·
Richard Gill
MA in Mathematics&Statistics (academic discipline), University of Cambridge, England (Graduated 1974) · Author has 398 answers and 290.5K answer views
·5y
cos(2x) = 2 cos(x)^2 -1
I suggest you define c = cos(x) and determine the solutions, if any, of 2 c^2 - c - 1 = 0
You could also try drawing graphs of cos(2x) and cos(x) and getting an idea what is going on here. After you “see” the answer with your eyes, you will be able to figure out a formal proof with words and formulas.
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David Goldsmith
Bachelors in Math from Brown, Masters in Math from Univ. of Montana · Author has 1.3K answers and 990K answer views
·5y
If your graphing calculator app (you do have a GC app, yes? If not, try EduCalc or GeoGebra, as free examples) has a mode which automatically uses multiples of π π to hash out the abscisa (and even if it doesn't, if you understand that the most likely exact solutions of an equation like this are rational multiples of π π) then the graph of the function can be very "suggestive":
Do you see that this graph suggests that, before you do any algebra, you check to see if 2 π 3,4 π 3,2 π 3,4 π 3, and of course 0, are the solutions sought?
which is what this graph was generated with.
Continue Reading
If your graphing calculator app (you do have a GC app, yes? If not, try EduCalc or GeoGebra, as free examples) has a mode which automatically uses multiples of π π to hash out the abscisa (and even if it doesn't, if you understand that the most likely exact solutions of an equation like this are rational multiples of π π) then the graph of the function can be very "suggestive":
Do you see that this graph suggests that, before you do any algebra, you check to see if 2 π 3,4 π 3,2 π 3,4 π 3, and of course 0, are the solutions sought?
which is what this graph was generated with.
Upvote ·
Alex Sadovsky
Studied Mathematics&Biomechanics at University of California, Irvine · Author has 13.4K answers and 7.8M answer views
·5y
Using the cosine double angle formula, we have:
0=cos(2 x)−cos(x)=2 cos 2(x)−1−cos(x)0=cos(2 x)−cos(x)=2 cos 2(x)−1−cos(x).
How let u=cos(x)u=cos(x). Our equation becomes the quadratic
0=2 u 2−u−1 0=2 u 2−u−1.
Solve it for u u (you will get at most two roots), then
solve
cos(x)=u cos(x)=u
for x x.
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Edi Thithit
Author has 129 answers and 73.6K answer views
·5y
cos 2x - cos x = 0 <＝> 2cos²x-cosx-1=0 <＝> (2cosx+1) (cosx-1) = 0 ＝> cosx = -1/2 ∨ cosx=1
cos x =-1/2 <＝> x=2∏/3 ; x=4∏/3
cos x =1 <＝> x=0
(2∏ is not included in the interval [0,2∏).)
Upvote ·
Leo Harten
BS, MS in Physics&Mathematics, Massachusetts Institute of Technology (Graduated 1977) · Author has 3.9K answers and 2.3M answer views
·5y
Use cos(2x)=2cos(x)^2–1 and you now have a quadratic eqn in cos(x):
2cos(x)^2–1-cos(x)=0 so cos(x)={1,-1/2}. On the specified interval, cos(x)=1 is x=0 since 2%pi is excluded. And cos(x)=-1/2 is x=2%pi/3 and x=4%pi/3, so we have x={0,2%pi/3,4%pi/3}. Each verifies.
Upvote ·
Dean Rubine
Been doing high school math since high school, circa 1975 · Author has 10.6K answers and 23.7M answer views
·1y
Related
How do I solve this equation (answer in degrees) : 2 sin x + cos x=0?
2 sin x=−cos x 2 sinx=−cosx
sin x cos x=−1 2 sinx cosx=−1 2
tan x=−1 2 tanx=−1 2
x=−arctan 1 2+180∘k,x=−arctan1 2+180∘k, integer k k
arctan 1 2 arctan1 2 is an important angle that our trigonometry obscures; it’s essentially impossible to write down the exact measure in radians or degrees beyond what we’ve written. It’s a baby angle, the small angle formed by the diagonal of the 1x2 rectangle. It’s actually half the large acute angle of the 3/4/5 right triangle, because (2+i)2=3+4 i.(2+i)2=3+4 i. The AP Physics test cringingly gives that angle arctan 4 3 arctan4 3 as 53∘;53∘; I’ll add an image to reach more people.
S
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2 sin x=−cos x 2 sinx=−cosx
sin x cos x=−1 2 sinx cosx=−1 2
tan x=−1 2 tanx=−1 2
x=−arctan 1 2+180∘k,x=−arctan1 2+180∘k, integer k k
arctan 1 2 arctan1 2 is an important angle that our trigonometry obscures; it’s essentially impossible to write down the exact measure in radians or degrees beyond what we’ve written. It’s a baby angle, the small angle formed by the diagonal of the 1x2 rectangle. It’s actually half the large acute angle of the 3/4/5 right triangle, because (2+i)2=3+4 i.(2+i)2=3+4 i. The AP Physics test cringingly gives that angle arctan 4 3 arctan4 3 as 53∘;53∘; I’ll add an image to reach more people.
So we can say:
Answer: x≈−26.5∘+180∘k x≈−26.5∘+180∘k, for any integer k k
Upvote ·
9 8
John K WilliamsSon
Accredited (MS Educ) nerd who loves talking about math · Author has 9K answers and 23.4M answer views
·3y
Related
How do you solve the equation for exact solutions over the interval [0, 2π]. 2–√cos(2 x)=−1 2–cos(2 x)=−1?
Look very carefully at your equation:
2–√cos(2 x)=−1 2–cos(2 x)=−1
What is the largest possible value of c o s(2 x)c o s(2 x)?
The range of the cosine function is [-1, 1], so the largest possible value of c o s(2 x)c o s(2 x) is 1.
Can 2–√cos(2 x)=−1 2–cos(2 x)=−1 ever be true? 2 minus the largest possible value of c o s(2 x)c o s(2 x) is equal to 1. It can never be equal to -1.
This eq...
Upvote ·
9 1
Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983) · Author has 3.9K answers and 4M answer views
·2y
Related
How do I solve the equation cos^2 x - sin^2x = sin x for 0 deg <= x <= 360?
cos 2(x)−sin 2(x)=sin(x)cos 2(x)−sin 2(x)=sin(x)
[1−sin 2(x)]−sin 2(x)−sin(x)=0[1−sin 2(x)]−sin 2(x)−sin(x)=0
1−2 sin 2(x)−sin(x)=0 1−2 sin 2(x)−sin(x)=0
2 sin 2(x)+sin(x)−1=0 2 sin 2(x)+sin(x)−1=0
(2 sin(x)−1)(sin(x)+1)=0(2 sin(x)−1)(sin(x)+1)=0
Either:
sin(x)+1=0 sin(x)+1=0
sin(x)=−1 sin(x)=−1
x=270∘+360∘n x=270∘+360∘n
Or:
2 sin(x)−1=0 2 sin(x)−1=0
sin(x)=1 2 sin(x)=1 2
x=30∘+360∘n x=30∘+360∘n
x=150∘+360∘n x=150∘+360∘n
where n n is any integer.
Therefore, the solutions within the given interval, 0∘≤x≤360∘0∘≤x≤360∘, are:
x=30∘x=30∘, 150∘150∘, 270∘270∘
A plot of the function:
y=cos 2(x)−sin 2(x)−sin(x)y=cos 2(x)−sin 2(x)−sin(x)
whose roots solve the given equation, looks like this:
Continue Reading
cos 2(x)−sin 2(x)=sin(x)cos 2(x)−sin 2(x)=sin(x)
[1−sin 2(x)]−sin 2(x)−sin(x)=0[1−sin 2(x)]−sin 2(x)−sin(x)=0
1−2 sin 2(x)−sin(x)=0 1−2 sin 2(x)−sin(x)=0
2 sin 2(x)+sin(x)−1=0 2 sin 2(x)+sin(x)−1=0
(2 sin(x)−1)(sin(x)+1)=0(2 sin(x)−1)(sin(x)+1)=0
Either:
sin(x)+1=0 sin(x)+1=0
sin(x)=−1 sin(x)=−1
x=270∘+360∘n x=270∘+360∘n
Or:
2 sin(x)−1=0 2 sin(x)−1=0
sin(x)=1 2 sin(x)=1 2
x=30∘+360∘n x=30∘+360∘n
x=150∘+360∘n x=150∘+360∘n
where n n is any integer.
Therefore, the solutions within the given interval, 0∘≤x≤360∘0∘≤x≤360∘, are:
x=30∘x=30∘, 150∘150∘, 270∘270∘
A plot of the function:
y=cos 2(x)−sin 2(x)−sin(x)y=cos 2(x)−sin 2(x)−sin(x)
whose roots solve the given equation, looks like this:
Upvote ·
9 8
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4477 | https://math.meta.stackexchange.com/questions/26187/which-if-any-inequalities-with-real-numbers-should-have-separate-tags | Which (if any) inequalities with real numbers should have separate tags? - Mathematics Meta Stack Exchange
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Which (if any) inequalities with real numbers should have separate tags?
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Asked 8 years, 5 months ago
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There are certainly many inequalities which are rather important and useful and which appear frequently in various areas in mathematics (AM-GM, Jensen, Cauchy-Schwarz, etc.) The question I want to ask is whether some of them would be also useful as tags on this site. And if yes, for which of them tags should be created.
To avoid making this question too general, let's deal just with the inequalities concerning real numbers. So for the purpose of this question let us leave inequalities from probability (such as Chebyshev's inequality), analysis (Harnack's inequality, Grönwall's inequality, etc.) and various other areas aside.
If we decide that some the inequalities should have their own tags, we should also discuss when these tags are supposed to be used. Are they only for the questions about these inequalities? Should they be added to the questions where the OP explicitly says that they want to see solution using this particular inequality? Or if some an answer is given which uses some inequality, should the tag be added based on the answer?
One more thing to keep in mind is that question can have at most five tags. So if we create too many very specific tags, it might happen that on some question we will not have enough space to add all tags which might be suitable there.
On the other hand, I can see that inequalities tag might start to resemble a big monolith of questions which are not divided into subcategories, which make questions about inequalities rather difficult to search. (Although I am not sure to which extent they can be reasonably categorized. And whether tags could actually improve searching among these question, or whether it is simpler to search for the exact formula.) It would be great if some of the users who are active in the inequality tag could comment on whether they think that inequalities tag needs to become more organized.
The reason why I have decided to post the question now was that not so long ago two tags of this nature have been created. Namely a.m.-g.m.-inequality was created in January and young-inequalitycreated very recently. (Another similar tag which existed, although only for a very brief period, was cauchy-schwarz.)
In both cases the tags start growing relatively fast. Therefore I considered asking about community opinion as a reasonable thing to do. If the community consensus is that those tags should be removed, it is better to find out before the tags contain too many questions and removing the tags is a lot of work. (Although moderators can remove tags without bumping - see here and here - but this only works if there is another tag where all currently tagged question would fit. Possibly in the case of these tags, inequalities might be a good fit for most questions, so it is possible that in this case we will not have to do manual retagging of all questions if the tags are removed.)
I have previously asked about a.m.-g.m.-inequality in the tag management thread. But since some other similar tags started appearing, it is probably better to discuss them in general.
EDIT: Recently (end of July 2017) two new inequality-related tags were created, namely cauchy-schwarz-inequality and holder-inequality. The list of questions currently having these tags can be found here.
EDIT2: And the tags rearrangement-inequality, jensen-inequality, convexity-inequality, muirhead-inequality have been created a few days later.
EDIT3: The tags karamata-inequality and tangent-line-method were added today. The latter is a method for proving inequalities rather than a specific inequality - the same is true about the tags mentioned in Jyrki's answer. This kind of moved the question a bit into contest-math.
I guess the above edits show that number of inequality-related tags is still growing - so I will not post further updates when new such tags appear. (I guess the above are sufficient as examples.)
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edited Aug 8, 2017 at 10:39
Martin SleziakMartin Sleziak
asked Apr 14, 2017 at 11:06
Martin SleziakMartin Sleziak
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7 The young-inequality tag is very young you say?Asaf Karagila –Asaf KaragilaMod 2017-04-14 12:44:21 +00:00 Commented Apr 14, 2017 at 12:44
3 @Asaf otherwise it would be holder-ineaquality, wouldn't it?quid –quid 2017-04-14 16:40:53 +00:00 Commented Apr 14, 2017 at 16:40
1 And the downside of creating a new tag is...?user541686 –user541686 2017-04-19 21:49:40 +00:00 Commented Apr 19, 2017 at 21:49
3 @Mehrdad: math.meta.stackexchange.com/questions/23513/…Asaf Karagila –Asaf KaragilaMod 2017-04-22 17:04:40 +00:00 Commented Apr 22, 2017 at 17:04
1 @Mehrdad The answer Asaf has linked to addresses your question. And I have also tried to write something about this here.Martin Sleziak –Martin Sleziak 2017-04-23 06:21:44 +00:00 Commented Apr 23, 2017 at 6:21
1 Holy cow, that's a long answer to such a short question. =P I wasn't trying to ask what the downside is in general; I was trying to ask what the downside is for the inequalities you mention. "Inequalities" is hardly appropriate for something about AM-GM, Cauchy-Schwartz, or Jensen's inequality, and something like "metric spaces" or "convex analysis" would be pretty overkill (though perhaps not entirely inappropriate). My view is that they're all used commonly enough that they would pretty obviously flourish on their own.user541686 –user541686 2017-04-23 07:23:27 +00:00 Commented Apr 23, 2017 at 7:23
1 @Mehrdad So I misunderstood your question. But the general answer also answer the specific question. Too many too specific tags cause problems. So it is good to discuss which inequalities should have their own tag, so that we do not end up with many tags which are not really used. (BTW I think that (inequalities) tag is definitely the most suitable for the questions about AM-GM etc. The question is whether to create additional tags.)Martin Sleziak –Martin Sleziak 2017-04-23 07:28:57 +00:00 Commented Apr 23, 2017 at 7:28
@Mehrdad Perhaps we could say that the question in your comment is very similar to what I am asking in the question. Anyway, if you think that for example the three equations you mention deserve to have a separate tag, you can post that as an answer - this is exactly what I am asking and from reaction of other users to your answer we would see whether they agree with you.) Also, if you wish, we can continue this discussion (general or specific to inequalities) in chat.Martin Sleziak –Martin Sleziak 2017-04-23 07:31:15 +00:00 Commented Apr 23, 2017 at 7:31
@Martin Sleziak Why you did not ask me about these things? Why are you doing this behind my back? I think it's not fair.Michael Rozenberg –Michael Rozenberg 2017-08-07 10:56:33 +00:00 Commented Aug 7, 2017 at 10:56
4 @MichaelRozenberg I do not think that what you are saying is accurate. The whole discussion started by the tags (a.m.-g.m.-inequality) created by Harsh Kumar, and (young-inequality) created by Leila. (I believe I notified Harsh Kumar about my post in tag management thread. So the tag creator was aware of this.) The post was publicly visible here on meta, and it appeared even in community bulletin. Anyway, if needed, we can discuss this in chat.Martin Sleziak –Martin Sleziak 2017-08-07 12:11:42 +00:00 Commented Aug 7, 2017 at 12:11
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I think all of those that you mentioned (AM-GM, Jensen, Cauchy-Schwarz) are just fine for tags.
I would even say Chebyshev's and Markov's inequalities are fine for tagging, as well as even the union bound, though not the other two you mentioned.
My rule of thumb would be that an inequality deserves a tag if:
Mathematically-inclined students would learn it before college (e.g. AM-GM, triangle), or
It is very likely to be taught in advanced undergraduate or introductory graduate courses outside of a math/physic/statistics department (e.g. Jensen's, Cauchy-Schwarz, Chebyshev)
The rationale would be that the audience would likely be large enough in these scenarios that the tags are very likely to flourish and help the question get better views.
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edited Apr 27, 2017 at 0:14
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4 What exactly would be tagged with say triangle-inequality and who should care to follow that tag specifically to see questions like: "Is there a triangle with side-length 3, 11, 29? I believe I need to use the triangle inequality." and "How can I show that the ring of p-adic integers is an ultra-metric space? I know that in an ultrametric space a stronger form of the triangle inequality needs to hold." side by side?quid –quid 2017-04-23 09:51:05 +00:00 Commented Apr 23, 2017 at 9:51
@quid: Same kind of people who presumably already care to follow "convex analysis" and see questions like "Proof that exponential function is convex" and "Legendre Transformation of a Lagrangian in Classical Mechanics" side-by-side?user541686 –user541686 2017-04-23 09:58:20 +00:00 Commented Apr 23, 2017 at 9:58
While this is somewhat similar. on the one hand the situation for triangle-inequality would be more extreme, and on the other hand, I for one would in fact argue that it'd be better if the first type of question would not be tagged convex-analysis. (Likely it also never was actively tagged convex-analysis; alas there are synonyms, which I do not find all that useful, and the wiki somewhat allows that use.) Note the first question initially also was (mis)tagged set-theory, and certainly those that follow set-theory do not follow it for this type of question.quid –quid 2017-04-23 10:11:27 +00:00 Commented Apr 23, 2017 at 10:11
3 To put this differently while there obviously are people that are experts in convex analysis, I have some difficulty to imagine what an expert on the the triangle inequality should be.quid –quid 2017-04-23 10:14:48 +00:00 Commented Apr 23, 2017 at 10:14
@quid: I would expect there to exist experts on triangle inequality about as much as there would on Cauchy-Schwartz, and you seem to be okay with the latter, so...user541686 –user541686 2017-04-23 10:23:52 +00:00 Commented Apr 23, 2017 at 10:23
1 I am not okay with C-S either. I just focused on one example you gave, admittedly the one that I find the worst. By the way there is no T there, just Schwarz.quid –quid 2017-04-23 10:31:17 +00:00 Commented Apr 23, 2017 at 10:31
1 @quid: Whoops, I used to spell it right but sometime in the past few months I started spelling it wrong again. Well, in that case, maybe you should first propose a common ground we can both stand on. Otherwise I'm just wasting my time rebutting against your arguments when you just keep rejecting everything and I have no idea what your stance on anything is. To me it's plain as day that Cauchy-Schwarz and Jensen deserve their own own tags, and the triangle inequality is a corollary...user541686 –user541686 2017-04-23 10:58:39 +00:00 Commented Apr 23, 2017 at 10:58
@quid I know that several users consider this as one of the test whether a tag might be useful: "Can you imagine that somebody ads this tag among favorited or ignored tags?" (I was able to find one mention of this here.) Personally, I would consider a tag useful if it greatly enhances searching for questions, even in the case where nobody would follow it or when it is unclear who would be expert in given tag. (I am mentioning this since you seem to be using this test implicitly.)Martin Sleziak –Martin Sleziak 2017-04-23 11:26:11 +00:00 Commented Apr 23, 2017 at 11:26
As a side note, to me the possible (triangle-inequality) tag seems closer to general topology, metric spaces and normed spaces. So it is slightly outside of the originally suggested scope of the question, which was inequalities for real numbers. (Of course, it is true that Cauchy-Schwarz also make sense in general inner product spaces. But for C-S I would expect more questions about real numbers. If a tag triangle-inequality is created, my expectation is that it will mostly appear in questions related to metric and normed spaces.)Martin Sleziak –Martin Sleziak 2017-04-23 11:28:58 +00:00 Commented Apr 23, 2017 at 11:28
@MartinSleziak: I can totally imagine these being among the favorite or ignored tags. In fact, I would have probably put most of these metric-space inequalities in my own ignored tags if I had been regular enough of a user of Math.SE to have any favorite or ignored tags at all. On the other hand I could see myself putting something like the minimax (aka min-max) inequality in my favorite tags. And yes, I totally agree with you on searchability; that's my main criterion for a tag as well. And tagging something as C-S most definitely would help in that regard.user541686 –user541686 2017-04-23 12:19:30 +00:00 Commented Apr 23, 2017 at 12:19
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Drawing attention to a couple more recently created related tags. Namely
sos, and
uvw.
As far as I can tell those two are methods for proving, for example, olympiad style inequalities.
I have mild misgivings about these tags for they seem to very specific. On the other hand:
The tags may come in handy for someone preparing for a math-contest looking for training material,
The probable creator (notified), to their credit, has written decent tag-wikis. This is a big plus in my eyes, because too often new tags have been created by a well-meaning but misguided badge hunter.
So I guess I am undecided whether these are good tags or not. Posting this partly to draw the creator's attention to this.
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edited Aug 6, 2017 at 18:50
answered Aug 6, 2017 at 13:57
Jyrki LahtonenMod Jyrki Lahtonen
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Re: Posting this partly to draw the creator's attention to this. I've noticed both creation of uvw and also sos and buffalo-way. Of course, I am grateful for letting me know, but it is probably more important what community as a whole thinks about the issue of specific-inequality tags (and some related tags).Martin Sleziak –Martin Sleziak 2017-08-06 19:46:19 +00:00 Commented Aug 6, 2017 at 19:46
1 @MartinSleziak I am not as well informed about our criteria of when to create new tags as you are. A goal was to bring Michael Rozenberg here. To educate him about our ways of clearing tag creations in advance, and also to give him a chance to speak up. Sorry, if I implied that you would not have noticed. I know that you pay attention to tags.Jyrki Lahtonen –Jyrki LahtonenMod 2017-08-06 21:01:58 +00:00 Commented Aug 6, 2017 at 21:01
1 I certainly take that as a compliment, although I am not sure whether anybody know criteria for creating new tags, since different users have different opinion on this. I agree with the point you made elsewhere that it is quite natural that subcommunities around certain tags naturally form on this site. So I think that voice of users who are active in the (inequalities) tag is important here. So I am grateful that you notified Michael Rozenberg of this discussion.Martin Sleziak –Martin Sleziak 2017-08-07 14:42:52 +00:00 Commented Aug 7, 2017 at 14:42
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I want to create a new tag EV-Method, but I want to know before, what Community thinks about it.
This method is very useful for the proof of hard symmetric inequalities with n n variables. I think if our user will want to learn this method he'll can click this tag and see many examples, how to use this method. There is a big problem with examples and by creating of this tag we'll solve this problem.
Without this tag I think it's impossible to find these examples in the tag "inequality".
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answered Aug 7, 2017 at 3:57
Michael RozenbergMichael Rozenberg
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3 Don't have any strong opinion on this (and btw I did not cast any votes in this thread). However, I think it might make sense to insert something related to inequality in the tag, like a.m.-g.m.-inequality or cauchy-schwarz-inequality have now. This would make it easier for both posters and later readers to find the right tag. Just sos or ev-method is a bit too cryptic for anyone who doesn't know already what they are looking for.dxiv –dxiv 2017-08-07 04:37:09 +00:00 Commented Aug 7, 2017 at 4:37
1 To which extent can EV-method be considered standard (i.e., commonly used) name. For example, I do not see this name in Cvetkovski's book.Martin Sleziak –Martin Sleziak 2017-08-08 05:34:00 +00:00 Commented Aug 8, 2017 at 5:34
@Martin Sleziak In my opinion, this is a book for beginners.Michael Rozenberg –Michael Rozenberg 2017-08-08 05:40:36 +00:00 Commented Aug 8, 2017 at 5:40
4 @MichaelRozenberg Ok, nice put-down :-) Still the question remains - is the name EV-method standard? Is it used somewhere?Martin Sleziak –Martin Sleziak 2017-08-08 05:44:26 +00:00 Commented Aug 8, 2017 at 5:44
@Martin Sleziak We can not use this method in IMO without proof, but in IMC it's obligatory! This method found Vasile Cirtoje 10 years ago and it's very usefull.Michael Rozenberg –Michael Rozenberg 2017-08-08 05:53:26 +00:00 Commented Aug 8, 2017 at 5:53
2 To address @dxiv's objection, maybe equal-variable-method or equal-variable-theorem would be a bit more descriptive name, if the tag is going to be created. (After all, ev-method can be added as a synonym.)Martin Sleziak –Martin Sleziak 2017-08-08 06:36:40 +00:00 Commented Aug 8, 2017 at 6:36
@Martin Sleziak I think the first is better.Michael Rozenberg –Michael Rozenberg 2017-08-08 06:37:55 +00:00 Commented Aug 8, 2017 at 6:37
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I think these tags (uvw and sos) are useful for the forum. User, which looks for to learn these methods, can click these tags and see many examples, how he can prove inequalities by these methods.
For example.
Let we need to prove that
a a 2+b c−−−−−−√+b b 2+a c−−−−−−√+c c 2+a b−−−−−−√≥2(a 2+b 2+c 2)(a b+a c+b c)−−−−−−−−−−−−−−−−−−−−−−−√a a 2+b c+b b 2+a c+c c 2+a b≥2(a 2+b 2+c 2)(a b+a c+b c)
for non-negatives a a, b b and c c.
A proof by SOS:
We need to prove that:
∑c y c a a 2+b c−−−−−−√≥2(a 2+b 2+c 2)(a b+a c+b c)−−−−−−−−−−−−−−−−−−−−−−−√∑c y c a a 2+b c≥2(a 2+b 2+c 2)(a b+a c+b c)
or
∑c y c(a 4+a 2 b c+2 a b(a 2+b c)(b 2+a c)−−−−−−−−−−−−−−√)≥∑c y c(2 a 3 b+2 a 3 c+2 a 2 b c)∑c y c(a 4+a 2 b c+2 a b(a 2+b c)(b 2+a c))≥∑c y c(2 a 3 b+2 a 3 c+2 a 2 b c)
or
∑c y c(a 4−a 3 b−a 3 c+a 2 b c)≥∑c y c(a 3 b+a 3 c+2 a 2 b c−2 a b(a 2+b c)(b 2+a c)−−−−−−−−−−−−−−√)∑c y c(a 4−a 3 b−a 3 c+a 2 b c)≥∑c y c(a 3 b+a 3 c+2 a 2 b c−2 a b(a 2+b c)(b 2+a c))
or
1 2∑c y c(a−b)2(a+b−c)2≥∑c y c a b(a 2+b c+b 2+a c−2(a 2+b c)(b 2+a c)−−−−−−−−−−−−−−√)1 2∑c y c(a−b)2(a+b−c)2≥∑c y c a b(a 2+b c+b 2+a c−2(a 2+b c)(b 2+a c))
or
∑c y c(a−b)2(a+b−c)2≥2∑c y c a b(a 2+b c−−−−−−√−b 2+a c−−−−−−√)2∑c y c(a−b)2(a+b−c)2≥2∑c y c a b(a 2+b c−b 2+a c)2
or
∑c y c(a−b)2(a+b−c)2⎛⎝⎜⎜1−2 a b(a 2+b c−−−−−−√+b 2+a c−−−−−−√)2⎞⎠⎟⎟≥0,∑c y c(a−b)2(a+b−c)2(1−2 a b(a 2+b c+b 2+a c)2)≥0,
which is obvious.
Now, try to find another examples for using SOS in the forum. For which it's enough to click the tag SOS and you'll see a full picture. I think it's impossible to see this picture without tag SOS.
Since the uvw method is much more useful than SOS (I think that particularly because of this method, inequalities were removed from IMO in the last six years), this tag also would be very useful for the forum.
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edited Aug 8, 2017 at 9:43
Martin Sleziak
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Michael RozenbergMichael Rozenberg
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4 I haven't voted on this post, but I suspect that the downvotes are due to you not discussing the usefulness of uvw/sos as TAGS rather than as techniques of doing math.Jyrki Lahtonen –Jyrki LahtonenMod 2017-08-07 04:15:16 +00:00 Commented Aug 7, 2017 at 4:15
5 Michael, my advice would be that before creating new tags you should read all of this. I am not the only one who would be willing to grant Martin Sleziak dictatorial powers re tagging. He knows what he's doing, and has earned community respect (not only) in this matter.Jyrki Lahtonen –Jyrki LahtonenMod 2017-08-07 07:01:19 +00:00 Commented Aug 7, 2017 at 7:01
1 To add to Jyrki's nice comments: you might want to read this and this, and then think carefully about whether tagging a question based on the techniques of only one of the answers in it is sensible.J. M. ain't a mathematician –J. M. ain't a mathematician 2017-08-07 07:44:38 +00:00 Commented Aug 7, 2017 at 7:44
@J. M. isn't a mathematician I just read it now and I think it exactly which I am doing, when I am adding or I am creating a new tag. Can you explain me where do you see a problem?Michael Rozenberg –Michael Rozenberg 2017-08-07 08:12:06 +00:00 Commented Aug 7, 2017 at 8:12
1 @MichaelRozenberg I am grateful that you found time to join the discussion - it is good to hear from people active in the inequalities tag (perhaps also contest-math). And also thanks for editing this post. (So far I did not vote on any of the answers here, I'll take some time thinking about these new tags before doing so.)Martin Sleziak –Martin Sleziak 2017-08-07 14:48:23 +00:00 Commented Aug 7, 2017 at 14:48
I haven't said there was a problem, Michael (I haven't voted any of your answers either). I only linked to those as a reminder. If, as you say, you can fully justify these tags to everybody else, then I've no objections.J. M. ain't a mathematician –J. M. ain't a mathematician 2017-08-07 14:57:33 +00:00 Commented Aug 7, 2017 at 14:57
Re: For which it's enough to click the tag SOS and you'll see a full picture. I think it's impossible to see this picture without tag SOS. Certainly, there are many other methods how to create a collection of examples for some methods than using tags.Martin Sleziak –Martin Sleziak 2017-08-08 07:30:23 +00:00 Commented Aug 8, 2017 at 7:30
@Martin Sleziak What do you mean? Give me example.Michael Rozenberg –Michael Rozenberg 2017-08-08 07:31:42 +00:00 Commented Aug 8, 2017 at 7:31
For example, various lists of problems. With additional advantage that they can be better organized and comments can be included. They can be created outside of math.SE, sometimes they are tolerated also here. See, for example, this question and other posts linked there.Martin Sleziak –Martin Sleziak 2017-08-08 07:37:17 +00:00 Commented Aug 8, 2017 at 7:37
@Martin Sleziak I think it's very bad idea for two reasons. 1. Adding question in live it's the best thing! It much better than a book. This thing we have because there is tagging. 2. We can else to make these comments and links. I think it's a big and unnecessary work.Michael Rozenberg –Michael Rozenberg 2017-08-08 07:40:39 +00:00 Commented Aug 8, 2017 at 7:40
@MichaelRozenberg If you wish to discuss question collections in more detail, we can move this to chat - because we digressed from the original topic which is tagging. (Although I am not sure I have much to add to what is written in the posts I linked above.)Martin Sleziak –Martin Sleziak 2017-08-08 09:33:53 +00:00 Commented Aug 8, 2017 at 9:33
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Pronounced Right or Wrong
What word when pronounced wrong is right but when pronounced right is wrong?
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The answer is the word “wrong”. When it’s pronounced “wrong” (rhyming with song) that’s the correct pronunciation. However, if you were to pronounce the word as “right” (rhyming with night) that would be an incorrect pronunciation and be wrong.
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11 Comments on "Pronounced Right or Wrong"
Richard nelson says February 25, 2014 @ 12:46
Thank you…very good
Ryan says July 1, 2014 @ 19:38
which word pronounced wrong is right and pronounced right is wrong? the answer is not “wrong”
Dan says July 1, 2014 @ 20:10
@Ryan I’ve been noodling on this one and haven’t come up with anything. It could be that the answer “wrong” is clouding my thinking. Would you care to share the answer, or provide a hint?
Brooke says April 7, 2015 @ 17:08
Wrong because it is
richard says March 17, 2016 @ 18:54
Richard u my namesake
SAM says April 2, 2016 @ 13:24
The answer is”wrong”
berenise says June 18, 2016 @ 15:14
What has two right but is still wrong?
wrong says May 1, 2017 @ 10:44
wrong is the awnser
Satam sonio says August 13, 2018 @ 10:28
Heads up to you…
Suparno says October 13, 2019 @ 23:57
WRONG
Darshana says March 26, 2020 @ 14:03
Wrong is the answer
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The Anatomical Pathogenesis of Stress Urinary Incontinence in Women
Xunguo Yang
Xingqi Wang
Zhenhua Gao
Ling Li
Han Lin
Haifeng Wang
Hang Zhou
Daoming Tian
Quan Zhang
Jihong Shen
Correspondence: kmsjh99@aliyun.com; Tel.: +86-135-7700-9705
These authors contributed equally to this work.
Roles
Received 2022 Dec 9; Revised 2022 Dec 12; Accepted 2022 Dec 15; Collection date 2023 Jan.
Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
Abstract
Stress urinary incontinence is a common disease in middle-aged and elderly women, which seriously affects the physical and mental health of the patients. For this reason, researchers have carried out a large number of studies on stress urinary incontinence. At present, it is believed that the pathogenesis of the disease is mainly due to changes related to age, childbirth, obesity, constipation and other risk factors that induce changes in the urinary control anatomy, including the anatomical factors of the urethra itself, the anatomical factors around the urethra and the anatomical factors of the pelvic nerve. The combined actions of a variety of factors lead to the occurrence of stress urinary incontinence. This review aims to summarize the anatomical pathogenesis of stress urinary incontinence from the above three perspectives.
Keywords: SUI, anatomy, pathogenesis, female
1. Introduction
Stress urinary incontinence (SUI) refers to the involuntary leakage of urine from the external urethral orifice when abdominal pressure increases due to actions such as sneezing, coughing and laughing. The prevalence of SUI among adult women in China is as high as 18.9% , and it seriously affects the physical and mental health of the patients and also represents a significant medical and economic burden on society [1,2]. To this end, researchers have conducted a large number of studies. Kelly observed an open vesical neck in women with SUI . Enhorning proposed the theory of pressure transmission, suggesting that the intra-abdominal pressure is evenly applied to the bladder and proximal urethra, and noticed that the transmission was reduced in women with SUI . Petros and Ulmsten presented an integral theory explaining why the reconstruction and restoration of urethral support can improve continence [5,6]. Delancey proposed the “hammock theory”, suggesting that the levator ani muscle, the anterior vaginal wall, the pelvic fascia and the pubourethral ligament, together, form a hammock structure to perform the function of urine control . Based on these theories, a number of treatment methods for stress urinary incontinence have been proposed, and certain therapeutic effects have been achieved, but there are still many complications, including a high recurrence rate and other problems. This situation requires us to further explore the anatomic pathogenesis of stress urinary incontinence. Currently, it is believed that the pathogenesis of stress urinary incontinence is mainly due to changes related to age, childbirth, obesity, constipation and other risk factors [8,9] that induce changes in the urinary control anatomical factors, including the anatomical factors of the urethra itself, the anatomical factors of periurethra and the anatomical factors of the pelvic nerve . Finally, stress incontinence occurs under the combined effects of a variety of change factors. Therefore, this review summarizes the anatomical pathogenesis of SUI from the above three perspectives so as to facilitate a better understanding of the occurrence of SUI and provide new ideas for the clinical treatment of SUI.
2. Anatomical Factors of the Urethra Itself
2.1. The Sealing Effect of the Urethral Mucosa Is Weakened
The urethral mucosa has a sealing effect. The normal urethral mucosa can produce mucus-like secretions, which create a strong sealing effect with the assistance of the submucosal blood vessels. The submucosa of the urethra is rich in blood vessels, providing a rich blood supply to the mucous membrane, promoting the proliferation of mucous membrane cells, thickening the mucous membrane, promoting mucous secretion by the mucous membrane cells and increasing the closure performance of the urethral lumen . When the blood vessels in the submucosa are filled, the urethral lumen can become compressed to strengthen the sealing effect of the urethral mucosa . Estrogen can act on the submucosal blood vessels of the urethra, promoting their dilation and filling and causing them to become highly vascularized in suburethra mucosa , thereby finally increasing the blood supply to the mucosa and strengthening the mucosal sealing effect of the urethral mucosa [13,14]. When the estrogen levels are low, the submucosal blood vessels of the urethra are reduced, the mucosal blood supply is insufficient, causing the mucosa atrophies and the urethral mucosal sealing effect to be weakened, and, finally, urinary incontinence is likely to occur . In chronic urethritis, long-term inflammatory stimulation leads to urethral mucosal fibrosis, urethral mucosal atrophy, a reduction in mucus secretion and a reduction in the submucosal blood vessels, which affects the mucosal sealing effect of the urethral mucosa and easily leads to the occurrence of SUI (Figure 1).
Figure 1.
The normal transverse section of the urethral anatomy.
2.2. Dysfunction or Defect of the Urethral Sphincter
The muscular layer of the urethra is formed of urethral smooth muscle and striated muscle (external urethral sphincter) [11,17,18]. The urethral smooth muscle is divided into two layers . The inner layer is a longitudinal smooth muscle fiber, which is arranged longitudinally with the urethral lumen at the center . Additionally, its elastic contraction increases the diameter of the longitudinal smooth muscle bundle, effectively narrowing the urethral lumen and increasing the resistance to urination. The outer layer is a circular smooth muscle fiber that completely envelops the urethra, which is a slow-contraction muscle fiber with anti-fatigue properties . The inner layer of urethra smooth muscle fibers is arranged longitudinally and circularly surrounded by the outer layer of urethra smooth muscle fibers . This kind of unique tissue arrangement is very important for continence in the context of urination. The outer circular smooth muscle fiber contracts with the inner longitudinal smooth muscle as a central filler, producing continuous tension , narrowing the lumen diameter of the urethra and maintaining static urethral tension [11,17] (Figure 1).
The external urethral sphincter in women is Ω-shaped . It is mainly distributed on the ventral surface and both sides of the urethra and does not completely envelop the urethra, forming the outermost layer of the urethra and covering about 80% of the total length of the urethra [23,24,25]. The external urethral sphincter uses the anterior vaginal wall as a “plate” to compress the urethra in the direction of the posterior urethral wall to achieve continence when the external urethral sphincter contracts [18,26]. Morgan et al. studied the components of the external urethral sphincter fibers and found that female external urethral sphincter muscle fibers can be divided into slow-contractile muscle fibers and fast-contractile muscle fibers . Slow-contractile muscle fibers have anti-fatigue properties, and they can continuously contract to generate tension and maintain a resting urethral pressure. When the abdominal pressure increases, the fast-contracting muscle fibers contract rapidly, compressing the urethra in the direction of the posterior wall of the urethra, preventing urine leakage and maintaining urethral pressure during stress periods . Frauscher et al. used ultrasound to visualize the external urethral sphincters of SUI patients and normal women. It was found that the external urethral sphincter in SUI patients was weak. Defects in, or the dysfunction of, the urethral sphincter can be divided into congenital and acquired categories. Congenital urethral sphincter disorders are mainly caused by congenital central nervous system diseases, while acquired urethral sphincter disorders are mainly caused by childbirth, surgical treatments and radiotherapy . Incontinence easily occurs when the urethral sphincter cannot contract effectively .
2.3. Decreased Elasticity of the Urethral Wall
Normally, the female urethra is a soft tubular structure. The urethral wall is rich in loose connective tissue, elastic fiber, collagen and other components, so that the urethral wall has good elasticity and flexibility. Under the actions of external forces, it can effectively deform and close the urethra, ensuring the tightness of the urethral closure . Zinner et al. showed that, in the mechanical model, when the lumen surface is filled with lubricant, the elasticity of the pipe wall is better, and the flow resistance of the pipe cavity is greater . Thus, the better the elasticity of the urethral wall is, the greater the flow resistance will be when the surface of the urethral lumen is filled with mucus. If the urethra undergoes multiple radiotherapy or surgery, the urethral wall will become stiff and the elasticity will be weakened, which will lead to a decrease in urethral closure capability, and urinary incontinence is prone to occur. The study found that the content of urethral elastic fibers and collagen in SUI patients was significantly lower than that of normal women . Estrogen can selectively act on the urethral epithelium , promoting the growth and maturation of the urethral epithelium cells and the synthesis of collagen . In the postmenopausal period, the decrease in the estrogen levels contributes to the decrease in the synthesis of elastic fibers and collagen in the urethral wall and the urethral closure function disorder, which can easily lead to the occurrence of urinary incontinence [13,14].
2.4. Shortened Length of the Functional Urethra
Ensuring the proper functional urethral length is the key factor in female continence. The thickest part of the urethral sphincter in the whole of the urethra is the middle part, which is the part mainly controlling urinary continence [37,38,39]. Tension-free middle urethral suspension for the treatment of SUI aims to ensure the effective functional length of the middle of the urethra when abdominal pressure increases and effectively improve the patient’s urinary continence function . Pelsang et al. showed that when the abdominal pressure of SUI patients increased, the bladder neck and the proximal urethra moved downward, forming a funnel shape and leading the relative length of the functional urethra to become shorter, thus causing urethral resistance, and the hydrostatic pressure of the urethra decreased, which can easily contribute to the occurrence of urinary incontinence. In the mechanical model, the longer the effective length of the lumen is, the greater the fluid resistance is [33,35]. Therefore, properly increasing the length of the functional urethra can increase the urethral resistance and improve the capacity for urinary continence.
3. Anatomical Factors Affecting the Urethra (Figure 2)
3.1. Weak Supporting Structure of the Bladder Neck
Under normal conditions, the bladder base is close to the horizontal level, and the bladder neck is closed. The bladder neck is located at the junction of the middle and lower third of the pubic symphysis, usually being higher than the pubococcygeal line (the line between the lower edge of the pubic symphysis and the coccyx tip) [40,41]. The angle between the proximal urethral axis and the horizontal tangent line of the bladder base is the posterior vesicourethral angle [42,43], which is normally between 90° and 110°. When the bladder neck is closed at a right angle, the proximal urethra has the strongest closing performance and the greatest resistance. Enhorning proposed the theory of pressure transmission, suggesting that the intra-abdominal pressure is evenly applied to the bladder neck and the proximal urethra in order to ensure that the intra-urethral pressure is always higher than the intra-bladder pressure, regardless of whether resting or increased abdominal pressure is applied, so as to maintain urine control . The weakness of the supporting structure of the bladder neck can lead to the inadequate closure of the bladder neck, so that the posterior wall of the bladder neck collapses, the bladder neck moves downward, and the posterior vesicourethral angle increases or even disappears. Furthermore, the proximal urethra becomes a part of the bladder neck, and the intra-abdominal pressure cannot be evenly transmitted to the bladder and the proximal urethra, resulting in lower intraurethral pressure compared to the bladder pressure . Eventually, these changes decrease the capacity for urinary continence, and urinary incontinence easily occurs when the abdominal pressure increases. In cystourethrography, it is found that the bladder neck of most SUI patients is funnel-shaped [44,45]. The bladder neck moves downward, reaching lower than the pubococcygeal line, and the posterior vesicourethral angle increases or even disappears [40,46,47]. McKinnie et al. showed that the lack of estrogen leads to the reduction in the supporting strength of the lower urinary tract and the increase in the bladder neck activity, eventually leading to urinary incontinence [13,14] (Figure 3).
Figure 3.
The female cystourethrography. (a) Normal female bladder: the bladder base is close to the horizontal level, and the bladder neck is closed. The bladder neck is located at the junction of the middle and lower 1/3 of the pubic symphysis, usually higher than the pubococcygeal line (the line between the lower edge of the pubic symphysis and the coccyx tip); (b) SUI patient: in cystourethrography, it is found that the bladder neck of most SUI patients is funnel-shaped. The bladder neck moves downward, reaching lower than the pubococcygeal line, and the posterior vesicourethral angle increases or even disappears.
Figure 2.
The normal pelvic floor anatomy. UB: urinary bladder, UT: uterus, V: vagina, U: urethra, R: rectum, S: sacrum, C: coccyx, PS: symphysis pubis, P: pubis PB: perineum body, LAM: levator ani muscle, PUL: pubic urethral ligament, ① upper abdominal pressure. ② Supporting force from the pubic bone. ③ supporting force from the sacrococcygeal bone. ④ Pressure from the bottom of the bladder against the upper part of the vagina ⑤ pressure from the urethra against the middle and lower parts of the vagina. Red dotted line: the angle formed by the middle and lower parts of the vagina.
3.2. Defective Nature or Prolapse of the Anterior Vaginal Wall Support
The middle and lower segments of the urethra cling to the anterior vaginal wall, which provides a stable “backing plate” for the middle and lower segments of the urethra. The pubococcygeal muscle contracts to support the anterior vaginal wall and squeeze the posterior urethral wall, effectively maintaining the urethral closure pressure when the abdominal pressure increases and playing a urinary control role . The stability of the supporting structure of the anterior vaginal wall directly affects urine control function in women. When the anterior vaginal wall support is defective or prolapses, the posterior urethral wall support is weakened, and the patients often have different degrees of urinary incontinence when the abdominal pressure is increased [49,50].
3.3. The Continuity and Integrity of the Pelvic Fascia and Pelvic Fascial Tendon Arch (ATFP) Are Impaired
The intrapelvic fascia is the fibrous connective tissue that covers the surfaces of the organs in the pelvis and connects these organs with the pelvic muscles and bones . It maintains the stability of the pelvic organs and allows the pelvic organs to have a certain degree of activity . In particular, the intrapelvic fascia, which wraps around the side of the vagina and connects with the pelvic fascia tendinous arch, plays an important role in maintaining the stability of the vagina and urethra .
The pelvic fascial tendon arch (ATFP) is a broad-band aponeurosis structure originating from the ischial spine and ending at the ventral side of the pubis [54,55]. Its function is to suspend and fix the urethra to the anterior wall of the vagina, similar to the cable of a hammock, and maintain the stability of the urethra and vagina . Pit et al. found that the pelvic fascia and the pelvic fascia tendinous arch play important roles in inhibiting the backward movement of the anterior vaginal wall and the proximal urethra . Delance et al. believe that the anterior side wall of the vagina is connected with the pelvic fascia tendinous arch and levator ani muscle through the intrapelvic fascia. When the levator ani muscle contracts, the anterior wall of the vagina is pulled to squeeze the posterior wall of the urethra through the intrapelvic fascia, causing the urethral pressure to rise and then exerting the urinary continence function . If the continuity and integrity of the pelvic fascia are destroyed, the stability of the pelvic organs such as the bladder, urethra and vagina will be weakened, and the activity of the urethra will be increased, which will easily lead to urinary incontinence.
3.4. Weak Pubic Urethral Ligaments
The pubourethral ligament is a dense fibrous connective tissue that connects the lower edge of the pubic symphysis with the bilateral walls of the urethra [38,56,57]. Its main function is to suspend and fix the urethra to the pubic bone, pelvic fascia tendon arch, puborectal muscle and other structures, providing a “hinge” support in the high-pressure urethral area and maintaining the stability of the urethra when the resting and abdominal pressure increases . If the pubourethral ligament is weak or defective, the activity of the urethra will increase under the condition of the increase in abdominal pressure, which can easily cause urinary incontinence . Kefer et al. showed that in a mouse model, when the pubourethral ligament of the mouse was injured or removed, urinary incontinence occurred in the mice under the condition of the increase in abdominal pressure.
3.5. Levator Ani Muscle Weakness or Dysfunction
The levator ani muscle includes the iliococcygeal muscle, pubococcygeal muscle and puborectal muscle [59,60,61]. The levator ani muscle fibers are mainly composed of type I muscle fibers (slow-contraction muscle fibers) and type II muscle fibers (fast-contraction muscle fibers) . The sustained contraction of the type I muscle fibers, as in the case of the “posture muscle” of the spine, pulls the middle and lower urethra, distal vagina and rectum toward the pubis , providing elastic support for the urethra and maintaining the normal shape and position of the urethra. Type II muscle fibers can rapidly contract when the abdominal pressure increases, causing the urethra to bend at the pelvic diaphragm, effectively closing the urethra and exerting the urinary continence function . Delancey pointed out the “hammock”-like structure formed by the continuity of the intrapelvic fascia, vaginal wall and levator ani muscle. This structure can maintain urethral stability and the urethral closure pressure when the abdominal pressure increases . When the levator ani muscle is weak or dysfunctional, the “hammock”-like structure cannot play an effective role, and the patient will be prone to urinary incontinence. Studies have shown that during vaginal delivery, when the fetal head passes through the urogenital hiatus, it produces great traction and shear force on the levator ani muscle . This leads to the rupture and injury of the levator ani muscle, which can easily cause postpartum urinary incontinence and, later, SUI [63,64]. Stoker J et al. also believe that SUI is related to levator ani muscle injury. In clinical practice, mild urinary incontinence is often treated by bioelectrical stimulation and pelvic floor muscle training so as to improve the contractility of the levator ani muscle.
4. Anatomical Factors of the Pelvic Floor Nerves
Pelvic Floor Neuromuscular Injury
The pelvic floor neuro-regulation of female continence is a complex process effected by the interaction of sympathetic, parasympathetic and somatic nerves [11,66]. The parasympathetic nerve originates from S2–S4 and mainly dominates the bladder detrusor and urethral smooth muscle. When it is excited, the nerve endings release ACH, causing the bladder detrusor to contract and the smooth muscle of the urethral and bladder neck to relax . The sympathetic nerve originates from T10–L2 and also controls the bladder detrusor and smooth muscle of the bladder and urethral neck. When it is excited, the nerve endings release NE, which can act on the β receptors of the bladder detrusor and α receptors of the smooth muscle of the bladder neck and urethra, causing the relaxation of the bladder detrusor and contraction of the smooth muscle of the bladder neck and urethra . The main somatic nerve involved in urination at the site of the pelvic floor is the pudendal nerve, which is issued by S2–S4 and mainly controls the external urethral sphincter. When the nerve is excited, its endings release ACH, causing the contraction of the external urethral sphincter, and when its excitability is inhibited, this causes the relaxation of the external urethral sphincter [69,70]. When nerves are damaged in any aspect, neuromuscular changes will occur, affecting normal urinary control function (Figure 4).
Figure 4.
The main pelvic floor neuro-regulation of female continence.
The pelvic floor nerves are often injured by external mechanical forces. For example, when the second stage of labor is prolonged during vaginal delivery, the pelvic nerve fibers are pulled and compressed for a long time, which can easily lead to hypoxia damage. Medical procedure complications, such as the improper angle of lateral episiotomy during vaginal delivery, may cause the branches of the pudendal nerves to become damaged. After pelvic nerve injury, neuromuscular changes occur. That is, the duration of the potential nerve action is prolonged, the amplitude is reduced, the relative refractory period of potential action is prolonged, and the number of nerve fibers is significantly reduced . The denervated muscle innervated by the pelvic floor nerve also gradually appears, which is characterized by muscular atrophy, a decrease in the number and density of the muscle fibers, a decrease in the diameter of the muscle fibers and a change in the proportion of muscle fiber types . When the pelvic nerve is injured, the smooth muscle of the urethra, sphincter and levator ani muscle innervated by it will be denervated and atrophied, and the contractility will be weakened. When the abdominal pressure increases, the affected muscles cannot effectively contract to close the urethra and maintain the urethral closure pressure, causing urinary incontinence.
5. The Key Anatomical Pathogenesis and Operation Improvement
The anatomical pathogenesis of SUI is very complex, involving the anatomical factors of urethra itself, the periurethra and the pelvic floor nerve . Among the many anatomical factors, the levator ani muscle and external urethral sphincter are particularly important. In a normal resting state, the contraction of the type I muscle fibers of the levator ani muscle pulls the rectum and the distal end of the vagina towards the pubic bone . The levator ani muscle pulls up the anterior wall of the vagina to compress the posterior wall of the urethra under the synergistic effect of the pelvic fascia, providing a lasting elastic support to the urethra . The type I muscle fibers of the external urethral sphincter contract, squeezing the urethra toward the posterior wall of the urethra and forming a force that opposes the contraction of the type I muscle fibers of the levator ani muscle . The middle urethral sphincter is the thickest and has the strongest contractility. Under the joint contraction of the type I muscle fibers of the levator ani muscle and external urethral sphincter, a bend angle is formed between the middle and lower segments of the urethra, effectively maintaining the urethral closure pressure at rest and exerting the function of urinary control [19,37]. When the stress state or abdominal pressure increases, the type II muscle fibers of the levator ani muscle and the external urethral sphincter rapidly contract (the direction of the contraction is the same as that at rest) to provide strong muscle tension, leading to the upward movement of the pelvic diaphragm, the reduction in the urethral bend angle and the increase in the urethral closure pressure. Eventually, the urethra is forcefully closed to prevent urine leakage . When the levator ani muscle and external sphincter of the urethra are dysfunctional, the urethra cannot be closed forcefully, resulting in urinary incontinence . Therefore, in the treatment of urinary incontinence, it is very important to pay attention to the repair of the levator ani muscle and external urethral sphincter. For mild SUI, pelvic floor muscle training and bioelectrical stimulation can be used to repair the function of the levator ani muscle and external urethral sphincter. For moderate or severe stress urinary incontinence, the authors’ research team performed posterior pelvic floor reconstruction while performing anti-incontinence surgery. On both sides of Deno’s space, fishbone sutures were used to suture the levator muscle plate, bilateral vaginal wall and damaged levator ani muscle in a layer-by-layer manner from deep to shallow. This not only restores the contractility of the levator ani muscle but also effectively reduces the hiatus of the levator ani muscle, achieving a good therapeutic effect.
The supporting structure of the bladder neck in patients with SUI is weak. During cystourethrography for patients with SUI, we found that the posterior wall of the bladder neck was collapsed, while there was inadequate closure of the bladder neck, and the bladder neck had moved downward [40,46]. Furthermore, in SUI, the proximal urethra becomes a part of the bladder neck, showing a funnel shape [44,45]. The authors’ research team attempted to design an incontinence sling with an “inverted T” shape (Figure 5), placing it over the area from the posterior of the bladder neck to the posterior of the middle urethra with as little tension as possible. When the abdominal pressure increased, it effectively strengthened the bladder neck and middle urethra support and increased the length of the functional urethra. This new sling achieved good results in the treatment of SUI.
Figure 5.
The “inverted T” shape sling: ① and ② are the central positioning lines, and ③ and ④ are the suspension guidelines on both sides.
6. Conclusions
To sum up, the anatomical factors of the urethra itself, the periurethra and pelvic floor nerves are all very important in the anatomical pathogenesis of SUI. However, among the various anatomical factors, we believe that the joint contraction of the levator ani muscle and the external urethral sphincter, leading to the formation of the urethral bend angle and to the urethra being forcefully closed, plays a key role in urinary continence. Therefore, in the treatment of SUI, we should pay attention to the repair and reconstruction of the levator ani muscle and external urethral sphincter on the basis of the current theory of mid-urethral suspension. Secondly, the surgical approach should be improved according to the anatomical pathogenesis of SUI, and more appropriate slings, such as the “inverted T”-shaped sling (as Figure 5), should be designed to significantly improve the therapeutic effect of SUI treatments.
Author Contributions
X.Y. wrote the manuscript. X.W., Z.G., H.W., H.L. and L.L. contributed to the conclusion and reviewed/edited the manuscript. H.Z., D.T. and Q.Z. visualization. The guarantor for this work is J.S. All authors have read and agreed to the published version of the manuscript.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Not applicable.
Conflicts of Interest
The authors declare no conflict of interest.
Funding Statement
This research was funded by the National Natural Science Foundation of China, grant numbers 82260297 and 81960133, and the Special Project of Yunnan Chronic Kidney Disease Clinical Medical Research Center, grant number: 202102AA100060.
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4481 | https://terpconnect.umd.edu/~jpfisher/index_files/lecture9.pdf | 1 © Copyright 2012, John P. Fisher, All Rights Reserved Respiration: Ventilation, Circulation, & Transport Adapted From: Textbook Of Medical Physiology, 11th Ed. Arthur C. Guyton, John E. Hall Chapters 37, 38, 39, & 40 John P. Fisher © Copyright 2012, John P. Fisher, All Rights Reserved Overview of Respiration Introduction • Respiration provides oxygen to tissues for the metabolism of fats, carbohydrates, and proteins, while removing the waste carbon dioxide that results from these metabolic reactions • Respiration can be divided into four functions • Pulmonary ventilation - the inflow and outflow of air between the atmosphere and lung alveoli • Gas diffusion in the lung - the transport of oxygen and carbon dioxide between the alveoli and the blood • Gas diffusion in the tissues - the transport of oxygen and carbon dioxide between the blood and tissues • Regulation of ventilation - the control systems that regulate these mechanisms 2 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Alveolar Ventilation • There are about 16 generations of conducting airways • The trachea (Z = 0), which bifurcate into the two main bronchi • The bronchi (Z = 1), which divide into the lobar, segmental and 4 to 5 further divisions of cartilaginous bronchi • The next eight generations (Z = 8 - 15) constitute progressively smaller, ciliated noncartilaginous bronchioles, the last of which is the terminal bronchiole which is about 0.5 mm in diameter • Total number of airways is approximately 216, or 65,000 • An adult contains 300 million to 1 billion alveoli, with a total alveolar surface area of 70 m2 • The tracheal cross sectional area is about 2.5 cm2 and this cross sectional area is approximatley constant until the third generation • Cross sectional area then increases by approximately 7/5 per generation to 180 cm2 at the 16th generation © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Alveolar Ventilation • Up to and including the terminal bronchioles, the air passages are lined with cuboidal or columnar epithelium - no gas exchange occurs here, and this volume is referred to as that anatomical dead space • Typically, anatomical dead space is 150 ml • The anatomical dead space acts to • Warm the air to body temperature • Humidify the air to saturation • Remove dust by nasal hairs of mucosal surfaces • Facilitate coughing due to the flexible dorsal membrane of the trachea 3 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Mechanics • Lungs can be expanded and contracted in two ways • Downward and upward movement of the diaphragm, changing chest cavity length • This mechanism is utilized in normal breathing • Contraction of diaphragm pulls the lungs downward, while relaxation allows the lungs to compress • Elevation and depression of the ribs, changing chest cavity diameter • As a consequence of the movements tending to expand the thorax and lungs, the pressure of the intrapulmonary gas is momentarily reduced and thus causing an influx of air Guyton & Hall. T extbook of Medical Ph ysiology , 11 th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Mechanics • The lung is elastic, requiring a force to keep it expanded and thus collapsing in the absence of this force • All muscle contraction occurs during inspiration • Expiration is almost entirely a passive process caused by the elastic recoil of the lungs and chest cage • The are few attachments between the lung and the chest cavity walls, rather the lung “floats” in the thoracic cavity surrounded by pleural fluid • The movement of pleural fluid into the lymphatic channels provides a suction force that holds the lungs to the thoracic wall • This pleural pressure is slightly negative, -5 cmH2O, thus this approximate pressure is required to keep the lungs open at rest • During inspiration, expansion of the chest creates an even greater negative pressure, about -7.5 cmH2O Guyton & Hall. Textbook of Medical Physiology, 11th Edition 4 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Mechanics • When the glottis is open and no air is flowing, the pressures in the entire respiratory tree of the lung is equal, and equal to atmospheric pressure • This gauge pressure is defined as 0 cmH2O • Alveolar pressure is the pressure of the air inside the lung alveoli • Alveolar pressure must become negative, about -1 cmH2O, for air inspiration, and positive, about +1 cmH2O, for air expiration • These pressures causes approximately 0.5 l of air to be taken into the lungs in 2 sec, and 0.5 l of air to be expelled out in 2 to 3 sec • The difference between alveolar pressure and pleural pressure is defined as transpulmonary pressure, and is a measure of the elastic forces of the lung Guyton & Hall. Textbook of Medical Physiology, 11th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Mechanics • Lung compliance is defined as the extent of lung volume increase due to an increase in transpulmonary pressure • Adult, normal lungs have a compliance of approximately 200 ml air / cmH2O • A compliance diagram describes lung volume change as a function of pleural pressure, for inspiration and expiration • Characteristics are determined by elastic forces of the lung tissue and elastic forces of the fluid that lines the inside walls of the alveoli • The surface tension of the alveolar fluid contributes 2/3 of the total lung elasticity and are dominated by the existence and composition of surfactants in this fluid Guyton & Hall. Textbook of Medical Physiology , 11th Edition 5 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Mechanics • Similar to the surface tension in raindrops or meniscus, the surface tension created by the alveolar fluid provides a contracting force in the alveoli - a surface tension induced elastic force • Surfactants, typically amphiphilic molecules, reduce water surface tension • In the lung, surfactants are release by type II alveolar epithelial cells • The released surfactant contains phospholipids, proteins, and ions - specifically dipalmitoylphosphatidylcholine, surfactant apoproteins, and calcium ions • A portion of the surfactant dissolves, while the remainder spreads over the alveolar fluid, reducing surface tension up to 90%, from 50 dynes/cm (alveolar fluid) to between 5 and 30 dynes/cm (alveolar fluid with surfactant) • In an occluded alveoli, the amount of pressure generated from the collapsing alveoli may be estimated by: • Pressure = 2 x surface tension / alveolar radius • Thus, the decrease is surface tension provided by surfactants plays a key role in keeping alveoli open • Also, note that smaller alveoli observe significantly higher pressures © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Pulmonary Volumes • Spirometry is the method by which the volume of air movement into and out of the lungs is recorded • Traditionally, a spirometer consisted of a drum inverted over a chamber of water, with the drum counter balanced by a weight • The drum contains breathable gas, such as air or oxygen • As one breathes into and out of the chamber, the drum rises and falls and a recording of this volume change is made Guyton & Hall. T extbook of Medical Physiology , 11 th Edition 6 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Pulmonary Volumes • Four volumes may be measured by spirometry • Tidal Volume (VT)- the volume of air inspired or expired with each normal breath (typically 500 ml) • Inspiratory Reserve Volume (IRV) - the extra volume of air that may be inspired in excess of the tidal volume (typically 3000 ml) • Expiratory Reserve Volume (ERV) - the extra volume of air that may be expired in excess of the tidal volume (typically 1100 ml) • Residual Volume (RV) - the volume of air remaining in the lungs after forceful expiration (typically 1200 ml) • Also, Minute Respiratory Volume is the tidal volume times the respiratory rate • Typically, 500 ml/b x 12 b/min = 6 l/min Guyton & Hall. T extbook of Medical Physiology , 11 th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Pulmonary Volumes • Four capacities may be calculated from spirometry as well • Inspiratory Capacity (IC) - tidal volume plus inspiratory reserve volume • Functional Residual Capacity (FRC) - expiratory reserve volume plus residual volume • Vital Capacity (VC) - inspiratory reserve volume plus tidal volume plus expiratory reserve volume • Total Lung Capacity (TLC) - vital capacity plus residual volume • Thus, • VC = IRV + VT + ERV = IC + ERV • TLC = VC + RV = IC + FRC • FRC = ERV + RV Guyton & Hall. T extbook of Medical Physiology , 11 th Edition 7 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Pulmonary Volumes • Measurement of Residual Volume • Residual volume is measured using spirometry with air containing helium • After expiration, helium is inspired during normal breathing • Helium mixes with the residual volume, allowing the functional residual capacity to be calculated as • FRC = (CiHe / CfHe - 1)Vinspir • CiHe = initial [He] in spirometer • CfHe = final [He] in spirometer • Vinspir = initial volume of spirometer • Then, RV may be calculated as • RV = FRC - ERV Guyton & Hall. T extbook of Medical Physiology , 11 th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Alveolar Ventilation • The rate at which air reaches areas of gas exchange, namely the alveoli, alveolar sacs, alveolar ducts, and respiratory bronchioles, is defined as the alveolar ventilation (VA) • Anatomical dead space air may be measured by suddenly inspiring a volume of pure oxygen after normal breathing, and then monitoring the emergence of nitrogen during rapid expiration • Anatomical dead space (VD) is the volume fraction of pure oxygen expelled times the total volume of expelled air • Thus, VA = RR x (VT - VD), where RR is the respiration rate • Normally, VA = 4200 ml/min Guyton & Hall. Textbook of Medical Physiology, 11th Edition 8 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Alveolar Ventilation • Air is distributed to the lungs through the trachea, bronchi, and bronchioles • To keep the channels from collapsing during inspiration, cartilage rings encircle the trachea and bronchi • Bronchioles remain open due to the positive transpulmonary pressure • Resistance to airflow, normally very low, does not occur in the alveoli, but in the larger bronchioles and bronchi • There are few bronchi compared to the 65,000 terminal bronchioles Guyton & Hall. T extbook of Medical Physiology , 11th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Ventilation Alveolar Ventilation • All respiratory passages are kept moist by a layer of mucus secreted, in part, by mucous goblet cells in the epithelial lining • Mucous lining also entraps air particles • The ciliated epithelium also lines the respiratory passages, with about 200 cilia per epithelial cell • The cilia beat at a rate of 10 to 20 times per sec towards the pharynx - down from the nose and up from the lungs • Thus, the mucus flows slowly towards the pharnyx, where it is finally removed by coughing or swallowed 9 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Circulation Physiology • The pulmonary artery extends 5 cm beyond the apex of the right ventricle, and then divides to supply blood to the two lungs • Pulmonary artery walls are thin, 1/3 that of the aorta • Pulmonary arterial branches are short • Most pulmonary vessels have diameters larger than those in the systemic system • Thus, pulmonary arteries have a high compliance, allowing the system to accommodate the stroke volume output of the right ventricle • Lung tissue is nourished by small bronchial arteries originating from the systemic system and emptying into the pulmonary veins © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Circulation Physiology • Pressure pulses in the right ventricle and pulmonary artery are significantly lower than those observed in the aorta • During systole, pulmonary artery and right ventricular pressure are nearly equal • During diastole, right ventricular pressure falls quickly, and pulmonary arterial pressure falls slowly as blood moves through the pulmonary capillary bed • Blood volume in the lungs is about 450 ml, 9% of total blood volume • Lung blood volume can increase from high lung pressures or systemic blood flow resistance Guyton & Hall. Textbook of Medical Physiology, 11th Edition 10 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Circulation Blood Distribution • Under most conditions, pulmonary vessels act as passive, distensible tubes that enlarge with increasing pressure and narrow with decreasing pressure • It is critical that blood flows to the regions of the lung where alveoli are most oxygenated • Normal concentration of oxygen in the air of the alveoli is 73 mmHg (PO2) • When concentrations falls below this level, the adjacent blood vessels constrict • This response is in opposition to that observed in the systemic system, where blood vessels would be expected to dilate is response to low oxygen • This constrictions prevents blood flow to the low oxygenated areas, thus increasing blood flow to the high oxygenated areas • Constriction is thought to occur by the release of a vasoconstrictor signaling molecule, although this molecule has not yet been identified © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Circulation Blood Distribution • Due to hydrostatic pressure, there is a significant difference in blood flow to different areas of the lung • There is about a 23 mmHg pressure difference between the top and bottom of the lung • Approximately 15 mmHg of this pressure is above the heart, and 8 mmHg of this pressure is below the heart • Also, the pressure difference between the alveolar air pressure and alveolar capillary pressure regulate blood flow • If alveolar air pressure is greater than capillary pressure, then the capillary collapses and there is no blood flow Guyton & Hall. Textbook of Medical Physiology, 11th Edition 11 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Circulation Blood Distribution • Thus, 3 zones of blood flow have been described • Zone 1: No blood flow during the entire cardiac cycle, as alveolar capillary pressure is never greater than alveolar air pressure • Zone 2: Intermittent blood flow, as only systolic pressures are high enough to push blood through the alveolar air pressure • Zone 3: Continuous blood flow, as alveolar capillary pressure is always higher than alveolar air pressure • A healthy, standing person has zone 2 flow in the apex and zone 3 flow in the lung base • A healthy person lying down has entirely zone 3 flow • During exercise, pulmonary vascular pressure rise high enough to ensure only zone 3 flow Guyton & Hall. Textbook of Medical Physiology, 11th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Circulation Blood Distribution • During extreme exercise, blood flow through the lungs increases 4 to 7 times and is accommodated by the following • Increasing the number of open capillaries • Distending all capillaries • Increasing capillary flow rate Guyton & Hall. Textbook of Medical Physiology, 11th Edition 12 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Circulation Blood Distribution • Notes about pulmonary capillary dynamics • Capillaries in the lung are so dense that they almost touch one another side by side, leading to blood flow in the alveolar walls to appear as nearly a sheet rather than through individual capillaries • Pulmonary capillary pressure has been estimated at 7 mmHg, between left atrial pressure (2 mmHg) and pulmonary arterial pressure (15 mmHg) • Blood flow rate through the pulmonary capillaries is typically 0.8 sec, but can be as little as 0.3 sec © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Circulation Blood Distribution • The movement of substances through the lung capillary membranes is similar to that observed in the peripheral tissues however • Pulmonary capillary pressure is low (7 mmHg) • Interstitial fluid pressure is lower than in the peripheral tissue • Pulmonary capillaries are leaky, so colloid osmotic pressure is twice that observed in peripheral tissue, approaching 14 mmHg • Alveolar walls are extremely thin • When fluid forces are summed, a +1 mmHg filtration pressure is found, meaning that there is a slight movement of fluid from the capillaries into the lymphatic system Guyton & Hall. Textbook of Medical Physiology, 11th Edition 13 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Circulation Blood Distribution • A critical concept centers on why the alveolus of the lung does not fill with fluid • Here, the slight negative interstitial fluid pressure pulls fluid from the alveolus into the interstitial space • This fluid is then either transported into the capillary or the lymphatic system • Pulmonary edema occurs when this negative pressure is lost, occurring is cases of • Cardiac conditions leading to increases in pulmonary venous pressure or capillary pressure • Damage to the pulmonary capillary membrane • Typically, pulmonary capillary pressure must rise to approximately 28 mmHg for edema to occur Guyton & Hall. Textbook of Medical Physiology, 11th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Circulation Pleural Cavity Fluid • Remember, the lungs lie within the pleural cavity and are lined with a thin layer of mucous • The total fluid volume in the pleural cavity is small (3 - 4 ml) and excess fluid is drained into the lymphatic system • The negative pressure of the pleural cavity (-7 mmHg) provides the force that keeps the lungs expanded (whose collapse pressure is about -4 mmHg), and this negative pressure is maintained by this pumping of fluid into the lymphatics • Pleural effusion, or edema in the pleural cavity, can occur from improper lymphatic drainage, cardiac failure, reduced plasma osmotic pressure, or inflammation of the surfaces of the pleural cavity Guyton & Hall. Textbook of Medical Physiology, 11th Edition 14 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Gas Exchange Introduction • After alveoli are ventilated with air, the next step in the respiratory process is the diffusion of oxygen from the alveoli into the pulmonary blood, with carbon dioxide diffusion out of the blood and into the alveoli • Diffusion of gas molecules is due to their thermal (Brownian) motion • Diffusion is driven by a concentration gradient, from areas of high concentration to areas of low concentration Guyton & Hall. T extbook of Medical Physiology, 11th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Gas Exchange Introduction • As inspired air is a mixture of gases, the rate of diffusion of each gas is directly proportional to the pressure caused by that gas, or the partial pressure of that gas • Partial pressure is defined by Henry’s Law: • The solubility coefficients for atmospheric gases are: • Oxygen 0.024 ( atm/ (vol/vol) ) • Carbon Dioxide 0.570 • Carbon Monoxide 0.018 • Nitrogen 0.012 • Helium 0.008 • The relative difference in partial pressures between the alveolar gas and pulmonary blood determines the net flux of gas molecules • If the partial pressure is higher in the alveolar gas, the molecule will tend to move towards the pulmonary blood • If the partial pressure is higher in the pulmonary blood, the molecule will tend to move towards the alveolar gas Partial Pressure = Concentration of Dissolved Gas Solubility Coefficient 15 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Gas Exchange Introduction • As air is inspired, water immediately evaporates from the surfaces of the airways and humidifies the air • Water evaporation is also driven by differences in water partial pressure in the atmospheric air and inspired air • The partial pressure exerted by water is called the vapor pressure • At physiological conditions, the vapor pressure of water is 47 mmHg • Thus, once air is inspired, its is humidified to relative saturation to 47 mmHg © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Gas Exchange Introduction • The rate of diffusion may be described by • where D = diffusion rate ΔP = partial pressure difference A = cross sectional area S = gas solubility d = distance MW = molecule’s molecular weight • Here, the diffusion coefficient (D) of the gas is proportional to S/(MW)1/2 • Diffusion coefficients for some gases in body fluids are • Oxygen 1.00 (D/D O2) • Carbon Dioxide 20.30 • Carbon Monoxide 0.81 • Nitrogen 0.53 • Helium 0.95 MW d S A P D Δ ∝ 16 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Gas Exchange Composition of Alveolar Air • The changing composition of air as it moves through respiration is provided below • Notes • Atmospheric air is mostly nitrogen and oxygen • Inspired air is almost completely humidified (relative humidity) • Humidification dilutes other gases’ concentrations • Alveolar air is low in oxygen and high in carbon dioxide compared to both humidified and expired air Atmospheric Air Humidified Air Alveolar Air Expired Air Nitrogen, N2 597.0 mmHg 78.62% 563.4 mmHg 74.09% 569.0 mmHg 74.90% 566.0 mmHg 74.50% Oxygen, O2 159.0 mmHg 20.84% 149.3 mmHg 19.67% 104.0 mmHg 13.60% 120.0 mmHg 15.70% Carbon Dioxide, CO2 0.3 mmHg 0.04% 0.3 mmHg 0.04% 40.0 mmHg 5.30% 27.0 mmHg 3.60% Water, H2O 3.7 mmHg 0.50% 47.0 mmHg 6.20% 47.0 mmHg 6.20% 47.0 mmHg 6.20% Total 760.0 mmHg 100.0% 760.0 mmHg 100.0% 760.0 mmHg 100.0% 760.0 mmHg 100.0% © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Gas Exchange Composition of Alveolar Air • The functional residual capacity of the lungs is 2300 ml, yet only 350 ml of new air is inspired with each breath, therefore • The volume of alveolar air replaced by new air is only 1/7 of the total volume • Multiple breaths are required for exchange of most of the alveolar air • This slow exchange of alveolar air ensures that gas concentrations in the blood do not suddenly change • Buffers oxygen concentration, carbon dioxide concentration, and pH Guyton & Hall. Textbook of Medical Physiology, 11th Edition 17 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Gas Exchange Composition of Alveolar Air • Oxygen concentration in the alveoli is determined by • Rate of entry of oxygen by ventilation • Rate of absorption of oxygen into the blood • Normal respiration sees an alveolar ventilation of 4.2 l/min, an alveolar oxygen partial pressure of 104 mmHg, and oxygen consumption of 250 ml/min • Increased oxygen consumptions requires a significant increase in ventilation to maintain oxygen partial pressure • Normal respiration and ventilation also sees an alveolar carbon dioxide partial pressure of 40 mmHg and carbon dioxide excretion of 200 ml/min • Again, increased carbon dioxide excretion requires a significant increase in ventilation Guyton & Hall. Textbook of Medical Physiology, 11th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Gas Exchange Diffusion Through the Respiratory Membrane • The respiratory unit is composed of a respiratory bronchiole, alveolar ducts, atria, and alveoli • There are about 300 million alveoli in the two lungs • Each alveoli has a diameter of approximately 0.2 mm • Due to the dense capillary bed and the thin alveoli walls, the alveolar gases are in very close proximity to the blood of the pulmonary capillaries • Gas exchange occurs through the membranes of all the terminal portions of the lung (not just alveoli), and are described as the respiratory membrane or pulmonary membrane Guyton & Hall. Textbook of Medical Physiology , 11th Edition 18 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Gas Exchange Diffusion Through the Respiratory Membrane • The layers of the respiratory membrane include • A fluid layer lining the alveolus • Alveolar epithelium • Epithelial basement membrane • Interstitial space • Capillary basement membrane • Capillary endothelial membrane • Note: gases must also pass through a thin plasma layer and the red blood cell membrane • The respiratory membrane averages 0.6 microns in thickness • Total respiratory membrane surface area is 70 m2 • Total blood volume in the pulmonary capillaries is 60 - 140 ml Guyton & Hall. Textbook of Medical Physiology, 11th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Gas Exchange Diffusion Through the Respiratory Membrane • Factors that determine the rate of gas diffusion through the respiratory membrane include • Thickness of the respiratory membrane - increases in thickness, due to edema or fibrosis, can reduce transport of oxygen • Surface area of the respiratory membrane - decreases in area, due to necrosis or emphysema where alveoli coalesce, will dramatically reduce transport of oxygen • Diffusion coefficient of the gases • Pressure difference across the respiratory membrane - thus decreased atmospheric oxygen concentrations or decreased bodily oxygen consumption will reduce transport of oxygen Guyton & Hall. Textbook of Medical Physiology, 11th Edition 19 © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Gas Exchange Diffusion Through the Respiratory Membrane • The diffusing capacity of oxygen is 21 ml/min/mmHg • Under normal conditions the resting oxygen pressure difference across the respiratory membrane, 11 mmHg • Thus the total amount of oxygen moving across the respiratory membrane is 21x11 = 230 ml each minute • In exercise, diffusing capacity can increase to 65 ml/min/mmHg • Opening of pulmonary capillaries • “Matching” of alveolar ventilation and alveolar capillary flow • The diffusing capacity of carbon dioxide has been estimated at 400 - 450 ml/min/mmHg Guyton & Hall. Textbook of Medical Physiology, 11th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Pulmonary Gas Exchange Diffusion Through the Respiratory Membrane • Some respiratory units lack for adequate blood flow, while others lack for adequate air flow - even though total ventilation and total pulmonary blood flow is normal • This imbalance may be described by the ventilation-perfusion ratio • Here, alveolar ventilation (VA) and blood flow (Q) are compared • VA/Q = 0 means no alveolar ventilation • VA/Q = ∞ means no blood flow • As VA/Q falls below normal, inadequate ventilation is observed: physiological shunt • As VA/Q rises above normal, inadequate blood flow is observed: physiological dead space • This takes anatomical dead space into account Guyton & Hall. Textbook of Medical Physiology, 11th Edition 20 © Copyright 2012, John P. Fisher, All Rights Reserved Transport of Gas in Blood and Tissues Introduction • Oxygen moves from the alveoli into the blood due to the low oxygen partial pressure in the blood and high partial pressure in the alveoli • Most of the oxygen exchange occurs in the first third of the capillary bed - allowing for a significant safety factor in oxygen exchange • In times of exercise when blood flow rates are high and oxygen consumption is high, blood still leaves the pulmonary capillary bed fully oxygenated • Blood entering the left atrium has a PO2 of about 95 mmHg, rather than the saturated level of 104 mmHg, due to the mixing in of “unoxygenated” blood which has just nourished the lungs Guyton & Hall. Textbook of Medical Physiology, 11th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Transport of Gas in Blood and Tissues Transport of Oxygen from Lungs to Tissues • Once in the systemic circulation and capillary bed, blood PO2 falls from 95 mmHg to 40 mmHg as the interstitial fluid takes up oxygen through pressure driven transport • Transport is a function of blood flow rate, with increasing rates increasing oxygen transport into the interstitial fluid up to the 95 mmHg limit • Transport is also a function of tissue consumption, with increasing consumption decreasing interstitial fluid PO2 or requiring a significant increase in flow to maintain interstitial fluid PO2 • Cellular PO2 is low due to the cell’s consumption of oxygen, but still in great excess to the cell’s minimum level of 1 to 3 mmHg Guyton & Hall. Textbook of Medical Physiology, 11th Edition 21 © Copyright 2012, John P. Fisher, All Rights Reserved Transport of Gas in Blood and Tissues Transport of Oxygen from Lungs to Tissues • Carbon dioxide moves in opposition to oxygen, but diffuses approximately 20 times as quickly • Thus small pressure differences in carbon dioxide are observed between cellular and interstitial space, interstitial space and capillary blood, and arterial and venous blood • Again, only one third of the capillary length is needed for “complete” carbon dioxide transport • Again, a decrease in blood flow increases interstitial fluid carbon dioxide pressure - and an increase in blood flow decreases interstitial fluid carbon dioxide pressure • Again, a increase in metabolism increases interstitial fluid carbon dioxide pressure and a decrease in metabolism decreases interstitial fluid carbon dioxide pressure Guyton & Hall. Textbook of M e d i c a l P h y s i o l o g y , 1 1 t h E d i t i o n © Copyright 2012, John P. Fisher, All Rights Reserved Transport of Gas in Blood and Tissues Transport of Oxygen from Lungs to Tissues • Normally, 97% of oxygen is carried by hemoglobin, while 3% is dissolved in the water solvent in blood • As blood PO2 increases, hemoglobin saturation increases, typically resting at about 97% under normal conditions • Normally, 100 ml of blood carries 15 gm hemoglobin which can bind oxygen at 1.34 ml / gm • Thus, 100 ml of blood carries 20 ml of oxygen bound to hemoglobin • This is described as 20 volumes % • Upon passing through the capillaries, blood “gives up” about 5 ml of oxygen to the tissues • 25% utilization • During heavy exercise, up to 15 ml of oxygen is transported from the blood to the tissues • 75% utilization Guyton & Hall. Textbook of Medical Physiology, 11th Edition 22 © Copyright 2012, John P. Fisher, All Rights Reserved Transport of Gas in Blood and Tissues Transport of Oxygen from Lungs to Tissues • Hemoglobin maintains constant PO2 in the tissues, as the 5 ml of oxygen consumed by the tissues requires that tissue PO2 be approximately 40 mmHg so that sufficient quantities of oxygen may be obtained from hemoglobin • Small decreases in tissue PO2, and thus blood PO2, cause significantly more oxygen to be released from hemoglobin • During exercise, increased oxygen is made available by only modest decreases in blood PO2 • Furthermore, since blood oxygen is 97% saturated at normal, atmospheric levels of air oxygen content, increases in inspired oxygen concentration do not significantly affect oxygen transport to tissues • Similarly, since blood oxygen remains relatively high despite reduced inspired oxygen content, down to 50 to 60 PO2, low levels of inspired oxygen concentration do not significantly affect oxygen transport to tissues Guyton & Hall. Textbook of Medical Physiology, 11th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Transport of Gas in Blood and Tissues Transport of Oxygen from Lungs to Tissues • The oxygen-hemoglobin saturation curve can be “shifted” to the left or right due to abnormal physiological states • The curve shifts to the right due to • Decreased blood pH • Increased CO2 • Increased temperature (exercise) • Increased 2,3-biphosphoglycerate (BPG) concentration • Bohr Effect • Increased blood CO2, increases blood H2CO3 (carbonic acid), and decreases blood pH - the curve shifts to the right and more oxygen is delivered to the tissues Guyton & Hall. Textbook of Medical Physiology, 11th Edition 23 © Copyright 2012, John P. Fisher, All Rights Reserved Transport of Gas in Blood and Tissues Transport of Oxygen from Lungs to Tissues • Under normal physiological conditions, oxygen concentration is not limiting cellular metabolism as PO2 of only 1 mmHg is required to supply sufficient oxygen, but ADP concentration is limiting • Most cells are within 50 microns of a blood perfused capillary, supplying enough oxygen to support cellular metabolism • Lacking this perfusion, cells may be diffusion limited • If blood flow through a tissue falls to such a low level, then the usage of oxygen may be blood flow limited Guyton & Hall. Textbook of Medical Physiology, 11th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Transport of Gas in Blood and Tissues Transport of Carbon Dioxide • Due to carbon dioxide’s ability to diffuse about 20 times as quickly as oxygen, its transport is much more easily accomplished when compared to oxygen • Under normal conditions, 100 ml of blood carries 4 ml of carbon dioxide, however, the carbon dioxide is transported in many forms Guyton & Hall. Textbook of Medical Physiology, 11th Edition 24 © Copyright 2012, John P. Fisher, All Rights Reserved Transport of Gas in Blood and Tissues Transport of Carbon Dioxide • PCO2 of venous blood is 45 mmHg and contains 2.7 ml CO2/dl blood, while PCO2 of arterial blood is 40 mmHg and contains 2.4 ml CO2/dl blood • Thus, for 100 ml of blood, only 0.3 ml of carbon dioxide is transported in the dissolved form, or about 7% of total carbon dioxide • RBC produce an enzyme, carbonic hydrolase, which catalyzes the reaction of carbon dioxide with water, forming carbonic acid • Carbonic acid dissociates to H+ and HCO3 - • H+ bind with hemoglobin • HCO3 - diffuse into the plasma, exchanging with Cl- • Carbonic acid mechanism accounts for 70% of normal carbon dioxide transport • Carbon dioxide also binds with amine radicals of hemoglobin, forming carbaminohemoglobin • Carbamino transport accounts for about 20% of normal carbon dioxide transport Guyton & Hall. Textbook of Medical Physiology, 11th Edition © Copyright 2012, John P. Fisher, All Rights Reserved Transport of Gas in Blood and Tissues Transport of Carbon Dioxide • The Haldane Effect • Binding of oxygen with hemoglobin tends to displace carbon dioxide from the blood • Binding of oxygen causes hemoglobin to become more acidic, resulting in • Decreased formation of carbaminohemoglobin • Release of H+, forming carbonic acid, and subsequently carbon dioxide Guyton & Hall. Textbook of Medical Physiology, 11th Edition |
4482 | https://math.jhu.edu/~mathclub/problems/problems2012/shuffling-solutions.pdf | JHMT 2012 Power Round and Solutions February 18, 2012 Introduction Shuffling a deck of playing cards is a very important life skill. The standard riffle shuffle goes like this: you take a stack of cards, split it into two piles, hold one pile in your left hand and the other in your right, and drop cards from each hand onto a common pile on the table in random order. After repeating this several times, your card deck is hopefully fairly well-mixed and ready for a game or maybe a magic trick. (Of course, in practice, the easiest way to perform the riffle shuffle is to allow the cards to interleave without dropping them, but this is mathematically equivalent.) People often perform only three or four riffle shuffles in a row before using a standard deck of 52 playing cards. One might ask if this is really enough. It’s not hard to see that after one riffle shuffle, many orderings of the cards are impossible to reach and many others are much more likely than they ought to be in a uniformly random probability distribution. How many shuffles in a row do you really need to approximate uniform randomness? This Power Round builds up some of the basic ideas you need to answer this question.
In order to understand how to mathematize shuffling, we first discuss the basic concept of a permutation and some specific properties that we will need. Next, we give a mathematically precise definition of the Gilbert-Shannon-Reeds riffle shuffle, which has been shown in experiments to be a good model for how real people shuffle, and develop the theory of the probabilities it generates. Unfortunately, actually computing the necessary number of shuffles for approximate uniform randomness is beyond the scope of this test, but hopefully you will come to believe that such a number is indeed computable. Have fun!
Permutation Enumeration One of the key tools that we will use to analyze shuffles is the permutation. A permutation of a set S is defined as a listing of elements of S in some order (with each element appearing precisely once); for example, permutations of S = {1, 2, 3, 4, 5} include (4, 2, 3, 5, 1) or (3, 2, 1, 4, 5).
1.
(a) List all permutations of {1, 2, 3}.
(b) Give an expression for the number of permutations of {1, 2, 3, . . . , n} in terms of n. Compute the number for n = 5.
Solution to Problem 1: (a) (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) (b) n!. 120.
We can also think of a permutation as an operation we perform on some ordered listing to get another ordered listing. For instance, with S = {1, 2, 3}, we can think of the permutation (2, 1, 3) as the operation of swapping the first and second elements in an ordered listing of three elements, and leaving the third in place. In general, we interpret a permutation (σ(1), σ(2), . . . , σ(n)) as the operation that sends the σ(1)th element to the first position, the σ(2)th element to the second position, and so on. The listings we previously wrote down are just what we get when we apply the permutation to (1, 2, . . . , n). Note that the permutation whose listing is itself (1, 2, . . . , n) corresponds to doing nothing at all—for this reason, we call it the identity permutation, and write it as 1.
Given this interpretation, we define the notions of composition and inverse. The composition σ ◦τ of two permutations σ and τ is the operation of performing τ first, then σ. The inverse σ−1 of a permutation σ is the permutation such that σ−1 ◦σ = 1.
2.
(a) Compute the composition σ ◦τ of permutations σ = (1, 5, 4, 3, 6, 2) and τ = (2, 4, 6, 3, 1, 5).
(b) Compute the inverse of (3, 1, 4, 2) and the inverse of (2, 4, 6, 3, 1, 5).
(c) Show that (σ ◦τ)−1 = τ −1 ◦σ−1 for all permutations σ and τ of {1, 2, . . . , n}.
Solution to Problem 2: (a) (2, 1, 3, 6, 5, 4) (b) (2, 4, 1, 3), (5, 1, 4, 2, 6, 3).
JHMT 2012 Power Round and Solutions February 18, 2012 (c) We can show that τ −1 ◦σ−1 is the inverse of (σ ◦τ) as follows: (τ −1 ◦σ−1) ◦(σ ◦τ) = τ −1 ◦(σ−1 ◦σ) ◦τ = τ −1 ◦τ = 1.
When talking about shuffles, both of these interpretations of permutations have a natural meaning.
The listing interpretation corresponds to a state of the deck, and the process interpretation corresponds to shuffling the deck form one state to another.
3. Suppose that a process shuffles a deck of σ into τ.
Which permutation will be produced when (1, 2, . . . , n) is shuffled by that process? Justify.
Solution to Problem 3: τ ◦σ−1. Since a permutation µ changes σ to µ ◦σ, µ should be τ ◦σ−1 in order for µ ◦σ to be τ.
So far, we’ve talked about permutations as deterministic processes that always produce the same result on the same input. But in real shuffling, people usually don’t produce the same result every time (unless they’re trained magicians!). Thus, we will mainly focus on shuffling processes that are random processes, i.e. different outcomes occur with certain probabilities. The probability that a random shuffle turns a deck σ into a deck τ is called the transition probability from σ to τ.
4. For any random shuffle, show that the transition probability from σ to τ is the same as the transition probability from 1 to τ ◦σ−1.
Solution to Problem 4: The probability that a random shuffle takes σ to µ ◦σ depends only on µ, since changing the labels on the cards does not affect the shuffle. So we can replace σ by 1 and, according to problem 3, therefore replace τ by µ = τ ◦σ−1 without changing the transition probability.
Now, let’s return to permutations. An ascent of a permutation σ of {1, 2, . . . , n} is any position 1 ≤i < n such that σ(i) < σ(i+1). For example, the permutation (2, 7, 1, 3, 5, 4, 8, 6) has ascents at positions 1, 3, 4, 6.
Similarly, a descent of a permutation σ is any position where σ(i) > σ(i + 1). In our example, the descents occur at positions 2, 5, 7. Note that every position i < n is either an ascent or a descent.
5.
(a) List the ascents and descents of (9, 2, 7, 6, 3, 1, 8, 4, 5).
(b) Compute the number of permutations of {1, 2, 3} with exactly one descent.
(c) There are 11 permutations of {1, 2, 3, 4} with exactly two ascents. List them.
No explanations required.
Solution to Problem 5: (a) Ascents at 2, 6, 8; descents at 1, 3, 4, 5, 7.
(b) 4. They are (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2).
(c) (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (2, 1, 3, 4), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (3, 1, 2, 4), (3, 4, 1, 2), (4, 1, 2, 3).
Define the Eulerian number n k as the number of permutations of {1, 2, . . . , n} with k ascents. For example, as given in the preceding problem, 4 2 = 11.
6. Prove the symmetry property of Eulerian numbers: n k = n n −k −1 .
Solution to Problem 6: We find a bijection (one-to-one correspondence) between the permu-tations with k ascents and the permutations with n −k −1 ascents.
Indeed, if a permutation σ = (σ(1), σ(2), . . . , σ(n)) has k ascents, then it has n −k −1 descents because each position i < n is either an ascent or a descent. Reversing the permutation swaps ascents and descents, so therefore the permutation (σ(n), σ(n −1), . . . , σ(1)) has n −k −1 ascents and k descents.
JHMT 2012 Power Round and Solutions February 18, 2012 7. Prove that the Eulerian numbers satisfy the recurrence n k = (k + 1) n −1 k + (n −k) n −1 k −1 .
Solution to Problem 7: Consider any permutation σ of {1, 2, . . . , n} with k ascents.
We have σ(i) = n for some 1 ≤i ≤n, and removing this σ(i) yields a permutation σ′ of {1, 2, . . . , n −1} with either k or k −1 ascents.
Every permutation of {1, 2, . . . , n} with k ascents is therefore built from a permutation of {1, 2, . . . , n− 1} with k or k −1 ascents by inserting n. There are now two cases.
Given a permutation of {1, 2, . . . , n−1} with k −1 ascents, we gain an ascent by inserting n only when we do so at a descent or at the end of the permutation. There are n −k −1 descents, so each such permutation produces n −k permutations of {1, 2, . . . , n} with k ascents.
Similarly, given a permutation of {1, 2, . . . , n −1} with k ascents, we want to preserve the number of ascents when inserting n. To do this, the insertion must happen at one of the k ascents, or at the beginning of the permutation. Each such permutation therefore produces k + 1 permutations of {1, 2, . . . , n} with k ascents.
Combining these two cases yields the desired recurrence.
Using these results, it is possible to prove Worpitzky’s Identity: xn = n X k=0 n k x + k n .
To ensure that the binomial coefficient makes sense, assume that x is an integer and x ≥n.1 A rising sequence of a permutation σ is a maximal sequence of consecutive numbers appearing as a subsequence of (not necessarily adjacent entries of) σ. (Here, “maximal” means that we cannot add more numbers to the rising sequence; (1, 2) and (1, 2, 3) cannot both be rising sequences.) Every permutation decomposes into disjoint rising sequences. For example, the permutation (6, 1, 2, 4, 7, 5, 3) decomposes into three rising sequences: (1, 2, 3), (4, 5), and (6, 7).
Here, (1, 2, 3) is a rising sequence of (6, 1, 2, 4, 7, 5, 3) because the numbers 1, 2, 3 appear in order in σ but 1, 2, 3, 4 do not.
8. Recall the definition of the inverse of a permutation from the text before problem 2. Show that the number of rising sequences of a permutation σ is equal to one more than the number of descents of σ−1. That is, show #{rising sequences of σ} = #{descents of σ−1} + 1.
Solution to Problem 8: If k and k + 1 are in the same rising sequence, then their positions σ−1(k) and σ−1(k + 1) must satisfy σ−1(k) < σ−1(k + 1). This means that k + 1 starts a new rising sequence if σ−1(k) > σ−1(k + 1), i.e. if σ−1 has a descent at position k. Also, there is an additional rising sequence that starts at 1; this does not correspond to a descent. Hence, the number of rising sequences of σ is one more than the number of descents of σ−1.
The Gilbert-Shannon-Reeds shuffle As remarked in the introduction, the Gilbert-Shannon-Reeds (GSR) shuffle is a mathematical model which has been shown in experiments to fit the way real people shuffle real card decks. Here we will develop this model and a number of its interesting properties.
First we introduce some standard notation. Let j1, j2, . . . , ja be nonnegative integers so that j1 + j2 + · · · + ja = n. We define n j1, j2, . . . , ja = n!
j1!j2! · · · ja!.
1This actually works in greater generality. We can define generalized binomial coefficients a b for any real numbers a and b, and Worpitzky’s identity holds in this more general context.
JHMT 2012 Power Round and Solutions February 18, 2012 This number is called a multinomial coefficient. Of course, when a = 2, this is just a binomial coefficient, which we will usually refer to as n j1 .
9. Compute (no explanations required): (a) 7 3,2,2 , (b) 8 2,2,2,2 , and (c) 100 99,1,0,0,0 .
Solution to Problem 9: (a) 210 (b) 2520 (c) 100 The standard GSR shuffle works like this. Take a deck of n cards and cut it into a left pile and a right pile containing the bottom x cards and top y cards respectively (so x + y = n), in such a manner that the probability of putting x cards into the left pile is n x /2n. Drop cards from the bottom of either the left or the right pile one at a time, in such a manner that if at any point you’re holding X cards on the left and Y cards on the right, the probability that the next card dropped comes from the left is X/(X + Y ).
10. Take a stack of three cards labeled 1, 2, 3 from bottom to top and apply the GSR shuffle once. Consider the resulting pile, from bottom to top, as a permutation of 1, 2, 3.
(a) Are any permutations impossible to get? If so, list them.
(b) Compute the probability of putting (i) 0, (ii) 1, (iii) 2, (iv) 3 cards into the left pile during the cut.
(c) Compute the probability of the final permutation being (i) 3, 1, 2, (ii) 1, 2, 3.
No explanations required.
Solution to Problem 10: (a) Yes, 3, 2, 1 only.
(b) (i) 1/8, (ii) 3/8, (iii) 3/8, (iv) 1/8.
(c) (i) 1/8, (ii) 1/2.
11.
(a) In the general case with n cards, why do the given probabilities of cutting 0, 1, . . . , n cards into the left pile always actually add up to 1? That is, show that ( n 0) 2n + ( n 1) 2n + · · · + ( n n) 2n = 1.
(b) Take a standard deck of 52 cards and perform one GSR shuffle. Show that the probability of cutting 0 cards into one of the piles is less than one in one trillion (10−12).
Solution to Problem 11: (a) Standard. Can be done by either quoting binomial theorem or providing combinatorial explana-tions.
(b) 2 252 < 1 248 < 1 1612 < 1 1012 .
Now we’re ready to describe the GSR a-shuffle, which is exactly like the standard GSR shuffle except with a piles. That is, take your deck of n cards, cut it into piles of size j1, . . . , ja with j1 + · · · + ja = n so that the probability of getting precisely those sizes in that order is n j1,...,ja 1 an (we will refer to this as the cutting stage), and drop cards from the piles, one at a time, so that whenever you are holding piles of size J1, . . . , Ja respectively, the probability of dropping the next card from the kth pile is Jk/(J1 + · · · + Ja) (this is the dropping stage). In the future, we will consistently assume the following: 1. The cards start out numbered 1 to n from bottom to top. 2. The order of the cards after the dropping stage, from bottom to top, will be considered as a permutation of 1, 2, . . . , n.
JHMT 2012 Power Round and Solutions February 18, 2012 12.
(a) Take a 4-card deck and perform one 3-shuffle. Compute the probability that after the cutting stage, the pile sizes will be 1, 1, 2 in some order.
(b) Now suppose the same 4-card deck has already been cut into piles of size 1, 1, 2 from left to right (so the leftmost pile has the card numbered 1, the middle pile has card 2, and the rightmost pile has cards 3 and 4). Perform the dropping stage.
(i) How many permutations of 1, 2, 3, 4 are possible results?
(ii) Compute the probability (given this initial cut) that the final permutation is 2, 3, 4, 1.
(iii) Compute the probability that it is 3, 2, 4, 1.
No explanations required.
Solution to Problem 12: (a) 4/9 (b) (i) 12 (ii) 1/12 (iii) 1/12 13.
(a) Prove that the probabilities we’ve given for every possible way to cut the cards during the cutting stage really do add up to 1.
(b) Take an n-card deck which has already been cut into a piles of size j1, . . . , ja. After the dropping stage, how many permutations of 1, . . . , n are possible? Justify.
(c) Prove that, given this initial cut, every permutation of 1, . . . , n which is possible after the dropping stage occurs with equal probability. Show that therefore every possible path of operation, from deck to cut piles to final final permutation, occurs with probability exactly 1/an. Conclude that the transition probability of the GSR a-shuffle from 1 to σ is the same as the number of paths leading to σ divided by an. Refer to the definitions from after problem 4.
Solution to Problem 13: (a) Standard. Possible solutions are: quoting multinomial theorem, explaining combinatorial mean-ing, or working with induction on a.
(b) n j1,...,ja .
The order of the cards within each pile cannot change, so they can be considered indistinguishable.
(c) Induct on n. Clear when all piles 0. Given some possible permutation, suppose WLOG the first card comes from pile 1. The probability of this permutation is then j1 n times the probability of the permutation with the first card removed given piles of size j1 −1, j2, . . . , ja, which by the inductive hypothesis is 1/ n−1 j1−1,...,ja . This is 1/ n j1,...,ja as desired.
(Alternatively, directly compute that regardless of dropping order, the numerator must be j1! · · · ja!
and the denominator must be n!.) We will now describe a few apparently different shuffles which turn out to be the GSR a-shuffle in disguise, or related. The diversity of these descriptions shows just how mathematically rich the GSR shuffle is!
14.
(a) A “maximum entropy a-shuffle” is any shuffle in which you cut an n-card deck into a (possibly empty) piles and then drop cards from the piles one by one, with the stipulation that every possible path from deck to piles to final permutation should be equally likely. Prove that (i) the GSR a-shuffle satisfies this property and (ii) the only way to satisfy this property is to use the same probabilities as in the GSR a-shuffle.
(b) A “sequential a-shuffle” works as follows. First you cut an n-card deck into a piles according to the GSR probability distribution (i.e. getting piles of size j1, . . . , ja occurs with probability n j1, . . . , ja ). Then you shuffle pile 1 and 2 together using the dropping stage of the standard GSR 2-shuffle. Having done this, you shuffle the combined pile with pile 3, take the result and shuffle with pile 4, and so on until you have only one pile left. Prove that the probability of getting any particular permutation at the end is the same as with the standard a-shuffle.
JHMT 2012 Power Round and Solutions February 18, 2012 (c) An “inverse a-shuffle” works as follows. Take your n-card deck and, dealing from the bottom, place each card on one of a piles uniformly at random (that is, choose each pile with probability 1/a).
Once you’re done, stack the piles together in order from left to right.
(i) Prove that any possible path of operation reachable by an inverse a-shuffle—from deck to randomly dealt piles to final permutation—appears with probability 1/an.
(ii) Show that inverse a-shuffle is not equivalent to the standard a-shuffle in general by exhibiting a permutation of 4 cards reachable by an inverse 2-shuffle which is not reachable by a standard 2-shuffle. You do not need to justify.
(iii) Show that the transition probability from σ to τ of the inverse a-shuffle is the same as the transition probability τ →σ of the standard a-shuffle. Refer to the definitions from after problem 4.
Solution to Problem 14: (a) (i) Clear from earlier calculations.
(ii) Given a deck cut into piles of size j1, . . . , ja, there are necessary n j1,...,ja outcomes which must be equally likely. Therefore the probability of getting piles of size j1, . . . , ja must be proportional to n j1,...,ja . Or it is just enough to note that probability of each individual path determines the probability for each middle stage of process tree.
(b) It is enough to show that the sequential a-shuffle also satisfies the property of 15(c): every possible permutation occurs with equal probability given the initial cut, as it characterizes the maximum entropy a-shuffle.
Consider the relative location of cards in pile 1 and 2. After shuffling pile 1 and 2 together, order of cards within those piles do not change anymore. So shuffling of pile 1 and 2 together should be uniquely determined if final permutation is given. Similarly we can show that in each 2-shuffles in the sequence should follow certain path to reach the final permutation. Thus the probability for all permutations should be same.
(c) (i) Clear.
(ii) 2413 is unique answer.
(iii) Both for the standard a-shuffle and inverse a-shuffle, we showed that the transition proba-bility of a permutation is the number of possible paths to the permutation divided by an.
Hence it suffices to show that the path from σ to the set of piles P to τ exists under the inverse a-shuffle if and only if the path from τ to P to σ exists under the standard a-shuffle.
In an inverse shuffle, the path (σ, P, τ) is possible if and only if merging P gives the target permutation τ and the cards in each pile of P are in order within σ. But in the standard shuffle, the path (τ, P, σ) is possible if and only if merging P gives the original permutation τ and the cards in each pile of P are in order within σ. The two conditions coincide exactly.
15.
(a) Prove that an inverse a-shuffle followed by an inverse b-shuffle gives rise to permutations with the same probabilities as an inverse ab-shuffle. (This is called the product rule.) (b) Explain why this property of an a-shuffle followed by a b-shuffle being the same as an ab-shuffle must also hold when carrying out the standard (AKA maximal entropy) and sequential forms of the GSR shuffle. Justify rigorously.
Solution to Problem 15: (a) Label each card with two numbers according to the piles it landed in during the a-shuffle and the b-shuffle. Those cards with the same label form a pile in an ab-shuffle.
(b) Let Pa(σ →τ) and P a(σ →τ) be the transition probabilities from σ to τ for the standard a-shuffle and inverse a-shuffle respectively. The probability of obtaining τ from σ after an a-shuffle and a b-shuffle is given by X µ Pa(σ →µ)Pb(µ →τ) JHMT 2012 Power Round and Solutions February 18, 2012 where the sum is taken over all permutations µ. Meanwhile problem 16(c) gives Pa(σ →µ) = P a(µ →σ), so this sum is the same as P µ P a(µ →σ)P b(τ →µ). Now this can be interpreted as the probability of obtaining σ from τ after an inverse b-shuffle and inverse a-shuffle, and according to 17(a) this is the same as P ba(τ →σ). Applying problem 16(c) again gives P ba(τ → σ) = Pab(σ →τ), so we are done.
16.
(a) Suppose σ is a permutation with r rising sequences. Prove that the transition probability from 1 to σ for GSR a-shuffle of an n-card deck is a+n−r n an .
(b) Use this to give another proof of Worpitzky’s identity.
(c) Use part a of this problem and Problem 17 to show that if we repeat an a-shuffle k times on the same deck, the probability of any one permutation σ appearing after the last shuffle approaches 1/n! as k approaches infinity.
Solution to Problem 16: (a) We need to count the number of different ways to cut the deck into piles which have σ as a possible resulting permutation. We will make a −1 cuts which can be in any of n + 1 locations in the deck. The rising sequences determine r −1 of these cuts, but the remaining a −r can be assigned arbitrarily. This gives a+n−r n ways in which the deck can be cut. There are an possible final permutations (counting repeats), giving the desired probability.
(b) Immediate.
(c) The probability is ak+n−r n 1 akn where r is the number of rising sequences of σ. This is ak + n −r ak · · · ak + 1 −r ak · 1 n!
and each multiplicative factor can be rewritten as 1 + n−i−r ak , which approaches 1 as k approaches ∞. Thus the whole expression approaches 1/n!.
Using the GSR model and some analysis too advanced to explain here, one can show that an n-card deck must be shuffled at least 3 2 log2 n times before the probability distribution of the resulting permutations begins to approach uniformly random. This number is 7 for a 52-card normal playing deck and 9 for an 81-card SET deck. We here at the Johns Hopkins Math Tournament consider it very important that you take this knowledge into account the next time you play a card game! |
4483 | https://pressbooks.library.torontomu.ca/ohsmath/chapter/2-4-applications/ | Skip to content
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2.4. Applications
Applications
Now that we have learned to determine equations of lines, we can apply these ideas to real-life equations.
Example 2.4.1
A taxi service charges $0.50 per mile plus a $5 flat fee. What will be the cost of traveling 20 miles? What will be cost of traveling x miles?
Solution
The cost of traveling 20 miles = y = (.50)(20) + 5 = 10 + 5 = 15
The cost of traveling x miles = y = (.50)(x) + 5 = .50x + 5
In the above problem, $0.50 per mile is referred to as the variable cost, and the flat charge $5 as the fixed cost. Now if we look at our cost equation y = .50x + 5, we can see that the variable cost corresponds to the slope and the fixed cost to the y-intercept.
Example 2.4.2
The variable cost to manufacture a product is $10 and the fixed cost $2500. If x represents the number of items manufactured and y the total cost, write the cost function.
Solution
The variable cost represents the slope and the fixed cost represents the y-intercept. Therefore, m = 10 and y = 2500.
The cost equation is y = 10x + 2500.
Example 2.4.3
It costs $750 to manufacture 25 items, and $1000 to manufacture 50 items. Assuming a linear relationship holds, find the cost equation, and use this function to predict the cost of 100 items.
Solution
We let x = the number of items manufactured, and let y = the cost.
Solving this problem is equivalent to finding an equation of a line that passes through the points (25, 750) and (50, 1000).
Therefore, the partial equation is y = 10x + b.
By substituting one of the points in the equation, we get b = 500.
Therefore, the cost equation is y = 10x + 500.
Now to find the cost of 100 items, we substitute x = 100 in the equation y = 10x + 500.
So the cost = y = 10(100) + 500 = 1500.
Example 2.4.4
The freezing temperature of water in Celsius is 0 degrees and in Fahrenheit 32 degrees. And the boiling temperatures of water in Celsius, and Fahrenheit are 100 degrees, and 212 degrees, respectively. Write a conversion equation from Celsius to Fahrenheit and use this equation to convert 30 degrees Celsius into Fahrenheit.
Solution
Let us look at what is given.
| | |
--- |
| Centigrade | Fahrenheit |
| 0 | 32 |
| 100 | 212 |
Again, solving this problem is equivalent to finding an equation of a line that passes through the points (0, 32) and (100, 212).
Since we are finding a linear relationship, we are looking for an equation y = mx + b, or in this case F = mC+ b, where x or C represent the temperature in Celsius, and y or F the temperature in Fahrenheit.
slope
The equation is
Substituting the point (0, 32), we get
Now to convert 30 degrees Celsius into Fahrenheit, we substitute C = 30 in the equation
Example 2.4.5
The population of Canada in the year 1970 was 18 million, and in 1986 it was 26 million. Assuming the population growth is linear, and x represents the year and y the population, write the function that gives a relationship between the time and the population. Use this equation to predict the population of Canada in 2010.
Solution
The problem can be made easier by using 1970 as the base year, that is, we choose the year 1970 as the year zero. This will mean that the year 1986 will correspond to year 16, and the year 2010 as the year 40.
Now we look at the information we have.
Solving this problem is equivalent to finding an equation of a line that passes through the points (0, 18) and (16, 26).
The equation is .
Substituting the point (0, 18), we get:
Now to find the population in the year 2010, we let x = 40 in the equation:
So the population of Canada in the year 2010 is estimated as 38 million.
| | |
--- |
| Year | Population |
| 0 (1970) | 18 million |
| 16 (1986) | 26 million |
More Applications
In this section, you will learn to:
Solve a linear system in two variables.
Find the equilibrium point when a demand and a supply equation are given.
Find the break-even point when the revenue and the cost functions are given.
In this section, we will do application problems that involve the intersection of lines. Therefore, before we proceed any further, we will first learn how to find the intersection of two lines.
Example 2.4.6
Find the intersection of the line y = 3x − 1 and the line y = −x + 7.
Solution
We graph both lines on the same axes, as shown below, and read the solution (2, 5).
Finding the intersection of two lines graphically is not always easy or practical; therefore, we will now learn to solve these problems algebraically.
At the point where two lines intersect, the x and y values for both lines are the same. So in order to find the intersection, we either let the x-values or the y-values equal.
If we were to solve the above example algebraically, it will be easier to let the y-values equal. Since y = 3x − 1 for the first line, and y = −x + 7 for the second line, by letting the y-values equal, we get:
By substituting x = 2 in any of the two equations, we obtain y = 5. Hence, the solution (2, 5).
One common algebraic method used in solving systems of equations is called the elimination method. The object of this method is to eliminate one of the two variables by adding the left and right sides of the equations together. Once one variable is eliminated, we get an equation that has only one variable for which it can be solved. Finally, by substituting the value of the variable that has been found in one of the original equations, we get the value of the other variable. The method is demonstrated in the example below.
Example 2.4.7
Find the intersection of the lines 2x + y = 7 and 3x − y = 3 by the elimination method.
Solution
We add the left and right sides of the two equations.
Now we substitute x = 2 in any of the two equations and solve for y.
Therefore, the solution is (2, 3).
Example 2.4.8
Solve the system of equations x + 2y = 3 and 2x + 3y = 4 by the elimination method.
Solution
If we add the two equations, none of the variables are eliminated. But the variable x can be eliminated by multiplying the first equation by -2, and leaving the second equation unchanged.
Substituting y = 2 in x + 2y = 3, we get
Therefore, the solution is (-1, 2).
Example 2.4.9
Solve the system of equations 3x − 4y = 5 and 4x − 5y = 6.
Solution
This time, we multiply the first equation by -4 and the second by 3 before adding. (The choice of numbers is not unique.)
By substituting y = -2 in any one of the equations, we get x = -1. Hence the solution (-1, -2).
Supply, Demand and the Equilibrium Market Price
In a free market economy the supply curve for a commodity is the number of items of a product that can be made available at different prices, and the demand curve is the number of items the consumer will buy at different prices. As the price of a product increases, its demand decreases and supply increases. On the other hand, as the price decreases the demand increases and supply decreases. The equilibrium price is reached when the demand equals the supply.
Example 2.4.10
The supply curve for a product is y = 1.5x + 10 and the demand curve for the same product is y = -2.5x + 34, where x is the price and y the number of items produced. Find the following:
a. How many items will be supplied at a price of $10?
b. How many items will be demanded at a price of $10?
c. Determine the equilibrium price.
d. How many items will be produced at the equilibrium price?
Solution
a. We substitute x = 10 in the supply equation, y = 1.5x + 10, and the answer is y = 25.
b. We substitute x = 10 in the demand equation, y = -2.5x + 34, and the answer is y = 9.
c. By letting the supply equal the demand, we get:
d. We substitute x = 6 in either the supply or the demand equation and we get y = 19. The graph below shows the intersection of the supply and the demand functions and their point of intersection, (6, 19).
Break-Even Point
In a business, profit is generated by selling products. If a company sells x number of items at a price P, then the revenue R is P times x , i.e., R = P · x. The production costs are the sum of the variable costs and the fixed costs, and are often written as C = mx + b, where x is the number of items manufactured.
A company makes a profit if the revenue is greater than the cost, and it shows a loss if the cost is greater than the revenue. The point on the graph where the revenue equals the cost is called the break-even point.
Example 2.4.11
If the revenue function of a product is R = 5x and the cost function is y = 3x + 12, find the following:
a. If 4 items are produced, what will the revenue be?
b. What is the cost of producing 4 items?
c. How many items should be produced to break-even?
d. What will be the revenue and the cost at the break-even point?
Solution
a. We substitute x = 4 in the revenue equation R = 5x, and the answer is R = 20.
b. We substitute x = 4 in the cost equation C = 3x + 12, and the answer is C = 24.
c. By letting the revenue equal the cost, we get:
d. We substitute x = 6 in either the revenue or the cost equation, and we get R = C = 30. The graph below shows the intersection of the revenue and the cost functions and their point of intersection, (6, 30).
Practice questions
1. The variable cost to manufacture an item is $20, and it costs a total of $750 to produce 20 items. If x represents the number of items manufactured and y the cost, write the cost function.
2. A person who weighs 150 pounds has 60 pounds of muscles, and a person that weighs 180 pounds has 72 pounds of muscles. If x represents the body weight and y the muscle weight, write an equation describing their relationship. Use this relationship to determine the muscle weight of a person that weighs 170 pounds.
3. In 2005, an average house in Greater Toronto Area cost $335,907 and the average house in 2018 cost $787,300. Assuming a linear relationship, predict the price of a similar house in the year 2025.
4. In 2010 there were 11,386 laboratory-confirmed cases of gonorrhea reported in Canada. In 2015, the number of cases increased to 19,845. Assuming a linear relationship, how many cases of gonorrhea might we expect in 2030?
5. The supply curve for a product is y = 2000x + 13000, and the demand curve is y = -1000x+ 28000, where x represents the price and y the number of items. At what price will the supply equal demand, and how many items will be produced at that price?
6. A company that produces toys has a fixed cost of $10,725, and variable cost of 20 cents a toy. Find the break-even point if the toys sell for $1.50 each.
License
Mathematics for Public and Occupational Health Professionals Copyright © 2019 by Ian Young is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License, except where otherwise noted. |
4484 | https://www.youtube.com/watch?v=kZ_V3cvi4iE | Mass and Density - Integral Calculus
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Posted: 20 Apr 2011
Free lecture about Mass and Density for Calculus students.
Integral Calculus - Chapter 3: Applications of Integration (Section 3.7: Mass and Density)
This was produced and recorded at the Worldwide Center of Mathematics in Cambridge, MA.
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Transcript:
in this section of the book I want to talk about mass and density um this is a more kind of serious application than maybe volumes and surface areas that you may find a little difficult to believe you would ever use to calculate anything with we're going to have to use some Concepts from multivariable calculus not so calculus with many variables not we're not really going to use them hopefully you'll have some intuition for them but I'm going to state things the way you would in a multivariable class initially because we need to talk about um the density of solid objects and those that a solid object has three dimensions and that's naturally a topic for multivariable calculus but I think you'll be all right we'll um just going to say it very intuitively write some notation that looks like integrals but and hopefully I think you'll be fine um so I you probably have an intuitive idea of what density is um density you know what it means you know what it means for an object one object to be more dense than the other it kind of means oh it's heavier for its size yeah density is mass per volume so it's mass divided by volume so that if you fix the mass and make the volume very big the density would be low yeah so you know something that's big that weighs a lot isn't as dense I said weighs a lot there is a difference between mass and weight and I want to say something about that but um that's so density mass per volume but you may never have thought about instantaneous density that objects or liquids or just various things can have densities that are different at different points in the object like humans you know density at various points in your body is very different so what is instantaneous density well you you might be able to guess if you have some solid object regardless of how weird the shape is what should the density at a point mean well you take a very small volume around that point so think of a a rectangular solid you know take a little rectangular chunk of your solid around that point very small um so take a rectangular solid chunk around that point and take the mass of that chunk and divide by the volume of that rectangular solid which is its length times its width times its height and you do this um you do this for smaller you think take the limit you it's the density would be the limit as you take small smaller and smaller um volumes smaller and smaller rectangular solids around that point and take the mass divide by the volume and you get an instantaneous density uh we like to use lowercase Delta for density Delta P the density at p and it's defined by a limiting process as you might expect but it's it's a multivariable limit because are three dimensions they're getting small but hopefully that's intuitively clear and what it means is you you should think of it as the instantaneous rate of change of mass so this is mass with respect to volume so yeah density is mass per volume if the density is constant you just take the total mass divided by the total volume but in a Calculus class you take instantaneous density the density at each point defined in this limiting way if you write this in differential notation or think infinim this says that the infimal amount of mass at a point p is the density times the infimal volume so that if you were given a density function and you knew how to do integrals in three three dimensions what this would tell you is that the total mass so what you would get is that the total mass would equal the integral of well the continuous sum of all the infantes little Blobs of mass um what should you write for like limits of integration well some notation that means you're integrating over the entire solid so I'm just this notation I'm just I'm just reading it integrate so take a continuous sum of the mass of DM little infant tal Blas mass as your points move throughout the entire solid so this notation uh isn't helpful for calculating it's just telling you what we're doing and so you would do this kind of thing oh I DM is the density time DV so I need to inte so take a continuous sum of all the infimal contributions you get as your point P moves through all the points in your solid s and you take the instantaneous density times little infantes chunks of volume great this you can you can say all this but this this is of course in single variable calculus and so the question is where do we get a one variable integral from where where how do we set things up so that we can do something like this um just in terms of a single variable and it just means that our our solids have to be fairly special um but we can still do interesting problems so here's our solid what we're going to assume is that somewhere you've got an axis and I'll call it the x- axis and that your solid lies between the perpendicular cross-sections perpendicular plans to X at xal A and xal B so your your solid lies between the cross-sections where xals A and xals B and that for each X between A and B we're going to assume two things one of them is what we assumed when we looked at volumes namely we're going to assume at each point x along here that we know ax the area of the cross-section at X of the say x cross-section and we want this area function this cross-sectional area function to be continuous so that we can integrate it um and what else do we have to assume well we want to calculate mass so we're going to have that little crosssection sectional area will give us a DV because we'll take ah as hopefully you recall from the section on on volumes we'll just take ax times an infimal little change in X that remember that thickens up this slice so that you get that thickens up this cross-section so that you get an infinitesimal little chunk of volume so that'll give us this DV but we would need a density that depends just on x also and what that means is we need to assume that while the density throughout the object can change is X changes that if you fix an X if you take a specific x value that all the points in that X cross-section have the same density so we also need to assume that we have a density function that depends solely on X so and once we do that we'll be able to do these problems or some interesting problems s so we also assume we have a continuous function Delta X which equals the density at each point in our solid in the X cross-section so I am assuming that that the density of the solid does not change in a given X cross-section so all these points would have the same density but as your X changes then the density of your solid can change all right um once we have that then this Mass integral will get a lot easier we will integrate as X goes from A to B will add up so as we go from here to here um that will take us by looking at cross-sections that'll take us through every point in our solid and so we'll take the integral so X goes from A to B of now we'll have a Delta X time DV but we know DV is a a of x DX it really does help to break the problem up into pieces and when you're thinking about physical applications of integration you really should think oh the total mass I add up all the infinitesimal pieces of mass uh how do I get how do I get a little chunk of mass oh I take density times ah something I've written that's wrong hopefully notice this you take right this should be a DX you take the density times a little chunk of volume but then you think oh but a little chunk of volume is the cross-sectional area times DX or whatever your variable is and so yeah you split up the problem into pieces okay so this is what we are going to do but first I I want to say something about the difference between um weight and mass so um you're probably familiar with mass in the the metric system the standard unit of mass in the metric system is the kilogram um if you you know people often when they're speaking about it just call it a kilo it's short for kilogram um that's a unit of mass that is not weight uh that distinction gets blurred in common speech because uh many people from Europe will say that oh I weigh some number of kilograms so you know what do I weigh I weigh 70 kilos they don't really mean that's what they weigh they mean that's their Mass there's just no good verb form of the word mass but what you weigh weight is the force of gravity that the force of gravity exerts on a mass so it's the mass times the acceleration of gravity um so weight is is a force and um 1 kilogram to get its weight you would actually multiply times the acceleration of gravity so it's 9.8 roughly 9.8 m/s squared so the 1 kilogram weighs nine roughly 9.8 Newtons now nobody talks about weighing a certain number of Newtons they I mean in theory they should but they don't um the situation with in the English system with pounds and feet is um complicated in another Direction because almost nobody talks about the unit of mass in the English system it's um so let me let me come over here and write that one kilogram if you if you assumed you had Newtons first so the unit of force first 1 kilogram is 1 Newton per m/s squared all right that's a kilogram in complete analogy with this the unit of mass in English system um one of them the standard unit of mass is well you you know the force it's kind of weird most people don't talk about weighing Newtons but you do and they do know the unit of mass it's kilograms but in the English system it's the reverse you know the unit of of weight it's a the standard is a pound which is abbreviated lb and we want to give a name to a pound per foot per second squared and the name of this apparently for a long time this didn't have a name this is called a slug um slugs I are not very well liked in the literature uh I've seen many students take a problem given to them in slugs pounds feet seconds and they convert everything into the metric system into um Newtons and kilogram and meters solve the problem there and then convert back people hate sug so much you really shouldn't be afraid of they're completely analogous to um what you do in the in the metric system um it's just different anyway the unit of mass is a slug some people talk about a pound Mass I will not um a pound mass is the amount of mass that would weigh one pound um I will not deal with a pound Mass um so this means that you know in complete analogy with with um the metric system how much does a slug weigh so if you take one Slug and you have it at sea level on Earth you would multiply times the acceleration of gravity which is roughly 32 ft per second squared right to get that it weighs weighs roughly 32 lb oh squared to get so just like one kilogram weighs 9.8 Newtons one slug weighs 32 lb um which means a pound mass is 132 of a slug so anyway all right I wanted to get that out of the way but now let's do some problems I want to look at a at a hemisphere of radius R and I want to assume that it's filled with some highly compressable material so that the weight of the stuff at the top smushes the stuff at the bottom and makes it more dense so um things are going to be set up where our variable instead of being X will be Z and we'll see what the limits of integration are but we're definitely going to do things um in terms of Z not X in this first problem so so you have a hemisphere of radius R I'm going to assume it's centered at the origin um this is call this the z-axis this is z um a hemisphere of radius R centered at the origin an equation for it zal the square < TK of r^ 2 - x^2 - y^ 2 and I'm going to assume that the um that the mass the the mass the density the density is a function of Z let's see there was a density function I wanted is 1,00 2 r - Z is 1,000 2 r - Z kilg per cubic meter so what does this mean it means when Z is zero your density is 2,000 R whatever R is so when Z is zero when you're down here at the bottom of the hemisphere your density is 2 R when you're up at the top of the hemisphere where Z is z is R you would get half of that because then you'll get a th times just R so the whatever this hemisphere is filled with we're assuming it's twice as dense down here as it is up here and the question is so this is the density um the question what is the mass of this filled hemisphere well you take density times crosssectional area multiply times DZ and you integrate as Z goes from will go from zero to R so yes Z will go from Z to R we have our density function our our question now is what's a of Z well we actually looked at this and we did volumes but I don't expect you to remember exactly everything we've done so if you're at some z-coordinate the question is if you take a cross-section perpendicular to that z-coordinate you get a circle or actually a disc a filled-in circle and the question is we what's this radius as a function of Z because if we knew that radius then the area of the cross-section would be Pi the radius squar so um how do you figure out that radius well it's it's easy this here's the origin this distance out to here is r and this is z and then you use the Pythagorean theorem so you get that z^2 + R2 equal Capital R2 so that first R was a little r so we get that the mass oh well let me go ahead and right so what's a of Z it's going to be Pi R 2 where what's r r of Z what we just said is R of Z squar so R is R of Z and what we just found said was that z^2 plus this R 2 should be little r should be capital r 2 which means that little R 2 is capital r² minus z^ 2 so that this cross-sectional area R little R 2 is capital r 2us z s so this is pi it's right here it's already squared for us Pi r^ 2us z^2 so that's the crosssectional area and as I said you should think of these things in pieces so the little chunks of volume be the cross-sectional area times times a little thickening so you get Pi r^ 2 - Z ^ 2 DZ and that's DV and then DM a little blob of mass is the density times a little chunk of volume we know the dens we we're handed the density function it's 1,00 2R minus Z times DV but we put in that DV is this so we get that times < r^ 2 - z^ 2 DZ and then yeah we want to add so that's a Time symbol supposed to be um and then we want to add up all of these as Z goes from0 to R so we integrate um this won't be too bad it's not particularly attractive but it's not so bad so what do we get we get the mass is the integral from 0 to R of I'm going to put the constants out front we get a thou oh I'm dropping all the units we know what they are um density was in kilograms per cubic meter the volume is going to come out in cubic meters um well I didn't say that I was measuring all the distances in meters but I was um so that our mass will come out in kilograms um I'll just suppress the units Until the End hopefully I won't forget to put them in then we get 1,000 pi times 2 r - Z r^ 2 - z^ 2 d z how do you integrate this well it's it's easy you pull out the th Pi it's a constant just the easiest thing to do is just go ahead and multiply this out you get a polinomial you use the power rule um and you get whatever you get you get 1,00 Pi time integral from 0 to R of all right you get a 2 R cubed and then a minus 2 R z^ 2 A minus r 2 Z and a plus z cubed DZ you get that and then you use the power rule repeatedly so remember that R is a constant so this is just some constant integrated with respect to Z so there we just get 2 R cub Z but then here you get a min-2 r power rule done to z^2 you add one to the exponent divid by the new exponent minus r^ 2 power rule here the power exponent is a one you add one to the exponent divide by the new exponent same thing here Z to 4th over 4 and you evaluate as Z goes from 0 to R and certainly when we we plug in Z is zero every one of these terms becomes zero so what we get is this quantity with Z replaced by R so what we get is 1,00 pi times when you put in Z as R we'll get a 2 R 4- 2/3 - 2/3 R 4 um minus r 4 / 2 plus r 4 over 4 we could certainly get a common denominator of 12ths um I guess I won't bother but you you could and this will come out in kilograms all right if you're wondering why all of these had to come out with R to the 4th in them well our units to to work out right we better kind of have the same power of R here each time or something's not not right um because if you had an extra power of our one or lesser power R then one of these things wouldn't be multiplying correctly to give us to make us end up with kilograms all right um so that's how you can do a kind of a three-dimensional solid density and mass problem with just one variable calculus you you have a formula for the cross-sectional area and you assume that that in each cross-section the density is constant in a given cross-section but can change as you change the crosssections there is because we found volumes another way there is another way you could get a one an interesting one variable problem out of a solid object and that's we could generate our solid as a solid of Revolution and use cylindrical shells then we have to make it we once again will have to assume that our density is constant in each shell and so that means we're assuming something special so I want to go back to a region we looked at in the volume section um so let's look [Applause] at this example where we have Y = 2 - x here's Y is the square root of x these intersect it when X is one um and Y is one and this is two and we want to look at this region and wrote revolve that revolve that around revolve that around the Y AIS and generate some I don't know looks kind of like a cool top so you know a top that you spin um you revolve that around the Y AIS and we'd like to get a a a mass problem out of this we found the volume of this before we use cylindrical shells so what you do is you take instead of taking crosssections that are perpendicular to your axis of Revolution you take a crosssection that's parallel so a cross-section of the original region that's parallel to your axis and when you rotate or revolve that around you get a cylinder but then you thicken the cylinder a little bit which means you can think of that as starting with a little thickened rectangle in the first place and sweeping that around that means you'll have an infinim little change in X right for the thickness of your cylinder and then what did we do you'll need you take if you call that the radius the distance from each cross-section to the axis that you're revolving around um call that R and you call the height call this the height the height of this cylinder well the formula for the area of a cylinder 2 pi RH the circumference of a circle times the height but then you give it a little infantes thickness and this will give you a little chunk of volume this is what we did when we looked at cylindrical shells since our axis of revolution is the the y- AIS the distance from what you're revolving to the Y AIS so the radius is just the x coordinate so in in this particular case r will just be X the height is well it's the the difference between the y-coordinate on this curve and the y-coordinate on that curve so the upper y coordinate is given by Y = 2 - x so H is 2 - x minus this y-coordinate and that y-coordinate is the square root of x so this so for our little chunks of volume by cylindrical shells what we get is 2kx 2 - x - the < TK X DX great how do we turn this into a mass problem what we need to assume now is if we want to just do one variable integrals which is what we want to do we need to assume that our density just depends on X so as X goes from 0 to one so but that means as X goes from 0 to one what you're getting are you'll get the shell that corresponds to this x coordinate you'll get the shell that corresponds to this x coordinate and you'll get the the shell that corresponds to this as your X goes from 0 to one you don't just get these pieces that it's just telling you the x coordinate that you then revolve to and calculate the volume of the corresponding shell this way so your X coordinat don't think oh I should go from minus one to one and I need to get all this revolves we've already encoded what you get by taking something at this x coordinate and revolving it now you just need to let X go from 0 to one to pick up what happens at each of the points in this region but we need to assume that our density is just in the solid that we generate that Delta just depends on R the distance to the axis of Revolution so that I I'm going to assume I just picked a function 200 1 + r kg per cubic meter I just made this up and so what this means is when you're right at the axis of Revolution so that R is zero the density at along points there it's 200 kg per cubic meter as you get out here where you're a distance of one away from the axis of Revolution and you would be at all the points that this you should picture this is some kind of circular look or top looking thing this this is swung around it revolved around so there's this whole circle of points that you get from here when you swing this around so when that Outer Circle Edge uh that would be where R is one and the density out there we're assuming is 200 kilg per cubic meter but we need that it depends that all the points in each cylindrical shell have the same density otherwise we're not going to be able to do this and that means that we need for the density to just be a function of the distance to the axis of Revolution um in this case as as I said before R the distance to the axis of revolution is just X and so we get this and then it's a one variable problem now and I'm not going to finish this integral but let me set it up in a lot of these application problems the the interesting part is setting up the correct integral actually evaluating it is um it is what it is so it's not uncommon for or it's fairly common for instructors to make test slightly shorter by just having people set up some integrals um and not actually evaluate them but you of course should pay attention to whatever instructions you're given so the mass the mass will be the integral as X goes from 0 to one 0 to one you'll add up all the infimal Blobs of mass what's an infimal blob of mass an infimal blob of mass well it's density times an infimal blob of volume [Music] um okay so what's an ininal chunk of volume we're using cylindrical shells so we got I'll put in the density function too so we got 200 1 + x for the density and then um the volume using cylindrical shells we got 2 pi x 2 - x - < TK of X DX as X goes from 0 to 1 this again I've set this up in kilogram and cubic meters and and again I didn't say that all our distances were measured in meters but I meant they were um and this will come out in kilogram the total mass uh how hard is this integral that's just the power rule over and over again uh you'll pull out a 400 Pi multiply these three quantities together you just have powers of x times constants just use the the power rule all right um that's those are two examples of how you find uh mass given density of of solid objects but in one variable calculus you're frequently given um kind of simplified problems that are more clearly uh that more clearly involve just a single variable like X so I want to I want to talk about length density and area density these are um kind of idealized densities um we can make sense of them in terms of real actual three-dimensional volume densities but we don't usually talk about them that way so suppose you've got a thin wire so think thin wire and what we really mean is we have an idealized one-dimensional object that it only has one dimension and that's its length then so maybe it goes from x equals call suppose it's laid out along the x axis um and it's between xal A and xal B okay so You' got this thin wire then what people normally talk about in this case is length density and I I'll write a Delta but with a sub L for length um the length density and this is mass per length so what that means and it's the infimal so if infimal mass divided by infimal length so it's it's dmdx if if you're measuring if you're got things laid out along the x-axis and it can change as X changes so I'm write this as a function of X and so reading this in terms of infin decimals and writing it in terms of differentials what you get is an infimal blob of mass is the length density times little changes in length what do you mean the mass per per length all right this is how people talk about um the mass of a wire that it has some mass per foot or per meter it's um you could set this up as as a three-dimensional problem and so if you want to think of it as well yeah what does that mean well yes it really you can picture the wies being fatter blow it up so look at it and go oh it's not you know it's not a onedimensional thing it's a three dimens dimensional thing if you want to do that then you need to set things up you would set things up the way we set them up before you would have to assume that you have a cross-sectional area function and that in each cross-section the the density is the same for a fixed cross-section the density at all points in the cross-section are the same but the the density can change as X changes so that we would do what we did before and get o so n DM would be Delta of x times little chunk of volume and a little chunk of volume as we so this is the normal three-dimensional density so mass per volume and then you'd write that DV is ax DX and what that means is we get little chunks of mass given by this expression instead of by this expression well that means that if you want to think of it as three-dimensional in terms of three-dimensional densities then what you're doing is defining the length density to be the everyday ordinary density so volume density times the cross-sectional area and you're assuming that the density doesn't change in each cross-section so I suggest you not think of it that way it's if you're trying to make sense of oh but link density doesn't really exist yes it does we can do that but this is how everybody talks about it and this is how I'll talk about it um for the rest of the time it's also true that you get an area there's you can talk about an area density in the same way you can make this rigorous in terms of volume densities just as I did but the way people talk about area densities you've got some idealized two-dimensional object so um just one of our examples will be a metal a triangular metal plate so think of a thin metal plate um there's a vocabulary word that you can use for really thin plates uh a lamina I probably won't say that often I prefer to say a thin metal plate um but we're thinking of it as an idealized two-dimensional object that it really for all for most intens and purposes it only has two dimensions and then of course we'll talk about area density and area density which I'll write as Delta sub a is mass per area but of course we mean this infinitesimally so we'll have a a DM da and we'll need it as a function of a single variable so in this kind of problem we would have to assume that either we're doing things in terms of X and the density is constant in each X cross-section or that we're doing we're looking at this in terms of Y and the density is constant in each y cross-section but one way or the other we have to turn it into one variable problem and then of course we'll read this as the infimal contribution to the Mass is the area density function times area Okay so let's just do one problem of each kind and then we'll be done with this section so no it could say most anything so let's suppose we've got a wire so an example suppose we have a wire laid out along the x-axis um between oh I think I want xal Z and xal 4 and I guess I'll switch to English system and use feet suppose the length density so Delta sub L of X is is I want eus X slugs per foot way as far as I know there's no abbreviation for the unit slugs you just keep writing slug um so the the density when X is zero is one slug per foot but when X is 4 is e to Theus 4 so 1 over e to the 4th slugs per foot so it's much less dense on the right and on the left um and for we could do we could ask lots of questions but let's just you know figure out let's just go ahead and figure out let's figure out the mass between 0 and two and the mass between 2 and four of course the mass between 0 and two is going to be bigger why because the there's a lot the density is a lot higher down close to where X is zero and gets much smaller down where X is four how how much more dense or sorry how much more mass does the left half have than the right half well let's just see so the mass of the left half so I mean half in terms of distance certainly not half in terms of mass mass of the left half this will be the integral as X goes from 0 to two of your length density times little chunks of length right I I guess to be consistent with writing it in pieces like I was doing before I would first write to find the mass you want to add up all the little Blobs of mass but now since we've got an idealized onedimensional object and are dealing with length densities that little blob of mass is just the length density times little chunks of length um so these problems inherently easier than the ones where we had a solid object you get the integral from 0 to two our density function e- X DX either by making the substitution U is min - x or just by you know you can think in your head uh what's something whose derivative is e- X well it's almost e- X but then you pick up an extra minus sign one way or the other you should find that an anti-derivative of e- X is minus e- X and of course now that's easy to verify because you can just take the derivative of this see that you get this and you evaluate from 0 to two this gives us minus E to the minus 2 minus what you get at zero which is min-1 so we get 1 minus E to-2 is 1 over e^2 I think I'll write it that way this is slugs if you wanted to get the weight of the left half you would now multiply this by 32t per second per second so multiply by 32 and that would give you the weight well approximately because 32 is only approximately the acceleration gravity but um in pounds slugs feet seconds pounds so if you multiply this by 32 you'd get the weight of the left half instead of its mass and that would make a lot of people more comfortable but once the the mass of the right half so uh the mass of the right half which should be a lot smaller now you do exactly the same thing except now you integrate from 2 to 4 so we'll just integrate from 2 to 4 e- x DX so you get minus E to the- X evaluated from 2 to 4 so we get minus E to- 4 minus what you get at 2 which is eus 2 so minus minus that's a plus so we get a 1 over e^2 minus 1 over e to 4th and this is in slugs and yeah this is this is significantly smaller than what we found before which was 1 - 1 e^2 um so the calculation is easy why is the calculation easy because I picked an easy um length density function I want to uh do one area density problem and then then we'll be finished with this section um so let me take um let me take a thin metal plate so an idealize two-dimensional object so let's let's take Y = 2 - x and assume that we have this triangular plate right here so think thin metal plate and I'm going to assume that it has an area density function it's a function of Y and so this means that I'm going to need I'm be assuming that the density in this play is constant along horizontal lines so for each horizontal line each horizontal cross-section of the metal plate the density is constant everywhere the the area density and I'm going to assume that that area density is a function of Y is just to continue picking on eus X or e to the minus y I'm going to assume it's this but now in say kilograms per square meter and I'm assuming that this is measured in all the lengths here are measured in meters so this is 2 m this is at 2 m okay then what's the find the mass of the plate well the mass then you would integrate as y goes from 0 to two you don't have a choice here if we were finding area and hopefully you'd know the area of a triangle and you wouldn't integrate to find the area but if we were using integrals to find the area you could do integrals with respect to Y think of little infant tessal rectangles horizontal rectangles or you could do integrals with respect to X and think of vertical rectangles we don't have that option now because our density function is a function of Y and if we tried to do things in terms of X then we wouldn't know the density's constant in each of our cross-sectional things and we wouldn't be able to get anywhere so um you integrate as why goes you add up all the little Blobs of mass as y goes from 0 to two but a little blob of mass is the area density times a little blob of area and put in the area density function that we were given so you get the integral from 0 to two of E e to Theus Y and then da da if we're looking at things in terms of Y at a given y-coordinate a little chunk of area it's it's this length times a little thickness so a little Dy to give yourself this little chunk of area of this rectangle so what is this well this length is the x coordinate on that curve but in terms of Y that x coordinate you put the X over here the Y over there in terms of Y that x coordinate is 2 - y okay so we get there's the area density function and then a little chunk of area you get that it's this length which is 2 - y times times the infimal little thickness Dy and this is all going to come out kilogram now I'm not going to finish this integral but you should and it's in the book it's um how do you how do you do this integral how do you find an anti-derivative of this well if you look at it for a few minutes hopefully it would occur to you what you do um to calculate that integral you need to split it into pieces so we would have to you want the integral from 0 to 2 of eus y 2 - y Dy you multiply the e to the Y the E Theus y times both parts you get 2 eus y - y e e to Theus y Dy you can split this integral up you know how to anti-differentiate that part right this is so you could split this integral into these two pieces this first part is simple we've in fact we've just done it a couple of times but it's this second part that might look bad so this is easy you an anti-derivative eus y minus E to the minus y you just pull the two out you evaluate from 0 to2 so that part's easy how do you do this integral hopefully if you look at this for a few minutes you'll remember our techniques of integration you do this by Parts by parts so you would let u y and that would leave e to the y e to the minus y DY for DV so DV then would be e to Theus y Dy and then you use integration by parts which I remind you is in terms of u and v is the integral of U DV is U V minus the integral of V du and this choice will enable you to do this you now find that du is dy you find that V is e to minus eus y and you apply integration by parts and you can do this integr but I'm going to leave that for you to do um in the next section of the book we're going to look at centers of mass um originally I had uh this material in the same section as the mass and density and I realized it was getting very long but the center of mass is a is a cool concept and you have but you have to have mass and density under your belt first before you can talk about the center of mass of an object which is of um a point that you think of is kind of for a lot of purposes you can think of your mass as being concentrated at that single point in space it's kind of cool we'll do that in the next section |
4485 | https://physics.stackexchange.com/questions/579140/deriving-ideal-gas-law-from-boyle-and-charles | thermodynamics - Deriving ideal gas law from Boyle and Charles - Physics Stack Exchange
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Deriving ideal gas law from Boyle and Charles
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My textbook states
Notice that since P V=constant P V=constant and V T=constant V T=constant for a given quantity of gas, then P V T P V T should also be a constant.
I tried to prove this, but no success:
P V=a P V=a
V T=b V T=b
P V 2 T=a b P V 2 T=a b
P T=a b P T=a b
But I am not able to cook up P V T P V T... Any help?
thermodynamics
pressure
temperature
ideal-gas
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edited Sep 12, 2020 at 18:44
Thomas Fritsch
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asked Sep 12, 2020 at 9:17
acrossacross
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Related in ChemSE: chemistry.stackexchange.com/a/127772/32355Verktaj –Verktaj 2020-09-12 18:59:58 +00:00 Commented Sep 12, 2020 at 18:59
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This formulation of Boyle's law
P V=const P V=const
is very misunderstandable. Actually the constant on the right side is only meant to be independent of P P and V V. But it may still depend on other parameters, like T T (temperature) and N N (number of molecules). So a better way to write this law is
P V=a(T,N)(1)(1)P V=a(T,N)
where a(T,N)a(T,N) is some unknown function of T T and N N.
Likewise this formulation of Charles's law
V T=const V T=const
is misunderstandable in the same way. A better way to write it is
V T=b(P,N)(2)(2)V T=b(P,N)
where b(P,N)b(P,N) is some unknown function of P P and N N.
Now we can divide equation (1) by T T and multiply equation (2) by P P to get
P V T=a(T,N)T=P b(P,N).P V T=a(T,N)T=P b(P,N).
The only way for this to hold true while varying P P and T T is that a(T,N)T a(T,N)T is independent of T T, and P b(P,N)P b(P,N) is independent of P P. Hence it only depends on N N, and we can call this function c(N)c(N).
So finally we arrived at the combined gas law
P V T=c(N)(3)(3)P V T=c(N)
where c(N)c(N) is some unknown function of N N only.
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edited Sep 12, 2020 at 20:02
answered Sep 12, 2020 at 10:24
Thomas FritschThomas Fritsch
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1 The missing step between (2)(2) and (3)(3) is dividing (1)(1) by T T and multiplying (2)(2) by P P to get P V T=a(T,N)T=P∗b(P,N)P V T=a(T,N)T=P∗b(P,N). The only way for this to hold true while varying P P and T T is if they cancel both out, meaning a(T,N)=T∗c(N)a(T,N)=T∗c(N) and b(P,N)=c(N)P b(P,N)=c(N)P BlueRaja - Danny Pflughoeft –BlueRaja - Danny Pflughoeft 2020-09-12 18:04:07 +00:00 Commented Sep 12, 2020 at 18:04
@BlueRaja-DannyPflughoeft You're right. Thanks for filling the gap in the logic.Thomas Fritsch –Thomas Fritsch 2020-09-12 18:21:00 +00:00 Commented Sep 12, 2020 at 18:21
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P V P V is constant for fixed T T, and V/T V/T is constant for fixed P P. Hence
P V=f(T)P V=f(T) and V/T=g(P)V/T=g(P).
From these we can write
V=f(T)/P=T×g(P)V=f(T)/P=T×g(P).
This implies that
f(T)=k T f(T)=k T and g(P)=k/P g(P)=k/P for some constant k k.
Hence P V/T=k P V/T=k (constant, actually n R n R) is the required answer.
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edited Sep 14, 2020 at 0:45
answered Sep 12, 2020 at 9:33
PeterPeter
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Nice! Pretty sure you meant g(P)=k/P g(P)=k/P in the last but one line...across –across 2020-09-12 18:12:32 +00:00 Commented Sep 12, 2020 at 18:12
Of course I did ;) Thanks - fixed now.Peter –Peter 2020-09-14 00:45:47 +00:00 Commented Sep 14, 2020 at 0:45
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You can't derive it like that because the proportionality relations hold only when the third parameter is kept constant.
However, you can derive the ideal gas law by noting that for high temperature, we get a limit as shown below:
lim p→0 p V¯¯¯¯=f(T)lim p→0 p V¯=f(T)
So, the limit of the product as pressure drops to zero is a unique function f(T)f(T) for all gases independent of the substance used. We can use this to define the linear kelvin scale. Using the triple point of water and absolute zero as our reference,
f(T)=f(T t r i p−p o i n t)273.16 K T f(T)=f(T t r i p−p o i n t)273.16 K T
Where f(T t r i p−p o i n t)f(T t r i p−p o i n t) is the value of the limit at the triple point, using this and our first equation, we can write,
lim p→0 p V¯¯¯¯=f(T t r i p−p o i n t)273.16 K T lim p→0 p V¯=f(T t r i p−p o i n t)273.16 K T
and now, the universal gas constant is defined as follows:
R=f(T t r i p−p o i n t)273.16 K R=f(T t r i p−p o i n t)273.16 K
Which leads us to:
lim p→0 p V¯¯¯¯=R T lim p→0 p V¯=R T
Now, we call an ideal gas is one which obeys the above relation even when the limit is not there.
p V¯¯¯¯=R T p V¯=R T
Reference: from 10:46 of this video
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edited Sep 12, 2020 at 9:42
answered Sep 12, 2020 at 9:37
Clemens BartholdyClemens Bartholdy
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pretty sure you meant ... even when the limit is there. in the last line. thank you for pointing me to that mit course XD across –across 2020-09-12 09:41:12 +00:00 Commented Sep 12, 2020 at 9:41
Ahhh I fixed it now :D Btw are you Indian? If so check out my profile , you may find something there :D Clemens Bartholdy –Clemens Bartholdy 2020-09-12 09:45:25 +00:00 Commented Sep 12, 2020 at 9:45
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Boyle’s Law: V ∝ (1/P) (constant T, n)
Charles’s Law: V ∝ T (constant P, n)
Avogadro’s Hypothesis: V ∝ n (constant T, P)
The combination of the three laws for ideal gases yields to
V ∝ nT/P
you can pass from proportionality to equality by introducing a constant R
V=R n T/P V=R n T/P
and so you have that P V/T=R n P V/T=R n
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answered Sep 12, 2020 at 9:32
gioretiktogioretikto
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If you can just combine the proportionalities like that, then why didn't op's method work?Clemens Bartholdy –Clemens Bartholdy 2020-09-12 10:09:47 +00:00 Commented Sep 12, 2020 at 10:09
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3(adjective)in the sense of shining
a highly polished surface
Synonyms
shining
shining brass buttons
bright
I was convinced that he was brighter than average.
smooth
The flagstones were worn smooth by centuries of use.
gleaming
a gleaming new car
glossy
glossy black hair
The leaves were dark and glossy.
slippery
The floor was wet and slippery.
burnished
glassy
There was a remote, glassy look in his eyes.
furbished
See examples for synonyms
Opposites
dark, rough, matt, dull
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Additional synonyms
in the sense of adept
Definition
proficient in something requiring skill
He is an adept guitar player.
Synonyms
skilful,
able,
skilled,
expert,
masterly,
practised,
accomplished,
versed,
tasty (British, informal),
masterful,
proficient,
adroit,
dexterous
in the sense of bright
I was convinced that he was brighter than average.
Synonyms
intelligent,
smart,
clever,
knowing,
thinking,
quick,
aware,
sharp,
keen,
acute,
alert,
rational,
penetrating,
enlightened,
apt,
astute,
brainy (informal),
wide-awake,
clear-headed,
perspicacious (formal),
quick-witted
in the sense of civilized
All truly civilized countries must deplore torture.
Synonyms
cultured,
educated,
sophisticated,
enlightened,
humane
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4487 | https://cran.r-project.org/web/packages/sets/vignettes/sets.pdf | Generalized and Customizable Sets in R David Meyer FH Technikum Wien Kurt Hornik WU Wirtschaftsuniversität Wien Abstract This introduction to the R package sets is a (slightly) modified version of Meyer and Hornik (2009a), published in the Journal of Statistical Software.
We present data structures and algorithms for sets and some generalizations thereof (fuzzy sets, multisets, and fuzzy multisets) available for R through the sets package. Fuzzy (multi-)sets are based on dynamically bound fuzzy logic families.
Further extensions include user-definable iterators and matching functions.
Keywords: R, set, fuzzy logic, multiset, fuzzy set.
1. Introduction Only few will deny the importance of sets and set theory, building the fundamentals of mod-ern mathematics. For theory-building typically axiomatic approaches (e.g., Zermelo 1908; Fraenkel 1922) are used.
However, even the primal, “naive” concept of sets representing “collections of distinct objects” (Cantor 1895) discarding order and count information seems both natural and practical. The main operation being “is-element-of”, sets alone are of lim-ited practical use—they most of the times serve as basic building blocks for more complex structures such as relations and generalized sets. A common way is to consider pairs (X, m) with set X (“universe”) and membership function m : X →D mapping each member to its “grade”. The subset of X of elements with non-zero membership is called “support”. In multisets, elements may appear more than once, i.e., D = N0 (m is also called the multiplic-ity function). There are many applications in computer science and other disciplines (for a survey, see, e.g., Singh, Ibrahim, Yohanna, and Singh 2007). In statistics, multisets appear as frequency tables. Fuzzy sets have become quite popular since their introduction by Zadeh (1965). Here, the membership function maps into the unit interval. An interesting charac-teristic of fuzzy sets is that the actual behavior of set operations depends on the underlying fuzzy logic employed, which can be chosen according to domain-specific needs. Fuzzy sets are actively used in fields such as machine learning, engineering, medical science, and artificial intelligence (Dubois, Prade, and Yager 1996). Fuzzy multisets (Yager 1986) combine both approaches by allowing each element to map to more than one fuzzy membership grade, i.e., D is the power set of multisets over the unit interval. Examples for the application of fuzzy multisets can be found in the field of information retrieval (e.g., Matthé, Caluwe, de Tré, Hallez, Verstraete, Leman, Cornelis, Moelants, and Gansemans 2006).
The use of sets and variants thereof is common in modern general purpose programming languages: Java and C++ provide corresponding abstract data types (ADTs) in their class libraries, Pascal and Python offer sets as native data type. Indeed, since set elements are order-2 Generalized and Customizable Sets in R invariant and unique, lookup mechanisms can be implemented very efficiently (for example via hashing, resulting in nearly constant run-time complexity, compared to linear search, requiring n/2 steps on the average for n elements). Surprisingly enough, sets are not standard in many mathematical programming environments such as MATLAB and Mathematica, and also R. Although the two latter offer set operations such as union and intersection, these are applied to linearly indexable structures (lists and vectors, respectively), interpreting them as sets. When it comes to R, this emulation is far from complete, and occasionally leads to inconsistent behavior. First of all, the existing infrastructure has no clear concept of how to compare elements, leading to possibly confusing results when different data types are involved in computations: R> s <- list(1, "1") R> union(s, s) 1 "1" R> intersect(s, s) 1 "1" The reason is that most of the existing operations rely on match() which automatically performs type conversions disturbing in this context. Also, quite a few other basic operations such as the Cartesian product, the power set, the subset predicate, etc., are missing, let alone more specialized operations such as the closure under union or intersection.
Then, the current facilities do not make use of a class system, making extensions hard (if not impossible).
Another consequence is that no distinction can be made between sequences (ordered collections of objects) and sets (unordered collections of objects), which is key for the definition of complex structures where both concepts are combined such as relations. Also, there is no support in base R for extensions such as fuzzy sets or multisets.
A few extension packages available from the Comprehensive R Archive Network deal with fuzzy concepts: Package fuzzyFDR (Lewin 2007) calculates fuzzy decision rules for multiple testing, but does not provide any explicit data structures for fuzzy sets. The main functions in fso (Roberts 2007) for fuzzy set ordination compute and return, among other information, membership values represented by numeric matrices for some variables of the the input data.
fuzzyRankTests (Geyer 2007) provides statistical tests based on fuzzy p values and fuzzy confidence intervals, the latter being returned as two separate numeric vectors for values and memberships. The gcl package (Vinterbo 2007) infers fuzzy rules from the input data, encap-sulated in a classifying function returned by the training function. The rules are composed of David Meyer, Kurt Hornik 3 triangular fuzzy sets, represented by triples describing the triangles’ corner points for which the memberships become (0, 1, 0), respectively. Similarly, the FKBL package for fuzzy knowl-edge base learning (Alvarez 2007) uses sequences of triangular fuzzy sets, defined by a vector of corner points. Finally, fuzzyOP (Aklan, Altindas, Macit, Umar, and Unal 2008) provides support for fuzzy numbers: A set of n numbers is represented by a k × 2n numeric matrix, where two consecutive columns represent (at most k) supporting points and memberships, respectively, of the corresponding piecewise linear membership function. If some numbers have fewer supporting points than others, the remaining cells are filled with missing values (NAs).
The sets package (Meyer and Hornik 2009b) presented here provides a flexible and customiz-able basic infrastructure for finite sets and the generalizations mentioned above, including basic operations for fuzzy logic. Apart from complementing the data structures implemented in base R, extension packages like the ones mentioned above could gain in flexibility from building on a common infrastructure, facilitating data exchange and leveraging synergies.
The remainder of the paper is structured as follows.
In Section 2, we discuss the design rationale of data structures and core algorithms. Section 3 introduces the most important set operations. Section 4 starts with constructors and specific methods for generalized sets, followed by a more focused presentation of the fuzzy logic infrastructure, and of functional-ity for handling and visualizing membership information. Section 5 shows how generalized sets can further be customized by specifying user-definable matching functions and iterators.
Section 6 presents three examples before Section 7 concludes.
2. Design issues There are many ways of implementing sets.
Choice and efficiency largely depend on the domain range (i.e., the number of possible values for each element). If the domain is relatively small, i.e. in the range of integral data types such as byte, integer, word etc., the probably most efficient representation is an array of bits representing the domain elements like in Pascal (Wirth 1983). Operations such as union and intersection can then straightforwardly be implemented using logical OR and AND, respectively.
This approach obviously fails for intractably large domains (e.g., strings or recursive objects). Without further application knowledge, one needs to resort to generic container ADTs with efficient element access such as hash tables or search trees (for unique elements). Operations can then be implemented following the classical element-based definitions: Union by inserting all elements of the smaller set into the larger one; intersection by creating a new set with all elements of the smaller set also contained in the larger one; etc.
Clearly, set comparison must be permutation invariant. Some care is needed for nested sets.
Assume, e.g., the comparison of A = {1, {2, 3}} and B = {1, {3, 2}} which clearly are identical.
To implement set equality, a matching operator would be used to check if all elements of A are contained in B. If elements were internally stored in this order during creation, the objects representing {2, 3} and {3, 2} would be different. Comparing two set elements for equality would thus require to recursively compare all elements down the nested structures, which can quickly become infeasible computationally. We avoid this by using a canonical ordering during set creation, guaranteeing that identical sets have identical physical representation as well. We chose to sort elements using the natural order for numeric values, the Unicode 4 Generalized and Customizable Sets in R character representation for strings, and the serialization byte sequence (as strings) for other objects. Eventually, the ordered elements are stored in a list.
For the sets package, further limitations are imposed by the extensions presented in Sections 4 and 5: Generalized sets require, for each element, the membership information, and we also support user-defined, high-level matching functions for comparing elements. Since operations defined for generalized sets basically operate on the memberships, it seems appropriate to store these as (generic) vectors in the same order than the corresponding elements. Thus, memberships of separate sets can simply be combined element-wise.
Many operations (e.g., testing for equality, subsetting, intersection, etc.) are based on match-ing elements of the sets involved. This is implemented by inserting the elements of the larger one into a hash table (we use hashed environments), and to look up the elements of the smaller set in this table (Knuth 1973, p. 391). As hash key, we use the elements’ character representation. Since different objects might map to the same hash key, we actually store the indexes of the list elements, and match the actual objects using a simple linear search. (Note that since the element list is sorted, elements with same representation are grouped, so the search will typically be fast.) The implementation is based on R’s S3 class system, allowing the definition of generic func-tions, dispatching appropriate methods depending on the first argument’s class information.
Objects for sets, generalized sets, and customizable sets have classes ‘set’, ‘gset’, and ‘cset’, respectively, with ‘set’ inheriting from ‘gset’ in turn inheriting from ‘cset’. Suitable opera-tors (such as & for intersection) are then “overloaded” to dispatch the right internal function corresponding to the operands’ classes by defining corresponding methods for group generics.
Additionally, all operations can directly be accessed using the corresponding name combined with a set_, gset_, or cset_ prefix to give the user the choice of up- or downcasts when objects of different class levels are involved in one computation. For example, consider the union of the set {1} and the fuzzy set {2/0.5}: using the generic operator will give an error since the operands’ classes differ. The user needs, in fact, resolve the semantic ambiguity by explicitly choosing the intended operation: If the result should be a generalized (fuzzy) set, gset_union() should be used. To make the result a set (stripping membership information), s/he employs set_union() instead.
3. Sets The basic constructor for creating sets is the set() function accepting any number of R objects as arguments.
R> s <- set(1L, 2L, 3L) R> print(s) {1L, 2L, 3L} For elements that are not sets or atomic vectors of length 1, the print method for sets will use labels indicating the class (and length for vectors): R> set("test", c, set("a", 2.5), list(1, 2)) David Meyer, Kurt Hornik 5 {"test", <>, {"a", 2.5}, <>} Mainly for cosmetic reasons, there is also a tuple class that can be used for vectors: R> set(1, pair(1,2), tuple(1L, 2L, 3L)) {1, (1, 2), (1L, 2L, 3L)} In addition, there is a generic as.set() function coercing suitable objects to sets.
R> s2 <- as.set(2:4) R> print(s2) {2L, 3L, 4L} There are some basic predicate functions (and corresponding generic operators) defined for the (in)equality (!=, ==), (proper) subset (<, <=), (proper) superset (>, >=), and element-of (%e%) operations: R> set_is_empty(set()) TRUE R> set(1) <= set(1,2) TRUE Note that all predicate functions are vectorized for convenience: R> 1:4 %e% set(1L, 2L, 3L) TRUE TRUE TRUE FALSE The sequence 1:4 as one element would be looked up by using list(1:4) on the left-hand side.
The class-specific functions dispatched by the generic operators are set_contains_element(), set_is_equal(), etc. Other than these predicate functions, one can use length() for the cardinality: R> length(s) 3 c() and | for the union, & for the intersection, %D% for the symmetric difference: R> s | set("a") {"a", 1L, 2L, 3L} 6 Generalized and Customizable Sets in R R> s & s2 {2L, 3L} R> s %D% s2 {1L, 4L} and ˆn for the (n-fold) Cartesian product (yielding a set of n-tuples): R> s s2 {(1L, 2L), (1L, 3L), (1L, 4L), (2L, 2L), (2L, 3L), (2L, 4L), (3L, 2L), (3L, 3L), (3L, 4L)} R> s ^ 2L {(1L, 1L), (1L, 2L), (1L, 3L), (2L, 1L), (2L, 2L), (2L, 3L), (3L, 1L), (3L, 2L), (3L, 3L)} and 2ˆ for the power set: R> 2 ^ s {{}, {1L}, {2L}, {3L}, {1L, 2L}, {1L, 3L}, {2L, 3L}, {1L, 2L, 3L}} The class-specific functions set_union(), set_intersection(), and set_symdiff() accept more than two arguments.1 It is also possible to compute the relative complement of a set X in Y , basically removing the elements of X from Y : R> set_complement(set(1), set(1,2,3)) {2, 3} Note, however, that for sets (as opposed to generalized sets), the concept of a “universe” is not necessarily required, and therefore the absolute complement of a set not a well-defined operation. In the sets package, objects of class ‘set’ are special cases of generalized sets. To stay faithful to the simplicity of the original set concept, we define a ‘set’ object’s universe to be identical to the set itself. The absolute complement of a ‘set’ object is therefore always the empty set: R> !set(1) {} 1The n-ary symmetric difference of a collection of sets consists of all elements contained in an odd number of the sets in the collection.
David Meyer, Kurt Hornik 7 set_combn() returns the set of all subsets of specified length: R> ## subsets R> set_combn(s, 2L) {{1L, 2L}, {1L, 3L}, {2L, 3L}} closure() and reduction() compute the closure and reduction under union or intersection for a set family (i.e., a set of sets): R> cl <- closure(set(set(1), set(2), set(3)), "union") R> print(cl) {{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} R> reduction(cl, "union") {{1}, {2}, {3}} The Summary() group methods will also work if defined for the elements: R> sum(s) 6 R> range(s) 1 3 Because set elements are unordered, it is not allowed to use positional subscripting. However, sets can be subset and elements be replaced by using the elements as index themselves: R> s2 <- set(1,2, c, list(1,2)) R> print(s2) {<>, 1, 2, <>} R> s2 <- "foo" R> s2 <- "bar" R> print(s2) {"bar", "foo", 1, 2} R> s2[list("foo", 1)] {"foo", 1} 8 Generalized and Customizable Sets in R Further, iterations over all elements can be carried out using for() and lapply()/sapply(): R> sapply(s, sqrt) 1.000000 1.414214 1.732051 R> for (i in s) print(i) 1 2 3 Note that for() only works because the underlying C code ignores the class information, and directly processes the low-level list representation instead. This will be replaced by a more intelligent “foreach” mechanism as soon as it exists in base R. sapply() and lapply() call the generic as.list() function before iterating over the elements. Since a corresponding method exists for sets objects, this is “safer” than using for().
Using set_outer(), it is possible to apply a function on all factorial combinations of the elements of two sets. If only one set is specified, the function is applied on all pairs of this set.
R> set_outer(set(1,2), set(1,2,3), "/") 1 2 3 1 1 0.5 0.3333333 2 2 1.0 0.6666667 4. Generalized sets There are several extensions of sets such as fuzzy sets and multisets. Both can be be seen as special cases of fuzzy multisets. All have in common that they are defined on some universe of elements, and that each element maps to some membership information. We present how generalized sets are constructed, and demonstrate the effect of choosing different fuzzy logic families.
4.1. Constructors and specific methods Generalized sets are created using the gset() function. The required arguments depend on whether membership information is specified extensionally (listing members) or intensionally (giving a rule for membership): 1. Extensional specification: a) Specify support and memberships as separate vectors. If memberships are omitted, they are assumed to be 1.
David Meyer, Kurt Hornik 9 b) Specify a set of elements along with their individual membership grades, using the element function (e()).
2. Intensional specification: Specify universe and membership function.
Note that for efficiency reasons, gset() will not store elements with zero memberships grades, and the specification of a universe is only required with membership functions. For conve-nience (and storage efficiency), a default universe can be defined using sets_options(). Set-specific universes supersede the default universe, if any. If no universe (general or specific) is defined, the support of a set will be interpreted as its universe. For multisets, the definition of a (general or set-specific) universe can be complemented by a maximum multiplicity or bound.
Without membership information, gset() creates a set (the support is converted to a set internally): R> X <- c("A", "B", "C") R> gset(support = X) {"A", "B", "C"} Note, however, that unlike for ‘set’ objects, it is possible to define a universe that differs from (i.e., is a proper superset of) the support: R> gset(support = X, universe = LETTERS[1:10]) {"A", "B", "C"} A multiset requires an integer membership vector: R> multi <- 1:3 R> gset(support = X, memberships = multi) {"A" , "B" , "C" } For fuzzy sets, the memberships need to be out of the unit interval: R> ms <- c(0.1, 0.3, 1) R> gset(support = X, memberships = ms) {"A" [0.1], "B" [0.3], "C" } Alternatively to separate support/membership specification, each elements can be paired with its membership value using e(): R> gset(elements = list(e("A", 0.1), e("B", 0.2), e("C", 0.3))) {"A" [0.1], "B" [0.2], "C" [0.3]} 10 Generalized and Customizable Sets in R Fuzzy sets can, additionally, be created using a membership function, applied to a specified (or the default) universe: R> f <- function(x) switch(x, A = 0.1, B = 0.2, C = 1, 0) R> gset(universe = X, charfun = f) {"A" [0.1], "B" [0.2], "C" } For fuzzy multisets, the membership argument expects a list of membership grades, either specified as vectors, or as multisets: R> ms2 <- list(c(0.1, 0.3, 0.4), c(1, 1), + gset(support = ms, memberships = multi)) R> gset(support = X, memberships = ms2) {"A" [{0.1, 0.3, 0.4}], "B" [{1 }], "C" [{0.1 , 0.3 , 1 }]} gset_cardinality() returns the (relative) cardinality of a generalized set, computed as the sum (mean) of all memberships. gset_support(), gset_memberships(), gset_height() and gset_core() can be used to retrieve support, memberships, height (maximum member-ship degree), and the core (elements with membership 1), respectively, of a generalized set.
gset_charfun() returns a (point-wise defined) characteristic function for a given gset. Note that in general, this will be different from the characteristic function possibly used for the creation.
As for sets, the usual operations such as union and intersection are available: R> X <- gset(c("A", "B", "C"), 4:6) R> Y <- gset(c("B", "C", "D"), 1:3) R> X | Y {"A" , "B" , "C" , "D" } R> X & Y {"B" , "C" } Additionally, the product (gset_product()), sum (+), and difference (-) of sets are defined, which multiply, add, and subtract multiplicities (or memberships for fuzzy sets): R> X + Y {"A" , "B" , "C" , "D" } R> X - Y {"A" , "B" , "C" } David Meyer, Kurt Hornik 11 R> gset_product(X, Y) {"B" , "C" } For fuzzy (multi-)sets, not only the relative, but also the absolute complement (!) is defined: R> !gset(1, 0.3) {1 [0.7]} R> X <- gset("a", universe = letters[1:3]) R> !X {"b", "c"} R> !!X {"a"} R> !gset(1L, 2, universe = 1:3, bound = 3) {1L , 2L , 3L } gset_mean() creates a new set by averaging corresponding memberships using the arithmetic, geometric or harmonic mean. Note that missing elements have 0 membership degree: R> x <- gset(1:3, 1:3/3) R> y <- gset(1:2, 1:2/2) R> gset_mean(x, y) {1L [0.4166667], 2L [0.8333333], 3L [0.5]} R> gset_mean(x, y, "harmonic") {1L [0.4], 2L [0.8]} R> gset_mean(set(1), set(1, 2)) {1 , 2 [0.5]} The membership vector of a generalized set can be transformed via gset_transform_memberships(), applying any vectorized function to the memberships: R> x <- gset(1:10, 1:10/10) R> gset_transform_memberships(x, pmax, 0.5) 12 Generalized and Customizable Sets in R {1L [0.5], 2L [0.5], 3L [0.5], 4L [0.5], 5L [0.5], 6L [0.6], 7L [0.7], 8L [0.8], 9L [0.9], 10L } Note that for multisets, an element’s membership (multiplicity) m is interpreted as a one-vector of length m, yielding possibly unexpected results: R> x <- gset(1, 2) R> gset_transform_memberships(x, , 0.5) {1 [{0.5 }]} For multisets, the rep() function is a more natural choice for membership transformations: R> rep(x, 0.5) {1} In addition, three convenience functions are defined for fuzzy (multi-)sets: gset_concentrate() and gset_dilate() apply the square and the square root function, and gset_normalize() normalizes the memberships to a specified maximum: R> gset_dilate(y) {1L [0.7071068], 2L } R> gset_concentrate(y) {1L [0.25], 2L } R> gset_normalize(y, 0.5) {1L [0.25], 2L [0.5]} (Note that these functions clearly have no effect for multisets.) 4.2. Fuzzy logic and fuzzy sets For fuzzy (multi-)sets, the user can choose the logic underlying the operations using the fuzzy_logic() function.
Fuzzy logics are represented as named lists with four compo-nents N, T, S, and I containing the corresponding functions for negation, conjunction (“t-norm”), disjunction (“t-conorm”), and (residual) implication (Klement, Mesiar, and Pap 2000). The fuzzy logic is selected by calling fuzzy_logic() with a character string spec-ifying the fuzzy logic “family”, and optional parameters.
The exported functions .N.(), .T.(), .S.(), and .I.() reflect the currently selected bindings. Available families include: "Zadeh" (default), "drastic", "product", "Lukasiewicz", "Fodor", "Frank", "Hamacher", "Schweizer-Sklar", "Yager", "Dombi", "Aczel-Alsina", "Sugeno-Weber", "Dubois-Prade", and "Yu" (see Appendix A). A call to fuzzy_logic() without arguments returns the current logic.
David Meyer, Kurt Hornik 13 R> x <- 1:10 / 10 R> y <- rev(x) R> .S.(x, y) 1.0 0.9 0.8 0.7 0.6 0.6 0.7 0.8 0.9 1.0 R> fuzzy_logic("Fodor") R> .S.(x, y) 1 1 1 1 1 1 1 1 1 1 Fuzzy set operations automatically use the active fuzzy logic setting: R> X <- gset(c("A", "B", "C"), c(0.3, 0.5, 0.8)) R> print(X) {"A" [0.3], "B" [0.5], "C" [0.8]} R> Y <- gset(c("B", "C", "D"), c(0.1, 0.3, 0.9)) R> print(Y) {"B" [0.1], "C" [0.3], "D" [0.9]} First, we try the Zadeh logic (default): R> fuzzy_logic("Zadeh") R> X & Y {"B" [0.1], "C" [0.3]} R> X | Y {"A" [0.3], "B" [0.5], "C" [0.8], "D" [0.9]} R> gset_complement(X, Y) {"B" [0.1], "C" [0.2], "D" [0.9]} The results are different by switching to the Fodor logic: R> fuzzy_logic("Fodor") R> X & Y {"C" [0.3]} R> X | Y 14 Generalized and Customizable Sets in R {"A" [0.3], "B" [0.5], "C" , "D" [0.9]} R> gset_complement(X, Y) {"D" [0.9]} The cut() method for generalized sets “filters” all elements with memberships exceeding a specified level (α-cuts)—the result, thus, is a crisp (multi)set: R> cut(X, 0.5) {"B", "C"} The method can also be used for ν-cuts, selecting elements according to their multiplicity.
4.3. Characteristic functions and their visualization The sets package provides several generators of characteristic functions to be used as templates for the creation of fuzzy sets, including the following shapes: gaussian curve (fuzzy_normal()), double gaussian curve (fuzzy_two_normals()), bell curve (fuzzy_bell()), sigmoid curve (fuzzy_sigmoid()), Π-like curves (fuzzy_pi3(), fuzzy_pi4()), trapezoid (fuzzy_trapezoid()), and triangle (fuzzy_triangular(), fuzzy_cone()). For example, a fuzzy normal function and a corresponding fuzzy set are created using: R> N <- fuzzy_normal(mean = 0, sd = 1) R> N(-3:3) 0.0111090 0.1353353 0.6065307 1.0000000 0.6065307 0.1353353 0.0111090 R> gset(charfun = N, universe = -3:3) {-3L [0.011109], -2L [0.1353353], -1L [0.6065307], 0L , 1L [0.6065307], 2L [0.1353353], 3L [0.011109]} For convenience, we also provide wrappers that directly generate corresponding sets, given a specified universe: R> fuzzy_normal_gset(universe = -3:3) {-3L [0.011109], -2L [0.1353353], -1L [0.6065307], 0L , 1L [0.6065307], 2L [0.1353353], 3L [0.011109]} (If no universe is specified, the default universe is used; if this is also missing, the universe currently defaults to seq(0, 20, 0.1).) It is also possible to create function generators for characteristic functions from other functions (such as distribution functions): R> fuzzy_poisson <- charfun_generator(dpois) R> gset(charfun = fuzzy_poisson(10), universe = seq(0, 20, 2)) David Meyer, Kurt Hornik 15 {0 [0.00036288], 2 [0.018144], 4 [0.1512], 6 [0.504], 8 [0.9], 10 , 12 [0.7575758], 14 [0.4162504], 16 [0.1734377], 18 [0.05667898], 20 [0.01491552]} fuzzy_tuple() generates a sequence (tuple) of sets based on any of the generating functions (except fuzzy_trapezoid() and fuzzy_triangular()). The chosen generating function is called with different values (chosen along the universe) passed to the first argument, thus varying the position or the resulting graph: R> ## creating a series of fuzzy normal sets R> fuzzy_tuple(fuzzy_normal, 5) (<>, <>, <>, <>, <>) (<> denotes an object of class ‘gset’ with 201 elements—the size of the default universe). The sets package provides support for visualizing the membership functions of generalized sets, and in particular fuzzy sets. For (fuzzy) multisets, the plot method produces a (grouped) barplot for the membership vector (see Figure 1, top left): R> ## a fuzzy multiset R> X <- gset(c("A", "B"), list(1:2/2, 0.5)) R> plot(X) Characteristic function generators can directly be plotted using a default universe (see Fig-ure 1, top right): R> plot(fuzzy_bell) There is a plot method for tuples for visualizing a sequence of sets (see Figure 1, bottom left): R> ## creating a sequence of sets R> plot(fuzzy_tuple(fuzzy_cone, 10), col = gray.colors(10)) Plots of several sets can be superposed using the line method (see Figure 1, bottom right): R> x <- fuzzy_normal_gset() R> y <- fuzzy_trapezoid_gset(corners = c(5, 10, 15, 17), height = c(0.7, 1)) R> plot(tuple(x, y), lty = 3) R> lines(x | y, col = 2) R> lines(gset_mean(x, y), col = 3, lty = 2) Finally, we note that the sets package provides basic infrastructure for fuzzy inference (fuzzy_inference()), involving the definition of linguistic variables (fuzzy_variable(), fuzzy_partition()) and fuzzy rules (fuzzy_rule()) to build a fuzzy system (fuzzy_system()). A few methods for defuzzification (gset_defuzzify()) are also provided.
16 Generalized and Customizable Sets in R "A" "B" 0.0 0.2 0.4 0.6 0.8 1.0 Universe Membership Grade 0.0 0.2 0.4 0.6 0.8 1.0 0 5 10 15 20 Universe Membership Grade 0.0 0.2 0.4 0.6 0.8 1.0 0 5 10 15 20 Universe Membership Grade 0.0 0.2 0.4 0.6 0.8 1.0 0 5 10 15 20 Figure 1: Membership plots for fuzzy sets. Top left: grouped barplot for a fuzzy multiset.
Top right: graph of a bell curve. Bottom left: sequence of triangular functions. Bottom right: two combinations of a normal and a trapezoid function (dotted lines: basic shapes; solid (red) line: union; dashed (green) line: arithmetic mean).
5. User-definable extensions We added customizable sets extending generalized sets in two ways: First, users can control the way elements are matched, i.e., define equivalence classes of elements. Second, arbitrary iteration orders can be specified.
5.1. Matching functions By default, sets and generalized sets use identical() to match elements which is maximally restrictive. Note that this differs from the behavior of R’s equality operator or match() which perform implicit type conversions and thus confound, e.g., 1, 1L and "1". In the following example, note that on most computer systems, 3.3 −2.2 will not be identical to 1.1 due to David Meyer, Kurt Hornik 17 numerical issues.
R> x <- set("1", 1L, 1, 3.3 - 2.2, 1.1) R> print(x) {"1", 1L, 1, 1.1, 1.1} R> y <- set(1, 1.1, 2L, "2") R> print(y) {"2", 2L, 1, 1.1} R> 1L %e% y FALSE R> x | y {"1", "2", 1L, 2L, 1, 1.1, 1.1} Customizable sets can be created using the cset() constructor, specifying the generalized set and some matching function.
R> X <- cset(x, matchfun = match) R> print(X) {"1", 1.1} R> Y <- cset(y, matchfun = match) R> print(Y) {"2", 1, 1.1} R> 1L %e% Y TRUE R> X | Y {"1", "2", 1.1} Matching functions take two vector arguments, say, x and table, with table being a vector where the elements of x are looked up. The function should be vectorized in the x, i.e. return the first matching position for each element of x. In order to make use of non-vectorized predicates such as all.equal(), the sets package provides matchfun() to generate one: 18 Generalized and Customizable Sets in R R> FUN <- matchfun(function(x, y) isTRUE(all.equal(x, y))) R> X <- cset(x, matchfun = FUN) R> print(X) {"1", 1L, 1.1} R> Y <- cset(y, matchfun = FUN) R> print(Y) {"2", 2L, 1, 1.1} R> 1L %e% Y TRUE R> X | Y {"1", "2", 1L, 2L, 1.1} sets_options() can be used to conveniently switch the default match and/or order function if a number of ‘cset’ objects need to be created: R> sets_options("matchfun", match) R> cset(x) {"1", 1.1} R> cset(y) {"2", 1, 1.1} R> cset(1:3) <= cset(c("1", "2", "3")) TRUE 5.2. Iterators In addition to specifying matching functions, it is possible to change the order in which itera-tors such as as.list() (but not for()—see end of Section 3) process the elements. Note that the behavior of as.list() influences the labeling and print methods for customizable sets.
Sets and generalized sets use the canonical internal ordering for iterations. With customizable sets, a “natural” ordering of elements can be kept by specifying either a permutation vector or an order function: R> cset(letters[1:5], orderfun = 5:1) David Meyer, Kurt Hornik 19 {"e", "d", "c", "b", "a"} R> FUN <- function(x) order(as.character(x), decreasing = TRUE) R> Z <- cset(letters[1:5], orderfun = FUN) R> print(Z) {"e", "d", "c", "b", "a"} R> as.character(Z) "e" "d" "c" "b" "a" Note that converters for ordered factors keep the order: R> o <- ordered(c("a", "b", "a"), levels = c("b", "a")) R> as.set(o) {a, b} R> as.gset(o) {a , b } R> as.cset(o) {b , a } Converters for other data types will use the order information only if elements are unique: R> as.cset(c("A", "quick", "brown", "fox")) {"A", "quick", "brown", "fox"} R> as.cset(c("A", "quick", "brown", "fox", "quick")) {"A" , "brown" , "fox" , "quick" } 6. Examples In the following, we present two examples for the use of multisets and fuzzy multisets.
6.1. Multisets Multisets are frequent in statistics since they can be seen as frequency tables of some objects.
Using the sets package, a “generalized” table can easily be constructed from a list of R objects using the as.gset() coercion function. Assume, e.g., that one samples a number of fourfold tables given the margins using r2dtable(): 20 Generalized and Customizable Sets in R R> set.seed(4711) R> l <- r2dtable(1000, r = 1:2, c = 2:1) Since the sum of the first row (and second column) are constrained to 1, the top left cell entry can only be 0 or 1. Also, given the marginals, there is only one degree of freedom in fourfold tables, so the value of this first cell determines the others, and thus only two possible tables exist: R> l[1:2] [,1] [,2] [1,] 0 1 [2,] 2 0 [,1] [,2] [1,] 1 0 [2,] 1 1 To count them, we can simply use as.gset() that will construct a multiset from the list: R> s <- as.gset(l) R> print(s) {<<2x2 matrix>> , <<2x2 matrix>> } Replace the matrices by the first cells’ values: R> for (i in s) s <- i R> print(s) {0L , 1L } The estimated probabilities of having 0 or 1 in the first cell can thus be obtained by: R> gset_memberships(s) / 1000 1 2 0.33 0.67 The probability for 0 clearly corresponds to the p value of the corresponding Fisher test: R> fisher.test(l)$p.value 0.3333333 David Meyer, Kurt Hornik 21 6.2. Fuzzy multisets Fuzzy multisets can be used to represent objects appearing several times with different mem-bership grades (e.g., weights, degrees of credibility, . . . ). Mizutani, Inokuchi, and Miyamoto (2008) describe an interesting application of fuzzy multisets to text mining: The occurrences of some terms of interest (“neural network”, “fuzzy”, “image”) in titles, abstracts, and keywords of 30 documents on fuzzy theory are represented by fuzzy multisets, with varying member-ships depending on whether a term occurs in the title (degree 1), the keywords (degree 0.6), and/or the abstract (degree 0.2).
R> data("fuzzy_docs") R> print(fuzzy_docs[8:9]) $x8 {"fuzzy" [{0.2, 0.6}], "neural network" [{0.2, 1}]} $x9 {"fuzzy" [{0.2, 0.6, 1}], "image" [{0.6}], "neural network" [{0.2, 0.6, 1}]} This information is then used to compute distances between documents, and ultimately to compare several (non-linear) clustering methods regarding their abilities of recovering the true underlying structure. In fact, it is known that the first 12 documents are related to neural networks, and the remaining 18 to image processing.
In the following, we will perform simple hierarchical clustering. We start by computing a distance matrix for the 30 documents. The sets package implements the Jaccard dissimilarity, defined for two generalized sets X and Y as 1 −|X ∩Y |/|X ∪Y | where | · | denotes the cardinality for generalized sets. A corresponding dissimilarity matrix can be obtained using, e.g., the proxy package (Meyer and Buchta 2009): R> library("proxy") R> d <- dist(fuzzy_docs, gset_dissimilarity) We then apply Ward’s clustering method: R> cl1 <- hclust(d, "ward") resulting in a clustering depicted in Figure 2. Clearly, the neural network-related documents (#1–#12) are separated from the image processing papers: R> labs1 <- cutree(cl1, 2) R> print(labs1) x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 x15 x16 x17 x18 x19 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 x20 x21 x22 x23 x24 x25 x26 x27 x28 x29 x30 2 2 2 2 2 2 2 2 2 2 2 22 Generalized and Customizable Sets in R x11 x1 x10 x9 x7 x4 x3 x5 x2 x6 x8 x12 x25 x28 x27 x29 x30 x14 x16 x18 x21 x13 x20 x26 x22 x15 x24 x17 x19 x23 0 1 2 3 4 Cluster Dendrogram hclust (, "ward.D") d Height Figure 2: Dendrogram for the fuzzy_docs data, clustered using Ward’s method on Jaccard distances computed from fuzzy multisets.
Note that for this data, using different weightings for terms in titles, keywords and ab-stracts are key to recover the subgroups. Naive text mining approaches operate on “classical” term-document-matrices only, counting term occurrences without further information on their relevance: R> tdm <- set_outer(c("neural networks", "fuzzy", "image"), + fuzzy_docs, %in%) R> print(tdm) x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 neural networks FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE fuzzy TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE image FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE x11 x12 x13 x14 x15 x16 x17 x18 x19 x20 neural networks FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE fuzzy TRUE TRUE TRUE FALSE TRUE FALSE TRUE TRUE FALSE TRUE image TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE x21 x22 x23 x24 x25 x26 x27 x28 x29 x30 neural networks FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE fuzzy TRUE TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE David Meyer, Kurt Hornik 23 image TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE By again computing Jaccard distances R> d <- dist(tdm, "Jaccard", by_rows = FALSE) and the corresponding hierarchical clustering, visualized in Figure 3, R> cl2 <- hclust(d, "ward") we can see that the “standard” text mining approach fails to correctly assign four documents (#9–#12): R> labs2 <- cutree(cl2, 2) R> print(labs2) x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 x15 x16 x17 x18 x19 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 x20 x21 x22 x23 x24 x25 x26 x27 x28 x29 x30 2 2 2 2 2 2 2 2 2 2 2 R> table(labs1, labs2) labs2 labs1 1 2 1 8 4 2 0 18 7. Conclusion In this paper, we described the sets package for R, providing infrastructure for sets and gener-alizations thereof such as fuzzy sets, multisets and fuzzy multisets. The fuzzy variants make use of a dynamic fuzzy logic infrastructure offering several fuzzy logic families. Generalized sets are further extended to allow for user-defined iterators and matching functions. Current work focuses on data structures and algorithms for relations, an important application of sets.
References Aklan S, Altindas E, Macit R, Umar S, Unal H (2008). fuzzyOP: Fuzzy Numbers and the Main Mathematical Operations. R package version 1.0, URL
org/package=fuzzyOP.
Alvarez AG (2007). FKBL: Fuzzy Knowledge Base Learning. R package version 0.50-4, URL
24 Generalized and Customizable Sets in R x8 x7 x6 x5 x4 x3 x1 x2 x26 x25 x24 x23 x22 x21 x20 x18 x17 x15 x13 x12 x11 x9 x10 x30 x29 x28 x27 x19 x14 x16 0 1 2 3 4 5 6 Cluster Dendrogram hclust (, "ward.D") d Height Figure 3: Dendrogram for the fuzzy_docs data, using Ward’s method on a term-document-matrix generated for the data.
Cantor G (1895). “Beiträge zur Begründung der transfiniten Mengenlehre.” In Mathematische Annalen, volume 46, pp. 481–512. Springer-Verlag.
Dubois D, Prade H, Yager RY (eds.) (1996). Fuzzy Information Engineering: A Guided Tour of Applications. John Wiley & Sons, New York.
Fraenkel AA (1922). “Über die Grundlagen der Cantor-Zermeloschen Mengenlehre.” In Math-ematische Annalen, volume 86, pp. 230–237. Springer-Verlag.
Geyer CJ (2007). fuzzyRankTests: Fuzzy Rank Tests and Confidence Intervals. R package version 0.3-2, URL
Klement EP, Mesiar R, Pap E (2000). Triangular Norms. Springer-Verlag, Dordrecht.
Knuth DE (1973). The Art of Computer Programming, volume 3. Addison-Wesley, Reading.
Lewin A (2007). fuzzyFDR: Exact Calculation of Fuzzy Decision Rules for Multiple Testing.
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Matthé T, Caluwe RD, de Tré G, Hallez A, Verstraete J, Leman M, Cornelis O, Moelants D, Gansemans J (2006). “Similarity Between Multi-valued Thesaurus Attributes: Theory and Application in Multimedia Systems.” In Flexible Query Answering Systems, Lecture Notes in Computer Science, pp. 331–342. Springer-Verlag.
David Meyer, Kurt Hornik 25 Meyer D, Buchta C (2009). proxy: Distance and Similarity Measures. R package version 0.4-3, URL
Meyer D, Hornik K (2009a). “Generalized and Customizable Sets in R.” Journal of Statistical Software, 31(2), 1–27. doi:10.18637/jss.v031.i02.
Meyer D, Hornik K (2009b). sets: Sets, Generalized Sets, and Customizable Sets. R package version 1.0, URL
Mizutani K, Inokuchi R, Miyamoto S (2008). “Algorithms of Nonlinear Document Clustering Based on Fuzzy Multiset Model.” International Journal of Intelligent Systems, 23, 176–198.
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“An Overview of the Applications of Multisets.” Novi Sad Journal of Mathematics, 37(3), 73–92.
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A. Available fuzzy logic families Let us refer to N(x) = 1 −x as the standard negation, and, for a t-norm T, let S(x, y) = 1 −T(1 −x, 1 −y) be the dual (or complementary) t-conorm. Available specifications and corresponding families are as follows, with the standard negation used unless stated otherwise.
"Zadeh" Zadeh’s logic with T = min and S = max. Note that the minimum t-norm, also known as the Gödel t-norm, is the pointwise largest t-norm, and that the maximum t-conorm is the smallest t-conorm.
"drastic" The drastic logic with t-norm T(x, y) = y if x = 1, x if y = 1, and 0 otherwise, and complementary t-conorm S(x, y) = y if x = 0, x if y = 0, and 1 otherwise. Note that the drastic t-norm and t-conorm are the smallest t-norm and largest t-conorm, respectively.
"product" The family with the product t-norm T(x, y) = xy and dual t-conorm S(x, y) = x + y −xy.
"Lukasiewicz" The Łukasiewicz logic with t-norm T(x, y) = max(0, x + y −1) and dual t-conorm S(x, y) = min(x + y, 1).
26 Generalized and Customizable Sets in R "Fodor" The family with Fodor’s nilpotent minimum t-norm given by T(x, y) = min(x, y) if x + y > 1, and 0 otherwise, and the dual t-conorm given by S(x, y) = max(x, y) if x + y < 1, and 1 otherwise.
"Frank" The family of Frank t-norms Tp, p ≥0, which gives the Zadeh, product and Łukasiewicz t-norms for p = 0, 1, and ∞, respectively, and otherwise is given by T(x, y) = logp(1 + (px −1)(py −1)/(p −1)).
"Hamacher" The three-parameter family of Hamacher, with negation Nγ(x) = (1 −x)/(1 + γx), t-norm Tα(x, y) = xy/(α + (1 −α)(x + y −xy)), and t-conorm Sβ(x, y) = (x + y + (β −1)xy)/(1 + βxy), where α ≥0 and β, γ ≥−1. This gives a deMorgan triple (for which N(S(x, y)) = T(N(x), N(y)) iffα = (1 + β)/(1 + γ).
The following parametric families are obtained by combining the corresponding families of t-norms with the standard negation and complementary t-conorm.
"Schweizer-Sklar" The Schweizer-Sklar family Tp, −∞≤p ≤∞, which gives the Zadeh (minimum), product and drastic t-norms for p = −∞, 0, and ∞, respectively, and otherwise is given by Tp(x, y) = max(0, (xp + yp −1)1/p).
"Yager" The Yager family Tp, p ≥0, which gives the drastic and minimum t-norms for p = 0 and ∞, respectively, and otherwise is given by Tp(x, y) = max(0, 1 −((1 −x)p + (1 − y)p)1/p).
"Dombi" The Dombi family Tp, p ≥0, which gives the drastic and minimum t-norms for p = 0 and ∞, respectively, and otherwise is given by Tp(x, y) = 0 if x = 0 or y = 0, and Tp(x, y) = 1/(1 + ((1/x −1)p + (1/y −1)p)1/p) if both x > 0 and y > 0.
"Aczel-Alsina" The family of t-norms Tp, p ≥0, introduced by Aczél and Alsina, which gives the drastic and minimum t-norms for p = 0 and ∞, respectively, and otherwise is given by Tp(x, y) = exp(−(| log(x)|p + | log(y)|p)1/p).
"Sugeno-Weber" The family of t-norms Tp, −1 ≤p ≤∞, introduced by Weber with dual t-conorms introduced by Sugeno, which gives the drastic and product t-norms for p = −1 and ∞, respectively, and otherwise is given by Tp(x, y) = max(0, (x+y−1+pxy)/(1+p)).
"Dubois-Prade" The family of t-norms Tp, 0 ≤p ≤1, introduced by Dubois and Prade, which gives the minimum and product t-norms for p = 0 and 1, respectively, and otherwise is given by Tp(x, y) = xy/ max(x, y, p).
"Yu" The family of t-norms Tp, p ≥−1, introduced by Yu, which gives the product and drastic t-norms for p = −1 and ∞, respectively, and otherwise is given by T(x, y) = max(0, (1 + p)(x + y −1) −pxy).
David Meyer, Kurt Hornik 27 Affiliation: David Meyer Department of Information Systems and Operations WU Wirtschaftsuniversität Wien Augasse 2–6 1090 Wien, Austria E-mail: David.Meyer@wu.ac.at URL: Kurt Hornik Department of Statistics and Mathematics WU Wirtschaftsuniversität Wien Augasse 2–6 1090 Wien, Austria E-mail: Kurt.Hornik@wu.ac.at URL: |
4488 | http://mathcentral.uregina.ca/QQ/database/QQ.09.03/fazia1.html | The product of any n consecutive integers
Quandaries and Queries
I have to prove that the product of any n consecutive integers is divisible by n!(eg:the product of five consecutive integers is divisible by 5!) It's easy enough to plug in numbers and prove it is true, but i'm finding it difficult to come up with a proper proof. Can someone help me please?
Many Thanks
Fazia
(university mathematics student)
Let M be the largest of the n numbers.
First ask yourself: In how many ways can you choose a n-member committee consisting of a president, a vice-president, a vice-vice-president (= vice 2-president), ..., and a vice-vice-...-vice-president (= vice n-1-president) among M people.
Then ask yourself: By how much should you divide that number to get the number of n-member committees chosen among M people, if the committee members have no title.
The answer to 2) is relevant to your question, but most importantly
it is an integer!
Claude
Go to Math Central |
4489 | https://math.stackexchange.com/questions/2380261/what-is-the-tangent-line-to-y-e-fracx2-through-0-0 | wolfram alpha - What is the tangent line to $y=e^{^{\frac{x}{2}}}$ through (0,0)? - Mathematics Stack Exchange
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What is the tangent line to y=e x 2 y=e x 2 through (0,0)?
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I'm trying to solve
y=e x 2 y=e x 2
The derivative:
e x√2 e x 2
So, I need to find the "slope" to the linear function y−y 1=m(x−x 1)y−y 1=m(x−x 1), solving the derivative by replacing x x by 0 0 is m=1 2 m=1 2, so the answer is:
y−y 1=m(x−x 1),y−0=1 2(x−0),y=1 2 x y−y 1=m(x−x 1),y−0=1 2(x−0),y=1 2 x
But the answer that Wolfram Alpha gives me is:
e x 2 e x 2
So, does this problem requires another formula or process to be solved? Or did I just fail in the process?
Greetings!
Feel free to edit the post if there are any English issues in the post, I appreciate it so much!
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edited Aug 2, 2017 at 17:00
Simply Beautiful Art
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asked Aug 2, 2017 at 16:52
Andrés David Hernández SánchezAndrés David Hernández Sánchez
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If you put x=0 x=0 in f′=1 2 e x 2 f′=1 2 e x 2 slope is 1 2 e 0=1 2 1 2 e 0=1 2 Khosrotash –Khosrotash 2017-08-02 16:55:00 +00:00 Commented Aug 2, 2017 at 16:55
What did you put into WA?Simply Beautiful Art –Simply Beautiful Art 2017-08-02 16:59:48 +00:00 Commented Aug 2, 2017 at 16:59
Anyone else get the strong feeling answerers will probably misread the entire question? Perhaps the WA tag might ring some bells as to what seems to be the real question.Simply Beautiful Art –Simply Beautiful Art 2017-08-02 17:00:51 +00:00 Commented Aug 2, 2017 at 17:00
@SimplyBeautifulArt I don't understand your comment. My answer agrees with WA. And it is the correct way to solve the question as it is formulated.mfl –mfl 2017-08-02 17:05:05 +00:00 Commented Aug 2, 2017 at 17:05
@mfl Sorry, and yes, I did see your answer (+1 to you)Simply Beautiful Art –Simply Beautiful Art 2017-08-02 17:05:47 +00:00 Commented Aug 2, 2017 at 17:05
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Firs of all note that (0,0)(0,0) is not a point of the graph of the function. So fix x 0∈R.x 0∈R. The tangent line at (x 0,y 0)(x 0,y 0) is given by
y−e x 0/2=1 2 e x 0/2(x−x 0).y−e x 0/2=1 2 e x 0/2(x−x 0).
(Note that f′(x)=1 2 e x/2 f′(x)=1 2 e x/2.) If this line contains the point (0,0)(0,0) then we have
e x 0/2=1 2 e x 0/2 x 0.e x 0/2=1 2 e x 0/2 x 0.
Since e x 0/2≠0 e x 0/2≠0 we get that x 0=2.x 0=2. Thus the tangent line is
y−e=e 2(x−2)y−e=e 2(x−2)
That is
y=e 2 x.y=e 2 x.
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answered Aug 2, 2017 at 16:57
mflmfl
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So many thanks @mfl! I have a question.. Why this e x 0 2≠0 e x 0 2≠0? And why x 0=2 x 0=2?Andrés David Hernández Sánchez –Andrés David Hernández Sánchez 2017-08-02 17:08:12 +00:00 Commented Aug 2, 2017 at 17:08
@AndrésDavidHernándezSánchez x 0=2 x 0=2 was just an example. You didn't specify which point you wanted to take the tangent line at in WA probably.Simply Beautiful Art –Simply Beautiful Art 2017-08-02 17:09:46 +00:00 Commented Aug 2, 2017 at 17:09
We have that e x>0 e x>0 for all real x.x. So it is ≠0.≠0. Thus we can divide both terms of the equality e x 0/2=1 2 e x 0/2 x 0 e x 0/2=1 2 e x 0/2 x 0 by the nonzero number e x 0/2.e x 0/2. Thus we get 1=1 2 x 0.1=1 2 x 0. That is, x 0=2.x 0=2.mfl –mfl 2017-08-02 17:09:55 +00:00 Commented Aug 2, 2017 at 17:09
@SimplyBeautifulArt You don't need to say the point where you take the tangent. See wolframalpha.com/input/…mfl –mfl 2017-08-02 17:12:53 +00:00 Commented Aug 2, 2017 at 17:12
No everything have sence! So many thanks @mfl, I never expect to do this analisys for e x>0 e x>0...Andrés David Hernández Sánchez –Andrés David Hernández Sánchez 2017-08-02 17:16:48 +00:00 Commented Aug 2, 2017 at 17:16
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How did you take the derivative here? Remember that d d x(e x)=e x d d x(e x)=e x and try using the chain rule.
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answered Aug 2, 2017 at 16:53
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4490 | https://en.wikipedia.org/wiki/Sum_of_squares | Jump to content
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From Wikipedia, the free encyclopedia
In mathematics, statistics and elsewhere, sums of squares occur in a number of contexts:
Statistics
[edit]
For partitioning of variance, see Partition of sums of squares
For the "sum of squared deviations", see Least squares
For the "sum of squared differences", see Mean squared error
For the "sum of squared error", see Residual sum of squares
For the "sum of squares due to lack of fit", see Lack-of-fit sum of squares
For sums of squares relating to model predictions, see Explained sum of squares
For sums of squares relating to observations, see Total sum of squares
For sums of squared deviations, see Squared deviations from the mean
For modelling involving sums of squares, see Analysis of variance
For modelling involving the multivariate generalisation of sums of squares, see Multivariate analysis of variance
Number theory
[edit]
For the sum of squares of consecutive integers, see Square pyramidal number
For representing an integer as a sum of squares of 4 integers, see Lagrange's four-square theorem
Legendre's three-square theorem states which numbers can be expressed as the sum of three squares
Jacobi's four-square theorem gives the number of ways that a number can be represented as the sum of four squares.
For the number of representations of a positive integer as a sum of squares of k integers, see Sum of squares function.
Fermat's theorem on sums of two squares says which primes are sums of two squares.
The sum of two squares theorem generalizes Fermat's theorem to specify which composite numbers are the sums of two squares.
Pythagorean triples are sets of three integers such that the sum of the squares of the first two equals the square of the third.
A Pythagorean prime is a prime that is the sum of two squares; Fermat's theorem on sums of two squares states which primes are Pythagorean primes.
Pythagorean triangles with integer altitude from the hypotenuse have the sum of squares of inverses of the integer legs equal to the square of the inverse of the integer altitude from the hypotenuse.
Pythagorean quadruples are sets of four integers such that the sum of the squares of the first three equals the square of the fourth.
The Basel problem, solved by Euler in terms of , asked for an exact expression for the sum of the squares of the reciprocals of all positive integers.
Rational trigonometry's triple-quad rule and triple-spread rule contain sums of squares, similar to Heron's formula.
Squaring the square is a combinatorial problem of dividing a two-dimensional square with integer side length into smaller such squares.
Algebra, algebraic geometry, and optimization
[edit]
Polynomial SOS, polynomials that are sums of squares of other polynomials
The Brahmagupta–Fibonacci identity, representing the product of sums of two squares of polynomials as another sum of squares
Hilbert's seventeenth problem on characterizing the polynomials with non-negative values as sums of squares
Sum-of-squares optimization, nonlinear programming with polynomial SOS constraints
The sum of squared dimensions of a finite group's pairwise nonequivalent complex representations is equal to cardinality of that group.
Euclidean geometry and other inner-product spaces
[edit]
The Pythagorean theorem says that the square on the hypotenuse of a right triangle is equal in area to the sum of the squares on the legs. The sum of squares is not factorable.
The squared Euclidean distance between two points, equal to the sum of squares of the differences between their coordinates
Heron's formula for the area of a triangle can be re-written as using the sums of squares of a triangle's sides (and the sums of the squares of squares)
The British flag theorem for rectangles equates two sums of two squares
The parallelogram law equates the sum of the squares of the four sides to the sum of the squares of the diagonals
Descartes' theorem for four kissing circles involves sums of squares
The sum of the squares of the edges of a rectangular cuboid equals the square of any space diagonal
See also
[edit]
Sums of powers
Sum of reciprocals
Quadratic form (statistics)
Reduced chi-squared statistic
Index of articles associated with the same name
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Binomial
A binominal is a polynominal with two terms, the sum of two monominals. It is common practice to bound binominals by brackets or parenthesis when operated upon.
Simple Operations
The binomial can be factored as a product of two other binomials, and .
The binomial can be factored as the product of two complex numbers, and .
A binomial to the nth power can be expanded using the binomial theorem or Pascal's triangle.
See Also
Binomial Theorem
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4492 | https://handwiki.org/wiki/Telescoping_series | Telescoping series - HandWiki
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Short description: Series whose partial sums eventually only have a fixed number of terms after cancellation
In mathematics, a telescoping series is a series whose general term t n is of the form t n=a n+1−a n, i.e. the difference of two consecutive terms of a sequence (a n).
As a consequence the partial sums only consists of two terms of (a n) after cancellation. The cancellation technique, with part of each term cancelling with part of the next term, is known as the method of differences.
For example, the series
∑n=1∞1 n(n+1)
(the series of reciprocals of pronic numbers) simplifies as
∑n=1∞1 n(n+1)=∑n=1∞(1 n−1 n+1)=lim N→∞∑n=1 N(1 n−1 n+1)=lim N→∞[(1−1 2)+(1 2−1 3)+⋯+(1 N−1 N+1)]=lim N→∞[1+(−1 2+1 2)+(−1 3+1 3)+⋯+(−1 N+1 N)−1 N+1]=lim N→∞[1−1 N+1]=1.
An early statement of the formula for the sum or partial sums of a telescoping series can be found in a 1644 work by Evangelista Torricelli, De dimensione parabolae.
[x]
Contents
1 In general
2 More examples
3 An application in probability theory
4 Similar concepts
4.1 Telescoping product
5 Other applications
6 References
In general
A telescoping series of powers. Note in the summation sign, ∑, the index n goes from 1 to m. There is no relationship between n and m beyond the fact that both are natural numbers.
Telescoping sums are finite sums in which pairs of consecutive terms cancel each other, leaving only the initial and final terms.
Let a n be a sequence of numbers. Then,
∑n=1 N(a n−a n−1)=a N−a 0
If a n→0
∑n=1∞(a n−a n−1)=−a 0
Telescoping products are finite products in which consecutive terms cancel denominator with numerator, leaving only the initial and final terms.
Let a n be a sequence of numbers. Then,
∏n=1 N a n−1 a n=a 0 a N
If a n→1
∏n=1∞a n−1 a n=a 0
More examples
Many trigonometric functions also admit representation as a difference, which allows telescopic canceling between the consecutive terms. ∑n=1 N sin(n)=∑n=1 N 1 2 csc(1 2)(2 sin(1 2)sin(n))=1 2 csc(1 2)∑n=1 N(cos(2 n−1 2)−cos(2 n+1 2))=1 2 csc(1 2)(cos(1 2)−cos(2 N+1 2)).
Some sums of the form ∑n=1 N f(n)g(n) where f and g are polynomial functions whose quotient may be broken up into partial fractions, will fail to admit summation by this method. In particular, one has ∑n=0∞2 n+3(n+1)(n+2)=∑n=0∞(1 n+1+1 n+2)=(1 1+1 2)+(1 2+1 3)+(1 3+1 4)+⋯⋯+(1 n−1+1 n)+(1 n+1 n+1)+(1 n+1+1 n+2)+⋯=∞. The problem is that the terms do not cancel.
Let k be a positive integer. Then ∑n=1∞1 n(n+k)=H k k where H k is the k th harmonic number. All of the terms after 1/(k − 1) cancel.
Let k,m with k≠m be positive integers. Then ∑n=1∞1(n+k)(n+k+1)…(n+m−1)(n+m)=1 m−k⋅k!m!
An application in probability theory
In probability theory, a Poisson process is a stochastic process of which the simplest case involves "occurrences" at random times, the waiting time until the next occurrence having a memorylessexponential distribution, and the number of "occurrences" in any time interval having a Poisson distribution whose expected value is proportional to the length of the time interval. Let X t be the number of "occurrences" before time t, and let T x be the waiting time until the x th "occurrence". We seek the probability density function of the random variableT x. We use the probability mass function for the Poisson distribution, which tells us that
Pr(X t=x)=(λ t)x e−λ t x!,
where λ is the average number of occurrences in any time interval of length 1. Observe that the event {X t ≥ x} is the same as the event {T x ≤ t}, and thus they have the same probability. Intuitively, if something occurs at least x times before time t, we have to wait at most t for the x t h occurrence. The density function we seek is therefore
f(t)=d d t Pr(T x≤t)=d d t Pr(X t≥x)=d d t(1−Pr(X t≤x−1))=d d t(1−∑u=0 x−1 Pr(X t=u))=d d t(1−∑u=0 x−1(λ t)u e−λ t u!)=λ e−λ t−e−λ t∑u=1 x−1(λ u t u−1(u−1)!−λ u+1 t u u!)
The sum telescopes, leaving
f(t)=λ x t x−1 e−λ t(x−1)!.
Similar concepts
Telescoping product
A telescoping product is a finite product (or the partial product of an infinite product) that can be cancelled by method of quotients to be eventually only a finite number of factors.
For example, the infinite product
∏n=2∞(1−1 n 2)
simplifies as
∏n=2∞(1−1 n 2)=∏n=2∞(n−1)(n+1)n 2=lim N→∞∏n=2 N n−1 n×∏n=2 N n+1 n=lim N→∞[1 2×2 3×3 4×⋯×N−1 N]×[3 2×4 3×5 4×⋯×N N−1×N+1 N]=lim N→∞[1 2]×[N+1 N]=1 2×lim N→∞[N+1 N]=1 2×lim N→∞[N N+1 N]=1 2.
Other applications
For other applications, see:
Grandi's series;
Proof that the sum of the reciprocals of the primes diverges, where one of the proofs uses a telescoping sum;
Fundamental theorem of calculus, a continuous analog of telescoping series;
Order statistic, where a telescoping sum occurs in the derivation of a probability density function;
Lefschetz fixed-point theorem, where a telescoping sum arises in algebraic topology;
Homology theory, again in algebraic topology;
Eilenberg–Mazur swindle, where a telescoping sum of knots occurs;
Faddeev–LeVerrier algorithm.
References
↑Apostol, Tom (1967). Calculus, Volume 1 (Second ed.). John Wiley & Sons. p.386.
↑Tom M. Apostol, Calculus, Volume 1, Blaisdell Publishing Company, 1962, pages 422–3
↑Brian S. Thomson and Andrew M. Bruckner, Elementary Real Analysis, Second Edition, CreateSpace, 2008, page 85
↑"Prehistory of the zeta-function". Number Theory, Trace Formulas and Discrete Groups: Symposium in Honor of Atle Selberg, Oslo, Norway, July 14–21, 1987. Boston, Massachusetts: Academic Press. 1989. pp.1–9. doi:10.1016/B978-0-12-067570-8.50009-3.
↑Weisstein, Eric W.. "Telescoping Sum" (in en). Wolfram.
↑ Jump up to: 6.06.1"Telescoping Series - Product" (in en-us). Brilliant.org.
↑Bogomolny, Alexander. "Telescoping Sums, Series and Products".
| Collapse v t e Sequences and series |
| Integer sequences | | Basic | Arithmetic progression Geometric progression Harmonic progression Square number Cubic number Factorial Powers of two Powers of 10 | | Advanced (list) | Complete sequence Fibonacci numbers Figurate number Heptagonal number Hexagonal number Lucas number Pell number Pentagonal number Polygonal number Triangular number | | |
| Properties of sequences | Cauchy sequence Monotone sequence Periodic sequence |
| Properties of series | Convergent series Divergent series Conditional convergence Absolute convergence Uniform convergence Alternating series Telescoping series |
| Explicit series | | Convergent | 1/2 − 1/4 + 1/8 − 1/16 + ⋯ 1/2 + 1/4 + 1/8 + 1/16 + ⋯ 1/4 + 1/16 + 1/64 + 1/256 + ⋯ 1 + 1/2 s+ 1/3 s + ... (Riemann zeta function) | | Divergent | 1 + 1 + 1 + 1 + ⋯ 1 + 2 + 3 + 4 + ⋯ 1 + 2 + 4 + 8 + ⋯ 1 − 1 + 1 − 1 + ⋯ (Grandi's series) Infinite arithmetic series 1 − 2 + 3 − 4 + ⋯ 1 − 2 + 4 − 8 + ⋯ 1 + 1/2 + 1/3 + 1/4 + ⋯ (harmonic series) 1 − 1 + 2 − 6 + 24 − 120 + ⋯ (alternating factorials) 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ⋯ (inverses of primes) | |
| Kinds of series | Taylor series Power series Formal power series Laurent series Puiseux series Dirichlet series Trigonometric series Fourier series Generating series |
| Hypergeometric series | Generalized hypergeometric series Hypergeometric function of a matrix argument Lauricella hypergeometric series Modular hypergeometric series Riemann's differential equation Theta hypergeometric series |
| Book Category |
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4493 | https://library.rangercollege.edu/scientificmethod/variables | Skip to Main Content
“The mission of Ranger College is to transform lives and give students the skills to be a positive influence in their communities.”
Scientific Method
Scientific Method
What is scientific research
Qualitative vs. Quantitative
Independent and Dependent Variables
Independent and Dependent Variables
In an experiment, the independent variable is the variable that is varied or manipulated by the researcher.
The dependent variable is the response that is measured. One way to think about it is that the dependent variable dependson the change in the independent variable. In theory, a change in the independent variable will lead to a change in the dependent variable.
Example 1:
In a study of how different doses of a drug affect the severity of symptoms, a researcher could compare the frequency and intensity of symptoms when different doses are administered.
Here the independent variable is the dose and the dependent variable is the frequency/intensity of symptoms.
Example 2:
The rudder on a boat directs the course of the boat. By changing the position of the rudder (turning it left or right), the rudder moves a certain way in the water, and that movement changes the trajectory of the boat.
Here the independent variable is the rudder, while the dependent variable is the trajectory of the boat.
Tips:
Independent and dependent variables are often referred to in other ways. For instance, independent variables are sometimes called experimental variables or predictor variables. Dependent variables are sometimes called outcome variables.
One way to differentiate between whether a variable is independent or dependent is to consider when each variable occurred. Typically, the independent variable will be the variable that happened earlier. Meaning, if I am looking at a dataset that has a variable for the year someone was born and a variable for their level of happiness in 2019, it’s a good bet that the birth year is the independent variable because it happened before the current measure of happiness in 2019 (assuming we are not surveying newborn babies). In effect, this question would be measuring whether someone’s year of birth (maybe translated as generation affiliation) relates to how happy they are in 2019.
<< Previous: Qualitative vs. Quantitative
Please contact us at 254-647-1414 or library@rangercollege.edu. |
4494 | https://www.ambrell.com/induction-heating-applications/shrink-fitting | Induction Shrink Fitting - Ambrell
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How it's Used Processes All Processes Aluminum Brazing Annealing Atmosphere Controlled Brazing Automotive Related Notes Bonding Brazing Cap Sealing Processes:More Carbide Heating Catheter Tipping Curing Fastener Manufacturing Fiber Optic Sealing Fluid Heating Forging Getter Firing Processes:More Hardening Heating Heat Staking Hot Forming Levitation Melting Material Testing Medical Device Manufacturing Melting Processes:More Metal-to-Glass Sealing Other Heating Processes Packaging Shrink Fitting Soldering Susceptor Heating Application Videos
Aluminum Brazing
Annealing
Atmosphere Controlled Brazing
Automotive Applications
Bonding
Brazing
Cap Sealing
Carbide Heating
Catheter Tipping
Curing
Fastener Manufacturing
Fiber Optic Heating
Fluid Heating
Forging
Getter Firing
Hardening
Heating
Heat Staking
Hot Forming
Levitation Melting
Material Testing
Medical Device Manufacturing
Melting
Metal-to-Glass
Other Heating Processes
Packaging
Shrink Fitting
Soldering
Susceptor Heating
Where it's Used Industries: All Industries Aerospace & Defense Aluminum Brazing Automotive Industry Brake Rotor Heating Industries:More Brazing Drill Bits Crystal Growing Electric Vehicle Production Fastener Heating Industries:More Forging Industry Heat Staking Heat Treating Medical Device Manufacture Industries:More Nanoparticle Heating Packaging Shell Annealing Tube and Pipe Heating
Aerospace & Defense
Aluminum Brazing
Automotive Industry
Brake Rotor Heating
Crystal Growing
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Electric Vehicle Production
Fastener Heating
Forging
Heat Staking
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Medical Applications
Nanoparticle Heating
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Information Learn: Learn About Induction What is Induction? Green Technology Green Energy Calculator Coil Design Guide Pro Skills Webinar Brazing Guide Learn:More Application Notes Application Videos Training Videos Industry 4.0 Technical Articles Our YouTube Channel FAQs About: About Us Careers Mission & Quality Principles Trade Shows Our Sales Team Our Channel Partners Find a Distributor About:More Gov't Contracting Info Newsroom Testimonials Feedback Patents ISO 9001 Certificate Sitemap
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How it's Used Processes All Processes Aluminum Brazing Annealing Atmosphere Controlled Brazing Automotive Related Notes Bonding Brazing Cap Sealing Processes:More Carbide Heating Catheter Tipping Curing Fastener Manufacturing Fiber Optic Sealing Fluid Heating Forging Getter Firing Processes:More Hardening Heating Heat Staking Hot Forming Levitation Melting Material Testing Medical Device Manufacturing Melting Processes:More Metal-to-Glass Sealing Other Heating Processes Packaging Shrink Fitting Soldering Susceptor Heating Application Videos
Aluminum Brazing
Annealing
Atmosphere Controlled Brazing
Automotive Applications
Bonding
Brazing
Cap Sealing
Carbide Heating
Catheter Tipping
Curing
Fastener Manufacturing
Fiber Optic Heating
Fluid Heating
Forging
Getter Firing
Hardening
Heating
Heat Staking
Hot Forming
Levitation Melting
Material Testing
Medical Device Manufacturing
Melting
Metal-to-Glass
Other Heating Processes
Packaging
Shrink Fitting
Soldering
Susceptor Heating
Where it's Used Industries: All Industries Aerospace & Defense Aluminum Brazing Automotive Industry Brake Rotor Heating Industries:More Brazing Drill Bits Crystal Growing Electric Vehicle Production Fastener Heating Industries:More Forging Industry Heat Staking Heat Treating Medical Device Manufacture Industries:More Nanoparticle Heating Packaging Shell Annealing Tube and Pipe Heating
Aerospace & Defense
Aluminum Brazing
Automotive Industry
Brake Rotor Heating
Crystal Growing
Drill Bit Brazing
Electric Vehicle Production
Fastener Heating
Forging
Heat Staking
Heat Treating
Medical Applications
Nanoparticle Heating
Packaging
Shell Annealing
Tube & Pipe Heating
Products & Services Products: Induction Systems New: EKOHEAT 2 Power Systems Workheads Accessories Cooling Systems Products:More Rental Plan Trade In Program Get a Quote FAQs Services: All Services Applications Lab Free Consultation Coil Design & Repair SmartCare Service Institutional Incentives Services:More Training Videos Lab Service Request Document Support FAQs Help Tickets
Induction Systems
Workheads
EKOHEAT 2 Power Systems
Accessories
Cooling Systems
Rental Plans
Trade-in Program
Get a Quote
FAQs
Services
Applications Lab
Free Consultation
SmartCARE Service
Institutional Incentives
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Document Support
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FAQs
Information Learn: Learn About Induction What is Induction? Green Technology Green Energy Calculator Coil Design Guide Pro Skills Webinar Brazing Guide Learn:More Application Notes Application Videos Training Videos Industry 4.0 Technical Articles Our YouTube Channel FAQs About: About Us Careers Mission & Quality Principles Trade Shows Our Sales Team Our Channel Partners Find a Distributor About:More Gov't Contracting Info Newsroom Testimonials Feedback Patents ISO 9001 Certificate Sitemap
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Induction Shrink Fitting
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How It's Used
Shrink Fitting
How does Induction Shrink Fitting Work?
Induction shrink fitting is a process in which a precise electromagnetic field is used to heat and thermally expand an electrically conductive material prior to assembly, resulting in an interference fit joint upon return to ambient temperature.
Can we help?
shrink fitting Application Notes
Select from our collection of shrink fitting notes, developed over 39 years supporting our customers. Read how we helped to solve their process heating challenges!
Reverse Shrink Fitting
Heat customer supplied parts to 300-400 °F for a shrink fitting and a reverse shrink fitting application
Shrink Fitting a Water Pump Shaft
Shrink fit customer supplied water pump parts to 800 °F
Shrink Fitting Steel Pipe Assemblies
To shrink fit steel pipe assemblies; the end product is a glass processing system
Press Fitting Steel Sleeves
Induction is typically a faster heating method for shrink fitting than alternatives like oven heating, offering instant on/instant off heating and
localized heating, making it an efficient option.
Shrink Fitting Magnetic Steel Pistons
Using induction for shrink-fitting magnetic steel pistons onto a chrome shaft; the client was using an unreliable handheld induction system and wanted a higher quality induction shrink-fitting solution.
Shrink Fitting Hammer Bits
Shrink-fitting hammer bits with induction for the insertion of carbide buttons; the end product is a drilling tool for the oil and gas industry.
Shrink Fitting a Magnetic Steel Gear
Induction shrink-fitting with EASYHEAT takes 45 minutes to heat the sample to the required temperature. The current oven process takes over two hours.
Shrink Fitting of Roller Bearings
A multi-turn helical coil delivers uniform heat to the entire range of the bearing sizes in the transverse mode.
Shrink Fitting an Aluminum Tube
With an EASYHEAT 2 kW induction shrink-fitting system, the aluminum tube heated to the required temperature within 30 seconds. Shrink fitting then took place
Shrink fitting an automotive aluminum motor housing
Induction shrink-fitting is fast, presents significant energy savings over an electric oven, requires a more modest footprint than an oven and can be easily integrated into an automated process
Inserting a Steel Bushing to an Aluminum Hub
The client currently uses an electric oven and the heating time is two hours, so at 60 seconds, the time savings with induction shrink-fitting is very significant
Shrink Fitting A Gear to a Shaft
The customer was using a torch, which can lead to inconsistent part quality. Induction's precise heating means the client can count on consistent results in their process
Shrink Fitting an Aluminum Motor Housing
The customer was using a cold press, but it was creating part defects. This was resolved with induction heating: the process took just two minutes compared to 40 minutes and they were able to achieve their targeted production rate.
Shrink Fitting Stainless-Steel Sleeve & Shaft
The customer currently uses electric ovens that run 24 hours a day, 5 days per week and their primary concern is to save on the energy cost of heating the parts in the ovens.
Shrinkfitting a steel mud pump liner
A twenty-turn helical coil is used to heat the chain. The chain is fed through the coil at a rate of 1 meter per minute to reach the desired 1760 °F (960 °C) for the tempering process...
Shrink fit a steel gear onto a steel gear motor shaft
A four-turn helical internal coil is used to heat the gear bore. The coil is inserted into the gear bore and power is applied for 90 seconds to reach the required 400 °F (204 °C) and expand the gear bore...
shrink-fitting a carbide ring into a valve seat
A three turn helical coil is used to heat the steel valve seat. The steel valve seat is placed in the coil and heated for 50 seconds to enlarge the center hole & drop the carbide ring in for the shrink-fitting process.
Shrink-fitting auto turbo charger impeller blades onto an aluminum shaft
Induction heating provides repeatable results, reduced cycle time, lower consumables cost and even distribution of heating
Shrink fitting aluminum pulley to insert inner bearing
A three turn helical coil is used to heat the aluminum pulley. The pulley is heated to 464 °F (240 °C) in 20 seconds to expand the inner diameter and then the inner bearing is inserted to form the completed part.
shrink-fitting an assembled wrist pin into a connecting rod
Induction heating provides more accurate control of heat vs a flame burner, it heats only the knuckle, not the whole part, it prevents discoloring due to lower temperature used and increases productivity due to repeatability & ease of operation. A foot pedal & timer is used.
shrink-fitting a cast iron rocker arm assemblies
A four-turn helical coil heats the ring at one end of the assembly. The coil is designed to concentrate the field towards the center of the assembly where the thermal mass is greatest.
Shrink fit a motor shaft and roller
Processing with induction heat saves power and time. The complete tube does not have to reach the desired temperature as it does when heating with an oven. Being able to selectively heat a zone allows for a much quicker transfer of heat.
Shrink-fitting a Camshaft Gear
Heating a camshaft gear with a bore size of 1.630 inch to shrink fit over a steel shaft that has a diameter of 1.632 inch. A temperature of 500F is required for the gear to expand 0.002 inch in order to slip over the shaft.
shrink-fitting a Fuel Pump Housing and Inserts
To heat an aluminum fuel pump housing measuring 8 x 45 x 3.5 in to 375 °F, allowing steel parts to be inserted.
Shrink fit a carbon graphite ring insert into an outer steel band
To heat a steel band to 1000 °F (538 °C) and insert a carbon graphite ring into the center of the steel band.
Shrink Fitting Cam Shaft Gears
To heat aluminum and steel camshaft gears to over 500 °F within 4 minutes for a shrink fitting application.
Shrink Fitting a Steel Part & Casing
A specially designed multiple turn internal helical coil was used to provide the heat to the various steel parts.
Benefits of Shrink Fitting with Induction
Process repeatability, accuracy, energy efficiency and speed are four hallmarks of induction heating for virtually any application. Additionally, induction heating delivers heat to the targeted part, not the atmosphere around it, so there is no risk of distortion.
Safety is another considerable benefit, as there is no open flame, which makes it a viable option for almost any manufacturing environment.
Frequently Asked Questions
Frequently Asked Questions
Other Resources
Why use induction for shrink fitting?
Induction delivers heat to the targeted part, not the atmosphere around it, so there is no risk of distortion and temperature can be controlled in a precise manner. Safety is improved, since it does not use an open flame.
Can induction be used for insertion & disassembly?
With induction shrink fitting, you use thermal expansion to fit or remove parts. A metal component is heated to 150-300 °C (305-572 °F), and that causes it to expand. This allows for the removal or insertion of a part.
10 Shrink Fitting Notes
We have collected these 10 popular application notes to help you understand the many ways induction heating can improve your shrink fitting processes.
10 Automotive Application Notes
We have collected these 10 popular Application Notes to help you understand the many ways induction heating can improve your precision automotive manufacturing processes.
From our blog
Press Fitting Steel Sleeves with Induction
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4495 | https://www.youtube.com/watch?v=4aH-NeqDwgc | General Solution: CES Elasticity of Substitution
Economics in Many Lessons
78200 subscribers
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Posted: 27 Jan 2020
Solving for the CES production function's elasticity of substitution.
Any channel donations are greatly appreciated:
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Transcript:
hello this video will solve for the elasticity of substitution from a generally specified CES production function so here we're going to solve for the CES production functions elasticity of substitution so in general form the CES production function looks like this K is the units of capital and L is units of labour the first thing we're gonna do is get the marginal product of labor the partial derivative of Q with respect so this exponent over here this minus 1 divided by beta I'm bringing that down in front and then I'll also subtract 1 from that exponent and then the next step we're going to take the derivative of what's in brackets here with respect to L so I'm gonna bring down this minus beta so that's where this minus beta is coming from and then minus beta minus 1 and everything else is unchanged can be simplified you notice right here minus 1 over beta and minus beta that'll just cancel so I get rid of those beta terms now and now the marginal product of capital taking the partial derivative of the production function with respect to K once again we bring down this exponent here this minus 1 divided by beta that's out in front we are going to subtract 1 from that exponent as well so we have this part here then we're gonna look in the brackets and we're gonna find the K term I'm gonna take the derivative of that so we're gonna bring down this minus beta in front here from the K exponent then we're gonna subtract 1 from this K exponent once again this minus 1 divided by beta and this minus beta will cancel and we're left with the following moving on rewriting the marginal product of labor rewriting the marginal product of capital and then writing the optimal input mix which is the marginal product of labor divided by the marginal of marginal product of capital will equal the ratio of the price of labor to the price of capital so making your substitution for the marginal product of labor and the marginal product of capital in the denominator we have the following and everything in red here will cancel so I'm just highlighting this above step here putting those things in red and that will all cancel there are identical identical terms in the numerator and denominator so after cancelling those terms we have this now on the left hand side and what we want to do is we want to solve for K divided by L so I'm going to follow the rules of exponents here I'm gonna bring this L term down to the denominator and I'm gonna move this K term up into the numerator so following the rules of exponents the signs on the exponents will change here and we're left with this result and then the next step here this one minus alpha divided by alpha I'm just gonna multiply through by its reciprocal so moving that over to the right hand side we get this result and as I said I want to solve for K divided by L so I am going to raise both sides to the 1 divided by beta plus 1 so raising both sides to 1 divided by beta plus 1 we get this simplification the next step is we're gonna define this letter V to equal K divided by L so let V equal K divided by L and then we're gonna let Z equal W divided by R so V just K divided by L just equals this right here ok and notice W divided by R I substitute that I substituted out Z okay so again K divided by L becomes V W divided by R becomes Z so we have this expression now the elasticity of substitution it's going to be given by this equation here we're going to take the derivative of the V equation with respect to Z and then we're going to multiply it by Z divided by V so we'll do that on the next slide so here again is our V equation and here again is how we're going to define the elasticity of substitution so we're going to first take the derivative of the V equation with respect to Z so this one divided by beta plus one comes down in front and then all we have to do is subtract one from that exponent here and we're done so the elasticity of substitution Sigma is this derivative result that we just found and then at Z divided by V which I have over here one thing you'll note here is that this Z and this Z over here this plus 1 and this minus 1 will cancel okay so we got Z raised to the 1/2 plus 1 minus 1 then we got Z race to the plus 1 that's just gonna cancel so I just have 1 here now and then this 1 plus beta plus 1 there's no longer minus 1 there so this Z and minus 1 here gonna cancel all right the next step recall what V is as I have over here here's V so we caught what V is and substitute it in to this expression this above expression so making that substitution for V you'll notice that in the denominator we have a like term in the numerator and that's just gonna cancel so this and this will cancel and you'll get the elasticity of substitution equal to 1 divided by beta plus 1 if beta is greater than minus 1 but less than 0 the elasticity of substitution will exceed 1 if beta is greater than 0 but less than infinity last Asiya substitution will be less than 1 and if beta equals 0 the elasticity of substitution will exactly equal 1 ok that's it I hope you found this video helpful |
4496 | https://blob.wenxiaobai.com/article/18326a7e-59c1-f237-2588-030a6fe86ea8 | 梯形的定义以及性质是什么
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梯形的定义以及性质是什么
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梯形的定义以及性质是什么
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梯形的定义是：一组对边平行的四边形。其性质包括：1. 只有一组对边平行；2. 不平行的两边为腰，平行的两边为底；3. 两底之间的距离称为梯形的高；4. 等腰梯形的两腰相等。梯形是一种特殊的四边形，具有平行和非平行边的组合特性。其定义和性质有助于我们更好地理解和应用梯形在几何问题中的作用。
梯形是一种特殊的四边形，其特征在于只有一组对边是平行的。这种几何形状在数学和工程学中有着广泛的应用。本文将详细介绍梯形的性质、判定方法以及分类，帮助读者更深入地理解这一几何概念。
梯形的性质
梯形具有以下基本性质：
上底与下底平行，这是梯形定义的核心特征。
梯形的中位线不仅平行于两底，而且其长度等于上下底之和的一半，这一性质在计算中位线长度时非常有用。
梯形的判定方法
要确定一个四边形是否为梯形，可以通过以下方法：
如果四边形中只有一组对边平行，且另一组对边不平行，那么这个四边形就是梯形。
如果四边形的一组对边平行，并且这组对边的长度不相等，那么这个四边形也是梯形。
梯形的分类
梯形可以根据其特征进一步分类为：
一般梯形：没有特殊角度或边长限制的梯形。
特殊梯形：包括直角梯形和等腰梯形。
直角梯形：至少有一个角为直角的梯形。
等腰梯形：两腰（即非平行边）长度相等的梯形。
等腰梯形的特有性质
等腰梯形除了具备一般梯形的性质外，还有以下独特性质：
同一底上的两个内角相等，这是等腰梯形对称性的体现。
两条对角线的长度相等，进一步强化了等腰梯形的对称性。
等腰梯形是轴对称图形，其对称轴为通过两底中点的直线，但不是中心对称图形。
通过上述内容，我们对梯形有了全面的认识。梯形的定义、性质、判定方法以及分类，都是理解这一几何形状的关键要素。希望本文能够帮助读者更好地掌握梯形的相关知识。
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4497 | https://www.acc.org/Clinical-Topics/Anticoagulation-Management/Anticoagulation-Solution-Set | Anticoagulation For Stroke Prevention in AFib Solution Set - American College of Cardiology
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Anticoagulation Management
Anticoagulation For Stroke Prevention in AFib Solution Set
The Anticoagulation For Stroke Prevention in Atrial Fibrillation (AFib) Solution Set is a one-stop-shop for clinicians who manage patients with AFib in their everyday practice.
Check out the tools and resources below and find the best one for your needs:
Clinical Policy
ACC’s Expert Consensus Decision Pathways (ECDPs) provide practical guidance in areas where evidence may be limited, new and evolving.
Periprocedural Management of Anticoagulation:Provides guidance on whether and how to interrupt and bridge anticoagulation as part of periprocedural planning for patients with nonvalvular AFib.
Bleed Management in Anticoagulation:Offers practical guidance on how to manage bleeding in a patient treated with anticoagulants for any indication including the need for reversal/hemostatic agents and the appropriateness and time of restarting anticoagulation.
Visit the AFib Guideline Hub, to access additional resources and information, including the Guidelines for the Management of Patient With AFib.
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ACC’s Clinical App Collection includes the latest science and ACC policy and are intended to be used by clinicians to support conversations with their patients.
AnticoagEvaluator: Helps clinicians make informed decisions on antithrombotic therapy initiation for patients with AFib.
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CardioSmart Heart Explorer App: Uses high-resolution cardiac graphics and animations to help effectively discuss common heart problems and treatment options with patients.
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Statin Drug Interactions With Immunosuppressants
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ACC’s shared decision-making tools are an essential part of the clinician-patient relationship. Considering the current evidence, they discuss the potential benefits and risks of a patient's options, improving accuracy of the patient's risk perception and clinician satisfaction. Done properly, these decision aids help clinicians navigate patients' wide-ranging therapy options along with consideration of the patients' goals.
Anticoagulation Shared Decision-Making Tool: Includes a personal risk calculator and interactive decision aids that help clinicians and patients choose the best option regarding anticoagulation therapy for individual patients with AFib.
Additional Decision Aids are available at CardioSmart.org/DecisionAids.
Patient Resources
Educational materials and tools developed to increase patient engagement and improve the patient-clinician dialogue are available in the CardioSmart AFib Hub.
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4498 | https://jacobimed.org/public/Docs/Bill's%20Papers/AORTIC%20STENOSIS%20NEJM.pdf | N Engl J Med, Vol. 346, No. 9 · February 28, 2002 · www.nejm.org · 677 CLINICAL PRACTICE Clinical Practice This Journal feature begins with a case vignette highlighting a common clinical problem. Evidence supporting various strategies is then presented, followed by a review of formal guidelines, when they exist. The article ends with the author’s clinical recommendations. A ORTIC S TENOSIS B LASE A. C ARABELLO , M.D. From the Department of Medicine, Baylor College of Medicine, and the Veterans Affairs Medical Center, Houston. Address reprint requests to Dr.
Carabello at the Veterans Affairs Medical Center Medical Service (111), 2002 Holcombe Blvd., Houston, TX 77030, or at blaseanthony.carabello@ med.va.gov. A 60-year-old man is evaluated for a heart mur-mur. He jogs 3 mi (5 km) per day and is asymp-tomatic. Physical examination reveals a delayed carotid upstroke, a 3/6 late-peaking systolic ejec-tion murmur that radiates to the neck, and a sin-gle S 2 . An echocardiogram shows normal systol-ic function and a heavily calcified aortic valve.
The patient’s peak Doppler transvalvular gradient is 64 mm Hg, with a mean gradient of 50 mm Hg.
His calculated valve area is 0.7 cm 2 . How should this patient be treated? THE CLINICAL PROBLEM Aortic stenosis is the most common cardiac-valve lesion in the United States. Two factors account for its common occurrence: approximately 1 to 2 per-cent of the population is born with a bicuspid aortic valve, which is prone to stenosis; and aortic stenosis develops with age, and the population is aging.
The clinician is usually first alerted to the presence of aortic stenosis by the finding of a systolic ejection murmur at the right upper sternal border that radiates to the neck. Clues that the disease is at least moderate in severity are peaking of the murmur late in systole, palpable delay of the carotid upstroke, and a soft sin-gle second heart sound, because the aortic component of S 2 disappears when the valve no longer opens or closes well.
Although once thought of as a degenerative lesion, calcific aortic stenosis has many features in common with coronary disease. 1 Both conditions are more com-mon in men, older persons, and patients with hyper-cholesterolemia, and both derive in part from an ac-tive inflammatory process. 2 Aortic stenosis is distinguished from aortic sclero-sis by the degree of valve impairment. In aortic scle-rosis, the valve leaflets are abnormally thickened, but obstruction to outflow is minimal, whereas in aortic stenosis, the functional area of the valve has decreased enough to cause measurable obstruction of outflow.
Little hemodynamic disturbance occurs as the valve area is reduced from the normal 3 to 4 cm 2 to 1.5 to 2 cm 2 . However, as shown in Table 1, an addition-al reduction in the valve area from half its normal size to one quarter of its normal size produces severe obstruction to flow and a progressive pressure over-load on the left ventricle. The concentric hypertrophy that develops in response to this overload is both adap-tive and maladaptive. Whereas the increased muscle mass allows the ventricle to generate the increased force necessary to propel blood past the obstruction, the hypertrophied myocardium has decreased coro-nary blood flow reserve 3 (even in the presence of nor-mal epicardial coronary arteries) and can also cause both diastolic and systolic left ventricular dysfunction, producing the symptoms of congestive heart failure. 4,5 The obvious question for the physician is when is the optimal time for clinical intervention? Although aor- Data were derived with the Gorlin formula: aortic-valve area = cardiac output÷(systolic ejection period¬heart rate) 44.3 √ mean gradient where the cardiac output was assumed to be 6 liters per min-ute, the systolic ejection period was assumed to be 0.33 sec-ond, and the heart rate was assumed to be 80 beats per minute. T ABLE 1. R ELATION OF THE A ORTIC -V ALVE A REA TO THE M EAN G RADIENT . A ORTIC -V ALVE A REA M EAN G RADIENT cm 2 mm Hg 4 1.7 3 2.9 2 6.6 1 26 0.9 32 0.8 41 0.7 53 0.6 73 0.5 105 , 678 · N Engl J Med, Vol. 346, No. 9 · February 28, 2002 · www.nejm.org The New England Journal of Medicine tic stenosis often coexists with other valvular diseases, this review focuses on isolated aortic stenosis. STRATEGIES AND EVIDENCE There is no effective medical therapy for severe aortic stenosis; aortic stenosis is a mechanical obstruc-tion to blood flow that requires mechanical correction.
In children with congenital aortic stenosis, the valve leaflets are merely fused, and balloon valvotomy may offer substantial benefit. 6 In adults with calcified valves, however, balloon valvotomy only temporarily relieves symptoms and does not prolong survival. 7 Thus, the intervention required in adults, other than standard prophylactic antibiotics against infective en-docarditis, 8 is the replacement of the valve. The risks of replacing that valve must be weighed against the risks of delaying the procedure. The procedure can usually be delayed until symptoms develop. 9,10 Stud-ies of aortic stenosis uniformly demonstrate that once angina, syncope, dyspnea, or other symptoms of heart failure develop and are found to be due to aortic ste-nosis, the patient’s life span is drastically shortened un-less the valve is replaced (Fig. 1). Of the 35 percent of patients with aortic stenosis who present with an-gina, half will die within five years in the absence of aortic-valve replacement. Of the 15 percent of pa-tients who present with syncope, half will die within three years, and of the 50 percent of patients who present with dyspnea, half will die within two years, unless the aortic valve is replaced. In contrast, 10-year age-corrected rates of survival among patients who have undergone aortic-valve replacement approach the rate in the normal population. 11,12 The striking contrast between the excellent prog-nosis after aortic-valve replacement and the dismal prognosis in the absence of replacement in sympto-matic patients makes the presence or absence of symp-toms the crucial factor with respect to management.
In general, the clinician can be confident that, in a giv-en patient, the symptoms are due to aortic stenosis if the mean aortic-valve gradient exceeds 50 mm Hg or if the aortic-valve area is no larger than 1 cm 2 . I will use these as criteria for severe disease, although there is no universally accepted definition that relies on valve area or gradient. AREAS OF UNCERTAINTY Management of Severe Asymptomatic Aortic Stenosis There is overwhelming evidence that patients with severe aortic stenosis who become symptomatic re-quire prompt aortic-valve replacement. Conversely, asymptomatic patients, even those with severe disease, generally have an excellent prognosis without aortic-valve replacement. Unfortunately, approximately 1 to 2 percent of asymptomatic patients die suddenly or have a very rapid rate of progression to the sympto-matic state and then to sudden death. 9,10,13,14 Thus, the question arises whether patients with severe asymp-tomatic aortic stenosis should undergo aortic-valve re-placement to protect them from sudden death. Al-though some experts advocate this approach, this strategy exposes the entire group of asymptomatic pa-tients with severe aortic stenosis to the risk of peri-operative and valve-related complications and death.
Even in the best of circumstances, the surgical mor-tality rate is approximately 1 percent and the risk of a valve-related complication (including thromboem-bolism; bleeding during therapy with anticoagulants; deterioration of the prosthetic valve, requiring reoper-ation; and infective endocarditis) is 1 percent per year. 15 Echocardiography and exercise testing may identify asymptomatic patients who are likely to benefit from surgery. Otto et al. 16 found that patients with asymp-tomatic aortic stenosis whose peak transaortic blood flow velocity exceeded 4 m per second (a peak gradi-ent of 64 mm Hg) had a risk of becoming symptomat-ic and requiring aortic-valve replacement of 70 percent within two years. None of the patients in this mod-erate-sized series died suddenly, and all underwent exercise testing.
Although exercise testing is unwarranted and dan-gerous in patients with symptomatic aortic stenosis, it has proved safe in patients with moderate-to-severe asymptomatic aortic stenosis. 16,17 A preliminary report suggests that exercise testing may identify some pa-tients with latent symptoms or exercise-induced he-modynamic instability, facilitating timely aortic-valve replacement. In a study of 58 asymptomatic patients, 18 21 had symptoms for the first time during the exer-cise test. Most likely, these patients had failed to rec-ognize the symptoms previously or had not engaged in activities that would have precipitated symptoms.
Although patients with asymptomatic aortic stenosis can exercise safely under a physician’s scrutiny, it seems most unwise to permit patients with moderate-to-severe aortic stenosis to engage in vigorous, unmon-itored exercise in view of the limitations imposed by left ventricular hypertrophy on coronary blood flow. Treatment of the Patient with a Low Gradient and Reduced Ejection Fraction In patients with left ventricular dysfunction who have a substantial transvalvular gradient (a mean gra-dient of more than 40 mm Hg), the outcome of sur-gery is excellent despite the presence of a reduced ejec-tion fraction preoperatively. 19 In these patients, the excessive afterload generated by the obstructing valve is a prime contributor to the left ventricular dysfunc-tion. Once the obstruction is removed and the after-load is reduced, left ventricular function returns to or approaches normal.
Patients with a reduced ejection fraction and a small CLINICAL PRACTICE N Engl J Med, Vol. 346, No. 9 · February 28, 2002 · www.nejm.org · 679 transvalvular gradient (less than 30 mm Hg) 19,20 have a high operative risk, and only half such patients are alive three to four years after surgery. 21 The poor out-come in these patients is related to the presence of both severely depressed myocardial contractility and the excessive afterload. 19 Although the overall prognosis for this group of pa-tients is poor, some patients in this category, pre-sumably those with severe valve obstruction, benefit from surgery. 21,22 In other patients, the calculated valve area may be severely reduced because cardiomyopathy inhibits the left ventricle from completely opening a mildly but not a severely stenotic valve. The presence of low output may lead clinicians to the false conclu-sion that the valve is severely stenotic (aortic pseudo-stenosis). 23-25 The best method for distinguishing these two conditions is to increase cardiac output during Doppler echocardiography or cardiac catheterization Figure 1. Survival among Patients with Severe Symptomatic Aortic Stenosis Who Underwent Valve Replacement and Similar Patients Who Declined to Undergo Surgery.
The overall and individual P values are shown, as is the overall chi-square value. Reprinted from Schwarz et al. 11 with the permission of the publisher.
0 100 0 5 20 40 60 80 1 2 3 4 Year Valve replacement Chi-square=23.5 P<0.001 No surgery P<0.001 P<0.05 Survival (%) NO. AT RISK Valve replacement No surgery 125 019 87 08 51 02 35 01 9 0 0 0 680 · N Engl J Med, Vol. 346, No. 9 · February 28, 2002 · www.nejm.org The New England Journal of Medicine and to use the new data to recalculate the valve area.
In the case of pseudostenosis, increased output causes a large increase in the calculated valve area, often to more than 1 cm 2 . In this group of patients, valve re-placement is unlikely to be beneficial. Conversely, pa-tients in whom increased cardiac output produces a substantial increase in gradient have a true outflow obstruction and may benefit from surgery. Patients with a low gradient who have no response to ino-tropic stimulation have a poor outcome, presumably because the myocardial damage is so far advanced. 26 Safety of Cardiac Surgery in Patients with Mild-to-Moderate Aortic Stenosis Controversy also exists regarding the optimal ap-proach to patients with mild-to-moderate aortic steno-sis (as indicated by a transvalvular gradient in the range of 10 to 30 mm Hg and a valve area of more than 1 cm 2 ) who require cardiac surgery for some other cause, usually coronary revascularization. Concomitant aortic-valve replacement increases the risk of both per-ioperative death and complications related to the pros-thetic valve. However, if the native valve is left in place, aortic stenosis may progress so rapidly that another cardiac surgery is required despite the fact that the bypass grafts are functioning normally. 27 Unfortunate-ly, there is wide individual variability in the rate of dis-ease progression, 27-29 virtually precluding prognosti-cation of the course of a given patient. Concomitant aortic-valve replacement is considered unwise if it is likely to increase the gradient even further. Safety of Noncardiac Surgery in Patients with Severe Asymptomatic Aortic Stenosis On the basis of a small number of adverse events, one study indicated that noncardiac surgery posed an increased risk among patients with aortic stenosis. 30 However, O’Keefe et al. examined the course of 48 patients with severe aortic stenosis who underwent noncardiac surgery 31 and found that there were no complications in the 25 patients who had local anes-thesia and only one complication in the 23 patients who received general anesthesia. Thus, although intra-operative hemodynamics must be closely monitored during noncardiac surgery in patients with asympto-matic aortic stenosis, there is no apparent need for concomitant aortic-valve intervention. Ability to Slow or Halt Progression of the Valve Lesion As noted above, the lesion of aortic stenosis shares many features with coronary disease. Although recent data suggest that hydroxymethylglutaryl coenzyme A reductase inhibitors (referred to as “statins”) can retard the progression of aortic stenosis, 32 in my opinion it is a bit too early to begin prescribing statins for this purpose. GUIDELINES Guidelines for the use of valve replacement in pa-tients with aortic stenosis are provided in Table 2. In 1998 the American Heart Association and American College of Cardiology issued guidelines for the treat-ment of valvular heart disease (available at http:// www.americanheart.org). 8 These guidelines recom-mend the use of standard antibiotic prophylaxis against infective endocarditis. Doppler echocardiography is recommended for the initial diagnosis and assessment of the severity of aortic stenosis and the function and hemodynamics of the left ventricle, as well as for the reevaluation of patients whose symptoms and signs are changing and of patients known to have severe asymptomatic aortic stenosis. CONCLUSIONS AND RECOMMENDATIONS Doppler echocardiography is indicated for the ini-tial evaluation in all patients suspected of having aortic stenosis, as well as in patients with established disease if symptoms develop or physical signs change. The de-velopment of angina, syncope, or dyspnea in a patient with severe aortic stenosis constitutes a grave medical condition, requiring prompt aortic-valve replacement.
For patients with severe asymptomatic disease, such as the patient described in the clinical vignette, the presence of an aortic-jet velocity of at least 4 m per second on Doppler echocardiography indicates the need for close scrutiny. I routinely recommend ex-ercise testing in this group (but never in symptomat-ic patients), since it may help identify which of these patients are at high enough risk to warrant undergo-ing surgery (Fig. 2).
The postoperative prognosis of patients with a re-duced ejection fraction is good if the mean transval-vular pressure gradient exceeds 40 mm Hg. Thus, a low ejection fraction alone should never be an abso- Aortic-valve replacement is not indicated to prevent sudden death in asymptomatic patients who have none of the findings listed. T ABLE 2. R ECOMMENDATIONS FOR THE U SE OF A ORTIC -V ALVE R EPLACEMENT IN P ATIENTS WITH A ORTIC S TENOSIS . Aortic-valve replacement indicated Patients with severe aortic stenosis and any of its classic symptoms (angina, syncope, or dyspnea) Patients with severe aortic stenosis who are undergoing coronary-artery by-pass surgery Patients with severe aortic stenosis who are undergoing surgery on the aor-ta or other heart valves Aortic-valve replacement possibly indicated Patients with only moderate aortic stenosis who require coronary-artery by-pass surgery or surgery on the aorta or other heart valves Asymptomatic patients with severe aortic stenosis and at least one of the following: an ejection fraction of no more than 0.50, hemodynamic in-stability during exercise (e.g., hypotension), or ventricular tachycardia CLINICAL PRACTICE N Engl J Med, Vol. 346, No. 9 · February 28, 2002 · www.nejm.org · 681 lute contraindication to surgery. However, the prog-nosis of patients with a low gradient and a low ejec-tion fraction is worse. For such patients I recommend hemodynamic manipulation in the catheterization or echocardiographic laboratory to help determine which patients are more likely to benefit from surgery.
Preliminary evidence indicates that patients who have moderate aortic stenosis (as indicated by a gra-dient of more than 20 mm Hg and a valve area of less than 1.2 cm 2 ) but who require heart surgery for oth-er diseases should probably undergo concomitant aor-tic-valve replacement. Further study of this approach is required. REFERENCES 1. Otto CM, Kuusisto J, Reichenbach DD, Gown AM, O’Brien KD. Char-acterization of the early lesion of “degenerative” valvular aortic stenosis: histological and immunohistochemical studies. Circulation 1994;90:844-53. 2. Otto CM, Lind BK, Kitzman DW, Gersh BJ, Siscovick DS. Association of aortic-valve stenosis with cardiovascular mortality and morbidity in the elderly. N Engl J Med 1999;341:142-7 . 3. Marcus ML, Doty DB, Hiratzka LF, et al. Decreased coronary reserve: a mechanism for angina pectoris in patients with aortic stenosis and normal coronary arteries. N Engl J Med 1982;307:1362-7 . 4. Hess OM, Ritter M, Schneider J, Grimm J, Turina M, Krayenbuehl HP . Diastolic stiffness and myocardial structure in aortic valve disease before and after valve replacement. Circulation 1984;69:855-65. 5. Huber D, Grimm J, Koch R, Krayenbuehl HP . Determinants of ejec-tion performance in aortic stenosis. Circulation 1981;64:126-34. 6. McCrindle BW. Independent predictors of immediate results of percu- Figure 2. Algorithm for the Management of Aortic Stenosis.
Aortic stenosis without symptoms present on physical examination Severe aortic stenosis and symptoms present on physical examination Perform Doppler echocardiography If echocardiography shows mild-to-moderate aortic stenosis, monitor patient for the development of symptoms If echocardiography shows severe aortic stenosis, perform exercise testing If results are normal, monitor patient closely If symptoms develop, repeat echocardi- ography If aortic stenosis becomes severe, perform coronary arteriography and aortic- valve replacement If patient remains asymptomatic but findings on examination change, repeat echocardi- ography If aortic stenosis remains mild to moderate, continue follow-up If results are abnormal, perform coronary arteriography and aortic-valve replacement Mild-to-moderate aortic stenosis and symptoms present on physical examination Search for other causes of symptoms Perform coronary arteriography and aortic-valve replacement 682 · N Engl J Med, Vol. 346, No. 9 · February 28, 2002 · www.nejm.org The New England Journal of Medicine taneous balloon aortic valvotomy in children. Am J Cardiol 1996;77:286-93. 7. Otto CM, Mickel MC, Kennedy JW, et al. Three-year outcome after balloon aortic valvuloplasty: insights into prognosis of valvular aortic ste-nosis. Circulation 1994;89:642-50. 8. ACC/AHA Guidelines for the management of patients with valvular heart disease: a report of the American College of Cardiology/American Heart Association Task Force on Practice Guidelines (Committee on Man-agement of Patients with Valvular Heart Disease). J Am Coll Cardiol 1998; 32:1486-588. 9. Ross J Jr, Braunwald E. Aortic stenosis. Circulation 1968;38:Suppl V: V-61–V-67 . 10. Pellikka PA, Nishimura RA, Bailey KR, Tajik AJ. The natural history of adults with asymptomatic, hemodynamically significant aortic stenosis. J Am Coll Cardiol 1990;15:1012-7 . 11. Schwarz E, Baumann P, Manthey J, et al. The effect of aortic valve re-placement on survival. Circulation 1982;66:1105-10. 12. Lindblom D, Lindblom U, Qvist J, Lundstrom H. Long-term relative survival rates after heart valve replacement. J Am Coll Cardiol 1990;15: 566-73. 13. Kelly TA, Rothbart RM, Cooper CM, Kaiser DL, Smucker ML, Gib-son RS. Comparison of outcome of asymptomatic to symptomatic patients older than 20 years of age with valvular aortic stenosis. Am J Cardiol 1988; 61:123-30.
14. Pellikka PA, Nishimura RA, Bailey KR, et al. Natural history of 610 adults with asymptomatic hemodynamically significant aortic stenosis over prolonged follow-up. J Am Coll Cardiol 2001;37:Suppl A:489A. abstract.
15. Thai HM, Gore JM. Prosthetic heart valves. In: Alpert JS, Dalen JE, Rahimtoola SH, eds. Valvular heart disease. 3rd ed. Philadelphia: Lippin-cott Williams & Wilkins, 2000:393-407 .
16. Otto CM, Burwash IG, Legget ME, et al. Prospective study of asymp-tomatic valvular aortic stenosis: clinical, echocardiographic, and exercise predictors of outcome. Circulation 1997;95:2262-70.
17. Linderholm H, Osterman G, Teien D. Detection of coronary artery disease by means of exercise ECG in patients with aortic stenosis. Acta Med Scand 1985;218:181-8.
18. Goldman L, Caldera DL, Nussbaum SR, et al. Multifactorial index of cardiac risk in noncardiac surgical procedures. N Engl J Med 1977;297: 845-50.
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Copyright © 2002 Massachusetts Medical Society.
IMAGES IN CLINICAL MEDICINE The Journal welcomes consideration of new submissions for Images in Clinical Medicine. Instructions for authors and procedures for submissions can be found on the Journal’s Web site at http:/ /www.nejm.org. At the discretion of the editor, images that are accepted for publication may ap-pear in the print version of the Journal, the electronic version, or both. |
4499 | https://www2.latech.edu/~deddy/chem103/103Hydrate.htm | CHEMISTRY 103: PERCENT WATER IN A HYDRATE
Hydrates are compounds that incorporate water molecules into their fundamental solid structure. In a hydrate (which usually has a specific crystalline form), a defined number of water molecules are associated with each formula unit of the primary material.
Gypsum is a hydrate with two water molecules present for every formula unit of CaSO4. The chemical formula for gypsum is CaSO4 2H2O and the chemical name is calcium sulfate dihydrate. Note that the dot in the formula (or multiplication sign) indicates that the waters are there. Other examples of hydrates are: lithium perchlorate trihydrate - LiClO4 3H2O; magnesium carbonate pentahydrate - MgCO3 5H2O; and copper(II) sulfate pentahydrate - CuSO4 5 H2O.
The water in the hydrate (referred to as "water of hydration") can be removed by heating the hydrate. When all hydrating water is removed, the material is said to be anhydrous and is referred to as an anhydrate.
CuSO4 5 H2O(s) + HEAT ---> CuSO4 (s) + 5 H2O (g) hydrate anhydrate
Experimentally measuring the percent water in a hydrate involves first heating a known mass of the hydrate to remove the waters of hydration and then measuring the mass of the anhydrate remaining. The difference between the two masses is the mass of water lost. Dividing the mass of the water lost by the original mass of hydrate used is equal to the fraction of water in the compound. Multiplying this fraction by 100 gives the percent water.
| |
| EXAMPLE 1 When a 1.000 g sample of CuSO4 5 H2O(s) was heated so that the waters of hydration were driven off, the mass of the anhydrous salt remaining was found to be 0.6390 g. What is the experimental value of the percent water of hydration? CuSO4 5 H2O(s) + HEAT ----> CuSO4 (s) + 5 H2O (g) 1.000 g 0.6390 g 1. The difference between the hydrate mass and anhydrate mass is the mass of water lost. 1.000 g - 0.6390 g = 0.3610 g 2. Divide the mass of the water lost by the mass of hydrate and multiply by 100. (0.3610 g /1.000 g)(100) = 36.10% |
The theoretical (actual) percent hydration (percent water) can be calculated from the formula of the hydrate by dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and multiplying by 100.
| |
| EXAMPLE 2 What is the percent water in copper(II) sulfate pentahydrate, CuSO4 5 H2O? 1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of hydration must be included. (1 Cu)(63.55 g/mol) + (1 S)(32.07 g/mol) + (4 O)(16.00 g/mol) = 159.62 g/mol Formula mass = 159.62 g/mol + (5 H20)( 18.02 g H20/mol) = 249.72 g/mol 2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and multiply this fraction by 100. Percent hydration = (90.10 g /249.72 g)(100) = 36.08% |
CHEMISTRY 103: PERCENT WATER IN A HYDRATE
Name___________________________________ Hood No.________ Date_____
Put on your CHEMICAL SPLASH-PROOF SAFETY GOGGLES! Attach a second sheet and clearly show all calculations.
PROCEDURE 1. Accurately weigh a clean, dry crucible. Record this mass to +0.01 g. 2. Transfer approximately 3 grams of barium chloride dihydrate, BaCl2 2 H2O, into the weighed crucible and weigh the crucible and its contents. Record this mass to +0.01 g. 3. Place the crucible on a ring stand using a ring and clay triangle and heat gently for 10 minutes. Then heat the sample more strongly for 10 more minutes by bringing the flame of the bunsen burner directly under the dish. The residue should be almost pure white. Allow the crucible to cool, then weigh it. Record this mass to +0.01 g. 4. Heat the crucible for another 5 minutes, cool, then weigh. If all the water has been driven off, the two masses should agree. Record this mass to +0.01 g. 5. Dispose of the barium chloride in the container provided.
DATA 1. Mass of empty crucible _______________g
2. Mass of crucible & BaCl2 2 H2O _______________g
3. Mass of BaCl2 2 H2O _______________g (#2 - #1)
4. Mass of crucible & BaCl2 after first heating _______________g
5. Mass of BaCl2 after first heating _______________g (#4 - #1)
6. Mass of crucible & BaCl2 after second heating _______________g
7. Mass of BaCl2 after second heating _______________g (#6 - #1)
8. Mass of water lost _______________g H2O
9. Percent hydration _______________% H2O
10. Theoretical value _______________% H2O
11. Percent error _______________%
Atomic masses: H = 1.01; O = 16.00; Cl = 35.45; Ba = 137.33
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| Reminders: 1. Barium chloride is toxic. Use care when handling. 2. The used barium chloride should be put in the waste container provided. It is very important that the evaporating dish cools to room temperature before weighing. If it is not cool, convection currents will be set up that will lower the mass. 3. The ring stand, ring, and crucible are hot. BE CAREFUL!!!!! | |
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