url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://www.reddit.com/user/Unlistedd | 1,553,485,131,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203548.81/warc/CC-MAIN-20190325031213-20190325053213-00145.warc.gz | 867,446,697 | 60,724 | Press J to jump to the feed. Press question mark to learn the rest of the keyboard shortcuts
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2
Posted by9 days ago
I downloaded a module that uses .pyd instead of .py, when i open the file i get a message that says: "The file is not displayed in the editor because it is either binary or uses an unsupported text encoding", i open it in the editor and a bunch of odd symbols shows up. How do i get access to the actual code?
3 points · 9 days ago
.pyd files are compiled native code, not Python code. You can't look at the source.
see more
Original Poster1 point · 9 days ago
Not at all? Is ther no way of knowing whats inside?
1
Posted by10 days ago
Dirichlet processes are almost always applied in big datasets, my question is wether or not there is a "cap", a requirment of datapoints, i.e. do i need 20, 50 or over 100 coordinates in order to get solid clusters of predictable Gaussians? Does the Dirichlet process work with fewer than 20 datapoints?
1
Posted by12 days ago
I want to be able to measure the uncertainty/riskiness of a business, how would one go about solving such a task?
I am familiar with the WACC, CAPM, APM, etc. The thing is that price-based risk is irrational, and so is volatility. How would one go about measuring the REAL risk of a business?
1 point · 12 days ago
What do you mean by "real"? The market does it's best through pricing to determine what is real.
see more
Original Poster2 points · 12 days ago
"real" meaning not rooted in betas and volatility. Would it be possible to reverse engineer the npv-formula in order to get the discount rate? Do i need to predict its future cash flows or am i able to use the cash flows up until this point in order to calculate the discount rate? I dont see how a investor is suppose to value risk.
2
Posted by13 days ago
Does the discount rate = risk premium + bond return?
1 point · 13 days ago
see more
Original Poster1 point · 13 days ago
It equals expected returns?
10
Posted by17 days ago
If i download a module its automatically accessible, but what if i want to dig into the modules and modify them? Is it possible to see the code within these modules? If so how?
1
Posted by1 month ago
Is it possible to apply a mathematicalprocess/model/equation to a indirect measure, in order to caculate the direct measure? If so, how?
What if i want to calculate the direct measure of the gradient from a light source, but i can only observe indirect measures of the light source itself. Can i still calculate the specific gradient?
Are there any general model(s) for this type of problem?
10
Posted by2 months ago
Why does this work:
Ph=['J&J', 'Bio']
for index, Ph in enumerate (Ph, start=1):
print(index, Ph)
But not this:
Ph=['J&J', 'Bio']
def Pha():
for index, Ph in enumerate (Ph, start=1):
print(index, Ph)
if __name__ == '__main__':
Pha()
And how do i fix it?
7 points · 2 months ago
What error do you get? The obvious thing is that your print(index, Ph) line is not indented properly.
see more
Original Poster-1 points · 2 months ago
It says that the error is on row 9 which is the for index, Ph in enumerate (Ph, start=1).
4 points · 2 months ago
Always post the full stack trace when requesting help.
see more
Original Poster1 point · 2 months ago
Traceback (most recent call last):
File "c:/Users/.../", line 13, in <module>
Pha()
File "c:/Users/.../", line 9, in Pha
for index, Ph in enumerate (Ph, start=1):
UnboundLocalError: local variable 'Ph' referenced before assignment
u/Unlistedd
29
##### Cake day
December 30, 2018
Trophy Case (1) | 908 | 3,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-13 | latest | en | 0.906729 |
https://www.quantitativeskills.com/sisa/distributions/binomial.php?Exp=00.050&Obs=005&N=0010 | 1,670,144,223,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00014.warc.gz | 982,279,103 | 2,566 | Calculate Binomial probabilities Expected: 0.05 Observed: 5 Sample Size: 10 Expected: 0.5 (=0.05*10)Variance Exp= 0.475; sd= 0.6892 BINOMIAL PROBABILITIES single; cumulative p(=0): 0.598737; p(>0): 0.401263 p(=1): 0.315125; p(>1): 0.086138p(=2): 0.074635; p(>2): 0.011504p(=3): 0.010475; p(>3): 0.001028p(=4): 0.000965; p(>4): 6.4E-5p(=5): 6.1E-5; p(>5): 3.0E-6(a=-0.052632,b=0.578947,0) *** summary ***Point probability= 6.1E-5 p(Obs>=5): 0.0001; (<5): 0.9999p(Obs>5): 0; (<=5): 1Mid-p: 0; 1-p: 1; 2*p: 0 Two sided probabilities(sum of small p's): abs(Exp-Obs)=abs(0.5-5)=4.5Pointpr(Exp-Obs=4.5): 6.1E-5p(Exp-Obs>=4.5)= 6.0E-5 p(Exp-Obs>4.5)= -0 Mid-p= 3.0E-5 For help go to SISA. More: Calculate CIaround obs= 5T-test the differenceExp-Obs=0.5-5=5.Calculate the minimum Sample Size required to see ifthe difference=5 isstatistically significantStudy the probability of N=10given exp=0.5 and obs=5 | 417 | 901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-49 | latest | en | 0.41755 |
http://easysurf.cc/masterj.htm | 1,623,998,096,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487635724.52/warc/CC-MAIN-20210618043356-20210618073356-00145.warc.gz | 15,726,444 | 4,435 | Easysurf Home Site Map E-Mail Help
Jotto - MASTERMIND Menu Directions Word List
Number TestWord Numberof Jots CorrectPosition
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Eliminating Alphabet
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Yes No
Jotto Directions
Word List
MasterMind Numerals 1-9
Letter Drop
Around the Square
Boggle Sixteen letter cubes - 4 x 4
Boggle with a challenge cube
Big Boggle 25 letter cubes - 5 x 5 - four or more letter words
Super Boggle 30 letter cubes - 5 x 6 - four or more letter words
Crossword Cubes
JOTTO
Letter Replacement
# Directions
Directions for MasterMind
Jotto MasterMind is a word game. The computer will select a secret English word with five different (unique) letters. The object of the game is to guess that secret word. Each guess is answered by the number of letters (Jots) in the guess (test) word that match or occur in the secret word. You will also be told how many of the letters are in the correct position in the secret word. Through a process of elimination, you should be able to deduce the correct letters using logic.
1. To start: Click inside of the Test Word entry box.
2. Type your five letter test or guess word. Letters may be in upper or lower case or a combination of upper and lower case.
3. After you have entered your guess word, click the "Guess" button.
4. See how many of the letters in your guess word occur in the secret word.
For example:
If the secret word is "solar"
- you enter "sslar"
The computer will report "Number of Jots = 4" and "Correct Position = 4"
- you enter "sssss"
The computer will report "Number of Jots = 1" and "Correct Position = 1"
5. Click inside of the next "Test Word" entry box.
6. Type your next guess word and then click the "Guess" button.
7. Continue in this manner until you correctly guess the secret word or your 20th guess. If you correctly guess the secret word, the computer will print "Correct Score =..."
8. To see the secret word, Click the "Answer" button at the bottom of the page.
9. To start a new game click "Click to Start a New Game"
10. An Eliminating Alphabet form is at the bottom of the page.
If you have ruled a letter in, place a "Y" in the Yes box under the digit.
If you have ruled a letter out, place an "N" in the No box under the digit. | 626 | 2,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-25 | latest | en | 0.830341 |
https://nz.education.com/resources/third-grade/two-digit-numbers/CCSS/ | 1,603,279,384,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107876307.21/warc/CC-MAIN-20201021093214-20201021123214-00475.warc.gz | 456,169,115 | 27,992 | # Search Our Content Library
20 filtered results
20 filtered results
Two Digit Numbers
Common Core
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Number Sequences
Worksheet
Number Sequences
Learn about different number patterns and relationships! Review basic math skills as to figure out the missing number in each sequence.
Math
Worksheet
Computation Station
Workbook
Computation Station
Improve your pattern-recognition and estimation skills as you practice multiplication and division. Children use techniques like breaking down numbers and fast addition to arrive at the correct answer — and to understand the concepts behind the process!
Math
Workbook
Write It & Round It
Worksheet
Write It & Round It
This exercise will help your students strengthen their number sense by reviewing the basics of place value to round numbers correctly.
Math
Worksheet
Lesson Plan
Analyzing and discussing arithmetic patterns builds a strong number sense in your students! Use this as a stand alone lesson or as a pre-lesson for *Boom, Clap! Patterns in the Multiplication Table.*
Math
Lesson Plan
Decomposing Numbers
Lesson Plan
Decomposing Numbers
Why make things harder on ourselves? Teach your students to decompose numbers to make math easier! Use this as a stand alone lesson or a pre-lesson for *Decompose to Multiply: 6, 7, 8, and 9.*
Math
Lesson Plan
Round 'Em Up
Worksheet
Round 'Em Up
Math
Worksheet
Input Output Math Tables
Worksheet
Input Output Math Tables
Find the hidden rule in these puzzling math tables! Compare your "in" number to the corresponding "out" number to find the pattern in each table.
Math
Worksheet
Multiplying by 3 Using Patterns
Worksheet
Multiplying by 3 Using Patterns
Multiplying by 3 is easier than multiplying by other numbers because of a certain pattern. When you multiply any number by 3, the digits of the answer must add up to a multiple of 3.
Math
Worksheet
Rounding to the Nearest Hundred
Worksheet
Rounding to the Nearest Hundred
Get your lasso ready, it's time to round up the hundreds herd! Your rounding rancher will practice estimating numbers to the nearest hundred.
Math
Worksheet
Mixed Problems
Worksheet
Mixed Problems
This worksheet reviews the multiplication patterns for the numbers two, three, and five, and the division pattern for the number two. Once you students familiarize themselves with these patterns, these math problems will become a breeze!
Math
Worksheet
Multiplying by 6 Using Patterns
Worksheet
Multiplying by 6 Using Patterns
With this worksheet, your third graders will learn a nifty trick about multiplying numbers by six. Any even number you multiply by six will be in the answer! For example, 6 x 2 = 12.
Math
Worksheet
Multiplying by 9 Using Patterns
Worksheet
Multiplying by 9 Using Patterns
With this worksheet, multiplying by nine got a lot easier for your third graders! Teach them the trick and they will be on the way to mastering their multiples of nine.
Math
Worksheet
Multiplying by 10, 100, or 1,000!
Worksheet
Multiplying by 10, 100, or 1,000!
Multiplying any number by 10, 100, or 1,000 will be easy for your third grader once they learn the trick! This worksheet will give them the knowledge they need to master this type of multiplication.
Math
Worksheet
Worksheet
Math
Worksheet
Multiplying by 2 and 5 Using Patterns
Worksheet
Multiplying by 2 and 5 Using Patterns
Math
Worksheet
Dividing by 2 and 3 Using Patterns
Worksheet
Dividing by 2 and 3 Using Patterns
With this worksheet, your students will learn a helpful trick about dividing by two and three. Dividing a number by two, is the same as one-half that number. Dividing by three is the same as one-third!
Math
Worksheet
Relating Single Digit Addition to a Double Digit
Worksheet
Relating Single Digit Addition to a Double Digit
With this worksheet, your third graders will be adding up two digit and three digit numbers with ease.
Math
Worksheet
Worksheet
Use these vocabulary cards with the EL Support Lesson: Talk About Patterns.
Math
Worksheet | 891 | 3,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2020-45 | latest | en | 0.841656 |
https://jp.mathworks.com/matlabcentral/profile/authors/7156805?s_tid=cody_local_to_profile | 1,611,313,673,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529179.46/warc/CC-MAIN-20210122082356-20210122112356-00210.warc.gz | 398,005,021 | 19,616 | Community Profile
# Shaun Kuo
9 2016 年以降の合計貢献数
#### Shaun Kuo's バッジ
Determine if input is odd
Given the input n, return true if n is odd or false if n is even.
4年以上 前
Select every other element of a vector
Write a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s...
4年以上 前
Is my wife right?
Regardless of input, output the string 'yes'.
4年以上 前
Given a and b, return the sum a+b in c.
4年以上 前
Sum all integers from 1 to 2^n
Given the number x, y must be the summation of all integers from 1 to 2^x. For instance if x=2 then y must be 1+2+3+4=10.
4年以上 前
Pizza!
Given a circular pizza with radius _z_ and thickness _a_, return the pizza's volume. [ _z_ is first input argument.] Non-scor...
4年以上 前
Find the sum of all the numbers of the input vector
Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...
4年以上 前
Make the vector [1 2 3 4 5 6 7 8 9 10]
In MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s...
4年以上 前
Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
4年以上 前 | 398 | 1,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-04 | latest | en | 0.449143 |
https://www.esaral.com/q/a-circular-park-is-surrounded-by-a-road-21-m-wide-54590 | 1,726,621,139,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00323.warc.gz | 702,702,016 | 11,683 | # A circular park is surrounded by a road 21 m wide.
Question:
A circular park is surrounded by a road 21 m wide. If the radius of theĀ park is 105 m, then find the area of the road.
Solution:
Given that, a circuiar park is surrounded by a road.
Width of the road $=21 \mathrm{~m}$
Radius of the park $(r)=105 \mathrm{~m}$
$\therefore$ Radius of whole circular portion (park + road),
Now, area of road = Area of whole circular portion
- Area of circular park
$=\pi r_{8}^{2}-\pi r_{i}^{2} \quad\left[\therefore\right.$ area of circle $\left.=\pi r^{2}\right]$
$=\pi\left(t_{e}^{2}-r_{i}^{2}\right)$
$=\pi\left\{\left(126^{2}-(105)^{2}\right\}\right.$
$=\frac{22}{7} \times(126+105)(126-105)$
$=\frac{22}{7} \times 231 \times 21 \quad\left[\because\left(a^{2}-b^{2}\right)=(a-b)(a+b)\right]$
$=66 \times 231$
$=15246 \mathrm{~cm}^{2}$
Hence, the required area of the road is $15246 \mathrm{~cm}^{2}$. | 328 | 917 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-38 | latest | en | 0.700928 |
https://www.grc.nasa.gov/www/k-12/airplane/kitelift.html | 1,670,030,813,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710918.58/warc/CC-MAIN-20221203011523-20221203041523-00853.warc.gz | 837,914,722 | 5,073 | + Text Only Site
+ Non-Flash Version
+ Contact Glenn
An excellent way for students to gain a feel for aerodynamic forces is to fly a kite. Kites fly because of forces acting on the parts of the kite. Though kites come in many shapes and sizes, the forces which act on the kite are the same for all kites. You can compare these forces to the forces that act on an airliner in flight and you will find that, with the exception of thrust, they are exactly the same. The similarity in forces allowed the Wright brothers to test their theories of flight by flying their aircraft as kites from 1900 to 1902. On this slide we show the equations which would describe the lift of a flying kite. The graphic shows a side view of the flying kite with the aerodynamic lift shown by the blue vector. The wind is blowing parallel to the ground and the lift direction is perpendicular to the wind. Since the forces on a kite are the same as the forces on an airplane, we can use the mathematical equations developed to predict airplane performance to predict the aerodynamic performance of a kite. In particular, the lift equation shown on the upper right side of the figure has been developed for aircraft. The lift L is equal to a lift coefficient Cl times the projected surface area A times the air density r times one half the square of the wind velocity V. L = Cl * A * r * .5 * V^2 The lift depends on two properties of the air; the density and velocity. In general, the density depends on your location on the earth. The higher the elevation, the lower the density. The standard value for air density r at sea level conditions is given as r = 1.229 kg/m^3 or .00237 slug/ft^3. The variation of lift with altitude is described on a separate page. The air velocity is the relative speed between the kite and the air. When the kite is held fixed by the control line, the relative air velocity is the wind speed. If the line breaks, or if you let out line, the velocity is something less than the wind speed; if you pull on the control line the velocity is the wind speed plus the speed of your pull. The lift changes with the square of the velocity. The aerodynamic lift of your kite depends directly on the surface area of the kite. You first learn how to compute the area for a geometric shape while you are in middle school. The surface area depends on the particular design of your kite. The lift depends on the lift coefficient, Cl, which depend on geometric properties of the kite and the angle between the kite surfaces and the wind. Lift coefficients are usually determined experimentally for aircraft, but the aerodynamic surfaces for most kites are simple, thin, flat plates. So we can use some experimental values of the lift coefficients for flat plates to get a first order idea of our kite performance. For a thin flat plate at a low angle of attack , the lift coefficient Clo is equal to 2.0 times pi (3.14159) times the angle a expressed in radians (180 degrees equals pi radians): Clo = 2 * pi * a We use Clo for the lift coefficient because there is another aerodynamic effect present on most kites. If we think of a kite as an aircraft wing, and use the terminology associated with aircraft wings, most kites have a low wing span (length from side to side) relative to the surface area. Most kites therefore have a low aspect ratio AR which is defined to be span s squared divided by the area A. AR = s^2 / A Near the tips of a wing the flow spills from the underside to the top side because of the difference in pressure. This creates a downwash which changes the effective angle of attack of the flow over a portion of the wing. For low aspect ratio wings, the portion of the wing affected by the downwash is greater than for high aspect ratio wings. Since most kites have a low aspect ratio AR, we have to include the effect of the downwash on the lift coefficient. The equation for the correction is given at the bottom right of the slide: Cl = Clo / (1 + Clo / (pi * AR) ) With these equations you can make a first prediction of the lift of your kite. You can use the KiteModeler program to further study how kites work and to design your own kites. Activities: Guided Tours Forces on a Kite KiteModeler Navigation .. Beginner's Guide Home Page
+ Inspector General Hotline + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act + Budgets, Strategic Plans and Accountability Reports + Freedom of Information Act + The President's Management Agenda + NASA Privacy Statement, Disclaimer, and Accessibility Certification Editor: Nancy Hall NASA Official: Nancy Hall Last Updated: May 13 2021 + Contact Glenn | 1,000 | 4,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2022-49 | latest | en | 0.935765 |
https://morioh.com/p/c52c5f1339ac | 1,603,141,868,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107866404.1/warc/CC-MAIN-20201019203523-20201019233523-00471.warc.gz | 444,830,165 | 3,567 | # Essential Mathematical Concepts for Machine Learning-Linear Algebra
We will start with Linear Algebra basics. I might be posting 2 or 3 more articles on Linear Algebra to cover those topics which are relevant to machine learning. The topics to be discussed are listed as follows:-
In this particular article, we will start with Linear Algebra basics. I might be posting 2 or 3 more articles on Linear Algebra to cover those topics which are relevant to machine learning. The topics to be discussed are listed as follows:-
1. Vectors
2. Vector Spaces and Subspaces
3. Linear Span
4. Column Space
5. Null Space
*Vectors:- *It is an object that has both magnitudes as well as the direction. Geometrically, we can visualize a vector as a directed line segment whose length gives the magnitude of the vector and arrow gives the direction.
In machine learning, we use vectors to represent features and labels of the input data like age, salary, etc. Here in this example, V is a two- dimensional vector from the field of Real Numbers.
*Vector Space:- *A vector space is a set V of vectors on which the vector addition and vector multiplication are defined such that,
1. (V,+) is an abelian group.
2. The dot product between real numbers and vector v is defined in such a way that for all “a” belongs to Real numbers and v belongs to the vector space V such that a.v should belong to the vector space V.
3. For all ‘a’ belongs to real numbers (R) and v,w belongs to V(vector space), a(v+w)=a.v+a.w
4. For all a,b belongs to R and v belongs to V, (a+b).v = a.v+b.v
5. For all a,b belongs to R and v belongs to V, (ab).v=a.(b.v)
6. (Unitary law) 1 belongs to R and v belongs to V, 1.v belongs to V.
## Linear Algebra: The hidden engine of machine learning
The hidden engine of machine learning. Algebra is firstly taken from a book, written by Khwarizmi(780-850 CE), which is about calculation and equations.
## Linear Algebra Concepts (For Machine Learning)
Basic Mathematical Concepts for Machine Learning. Machine learning at its base is complete and pure mathematics and logic.
## What is Supervised Machine Learning
What is neuron analysis of a machine? Learn machine learning by designing Robotics algorithm. Click here for best machine learning course models with AI
## Pros and Cons of Machine Learning Language
AI, Machine learning, as its title defines, is involved as a process to make the machine operate a task automatically to know more join CETPA
## Appendix: Linear Algebra
These notes are intended to provide a “fast” collection of theoretical definitions, theorems, and concepts for Linear Algebra as a refresher only | 599 | 2,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2020-45 | latest | en | 0.932427 |
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# How do you teach math concepts?
## How do you teach math concepts?
7 Effective Strategies for Teaching Elementary Math
1. Make it hands-on.
2. Use visuals and images.
3. Find opportunities to differentiate learning.
4. Ask students to explain their ideas.
5. Incorporate storytelling to make connections to real-world scenarios.
6. Show and tell new concepts.
7. Let your students regularly know how they’re doing.
## How do you teach math stories?
1. Use stories to develop number line skills: One of the biggest contributors to math achievement, even into adulthood, is a strong grasp of the number line. While we might be tempted to think the number line is just about counting, it’s also about sequence and spatial relationship.
## How do I teach my kindergarten number line?
Basic Number Line Skills
1. Find the dot marked on a number (and name the dot)
2. Place a dot at the desired location.
3. Find the desired number below the number line’s tick mark*
4. Add the missing number.
5. Learn the numbers to the right are higher than numbers to left.
6. Determine which number is greater than, less than or equals.
## How do you add and subtract on a number line?
Adding and Subtracting Positive and Negative Numbers using the Number Line
1. To add a positive number means that we move the point to the right of the number line.
2. Similarly, to add a negative number implies that we move the point to the left of the number line.
## How do you add without using number lines?
Hint: Adding one number to the other is just counting starting from the first number as many times the second number. When the second number is negative, the addition sign changes to subtraction, and the negative number is written as positive.
## How do you add a number line?
Add line numbers to a section or to multiple sections
1. Click in a section or select multiple sections.
2. On the Layout tab, in the Page Setup group, click Line Numbers.
3. Click Line Numbering Options, and then click the Layout tab.
4. In the Apply to list, click Selected sections.
5. Click Line Numbers.
## What is number line strategy?
A number line with no numbers or markers, essentially the empty number line is a visual representation for recording and sharing students’ thinking strategies during the process of mental computation. Before using an empty number line students need to show a secure understanding of numbers to 100.
## Why can you use adding up on an open number line to find the difference?
About Open Number Lines An open number line reinforces the idea that the answer to a subtraction problem is the difference between two numbers. It supports using the inverse relationship between addition and subtraction by counting up instead of back. It engages students in decomposing numbers and reasoning.
## What is 3/4 on a number line?
It’s the top number. This means that the fraction 3/4 is 3 out of 4 equal parts. So, count 3 parts from 0 and you will get 3/4. That’s it!
## Where would 2/3 Be on a number line?
1 Answer. On the real number line, it is between 1 and 2. To be more precise, the distance between 23 and 0 is twice the distance between 23 and 1 .
1/4
## What percent is 5 hours in a day?
20.83%
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https://mkihforums.com/articles/what-is-geometric-sequence-in-math-grade-10 | 1,675,150,616,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499845.10/warc/CC-MAIN-20230131055533-20230131085533-00515.warc.gz | 413,460,045 | 16,848 | # What is geometric sequence in math grade 10? (2023)
## What is geometric sequence in math grade 10?
A geometric sequence is a sequence of numbers in which each new term (except for the first term) is calculated by multiplying the previous term by a constant value called the constant ratio (r). This means that the ratio between consecutive numbers in a geometric sequence is a constant (positive or negative).
(Video) Geometric Series and Geometric Sequences - Basic Introduction
(The Organic Chemistry Tutor)
What is geometric sequence short answer?
A geometric sequence is a special progression, or a special sequence, of numbers, where each successive number is a fixed multiple of the number before it. Let me explain what I'm saying. So let's say my first number is 2 and then I multiply 2 by the number 3. So I multiply it by 3, I get 6.
(Video) GEOMETRIC SEQUENCE || GRADE 10 MATHEMATICS Q1
(WOW MATH)
What is a geometric sequence in math definition?
A geometric sequence is a sequence in which each term is found by multiplying the preceding term by the same value. Its general term is. a n= a 1 r n 1. The value r is called the common ratio. It is found by taking any term in the sequence and dividing it by its preceding term.
(Video) Grade 10 Math: Finding the next Term of a Geometric Sequence
(Prof Math Wizard)
What are 2 examples of geometric sequence?
{2,6,18,54,162,486,1458,...} is a geometric sequence where each term is 3 times the previous term. Example 2: {12,−6,3,−32,34,−38,316,...}
(Video) Introduction to geometric sequences | Sequences, series and induction | Precalculus | Khan Academy
Can you give me an example of geometric sequence?
An example of a Geometric sequence is 2, 4, 8, 16, 32, 64, …, where the common ratio is 2.
(Video) Illustrating Geometric Sequence Grade 10 - Online Class
(Melchor Paguirigan)
How do you find the geometric mean of 10 numbers?
Geometric mean formula is obtained by multiplying all the numbers together and taking the nth root of the product.
(Video) GRADE 10 MATHEMATICS QUARTER 1 WEEK 4 GEOMETRIC SEQUENCE, MEANS AND SERIES
What is sequence easy words?
A sequence of events or things is a number of events or things that come one after another in a particular order.
(Video) Grade 10 Week 3 Geometric Sequences
(Math Nacian)
How do you explain sequence?
A sequence is an ordered list of numbers . The three dots mean to continue forward in the pattern established. Each number in the sequence is called a term. In the sequence 1, 3, 5, 7, 9, …, 1 is the first term, 3 is the second term, 5 is the third term, and so on.
(Video) [TAGALOG] Grade 10 Math Lesson: SOLVING GEOMETRIC SEQUENCE (Part I) - FINDING THE nth TERM
(YouMore PH)
What is the best definition of geometric?
: using straight or curved lines in designs or outlines. : of or relating to art based on simple geometric shapes (such as straight lines, circles, or squares)
(Video) How to Find the Geometric Means? Geometric Sequence - Grade 10 Math
(Math Teacher Gon)
Why is it called geometric sequence?
Apparently, the expression “geometric progression” comes from the “geometric mean” (Euclidean notion) of segments of length a and b: it is the length of the side c of a square whose area is equal to the area of the rectangle of sides a and b.
(Video) GEOMETRIC SEQUENCE GRADE 10 MATH
(JHGCA Main_Branch)
## How do you find the next number in a geometric sequence?
A geometric sequence is a sequence of numbers that follows a pattern were the next term is found by multiplying by a constant called the common ratio, r.
(Video) Grade 10 - Geometric Sequence
(Ma'am Mariel)
What is geometry formula?
Geometry formulas are used for finding dimensions, perimeter, area, surface area, volume, etc. of the geometric shapes. Geometry is a part of mathematics that deals with the relationships of points, lines, angles, surfaces, solids measurement, and properties.
What is an example of a geometric?
Circles, squares, triangles, and rectangles are all types of 2D geometric shapes. Check out a list of different 2D geometric shapes, along with a description and examples of where you can spot them in everyday life. Keep in mind that these shapes are all flat figures without depth.
What are 3 examples of geometric forms?
List of Geometric Shapes
• Triangle.
• Circle.
• Semi-Circle.
• Square.
• Rectangle.
• Parallelogram.
• Rhombus.
• Trapezium.
8 Jan 2021
What is the easiest way to find the formula for a sequence?
Solution: To find a specific term of an arithmetic sequence, we use the formula for finding the nth term. Step 1: The nth term of an arithmetic sequence is given by an = a + (n – 1)d. So, to find the nth term, substitute the given values a = 2 and d = 3 into the formula.
Which is an example of a sequence?
A sequence is an ordered list of elements with a specific pattern. For example, 3, 7, 11, 15, ... is a sequence as there is a pattern where each term is obtained by adding 4 to its previous term.
What is the easiest way to calculate geometric mean?
To calculate the geometric mean of two numbers, you would multiply the numbers together and take the square root of the result.
What is geometric series with example?
geometric series, in mathematics, an infinite series of the form a + ar + ar2 + ar3+⋯, where r is known as the common ratio. A simple example is the geometric series for a = 1 and r = 1/2, or 1 + 1/2 + 1/4 + 1/8 +⋯, which converges to a sum of 2 (or 1 if the first term is excluded).
How do you solve mean?
To calculate the mean, you first add all the numbers together (3 + 11 + 4 + 6 + 8 + 9 + 6 = 47). Then you divide the total sum by the number of scores used (47 / 7 = 6.7). In this example, the mean or average of the number set is 6.7.
What is sequence in one word?
sequence noun (ORDERED SERIES)
a series of related things or events, or the order in which they follow each other: The first chapter describes the strange sequence of events that led to his death.
## What is sequence in a sentence?
Noun He listened to the telephone messages in sequence. a chase sequence in a spy movie I enjoyed the movie's opening sequence.
What are the 3 types of sequences?
There are mainly three types of sequences: Arithmetic Sequences. Geometric Sequence. Fibonacci Sequence.
Why is it important to sequence?
Why is sequencing important? We sequence all day long—we divide our time into what we need to do first, second, and last; we understand events in our lives by understanding the order in which they occur. For some children, sequencing can be a hard concept to grasp, especially when they are trying to tell a story.
Why are sequences so important?
Sequences are useful in a number of mathematical disciplines ... In particular, sequences are the basis for series, which are important in differential equations and analysis. Sequences are also of interest in their own right and can be studied as patterns or puzzles, such as in the study of prime numbers.
How do you create a sequence explain with example?
Example 1:
CREATE SEQUENCE sequence_1 start with 1 increment by 1 minvalue 0 maxvalue 100 cycle; Above query will create a sequence named sequence_1. Sequence will start from 1 and will be incremented by 1 having maximum value 100. Sequence will repeat itself from start value after exceeding 100.
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Enseignant: Thiran Jean-Philippe
Langue: Français
## Summary
This class teaches the theory of linear time-invariant (LTI) systems. These systems serve both as models of physical reality (such as the wireless channel) and as engineered systems (such as electrical circuits, filters and control strategies).
## Keywords
Systems, Circuits, Signals, Frequency Response, Transfer Function, Fourier Transform, Laplace Transform, Z Transform, Stability, Causality, Sampling
## Required courses
Analysis I, II, III. Linear algebra I.
## Recommended courses
Linear algebra II
## Learning Outcomes
By the end of the course, the student must be able to:
• Describe properties of LTI systems
• Solve for poles and zeros of LTI systems
• Recall properties of CT Fourier transform
• Analyze LTI systems by spectral analysis
• Operate with Fourier transform tools
• Work out / Determine impulse response of CT LTI
## Teaching methods
• Classroom lectures
• Written exercises
## Expected student activities
• Read course book in english (the course is taught in english)
## Assessment methods
Homeworks and written mid-term exam and final exams
## Bibliography
The following is a recommended (but not required) book:
A. V. Oppenheim and A. S. Willsky, with S. Hamid Nawab, Signals and Systems. Upper Saddle River, NJ: Prentice Hall, 2nd ed., 1996.
## Dans les plans d'études
• Semestre: Printemps
• Forme de l'examen: Ecrit (session d'été)
• Matière examinée: Signaux et systèmes (pour EL)
• Cours: 2 Heure(s) hebdo x 14 semaines
• Exercices: 2 Heure(s) hebdo x 14 semaines
• Semestre: Printemps
• Forme de l'examen: Ecrit (session d'été)
• Matière examinée: Signaux et systèmes (pour EL)
• Cours: 2 Heure(s) hebdo x 14 semaines
• Exercices: 2 Heure(s) hebdo x 14 semaines
## Semaine de référence
Lu Ma Me Je Ve 8-9 9-10 10-11 11-12 12-13 13-14 14-15 15-16 16-17 17-18 18-19 19-20 20-21 21-22 | 551 | 1,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-40 | longest | en | 0.650375 |
https://stats.stackexchange.com/questions/121852/how-to-choose-between-t-test-or-non-parametric-test-e-g-wilcoxon-in-small-sampl/372636 | 1,643,220,933,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304959.80/warc/CC-MAIN-20220126162115-20220126192115-00163.warc.gz | 588,750,305 | 58,311 | # How to choose between t-test or non-parametric test e.g. Wilcoxon in small samples
Certain hypotheses can be tested using Student's t-test (maybe using Welch's correction for unequal variances in the two-sample case), or by a non-parametric test like the Wilcoxon paired signed rank test, the Wilcoxon-Mann-Whitney U test, or the paired sign test. How can we make a principled decision about which test is most appropriate, particularly if the sample size is "small"?
Many introductory textbooks and lecture notes give a "flowchart" approach where normality is checked (either – inadvisedly – by normality test, or more broadly by QQ plot or similar) to decide between a t-test or non-parametric test. For the unpaired two-sample t-test there may be a further check for homogeneity of variance to decide whether to apply Welch's correction. One issue with this approach is the way the decision on which test to apply depends on the observed data, and how this affects the performance (power, Type I error rate) of the selected test.
Another problem is how hard checking normality is in small data sets: formal testing has low power so violations may well not be detected, but similar issues apply eyeballing the data on a QQ plot. Even egregious violations could go undetected, e.g. if the distribution is mixed but no observations were drawn from one component of the mixture. Unlike for large $n$, we can't lean on the safety-net of the Central Limit Theorem, and the asymptotic normality of the test statistic and t distribution.
One principled response to this is "safety first": with no way to reliably verify the normality assumption in a small sample, stick to non-parametric methods. Another is to consider any grounds for assuming normality, theoretically (e.g. variable is sum of several random components and CLT applies) or empirically (e.g. previous studies with larger $n$ suggest variable is normal), and using a t-test only if such grounds exist. But this usually only justifies approximate normality, and on low degrees of freedom it's hard to judge how near normal it needs to be to avoid invalidating a t-test.
Most guides to choosing a t-test or non-parametric test focus on the normality issue. But small samples also throw up some side-issues:
• If performing an "unrelated samples" or "unpaired" t-test, whether to use a Welch correction? Some people use a hypothesis test for equality of variances, but here it would have low power; others check whether SDs are "reasonably" close or not (by various criteria). Is it safer simply to always use the Welch correction for small samples, unless there is some good reason to believe population variances are equal?
• If you see the choice of methods as a trade-off between power and robustness, claims about the asymptotic efficiency of the non-parametric methods are unhelpful. The rule of thumb that "Wilcoxon tests have about 95% of the power of a t-test if the data really are normal, and are often far more powerful if the data is not, so just use a Wilcoxon" is sometimes heard, but if the 95% only applies to large $n$, this is flawed reasoning for smaller samples.
• Small samples may make it very difficult, or impossible, to assess whether a transformation is appropriate for the data since it's hard to tell whether the transformed data belong to a (sufficiently) normal distribution. So if a QQ plot reveals very positively skewed data, which look more reasonable after taking logs, is it safe to use a t-test on the logged data? On larger samples this would be very tempting, but with small $n$ I'd probably hold off unless there had been grounds to expect a log-normal distribution in the first place.
• What about checking assumptions for the non-parametrics? Some sources recommend verifying a symmetric distribution before applying a Wilcoxon test (treating it as a test for location rather than stochastic dominance), which brings up similar problems to checking normality. If the reason we are applying a non-parametric test in the first place is a blind obedience to the mantra of "safety first", then the difficulty assessing skewness from a small sample would apparently lead us to the lower power of a paired sign test.
With these small-sample issues in mind, is there a good - hopefully citable - procedure to work through when deciding between t and non-parametric tests?
There have been several excellent answers, but a response considering other alternatives to rank tests, such as permutation tests, would also be welcome.
• I should explain what a "method for choosing a test" might be - introductory texts often use flowcharts. For unpaired data, maybe: "1. Use some method to check if both samples are normally distributed (if not go to 3), 2. Use some method to check for unequal variances: if so, perform two-sample t-test with Welch's correction, if not, perform without correction. 3. Try transforming data to normality (if works go to 2 else go to 4). 4. Perform U test instead (possibly after checking various assumptions)." But many of these steps seem unsatisfactory for small n, as I hope my Q explains! Oct 29 '14 at 16:01
• Interesting question (+1) and a brave move to set up a bounty. Looking forward for some interesting answers. By the way, what I often see applied in my field is a permutation test (instead of either t-test or Mann-Whitney-Wilcoxon). I guess it could be considered a worthy contender as well. Apart from that, you never specified what you mean by "small sample size". Nov 4 '14 at 1:09
• It is not clear to me why one would assert that nonparametric tests (rank sum or sign rank) would require symmetry? Nov 5 '14 at 22:36
• @Silverfish " if the results are seen as a statement about location" That is an important caveat, as these tests are most generally statements about evidence for H$_{0}: P(X_{A} > X_{B}) =0.5$. Making additional distributional assumptions narrows the scope of inference (e.g. tests for median difference), but are not generally requisites for the tests. Nov 6 '14 at 3:09
• It might be worth exploring just how "flawed" the "95% power for the Wilcoxon" reasoning is for small samples (in part it depends on what, exactly, one does, and how small is small). If for example, you're happy to conduct tests at say 5.5% instead of 5%, should that be the nearest suitable achievable significance level, the power often tends to hold up fairly well. Once can of course - at the "power calculation" stage before you collect data - figure out what the circumstances may be and get a sense of what the properties of the Wilcoxon are at the sample sizes you're considering. Oct 22 '15 at 22:07
I am going to change the order of questions about.
I've found textbooks and lecture notes frequently disagree, and would like a system to work through the choice that can safely be recommended as best practice, and especially a textbook or paper this can be cited to.
Unfortunately, some discussions of this issue in books and so on rely on received wisdom. Sometimes that received wisdom is reasonable, sometimes it is less so (at the least in the sense that it tends to focus on a smaller issue when a larger problem is ignored); we should examine the justifications offered for the advice (if any justification is offered at all) with care.
Most guides to choosing a t-test or non-parametric test focus on the normality issue.
That’s true, but it’s somewhat misguided for several reasons that I address in this answer.
If performing an "unrelated samples" or "unpaired" t-test, whether to use a Welch correction?
This (to use it unless you have reason to think variances should be equal) is the advice of numerous references. I point to some in this answer.
Some people use a hypothesis test for equality of variances, but here it would have low power. Generally I just eyeball whether the sample SDs are "reasonably" close or not (which is somewhat subjective, so there must be a more principled way of doing it) but again, with low n it may well be that the population SDs are rather further apart than the sample ones.
Is it safer simply to always use the Welch correction for small samples, unless there is some good reason to believe population variances are equal? That’s what the advice is. The properties of the tests are affected by the choice based on the assumption test.
Some references on this can be seen here and here, though there are more that say similar things.
The equal-variances issue has many similar characteristics to the normality issue – people want to test it, advice suggests conditioning choice of tests on the results of tests can adversely affect the results of both kinds of subsequent test – it’s better simply not to assume what you can’t adequately justify (by reasoning about the data, using information from other studies relating to the same variables and so on).
However, there are differences. One is that – at least in terms of the distribution of the test statistic under the null hypothesis (and hence, its level-robustness) - non-normality is less important in large samples (at least in respect of significance level, though power might still be an issue if you need to find small effects), while the effect of unequal variances under the equal variance assumption doesn’t really go away with large sample size.
What principled method can be recommended for choosing which is the most appropriate test when the sample size is "small"?
With hypothesis tests, what matters (under some set of conditions) is primarily two things:
• What is the actual type I error rate?
• What is the power behaviour like?
We also need to keep in mind that if we're comparing two procedures, changing the first will change the second (that is, if they’re not conducted at the same actual significance level, you would expect that higher $$\alpha$$ is associated with higher power).
(Of course we're usually not so confident we know what distributions we're dealing with, so the sensitivity of those behaviors to changes in circumstances also matter.)
With these small-sample issues in mind, is there a good - hopefully citable - checklist to work through when deciding between t and non-parametric tests?
I will consider a number of situations in which I’ll make some recommendations, considering both the possibility of non-normality and unequal variances. In every case, take mention of the t-test to imply the Welch-test:
• n medium-large
Non-normal (or unknown), likely to have near-equal variance:
If the distribution is heavy-tailed, you will generally be better with a Mann-Whitney, though if it’s only slightly heavy, the t-test should do okay. With light-tails the t-test may (often) be preferred. Permutation tests are a good option (you can even do a permutation test using a t-statistic if you're so inclined). Bootstrap tests are also suitable.
Non-normal (or unknown), unequal variance (or variance relationship unknown):
If the distribution is heavy-tailed, you will generally be better with a Mann-Whitney
• if inequality of variance is only related to inequality of mean - i.e. if H0 is true the difference in spread should also be absent. GLMs are often a good option, especially if there’s skewness and spread is related to the mean. A permutation test is another option, with a similar caveat as for the rank-based tests. Bootstrap tests are a good possibility here.
Zimmerman and Zumbo (1993)$$^{[1]}$$ suggest a Welch-t-test on the ranks which they say performs better that the Wilcoxon-Mann-Whitney in cases where the variances are unequal.
• n moderately small
rank tests are reasonable defaults here if you expect non-normality (again with the above caveat). If you have external information about shape or variance, you might consider GLMs . If you expect things not to be too far from normal, t-tests may be fine.
• n very small
Because of the problem with getting suitable significance levels, neither permutation tests nor rank tests may be suitable, and at the smallest sizes, a t-test may be the best option (there’s some possibility of slightly robustifying it). However, there’s a good argument for using higher type I error rates with small samples (otherwise you’re letting type II error rates inflate while holding type I error rates constant). Also see de Winter (2013)$$^{[2]}$$.
The advice must be modified somewhat when the distributions are both strongly skewed and very discrete, such as Likert scale items where most of the observations are in one of the end categories. Then the Wilcoxon-Mann-Whitney isn’t necessarily a better choice than the t-test.
Simulation can help guide choices further when you have some information about likely circumstances.
I appreciate this is something of a perennial topic, but most questions concern the questioner's particular data set, sometimes a more general discussion of power, and occasionally what to do if two tests disagree, but I would like a procedure to pick the correct test in the first place!
The main problem is how hard it is to check the normality assumption in a small data set:
It is difficult to check normality in a small data set, and to some extent that's an important issue, but I think there's another issue of importance that we need to consider. A basic problem is that trying to assess normality as the basis of choosing between tests adversely impacts the properties of the tests you're choosing between.
Any formal test for normality would have low power so violations may well not be detected. (Personally I wouldn't test for this purpose, and I'm clearly not alone, but I've found this little use when clients demand a normality test be performed because that's what their textbook or old lecture notes or some website they found once declare should be done. This is one point where a weightier looking citation would be welcome.)
Here’s an example of a reference (there are others) which is unequivocal (Fay and Proschan, 2010$$^{[3]}$$):
The choice between t- and WMW DRs should not be based on a test of normality.
They are similarly unequivocal about not testing for equality of variance.
To make matters worse, it is unsafe to use the Central Limit Theorem as a safety net: for small n we can't rely on the convenient asymptotic normality of the test statistic and t distribution.
Nor even in large samples -- asymptotic normality of the numerator doesn’t imply that the t-statistic will have a t-distribution. However, that may not matter so much, since you should still have asymptotic normality (e.g. CLT for the numerator, and Slutsky’s theorem suggest that eventually the t-statistic should begin to look normal, if the conditions for both hold.)
One principled response to this is "safety first": as there's no way to reliably verify the normality assumption on a small sample, run an equivalent non-parametric test instead.
That’s actually the advice that the references I mention (or link to mentions of) give.
Another approach I've seen but feel less comfortable with, is to perform a visual check and proceed with a t-test if nothing untowards is observed ("no reason to reject normality", ignoring the low power of this check). My personal inclination is to consider whether there are any grounds for assuming normality, theoretical (e.g. variable is sum of several random components and CLT applies) or empirical (e.g. previous studies with larger n suggest variable is normal).
Both those are good arguments, especially when backed up with the fact that the t-test is reasonably robust against moderate deviations from normality. (One should keep in mind, however, that "moderate deviations" is a tricky phrase; certain kinds of deviations from normality may impact the power performace of the t-test quite a bit even though those deviations are visually very small - the t-test is less robust to some deviations than others. We should keep this in mind whenever we're discussing small deviations from normality.)
Beware, however, the phrasing "suggest the variable is normal". Being reasonably consistent with normality is not the same thing as normality. We can often reject actual normality with no need even to see the data – for example, if the data cannot be negative, the distribution cannot be normal. Fortunately, what matters is closer to what we might actually have from previous studies or reasoning about how the data are composed, which is that the deviations from normality should be small.
If so, I would use a t-test if data passed visual inspection, and otherwise stick to non-parametrics. But any theoretical or empirical grounds usually only justify assuming approximate normality, and on low degrees of freedom it's hard to judge how near normal it needs to be to avoid invalidating a t-test.
Well, that’s something we can assess the impact of fairly readily (such as via simulations, as I mentioned earlier). From what I've seen, skewness seems to matter more than heavy tails (but on the other hand I have seen some claims of the opposite - though I don't know what that's based on).
For people who see the choice of methods as a trade-off between power and robustness, claims about the asymptotic efficiency of the non-parametric methods are unhelpful. For instance, the rule of thumb that "Wilcoxon tests have about 95% of the power of a t-test if the data really are normal, and are often far more powerful if the data is not, so just use a Wilcoxon" is sometimes heard, but if the 95% only applies to large n, this is flawed reasoning for smaller samples.
But we can check small-sample power quite easily! It’s easy enough to simulate to obtain power curves as here.
(Again, also see de Winter (2013)$$^{[2]}$$).
Having done such simulations under a variety of circumstances, both for the two-sample and one-sample/paired-difference cases, the small sample efficiency at the normal in both cases seems to be a little lower than the asymptotic efficiency, but the efficiency of the signed rank and Wilcoxon-Mann-Whitney tests is still very high even at very small sample sizes.
At least that's if the tests are done at the same actual significance level; you can't do a 5% test with very small samples (and least not without randomized tests for example), but if you're prepared to perhaps do (say) a 5.5% or a 3.2% test instead, then the rank tests hold up very well indeed compared with a t-test at that significance level.
Small samples may make it very difficult, or impossible, to assess whether a transformation is appropriate for the data since it's hard to tell whether the transformed data belong to a (sufficiently) normal distribution. So if a QQ plot reveals very positively skewed data, which look more reasonable after taking logs, is it safe to use a t-test on the logged data? On larger samples this would be very tempting, but with small n I'd probably hold off unless there had been grounds to expect a log-normal distribution in the first place.
There’s another alternative: make a different parametric assumption. For example, if there’s skewed data, one might, for example, in some situations reasonably consider a gamma distribution, or some other skewed family as a better approximation - in moderately large samples, we might just use a GLM, but in very small samples it may be necessary to look to a small-sample test - in many cases simulation can be useful.
Alternative 2: robustify the t-test (but taking care about the choice of robust procedure so as not to heavily discretize the resulting distribution of the test statistic) - this has some advantages over a very-small-sample nonparametric procedure such as the ability to consider tests with low type I error rate.
Here I'm thinking along the lines of using say M-estimators of location (and related estimators of scale) in the t-statistic to smoothly robustify against deviations from normality. Something akin to the Welch, like:
$$\frac{\stackrel{\sim}{x}-\stackrel{\sim}{y}}{\stackrel{\sim}{S}_p}$$
where $$\stackrel{\sim}{S}_p^2=\frac{\stackrel{\sim}{s}_x^2}{n_x}+\frac{\stackrel{\sim}{s}_y^2}{n_y}$$ and $$\stackrel{\sim}{x}$$, $$\stackrel{\sim}{s}_x$$ etc being robust estimates of location and scale respectively.
I'd aim to reduce any tendency of the statistic to discreteness - so I'd avoid things like trimming and Winsorizing, since if the original data were discrete, trimming etc will exacerbate this; by using M-estimation type approaches with a smooth $$\psi$$-function you achieve similar effects without contributing to the discreteness. Keep in mind we're trying to deal with the situation where $$n$$ is very small indeed (around 3-5, in each sample, say), so even M-estimation potentially has its issues.
You could, for example, use simulation at the normal to get p-values (if sample sizes are very small, I'd suggest that over bootstrapping - if sample sizes aren't so small, a carefully-implemented bootstrap may do quite well, but then we might as well go back to Wilcoxon-Mann-Whitney). There's be a scaling factor as well as a d.f. adjustment to get to what I'd imagine would then be a reasonable t-approximation. This means we should get the kind of properties we seek very close to the normal, and should have reasonable robustness in the broad vicinity of the normal. There are a number of issues that come up that would be outside the scope of the present question, but I think in very small samples the benefits should outweigh the costs and the extra effort required.
[I haven't read the literature on this stuff for a very long time, so I don't have suitable references to offer on that score.]
Of course if you didn't expect the distribution to be somewhat normal-like, but rather similar to some other distribution, you could undertake a suitable robustification of a different parametric test.
What if you want to check assumptions for the non-parametrics? Some sources recommend verifying a symmetric distribution before applying a Wilcoxon test, which brings up similar problems to checking normality.
Indeed. I assume you mean the signed rank test*. In the case of using it on paired data, if you are prepared to assume that the two distributions are the same shape apart from location shift you are safe, since the differences should then be symmetric. Actually, we don't even need that much; for the test to work you need symmetry under the null; it's not required under the alternative (e.g. consider a paired situation with identically-shaped right skewed continuous distributions on the positive half-line, where the scales differ under the alternative but not under the null; the signed rank test should work essentially as expected in that case). The interpretation of the test is easier if the alternative is a location shift though.
*(Wilcoxon’s name is associated with both the one and two sample rank tests – signed rank and rank sum; with their U test, Mann and Whitney generalized the situation studied by Wilcoxon, and introduced important new ideas for evaluating the null distribution, but the priority between the two sets of authors on the Wilcoxon-Mann-Whitney is clearly Wilcoxon’s -- so at least if we only consider Wilcoxon vs Mann&Whitney, Wilcoxon goes first in my book. However, it seems Stigler's Law beats me yet again, and Wilcoxon should perhaps share some of that priority with a number of earlier contributors, and (besides Mann and Whitney) should share credit with several discoverers of an equivalent test.[4][5] )
References
[1]: Zimmerman DW and Zumbo BN, (1993),
Rank transformations and the power of the Student t-test and Welch t′-test for non-normal populations,
Canadian Journal Experimental Psychology, 47: 523–39.
[2]: J.C.F. de Winter (2013),
"Using the Student’s t-test with extremely small sample sizes,"
Practical Assessment, Research and Evaluation, 18:10, August, ISSN 1531-7714
http://pareonline.net/getvn.asp?v=18&n=10
[3]: Michael P. Fay and Michael A. Proschan (2010),
"Wilcoxon-Mann-Whitney or t-test? On assumptions for hypothesis tests and multiple interpretations of decision rules,"
Stat Surv; 4: 1–39.
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2857732/
[4]: Berry, K.J., Mielke, P.W. and Johnston, J.E. (2012),
"The Two-sample Rank-sum Test: Early Development,"
Electronic Journal for History of Probability and Statistics, Vol.8, December
pdf
[5]: Kruskal, W. H. (1957),
"Historical notes on the Wilcoxon unpaired two-sample test,"
Journal of the American Statistical Association, 52, 356–360.
• A couple of things I'd like clarification on. There's several points where you mention e.g. "If the distribution is heavy-tailed, ..." (or skewed etc) - presumably this should be read as "if it's reasonable to assume that the distribution will be heavy-tailed" (from theory/previous studies/whatever) rather than "if the sample is heavy-tailed", otherwise we are back at multi-step testing again which is the thing we are trying to avoid? (It seems to me that a central issue in this topic is how to justify beliefs or assumptions about distributions, without reading too much into the sample.) Nov 10 '14 at 16:51
• Yes, that should be understood as "population is either known to be heavy-tailed, or may reasonably expected to be heavy tailed". That certainly includes things like theory (or sometimes even general reasoning about the situation that doesn't quite reach the status of theory), expert knowledge, and previous studies. It's not suggesting testing for heavy-tailedness. In situations where it's simply unknown, it may be worth investigating how bad things might be under various distributions which might be plausible for the specific situation you have. Nov 10 '14 at 17:13
• Any chance that this already excellent answer could incorporate a little more detail on what options there might be to "robustify" the t-test? Nov 10 '14 at 18:59
• Silverfish - I'm not sure if I sufficiently addressed your question asking for detail on robustifying. I'll add a little more now. Jan 28 '15 at 1:33
• Many thanks for the addition, I thought that added a lot to the quality of this answer. Now this question has settled down a bit, and generated a good set of responses, I'd like to give the original question a good copy-edit and remove anything which might be misleading (for the benefit of readers who don't read past the question!). Is it okay when I do so for me to make appropriate edits to your response so quotes match with the reorganized question? Feb 17 '15 at 17:21
In my view the principled approach recognizes that (1) tests and graphical assessments of normality have insufficient sensitivity and graph interpretation is frequently not objective, (2) multi-step procedures have uncertain operating characteristics, (3) many nonparametric tests have excellent operating characteristics under situations in which parametric tests have optimum power, and (4) the proper transformation of $Y$ is not generally the identity function, and nonparametric $k$-sample tests are invariant to the transformation chosen (not so for one-sample tests such as the Wilcoxon signed rank test). Regarding (2), multi-step procedures are particularly problematic in areas such as drug development where oversight agencies such as FDA are rightfully concerned about possible manipulation of results. For example, an unscrupulous researcher might conveniently forget to report the test of normality if the $t$-test results in a low $P$-value.
Putting all this together, some suggested guidance is as follows:
1. If there is not a compelling reason to assume a Gaussian distribution before examining the data, and no covariate adjustment is needed, use a nonparametric test.
2. If covariate adjustment is needed, use the semiparametric regression generalization of the rank test you prefer. For the Wilcoxon test this is the proportional odds model and for a normal scores test this is probit ordinal regression.
These recommendations are fairly general although your mileage may vary for certain small sample sizes. But we know that for larger samples the relative efficiency of the Wilcoxon 2-sample test and signed rank tests compared to the $t$-test (if equal variance holds in the 2-sample case) is $\frac{3}{\pi}$ and that the relative efficiency of rank tests is frequently much greater than 1.0 when the Gaussian distribution does not hold. To me, the information loss in using rank tests is very small compared to the possible gains, robustness, and freedom from having to specify the transformation of $Y$.
Nonparametric tests can perform well even if their optimality assumptions are not satisfied. For the $k$-sample problem, rank tests make no assumptions about the distribution for a given group; they only make assumptions for how the distributions of the $k$ groups are connected to each other, if you require the test to be optimal. For a $-\log-\log$ link cumulative probability ordinal model the distributions are assumed to be in proportional hazards. For a logit link cumulative probability model (proportional odds model), the distributions are assumed to be connected by the proportional odds assumptions, i.e., the logits of the cumulative distribution functions are parallel. The shape of one of the distributions is irrelevant. Details may be found in http://biostat.mc.vanderbilt.edu/CourseBios330 in Chapter 15 of Handouts.
There are two types of assumptions of a frequentist statistical method that are frequently considered. The first is assumptions required to make the method preserve type I error. The second relates to preserving type II error (optimality; sensitivity). I believe that the best way to expose the assumptions needed for the second are to embed a nonparametric test in a semiparametric model as done above. The actual connection between the two is from Rao efficient score tests arising from the semiparametric model. The numerator of the score test from a proportional odds model for the two-sample case is exactly the rank-sum statistic.
• Thanks for this, I'm very sympathetic to the philosophy of this answer - for instance, lots of sources suggest I should at least eyeball-check data for normality before deciding on a test. But this sort of multi-step procedure clearly, albeit subtly, influences how the tests operate. Nov 4 '14 at 14:06
• Some queries: (1) suppose there's good reason to assume a Gaussian distribution a priori (eg previous studies) so we prefer a t-test. For tiny $n$ there's no point trying to assess normality - there'd be no way to detect its breach. But for $n=15$ or so, a QQ plot may well show up eg if there's severe skew. Does the philosophy of avoiding multi-step procedures mean we should simply justify our normality assumption, then proceed without checking the apparent distribution of our data? Similarly, in the k sample case, should we by default assume unequal variances rather than try to check it? Nov 4 '14 at 14:19
• (+1) I am wondering what is your take on Mann-Whitney-Wilcoxon vs. permutation tests (I am referring to Monte Carlo permutation test, when group labels are shuffled e.g. $10\,000$ times and $p$-value is computed directly as the number of shuffles resulting in a larger group difference)? Nov 4 '14 at 15:55
• Permutation tests are ways to control type I error but do not address type II error. A permutation test based on suboptimal statistics (e.g., ordinary mean and variance when the data come from a log-Gaussian distribution) will suffer in terms of power. Nov 4 '14 at 16:26
• Yes Chapter 15 in the Handouts is expanded into a new chapter in the upcoming 2nd edition of my book which I'll submit to the publisher next month. Nov 6 '14 at 12:40
Rand Wilcox in his publications and books make some very important points, many of which were listed by Frank Harrell and Glen_b in earlier posts.
1. The mean is not necessarily the quantity we want to make inferences about. There maybe other quantities that better exemplifies a typical observation.
2. For t-tests, power can be low even for small departures from normality.
3. For t-tests, observed probability coverage can be substantially different than nominal.
Some key suggestions are:
1. A robust alternative is to compare trimmed means or M-estimators using the t-test. Wilcox suggests 20% trimmed means.
2. Empirical Likelihood methods are theoretically more advantageous (Owen, 2001) but not necessarily so for medium to small n.
3. Permutations tests are great if one needs to control Type I error, but one cannot get CI.
4. For many situations Wilcox proposes the bootstrap-t to compare trimmed means. In R, this is implemented in the functions yuenbt, yhbt in the WRS package.
5. Percentile bootstrap maybe better than percentile-t when amount of trimming is >/=20%. In R this is implemented in the function pb2gen in the aforementioned WRS package.
Two good references are Wilcox (2010) and Wilcox (2012).
Bradley, in his work Distribution-Free Statistical Tests (1968, pp. 17–24), brings thirteen contrasts between what he calls "classical" and "distribution-free" tests. Note that Bradley differentiates between "non-parametric" and "distribution-free," but for the purposes of your question this difference is not relevant. Included in those thirteen are elements that relate not just to the derivatinos of the tests, but their applications. These include:
• Choice of significance level: Classical tests have continuous significance levels; distribution-free tests usually have discrete observations of the significance levels, so the classical tests offer more flexibility in setting said level.
• Logical validity of rejection region: Distribution-free test rejection regions can be less intuitively understandable (neither necessarily smooth nor continuous) and may cause confusion as to when the test should be considered to have rejected the null hypothesis.
• Type of statistics which are testable: To quote Bradley directly: "Statistics defined in terms of arithmetical operations upon observation magnitudes can be tested by classical techniques, wheras thse defined by order relationships (rank) or category-frequencies, etc. can be tested by distribution-free methods. Means and variances are examples of the former and medians and interquartile ranges, of the latter." Especially when dealing with non-normal distributions, the ability to test other statistics becomes valuable, lending weight to the distribution-free tests.
• Testability of higher-order interactions: Much easier under classical tests than distribution-free tests.
• Influence of sample size: This is a rather important one in my opinion. When sample sizes are small (Bradley says around n = 10), it may be very difficult to determine if the parametric assumptions underlying the classical tests have been violated or not. Distribution-free tests do not have these assumptions to be violated. Moreover, even when the assumptions have not been violated, the distribution-free tests are often almost as easy to apply and almost as efficient of a test. So for small sample sizes (less than 10, possible up to 30) Bradley favors an almost routine application of distribution-free tests. For large sample sizes, the Central Limit Theorem tends to overwhelm parametric violations in that the sample mean and sample variance will tend to the normal, and the parametric tests may be superior in terms of efficieny.
• Scope of Application: By being distribution-free, such tests are applicable to a much larger class of populations than classical tests assuming a specific distribution.
• Detectibility of violation of assumption of a continuous distribution: Easy to see in distributio-free tests (e.g. existence of tied scores), harder in parametric tests.
• Effect of violation of assumption of a continuous distribution: If the assumption is violated the test becomes inexact. Bradley spends time explaining how the bounds of the inexactitude can be estimated for distribution-free tests, but there is no analogous routine for classical tests.
• Thank you for the citation! Bradley's work seems quite old so I suspect it does not have much work on modern simulation studies to compare efficiencies and Type I/II error rates in various scenarios? I'd also be interested in what he suggests about Brunner-Munzel tests - should they be used instead of a U test if variances in the two groups are not known to be equal? Nov 5 '14 at 22:11
• Bradley does discuss efficiencies, although most of the time, it is in the context of asymptotic relative efficiency. He brings sources sometimes for statements about finite sample-size efficiency, but as the work is from 1968, I'm sure much better analyses have been done since then. Speaking of which, If I have it right, Brunner and Munzel wrote their article in 2000, which explains why there is no mention of it in Bradley. Nov 5 '14 at 23:23
• Yes that would indeed explain it! :) Do you know if there is a more up to date survey than Bradley? Nov 5 '14 at 23:25
• A brief search shows that there are a lot of recent texts on non-parametric statistics. For example: Nonparametric Statistical Methods (Hollander et al, 2013), Nonparametric Hypothesis Testing: Rank and Permutation Methods with Applications in R (Bonnini et al, 2014), Nonparametric Statistical Inference, Fifth Edition (Gibbons and Chakraborti, 2010). There are many others which come up in various searches. As I don't have any, I cannot make any recommendations. Sorry. Nov 6 '14 at 0:20
Starting to answer this very interesting question.
For non-paired data:
Performance of five two-sample location tests for skewed distributions with unequal variances by Morten W. Fagerland, Leiv Sandvik (behind paywall) performs a series of experiments with 5 different tests (t-test, Welch U, Yuen-Welch, Wilcoxon-Mann-Whitney and Brunner-Munzel) for different combinations of sample size, sample ratio, departure from normality, and so on. The paper ends up suggesting Welch U in general,
But appendix A of the paper lists the results for each combination of sample sizes. And for small sample sizes (m=10 n=10 or 25) the results are more confusing (as expected) - in my estimation of the results (not the authors') Welch U, Brunner-Munzel seems to perform equally well, and t-test also well in m=10 and n=10 case.
This is what I know so far.
For a "fast" solution, I used to cite Increasing Physicians’ Awareness of the Impact of Statistics on Research Outcomes: Comparative Power of the t-test and Wilcoxon Rank-Sum Test in Small Samples Applied Research by Patrick D Bridge and Shlomo S Sawilowsky (also behind paywall) and go straight to Wilcoxon no matter the sample size, but caveat emptor, for example Should we always choose a nonparametric test when comparing two apparently nonnormal distributions? by Eva Skovlund and Grete U. Fensta.
I have not yet found any similar results for paired data
• I appreciate the citations! For clarification, is the "Welch U" being referred to, the same test also known as the "Welch t" or "Welch-Aspin t" or (as I perhaps impropery called it in the question) "t test with Welch correction"? Nov 4 '14 at 13:51
• As far as I understand from the paper, Welch U is not the usual Welch-Aspin - it does not use the Welch–Satterthwaite equation for the degrees of freedom, but a formula that has a difference of the cube and the square of the sample size. Nov 4 '14 at 15:24
• Is it still a t-test though, despite its name? Everywhere else I search for "Welch U" I seem to find it's referring to the Welch-Aspin, which is frustrating. Nov 5 '14 at 22:27
Is normality testing 'essentially useless'?
Need and best way to determine normality of data
To simplify things, since non-parametric tests are reasonably good even for normal data, why not use them always for small samples.
# Simulating the difference of means of Gamma populations
Comparing the t-test and the Mann Whitney test
## Summary of results
• When the variance of the two populations is the same, the Mann Whitney test has greater true power but also greater true type 1 error than the t-test.
• For large sample N = 1000, the minimum true type 1 error for the Mann whitney test is 9%, whereas the t-test has true Type 1 of 5% as required by the experiment setup (reject $$H_0$$ for p values below 5%)
• When the variance of two populations is different, then the Mann Whitney test leads to large type 1 error, even when the means are the same. This is expected since the Mann Whitney tests for difference in distributions, not in means.
• The t test is robust to differences in variance but identical means
# Experiment 1) Different means, same variance
Consider two gamma distributions parametrized using k (shape) and scale $$\theta$$, with parameters
• $$X_1$$: gamma with $$k = 0.5$$ and $$\theta = 1$$ hence mean $$E[X_1] = k\theta = 0.5$$ and variance $$Var[X_1] = k\theta^2 = 0.5$$
• $$X_2$$: gamma with $$k = 1.445$$ and $$\theta = 0.588235$$ $$E[X_2] = .85$$ and variance $$Var[X_2] = .5$$
We will be testing for a difference in means of samples from $$X_1$$ and $$X_2$$. Here the setup is chosen such that $$X_1$$ and $$X_2$$ have the same variance, hence the true cohen d distance is 0.5
$$d = (.85 - .5) / \sqrt{.5} = 0.5$$
We will compare two testing methods: the two sample t-test and the Mann Whitney non parametric test, and simulate the true Type I and Power of these tests for different sample size (assuming we reject null hypothesis for $$p$$ value < 0.05)
• $$H_0: \mu_{X_1} = \mu_{X_2} = 0.5$$
• $$H_1: \mu_{X_1} \neq \mu_{X_2}$$
The true type 1 error is calculated as: $$P(\text{reject} | H_0)$$ and the true power is calculated as: $$P(\text{reject} | H_1)$$. We simulate thousands of experiment using the true distribution of $$H_0$$ and $$H_1$$
Sources:
### Discussion
• As expected, the sample mean is not normally distributed for small sample size ($$N = 10$$) as shown by the distribution skew and kurtosis. For larger sample size, the distribution is approximately normal
• For all sample sizes, the Mann Whitney test has more power than the t-test, and in some cases by a factor of 2
• For all samples sizes, the Mann Whitney test has greater type I error, and this by a factor or 2 - 3
• t-test has low power for small sample size
Discussion: when the variance of the two populations are indeed the same, the Mann Whitney test greatly outperforms the t-test in terms of power for small sample size, but has a higher Type 1 error rate
# Experiment 2: Different variances, same mean
• $$X_1$$: gamma with $$k = 0.5$$ and $$\theta = 1$$ hence mean $$E[X_1] = k\theta = .5$$ and variance $$Var[X_1] = k\theta^2 = .5$$
• $$X_2$$: gamma with $$k = 0.25$$ and $$\theta = 2$$ $$E[X_2] = .5$$ and variance $$Var[X_2] = 1$$
Here we won't be able to computer the power because the simulation does not contain the true $$H_1$$ scenario. However we can compute the type 1 error when $$Var[X_1] = Var[X_2]$$ and when $$Var[X_1] \neq Var[X_2]$$
Discussion Results from the simulation show that the t-test is very robust to different variance, and the type I error is close to 5% for all sample sizes. As expected, the Mann Whitney test performs poorly in this case since it is not testing for a difference in means but for a difference in distributions | 9,659 | 43,270 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 49, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-05 | latest | en | 0.904565 |
https://ey.westside66.org/34-math-minute-pizza-math/ | 1,723,077,760,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713903.39/warc/CC-MAIN-20240808000606-20240808030606-00751.warc.gz | 191,131,587 | 13,373 | # #34: Pizza Math
##### Spark your math thinking!
1. Set up your math mini spark recording page: #34: Pizza Math
2. Watch the videos, Pizza Pi and The Twitter Pizza Equation Explained and add details to your recording page showing the math behind the pizza sizes.
3. Omni has a great site with information on pizza math and an pizza size calculator. On the left side there are a few articles about pizza. Record the title of each article and 3 details from each one on your note taking sheet.
4. Now try out the calculator. You can use the measurements from one of the videos or create your own and see what the comparisons are. Record what you find.
5. OPTIONAL: Calculate the area, circumference, and price per square inch of the next pizza you order. Make a display of the math/calculations.
6. Share your math mini spark recording page with your teacher/EY coordinator.
## 3 thoughts on “#34: Pizza Math”
1. Liam P., Sunset says:
I didn’t know that 1 large pizza might be less money than 2 smaller pizzas.
2. Sicily says:
Each 2 inches the 2 more \$?
3. Kadence D. Rockbrook says:
Haddley and Worsley discovered a way to slice curvy pieces with nearly an infinite number of sides (as long as it’s an odd number of sides) | 296 | 1,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-33 | latest | en | 0.830609 |
https://ncertsolution.org/ncert-class-6-science-solutions-chapter-7-motion-and-measurement-of-distances/ | 1,719,019,081,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862189.36/warc/CC-MAIN-20240621223321-20240622013321-00004.warc.gz | 359,443,729 | 49,512 | NCERT Class 6 Science Solutions Chapter 7: Motion and Measurement of Distances
1. Give two examples each, of modes of transport used on land, water and air.
Solution:
Land – Train, Bus
Water – Ship, Boat
Air – Helicopter, Aeroplane
2. Fill in the blanks:
(i) One metre is ______________ cm.
(ii) Five kilometre is ______________ m.
(iii) Motion of a child on a swing is ______________.
(iv) Motion of the needle of a sewing machine is ______________.
(v) Motion of wheel of a bicycle is______________.
Solution:
(i) One metre is 100 cm.
(ii) Five kilometres is 5000 m.
(iii) Motion of a child on a swing is periodic.
(iv) Motion of the needle of a sewing machine is periodic.
(v) Motion of the wheel of a bicycle is circular.
3. Why can a pace or a footstep not be used as a standard unit of length?
Solution:
Pace or a footstep cannot be used as a standard unit of length because it varies from person to person.
4. Arrange the following lengths in their increasing magnitude: 1 metre, 1 centimetre, 1 kilometre, 1 millimetre.
Solution:
1 millimetre, 1 centimetre, 1 metre, 1 kilometre
5. The height of a person is 1.65 m. Express it into cm and mm.
Solution:
1.65= 165 cm = 1650 mm
6. The distance between Radha’s home and her school is 3250 m. Express this distance in km.
Solution:
1km = 1000 m
Hence, 3250 m = 3.25 kms
7. While measuring the length of a knitting needle, the reading of the scale at one end is 3.0 cm and at the other end is 33.1 cm. What is the length of the needle?
Solution:
Length of needle = 33.1 – 3 = 30.1 cm
8. Write the similarities and differences between the motion of a bicycle and a ceiling fan that has been switched on.
Solution:
Similarities – The blades of a fan and the wheels of a bicycle show circular motion
Differences – Bicycles vs. Fans
While bicycles move in a straight line, fans do not move in a rectilinear motion.
9. Why would you not like to use a measuring tape made of an elastic material like rubber to measure distance? What would be some of the problems you would meet in telling someone about a distance you measured with such a tape?
Solution:
Accurate measurements cannot be obtained using an elastic measuring tape as its length stretches and size reduces when pulled. Consequently, when using elastic tape, it is necessary to specify whether it was stretched and by how much to express measurements accurately. Due to these factors, measurements taken from elastic tape can be challenging to interpret accurately.
10. Give two examples of periodic motion.
Solution:
a) A needle of a sewing machine
b) Pendulum
Experts Ncert Solution
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WhatsApp API is now publicly available! | 666 | 2,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-26 | latest | en | 0.890208 |
https://www.kylesconverter.com/frequency/yottahertz-to-degrees-per-minute | 1,708,681,350,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474377.60/warc/CC-MAIN-20240223085439-20240223115439-00559.warc.gz | 888,851,648 | 5,603 | # Convert Yottahertz to Degrees Per Minute
### Kyle's Converter > Frequency > Yottahertz > Yottahertz to Degrees Per Minute
Yottahertz (YHz) Degrees Per Minute (deg/m)* Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Degrees Per Minute to Yottahertz
(or just enter a value in the "to" field)
#### Please share if you found this tool useful:
Unit Descriptions
1 Yottahertz:
1 Yottahertz is exactly one septillion Hertz. 1 YHz = 1 x 1024 Hz. 1 YHz = 1000000000000000000000000 Hz.
1 Degree per Minute:
1 Degree per minute is comparative to 0.000046296296296296 Hertz. Degrees per minute is a measure of angular frequency, it can be compared to Hertz or other angular units. Approximation of PI used. 1 deg/m is approximately 0.000046296296296296 Hz.
Conversions Table
1 Yottahertz to Degrees Per Minute = 2.16E+2870 Yottahertz to Degrees Per Minute = 1.512E+30
2 Yottahertz to Degrees Per Minute = 4.32E+2880 Yottahertz to Degrees Per Minute = 1.728E+30
3 Yottahertz to Degrees Per Minute = 6.48E+2890 Yottahertz to Degrees Per Minute = 1.944E+30
4 Yottahertz to Degrees Per Minute = 8.6400000000001E+28100 Yottahertz to Degrees Per Minute = 2.16E+30
5 Yottahertz to Degrees Per Minute = 1.08E+29200 Yottahertz to Degrees Per Minute = 4.32E+30
6 Yottahertz to Degrees Per Minute = 1.296E+29300 Yottahertz to Degrees Per Minute = 6.48E+30
7 Yottahertz to Degrees Per Minute = 1.512E+29400 Yottahertz to Degrees Per Minute = 8.6400000000001E+30
8 Yottahertz to Degrees Per Minute = 1.728E+29500 Yottahertz to Degrees Per Minute = 1.08E+31
9 Yottahertz to Degrees Per Minute = 1.944E+29600 Yottahertz to Degrees Per Minute = 1.296E+31
10 Yottahertz to Degrees Per Minute = 2.16E+29800 Yottahertz to Degrees Per Minute = 1.728E+31
20 Yottahertz to Degrees Per Minute = 4.32E+29900 Yottahertz to Degrees Per Minute = 1.944E+31
30 Yottahertz to Degrees Per Minute = 6.48E+291,000 Yottahertz to Degrees Per Minute = 2.16E+31
40 Yottahertz to Degrees Per Minute = 8.6400000000001E+2910,000 Yottahertz to Degrees Per Minute = 2.16E+32
50 Yottahertz to Degrees Per Minute = 1.08E+30100,000 Yottahertz to Degrees Per Minute = 2.16E+33
60 Yottahertz to Degrees Per Minute = 1.296E+301,000,000 Yottahertz to Degrees Per Minute = 2.16E+34 | 781 | 2,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-10 | latest | en | 0.668916 |
https://legacy.sawtoothsoftware.com/forum/25633/conditional-pricing-and-wtp-analysis?show=25643 | 1,603,453,195,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881369.4/warc/CC-MAIN-20201023102435-20201023132435-00396.warc.gz | 411,048,210 | 9,003 | # Conditional pricing and WTP analysis
Hi,
I already have seen similar topics on this question, but I still do not quite understand. I have used conditional pricing (a standard increase in price per increment in level per attribute + price as a separate attribute with the levels: low (-30%), medium or high (+30%)).
Now I want to calculate the WTP per attribute/level. I have used the HB analysis to estimate utilities as this was recommended. But how do I get my WTP through your lighthouse software? Should I define price as linear in the analysis instead of path-worth? And then just do -1*(attribute coefficient/price coefficient)
Kind regards,
Irene
related to an answer for: Conditional Pricing and Willingness to Pay
With conditional pricing, each brand or SKU (if you made price conditional on another attribute such as brand or SKU) has a different set of prices.
For computing WTP, we strongly advise to use the competitive simulation approach rather than estimating the algebraic approach. When using conditional pricing, this is even more a concern.
The competitive simulation approach involves putting a test product (for which WTP will be estimated) into the market simulator in competition with a large set of realistic competitive products. Perhaps that's 8 or 10 products, plus an additional "product" in the None alternative.
Let's imagine you want to estimate the WTP for "has sunroof" versus "doesn't have sunroof". You first specify your product (the product for which WTP will be estimated) in competition with the realistic set of competitors as would be seen in the marketplace where your product does not have a sunroof. Write down the share of preference for the test product (perhaps it is 12%).
Next, change the test product to have a sunroof. Keep the competition constant. Run the simulation and write down the share of preference, perhaps it is now 15%.
Next, increase the price for the test product with the sunroof until its share is driven back down to 12%. That difference in price that leads to indifference in share of preference for having or not having a sunroof is the estimated WTP for sunroof.
This approach takes into account the conditional pricing table, as the market simulator knows how to map actual prices to the conditional pricing table for simulating share of preference for products.
answered Jun 2 by Platinum (177,015 points)
selected Jun 7 by Ireen92
Thank you for your response. I understand your approach but I am not sure how to run this competitive simulation approach in the lighthouse studio after I already did my CBC test?
If you are working in Lighthouse Studio, you should click Analysis + Analysis Manager. Then, to run HB estimation, click the "Add" icon, then change the "Analysis Types" in the Analysis Manager from Counts to "HB". You can use the default settings. Click Run. An HB run is created called "Analysis Run 1" by default.
After HB estimation concludes, you can exit the Analysis Manager to get back to the main Lighthouse Studio menu. Now, the set up and run Market Simulations, click Analysis + Choice Simulator.
In the choice simulator, add your competitive scenario and run market simulations.
If you have further questions, there is documentation in the Help manual for Lighthouse Studio regarding the simulator. Or, you can also call our tech support team at +1 801 477 4700.
Perfect! Thank you so much | 736 | 3,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-45 | latest | en | 0.948427 |
https://community.oracle.com/message/10745904 | 1,394,745,014,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678682078/warc/CC-MAIN-20140313024442-00041-ip-10-183-142-35.ec2.internal.warc.gz | 637,230,370 | 38,145 | This discussion is archived
10 Replies Latest reply: Dec 13, 2012 8:56 AM by hitgon
If with multiple conditon "OR"
Currently Being Moderated
Hello ,
I am not sure what i am doing wrong in this should be some silly mistake but i am not getting around this. :(
I just used reverse condition to suit my situation but would like to know what casing this.
``````declare
rt varchar2(20);
begin
rt := 9;
if (rt != 9) OR (rt != 0) then
rt := 'Error -'||rt;
dbms_output.put_line(rt);
end if;
end;``````
output when rt := 9 then Error -9
output when rt := 11 then Error -11
Thanks
• 1. Re: If with multiple conditon "OR"
Currently Being Moderated
The condition (rt!=0) is true, so rt will become 'Error -'||rt
hm
• 2. Re: If with multiple conditon "OR"
Currently Being Moderated
There are some SQL functions that you could use here like DECODE or NVL
• 3. Re: If with multiple conditon "OR"
Currently Being Moderated
Basic Boolean principle:
If any one (or more) conditions in an OR is true, then the result is true.
In your case, with rt equal to 9:
+(rt != 9) OR (rt != 0)+
Is:
+(FALSE) OR (TRUE)+
Result:
TRUE
The condition you have however does not make sense - and AND would result in a more logical and send sensible condition.
• 4. Re: If with multiple conditon "OR"
Currently Being Moderated
hitgon wrote:
There are some SQL functions that you could use here like DECODE or NVL
Such functions do not automagically fix basic logic errors.
Get the logic right first. And then select the functions to apply that logic programatically.
• 5. Re: If with multiple conditon "OR"
Currently Being Moderated
Here's why your condition test is not sensible - as one of the predicates will always be true... and thus make the condition check always true.
``````SQL> with test_data( r ) as(
2 select level-1 from dual
3 connect by level <= 10
4 )
5 select
6 r,
7 case when
8 (r != 9 ) then 'TRUE'
9 else
10 'FALSE'
11 end as "(r != 9)",
12 case when
13 (r != 0) then 'TRUE'
14 else
15 'FALSE'
16 end as "(r != 0)",
17 case when
18 (r != 9 ) or (r != 0) then 'TRUE'
19 else
20 'FALSE'
21 end as "Result"
22 from test_data
23 order by r;
R (r != 9) (r != 0) Result
---------- ---------- ---------- ----------
0 TRUE FALSE TRUE
1 TRUE TRUE TRUE
2 TRUE TRUE TRUE
3 TRUE TRUE TRUE
4 TRUE TRUE TRUE
5 TRUE TRUE TRUE
6 TRUE TRUE TRUE
7 TRUE TRUE TRUE
8 TRUE TRUE TRUE
9 FALSE TRUE TRUE
10 rows selected.
SQL> ``````
• 6. Re: If with multiple conditon "OR"
Currently Being Moderated
• 7. Re: If with multiple conditon "OR"
Currently Being Moderated
What is it you want.
Perhaps you mean?
``if rt NOT IN (9,0)``
Or similarly:
``if NOT ((rt = 9) or (rt=0)) then``
Also consider NULLs.
• 8. Re: If with multiple conditon "OR"
Currently Being Moderated
Thanks Guys for all your suggestion/help i think i got an what was wrong here :) thanks!!!
• 9. Re: If with multiple conditon "OR"
Currently Being Moderated
Ramesh Penti wrote:
No it is not.
Explain what is correct to code an IF condition that is always true?
This is extra code and just silly:
``````IF true IS true THEN
DO x
END IF``````
As the following does the EXACT same thing with a SINGLE statement:
``DO x``
Code is never correct simply because the syntax is correct and it compiles without an error.
• 10. Re: If with multiple conditon "OR"
Currently Being Moderated
Thanks Billy Verreynne for great explanation
Legend
• Correct Answers - 10 points | 1,013 | 3,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2014-10 | longest | en | 0.850805 |
https://www.mathsplatter.com/post/how-to-find-karl-pearson-s-coefficient-of-skewness | 1,685,311,266,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644571.22/warc/CC-MAIN-20230528214404-20230529004404-00775.warc.gz | 988,527,797 | 163,737 | top of page
# How to find Karl Pearson's Coefficient of Skewness?
## Question:
Find Karl Pearson's Coefficient of Skewness when Variance = 900, Median = 75 and Mean = 60.
## Solution:
Step 1: Find Standard deviation from Variance
Standard Deviation = √Variance
∴ Standard Deviation = √900
∴ Standard Deviation = 30
Step 2: Find Karl Pearson's Coefficient of Skeweness
Sₖₚ = 3 x [Mean - Median] / Standard Deviation
∴ Sₖₚ = 3 x [60 - 75] / 30
∴ Sₖₚ = 3 x [-15] / 30
∴ Sₖₚ = -15 / 10
∴ Sₖₚ = -1.5 | 188 | 508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-23 | latest | en | 0.670897 |
https://studydaddy.com/question/suppose-that-the-demand-for-a-company-s-product-in-weeks-1-2-and-3-are-each-norm-1 | 1,580,219,086,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251778272.69/warc/CC-MAIN-20200128122813-20200128152813-00044.warc.gz | 662,901,433 | 7,452 | Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
QUESTION
# Suppose that the demand for a company's product in weeks 1, 2, and 3 are each normally distributed and the mean demand during each of these three...
Suppose that the demand for a company's product in weeks 1, 2, and 3 are each normally distributed and the mean demand during each of these three weeks is 50, 45, and 65, respectively. Suppose the standard deviation of the demand during each of these three weeks is known to be 10, 5, and 15, respectively. It turns out that if we can assume that these three demands are probabilistically independent then the total demand for the three week period is also normally distributed. And, the mean demand for the entire three week period is the sum of the individual means. Likewise, the variance of the demand for the entire three week period is the sum of the individual weekly variances. But be careful! The standard deviation of the demand for the entire 3 week period is not the sum of the individual standard deviations. Square roots don't work that way! | 244 | 1,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-05 | latest | en | 0.970387 |
https://www.felderbooks.com/papers/psinotes.html | 1,721,085,527,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514724.0/warc/CC-MAIN-20240715224905-20240716014905-00375.warc.gz | 680,115,281 | 9,844 | Complex Numbers and Magnitudes
If you take any real number and square it, you get a positive number (or zero). There is no number that, when you square it, gives you a negative number. So at some point, someone just made one up, and designated it by the letter i (which stands for "imaginary"): i2=–1, by definition. You can readily see that (–i)2=–1, that (2i)2=–4, that i3=–i, and so on.
A real number is defined as one that has no i in it. An imaginary number is a real number times i (such as 4i). A complex number has both real and imaginary parts. Any complex number z can be written in the form x+iy where x and y are real numbers. x is the "real part" and y is the "imaginary part." For instance, 2+3i is a complex number.
The complex conjugate of a complex number, represented by a star next to the number, is the same real part but the negative imaginary part. So if z=2+3i then z*=2–3i. Note that the reverse is also true: if z=2–3i then z*=2+3i. So you always have two numbers that are complex conjugates of each other.
Real numbers can be graphed on a line (the number line). Complex numbers, since they have two separate components, are graphed on a plane called the "complex plane" where the real part is mapped to the x-axis, the imaginary part the y-axis. Every complex number exists at exactly one point on the complex plane: a few examples are given below.
Every complex number has a magnitude which gives its distance from the origin (0) on the complex plane. So 1 and i both have magnitude 1. The magnitude is written with absolute value signs. So we can say that if z=2+3i then |z|2=22+32.
If you multiply any number by its own complex conjugate, you get the magnitude of the number squared (you should be able to convince yourself of this pretty easily). So we can write that, for any z, z*z=|z|2.
[close]
Expectation Values
Suppose you roll a 4-sided die. What number do you expect to get? Answer: you have no idea. Any number is as likely as any other, presumably.
OK, but on average what number will you get? For instance, if you rolled the die a million times, and averaged all the results (divided the sum by a million), what would the answer probably be? This is called the "expectation value." You can probably see, intuitively, that the answer in this case is 2½. Never mind that an individual die roll can never come out 2½: on the average, we will land right in the middle of the 1-to-4 range, and that will be 2½.
But you can't guess the answer quite so easily if the odds are uneven. For instance, suppose the die is weighted so that we have a 1/2 chance of rolling a 1, a 1/4 chance of rolling a 2, a 1/8 chance of rolling a 3, and a 1/8 chance of rolling a 4? Your intuition should tell you that our average roll will now be a heck of a lot lower than 2½, since the lower numbers are much more likely. But exactly what will it be?
Well, let's roll a million dice, add them, and then divide by a million. How many 1s will we get? About 1/2 million. How many 2s? About 1/4 million. And we will get about 1/8 million each of 3s and 4s. So the total, when we add, will be (1/2 million)*1+(1/4 million)*2+(1/8 million)*3+(1/8 million)*4. And after we add—this is the kicker—we will divide the total by a million, so all those "millions" go away, and leave us with (1/2)1+(1/4)2+(1/8)3+(1/8)4=1 7/8.
The moral of the story is: to get the expectation value, you do a sum of each possible result multiplied by the probability of that result. This tells you what result you will find on average.
[close]
Fourier Transforms
Before explaining Fourier transforms, we will start by quickly reviewing a related topic you may be more familiar with, namely Fourier series. If you have any function f(x) that is periodic with period L you can write it as a sum of sine and cosine terms:
For each sine or cosine wave in this series the wavelength is given by L/m or L/n respectively. In other words the indices m and n indicate the wave number (one over wavelength) of each wave. Using the formula
eix=cos x + i sin x
you can rewrite the Fourier series in the equivalent form
The function e2pipx/L is called a plane wave. Like a sine or cosine wave it has a value that oscillates periodically, only with plane waves this value is complex. That means that even if f(x) is a real function the coefficients cp will in general be complex. Either way you represent it, however, a Fourier series is simply a way of rewriting a periodic function f(x) as an infinite series of simple periodic functions with numerical coefficients.
A Fourier transform is the exact same thing, only f(x) doesn't have to be periodic. This means that the index p can now take on any real value instead of just integer values, and that the expansion is now an integral rather than a sum.
The factor of is a matter of convention and is set differently by different authors. Likewise some people add a factor (usually ±2p) in the exponential. The basic properties of the Fourier transform are unchanged by these differences. Note that this formula has nothing to do with position and momentum. The variable x could be position, or time, or anything else, and the variable p tells you the wave number (one over wavelength) of each plane wave in the expansion. (If x is time then the wave number is the same thing as the frequency of the wave.) To invert this formula, first multiply it by e—ip'x (where p' is just an arbitrary number) and then integrate with respect to x.
It is possible to evaluate the integrals on the right hand side of this equation, but doing so requires a familiarity with Dirac delta functions. If you are not familiar with these functions you can just skip the next paragraph and jump straight to the result. (We discuss Dirac delta functions in a later footnote, but if you're not already somewhat familiar with them trying to follow the argument below may be an exercise in frustration.)
The only x dependence in this integral comes in the exponential, which is an oscillatory function. If p-p' is anything other than zero then this exponential will have equal positive and negative contributions and the integral will come out to zero. (Actually eipx is complex so the equal contributions come from all directions in the complex plane. If you can't picture this just take our word for it.) If p-p' is zero, however, then the exponential is simply one everywhere and the x integral is infinite. Thus the x integral is a function of p-p' that is infinite at p-p', and zero everywhere else, which is to say a delta function. Of course this isn't a rigorous proof. It can be proven rigorously that the integral of a plane wave is a delta function, however, and when you do the math carefully you find that the integral produces a delta function with a coefficient of 2p. So
where the integration over p uses the defining characteristic of a delta function. Replacing the dummy variable name p' with p we finally arrive at the formula
We should emphasize once more that mathematically x is just the argument to the function f(x) and p is just the wave number of the plane waves in the expansion. The fact that the relationship between Y(x) and f(p) in quantum mechanics takes the form of a Fourier transform means that for a given wavefunction Y(x) the momentum happens to be inversely related to the wavelength. (Once again, though, a realistic wavefunction will be made of many waves with different wavelengths.)
We said before that the numerical coefficients in the Fourier transform definition were arbitrary. In quantum mechanics these coefficients pick up an extra factor of , so the full relation between Y(x) and f(p) is
Finally, we should note one convenient theorem about Fourier transforms. Consider the normalization of the wavefunction, which is determined by the integral of its squared magnitude. Writing that in terms of the Fourier expansion we find
This result can be proven by taking the complex conjugate of the expression for Y(x) above, multiplying it by Y(x), and integrating over x using the same delta function trick we used above. We won't bother to show this proof here. Recall that f(p) is the coefficient of the momentum basis state with momentum p, so what this result says is that if the wavefunction is properly normalized so that the total probability of the position being anywhere is one, then it will automatically be normalized so that the total probability of the momentum having any value is one.
[close]
The Uncertainty Principle
In classical physics a particle at any given moment has a position and a momentum. We may not know what they are, but they exist with some exact values. In quantum mechanics a particle has a wavefunction which gives it some probability distributions for position and momentum. The uncertainty principle says that the more definite position is the less definite the momentum, and vice-versa.
We can see how this comes out of the rules we've given so far by considering the basis states of momentum, which are of the form eipx/, which is equal to cos(px/)+i sin(px/). In other words the basis states of momentum are waves whose wavelength is determined by the value p. (Specifically, the wavelength is 2p/p.) When you expand a wavefunction in momentum basis states you are writing Y(x) as a superposition of these waves.
Those familiar with Fourier expansions know that a localized wave is made up of components with many different wavelengths. At the other extreme a function made up of a single wave isn't localized at all. If a particle is in a momentum eigenstates Y=eipx/ then its position probabilities are equally spread out over all space. The bigger the range of wavelengths contained in the superposition the more localized the function Y(x) can be. From this we can see that a particle highly localized in momentum (a small range of wavelengths for Y) will have a very uncertain position (Y(x) spread out over a large area), and a particle highly localized in position (Y(x) concentrated near a point) will have a very uncertain momentum (many wavelengths contributing to the Fourier Transform of Y). You can certainly cook up a wavefunction with a large uncertainty in both position and momentum, or one that's highly localized in one but uncertain in the other, but you can't make one that's highly localized in both.
It's possible to make a mathematical definition of what you mean by the uncertainty in position or momentum and use the properties of Fourier Transforms to make this rule more precise. The rule ends up being that the product of the position uncertainty times the momentum uncertainty must always be at least /2.
[close]
Squaring an Operator
What does it mean to square an operator? It means "to do that operator twice." Why does it mean that? Really, it's just a matter of definition. So if D is the operator d/dx, then we say that D2f=d2f/dx2. It could just as easily have been defined some other way (for instance, as (df/dx)2) but it isn't: squaring an operator means doing it twice.
OK, you say, but if it's that arbitrary, then why does it work? Why can we say that kinetic energy is p2/2m, so the kinetic energy operator will be found by squaring the momentum operator and then dividing by 2m? Well, to some extent, you have to accept the operators as postulates of the system. But it does make sense if you consider the special case of an operator corresponding to some observable acting on an eigenfunction. If Yp is an eigenfunction of momentum with eigenvalue p that means that a particle whose wavefunction is Yp must have a momentum of exactly p. In that case, though, its momentum squared must be exactly p2, which is the eigenvalue you get by acting on Yp with the momentum operator twice. So in the end, it's not quite as arbitrary as it first appears. (Of course, this is not a proof of anything, but hopefully it's a helpful hand-wave.)
[close]
Where did the formula <O>=Y*OYdx come from?
To see why this works consider first the case where Y happens to be an eigenfunction of O with eigenvalue v. We know that a measurement of the quantity O will yield exactly v, so the expectation value of O must be v. We also know, however, that OY= vY, so the formula gives us
where in the last step we've used the fact that the wavefunction must be normalized. For this special case we can see that our formula worked. What happens when Y is not an eigenfunction of O? For simplicity let's consider a wavefunction made up of just two basis states:
Y=C1Y1+C2Y2
We know for this case the correct answer for the expectation value of O is v1|C1|2+v2 |C2|2. To see how our formula gives us this result we act on Y with the operator O to get
OY=C1v1Y1+C2v2Y2
Our formula then tells us that
The first two terms look like exactly what we need since each basis state Yi is itself a valid wavefunction that must be properly normalized. (This is a bit more complicated for continuous variables, where the basis states are not normalized in this way, but a similar argument applies.) The last two terms seem to spoil the result, though. Here's the hand-waving part: for any two different basis states Y1 and Y2, the integral
will always be zero. The technical way to say this is that the basis states of all operators representing observables are orthogonal. This property, which we're not going to prove here, is the last necessary step to show why the expectation value formula works. (We do make an intuitive argument why this is true for the momentum basis states in our footnote on Fourier transforms.)
[close]
The Basis State of Position
What is the basis function for position?
The best way to answer this is actually not mathematically, but physically. Remember that a basis function represents a state where the particle has exactly one value, and no probability of being any other value. But we know that you find probabilities of position by squaring the wavefunction. So if we want a function that says "the position must be exactly 5" then clearly this wavefunction must equal 0 at every point but x=5. But if the probability distribution is completely concentrated at one point then it must be infinitely high at that point!
This function—infinitely high at one point, zero everywhere else—is known as the Dirac delta function and is written as the Greek letter delta. So d(x-5) is the function that is infinitely high at x=5, and zero everywhere else, such that the total integral is 1. To make this definition mathematically rigorous, you have to define it as a limiting case of a function that gets thinner-and-taller while keeping its integral at 1.
We said earlier that the position operator is x (meaning "multiply by x"). Now we're saying that the delta function is the position basis function: in other words, the eigenfunction for that operator. So it should be true, for instance, that xY=5Y when Y=d(5). In one sense we can see that this is true. Since d(x-5)=0 everywhere except x=5 multiplying it by x can only have an effect at the point x=5. So we can say that xd(x-5)=5d(x-5). At the same time, it seems kind of silly to multiply an infinitely large number by 5. What it really means is that the integral of the delta function, which was 1 by definition, will now be equal to 5.
Of course the above explanation isn't at all rigorous mathematically. In fact the "definition" we gave of the Dirac delta function was pretty hand-waving. The more rigorous definition involves taking the limit of a series of functions that get thinner and taller around x=5 while keeping their total integral equal to 1. If you think about multiplying one of these tall, narrow functions by x and then integrating you should be able to convince yourself that you get the same narrow function, only five times taller.
[close]
Partial Differential Equations
The simplest type of differential equation is an equation involving some function f(x) and its derivatives f'(x), f''(x), etc. Consider for example Newton's second law F=ma. In a typical situation you will know the force F acting on a particle as a function of that particle's position. Say the particle is a mass on a spring so that F=-kx. Then Newton's second law becomes the differential equation
x''(t)=-k/m x(t)
An equation like this is called an ordinary differential equation (ODE) because the dependent variable, namely the function x that you are solving for, depends only on one independent variable, t. You can write down a general solution to this equation, but it will necessarily have undetermined constants in it. Because this is a second order ODE it has two undetermined constants. The full solution of this particular equation is
x(t)=C1sin(t)+C2cos(t)
The undetermined constants C1 and C2 indicate that you can put any number there, and you will still have a solution: for instance, 2sin(t)-¾cos(t) is one perfectly valid solution. To find the constants that should be used for a particular situation, you must specify two specific data, such as the initial conditions x(0) and x'(0). In other words if you know the state of your particle (position and velocity) at t=0 then you can solve for its state at any future time.
In quantum mechanics we have to deal with a partial differential equation (PDE), which is an equation where the dependent variable (Y in this case) depends on two independent variables (x and t). As a simple example of a PDE (simpler in some ways than Schrödinger's Equation) we can consider the wave equation
Here f(x,t) is some function of space and time (eg the height of a piece of string). Once again we can write down the general solution to this equation, but now instead of a couple of undetermined constants we have some entire undetermined functions!
f(x,t)=g1(x-t)+g2(x+t)
You can (and should) check for yourself that this solution works no matter what functions you put in for g1 and g2. You could write for example
f(x,t)=sin2(x-t)+ln(x+t)
and you would find that f satisfies the wave equation. How can you determine these functions? For the ODE above it was enough to specify two initial values, x(0) and x'(0). For this PDE we must specify two initial functions f(x,0) and f'(x,0) (where the prime here refers to a time derivative). For a piece of string, for instance, I must specify the height and velocity of each point on the string at time t=0, and then my solution to the wave equation can tell me the state of my system at any future time.
How does this apply to Schrödinger's Equation? Schrödinger's Equation is only first order in time derivatives, which means you can write the general solution in terms of only one undetermined function. In other words if I specify Y(x,0) that is enough to determine Y for all future times. (One way you can see this is by noting that if you know Y(x,0) Schrödinger's Equation tells you Y'(x,0), so it wouldn't be consistent to specify both of them separately.) So to solve for the evolution of a quantum system I must specify my initial wavefunction and then solve Schrödinger's Equation to see how it evolves in time.
[close] | 4,379 | 19,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-30 | latest | en | 0.937751 |
http://www.jiskha.com/display.cgi?id=1248447851 | 1,498,164,779,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319902.52/warc/CC-MAIN-20170622201826-20170622221826-00250.warc.gz | 569,901,834 | 3,681 | # algebra
posted by on .
The formula R= - 0.075 + 3.85 can be used to predict the world record in the 1500 meter run, t years after 1930. Determine an inequality that identifies the years in which the world record will be less than 3.4 minutes.
Solve for T
t >
(Round to the nearest whole number)
• algebra - TYPO - ,
The function
R(t)= - 0.075 + 3.85
does not have a variable t in the expression. Could you double check the formula? | 122 | 437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-26 | latest | en | 0.819358 |
https://math.stackexchange.com/questions/2901581/probability-of-getting-2-questions-correct-on-a-test-by-students-of-different-sk | 1,561,443,181,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999800.5/warc/CC-MAIN-20190625051950-20190625073950-00335.warc.gz | 510,262,772 | 35,096 | # Probability of getting 2 questions correct on a test by students of different skill levels
Out of the 14 students taking a test, 5 are well prepared, 6 are adequately prepared and 3 are poorly prepared. There are 10 questions on the test paper. A well prepared student can answer 9 questions correctly, an adequately prepared student can answer 6 questions correctly and a poorly prepared student can answer only 3 questions correctly.
(a) If a randomly chosen student is asked two distinct randomly chosen questions from the test, what is the probability that the student will answer both questions correctly? Note: The student and the questions are chosen independently of each other. “Random” means that each individual student/each pair of questions is equally likely to be chosen. (b) Now suppose that a student is chosen at random and asked two randomly chosen questions from the exam, and moreover ݀݅݀ answer both questions correctly. Find the probability that the chosen student was well prepared.
I started by making 3 cases
Case 1: The student is well prepared.
The probability of picking a well prepared student is $\frac{5}{14}$, the probability that he got the first question right was $\frac{9}{10}$ and the probability of getting the second question correct will be $\frac{8}{9}$ Multiplying them we get $\frac{2}{7}$
I similarly proceeded with the other cases and added them up to get the answer as $\frac{31}{70}$ for the first part and $\frac{20}{31}$ for the second one.
However none of my answers match the correct ones... Please tell me where I went wrong and what is the correct answer, I was thinking in terms of the distinguishability of the 2 questions in the first one for which we might have to divide further by $2!$
How to do this question correctly?
• I think the question is ambiguous. Does it mean, as you seem to imagine, that (for the prepared student, say) there are $9$ questions he can answer and $1$ that he can not or does it mean that he has a $.9$ probability of getting any given question right? These are similar but not the same. – lulu Sep 1 '18 at 11:14
• What exactly is meant with "can answer"? – drhab Sep 1 '18 at 11:19
• I believe it is the former as it says that he can answer 9 question correctly. – K. Chopra Sep 1 '18 at 11:20
• Possibly the ability to answer only a certain number of questions. At least that's what I assumed. – K. Chopra Sep 1 '18 at 11:20
• Well, you also say that your answer is incorrect. What is the "official" answer? Is it consistent with the second interpretation I give? – lulu Sep 1 '18 at 11:21 | 615 | 2,589 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-26 | latest | en | 0.964523 |
https://rellek.net/book-2016.1/s_recurrence_rigorous.html | 1,603,161,593,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107869785.9/warc/CC-MAIN-20201020021700-20201020051700-00664.warc.gz | 509,246,821 | 11,281 | # Section9.5Formalizing our approach to recurrence equations¶ permalink
So far, our approach to solving recurrence equations has been based on intuition, and we've not given a lot of explanation for why the solutions we've given have been the general solution. In this section, we endeavor to remedy this. Some familiarity with the language of linear algebra will be useful for the remainder of this section, but it is not essential.
Our techniques for solving recurrence equations have their roots in a fundamentally important concept in mathematics, the notion of a vector space. Recall that a vector space 1 consists of a set $V$ of elements called vectors; in addition, there is a binary operation called addition with the sum of vectors $x$ and $y$ denoted by $x+y\text{;}$ furthermore, there is an operation called scalar multiplication which combines a scalar (real number) $\alpha$ and a vector $x$ to form a product denoted $\alpha x\text{.}$ These operations satisfy the following properties:
1. $x+y=y+x$ for every $x,y,\in V\text{.}$
2. $x+(y+z) = (x+y)+z\text{,}$ for every $x,y,z\in V\text{.}$
3. There is a vector called zero and denoted $0$ so that $x+0=x$ for every $x\in V\text{.}$ Note: We are again overloading an operator and using the symbol $0$ for something other than a number.
4. For every element $x\in V\text{,}$ there is an element $y\in V\text{,}$ called the additive inverse of $x$ and denoted $-x$ so that $x+(-x)=0\text{.}$ This property enables us to define subtraction, i.e., $x-y= x+(-y)\text{.}$
5. $1x=x$ for every $x\in X\text{.}$
6. $\alpha(\beta x) = (\alpha\beta)x\text{,}$ for every $\alpha,\beta\in\reals$ and every $x\in V\text{.}$
7. $\alpha(x+y)=\alpha x + \alpha y$ for every $\alpha\in\reals$ and every $x,y\in V\text{.}$
8. $(\alpha +\beta) x = \alpha x + \beta x\text{,}$ for every $\alpha,\beta\in\reals$ and every $x\in V\text{.}$
When $V$ is a vector space, a function $\phi\colon V\rightarrow V$ is called an linear operator, or just operator for short, when $\phi(x+y)=\phi(x)+\phi(y)$ and $\phi(\alpha x)=\alpha\phi(x)\text{.}$ When $\phi\colon V\rightarrow V$ is an operator, it is customary to write $\phi x$ rather than $\phi(x)\text{,}$ saving a set of parentheses. The set of all operators over a vector space $V$ is itself a vector space with addition defined by $(\phi+\rho)x = \phi x +\rho x$ and scalar multiplication by $(\alpha\phi)x=\alpha(\phi x)\text{.}$
In this chapter, we focus on the real vector space $V$ consisting of all functions of the form $f\colon\ints\rightarrow\reals\text{.}$ Addition is defined by $(f+g)(n)= f(n)+g(n)$ and scalar multiplication is defined by $(\alpha f)(n)=\alpha(f(n))\text{.}$
Here is the basic theorem about solving recurrence equations (stated in terms of advancement operator equations)—and while we won't prove the full result, we will provide enough of an outline where it shouldn't be too difficult to fill in the missing details.
The conclusion that the set $W$ of all solutions is a subspace of $V$ is immediate, since \begin{equation*} p(A)(f+g)=p(A)f+p(A)g\quad\text{ and } \quad p(a)(\alpha f)=\alpha p(A)(f). \end{equation*} What takes a bit of work is to show that $W$ is a $k$-dimensional subspace. But once this is done, then to solve the advancement operator equation given in the form of Theorem 9.18, it suffices to find a basis for the vector space $W\text{.}$ Every solution is just a linear combination of basis vectors. In the next several subsections, we outline how this goal can be achieved.
The development proceeds by induction (surprise!) with the case $k=1$ being the base case. In this case, we study a simple equation of the form $(c_0A+c_1)f=0\text{.}$ Dividing by $c_0$ and rewriting using subtraction rather than addition, it is clear that we are just talking about an equation of the form $(A-r)f=0$ where $r\neq0\text{.}$
Using the preceding two results, we can now provide an outline of the inductive step in the proof of Theorem 9.18, at least in the case where the polynomial in the advancement operator has distinct roots.
##### Proof
Combining the results in the preceding sections, we can quickly write the general solution of any homogeneous equation of the form $p(A)f=0$ provided we can factor the polynomial $p(A)\text{.}$ Note that in general, this solution takes us into the field of complex numbers, since the roots of a polynomial with real coefficients are sometimes complex numbers—with non-zero imaginary parts.
We close this section with one more example to illustrate how quickly we can read off the general solution of a homogeneous advancement operator equation $p(A)f=0\text{,}$ provided that $p(A)$ is factored. | 1,286 | 4,695 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 8, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2020-45 | latest | en | 0.870669 |
https://stat.ethz.ch/pipermail/r-help/2020-January/465420.html | 1,701,870,576,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100599.20/warc/CC-MAIN-20231206130723-20231206160723-00672.warc.gz | 606,963,133 | 2,052 | # [R] Binary Number To Two´s Complement Representation
Paul Bernal p@u|bern@|07 @end|ng |rom gm@||@com
Mon Jan 20 15:57:56 CET 2020
```Dear Rui,
Based on the rules given in the link below, I want to transform the binary
numbers into latitude and longitude coordinates (in degrees and minutes),
so that is basically what I am trying to accomplish. The first integer
gives the sign (positive or negative) of the number, and the rest n-1
digits, give the magnitude of the number.
So for example: 01111001, the first integer indicates that the decimal is
positive, and, in the case of 1001011. the first integer starts with "1",
so that means the number is negative, then the rest of the integers give
the actual number, while the first integer (from left to right) give the
sign, with 0 meaning it´s a positive integer, and 1 meaning is a negative
integer.
Best regards!
https://www.electronics-tutorials.ws/binary/signed-binary-numbers.html
[[alternative HTML version deleted]]
``` | 252 | 987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-50 | latest | en | 0.80037 |
https://wiki-helper.com/a-bo-contains-60-identical-marbles-where-of-them-are-red-while-the-rest-are-white-if-a-marble-id-39833149-20/ | 1,726,117,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00229.warc.gz | 567,353,899 | 30,285 | # a box contains 60 identical marbles where x of them are red while the rest are white if a marble id drawn and the probability th
a box contains 60 identical marbles where x of them are red while the rest are white if a marble id drawn and the probability that it is ehite is find the value of x | 70 | 298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-38 | latest | en | 0.928094 |
https://math.stackexchange.com/questions/2092882/skyscraper-sheaves-cohomology | 1,716,681,305,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058858.81/warc/CC-MAIN-20240525222734-20240526012734-00563.warc.gz | 329,770,568 | 38,114 | # Skyscraper sheaves cohomology
Let $X$ be a topological space and $G$ an abelian group. Denote by $\mathcal{S}$ the skyscraper sheaf with group $G$ at the point $x\in X$. How I can prove that $\mathcal{S}$ has not cohomology, i.e $H^i(X,\mathcal{S})=0, \: \forall i>0$ ?
• Do you assume $G$ is an abelian group? Jan 11, 2017 at 5:39
• Yes. Thanks fo the remark :) Jan 11, 2017 at 18:06
• Can I have a moment to understand the answers? Jan 11, 2017 at 19:13
• Yes! -:) ${}{}{}$ Jan 11, 2017 at 20:23
• There is also a cheaper way for showing this : you can build a discrete partition of unity adapted to your sheaf. This is done in the book of Miranda, Algebraic curves and Riemann surfaces.
– user171326
Jan 16, 2017 at 1:22
A skyscraper sheaf is flasque, hence has no cohomology: Hartshorne Chap. III, Prop.2.5, page 208.
You can also see this using Čech cohomology:
Consider an open cover $\mathfrak{U}=(U_i)_{i\in I}$ of your space $X$. You can always refine this cover so that only one of the sets $U_i$ contains the point $x$: Pick a set $U_0$ containing $x$ and consider the cover $\mathfrak{U}'$ consisting of $U_0$ and $U_i\setminus\{x\}$ for all $i\in I$. Then $\mathfrak{U}'$ is a refinement of $\mathfrak{U}$ and $x$ is only contained in the set $U_0$.
In particular, $x$ is not contained in any intersection of two or more distinct sets in $\mathfrak{U}'$, so the scyscraper sheaf has no sections over these. Hence all higher Čech-cocylces are $0$ and all higher cohomology groups vanish.
(I believe this approach can also be found in Forster's book on Riemann Surfaces.)
• I got a regarding the equivalence of Cech cohomology and the normal cohomology group H^i. In Hartshorne III 4.5, it said the two cohomology are equivalent only if we assume the open covering used for the Cech consideration is affine. How can you guarantee your specially chosen open covering to be affine? (if I am considering from the algebraic geometry perspective) Nov 26, 2019 at 6:45
• Dear @IvanSo, the question as asked was in the setting of topological spaces, where there is no notion of affine. If you are in the situation of, say, schemes, then you may mimic this approach by starting with an affine cover (which exists by definition of a scheme) and refine this as described, noting that intersections and open subsets of affines remain affine. Nov 26, 2019 at 12:42
• But I suppose affine intersect affine is affine is only true for X being seperated (hausdorff)? Nov 26, 2019 at 13:38
• @IvanSo You are right, there are subtleties involved. One should probably require Noetherian schemes, as Hartshorne does (if I am not mistaken). Maybe qcqs is enough, though... Nov 26, 2019 at 14:07
These notes probably have enough detail to give you what you want. I'll assume you mean $G$ to be an abelian group. Basically the argument goes like this: if $G$ is an abelian group, take an injective resolution $G \to I^\bullet$ of abelian groups. Then $\pi_*G \to \pi_*I^\bullet$ is an injective resolution, where $\pi:{\ast} \to X$ is the inclusion of a point. Note that $\pi_*G$ is your skyscraper sheaf. To compute $H^i(X,G)$ take global sections of $\pi_*G \to \pi_*I^\bullet$, which just gives back the resolution $G\to I^\bullet$, which shows that the higher cohomology of $\pi_*G$ vanishes.
• The link is not working. Do you have a new link for the notes?
– FNH
Dec 26, 2020 at 10:34 | 1,007 | 3,389 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.885851 |
http://www.researchgate.net/publication/220189740_Automorphisms_of_cubic_Cayley_graphs_of_order_2pq | 1,448,678,477,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398450745.23/warc/CC-MAIN-20151124205410-00160-ip-10-71-132-137.ec2.internal.warc.gz | 663,306,816 | 41,890 | Article
# Automorphisms of cubic Cayley graphs of order
Department of Mathematics, Beijing Jiaotong University, Beijing 100044, PR China
(Impact Factor: 0.56). 05/2009; 309(9):2687-2695. DOI: 10.1016/j.disc.2008.06.023
Source: DBLP
ABSTRACT
In this paper the automorphism groups of connected cubic Cayley graphs of order 2pq for distinct odd primes p and q are determined. As an application, all connected cubic non-symmetric Cayley graphs of order 2pq are classified and this, together with classifications of connected cubic symmetric graphs and vertex-transitive non-Cayley graphs of order 2pq given by the last two authors, completes a classification of connected cubic vertex-transitive graphs of order 2pq.
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• Source
• "One difficult problem in Algebraic Graph Theory is to determine the automorphism groups of Cayley graphs. Although there are some nice results on the automorphism groups of Cayley graphs (see [6] [7] [8] [10] [13] [23] [24] [25] "
##### Article: Maximum-Size Independent Sets and Automorphism Groups of Tensor Powers of the Even Derangement Graphs
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ABSTRACT: Let $A_n$ be the alternating group of even permutations of $X:=\{1,2,...,n\}$ and ${\mathcal E}_n$ the set of even derangements on $X.$ Denote by $A\T_n^q$ the tensor product of $q$ copies of $A\T_n,$ where the Cayley graph $A\T_n:=\T(A_n,{\mathcal E}_n)$ is called the even derangement graph. In this paper, we intensively investigate the properties of $A\T_n^q$ including connectedness, diameter, independence number, clique number, chromatic number and the maximum-size independent sets of $A\T_n^q.$ By using the result on the maximum-size independent sets $A\T_n^q$, we completely determine the full automorphism groups of $A\T_n^q.$
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• "Feng and Xu [7] determined the automorphism groups of tetravalent Cayley graphs on regular p-groups. Recently, Zhang et al. [22] determined the automorphism groups of cubic Cayley graphs of order 2pq. For other results on the automorphism groups of Cayley graphs, we refer the readers to [5] [6] [11] [12] [18] [20] [21]. "
##### Article: Automorphism Group of the Derangement Graph.
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ABSTRACT: We prove that the full automorphism group of the derangement graph Γ n (n≥3) is equal to (R(S n )⋊Inn(S n ))⋊Z 2 , where R(S n ) and Inn(S n ) are the right regular representation and the inner automorphism group of S n respectively, and Z 2 =〈φ〉 with the mapping φ:σ φ =σ -1 , ∀σ∈S n . Moreover, all orbits on the edge set of Γ n (n≥3) are determined.
The electronic journal of combinatorics 01/2011; 18. · 0.49 Impact Factor
• Source
• "Let p and q be two primes. In [25] [26] [27], all connected cubic non-normal Cayley graphs of order 2pq are determined. Wang and Xu [23] determined all tetravalent non-normal 1- regular Cayley graphs on dihedral groups. "
##### Article: Tetravalent Non-Normal Cayley Graphs of Order 4p.
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ABSTRACT: A Cayley graph Cay(G, S) on a group G is said to be normal if the right regular representation R(G) of G is normal in the full automorphism group of Cay(G, S). In this paper, all connected tetravalent non-normal Cayley graphs of order 4p are constructed explicitly for each prime p. As a result, there are fifteen sporadic and eleven infinite families of tetravalent non-normal Cayley graphs of order 4p.
The electronic journal of combinatorics 09/2009; 16(1). · 0.49 Impact Factor | 944 | 3,411 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2015-48 | latest | en | 0.823218 |
https://socratic.org/questions/what-is-the-derivative-of-y-e-1#100303 | 1,695,851,673,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510326.82/warc/CC-MAIN-20230927203115-20230927233115-00592.warc.gz | 591,125,766 | 5,776 | # What is the derivative of y=e^(-1)?
Aug 5, 2014
y = ${e}^{- 1}$ is a constant function.
See, ${e}^{- 1} = \frac{1}{e}$ and there's no x in the expression.
The derivative of any constant is zero.
So $\frac{\mathrm{dy}}{\mathrm{dx}} = 0.$
Unless… the letter e has been defined as a variable, and the question is to find $\frac{\mathrm{dy}}{\mathrm{de}} .$ Then the derivative would be $- \frac{1}{{e}^{2}}$. But that seems unlikely.
\dansmath to the rescue!/ | 149 | 464 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-40 | latest | en | 0.8805 |
https://sublinear.info/index.php?title=Open_Problems:3&oldid=610 | 1,568,921,859,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573570.6/warc/CC-MAIN-20190919183843-20190919205843-00390.warc.gz | 691,963,204 | 6,726 | # Problem 3: $L_\infty$ Estimation
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Suggested by Graham Cormode Kanpur 2006 https://sublinear.info/3
One of the earliest results shown in data streaming is that approximating $L_ \infty$ of a stream of values requires space proportional to the dimensionality of the stream. The hard case used to prove this is when most items in the stream have frequency of occurrence 1, and approximating $L_ \infty$ is equivalent to testing whether any item has frequency two or higher. However, a variation of this problem is routinely studied under the name “heavy hitters.” Here, the lower bound is avoided by asking to find all items whose frequencies are greater than some fixed fraction $\phi$ of the total stream length, and tolerating approximation error $\epsilon$. Bounds are then provided which are polynomial in $(1/\phi)$ or $(1/\epsilon)$. A side effect of these algorithms is to estimate $L_ \infty$ of the stream with error proportional to $\epsilon$ times the $L_1$ or $L_2$ norm of the stream. Let the stream consist of items specified in $\log m$ bits. For insert only streams, the best space bound is $O(\epsilon^{-1}(\log m + \log L_1) )$ [MisraG-82,MetwallyAA-05], for computing on the difference between two streams the bounds are $O(\epsilon^{-1}\log m (\log m + \log L_1))$ [CormodeM-05a]. These algorithms approximate the $L_\infty$ distance in the sense above, but additionally identify a set of items which contribute significantly to the distance.
The open question is whether it is possible to approximate $L_\infty$ with additive error in terms of $\epsilon$ times $L_1$ or $L_2$ with less space. In particular, is it possible to reduce the dependency on $m$, since this is not needed in the output? One possible direction is to analyze data structures such as the Count-Min sketch, from which items frequencies can be estimated and in which $m$ does not occur in the (word) space complexity [CormodeM-05].[1]
## Notes
1. Formally, $\log m$ does affect the bit space complexity in two places: the data structure consists of $O(\log 1/\delta)$ hash functions whose specification requires $O(\log m)$ bits; and $O(\epsilon^{-1} \log 1/\delta)$ counters which in the worst case may count to the $L_1$ norm of the whole stream—this may perhaps be addressed by using approximate counters. | 594 | 2,407 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-39 | latest | en | 0.886694 |
https://nrich.maths.org/8109/clue?nomenu=1 | 1,511,291,080,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806422.29/warc/CC-MAIN-20171121185236-20171121205236-00111.warc.gz | 684,878,539 | 1,925 | What do you get if you square $a+ib$?
if $(a+ib)^2$ is real, what relationships must $a$ and $b$ satisfy?
if $(a+ib)^2$ is imaginary, what relationships must $a$ and $b$ satisfy? | 60 | 178 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-47 | latest | en | 0.845742 |
https://de.zxc.wiki/wiki/Bandbreite | 1,709,044,601,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474676.26/warc/CC-MAIN-20240227121318-20240227151318-00556.warc.gz | 201,487,576 | 15,068 | # Bandwidth
The bandwidth is a parameter in signal processing that defines the width of the interval in a frequency spectrum in which the dominant frequency components of a signal to be transmitted or stored are located. The bandwidth is characterized by a lower and an upper limit frequency , whereby different definitions of the two limit values exist depending on the application and thus different bandwidths exist as characteristic values depending on the context. The term is used to describe signal transmission systems in various areas such as communications technology , radio technology or acoustics .
## Determinations
Each transmission channel has - depending on its physical properties - a lower and an upper limit frequency. The lower limit frequency can also be zero; in this case one speaks of baseband position , otherwise of bandpass position. The difference in amount between the two limit frequency values is called the bandwidth. The cut-off frequencies are either in the unit Hertz (Hz) and usually abbreviated with f or by means of the angular frequency in the unit s −1 and referred to as . ${\ displaystyle \ omega}$
Different definitions are used to define the cut-off frequencies and thus the bandwidth, depending on the application and reference. With identical physical properties, these different definitions can lead to different bandwidth specifications. Some common definitions of the bandwidth are described below.
### Strict band limitation
Magnitude frequency response in the baseband with bandwidth ${\ displaystyle B}$
A signal is strictly band-limited if the magnitude frequency response , with the parameter as the angular frequency , is equal to 0 outside the range of the bandwidth. In real terms, this is only possible in approximation and the type of bandwidth definition serves as a simplified model in the context of signal theory . ${\ displaystyle | X (\ mathrm {j} \ omega) |}$${\ displaystyle \ omega}$
In the case of baseband signals with strict band limitation, the bandwidth is limited by an ideal low pass . Real signals in the baseband position always have negative frequency components , the so-called mirror spectrum , as shown by way of example in the adjacent figure on the absolute frequency profile of a real-valued signal. Without the negative frequency components, the bandwidth is defined as:
${\ displaystyle B = \ omega _ {g}}$
Magnitude frequency response in bandpass position with bandwidth ${\ displaystyle B '}$
For signals in the so-called bandpass position, the band is limited by a bandpass filter . Signals in the bandpass position are created, for example, by modulating a baseband signal; they occur, among other things, in intermediate frequency levels in radio devices. The modulation shifts the center frequency of the baseband signal from zero to the carrier frequency , which means that the lower limit frequency has a positive value: ${\ displaystyle \ omega _ {m}}$${\ displaystyle \ omega _ {u}}$
${\ displaystyle B '= \ omega _ {g} - \ omega _ {u}}$
As in the baseband, negative spectral components are not added to the bandwidth. It should be noted that the mirror spectrum of the real-valued baseband signal leads to a doubling of the bandwidth in the bandpass position due to the frequency shift in the case of linear modulation, since the modulation shifts the negative frequencies into the positive spectral range:
${\ displaystyle B '= 2 \ cdot B}$
The two positive partial spectra below and above around the center frequency are also referred to as the lower and upper sideband and have the same information content in real-valued baseband signals. For baseband signals that do not have negative frequencies, this is the case with an analytical signal , the bandwidth is identical both in the baseband and in the bandpass position - analytical signals can only be represented as complex signals in the baseband . Technically, this property is implemented in various ways, such as with single sideband modulation . ${\ displaystyle \ omega _ {m}}$
With non-linear modulation techniques such as frequency modulation , there is no direct connection between the bandwidth of the baseband signal and the required bandwidth in the bandpass position. The bandwidth is expressed approximately by the frequency deviation in the Carson formula .
### 3 dB bandwidth
Power density spectrum of a signal in the baseband position with 3 dB bandwidth ${\ displaystyle B}$
Power density spectrum of a signal in the bandpass position with 3 dB bandwidth ${\ displaystyle B '}$
In real systems, due to the finite attenuation of filters, spectral components are distributed over the entire spectrum; with a strict definition, the parameter of the bandwidth would be infinitely large and therefore not very meaningful. The usual practical parameter is the 3 dB bandwidth, which is defined by the power density spectrum around the maximum amount . The cut-off frequencies are set at half the maximum power value, which corresponds to a reduction to a rounded 3 dB : ${\ displaystyle | X (\ mathrm {j} \ omega) | ^ {2}}$${\ displaystyle A}$${\ displaystyle A / 2}$
${\ displaystyle 10 \ cdot \ log {\ frac {1} {2}} \ approx -3 {,} 01 \, \ mathrm {dB}}$
At the cutoff frequency, this corresponds to a reduction in amplitude by the factor . ${\ displaystyle 1 / {\ sqrt {2}}}$
The bandwidth is thus in the baseband position
${\ displaystyle B = \ omega _ {g}}$
and in bandpass position
${\ displaystyle B '= \ omega _ {g} - \ omega _ {u}}$
set.
For example, in the system of a low-pass filter (1st order), the 3 dB bandwidth corresponds exactly to the bandwidth of the filter; it is therefore also referred to as the 3 dB cutoff frequency . ${\ displaystyle B}$
For a series or parallel resonant circuit , the dimensionless relative bandwidth is the ratio of the 3 dB bandwidth and the center frequency : ${\ displaystyle d}$${\ displaystyle B '}$${\ displaystyle \ omega _ {m}}$
${\ displaystyle d = {\ frac {B '} {\ omega _ {m}}} = {\ frac {1} {Q}}}$
The relative bandwidth is identical to the loss factor and reciprocal to the quality factor Q .
### Carson bandwidth
The Carson bandwidth, named after John Renshaw Carson, is used for angle modulations such as frequency modulation or phase modulation . The 10% Carson bandwidth is the bandwidth that spans the spectral lines that make up 90% of the power of the signal; the rarely used 1% Carson bandwidth is the bandwidth in which the spectral lines lie that make up 99% of it. ${\ displaystyle B_ {10 \, \%}}$${\ displaystyle B_ {1 \, \%}}$
### Nyquist bandwidth
The Nyquist-Shannon sampling theorem has a central position in the theory of digital signal processing . It states that a time-continuous signal can then be reconstructed as precisely as required from the sampled, time-discrete sequence if the bandwidth of the signal is a maximum of half the sampling frequency . This maximum is called the Nyquist bandwidth.
The naive reconstruction as a step function is crude: The rectangular pulses that make up the step function in time, have the sinc function as a spectrum, i.e. an infinite bandwidth. But the bandwidth within the first two zeros of the sinc function (for positive and negative frequencies) is precisely the Nyquist bandwidth. Their product with the sampling period is 1, see time-bandwidth product .
### Antenna technology
In the field of antenna technology , among other things, relative, i.e. H. dimensionless, bandwidth specifications used. For narrowband antennas, these are antennas whose frequency response is approximately constant, a percentage bandwidth specification is used:
${\ displaystyle B _ {\%} = 2 \ cdot {\ frac {\ omega _ {g} - \ omega _ {u}} {\ omega _ {g} + \ omega _ {u}}}}$
The theoretical maximum value of the percentage bandwidth is 200% when the lower limit frequency is zero.
For broadband antennas whose magnitude frequency response is not constant, the two limit frequencies of the antenna are set in relation to each other as a relative bandwidth specification and expressed in the form : ${\ displaystyle B _ {\ text {N}}: 1}$
${\ displaystyle B _ {\ text {N}} = {\ frac {\ omega _ {g}} {\ omega _ {u}}}}$
### Legal bandwidth specifications
#### Occupied bandwidth
The term occupied bandwidth, which defines a frequency range that is used by a radio transmission, is particularly common in frequency management . The definition to be seen in the legal context according to Article 1.153 of the Implementing Regulations for the Radio Service (VO Funk) of the International Telecommunication Union (ITU) takes place as integration via the spectral power density , with 99.0% of the total transmitted power within the range between the lower and upper Frequency limit is. The remainder of 1.0% (2 x 0.5%) of the radiated power is outside of this fixed band.
#### Required bandwidth
The legal definition of the required bandwidth in accordance with Article 1.152 of the Radio Regulations (VO Funk) of the International Telecommunication Union (ITU) is the bandwidth that is just sufficient for a type of transmission at a given speed to ensure the transmission of the message.
## Examples of bandwidths
As a first approximation, the transmission system of a telephone has a lower limit frequency of 300 Hz and an upper limit frequency of 3400 Hz, which corresponds to a bandwidth of 3100 Hz and is sufficient for intelligible voice transmission. Frequency components in speech that are below or above the limit frequency are suppressed in a telephone system by means of band limitation and are not transmitted.
application approximate bandwidth
Nuclear magnetic resonance spectroscopy 0.1 Hz
Longitudinal wave 1 Hz
Electrocardiogram (EKG) 40 Hz
Telephone , slow scan television 3.1 kHz
analog FM broadcast (audio) 15 kHz
Audio CD 22 kHz
Cellular radio ( GSM ), signal in band pass position 200 kHz
VHF radio signal including additional services, in band pass position 300 kHz
analog AM television signal including sound 7 MHz
digital DVB-T multiplex signal (usually contains four programs)
WLAN according to IEEE-802.11 a / b, bandpass layer 22 MHz
Front side bus in the computer 400 MHz
Fiber optic ethernet up to 50 GHz
## literature
• Martin Werner: Signals and Systems . 3. Edition. Vieweg Teubner, 2008, ISBN 978-3-8348-0233-0 .
• Karl-Dirk Kammeyer: message transmission . 4th edition. Vieweg Teubner, 2008, ISBN 978-3-8351-0179-1 .
• Michael Dickreiter: Handbook of the recording studio technology . 6th edition. KG Saur Verlag KG, Munich 1997, ISBN 3-598-11320-X .
## Individual evidence
1. ^ John R. Carson: Notes on the Theory of Modulation . In: Proceedings of the IRE . tape 10 , no. 1 , 1922, pp. 57-64 .
2. ^ Warren L. Stutzman and Gary A. Theiele: Antenna Theory and Design . 2nd Edition. New York 1998, ISBN 0-471-02590-9 .
3. ^ VO Funk, Edition 2012, Article 1.153, Definition: occupied (frequency) bandwidth / occupied (frequency) bandwidth
4. ^ VO Funk, 2012 edition, Article 1.152, Definition: necessary bandwidth / required bandwidth | 2,460 | 11,085 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 30, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-10 | latest | en | 0.893092 |
https://www.topperlearning.com/forums/home-work-help-19/the-sum-of-reciprocals-of-two-consecutive-numbers-is-13-132-mathematics-quadratic-equations-61056/reply | 1,508,771,350,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826114.69/warc/CC-MAIN-20171023145244-20171023165244-00394.warc.gz | 1,025,352,169 | 40,625 | Question
Fri November 30, 2012 By:
# the sum of reciprocals of two consecutive numbers is 13/132 find the numbers.
Sat December 01, 2012
Let the number be X & X+1
1/X+1/(X+1)=13/132
or, taking LCM
(X+1+X)/(X^2+X)=13/132
Cross multiplying
132+264X=13X^2+13X
or, 13X^2-251X-132=0
this equation does not have ingral roots as the discriminant is not a perfect square ther such a condition is not possible .
NOTE: If in place of 13/132 it is 23/132 then the numbers will be 12 &13.
Related Questions
Wed September 13, 2017
# IF SIN ALPHA AND COS ALPHA ARE THE ROOTS OF THE EQUATION ax^2+bx+c=0.THEN PROVE THAT:a^2+2ac=b^2
Tue September 12, 2017 | 228 | 646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2017-43 | latest | en | 0.858784 |
https://numberworld.info/983319 | 1,610,736,322,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00483.warc.gz | 543,981,018 | 3,853 | # Number 983319
### Properties of number 983319
Cross Sum:
Factorization:
3 * 23 * 14251
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
f0117
Base 32:
u08n
sin(983319)
0.47892207376646
cos(983319)
0.87785741852492
tan(983319)
0.54555792735818
ln(983319)
13.79868886327
lg(983319)
5.9926944308182
sqrt(983319)
991.62442487063
Square(983319)
### Number Look Up
Look Up
983319 which is pronounced (nine hundred eighty-three thousand three hundred nineteen) is a unique number. The cross sum of 983319 is 33. If you factorisate 983319 you will get these result 3 * 23 * 14251. The figure 983319 has 8 divisors ( 1, 3, 23, 69, 14251, 42753, 327773, 983319 ) whith a sum of 1368192. 983319 is not a prime number. The figure 983319 is not a fibonacci number. The figure 983319 is not a Bell Number. The number 983319 is not a Catalan Number. The convertion of 983319 to base 2 (Binary) is 11110000000100010111. The convertion of 983319 to base 3 (Ternary) is 1211221212020. The convertion of 983319 to base 4 (Quaternary) is 3300010113. The convertion of 983319 to base 5 (Quintal) is 222431234. The convertion of 983319 to base 8 (Octal) is 3600427. The convertion of 983319 to base 16 (Hexadecimal) is f0117. The convertion of 983319 to base 32 is u08n. The sine of the number 983319 is 0.47892207376646. The cosine of 983319 is 0.87785741852492. The tangent of 983319 is 0.54555792735818. The root of 983319 is 991.62442487063.
If you square 983319 you will get the following result 966916255761. The natural logarithm of 983319 is 13.79868886327 and the decimal logarithm is 5.9926944308182. You should now know that 983319 is impressive number! | 626 | 1,809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-04 | latest | en | 0.724743 |
https://tipings.com/en/pages/60801 | 1,642,655,588,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301720.45/warc/CC-MAIN-20220120035934-20220120065934-00318.warc.gz | 636,480,118 | 6,031 | # Total capitalization rate and its calculation
Different sources quite a lot of attention paid to what the cap rate and how is it calculated.However, the category of "overall capitalization rate" needs some further explanation.
It is calculated by dividing the value of the operating profit in the value of the total sale price of the entire company or produced by the enterprise.This figure includes the value and return of investment, and the value of their yield.Defined by this method it excludes debt - so the assumption is that the company has no long-term debt.This value is then summed with the total market value.This is done as follows: it is assumed that part of long-term debt in favor of equity of the company.After this, the net value generated by an enterprise or by product (calculated according to the values before tax) is added to depreciation, as well as those expenses that the company incurred in the payment of interest.
Long-term debt is added to the value of net worth in assets.Further, according to the same procedure, the value is added to the profit interest accrued on the totality of the amount of the debt.These articles are perfectly permissible exceptions (deductions), and therefore does not appear sufficient, and the more binding basis for return on investment.That eventually turns out such a general capitalization rate, which reflects the value of the total return (arising due to amortization and depreciation), as well as the value of the total income (including interest) on the amount of equity capital of the enterprise or company and borrowed funds.
To illustrate how a capitalization rate, the calculation of which is produced in this manner, we assume that the data for the calculation of selected information on.Imagine this technique step by step form.
Step 1. Here is the detection of the total cost of the shares of the enterprise or company.It uses the mean value of the period, which is most revealing in terms of the stability of market factors.This average price of the asset multiplied by the number of ordinary shares that are released into circulation for the selected period.Furthermore, it should consider the possibility of making some corrections in the calculation taking into account the preferred shares.The resulting value is the total value of assets of the enterprise.
Step 2: At this stage of the calculation are added values of long-term debt for a given period to the total price of the ordinary shares.
Step 3. Here, the company's net profit, calculated to the payment of taxes, is added to the value of depreciation.
Step 4. At this stage the amount of net income and depreciation expenses divided by the sum, which is obtained by adding the market value of assets and long-term debt.As a result, we obtain the indicators that characterize the overall capitalization rate.
Step 5. It is estimated the net profit before tax and value of depreciation and interest deductions.
Step 6. The value obtained in the previous calculation of the shares in the consolidated rate, the rate of which is determined on the basis of information from the database of the enterprise.In the absence of such, or if they are insufficient, an alternative method by which the capitalization rate is determined.Real estate, which brings income to the enterprise as the subject of the calculation in this case is also excluded.Based on this alternative method of summing the sequential procedure.
Step 7. This is done by dividing the amount of net income and depreciation on the value of the total bet.The result is the full price of equity or venture company based on the value of borrowed funds.
It should be noted that, in the calculations, it was assumed that long-term debt was taken as part of the equity.Naturally, the calculation for the target firm will need to make the subtraction value of long-term debt from equity price index. | 760 | 3,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-05 | latest | en | 0.949315 |
solatatech.com | 1,723,243,479,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782236.55/warc/CC-MAIN-20240809204446-20240809234446-00813.warc.gz | 420,284,114 | 18,366 | # Selling Price Formula - What is the Selling Price Formula? Examples, Derivation, Formula (2024)
The selling price formula is used to calculate the selling price, which is the price at which a product is sold. If we compare the selling price and the cost price of any article, we can find the profit or loss incurred in the transaction. There are different formulas with the help of which the selling price can be calculated. Let us understand the different selling price formulas in the lesson.
## What is the Selling Price Formula?
The basic formula that is used to calculate the selling price of a product is: Selling price (S.P.) = Cost Price (C.P.) + Profit. Selling price can be calculated by using different formulas. In order to understand the other formulas, we need to know the terms related to them.
Cost Price (C.P.): Cost price is the price at which a product is purchased.
Profit (Gain): Profit is the amount that is gained or earned in the transaction. In other words, if the selling price is greater than the cost price, then the difference between them is called profit.
Loss: Loss is the amount that is lost in the transaction. This means if the selling price is less than the cost price, then the difference between them is called loss.
Formula 1: Selling Price Formula = {(100 + Gain%)/100} × CP
If we observe the first formula, we see that when the Cost price and gain percentage is given, we can easily calculate the selling price.
Example: If the cost price of an article is \$40 and there is a gain of 20% in the transaction, find its selling price.
Substituting the given values in the formula: Selling Price = {(100 + Gain%)/100} × CP = {(100 + 20)/100} × 40 = (120/100) × 40 = 48. Therefore, the Selling price of the article is \$48.
Formula 2: Selling Price Formula = {(100 – Loss%)/100} × CP
If we observe the second formula, we see that when the Cost price and loss percentage is given we can calculate the selling price.
Example: If the cost price of an article is \$30 and there is a loss of 10% when it is sold, find the selling price.
Substituting the given values in the formula: Selling Price = {(100 – Loss%)/100} × CP = {(100 – 10)/100} × 30 = 27. Therefore, the Selling price of the article is \$27.
Formula 3: Selling Price Formula = CP + Profit
If we observe the third formula, we see that when the Cost price and profit (\$) is given we can calculate the selling price.
Example: If the cost price of an article is \$25 and it is sold at a profit of \$5, find the selling price.
Substituting the given values in the formula: Selling Price = CP + Profit = 25 + 5 = 30. Therefore, the Selling price of the article is \$30.
Formula 4: Selling Price Formula = CP – Loss
In the fourth formula, we see that when the cost price and loss (\$) is given we can find the selling price.
Example: If the cost price of an article is \$35 and it is sold at a loss of \$3, find the selling price.
Substituting the given values in the formula: Selling Price = CP - loss = 35 - 3 = 32. Therefore, the Selling price of the article is \$32.
## Difference Between Selling Price and Marked Price
In order to be at par with the competition in business and to increase the sale of goods, shopkeepers offer some rebates to customers. It should be noted that there are two more terms related to this concept - the marked price (list price), and discount. Marked price is a price on which the seller offers a discount. After the discount is applied to the Marked price, it is sold at a reduced price known as the selling price.
Marked price: Marked price is the price set by the seller on the label of the article.
Discount: The rebate given by the shopkeepers to attract the customers is called a discount. Discount is always calculated on the Marked price of the article.
Therefore, we have another formula to calculate the selling price:
Selling Price = Marked Price - Discount.
Example: If the marked price of an article is \$300, and there is a 20% discount on it, find the price at which it is sold.
In this case, there is a 20% discount on the marked price.
20% discount on marked price = (20/100) × 300
Discount (\$)= 6000/100 = \$60
Now, using the selling price formula, Selling Price = Marked Price - Discount (\$) = \$300 − \$60 = \$240. Therefore, the selling price of the article is \$240.
Let us have a look at a few more solved examples to understand the selling price formula better.
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## Selling Price Formula Examples
Example1: Ryan buys a calculator for \$720 and sells it at a loss of 6(2/3)%. Find the selling price of the calculator.
Solution: Given, CP= \$720; Loss= 6(2/3)% =20/3%
Using the selling price formula,
SP = {(100 – Loss %)/100} × CP
Substituting the values,
SP = {(100-(20/3))/100} × 720
SP = {280/300} × 720
SP = \$672
Answer: The selling price of the calculator is \$672.
Example 2: James bought a bicycle for \$600 and sold it at a profit of \$100. Find the selling price of the bicycle.
Solution: Given, CP = \$600, Profit = \$100
Using the selling price formula,
SP = CP + Profit
Substituting the values,
SP = 600 + 100
SP = 700
Answer: The selling price of the bicycle is \$700.
Example 3: If the marked price of an article is \$400, and there is a 15% discount on it, find the selling price.
Solution:
In this case, there is a 15% discount on the marked price. 15% discount on marked price = (15/100) × 400 = 60
Discount (\$)= 6000/100 = \$60
Therefore, using the selling price formula, Selling Price = Marked Price - Discount (\$) = \$400 − \$60 = \$340
Answer: The selling price of the article is \$340.
## FAQs on Selling Price Formula
### What is the Selling Price Formula When Gain Percentage is Given?
Gain is the profit earned in a transaction and sometimes it is given in terms of percentage. Cost price is the price at which a product is purchased. When the gain percentage and the cost price is given, we calculate the selling price using the formula, Selling price (SP) = {(100 + Gain%)/100} × Cost Price
### What is the Selling Price Formula When Loss Percentage is Given?
When the loss percentage and the cost price is given, we calculate the selling price using the formula, Selling price(S.P.) = {(100 – Loss%)/100} × Cost Price. It should be noted that cost price is the price at which a product is purchased. Loss is incurred in a transaction when the selling price is less than the cost price. It is also expressed in the form of a percentage.
### What is the Selling Price Formula When Cost Price and Profit is Given?
When the cost price and profit is given, we calculate the selling price using the formula, Selling price = Cost Price + Profit. We know that Cost price is the price at which a product is purchased and profit is the amount that is gained or earned in the transaction.
### What is the Selling Price Formula When Cost Price and Loss is Given?
In a transaction, a loss occurs when a product is sold at a lesser price than its cost. Cost price is the price at which a product is purchased. When the cost price and loss is given, we calculate the selling price using the formula, Selling Price = Cost Price – Loss.
### What is the Selling Price Formula When Marked Price and Discount is Given?
When the marked price and discount on an article is given, we calculate the selling price using the formula, Selling Price = Marked Price - Discount. it should be noted that Marked price is the price set by the seller on the label of the article and discount is the rebate given by the shopkeepers to attract the customers. For example, the marked price (list price) of an article is \$50, and there is a \$5 discount on it, we can find its selling price using the formula, Selling Price = Marked Price - Discount. Substituting the values in the formula, Selling Price = Marked Price - Discount = 50 - 5 = 45. Therefore, the selling price of the article is \$45.
## FAQs
### What is the formula to calculate selling price? ›
Determine the total cost of all units purchased. Divide the total cost by the number of units purchased to get the cost price. Use the selling price formula to calculate the final price: Selling Price = Cost Price + Profit Margin.
What is the selling price? ›
The selling price of something is the price for which it is sold. The difference between buying and selling prices is called the spread.
How do you calculate average selling price example? ›
Calculating average sales price is rather simple compared to other inventory management formulas. To figure out your ASP, divide the total revenue earned by the total number of units sold. This looks like: average selling price = [total revenue earned ÷ number of units sold].
How do you find the selling price without a calculator? ›
Rules for finding the sale price given the original price and percent discount
1. At first, we consider the original price and discount rate.
2. The rate is usually given as a percent.
3. To find the discount, multiply the rate by the original price.
4. To find the sale price, subtract the discount from original price.
What is sale price example? ›
Thus, a 30% discount on a \$100 list price will result in a sale price of \$70. Another interpretation of the term is that it is simply the price at which something sells. For example, if a work of art is bid up to a record price of \$10 million, then that is its sale price.
What is an example of a selling cost? ›
Selling expenses can include: Distribution costs such as logistics, shipping and insurance costs. Marketing costs such as advertising, website maintenance and spending on social media. Selling costs such as wages, commissions and out-of-pocket expenses.
How do you calculate selling price and margin? ›
Calculate a retail or selling price by dividing the cost by 1 minus the profit margin percentage. If a new product costs \$70 and you want to keep the 40 percent profit margin, divide the \$70 by 1 minus 40 percent – 0.40 in decimal. The \$70 divided by 0.60 produces a price of \$116.67.
How do you calculate average buy and sell price? ›
To calculate the average cost, divide the total purchase amount (\$2,750) by the number of shares purchased (56.61) to figure the average cost per share = \$48.58. Cost Basis = Average cost per share (\$48.58) x # of shares sold (5) = \$242.90.
What is the formula of average with example? ›
Average This is the arithmetic mean, and is calculated by adding a group of numbers and then dividing by the count of those numbers. For example, the average of 2, 3, 3, 5, 7, and 10 is 30 divided by 6, which is 5.
How do you calculate average product with example? ›
The Basic Calculation
Divide the total product by the input of labor to find the average product. For example, a factory that produces 100 widgets with 10 workers has an average product of 10. Average product is useful for defining production capabilities at a specific level of input.
### What is the example of price? ›
Noun You paid a high price for the car. We bought the house at a good price. The price of milk rose. What is the difference in price between the two cars?
What is selling give an example? ›
Selling is any transaction in which money is exchanged for a good or service. During a sales negotiation, the seller attempts to convince or “sell” the buyer on the benefits of their offer.
What does selling price mean in math? ›
Selling price = (cost) + (desired profit margin) In the formula, the revenue is the selling price, the cost represents the cost of goods sold (the expenses you incur to produce or purchase goods to sell) and the desired profit margin is what you hope to earn.
What are selling and distribution costs examples? ›
Sales and Distribution Expenses Sales and distribution expenses mainly consist of salaries and benefits for staff, transportation and insurance costs, maintenance and repair expenses, travelling expenses, office utility expenses, business entertainment and marketing expenses and depreciation costs.
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Introduction: My name is Mr. See Jast, I am a open, jolly, gorgeous, courageous, inexpensive, friendly, homely person who loves writing and wants to share my knowledge and understanding with you. | 2,983 | 12,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-33 | latest | en | 0.930722 |
https://tbc-python.fossee.in/convert-notebook/High_Voltage_Engineering_Theory_and_Practice_by_M._A._Salam,_H._Anis,_A._El_Morshedy_and_R._Radwan/Chapter_2_Electric_Fields.ipynb | 1,709,276,740,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475203.41/warc/CC-MAIN-20240301062009-20240301092009-00532.warc.gz | 543,403,242 | 37,994 | # Chapter 2 Electric Fields¶
## Example 2.8.5 pgno:65¶
In [1]:
#Calculate the maximum field at the sphere surface
#Calulating Field at surface E based on figure 2.31 and table 2.3
from math import pi
Q1 = 0.25
e0 = 8.85418*10**-12 #Epselon nought
RV1= ((1/0.25**2)+(0.067/(0.25-0.067)**2)+(0.0048/(0.25-0.067)**2))
RV2= ((0.25+0.01795+0.00128)/(0.75-0.067)**2)
RV= RV1+RV2
E = (Q1*RV)/(4*pi*e0*10**10)
print"Maximum field = ",E,"V/m per volt"
#Answers vary due to round off error
Maximum field = 4.20643156401 V/m per volt
## Example 2.8.6 pgno:66¶
In [2]:
#calculation based on figure 2.32
from math import log
from math import pi
#(a)Charge on each bundle
print"Part a\t"
req = (0.0175*0.45)**0.5
print"Equivalent radius = %f m "%req
V = 400*10**3 #Voltage
H = 12. #bundle height in m
d = 9. #pole to pole spacing in m
e0 = 8.85418*10**-12 #Epselon nought
Hd = ((2*H)**2+d**2)**0.5#2*H**2 + d**2
Q = V*2*pi*e0/(log((2*H/req))-log((Hd/d)))
q = Q/2
print"Charge per bundle = %f uC/m "%Q #micro C/m
print"Charge per sunconducter = %f uC/m "%q #micro C/m
#(b part i)Maximim & average surface feild
print"\tPart b"
print"\tSub part 1\t"
R = 0.45 #conductor to subconductor spacing
MF = (q/(2*pi*e0))*((1/r)+(1/R)) # maximum feild
print"Maximum feild = %f V/m \t"%MF
MSF = (q/(2*pi*e0))*((1/r)-(1/R)) # maximum surface feild
print"Maximum feild = %f V/m \t"%MSF
ASF = (q/(2*pi*e0))*(1/r) # Average surface feild
print"Maximum feild = %f V/m \t"%ASF
#(b part ii) Considering the two sunconductors on the left
print"\tSub part 2\t"
#field at the outer point of subconductor #1
drO1 = 1/(d+r)
dRrO1 = 1/(d+R+r)
EO1 = MF -((q/(2*pi*e0))*(drO1+dRrO1))
print"EO1 = %f V/m \t"%EO1
#field at the outer point of subconductor #2
drO2 = 1/(d-r)
dRrO2 = 1/(d-R-r)
EO2 = MF -((q/(2*pi*e0))*(dRrO2+drO2))
print"EO2 = %f V/m \t"%EO2
#field at the inner point of subconductor #1
drI1 = 1/(d-r)
dRrI1 = 1/(d+R-r)
EI1 = MSF -((q/(2*pi*e0))*(drI1+dRrI1))
print"EI1 = %f V/m \t"%EI1
#field at the inner point of subconductor #2
drI2 = 1/(d+r)
dRrI2 = 1/(d-R+r)
EI2 = MSF -((q/(2*pi*e0))*(dRrI2+drI2))
print"EI2 = %f V/m \t"%EI2
#(part c)Average of the maximim gradient
print"\tPart c\t"
Eavg = (EO1+EO2)/2
print"The average of the maximum gradient = %f V/m \t"%Eavg
#Answers might vary due to round off error
Part a
Charge per bundle = 0.000005 uC/m
Charge per sunconducter = 0.000002 uC/m
Part b
Sub part 1
Maximum feild = 2607466.950170 V/m
Maximum feild = 2412255.520745 V/m
Maximum feild = 2509861.235457 V/m
Sub part 2
EO1 = 2597956.835582 V/m
EO2 = 2597429.477437 V/m
EI1 = 2402709.212726 V/m
EI2 = 2402258.056301 V/m
Part c
The average of the maximum gradient = 2597693.156510 V/m
## Example 2.8.7 pgno:69¶
In [3]:
#Electric feild induced at x
from math import pi
e0 = 8.85418*10**-12 #Epselon nought
q = 1 # C/m
C = (q/(2*pi*e0))
#Based on figure 2.33
E = C-(C*(1./3.+1./7.))+(C*(1+1./5.+1./9.))+(C*(1./5.+1./9.))-(C*(1./3.+1./7.))
print"Electric Feild = %f V/m \t"%E
#Answers might vary due to round off error
Electric Feild = 30015596280.410564 V/m
## Example 2.8.8 pgno:70¶
In [5]:
#Calculate the volume of the insulator
from math import e
from math import pi
V = 150*(2)**0.5
Ebd = 50
T = V/Ebd
print"\nThickness of graded design= %f cm "%T
#Based on figure 2.24
r = 2 # radius of the conductor
l = 10 #length of graded cylinder; The textbook uses 10 instead of 20
zr = l*(T+r)
print"Curve = %f cm**2 "%zr
V1 = 4*pi*zr*(zr-r)
print"V1 = %f cm**3 "%V1 #Unit is wrong in the textbook
#Thickness of regular design as obtained form Eq.2.77
pow = V/(2*Ebd)
t = 2*(e**pow-1)
print"Thickness of regular design = %f cm "%t
#Volume of regular design V2
V2 = pi*((2+t)**2-4)*10
print"V2 = %f cm**3 "%round(V2,2)#unit not mentioned in textbook
#Answers may vary due to round off error
Thickness of graded design= 4.242641 cm
Curve = 62.426407 cm**2
V1 = 47402.906725 cm**3
Thickness of regular design = 14.684289 cm
V2 = 8619.450000 cm**3
## Example 2.8.11 pgno:75¶
In [6]:
#Calculate the potential within the mesh
#Based on figure 2.38(b)
#equations are obtained using Eq.2.46
import numpy
from numpy import linalg
A1 = 1/2*(0.54+0.16)
A2 = 1/2*(0.91+0.14)
S = numpy.matrix([[0.5571, -0.4571, -0.1],[-0.4751, 0.828, 0.3667],[-0.1, 0.667, 0.4667]])
#By obtaining the elements of the global stiffness matrix(Sadiku,1994)
#and by emplying the Eq.2.49(a)
S1 = numpy.matrix([[1.25, -0.014],[-0.014, 0.8381]])
S2 = numpy.matrix([[-0.7786, -0.4571],[-0.4571, -0.3667]])
Phi13 = numpy.matrix([[0], [10]])
val1 = S2*Phi13
Phi24 = val1/S1
print"The values of Phi2 and Phi4 are: \n",-Phi24
#Answers may vary due to round of error
The values of Phi2 and Phi4 are:
[[ 3.6568 -326.5 ]
[-261.92857143 4.37537287]] | 1,985 | 4,785 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-10 | longest | en | 0.555668 |
https://codedump.io/share/UAdnVeMZO3JD/1/how-to-retrieve-a-total-pixel-value-above-an-average-based-threshold-in-python | 1,480,748,263,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540909.75/warc/CC-MAIN-20161202170900-00509-ip-10-31-129-80.ec2.internal.warc.gz | 844,808,666 | 9,235 | Damien Paul - 2 months ago 14
Python Question
How to retrieve a total pixel value above an average-based threshold in Python
Currently, I am practicing with retrieving the total of the pixel values above a threshold based on the mean of the whole image. (I am very new to Python). I am using Python 3.5.2, and the above code was copied from the Atom program I am using to write and experiment with the code.
For the time being, I am just practicing with the red channel - but eventually, I will need to individually analyse all colour channels.
The complete code that I am using so far:
``````import os
from skimage import io
from tkinter import *
def callback():
M = askopenfilename() #to select a file
red = image[:,:,0] #selecting the red channel
red_av = red.mean() #average pixel value of the red channel
threshold = red_av + 100 #setting the threshold value
red_val = red > threshold
red_sum = sum(red_val)
print(red_sum)
Button(text = 'Select Image', command = callback).pack(fill = X)
mainloop()
``````
Now, everything works so far, except when I run the program, red_sum comes out to be the number of pixels above the threshold, not the total of the pixels.
What I am I missing? I am thinking that my (possible naive) way of declaring the
`red_val`
variable has something to do with it.
But, how do I retrieve the total pixel value above the threshold?
When you did (red > threshold) you got a mask such that all the pixels in red that are above the thrshold got the value 1 and 0 other wise. Now to get the values you can just multiply the mask with the red channel. The multiplcation will zero all the values that are less than the threshold and will leave the values over the threshold unchanged.
The code:
``````red_val = (red > threshold)*red
red_sum = sum(red_val)
`````` | 433 | 1,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-50 | longest | en | 0.853937 |
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# Qwizdom year 7 maths - bar charts
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### Qwizdom year 7 maths - bar charts
1. 1. YEAR 7 MATHS: BAR CHARTS AN INTERACTIVE QWIZDOM LESSON
2. 2. True or False <ul><li>Hair colour is an example of Discrete Data </li></ul>
3. 3. True or False <ul><li>Height is an example of Continuous data. </li></ul>
4. 4. True or False <ul><li>Favourite type of food is </li></ul><ul><li>an example of Continuous data </li></ul>
5. 5. True or False <ul><li>Shoe size is an example of Discrete data </li></ul>
6. 6. True or False <ul><li>Length of time it takes you to walk to school each morning is an example of Continuous data. </li></ul>
7. 7. Fill in the gap
8. 8. Fill in the gap
9. 9. How many purple?
10. 10. How many green?
11. 11. What’s the difference between red and blue?
12. 12. Which is a type of data which is measured? <ul><li>Discrete </li></ul><ul><li>Continuous </li></ul><ul><li>Frequency </li></ul>
13. 13. Which is the correct spelling? <ul><li>Contenuous </li></ul><ul><li>Continuous </li></ul><ul><li>Continous </li></ul><ul><li>Continueous </li></ul> | 497 | 1,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2016-44 | latest | en | 0.642177 |
https://econometrics.com/comdata/greene7/ex_4.1.sha | 1,660,654,178,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-33/segments/1659882572304.13/warc/CC-MAIN-20220816120802-20220816150802-00066.warc.gz | 239,856,067 | 1,593 | * Sampling Distribution of a Least Squares Estimator * * Keywords: * regression, ols, sampling, distribution, monte carlo, histogram * * Description: * We illustrate how to simulate a sampling distribution of a Least Squares * Estimator by estimating 500 regressions from 100 generated datapoints * each and plotting its Histogram * * Author(s): * Noel Roy * Skif Pankov * * Source: * William H. Greene, Econometric Analysis - 7th Edition * Pearson International Edition, Chapter 4, Example 4.1 (page 94) * * Defining the variable b and setting it equal to 500 dim b 500 * Specifying the sample size - in our case, this determines how many datapoints * to generate for each variable sample 1 100 * Specifying not to print out any commands during a do-loop set nodoecho * Initiating a do-loop with 500 iterations do #=1, 500 * Generating variables w and x by drawing from a Normal distribution with * a unit variance genr w=nor(1) genr x=nor(1) * Generating remaining variables e and y genr e=0.5*w genr y=0.5+0.5*x+e * Running an OLS regression of newly generated variables y on x, speficying to * store the estimated coefficients in a vector c. "?" in front of the ols command * suppresses any output we'd normally get ?ols y x / coef = c * The vector C has two elements, the estimate of the slope and the * estimate of the constant. The former is denoted by C(1). We want * to save this in the B vector. We do this using the GEN1 command. This * command is like the GENR command, but generates a constant rather than a * series. * Saving the element 1 from vector c (estimated slope) as an #'th element of * the vector b gen1 b(#)=c(1) * Stopping the loop endo * Redefining a sample for plotting a histogram sample 1 500 * Plotting a historgram of b, specifying to split the data into 30 groups graph b / histo groups=30 stop * Comment: * Because of the random nature of estimation, the exact histogram * produced may differ from the one in the book in some details, but * the key feature (shape) will remain the same | 508 | 2,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-33 | latest | en | 0.823527 |
https://www.mathworks.com/help/optim/ug/deblur-solver-based.html | 1,680,054,004,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948900.50/warc/CC-MAIN-20230328232645-20230329022645-00507.warc.gz | 988,587,009 | 19,446 | Large-Scale Constrained Linear Least-Squares, Solver-Based
This example shows how to recover a blurred image by solving a large-scale bound-constrained linear least-squares optimization problem. The example uses the solver-based approach. For the problem-based approach, see Large-Scale Constrained Linear Least-Squares, Problem-Based.
The Problem
Here is a photo of people sitting in a car having an interesting license plate.
```load optdeblur [m,n] = size(P); mn = m*n; imshow(P) title(sprintf('Original Image, size %d-by-%d, %d pixels',m,n,mn))```
The problem is to take a blurred version of this photo and try to deblur it. The starting image is black and white, meaning it consists of pixel values from 0 through 1 in the m x n matrix P.
Simulate the effect of vertical motion blurring by averaging each pixel with the 5 pixels above and below. Construct a sparse matrix `D` to blur with a single matrix multiply.
```blur = 5; mindex = 1:mn; nindex = 1:mn; for i = 1:blur mindex=[mindex i+1:mn 1:mn-i]; nindex=[nindex 1:mn-i i+1:mn]; end D = sparse(mindex,nindex,1/(2*blur+1));```
Draw a picture of D.
```cla axis off ij xs = 31; ys = 15; xlim([0,xs+1]); ylim([0,ys+1]); [ix,iy] = meshgrid(1:(xs-1),1:(ys-1)); l = abs(ix-iy)<=5; text(ix(l),iy(l),'x') text(ix(~l),iy(~l),'0') text(xs*ones(ys,1),1:ys,'...'); text(1:xs,ys*ones(xs,1),'...'); title('Blurring Operator D (x = 1/11)')```
Multiply the image P by the matrix D to create a blurred image G.
```G = D*(P(:)); figure imshow(reshape(G,m,n)); title('Blurred Image')```
The image is much less distinct; you can no longer read the license plate.
Deblurred Image
To deblur, suppose that you know the blurring operator D. How well can you remove the blur and recover the original image P?
The simplest approach is to solve a least squares problem for x:
$\mathrm{min}\left(‖Dx-G{‖}^{2}\right)$ subject to $0\le x\le 1$.
This problem takes the blurring matrix D as given, and tries to find the x that makes Dx closest to G = DP. In order for the solution to represent sensible pixel values, restrict the solution to be from 0 through 1.
```lb = zeros(mn,1); ub = 1 + lb; sol = lsqlin(D,G,[],[],[],[],lb,ub);```
```Minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance. ```
```xpic = reshape(sol,m,n); figure imshow(xpic) title('Deblurred Image')```
The deblurred image is much clearer than the blurred image. You can once again read the license plate. However, the deblurred image has some artifacts, such as horizontal bands in the lower-right pavement region. Perhaps these artifacts can be removed by a regularization.
Regularization
Regularization is a way to smooth the solution. There are many regularization methods. For a simple approach, add a term to the objective function as follows:
$\mathrm{min}\left(‖\left(D+\epsilon I\right)x-G{‖}^{2}\right)$ subject to $0\le x\le 1$.
The term$\epsilon I$ makes the resulting quadratic problem more stable. Take $\epsilon =0.02$ and solve the problem again.
```addI = speye(mn); sol2 = lsqlin(D+0.02*addI,G,[],[],[],[],lb,ub);```
```Minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance. ```
```xpic2 = reshape(sol2,m,n); figure imshow(xpic2) title('Deblurred Regularized Image')```
Apparently, this simple regularization does not remove the artifacts. | 989 | 3,712 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-14 | latest | en | 0.821455 |
http://www.topperlearning.com/forums/ask-experts-19/an-object-is-at-a-distance-of-a-2x-b-3-2x-from-a-concave-physics-light-reflection-and-refraction-52978/reply | 1,487,657,038,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170651.78/warc/CC-MAIN-20170219104610-00420-ip-10-171-10-108.ec2.internal.warc.gz | 642,692,091 | 38,208 |
Question
Fri February 10, 2012 By:
# An object is at a distance of a) 2x b) 3/2x , from a concave lens having a focal length of magnitude x. draw ray diagrams showing formation of the image in both the cases ?
Sun February 12, 2012
### The object is located at 2F
When the object is located at the 2F point, the image will also be located at the 2F point on the other side of the lens. In this case, the image will be inverted (i.e., a right side up object results in an upside-down image). The image dimensions are equal to the object dimensions. A six-foot tall person would have an image that is six feet tall; the absolute value of the magnification is exactly 1. Finally, the image is a real image. Light rays actually converge at the image location. As such, the image of the object could be projected upon a sheet of paper.
### The object is located between 2F and F
When the object is located in front of the 2F point, the image will be located beyond the 2F point on the other side of the lens. Regardless of exactly where the object is located between 2F and F, the image will be located in the specified region. In this case, the image will be inverted (i.e., a right side up object results in an upside-down image). The image dimensions are larger than the object dimensions. A six-foot tall person would have an image that is larger than six feet tall. The absolute value of the magnification is greater than 1. Finally, the image is a real image. Light rays actually converge at the image location. As such, the image of the object could be projected upon a sheet of paper.
Related Questions
Wed February 15, 2017
# in an experiment the focal length of a convex lens a student obtained a sharp and inverted image of a distant tree on the screen behind the lens.she then removed the screen and looked through lens in the direction of the object .she will see a) an inverted image of the tree at the focus of the lens b)no image as the screen is removed c) a blurered image on the wall of laboratory d0 an erect image of the tree on the lens pls explain me the answer elaborately
Sun January 29, 2017
| 511 | 2,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-09 | longest | en | 0.943045 |
https://mail.haskell.org/pipermail/haskell-cafe/2021-June/134094.html | 1,713,514,644,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817382.50/warc/CC-MAIN-20240419074959-20240419104959-00586.warc.gz | 348,244,734 | 3,910 | # [Haskell-cafe] Pearl! I just proved theorem about impossibility of monad transformer for parsing with (1) unbiased choice and (2) ambiguity checking before running embedded monadic action (also, I THREAT I will create another parsing lib)
Fri Jun 11 22:18:30 UTC 2021
```Hi. I hope people here, at haskell-cafe, like various theorem, pearls on parsing, monads, monad transformers and anything like that, right? So, I just proved something interesting for you!
Theorem about impossibility of monad transformer for parsing with (1) unbiased choice and (2) ambiguity checking before running embedded monadic action.
I already presented it on #haskell in Libera chat, now let me send it here. This is not quite theorem, i. e. I would not say that it keeps with mathematical standards of rigor.
---------------------------
I want monad transformer (let's name it ParserT) with following properties (hint: I proved such transformer doesn't exist):
- it has unbiased choice (i. e. (<|>) :: ParserT m a -> ParserT m a -> ParserT m a) (this means running parser can give you multiple results)
- you can run parser without executing embedded monadic actions. i. e. "run" function should look like this:
run :: ParserT m a -> String -> [m a] {- Not m [a] -}
I want such function, because I want to do this: run parser, then count number of possible parses, then (if count is equal to one) actually execute embedded monadic action. I. e. I want something like this:
parser :: ParserT (Either String) ()
parser = .....
main :: IO ()
main = do {
input <- ....;
case run parser input of {
[res] -> do {
-- Cool! we got exactly one result! Let's run it
case res of {
Left e -> putStr \$ "Parsing succeed, but we have semantic error: " ++ e;
Right () -> putStr "Everything OK";
};
};
_ -> putStr "Cannot parse. 0 parse trees or >1 parse trees";
};
}
Now let me state theorem.
Theorem. Such transformer doesn't exist. Moreover, even if I replace arbitrary monad "m" with particular monad "Maybe" (or "Either String") the task is still unsolvable. But the task seems achievable using arrows
-------------------------------------
Proof. Let's imagine such "ParsecT Maybe" exists. Let's imagine we have such function:
char :: Char -> ParserT Maybe Char {- matches supplied char -}
Also, keep in mind we have "lift":
lift :: Maybe a -> ParserT Maybe a
Now let's write this:
parseA = char 'a'
parseAorA = char 'a' <|> char 'a'
contr = do {
a <- lift \$ Nothing;
if a > 5 then parseA else parseAorA;
return ();
}
Now, please, say, what is (run contr "a")? How many elements will (run contr "a") have? If "a > 5", then (run contr "a") will contain one element. If "a <= 5", then (run contr "a") will contain two elements. But we cannot know whether "a > 5", because we cannot extract "a" from "Nothing". So, we got contradiction. QED
-----------------
In other words, function "run" with type "ParserT m a -> String -> [m a]" means that we should somehow be able to know count of parses before executing monadic action. But this is impossible, because our parser can contain code like this: (do { a <- lift \$ ...; if a then ... else ...; }). I. e. count of parses can depend on value, embedded in inner monad, so we cannot know count of parses before we run embedded inner monadic action.
This theorem is not simply out of curiosity. I actually need parsing lib with this properties.
1. I need unbiased choice, because it is compatible with CFGs, and I love CFGs (as opposed to PEG)
2. I need embedding monads, because this allows me to check semantic errors (such as "type mismatch") in time of parsing. I. e. I can simply embed (Either String) monad into parsing monad, and use this (Either String) monad for checking semantic errors
3. I need to know number of parse trees before executing monadic actions (i. e. i need ([m a]) as opposed to (m [a])), because I want to check that number of parses is equal to 1 (if not, then print error), and then extract inner monad (which will contain information about semantic errors). In other words, I want check the input for ambiguity (and print error if it is ambiguous), and if it is not ambiguous, then print possible semantic errors (if any). The possibility of ambiguity detection is whole point of using CFGs as opposed to PEGs (which are always unambiguous, but this unambiguity is simply trick).
4. Every terminal and nonterminal should carry location info (i. e. begin and end of this string). And this location info should be passed implicitly. And it should not be embedded deeply in ASTs I will use. (This means that "happy" is not for me. As well as I understand it cannot implicitly pass location info. In https://github.com/ghc/ghc/blob/master/compiler/GHC/Parser.y I see explicit location management all the time.)
5. Left recursion is not needed
So, does such lib exist? Well, I already know such lib should be arrow based (as opposed to monad). So, is there such arrow parsing lib? Embedding arbitrary inner monad is not needed, just embedding (Either String) will go.
If there is no such lib, I plan to create my own. This is threat! :)
Also, I wrote letter long ago asking for parsing lib with different properties. I included desire for reversibility, among others. Well, this was other task. For my current task I don't need reversibility.
== | 1,297 | 5,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-18 | latest | en | 0.821779 |
https://jumpsaildive.com/2014/03/21/oxygen-toxicity/ | 1,686,045,371,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652494.25/warc/CC-MAIN-20230606082037-20230606112037-00740.warc.gz | 363,662,324 | 33,263 | # Oxygen Toxicity
When diving Enriched Air Nitrox there are two things that you have to be careful about. The first is the maximum operating depth which is linked to CNS, and the second is oxygen exposure (Pulmonary Toxicity).
### How to calculate the Maximum Operating Depth
In a previous post about partial pressures I touched upon the maximum operating depth but not how to calculate it. There are two depths that you calculate after you have analysed your cylinder. The maximum operating depth (MOD) and the contingency operating depth (COD). The MOD is calculated at a PO2 1.4 ata / bar and the COD is calculated at PO2 1.6 ata / bar. The COD is always a little deeper than the MOD, this is to give you a little wiggle room so that in case you go below your MOD you should still be OK.
Now to calculate the MOD or the COD we need to know the blend of EANx that we plan to use. For example there are two common blends NOAA1 which is 32% and NOAA2 which is 36%. We can use either of the following formulae depending on whether you are working in feet or metres.
Formula for calculating the MOD in metres
Formula for calculating the MOD in feet
Lets work out the MOD and the COD for NOAA1 which is 32% Nitrox in both feet and metres. If you look at the formulae for the MOD the only difference between the formula for feet and the one for metres is what you multiply by.
So we take the partial pressure of oxygen at the MOD which is PO2 1.4 ata / bar and we divide this by the blend of Nitrox that we are using 32% (as a decimal so it would be 0.32). So we do 1.4 divided by 0.32, then subtract 1 from this and multiply the answer be either 33 or 10 depending if you are using feet or metres respectively.
MOD in Feet = ((1.4 / 0.32) – 1) * 33 = 111.375 feet
MOD in Metres = ((1.4 / 0.32) – 1) * 10 = 33.75 metres
It’s a straight forward calculation that you must always do when you check the fill in your tank. However, most dive computers will tell you the MOD and the COD when you put the blend of Nitrox into them. So you can avoid the really complex maths. lol. Remember to mark the MOD on your tank and the percentage of the Oxygen so that you and/or the other divers can easily see it.
### Ok but my computer says it’s less than this. Why?
You might find that when you put the blend into your dive computer it says that the MOD is slightly less than this. For example the SUUNTO brand of dive computers are more conservative and as such calculate their MOD as if you were using 1% more Oxygen in the blend. (Don’t quote me on this, it is from personal experience from using my computer and and comparing the calculations). I would rather have a computer that was conservative and helped keep me safe rather than one that put me in danger. Err on the side of caution is always a good option.
### What happens if I go below the MOD or the COD?
You should try not to go below the MOD but you should definitely not go below the COD. The risk of Central Nervous Shockincreases rapidly and you could suffer a convulsion. Please stay about the MOD.
### Central Nervous Shock
Going below the COD means that you are entering the realm where oxygen is becoming toxic. There are symptoms that may or may not appear before a CNS convulsion happens these are:
1. visual disturbances (including tunnel vision)
2. tinnitus
3. nausea
4. twitching or muscles spasms (particularly in the face)
5. irritability
6. restlessness
7. euphoria or anxiety
8. dizziness
Some of these may occur, you may get none of them. The worry about CNS is that you will probably convulse and spit your regulator out. There is a good chance that you will drown if this happens.
### Pulmonary Toxicity
This is caused by prolonged exposure to high partial pressures of oxygen. So the longer that you are exposed to partial pressures of oxygen greater than PO2 1 ata / bar, you risk injuring your lungs. So to avoid this, we must track our exposure. To do this we can either use our computers to track the exposure or we can use tables. My SUUNTO D9tx will actually tell me how much as a percentage I have used of my available oxygen exposure, this is really handy. My Vyper just had a bar that increased with your exposure. Remember read the manual to your dive computer and follow the manufacturers recommendation.
The most important thing to note is that you should not exceed 100% exposure in a 24 hour period. If you do this you risk reaching toxicity. It is recommended that you be conservative and do not exceed 90% wherever possible. If you reach 100% of your allowable exposure, wait 12 hours before your next dive. If you go below the contingency depth treat the exposure as 100% and do not dive for 12 hours.
So it’s really easy to work out your exposure. All you need to know is the maximum partial pressure of the oxygen on the dive and use the DSAT Oxygen Exposure Table. The table (shown below but blanked out because you should buy one) is very easy to use. You take the partial pressure of the oxygen and then go down until you find the closest time and then read across to the right and this gives you the exposure as a percentage.
The PADI DSAT Oxygen Exposure Table
Example:
You make a dive where the partial pressure of oxygen is 1.12 ata / bar for 37 minutes. We need to round the partial pressure up to the nearest which is 1.2 ata / bar. We then go down and find the closest number of minutes, because 37 is not in the table we round up to the next time available which is 42 minutes. Then reading across to the right tells us that the exposure is 20%.
It’s that simple, just remember to keep track of your oxygen exposure between each dive.
Please note the information in the blog is for illustration purposes only. You must get proper training from a qualified diving professional. Only certified Nitrox divers should dive using gas blends with oxygen higher then 21%. Most diving agencies have courses that you can take. | 1,364 | 5,943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-23 | latest | en | 0.937187 |
https://cracku.in/12-which-of-the-four-companies-experienced-the-highes-x-cat-2022-slot-2-lrdi | 1,680,149,436,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949097.61/warc/CC-MAIN-20230330035241-20230330065241-00324.warc.gz | 221,642,066 | 20,788 | ### CAT 2022 Question Paper (Slot 2) - LRDI Question 12
Instructions
The two plots below show data for four companies code-named A, B, C, and D over three years - 2019, 2020, and 2021.
The first plot shows the revenues and costs incurred by the companies during these years. For example, in 2021, company C earned Rs.100 crores in revenue and spent Rs.30 crores. The profit of a company is defined as its revenue minus its costs
The second plot shows the number of employees employed by the company (employee strength) at the start of each of these three years, as well as the number of new employees hired each year (new hires). For example, Company B had 250 employees at the start of 2021, and 30 new employees joined the company during the year.
Question 12
# Which of the four companies experienced the highest annual loss in any of the years?
Solution
For all the companies in all three years, cost incurred is less than Revenue except for D in 2020.
Revenue is 20 and cost incurred is 50
Company D experienced the highest annual loss in 2020. | 258 | 1,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-14 | latest | en | 0.970251 |
https://pressbooks.bccampus.ca/douglasphys1107/chapter/10-6-the-simple-pendulum/ | 1,726,075,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651390.33/warc/CC-MAIN-20240911152031-20240911182031-00582.warc.gz | 444,004,451 | 25,985 | Chapter 10 Oscillatory Motion and Simple Harmonic Motion
# 10.6 The Simple Pendulum
### Summary
• Measure acceleration due to gravity.
• State what the period for a simple pendulum depends on.
Pendulums are in common usage. Some have crucial uses, such as in clocks; some are for fun, such as a child’s swing; and some are just there, such as the sinker on a fishing line. For small displacements, a pendulum is a simple harmonic oscillator. A simple pendulum is defined to have an object that has a small mass, also known as the pendulum bob, which is suspended from a light wire or string, such as shown in Figure 1. Exploring the simple pendulum a bit further, we can discover the conditions under which it performs simple harmonic motion, and we can derive an interesting expression for its period.
We begin by defining the displacement to be the arc length s. We see from Figure 1 that the net force on the bob is tangent to the arc and equals –mg sin θ. (The weight mg has components mg cos θ along the string and mg sin θ tangent to the arc.) Tension in the string exactly cancels the component mg cos θ parallel to the string. This leaves a net restoring force back toward the equilibrium position at θ = 0.
Now, if we can show that the restoring force is directly proportional to the displacement, then we have a simple harmonic oscillator. In trying to determine if we have a simple harmonic oscillator, we should note that for small angles (less than about 15°), sin θ is almost equal to θ ( sin θ and θ differ by about 1% or less at smaller angles). Thus, for angles less than about 15°, the restoring force F is
$F\approx{-}mg\theta.$
The displacement s is directly proportional to θ. When θ is expressed in radians, the arc length in a circle is related to its radius (L in this instance) by:
$s=L\theta,$
so that
$\theta=\dfrac{s}{L}.$
For small angles, then, the expression for the restoring force is:
$F\approx{-}\dfrac{mg}{L}s$
This expression is of the form:
$F=-kx,$
where the force constant is given by k = mg/L and the displacement is given by x = s. For angles less than about 15°, the restoring force is directly proportional to the displacement, and the simple pendulum is a simple harmonic oscillator.
Using this equation, we can find the period of a pendulum for amplitudes less than about 15°. For the simple pendulum:
$T=2\pi\sqrt{\dfrac{m}{k}}=2\pi\sqrt{\dfrac{m}{mg/L}}.$
Thus,
$T=2\pi\sqrt{\dfrac{L}{g}}$
for the period of a simple pendulum. This result is interesting because of its simplicity. The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass. As with simple harmonic oscillators, the period T for a pendulum is nearly independent of amplitude, especially if θ is less than about 15°. Even simple pendulum clocks can be finely adjusted and accurate.
Note the dependence of T on g. If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity. Consider the following example.
### Example 1: Measuring Acceleration due to Gravity: The Period of a Pendulum
What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?
Strategy
We are asked to find g given the period T and the length L of a pendulum. We can solve $T=2\pi\sqrt{\dfrac{L}{g}}$ for g, assuming only that the angle of deflection is less than 15°.
Solution
Square $T=2\pi\sqrt{\dfrac{L}{g}}$ and solve for g:
$g=4\pi^2\dfrac{L}{T^2}.$
Substitute known values into the new equation:
$g=4\pi^2\dfrac{0.75000~\text{m}}{(1.7357~\text{s})^2}.$
Calculate to find g:
$g=9.8281~\text{m/s}^2.$
Discussion
This method for determining g can be very accurate. This is why length and period are given to five digits in this example. For the precision of the approximation sin θ almost equal to θ to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about 0.5°.
### MAKING CAREER CONNECTIONS
Knowing g can be important in geological exploration; for example, a map of g over large geographical regions aids the study of plate tectonics and helps in the search for oil fields and large mineral deposits.
### TAKE-HOME EXPERIMENT: DETERMINING g
Use a simple pendulum to determine the acceleration due to gravity g in your own locale. Cut a piece of a string or dental floss so that it is about 1 m long. Attach a small object of high density to the end of the string (for example, a metal nut or a car key). Starting at an angle of less than 10°, allow the pendulum to swing and measure the pendulum’s period for 10 oscillations using a stopwatch. Calculate g. How accurate is this measurement? How might it be improved?
An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10 kg. Pendulum 2 has a bob with a mass of 100 kg. Describe how the motion of the pendula will differ if the bobs are both displaced by 12°.
### PHET EXPLORATIONS: PENDULUM LAB
Play with one or two pendulums and discover how the period of a simple pendulum depends on the length of the string, the mass of the pendulum bob, and the amplitude of the swing. It’s easy to measure the period using the photogate timer. You can vary friction and the strength of gravity. Use the pendulum to find the value of$g$on planet X. Notice the anharmonic behavior at large amplitude.
# Section Summary
• A mass m suspended by a wire of length L is a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15°.
• The period of a simple pendulum is
$T=2\pi\sqrt{\dfrac{L}{g}},$
where L is the length of the string and g is the acceleration due to gravity.
# Glossary
simple pendulum
an object with a small mass suspended from a light wire or string
### Conceptual Questions
1: Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer.
### Problems & Exercises
As usual, the acceleration due to gravity in these problems is taken to be g = 9.80 m/s2, unless otherwise specified.
1: What is the length of a pendulum that has a period of 0.500 s?
2: Some people think a pendulum with a period of 1.00 s can be driven with “mental energy” or psycho kinetically, because its period is the same as an average heartbeat. True or not, what is the length of such a pendulum?
3: What is the period of a 1.00-m-long pendulum?
4: How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?
5: The pendulum on a cuckoo clock is 5.00 cm long. What is its frequency?
6: Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?
7: (a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is 9.79 m/s2 is moved to a location where it the acceleration due to gravity is 9.82 m/s2. What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity.
8: A pendulum with a period of 2.00000 s in one location (g=9.80 m/s2) is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?
9: (a) What is the effect on the period of a pendulum if you double its length?
(b) What is the effect on the period of a pendulum if you decrease its length by 5.00%?
10: Find the ratio of the new/old periods of a pendulum if the pendulum were transported from Earth to the Moon, where the acceleration due to gravity is 1.63 m/s2.
11: At what rate will a pendulum clock run on the Moon, where the acceleration due to gravity is 1.63 m/s2, if it keeps time accurately on Earth? That is, find the time (in hours) it takes the clock’s hour hand to make one revolution on the Moon.
12: Suppose the length of a clock’s pendulum is changed by 1.000%, exactly at noon one day. What time will it read 24.00 hours later, assuming it the pendulum has kept perfect time before the change? Note that there are two answers, and perform the calculation to four-digit precision.
13: If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect time?
# Solutions
1: The movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum’s length) and by the acceleration due to gravity.
Problems & Exercises
1: 6.21 cm
3: 2.01 cm
5: 2.23 Hz
7: (a) 2.99541 s (b) Since the period is related to the square root of the acceleration of gravity, when the acceleration changes by 1% the period changes by$(0.01)^2=0.01\%$so it is necessary to have at least 4 digits after the decimal to see the changes.
9: (a) Period increases by a factor of 1.41 ($\sqrt{2}$) (b) Period decreases to 97.5% of old period
11: Slow by a factor of 2.45
13: length must increase by 0.0116%. | 2,430 | 9,537 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-38 | latest | en | 0.906505 |
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Half Title
Series Page
Title Page
Contents
Preface
Authors
1 Logics and Proofs
1.1 Introduction
1.2 Proposition
1.3 Compound Propositions
1.4 Truth Table
1.5 Logical Operators
1.5.1 Negation
1.5.2 Conjunction
1.5.3 Disjunction
1.5.4 Molecular Statements
1.5.5 Conditional Statement [If then] [ → ]
1.5.6 Biconditional [If and only if or iff] [↔ or ⇌]
1.5.7 Solved Problems
1.5.8 Tautology
1.5.10 Contingency
1.5.11 Equivalence Formulas
1.5.12 Equivalent Formulas
1.5.13 Duality Law
1.5.14 Tautological Implication
1.5.15 Some More Equivalence Formulas
1.5.16 Solved Problems
1.6 Normal Forms
1.6.1 Principal Disjunctive Normal Form or Sum of Products Canonical Form
1.6.2 Principal Conjunctive Normal Form or Product of Sum Canonical Form
1.6.3 Solved Problems
1.7 Inference Theory
1.7.1 Rules of Inference
1.7.2 Solved Problems
1.8 Indirect Method of Proof
1.8.2 Solved Problems
1.9 Method of Contrapositive
1.9.1 Solved Problems
1.10 Various Methods of Proof
1.10.1 Trivial Proof
1.10.2 Vacuous Proof
1.10.3 Direct Proof
1.11 Predicate Calculus
1.11.1 Quantifiers
1.11.2 Universe of Discourse, Free and Bound Variables
1.11.3 Solved Problems
1.11.4 Inference Theory for Predicate Calculus
1.11.5 Solved Problems
2 Combinatorics
2.1 Introduction
2.2 Mathematical Induction
2.2.1 Principle of Mathematical Induction
2.2.2 Procedure to Prove that a Statement P(n) is True for all Natural Numbers
2.2.3 Solved Problems
2.2.4 Problems for Practice
2.2.5 Strong Induction
2.2.6 Well-Ordering Property
2.3 Pigeonhole Principle
2.3.1 Generalized Pigeonhole Principle
2.3.2 Solved Problems
2.3.3 Another Form of Generalized Pigeonhole Principle
2.3.4 Solved Problems
2.3.5 Problems for Practice
2.4 Permutation
2.4.1 Permutations with Repetitions
2.4.2 Solved Problems
2.4.3 Problems for Practice
2.5 Combination
2.5.1 Solved Problems
2.5.2 Problems for Practice
2.5.3 Recurrence Relation
2.5.4 Solved Problems
2.5.5 Linear Recurrence Relation
2.5.6 Homogenous Recurrence Relation
2.5.7 Recurrence Relations Obtained from Solutions
2.6 Solving Linear Homogenous Recurrence Relations
2.6.1 Characteristic Equation
2.6.2 Algorithm for Solving k[sup(th)]-order Homogenous Linear Recurrence Relations
2.6.3 Solved Problems
2.7 Solving Linear Non-homogenous Recurrence Relations
2.7.1 Solved Problems
2.7.2 Problems for Practice
2.8 Generating Functions
2.8.1 Solved Problems
2.8.2 Solution of Recurrence Relations Using Generating Function
2.8.3 Solved Problems
2.8.4 Problems for Practice
2.9 Inclusion—Exclusion Principle
2.9.1 Solved Problems
2.9.2 Problems for Practice
3 Graphs
3.1 Introduction
3.2 Graphs and Graph Models
3.3 Graph Terminology and Special Types of Graphs
3.3.1 Solved Problems
3.3.2 Graph Colouring
3.3.3 Solved Problems
3.4 Representing Graphs and Graph Isomorphism
3.4.1 Solved Problems
3.4.2 Problems for Practice
3.5 Connectivity
3.5.1 Connected and Disconnected Graphs
3.6 Eulerian and Hamiltonian Paths
3.6.1 Hamiltonian Path and Hamiltonian Circuits
3.6.2 Solved Problems
3.6.3 Problems for Practice
4 Algebraic Structures
4.1 Introduction
4.2 Algebraic Systems
4.2.1 Semigroups and Monoids
4.2.2 Solved Problems
4.2.3 Groups
4.2.4 Solved Problems
4.2.5 Subgroups
4.2.6 Cyclic Groups
4.2.7 Homomorphisms
4.2.8 Cosets and Normal Subgroups
4.2.9 Solved Problems
4.2.10 Permutation Functions
4.2.11 Solved Problems
4.2.12 Problems for Practice
4.2.13 Rings and Fields
4.2.14 Solved Problems
4.2.15 Problems for Practice
5 Lattices and Boolean Algebra
5.1 Introduction
5.2 Partial Ordering and Posets
5.2.1 Representation of a Poset by Hasse Diagram
5.2.2 Solved Problems
5.2.3 Problems for Practice
5.3 Lattices, Sublattices, Direct Product, Homomorphism of Lattices
5.3.1 Properties of Lattices
5.3.2 Theorems on Lattices
5.3.3 Solved Problems
5.3.4 Problem for Practice
5.4 Special Lattices
5.4.1 Solved Problems
5.4.2 Problems for Practice
5.5 Boolean Algebra
5.5.1 Solved Problems
5.5.2 Problems for Practice
Bibliography
Index
##### Citation preview
Authored by B. V. Senthil Kumar and Hemen Dutta
Discrete Mathematical Structures
Mathematics and Its Applications: Modelling, Engineering, and Social Sciences Series Editor: Hemen Dutta Discrete Mathematical Structures A Succinct Foundation B. V. Senthil Kumar and Hemen Dutta Concise Introduction to Logic and Set Theory Iqbal H. Jebril and Hemen Dutta Tensor Calculus and Applications Simplified Tools and Techniques Bhaben Kalita
Authored by B. V. Senthil Kumar and Hemen Dutta
Discrete Mathematical Structures A Succinct Foundation
CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 c 2020 by Taylor & Francis Group, LLC
Contents
Preface Authors 1 Logics and Proofs 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 1.2 Proposition . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Compound Propositions . . . . . . . . . . . . . . . . . 1.4 Truth Table . . . . . . . . . . . . . . . . . . . . . . . 1.5 Logical Operators . . . . . . . . . . . . . . . . . . . . 1.5.1 Negation . . . . . . . . . . . . . . . . . . . . . 1.5.2 Conjunction . . . . . . . . . . . . . . . . . . . . 1.5.3 Disjunction . . . . . . . . . . . . . . . . . . . . 1.5.4 Molecular Statements . . . . . . . . . . . . . . 1.5.5 Conditional Statement [If . . . then] [ → ] . . . 1.5.6 Biconditional [If and only if or iff] [ ↔ or ] . 1.5.7 Solved Problems . . . . . . . . . . . . . . . . . 1.5.8 Tautology . . . . . . . . . . . . . . . . . . . . . 1.5.9 Contradiction . . . . . . . . . . . . . . . . . . . 1.5.10 Contingency . . . . . . . . . . . . . . . . . . . 1.5.11 Equivalence Formulas . . . . . . . . . . . . . . 1.5.12 Equivalent Formulas . . . . . . . . . . . . . . . 1.5.13 Duality Law . . . . . . . . . . . . . . . . . . . . 1.5.14 Tautological Implication . . . . . . . . . . . . . 1.5.15 Some More Equivalence Formulas . . . . . . . . 1.5.16 Solved Problems . . . . . . . . . . . . . . . . . 1.6 Normal Forms . . . . . . . . . . . . . . . . . . . . . . 1.6.1 Principal Disjunctive Normal Form or Sum of Products Canonical Form . . . . . . . . . . . 1.6.2 Principal Conjunctive Normal Form or Product of Sum Canonical Form . . . . . . . . . . . . . 1.6.3 Solved Problems . . . . . . . . . . . . . . . . . 1.7 Inference Theory . . . . . . . . . . . . . . . . . . . . . 1.7.1 Rules of Inference . . . . . . . . . . . . . . . . 1.7.2 Solved Problems . . . . . . . . . . . . . . . . . 1.8 Indirect Method of Proof . . . . . . . . . . . . . . . . 1.8.1 Method of Contradiction . . . . . . . . . . . . 1.8.2 Solved Problems . . . . . . . . . . . . . . . . .
ix xi
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1 1 1 1 2 2 2 2 3 3 3 4 4 6 6 6 6 7 7 7 7 8 9
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Contents 1.9
Method of Contrapositive . . . . . . . . . . . . . . . . . 1.9.1 Solved Problems . . . . . . . . . . . . . . . . . . 1.10 Various Methods of Proof . . . . . . . . . . . . . . . . . 1.10.1 Trivial Proof . . . . . . . . . . . . . . . . . . . . 1.10.2 Vacuous Proof . . . . . . . . . . . . . . . . . . . 1.10.3 Direct Proof . . . . . . . . . . . . . . . . . . . . 1.11 Predicate Calculus . . . . . . . . . . . . . . . . . . . . . 1.11.1 Quantifiers . . . . . . . . . . . . . . . . . . . . . 1.11.2 Universe of Discourse, Free and Bound Variables 1.11.3 Solved Problems . . . . . . . . . . . . . . . . . . 1.11.4 Inference Theory for Predicate Calculus . . . . . 1.11.5 Solved Problems . . . . . . . . . . . . . . . . . . 1.12 Additional Solved Problems . . . . . . . . . . . . . . .
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21 21 21 21 22 22 22 23 23 24 28 28 32
2 Combinatorics 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . 2.2.1 Principle of Mathematical Induction . . . . . . . . . . 2.2.2 Procedure to Prove that a Statement P (n) is True for all Natural Numbers . . . . . . . . . . . . . . . . . . . 2.2.3 Solved Problems . . . . . . . . . . . . . . . . . . . . . 2.2.4 Problems for Practice . . . . . . . . . . . . . . . . . . 2.2.5 Strong Induction . . . . . . . . . . . . . . . . . . . . . 2.2.6 Well-Ordering Property . . . . . . . . . . . . . . . . . 2.3 Pigeonhole Principle . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Generalized Pigeonhole Principle . . . . . . . . . . . . 2.3.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . 2.3.3 Another Form of Generalized Pigeonhole Principle . . 2.3.4 Solved Problems . . . . . . . . . . . . . . . . . . . . . 2.3.5 Problems for Practice . . . . . . . . . . . . . . . . . . 2.4 Permutation . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Permutations with Repetitions . . . . . . . . . . . . . 2.4.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . 2.4.3 Problems for Practice . . . . . . . . . . . . . . . . . . 2.5 Combination . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . 2.5.2 Problems for Practice . . . . . . . . . . . . . . . . . . 2.5.3 Recurrence Relation . . . . . . . . . . . . . . . . . . . 2.5.4 Solved Problems . . . . . . . . . . . . . . . . . . . . . 2.5.5 Linear Recurrence Relation . . . . . . . . . . . . . . . 2.5.6 Homogenous Recurrence Relation . . . . . . . . . . . . 2.5.7 Recurrence Relations Obtained from Solutions . . . .
39 39 39 39 40 40 56 57 57 58 58 58 60 61 69 70 71 72 79 80 81 85 87 87 88 88 89
Contents 2.6
2.7
2.8
2.9
vii
Solving Linear Homogenous Recurrence Relations . . 2.6.1 Characteristic Equation . . . . . . . . . . . . . 2.6.2 Algorithm for Solving k th -order Homogenous Recurrence Relations . . . . . . . . . . . . . . . 2.6.3 Solved Problems . . . . . . . . . . . . . . . . . Solving Linear Non-homogenous Recurrence Relations 2.7.1 Solved Problems . . . . . . . . . . . . . . . . . 2.7.2 Problems for Practice . . . . . . . . . . . . . . Generating Functions . . . . . . . . . . . . . . . . . . 2.8.1 Solved Problems . . . . . . . . . . . . . . . . . 2.8.2 Solution of Recurrence Relations Using Generating Function . . . . . . . . . . . . . . . 2.8.3 Solved Problems . . . . . . . . . . . . . . . . . 2.8.4 Problems for Practice . . . . . . . . . . . . . . Inclusion–Exclusion Principle . . . . . . . . . . . . . . 2.9.1 Solved Problems . . . . . . . . . . . . . . . . . 2.9.2 Problems for Practice . . . . . . . . . . . . . .
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90 91 91 92 95 96 102 103 103
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106 106 116 117 118 131
3 Graphs 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . 3.2 Graphs and Graph Models . . . . . . . . . . . . . 3.3 Graph Terminology and Special Types of Graphs 3.3.1 Solved Problems . . . . . . . . . . . . . . . 3.3.2 Graph Colouring . . . . . . . . . . . . . . . 3.3.3 Solved Problems . . . . . . . . . . . . . . . 3.4 Representing Graphs and Graph Isomorphism . . 3.4.1 Solved Problems . . . . . . . . . . . . . . . 3.4.2 Problems for Practice . . . . . . . . . . . . 3.5 Connectivity . . . . . . . . . . . . . . . . . . . . . 3.5.1 Connected and Disconnected Graphs . . . . 3.6 Eulerian and Hamiltonian Paths . . . . . . . . . . 3.6.1 Hamiltonian Path and Hamiltonian Circuits 3.6.2 Solved Problems . . . . . . . . . . . . . . . 3.6.3 Problems for Practice . . . . . . . . . . . . 3.6.4 Additional Problems for Practice . . . . . .
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4 Algebraic Structures 4.1 Introduction . . . . . . . . . . . 4.2 Algebraic Systems . . . . . . . . 4.2.1 Semigroups and Monoids 4.2.2 Solved Problems . . . . . 4.2.3 Groups . . . . . . . . . . 4.2.4 Solved Problems . . . . . 4.2.5 Subgroups . . . . . . . . . 4.2.6 Cyclic Groups . . . . . . .
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viii
Contents 4.2.7 4.2.8 4.2.9 4.2.10 4.2.11 4.2.12 4.2.13 4.2.14 4.2.15
Homomorphisms . . . . . . . . Cosets and Normal Subgroups Solved Problems . . . . . . . . Permutation Functions . . . . . Solved Problems . . . . . . . . Problems for Practice . . . . . Rings and Fields . . . . . . . . Solved Problems . . . . . . . . Problems for Practice . . . . .
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5 Lattices and Boolean Algebra 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 5.2 Partial Ordering and Posets . . . . . . . . . . . . . 5.2.1 Representation of a Poset by Hasse Diagram 5.2.2 Solved Problems . . . . . . . . . . . . . . . . 5.2.3 Problems for Practice . . . . . . . . . . . . . 5.3 Lattices, Sublattices, Direct Product, Homomorphism of Lattices . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Properties of Lattices . . . . . . . . . . . . . 5.3.2 Theorems on Lattices . . . . . . . . . . . . . 5.3.3 Solved Problems . . . . . . . . . . . . . . . . 5.3.4 Problem for Practice . . . . . . . . . . . . . . 5.4 Special Lattices . . . . . . . . . . . . . . . . . . . . 5.4.1 Solved Problems . . . . . . . . . . . . . . . . 5.4.2 Problems for Practice . . . . . . . . . . . . . 5.5 Boolean Algebra . . . . . . . . . . . . . . . . . . . . 5.5.1 Solved Problems . . . . . . . . . . . . . . . . 5.5.2 Problems for Practice . . . . . . . . . . . . .
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. . . . . . . . . . .
. . . . . . . . . . .
231 231 233 236 240 240 242 247 248 252 255
Bibliography
257
Index
259
Preface
The aim of this book is to provide a concise introduction to some significant topics covered in the subject Discrete Mathematics. Various themes of discrete mathematics have extensive applications in several courses in disciplines like Computer Science, Engineering, and Information Technology. This motivated us to provide a resource in the form of a textbook for undergraduate and postgraduate students adopting relevant courses in Computer Science, Mathematics, Engineering, Information Technology, etc. While in some courses this book can be adopted as a text book, there are several other courses where this book will serve the need of a reference book. This book also will serve as a Handbook for the teachers in class room teaching as it contains many solved problems and problems for practice which are included at the end of each subtopic. In order to meet the needs of the learners, we have included essential topics in this book, such as Logics and Proofs, Combinatorics, Graphs, Algebraic Structures, Lattices and Boolean Algebra. In Chapter 1: Logics and Proofs, vital concepts from basic level of logic and proofs are provided. In Chapter 2: Combinatorics, we have dealt with the idea of mathematical induction with more number of solved problems for better understanding of the reader. Also, various other topics in combinatorics like pigeonhole principle, permutations and combinations, and recurrence relation are discussed in meticulous approach. In Chapter 3: Graphs, fundamental notion of graph theory and various types of graphs are introduced with lot of illustrations. In Chapter 4: Algebraic Structures, the elementary tools of discrete mathematics such as algebraic structure, semigroup, monoid, abelian group, subgroup, cosets, Lagrange’s theorem, normal subgroup, homomorphism of groups, rings and fields are provided with sufficient theorems with proofs and examples. In Chapter 5: Lattices and Boolean Algebra, the concepts of partially ordered sets, lattices, Boolean algebra and their properties are focussed. These concepts are useful in many different types of computational circuits. We thank all the authors who have contributed extensively in the field of Discrete Mathematics and our family members and friends who have motivated us to bring our objective in the form of a textbook. We are indebted to the Management of Nizwa College of Technology, Nizwa, Oman for their constant support and encouragement during the preparation of this book. We would like to thank the editors in charge for this book project and supporting
ix
x
Preface
staff at CRC Press, Taylor & Francis Group, for their timely cooperation in publishing this book. We also welcome productive suggestions and comments to improve the quality of the book for next edition. B. V. Senthil Kumar, Nizwa, Oman Hemen Dutta, Guwahati, India
Authors
B. V. Senthil Kumar is serving in the Section of Mathematics, Department of Information Technology, Nizwa College of Technology, Nizwa, Oman. His areas of interest are solution and stability of functional, differential, and difference equations; operations research, statistics; and discrete mathematics. He obtained his Ph.D. Degree in 2015 and has 18 years of teaching and research experience. He has published more than 40 research papers in National and International reputed journals. He has authored the books titled Functional Equations and Inequalities: Solution and Stability Results (published by World Scientific Publishing Company) and Probability & Queueing Theory (published by KKS Publishers, Chennai, India) to his credit. He has also contributed some book chapters. He has delivered invited talks in various institutions and also organised many academic and non-academic events. He is a member of many mathematical societies. Hemen Dutta is a faculty member at the Department of Mathematics, Gauhati University, India. He did his Master of Science in Mathematics, Post Graduate Diploma in Computer Application, and Ph.D. in Mathematics from Gauhati University, India. He received his M.Phil in Mathematics from Madurai Kamaraj University, India. He currently teaches subjects like real analysis, functional analysis, algebra, mathematical logic, computer applications, etc. His primary research interest includes areas of mathematical analysis. He has to his credit several research papers, some book chapters, and few books. He has delivered talks at different institutions and organised a number of academic events. He is a member of several mathematical societies.
xi
1 Logics and Proofs
1.1
Introduction
In this chapter, we discuss propositional logic and various methods of proving validity of propositions. The concept of logic has many applications in computer science to develop computer programs, to verify the logic of program and also in electronics to design circuits.
1.2
Proposition
A proposition (or statement) is a declarative sentence which is true or false, but not both. Consider, for example, (i) The year 2000 is a leap year. (ii) 5 + 3 = 7. (iii) x = 1 is a solution of x3 = 1. (iv) Close the door. In the above, (i)–(iii) are propositions, whereas (iv) is not a proposition. Moreover, (i) and (iii) are true, while (ii) is false.
1.3
Compound Propositions
Many propositions are composite, that is, composed of subpropositions and various connectives discussed in the next section. Such composite propositions are called compound propositions. A proposition is said to be primitive if it cannot be broken into smaller propositions, that is, if it is not composite. Examples: (i) Apples are red, and milk is white. (ii) Jack is brilliant or is a hardworking student. 1
2
Discrete Mathematical Structures
Note: The truth value of a compound proposition is obtained by the truth values of its subpropositions together with the way in which they are connected to form the compound propositions.
1.4
Truth Table
A truth table lists all possible combinations of truth values of the propositions in the left most column and the truth values of the resulting propositions in the right most column.
1.5 1.5.1
Logical Operators Negation
If P is a statement, then negation of P written as ¬P or ∼ P is read as “Not P ”. The truth table for the operator “negation” is shown below. Negation P ¬P T F F T Example: P : Apple is red. ¬P : Apple is not red.
1.5.2
Conjunction
The conjunction of two statements P and Q is the statement P ∧ Q which is read as “P and Q”. The statement P ∧ Q has a truth value T whenever both P and Q have the truth value T ; otherwise, it has a truth value F . The conjunction is defined by the truth table below. Conjunction P Q P ∧Q T T T T F F F T F F F F
Logics and Proofs
3
Example: P : John worked hard. Q: John passed the examination. P ∧ Q: John worked hard, and he passed the examination.
1.5.3
Disjunction
The disjunction of two statements P and Q is the statement P ∨ Q and has a truth value F only when both P and Q have truth value F ; otherwise, it has a truth value T . The disjunction is defined by the truth table shown below. Disjunction P Q P ∨Q T T T T F T F T T F F F Example: P : 2 + 4 = 6 (T ). Q: 2 > 10 (F ). P ∨ Q: 2 + 4 = 6 or 2 > 10 is true.
1.5.4
Molecular Statements
The statements that contain one or more atomic statements and some connectives are called molecular statements. Examples: ¬P , P ∧ ¬Q, ¬P ∨ ¬Q, etc.
1.5.5
Conditional Statement [If . . . then] [ → ]
If P and Q are any two statements, then the statement P → Q which is read as “If P then Q” is called a conditional statement. Here, P is called “antecedent”, and Q is called “consequent”. The truth table is shown below.
P T T F F
If . . . then Q P →Q T T F F T T F T
Note: P → Q has a truth value F if P has the truth value T and Q has the truth value F . In all the remaining cases, it has the truth value T .
4
Discrete Mathematical Structures
Example: P : It is hot. Q: 2 + 3 = 5. P → Q: If it is hot, then 2 + 3 = 5.
1.5.6
Biconditional [If and only if or iff] [ ↔ or ]
If P and Q are any two statements, then the statement P ↔ or P Q which is read as “P if and only if Q” is called biconditional statement. The statement P ↔ Q has the truth value T whenever both P and Q have identical truth values. The truth table is shown below. If and only if P Q P ↔Q T T T T F F F T F F F T Example: P : John is rich. Q: John is happy. P ↔ Q: John is rich if and only if he is happy.
1.5.7
Solved Problems
1. Give the contrapositive statement of the statement “If there is rain, then I buy an umbrella”. Solution. Let P : “There is rain” and Q: “I buy an umbrella”. Then the given statement is P → Q. Its contrapositive is ¬Q → ¬P . 2. Construct the truth table for P → ¬Q. Solution. The truth table is shown below. Truth Table for P → ¬Q P Q ¬Q P → ¬Q T T F F T F T T F T F T F F T T
Logics and Proofs
5
3. Find the truth table for P → Q. Solution. The truth table is shown below. Truth Table for P → Q P Q P →Q T T T T F F F T T F F T 4. Construct the truth table for the compound proposition (P → Q) ↔ (¬P → ¬Q). Solution. The truth table is shown below.
P T T F F
Q T F T F
Truth Table for (P P → Q ¬P ¬Q ¬P T F F F F T T T F T T T
→ Q) ↔ (¬P → ¬Q) → ¬Q (P → Q) ↔ (¬P → ¬Q) T T T F F F T T
5. What are the contrapositive, the converse, and the inverse of the following conditional statement? “If you work hard, then you will be rewarded”. Solution. P : You work hard. Q: You will be rewarded. ¬P : You will not work hard. ¬Q: You will not be rewarded. Converse: Q → P : If you will be rewarded, then you work hard. Contrapositive: ¬Q → ¬P : If you will not be rewarded, then you will not work hard. Inverse: ¬P → ¬Q: If you will not work hard, then you will not be rewarded. 6. Construct a truth table for the compound proposition (P → Q) → (Q → P ). Solution. The truth table is shown below.
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Discrete Mathematical Structures
P T T F F
1.5.8
Q T F T F
Truth Table for (P → Q) → (Q → P ) P → Q Q → P (P → Q) → (Q → P ) T T T F T T T F F T T T
Tautology
A statement formula which is true regardless of the truth values of the statements which replace the variables in it is called a tautology or a universally valid formula or a logical truth. Example: P ∨ ¬P is a tautology.
1.5.9
A statement formula which is false regardless of the truth values of the statements which replace variables in it is called a contradiction. Example: P ∧ ¬P is a contradiction.
1.5.10
Contingency
A statement formula which is neither tautology nor contradiction is called contingency. Example: P → Q is a contingency. Note: To determine whether a given formula is a tautology or a contradiction, construct the truth table. But this process is very lengthy since the truth table will have 2n rows for n statements.
1.5.11
Equivalence Formulas
Let A and B be two statement formulas, and let p1 , p2 , . . . , pn denote all the variables occurring in both A and B. If the truth value of A is equal to the truth value of B for every one of the 2n possible sets of truth values assigned to p1 , p2 , . . . , pn , then A and B are said to be equivalent. Assuming that the variables and assignment of truth values to the variables appear in the same order in the truth tables of A and B, the final columns in the truth tables for A and B are identical if A and B are equivalent. Examples: (i) ¬¬P is equivalent to P . (ii) P ∨ ¬P is equivalent to Q ∨ ¬Q.
Logics and Proofs
7
Remark: We know that A B is true whenever A and B have identical truth values. This means A is equivalent to B (⇔) if and only if A B is a tautology.
1.5.12
Equivalent Formulas
(1) P ∨ ¬P , P ∧ ¬P [Idempotent laws] (2) (P ∨ Q) ∨ R ⇔ P ∨ (Q ∨ R), [Associative laws]
(P ∧ Q) ∧ R ⇔ P ∧ (Q ∧ R)
(3) P ∨ Q ⇔ Q ∨ P , P ∧ Q ⇔ Q ∧ P [Commutative laws] (4) P ∨ (Q ∧ R) ⇔ (P ∨ Q) ∧ (P ∨ R), P ∧ (Q ∨ R) ⇔ (P ∧ Q) ∨ (P ∧ R) [Distributive laws] (5) ¬(P ∨ Q) ⇔ ¬P ∧ ¬Q, ¬(P ∧ Q) ⇔ ¬P ∨ ¬Q [De Morgan’s laws]
1.5.13
Duality Law
Two formulas A and A? are duals of each other if either one can be obtained from the other by replacing ∧ by ∨ and ∨ by ∧. The connectives ∨ and ∧ are also called duals of each other. Examples: Write the duals of (i)(P ∨ Q) ∧ R, (ii)(P ∧ Q) ∨ T. The duals are (i)(P ∧ Q) ∨ R, (ii)(P ∨ Q) ∧ T.
1.5.14
Tautological Implication
A statement A is said to tautologically imply a statement B if A → B is a tautology. We use the notation ⇒. Remark: To prove P ⇒ Q, we assume P to be true and prove Q to be true. Otherwise, assume Q to be false, and prove P to be false also. Construction of truth table is another method for proving the implication.
1.5.15
Some More Equivalence Formulas
1. P ∧ ¬P ⇔ F , P ∨ ¬P ⇔ T
[Complement laws]
2. P ∨ T ⇔ T , P ∧ F ⇔ F
[Dominance laws]
3. P ∧ T ⇔ P , P ∨ F ⇔ P
[Identity laws]
4. P ∨ (P ∧ Q) ⇔ P , P ∧ (P ∨ Q) ⇔ P 5. ¬(¬P ) ⇔ P
[Absorption laws]
[Double Negation law]
6. P → Q ⇔ ¬Q → ¬P
[Contrapositive law]
7. P → Q ⇔ ¬P ∨ Q [Conditional as disjunction] 8. P Q ⇔ (P → Q) ∧ (Q → P )
[Biconditional as conjunction]
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Discrete Mathematical Structures
1.5.16
Solved Problems
1. Using truth table, show that the proposition P ∨ ¬(P ∧ Q) is a tautology. Solution. The truth table is shown below. Since all the entries in the last column are T , the given proposition is a tautology. P T T F F
Truth Table of P ∨ ¬(P ∧ Q) Q P ∧ Q ¬(P ∧ Q) P ∨ ¬(P ∧ Q) T T F T F F T T T F T T F F T T
2. Express A ↔ B in terms of the connectives {∧, ¬}. Solution. A ↔ ⇔ (A → B) ∧ (B → A) ⇔ (¬A ∨ B) ∧ (¬B ∨ A). 3. Show that (p → r)∧(q → r) and (p∨q) → r are logically equivalent. Solution. (p → r) ∧ (q → r) ⇔(¬p ∨ r) ∧ (¬q ∨ r) ⇔(¬p ∧ ¬q) ∨ r
(conditional as disjunction)
(Distributive law)
⇔¬(p ∨ q) ∨ r
(De Morgan’s law)
⇔(p ∨ q) → r
(conditional as disjunction).
4. Is (¬p ∧ (P ∨ q)) → q is a tautology. Solution. (¬p ∧ (p ∨ q)) → q ⇔(¬p ∧ p) ∨ (¬p ∧ q) → q ⇔F ∨ (¬p ∧ q) → q ⇔(¬p ∧ q) → q ⇔¬(¬p ∧ q) ∨ q ⇔(p ∨ ¬q) ∨ q
(Distributive law) [p ∧ ¬p ⇔ F ]
[p ∨ F ⇔ p] [p → q ⇔ ¬p ∨ q] (De Morgan’s law)
⇔p ∨ q ∨ ¬q
(Associative law)
⇔p ∨ T
[p ∨ ¬p ⇔ T ]
⇔T. Therefore, the given statement is a tautology.
Logics and Proofs
9
5. Show that the propositions p → q and ¬p∨q are logically equivalent. Solution. From the truth table, p → q and ¬p ∨ q are equivalent. Truth Table of p → q and ¬p ∨ q p q ¬p p → q ¬p ∨ q T T F T T T F F F F F T T T T F F T T T
1.6
Normal Forms
1.6.1
Principal Disjunctive Normal Form or Sum of Products Canonical Form
Consider two statements P and Q. Consider a possible formula using conjunction as follows: P ∧ Q, ¬P ∧ Q, P ∧ ¬Q, ¬P ∧ ¬Q (Duplication is not allowed and only distinct formulas are considered). We call the above terms as “minterms”. For a given formula, an equivalent formula consisting of disjunction of minterms only is known as Principal Disjunctive Normal Form (PDNF) or Sum of Products Canonical Form. Procedure I: For every truth value T in the truth table of the given formula, select the minterm which also has the value T for the same combination of the truth values of P and Q. The disjunction of these minterms will then be equivalent to the given formula. From the table below, we observe that
P T T F F
Truth Q P ∧ Q ¬P T T F F F F T F T F F T
Table showing Disjunctions of ¬Q ¬P ∧ Q P ∧ ¬Q ¬P F F F T F T F T F T F F
P and Q ∧ ¬Q P → Q P ∨ Q F T T F F T F T T T T F
P → Q ⇔ (P ∧ Q) ∨ (¬P ∧ Q) ∨ (¬P ∧ ¬Q) P ∨ Q ⇔ (P ∧ Q) ∨ (P ∧ ¬Q) ∨ (¬P ∧ Q).
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Discrete Mathematical Structures
Procedure II: This is explained in the following example: P ∨ Q ⇔ ¬P ∧ (Q ∨ ¬Q) ∨ (Q ∧ (P ∨ ¬P )) [since A ∧ T ⇔ A] ⇔ (¬P ∧ Q) ∨ (¬P ∧ ¬Q) ∨ (Q ∧ P ) ∨ (Q ∧ ¬P ) [Distributive laws] ⇔ (¬P ∧ Q) ∨ (¬P ∧ ¬Q) ∨ (P ∧ Q) [P ∨ P ⇔ P ] Note: 1. The number of minterms appearing in the normal form is the same as the number of entries with the truth value T in the truth table of the given formula. Thus, every formula which is not a contradiction has an equivalent PDNF. 2. If a formula is a tautology, then all the minterms will appear in its PDNF. 3. To show that two formulas are equivalent, obtain PDNFs of the two formulas. If the normal forms are identical, then both the formulas are equivalent. 4. Minterms of three variables are P ∧ Q ∧ R, P ∧ Q ∧ ¬R, P ∧ ¬Q ∧ R, ¬P ∧Q∧R, P ∧¬Q∧¬R, ¬P ∧Q∧¬R, ¬P ∧¬Q∧¬R, ¬P ∧¬Q∧¬R.
1.6.2
Principal Conjunctive Normal Form or Product of Sum Canonical Form
For a given number of variables, the “maxterms” consist of disjunctions in which each variable or its negation, but not both, appear only once. For a given formula, an equivalent formula consisting of conjunctions of maxterms only is known as its Principal Conjunctive Normal Form (PCNF) or Product of Sum of Canonical Form. Note: If the PDNF or PCNF of a given formula A consisting of n variables is known, then the PDNF or PCNF of ¬A will consist of the disjunction (or conjunction) of the remaining minterms (or maxterms) which do not appear in the PDNF or PCNF of A. From A ⇔ ¬¬A, one can obtain the PDNF or PCNF of A by repeated applications of De Morgan’s laws to the PDNF or PCNF of ¬A.
1.6.3
Solved Problems
1. Obtain the PDNF and PCNF of (P ∧ Q) ∨ (¬P ∧ R). Solution. (P ∧ Q) ∨ (¬P ∧ R) ⇔((P ∧ Q) ∧ T ) ∨ ((¬P ∧ R) ∧ T ) [since A ∧ T ⇔ A] ⇔((P ∧ Q) ∧ (R ∨ ¬R)) ∨ ((¬P ∧ R) ∧ (Q ∨ ¬Q)) [since P ∨ ¬P ⇔ T ]
Logics and Proofs
11
⇔(P ∧ Q ∧ R) ∨ (P ∧ Q ∧ ¬R) ∨ (¬P ∧ Q ∧ R) ∨ (¬P ∧ ¬Q ∧ R) [Distributive laws] is the required PDNF. The remaining minterms are P ∧ ¬Q ∧ R,
P ∧ ¬Q ∧ ¬R,
¬P ∧ Q ∧ ¬R,
¬P ∧ ¬Q ∧ ¬R.
The required PCNF is ¬((P ∧ ¬Q ∧ R) ∨ (P ∧ ¬Q ∧ ¬R) ∨ (¬P ∧ Q ∧ ¬R) ∨ (¬P ∧ ¬Q ∧ ¬R)) ⇔ (¬P ∨ Q ∨ ¬R) ∧ (¬P ∨ Q ∨ R) ∧ (P ∨ ¬Q ∨ R) ∧ (P ∨ Q ∨ R). 2. Obtain the PCNF and PDNF of (¬P → R) ∧ (Q ↔ P ) by using equivalences. Solution. (¬P → R) ∧ (Q ↔ P ) ⇔(P ∨ R) ∧ ((Q → P ) ∧ (P → Q)) [since P → Q ⇔ ¬P ∨ Q, P ↔ Q ⇔ (P → Q) ∧ (Q → P )] ⇔(P ∨ R) ∧ ((¬Q ∨ P ) ∧ (¬P ∨ Q)) ⇔((P ∨ R) ∨ (Q ∧ ¬Q)) ∧ ((¬Q ∨ P ) ∨ (R ∨ ¬R)) ∧ ((¬P ∨ Q) ∨ (R ∧ ¬R)) ⇔(P ∨ Q ∨ R) ∧ (P ∨ ¬Q ∨ R) ∧ (P ∨ ¬Q ∨ R) ∧ (P ∨ ¬Q ∨ ¬R) ∧ (¬P ∨ Q ∨ R) ∧ (¬P ∨ Q ∧ ¬R) ⇔(P ∨ Q ∨ R) ∧ (P ∨ ¬Q ∨ R) ∧ (P ∨ ¬Q ∨ ¬R) ∧ (¬P ∨ Q ∨ R) ∧ (¬P ∨ Q ∨ ¬R) which is the required PDNF. The remaining minterms are P ∨ Q ∨ ¬R,
¬P ∨ ¬Q ∨ R,
¬P ∨ ¬Q ∨ ¬R.
The required PCNF is ¬((P ∨ Q ∨ ¬R) ∧ (¬P ∨ ¬Q ∨ R) ∧ (¬P ∨ ¬Q ∨ ¬R)) ⇔(¬P ∧ ¬Q ∧ R) ∨ (P ∧ Q¬R) ∨ (P ∧ Q ∧ R). 3. Find the PDNF of (Q ∨ (P ∧ R)) ∧ ¬((P ∨ R) ∧ Q). Solution. (Q ∨ (P ∧ R)) ∧ ¬((P ∨ R) ∧ Q) ⇔(Q ∨ (P ∧ R)) ∧ ((¬P ∧ ¬R) ∨ ¬Q) [by De Morgan’s laws] ⇔Q ∧ (¬P ∧ ¬R) ∨ (Q ∧ ¬Q) ∨ (P ∧ R ∧ ¬P ∧ ¬R) ∨ (P ∧ R ∧ ¬Q) [using Distributive law]
12
Discrete Mathematical Structures ⇔(¬P ∧ Q ∧ ¬R) ∨ F ∨ (F ∧ R) ∨ (P ∧ ¬Q ∧ R) ⇔(¬P ∧ Q ∧ ¬R) ∨ (P ∧ ¬Q ∧ R). 4. Prove that ((P ∨Q)∧¬(¬P ∧(¬Q∨¬R))))∨(¬P ∧¬Q)∨(¬P ∧¬R) is a tautology. Solution. Consider ¬(¬P ∧ (¬Q ∨ ¬R)) ⇔ P ∨ ¬(¬Q ∨ ¬R) [De Morgan’s law] ⇔ P ∨ (Q ∧ R) [De Morgan’s law] ⇔ (P ∨ Q) ∧ (P ∨ R) [De Morgan’s law]. (1.1) Now, consider (¬P ∧ ¬Q) ∨ (¬P ∧ ¬R) ⇔(¬(P ∨ Q) ∨ ¬(P ∨ R) [De Morgan’s law] ⇔¬((P ∨ Q) ∧ (P ∨ R)) [De Morgan’s law].
(1.2)
From (1.1) and (1.2), we obtain ((P ∨ Q) ∧ (P ∨ Q) ∧ (P ∨ R)) ∨ ¬((P ∨ Q) ∧ (P ∨ R)) ⇔((P ∨ Q) ∧ (P ∨ R)) ∨ ¬((P ∨ Q) ∧ (P ∨ R)) ⇔T. Hence, the given statement formula is a tautology. 5. Prove that (P → Q) ∧ (Q → R) ⇒ (P → R). Solution. It is enough to prove (P → Q)∧(Q → R) ⇒ (P → R) is a tautology. Let S = (P → Q) ∧ (Q → R) ⇒ (P → R). Since the last column is T for all eight combinations as shown in the table below, the given statement formula is a tautology.
P T T T F T F F F
Q T T F T F T F F
Truth Table showing (P → Q) ∧ (Q → R) ⇒ (P → R) R P → Q Q → R P → R (P → Q) ∧ (Q → R) S T T T T T T F T F F F T T F T T F T T T T T T T F F T F F T F T F T F T T T T T T T F T T T T T
Logics and Proofs
13
6. Show that P ∨(Q∧R) and (P ∨Q)∧(P ∨R) are logically equivalent. Solution. From columns (6) and (8), we have P ∨(Q∧R) and (P ∨Q)∧(P ∨R) are logically equivalent. Truth Table showing the given Formulas are Logically Equivalent Q R P ∨ Q P ∨ R (P ∨ Q) ∧ (P ∨ R) Q ∧ R P ∨ (Q ∧ R) T T T T T T T T F T T T F T F T T T T F T T T T T T T T F F T T T F T T F T F F F F F T F T F F F F F F F F F F
P T T T F T F F F
7. Show that the propositions P → Q and ¬P ∨ Q are logically equivalent. Solution. From columns (3) and (5) in the table below, we have P → Q and ¬P ∨ Q are logically equivalent. Truth Table showing the given Formulas are Logically Equivalent P Q P → Q ¬P ¬P ∨ Q T T T F T T F F F F F T T T T F F T T T
1.7
Inference Theory
Given a set of premises H1 , H2 , . . . , Hm and a conclusion C, we want to show whether C logically follows from H1 , H2 , . . . , Hm . That is, we want to show (H1 ∧ H2 ∧ · · · ∧ Hm ) → C is a tautology. Procedure 1. We look for those rows of H1 , H2 , . . . , Hm which have a truth value T . If for every such row C also has a truth value T , then the conclusion C logically follows from H1 , H2 , . . . , Hm . Example 1: H1 : P , H2 : P → Q, C : Q. In the first row of the truth table,
14
Discrete Mathematical Structures Truth Table Showing P, P → Q ⇒ Q P Q P →Q T T T T F F F T T F F T
H1 : P is True, H2 : P → Q is True. Also, C : Q is True. Therefore, P, P → Q ⇒ Q.
1.7.1
Rules of Inference
1. Rule P: A premise can be introduced at any point of derivation. 2. Rule T: A formula can be introduced provided it is tautologically implied by previously introduced formulas in the derivation. 3. Rule CP: If S can be derived from R and a set of premises, then R → S can be derived from the set of premises alone.
Set of premises
R
R S
Rule CP
We use the following tables of implications and equivalences. Implications Table I1 : P ⇒ P ∨ Q I2 : Q ⇒ P ∨ Q I3 : P ∧ Q ⇒ P I4 : P ∧ Q ⇒ Q I5 : P, P → Q ⇒ Q I6 : ¬Q, P → Q ⇒ ¬P
S
Logics and Proofs
15
I7 : ¬P, P ∨ Q ⇒ Q I8 : P → Q, Q → R ⇒ P → R I9 : P, Q ⇒ P ∨ Q I10 : Q ⇒ P → Q I11 : P ∨ Q, Q → R ⇒ ¬P → R I12 : ¬P ⇒ P → Q. Equivalences Table E1 : ¬¬P ⇔ P E2 : P → Q ⇔ ¬P ∨ Q E3 : P → Q ⇔ ¬Q → ¬P E4 : (P Q) ⇔ (P → Q) ∧ (Q → P ) E5 : P → (Q → R) ⇔ (P ∧ Q) → R E6 : ¬(P ∧ Q) ⇔ ¬P ∨ ¬Q. Note: 1. Rule CP means rule of Conditional Proof. 2. Rule CP is also called the deduction theorem. 3. In general, whenever conclusion is of the form R → S (in terms of conditional), we should apply Rule CP . In such case, R is taken as an additional premise, and S can be derived from the given premises and R.
1.7.2
Solved Problems
1. Show that R∧(P ∨Q) is a valid conclusion from the premises P ∨Q, Q → R, P → M , and ¬M . Solution. Given premises are P ∨ Q, Q → R, P → M , ¬M Conclusion: R ∧ (P ∨ Q). {1} {2} {1, 2} {4} {4} {1, 2, 4} {7} {1, 2, 4, 7} {1, 2, 4, 7}
(1) P →M (2) ¬M (3) ¬P (4) P ∨Q (5) ¬P → Q (6) Q (7) Q→R (8) R (9) R ∧ (P ∨ Q)
Rule Rule Rule Rule Rule Rule Rule Rule Rule
P P T P T T P T T
[¬Q, P → Q ⇒ ¬P ] [P → Q ⇔ ¬P ∨ Q] [P, P → Q ⇒ Q] [P, P → Q ⇒ Q] [P, Q ⇒ P ∧ Q].
16
Discrete Mathematical Structures 2. Prove that the premises P → Q, Q → R, R → S, S → ¬R, and P ∧ S are inconsistent. Solution. {1} {2} {1, 2} {4} {4} {1, 2, 4} {1, 2, 4} {1, 2, 4} {9} {1, 2, 4, 9}
(1) P →Q (2) Q→R (3) P →R (4) S → ¬R (5) R → ¬S (6) P → ¬S (7) ¬P ∨ ¬S (8) ¬(P ∧ S) (9) P ∧S (10) (P ∧ S)∧ ¬(P ∧ S)
Rule Rule Rule Rule Rule Rule Rule Rule Rule Rule
P P T P T T T T P T
[P → Q, Q → R ⇒ P → R] [P → Q ⇔ ¬Q → ¬P ] [P → R, R → ¬S ⇒ P → ¬S] [P → Q ⇔ ¬P ∨ Q] [De Morgan’s law] [P, Q ⇒ P ∧ Q]
which is false. Therefore, the given set of premises are inconsistent. 3. Show that (p → q) ∧ (r → s), (q → t) ∧ (s → u), ¬(t ∧ u), and (p → r) ⇒ ¬p. Solution. {1} {1} {1} {4} {4} {4} {1, 4}
(1) (2) (3) (4) (5) (6) (7)
(p → q) ∧ (r → s) p→q r→s (q → t) ∧ (s → u) q→t s→u p→t
{1, 4}
(8)
r→u
{9} {1, 4, 9}
(9) (10)
p→r p→u
{1, 4, 9} {1, 4} {1, 4, 9}
(11) (12) (13)
¬u → ¬p ¬t → ¬p (¬t ∨ ¬u) → ¬p
{1, 4, 9}
(14)
¬(t ∧ u) → ¬p
Rule P Rule T [P ∧ Q ⇒ P ] Rule T [P ∧ Q ⇒ Q] Rule P Rule T [P ∧ Q ⇒ P ] Rule T [P ∧ Q ⇒ Q] Rule T [P → Q, Q → R ⇒ P → R] Rule T [P → Q, Q → R ⇒ P → R] Rule P Rule T [P → Q, Q → R ⇒ P → R] Rule T Rule T Rule T [P → Q, R → Q ⇒ (P ∨ R) → Q] Rule T [De Morgan’s law]
Logics and Proofs {15} {1, 4, 9, 15}
17 ¬(t ∧ u) ¬p
(15) (16)
Rule P Rule T [P, P → Q ⇒ Q].
4. Show that the hypotheses, “It is not sunny this afternoon and it is colder than yesterday”, “We will go swimming only if it is sunny”, “If we do not go swimming, then we will take a canoe trip”, and “if we take a canoe trip, then we will be home by sunset”, lead to the conclusion, “We will be home by sunset”. Solution. Let A: It is sunny B: It is colder than yesterday C: We will go swimming D: We will take a canoe trip E: We will be home by sunset. Then the given premises are (1) ¬A ∧ B (2) A → C (3) ¬C → D The conclusion is C. {1} {1} {2} {1, 2} {5} {1, 2, 5} {7} {1, 2, 5, 7}
(1) ¬A ∧ B (2) ¬A (3) A→C (4) ¬C (5) ¬C → D (6) D (7) D→E (8) E
(4) D → E.
Rule Rule Rule Rule Rule Rule Rule Rule
P T P T P T P T
[P ∧ Q ⇒ P ] [¬P, P → Q ⇒ ¬Q] [P, P → Q ⇒ Q] [P, P → Q ⇒ Q].
5. Prove that the following argument is valid. p → ¬q, r → q, r ⇒ ¬p. Solution. {1} {2} {1, 2} {4} {1, 2, 4}
(1) (2) (3) (4) (5)
r r→q q p → ¬q ¬p
Rule Rule Rule Rule Rule
P P T [P, P → Q ⇒ Q] P T [P → ¬Q, q ⇒ ¬P ].
6. Using indirect method of proof, derive p → ¬s from the premises p → (q ∨ r), q → ¬p, s → ¬r, and p.
18
Discrete Mathematical Structures Solution. Let us include ¬(p → ¬s) as an additional premise and prove this problem by the method of contradiction. Now, ¬(p → ¬s) = ¬(¬p ∨ ¬s) = p ∧ s. Therefore, the additional premise is p ∧ s. {1} {2} {3} {2, 3} {1, 2, 3} {6} {1, 2, 3, 6} {1, 2, 3, 6} {9} {1, 2, 3, 6, 9} {1, 2, 3, 6, 9}
(1) p∧s (2) p → (q ∨ r) (3) p (4) q∨r (5) s (6) s → ¬r (7) ¬r (8) q (9) q → ¬p (10) ¬p (11) p ∧ ¬p
Additional premise Rule P Rule P Rule T [P, P → Q ⇒ Q] Rule T [P ∧ Q ⇒ Q] Rule P Rule T [P, P → Q ⇒ Q] Rule T [P, P → Q ⇒ Q] Rule P Rule P Rule T [P, Q ⇒ P ∧ Q]
which is false. Therefore, by the method of contradiction, p → ¬s follows. 7. Show that R → S can be derived from the premises P → (Q → S), ¬R ∨ P , and Q. Solution. {1}
(1)
R
{2}
(2)
¬R ∨ P
Rule P
{2}
(3)
R→P
Rule T [P → Q ⇔ ¬P ∨ Q]
{1, 2}
(4)
P
{5}
(5) P → (Q → S)
Assumed premise
Rule T [P, P → Q ⇒ Q] Rule P
{1, 2, 5}
(6)
Q→S
Rule P [P, P → Q ⇒ Q]
{7}
(7)
Q
Rule P
{1, 2, 5, 7}
(8)
S
Rule T [P, P → Q ⇒ Q]
{1, 2, 5, 7}
(9)
R→S
Rule CP.
8. Prove that A → ¬D is a conclusion from the premises A → B ∨ C, B → ¬A, and D → ¬C by using conditional proof. Solution. {1}
(1)
A
{2}
(2)
A→B∨C
Assumed premise Rule P
Logics and Proofs
19
{1, 2}
(3)
B∨C
Rule T
{1, 2}
(4)
¬B → C
Rule T
{5}
(5)
B → ¬A
Rule P
{5}
(6)
A → ¬B
Rule T
{1, 2, 5}
(7)
A→C
Rule T
[P, P → Q ⇒ Q]
[P → Q ⇔ ¬Q → ¬P ]
[P → Q, Q → R ⇒ P → R] {8}
(8)
D → ¬C
Rule P
{8}
(9)
C → ¬D
Rule T
(10)
A → ¬D
Rule CP.
{1, 2, 5, 8}
1.8 1.8.1
[P → Q ⇔ ¬P ]
Indirect Method of Proof Method of Contradiction
In order to show that a conclusion C follows logically from the premises H1 , H2 , . . . , Hm , we assume that C is false and consider ¬C as an additional premise. If the new set of premises gives contradicting value, then the assumption ¬C is true does not hold simultaneously with H1 ∧ H2 ∧ · · · ∧ Hm being true. Therefore, C is true whenever H1 ∧ H2 ∧ · · · ∧ Hm is true. Thus, C logically follows from the premises H1 ∧ H2 ∧ · · · ∧ Hm .
1.8.2
Solved Problems
1. Show that
√ 2 is irrational.
Solution. √ Suppose 2 is irrational. Therefore, 2 = divisor. Therefore,
p2 q2
p q
for p, q ∈ Z, q 6= 0, p and q have no common
= 2 =⇒ p2 = 2q 2 .
Since p2 is an even integer, p is an even integer. Let p = 2m for some integer m. ∴
p2 = 2q 2 =⇒ (2m)2 = 2q 2 =⇒ q 2 = 2m2 .
Since q 2 is an even integer, q is an even integer. Let q = 2n for some integer n.
20
Discrete Mathematical Structures Hence, p and q are even integers. Hence, they can have a common factor 2 which is a contradiction to our assumption. √ 2 is irrational. ∴ 2. Prove that the following argument is valid: “If 7 is less than 4, then 7 is not a prime number; 7 is not less than 4; therefore 7 is a prime number”. Solution. Let A: 7 is less than 4. B: 7 is a prime number. The given premises are is B. {1} {2} {1, 2} {1, 2}
(1) (2) (3) (4)
(i) A → ¬B
¬A A → ¬B ¬(¬B) B
(ii) ¬A, and the conclusion
Rule P Rule P Rule T
[¬P, P → Q ⇒ ¬Q]
Therefore, the statements are valid. 3. Prove that the sum of an irrational number and a rational number is irrational. Solution. Let a be a rational number and b be an irrational number. Assume a + b = c is a rational number. ∴ b = c + (−a) is a rational number since the sum of two rational numbers is a rational number, which is a contradiction to our assumption. Hence, the sum of an irrational number and a rational number is an irrational number. 4. Show that the statement “Every positive integer is the sum of the squares of three integers” is false. Solution. Any number of the form 4m + 7, where m is non-negative integer, cannot be written as the sum of the squares of three integers. For example, 7, 11, 15, 19, . . . cannot be written as the sum of squares of three integers. ∴ “Every positive integer is the sum of the squares of three integers” is a false statement.
Logics and Proofs
1.9
21
Method of Contrapositive
In order to prove that H1 ∧ H2 ∧ · · · ∧ Hm ⇒ C, if we prove ¬C ⇒ ¬(H1 ∧H2 ∧· · ·∧Hm ), then the original problem follows. This procedure is called the method of contrapositive.
1.9.1
Solved Problems
1. Prove that if 3n + 2 is odd, then n is odd. Solution. Let us prove this problem by the method of contrapositive. Assume that n is even. Then n = 2k, for some integer k. Now, 3n + 2 = 3(2k) + 2 = 2(3k + 1) =⇒ 3n + 2 is an even number which contradicts that 3n + 2 is odd. ∴ n is odd. 2. Prove that if n is an integer and n3 + 5 is odd, then n is even. Solution. Given: n3 + 5 is odd. To prove: n is even. Assume that n is odd. ∴ n = 2k + 1 for some integer k. n3 + 5 = (2k + 1)3 + 5 = 8k 3 + 12k 2 + 6k + 6 = 2 4k 3 + 6k 2 + 3k + 3
=⇒ n3 + 5 is an even number which is a contradiction. ∴ n is even.
1.10 1.10.1
Various Methods of Proof Trivial Proof
In an implication p → q, if we can establish that q is true, then regardless of the truth value of p, the implication p → q will be true. Hence, to construct a trivial proof of p → q, we need to show that the truth value of q is true.
22
1.10.2
Discrete Mathematical Structures
Vacuous Proof
If the hypothesis p of an implication p → q is false, then p → q is true for any proposition q. Example: Prove the proposition p(0) where p(n) is the proposition: “If n is a positive integer greater than 1, then n2 > n”. Solution. Let p(0): If 0 is a positive integer greater than 1, then 02 > 0. Since 0 is not a positive integer greater than 1, the proposition is true.
1.10.3
Direct Proof
Suppose, the hypothesis p is true. Then, the implication p → q can be proved if we can prove that q is true by using the rules of inference and some other theorems. Example: Prove that the sum of two odd integers is even. Solution. Let a and b be two odd integers. ∴ a = 2m + 1, for some integer m. b = 2n + 1, for some integer n. ∴ a + b = 2m + 2n + 2 = 2(m + n + 1) =⇒ a + b is even. ∴ Sum of two odd integers is even.
1.11
Predicate Calculus
Consider the following statement: Amruta is a student. Suppose “Amruta” is taken as a and “is a student” as S, then the above statement can be symbolically written as S(a). Here, “is a student” is a predicate, and “Amruta” is a subject. Any statement “r is P ” can be written as P (r). This can be extended to two place predicate. For example, Amruta is taller than Arvindh. Similarly, a statement with three place predicate can be extended as “Amruta is younger than Boomika but elder than Arvindh”. If a statement function of one variable is defined to be an expression consisting of a predicate symbol and an individual variable, such a statement function becomes a statement when the variable is replaced by the name of the object. Example for statement function of two variables: A(x, y): x is taller than y. G(a, r): Amruta is taller than Arvindh where a is Amruta and r is Arvindh.
Logics and Proofs
1.11.1
23
Quantifiers
Consider the following example: All apples are red. This can be understood as “for any x, if x is an apple, then x is red”. If we denote A(x): x is an apple and R(x): x is red, then we can write the above statement as (x)(A(x) → R(x)). Here (x) is called “Universal Quantifier”. We use universal quantifier for those statements of the form “All P are Q”. Now, consider the following example: Some men are clever. This can be written as “There exists x; if x is a man, then he is clever”. If we denote M (x): x is a man and C(x): x is clever, then we can write the above statement as (∃x)(M (x) ∧ C(x)). Here (∃x) is called “Existential Quantifier”. Existential quantifier is used for those statements which are of the form “Some P are Q”.
1.11.2
Universe of Discourse, Free and Bound Variables
Consider the following statement: All men are giants. This can be symbolically written as (x)(M (x) → G(x)) where M (x): x is a man and G(x): x is giant. In the above example, if we restrict the class as the class of men, then the symbolic representation will be (x)G(x). Such a restricted class is called “Universe of Discourse”. In any formula, the part containing (x)P (x) or ∃xP (x) is called the x bound part of the formula. Any variable appearing in an x bound part of the formula is called bound variable. Otherwise, it is called free. Any formula immediately following (x) or (∃x) is called the scope of the quantifier. Example: (x)P (x) ∧ Q(x) In this, all x in P (x) is bound, whereas the x in Q(x) is free. The scope of (x) is P (x).
24
Discrete Mathematical Structures
1.11.3
Solved Problems
1. Symbolise the expression “x is the father of the mother of y”. Solution. Let P (x): x is a person F (x, y): x is the father of y M (x, y): x is the mother of y. Let z be the mother of y. Hence, the given statement can be written as “x is the father of z, and z is the mother of y”. In symbolic notation, (∃z)(P (z) ∧ F (x, z) ∧ M (z, y)). 2. Given P = {2, 3, 4, 5, 6}, state the truth value of the statement (∃x ∈ P )(x + 3 = 10). Solution. For x = 2, 3, 4, 5, 6, no x satisfies x + 3 = 10. ∴ (∃x ∈ P )(x + 3 = 10) is false. 3. Write the symbolic form, and negate the following statements: (i) Everyone who is healthy can do all kinds of work. (ii) Some people are not admired by everyone. (iii) Everyone should help his neighbours, or his neighbours will not help him. Solution. (i) Let H(x): x is healthy W (x): x can do all kinds of work. Symbolic form of the statement is (x)(H(x) → W (x)). Negation of this expression is ¬((x)(H(x) → W (x)) =⇒ ¬((x)(¬H(x) ∨ W (x))) =⇒ (∃x)(H(x) ∧ ¬W (x)). That is, someone is healthy and cannot do all kinds of work.
Logics and Proofs
25
(ii) Let A(x): x is admired. Then, the given statement can be written as “for some x, it is not a case that x is admired by everyone”. Symbolic form is (∃x)(¬A(x)). Negation of the above statement is ¬((∃x)¬A(x)) ⇒ (∀x)A(x). That is, all people are admired by everyone. (iii) This statement can be rewritten as “For all x, x is a person, x should help his neighbours, or his neighbours will not help him”. Let P (x): x is a person H(x): x helps his neighbour. The symbolic form is (x)((P (x) → H(x)) ∨ (H(x) → ¬P (x))). Negation of the above statement is (∃x)((H(x) → P (x)) ∧ (¬P (x) → H(x))). 4. Find a counterexample, if possible to these universally quantified statements, in which the universe of discourse for all variables consists of all integers. (i) ∀x ∀y (x2 = y 2 → x = y). (ii) ∀x ∀y (xy ≥ x). Solution. (i) Suppose x = −3 and y = 3, then x2 = y 2 = 9. But x 6= y. ∴ ∀x ∀y (x2 = y 2 → x = y) is false. (ii) Suppose x = −3 and y = 3, then xy = −9. −9 < −3 =⇒ xy < x. ∴ ∀x ∀y(xy > x) is false. 5. Establish this logical equivalence, where A is a proposition not involving any quantifiers. Show that (∀xp(x)) ∧ A ≡ ∀x(p(x) ∧ A) and (∃x p(x)) ∧ A ≡ ∃x(p(x) ∧ A). Solution. (a) Consider (∀xp(x)) ∧ A ≡ ∀x(p(x) ∧ A).
(1.3)
26
Discrete Mathematical Structures Case (i): Suppose A is true. Since P ∧T = P , equation (1.3) =⇒ (∀xp(x)) ∧ T = (∀x)p(x). Therefore, both sides are the same, and hence equation (1.3) is logically equivalent. Case (ii): Suppose A is false. Since P ∧ F = F , the left-hand side of equation (1.3) is false. Also, for every x, p(x) ∧ A is false. Hence, right-hand side of equation (1.3) is false. Therefore, both sides are the same, and hence, it is logically equivalent. (b) Now, consider (∃xp(x) ∧ A) = ∃x(p(x) ∧ A).
(1.4)
Case (i): Suppose A is true. Since P ∧T = P , equation (1.4) becomes ∃xp(x) = (∃x)p(x), and hence, equation (1.4) is logically equivalent. Case (ii): Suppose A is false. Since P ∧ F = F , the left-hand side of equation (1.4) is false. Also, for every x, p(x) ∧ A is false. Therefore, ∃x(p(x) ∧ A) is false. Hence, the two sides of equation (1.4) are the same, and hence equation (1.4) is logically equivalent. 6. Show that ∃xP (x) ∧ ∃xQ(x) and ∃x(P (x) ∧ Q(x)) are not logically equivalent. Solution. We prove the two statements are not logically equivalent by using the following counterexample. Let us assume that the universe of discourse is the set of integers. Let P (x): x is a positive integer Q(x): x is a negative integer. Then, ∃xP (x) ∧ ∃xQ(x) is true. But the truth value of ∃x(P (x) ∧ Q(x)) is false. ∴ Given two statements are not logically equivalent. 7. Use quantifiers and predicate to express the fact that lim f (x) does x→a not exist. Solution. Limit exists means “f (x) approaches L (where L ∈ R) as x approaches a, if given > 0, ∃ δ > 0 such that |f (x) − L| < whenever 0 < |x − a| < δ 00 .
Logics and Proofs
27
Limit does not exist means “for all real numbers L, lim f (x) 6= L”. x→a The above statement can be expressed as ¬(∀ > 0, ∃ δ > 0, ∀x(0 < |x − a| < δ) → |f (x) − L| < )) =⇒ ∀ L, ∃ > 0, ∀ δ > 0, (∃ x)() < |x − a| < δ ∧ |f (x) − L| < ) [∵ ¬(p → q) = p ∧ ¬q]. 8. Let H = {−1, 0, 1, 2} denote the universe of discourse. If p(x, y) : x + y = 1, find the truth value of (∃ x)(∃ y) p(x, y). Solution. When When When When
x = −1, ∃y = 2 such that − 1 + 2 = 1. x = 0, ∃y = 1 such that 0 + 1 = 1. x = 1, ∃y = 0 such that 1 + 0 = 1. x = 2, ∃y = −1 such that 2 − 1 = 1. ∴ (∀x)(∃y) p(x, y) is true.
9. What are the negations of the statements ∀x(x2 > x) and ∃x(x2 = 2)? Solution. (i) Given statement is ∀x(x2 > x). Its negation is ∃x(x2 ≤ x). (ii) Given statement is ∃x(x2 = 2). Its negation is ∀x(x2 6= 2). 10. Write the negation of the statement ∃x (∀y) p(x, y). Solution. Given statement is (∃x) (∀y) p(x, y). Its negation is ¬((∃x) (∀y) p(x, y)) =⇒ (∀x) (∃y)¬p(x, y). 11. Consider the statement “Given any positive integer, there is a greater positive integer”. Symbolize this statement using and without using the set of positive integers as the universe of discourse. Solution. Let G(x, y) : x is greater than y. If we use the universe of discourse as the set of positive integers, then we can write (x) (∃y) G(y, x). If we do not impose the restriction on the universe of discourse and if we write P (x) : x is a positive integer, then we can write as (x) (P (x) → (∃y) (P (y) ∧ G(y, x))).
28
Discrete Mathematical Structures 12. Indicate free and bound variables. Also indicate the scope of the quantifier in (i) (x) (P (x) ∧ R(x)) ⇒ (x) (P (x)) ∧ Q(x)). (ii) ((x)(P (x) Q(x) ∧ ∃(x) R(x))) ∧ S(x). Solution. (i) All occurrences of x in P (x) ∧ R(x) are bound occurrences. The occurrence of x in xP (x) is bound. The occurrence of x in Q(x) is free. The scope of (x) is P (x) ∧ R(x) and P (x). (ii) All occurrences of x in P (x) Q(x) ∧ (∃ x) R(x) are bound, and the occurrence of x in S(x) is free. The scope of (x) is P (x) Q(x) ∧ (∃x) R(x), and the scope of (∃x) is R(x).
1.11.4
Inference Theory for Predicate Calculus
We have seen implication table and equivalence table already. Those rules can be extended here also. For example, P, P → Q ⇒ Q can be extended as P (x), P (x) → Q(x) ⇒ Q(x). In addition, we use the following rules: 1. Universal Specification (US): (x)A(x) ⇒ A(y) 2. Universal Generalization (UG): A(y) ⇒ (x)A(x) 3. Existential Specification (ES): (∃x)A(x) ⇒ A(y) 4. Existential Generalization (EG): A(y) ⇒ (∃x)A(x). Remark. Let the universe of discourse be denoted by S = {a1 , a2 , . . . , an }. Then, (x)A(x) = A(a1 ) ∧ A(a2 ) ∧ · · · ∧ A(an ) (∃x)A(x) = A(a1 ) ∨ A(a2 ) ∨ · · · ∨ A(an ). Consider now
Similarly,
1.11.5
¬(x)A(x) ⇔¬(A(a1 ) ∧ A(a2 ) ∧ · · · ∧ A(an )) ⇔¬A(a1 ) ∨ ¬A(a2 ) ∨ · · · ∨ ¬A(an ) ⇔(∃x)¬A(x).
¬((∃x)A(x)) ⇔ (x)¬A(x)).
Solved Problems
1. Show that (x)(P (x) → Q(x)) ∧ (x)(Q(x) → R(x)) ⇒ (x)(P (x) → R(x)).
Logics and Proofs
29
Solution. {1} {2} {3} {3} {1, 3}
(1) (2) (3) (4) (5)
(x)(P (x) → Q(x)) P (y) → Q(y) (x)(Q(x) → R(x)) Q(y) → R(y) P (y) → R(y)
{1, 3}
(6)
(x)(P (x) → R(x))
Rule P Rule U S Rule P Rule U S Rule T [∵ P → Q, Q → R ⇒ P → R] Rule U G.
2. Show that ∀x(P (x) ∨ Q(x)) ⇒ ∀x P (x) ∨ ∃x Q(x) using indirect method. Solution. We use the method of contradiction. Assume ¬((x)P (x)∨(∃x)Q(x)) as an additional premise. {1} {1} {1} {1} {1} {1} {1} {1} {9} {9} {1, 9}
(1) ¬((x)P (x) ∨ (∃x)Q(x)) (2) (∃x)¬P (x) ∧ (x)¬Q(x) (3) (∃x)¬P (x) (4) (x)¬Q(x) (5) ¬P (y) (6) ¬Q(y) (7) ¬P (y) ∧ ¬Q(y) (8) ¬(P (y) ∨ Q(y)) (9) (x)(P (x) ∨ Q(x)) (10) P (y) ∨ Q(y) (11) (P (y) ∨ Q(y))∧ ¬(P (y) ∨ Q(y))
Assumed premise Rule T [De Morgan’s law] Rule T [P ∧ Q ⇒ P ] Rule T [P ∧ Q ⇒ Q] Rule ES Rule U S Rule T [P, Q ⇒ P ∧ Q] Rule T [De Morgan’s law] Rule P Rule U S Rule T [P, Q ⇒ P ∧ Q]
which is false. Therefore, by the method of contradiction, we have ∀x(P (x) ∨ Q(x)) ⇒ ∀x P (x) ∨ ∃x Q(x). 3. Show that ∀xP (x) ∧ ∃xQ(x) is equivalent to ∀x∃y(P (x) ∧ Q(y)). Solution. {1}
(1)
∀xP (x) ∧ ∃xQ(x)
{1}
(2)
∀xP (x)
{1}
(3)
P (m)
{1}
(4)
∃xQ(x)
Rule P Rule T [P ∧ Q ⇒ P ] Rule U S Rule T [P ∧ Q ⇒ Q]
30
Discrete Mathematical Structures {1}
(5)
Q(n)
{1}
(6)
P (m) ∧ Q(n)
{1}
(7)
∃y(P (m) ∧ Q(y))
Rule EG
{1}
(8)
(∀x)∃y(P (x) ∧ Q(y))
Rule U G
Rule ES Rule T
Hence, ∀xP (x) ∧ ∃xQ(x) and ∀x∃y(P (x) ∧ Q(y)) are logically equivalent. 4. Show that (x)(P (x) → Q(x)) ⇒ (x)P (x) → (x)Q(x). Solution. We use contrapositive method to prove this problem. {1} (1)
¬((x)P (x) → (x)Q(x))
{1} (2)
(x)P (x) ∧ ¬((x)Q(x))
{1} (3)
(x)P (x)
Rule T [P ∧ Q ⇒ P ]
{1} (4)
¬((x)Q(x))
Rule T [P ∧ Q ⇒ Q]
{1} (5)
(∃x)¬Q(x)
Rule T [apply ¬]
Assumed premise Rule T [¬(P → Q) ⇔ P ∧ ¬Q]
{1} (6)
P (y)
Rule U S
{1} (7)
¬Q(y)
Rule ES
{1} (8)
P (y) ∧ ¬Q(y)
{1} (9)
¬(P (y) → Q(y))
{1} (10) {1} (11)
(∃x)¬(P (x) → Q(x)) ¬((x)(P (x) → Q(x)))
Rule T [P, Q ⇒ P ∧ Q] Rule T [P ∧ ¬Q ⇔ ¬(P → Q)] Rule EG Rule T [apply ¬].
∴ By the method of contrapositive, we have (x)(P (x) → Q(x)) ⇒ (x)P (x) → (x)Q(x). 5. Show that ∃x(P (x) ∧ Q(x)) ⇒ (∃x)P (x) ∧ (∃x)Q(x). Is the converse true? Solution. {1} {1} {1} {1} {1} {1} {1}
(1) (2) (3) (4) (5) (6) (7)
(∃x)(P (x) ∧ Q(x)) P (y) ∧ Q(y) P (y) Q(y) ∃xP (x) ∃xQ(x) ∃xP (x) ∧ ∃xQ(x)
Rule Rule Rule Rule Rule Rule Rule
P ES T [P ∧ Q ⇒ P ] T [P ∧ Q ⇒ Q] EG EG T [P, Q ⇒ P ∧ Q].
Logics and Proofs
31
Converse is also true, since {1} {1} {1} {1} {1} {1} {1}
(1) (2) (3) (4) (5) (6) (7)
(∃x) P (x) ∧ (∃x)Q(x) (∃x) P (x) (∃x)Q(x) P (y) Q(y) P (y) ∧ Q(y) (∃x)(P (x) ∧ Q(x))
Rule Rule Rule Rule Rule Rule Rule
P T [P ∧ Q ⇒ P ] T [P ∧ Q ⇒ Q] ES ES T [P, Q ⇒ P ∧ Q] EG.
6. Verify the validity of the following argument. Every living thing is a plant or an animal. John’s gold fish is alive, and it is not a plant. All animals have hearts. Therefore, John’s gold fish has a heart. Solution. Let L(x): x is a living thing P (x): x is a plant A(x): x is an animal H(x): x has a heart. Given premises with their symbolic forms are (i) Every living thing is a plant or an animal. (x)(L(x) → P (x) ∨ A(x)). (ii) John’s gold fish is alive, and it is not a plant. L(y) ∧ ¬P (y). (iii) All animals have hearts. (x)(A(x) → H(x)). Conclusion is H(y). {1} {1} {3} {4} {5} {1, 3}
(1) (2) (3) (4) (5) (6)
(x)(L(x) → P (x) ∨ A(x)) L(y) → P (y) ∨ A(y) L(y) ∧ ¬P (y) L(y) ¬P (y) P (y) ∨ A(y)
Rule Rule Rule Rule Rule Rule
P P P T [P ∧ Q ⇒ P ] T [P ∧ Q ⇒ Q] T [P, P → Q ⇒ Q]
32
Discrete Mathematical Structures
{1, 3} {8} {8} {1, 3, 8}
(7) (8) (9) (10)
¬P (y) → A(y) (x)(A(x) → H(x)) A(y) → H(y) ¬P (y) → H(y)
{1, 3, 8}
(11)
H(y)
Rule T [P → Q ⇒ ¬P ∨ Q] Rule P Rule U S Rule T [P → Q, Q → R ⇒ P → R] Rule T [P → Q, P ⇒ Q].
Hence, the given statements are valid statements.
1.12
1. Let P (x) denote the statement x ≤ 4. Write the truth values of P (2) and P (4). Solution. P (x) : x ≤ 4 P (2) : 2 ≤ 4 is True P (6) : 6 ≤ 4 is False. 2. Give the converse and the contrapositive of the implication: “If it is raining, then I get wet”. Solution. Let P : It is raining Q: I get wet. Given statement is P → Q. Its inverse is Q → P : If I get wet, then it is raining. Its contrapositive is ¬Q → ¬P : If I do not get wet, then it is not raining. 3. Show that R ∨ S follows logically from the premises C ∨ D, C ∨ D → ¬H, ¬H → (A ∧ ¬B), and (A ∧ ¬B) → (R ∨ S). Solution. {1} {2} {1, 2} {4} {1, 2, 4}
(1) C ∨ D → ¬H (2) ¬H → (A ∧ ¬B) (3) C ∨ D → A ∧ ¬B (4) (A ∧ ¬B) → (R ∨ S) (5) C ∨D →R∨S
Rule Rule Rule Rule Rule
P P T P T
Logics and Proofs
33
{6} {1, 2, 4, 6}
C ∨D R∨S
(6) (7)
Rule P Rule T [P → Q, P ⇒ Q].
4. Show that the following premises are inconsistent. (i) If Jack misses many classes through illness, then he fails high school. (ii) If Jack fails high school, then he is uneducated. (iii) If Jack reads a lot of books, then he is not uneducated. (iv) Jack misses many classes through illness and reads a lot of books. Solution. E : Jack S : Jack A : Jack H : Jack
misses many classes fails high school reads a lot of books is uneducated
The given premises are E → S, S → H, A → ¬H, and E ∧ A. {1} {2} {1, 2} {4} {4} {1, 2, 4} {1, 2, 4} {1, 2, 4} {9} {1, 2, 4, 9}
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
E→S S→H E→H A → ¬H H → ¬A E → ¬A ¬E ∧ ¬A ¬(E ∧ A) E∧A (E ∧ A) ∧ (¬(E ∧ A))
Rule Rule Rule Rule Rule Rule Rule Rule Rule Rule
which is false. Hence, the given premises are inconsistent. 5. Write the dual of (P ∧ Q) ∨ T . Solution. (P ∨ Q) ∧ F . 6. Negate the following statements. (i) Ottawa is a small town. (ii) Every city in Canada is clean.
P P T P T T T T P T
34
Discrete Mathematical Structures Solution. (i) Ottawa is not a small town. (ii) Every city in Canada is not clean. 7. Construct the truth table for P ∧ (P ∨ Q). Solution. The truth table is shown below. Truth Table for P ∧ (P ∨ Q) P Q P ∨ Q P ∧ (P ∨ Q) T T T T T F T T F T T F F F F F 8. Write the following in symbolic form: If John takes calculus or Peter takes analytical geometry, then Mohan will take English. Solution. Let J: John takes calculus P : Peter takes analytical geometry M : Mohan will take English. Then the symbolic form is (J ∨ P ) → M . 9. Symbolise the expression: “All the world loves a lover”. Solution. Let P (x): x is a person L(x): x is a lover R(x, y): x loves y. The symbolic form is (x)(P (x) → (y)(P (y) ∧ L(y) → R(x, y)). 10. Obtain the PDNF and PCNF of P → ((P → Q) ∧ ¬(¬Q ∨ ¬P )). Solution. Let
A ⇔ P → ((P → Q) ∧ ¬(¬Q ∨ ¬P )) ⇔ ¬P ∨ ((¬P ∨ Q) ∧ (Q ∧ P )) ↔ ¬P ∨ ((¬P ∧ (Q ∧ P )) ∨ (Q ∧ (Q ∧ P )) ⇔ ¬P ∨ F ∨ (P ∧ Q) ⇔ ¬P ∨ (P ∧ Q)
Logics and Proofs
35 ⇔ ∨(P ∧ Q) ⇔ (¬P ∧ (Q ∨ ¬Q)) ∨ (P ∧ Q) ⇔ (¬P ∧ Q) ∨ (¬P ∧ ¬Q) ∨ (P ∧ Q)
which is the required PDNF. Its PCNF is ¬¬A ⇔ ¬P ∨ Q. 11. Obtain the PCNF of S : (¬P → R) ∧ (Q P ). Hence, obtain PDNF. Solution. S : (¬P → R) ∧ (Q P ) ⇔ (¬P → R) ∧ ((Q → P ) ∧ (P → Q)) ⇔ (P ∨ R) ∧ (¬Q ∨ P ) ∧ (¬P ∨ Q) ⇔ ((P ∨ R) ∨ F ) ∧ ((¬Q ∨ P ) ∨ F ) ∧ ((¬P ∨ Q) ∨ F ) ⇔ ((P ∨ R) ∨ (Q ∧ ¬Q)) ∧ ((¬Q ∨ P ) ∨ (R ∧ ¬R)) ∧ ((¬P ∨ Q) ∨ (R ∧ ¬R)) ⇔ (P ∨ Q ∨ R) ∧ (P ∨ ¬Q ∨ R) ∧ (P ∨ ¬Q ∨ R) ∧ (P ∨ ¬Q ∨ ¬R) ∧ (¬P ∨ Q ∨ R) ∧ (¬P ∨ Q ∨ ¬R) ⇔ (P ∨ Q ∨ R) ∧ (P ∨ ¬Q ∨ R) ∧ (P ∨ ¬Q ∨ ¬R) ∧ (¬P ∨ Q ∨ R) ∧ (¬P ∨ Q ∨ ¬R) which is the required PCNF. Now, ¬S : (P ∨ Q¬R) ∧ (¬P ∨ ¬Q ∨ R) ∧ (¬P ∨ ¬Q¬R) ¬¬S :¬((P ∨ Q¬R) ∧ (¬P ∨ ¬Q ∨ R) ∧ (¬P ∨ ¬Q¬R)) ⇔ (¬P ∧ ¬Q ∧ R) ∨ (P ∧ Q ∧ R) ∨ (P ∧ Q ∧ R) which is the required PDNF. 12. What is the contrapositive statement of “The home town wins whenever it is raining”? Solution. Let P : It is raining Q: The home town wins. Given statement is P → Q. Its contrapositive statement is ¬Q → ¬P : If the home town does not win, then it is not raining.
36
Discrete Mathematical Structures 13. Give the symbolic form of “Some men are giants”. Solution. The given statement can be written as “there is an x such that x is a man and x is giant”. Let M (x): x is a man G(x): x is a giant. ∴ The symbolic form is (∃x)(M (x) ∧ G(x)). 14. Find the PCNF of (P ∨ R) ∧ (P ∨ ¬Q). Also find its PDNF, without using truth table. Solution. Let A ⇔ (P ∨ R) ∧ (P ∨ ¬Q) ⇔ ((P ∨ R) ∨ F ) ∧ ((P ∨ ¬Q) ∨ F ) ⇔ ((P ∨ R) ∨ (Q ∧ ¬Q)) ∧ ((P ∨ ¬Q) ∨ (R ∧ ¬R)) ⇔ (P ∨ Q ∨ R) ∧ (P ∨ ¬Q ∨ R) ∧ (P ∨ ¬Q ∨ R) ∧ (P ∨ ¬Q ∧ ¬R) ⇔ (P ∨ Q ∨ R) ∧ (P ∨ ¬Q ∨ R) ∧ (P ∨ ¬Q ∨ ¬R) which is the required PCNF. Now, ¬A : (P ∨ Q ∨ ¬R) ∧ (¬P ∨ Q ∨ R) ∧ (¬P ∨ Q ∨ ¬R) ∧ (¬P ∨ ¬Q ∨ R) ∧ (¬P ∨ ¬Q ∨ ¬R) ¬¬A :¬ (P ∨ Q ∨ ¬R) ∧ (¬P ∨ Q ∨ R) ∧ (¬P ∨ Q ∨ ¬R) ∧ (¬P ∨ ¬Q ∨ R) ∧ (¬P ∨ ¬Q ∨ ¬R) ⇔ (¬P ∧ ¬Q ∧ R) ∨ (P ∧ ¬Q ∧ ¬R) ∨ (P ∧ ¬Q ∧ R) ∨ (P ∧ Q ∧ ¬R) ∨ (P ∧ Q ∧ R) which is the required PDNF. 15. Show that using rule CP, ¬P ∨ Q, ¬Q ∨ R, R → S ⇔ P → S. Solution. {1} {2} {2} {1, 2} {5} {5}
(1) (2) (3) (4) (5) (6)
P ¬P ∨ Q P →Q Q ¬Q ∨ R Q→R
Assumed premise Rule P Rule T [P → Q ⇔ ¬P ∨ Q] Rule T [P, P → Q ⇒ Q] Rule P Rule T [P → Q ⇔ ¬P ∨ Q]
Logics and Proofs {1, 2, 5} {8} {1, 2, 5, 8} {1, 2, 5, 8}
37 (7) (8) (9) (10)
R R→S S P →S
Rule Rule Rule Rule
T [P, P → Q ⇒ Q] P T [P, P → Q ⇒ Q] CP.
16. Show that (¬P ∧ (¬Q ∧ R)) ∨ (Q ∧ R) ∨ (P ∧ R) ⇔ R without using truth table. Solution. ¬P ∧ (¬Q ∧ R) ⇔ (¬P ∧ ¬Q) ∧ R (Associative law) ⇔ ¬(P ∨ Q) ∧ R (De Morgan’s law).
(1.5)
Now, (Q ∧ R) ∨ (P ∧ R) ⇔ (Q ∨ P ) ∧ R ⇔ (P ∨ Q) ∧ R
(Distributive law) (Commutative law).
From (1.5) and (1.6), we have (¬P ∧ (¬Q ∧ R)) ∨ (Q ∧ R) ∨ (P ∧ R) ⇔ (¬(P ∨ Q) ∧ R) ∨ ((P ∨ Q) ∧ R) ⇔ (¬(P ∨ Q) ∨ (P ∨ Q)) ∧ R (Distributive law) ⇔T ∧R ⇔ R. ∴ (¬P ∧ (¬Q ∧ R)) ∨ (Q ∧ R) ∨ (P ∧ R) ⇔ R.
(1.6)
2 Combinatorics
2.1
Introduction
In this chapter, we discuss about the technique of mathematical induction which is used for proving many standard results over natural numbers. Then, we discuss about the basis of counting, pigeonhole principle, permutations and combinations, and recurrence relation. These concepts are useful in the analysis of certain discrete time systems, analysis of algorithms, error-correcting code, etc. At the end of the chapter, we discuss about generating function which is used to solve linear recurrence relations.
2.2
Mathematical Induction
Mathematical induction is a method of finding the truth from a general statement for particular cases. A statement may be true with reference to more than hundred cases yet we cannot conclude it to be true in general. It is possible to disprove the statement by a counter-example. A statement need not be accepted to be true. Such a statement inferred from a particular case is called a conjecture. If the conjecture is a statement involving natural numbers, we can use the principle of the mathematical induction to prove the same.
2.2.1
Principle of Mathematical Induction
For a given statement involving a natural number n, if we can show that (1) The statement is true for n = 1 or n = n0 , (2) The statement is true for n = m + 1 under the assumption that the statement is true for n = m(m ≥ n0 ), then we can conclude that the statement is true for all natural numbers. Remark 2.2.1 In the above principle, (1) is usually referred to as the basis of induction, and (2) is usually referred to as the induction step. Also, the assumption that the statement is true for n = m in (2) is usually referred to as the induction hypothesis. 39
40
Discrete Mathematical Structures
2.2.2
Procedure to Prove that a Statement P (n) is True for all Natural Numbers
Step 1. We must prove that P (1) is true. Step 2. By assuming P (m) is true, we must prove that P (m + 1) is also true. In the sequel, we apply the principle of mathematical induction to prove statements involving natural numbers.
2.2.3
Solved Problems
1. Prove that 1 + 2 + 3 + · · · + n = Solution. Let P (n) : 1 + 2 + 3 + · · · + n =
n(n + 1) by induction principle. 2 n(n + 1) . 2
1(1 + 1) = 1 is true. 2 (2) Assume that P (m) is true. m(m + 1) That is, 1 + 2 + 3 + · · · + m = . 2 (3) Now, (1) P (1) : 1 =
m(m + 1) +m+1 2 m(m + 1) + 2(m + 1) = 2 (m + 1)(m + 2) . = 2
1 + 2 + 3 + · · · + m + (m + 1) =
∴ By mathematical induction, the given statement is true for all n. n(n + 1)(2n + 1) 2. Show that 12 + 22 + 32 + · · · + n2 = , n ≥ 1 by 6 mathematical induction. Solution. Let P (n) : 12 + 22 + 32 + · · · + n2 =
n(n + 1)(2n + 1) . 6
1(1 + 1)(2 + 1) = 1 is true. 6 (2) Assume P (m) is true. m(m + 1)(2m + 1) That is, 12 + 22 + 32 + · · · + m2 = . 6 (3) Now, (1) P (1) : 12 =
12 + 22 + 32 + · · · + m2 + (m + 1)2
Combinatorics
41 = = = = = = = = =
m(m + 1)(2m + 1) + (m + 1)2 6 m(m + 1)(2m + 1) + 6(m + 1)2 6 (m + 1)[m(2m + 1) + 6(m + 1)] 6 (m + 1)(2m2 + m + 6m + 6) 6 (m + 1)(2m2 + 7m + 6) 6 (m + 1)(2m2 + 4m + 3m + 6) 6 (m + 1)[2m(m + 2) + 3(m + 2)] 6 (m + 1)(m + 2)(m + 3) 6 (m + 1)[(m + 1) + 1][2(m + 1) + 1] . 6
∴ By mathematical induction, the given statement is true for all n ≥ 1. n2 (n + 1)2 3. Prove that 13 + 23 + 33 + · · · + n3 = , n ∈ N. 4 Solution. n2 (n + 1)2 Let P (n) : 13 + 23 + 33 + · · · + n3 = . 4 12 (1 + 1)2 (1) P (1) : 12 = is true. 4 (2) Assume P (m) is true. m2 (m + 1)2 . That is, 13 + 23 + 33 + · · · + m3 = 4 (3) Now, m2 (m + 1)2 13 + 23 + 33 + · · · + m3 + (m + 1)3 = + (m + 1)3 4 m2 (m + 1)2 + 4(m + 1)3 = 4 (m + 1)2 [m2 + 4(m + 1)] = 4 (m + 1)2 (m2 + 4m + 4) = 4 (m + 1)2 (m + 2)2 = 4 (m + 1)[(m + 1) + 1]2 = . 4
42
Discrete Mathematical Structures ∴ By mathematical induction, the statement is true for all n ∈ N. Pn 4. Prove that i=1 (2i − 1)2 = n2 , for all n ∈ N. (or) Prove that the sum of the first n odd integers is n2 for all integers n. Solution. Let P (n) : 1 + 3 + 5 + · · · + (2n − 1) = n2 . (1) P (1) : 1 = 12 is true. (2) Assume P (m) is true. That is, 1 + 3 + 5 + · · · + (2m − 1) = m2 . (3) Now, 1 + 3 + 5 + · · · + (2m − 1) + [2(m + 1) − 1] = m2 + [2(m + 1) − 1] = m2 + [2(m + 1) − 1] = m2 + 2m + 2 − 1 = m2 + 2m + 1 = (m + 1)2 . ∴ By mathematical induction, the given statement is true for all n ∈ N. (or) (1) P (1) : 1 = 12 is true. (2) Assume P (m) is true. Pm That is, i=1 (2i − 1) = m2 . (3) Now, m+1 X
m X (2i − 1) = (2i − 1) + [2(m + 1) − 1]
i=1
i=1
= m2 + 2m + 2 − 1 = m2 + 2m + 1 = (m + 1)2 . ∴ By mathematical induction, the given statement is true for all n ∈ N. 5. Using mathematical induction, prove that 2 + 5 + 8 + · · · + (3n − 1) = Solution. Let P (n) : 2 + 5 + 8 + · · · + (3n − 1) =
n(3n + 1) . 2
n(3n + 1) . 2
Combinatorics
43
1[3 · 1 + 1] = 2 is true. 2 (2) Assume P (m) is true.
(1) P (1) : 2 =
That is, 2 + 5 + 8 + · · · + (3m − 1) =
m(3m + 1) . 2
(3) Now, 2 + 5 + 8 + · · · + (3m − 1) + [3(m + 1) − 1] = = = = = = = =
m(3m + 1) + (3m + 3 − 1) 2 m(3m + 1) + 2(3m + 2) 2 3m2 + m + 6m + 4 2 3m2 + 7m + 4 2 3m2 + 3m + 4m + 4 2 3m(m + 1) + 4(m + 1) 2 (m + 1)(3m + 4) 2 (m + 1)[3(m + 1) + 1] . 2
∴ By mathematical induction, the given statement is true. 6. Prove that for n ≥ 0, 1 + 2 + 4 + · · · + 2n = 2n+1 − 1. Solution. Let P (n) : 1 + 2 + 4 + · · · + 2n = 2n+1 − 1. (1) P (1) : 1 + 2 = 21+1 − 1 is true. (2) Assume P (m) is true. That is, 1 + 2 + 4 + · · · + 2m = 2m+1 − 1. (3) Now, 1 + 2 + 4 + · · · + 2m + 2m+1 = (2m+1 − 1) + 2m+1 = 2 · 2m+1 − 1 = 2m+2 − 1 = 2(m+1)+1 − 1. ∴ By mathematical induction, the given statement is true. 7. Prove that if n ≥ 1, then 1(1!) + 2(2!) + · · · + n(n!) = (n + 1)! − 1.
44
Discrete Mathematical Structures Solution. Let P (n) : 1(1!) + 2(2!) + · · · + n(n!) = (n + 1)! − 1. (1) P (1) : 1(1!) = (1 + 1)! − 1 is true. (2) Assume P (m) is true. That is, 1(1!) + 2(2!) + · · · + m(m!) = (m + 1)! − 1. (3) Now, 1(1!) + 2(2!) + · · · + m(m!) + (m + 1)[(m + 1)!] = [(m + 1)! − 1] + (m + 1)[(m + 1)!] = (m + 1)! + (m + 1)(m + 1)! − 1 = (m + 1)![1 + (m + 1)] − 1 = (m + 1)!(m + 2) − 1 = (m + 2)! − 1. ∴ By mathematical induction, the given statement is true for all n ≥ 1. 8. Use mathematical induction to show that 1 1 1 n 1 + + + ··· + = , 1·2 2·3 3·4 n(n + 1) n+1
for all n ≥ 1.
Solution. Let P (n) :
1 1 1 1 n + + + ··· + = . 1·2 2·3 3·4 n(n + 1) n+1 1 1 (1) P (1) : = is true. 1·2 1+1 (2) Assume P (m) is true. 1 1 1 1 m That is, + + + ··· + = . 1·2 2·3 3·4 m(m + 1) m+1 (3) Now, 1 1 1 1 1 + + + ··· + + 1·2 2·3 3·4 m(m + 1) (m + 1)(m + 2) m 1 = + m + 1 (m + 1)(m + 2) m(m + 2) + 1 = (m + 1)(m + 2) m2 + 2m + 1 = (m + 1)(m + 2) (m + 1)2 = (m + 1)(m + 2) m+1 = . (m + 1) + 1 ∴ By mathematical induction, the given statement is true for all n.
Combinatorics
45
9. Use mathematical induction to prove that n3 − n is divisible by 3 whenever n is a positive integer. Solution. Let P (n) : n3 − n is divisible by 3. (1) P (1) : 11 − 1 = 0 is divisible by 3. (2) Assume P (m) is true. That is, m3 − m is divisible by 3. (3) Now, (m + 1)3 − (m + 1) = (m3 + 3m2 + 3m + 1) − (m + 1) = (m3 − m) + 3m2 + 3m = (m3 − m) + 3(m2 + m). Since both terms in this sum are divisible by 3, it follows that (m + 1)3 − (m + 1) is also divisible by 3. ∴ By mathematical induction, n3 − n is divisible by 3 for all n ∈ N. 10. Use mathematical induction to show that n3 + 2n is divisible by 3 whenever n is a non-negative integer. (or) 3
Prove that for all n ≥ 1, n + 2n is a multiple of 3. Solution. Let P (n) : n3 + 2n is a multiple of 3. (1) P (1) : 13 + 2 · 1 = 3 is a multiple of 3. (2) Assume P (m) is true. That is, m3 + 2m is a multiple of 3. (3) Now, (m + 1)3 + 2(m + 1) = m3 + 3m2 + 3m + 1 + 2m + 2 = (m3 + 2m) + 3(m2 + m + 1). Since both terms on the right-hand side are divisible by 3, it follows that (m + 1)3 + 2(m + 1) is a multiple of 3. Hence, by mathematical induction, n3 + 2n is a multiple of 3 for all n ∈ N.
46
Discrete Mathematical Structures 11. Use mathematical induction to show that 8n − 3n is a multiple of 5. Solution. Let P (n) : 8n − 3n is a multiple of 5. (1) P (1) : 8 − 3 = 5 is a multiple of 5. (2) Assume P (m) is true. That is, 8m − 3m is a multiple of 5. (3) Now, 8m+1 − 3m+1 = 8m (8) − 3m+1 = 8m (5 + 3) − 3m+1 = 5 · 8m + 3 · 8m − 3m+1 = 5 · 8m + 3(8m − 3m ). Since both terms on the right-hand side are multiples of 5, it follows that 8m+1 − 3m+1 is also a multiple of 5. ∴ By mathematical induction, 8n − 3n is a multiple of 5. 12. Use mathematical induction to show that n2 − 7n + 12 is non-negative whenever n is an integer greater than 3. Solution. Let P (n) : n2 − 7n + 12 be non-negative. (1) P (4) : 42 − 7(4) + 12 = 16 − 28 + 12 = 0, which is non-negative. (2) Assume P (m) is true. That is, m2 − 7m + 12 is non-negative. (3) Now, (m + 1)2 − 7(m + 1) + 12 = m2 + 2m + 1 − 7m − 7 + 12 = (m2 − 7m + 12) + (2m − 6). The first term is non-negative, and the second term is also non-negative for m > 3. Hence, (m + 1)2 − 7(m + 1) + 12 is non-negative. ∴ By mathematical induction, n2 − 7n + 12 is non-negative for n > 3. 13. Use mathematical induction to prove the inequality n < 2n for all positive integers n. Solution. Let P (n) : n < 2n . (1) P (1) : 1 < 21 is true. (2) Assume P (m) is true. That is, m < 2m .
Combinatorics
47
(3) Now, m + 1 < 2m + 1 < 2m + 2m < 2 · 2m
since 1 < 2m
< 2m+1 . ∴ By mathematical induction, the result is true for all n ∈ N. 14. Prove that n + 10 ≤ 2n for all n ∈ N and n ≥ 4. Solution. Let P (n) : n + 10 ≤ 2n , n ≥ 4. (1) P (4) : 4 + 10 ≤ 24 is true. (2) Assume P (m) is true. That is, m + 10 ≤ 2m . (3) Now, (m + 1) + 10 = (m + 10) + 1 ≤ 2m + 1 ≤ 2m + 2m since 1 < 2m ≤ 2 · 2m ≤ 2m+1 . ∴ By the principle of mathematical induction, the result is true for all n ∈ N and n ≥ 4. Note: In the above problem, the result is false for n = 1, 2 and 3. 15. Prove that 2n < n! for n ≥ 4 and n ∈ N. Solution. Let P (n) : 2n < n! for n ≥ 4. (1) P (4) : 24 < 4! is true (since 16 < 24). (2) Assume P (m) is true. That is, 2m < m! (3) Now, 2m+1 = 2m · 2 < 2m (m + 1) since 2 < m + 1 < m!(m + 1) by assumption < (m + 1)!
for m ≥ 4
∴ The result is true for all n ≥ 4 by the principle of mathematical induction.
48
Discrete Mathematical Structures 16. Suppose there are n people in a room (n ≥ 1) and all shake (n − 1)n hands with one another. Prove that handshakes will have 2 occurred. Solution.
(n − 1)n handshakes. 2 (1 − 1)1 = 0. There is no handshake when n = 1. (1) P (1) : 2 (2) Assume P (m) is true. (m − 1)m That is, there are handshakes. 2 (3) Now, if one more person enters the room, he will shake hands with m people. So, Let P (n) :
(m − 1)m m2 − m + 2m +m= 2 2 2 m +m = 2 m(m + 1) = 2 [(m + 1) − 1](m + 1) = . 2 ∴ By the principle of mathematical induction, the result follows. 17. Use mathematical induction to prove that 3n + 7n − 2 is divisible by 8, for n ≥ 1. Solution. Let P (n) : 3n + 7n − 2 is divisible by 8. (1) P (1) : 31 + 71 − 2 = 8 is divisible by 8, which is true. (2) Assume P (m) is true. That is, 3m + 7m − 2 is divisible by 8. (3) Now, 3m+1 + 7m+1 − 2 = 3 · 3m + 7 · 7m − 2 = 3 · 3m + (3 + 4) · 7m − 2 = 3 · 3m + 3 · 7m + 4 · 7m − 6 + 4 = 3(3m + 7m − 2) + 4(7m + 1). Since both terms on right-hand side are divisible by 8, 3m+1 + 7m+1 − 2 is also divisible by 8. ∴ By the principle of mathematical induction, the result follows.
Combinatorics
49
18. Show that an − bn is divisible by a − b. Solution. Let P (n) : an − bn be divisible by a − b. (1) P (1) : a1 − b1 is divisible by a − b. (2) Assume P (m) is true. That is, am − bm is divisible by a − b. ⇒
am − bm = k(a − b)
am = bm + k(a − b).
(2.1)
(3) Now, am+1 − bm+1 = am · a − bm · b [using (2.1)] = ak(a − b) + abm − b · bm = (a − b)ak + (a − b)bm = (a − b)[ak + bm ], which is a multiple of (a − b). Hence, am+1 − bm+1 is divisible by a − b. ∴ By the principle of mathematical induction, an − bn is divisible by a − b for all n ≥ 1. 19. Using mathematical induction, prove that 2 + 22 + 23 + · · · + 2n = 2n+1 − 2. Solution. Let P (n) : 2 + 22 + 23 + · · · + 2n = 2n+1 − 2. (1) P (1) : 21 = 21+1 − 2 = 2 is true. (2) Assume P (m) is true. That is, 2 + 22 + 23 + · · · + 2m = 2m+1 − 2 is true. (3) Now, 2 + 22 + 23 + · · · + 2m + 2m+1 = 2m+1 − 2 + 2m+1 = 2 · 2m+1 − 2 = 2m+2 − 2. ∴ By the principle of mathematical induction, the result is true for all n.
50
Discrete Mathematical Structures 20. Use mathematical induction to show that n! ≥ 2n+1 ; n = 5, 6, . . . Solution. Let P (n) : n! ≥ 2n+1 ; n = 5, 6 . . . (1) P (5) : 5! ≥ 25+1 is true since 120 ≥ 64. (2) Assume P (m) is true. m! ≥ 2m+1 ; m = 5, 6 . . .
That is,
(2.2)
(3) Multiplying both sides of (2.2) by 2, we have 2(m!) ≥ 2 · 2m+1 =⇒ =⇒
(m + 1)m! ≥ 2m+2 (m + 1)! ≥ 2m+2 .
∴ By mathematical induction, n! ≥ 2n+1 ; n = 5, 6, . . . 21. Show that
1 1 n 1 + + ··· + = . 1·2 2·3 n(n + 1) n+1
Solution. Let P (n) :
1 1 1 n + + ··· + = . 1·2 2·3 n(n + 1) n+1 1 1 = is true. (1) P (1) : 1·2 1+1 (2) Assume P (m) is true. 1 1 1 m That is, + + ··· + = . 1·2 2·3 m(m + 1) m+1 (3) Now, 1 1 1 1 + + ··· + + 1·2 2·3 m(m + 1) (m + 1)(m + 2) m 1 = + m + 1 (m + 1)(m + 2) m(m + 2) + 1 = (m + 1)(m + 2) m2 + 2m + 1 = (m + 1)(m + 2) (m + 1)2 = (m + 1)(m + 2) m+1 = m+2 m+1 = . (m + 1) + 1 ∴ By the principle of mathematical induction, the result follows.
Combinatorics
51
22. Using mathematical induction, prove that if n ≥ 1, then 1 · 1! + 2 · 2! + 3 · 3! + · · · + n · n! = (n + 1)! − 1,
n ≥ 1.
Solution. Let P (n) : 1 · 1! + 2 · 2! + 3 · 3! + · · · + n · n! = (n + 1)! − 1. (1) P (1) : 1 · 1! = (1 + 1)! − 1 is true. (2) Assume P (m) is true. That is, 1 · 1! + 2 · 2! + 3 · 3! + · · · + m · m! = (m + 1)! − 1. (3) Now, 1 · 1! + 2 · 2! + 3 · 3! + · · · + m · m! + (m + 1)(m + 1)! = [(m + 1)! − 1] + (m + 1)(m + 1)! = (m + 1)![1 + m + 1] − 1 = (m + 1)!(m + 2) − 1 = (m + 2)! − 1. ∴ By mathematical induction, 1 · 1! + 2 · 2! + 3 · 3! + · · · + n · n! = (n + 1)! − 1, 23. Using mathematical induction, prove that Solution. Let P (n) : 30 + 31 + · · · + 3n =
Pn
k=0
3k =
n ≥ 1. 3n+1 − 1 . 2
3n+1 − 1 . 2
31 − 1 2 = is true. 2 2 (2) Assume P (m) is true. 3m+1 − 1 That is, 30 + 31 + · · · + 3m = . 2 (3) Now, (1) P (0) : 30 =
30 + 31 + · · · + 3m + 3m+1 = = = = = ∴ By mathematical induction,
Pn
k=0
3m+1 − 1 + 3m+1 2 3m+1 − 1 + 2 · 3m+1 2 3 · 3m+1 − 1 2 3m+2 − 1 2 3(m+1)+1 . 2
3k =
3n+1 − 1 is true. 2
52
Discrete Mathematical Structures 24. Using mathematical induction, prove that √ 1 1 1 1 √ + √ + √ + · · · + √ > n for n ≥ 2. n 1 2 3 Solution. √ 1 1 1 1 Let P (n) : √ + √ + √ + · · · + √ > n for n ≥ 2. n 1 2 3 √ 1 1 (1) P (2) : √ + √ = 1.707 > 2 = 1.414 is true. 1 2 (2) Assume P (m) is true. √ 1 1 1 1 That is, √ + √ + √ + · · · + √ > m for m ≥ 2. m 1 2 3 (3) Now, 1 1 1 1 1 √ + √ + √ + ··· + √ + √ m m+1 1 2 3 √ 1 > m+ √ m+1 √ √ m m+1+1 √ > m+1 p m(m + 1) + 1 √ > m+1 √ m·m+1 > √ (∵ m + 1 > m) m+1 √ m2 + 1 > √ m+1 m+1 >√ m+1 √ > m + 1. ∴ By the principle of mathematical induction, √ 1 1 1 1 √ + √ + √ + · · · + √ > n for n ≥ 2. n 1 2 3 25. Using mathematical induction, prove that n3 + (n + 1)3 + (n + 2)3 is divisible by 9, for n ≥ 1. Solution. Let P (n) : n3 + (n + 1)3 + (n + 2)3 be divisible by 9, for n ≥ 1. (1) P (1) : 13 + (1 + 1)3 + (1 + 2)3 = 1 + 8 + 27 = 36 is divisible by 9.
Combinatorics
53
(2) Assume P (m) is true. That is, m3 + (m + 1)3 + (m + 2)3 is divisible by 9, for m ≥ 1. (3) Now, (m + 1)3 + [(m + 1) + 1]3 + [(m + 1) + 2]3 = (m + 1)3 + (m + 2)3 + (m + 3)3 = (m + 1)3 + (m + 2)3 + m3 + 9m2 + 27m + 27 [m3 + (m + 1)3 + (m + 2)3 ] + 9(m2 + 3m + 3). Both the terms on the right-hand side are divisible by 9, and hence, the terms on the left-hand side are also divisible by 9. ∴ By the principle of mathematical induction, the result follows. 26. Show that 32n + 4n+1 is divisible by 5, for n ≥ 0. Solution. Let P (n) : 32n + 4n+1 be divisible by 5, for n ≥ 0. (1) P (0) : 30 + 41 = 5 is divisible by 5. (2) Assume P (m) is true. 32m + 4m+1 is divisible by 5.
That is, ⇒ ⇒
32m + 4m+1 = 5k 2m
3
m+1
= 5k − 4
(where k is an integer) .
(2.3)
(3) Now, 32(m+1) + 4m+1+1 = 32m · 32 + 4m+2 = (5k − 4m+1 ) · 32 + 4m+1 · 4 = 5k · 9 − 4
m+1
= 5k · 9 − 4
m+1
2
·3 +4
m+1
·9+4·4
[using (2.3)]
·4
m+1
m+1
= 5k · 9 − 5 · 4
= 5(9k − 4m+1 )
which is a multiple of 5.
∴ By the principle of mathematical induction, 32n + 4n+1 is divisible by 5 for n ≥ 0. 27. Using mathematical induction, prove that H2n ≥ 1 +
n 2
where Hk = 1 +
1 1 1 + + ··· + . 2 3 k
54
Discrete Mathematical Structures Solution. 1 1 Let P (n) : H2n = 1 + + + · · · + 2 3 0 (1) P (0) : H20 = H1 = 1 ≥ 1 + . 2 That is, 1 ≥ 1 is true. (2) Assume P (m) is true. 1 That is, P (m) : H2m = 1 + + 2 (3) Now, H2m+1 = 1 +
n 1 ≥1+ . 2n 2
1 1 m + ··· + m ≥ 1 + . 3 2 2
1 1 1 1 1 1 + + ··· + m + m + + · · · + m+1 2 3 2 2 + 1 2m + 2 2
= H2m +
2m
1 1 1 + m + · · · + m+1 +1 2 +2 2
m 1 1 1 ≥ 1+ + m + m + · · · + m+1 2 2 +1 2 +2 2 1 m + 2m · m+1 since there are 2m terms ≥ 1+ 2 2 1 each not less than m+1 2 m 1 ≥ 1+ + 2 2 m+1 ≥1+ . 2 ∴ By the principle of mathematical induction, H2n ≥ 1 +
n . 2
28. Using mathematical induction, prove that 12 + 32 + 52 + · · · + (2n − 1)2 =
n(2n − 1)(2n + 1) . 3
Solution. Let P (n) : 12 + 32 + 52 + · · · + (2n − 1)2 =
n(2n − 1)(2n + 1) . 3
1(2 − 1)(2 + 1) = 1 is true. 3 (2) Assume P (m) is true.
(1) P (1) : 12 =
That is, P (n) : 12 + 32 + 52 + · · · + (2m − 1)2 =
m(2m − 1)(2m + 1) . 3
Combinatorics
55
(3) Now, 12 + 32 + 52 + · · · + (2m − 1)2 + [(2(m + 1) − 1]2 = = = = = = = = = =
m(2m − 1)(2m + 1) + (2m + 1)2 3 1 [m(2m − 1)(2m + 1) + 3(2m + 1)]2 3 2m + 1 [m(2m − 1) + 3(2m + 1)] 3 2m + 1 [2m2 − m + 6m + 3] 3 2m + 1 (2m2 + 5m + 3) 3 2m + 1 (2m2 + 2m + 3m + 3) 3 2m + 1 [2m(m + 1) + 3(m + 1)] 3 2m + 1 (2m + 3)(m + 1) 3 1 (m + 1)(2m + 1)(2m + 3) 3 1 (m + 1)[2(m + 1) − 1][2(m + 1) + 1]. 3
∴ By the principle of mathematical induction, 12 + 32 + 52 + · · · + (2n − 1)2 =
n(2n − 1)(2n + 1) . 3
29. Use mathematical induction to show that n3 − n is divisible by 3, for n ∈ N. Solution. Let P (n) : n3 − n be divisible by 3. (1) P (1) : 13 − 1 = 0 is divisible by 3. (2) Assume P (m) is true. That is, m3 − m is divisible by 3. (3) Now, (m + 1)3 − (m + 1) = m3 + 3m2 + 3m + 1 − m − 1 = m3 + 3m2 + 2m = m3 − m + 3m2 + 2m + m = (m3 − m) + 3(m2 + m).
56
Discrete Mathematical Structures Since both m3 − m and 3(m2 + m) are divisible by 3, (m + 1)3 − (m + 1) is also divisible by 3. Hence, by mathematical induction, n3 − n is divisible by 3.
2.2.4
Problems for Practice
Using the principle of mathematical induction, 1. Prove that 2 + 22 + 23 + · · · + 2n = 2n+1 − 2. 2. Show that the sum of first n even integers is n2 + n. 4n3 − n 3. Prove that 12 + 32 + 52 + · · · + (2n − 1)2 = . 3 1 1 1 1 1 4. Prove that + 2 + 3 + · · · + n = 1 − n . 2 2 2 2 2 5. Show that n4 − 4n2 is divisible by 3 for all n ∈ N. n(n + 1)(n + 2) . 6. Prove that 1 · 2 + 2 · 3 + · · · + n(n + 1) = 3 n(3n − 1) 7. Prove that 1 + 4 + 7 + · · · + (3n − 2) = . 2 1 1 1 n 1 + + + ··· + = . 8. Show that 1·3 3·5 5·7 (2n − 1)(2n + 1) 2n + 1 9. Prove that the sum of the cubes of three consecutive integers is divisible by 9. 10. Show that 22n − 1 is divisible by 3 for all n ∈ N. 11. Prove that 5n − 4n − 1 is exactly divisible by 16 for n ∈ N. 12. Show that 11n − 4n is divisible by 7 for n ∈ N. 13. Show that n + 1 < n2 for n ≥ 2. 14. Show that 2n < 3n for all n ∈ N. 15. Show that 2n < n3 for n ≥ 10. 16. Prove that n3 − n is divisible by 6. 17. Show that 52n − 25n is divisible by 7. 18. Prove that n5 − n is divisible by 5. 19. Show that 10n+1 + 10n + 1 is divisible by 3. 20. Show that n2 < 2n for n ≥ 4. 21. Show that 2n ≥ (2n + 1) for n ≥ 3. 22. Prove that 1· 2 · 3 + 2 · 3 · 4 + 3 · 4 · 5 + · · · + n(n + 1)(n + 2) =
n(n + 1)(n + 2)(n + 3) , n ≥ 1. 4
Combinatorics
2.2.5
57
Strong Induction
Strong induction is another form of mathematical induction which is very useful in proofs. In this form, we use the basic step similar to the principle of mathematical induction, and for inductive step, we assume that P (k) is true for k = 1, 2, . . . , m and then show that P (m + 1) is true based on this assumption. Strong induction is also called as the second principle of mathematical induction. Procedure for strong induction Basis step: P (1) has to be proved as true. Inductive step: [P (1) ∧ P (2) ∧ · · · ∧ P (m)] → P (m + 1) has to be proved as true by assuming P (k) is true for k = 1, 2, . . . , m. Solved Example: Show that if n is an integer greater than 1, then n can be written as the product of primes. Solution. Let P (n) : n be written as the product of primes. Basic step: P (2) is true since 2 = 1 × 2, product of primes. Inductive step: Assume that P (k) is true for all positive integers k with k ≤ m. To complete the inductive step, it must be shown that P (m + 1) is true under this assumption. Two cases arise, namely (i) when (m + 1) is prime (ii) when (m + 1) is composite. Case (i) : If (m + 1) is prime, it is obvious that P (m + 1) is true. Case (ii) : If (m + 1) is composite, then it can be written as a product of two positive integers a and b with 2 ≤ a < b ≤ m + 1. By the induction hypothesis, both a and b can be written as the product of primes. Thus, if (m + 1) is composite, it can be written as the product of primes, namely those primes in the factorisation of a and those in the factorisation of b.
2.2.6
Well-Ordering Property
The validity of mathematical induction follows from the following fundamental axioms about the set of integers. Every non-empty set of non-negative integers has a least element. The well-ordering property can often be used directly in the proof.
58
Discrete Mathematical Structures
Solved Example: What is wrong with this “proof” by strong induction? Theorem: For every non-negative integer n, 5n = 0. Proof. Basis step: 5 · 0 = 0. Induction step: Suppose that 5j = 0 for all non-negative integers j with 0 ≤ j ≤ m. Write m + 1 = i + j where i and j are natural numbers less than m + 1. By the induction hypothesis, 5(m + 1) = 5(i + 1) = 5i + 5j = 0 + 0 = 0.
2.3
Pigeonhole Principle
If n pigeonholes are occupied by n + 1 or more pigeons, then at least one pigeonhole is occupied by more than one pigeon. Example 1. Suppose a department contains 13 professors. Then, two of the professors (pigeons) were born in the same month (pigeonhole). Example 2. Suppose a laundry bag contains many red, white, and blue socks. Then, one needs to only grab four socks (pigeons) to be sure of getting a pair with the same colour (pigeonhole). Example 3. Find the minimum number of elements that one needs to take from the set S = {1, 2, . . . , 9} to be sure that two of the numbers add up to 10. Hence, the pigeonholes are the five sets {1, 9}, {2, 8}, {3, 7}, {4, 6}, {5}. Thus, any choice of six elements (pigeons) of S will guarantee that two of the numbers add up to 10.
2.3.1
Generalized Pigeonhole Principle
If n pigeonholes are occupied by kn + 1 or more pigeons, where k is a positive integer, then at least one pigeonhole is occupied by k + 1 or more pigeons.
2.3.2
Solved Problems
1. Find the minimum number of students in a class to be sure that three of them were born in the same month.
Combinatorics Solution. Here, n = 12 months are the pigeonholes and k + 1 = 3 or k = 2. Hence, among any kn + 1 = 25 students (pigeons), three of them were born in the same month. 2. Suppose a laundry bag contains many red, white, and blue socks. Find the minimum number of socks that one needs to choose in order to get two pairs (four socks) of the same colour. Solution. There are n = 3 colours (pigeonholes) and k + 1 = 4 or k = 3. Thus, among any kn + 1 = 10 socks (pigeons), four of them have the same colour. 3. Assume there are n distinct pairs of shoes in a closet. Show that if you choose n + 1 single shoes at random from the closet, you are certain to have a pair. Solution. The n distinct pairs constitute n pigeonholes. The n+1 single shoes correspond to n + 1 pigeons. Therefore, there must be at least one pigeonhole with two shoes, and thus you will certainly have drawn at least one pair of shoes. 4. Assume there are three men and five women at a party. Show that if these people are lined up in a row, at least two women will be next to each other. Solution. Consider the case where the men are placed so that no two men are next to each other and not at either end of the line. In this case, the three men generate four potential locations (pigeonholes) to place women (at either end of the line and two locations between men within the line). Since there are five women (pigeons), at least one slot will contain two women who must, therefore, be next to each other. If the men are allowed to be placed next to each other or at the end of the line, there are even fewer pigeonholes and, once again, at least two women will have to be placed next to each other. 5. Find the minimum number of students needed to guarantee that five of them belong to the same class (Freshman, Sophomore, Junior, Senior). Solution. Here, n = 4 classes are the pigeonholes and k+1 = 5 or k = 4. Thus, among any kn + 1 = 17 students (pigeons), five of them belong to the same class. 6. A student must take five classes from three areas of study. Numerous classes are offered in each discipline, but the student cannot take
59
60
Discrete Mathematical Structures more than two classes in any given area. Using pigeonhole principle, show that the student will take at least two classes in one area. Solution. The three areas are the pigeonholes, and the student must take five classes (pigeons). Hence, the student must take at least two classes in one area. 7. Let L be a list (not necessarily in alphabetical order) of the 26 letters in the English alphabet (which consists of 5 vowels, A, E, I, O, U, and 21 consonants). (i) Show that L has a sublist consisting of four or more consecutive consonants. (ii) Assuming L begins with a vowel, say A, show that L has a sublist consisting of five or more consecutive consonants. Solution. (i) The five letters partition L into n = 6 sublists (pigeonholes) of consecutive consonants. Here, k + 1 = 4 and so k = 3. Hence, nk + 1 = 6(3) + 1 = 19 < 21. Hence, some sublist has at least four consecutive consonants. (ii) Since L begins with a vowel, the remaining vowels partition L into n = 5 sublists. Here, k + 1 = 5 and so k = 4. Hence, kn + 1 = 21. Thus, some sublist has at least five consecutive consonants. 8. Find the minimum number n of integers to be selected from S = {1, 2, . . . , 9} so that (i) the sum of two of the n integers is even (ii) the difference of two of the n integers is 5. Solution. (i) The sum of two even integers or of two odd integers is even. Consider the subsets {1, 3, 5, 7, 9} and {2, 4, 6, 8} of S as pigeonholes. Hence, n = 3. (ii) Consider the five subsets {1, 6}, {2, 7}, {3, 8}, {4, 9}, {5} of S as pigeonholes. Then, n = 6 will guarantee that two integers will belong to one of the subsets and their difference will be 5.
2.3.3
Another Form of Generalized Pigeonhole Principle
If m pigeons occupy n holes (m > n), then at least one hole has more than m−1 + 1 pigeons. n Here, [x] denotes the greatest integer less than or equal to x, which is a real number.
Combinatorics
2.3.4
61
Solved Problems
1. Show that among 100 people, at least nine of them were born in the same month. Solution. Here, number of pigeons = m = number of people = 100. number of holes = n = number of months = 12. Then by generalized pigeonhole principle, at least 100 − 1 99 m−1 +1= +1= + 1 = 8 + 1 = 9 were born n 12 2 in the same month. 2. Show that if seven colours are used to paint 50 bicycles, at least eight bicycles will be the same colour. Solution. Here, number of pigeons = m = number of bicycles = 50. number of holes = n = number of colours = 7. Then by generalized pigeonhole principle, at least m−1 50 − 1 +1= + 1 = 7 + 1 = 8 bicycles will have the n 7 same colour. 3. Show that if 25 dictionaries in a library contain a total of 40235 pages, then one of the dictionaries must have at least 1,614 pages. Solution. Here, number of pigeons = m = number of bicycles = 50. number of holes = n = number of colours = 7. Then by generalized pigeonhole principle, at least m−1 40325 − 1 40324 +1= +1= + 1 = 1613 + 1 = 1614 n 25 25 pages. 4. Prove that in any group of six people, there must be at least three mutual friends or at least three mutual enemies. Solution. Let those six people be A, B, C, D, E and F. Fix A. The remaining five people can be accommodated into two groups, namely (i) Friends of A and (ii) Enemies of A. Now, by generalized pigeonhole principle, at least one of the groups 5−1 + 1 = 3 people. must contain 2
62
Discrete Mathematical Structures (i) If any two of these three people (B, C, D) are friends, then these two together with A form three mutual friends. (ii) If no two of these three people are friends, then these three people (B, C, D) are mutual enemies. In either case, we get the required conclusion. If the group of enemies of A contains three people, by the above similar argument, we get the required conclusion. 5. If we select ten points in the interior of an equilateral triangle of side 1, show that there must be at least two points whose distance 1 apart is less than . 3 Solution. Let ABC be the given equilateral triangle. Let D and E be the points of trisection of the side AB, F and G be the points of trisection of the side BC, and H and I be the points of trisection of AC, so that the triangle ABC is divided into nine equilateral 1 triangles each of side . 3 A
1 D
H 3
2
4
E
I 5
B
6
8 7
F
9 G
C
Equilateral triangle of side 1 unit Here, number of pigeons = m = number of interior points = 10. number of holes = n = number of triangles = 9. Then by generalized pigeonhole principle, at least one triangle contains 10 − 1 + 1 = 2 interior points. 2 1 , the distance between any two 3 1 interior points of any sub-triangle cannot exceed . 3 6. Find the minimum number of students needed to guarantee that five of them belong to the same subject, having majors as English, Maths, Physics, and Chemistry. Since each triangle is of length
Combinatorics
63
Solution. Number of pigeonholes = Number of subjects = n = 4 Let k be the number of students (pigeons) in each subject. Now, k + 1 = 5 ⇒ k = 4. Therefore, the total number of students = kn + 1 = 4(4) + 1 = 17. 7. Show that if any 11 numbers from 1 to 20 are chosen, then 2 of them will add up to 21. Solution. Construct the following sets with two numbers that add up to 21. A1 = {1, 20}, A5 = {5, 16}, A9 = {9, 12},
A2 = {2, 19}, A6 = {6, 15}, A10 = {10, 11}.
A3 = {3, 18}, A7 = {7, 14},
A4 = {4, 17}, A8 = {8, 13},
By pigeonhole principle, if any 11 numbers from 1 to 20 are chosen, then we must have to select all 2 elements from at least 1 set from the above 10 sets, which will give the sum as 21. 8. If we select any group of 1,000 students on campus, show that at least 3 of them must have the same birthday. Solution. The maximum number of days in a year is 366. Here, number of students = number of pigeons = m = 1,000. Number of days in a year = number of holes = n = 366. By generalized pigeonhole principle, at least 1000 − 1 m−1 +1 = + 1 = 2 + 1 = 3 students must have the n 366 same birthday. 9. How many students must be in a class to guarantee that at least two students receive the same score on the final exam, if exam is graded on a scale from 0 to 100 points. Solution. There are 101 possible scores as 0, 1, 2, . . . , 100. By pigeonhole principle, we have 102 students. Hence, there must be at least two students with the same score. Therefore, the class must contain minimum 102 students. 10. Show that among (n + 1) positive integers not exceeding 2n, there must be an integer that divides one of the other integers. Solution. Let the (n + 1) integers be a1 , a2 , . . . , an+1 .
64
Discrete Mathematical Structures Each of these numbers can be expressed as an odd multiple of a power of 2. ai = 2ki × mi ,
That is,
where ki is a non-negative integer, mi is an odd number; i = 1, 2, . . . , n + 1. Here, pigeon = odd positive integers m1 , m2 , . . . , mn+1 less than 2n. Pigeonhole = n odd positive integers less than 2n. Therefore, by pigeonhole principle, two of the integers must be equal. Let it be mi = mj . ai = 2ki
Now, ⇒
2ki ai = kj aj 2
and
aj = 2kj mj
(∵ mi = mj ).
Case (i): If ki < kj , then 2ki divides 2kj , and hence ai divides aj . Case(ii): If ki > kj , then aj divides ai . 11. Prove that in an equilateral triangle whose sides are of length 1 unit, if any five points are chosen, then at least two of them lie in a 1 triangle whose sides apart is less than . 2 Solution. Let D, E, and F be midpoints of the sides AB, BC, and AC, respectively, so that triangle ABC is divided into four equilateral 1 triangles each of side . 2 A
1 D
F 3
2
B
4
E
Equilateral triangle of side 1 unit
C
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65
Now, number of pigeon = m = number of interior points = 5. Number of pigeonholes = n = number of triangles = 4. By pigeonhole principle, at least one triangle has more than one point (or maximum two points). 1 Since each triangle side is , the distance between two interior 2 1 points of any subtriangle is less than . 2 12. Show that among 13 children, there are at least 2 children who were born in the same month. Solution. Assume the 13 children as pigeons and 12 months (from January to December) as the pigeonholes. Then, by the pigeonhole principle, there will be at least two children who were born in the same month. 13. Show that if any four numbers from 1 to 6 are chosen, then two of them will add up to 7. Solution. The following sets contain two numbers whose sum is 7. A1 = {1, 6},
A2 = {2, 5},
A3 = {3, 4}.
The numbers from 1 to 6 can be splitted into 3 sets above who sum add up to 7. Hence if any four numbers from 1 to 6 are chosen, then two of them will belong to any one of the above 3 sets whose sum is 7. 14. Show that among any group of five (not necessarily consecutive) integers, there are two with the same remainder when divided by 4. Solution. Take any group of five integers. When these are divided by 4, each has some remainder. Since there are five integers and four possible remainders when an integer is divided by 4, the pigeonhole principle implies that given five integers, at least two have the same remainder. 15. A bag contains 12 pairs of socks (each pair is in different colour). If a person draws the socks one by one at random, determine at most how many draws are required to get at least one pair of matched socks. Solution. Let n denote the number of draws. For n ≤ 12, it is possible that the socks drawn are of different colours, since there are 12 colours. For n = 13, all socks cannot have different colours, and at least two must have the same colour. Here 13 is the number of pigeons and
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Discrete Mathematical Structures 12 colours are 12 pigeonholes. Hence, at most 13 draws are required to have at least one pair of socks of the same colour. 16. Show that for every integer n there is a multiple of n that has only 0’s and 1’s in its decimal expansion. Solution. Let n be a positive integer. Consider the n + 1 integers 1, 11, 111, . . . There are n possible remainders when an integer is divided by n. Since there are n + 1 integers in this list, by the pigeonhole principle, there must be two with the same remainder when divided by n. The larger number of these integers minus the smaller one is a multiple of n, which has a decimal expansion consisting entirely of 0’s and 1’s. 17. Prove the statement: If m = kn+1 pigeons (where k ≥ 1) occupy n pigeonholes, then at least one pigeonhole must contain k + 1 or more pigeons. Solution. Let us assume that the conclusion of the given statement is false. Then, every pigeonhole contains k or less number of pigeons. Then, the total number of pigeons would be nk. This is a contradiction. Hence, the assumption made is wrong, and the given statement is true. 18. Let n1 , n2 , . . . , nr be r objects. Show that if n1 +n2 +· · ·+nr −r +1 objects are placed in r boxes, then for some i = 1, 2, . . . , r, the ith box contains at least ni objects. Solution. Assume that the conclusion part of the given statement is false. Here n1 , n2 , . . . , nr are pigeons, r boxes are pigeonholes. Then, for holes containing nj−1 or less number of pigeons, j = 1, 2, . . . , m. Then, the total number of pigeons would be less than or equal to (n1 − 1) + (n2 − 1) + · · · + (nr − 1) = n1 + n2 + · · · + nr − r = m − 1. This is a contradiction, since the number of pigeons is equal to m. Hence, the assumption made is wrong, and the given statement is true. 19. Seven members of a family have totally Rs. 2886 in their pockets. Show that at least one of them must have at least Rs. 416 in his pocket. Solution. Assume “members” as pigeonholes and “rupees” as pigeons. 2886 pigeons are to be assigned to seven pigeonholes.
Combinatorics By generalized pigeonhole principle, at least m−1 2886 − 1 +1= + 1 = 416 rupees in one member’s n 7 pocket. 20. If nine books are to be kept in four shelves, there must be at least one shelf which contains at least three books. Solution. Assume “books” as pigeons and “shelves” as pigeonholes. Nine pigeons are to be assigned to four shelves. By generalized pigeonhole principle, at least 9−1 m−1 +1= + 1 = 3 books in one shelf. n 4 21. How many people must you have to guarantee that at least nine of them will have birthdays in the same day of the week. Solution. Assume “days in a week” as pigeonholes and “people” as pigeons. We have to find the number of people (pigeons) to be assigned to seven pigeonholes. By generalized pigeonhole principle (given at least nine of them will have birthdays in the same week), m−1 +1=9 n m−1 +1=9 7 m−1+7 =9 7 m+6 =9 7 m = 57. Hence, there must be 57 people to guarantee that at least nine of them will have birthdays in the same day of the week. 22. Show that if 30 dictionaries in a library contain a total of 61327 pages, then one of the dictionaries must have at least 2045 pages. Solution. Assume “pages” as pigeons and “dictionaries” as pigeonholes. 61327 pages (pigeons) are to be assigned to 30 dictionaries (pigeonholes). By generalized pigeonhole principle, one dictionary must contain
67
68
Discrete Mathematical Structures
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69
Solution. Consider the following sets: A1 = {1, 8},
A2 = {2, 7},
A3 = {3, 6},
A4 = {4, 5}.
These are the only sets containing two numbers from 1 to 8, whose sum is 9. Since every number from 1 to 8 belongs to one of the above sets, each of the five numbers chosen must belong to one of the sets. Since there are only four sets, two of the five chosen numbers have to belong to the same set (by the pigeonhole principle). These two numbers have their sum equal to 9.
2.3.5
Problems for Practice
1. If m is an odd positive integer, then prove that there exists a positive integer n such that m divides 2n − 1. 2. A man hiked for 10 hours and covered a total distance of 45 km. It is known that he hiked 6 km in the first hour and only 3 km in the last hour. Show that he must have hiked at least 9 km within a certain period of two consecutive hours. 3. Consider a tournament in which each of n players plays against every other player and each player wins at least once. Show that there are at least two players having the same number of wins. 4. Show that any set of seven distinct integers includes two integers, x and y, such that either x + y or x − y is divisible by 10. 5. What is the minimum number of students, each of whom comes from one of the 50 states, who must be enrolled in a university to guarantee that there are at least 100 who come from the same state? 6. Show that if any eight positive integers are chosen, two of them will have the same remainder when divided by 7. 7. A drawer contains a dozen brown socks and a dozen black socks, all unmatched. A man takes socks out at random in the dark. (i) How least (ii) How least
many socks must he take out to be sure that he has at two socks of the same colour? many socks must he take out to be sure that he has at two black socks?
8. There are 38 different time periods during which classes at a university can be scheduled. If there are 677 different classes, how many different rooms will be needed? 9. Construct a sequence of 16 positive integers that has no increasing or decreasing subsequence of five terms.
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Discrete Mathematical Structures 10. Suppose there are 26 students and seven cars to transport them. Then, show that at least one car must have four or more passengers. 11. Show that in any set of eleven integers, there are two whose difference is divisible by 15. 12. Show that in any room of people who have been doing handshaking, there will always be at least two people who have shaken hands the same number of times. 13. Show that if nine colours are used to paint 100 houses, at least 12 houses will be of the same colour. 14. Show that if any five integers from 1 to 8 are chosen, then at least two of them will have a sum 9. 15. Prove that if any 30 people are selected, then we may choose a subset of five so that all five were born on the same day of the week. 16. Show that in any set of 11 integers, there are two whose difference is divisible by 15. 17. A drawer contains ten black and ten white socks. What is the last number of socks one must pull out to be sure to get a matched pair? 18. In a group of 13 children, show that there must be at least two children who were born in the same month. 19. Prove that every set of 37 positive integers contains at least two integers that leave the same remainder upon division by 36. 20. Let A be some fixed ten element set of {1, 2, 3, . . . , 50}. Show that A possesses two different five element subsets, the sum of whose elements are equal.
2.4
Permutation
Any arrangement of a set of n objects in a given order is called a permutation of the objects (taken all at a time). An arrangement of any r ≤ n of these objects in a given order is called an r-permutation or a permutation of the n objects taken r at a time. For example, consider the set of letters: a, b, c, and d. Then, (i) bdca, dcba, and acdb are permutations of the four letters (taken all at a time). (ii) bad adb, cbd, and bca are permutations of the four letters taken three at a time.
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(iii) ad, cb, da, and bd are permutations of the four letters taken two at a time. The number of permutations of n objects taken r at a time is denoted by nPr
or P (n, r)
or Pn,r
or Prn
or
(n)r .
We shall use nPr or P (n, r). Example: Find the number of permutations of six objects, say, A, B, C, D, E, and F taken three at a time. In other words, find the number of three-letter words using only the given six letters without repetition. Solution. Let the general three-letter words be represented by the following three boxes:
The first letter can be chosen in six different ways. Following this, the second letter can be chosen in five different ways, and, following this, the last letter can be chosen in four different ways. Write each number in its appropriate box as follows: 6 5 4 Therefore, by the fundamental principle of counting, there are 6 × 5 × 4 = 120 possible three-letter words without repetition from the six letters, or there are 120 permutations of six objects taken three at a time. 6P3 = P (6, 3) = 120. Formula for nP r : nPr =
n! (n − r)!
Remark: (i) When r = n, then nPn = n! (ii) There are n! permutations of n objects (taken all at a time). For example, there are 3! = 1 × 2 × 3 = 6 permutations of the three letters a, b, and c. They are abc, acb, bac, bca, cab, and cba.
2.4.1
Permutations with Repetitions
The number of permutations of n objects of which n1 is alike, n2 is alike,. . . , nr is alike is n! P (n; n1 , n2 , . . . , nr ) = . n1 !n2 ! . . . nr !
72
Discrete Mathematical Structures
2.4.2
Solved Problems
1. How many seven-letter words can be formed using the letters of the word “BENZENE”? Solution. There are three E’s and two N’s in the given word. Therefore, here n = 7, n1 = 3, n2 = 2. P (n; n1 , n2 ) = P (7; 3, 2) =
n! n1 !n2 !
7! 7×6×5×4×3×2×1 = = 420. 3! × 2! 6×2
2. How many different signals, each consisting of eight flags hung in a vertical line, can be formed from a set of four indistinguishable red flags, three indistinguishable white flags, and a blue flag? Solution. Here n = 8, n1 = 4, n2 = 3. P (n; n1 , n2 ) = P (8; 4, 3) =
n! n1 !n2 !
8! = 280. 4! × 3!
3. There are four bus lines between A and B, and three bus lines between B and C. In how many ways can a man travel (i) by bus from A to C by way of B? (ii) round-trip by bus from A to C by way of B? (iii) round-trip by bus from A to C by way of B, if he does not want to use a bus line more than once? Solution. (i) There are four ways to go from A to B and three ways to go from B to C. Hence, there are 4 × 3 = 12 ways to go from A to C by way of B. (ii) There are 12 ways to go from A to C by way of B and 12 ways to return. Hence, there are 12 × 12 = 144 ways to travel round-trip. (iii) The men will travel from A to B to C to B to A. Enter these letters connecting arrows as follows: A −→ B −→ C −→ B −→ A. The man can travel four ways from A to B and three ways from B to C, but he can only travel two ways from C to B and three
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ways from B to A since he does not want to use a bus line more than once. Enter these numbers above the corresponding arrows as follows: 4
3
2
3
A −→ B −→ C −→ B −→ A. Therefore, there are 4 × 3 × 2 × 3 = 72 ways to travel round-trip without using the same bus line more than once. 4. Suppose repetitions are not permitted. (i) How many three-digit numbers can be formed from the six digits 2, 3, 5, 6, 7, and 9? (ii) How many of these numbers are less than 400? (iii) How many are even? Solution. to represent an arbitrary In each case, draw three boxes number, and then write in each box the number of digits that can be placed there. (i) The box on the left can be filled in six ways. Following this, the middle box can be filled in five ways. Lastly, the box on the 5 4 . Therefore, there right can be filled in four ways: 6 are 6 × 5 × 4 = 120 numbers. (ii) The box on the left can be filled only in two ways by 2 or 3, since each number must be less than 400. The middle box can be filled in five ways. Lastly, the box on the right can be filled in four ways. Therefore, there are 2 × 5 × 4 = 40 numbers. (iii) The box on the right can be filled in only two ways by two or six, since the numbers must be even. The box on the left can be filled in five ways, and lastly, the middle box can be filled in 5 4 2 . Therefore, there are 5 × 4 × 2 = 40 four ways: numbers. 5. Find the number of ways in which a party of seven persons can arrange themselves: (i) in a row of seven chairs (ii) around a circular table. Solution. (i) The seven persons can arrange themselves in a row in 7 × 6 × 5 × 4 × 3 × 2 × 1 = 7! ways. (ii) One person can sit at any place in the circular table. The other six persons can then arrange themselves in 6 × 5 × 4 × 3 × 2 × 1 = 6! ways around the table.
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Discrete Mathematical Structures Remark. This is an example of a circular permutation. In general, n objects can be arranged in a circle in (n − 1) × (n − 2) × · · · × 3 × 2 × 1 = (n − 1)! ways. 6. Find the number of distinct permutations that can be formed from all the letters of the word: (i) RADAR (ii) UNUSUAL. Solution. 5! = 30, since there are five letters of which two are R and (i) 2! × 2! two are A. 7! = 840, since there are seven letters of which three are U. (ii) 3! 7. In how many ways can four mathematics books, three history books, three chemistry books, and two sociology books be arranged on a shelf so that all books of the same subject are together? Solution. First, the books must be arranged on the shelf in four units according to subject matter: . The box on the left can be filled by any of the four subjects, the next by any three remaining subjects, the next by any two remaining subjects, 4 3 2 1 . and the box on the right by the last subject: Therefore, there are 4 × 3 × 2 × 1 = 4! ways to arrange the books on the shelf according to subject matter. Now, in each of the above cases, the mathematics books can be arranged in 4! ways, the history books in 3! ways, the chemistry books in 3! ways, and the sociology books in 2! ways. Thus, altogether, there are 4! × 4! × 3! × 3! × 2! = 41472 arrangements. 8. Find n if (i) P (n, 2) = 72 (ii) P (n, 4) = 42P (n, 2) (iii) 2P (n, 2) + 50 = P (2n, 2). Solution. (i) P (n, 2) = n(n − 1) = n2 − n ∴
n2 − n = 72 ⇒ n2 − n − 72 = 0 ⇒ (n − 9)(n + 8) = 0 ⇒ n = 9, −8 ⇒ n = 9 since n is positive.
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(ii) P (n, 4) = n(n − 1)(n − 2)(n − 3) and P (n, 2) = n(n − 1) Therefore, n(n − 1)(n − 2)(n − 3) = 42n(n − 1). If n 6= 0, n 6= 1, (n − 2)(n − 3) = 42 ⇒ n2 − 5n + 6 = 42 ⇒ n2 − 5n − 36 = 0 ⇒ (n − 9)(n + 4) = 0 ⇒ n = 9 since n is positive. (iii) P (n, 2) = n(n − 1) = n2 − n P (2n, 2) = 2n(2n − 1) = 4n2 − 2n Therefore, 2(n2 − n) + 50 = 4n2 − 2n ⇒ 2n2 − 2n + 50 = 4n2 − 2n ⇒ 2n2 = 50 ⇒ n2 = 25 ⇒ n = 5 since n must be positive. 9. In how many ways can six persons occupy three vacant seats? Solution. Total number of ways = P (6, 3) = 6 × 5 × 4 = 120 ways. 10. How many permutations of the letters A, B, C, D, E, F, G, H contain the string ABC? Solution. Since the letters A, B, and C must occur as block, we can find the answer by finding number of permutations of six objects, namely the block ABC and individual letters D, E, F, G, and H. Therefore, there are 6! = 720 permutations of the letters A, B, C, D, E, F, G, H in which ABC occurs. 11. If P (12, r) = 1320, find r. Solution. P (12, r) = 12 × 11 × 10 . . . r factors ⇒ 1320 = 12 × 11 × 10 . . . r factors ⇒ r = 3. 12. In how many of the permutations of ten things taken four at a time will (i) one thing always occur (ii) one thing never occur.
76
Discrete Mathematical Structures Solution. (i) We can keep aside the particular thing which will always occur; the number of permutations of nine things taken three at a time is P (9, 3). Now, this particular thing can take up any one of the 4 places, and 50 can be arranged in four ways. Therefore, the total number of permutations = P (9, 3) × 4 = 9 × 8 × 7 × 4 = 2016. (ii) If we are keeping the particular thing aside as never to occur, the number of permutations of nine things (10 − 1 = 9) taken four at a time is P (9, 4) = 9 × 8 × 7 × 6 = 3026. 13. In how many ways can six boys and four girls be arranged in a straight line so that no two girls are ever together. Solution. The arrangement may be done in two operations. (i) First, we fix the positions of six boys. Their positions are indicated by B1 , B2 , . . . , B6 . That is, X B1 X B2 X B3 X B4 X B5 X B6 . This can be done in 6! ways. (ii) If the positions of girls are fixed at places including those at the two ends as shown by the crosses, the four girls will never come together. In any one of these arrangements, there are seven places for four girls, and so the girls can sit in 7P4 ways. ∴ The number of ways of seating six boys and four girls = 7P4 × 6! = 7 × 6 × 5 × 4 × 6 × 5 × 4 × 3 × 2 × 1 = 604800. 14. There are six books on Economics, three on Commerce and two on History. In how many ways can these be placed on a shelf of books if the same subjects are to be together? Solution. Six Economics books can be arranged in 6P6 ways or 6! ways. Three commerce books can be arranged in 3P3 ways or 3! ways. Two history books can be arranged in 2P2 ways or 2! ways. The three subject books, Economics, Commerce, and History books, can be arranged in 3P3 ways or 3! ways. ∴ The total number of required arrangements = 6! × 3! × 2! × 3! ways = 51840 ways. 15. Suppose there are six boys and five girls. (i) In how many ways can they sit in a row? (ii) In how many ways can they sit in a row if the boys and girls each sit together? (iii) In how many ways can they sit in a row if the girls are to sit together and the boys are not to sit together?
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77
(iv) How many seating arrangements are there with no two girls sitting together? Solution. (i) There are 6 + 5 = 11 persons, and they can sit in 11P11 ways. That is, 11P11 = 11! ways. (ii) The boys among themselves can sit in 6! ways, and girls among themselves can sit in 5! ways. They can be considered as two units and can be permuted in 2! ways. Thus, the required seating arrangements can be done in = 2! × 6! × 5! ways = 2 × 720 × 120 ways = 172800 ways. (iii) The boys can sit in 6! ways and girls in 5! ways. Since girls have to sit together, they are considered as one unit. Among the six boys, either 0 or 1 or 2 or 3 or 4 or 5 or 6 have to sit to the left of the girls’ units. Of these seven ways, 0 and 6 cases have to be omitted as the boys do not sit together. Thus, the required number of arrangements = 5 × 6! × 5! ways = 5 × 720 × 120 ways = 432000 ways. (iv) The boys can sit in 6! ways. There are seven places where the girls can be placed. Thus, the total arrangements are = 7P5 × 6! ways 7! × 720 = 2! = 2520 × 720 = 1814400 ways. 16. Find the number of ways in which five boys and five girls can be seated in a row if the boys and girls are to have alternate seats. Solution. Case (i): Boys can be arranged among themselves in 5! ways. B
B
B
B
B
There are six places for girls. Hence, there are 6P5 × 5! arrangements. Case (ii): Girls can be arranged in 5! ways. G
G
G
G
G
There are six places for boys. Hence, there are 6P5 × 5! ways. Hence, taking two cases into account, there are 2 × 6P5 × 5! arrangements in total. ∴ There are 2 × 120 × 6 = 240 ways. 17. How many permutations of {a, b, c, d, e, f, g} (i) end with a (ii) begin with c
78
Discrete Mathematical Structures (iii) begin with c and end with a (iv) have c and a occupying the end places? Solution. (i) The last position can be filled in only one way. The remaining six letters can be arranged in 6! ways. ∴ The total number of permutations ending with a = 6! × 1 = 720 ways. (ii) The first position can be filled in only one way. The remaining six letters can be arranged in 6! ways. ∴ The total number of permutations starting with c = 1 × 6! ways = 720 ways. (iii) The first position can be filled in only one way, and the last position can be filled in only one way. The remaining five letters can be arranged in 5! ways. ∴ The total number of permutations begin with c and end with a is = 1 × 5! × 1 ways = 120 ways. (iv) c and a occupy end positions in 2! ways, and the remaining five letters can be arranged in 5! ways. ∴ The total number of permutations = 5! × 2! ways = 240 ways. 18. How many bit strings of length 10 contain (i) (ii) (iii) (iv)
exactly four 1’s at most four 1’s at least four 1’s an equal number of 0’s and 1’s?
Solution. (i) A bit string of length 10 can be considered to have ten positions and should be filled with four 1’s and six 0’s. 10! ∴ Required number of bit strings = = 210. 4! × 6! (ii) Required number of bit strings =
10! 10! 10! 10! 10! + + + + = 386. 0! × 10! 1 × 9! 2! × 8! 3! × 7! 4! × 6!
(iii) Required number of bit strings =
10! 10! 10! 10! 10! 10! 10! + + + + + + = 848. 4! × 6! 5! × 5! 6! × 4! 7! × 3! 8! × 2! 9! × 1! 10! × 0!
(iv) Required number of bit strings =
10! = 252. 5! × 5!
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19. Suppose that there are 9 faculty members in the mathematics department and 11 in the computer science department. How many ways are there to select a committee to develop a discrete mathematics course in a school if the committee is to consist of three faculty members from the mathematics department and four from the computer science department? Solution. By the product rule, the answer is the product of the number of 3-combinations of a set with nine elements and the number of 4-combinations of a set with 11 elements. The number of ways to select the committee 11! 9! × = 84 × 330 = 27720. = 9C3 × 11C4 = 3! × 6! 4! × 7! 20. How many possibilities are there for the win, place, and show (first, second, and third) positions in a horse race with 12 horses if all orders of finish are possible? Solution. The number of ways to pick the three winners is the number of ordered selections of three elements from 12. ∴ The required number of possibilities = 12P3 = 12 × 11 × 10 = 1320.
2.4.3
Problems for Practice
1. How many automobile license plates can be made if each plate contains two different letters followed by three different digits? Solve the problem if the first digit cannot be 0. 2. There are six roads between A and B and four roads between B and C. Find the number of ways in which one can drive (i) from A to C by way of B (ii) round-trip from A to C by way of B (iii) round-trip from A to C by way of B without using the same road more than once. 3. Find the number of ways in which six people can ride a toboggan if one of a subset of three must drive. 4.
(i) Find the number of ways in which five persons can sit in a row. (ii) How many ways are there if two of the persons insist on sitting next to one another? (iii) Solve (i) assuming they sit around a circular table. (iv) Solve (ii) assuming they sit around a circular table.
5. Find the number of ways in which five large books, four medium-size books, and three small books can be placed on a shelf so that all books of the same size are together.
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Discrete Mathematical Structures 6.
(i) Find the number of permutations that can be performed from the letters of the word ELEVEN. (ii) How many of them begin and end with E? (iii) How many of them have three E’s together? (iv) How many begin with E and end with N?
7.
(i) In how many ways can three boys and two girls sit in a row? (ii) In how many ways can they sit in a row if the boys and girls are each to sit together? (iii) In how many ways can they sit in a row if just the girls are to sit together?
8. Show that (i) P (n, 0) + P (n, 1) + P (n, 2) + · · · + P (n, n) = 2n . (ii) P (n, 0) − P (n, 1) + P (n, 2) − P (n, 3) + · · · + P (n, n) = 0. 9. How many bit strings of length 10 contain at least three 1’s and at least three 0’s? 10. How many ways are there for eight men and five women to stand in a line so that no two women stand next to each other? 11. The English alphabet contains 21 consonants and five vowels. How many strings of six lowercase letters of the English alphabet contain (i) (ii) (iii) (iv)
exactly one vowel exactly two vowels at least one vowel at least two vowels?
12. A committee of 11 members sit at a round table. In how many ways can they be seated if the “president” and “secretary” choose to sit together? 13. In an examination, six papers are set of which two are mathematics. In how many ways can the examination be arranged if the mathematics papers are not to be together? 14. In how many ways can eight people sit around a table? 15. How many numbers are there in all which consist of five digits? 16. How many odd numbers of three digits can be formed with 1, 2, 3, 4, and 5?
2.5
Combination
Suppose we have a collection of n objects. A combination of these n objects taken r at a time is any selection of r of the objects where order does not
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81
count. In other words, an r-combination of a set of n objects is any subset of r elements. For example, the combinations of the letters a, b, c, d taken three at a time are {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d} or simply abc, abd, acd, bcd, respectively. It can be noted that the following combinations are equal: abc, acb, bac, bca, cab
and cba.
That is, each denote the same set {a, b, c}. The number of combinations of n objects taken r at a time is denoted by C(n, r). The symbols nCr , Cn,r and Crn can also be used. Formula for nC r : n! nCr = r!(n − r)!
2.5.1
Solved Problems
1. How many committees of three can be formed from eight people? Solution. Each committee is a combination of eight people taken three at a time. Therefore, the number of committees that can be formed is 8! 8C3 = = 56. 3! × 5! 2. A farmer buys three cows, two pigs, and four hens from a man who has six cows, five pigs, and eight hens. How many choices does the farmer have? Solution. The farmer can choose the cows in 6C3 ways, the pigs in 5C2 ways, and the hens in 8C4 ways. Hence, altogether he can choose the animals in 6C3 × 5C2 × 8C4 = 20 × 10 × 70 = 14000 ways. 3. In how many ways can a committee consisting of three men and two women be chosen from seven men and five women? Solution. The three men can be chosen from the seven men in 7C3 ways, and the two women can be chosen from the five women in 5C2 ways. Hence, the committee can be chosen in 7C3 × 5C2 = 350 ways. 4. How many committees of five with a given chairperson can be selected from 12 persons? Solution. The chairperson can be chosen in 12 ways, and, following this, the other four on the committee can be chosen from the 11 remaining
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Discrete Mathematical Structures in 11C4 ways. There are 12 × 11C4 = 12 × 330 = 3960 such committees. 5. A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if (i) they can be any colour (ii) two must be white and two red (iii) they must all be of the same colour. Solution. (i) The four marbles (of any colour) can be chosen from the 11 marbles in 11C4 = 330 ways. (ii) The two white marbles can be chosen in 6C2 ways, and the two red marbles can be chosen in 5C2 ways. Thus, there are 6C2 × 5C2 = 150 ways of drawing two white marbles and two red marbles. (iii) There are 6C4 = 15 ways of drawing four white marbles and 5C4 = 5 ways of drawing four red marbles. Thus, there are 15 + 5 = 20 ways of drawing four marbles of the same colour. 6. In how many ways can a set of five letters be selected from the English alphabet? Solution. The number of ways to select five letters from 26 alphabets is 26C5 = 65780. 7. How many bit strings of length n contain exactly r 1’s? Solution. The positions of r 1’s in a bit string of length n form r-combination of the set {1, 2, . . . , n}. Hence, there are nCr bit strings of length n that contain exactly r 1’s. Note: (i) (ii) (iii) (iv)
nCn = nC0 = 1. nCr = nCn−r . r nCr = nP r! . nCx = nCy ⇒ n = x + y
or x = y.
8. Find the value of n if 20Cn+2 = 20C2n−1 . Solution. Given:
20Cn+2 = 20C2n−1 ⇒ n + 2 = 2n − 1 (∵ ncx = ncy ⇒ x = y) ⇒ n = 3.
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9. How many ways are there to form a committee, if the committee consists of 3 educationalists and 4 socialists, if there are 9 educationalists and 11 socialists. Solution. Three educationalists can be chosen from nine educationalists in 9C3 ways. Four socialists can be chosen from 11 socialists in 11C4 ways. Hence, by product rule, the number of ways to select the committee = 9C3 × 11C4 =
11! 9! × = 27720 ways. 3! × 6! 4! × 7!
10. A team of 11 players is to be chosen from 15 members. In how many ways can this be done if (i) one particular player is always included (ii) two such players have to be always included? Solution. (i) Let one player be fixed. The remaining players are 14. Out of these 14 players, we have to select ten players in 14C10 = 1001 ways. (ii) Let two players be fixed. The remaining players are 13. Out of these 13 players, we have to select nine players in 13C9 = 715 ways. 11. Find the number of diagonals that can be drawn by joining the angular points of a heptagon. Solution. A heptagon has seven angular points and seven sides. The join of two angular points is either a side or a diagonal. 7×6 The number of lines joining the angular points = 7C2 = = 21. 1×2 But the number of sides = 7. ∴ The number of diagonals = 21 − 7 = 14. 12. There are five questions in a question paper. In how many ways can a boy solve one or more questions? Solution. The boy can dispose of each question in two ways. He may either solve it or leave it. Thus, the number of ways of disposing all the questions = 25 .
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Discrete Mathematical Structures But this includes the case in which he has left all the questions unsolved. ∴ The total number of ways of solving the paper = 25 − 1 = 31. 13. Find the value of r if 20Cr = 20Cr+2 . Solution.
⇒ ⇒ ⇒ ⇒
20Cr = 20Cr+2 20Cr = 20C20−(r+2) r = 20 − (r + 2) (∵ r = r + 2 ⇒ 2 = 0 2r = 18 r = 9.
is not possible)
14. If nC5 = 20 · nC4 , find n. Solution. nC5 = 20 × C4 n(n − 1)(n − 2)(n − 3) n(n − 1)(n − 2)(n − 3)(n − 4) = 20 × 1×2×3×4×5 1×2×3×4 n−4 = 20 5 n − 4 = 100 n = 104. 15. From a committee consisting of six men and seven women, in how many ways can we select a committee of (i) (ii) (iii) (iv) (v)
three men and four women four members that has at least one woman four persons that has at most one man four persons of both genders four persons in which Mr and Mrs Joseph are not included.
Solution. (i) Three men can be selected from six men in 6C3 ways. Four women can be selected from seven women in 7C4 ways. ∴ By product rule, the committee of three men and four women can be selected in 6C3 × 7C4 = 700 ways. (ii) For the committee of at least one woman, we have the following possibilities:
Combinatorics (i) One woman and three men (ii) Two women and two men (iii) Three women and one man (iv) Four women and zero men. Hence, the selection can be done in = 7C4 × 6C3 + 7C2 × 6C2 + 7C3 × 6C1 + 7C4 × 6C0 = 700 ways. (iii) For the committee of at most one man, we have the following possibilities: (i) One man and three women (ii) Zero men and four women. Hence, the selection can be done in = 6C1 × 7C3 + 6C0 × 7C4 = 245 ways. (iv) For the committee of both genders, we have the following possibilities: (i) One man and three women (ii) Two men and two women (iii) Three men and one woman which can be done in 6C1 × 7C3 + 6C2 × 7C2 + 6C3 × 7C1 = 665 ways. (v) Since the committee does not consist of Mr. and Mrs. Joseph, we have five men and six women in the committee. Now, we can select 4 members from 11 members in 11C4 = 330 ways.
2.5.2
Problems for Practice
1. A woman has 11 close friends. (i) In how many ways can she invite five of them to dinner? (ii) In how many ways if two of the friends are married and will not attend separately? (iii) In how many ways if two of them are not on speaking terms and will not attend together? 2. A woman has 11 close friends of whom six are also women. (i) In how many ways can she invite three or more to a party? (ii) In how many ways can she invite three or more of them if she wants the same number of men and women (including herself)? 3. A student is to answer 10 out of 13 questions in an exam. (i) How many choices does he have? (ii) How many if he must answer the first two questions?
85
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Discrete Mathematical Structures (iii) How many if he must answer the first or second question not both? (iv) How many if he must answer exactly three out of the first five questions? (v) How many if he must answer at least three of the first five questions? 4. How many diagonals are there in a polygon of ten sides? 5. A committee is to consist of two men and three women. How many different committees are possible if five men and seven women are eligible. 6. How many different groups can be selected for playing tennis out of four ladies and three gentlemen, there being one lady and one gentleman on each side? 7. From a committee of five women and seven men, in how many ways can a subcommittee of four be chosen so as to contain one particular man? 8. In how many ways can a selection be made out of five oranges, eight apples, and seven plantains? 9. In how many ways can 20 students be divided into four equal groups? 10. How many bit strings of length 10 have (i) (ii) (iii) (iv)
exactly three 0’s at least three 1’s more 0’s than 1’s an odd number of 0’s?
11. How many bit strings of length 12 contain (i) exactly three 1’s (ii) at least three 1’s (iii) an equal number of 1’s and 0’s? 12. In how many ways can a party of 16 people can be conveyed in two vehicles, one of which will not hold more than eight and the other not more than ten? 13. In how many ways can a committee of 8 be chosen from 12 socialists and 9 conservatives to give a socialist majority with at least 2 conservatives included? 14. A committee of 12 is to be selected from 10 men and 10 women. In how many ways can the selection be carried out if (i) there are no restrictions (ii) there must be equal number of men and women
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(iii) there must be an even number of women (iv) there must be more women than men (v) there must be at least eight men?
2.5.3
Recurrence Relation
A recurrence relation for the sequence {fn } is a formula that expresses fn in terms of one or more of the previous terms of the sequence, namely f0 , f1 , . . . , fn−1 , for all integers n with n ≥ n0 , where n0 is non-negative integer. A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation.
2.5.4
Solved Problems
1. Determine whether the sequence {fn } = {3n} is a solution of the recurrence relation: fn = 2fn−1 − fn−2 , for n = 2, 3, 4, . . . Solution. Suppose fn = 3n. Then for n ≥ 2, fn = 2fn−1 − fn−2 = 2[3(n − 1)] − 3(n − 2) since fn = 3n = 6n − 6 − 3n + 6 = 3n. ∴ {fn }, where fn = 3n, is a solution of the recurrence relation. 2. Show that the sequence {fn } is a solution of the recurrence relation fn = −3fn−1 + 4fn−2 if fn = 2(−4)n + 3. Solution. fn = −3fn−1 + 4fn−2 = −3 2(−4)n−1 + 3 + 4 2(−4)n−2 + 3 = −6(−4)n−1 − 9 + 8(−4)n−2 + 12 = −6(−4)n−1 + 8(−4)n−2 + 3 = −6(−4)n−1 − 2(−4)n−1 + 3 = 2(−4)n + 3. ∴ fn = 2(−4)n + 3 is a solution of the recurrence relation. Now, we discuss about a class of recurrence relations known as linear recurrence relations with constant coefficients.
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2.5.5
Linear Recurrence Relation
A recurrence relation of the form a0 fn + a1 fn−1 + a2 fn−2 + · · · + ak fn−k = f (n)
(2.4)
where ai ’s are constants, is called a linear recurrence relation with constant coefficients. The recurrence relation (2.4) is known as a k th -order recurrence relation, provided both a0 and ak are non-zero. Note: The phrase “k th -order” means that each term in the sequence depends only on the k previous terms. Example 1: Consider the Fibonacci sequence defined by the recurrence relation fn = fn−1 + fn−2 , n ≥ 2 and the initial conditions f0 = 0 and f1 = 1. The recurrence relation is called a second-order relation because fn depends on the two previous terms of fn . Example 2: Consider the recurrence relation f (k) − 5f (k − 1) + 6f (k − 2) = 4k + 10 defined for k ≥ 2, together with the initial conditions f (0) = 73 and f (1) = 5. Clearly, it is a second-order linear recurrence relation.
2.5.6
Homogenous Recurrence Relation
th
A k -order linear relation is a homogenous recurrence relation if f (n) = 0 for all n. Otherwise, it is called non-homogenous. Example 1: Consider the recurrence relation C(k)−5C(k−1)+8C(k−2) = 0 together with the initial conditions C(0) = 5 and C(1) = 2. It is a second-order homogenous recurrence relation. Example 2: Which of the following recurrence relations are homogenous and which of them are non-homogenous? (i) fn = fn−2 . (ii) an = an−1 + an−3 . (iii) bn = bn−1 + 2. (iv) s(n) = s(n − 2) + s(n − 4). Solution. The relations fn = fn−2 , an = an−1 + an−3 , s(n) = s(n − 2) + s(n − 4) are all homogenous, and the relation bn = bn−1 + is non-homogenous.
Combinatorics
2.5.7
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Recurrence Relations obtained from Solutions
Before giving an algorithm for solving a recurrence relation, we will examine a few recurrence relations that arise from certain closed form expressions. The procedure is illustrated by the following examples. 1. Form the recurrence relation given fn = 3 · 5n , n ≥ 0. Solution. If n ≥ 1, then fn = 3 · 5n = 3 · 5 · 5n−1 = 5 · 3 · 5n−1 = 5fn−1 . ∴ The recurrence relation is fn = 5fn−1 with f0 = 3. 2. Find the recurrence relation satisfying yn = A(3)n + B(−2)n . Solution. Given yn = A(3)n + B(−2)n . ∴
yn+1 = A(3)n+1 + B(−2)n+1 = 3A(3)n − 2B(−2)n yn+2 = A(3)n+2 + b(−2)n+2 = 9A(3)n + 4B(−2)n
Eliminating A and B from the yn yn+1 yn+2
above equations, 1 1 3 −2 = 0 9 4
Expanding along column 1, or or or
yn (12 + 18) − yn+1 (4 − 9) + yn+2 (−2 − 3) = 0 30yn + 5yn+1 − 5yn+2 = 0 6yn + yn+1 − yn+2 = 0 yn+2 − yn+1 − 6yn = 0
which is the required recurrence relation. 3. Find the recurrence relation satisfying yn = A(3)n + B(−4)n . Solution. Given yn = A(3)n + B(−4)n . ∴
yn+1 = 3A(3)n − 4B(−4)n yn+2 = 9A(3)n + 16B(−4)n
Eliminating A and B from the yn yn+1 yn+2
above equations, 1 1 3 −4 = 0 9 16
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Discrete Mathematical Structures Expanding along column 1,
or or or
yn (48 + 36) − yn+1 (16 − 9) + yn+2 (−4 − 3) = 0 84yn − 7yn+1 − 7yn+2 = 0 12yn − yn+1 − yn+2 = 0 yn+2 + yn+1 − 12yn = 0
which is the required recurrence relation. 4. Find the recurrence relation satisfying yn = (A + Bn)4n . Solution. Given yn = (A + Bn)4n = A(4)n + Bn(4)n . ∴
yn+1 = 4A(4)n + 4B(n + 1)(4)n yn+2 = 16A(4)n + 16B(n + 2)(4)n
Eliminating A and B from yn yn+1 yn+2
the above equations, 1 n 4 4(n + 1) = 0 16 16(n + 2)
Expanding along column 1,
or or
yn [64(n + 2) − 64(n + 1)] − yn+1 [16n + 32 − 16n] + yn+2 [4n + 4 − 4n] = 0 64yn − 32yn+1 + 4yn+2 = 0 yn+2 − 8yn+1 + 16yn = 0
which is the required recurrence relation.
2.6
Solving Linear Homogenous Recurrence Relations
Consider a linear homogenous recurrence relation of degree k with constant coefficients fn = a1 fn−1 + a2 fn−2 + · · · + ak fn−k where a1 , a2 , . . . , ak are real numbers and ak 6= 0. The basic approach for solving linear homogenous recurrence relations is to look for solutions of the form fn = rn , where r is a constant. Note that fn = rn is a solution of the recurrence relation fn = a1 fn−1 + a2 fn−2 + · · · + ak fn−k if and only if rn = c1 rn−1 + c2 rn−2 + · · · + ck rn−k .
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When both sides of this equation are divided by rn−k and the right-hand side is subtracted from the left, we obtain rk − c1 rk−1 − c2 rk−2 − · · · − ck−1 r − ck = 0. Consequently, the sequence {fn } with fn = rn is a solution if and only if r is a solution of this last equation.
2.6.1
Characteristic Equation
The characteristic equation of the homogenous k th -order linear recurrence relation fn + a1 fn−1 + a2 fn−2 + · · · + ak fn−k = 0 is the k th -degree polynomial equation rk + a1 rk−1 + a2 rk−2 + · · · + ak−1 rk−1 + ak = 0. The solutions of this equation are called the characteristic roots of the recurrence relation. Examples: 1. The characteristic equation of Q(k) + 2Q(k − 1) − 3Q(k − 2) − 6Q(k-4) = 0 is r4 + 2r3 − 3r2 − 6 = 0. Note that the absence of Q(k − 3) term implies that there is no r4−3 = r term in the characteristic equation. 2. The characteristic equation of T (k) − 7T (k − 2) + 6T (k − 3) = 0 is r3 − 7r + 6 = 0, i.e. r3 − 7r + 6 = 0, and 1, 2, and −3 are the characteristic roots.
2.6.2
Algorithm for Solving k th -order Homogenous Linear Recurrence Relations
Step 1: If fn + a1 fn−1 + a2 fn−2 + · · · + ak fn−k = 0 is a given recurrence relation, then write its characteristic equation as rk +a1 rk−1 +a2 rk−2 +· · ·+ak−1 r +ak = 0. Step 2: Find all the characteristic roots of this equation. Step 3: Case (i): If there are k distinct roots, say c1 , c2 , . . . , ck , then the general solution of the recurrence relation is fn = A1 ck1 + A2 ck2 + · · · + Ak Ckk . Case (ii): Suppose that c1 is a root of multiplicity m. Then, the corresponding solution is fn A1 rm−1 + A2 rm−2 + · · · + Am−2 r2 + Am−1 r + Am cr1 . Step 4: Use the boundary conditions to determine the constants A1 , A2 , . . . , Ak .
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2.6.3
Solved Problems
1. Solve the Fibonacci sequence {fn } defined by fn = fn−1 + fn−2 for n ≥ 2 with the initial conditions f0 = 0 and f1 = 1. Solution. The characteristic equation of the given recurrence relation is r2 − r − 1 = 0. Solving this equation, we get √ √ 1± 5 1± 1+4 = . r= 2 2 √ √ 1+ 5 1− 5 ∴ c1 = , c2 = . 2 2 The general solution is fn = A1 cn1 + A2 cn2 where A1 and A2 are constants. Given: f0 = 1 ⇒ A1 + A2 = 0. f1 = 1 ⇒ A1 c1 + A2 c2 = 1 √ ! 1+ 5 ⇒ A1 + A2 2
√ ! 1− 5 = 1. 2
(2.5)
(2.6)
Solving (2.5) and (2.6), we get 1 A1 = √ 5 ∴
and
1 A2 = − √ . 5
The solution is 1 fn = √ 5
"
√ !n 1+ 5 − 2
√ !n # 1− 5 . 2
2. If the recurrence relation is un+1 − 2un = 0, find the closed form expression (solution) for un . Solution. The characteristic equation is r−2=0 ⇒ r = 2. The general solution is un = A · 2 n where A is a constant.
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3. Find f (n) if f (n) = 7f (n − 1) − 10f (n − 2), given that f (0) = 4 and f (1) = 17. Solution. The characteristic equation is r2 − 7r + 10 = 0 ⇒ r = 2, 5 c1 = 2, c2 = 5.
∴ The general solution is
f (n) = A1 cn1 + A2 cn2 = A1 2n + A2 5n . Given: f (0) = 4 ⇒ A1 + A2 = 4.
(2.7)
f (1) = 17 ⇒ 2A1 + 5A2 = 17.
(2.8)
Solving (2.7) and (2.8), we get A1 = 3
and
A2 = 3.
f (n) = 2n + 3(5)n .
4. Find T (k) if T (k) − 7T (k − 2) + 6T (k − 3) = 0, where T (0) = 8, T (1) = 6, and T (2) = 22. Solution. The characteristic equation is r3 − 7r + 6 = 0. 1 1 0
0 −7 1 1
6 6
2 1 0
1 −6 2 6
0
1
3
0
The characteristic roots are c1 = 1,
c2 = 2,
c3 = −3.
The general solution is T (k) = A1 ck1 + A2 ck2 + A3 ck3 T (k) = A1 + A2 (2)k + A3 (−3)k .
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Discrete Mathematical Structures Given: T (0) = 8 ⇒ A1 + A2 + A3 = 8.
(2.9)
T (1) = 6 ⇒ A1 + 2A2 − 3A3 = 6.
(2.10)
T (2) = 22 ⇒ A1 + 4A2 + 9A3 = 22.
(2.11)
Solving (2.9), (2.10) and (2.11), we get A1 = 5,
A2 = 2,
A3 = 1.
T (k) = 5 + 2(2)k + 1(−3)k
T (k) = 5+2k+1 +(−3)k .
or
5. Solve fk − 8fk−1 + 16fk−2 = 0 where f2 = 16 and f3 = 80. Solution. The characteristic equation is r2 − 84 + 16 = 0 ⇒ r = 4, 4 (repeated). The general solution is fk = (A1 + A2 k)4k . Given: f2 = 16 ⇒ (A1 + 2A2 )16 = 16 ⇒ A1 + 2A2 = 1.
(2.12)
f3 = 80 ⇒ (A1 + 3A2 )64 = 80 ⇒ 4(A1 + 3A2 ) = 5 ⇒ 4A1 + 12A2 = 5.
(2.13)
Solving (2.12) and (2.13), we get A1 = ∴
1 , 2
A2 =
1 . 4
The solution is fk =
1 1 + k 4k = (2 + k)4k−1 . 2 4
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6. Find a solution to the recurrence relation Cn = −3Cn−1 − 3Cn−2 − Cn−3 for n ≥ 3 with initial conditions C0 = 1, C1 = −2, and C2 = 1. Solution. The characteristic equation is r3 + 3r2 + 3r + 1 = 0 ⇒ (r + 1)3 = 0. ∴ r = −1 is a characteristic root of multiplicity 3. The general solution is Cn = A1 + A2 n + A3 n2 (−1)n . Given: C0 = 1 ⇒ A1 = 1.
(2.14)
C1 = −2 ⇒ −(A1 + A2 + A3 ) = −2.
(2.15)
C2 = 1 ⇒ A1 + 2A2 + 4A3 = 1.
(2.16)
Solving (2.14), (2.15), and (2.16), we get A1 = 1,
A2 = 2,
A3 = −1.
∴ The solution is Cn = 1 + 2n − n2 (−1)n .
2.7
Solving Linear Non-homogenous Recurrence Relations
The solution of a linear non-homogenous recurrence relation with constant coefficients is the sum of the two parts, the homogenous solution, which satisfies the recurrence relation when the right-hand side of the equation is set to 0, and the particular solution, which satisfies the difference equation with f (n) on the right-hand side. There is no general procedure for determining the particular solution of a difference equation. However, in simple cases, this solution can be obtained by the method of inspection. To determine the particular solution, we use the following rules: Rule 1: When f (n) is of the form of a polynomial of degree m in n, k0 + k1 n + k2 n2 + k3 n3 + · · · + km−1 nm−1 + km nm ,
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Discrete Mathematical Structures
the corresponding particular solution will be of the form Q0 + Q1 n + Q2 n2 + Q3 n3 + · · · + Qm−1 nm−1 + Qm nm . Rule 2: When f (n) is of the form k0 + k1 n + k2 n2 + · · · + km−1 nm−1 + km nm an , the corresponding particular solution is of the form Q0 + Q1 n + Q2 n2 + · · · + Qm−1 nm−1 + Qm nm an if a is not a characteristic root of the recurrence relation. Rule 3: If a is a characteristic root of multiplicity r − 1, when f (n) is of the form k0 + k1 n + k2 n2 + · · · + Km−1 nm−1 + K − mnm an , the corresponding particular solution is of the form rn−1 Q0 + Q1 n + Q2 n2 + · · · + Qm−1 nm−1 + Qm nm an . Note: The general solution of the recurrence relation is the sum of the homogenous solution and particular solution. If no initial conditions are given, then you have finished. If m initial conditions are given, obtain m linear equations in m unknowns and solve the system, if possible, to get a complete solution.
2.7.1
Solved Problems
1. Solve S(k)−S(k−1)−6S(k−2) = −30 where S(0) = 20, S(1) = −5. Solution. The associated homogenous relation is S(k) − S(k − 1) − 6S(k − 2) = 0. The characteristic equation is r2 − r − 6 = 0. The characteristic roots are r = −2, 3. The homogenous solution is A1 (−2)k + A2 (3)k . Since the right-hand side of S(k) − S(k − 1) − 6S(k − 2) = −30
(2.17)
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is a constant, by Rule 1, the particular solution will be a constant, say Q. Substituting Q into (2.17), we obtain Q − Q − 6Q = −30 ⇒ Q = 5. ∴
The general solution is S(k) = A1 (−2)k + A2 (3)k + 5.
Using the initial conditions, we have S(0) = 20 ⇒ A1 + A2 + 5 = 20.
(2.18)
S(1) = −5 ⇒= 2A1 + 3A2 + 5 = −5.
(2.19)
Solving (2.18) and (2.19), we get A1 = 11, A2 = 4. ∴
The complete solution is S(k) = 11(−2)k + 4(3)k + 5.
2. Solve the recurrence relation fn − 5fn−1 + 6fn−2 = 1. Solution. The associated homogenous relation is fn − 5fn−1 + 6fn−2 = 0. The characteristic equation is r2 − 5r + 6 = 0. The characteristic roots are r = 2, 3. The homogenous solution is A2 (2)n + A2 (3)n . Since the right-hand side of the given relation is 1 (a constant), by Rule 1, the particular solution will also be a constant, say Q. Q − 5Q + 6Q = 1 1 ⇒ Q= . 2 ∴
The complete solution is 1 fn = A1 (2)n + A2 (3)n + . 2
3. Find the particular solution of the recurrence relation f (n) + 5f (n − 1) + 6f (n − 2) = 3n2 − 2n + 1.
(2.20)
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Discrete Mathematical Structures Solution. By rule 2, the particular solution is of the form Q0 + Q1 n + Q2 n2 .
(2.21)
Substituting (2.21) in (2.20), we get Q0 + Q1 n + Q2 n2 + 5 Q0 + Q1 (n − 1) + Q2 (n − 1)2 + 6 Q0 + Q1 (n − 2) + Q2 (n − 2)2 = 3n2 − 2n + 1 which simplifies to (12Q0 − 17A1 + 29Q2 ) + (12Q1 − 34Q2 )n + 12Q2 n2 = 3n2 − 2n + 1.
(2.22)
Comparing both sides of (2.22), we obtain 12Q2 = 3;
12Q1 − 34Q2 = −2;
12Q0 − 17Q1 + 29Q2 = 1
which gives 13 1 ; Q1 = ; 4 24 The particular solution is Q2 =
Q0 =
71 . 288
71 13 1 + n + n2 . 288 24 4 4. Solve ar + 5ar−1 = 9 with initial condition a0 = 6. Solution. The associated homogenous relation is ar + 5ar−1 = 0. The characteristic equation is r + 5 = 0. The characteristic root is r = −5. The homogenous solution is A(−5)r . Since the right-hand side of the given relation is a constant, the particular solution will also be a constant Q. Substituting in the relation, we get Q + 5Q = 9 3 ⇒ Q= . 2 ∴
The general solution is 3 Ar = A(−5)r + . 2
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Given: 3 =6 2 9 ⇒A= . 2
a0 = 6 ⇒ A +
The complete solution is ar =
9 3 (−5)r + . 2 2
5. Solve the recurrence relation f (n) − 7f (n − 1) + 10f (n − 2) = 6 + 8n with f (0) = 1 and f (1) = 2. Solution. The characteristic equation is r2 − 7r + 10 = 10 ⇒ r = 2, 5. Homogenous solution is A1 (2)n + A2 (5)n . By Rule 1, the particular solution is of the form Q0 + Qn1 . Substituting in the given relation, we obtain (Q0 + Qn1 ) − 7[Q0 + Q1 (n − 1)] + 10[Q0 + Q1 (n − 2)] = 6 + 8n. Comparing both sides, we obtain 4Q0 − 13Q1 = 6
and
4Q1 = 8
which yield Q0 = 8 and Q1 = 2. ∴
The particular solution is 8 + 2n.
The general solution is f (n) − A1 (2)n + A2 (5)n + 8 + 2n. Given: f (0) = 1 ⇒ A1 + A2 + 8 = 1.
(2.23)
f (1) = 2 ⇒ 2A1 + 5A2 + 10 = 2.
(2.24)
Solving (2.23) and (2.24), we get A1 = −9, ∴
A2 = 2.
The complete solution is f (n) = −9(2)n + 2(5)n + 8 + 2n.
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Discrete Mathematical Structures 6. Find the particular solution of the recurrence relation an + 5an−1 + 6an−2 = 42(4)n . Solution. The characteristic equation is r2 + 5r + 6 = 0. The characteristic roots are r = −2, −3. Since 4 is not a characteristic root, by Rule 2, we assume that the general form of the particular solution is Q · (4)n . Substituting in the given relation, we obtain Q · (4)n + 5Q · (4)n−1 + 6Q · (4)n−2 = 42(4)n
Q · 4n−2 [16 + 20 + 6] = 42(4)n
⇒ ⇒
Q · 4n−2 (42) = 42(4)n Q = 16.
The particular solution is 16(4)n = 4n+2 .
7. Find the particular solution of the recurrence relation fn + fn−1 = 3n2n . Solution. The characteristic equation is r + 1 = 0. The characteristic root is r = −1. Since 2 is not a characteristic root, by Rule 2, the general form of the particular solution is (Q0 + Q1 n)2n . Substituting in the given relation, we obtain (Q0 + Q1 n)2n + [Q0 + Q1 (n − 1)]2n − 1 = 3n2n which simplifies to 1 1 1 Q0 2n + Q1 n2n + Q0 2n + Q1 n2n − Q1 n2n = 3n2n 2 2 2 3 1 3 n n ⇒ Q0 − Q1 2 + Q1 n2 = 3n2n 2 2 2 3 1 Q0 − Q1 = 0 ⇒ 2 2 3 and Q1 = 3. 2 Solving, we get Q0 =
2 3
and Q1 = 2. 2 ∴ The particular solution is + 2n 2n . 3 8. Find the particular solution of the recurrence relation f (n) − 2f (n − 1) = 3 · 2n .
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101
Solution. The characteristic equation is r − 2 = 0. The characteristic equation is r = 2. Since r = 2 is the characteristic root of multiplicity 1, by Rule 3, the general form of the particular solution is Qn · 2n . Substituting in the given relation, we obtain Qn2n − 2 Q · (n − 1)2n−1 = 3 · 2n ⇒ Qn2n − Qn2n + Q2n = 3 · 2n ⇒ Q = 3. The particular solution is 3n(2n ).
9. Find the general solution of f (n) − 3f (n − 1) − 4f (n − 2) = 4n .
(2.25)
Solution. The associated homogenous relation is f (n) − 3f (n − 1) − 4f (n − 2) = 0. The characteristic equation is r2 − 3r − 4 = 0. The characteristic roots are r = −1, 4. The homogenous solution is A1 (−1)n + A2 (4)n . Since 4 is a characteristic root, by Rule 3, we assume that the general form of the particular solution is Qn4n . Substituting in (2.25), we obtain Qn4n − 3Q · (n − 1)4n−1 − 4Q · (n − 2)4n−2 = 4n ⇒
Qn4n − 3Qn4n−1 + 3Q4n−1 − 4Qn4n−2 + 8Q4n−2 = 4n
⇒ ⇒
(16Qn − 12Qn + 12Q − 4Qn + 8Q)4n−2 = (16)4n−2 20Q = 16 4 Q= . 5
4 n(4)n . 5 The general solution of the given recurrence relation is ∴
The particular solution is
4 f (n) = A1 (−1)n + A2 (4)n + n4n . 5
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Discrete Mathematical Structures
Remark: What if the characteristic equation gives rise to complex roots? Here, our methods are still valid, but the method for expressing the solutions of the recurrence relations is different. Since an understanding of these representations require some background in complex numbers, we suggest that an interested reader refer to a more advanced treatment of recurrence relations.
2.7.2
Problems for Practice
1. Find the general solution of the following recurrence relations. (i) (ii) (iii) (iv)
fn − 3fn−1 − 10fn−2 = 0 fn+2 + 6fn+1 + 9fn = 0 2fn + 2fn−1 − fn−2 = 0 fn − 3fn−1 − 4fn−2 .
2. Solve the following recurrence relations. f (n) − 10f (n − 1) + 9f (n − 2) = 0; f (0) = 3; f (1) = 11 f (n) − 9f (n − 1) + 18f (n − 2) = 0; f (0) = 1; f (1) = 4 f (n + 2) − 8f (n + 1) + 16f (n) = 0; f (0) = 0; f (1) = 8 f (n) − 3f (n − 1) + 3f (n − 2) − f (n − 3) = 0; f (1) = 0; f (2) = 1; f (3) = 0 (v) f (n + 2) − 2f (n + 1) + f (n) = 0; f (0) = 1; f (1) = 2 (vi) f (n) − 20f (n − 1) + 100f (n − 2) = 0; f (0) = 2; f (1) = 30.
(i) (ii) (iii) (iv)
3. Find the recurrence relation satisfying (i) (ii) (iii) (iv) (v)
yn yn yn yn yn
= A(3)n + B(8)n = (A + Bn)(−2)n = (A + Bn)(6)n = A(3)n + B(5)n = 2(3)n .
4. Solve the following set of recurrence relations with the initial conditions. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)
yn − 2yn−1 = 6n; y1 = 2 yn+2 + 2yn+1 − 15yn = 6n + 10; y0 = 1; y1 = − 21 yn+1 + 2yn = 3 + 4n ; y0 = 2 yn+2 − 2yn+1 + yn = 1; y0 = 1; y1 = 12 yn+1 + yn = 5; y0 = 1 yn − 3yn−1 + 2yn−2 − n2 ; y0 = 0; y1 = 0 yn − 4yn−1 + 4yn−2 = 3n + 2n ; y0 = 1; y1 = 1 yn − 5yn−1 = 5n ; y0 = 3.
Combinatorics
2.8
103
Generating Functions
In this section, we will show how recurrence relations can be solved using the powerful generating function method. Generating function is an important tool in discrete mathematics, and its use is by no means confined to the solution of recurrence relations. If a0 , a1 , a2 , . . . , an is a finite sequence of numbers, the generating function for the an ’s is the polynomial G(z) =
n X
ak z k = a0 + a1 z + a2 z 2 + · · · + an z n
k=0
where z is an indeterminate (that is, an abstract) symbol. If a0 , a1 , a2 , . . . , an , . . . is an infinite sequence of numbers, its generating function is defined to be G(z) =
∞ X
ak z k = a0 + a1 z + a2 z 2 + . . . .
k=0
The symbol z is just the name given to a variable and has no special significance. For any sequence {an }, we write G(z) to denote the generating function of {an }. Clearly, given a sequence, we can easily obtain its generating function and its converse. For example, the generating function of an = αn , n ≥ 0 is α0 + αz + α2 z 2 + α3 z 3 + . . . (2.26) 1 We note that the infinite series (2.26) can be written in closed form as 1 − αz which is a rather compact way to represent the sequence {an } or (a, α, α2 , . . . ).
2.8.1
Solved Problems
1. Find the generating function for the sequence 1, 1, 1, 1, 1, 1. Solution. By definition, the generating function of 1, 1, 1, 1, 1, 1 is G(z) = 1 + z + z 2 + z 3 + z 4 + z 5 =
z6 − 1 . z−1
2. Let n be a positive integer. Let ak = C(n, k) for k = 0, 1, 2, . . . , n. Find the generating function for the sequence a0 , a1 , . . . , an . Solution. The generating function for this sequence is G(z) = C(n, 0) + C(n, 1)z + C(n, 2)z 2 + · · · + c(n, n)z n = (1 + z)n , by binomial theorem.
104
Discrete Mathematical Structures 3. Find the generating function for the infinite sequence 1, α, α2 , α3 , . . . , where α is a fixed constant. Solution. The generating function for this sequence is G(z) = 1 + αz + α2 z 2 + α3 z 3 + . . . 1 = . 1 − αz 4. Find the generating function for fn = 3n , n ≥ 0 in closed form. Solution. The generating function for this sequence is G(z) = 1 + 3z + (3z)2 + (3z)3 + . . . 1 . = 1 − 3z 5. Find the generating function (in closed form) of the Fibonacci sequence {fn } defined by fn = fn−1 + fn−2 ; f0 = 0; f1 = 1. Solution. The generating function is G(z) = f0 + f1 z + f2 z 2 + f3 z 3 + · · · =
∞ X
fn z n .
(2.27)
n=0
Consider fn = fn−1 + fn−2 ; n ≥ 0. Multiplying both sides by z n and summing over all n ≥ 2, we get ∞ X
fn z n =
n=2
∞ X
fn−1 z n +
n=2
∞ X
fn−2 z n .
n=2
Consider the first sum ∞ X
fn z n = f2 z 2 + f3 z 3 + . . .
n=2
= G(z) − f0 − f1 z
[using (2.28)] .
Similarly, ∞ X
fn−1 z n = f1 z 2 + f2 z 3 + . . .
n=2
= z(f1 z + f2 z 2 + . . . ) = z[G(z) − f0 ] [using (2.28)]
(2.28)
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105
and ∞ X
fn−2 z n = f0 z 2 + f1 z 3 + f2 z 4 + . . .
n=2
= z 2 (f0 + f1 z + f2 z 2 + . . . ) = z 2 [G(z)]
[using (2.28)] .
Substituting these expressions in (2.28), we obtain G(z) − f0 − f1 z = z[G(z) − f0 ] + z 2 G(z). Since f0 = 0 and f1 = 1, we get G(z) − z = zG(z) + z 2 G(z) ⇒ ⇒
G(z)(1 − z − z 2 ) = z z G(z) = 1 − z − z2
which is the required generating function. 6. Find the generating function of the sequence y0 , y1 , . . . , yn defined as follows: yn + 2yn−1 − 15yn−2 = 0
for n ≥ 2
with y0 , y1 = 1. (2.29)
Solution. The generating function is G(z) = y0 + y1 z + y2 z 2 + · · · =
∞ X
yn z n .
n=2
Multiplying (2.29) by z n and summing over all n ≥ 2, we get ∞ X
n
yn z + 2
n=2
G(z) −
∞ X
n=2 y0 − y12
n
yn−1 z − 15
∞ X
yn−2 z n = 0
n=2
+ 2 [z{G(z) − y0 }] − 15z 2 G(z) = 0.
Since y0 = 0, y1 = 1, we get [G(z) − 1] + 2[zG(z) − 15z 2 G(z)] = 0 ⇒ ⇒ ⇒
G(z)(1 + 2z − 15z 2 ) = z z G(z) = 1 + 2z − 15z 2 z G(z) = . (1 − 3z)(1 + 5z)
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Discrete Mathematical Structures
2.8.2
Solution of Recurrence Relations Using Generating Function
We can find the solution to a recurrence relation with its initial conditions by finding an explicit formula for the associated generating function. This is illustrated in the following examples. The following are some important fundamental results useful for solved examples presented below: P∞ xr 1. (1 + x)n = r=0 n(n−1)...(n−r+1) r! P∞ 2. (1 + x)−n = r=0 (−1)r n(n−1)...(n−r+1) r! P∞ 3. (1 + x)−1 = r=0 (−1)r xr = 1 + x + x2 + . . . P∞ 4. (1 + x)−2 = r=0 (−1)r (r + 1)xr = 1 − 2x + 3x2 − 4x3 + . . . P∞ 5. (1 − x)−1 = r=0 xr = 1 + x + x2 + . . . P ∞ 6. (1 − x)−2 = r=0 (r + 1)xr = 1 + 2x + 3x2 + 4x3 + . . . P∞ r 2 3 x + x2! + x3! + . . . 7. ex = r=0 xr! = 1 + 1!
2.8.3
Solved Problems
1. Using the generating function, solve the recurrence relation fn = 3fn−1 , for n = 1, 2, 3 . . . and initial condition f0 = 2. Solution. Let the generating function be G(z) =
∞ X
fn z n .
(2.30)
n=0
Multiplying the given relation by z n and summing for all n ≥ 1, we obtain ∞ ∞ X X fn z n = 3 fn−1 z n . (2.31) n=1
n=1
The first sum ∞ X
fn z n = f1 z + f2 z 2 + · · · = G(z) − f0
[using (2.30)]
n=1
and the second sum ∞ X
fn−1 z n = f0 z + f1 z 2 + f2 z 3 + . . .
n=1
= z[f0 + f1 z + f2 z 2 + . . . ] = zG(z) [using (2.30)] .
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107
Hence, (2.31) becomes
G(z) − f0 = 3zG(z) G(z) − 2 = 3zG(Z)
G(z) = =
∞ X
∞ X 2 = 2(1 − 3z)−1 = 2 (3z)n 1 − 3z n=0
2 · 3n z n .
(2.32)
n=0
Comparing (2.30) and (2.32), we get fn = 2 · 3n which is the required solution. 2. Using generating function, solve the recurrence relation yn = 3yn−1 + 2; n ≥ 1 with y0 = 1. Solution. Let the generating function be ∞ X
yn z n .
(2.33)
yn = 3yn−1 + 2.
(2.34)
G(z) =
n=0
Given: Multiplying both sides of (2.34) by z n and summing for n ≥ 1, we get ∞ ∞ ∞ X X X yn z n = 3 yn−1 z n + 2 zn. (2.35) n=1
n=1
n=1
Consider the first sum ∞ X
yn z n = y1 z + y2 z 2 + · · · = G(z) − y0
[using (2.33)] ,
n=1
the second sum ∞ X
yn−1 z n = y0 z + y1 z 2 + y2 z 3 + . . .
n=1
= z[y0 + y1 z + y2 z 2 + . . . ] = zG(z) [using (2.33)] ,
108
Discrete Mathematical Structures and the third sum ∞ X
= z + z2 + z3 + . . .
n=1
= z(1 + z + z 2 + . . . ) = z(1 − z)−1 z = . 1−z Hence, (2.35) becomes G(z) − y0 = 3zG(z) +
2z . 1−z
Using y0 = 1, G(z)(1 − 3z) = 1 + ⇒
G(z) =
2z 1+z = . 1−z 1−z
1+z . (1 − z)(1 − 3z)
Using partial fraction,
A B 1+z = + (1 − z)(1 − 3z) 1 − 3z 1−z 1 + z = A(1 − z) + B(1 − 3z).
Put z = 1 =⇒ 2 = −2B =⇒ B = −1. 4 2 1 Put z = =⇒ = A =⇒ A = 2. 3 3 3 ∴
G(z) =
2 1 − = 2(1 − 3z)−1 − (1 − z)−1 1 − 3z 1−z ∞ ∞ X X =2 (3z)n − zn n=0
=
∞ X
n=0
(2 · 3n − 1) z n .
n=0
Comparing (2.33) and (2.36), we get yn = 2 · 3n − 1. 3. Using generating function, solve the difference equation yn+2 − 4yn+1 + 3yn = 0; y0 = 2; y1 = 4.
(2.36)
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109
Solution. Let the generating function be G(z) =
∞ X
yn z n .
(2.37)
n=0
Multiplying the given relation by z n and summing for n ≥ 0, we get ∞ ∞ ∞ X X X yn+2 z n − 4 yn+1 z n + 3 yn z n = 0. (2.38) n=0
n=0
n=0
Consider the first sum ∞ X
yn+2 z n = y2 + y3 z + y4 z 2 + y5 z 3 + . . .
n=2
1 y2 z 2 + y3 z 3 + y4 z 4 + . . . z2 1 = [G(z) − y0 ] [using (2.37)] , 2
=
the second sum ∞ X
= y1 + y2 + y3 z 2 + y4 z 3 + . . .
n=0
1 [y1 z + y2 z 2 + y3 z 3 + . . . ] z 1 = [G(z) − y0 ] [using (2.37)] , z =
and the third sum ∞ X
yn z n = G(z)
[using (2.37)] .
n=0
(2.38) becomes 1 4 [G(z) − y0 − y1 z] − [G(z) − y0 ] + 3G(z) = 0. 2 z z
Since y0 = 2, y1 = 4,
=⇒ =⇒ =⇒
1 4 [G(z) − 2 − 4z] − [G(z) − 2] + 3G(z) = 0 z2 z [G(z) − 4z − 2] − 4z[G(z) − 2] + 3z 2 G(z) = 0 G(z)[1 − 4z + 3z 2 ] = 2 − 4z 2 − 4z 2 − 4z G(z) = = . 1 − 4z + 3z 2 (1 − z)(1 − 3z)
110
Discrete Mathematical Structures Using partial fractions,
=⇒
2 − 4z A B = + (1 − z)(1 − 3z) 1−z 1 − 3z 2 − 4z = A(1 − 3z) + B(1 − z).
Put z = 1 =⇒ −2 = −2A =⇒ A = 1. Put z = 13 =⇒ 23 = 32 B =⇒ B = 1. ∴ G(z) =
1 1 + = (1 − z)−1 + (1 − 3z)−1 1−z 1 − 3z ∞ ∞ X X = zn + (3z)n =
n=0 ∞ X
n=0
(1 + 3n )z n .
(2.39)
n=0
Comparing (2.37) and (2.39), the required solution is yn = 1 + 3n . 4. Using generating function, solve the difference equation yn+2 − 6yn+1 + 8yn = 0, y0 = 1, y1 = 4. Solution. Let the generating function be G(z) =
∞ X
yn z n .
(2.40)
n=0
Multiplying the given equation by z n and summing for n ≥ 0, we get ∞ ∞ ∞ X X X n n yn+2 z − 6 yn+1 z + 8 yn z n = 0. (2.41) n=0
n=0
n=0
Consider the first sum ∞ X
yn+2 z n = y2 + y3 z + y4 z 2 + y5 z 3 + . . .
n=0
1 y2 z 2 + y3 z 3 + y4 z 4 + . . . 2 z 1 = 2 [G(z) − y0 − y1 z] [using (2.40)] , z =
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111
the second sum ∞ X
yn+1 z n = y1 + y2 z + y3 z 2 + . . .
n=0
1 (y1 z + y2 z 2 + y3 z 3 + . . . ) z 1 = [G(z) − y0 ] [using (2.40)] , z
=
and the third sum ∞ X
yn z n = G(z)
[using (2.40)] .
n=0
Hence, (2.41) becomes
=⇒ =⇒ =⇒
6 1 [G(z) − y0 − y1 z] − [G(z) − y0 ] + 8G(z) = 0 2 z z [G(z) − y0 − y1 z] − 6z[G(z) − y0 ] + 8z 2 G(z) = 0 G(z)[1 − 6z + 8z 2 ] = 1 − 2z (∵ y0 = 1, y1 = 4) 1 − 2z 1 − 2z G(z) = = 1 − 6z + 8z 2 (1 − 2z)(1 − 4z) ∞ X −1 = (1 − 4z) = 4n z n . (2.42) n=0
Comparing (2.40) and (2.42), the solution is yn = 4n . 5. Solve S(k) − 7S(k − 2) + 6S(k − 3) = 0, S(0) = 8, S(1) = 6, and S(2) = 22. Solution. Let the generating function be G(z) =
∞ X
S(k)z k .
(2.43)
k=0
Multiplying the given equation by z k and summing for k ≥ 3, we get ∞ X k=3
S(k)z k − 7
∞ X k=3
S(k − 2)z k + 6
∞ X
S(k − 3)z k = 0.
(2.44)
k=3
Consider the first sum ∞ X
S(k)z k = S(3)z 3 + s(4)z 4 + s(5)z 5 + . . .
k=3
= G(z) − S(0)S(1)z − S(2)z 2
[using (2.43)] ,
112
Discrete Mathematical Structures the second sum ∞ X
s(k − 3)z k = S(0)z 3 + S(1)z 4 + S(2)z 5 + . . .
k=3
= z 3 [S(0) + S(1)z + S(2)z 2 + . . . ] = z 3 G(z)
[using (2.43)] ,
and the third sum ∞ X
s(k − 3)z k = S(0)z 3 + S(1)z 4 + S(2)z 5 + . . .
k=3
= z 3 [S(0) + S(1)z + S(2)z 2 + . . . ] = z 3 G(z)
[using (2.43)] .
(2.44) becomes
[G(z) − s(0) − s(1)z − s(2)z 2 ] − 7z 2 [G(z) − s(0)] + 6z 3 G(z) = 0. Since S(0) = 8, S(1) = 6, S(2) = 22, [G(z) − 8 − 6z − 22z 2 ] − 7z 2 [G(z) − 8] + 6z 3 G(z) = 0 =⇒
G(z)[1 − 7z 2 + 6z 3 ] = 8 + 6z − 34z 2
=⇒
G(z) =
8 + 6z − 34z 2 1 − 7z 2 + 6z 3 8 + 6z − 34z 2 . G(z) = (1 − z)(1 − 2z)(1 − 3z)
=⇒
Using partial fractions, 8 + 6z − 34z 2 A B C = + + (1 − z)(1 − 2z)(1 − 3z) 1−z 1 − 2z 1 + 3z =⇒
8 + 6z − 34z 2 = A(1 − 2z)(1 + 3z) + B(1 − z)(1 + 3z) + C(1 − z)(1 − 2z).
Put z = 1 =⇒ −20 = −4A =⇒ A = 5. 1 10 5 Put z = =⇒ = B =⇒ B = 2. 2 4 4 1 20 20 Put z = − =⇒ = C =⇒ C = 1. 3 9 9 ∴
5 2 1 + + 1−z 1 − 2z 1 + 3z = 5(1 − z)−1 + 2(1 − 2z)−1 + (1 + 3z)−1 ∞ ∞ ∞ X X X =5 zk + 2 (2z)k + (−3)k z k
G(z) =
k=0
k=0
k=0
Combinatorics
113 =
∞ X 5 + 2k+1 + (−3)k z k .
(2.45)
k=0
Comparing (2.43) and (2.45), the solution is S(k) = 5 + 2k+1 + (−3)k . 6. Suppose that a valid code word is an n-digit number in decimal notation containing an even number of 0’s. Let an denote the number of valid code words of length n. The sequence {an } satisfies the recurrence relation an = 8an−1 +10n−1 and the initial condition a1 = 9. Use generating function to find an explicit formula for an . Solution. To make our work with generating function simpler, we extend this sequence by setting a0 = 1 so that a1 = 8a0 + 100 = 9. Let the generating function be G(z) =
∞ X
an z n .
(2.46)
n=0
Multiplying both sides of an = 8an−1 + 10n−1 by z n and summing for all n ≥ 1, we get ∞ X
an z n = 8
∞ X
an−1 z n +
n=1
n=1
∞ X
10n−1 z n .
Consider the first sum ∞ X an z n = a1 z + a2 z 2 + a3 z 3 + · · · = G(z) − a0
[using (2.47)] ,
n=1
the second sum ∞ X
an−1 z n = a0 z + a1 z 2 + a2 z 3 + . . .
n=1
= z(a0 + az + a2 z 2 + . . . ) = zG(z) [using (2.46)] , and the third sum ∞ ∞ X X 10n−1 z n = z (10z)n n=1
n=1
= z(1 − 10z)−1 = ∴
(2.47)
n=1
z . 1 − 10z
(2.47) becomes [G(z) − a0 ] = 8zG(z) +
z . 1 − 10z
114
Discrete Mathematical Structures Using a0 = 1, z 1 − 9z = 1 − 10z 1 − 10z 1 − 9z G(z) = . (1 − 8z)(1 − 10z) G(z)[1 − 8z] = 1 + =⇒
Using partial fraction,
=⇒
A B 1 − 9z = + (1 − 8z)(1 − 10z) 1 − 8z 1 − 10z 1 − 9z = A(1 − 10z) + B(1 − 8z). 1 5 1 z = =⇒ − = A 1 − 8 8 4 1 1 =⇒ − = − A 8 4 1 =⇒ A = . 2 z=
1 1 1 =⇒ = B 10 10 5 1 =⇒ B = . 2
1 1 (1 − 8z)−1 + (1 − 10z)−1 2 2 ∞ ∞ X 1 1X n = (8z) + (10z)n 2 n=0 2 n=0
G(z) =
=
∞ X 1 n (8 + 10n ) z n . 2 n=0
(2.48)
Comparing (2.46) and (2.48), the solution is an =
1 n (8 + 10n ) . 2
7. Solve S(n) − 2S(n − 1) − 3S(n − 2) = 0, n ≥ 2 with S(0) = 3 and S(1) = 1 using generating function. Solution. Let the generating function be G(z) =
∞ X n=0
S(n)z n .
(2.49)
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115
Multiplying the given equation by z n and summing for n ≥ 2, we get ∞ X
S(n)z n − 2
n=2
∞ X
S(n − 1)z n − 3
n=2
∞ X
S(n − 2)z n = 0.
(2.50)
n=2
Consider the first sum ∞ X
S(n)z n = S(2)z 2 + S(3)z 3 + S(4)z 4 + . . .
n=2
= G(z) − S(0) − S(1)z
[using (2.49)] ,
the second sum ∞ X
S(n − 1)z n = S(1)z 2 + S(2)z 3 + S(3)z 4 + . . .
n=2
= z[S(1)z + S(2)z 2 + S(3)z 3 + . . . ] = z[G(z) − S(0)] [using (2.49)] , and the third sum ∞ X
S(n − 2)z n S(0)z 2 + S(1)z 3 + S(2)z 4 + . . .
n=2
= z 2 [S(0) + S(1)z + S(2)z 2 + . . . ] = z 2 G(z). ∴
(2.50) becomes [G(z) − S(0) − S(1)z] − 2z[G(z) − S(0)] − 3z 2 G(z) = 0.
Using S(0) = 3 and S(1) = 1, we get [G(z) − 3 − z] − 2z[G(z) − 3] − 3z 2 G(z) = 0 =⇒ =⇒
G(z)[1 − 2z − 3z 2 ] = 3 − 5z 3 − 5z 3 − 5z . = G(z) = 2 1 − 2z − 3z (1 − 3z)(1 + z)
Using partial fractions,
=⇒
3 − 5z A B = + (1 − 3z)(1 + z) 1 − 3z 1+z 3 − 5z = A(1 + z) + B(1 − 3z).
Put z = −1 =⇒ 8 = 4B =⇒ B = 2.
116
Discrete Mathematical Structures Put z =
1 4 4 =⇒ = A =⇒ A = 1. 3 3 3 ∴
1 2 + 1 − 3z 1+z = (1 − 3z)−1 + 2(1 + z)−1 ∞ ∞ X X = (3z)n + 2 (−z)n
G(z) =
=
n=0 ∞ X
n=0
[3n + 2(−1)n ] z n .
(2.51)
n=0
Comparing (2.49) and (2.51), the solution is S(n) = 3n + 2(−1)n .
2.8.4
Problems for Practice
1. Find the generating function of the following sequences. (i) (ii) (iii) (iv)
2, 2, 2, 2, 2, 2, . . . an = (−2)n 0, 0, 1, 1, 1, 1, . . . 1, 0, −1, 0, 1, 0, 0, −1, 0, 1, 0, −1, 0, . . .
2. For the following expressions, identify the sequences having the expression as a generating function. [Hint: Use partial fractions.] 5 + 2z 1 − 4z 2 6 − 29z (ii) 1 − 11z + 30z 2 32 − 22z (iii) 2 − 3z + z 2 3 + 7z (iv) 1 + 3z − 4z 2 3 + 5z . (v) 1 − 2z − 3z 2 3. Find the generating functions for the following sequences satisfying the given initial conditions. (i)
(i) (ii) (iii) (iv) (v)
yn+2 + 2yn+1 − 15yn = 0; y0 =, y1 = 1. yn+1 − yn−1 = 0; y0 = 0, y1 = 1. yn+2 − 2yn+1 − 3yn = 0; y0 = 0, y1 = 1. S(n) − 2S(n − 1) − 3S(n − 2) = 0; S(0) = 3, S(1) = 1. S(k) + 3S(k − 1) − S(k − 2) = 0; S(0) = 3, S(1) = 2.
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4. Using generating function, solve the following recurrence relations. (i) (ii) (iii) (iv) (v) (vi) (vii)
yn = 7yn−1 ; y0 = 5. yn = 3yn−1 + 4n−1 ; y0 = 1. yn + 2yn−1 − 15yn−2 = 0; y0 = 0, y1 = 1. yn+2 − 8yn+1 + 16yn = 0; y0 = 0, y1 = 8. yn+2 − 2yn+1 + yn = 0; y0 = 2, y1 = 1. yn+2 − yn+1 − 6yn = 0; y0 = 2, y1 = 1. yn+2 − 5yn+1 + 6yn = 0; y0 = 1, y1 = 3.
5. Use a generating function to find an explicit formula for the Fibonacci numbers.
2.9
Inclusion–Exclusion Principle
Let A and B be any finite sets. Then, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). In other words, to find the number n(A ∪ B) of elements in the union A ∪ B, we add n(A) and n(B), and then we subtract n(A ∩ B); that is, we “include” n(A) and n(B), and we “exclude” n(A ∩ B). This follows from the fact that, when we add n(A) and n(B), we have counted the elements of A ∩ B twice. This principle holds for any number of sets. We first state it for three sets. Theorem 2.9.1 For any finite sets A, B, and C, we have n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C). That is, we “include” n(A), n(B), n(C), we “exclude” n(A ∩ B), n(B ∩ C), n(A ∩ C), and we “include” n(A ∩ B ∩ C). Example: Find the number of mathematics students at a college taking at least one of the languages French, German, and Russian given the following data: 65 study French, 20 study French and German, 45 study German, 25 study French and Russian, 42 study Russian, 15 study German and Russian, and 8 study all three languages.
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Discrete Mathematical Structures
We want to find n(F ∪ G ∪ R), where F , G, and R denote the sets of students studying French, German, and Russian, respectively. By the inclusion–exclusion principle, n(F ∪ G ∪ R) = n(F ) + n(G) + n(R) − n(F ∩ G) − n(F ∩ R) − n(G ∩ R) + n(F ∩ G ∩ R) = 65 + 45 + 42 − 20 − 25 − 15 + 8 = 100. ∴ 100 students study at least one of the languages. Note: Principle of inclusion–exclusion can also be denoted as (i) |A ∪ B| = |A| + |B| − |A ∩ B| if A and B are not disjoint sets (ii) |A ∪ B| = |A| + |B| if A and B are disjoint sets where |A| = n(A) = cardinality of A = number of elements in A.
2.9.1
Solved Problems
1. Find the number of positive integers not exceeding 100 that are divisible by 7 or by 11. Solution. Let A be the set of positive integers not exceeding 100 that are divisible by 7. Let B be the set of positive integers not exceeding 100 that are divisible by 11. Then, A∪B is the set of positive integers not exceeding 100 that are divisible by either 7 or 11, and A ∩ B is the set of positive integers not exceeding 100 that are divisible by both 7 and 11. We knowthat among the positive integers not exceeding 100, there 100 100 are integers divisible by 7 and integers divisible by 11. 7 11 Since 7 and 11 are relatively prime, the integers divisible by both 7 and 11 are those divisible by 7 and 11. 100 There are positive integers not exceeding 100 that are 7 × 11
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119
divisible by both 7 and 11. ∴
|A ∩ B| = |A| + |B| − |A ∩ B| 100 100 100 = + − 7 11 7 × 11 = 14 + 9 − 1 = 22.
2. Among the first 1000 positive integers, determine the integers which are not divisible by 5, nor by 7, nor by 9. Solution. Let A = set of integers divisible by 5 B = set of integers divisible by 7 C = set of integers divisible by 9. 1000 1000 1000 ∴ |A| = = 200; |B| = = 142; C = = 111. 5 7 9 1000 1000 |A ∩ B| = = = 28 LCM (5, 7) 35 1000 1000 |B ∩ C| = = = 15 LCM (7, 9) 63 1000 1000 |A ∩ C| = = = 22 LCM (5, 9) 45 1000 1000 1000 = = = 3. |A ∩ B ∩ C| = LCM (5, 7, 9) 5×7×9 315 The number of integers divisible by 5, 7, and 9 is |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |B ∩ C| − |A ∩ C| + |A ∩ B ∩ C| = 200 + 142 + 111 − 28 − 15 − 22 + 3 = 391. The number of integers not divisible by 5 nor 7 nor 9 = Total number of integers - integers divisible by 5, 7, and 9 = 1000 − 391 = 609. 3. In a survey of 300 students, 64 had taken a Mathematics course, 94 had taken an English course, 58 had taken a Computer course, 28 had taken both Mathematics and Computer courses, 26 had taken both English and Mathematics courses, 22 had taken both
120
Discrete Mathematical Structures English and Computer courses, and 14 had taken all three courses. How many students were surveyed who had taken none of the three courses? Solution. Given: |M | = 64; |M ∩ C| = 28;
|E| = 94;
|M ∩ E| = 26;
|C| = 58; |E ∩ C| = 22;
|M ∩ C ∩ E| = 14.
|M ∪ E ∪ E| = |M | + |E| + |C| − |M ∩ E| − |M ∩ C| − |E ∩ C| + |M ∩ E ∩ C| = 64 + 94 + 58 − 26 − 28 − 22 + 14 = 154. ∴ Students who had taken none of the courses = 300−154 =146. 4. How many solutions does x1 + x2 + x3 = 13 have, where x1 , x2 , and x3 are non-negative integers with x1 < 6, x2 < 6, and x3 < 6? Solution. To apply the principle of inclusion–exclusion, let a solution have property P1 if x ≥ 6, property P2 if x2 ≥ 6, and property P3 if x3 ≥ 6. The number of solutions satisfying the inequalities x1 < 6, x2 < 6, and x3 < 6 is N (P10 P20 P30 ) = N − N (P1 ) − N (P2 ) − N (P3 ) + N (P1 P2 ) + N (P1 P3 ) + N (P2 P3 ) − N (P1 P2 P3 ) (2.52) where N = total number of solutions = C(3 + 13 − 1, 13) = C(15, 13) = 105. N (P1 ) = number of solutions with x1 ≥ 6 = C(3 + 7 − 1, 7) = C(9, 7) = 36 N (P2 ) = number of solutions with x2 ≥ 6 = C(3 + 7 − 1, 7) = C(9, 7) = 36 N (P3 ) = number of solutions with x3 ≥ 6 = C(3 + 7 − 1, 7) = C(9, 7) = 36 N (P1 P2 ) = number of solutions with x1 ≥ 6 and x2 ≥ 6 = C(3 + 1 − 1, 1) = C(3, 1) = 3 N (P1 P3 ) = number of solutions with x1 ≥ 6 and x3 ≥ 6 = C(3 + 1 − 1, 1) = C(3, 1) = 3 N (P2 P3 ) = number of solutions with x2 ≥ 6 and x3 ≥ 6 = C(3 + 1 − 1, 1) = C(3, 1) = 3 N (P1 P2 P3 ) = number of solutions with x ≥ 6, x2 ≥ 6, and x3 ≥ 6 = 0.
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121
Inserting these quantities into the formula N (P10 P20 N30 ) shows that the number of solutions with x1 ≤ 6, x2 ≤ 6, and x3 ≤ 6 equals [implying from (2.52)] N (P10 P20 N30 ) = 105 − 36 − 36 − 36 + 3 + 3 + 3 − 0 = 6. 5. A survey of 500 students from a school produced the following information. 200 play volleyball, 120 play hockey, 60 play both volleyball and hockey. How many are not playing either volleyball or hockey? Solution. Let A be the set of students who play volleyball. Let B be the set of students who play hockey. Given: |A| = 200,
|B| = 120,
|A ∩ B| = 60.
By the principle of inclusion–exclusion, the number of students playing either volleyball or hockey is |A ∪ B| = |A| + |B| − |A ∩ B| = 200 + 120 − 60 = 260. ∴ The number of students not playing either volleyball or hockey = 500 − 260 = 240. 6. A total of 1232 students have taken a course in Russian, 879 have taken a course in German, and 114 have taken a course in French. Further 103 have taken a course in both Russian and German, 23 have taken a course in Russian and French, and 14 have taken a course in German and French. If 2092 students have taken at least one of the courses Russian, German, and French, how many students have taken a course in all three languages. Solution. Let A be the set of students who have taken a course in Russian. Let B be the set of students who have taken a course in German. Let C be the set of students who have taken a course in French. Given: |A| = 1232, |A ∩ B| = 103,
|B| = 879,
|A ∩ C| = 23,
|C| = 114, |B ∩ C|14,
|A ∩ B ∩ C| = 2092.
By the principle of inclusion–exclusion, we have
=⇒ =⇒
|A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |B ∩ C| − |A ∩ C| + |A ∩ B ∩ C| 2092 = 1232 + 879 + 114 − 103 − 23 − 14 + |A ∩ B ∩ C| |A ∩ B ∩ C| = 7.
122
Discrete Mathematical Structures ∴ There are seven students who have taken a course in Russian, German, and French. 7. In a survey of 100 students, it was found that 30 studied Mathematics, 54 studied Statistics, 25 studied Operations Research, 1 studied all the three subjects, 20 studied Mathematics and Statistics, 3 studied Mathematics and Operations Research, and 15 studied Statistics and Operations Research. (i) How many students studied none of these subjects? (ii) How many students studied only Mathematics? Solution. Let A denote the set of students who studied Mathematics. Let B denote the set of students who studied Statistics. Let C denote the set of students who studied Operations Research. Given: |A| = 30, |A ∩ B| = 20,
|B| = 54,
|A ∩ C| = 3,
C = 25, |B ∩ C| = 15,
|A ∩ B ∩ C| = 1.
(i) By the principle of inclusion–exclusion, the number of students who studied any one of the subjects is |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |B ∩ C| − |A ∩ C| + |A ∩ B ∩ C| = 30 + 54 + 25 − 2003 − 15 + 1 = 72. ∴ Number of students who studied none of these subjects = 100 − 72 = 28. (ii) Number of students who studied only Mathematics and Statistics = |A ∩ B| − |A ∩ B ∩ C| = 20 − 1 = 19. Number of students who studied only Mathematics and Operations Research = |A ∩ C| − |A ∩ B ∩ C| = 3 − 1 = 2. ∴ Number of students who studied only Mathematics = 30 − 19 − 2 − 1 = 8. 8. How many positive integers not exceeding 1000 are divisible by 7 or 11? Solution. Let A denote the set of positive integers not exceeding 1000 that are divisible by 7. Let B denote the set of positive integers not exceeding 1000 that are divisible by 11. Then,
Combinatorics 123 1000 1000 1000 |A| = = 142, |B| = = 90, |A ∩ B| = = 12. 7 11 7 × 11 The number of positive integers not exceeding 1000 that are divisible by either 7 or 11 is |A ∪ B|. By the principle of inclusion–exclusion, |A ∪ B| = |A| + |B| − |A ∩ B| = 142 + 90 − 12 = 220. 9. Determine n such that 1 ≤ n ≤ 100 and it is not divisible by 5 or 7. Solution. Let A denote the number n, 1 ≤ n ≤ 100, which is divisible by 5. Let B denote the number n, 1 ≤ n ≤ 100, which is divisible by 7. Then, 100 100 100 |A| = = 20, |B| = = 14, |A ∩ B| = = 2. 5 7 5×5 By the principle of inclusion–exclusion, the number n, 1 ≤ n ≤ 100, which is divisible by either 5 or 7 is |A ∪ B|. |A ∪ B| = |A| + |B| − |A ∩ B| = 20 + 14 − 2 = 32. ∴ The number n, 1 ≤ n ≤ 100, which is not divisible by 5 and 7 is = 100 − 32 = 68. 10. A survey among 100 students shows that of the three ice cream flavours vanilla, chocolate, and strawberry, 50 students like vanilla, 43 like chocolate, 28 like strawberry, 13 like vanilla and chocolate, 11 like chocolate and strawberry, 12 like strawberry and vanilla, and 5 like all of them. Find the number of students surveyed who like the following flavours: (i) chocolate but not strawberry (ii) chocolate and strawberry but not vanilla (iii) vanilla or chocolate but not strawberry. Solution. Let A denote the set of students who like vanilla. Let B denote the set of students who like chocolate. Let C denote the set of students who like strawberry. Since five students like all flavours, |A ∩ B ∩ C| = 5.
124
Discrete Mathematical Structures Twelve students like both strawberry and vanilla =⇒
|A ∩ C| − 5 = 12 − 5 = 7.
But five of them like chocolate also =⇒ Six of them like vanilla =⇒
|A ∩ C| = 12.
|B ∩ C| − 6 = 12 − 6 = 6.
Out of 28 students who like strawberry, we have already accounted for 7 + 5 + 6 = 18. ∴ The remaining ten students belong to the set C − (A ∪ B). Similarly, |A − (B ∪ C)| = 30
and
|B − (A ∪ C)| = 24.
Hence, we have accounted for 90 of the 100 students. The remaining ten students like outside the region A ∪ B ∪ C. Now, (i) |B − C| = 24 + 8 = 32. ∴ 32 students like chocolate but not strawberry. (ii) |(B ∩ C) − A| = 6. ∴ Six students like both chocolate and strawberry but not vanilla. (iii) |(A ∪ B) − C| = 30 + 8 + 24 = 62. ∴ 62 students like vanilla or chocolate, but not strawberry. 11. Find the number of integers between 1 and 250 that are not divisible by any of the integers 2, 3, 5, and 7. Solution. Let A denote the set of integers between 1 and 250 that are divisible by 2. Let B denote the set of integers between 1 and 250 that are divisible by 3. Let C denote the set of integers between 1 and 250 that are divisible by 5. Let D denote the set of integers between 1 and 250 that are divisible by 7. Now, 250 = 125, |A| = 2 250 |C| = = 5, 5
250 |B| = = 83 3 250 |D| = = 35. 7
Number of integers between 1 and 250 that are divisible by 2 and 3 250 = |A ∩ B| = = 41 2×3
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125
Number of integers between 1 and 250 that are divisible by 2 and 5 250 = |A ∩ C| = = 25. 2×5 Similarly,
250 |A ∩ D| = = 17 2×7 250 |B ∩ C| = = 16 3×5 250 |B ∩ D| = = 11 3×7 250 = 7. |C ∩ D| = 5×7 Number of integers between 1 and 250 that are divisible by 2, 3, and 5 250 = |A ∩ B ∩ C| = = 8. 2 × 3 × 50 Similarly,
250 =5 2×3×7 250 |A ∩ C ∩ D| = =3 2×5×7 250 |B ∩ C ∩ D| = =2 3×5×7 250 = 1. |A ∩ B ∩ C ∩ D| = 2×3×5×7 |A ∩ B ∩ D| =
The number of integers between 1 and 250 that are divisible by 2, 3, 5, and 7 is |A ∪ B ∪ C ∪ D|. By principle of inclusion–exclusion, |A ∪ B ∪ C ∪ D| = |A| + |B| + |C| + |D| − |A ∩ B| − |B ∩ C| − |C ∩ D| − |A ∩ C| − |B ∩ D| − |A ∩ D| + |A ∩ B ∩ C| + |A ∩ B ∩ D| + |A ∩ C ∩ D| + |B ∩ C ∩ D| − |A ∩ B ∩ C ∩ D| = 125 + 83 + 50 + 35 − 41 − 25 − 17 − 16 − 11 − 7 +8+5+3+2−1 = 193.
126
Discrete Mathematical Structures ∴ Number of integers between 1 and 250 that are not = 250 − 193 = 57. divisible by 2, 3, 5, and 7
12. A survey shows that 57% of Indians like coffee whereas 75% like tea. What can you say about the percentage of Indians who like both coffee and tea. Solution. Let A denote the set of Indians who like coffee. Let B denote the set of Indians who like tea. Assume the total population is 100. ∴ |A| = 57;
|B| = 75. Now, |A ∪ B| = |A| + |B| − |A ∩ B| = 57 + 75 − |A ∩ B| = 132 − |A ∩ B|.
Since |A ∪ B| ≤ 100, it follows that |A ∩ B| ≥ 32.
(2.53)
Since A ∩ B ⊆ A and A ∩ B ⊆ B, we have |A ∩ B| ≤ |A| ∴
|A ∩ B| ≤ 57 =⇒
and
|A ∩ B| ≤ |B|.
and
|A ∩ B| ≤ 75.
|A ∩ B| ≤ 57.
(2.54)
From (2.53) and (2.54), the percentage of Indians who like both coffee and tea lies between 32 and 57. 13. Out of 100 students in a college, 38 play tennis, 57 play cricket, 31 play hockey, 9 play cricket and hockey, 10 play hockey and tennis, and 12 play tennis and cricket. How many play (i) all three games (ii) just one game (iii) tennis and cricket but not hockey. Assume that each student plays at least one game. Solution. Let T, C, and H denote the set of students playing tennis, cricket, and hockey, respectively (Figure 2.1). Given: |T | = 38, |C| = 57, |H| = 31
Combinatorics
127 C
T
21
41
7
5 5
4
17
H
FIGURE 2.1 Venn diagram |T ∩ C| = 12, |T ∩ H| = 10, |C ∩ H| = 9, |T ∪ C ∪ H| = 100. Number of students who play all three games = |T ∩ C ∩ H|. By principle of inclusion–exclusion, we have
=⇒ =⇒
|T ∩ C ∩ H| = |T | + |C| + |H| − |T ∩ C| − |C ∩ H| − |T ∩ H| + |T ∩ C ∩ H| 100 = 38 + 57 + 31 − 12 − 9 − 10 + |T ∩ C ∩ H| |T ∩ C ∩ H| = 100 − 126 + 31 = 5.
∴ Number of students who play all three games = 5. From the given data, we have Number of students playing just one game = Number of students playing tennis only + Number of students playing cricket only + Number of students playing hockey only = 21 + 41 + 17 = 79. Number of students playing tennis and cricket but not hockey = |T ∩ C| − |T ∩ C ∩ H| = 12 − 5 = 7. 14. How many integers between 1 and 100 are (i) not divisible by 7, 11, or 13 (ii) divisible by 3 but not by 7?
128
Discrete Mathematical Structures Solution. Let A, B, and C be the set of integers between 1 and 100 that are divisible by 7, 11, and 13, respectively. 100 100 100 ∴ |A| = = 14; |B| = ; |C| = = 7; 7 11 13 100 100 = 1; |A ∩ C| = = 1; |A ∩ B| = 7 × 11 7 × 13 100 100 |B ∩ C| = = 0; = 0. 11 × 13 7 × 11 × 13 Number of integers between 1 and 100 that are divisible by 7, 11, or 13 is |A ∪ B ∪ C|. By principle of inclusion–exclusion, we have |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |B ∩ C| − |A ∩ C| + |A ∩ B ∩ C| = 14 + 9 + 7 − 1 − 0 − 1 + 0 = 28. (i) Number of integers between 1 and 100 that are not divisible by 7, 11, or 13 = 100 − 28 = 72. (ii) Let U and V denote the set of integers between 1 and 100 that are divisible by 3 and 7, respectively. 100 100 100 |U | = = 33; |V | = = 14; |U ∩ V | = = 4. 3 7 3×7 ∴
Number of integers divisible by 3 but not by 7 = |U | − |U ∩ V | = 33 − 4 = 29.
15. Find the number of integers between 1 and 100 that are divisible by (i) 2, 3, 5, or 7 (ii) 2, 3, 5 but not by 7. Solution. Let A, B, C, and D denote the set of positive integers between 1 and 100 that are divisible by 2, 3, 5, and 7, respectively. 100 100 = 50; |B| = = 33; ∴ |A| = 2 3 100 100 |C| = = 20; |D| = = 14; 5 7 100 100 |A ∩ B| = = 16; |A ∩ C| = = 10; 2×3 2×5
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129
100 100 = 7; |B ∩ C| = = 7; 2×7 3×5 100 100 |B ∩ D| = = 4; |C ∩ D| = = 2; 3×7 5×7 100 100 |A ∩ B ∩ C| = = 3; |A ∩ B ∩ D| = = 2; 2×3×5 2×3×7 100 100 = 1; |B ∩ C ∩ D| = = 0; |A ∩ C ∩ D| = 2×5×7 3×5×7 100 |A ∩ B ∩ C ∩ D| = = 0. 2×3×5×7 |A ∩ D| =
(i) By the principle of inclusion–exclusion, we have |A ∪ B ∪ C ∪ D| = |A| + |B| + |C| + |D| − |A ∩ B| − |A ∩ C| − |A ∩ D| − |B ∩ C| − |B ∩ D| + |A ∩ B ∩ C| + |A ∩ B ∩ D| + |A ∩ C ∩ D| + |B ∩ C ∩ D| − |A ∩ B ∩ C ∩ D| = 50 + 33 + 20 + 14 − 16 − 10 − 7 − 7 − 4 − 2 +3+2+1+0−0 = 117 − 46 + 6 = 123 − 46 = 77. (ii) The number of integers between 1 and 100 that are divisible by 2, 3, 5 but not by 7 = |A ∩ B ∩ C| − |A ∩ B ∩ C ∩ D| = 3 − 0 = 3. 16. How many prime numbers not exceeding 100 are there? Or determine a prime number n, where 1 ≤ n ≤ 100. Solution. To find the number of primes not exceeding 100, first note that a composite integer not exceeding 100 must have a prime factor not exceeding 10. The primes not exceeding 100 are 2, 3, 5, and 7 and the numbers that are divisible by none of 2, 3, 5, or 7. Let P1 be the property that an integer is divisible by 2. Let P2 be the property that an integer is divisible by 3. Let P3 be the property that an integer is divisible by 5. Let P4 be the property that an integer is divisible by 7. Number of primes not exceeding = 4 + N (P10 P20 P30 P40 ).
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Discrete Mathematical Structures Now,
N (P10 P20 P30 P40 ) = 99 − N (P1 ) − N (P2 ) − N (P3 ) − N (P4 ) + N (P1 P2 ) + N (P1 P3 ) + N (P1 P4 ) + N (P2 P3 ) + N (P2 P4 ) + N (P3 P4 ) − N (P1 P2 P3 ) − N (P1 P2 P4 ) − N (P1 P3 P4 ) − N (P2 P3 P4 ) − N (P1 P2 P3 P4 ) (∵ there are 99 integers > 1 and not exceeding 100). N (P10 P20 P30 P40 )
100 100 100 100 − − − = 99 − 2 3 5 7 100 100 100 + + + 2×3 2×5 2×7 100 100 100 + + + 3×5 3×7 5×7 100 100 100 − − − 2×3×5 2×3×7 2×5×7 100 100 − + 3×5×7 2×3×5×7 = 99 − 50 − 33 − 20 − 14 + 16 + 10 + 7 + 6 +4+2−3−2−1−0+0 = 21.
∴ There are 4 + 21 = 25 primes. 17. How many solutions does x1 + x2 + x3 = 11 have, where x1 , x2 , and x3 are non-negative integers with x1 ≤ 3, x2 ≤ 4, and x3 ≤ 6? Solution. Let P1 be the property that x1 > 3. Let P2 be the property that x2 > 4. Let P3 be the property that x3 > 6. Now, the number of solutions satisfying the inequalities x1 ≤ 1, x2 ≤ 4, and x3 ≤ 6 is N (P10 P20 P30 ). By principle of inclusion–exclusion, we have N (P10 P20 P30 )N − N (P1 ) − N (P2 ) − N (P3 ) + N (P1 P2 ) + N (P1 P3 ) + N (P2 P3 ) − N (P1 P2 P3 ). Now, N = Total number of solutions = C(3 + 11 − 1, 11) = 78
Combinatorics
131
[since the number of r-combinations from a set with n elements when repetitions are allowed is (n+r −1)Cr ways or C(n+r −1, r)]. N (P1 ) = Number of solutions with x1 ≥ 4 = C(3 + 7 − 1, 7) = C(9, 7) = 36. N (P2 ) = Number of solutions with x2 ≥ 5 = C(3 + 6 − 1, 6) = C(8, 6) = 28. N (P3 ) = Number of solutions with x3 ≥ 7 = C(3 + 4 − 1, 4) = C(6, 4) = 15. N (P1 P2 ) = Number of solutions with x1 ≥ 4 and x2 ≥ 5 = C(3 + 2 − 1, 2) = C(4, 2) = 6. N (P1 P3 ) = Number of solutions with x1 ≥ 4 and x3 ≥ 7 = C(3 + 0 − 1, 0) = C(2, 0) = 1. N (P2 P3 ) = Number of solutions with x2 ≥ 5 and x3 ≥ 7 = 0. N (P1 P2 P3 ) = Number of solutions with x1 ≥ 4, x2 ≥ 5 and x3 ≥ 7 = 0. ∴
N (P10 P20 P30 ) = 78 − 36 − 28 − 15 + 6 + 1 + 0 + 0 = 6.
Hence, the equation x1 + x2 + x3 = 11 with respect to the given conditions has six solutions.
2.9.2
Problems for Practice
1. Find the number of elements in A1 ∪ A2 ∪ A3 if there are 100 elements in each set if (i) the sets are pairwise disjoint (ii) there are 50 common elements in each pair of sets and no element in all three sets (iii) the sets are equal. 2. How many elements are in the union of four sets if the sets have 50, 60, 70, and 80 elements, respectively, each pair of sets has five elements in common, each trio of the sets has one common element, and no element is in all four sets? 3. How many bit strings of length 8 do not contain six consecutive 0’s? 4. How many solutions does the equation x1 +x2 +x3 = 16 have where x1 , x2 , and x3 are non-negative integers less than 6? 5. How many ways are there to distribute six different toys to three different children such that each child gets at least one toy? 6. How many ways can the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 be arranged so that no even digit is in its original position?
132
Discrete Mathematical Structures 7. By using principle of inclusion–exclusion, find how many positive integers between 1 and 250 are (i) divisible by 2, 3, 5, or 7 (ii) not divisible by 2, 3, 5, and 7. 8. How many positive integers not exceeding 1000 are divisible by 7 or 11? 9. How many elements are in A1 ∪ A2 if there are 12 elements in A1 , 18 elements in A2 , and (i) |A1 ∩ A2 | = 1 (ii) |A1 ∩ A2 | = 6?
10. There are 2504 Computer Science students at a college. Of these, 1876 have taken a course in Pascal, 999 have taken a course in Fortran, and 345 have taken course in C. Further, 876 have taken courses in both Pascal and Fortran, 231 have taken courses in both Fortran and C, and 290 have taken courses in both Pascal and C. If 189 of these students have taken courses in Fortran, Pascal, and C, how many of these 2504 students have not taken a course in any of these 3 programming languages? 11. How many permutations of ten digits either begin with the three digits 987, contain the digits 45 in the fifth and sixth positions, or end with the three digits 123? 12. Find the probability that when four numbers from 1 to 100, inclusive, are picked at random with no repetitions allowed, either all are odd, all are divisible by 3, or all are divisible by 5. 13. Find the number of primes less than 200 using the principle of inclusion–exclusion. 14. How many derangements are there of a set with seven elements? 15. How many positive integers less than 10000 are not the second or higher powers of an integer? 16. How many integers between 1 and 300 are divisible by (i) at least one of 3, 5, 7 (ii) 3 and 5 but not by 7 (iii) 5 but not by 3 and 7. 17. A survey of households in the United States reveals that 96% have at least one television set, 98% have telephone service, and 95% have telephone service and at least one television set. What percentage of households in the United States have neither telephone service nor a television set?
Combinatorics
133
18. How many students are enrolled in a course either in calculus, discrete mathematics, data structures, or programming languages at a school if there are 507, 292, 312, and 344 students in these courses, respectively; 14 in both calculus and data structures; 213 in both calculus and programming languages; 211 in both discrete mathematics and data structures; 43 in both discrete mathematics and programming languages; and no student may take calculus and discrete mathematics, or data structures and programming languages, concurrently? 19. Find the number of positive integers not exceeding 100 that are either odd or the square of an integer. 20. Suppose in a bushel of 100 apples, there are 20 that have worms in them and 15 that have bruises. Only those apples with neither worms or bruises can be sold. If there are ten bruised apples that have worms in them, how many of the 100 apples can be sold? 21. In how many ways can seven different jobs be assigned to four different employees so that each employee is assigned at least one job and the most difficult job is assigned to the best employee?
3 Graphs
3.1
Introduction
Graphs are mathematical discrete structures which have major role in computer science (algorithms and computation), electrical engineering (communication networks and coding theory), operations research (scheduling), and in many fields of engineering and also in sciences such as chemistry, biochemistry (genomics), biology, linguistics, sociology, and other fields. For instance, graphs are encoded to represent the relationship between objects. Many real-world situations can conveniently be described by means of a diagram consisting of a set of points together with lines joining certain pairs of these points. For example, the points could represent people, with lines joining pairs of friends; or the points might be communication centres, with lines representing communication links. Notice that in such diagrams, one is mainly interested in whether or not two given points are joined by a line; the manner in which they are joined is immaterial. A mathematical abstraction of situations of this type gives rise to the concept of a graph. In this chapter, we focus on the terminology of graphs, its various types, connectivity of graphs, Eulerian path, and Hamiltonian path. Graph theory is introduced as an abstract mathematical system. The most common representation of a graph is by means of a diagram, in which the vertices are represented as points and each edge as a line segment joining its end vertices.
3.2
Graphs and Graph Models
Definition 3.2.1 Graph: A graph G = (V, E, φ) consists of a non-empty set V = {v1 , v2 , . . . } called the set of vertices of the graph and E = {e1 , e2 , . . . } called the set of edges of the graph, and φ is a mapping from the set of edges E to the set of ordered or unordered pair of elements of vertices.
135
136
Discrete Mathematical Structures v4
e3
v3
e4
e2
e1
v1
v2
Example of a graph Definition 3.2.2 Self-loop: An edge having same vertex as both its end vertices is called a self-loop. Example 3.2.3 Here e is a self-loop. v3
e
v1
v2
Example of a self-loop Definition 3.2.4 Parallel Edges: If more than one edge has the same pair of end vertices, then the edges are called parallel edges. v3
e1 e2 e3 v2
v1
e4
Example of parallel edges Example 3.2.5 In the graph in (i) e1 , e2 are incident with v1 . (ii) e2 , e3 are incident with v5 . Here, e3 and e4 are parallel edges.
Graphs
137
Definition 3.2.6 If vi is an end vertex of some edge ej , then ej is said to be incident at vi . v1
e1
v2 v3
e2 e4 v5
v4
e3
Example of incident edges on vertices Example 3.2.7 The edge e1 is incident at the vertex v1 . The edge e3 is incident at the vertex v5 . Definition 3.2.8 Adjacent Edges and Vertices: Two non-parallel edges are said to be adjacent if they are incident on a common vertex. Two vertices are said to be adjacent if they are the end vertices of the same edge. Example 3.2.9 Here, e1 is adjacent to e2 and e3 , and v1 is adjacent to v2 and v3 . v3
e1 e2 e3 v1
v2
Adjacent vertices and edges Definition 3.2.10 Simple Graph: A graph which has neither self-loops nor parallel edges is called a simple graph as shown below.
Examples of simple graphs Definition 3.2.12 Isolated Vertex: A vertex having no edge incident on it is called an isolated vertex as shown below.
138
Discrete Mathematical Structures v2
v1
v3
v4
Example of isolated vertex Definition 3.2.14 Directed Graph: A graph in which every edge is directed is called a directed graph or digraph as shown below.
Example of directed graph Definition 3.2.16 Undirected Graph: A graph in which every edge is undirected is called an undirected graph.
Example of undirected graph Definition 3.2.18 Mixed Graph: A graph in which some edges are directed and some are undirected is called a mixed graph. Definition 3.2.19 Multigraph: A graph which contains some parallel edges is called a multigraph. Definition 3.2.20 Pseudo Graph: A graph in which loops and parallel edges are allowed is called a pseudograph.
3.3
Graph Terminology and Special Types of Graphs
Definition 3.3.1 Degree of a vertex: The number of edges incident at the vertex vi in an undirected graph is called the degree of the vertex vi . But a loop at a vertex contributes twice to the degree of that vertex. The degree of the vertex vi is denoted by deg(vi ).
Graphs
139
Definition 3.3.2 In-degree and Out-degree of a vertex: In a directed graph, the in-degree of a vertex vi is denoted by deg − (vi ) and defined by the number of edges with vi as their terminal vertex. The out-degree of a vertex vj is denoted by deg + (vj ) and defined as the number of edges with their initial. Theorem 3.3.3 Handshaking Theorem: Let G = (V, E) be an undirected P graph with e edges. Then, v∈V deg(v) = 2e. Proof. Since every edge is incident with exactly two vertices, every edge contributes two to the sum of degrees of the vertex. Therefore, all the edges e contribute 2e to the sum of degrees of the vertex. P ∴ = 2e. v∈V Theorem 3.3.4 The maximum number of edges in a simple graph with n n(n − 1) . vertices is 2 Proof. From handshaking theorem, we have X deg(v) = 2e, v∈V
where e is the number of edges with n vertices in the graph G. That is, deg(v1 ) + deg(v2 ) + · · · + deg(vn ) = 2e.
(3.1)
We know that the maximum degree of each vertex in the graph G can be (n − 1). ∴ (3.1) =⇒ (n − 1) + (n − 1) + . . . ton terms = 2e =⇒ n(n − 1) = 2e =⇒ e =
n(n − 1) . 2
Hence, the maximum number of edges in any simple graph with n vertices is n(n − 1) . 2 Theorem 3.3.5 A simple graph with at least two vertices has at least two vertices of same degree. Proof. Let G be a simple graph with n ≥ 2 vertices. The graph G has no loop and parallel edges. Hence, the degree of each vertex is ≤ n − 1. Suppose that all the vertices of G are of different degrees. Following degrees 0, 1, 2, . . . , n − 1 are possible for n vertices of G.
140
Discrete Mathematical Structures
Let u be the vertex with degree 0. Then, u is an isolated vertex. Let v be the vertex with degree n − 1. Then, v has n − 1 adjacent vertices. Since v is not an adjacent vertex of itself, every vertex of G other than u is an adjacent vertex of G other than u. Hence, u cannot be an isolated vertex; this contradiction proves that a simple graph contains two vertices of same degree. Note: The converse of the above theorem is not true. Theorem 3.3.6 The number of odd degree vertices is always even. Proof. Let G = (V, E) be any graph with n number of vertices and e number of degrees. 0 be the Let v1 , v2 , . . . , vk be the vertices of odd degree and v10 , v20 , . . . , vm vertices of even degree. To prove, k is even. P We know that v∈V deg(v) = 2|E| = 2e. Pk Pm =⇒ deg(vi ) + j=1 deg(vj0 ) = 2e. i=1 Pm Clearly, j=1 deg(vj0 ) and 2e are even numbers. Pk That is, i=1 deg(vi ) = an even number. Since each term deg(vi ) is odd, the number of terms in the left-hand side sum must be even. =⇒ k is even. Hence, the theorem is proved.
3.3.1
Solved Problems
1. Can a simple graph exist with 15 vertices each of degree 5? Solution. We know that the number of odd degree vertices is even. Hence, the number of odd degree vertices to be odd is not possible. We cannot say that a simple graph exists with 15 vertices each of degree 5. 2. How many vertices does a regular graph of degree 4 with ten edges have? Solution. We know that X
deg(v) = 2e
v∈V
n × 4 = 2 × 10 20 n= 4 n = 5. ∴
The number of vertices is five.
Graphs
141
3. Is there a simple graph corresponding to the following degree sequences? (i) (1, 1, 2, 3) (ii) (2, 2, 4, 6) Solution. (i) There are odd number (3) of degree vertices 1, 1, and 3. Hence, there does not exist a graph corresponding to this degree sequence. (ii) The number of vertices in the graph is four, and the maximum degree of a vertex is 6 which is not possible as the maximum degree cannot be one less than the number of vertices. 4. Show that in a group, there must be two people who know the same number of other people in the group. Solution. Construct the simple graph model in which V is the set of people in the group, and there is an edge associated with (u, v) if u and v know each other. Then, the degree of vertex v is the number of people v knows. We know that there are two vertices with the same degree. Therefore, there are two people who know the same number of other people in the group. 5. Show that the degree of a vertex of a simple graph G of n vertices cannot exceed n − 1. Solution. Let v be a vertex of G because G is simple and no multiple edges or loops are allowed in G. Thus, v can be adjacent to at most all the remaining n − 1 vertices of G. Hence, v may be of maximum degree n − 1 in G. Then, 0 ≤ deg(v) ≤ n − 1, or all v ∈ V . 6. How many edges are there in a graph with ten vertices each of degree 6? Solution. Sum of the degrees of the ten vertices is (6) × (10) = 60 =⇒ 2e = 60 =⇒ e = 30. 7. Show that the sum of degrees of all the vertices in a graph G is even. Solution. Each edge contributes two degrees in a graph.
142
Discrete Mathematical Structures Also, each edge contributes one degree to each of the vertices on which it is incident. Hence, if there are N edges in G, then 2N = deg(v1 ) + deg(v2 ) + · · · + deg(vN ). Thus, 2N is always even.
Definition 3.3.7 Regular Graph: If every vertex of a simple graph has the same degree, then the graph is called a regular graph. If every vertex in a regular graph has degree k, then the graph is called k-regular graph. Note: 1. Every null graph is regular of degree 0. 2. The complete graph Kn is of degree n − 1. 3. If a graph G has n vertices and is regular of degree k, then G has rn edges. 2 Definition 3.3.8 Complete Graph: A simple graph with n vertices is said to be a complete graph if the degree of every vertex is n − 1. (or) In a graph G, if every vertex v is adjacent to all other vertices, then G is called a complete graph. The complete graph with n vertices is denoted by Kn .
K3
K4
K5
Examples of complete graphs
Example 3.3.9 Definition 3.3.10 Subgraph: A graph H = (V 0 , E 0 ) is called a subgraph of a graph G = (V, E) if V 0 ⊆ V and E 0 ⊆ E. Definition 3.3.11 Cycle Graph: A cycle graph of order n is a connected graph whose edges form a cycle of length n and is denoted by Cn .
Graphs
143
Example 3.3.12
C6
Example of cycle graph Note: 1. In a graph, a cycle that is not a loop must have length at least three, but there may be cycles of length two in a multigraph. 2. A simple digraph having no cycles is called a cyclic graph. 3. A cyclic graph cannot have any loops. 4. The cycle Cn , n ≥ 3, consists of n vertices 1, 2, . . . , n and edges {1, 2}, {2, 3}, . . . , {n − 1, n}. Definition 3.3.13 Wheel Graph: A wheel graph of order n is obtained by joining a new vertex called “Hub” to each vertex of a cycle graph of order n − 1, denoted by Wn .
Hub
W5
Example of wheel graph Example 3.3.14 Note: We obtain the wheel Wn when we add an additional vertex to the cycle Cn , for n ≥ 3, and connect this new vertex to each of the n vertices in Cn , by new edges. Definition 3.3.15 Bipartite Graph: A graph G is said to be bipartite if its vertex set V can be partitioned into two disjoint non-empty sets V1 and V2 such that V1 ∪ V2 and every edge in E has one end in V1 and the other end in V2 . Example 3.3.16 Let V = V1 ∪ V2 where V1 = {u1 , u3 , u5 , u7 } and V2 = {u2 , u4 , u6 , u8 }. Then, G is a bipartite graph shown below.
144
Discrete Mathematical Structures
u1
u2
u3
u4
u5
u6
u7
u8
V1
V2
Example of bipartite graph Definition 3.3.17 Complete Bipartite Graph: A bipartite graph G with the bipartition V1 and V2 is called complete bipartite if every vertex in V1 is adjacent to every vertex in V2 . A complete bipartite graph that may be partitioned into sets A and B as above such that |A| = a and |B| = b is denoted by Ka,b . Example 3.3.18 The graph K3,3 is a complete bipartite graph. V1
V2
V4
V5
V3
V6
K3,3
Example of complete bipartite graph Definition 3.3.19 Star Graph: Any graph that is K1,n is called a star graph. Example 3.3.20 The graph K1,6 is a star graph.
K1,6
Example of star graph
Graphs
3.3.2
145
Graph Colouring
The assignment of colours to the vertices of G, one colour to each vertex, so that adjacent vertices are assigned different colours is called the proper colouring of G or simply vertex colouring. If G has n colouring, then G is said to be n-colourable. Theorem 3.3.21 A simple graph is bipartite if and only if it is possible to assign one of two different colours to each vertex of the graph so that no two adjacent vertices are assigned the same colour. Proof. Let G = (V, E) be a bipartite simple graph. Then, V = V1 ∪ V2 , where V1 and V2 are disjoint sets and every edge in E connects a vertex in V1 and a vertex in V2 . If we assign one colour to each vertex in V1 and a second colour to each vertex in V2 , then no two adjacent vertices are assigned the same colour. Suppose that it is possible to assign colours to the vertices of the graph using just two colours. =⇒ No two adjacent vertices are assigned the same colour. Let V1 be the set of vertices assigned one colour and V2 be the set of vertices assigned the other colour. Then, V1 and V2 are disjoint and V = V1 ∪ V2 . That is, every edge connects a vertex in V1 and a vertex in V2 since no two adjacent vertices are either both in V1 or both in V2 . Consequently, G is bipartite.
3.3.3
Solved Problems
1. What is the degree sequence of Kn , where n is positive integer? Explain your answer. Solution. Each of the n vertices is adjacent to each of the other n − 1 vertices, so the degree sequence is n − 1, n − 1,. . . , n − 1 (n terms). 2. Determine whether each of the following sequences is a graph. For those that are, draw a graph having the given degree sequence. (i) 5, 4, 3, 2, 1 (ii) 3, 2, 2, 1, 0 (iii) 1, 1, 1, 1, 1 Solution. (i) No, since the sum of degrees = 5 + 4 + 3 + 2 + 1 = 15 which is odd. (ii) Yes. (iii) No, since the sum of degrees = 1 + 1 + 1 + 1 + 1 = 5 which is odd.
146
Discrete Mathematical Structures
Graph of the given sequence 3. How many vertices and edges are there in Kn ? Solution.
n(n − 1) edges. 2 4. Find the degree sequence of each of the following graphs. Kn has n vertices and
(i) K4 (ii) K5 (iii) K2 Solution. (i) 3, 3, 3, 3 (ii) 4, 4, 4, 4, 4 (iii) 1, 1 5. How many vertices and edges do the following graphs have? (i) Cn (ii) C8 (iii) Also find the degree sequence of C4 . Solution. (i) n vertices and n edges (ii) Eight vertices and eight edges (iii) 2, 2, 2, 2. 6. Show that C6 is a bipartite graph? Solution. The vertex set of C6 can be partitioned into the two sets V1 = {v1 , v3 , v5 } and V2 = {v2 , v4 , v6 }, and every edge of C6 connects a vertex in V1 and a vertex in V2 . Hence, C6 is a bipartite graph.
Graphs
147 v1
v2
v3 v6
v5
v4
Graph of C6
7. Is K3 bipartite? Solution. No, the complete graph K3 is not bipartite as shown below. v1
v3
v2
Graph of K3 If we divide the vertex set of K3 into two disjoint sets, one of the two sets must contain two vertices. If the graph is bipartite, these two vertices should not be connected by an edge, but in K3 each vertex is connected to every other vertex by an edge. ∴ K3 is not bipartite. 8. How many vertices and edges are there in a complete bipartite graph Km,n ? Solution. There are m + n vertices and mn edges. 9. Find the degree sequence of the graph K2,3 . Solution. 3, 3, 2, 2, 2. 10. For which values of m and n is Km,n regular? Solution. A complete bipartite graph Km,n is not regular if m 6= n. =⇒ If m = n, then Km,n is regular. 11. Prove that a graph which contains a triangle cannot be bipartite. Solution. At least two of the three vertices must lie in one of the bipartite sets because these two are joined by edge; thus, the graph cannot be bipartite.
148
Discrete Mathematical Structures
12. Show that if G is a bipartite simple graph with v vertices and e v2 edges, then e ≤ . 4 Solution. Let G be a complete bipartite graph with v vertices. Let v1 and v2 be the number of vertices in the partitions V1 and V2 of vertex set of G. Since G is complete bipartite, each vertex in V1 is joined to each vertex in V2 by exactly one edge. Thus, G has v1 v2 edges when v1 + v2 = v. But we know the maximum value of v1 v2 subject to v1 + v2 = v v2 . is 4 v2 . Thus, the maximum number of edges in G is 4 2 v That is, e ≤ . 4 13. Show that the graph G is bipartite. V2 V1
V3
V7
V6
V4 V5
Graph G Solution. Graph G is bipartite since its vertex set is the union of two disjoint sets {v1 , v2 , v3 } and {v4 , v5 , v6 , v7 } and each edge connects a vertex in one of these subsets to a vertex in the other subset. 14. Draw the complete bipartite graphs K2,3 , K3,3 , K3,5 , and K2,6 . Solution. The complete bipartite graphs are shown below.
Graphs
149
K2,3
K3,3
K3,5
K2,6
Given complete bipartite graphs 15. How many subgraphs with at least one vertex does K3 have? Solution. 17
3.4
Representing Graphs and Graph Isomorphism
We can represent a simple graph in the form of edge list or in the form of adjacency lists which may be useful in computer programming. Definition 3.4.1 Adjacency matrix of a simple graph: Let G = (V, E) be a simple graph with n vertices {v1 , v2 , . . . , vn }. Its adjacency matrix is denoted by A = (aij ) and defined by ( 1, if vi and vj are adjacent aij = 0, otherwise. Example 3.4.2 v1
v5
v2
v4
v3
Example of a graph with adjacency matrix
150
Discrete Mathematical Structures
The adjoint matrix for the graph in the figure above is
A=
v1 0 1 0 0 1
v1 v2 v3 v4 v5
v2 1 0 1 1 1
v3 0 1 0 1 1
v4 0 1 1 0 1
v5 1 1 1 1 0
Definition 3.4.3 Incidence Matrix: Let G = (V, E) be a graph with n vertices v1 , v2 , . . . , vn and m edges e1 , e2 , . . . , em . Then, the n × m matrix B = (bij ) where ( 1, if the edge ej is incident on vi bij = 0, otherwise. Example 3.4.4 v1
e1
e4 v4
v2 e2
e5 e3
v3
Example of a graph with incidence matrix The incidence matrix for the graph in the figure above is
B=
v1 v2 v3 v4
e1 1 1 0 0
e2 0 1 1 0
e3 0 0 1 1
e4 1 0 0 1
e5 1 0 1 0
Observations about the incidence matrix: 1. Since every edge is incident on exactly two vertices, each column of B has exactly two 1’s. 2. The number of 1’s in each row is equal to the degree of the corresponding vertex. 3. A row with all 0’s represents an isolated vertex. 4. Parallel edges in a graph produce identical columns in its incidence matrix.
Graphs
151
Definition 3.4.5 Isomorphic Graphs: The simple graphs G1 = (V1 , E1 ) and G2 = (V2 , E2 ) are isomorphic if there is a one-to-one and onto function f from V1 to V2 with the property that a and b are adjacent in G1 if and only if f (a) and f (b) are adjacent in G2 , for all a and b in V1 . Such a function f is called an isomorphism. In other words, two graphs G1 and G2 are isomorphic if there is a function f : V (G1 ) −→ V (G2 ) from the vertices of G1 to the vertices of G2 such that (i) f is one-to-one (ii) f is onto and (iii) for each pair of vertices u and v of G1 [u, v] ∈ E(G1 ) ⇔ [f (u), f (v)] ∈ E(G2 ). Any function f with the above three properties is called an isomorphism from G1 to G2 . Example 3.4.6 Consider the graphs G1 and G2 in the following figure. Let f : G1 −→ G2 be a function with f (u1 ) = v1 , f (u2 ) = v4 , f (u3 ) = v3 , f (u4 ) = v2 . Then, f is a one-to-one and onto function between G1 and G2 . Here, f preserves the adjacency. The adjacent vertices in G1 are u1 and u2 , u1 and u3 , u2 and u4 , and u3 and u4 , and each of the pairs f (u1 ) = v1 and f (u2 ) = v4 , f (u1 ) = v1 and f (u3 ) = v3 , f (u2 ) = v4 and f (u4 ) = v2 , and f (u3 ) = v3 and f (u4 ) = v2 are adjacent in G2 . Hence, the graphs G1 and G2 are isomorphic. u1
u2
v1
u3
u4
v3
v2
v4
G1
G2
Example of isomorphic graphs
3.4.1
Solved Problems
1. Write the adjacency matrix of the following graph. u1
u2
u3
u4
Given graph
152
Discrete Mathematical Structures Solution. The adjacency matrix is
u1 u2 u3 u4
A=
u u2 1 0 1 1 0 1 0 1 1
u3 1 0 0 1
u4 1 1 1 0
2. Write the incidence matrix for the following graph. e2
v1
v2
e4
v3
e3
e1
e6
e7
v4
e5 v5
e8
Given graph Solution. The incidence matrix is
B=
e1 e2 1 1 0 1 0 0 0 0 0 0
v1 v2 v3 v4 v5
e3 1 1 0 0 0
e4 0 1 1 0 0
e5 0 0 1 0 1
e6 0 1 0 0 1
e7 0 1 0 0 0
e8 0 0 0 0 0
3. What is the sum of the entries in a row of the incidence matrix for an undirected graph? Solution. Sum is 2 if the edge e is not a loop and 1 if the edge e is a loop. 4. Check whether the two graphs are isomorphic or not.
G1
G2
Given graphs
Graphs
153 Solution. In the graph G1 , vertices of degree 2 are not adjacent, while in the graph G2 , vertices of degree 2 are adjacent. Since isomorphism preserves adjacency of vertices, the graphs are not isomorphic.
5. Prove that the graphs G1 and G2 are isomorphic. v1
u1 v5
u2
u3
u4
v2
v4
u5
v3
G2
G1
Given graphs Solution. The two graphs have the same number of vertices, same number of edges, and same degree sequence. Consider the function f defined by f (u1 ) = v1 , f (u2 ) = v3 , f (u3 ) = v4 , f (u4 ) = v2 , f (u5 ) = v5 . Then, the adjacency matrices of the two graphs corresponding to f are u1 u2 u3 u4 u5 u1 0 1 1 0 0 1 0 0 0 1 u2 A(G1 ) = 1 0 0 1 0 u3 0 0 1 0 1 u4 0 1 0 1 0 u5
A(G2 ) =
v1 v2 v3 v4 v5
v1 0 1 1 0 0
v2 1 0 0 0 1
v3 1 0 0 1 0
v4 0 0 1 0 1
v5 0 1 0 1 0
Therefore, A(G1 ) = A(G2 ). Hence, G1 and G2 are isomorphic to each other. 6. Prove that any two simple connected graphs with n vertices all of degree 2 are isomorphic. Solution. We know that the total degree of a graph is given by n X i=1
deg(vi ) = 2|E|.
154
Discrete Mathematical Structures Then, |V | = number of vertices = n |E| = number of edges. Further, Pn the degree of each vertex is 2. ∴ i=1 2 = 2|E| =⇒ n = |E| ∴ Number of edges = number of vertices. Hence, the graphs are cycle graphs. Therefore, they are isomorphic. 7. Can a simple graph with seven vertices be isomorphic to its complement? Solution. A graph with seven vertices can have a maximum number of edges 7×6 7(7 − 1) = = 21 edges. = 2 2 But 21 edges cannot be splitted into two equal integers. Therefore, a graph and its complement cannot have equal number of edges. Hence, a graph with seven vertices cannot be isomorphic to its complement. 8. Let G be a simple graph, all of whose vertices have degree 3 and |E| = 2|V | − 3. What can be said about G? Solution. |V | X
deg(vi ) = 2|E|
i=1
3(|V | − 1 + 1) = 2|E| 3|V | = 2|E| = 2(2|V | − 3) = 4|V | − 6 =⇒ |V | = 6. The number of vertices in G is six. Hence, it can be concluded that G is isomorphic to K3,3 . 9. Show that isomorphism of simple graphs is an equivalence relation. Solution. (i) Reflexive: G is isomorphic to itself by the identity function. Hence, isomorphism is reflexive. (ii) Symmetric: Suppose that G is isomorphic to H. Then, there exists a one-to-one correspondence f from G to H that preserves adjacency and non-adjacency. From this, f −1 is a oneto-one correspondence from H to G that preserves adjacency and non-adjacency. Hence, isomorphism is symmetric. (iii) Transitive: If G is isomorphic to H and H is isomorphic to K, then there are one-to-one correspondences f and g from G to H and from H to K that preserve adjacency and nonadjacency. It follows that g ◦ f is a one-to-one correspondence
Graphs
155 from G to K that preserves adjacency and non-adjacency. Hence, isomorphism is transitive. From the above (i)–(iii), isomorphism is an equivalence relation.
3.4.2
Problems for Practice
1. Write the adjacency matrix of the following graph.
a
b
c
d
2. Draw the directed graph for the following adjacency matrix. 0 0 1 1 0 0 1 0 1 1 0 1 1 1 1 0 3. Draw an undirected graph for 1 2 0 1
the following adjacency matrix. 2 0 1 0 3 0 3 1 1 0 1 0
4. Find the adjacency matrix of the given directed graph.
v1 v2
v3
v4
156
Discrete Mathematical Structures 5. Check whether the graphs with the following adjacency matrices are isomorphic. 0 0 1 0 1 1 0 0 1 , 1 0 0 1 1 0 1 0 0 6. Determine whether the matrices are isomorphic. 0 1 1 0 1 0 0 1
graphs with the following adjacency 1 0 0 1
0 0 1 1 , 1 0 1 0
1 0 0 0
0 0 0 1
1 0 1 0
7. Find whether the following graphs are isomorphic to each other. u1
u2
u3
v1
u4 G1
v2
v4
v3 G2
8. Find a pair of non-isomorphic graphs with the same degree sequence such that one graph is bipartite but the other graph is not bipartite. 9. What is the product of the incidence matrix and its transpose for an undirected graph? 10. Draw the graph represented by 1 2 0
3.5
the following adjacency matrix. 2 1 0 0 2 2
Connectivity
Definition 3.5.1 Walk: A walk is defined as a finite alternating sequence of vertices and edges, beginning and ending with vertices such that each edge is incident with the vertex preceding and following it. (No edge appears more than once, and vertex may be repeated.)
Graphs
157
Example 3.5.2 Consider the graph G. In this graph, v1 e1 v4 e2 v3 e6 v5 is a walk. v1
e1
v4
e4
e3
e2
v2
e5
v5 e6
v3
Example of a walk Remarks: 1. A walk is also referred to as an edge train or a chain. 2. No edge appears more than once in a walk. 3. Every walk is a subgraph of G. Definition 3.5.3 Terminal Vertex: In a walk, the vertex that begins and ends the walk is called its terminal vertex. For example, in the walk v1 e3 v3 e5 v2 in the figure above, the terminal vertices are v1 and v2 . Definition 3.5.4 Closed Walk: A walk with same end vertices is called a closed walk. In the example above, v1 e3 v3 e5 v2 e4 v1 is a closed walk. Definition 3.5.5 Open Walk: A walk which is not closed is called an open walk. In the graph above, v1 e3 v3 e5 v2 is an open walk. Definition 3.5.6 Path or Simple Path or Elementary Path: An open walk in which no vertex appears more than once is called a path or simple path or an elementary path. For example, in the graph above, v1 e3 v3 e6 v5 is a path, but v1 e3 v3 e5 v2 e4 v1 e1 v4 is not a path, since v1 is repeated twice. Definition 3.5.7 Circuit or Cycle or Circular Path or Elementary Cycle: A closed walk in which no vertex appears more than once is called a circuit or a cycle or a circular path or an elementary cycle. In the graph above, v1 e3 v3 e5 v2 e4 v1 is a circuit. Remarks: 1. A path does not intersect itself. 2. A self-loop can be included in a walk but not a path.
158
Discrete Mathematical Structures 3. Every circuit is called as a non-intersecting walk. 4. Every self-loop is a circuit, but every circuit is not a self-loop. 5. Every vertex in a circuit is of degree 2.
3.5.1
Connected and Disconnected Graphs
Definition 3.5.8 Connected Graph: A graph G is a connected graph if there is at least one path between every pair of vertices in G. Otherwise G is a disconnected graph. Example 3.5.9 Consider the connected graph below. e4
v1
e1
v4
e3
e2
v2
e5
v5 e6
v3
Connected graph Definition 3.5.10 Component or Block: A disconnected graph consists of two or more connected graphs. Each of these connected subgraphs is called a component or block. Example 3.5.11
Disconnected graph with two components Theorem 3.5.12 A graph G is disconnected iff its vertex set V can be partitioned into two non-empty disjoint subsets V1 and V2 such that there exists no edge in G whose one end vertex is in the subset V1 and the other in the subset V2 . Proof. Let G be a disconnected graph. Let us fix a vertex a in G. Let V1 be the set of all vertices such that they are joined by paths to a.
Graphs
159
Now, consider V2 = V − V1 . Clearly, V2 is not empty (since G is disconnected, V1 does not contain all vertices of G). Also, no vertex in V1 is joined to any vertex in V2 by an edge. Conversely, suppose the partition exists. Consider any two arbitrary vertices a and b of G such that a ∈ V1 and b ∈ V2 . =⇒ There exists no path between a and b. Otherwise, there would be at least one edge whose one end vertex would be in V1 and other end in V2 . =⇒ G is not connected. Hence, the theorem is proved. Theorem 3.5.13 If a graph (connected or disconnected) has exactly two vertices of odd degree, then there must be a path joining these two vertices. Proof. Let G be a graph with exactly two odd degree vertices. Let them be v1 and v2 . Case (i): Let G be a connected graph. =⇒ There exists at least one path between v1 and v2 . Case (ii): Let G be a disconnected graph. =⇒ There exist at least two components. Without loss of generality, we can assume G contains two components g1 and g2 . If both v1 and v2 lie in the same component, then there exists a path joining between v1 and v2 . Suppose v1 lies in g1 and v2 lies in g2 . Then, g1 is a graph containing exactly one odd degree vertex, and g2 is a graph having exactly one odd degree vertex, which is a contradiction to the theorem that “the number of odd degree vertices is always even”. Therefore, v1 and v2 must be in the same component. Hence, the theorem is proved. Theorem 3.5.14 A simple graph with n vertices and k components can have (n − k)(n − k + 1) edges. at most 2 Proof. Let G be a simple graph with n vertices and k components, namely g1 , g2 , . . . , gk . Let ni be the number of vertices in the ith components gi , for all i = 1, 2, . . . , k. Clearly, Pkn1 + n2 + · · · + nk = n =⇒ i=1 ni = n, and ni ≥ 1. We know that the maximum number of edges in the ith component is ni (ni − 1) . 2
160
Discrete Mathematical Structures ∴ The maximum number of edges in G = the number of edges in g1 + the number of edges in g2 + · · · + the number of edges in gk = n1 (n1 − 1)/2 + n2 (n2 − 1)/2 + · · · + nk (nk − 1)/2 ! k k k k X X X X = ni (ni − 1)/2 = n2i − ni /2 = n2i − ni /2 i=1
i=1
k X
=
i=1
i=1
! n2i − n /2.
(3.2)
i=1
Let us consider k k k X X X 1 ni − (ni − 1) = i=1
i=1
i=1
= (n − k). Squaring on both sides, we get " k #2 X (ni − 1) = (n − k)2 i=1
(or)
[(n1 − 1) + (n2 − 1) + · · · + (nk − 1)]2 = (n − k)2 [(n1 − 1)2 + (n2 − 1)2 + · · · + (nk − 1)2 ] + (positive cross terms) = (n − k)2
(or)
[(n21 + 1 − 2n1 ) + (n22 + 1 − 2n2 ) + · · · + (n2k + 1 − 2nk )] + (positive cross terms) = (n − k)2
(or)
(n21 + n22 + · · · + n2k ) + (1 + 1 + · · · + 1) {z } | k times
− 2(n1 + n2 + · · · + nk ) + (positive cross terms) = (n − k)2 (or)
k X
n2i − k − 2
i=1
(or)
k X i=1
(or)
k X
k X
ni + (positive cross terms) = (n − k)2
i=1
n2i + k − 2
k X
ni ≤ (n − k)2 (by omitting the positive cross terms)
i=1
n2i + k − 2n ≤ (n − k)2
i=1
(or)
k X i=1
n2i ≤ (n − k)2 − k + 2n.
(3.3)
Graphs
161 Therefore, the maximum number of edges in G k
=
1X 2 n n − 2 i=1 i 2
[using (3.2)]
≤ [(n − k)2 − k + 2n − n]/2
[using (3.3)]
2
= [(n − k) − k + n]/2 = (n − k)(n − k + 1)/2. Hence, the theorem is proved.
3.6
Eulerian and Hamiltonian Paths
Definition 3.6.1 Eulerian Circuit: An Eulerian circuit in a graph G is a simple circuit containing every edge of G. Definition 3.6.2 Eulerian Trail: A trail in G is called an Eulerian trail if it includes each edge of G exactly once. Definition 3.6.3 Eulerian Path: An Eulerian path in G is a simple path containing every edge of G. Definition 3.6.4 Eulerian Graph: A closed walk which contains all edges of the graph G is called an Euler line, and the graph containing at least one Euler line is called an Eulerian graph. Example 3.6.5 The graphs are Eulerian graphs.
Star of David
Mohammed's Semi Stars
Examples of Eulerian graphs Theorem 3.6.6 A given connected graph G is an Eulerian graph if and only if all vertices of G are of even degree. Proof. Suppose G is an Eulerian graph.
162
Discrete Mathematical Structures
=⇒ G contains an Euler line. =⇒ G contains a closed walk covering all edges. To prove: All vertices of G are of even degree. In tracing the closed walk, every time the walk meets a vertex v, it goes through two new edges incident on v with one we “entered” and other “exited”. Since it is a closed walk, this is true for all vertices (intermediate and terminal vertices). Thus, the degree of every vertex is even. Conversely, suppose that all vertices of G are of even degree. To prove: G is an Eulerian graph. That is, to prove G contains an Euler line. Construct a closed walk starting at an arbitrary vertex v and going through the edges of G such that no edge is repeated. Since each vertex is of even degree, we can exit from each and every vertex where we enter, and the tracing can stop only at the vertex v. Name the closed walk as h. Case (i): If h covers all edges of G, then h becomes an Euler line, and hence G is an Eulerian graph. Case (ii): If h does not cover all edges of G, then remove all edges of h from G and obtain the graph G0 . Since both G and G0 have vertices which are of even degree, every vertex in G0 is also of even degree. Since G is connected, h will touch G0 in at least one vertex v 0 . Starting from v 0 , we can again construct a new walk h0 in G0 . This will terminate only at v 0 , since every vertex in G0 is also of even degree. Now, this walk h0 combined with h forms a closed walk, starts and ends at v, and has more edges than h. This process is repeated until we obtain a closed walk covering all edges of G. Thus, G is an Eulerian graph. Theorem 3.6.7 A connected multigraph with at least two vertices has an Eulerian circuit if and only if each of its vertices has even degree. Proof. Given: A connected multigraph G has an Eulerian path but not an Eulerian circuit. To prove: G has exactly two vertices of odd degree. Suppose that a connected multigraph has an Eulerian path from a to b but not an Eulerian circuit. The first edge of the path contributes 1 to the degree of a. A contribution of 2 to the degree of a is made every time the path passes through a. The last edge in the path contributes 1 to the degree of b. Every time the path goes through b, there is a contribution of 2 to its degree. Consequently, both a and b have odd degree. Every other vertex has even degree, because the path contributes 2 to the degree of a vertex whenever it passes through it. Theorem 3.6.8 A connected multigraph has an Eulerian path but not an Eulerian circuit if and only if it has exactly two vertices of odd degree. Proof. Given: The graph has exactly two vertices of odd degree.
Graphs
163
To prove: G has an Eulerian path. Suppose that a graph has exactly two vertices of odd degree, say a and b. Consider the larger graph made up of the original graph with the addition of an edge {a, b}. Every vertex of this larger graph has even degree, so there is an Eulerian circuit. The removal of the new edge produces an Eulerian path in the original graph. Chinese postman problem: If a postman can find an Eulerian path in the graph that represents the streets the postman needs to cover, this path produces a route that traverses each street of the route exactly once. If no Eulerian path exists, some streets will have to be traversed more than once. This problem is known as the Chinese postman problem. Theorem 3.6.9 A directed multigraph having no isolated vertices has an Eulerian path but not an Eulerian circuit if and only if the graph is weakly connected and the in-degree and out-degree of each vertex are equal for all but two vertices, one that has in-degree larger than its out-degree by 1 and the other that has out-degree larger than its in-degree by 1. Proof. If there is an Eulerian path, as we follow, each vertex except the starting and ending vertices must have equal in-degree and out-degree, since whenever we come to a vertex along an edge, we leave it along another edge. The starting vertex must have out-degree 1 larger than its in-degree, since we use one edge leading out of this vertex and whenever we visit it again, we use one edge leading into it and one leaving it. Similarly, the ending vertex must have in-degree 1 greater than its out-degree. Since the Eulerian path with directions erased produces a path between any two vertices, in the underlying undirected graph, the graph is weakly connected. Conversely, suppose the graph meets the degree conditions stated. If we add one more edge from the vertex of deficient out-degree to the vertex X with equal in-degree and out-degree. Because the graph is still weakly connected, by this new graph has an Eulerian circuit. Now, delete the added edge to obtain the Eulerian path.
3.6.1
Hamiltonian Path and Hamiltonian Circuits
Definition 3.6.10 Hamiltonian Path: A simple path in a graph G that passes through every vertex exactly once is called a Hamiltonian path. That is, the simple path x0 , x1 , x2 , . . . , xn−1 , xn in the graph G = (V, E) is a Hamiltonian path if V = {x0 , x1 , x2 , . . . , xn−1 , xn } and xi 6= xj for 0 ≤ i < j ≤ n. Definition 3.6.11 A simple circuit in a graph G that passes through every vertex exactly once is called a Hamiltonian circuit. And the simple
164
Discrete Mathematical Structures
circuit x0 , x1 , . . . , xn−1 , xn , x0 (with n > 0) is a Hamiltonian circuit if x0 , x1 , . . . , xn−1 , xn is a Hamiltonian path. Definition 3.6.12 Unicursal Graph: An open walk that includes all edges of G without retracing any edge is called a unicursal line or an open Euler line. A connected graph which has a unicursal line is called a unicursal graph. Example 3.6.13 In the graph, the unicursal line is v2 e1 v4 e2 v3 e3 v2 e4 v1 . v4
v1 e4 v2
e1
e2 e3
v3
Example of unicursal graph Remark: 1. Adding an edge between the initial and final vertices of unicursal line, we obtain an Euler line. 2. A connected graph is unicursal, if it has exactly two odd degree vertices. Theorem 3.6.14 In a connected graph G, with exactly 2k odd degree vertices, there exist k edge-disjoint subgraphs such that they together contain all edges of G and that each is a unicursal graph. Proof. Let the odd degree vertices of the given graph be named as v1 , v2 , . . . , vk and ω1 , ω2 , . . . , ωk in any arbitrary order. Add k edges (new edges) to G between the pair of vertices (v1 , ω1 ), (v2 , ω2 ),. . . , (vk , ωk ) to form a new graph G0 . In the resultant graph G0 , every vertex is of even degree. =⇒ G0 is an Eulerian graph. =⇒ G0 contains an Euler line, say P . If we remove k newly added edges from P , that will split P into k walks each of a unicursal line to itself. That is, the first removal will leave a single unicursal line. The second removal will split that into two unicursal lines, and each successive removal will split a unicursal line into two unicursal lines, until there are k of them. Hence, the theorem is proved.
3.6.2
Solved Problems
1. Does the graph given below have a Hamiltonian path? If so, find such a path. If it does not, give an argument to show why no such path exists.
Graphs
165 d
a c
f
b
e
Given graph Solution. a–b–c–f –d–e is a Hamiltonian path. 2. Does the graph below have a Hamiltonian path. If so, find such a path. If it does not, give an argument to show why no such path exists. a
b
c d e
f
Given graph Solution. f –e–d–a–b–c is a Hamiltonian path. 3. Does the graph given below have a Hamiltonian path? If so, find such a path. If it does not, give an argument to show why no such path exists. b
a
c
j
k
i d
o
q
p
n
h
l m
e
g f
Given graph Solution. No Hamiltonian path exists. There are eight vertices of degree 2, and only two of them can be end vertices of a path. For each of the other six, their two incident edges must be in the path. It is easy to see that if there is to be a Hamiltonian path, exactly one of the inside corner vertices must be an end and that this is impossible.
166
Discrete Mathematical Structures 4. Show that Kn has a Hamiltonian circuit whenever n ≥ 3. Solution. We can form a Hamiltonian circuit in Kn beginning at any vertex. Such a circuit can be built by visiting vertices in any order we choose, as long as the path begins and ends at the same vertex and visits each of the other vertices exactly once. It is possible since there are edges in Kn between any two vertices. 5. Give an example of a graph that has an Eulerian circuit and a Hamiltonian circuit, which are distinct. Solution. The graph having an Eulerian circuit and a Hamiltonian circuit which are distinct as shown below. The Eulerian circuit is a–c–b–c–d–b–a. b
a
d
c
Graph with Eulerian and Hamiltonian circuits The Hamiltonian circuit is a–b–d–c–a. 6. Give an example of a graph which has a Hamiltonian circuit but not an Eulerian circuit. Solution. The graph having a Hamiltonian circuit but not an Eulerian circuit as shown below. a
b
c
d
Graph with Hamiltonian circuit but no Eulerian circuit The Hamiltonian circuit is a–b–d–c–a. There is no Eulerian circuit. 7. Give an example of a graph which has an Eulerian circuit but not a Hamiltonian circuit.
Graphs
167 Solution. The graph having an Eulerian circuit but not a Hamiltonian circuit as shown below. b
d
a c
e
f
Graph with Eulerian circuit but no Hamiltonian circuit The Eulerian circuit is a–e–b–e–c–d–f –c–b–a. There is no Hamiltonian circuit. 8. Show that a bipartite graph with an odd number of vertices does not have a Hamiltonian circuit. Solution. Suppose that G = (V, E) is a bipartite graph with V = V1 ∪ V2 , where no edge connects a vertex in V1 and a vertex in V2 . Suppose that G has a Hamiltonian circuit. Such a circuit must be of the form a1 , b1 , a2 , b2 , . . . , ak , bk , where ai ∈ V1 and bi ∈ V2 for i = 1, 2, . . . , k. Since the Hamiltonian circuit visits each vertex exactly once, except for v1 , where it begins and ends, the number of vertices in the graph equals 2k, an even number. Hence, a bipartite graph with an odd number of vertices cannot have a Hamiltonian circuit. 9. For which values of m and n does the complete bipartite graph Km,n have a Hamiltonian circuit? Solution. m = n ≥ 2. n−1 10. In a complete graph with n vertices, show that there are 2 edge-disjoint Hamiltonian circuits, if n is an odd number ≥ 3. Solution. A complete graph G of n vertices has n(n − 1)/2 edges, and a Hamiltonian circuit in G consists of n edges. Therefore, the number of edge-disjoint Hamiltonian circuits in G cannot exceed (n − 1)/2. That is, there are (n − 1)/2 edge-disjoint Hamiltonian circuits, when n is odd. The subgraph (of a complete graph of n vertices) in the figure below is a Hamiltonian circuit. Keeping the vertices fixed on a circle, rotate the polygonal pattern clockwise by 360/(n − 1), 2 · 360/(n − 1),
168
Discrete Mathematical Structures 3 · 360/(n − 1),. . . , (n − 3)/2 · 360/(n − 1) degrees. Observe that each rotation produces a Hamiltonian circuit that has no edge in common with any of the previous ones. Thus, we have (n − 3)/2 new Hamiltonian circuits, all edgedisjoint from the one as shown below and also edge-disjoint among themselves. 5 n-2
3
n 2
1 n-1 4
n-3
Required graph for the problem 11. Explain Konigsberg bridge problem. Represent the problem by means of a graph. Does the problem have a solution? Solution. There are two islands A and B formed by a river. They are connected to each other and to the river banks C and D by means of seven bridges. The problem is to start from any one of the four land areas A, B, C, D, walk across each bridge exactly once, and return to the starting point. C
A
B
Konigsberg bridge and its graph
The situation is represented by a graph, with vertices representing the land areas and the edges and the bridges.
D
Graphs
169 This problem is the same as that of drawing the graph without lifting the pen from the paper and without retracing any line. In other words, the problem is to find whether there is an Eulerian circuit in the graph. But a connected graph has an Eulerian circuit if and only if each of its vertices is of even degree. In the present case, all the vertices are of odd degree. Hence, Konigsberg bridge problem has no solution.
3.6.3
Problems for Practice
1. Can someone cross all the bridges shown in this map exactly once and return to the starting point?
2. In Kaliningrad (the Russian name for Konigsberg), there are two additional bridges, besides the seven that were present in the 18th century. These new bridges connect regions B and C and regions B and D, respectively. Can someone cross all nine bridges in Kaliningrad exactly once and return to the starting point? 3. Show that a directed multigraph having no isolated vertices has an Eulerian circuit if and only if the graph is weakly connected and the in-degree and out-degree of each vertex are equal. 4. For which values of n do the following graphs have an Eulerian circuit? (a) Kn (b) Cn (c) Wn (d) Qn
3.6.4
1. Can you draw a graph of five vertices with degree sequence 1, 2, 3, 4, 5? 2. Show that there does not exist a graph with five vertices with degrees 1, 3, 4, 2, 3, respectively. 3. How many edges are there in a graph with ten vertices each of degree 5?
170
Discrete Mathematical Structures 4. Let G be a graph with ten vertices. If four vertices have degree 4 and six vertices have degree 5, then find the number of edges of G. 5. Draw the graph represented by the given adjacency matrix: 0 1 0 1 1 0 1 0 0 1 0 1 . 1 0 1 0 6. How do you find the number of different paths of length r from i to j in a graph G with adjacency matrix A? 7. Is the directed graph given below strongly connected? Why or why not? 1
2
4
3
8. Define a bipartite graph. 9. Draw the complete graph K5 . 10. Define isomorphism of directed graphs. 11. What do strongly connected components of a telephone call graph represent? 12. Define Hamiltonian path. 13. Give an example for a graph which is (i) Eulerian and Hamiltonian (ii) neither Eulerian nor Hamiltonian. 14. Describe a discrete structure based on a graph that can be used to model airline routes and their flight times. 15. Show that a simple graph G with n vertices is connected if it has more than (n − 1)(n − 2)/2 edges. 16. Show the isomorphism of simple graphs is an equivalence relation. 17. Derive an algorithm for constructing Eulerian path in directed graphs. 18. Are simple graphs with the following adjacency matrices isomorphic? 0 1 0 0 0 1 0 1 0 0 0 1 1 0 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 0 1 0 1 1 0 , 0 0 1 0 1 0 0 0 1 0 1 0 0 1 0 1 0 1 0 0 1 1 0 1 1 0 1 0 1 0 1 1 0 0 1 0
Graphs
171
19. Examine whether the two graphs G and G0 following adjacency matrices are isomorphic. 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 1 0 1 0 1 0 1 0 0 0 1 0 1 0 1 0 1 0 , 0 0 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 0 0 1 0
associated with the 0 0 1 0 1 0
1 0 0 1 0 1
0 0 1 0 1 0
20. Discuss the various graph invariants preserved by isomorphic graphs. 21. If G is a self-complementary graph, then prove that G has n ≡ 0 or 1 (mod 4) vertices. 22. If G is a connected simple graph with n vertices with n ≥ 3, such n that the degree of every vertex in G is at least , then prove that 2 G has Hamiltonian cycle. 23. In a round robin tournament, the team 1 beats team 2, team 3, and team 4; team 2 beats team 3 and team 4; and team 3 beats team 4. Model this outcome with a directed graph. 24. Show that the number of vertices of odd degree in an undirected graph is even. 25. If a graph, either connected or disconnected, has exactly two vertices of odd degree, prove that there is a path joining these two vertices. 26. Find an Eulerian path or Eulerian circuit if it exists in each of the following two graphs. B
B
A
C
A (i)
D
(ii)
C
D
E
E
F
27. Determine whether the following graphs G and H are isomorphic. Give reason. u2
v1
u1
u3
u5
u4 G
v2
v5
v3
v4 H
172
Discrete Mathematical Structures
28. Which of the following simple graphs have a Hamiltonian circuit, or if not, a Hamiltonian path? G1
G2
G3
a
b
a
b
e
c
d
c
a
b
g
d
c
e
f
d
29. Prove that a simple graph is bipartite if and only if it is possible to assign one of two different colours to each vertex of the graph so that no two adjacent vertices are assigned the same colour. 30. Show that every connected graph with n vertices has at least n − 1 edges. 31. Show that there does not exist a graph with five vertices with degrees 1, 3, 4, 2, 3, respectively. 32. Can you draw a graph of five vertices with degree sequence 1, 2, 3, 4, 5? 33. Draw a graph that is Eulerian but not Hamiltonian. 33. Let G be a graph with ten vertices. If four vertices have degree 4 and six vertices have degree 5, then find the number of edges of G. 34. How many edges are there in a graph with ten vertices each of degree 5? 35. Give an example for a graph which is (i) Eulerian and Hamiltonian (ii) neither Eulerian nor Hamiltonian. 36. How do you find the number of different paths of length r from i to j in a graph G with adjacency matrix A? 37. Establish the isomorphism of the following pairs of graphs. v6
v1
v2
v3
u6
v4
v5
u1
u2
u3
u4
u5
4 Algebraic Structures
4.1
Introduction
An algebraic system can be described as a set of objects together with some operations. These operations will impose a certain structure on the set. In this chapter, we study the axiomatic set theory, semigroups, groups, and monoids which are the basic tools of discrete mathematics.
4.2
Algebraic Systems
Definition 4.2.1 Binary Operation: Let A be any set. A mapping f : A × A −→ A is called a binary operation. Definition 4.2.2 Algebraic System or Algebra: A set together with a number of (binary) operations on the set is called an algebraic system or an algebra. Properties of Binary Operations: Let G be a set. (i) Closure Property: A binary operation ? : G × G −→ G is said to be closed, if for all a, b ∈ G, an element a ? b = x ∈ G. (ii) Associative Property: a ? (b ? c) = (a ? b) ? c, for all a, b, c ∈ G. (iii) Existence of Identity: There exists an element e ∈ G such that e ? a = a ? e = a, for all a ∈ G. The element e is called the identity element. (iv) Existence of Inverse: For a ∈ G, there exists an element b ∈ G such that a ? b = b ? a = e. The element b is called the inverse of a, and it is denoted by b = a−1 . (v) Commutativity: For all a, b ∈ G, if a ? b = b ? a, then ? is commutative.
173
174
Discrete Mathematical Structures
(vi) Distributive Properties: Let • be any other binary operation on G. Then, a ? (b • c) = (a ? b) • (a ? c) (b • c) ? a = (b ? a) • (c ? a)
(Left distributive law) (Right distributive law)
for all a, b, c ∈ G. (vii) Cancellation Property: For a, b, c ∈ G and a 6= 0, a ? b = a ? c =⇒ b = c. Definition 4.2.3 Algebraic Structure: The operations on a set G define a structure on the elements of G. Then, the algebraic system G is called an algebraic structure. Example 4.2.4 Let R be the set of real numbers. Consider the algebraic system (R, +, ×) where + and × are the operations of addition and multiplication on R.
4.2.1
Semigroups and Monoids
Definition 4.2.5 Semigroup: A non-empty set S together with the binary operation ? : S × S −→ S is said to be a semigroup if ? satisfies the following conditions, namely, the closure property and the associative property. We denote the semigroup as (S, ?). Example 4.2.6 Let N = {1, 2, 3, . . . } be the set of natural numbers. Then, (N, +) and (N, •) are semigroups under the binary operations of addition and multiplication, respectively. Definition 4.2.7 Monoid: A semigroup (M, ?) with an identity element e, with respect to the operation ? is called a monoid. In other words, a non-empty set M together with the binary operation ? : M × M −→ M is said to be a monoid if ? satisfies the closure property, associative property, and the identity property. Example 4.2.8 Let Z+ = {0, 1, 2, 3, . . . } be the set of all non-negative integers. Then, (Z+ , +) is a semigroup as well as a monoid. Definition 4.2.9 A semigroup (or monoid) (S, ?) is said to be commutative or abelian if a ? b = b ? a, for all a, b ∈ S. Example 4.2.10 The set of integers, the set of reals, the set of complex numbers are abelian semigroups (abelian monoids) under the usual operations of addition and multiplication. Definition 4.2.11 Idempotent element: Let (G, ?) be a group. An element a ∈ G is said to be idempotent if a ? a = a.
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Definition 4.2.12 Congruence Relation: Let (X, ?) be an algebraic system and E an equivalence relation on X. The relation E is called a congruence relation on (X, ?) if E satisfies the substitution property with respect to the operation ?. Remark 4.2.13 Substitution Property: Let (X, ?) be an algebraic system in which ? is a binary operation on X. Let us assume that E is an equivalence relation on X. The equivalence relation E is said to have the substitution property with respect to the operation ? if and only if for any x1 , x2 ∈ X, (x1 Ex01 ) ∧ (x2 Ex02 ) = (x1 ? x2 )E(x2 ? x02 ) where x01 , x02 ∈ X.
4.2.2
Solved Problems
1. Show that intersection of any two congruence relations on a set A is again a congruence relation on A. Solution. Let E1 and E2 be two congruence relations on (A, ?). =⇒ (a1 E1 a01 ) ∧ (a2 E1 a02 ) = (a1 ? a2 )E1 (a01 ? a02 ) and (a1 E2 a01 ) ∧ (a2 E2 a02 ) = (a1 ? a2 )E2 (a01 ? a02 ). Let E = E1 ∩ E2 . To prove: E is a congruence relation on A (a1 Ea01 ) ∧ (a2 Ea02 ) = [a1 (E1 ∩ E2 )a01 ] ∧ [a2 (E1 ∩ E2 )a02 ] = (a1 E1 a01 ) and (a1 E1 a01 ) ∧ (a2 E2 a02 ) and (a2 E2 a02 ) = (a1 E1 a01 ) ∧ (a2 E1 a02 ) and (a1 E2 a01 ) ∧ (a2 E2 a02 ) = (a1 ? a2 )E1 (a01 ? a02 ) and (a1 ? a2 )E2 (a01 ? a02 ) = (a1 ? a2 )(E1 ∩ E2 )(a01 ? a02 ) = (a1 ? a2 )E(a01 ? a02 ). Hence, E is a congruence relation on A. 2. Show that a semigroup with more than one idempotent cannot be a group. Give an example of a semigroup which is not a group. Solution. Let (S, ?) be a semigroup. Let a, b are two idempotents. ∴ a ? a = a and b ? b = b. Let us assume that (S, ?) is group. Then, each element has an inverse. Hence, (a ? a) ? a−1 = a ? (a ? a−1 ) (by associative property).
176
Discrete Mathematical Structures Now, (a ? a) ? a−1 = a ? a−1 = e
since a ? a = a.
(4.1)
Also, a ? (a ? a−1 ) = a ? e = a.
(4.2)
From (4.1) and (4.2), we get a = e. Similarly, we can prove that b = e. But in a group we cannot have two identities, and hence (S, ?) cannot be a group. This contradiction is due to an assumption that (S, ?) has two idempotents. Example: Let S = {a, b, c} under the operation ?. The composition table of (S, ?) is shown in the following table. Composition Table of (S, ?) ? a b c a a c a b c b a c b a c (S, ?) is a semigroup which is not a group. 3. Give an example of a semigroup which is not a monoid. Solution. Let D = {. . . , −4, −2, 0, 2, 4, . . . } (D, ·) is a semigroup but not a monoid since its multiplicative identity is 1, but 1 ∈ / D. 4. Give an example of a monoid which is not a group. Solution. (Z+ , ·) is a monoid which is not a group, since for all a ∈ Z+ , 1 ∈ / Z+ . a 5. What do you call a homomorphism of a semigroup into itself? Solution. A homomorphism of a semigroup into itself is called a semigroup endomorphism. 6. If (Z, +) and (E, +) are two semigroups, where Z is the set of all integers and E is the set of all even integers, show that the two semigroups (Z, +) and (E, +) are isomorphic.
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Solution. First, we define a function g : Z −→ E by g(a) = 2a, for all a ∈ Z. To prove g is one-to-one: Suppose g(a1 ) = g(a2 ), where a1 , a2 ∈ Z. Then, 2a1 = 2a2 =⇒ a1 = a2 . Therefore, g is one-to-one. To prove g is onto: Suppose b is an even integer. b Let a = . Then, a ∈ Z and 2 b b g(a) = g = 2 · = b. 2 2 That is, every element b ∈ E has a preimage in Z. Therefore, g is onto. To prove g is homomorphism: Let a, b ∈ Z. g(a + b) = 2(a + b) = 2a + 2b = g(a) + g(b). Hence, (Z, +) and (E, +) are isomorphic semigroups. 7. If ? is a binary operation on the set of R of real numbers defined by a ? b = a + b + 2ab, (i) show that (R, ?) is a semigroup. (ii) find the identity element if it exists. (iii) which elements has inverse and what are they? Solution. (i) (a ? b) ? c = (a + b + 2ab) + c + 2(a + b + 2ab)c = a + b + c + 2(ab + bc + ca) + 4abc and a ? (b ? c) = a + (b + c + 2bc) + 2a(b + c + 2bc) = a + b + c + 2(ab + bc + ca) + 4abc. Hence, (a ? b) ? c = a ? (b ? c). Therefore, ? is associative. (ii) If the identity element exists, let it be e. Then for any a ∈ R, or or
a?e=a a + e + 2ae = a e(1 + 2a) = 0.
Therefore, e = 0, since 1 + 2a 6= 0, for any a ∈ R.
178
Discrete Mathematical Structures (iii) Let a−1 be the inverse of an element a ∈ R. Then, a ? a−1 = e. That is, a + a−1 + 2a · a−1 = 0 a−1 · (1 + 2a) = −a. a Therefore, a−1 = − . 1 + 2a 1 a Hence, if a 6= , then a−1 = − . 2 1 + 2a 8. Show that a semigroup with more than one idempotent cannot be a group. Give an example of a semigroup which is not a group. Solution. Let (S, ?) be a semigroup. Let a, b ∈ S be two idempotents. Then, a ? a = a and b ? b = b. Let us assume that (S, ?) is a group. Then, each element has its inverse. Now, by associative property, we have (a ? a) ? a−1 = a ? (a ? a−1 ). (a ? a) ? a−1 = a ? a−1 = e. a ? (a ? a
−1
) = a ? e = a.
(4.3) (4.4)
From (4.3) and (4.4), we get a = e. Similarly, we can prove that b = e. But in a group, we cannot have two identities and hence (S, ?) cannot be a group. This contradiction is due to the assumption that (S, ?) has two idempotents. Example: Let S = {a, b, c} under the operation ?. The composition table of (S, ?) is given inthe following table. Composition Table of (S, ?) ? a b c a a c a b c b a c b a c (S, ?) is a semigroup but not a group. 9. Let (N, +) be the semigroup of natural numbers and (S, ?) be a semigroup where S = {e, 0, 1} with the operation ? given in the following table. Composition Table of (S, ?) ? e 0 1 e e 0 1 0 0 0 0 1 1 0 1
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A mapping g : N −→ S is defined by g(0) = 1 and g(j) = 0 for j 6= 0. Is g is a semigroup homomorphism? Solution. Though both (N, +) and (S, ?) are monoids with identities 0 and e, respectively, g is not a monoid homomorphism because g(0) 6= e. ∴ g is a semigroup homomorphism. 10. If ? is the operation defined on S = Q × Q, the set of ordered pair of rational numbers and given by (a, b) ? (x, y) = (ax, ay + b), show that (S, ?) is a semigroup. Is it commutative? Also, find the identity element of S. Solution. Given (a, b) ? (x, y) = (ax, ay + b).
(4.5)
To prove: (S ,?) is a semigroup, that is, to prove ? is associative. [(a, b) ? (x, y)] ? (c, d) = (ax + ay + b) ? (c, d) [using (4.5)] = (acx, adx + ay + b) [using (4.5)]. (4.6) (a, b) ? [(x, y) ? (c, d)] = (a, b) ? (cx, dx + y) = (acx, adx + ay + b).
(4.7)
From (4.6) and (4.7), ? is associative on S. To prove (S ,?) is not commutative: (x, y) ? (a, b) = (ax, bx + y).
(4.8)
(a, b) ? (x, y) = (ax, ay + b).
(4.9)
From (4.8) and (4.9), (a, b) ? (x, y) 6= (x, y) ? (a, b). Hence, (S, ?) is not commutative. To find the identity element of (S ,?): Let (e1 , e2 ) be the identity element of (S, ?). Then, for all (a, b) ∈ S, we have (a, b) ? (e1 , e2 ) = (a, b) =⇒ (ae1 , ae2 + b) = (a, b). =⇒ ae1 = a and ae2 + b = b. =⇒ e1 = 1 and ae2 = 0 or e2 = 0. Therefore, (1, 0) is the identity element of (S, ?). 11. Is it true that a semigroup homomorphism preserves identity? Justify your answer. Or prove by an example that semigroup homomorphism need not preserve an identity. Solution. To prove that semigroup homomorphism need not preserve an identity:
180
Discrete Mathematical Structures Let W = {0, 1, 2 . . . }. Then, (W, +) is a semigroup homomorphism with identity element 0. Let S = {e, 0, 1} and ? be the operation on S given by the table below. Composition Table of (S, ?) ? e 0 1 e e 0 1 0 0 0 0 1 1 0 1 Then, (S, ?) is a semigroup with identity e. Now, define a mapping g : W −→ S by g(0) = 1 and g(i) = 0 for i 6= 0. We can see that g(a + b) = g(a) ? g(b), for all a, b ∈ W . Thus, g is a semigroup homomorphism. But g(0) = 1 6= e. Thus, g does not preserve the identity.
12. Find all semigroups of Z6 , ×6 where Z6 = {[0], [1], [2], [3], ][4], [5]}. Solution. The composition table is given below. Composition ×6 [0] [1] [0] [0] [0] [1] [0] [1] [2] [0] [2] [3] [0] [3] [4] [0] [4] [5] [0] [5]
Table of (Z6 , ×6 ) [2] [3] [4] [5] [0] [0] [0] [0] [2] [3] [4] [5] [4] [0] [2] [4] [0] [3] [0] [3] [2] [0] [4] [2] [4] [3] [2] [1]
The semigroups are {[0]}, {[1], [2], [4]}, {[0], [3], [4]}, {[0], [4]}, {[0], [1], [2], [3]}.
{[0], [1]}, {[0], [1], [2], [4]}, {[1], [5]}, {[0], [1], [4]},
{[1]}, {[2], [4]}, {[0], [1], [5]}, {[0], [2], [3]},
13. Prove that (Z5 , ×5 ) is a commutative monoid, where ×5 is the multiplication modulo 5. Solution. Z5 = {[0], [1], [2], [3], [4]}. The composition table is given below.
Algebraic Structures
181 Composition Table of ×5 [0] [1] [2] [0] [0] [0] [0] [1] [0] [1] [2] [2] [0] [2] [4] [3] [0] [3] [1] [4] [0] [4] [3]
(Z5 , ×5 ) [3] [4] [0] [0] [3] [4] [1] [3] [4] [2] [2] [1]
(i) Closure Property: From the table above, it is clear that Z5 is closed under ×5 . (ii) Associative Property: It is also clear that the associative property holds from the table above. That is, [a] ×5 ([b] ×5 [c]) = ([a] ×5 [b]) ×5 [c], for all [a], [b], [c] ∈ Z5 . (iii) Existence of Identity: [1] is the identity element since [a] ×5 [1] = [a], for all [a] ∈ Z5 . (iv) Commutative Property: Clearly, from the table above, [a] ×5 [b] = [b] ×5 [a], for all [a], [b] ∈ Z5 . Hence, (Z5 , ×5 ) is a commutative monoid. 14. Let (M, ?, eM ) be a monoid and a ∈ M . If a is invertible, then show that its inverse is unique. Solution. Let b and c be inverse elements of a ∈ M such that a ? b = b ? a = e and a ? c = c ? a = e. Now, b = b ? e = b ? (a ? c) = (b ? a) ? c = e ? c = c. Therefore, its inverse is unique. 15. Show that the set N of natural numbers is a semigroup under the operation x ? y = max{x, y}. Is it a monoid? Solution. Let N = {1, 2, 3, . . . }. Define the operation x ? y = max{x, y} for x, y ∈ N. Clearly, (N, ?) is closed because x ? y = max{x, y} ∈ N and ? is associative since (x ? y) ? z = max{x ? y, z} = max{max{x, y}, z} = max{x, y, z} = max{x, max{y, z}} = max{x, y ? z} = x ? (y ? z). Therefore, (N, ?) is a semigroup. 16. Prove that monoid homomorphism preserves invertibility and monoid epimorphism preserves a zero element (if it exists).
182
Discrete Mathematical Structures Solution. Let (M, ?, eM ) and (T, ∆, eT ) be any two monoids, and let g : M −→ T be a monoid homomorphism. If a ∈ M is invertible, let a−1 be the inverse of a in M . We will now show that g(a−1 ) will be an inverse of g(a) in T . a ? a−1 = a−1 ? a = eM (by definition of inverse) So, g(a ? a−1 ) = g(a−1 ? a) = g(eM ). Hence, g(a)∆g(a−1 ) = g(a−1 )∆g(a) = g(eM ). (since g is a homomorphism) But g(eM ) = eT . (since g is a monoid homomorphism) Therefore, g(a)∆g(a−1 ) = g(a−1 )∆g(a) = eT . This means g(a−1 ) is an inverse of g(a). That is, g(a) is invertible. Thus, the property of invertibility is preserved under monoid homomorphism. Assume g is a monoid epimorphism. Let t = g(b) ∈ T . Then t∆g(z) = g(b)∆g(z) = g(b ? z) = g(z) and g(z)∆t = g(z)∆g(b) = g(z ? b) = g(z). Therefore, g(z) is the zero element of T .
17. The operation ? is defined by a ? b = a + b − ab, on the set Q of all rational numbers. Show that under this operation, Q is a commutative monoid. Solution. (i) Closure Property: Since a + b − ab is a rational number for all rational numbers a, b, the given operation ? is a binary operation on Q. (ii) Associative Property: For all a, b, c ∈ Q, (a ? b) ? c = (a + b − ab) ? c = (a + b − ab) + c − (a + b − ab)c = a + b − ab + c − ac − bc + abc = a + (b + c − bc) − a(b + c − bc) = a ? (b + c − bc) = a ? (b ? c). Hence, ? is associative. (iii) Existence of Identity: For any a ∈ Q, a?0=a+0−a·0=a and 0 ? a = 0 + a − 0 · a = a. Hence, 0 is the identity element in Q under the operation ?.
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(iv) Commutative Property: From the definition of the operation ?, it is clear that ? is commutative. Hence, under the operation ?, Q is a commutative monoid with 0 as the identity element. 18. Let V = {a, b} and A be the set of all sequences on V including ∧ beginning with a. Show that (A, ◦, ∧) is a monoid. Solution. Let V = {a, b} and A be the set of all sequences on V including ∧ beginning with a. Then, A = {∧, a, ab, aa, ab, aba, abb, . . . }. Let ◦ be a concatenation operation on the sequences in A. Clearly for any two elements α, β ∈ A, α ◦ β = αβ also belongs to A, and hence (A, ◦) is closed. Also ◦ is associative because (α ◦ β) ◦ γ = αβγ = α ◦ (βγ) = (α ◦ β ◦ γ). ∧ is the identity since ∧ ◦ α = α ◦ ∧ = α, for all α ∈ A. Therefore, (A, ◦, ∧) is a monoid. 19. Let V = {a, b}. Show that (V ? , ◦, ∧) is an infinite monoid. Solution. While defining the alphabet and set of strings V ? , we proved that (V ? , ◦, ∧) is a monoid, where ∧ is an empty string. So, it is enough to show that V ? is an infinite set. As a is an element of V , a, aa, aaa, aaaa,. . . ; b, bb, bbb, bbbb,. . . ; and ab, abb, abbb, . . . are the elements of V ? , and hence V ? contains infinitely many strings including an empty set.
4.2.3
Groups
Definition 4.2.14 Group: A non-empty set G together with a binary operation ?, that is (G, ?), is called a group if ? satisfies the following conditions: (i) Associative: For every a, b, c ∈ G, a ? (b ? c) = (a ? b) ? c. (ii) Existence of Identity: There exists an element e ∈ G called the identity element such that a ? e = e ? a = a, for all a ∈ G. (iii) Existence of Inverse: There exists an element a−1 ∈ G called the inverse of a such that a ? a−1 = a−1 ? a = e, for each a ∈ G.
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Discrete Mathematical Structures
Example 4.2.15 The set of all integers Z with the addition operation is a group. Example 4.2.16 The set of all non-zero real numbers R? under the multiplication operation is a group. Definition 4.2.17 Abelian Group or Commutative Group: A group (G, ?) is said to be an abelian group or commutative group if a ? b = b ? a, for all a, b ∈ G. Otherwise, it is a non-abelian group. The set of all integers Z with the addition operation is an abelian group. Properties of Groups 1. The identity of a group is unique. 2. The left and right cancellation laws are true. (i) a ? b = a ? c =⇒ b = c (left cancellation law) and (ii) b ? a = c ? a =⇒ b = c (right cancellation law). 3. The inverse of any element in a group is unique. −1 4. If a is an element of a group G, then a−1 = a. 5. For any two elements a, b in a group G, (a ? b)−1 = b−1 ? a−1 . 6. In a group, the solution for the equations a ? x = b and y ? b = a exists, and it is unique. Theorem 4.2.18 Every row or column in the composition table of a group (G, ?) is a permutation of the elements of G. Proof. Initially, we shall show that no row or column in the composition table can have an element of G more than once. Let us assume the contrary. Suppose that the row corresponding to an element a ∈ G has two entries which are both k. That is, assume that a ? b1 = a ? b2 = k, where k, b1 , b2 ∈ G and b1 6= b2 . Then by cancellation law, we have b1 = b2 which is a contradiction. A similar result holds for any column. Next we will show that every element of G appears in each row and column of the table of composition. Consider the row corresponding to the element a ∈ G, and let b be an arbitrary element of G. Since b = a ? (a−1 ? b), “b” must appear in the row corresponding to the element a ∈ G. The same argument applies to every column of the table. Thus, we obtain that no two rows or columns are identical. Hence, every row of the composition table is obtained by a permutation of the elements G and that each row is a distinct permutation. The same result applies to the columns of the composition table.
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Theorem 4.2.19 In a group (G, ?), an element a ∈ G such that a2 = e, a 6= e if and only if a = a−1 . Proof. Let us assume that a = a−1 . Then, a2 = a ? a = a ? a−1 = e. Conversely, assume that a2 = e with a 6= e. That is, a
a?a=e −1
? a ? a = a−1 ? e e ? a = a−1 a = a−1 .
Theorem 4.2.20 In a group (G, ?), (a−1 )−1 = a, for all a ∈ G. Proof. Let a−1 be the inverse of a. a ? a−1 = a−1 ? a = e =⇒ a is the inverse of a−1 . That is, (a−1 )−1 = a. Definition 4.2.21 Permutation Group or Symmetric Group: The set Pn of all permutations of n elements is a permutation group or a symmetric group under the composition of functions. That is, Pn = {f /f is a one-to-one and onto mapping from S to S} is a group under the composition operation of functions, where S is any non-empty set. Example 4.2.22 The set P3 of all permutations on S = {1, 2, 3} is a finite non-abelian group of order six with respect to composition of mappings. The composition table for P3 is given in the table below where, Composition Table ◦ f1 f2 f3 f4 f1 f1 f2 f3 f4 f2 f2 f1 f6 f5 f3 f3 f5 f1 f6 f4 f4 f6 f5 f1 f5 f5 f3 f4 f2 f6 f6 f4 f2 f3
of P3 f5 f6 f5 f6 f4 f3 f2 f4 f3 f2 f6 f1 f1 f5
f1 =
1 1
2 2
3 1 , f2 = 3 3
2 2
3 1 , f3 = 1 2
2 3
3 1
f4 =
1 3
2 1
3 1 , f5 = 2 2
2 1
3 1 , f6 = 3 1
2 3
3 . 2
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Discrete Mathematical Structures
4.2.4
Solved Problems n a 0 o : a 6= 0 ∈ R is an abelian group under 0 0 matrix multiplication.
1. Show that G = Solution.
(i) Closure Property: a 0 b 0 Let A = ,B = ∈ G. 0 0 0 0 ab 0 Then, AB = ∈ G since ab ∈ R, for all a, b ∈ R. 0 0 (ii) Commutative Property: AB = BA is true forall A, B ∈ G, since ab 0 AB = BA = [∵ ab = ba is true in R]. 0 0 (iii) Associative Property: Matrix multiplication is associative always. That is, A(BC) = (AB)C, for all A, B, C ∈ G. (iv) Existence of Identity: 1 0 E= ∈ G is the identity in G, since 0 0 a 0 1 0 a 0 AE = = = A, for all A ∈ G. 0 0 0 0 0 0 (v) Existence ofInverse: 1 a 0 0 −1 a ∈ G is the inverse of If A = ∈ G, then A = 0 0 0 0 1 1 0 A, since AA−1 = (∵ a 6= 0 ∈ R =⇒ 6= 0 ∈ R). 0 0 a Hence, G is an abelian group under matrix multiplication. n o a a 2. Examine whether G = : a 6= 0 ∈ R is a commutative a a group under matrix multiplication, where R is the set of all real numbers. Solution. (i) Closure Property: a a b b Let A = ,B = ∈ G. Then, a a b b 2ab 2ab AB = ∈ G since 2ab ∈ R, for all a, b ∈ R. 2ab 2ab (ii) Commutative Property: AB = BA is true for allA, B ∈ G, since 2ab 2ab AB = BA = [∵ 2ab = 2ba is true in R]. 2ab 2ab
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(iii) Associative Property: Matrix multiplication is associative always. That is, A(BC) = (AB)C, for all A, B, C ∈ G. (iv) Existence of ! Identity: E=
1 2 1 2
1 2 1 2
AE =
a a
∈ G is the identity in G, since 1 1! a a a 2 2 = = A, for all A ∈ G. 1 1 a a a 2
2
(v) Existence of Inverse: a a If A = ∈ G, then A−1 = a a ! of A, since AA−1 =
1 2 1 2
1 2 1 2
1 4a 1 4a
1 4a 1 4a
! ∈ G is the inverse
(∵ a 6= 0 ∈ R =⇒
1 4a
6= 0 ∈ R).
Hence, G is a commutative group under matrix multiplication. n 1 0 −1 0 1 0 −1 0 o 3. Prove that G = , , , 0 1 0 1 0 −1 0 −1 forms an abelian group under matrix multiplication. Solution. 1 0 −1 0 1 0 −1 0 Let A = ,B = ,C = ,D = . 0 1 0 1 0 −1 0 −1 The composition table is shown in the table below. Composition Table · A B A A B B B A C C D D D C
of (G, ·) C D C D D C A B B A
(i) Closure Property: Clearly, from the table above, we have xy ∈ G for all x, y ∈ G. Hence, closure property is satisfied. (ii) Commutative Property: We observe from the table above that xy = yx, for all x, y ∈ G. Hence, commutative property holds. (iii) Associative Property: Matrix multiplication is associative always. That is x(yz) = (xy)z, for all x, y, z ∈ G. (iv) Existence of Identity: 1 0 A= is the identity element in G since 0 1
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Discrete Mathematical Structures AA = A, AB = BA = B, AC = CA = C, and AD = DA = D. (v) Existence of Inverse: From Table 4.2, all elements in G are self-inverses. That is, inverse of A is A, inverse of B is B, inverse of C is C, inverse of D is D, since AA = A, BB = A, CC = A, DD = A. Hence, G forms an abelian group under matrix multiplication. 4. Show that (Q+ , ?) is an abelian group, where ? is defined by ab a ? b = , ∀ a, b ∈ Q+ . 2 Solution. (i) Closure Property: ab It is clear that for all a, b ∈ Q+ , a ? b ∈ Q+ , since ∈ Q+ . 2 Hence, closure property is satisfied. (ii) Commutative Property: a ? b = b ? a is true for all a, b ∈ Q+ , since ab ab ba a?b=b?a= [∵ = is true in Q+ ]. 2 2 2 (iii) Associative Property: a bc bc abc a ? (b ? c) = a ? . = 2 = 2 2 4 ab c ab abc ?c= 2 = . (a ? b) ? c = 2 2 4 Therefore, a ? (b ? c) = (a ? b) ? c, for all a, b, c ∈ Q+ . Hence, associative property is satisfied. (iv) Existence of Identity: e = 2 ∈ Q+ is the identity element, since a·2 a?e=a?2= = a, for all a ∈ Q+ . 2 (v) Existence of Inverse: 4 a−1 = ∈ Q+ is the inverse of a ∈ Q+ , since a a · a4 4 4a a ? a−1 = a ? = = = 2. a 2 2a Hence, Q+ is an abelian group under the operation ? defined in the problem. 5. Prove that the identity element of a group is unique. Solution. Let (G, ?) be a group. Let e1 and e2 be two identity elements in G.
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Then, e1 ? e 2 = e1 e1 ? e 2 = e2
[∵ e2 is the identity] [∵ e1 is the identity].
Thus, e1 = e2 . Hence, the identity is unique. 6. Prove that the identity element is the only idempotent element of a group. Solution. Let (G, ?) be a group. Since e ? e = e, e is the idempotent element. Let a be any idempotent element of G. Then, a ? a = a. Also, e ? a = a [∵ e is the identity element]. It follows that a ? a = e ? a. By the right cancellation law, we have a = e, and so e is the only idempotent element. 7. Prove that if every element in a group is its own inverse, then the group must be abelian. Or prove that for any group (G, ?), if a2 = e with a 6= e, then G is abelian. Solution. Given a = a−1 for all a ∈ G. Let a, b ∈ G. Then, a = a−1 and b = b−1 . Now,
(a ? b) = (a ? b)−1
=⇒
= b−1 ? a−1 = b ? a. G is abelian.
8. Prove for any element a in a group G, the inverse is unique. Solution. Let a be any element of a group G. If possible, let a0 and a00 be two inverses of a. Then a ? a 0 = a0 ? a = e a ? a00 = a00 ? a = e. 0 Now, a = a0 = a0 ? (a ? a00 ) = (a0 ? a) ? a00 = e ? a00 = a00 . Hence, the inverse is unique. 9. Prove that in a group (G, ?), (a ? b)−1 = b−1 ? a−1 . Solution. (a ? b)(b−1 ? a−1 ) = a ? (b ? b−1 ) ? a−1
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Discrete Mathematical Structures = a ? e ? a−1 = a ? a−1 = e and
(b−1 ? a−1 ) ? (a ? b) = b−1 ? a−1 ? a ? b = b−1 ? e ? b = b−1 ? b = e.
Hence, (a ? b)−1 = b−1 ? a−1 . 10. If a and b are any two elements of a group (G, ?), then show that G is abelian if and only if (a ? b)2 = a2 ? b2 . Solution. Necessary Part: Given that (G, ?) is an abelian group. =⇒
For all a, b ∈ G, a ? b = b ? a. 2
2
(4.10)
2
To prove: (a ? b) = a ? b . (a ? b)2 = (a ? b) ? (a ? b) = a ? (b ? a) ? b = a ? (a ? b) ? b [using (4.10)] = (a ? a) ? (b ? b) = a2 ? b2 . Sufficient Part: Given:
(a ? b)2 = a2 ? b2 .
(4.11)
To prove: a ? b = b ? a. (4.11) =⇒ =⇒ =⇒ =⇒ =⇒ =⇒ =⇒
(a ? b)2 = a2 ? b2 (a ? b) ? (a ? b) = (a ? b) ? (b ? b) a ? [b ? (a ? b)] = a ? [a ? (b ? b)] b ? (a ? b) = a ? (b ? b) (using left cancellation law) (b ? a) ? b = (a ? b) ? b (using associative property) b ? a = a ? b (using right cancellation law) G is abelian.
11. Show that every group of order four is abelian. Solution. Let (G, ?) be a group of order four where G = {e, a, b, c}. Since G is of even order, there exists at least one element, say a such that a−1 = a. Then, two cases arise:
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(i) b−1 = b, c−1 = c (ii) b−1 = c, c−1 = b. Case (i): e−1 = e, a−1 = a, b−1 = b, c−1 = c. That is, every element has its own inverse. Then, (G, ?) is abelian. Case (ii): a−1 = a, b−1 = c, c−1 = b. Therefore, a2 = e, b ? c = e, c ? b = e. Since (G, ?) is a group, its elements will appear in a row (column) only once. Since a, e appear in the second row and b appears in the third column, c will appear as (2, 3)th element. ∴ (2, 4)th element is b (3, 3)th element is a (3, 2)th element is c (4, 2)th element is b (4, 4)th element is a. Composition Table of (G, ?) ? e a b c e e a b c a a e c b b b c a e c c b e a 12. Show that the set S = {[1], [5], [7], [11]} is a group with respect to multiplication modulo 12. Solution. The composition table of S with respect to ×12 is given in the table below: Here, 5 ×12 7 = 35, which on division by 12 gives the Composition ×12 [1] [1] [1] [5] [5] [7] [7] [11] [11]
Table of (S, ×12 ) [5] [7] [11] [5] [7] [11] [1] [11] [7] [11] [1] [5] [7] [5] [1]
remainder 11, 11 ×12 7 = 77, which on division by 12 gives the remainder 5, etc. Hence, S is a group, in which [1] is the identity and each element of S is its own inverse. cos α − sin α 13. Show the set of matrices G = , a ∈ R forms a sin α cos α group under matrix multiplication.
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Discrete Mathematical Structures Solution. (i) Closure Property: cos α − sin α cos β − sin β Let Aα = ∈ G and Aβ = ∈ G. sin α cos α sin β cos β Then cos α − sin α cos β − sin β Aα Aβ = sin α cos α sin β cos β cos α cos β − sin α sin β −(cos α sin β + sin α cos β) = sin α cos β + cos α sin β cos α cos β − sin α sin β cos(α + β) − sin(α + β) = = Aα+β ∈ G. (4.12) sin(α + β) cos(α + β) (ii) Associative Property: We know that matrix multiplication is associative. (iii) Existence of Identity: 1 0 I0 = is the identity in G, since Aα I0 = I0 Aα = Aα for 0 1 Aα ∈ G. (iv) Existence of Inverse: A−α is the inverse of Aα for each Aα ∈ G, since Aα A−α = Aα+(−α) = A0 = I0 , using (4.12) Hence, G forms a group under matrix multiplication.
4.2.5
Subgroups
Definition 4.2.23 Subgroup: A non-empty subset H of a group G is said to be a subgroup of G, if H itself is a group under the same operation defined on G and with the same identity element. Example 4.2.24 The set of all integers Z is a subgroup of the set of all real numbers R under usual addition. That is, (Z, +) is a subgroup of (R, +). Theorem 4.2.25 The necessary and sufficient condition is that a non-empty subset H of a group (G, ?) is a subgroup iff for any a, b ∈ H, a ? b−1 ∈ H. Proof. Necessary condition: Assume that H is a subgroup of G. Since H itself is a group, we have a, b ∈ H =⇒ a ? b ∈ H (using closure property). Also, b ∈ H =⇒ b−1 ∈ H (using inverse property). ∴ a, b ∈ H =⇒ a, b−1 ∈ H =⇒ a ? b−1 ∈ H. Sufficient condition: Let a ? b−1 ∈ H, for all a, b ∈ H and H is a subset of G. We have to prove H is a subgroup of G.
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(i) Existence of Identity: Let a ∈ H =⇒ a ? a−1 ∈ H ⊆ G =⇒ e ∈ H. ∴ e is the identity element of H. (ii) Existence of Inverse: Let e ∈ H, a ∈ H =⇒ e ? a−1 ∈ H ⊆ G =⇒ a−1 ∈ H. ∴ Every element of H has an inverse in H. (iii) Closure Property: Let b ∈ H =⇒ b−1 ∈ H. ∴
For a, b ∈ H =⇒ a, b−1 ∈ H −1 =⇒ a ? b−1 ∈H⊆G =⇒ a ? b ∈ H.
∴ H is closed under the operation ?. (iv) Associative Property: Given that H ⊆ G. =⇒ The elements of H are also the elements of G. Since ? is associative in G, it must also be associative in H. Therefore, H itself is a group under the operation ? in G.
4.2.6
Cyclic Groups
Definition 4.2.26 Order of a group: Let (G, ?) be a group. The number of elements in G is called the order of the group G and is denoted by O(G). Note: If O(G) is finite, then G is called a finite group, otherwise it is called an infinite group. Definition 4.2.27 Cyclic Group: A group (G, ?) is said to be cyclic if there exists an element a ∈ G such that any x ∈ G can be written as either x = an or x = na, where n is some integer. This element a is called the generator of the cyclic group G, that is, the cyclic group generated by a, and we denote it by G = . Example 4.2.28 The multiplicative group, G = {1, −1, i, −i}, (i being the complex number) is cyclic. We can write 1 = i4 , −1 = i2 , i = i3 . That is all the elements of G can be expressed as integral powers of the element i. Therefore, G is a cyclic group generated by i. Since i is the generator of G, i−1 is also a generator of G. Hence, G is a cyclic group, and its generators are i and i−1 .
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Theorem 4.2.29 Every cyclic group is abelian. Proof. Let (G, ?) be a cyclic group generated by an element a ∈ G, that is G = . Then, for any two elements x, y ∈ G, we have x = an , y = am , where m, n are integers. Therefore, x ? y = an ? am = am+n = am ? an = y ? x. Thus, (G, ?) is abelian. Theorem 4.2.30 Let (G, ?) be a finite group generated by an element a ∈ G. If G is of order n, that is, O(G) = n, then an = e so that G = {a, a2 , . . . , an = e}. Further, n is the least positive integer for which an = e. Proof. Let us assume that, for some positive integer m < n, am = e. Since G is cyclic, any element of G can be written as ak , for some k ∈ Z. By division algorithm, we have k = mq + r, where q ∈ Z and 0 ≤ r ≤ m. Therefore, ak = amq+r = amq ? ar q = (am ) ? ar = eq ? a r = e ? ar = ar . Hence, every element of G can be expressed as ar , for some 0 ≤ r ≤ m. Therefore, G has at most m distinct elements. That is, O(G) = m < n, which is a contradiction. Hence, am = e, for m < n is not possible. We now proceed to show that the elements a, a2 , a3 , . . . , an are all distinct where an = e. If possible, let ai = aj , for i < j ≤ n. Therefore, ai ? a−j = aj ? a−j =⇒
ai−j = aj−j = e,
which is again a contradiction. Hence, ai 6= aj , for i, j ≤ n. Hence, the theorem is proved.
where i − j < n,
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Theorem 4.2.31 Every subgroup of a cyclic group is cyclic. Proof. Let G be a finite cyclic group of order n with generator a. That is, g = {e, a, a2 , . . . , an−1 }. Let H be a subgroup of G. Then, elements of H are of the form ak with 1 ≤ k < n. Let t be the smallest positive integer such that at ∈ H. We shall prove that H = . Indeed, let am ∈ H. By the division algorithm, there exist unique integers q and r such that m = tq + r where 0 ≤ r < t. It follows that am = (at )q ar or ar = am (at )−q . But am ∈ H and at ∈ H. Then by closure, ar ∈ H. Since t is the smallest positive integer such that at ∈ H, we must have r = 0. Hence, am = (at )q or q m ∈ . Clearly, ⊆ H since at ∈ H and H is a group. Theorem 4.2.32 Every group of prime order is cyclic and hence is abelian. Proof. Let G be a group with O(G) = p, a prime. Let a 6= e ∈ G and H = be the cyclic group of G generated by a. By Lagrange’s theorem, O(H)|p. So, O(H) = 1 or p. Since O(H) 6= 1 (as a 6= e and a, e ∈ H, O(H) ≥ 2), we have O(H) = p. So, G = H = is a cyclic group. Since every cyclic group is abelian, G is abelian.
4.2.7
Homomorphisms
Definition 4.2.33 Homomorphism: Let (G, ?) and (H, ∆) be any two groups. A mapping f : G −→ H is said to be a homomorphism if f (a ? b) = f (a)∆f (b), for a, b ∈ G. Example 4.2.34 Let G = (Z, +) and H = (nZ, +) be two groups (for a fixed integer n). The mapping f : G −→ H defined by f (m) = nm for m ∈ Z is a homomorphism from G into H. Definition 4.2.35 Kernel of a Homomorphism: Let f : G −→ G0 be a group homomorphism. The set of elements of G which are mapped into e0 (the identity of G0 ) is called the kernel of f and is denoted by ker(f ). That is, ker(f ) = {x ∈ G/f (x) = e0 , where e0 is the identity of G0 }. Theorem 4.2.36 The kernel of a homomorphism f from a group G into a group G0 is a subgroup of G. Proof. Let f : (G, ?) −→ (G0 , ?0 ) be any homomorphism. Then,
ker(f ) = {x ∈ G/f (x) = e0 , where e0 is the identity of G0 }.
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Since f (e) = e0 is true always, at least e ∈ ker(f ). Therefore, ker(f ) is a non-empty subset of G. Let a, b ∈ ker(f ) with f (a) = e0 and f (b) = e0 . Therefore, f (a ? b−1 ) = f (a) ?0 f (b−1 )(since f is a homomorphism) = f (a) ?0 (f (b))−1 = e0 ?0 e0 = e0 =⇒
a ? b−1 ∈ ker(f ).
That is, a, b ∈ ker(f ) =⇒ a ? b−1 ∈ ker(f ). Hence, ker(f ) is a subgroup of G. Definition 4.2.37 Endomorphism: A homomorphism f of a group into itself is called an endomorphism. Definition 4.2.38 Isomorphism: A mapping f from a group G to a group G0 is said to be an isomorphism if f is a one-to-one and onto homomorphism. Theorem 4.2.39 Cayley’s Representation Theorem: Every finite group of order n is isomorphic to a permutation group of degree n. Proof. Let G be any finite group of order n. For each a ∈ G, define a function fa : G −→ G such that fa (x) = ax, for every x ∈ G. Clearly, this function fa is bijective (one-to-one and onto). Consider G1 = {fa /a ∈ G}. This G1 becomes a group under the composition operation of functions. That is, (G1 , ◦) is the permutation group of order n. Now, define a function Φ : G −→ G1 such that Φ(a) = fa , for all a ∈ G. Claim 1: Φ is a homomorphism Φ(b) = fab = fa ◦ fb [since fab (x) = abx = a(bx) = fa (bx) = fa ◦ fb (x)] = Φ(a) ◦ Φ(b). Claim 2: Φ is bijective Clearly, Φ is one-to-one, since
=⇒ =⇒ =⇒ =⇒
Φ(a) = Φ(b) fa = fb fa (x) = fb (x), ax = bx a = b.
for every x ∈ G
Also, for every fa ∈ G, we have a ∈ G such that Φ(a) = fa .
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Therefore, Φ is onto. Hence, Φ is bijective. Thus, Φ : G −→ G1 becomes as an isomorphism. Hence, every finite group of order n is isomorphic to a permutation group of degree n. Theorem 4.2.40 Any cyclic group of order n is isomorphic to the additive group of residue classes of integers modulo n. Proof. Let G = {a, a2 , . . . , an = e} be a cyclic group of order n generated by a. We know that (Zn , +n ) is the additive group of residue classes modulo n =⇒ Zn = {[1], [2], . . . , [n] = [0]}. Let f : G −→ Zn be defined by f (ar ) = [r], for all ar ∈ G. For all [r] ∈ Zn , there exists ar ∈ G such that f (ar ) = [r] =⇒ f is onto. For r 6= s, [r] 6= [s] and hence f (ar ) 6= f (as ) =⇒ f is one-to-one. For all ar , as ∈ G, f (ar · as ) = f (ar+s ) = [r + s] = [r] + [s] = f (ar ) +n f (as ) =⇒ f is a homomorphism. Hence, (G, ·) is isomorphic to (Zn , +n ).
4.2.8
Cosets and Normal Subgroups
Definition 4.2.41 Left and Right Cosets: Let (H, ?) be a subgroup of a group (G, ?). (i) For any a ∈ G, the set a ? H defined by a ? H = {a ? h/h ∈ H} is called the left coset of H in G determined by the element a ∈ G. (ii) For any a ∈ G, the set H ? a defined by H? = {h ? a/h ∈ H} is called the right coset of H in G determined by the element a ∈ G. Example 4.2.42 Consider the multiplicative group G = {1, −1, i, −i} and a subgroup H = {1, −1}. Clearly, iH, −iH, 1H, and −1H are the left cosets. Definition 4.2.43 Index of a subgroup in a group: Let (H, ?) be a subgroup of a group (G, ?). Then, the number of different left (or right) cosets of H in G is called the index of H in G, and it is denoted by iG (H). Some important results: (i) If G is abelian, then a ? H = H ? a, for all a ∈ G.
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(ii) If H is a subgroup of G and e ∈ H, then e ? H = H ? e = H. (iii) Any two left or right cosets of H in G are either disjoint or identical. (iv) The union of all distinct left (or right) cosets of H in G is equal to G. Theorem 4.2.44 Let (H, ?) be a subgroup of a group (G, ?). The set of left cosets of H in G forms a partition of G. Also, every element of G belongs to one and only one left coset of H in G. Proof. To prove: Every element of G belongs to one and only one left coset of H in G. Let H be a subgroup of a group G. Let a ∈ G. Then, aH = H if and only if a ∈ H. Suppose a ∈ G and aH = H. Then, aH = H =⇒ ae ∈ H =⇒ a ∈ H
(since H is a subgroup and e ∈ H).
Conversely, assume that a ∈ H. Then ah ∈ H, for all h ∈ H. So, aH ⊆ H.
(4.13)
Given any y ∈ H, a−1 y ∈ H, and y = a(a−1 y) ∈ H. So, y ∈ aH, for all y ∈ H. That is, H ⊆ aH. (4.14) From (4.13) and (4.14), we have H = aH. Hence, every element of G belongs to one and only one left coset of H in G. To prove: The set of left cosets of H in G forms a partition of G. Let a, b ∈ G and H be a subgroup of G. If aH ∩ Ha 6= φ, then let c ∈ aH ∩ Ha. Since c ∈ aH, we have cH = aH. Let H be a subgroup of a group G. Let a, b ∈ G if b ∈ aH; then bH = aH. Since c ∈ bH, we have cH = bH. So aH = cH = bH. Thus, if aH ∩ bH 6= φ, then aH = bH. Therefore, any two distinct left cosets are disjoint. Hence, the set of all (distinct) left cosets of H in G forms a partition of G. Theorem 4.2.45 Lagrange’s Theorem: The order of each subgroup of a finite group is a divisor of the order of the group.
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Proof. Let G be a finite group and H be a subgroup of G. Let O(G) = n and O(H) = m. Let us consider all left cosets of H in G. Each coset has exactly m elements. ah1 = ah2 =⇒ h1 = h2 , for all a ∈ G. By result (iii), namely, G is decomposed into say r mutually disjoint subsets, each of order m. Therefore, n = rm. That is, O(G) = rO(H). Thus, O(H) divides O(G). Note: The converse of Lagrange’s theorem is not true in general. That is, if n is a divisor of a group G, then it does not necessarily follow that G has a subgroup of order n. Theorem 4.2.46 If (G, ?) is a finite group of order n, then for any a ∈ G, we must have an = e, where e is the identity of the group G. Proof. Let O(G) = n. Let a ∈ G. Then, the order of the subgroup is the order of the element a. If O()= m, then am = e, and by Lagrange’s theorem, we get m|n. Let n = mk. Then, an = amk = (am )k = ek = e. Definition 4.2.47 Normal Subgroup: A subgroup (H, ?) of a group (G, ?) is said to be a normal subgroup of G if for every x ∈ G and for every h ∈ H, xhx−1 ∈ H or xHx−1 ⊆ H. Example 4.2.48 Consider the group (Z, +). Clearly, (3Z, +) is a normal subgroup of (Z, +). Definition 4.2.49 Quotient Group or Factor Group: Let N be a normal subgroup of a group (G, ?) and the set of all right cosets of N in G be denoted by G/N = {N a|a ∈ G}. Now, define ⊗ as binary operation on G/N as N a ⊗ N b = N (a ? b). Then, (G/N, ⊗) will form a group called quotient group or factor group. Theorem 4.2.50 The kernel of a homomorphism is a normal subgroup. Proof. Let f : (G, ?) −→ (G0 , ?0 ) be any homomorphism. Then, ker(f ) = {x ∈ G/f (x) = e0 (where e0 is the identity element of G0 )} is a subgroup of G by Theorem 4.2.36. Let x ∈ ker(f ) and let g ∈ G. =⇒ f (x) = e0 .
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Consider f (g ? x ? g −1 ) = f (g) ?0 f (x ? g −1 ) = f (g) ?0 [f (x) ?0 f (g −1 )] = f (g) ?0 [e0 ?0 f (g −1 )] = f (g) ?0 f (g −1 ) = f (g ? g −1 ) = f (e) = e0 . Thus, f (g ? x ? g −1 ) = e0 . Therefore, g ? x ? g −1 ∈ ker(f ). Hence, ker(f ) is a normal subgroup. Theorem 4.2.51 Let (H, ?) be a subgroup of a group (G, ?). Then, (H, ?) is a normal subgroup if and only if a ? h ? a−1 = H, for all a ∈ G. Proof. Let H be a normal subgroup of G. Then by definition, a ? H = H ? a, for all a ∈ G. Hence, a ? H ? a−1 = a ? (a−1 ? H) = (a ? a−1 ) ? H =e?H = H. Conversely, let a−1 ? H ? a = H, for all a ∈ G. That is, a ? (a−1 ? H ? a) = a ? H or (a ? a−1 ) ? (H ? a) = a ? H or e ? (H ? a) = a ? H or H ? a = a ? H. Thus, H is a normal subgroup. Theorem 4.2.52 Let (G, ?) be a group. Let H = {a|a ∈ G & a ? b = b ? a, ∀ b ∈ G}. Then, H is a normal subgroup. Proof. H = {a|a ∈ G & a ? b = b ? a, ∀ b ∈ G}. Since e ? a = a ? e = a, ∀a ∈ G, we have e ∈ H. Therefore, H is non-empty. Let x, y ∈ H. Then a ? x = x ? a, ∀ x ∈ G, and a ? y = y ? a, ∀ y ∈ G. Claim: H is a normal subgroup. Consider a ? (x ? y) = (a ? x) ? y = (x ? a) ? y = x ? (a ? y)
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201 = x ? (y ? a) = (x ? y) ? a =⇒ x ? y ∈ H.
Let a ∈ H. Then, a ? x = x ? a, ∀x ∈ G. Then a−1 ? (a ? x) = a−1 ? (x ? a) =⇒ x = a−1 ? (x ? a) =⇒ x ? a−1 = a−1 ? (x ? a) ? a−1 = (a−1 ? x) ? (a ? a−1 ) = a−1 ? x =⇒ x ? a−1 = a−1 ? x, ∀x ∈ G =⇒ a−1 ∈ H. Thus, H is a subgroup. To prove: H is normal. Let x ∈ H, g ∈ G. Then, a ? x = x ? a, ∀a ∈ G. Then, g ? x ? g −1 = x ? g ? g −1 =⇒ x ∈ H. Thus, g ? x ? g −1 ∈ H =⇒ H is normal. Theorem 4.2.53 N is a normal subgroup of a group G if and only if gN g −1 = N , for every g ∈ G (or gN = N g). Show that the number of right and left cosets are equal in normal subgroups and every left coset is a right coset. Proof. Let N be a normal subgroup of G. Let x ∈ gN g −1 =⇒ x = gng −1 , for some n ∈ N . Therefore, x = gng −1 ∈ N (∵ N is a normal subgroup). Hence, gN g −1 ⊆ N . Now, g −1 N g = g −1 N (g −1 )−1 ⊆ N , since g −1 ∈ G, and g −1 ng ∈ N . Therefore, N = g(g −1 N g)g −1 ∈ gN g −1 Therefore, N ⊆ gN g −1 . Hence, N = gN g −1 . Conversely, let N g −1 = N , for every g ∈ G. That is, gN g −1 is the set of all gng −1 , for n ∈ N . Clearly, gN g −1 ⊆ N . Therefore, N is a normal subgroup. We get if N is a normal subgroup, then gN g −1 = N or gN = N g, that is, the left and right cosets are equal. Therefore, the right and left cosets are equal in number in normal subgroups, and every left coset is a right coset.
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Fundamental Theorem of Group Homomorphism Let H be a normal subgroup of a group G. Let G/H be the set of all left cosets of H in G. That is, G/H = {aH/a ∈ G}. Let us define an operation “·” as follows. For any a, b ∈ G, (ab)H = (aH)(bH). This is a binary operation under which G/H becomes a group. It is called the quotient group or factor group. Let f : G −→ G/H be defined as f (a) = aH, for any a ∈ G. Then, for any a, b ∈ G, f (ab) = (ab)H = (aH) · (bH) = f (a) · f (b). Therefore, f is a homomorphism of G into G/H. It is called the natural homomorphism or canonical homomorphism. Theorem 4.2.54 Let g be a homomorphism of a group G into a group G0 . Let K be the kernel of g and R be the image set of g in G0 . Then, G/K is isomorphic to R. Proof. We have already shown that K is a normal subgroup of G. Therefore, there exists a canonical homomorphism f : G −→ G/K given by f (a) = aK, for any a ∈ G. Now, let us define a mapping h : G/K −→ R such that h(aK) = g(a). The image set of h is the same as the image set of g, and hence h is onto. Further, for any a, b ∈ G such that aK = bK, we have ak1 = bk2 , for some k1 , k2 ∈ K. Therefore, g(ak1 ) = g(a)g(k1 ) = g(a)e0 = g(a) and g(bk2 ) = g(b)g(k2 ) = g(b) = e0 = g(b) so that aK = bK =⇒ g(a) = g(b). Therefore, h(aK) = g(a) = g(b) = h(bK). Also, f (a) = f (b). Hence, h is one-to-one and onto. Further, h(aKbK) = h(abK) == g(ab) = g(a)g(b) = h(aK)h(bK). Hence, h is an isomorphism of G/K to R.
4.2.9
Solved Problems
1. Prove that the intersection of any two subgroups of a group (G, ?) is again a subgroup of (G, ?). Solution. Let H and K be subgroups of G. Let a ∈ H =⇒ a, b ∈ H and a, b ∈ K =⇒ a ? b−1 ∈ H and a ? b−1 ∈ K (as H and K are subgroups) =⇒ a ? b−1 ∈ H ∩ K. Hence, H ∩ K is a subgroup of G.
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2. Is the union of two subgroups of a group a subgroup of G? Justify your answer. Solution. The union of two subgroups of a group G need not be a subgroup of G. For example, we know (Z, +) is a group. Let H = 3Z = {0, ±3, ±6, . . . } Let K = 2Z = {0, ±2, ±4, . . . } =⇒ H and K are subgroups of (Z, +) =⇒ 3 ∈ 3Z ⊆ 3Z ∪ 2Z = H ∪ K =⇒ 2 ∈ 2Z ⊆ 2Z ∪ 3Z = H ∪ K. But 3 + 2 = 5 ∈ / 2Z ∪ 3Z. ∴ H ∪ K is not a subgroup of (Z, +). 3. If H1 and H2 are subgroups of a group (G, ?), then prove that H1 ∪H2 is a subgroup of G if and only if either H1 ⊆ H2 or H2 ⊆ H1 . Solution. Given H1 and H2 are two subgroups of (G, ?) and H1 ⊆ H2 or H2 ⊆ H1 . If H1 ⊆ H2 , then H1 ∪ H2 = H2 which is a subgroup of G. If H2 ⊆ H1 , then H1 ∪ H2 = H1 which is a subgroup of G. Conversely, suppose H1 6⊂ H2 and H2 6⊂ H1 . Then, there exist a ∈ H1 and a ∈ / H2 , and there exist b ∈ H2 and b∈ / H1 . Now a, b ∈ H1 ∪ H2 . Since H1 ∪ H2 is a subgroup, it follows that a ? b ∈ H1 ∪ H2 . Hence, a ? b ∈ H1 or a ? b ∈ H2 . Case (i): If a ? b ∈ H1 , then a−1 ? (a ? b) ∈ H1 . That is, b ∈ H1 , which is a contradiction. Case(ii): If a ? b ∈ H2 . Then, (a ? b) ? b−1 ∈ H2 . That is, a ∈ H2 , which is a contradiction. Thus, either H1 ⊆ H2 or H2 ⊆ H1 . 4. Find all the subgroups of (Z9 , +9 ). Solution. Z9 = {[0], [1], [2], [3], [4], [5], [6], [7], [8]}. The binary operation is addition modulo 9 (or +9 ). Consider the subsets H1 = {[0], [2], [4], [6], [8]}, H2 = {[0], [3], [6]}, H3 = {[0], [4], [8]}, H4 = {[0], [5]}. The improper subgroups of (Z9 , +9 ) are ({[0]}, +9 ) and (Z9 , +9 ).
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Discrete Mathematical Structures +9 [0] [3] [6]
H2 is closed since
H1 is closed since
H3 is closed since
H4 is closed since
+9 [0] [2] [4] [6] [8]
[0] [0] [2] [4] [6] [8] +9 [0] [4] [8]
[0] [3] [6] [0] [3] [6] [3] [6] [0] [6] [0] [3] [2] [2] [4] [6] [8] [1]
[4] [4] [6] [8] [1] [3]
[6] [6] [8] [1] [3] [5]
[8] [8] [1] [3] [5] [7]
[0] [4] [8] [0] [4] [8] [4] [8] [3] [8] [3] [7] +9 [0] [5]
[0] [5] [0] [5] [5] [1]
The above composition tables show that H1 , H2 , H3 , and H4 are closed under +9 . Therefore, the possible subgroups of (Z9 , +9 ) are (H1 , +9 ), (H2 , +9 ), (H3 , +9 ), and (H4 , +9 ). 5. Find the left cosets of {[0], [3]} in the addition modular group (Z6 , +6 ). Solution. Let Z6 = {[0], [1], [2], [3], [4], [5]} be a group and H = {[0], [3]} be a subgroup of Z6 under +6 (addition mod 6). The left cosets of H are [0] +6 H = {[0], [3]} = H [1] +6 H = {[1], [4]} [2] +6 H = {[2], [5]} [3] +6 H = {[3], [6]} = {[3], [0]} = {[0], [3]} = H [4] +6 H = {[4], [7]} = {[4], [1]} = [1] +6 H [5] +6 H = {[5], [8]} = {[5], [2]} = [2] +6 H. ∴ [0] +6 H = [3] +6 H = H and [1] +6 H = [4] +6 H, [2] +6 H = [5] +6 H are the distinct left cosets of H in Z6 .
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6. Find the left cosets of {[0], [2]} in the group (Z4 , +4 ). Solution. Let Z4 = {[0], [1], [2], [3]} be a group and H = {[0], [2]} be a subgroup of Z4 under +4 (addition mod 4). The left cosets of H are [0] + H = {[0], [2]} = H [1] + H = {[1], [3]} [2] + H = {[2], [4]} = {[2], [0]} = {[0], [2]} = H [3] + H = {[3], [5]} = {[3], [1]} = {[1], [3]} = [1] + H. Therefore, [0] + H = [2] + H = H and [1] + H = [3] + H are the two distinct left cosets of H in Z4 . 7. Let G = {1, a, a2 , a3 } (a4 = 1) be a group and H = {1, a2 } be a subgroup of G under multiplication. Find all the cosets of H. Solution. The right cosets of H in G are H1 = {a, a2 } = H Ha = {a, a3 } Ha2 = {a2 , a4 } = {a2 , a} = H and Ha3 = {a3 , a5 } = {a3 , a} = Ha. Therefore, H1 = H = Ha2 = {1, a2 } and Ha = Ha3 = {a, a3 } are two distinct right cosets of H in G. Similarly, we can find the left cosets of H in G. 8. Prove that any two infinite cyclic groups are isomorphic to each other. Solution. Let G1 = and G2 = be two cyclic groups of infinite order. G1 = {an |n is an integer} and G2 = {bn |n is an integer}. Define a map f : G1 −→ G2 by f (an ) = bn . Let x, y ∈ G1 . Then, x = an , y = am for some integers n and m. Now, f (x)f (y) = f (an )f (am ) = bn bm = bn+m = f (an+m ) = f (an am ) = f (xy). Hence, f is a homomorphism. If f (x) = f (y), then f (an ) = f (am ) =⇒ bn = bm . Then, bn−m = e0 in G2 . As G2 is an infinite cyclic group generated by b, there is no non-zero integer k such that bk = e0 . Hence, from bn−m = e0 , we have n − m = 0, or n = m, and hence x = an = am = y. Thus, f is one-to-one.
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Discrete Mathematical Structures Let z ∈ G2 . Then, z = bn for some integer n. Now, take x = an . Then, f (x) = f (an ) = bn = z. So, the map f is onto. Now, f is one-to-one and onto homomorphism. Hence, it is an isomorphism. 9. Prove that the group homomorphism preserves the identity element. Solution. Let (G, ?) and (H, ◦) be two groups. Letf : G −→ H be a group homomorphism. Let eG be the identity element of G. Let eH be the identity element of H. To prove: f (eG ) = eH . Consider f (eG ) = f (eG ? eG ) (since eG is the identity in G) = f (eG ) ◦ f (eG ) (since f is a homomorphism) Multiplying bothsides by f (eG )−1 on the right side, we get f (eG ) ◦ f (eG )−1 = f (eG ) ◦ f (eG ) ◦ f (eG )−1 eH = f (eG ). Hence, the group homomorphism preserves the identity element.
10. Prove that the group homomorphism preserves the inverse element. Solution. Let (G, ?) and (H, ◦) be two groups. Let f : G −→ H be a group homomorphism. Let eG be the identity element of G. Let eH be the identity element of H. To Prove: [f (x)]−1 = f (x−1 ), for all x ∈ G. It is sufficient to prove that f (x) ◦ f (x−1 ) = eH . Now, for all x ∈ G, we can write f (x) ◦ f (x−1 ) = f (x ? x−1 ) (since f is a homomorphism) =⇒ f (x) ◦ f (x−1 ) = f (eG ) = eH . Hence, the group homomorphism preserves the inverse element. 11. If f : G −→ G0 is a group homomorphism from (G, ?) to (G0 , ∆), then prove that for any a ∈ G, f (a−1 ) = [f (a)]−1 . Solution. For all a, a−1 ∈ G, we have f (a ? a−1 ) = f (a)∆f (a−1 ) f (e) = f (a)∆f (a−1 )
or or
e0 = f (a)∆f (a−1 ).
(4.15)
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Similarly, f (a−1 ∆a) = f (a−1 )∆f (a) f (e) = f (a−1 )∆f (a)
or or
e0 = f (a−1 )∆f (a).
(4.16)
From (4.15) and (4.16), we get f (a)∆f (a−1 ) = f (a−1 )∆f (a) =⇒ f (a−1 ) is the inverse of f (a). That is, f (a−1 ) = [f (a)]−1 . 12. Let G be a group and a ∈ G. Let f : G −→ G be given by f (x) = axa−1 , for all x ∈ G. Prove that f is an isomorphism of G onto G. Solution. To show f is a homomorphism: If x, y ∈ G, then f (x)f (y) = (axa−1 )(aya−1 ) = ax(−1 a)ya−1 = axya−1 = a(xy)a−1 = f (xy). Therefore, f is a homomorphism. To show f is one-to-one: If f (x) = f (y), then axa−1 = aya−1 . Hence, by left cancellation law, we have xa−1 = ya−1 ; again by right cancellation law, we get x = y. Therefore, f (x) = f (y) =⇒ x = y. Hence, f is one-to-one. To show f is onto: Let y ∈ G; then a−1 ya ∈ G and f (a−1 ya) = a(a−1 ya)a−1 = (aa−1 )y(aa−1 ) = y. Therefore, f (x) = y, for some x ∈ G. Thus, f is an isomorphism. 13. Prove that the intersection of two normal subgroups is a normal subgroup. Solution. Let H and K be any two normal subgroups of a group G. We have to prove that H ∩ K is a normal subgroup of G.
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Discrete Mathematical Structures Since H and K are subgroups of G, e ∈ H and e ∈ H. Hence, e ∈ H ∩ K. Thus, H ∩ K is a non-empty set. Let a, b ∈ H ∩ K. Claim: ab−1 ∈ H ∩ K. Since a, b ∈ H ∩ K, both a and b belong to H and K. Since H and K are subgroups of G, ab−1 ∈ H and ab−1 ∈ K, so that ab−1 ∈ H ∩ K. Hence, H ∩ K is a subgroup of G, by a criterion for subgroup. To prove: H ∩ K is normal. Let x ∈ H ∩ K and g ∈ H. Since x ∈ H ∩ K, x ∈ H and x ∈ K. Since x ∈ H, g ∈ G =⇒ gxg −1 ∈ K (as H is normal). Similarly, x ∈ K, g ∈ G =⇒ gxg −1 ∈ K (as K is normal). Hence, x ∈ H ∩ K and g ∈ G =⇒ gxg −1 ∈ H ∩ K. Thus, H ∩ K is a normal subgroup of G.
14. Prove that every subgroup of an abelian group is a normal subgroup. Solution. Let (G, ?) be an abelian group and (N, ?) be a subgroup of G. Let g be any element in G, and let n ∈ N . Now, g ? n ? g −1 = (n ? g) ? g −1 = n ? (g ? g =n?e = n ∈ N.
−1
[∵ G is abelian]
)
Therefore, for all g ∈ G and n ∈ N , g ? n ? g −1 ∈ N . Hence, (N, ?) is a normal subgroup.
4.2.10
Permutation Functions
Definition 4.2.55 A bijection from a set A to itself is called a permutation of A. Example 4.2.56 Let A = R, and let f : A −→ A be defined by f (a) = 2a+1. Since f is one-to-one and onto, it follows that f is a permutation of A. Example 4.2.57 Let A = {1, 2, 3}. Then, 1 2 3 1 2 1A = , p1 = 1 2 3 1 3 1 2 3 1 2 p3 = , p4 = 2 3 1 3 1
all the permutations of A are 3 1 2 3 , p2 = , 2 2 1 3 3 1 2 3 , p5 = . 2 3 2 1
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Remark 4.2.58 In the above example, we can write the permutations as ordered pairs. For example, p4 = {(1, 3), (2, 1), (3, 2)} and p−1 4 = {(3, 1), (1, 2), (2, 3)}. Or, if the first component of each ordered pair is written in increasing order, then we have p−1 4 = {(1, 2), (2, 3), (3, 1)}. 1 2 3 −1 Thus, p4 = = p3 . 2 3 1 Remark 4.2.59 The function p2 takes 1 to 2, and p3 takes 2 to 3, so p3 ◦ p2 takes 1 to 3. Also, p2 takes 2 to 1, and p3 takes 1 to 2, so p3 ◦ p2 takes 2 to 2. Finally, p2 takes 3 to 3, and p3 takes 3 to 1, so p3 ◦ p2 takes 3 to 1. Thus, 1 2 3 p3 ◦ p2 = . 3 2 1 Remark 4.2.60 The process of forming p3 ◦p2 is shown below. It can be noted that p3 ◦ p2 = p5 . 1 2 3 ←− 1 2 3 = 1 2 3 = p5 ↑ ◦ ↓ p3 ◦ p2 = 2 ↓ 3 2 1 1 3 ←− 2 3 1 Theorem 4.2.61 If A = {a1 , a2 , . . . , an } is a set containing n elements, then there are n! = n · (n − 1) · · · 2 · 1 permutations of A. Definition 4.2.62 Cyclic Permutation: Let b1 , b2 , . . . , br be r distinct elements of the set A = {a1 , a2 , . . . , an }. The permutation p : A −→ A is defined by p(b1 ) = b2 p(b − 2) = b3 .. .. . . p(br−1 ) = br p(br ) = b1 . p(x) = x if x ∈ A, x ∈ / {b1 , b2 , . . . , br } is called a cyclic permutation of length r, or simply a cycle of length of r, and will be denoted by (b1 , b2 , . . . , br ). Example 4.2.63 Let A = {1, 2, 3, 4, 5}. permutation 1 2 3 4 3 2 5 4
The cycle (1, 3, 5) denotes the 5 . 1
Definition 4.2.64 Disjoint Cycles: Two cycles of a set A are said to be disjoint if no element of A appears in both cycles.
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Example 4.2.65 Let A = {1, 2, 3, 4, 5, 6}. Then, the cycles (1, 2, 5) and (3, 4, 6) are disjoint, whereas the cycles (1, 2, 5) and (2, 4, 6) are not. Theorem 4.2.66 A permutation of a finite set that is not the identity or a cycle can be written as a product of disjoint cycles of length ≥ 2. Definition 4.2.67 Transposition: A cycle of length 2 is called a transposition. That is, a transposition is a cycle p = (ai , aj ), where p(ai ) = aj and p(aj ) = ai . Remark 4.2.68 Note that if p = (ai , aj ) is a transposition of A, then p ◦ p = 1A , the identity permutation of A. Every cycle can be written as a product of transpositions. In fact, (b1 , b2 , . . . , br ) = (b1 , br ) ◦ (b1 , br−1 ) ◦ · · · ◦ (b1 , b3 ) ◦ (b1 , b2 ). This case can be verified by induction on r as follows. Basis Step: If r = 2, then the cycle is just (b1 , b2 ), which already has the proper form. Induction Step: We use P (m) to find P (m + 1). Let (b1 , b2 , . . . , bm , bm+1 ) be a cycle of length m + 1. Then, (b − 1, b − 2, . . . , bm , bm+1 ) = (b1 , bm+1 ) ◦ (b1 , b2 , . . . , bm ) as may be verified by computing the composition. Using P (m), (b1 , b2 , . . . , bm ) = (b1 , bk ) ◦ (b1 , bm−1 ) ◦ · · · ◦ (b1 , b2 ). Thus, by substitution, (b1 , b2 , . . . , bm+1 ) = (b1 , bm+1 ) ◦ (b1 , bm ) ◦ · · · ◦ (b1 , b3 ) ◦ (b1 , b2 ). This completes the induction step. Thus, by the principle of mathematical induction, the result holds for every cycle. For example, (1, 2, 3, 4, 5) = (1, 5) ◦ (1, 4) ◦ (1, 3) ◦ (1, 2). Corollary 4.2.69 Every permutation of a finite set with at least two elements can be written as a product of transpositions. Theorem 4.2.70 If a permutation of a finite set can be written as a product of an even number of transpositions, then it can never be written as a product of an odd number of transpositions, and the converse is also true. Definition 4.2.71 Even and Odd Permutations: A permutation of a finite set is called even if it can be written as a product of an even number of transpositions, and it is called odd if it can be written as a product of an odd number of transpositions. Remark 4.2.72 From the definition of even and odd permutations, we have the following: (a) The product of two even permutations is even. (b) The product of two odd permutations is even. (c) The product of an even and an odd permutation is odd.
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Theorem 4.2.73 Let A = {a1 , a2 , . . . , an } be a finite set with n elements, n! n! n ≥ 2. Then, there are even permutations and odd permutations. 2 2 Proof. Let An be the set of all even permutations of A, and let Bn be the set of all odd permutations. We shall define a function f : An −→ Bn , which we shall show is one-to-one and onto, and this will show that An and Bn have the same number of elements. Since n ≥ 2, we can choose a particular transposition q0 of A, say q0 = (an−1 , an). We now define the function f : An −→ Bn by f (p) = q0 ◦ p,
p ∈ An .
Observe that if p ∈ An , then p is an even permutation, so q0 ◦ p is an odd permutation, and thus f (p) ∈ Bn . Suppose now that p1 and p2 are in An and f (p1 ) = f (p2 ). Then q0 ◦ p1 = q0 ◦ p2 .
(4.17)
We now compose each side of equation (4.17) with q0 : q0 ◦ (q0 ◦ p1 ) = q0 ◦ (q0 ◦ p2 ); so by the associative property, (q0 ◦ q0 ) ◦ p1 = (q0 ◦ q0 ) ◦ p2 , or since q0 ◦ q0 = 1A , 1A ◦ p1 = 1A ◦ p2 p1 = p2 . Thus, f is one-to-one. Now, let q ∈ Bn . Then, q0 ◦ q ∈ An , and f (q0 ◦ q) = q0 ◦ (q0 ◦ q) = (q0 ◦ q0 ) = 1A ◦ q = q, which means that f is an onto function. Since f : An −→ Bn is one-to-one and onto, we conclude that An and Bn have the same number of elements. Note that An ∩ Bn = φ, since no permutation can be both even and odd. Also, by theorem, |An ∪ Bn | = n! n! = |An ∪ Bn | = |An | + |Bn | − |An ∩ Bn | = 2|An |. n! Hence, we have |An | = |Bn | = . 2
4.2.11
Solved Problems
1. Let A = {1, 2, 3, 4, 5, 6}. Compute (4, 1, 3, 5) ◦ (5, 6, 3) and (5, 6, 3) ◦ (4, 1, 3, 5).
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Solution. We have 1 2 (4, 1, 3, 5) = 3 2
3 5
4 1
5 4
6 6
1 Then, (4, 1, 3, 5)◦(5, 6, 3) = 3
and
(5, 6, 3) =
2 2
3 5
4 1
5 4
6 1 ◦ 6 1
and (5, 6, 3)◦(4, 1, 3, 5) =
1 1
1 3
2 2
3 4
4 1
5 6
1 1
2 2
3 5
4 4
5 6
6 1 ◦ 3 3
1 5
2 2
3 6
4 1
5 4
=
3 5
4 4
6 . 3
5 6
2 2
3 5
4 4
5 6
6 3
2 2
3 5
4 1
5 4
6 6
6 5
=
2 2
6 . 3
Observe that (4, 1, 3, 5) ◦ (5, 6, 3) 6= (5, 6, 3) ◦ (4, 1, 3, 5) and that neither product is a cycle. 2. Let A a set. Then, write the permutation = {1, 2, 3, 4, 5, 6, 7, 8} be 1 2 3 4 5 6 7 8 p= as a product of disjoint cycles. 3 4 6 5 2 1 8 7 Solution. We start with 1 and find that p(1) = 3, p(3) = 6, and p(6) = 1, so we have the cycle (1, 3, 6). Next, we choose the first element of A that has not appeared in a previous cycle. We choose 2, and we have P (2) = 4, p(4) = 5, and p(5) = 2, so we obtain the cycle (2, 4, 5). We now choose 7, the first element of A that has not appeared in a previous cycle. Since p(7) = 8 and p(8) = 7, we obtain the cycle (7, 8). We can then write p as a product of disjoint cycles as p = (7, 8) ◦ (2, 4, 5) ◦ (1, 3, 6). 1 2 3 4 5 6 7 3. Is the permutation p = even or odd? 2 4 5 7 6 3 1 Solution. We first write p as a product of disjoint cycles, obtaining p = (3, 5, 6) ◦ (1, 2, 4, 7). Next, we write each of the cycles as a product of transpositions: (1, 2, 4, 7) = (1, 7) ◦ (1, 4) ◦ (1, 2) (3, 5, 6) = (3, 6) ◦ (3, 5).
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Then, p = (3, 6) ◦ (3, 5) ◦ (1, 7) ◦ (1, 4) ◦ (1, 2). Since p is a product of an odd number of transpositions, it is an odd permutation. 1 2 3 4 5 6 4. Show that the permutation is odd, while the 5 6 2 4 1 3 1 2 3 4 5 6 permutation is even. 6 3 4 5 2 1 Solution. 1 2 3 5 6 2
4 4
5 1
6 = (1 5)(2 6 3) = (1 5)(2 6)(2 3). 3
The given permutation can be expressed as the product of an odd number of transpositions, and hence the permutation is odd. Again 1 2 3 4 5 6 = (1 6)(2 3 4 5) = (1 6)(2 3)(2 4)(2 5). 6 3 4 5 2 1 Since it is a product of even number of transpositions, the permutation is an even permutation. 1 2 3 4 5 6 5. Express the permutation as a product of 6 5 2 4 3 1 transpositions. Solution. 1 2 3 6 5 2
6 = (1 6)(2 5 3) = (1 6)(2 5)(2 3). 1 1 2 3 4 5 6. Find the inverse of the permutation . 2 3 1 5 4 4 4
5 3
Given
1 2
Solution.
2 3
3 1
4 5
5 . 4
5 . v 1 2 3 4 5 1 2 3 4 5 1 2 Then, = 2 3 1 5 4 x y z u v 1 2 1 2 3 4 5 1 2 3 4 5 =⇒ = y z x v u 1 2 3 4 5 Let the inverse permutation be
1 x
2 y
3 z
y = 1, z = 2, x = 3, v = 4, u = 5. 1 2 Hence, the inverse permutation is 3 1
4 u
=⇒
3 2
4 5
7. If A = (1 2 3 4 5), B = (2 3)(4 5), find AB.
5 . 4
3 3
4 4
5 5
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Discrete Mathematical Structures Solution. Given A = (1 2 3 4 5), 1 2 3 4 AB = 2 3 4 5 =
1 3
2 2
3 5
4 4
B = (2 3)(4 5). 5 1 2 3 4 5 1 1 3 2 5 4 5 1
= (1 3 5). 8. If A = {1, 2, 3, 4, 5, 6, 7, 8}, then express the following permutations as a product of disjoint cycles. 1 2 3 4 5 6 7 8 (i) p = 6 5 7 8 4 3 2 1 1 2 3 4 5 6 7 8 (ii) p = . 2 3 1 4 6 7 8 5 Solution. (i) p(1) = 6, p(6) = 3, p(3) = 7, p(7) = 2, p(2) = p(4) = 8, p(8) = 1. Therefore, p = (1 6 3 7 2 5 4 8). (ii) p(1) = 2, p(2) = 3, p(3) = 1 =⇒ (1 2 3) p(5) = 6, p(6) = 7, p(7) = 8, p(8) = 5 =⇒ (5 6 7 Therefore, p = (5 6 7 8)(1 2 3). 1 2 3 4 9. Let A = {1, 2, 3, 4, 5, 6} and p = 2 4 3 1 permutation of A. (i) (ii) (iii) (iv)
5, p(5) = 4,
8). 5 5
6 be a 6
Write p as a product of disjoint cycles. Compute p−1 . Compute p2 . Find the period of p, that is, the smallest positive integer k such that pk = 1A .
Solution. (i) Given p =
1 2
2 4
3 3
4 1
5 5
6 . 6
Since p(1) = 2, p(2) = 4, and p(4) = 1, we write p = (1, 2, 4) as the other elements are fixed. 2 4 3 1 5 6 1 2 3 4 5 6 −1 (ii) p = =p= . 1 2 3 4 5 6 4 1 3 2 5 6 1 2 3 4 5 6 (iii) p2 = p ◦ p = . 4 1 3 2 5 6
Algebraic Structures
215 1 2 3 4 5 6 (iv) p3 = p2 ◦ p = = 1A . 1 2 3 4 5 6 p4 = p, p5 = p2 , etc. Therefore, the period of p = 3. 1 2 3 4 1 2 3 4 10. If f = and g = are permutations, 3 2 1 4 2 3 4 1 prove that (g ◦ f )−1 = f −1 ◦ g −1 . Solution.
f
−1
f
−1
◦g
−1
3 1
2 2
1 3
1 4
2 3
3 2
1 2
2 3
3 4
1 4
2 3
3 2
= =
g◦f = (g ◦ f )
−1
=
4 1 2 3 4 −1 and g = . 4 4 1 2 3 4 1 4 1 2 3 4 1 2 3 ◦ = 1 3 2 1 4 4 3 2 4 . 1
Hence, (g ◦ f )−1 = f −1 ◦ g −1 . 1 2 3 4 5 6 7 1 11. Let p1 = and p2 = 7 3 2 1 4 5 6 6
2 3
4 1
3 2
4 1
5 5
6 4
1 2 3 4 5 6 7 1 2 3 4 (i) p1 ◦ p2 = ◦ 7 3 2 1 4 5 6 6 3 2 1 1 2 3 4 5 6 7 = . 5 2 3 7 4 1 6 1 2 3 4 5 6 7 (ii) p−1 1 = 4 3 2 5 6 7 1 . (iii) p1 = (1 7 6 5 4) ◦ (2 3) = (1 4) ◦ (1 5) ◦ (1 6) ◦ (1 7) ◦ (2 3) = product of odd number of transpositions.
5 5
6 4
7 7
7 . 7
(i) Compute p1 ◦ p2 . (ii) Compute p−1 1 . (iii) Is p1 an even or odd permutation? Explain. Solution.
Therefore, p1 is an odd permutation. 12. If x = (1 2 3), y = (2 4 3), and z = (1 3 4), then show that xyz = 1. Solution. Given x = (1 2 3) =
1 2
2 3
3 1
4 4
216
1 1 1 z = (1 3 4) = 3 1 2 3 Therefore, xyz = 2 3 1 1 2 3 = 4 2 1 1 2 3 = 1 2 3 y = (2 4 3) =
4.2.12
Discrete Mathematical Structures 2 3 4 4 2 3 2 3 4 . 2 4 1 4 1 2 3 4 1 2 3 4 ◦ ◦ 4 1 4 2 3 3 2 4 1 4 1 2 3 4 ◦ 3 3 2 1 4 4 = 1. 4
Problems for Practice
1. Which of the following functions f : Z −→ Z are permutations of Z? (i) f is defined by f (a) = a + 1. (ii) f is defined by f (a) = (a − 1)2 . 2. Which of the following functions f : R −→ R are permutations of R? (i) f is defined by f (a) = a3 . (ii) f is defined by f (a) = ea . 3. Which of the following functions f : R −→ R are permutations of R? (i) f is defined by f (a) = a − 1. (ii) f is defined by f (a) = a2 . 4. Which of the following functions f : Z −→ Z are permutations of Z? (i) f is defined by f (a) = a2 + 1. (ii) f is defined by f (a) = a3 − 3. 5. Let A = {a, b, c, d, e, f, g}. Write each of the following permutations as a product of disjoint cycles. a b c d e f g (i) g d b a c f e a b c d e f g (ii) d e a b g f c 6. Let A = {1, 2, 3, 4, 5, 6, 7, 8}. Write each of the following permutations as a product of transpositions. (i) (2 1 4 5 8 6) (ii) (3 1 6) ◦ (4 8 2 5) 7. Code the message “WHERE ARE YOU” by applying the permutation (1 7 3 5 11) ◦ (2 6 9) ◦ (4 8 10).
Algebraic Structures
217
8. Decode the message “ATEHAOMOMNTI”, which was encoded using the permutation (3 7 1 12) ◦ (2 5 8) ◦ (4 10 6 11 9). 9. Let A = {1, 2, 3, 4, 5, 6, 7, 8}. Determine whether the following permutations are even or odd. 1 2 3 4 5 6 7 8 (i) 4 2 1 6 5 8 7 3 1 2 3 4 5 6 7 8 (ii) 7 3 4 2 1 8 6 5 (iii) (6 4 2 1 5) (iv) (4 8) ◦ (3 5 2 1) ◦ (2 4 7 1) 10. Prove that the product of two even permutations is even. 11. Prove that the product of two odd permutations is even. 12. Prove that the product of an even and odd permutation is odd. 13. Let A = {1, 2, 3, 4, 5}. Let f = (5 2 3) and g = (3 4 1) be permutations of A. Compute each of the following, and write the result as the product of disjoint cycles: (i) f ◦ g (ii) f −1 ◦ g −1 .
4.2.13
Rings and Fields
Definition 4.2.74 Ring: An algebraic system (S, +·) is called a ring if the binary operations + and · on S satisfy the following three properties: 1. (S, +) is an abelian group. 2. (S, ·) is a semigroup. 3. The operation · is distributive over +; that is, for any a, b, c ∈ S, a · (b + c) = a · b + a · c
and
(b + c) · a = b · a + c · a.
Examples: 1. The set of all integers Z, the set of all rational numbers Q, the set of all real numbers R are rings under the usual addition and usual multiplication. 2. The set of all n × n matrices Mn is a ring under the matrix addition and matrix multiplication. 3. If n is a positive integer, then Zn = {[0], [1], . . . , [n − 1]} is a ring under +n , the addition modulo n, and ×n , the multiplication modulo n.
218
Discrete Mathematical Structures 4. Let (R, +, ·) be a ring and X be a non-empty set. Let A be the set of all functions from X to R. That is, A = {f |f : X −→ R is a function}. We define ⊕ and · on A as follows: (i) if f, g ∈ A, then f ⊕ g : X −→ R is given by (f ⊕ g)(x) = f (x) + g(x),
for all x ∈ X.
(ii) if f, g ∈ A, then f, g : X −→ R is given by (f · g)(x) = f (x) · g(x),
for all x ∈ X.
Definition 4.2.75 Integral Domain: A commutative ring (S, +, •) with identity and without divisors of zero is called an integral domain. Definition 4.2.76 Field: A commutative ring (S, +, •) which has more than one element such that every non-zero element of S has a multiplicative inverse in S is called a field. Definition 4.2.77 Subring: A subset R of a ring (S, +, •) is called a subring if (R, +, •) itself is a ring with the operations + and • restricted to R. Examples: 1. The set of integers Z is a subring of the ring of all rational numbers Q. 2. The set of all even integers is a subring of the ring of all integers Z. Definition 4.2.78 Ring Homomorphism: Let (R, +, •) and (S, ⊕, ) be rings. A mapping g : R −→ S is called a ring homomorphism from (R, +, •) to (S, ⊕, ) if for any a, b ∈ R, g(a + b) = g(a) ⊕ g(b) and g(a · b) = g(a) g(b). Examples: 1. The ring Mn of all non-zero matrices is not commutative and 0 non-zero divisors. For example, let n = 2; then if A = 0 1 0 0 0 0 1 and B = , then AB = and BA = . 0 0 0 0 0 0 AB 6= BA, and A is a non-zero divisor.
has 1 0 So,
2. The ring Q of rational numbers and the ring R of real numbers are fields. 3. The ring (Z7 , +7 , ×7 ) is a field.
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219
4. The ring (Z10 , +10 , ×10 ) is not an integral domain since 5 ×10 2 = 0, even though 5 6= 0, 2 6= 0 in Z10 . 5. The ring Z of all integers is an integral domain but not a field. Definition 4.2.79 Commutative Ring: A ring (R, +, ·) is said to be commutative if a · b = b · a, for all a, b ∈ R. Theorem 4.2.80 Every finite integral domain is a field. Proof. Let (R, +, •) be a finite integral domain. To prove: (R − {0}, •) is a group, that is, to prove (i) there exists an element 1 ∈ R such that 1 · a = a · 1 = a, for all a ∈ R
(since 1 ∈ R is an identity)
(ii) for every element of 0 6= a ∈ R, there exists an element a−1 ∈ R such that a · a−1 = a−1 · a = a. Let R − {0} = {a1 , a2 , a3 , . . . , an }. Let a ∈ R − {0}. Then, the elements aa1 , aa2 , . . . , aan are all in R − {0}, and they are all distinct. That is, if a · ai = a · aj , i 6= j, then a · (ai − aj ) = 0. Since R is an integral domain and a 6= 0, we must have ai − aj = 0 =⇒ ai = aj , which is a contradiction. Therefore, R − {0} has exactly n elements, and R is a commutative ring with cancellation law. Hence, we get a = a · ai0 , for some i0 (since a ∈ R − {0}). That is, a · ai0 = ai0 · a (since R is commutative). Thus, let x = a · ai for some ai ∈ R − {0}, and y · ai0 = a · ai0 = (ai · a)ai0 = ai · a = a · aj = y. Therefore, ai0 is unity in R − {0}. We write it as 1. Since 1 ∈ R − {0}, there exists an element aak ∈ R − {0} such that aak = 1. Therefore, ba = b = 1 (let ak = b). Hence, b is the inverse of a, and the converse is also true. Hence, (R, +, •) is a field. Theorem 4.2.81 Every field is an integral domain. Proof. Let (F, +, ·) be a field. That is, F is a commutative ring with unity. To prove F is an integral domain, it is enough to show that it has no zero divisor. Let a, b ∈ F such that a · b = 0.
220
Discrete Mathematical Structures
If a 6= 0, then a−1 ∈ F . Therefore, a · b = 0 =⇒ a−1 · (a · b) = a−1 · 0 =⇒ 1 · b = 0 =⇒ b = 0. Hence, the theorem is proved. Note: The converse of the above Theorem 4.2.81 need not be true. Theorem 4.2.82 A commutative ring (R, +, ·) is an integral domain if and only if the cancellation law holds in R. That is, for a 6= 0, a · b = a · c =⇒ b = c, for all a, b, c ∈ R. Proof. Let R be an integral domain and a · b = a · c and a 6= 0, for all a, b, c ∈ R. We have a · b − a · c = 0 =⇒ a · (b − c) = 0. Therefore, since R is an integral domain and a 6= 0, b − c = 0. (R has no zero divisor). Therefore, b = c. Hence, the cancellation law holds. Converse Part: Assume that the cancellation law holds in a ring R. Let a · b = 0, for a 6= 0 and b ∈ R. We have ab = 0 = a0 =⇒ b = 0. Thus, ab = 0 in R =⇒ a = 0 or b = 0. Therefore, R has no zero divisors. Therefore, R is an integral domain.
4.2.14
Solved Problems
1. Prove that the set Z4 = {[0], [1], [2], [3]} is a commutative ring with respect to the binary operations addition modulo 4 (+4 ) and multiplication modulo 4 (×4 ). Solution. Tables 4.1 and 4.2 are composition tables for addition modulo 4 (+4 ) and multiplication modulo 4 (×4 ), respectively. From Tables 4.1 and 4.2, we get the following:
TABLE 4.1 Composition Table for +4 +4 [0] [1] [2] [3] [0] [0] [1] [2] [3] [1] [1] [2] [3] [0] [2] [2] [3] [0] [1] [3] [3] [0] [1] [2]
Algebraic Structures
221
TABLE 4.2 Composition Table for ×4 ×4 [0] [1] [2] [3] [0] [0] [0] [0] [0] [1] [0] [1] [2] [3] [2] [0] [2] [0] [2] [3] [0] [3] [2] [1]
(i) All the entries in both the tables belong to Z4 . Therefore, Z4 is closed under the operations +4 and ×4 . (ii) In both the tables, Entries in the first Entries in the second Entries in the third Entries in the fourth
row = Entries row = Entries row = Entries row = Entries
in in in in
the the the the
first column second column third column fourth column.
Therefore, the operations +4 and ×4 are commutative in Z4 . (iii) Also, for any a, b, c ∈ Z4 , we have a +4 (b +4 c) = (a +4 b) +4 c and a ×4 (b ×4 c) = (a ×4 b) ×4 c since 0 +4 (1 +4 2) = 0 +4 3 = 3 and (0 +4 1) +4 2 = (1 +4 2) = 3. ∴ (0 +4 1) +4 2 = (0 +4 1) +4 2. Also, 1 ×4 (2 ×4 3) = 1 ×4 2 = 2 and (1 ×4 2) ×4 3 = 2 ×4 3 = 2. ∴ 1 ×4 (2 ×4 3) = (1 ×4 2) ×4 3. Thus, the operations +4 and ×4 are associative in Z4 . (iv) [0] is the additive identity of Z4 , and [1] is the multiplicative identity of Z4 . (v) Additive inverses of [0], [1], [2], [3] are, respectively, [0], [3], [2], [1]. Multiplicative inverses of the non-zero elements [1], [2], [3] are, respectively, [1], [2], [3]. (vi) If a, b, c ∈ Z4 , then a ×4 (b +4 c) = (a ×4 b) +4 (a ×4 c) and (a +4 b) ×4 c = (a ×4 c) +4 (b ×4 c). Thus, the operation ×4 is distributive over +4 in Z4 . Hence, (Z4 , +4 , ×4 ) is a commutative ring with unity. 2. Show that (Z, +, ×) is an integral domain where Z is the set of all integers.
222
Discrete Mathematical Structures Solution. We know the following: (Z, +) is an abelian group. (Z, ×) is a monoid. The operation × is distributive over +. (Z, ×) is commutative. (Z, +, ×) is without zero divisors. (Z, +, ×) is an integral domain. 3. Give an example of a ring which is not a field. Solution. The ring Z of all integers is an integral domain but not a field.
4.2.15
Problems for Practice
1. Discuss a ring and a field with suitable examples. 2. If (R, +, ·) is a ring, then prove that a · 0 = 0, for all a ∈ R, and 0 is the identity element in R under addition.
5 Lattices and Boolean Algebra
5.1
Introduction
In this chapter, we focus on partially ordered sets, lattices, Boolean algebra, and their properties. These structures are useful in set theory, algebra, sorting, and searching and in the construction of logical representation for computer science. The concept of the lattices is a special case of a partially ordered set. Boolean algebra is a special lattice.
5.2
Partial Ordering and Posets
Definition 5.2.1 Partial Order Relation: A relation R on a non-empty set P is called a partial order, if R is reflexive, antisymmetric, and transitive. That is, if R satisfies (i) xRx, for all x ∈ P
(reflexive)
(ii) xRy and yRx =⇒ x = y, for all x, y ∈ P (iii) xRy and yRz =⇒ xRz, for all x, y, z ∈ P
(antisymmetric) (transitive).
Example 5.2.2 Let P be the set of all positive integers. Define the relation ‘R’ such that xRy holds if and only if x ≤ y, for all x, y ∈ P . Clearly, “≤” relation is reflexive, antisymmetric, and transitive. Hence, “≤” relation on P is a partial order relation. Remark 5.2.3 Usually, the partially ordered relation is denoted by “≤”. Definition 5.2.4 Partially Ordered Set or Poset: A set P with the partial order relation “≤” is called a partially ordered set or simply a poset. It is denoted by hP, ≤i. Example 5.2.5 Consider P = {collection of all subsets of any set}. Clearly, “⊆” relation (set inclusion) is a partially ordered relation on P . Definition 5.2.6 Totally Ordered Set: Let hP, ≤i be a partially ordered set. If for every a, b ∈ P we have either a ≤ b or b ≤ a, then ≤ is called 223
224
Discrete Mathematical Structures
simple ordering or linear ordering on P , and hP, ≤i is called a totally ordered or simply ordered set or a chain. Example 5.2.7 The poset hZ, ≤i is totally ordered, since a ≤ b or b ≤ a whenever a and b are integers. Definition 5.2.8 Well-ordered Set: A partially ordered set is called wellordered if every non-empty subset of it has a least member.
5.2.1
Representation of a Poset by Hasse Diagram
A partially ordered relation “≤” on a set P can be represented by means of a diagram known as a Hasse diagram. In such a diagram, each element is represented by a small circle or a dot. The circle for an element x in P is drawn below the circle for y in P , if x < y, and a line is drawn between x and y, if y covers x. If x < y but y does not cover x, then x and y are not connected directly by a single line. Example 5.2.9 For example, let P = {1, 2, 3, 4} and “≤” be the relation “less than or equal to”. Then, the Hasse diagram is shown below. 4 3 2 1
Hasse diagram of P Example 5.2.10 Consider the set X = {2, 3, 6, 12, 24, 36} and the relation “≤” is defined as x ≤ y if and only if x divides y. The Hasse diagram of the poset hX, ≤i is shown below. 24
36 12
6
2
3
Hasse diagram of X
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225
Note: 1. Hasse diagram is named after the twentieth-century German mathematician Helmut Hasse. 2. In a digraph, if we apply the following rules, then we get Hasse diagram. (i) Each vertex of a poset P must be related to itself. So, the arrows from vertex to itself are not necessary. (ii) If a vertex b appears above vertex a and if vertex a is connected to vertex b by an edge, then we have aRb; so, direction arrows are not necessary. (iii) If vertex c is above a and if c is connected to a by a sequence of edges, then we have aRc. (iv) The vertices are denoted by points rather than by circles. Example 5.2.11 Let A = {a, b}. Let B = P (A) = {{φ}, {a}, {b}, {a, b}}. Then, ⊆ is a relation on A whose digraph and Hasse diagram are given in Figures 5.1 and 5.2.
{a,b}
{a}
{b}
FIGURE 5.1 Digraph of hB, ⊆i {a,b}
{a}
{b}
FIGURE 5.2 Hasse diagram of hB, ⊆i
226
Discrete Mathematical Structures
5.2.2
Solved Problems
1. Show that the “greater than or equal” relation (≥) is a partial ordering on the set of integers. Solution. (i) Since a ≥ a for every integer a, the relation ≥ is reflexive. (ii) If a ≥ b and b ≥ a, then a = b. Hence, ≥ is antisymmetric. (iii) The relation ≥ is transitive since a ≥ b and b ≥ c imply that a ≥ c. Hence, ≥ is a partial ordering on the set of integers, and hZ, ≥i is a poset. 2. Show that the inclusion relation ⊆ is a partial ordering on the power set of a set S. Solution. (i) Since A ⊆ A, whenever A is a subset of S, the relation ⊆ is reflexive. (ii) Since A ⊆ B and b ⊆ A imply that A = B, the relation ⊆ is antisymmetric. (iii) Since A ⊂ B and B ⊂ C imply that A ⊆ C, the relation ⊆ is transitive. Therefore, the relation ⊆ is a partial ordering on P (S), and hP (S), ⊆i is a poset. 3. Let R be a binary relation on the set of all positive integers such that R = {(a, b)/a = b2 }. Is R reflexive, symmetric, antisymmetric, transitive, an equivalence relation, or a partial order relation? Solution. (i) R = {(a, b)/a, b are positive integers and a = b2 }. For R to be reflexive, we should have aRa, for all positive integers a. But aRa holds only when a = a2 by hypothesis. Now, a = a2 is not true for all positive integers. In fact, only when a = 1, we have a = a2 . Hence, R is not reflexive. (ii) For R to be symmetric, if aRb holds, then we should have bRa. But aRb implies a = b2 . But a = b2 does not imply b = a2 always for positive integers. For instance, 16 = 42 , but 4 6= 162 . Hence, aRb does not imply bRa. Hence, R is not symmetric. (iii) For R to be antisymmetric, for positive integers a, b if aRb and bRa hold, then a = b. aRb implies a = b2 , and bRa implies b = a2 . Hence, if a = b2 and b = a2 , then a = b2 = (a2 )2 = a4 , that is, a4 − a = 0, that is, a(a3 − 1) = 0. Since a is a positive
Lattices and Boolean Algebra
227
integer, a 6= 0 so that a3 − 1 = 0, that is, a3 = 1 which implies a = 1. This means b = a2 = 1. Hence, aRb and bRa imply a = b = 1. Hence, R is antisymmetric. (iv) For R to be transitive, if aRb holds and bRc holds, then aRc should hold. That is, aRb implies a = b2 , and bRc implies b = c2 , so that a = b2 = c4 . Hence, aRc does not hold. For example, 256 = 162 and 16 = 42 but 256 6= 42 . Thus, R is not transitive. (v) R is not an equivalence relation since an equivalence relation is reflexive, symmetric, and transitive. (vi) R is not a partial order relation, since a partial ordering relation is reflexive, antisymmetric, and transitive. 4. Give examples of a relation which is both a partial ordering relation and an equivalence relation on a set. Solution. Equality and similarity of triangles are examples of a relation which is both a partial ordering relation and an equivalence relation. 5. Let S be a set. Determine whether there is a greatest element and a least element in the poset hP (S), ⊆i. Solution. The least element is the empty set since φ ⊆ T for any subset T of S. The set S is the greatest element in this poset. Hence T ⊆ S whenever T is a subset of S. 6. Is there a greatest element and a least element in the poset hZ+ , |i? Solution. The integer 1 is the least element since 1 divides n whenever n is a positive integer. Since there is no integer that is divisible by all positive integers, there is no greatest element. 7. Let A be a given finite set and P (A) its power set. Let ⊆ be the inclusion relation on the elements of P (A). Draw Hasse diagram of hP (A), ⊆i for (i) A = {a} (iii) A = {a, b, c}
(ii) A = {a, b} (iv) A = {a, b, c, d}.
228
Discrete Mathematical Structures Solution. {a,b,c} {a}
{a,b} {a,b} {a}
{a,c}
{b,c}
{b}
{c}
{b} {a}
(i)
(ii)
(iii)
Hasse diagrams of (i), (ii), and (iii) {a,b,c,d}
{a,b,c}
{a,b,d}
{a,c,d}
{a,c}
{b,c,d}
{a,d} {b,c}
{a,b}
{b,d} {a}
{c}
{b}
{c,d}
{d}
(iv)
Hasse diagram of (iv) 8. Which elements of the poset h{2, 4, 5, 10, 12, 20, 25}, |i are maximal, and which of them are minimal? Solution. The Hasse diagram is shown in Figure 5.3. From the Hasse diagram in Figure 5.3, this poset shows that the maximal elements are 12, 20, and 25 and the minimal elements are 12
20
4
10
2
5
25
FIGURE 5.3 Hasse diagram of the given poset
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229
2 and 5. As this example shows, a poset can have more than one maximal element and more than one minimal element. 9. Determine whether the posets represented by each of the Hasse diagrams in the following figure have a greatest element and a least element. b
c
d
d
d
e
d
c
c
c
b
a
a
a
(ii)
(i)
b
b (iii)
a
(iv)
Hasse diagrams of the given posets Solution. (i) The least element of the poset with Hasse diagram (i) is a. This poset has no greatest element. (ii) The poset with Hasse diagram (ii) has neither a least nor a greatest element. (iii) The poset with Hasse diagram (iii) has no least element. Its greatest element is d. (iv) The poset with Hasse diagram (iv) has the least element a and greatest element d. 10. Draw the Hasse diagram of the set of partitions of 5. Solution. 5
4+1
3+2
2+2+1
3+1+1 2+1+1+1
1+1+1+1+1
Hasse diagram of the set of partitions of 5
230
Discrete Mathematical Structures 5=5 5=4+1 5=3+2 5=3+1+1 5=2+2+1 5=2+1+1+1 5 = 1 + 1 + 1 + 1 + 1.
5.2.3
Problems for Practice
1. Let R be the relation on the set of people such that xRy holds if x is older than y. Show that R is not a partial ordering. 2. Show that hN, ≤i is a partially ordered set, where N is the set of all positive integers and ≤ is a relation defined by m ≤ n if and only if n − m is a non-negative integer. 3. Show that there are only five distinct Hasse diagrams for partially ordered sets that contain three elements. 4. Give an example of a set X such that hP (X), ⊆i is a totally ordered set. 5. Let S denote the set of all the partial ordering relations on a set P . Define a partial ordering relation on S, and interpret this relation in terms of the elements of P . 6. Let X = {1, 2, 3, 4, 6, 8, 12, 24} and R be a division relation defined on X. Find the Hasse diagram of the poset hX, Ri. 7. Draw the Hasse diagrams of the following sets under the partial ordering relation “divides”, and indicate those which are totally ordered: {2, 6, 24}, {3, 5, 15}, {1, 2, 3, 6, 12}, {2, 4, 8, 16}, {3, 9, 27, 54}. 8. If R is a partial ordering relation on a set X and A ⊆ X, show that R ∩ (A × A) is a partial ordering relation on A. 9. Let D30 = {1, 2, 3, 5, 6, 10, 15, 30}, and let the relation R be divisor on D30 . Find (i) (ii) (iii) (iv)
all the lower bounds of 10 and 15 the greatest lower bound of 10 and 15 all upper bounds of 10 and 15 the least upper bound of 10 and 15.
Also, draw the Hasse diagram. 10. Draw the Hasse diagram of hX, ≤i, where X = {2, 4, 5, 10, 12, 20, 25} and the relation ≤ be such that x ≤ y if x divides y.
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231
Definition 5.2.12 Linearly ordered set or Chain: A poset hP, ≤i is called a linearly ordered set or a chain if every pair of elements in a poset hP, ≤i are comparable. Example 5.2.13 Let Z+ be the set of all positive integers. The usual relation “≤” is a partial order relation on Z+ , since any two integers in Z+ can be comparable with respect to the relation “≤”. Thus, hZ+ , ≤i is a linearly ordered set. Definition 5.2.14 Upper bound and Lower bound: Let S be any subset of a poset hP, ≤i. An element x ∈ P is called an upper bound of S if y ≤ x, for all y ∈ S. An element z ∈ P is called a lower bound of S if z ≤ y, for all y ∈ S. Example 5.2.15 Let A = {a, b, c} be a given set and ρ(A) be its power set. Let “⊆” be the relation on ρ(A). Then clearly, hρ(A), ⊆i is a poset. For the subset S = {{a, b}, {a}, {b}, {c}} ⊆ ρ(A), the upper bounds are {a, b} and {a, b, c}, and its lower bound is φ.
5.3
Lattices, Sublattices, Direct Product, Homomorphism of Lattices
Definition 5.3.1 Lattice: A lattice is a poset hL, ≤i in which any subset {a, b} consisting of two elements has a least upper bound and a greatest lower bound. We denote LU B({a, b}) by a ⊕ b and call it as the join of a and b. Similarly, we denote GLB({a, b}) by a ? b and call it as the meet of a and b. Example 5.3.2 Let S be a set, and let L = ρ(S). Let “⊆” (set inclusion) be the relation on L. Clearly, hL, ⊆i is a lattice in which the meet and join are the same as the operations ∩ and ∪ on sets, respectively. That is, for any two elements A, B ∈ ρ(S), GLB({A, B}) = A ∩ B and LU B({A, B}) = A ∪ B. Definition 5.3.3 Dual Lattices: Let hL, ≤i be a poset, and let hL, ≥i be the dual poset (the symbol ‘≥’ used for the partial order is ≤0 ). If hL, ≤i is a lattice, we can show that hL, ≥i is also a lattice. In fact, for any a, b ∈ L, the LU B(a, b) in hL, ≤i is equal to GLB(a, b) in hL, ≥i. Similarly, the GLB(a, b) in hL, ≤i is equal to LU B(a, b) in hL, ≥i. Thus, the dual of hL, ≤i is hL, ≥i and vice-versa.
5.3.1
Properties of Lattices
In the following theorems, let hL, ≤i be a lattice.
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Discrete Mathematical Structures
Theorem 5.3.4 [Idempotent law] For any a, b, c ∈ L, we have a ? a = a and a ⊕ a = a. Proof. Let a, b, c ∈ L. Then by the definition of GLB of a and b, we have a?b≤a
(5.1)
a ≤ a ? b.
(5.2)
and if a ≤ a and a ≤ b, then Since a ≤ a, from (5.1) and (5.2), we have a ? a ≤ a and a ≤ a ? a. By the antisymmetric property, it follows that a = a ? a. Similarly, we can prove that a ⊕ a = a. Theorem 5.3.5 [Associative law] The operations of meet and join on hL, ≤i are associative. That is, for any a, b, c ∈ L, we have the following: (i) a ? (b ? c) = (a ? b) ? c (ii) a ⊕ (b ⊕ c) = (a ⊕ b) ⊕ c. Proof. To prove: a ? (b ? c) = (a ? b) ? c. Let a, b, c ∈ L, Then by the definition, we have (a ? b) ? c ≤ a ? b and (a ? b) ? c ≤ c. By the definition of GLB of a and b, we have a ? b ≤ a and a ? b ≤ b. Hence, by the transitive property of ≤, we have (a ? b) ? c ≤ a and (a ? b) ? c ≤ b. Since (a ? b) ? c ≤ b and (a ? b) ? c ≤ c, we see that (a ? b) ? c is a lower bound for b and c. From the definition of b ? c, it follows that (a ? b) ? c ≤ b ? c. Since (a ? b) ? c ≤ a and (a ? b) ? c ≤ b ? c, from the definition of a ? (b ? c), we have (a ? b) ? c ≤ a ? (b ? c). (5.3) Now, a ? (b ? c) ≤ a and a ? (b ? c) ≤ b ? c. Since b ? c ≤ b, by transitivity, we have a ? (b ? c) ≤ b. Since a ? (b ? c) ≤ a and a ? (b ? c) ≤ b, we have a ? (b ? c) ≤ a ? b. Since a ? (b ? c) ≤ b ? c ≤, we have a ? (b ? c) ≤ (a ? b) ? c. From (5.3), (5.4) and by antisymmetric property, it follows that a ? (b ? c) = (a ? b) ? c. Similarly, we can prove that a ⊕ (b ⊕ c) = (a ⊕ b) ⊕ c.
(5.4)
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Theorem 5.3.6 [Commutative law] The operations of meet and join on hL, ≤i satisfy commutative property. That is, for any a, b ∈ L, we have the following: (i) a ? b = b ? a (ii) a ⊕ b = b ⊕ a. Proof. Given: a, b ∈ L. Both a ? b and b ? a are GLB of a and b. By the uniqueness of GLB of a and b, we have a ? b = b ? a. Similarly, a ⊕ b = b ⊕ a holds good. Theorem 5.3.7 [Absorption law] For any a, b ∈ L, we have the following: (i) a ? (a ⊕ b) = a (ii) a ⊕ (a ? b) = a. Proof. Let a, b ∈ L. Then, a ≤ a and a ≤ a ⊕ b. So, a ≤ a ? (a ⊕ b). On the other hand, a ? (a ⊕ b) ≤ a. By antisymmetric property of ≤, we have a = a ? (a ⊕ b). Similarly, we have a ⊕ (a ? b) = a, for all a, b ∈ L.
5.3.2
Theorems on Lattices
Theorem 5.3.8 Let hL, ≤i be a lattice in which ? and ⊕ denote the operations of meet and join respectively. For any a, b ∈ L, a ≤ b ⇐⇒ a ? b = a ⇐⇒ a ⊕ b = b. Proof. First, let us prove that a ≤ b ⇐⇒ a ? b = a ⇐⇒ a ⊕ b = b. Let us assume that a ≤ b, and also, we know that a ≤ a. ∴
a ≤ a ? b.
(5.5)
But, from the definition of a ? b, we have a ? b ≤ a. Hence, a ≤ b =⇒ a ? b = a [using (5.5) and (5.6)]. Next, assume that a ? b = a. But it is only possible if a ≤ b. That is, a ? b = a =⇒ a ≤ b. Combining these two results, we get a ≤ b ⇐⇒ a ? b = a. Similarly, we can prove that a ≤ b ⇐⇒ a ⊕ b = b. From a ? b = a, we have b ⊕ (a ? b) = b ⊕ a = a ⊕ b. But b ⊕ (a ? b) = b. Hence, a ⊕ b = b follows that a ? b = a.
(5.6)
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Discrete Mathematical Structures
Theorem 5.3.9 Let hL, ≤i be a lattice. For any a, b ∈ L, the following are equivalent: (i) a ≤ b (ii) a ? b = a (iii) a ⊕ b = b. Proof. First, consider (i) ⇐⇒ (ii). We have a ≤ a. Assume a ≤ b. Therefore, a ≤ a ? b. By the definition of GLB, we have a ? b ≤ a. Hence, by antisymmetric property, a ? b = a. Assume that a ? b = a, but it is only possible if a ≤ b =⇒ a ? b = a =⇒ a ≤ b. Combining these two results, we have a ≤ b ⇐⇒ a ? b = a. Similarly, a ≤ b ⇐⇒ a ⊕ b = b. Now, consider (ii) ⇐⇒ (iii). Assume a ? b = a, we have b ⊕ (a ? b) = b ⊕ a = a ⊕ b, but by absorption, b ⊕ (a ? b) = b. Hence, a ⊕ b = b. By similar arguments, we can show that a ? b = a follows from a ⊕ b = b. (ii) ⇐⇒ (iii) Hence, the theorem is proved. Theorem 5.3.10 Let hL, ≤i be a lattice. For any a, b ∈ L, the following inequalities hold: (1) Distributive Inequalities (i) a ⊕ (b ? c) ≤ (a ⊕ b) ? (a ⊕ c) (ii) a ? (b ⊕ c) ≥ (a ? b) ⊕ (a ? c). (2) Modular Inequalities (i) a ≤ c ⇐⇒ a ⊕ (b ? c) ≤ (a ⊕ b) ? c (ii) a ≥ c ⇐⇒ a ? (b ⊕ c) ≥ (a ? b) ⊕ c. Proof. Since (ii) in (1) and (ii) in (2) are duals of (i) in (1) and (i) in (2) respectively, it is enough to prove (i) in (1) and (i) in (2) only. Consider (i) in (1). Let a, b, c ∈ L. Since a ≤ a ⊕ b and a ≤ a ⊕ c, we have a ≤ [(a ⊕ b) ? (a ⊕ c)]. Since b ? c ≤ b ≤ a ⊕ b and b ? c ≤ c ≤ a ⊕ c, we have (b ? c) ≤ (a ⊕ b) ? (a ⊕ c). Therefore, (a ⊕ b) ? (a ⊕ c) is an upper bound for a and b ? c, and hence a ⊕ (b ? c) ≤ (a ⊕ b) ? (a ⊕ c). Thus, (i) in (1) is proved.
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235
The inequality (i) in (2) is a special case of (i) in (1). If a ≤ c, then a ⊕ c = c, and from (i) in (1), we obtain a ⊕ (b ? c) ≤ (a ⊕ b) ? (a ⊕ c) = (a ⊕ b) ? c, which is inequality (i) in (2). Hence, the theorem is proved. Theorem 5.3.11 In a lattice hL, ≤i, for all a, b, c ∈ L, we have the following: (i) (a ? b) ⊕ (c ? d) ≤ (a ⊕ c) ? (b ⊕ d) (ii) (a ? b) ⊕ (b ? c) ⊕ (c ? a) ≤ (a ⊕ b) ? (b ⊕ c) ? (c ⊕ a). Proof. Let a, b, c ∈ L. Then a ? b ≤ a (or) b ≤ a ⊕ b.
(5.7)
a ? b ≤ a ≤ c ⊕ a.
(5.8)
a ? b ≤ b ≤ b ⊕ c.
(5.9)
Using (5.7), (5.8), and (5.9), we get a ? b ≤ (a ⊕ b) ? (b ⊕ c) ? (c ⊕ a). Similarly, b ? c ≤ (a ⊕ b) ? (b ⊕ c) ? (c ⊕ a), c ? a ≤ (a ⊕ b) ? (b ⊕ c) ? (c ⊕ a). This proves (ii). We have a ≤ a ⊕ c and b ≤ b ⊕ d. We know that c ≤ a ⊕ c. d ≤ b ⊕ d.
(5.10) (5.11)
Therefore, c ? d ≤ (a ⊕ c) ? (b ⊕ d). By (5.10) and (5.11), we have (a ? b) ⊕ (c ? d) ≤ (a ⊕ c) ? (b ⊕ d). This proves (i). Theorem 5.3.12 In a lattice hL, ≤i, prove that for a, b, c ∈ L, (i) (a ? b) ⊕ (a ? c) ≤ a ? (b ⊕ (a ? c)) (ii) (a ⊕ b) ? (a ⊕ c) ≥ a ⊕ (b ? (a ⊕ c)). Proof. We know that a ? b ≤ a, a ? c ≤ a. Therefore, Also, =⇒
(a ? b) ⊕ (a ? c) ≤ a ⊕ a = a.
(5.12)
a ? b ≤ b, a ? c ≤ a ? c (a ? b) ⊕ (a ? c) ≤ b ⊕ (a ? c).
From (5.12) and (5.13), we have (a ? b) ⊕ (a ? c) ≤ a ? (b ⊕ (a ? c)).
(5.13)
236
Discrete Mathematical Structures This proves (i). a ≤ a ⊕ b; a ≤ a ⊕ c
We know that
a = a ? a ≤ (a ⊕ b) ? (a ⊕ c).
=⇒
(5.14)
b ≤ a ⊕ b; a ⊕ c ≤ a ⊕ c
Further,
b ? (a ⊕ c) ≤ (a ⊕ b) ? (a ⊕ c).
=⇒
(5.15)
Using (5.14) and (5.15), we have a ⊕ (b ? (a ⊕ c)) ≤ (a ⊕ b) ? (a ⊕ c). This proves (ii). Theorem 5.3.13 In a lattice if a ≤ b ≤ c, show that (i) a ⊕ b = b ? c (ii) (a ? b) ⊕ (b ? c) = (a ⊕ b) ? (a ⊕ c) = b. Proof. Let a ≤ b ≤ c. a ≤ b =⇒ a ⊕ b = b, a ? b = a. b ≤ c =⇒ b ⊕ d = c, b ? c = b. a ≤ c =⇒ a ⊕ c = c, a ? c = a. Therefore, a ⊕ b = b = b ? c, which is (i). Now, (a ? b) ⊕ (b ? c) = a ⊕ b = b (a ⊕ b) ? (a ⊕ c) = b ? c = b, which is (ii).
5.3.3
Solved Problems
1. Determine whether the posets represented by each of the Hasse diagrams are lattices. Solution. f e
h e
d
c
e f
g
d c b
b
c
b
a
a
(ii)
(iii)
a (i)
Hasse diagrams of the given posets
d
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237
The posets represented by the Hasse diagrams in (i) and (iii) are both lattices because in each poset, every pair of elements has both a least upper bound and a greatest lower bound. On the other hand, the poset with the Hasse diagram shown in (ii) is not a lattice, since the elements b and c have no least upper bound. It is to be noted that each of the elements d, e, and f is an upper bound, but none of these three elements precede the other two with respect to the ordering of this poset. 2. Is the poset hZ+ , |i a lattice? Solution. Let a and b be two positive integers. The least upper bound and greatest lower bound of these two integers are the least common multiple and the greatest common divisor of these integers, respectively. Hence, it follows that this poset is a lattice. 3. Explain why the partially ordered sets of Figures 5.4 and 5.5 are not lattices.
(i)
(ii)
(iii)
FIGURE 5.4 Hasse diagrams of the given posets
f
e
e d
d b
b (i)
e
b
c
c
c a
d
a (ii)
FIGURE 5.5 Hasse diagrams of the given posets
a (iii)
238
Discrete Mathematical Structures Solution. Given: (i) does not represent a lattice since e ⊕ f does not exist. (ii) does not represent a lattice since b ⊕ c does not exist. (iii) does not represent a lattice because neither d ⊕ c nor b ? c exists. 4. Let the sets S0 , S1 , S2 , . . . , S7 be given by S0 S2 S4 S6
= {a, b, c, d, e, f }, = {a, b, c, d, f }, = {a, b, c}, = {a, c},
S1 S3 S5 S7
= {a, b, c, d, e}, = {a, b, c, e}, = {a, b}, = {a}.
Draw the diagram of hL, ⊆i where L = {S0 , S1 , S2 , . . . , S7 }. Solution. The diagram is shown below. S0
S1
S2 S3
S4 S5
S6
S7
Hasse diagram of hL, ⊆i 5. Show that every non-empty subset of a lattice has a least upper bound and a greatest lower bound. 6. Show that every totally ordered set is a lattice. 7. Let A = {1, 2, 5, 10} with the relation “divides”. Draw the Hasse diagram. Definition 5.3.14 Sublattice: Let hL, ?, ⊕i be a lattice, and let S ⊆ L be a subset of L. The algebra hS, ?, ⊕i is a sublattice of hL, ?, ⊕i if and only if S is closed under both operations ? and ⊕. Example 5.3.15 Let hL, ≤i be a lattice in which L = {a1 , a2 , . . . , a8 } and S1 , S2 , and S3 be the sublattices of L given by S1 = {a1 , a2 , a4 , a6 }, S2 = {a3 , a5 , a7 , a8 }, and S3 = {a1 , a2 , a4 , a8 }. The diagram of hL, ≤i is below.
Lattices and Boolean Algebra
239 a1
a2
a3
a4
a6 a7
a5
a8
Hasse diagram of hL, ⊆i Note that hS1 , ≤i and hS2 , ≤i are sublattices of hL, ≤i, but hS3 , ≤i is not a sublattice since a2 , a4 ∈ S3 but a2 ? a4 = a6 ∈ S3 . Also, note that hS3 , ≤i is a lattice. Definition 5.3.16 Direct Product of Lattices: Let hL, ?, ⊕i and hS, ∧, ∨i be two lattices. The algebraic system hL×S, ·, +i in which the binary operations “+” and “·” on L × S are such that for any (a1 , b1 ) and (a2 , b2 ) in L × S (a1 , b1 ) · (a2 , b2 ) = (a1 ? a2 , b1 ∧ b2 ) (a1 , b1 ) + (a2 , b2 ) = (a1 ⊕ a2 , b1 ∨ b2 ) is called the direct product of the lattices hL, ?, ⊕i and hS, ∧, ∨i. Definition 5.3.17 Lattice Homomorphism: Let hL, ?, ⊕i and hS, ∧, ∨i be two lattices. A mapping g : L −→ S is called a lattice homomorphism from the lattice hL, ?, ⊕i to hS, ∧, ∨i if for any a, b ∈ L, g(a ? b) = g(a) ∧ g(b) and g(a ⊕ b) = g(a) ∨ g(b). Remark 5.3.18 Observe that both the operations of meet and join are preserved. These may be mappings which preserve only one of the two operations. Such mappings are not lattice homomorphisms. Definition 5.3.19 Lattice Isomorphism: If a homomorphism g : L −→ S of two lattices hL, ?, ⊕i and hS, ∧, ∨i is bijective, that is, one-to-one and onto, then g is called an isomorphism. If there exists an isomorphism between two lattices, then the lattices are called isomorphic. Definition 5.3.20 Lattice Endomorphism: Let hL, ?, ⊕i be a lattice. A homomorphism g : L −→ L is called an endomorphism. Definition 5.3.21 Lattice Automorphism: Let hL, ?, ⊕i be a lattice. If g : L −→ L is an isomorphism, then g is called an automorphism. Remark 5.3.22 Let hL, ?, ⊕i be a lattice. If g : L −→ L is an endomorphism, then the image set of g is a sublattice of L.
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Definition 5.3.23 Order-Preserving Mapping: Let hP, ≤i and hQ, ≤0 i be two partially ordered sets. A mapping f : P −→ Q is said to be order-preserving relative to the ordering ≤ in P and ≤0 in Q if and only if for any a, b ∈ P such that a ≤ b, f (a) ≤0 f (b) in Q. Remark 5.3.24 If hP, ≤i and hQ, ≤0 i are lattices and g : P −→ Q is a lattice homomorphism, then g is order-preserving. Definition 5.3.25 Order-isomorphic Partially Ordered Sets: Two partially ordered sets hP, ≤i and hQ, ≤0 i are called order-isomorphic if there exists a mapping f : P −→ Q which is bijective and if both f and f −1 are order-preserving.
5.3.4
Problem for Practice
1. Let hL, ?, ⊕i and hS, ∧, ∨i be any two lattices with the partial orderings ≤ and ≤0 respectively. If g is a lattice homomorphism, then g preserves the partial ordering.
5.4
Special Lattices
Let hL, ?, ⊕i be a lattice and S ⊆ L be a finite subset of L where S = {a1 , a2 , . . . , an }. The greatest lower bound and the least upper bound of S can be expressed as GLB S = ?ni=1 ai where
?2i=1 ai = a1 ? a2
and LU B S = ⊕ni=1 ai
and
?ki=1 ai = ?k−1 i=1 (ai ? ak ), k = 3, 4, . . .
A similar representation can be given for ⊕ni=1 . In lieu of the associative property of the operations ? and ⊕, we can write ?ni=1 ai = a1 ? a2 ? · · · ? an and
⊕ni=1 ai = a1 ⊕ a2 ⊕ · · · ⊕ an .
Definition 5.4.1 Complete Lattice: A lattice is called complete if each of its non-empty subsets has a least upper bound and a greatest lower bound. Definition 5.4.2 Complement Element: In a bounded lattice hL, ?, ⊕, 0, 1i, an element b ∈ L is called a complement of an element a ∈ L if a?b=0
and
a ⊕ b = 1.
Definition 5.4.3 Complemented Lattice: A lattice hL, ?, ⊕, 0, 1i is said to be a complemented lattice if every element of L has at least one complement.
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241
Definition 5.4.4 Distributive Lattice: A lattice hL, ?, ⊕i is called a distributive lattice if for any a, b, c ∈ L, a ? (b ⊕ c) = (a ? b) ⊕ (a ? c) a ⊕ (b ? c) = (a ⊕ b) ? (a ⊕ c). In other words, in a distributive lattice, the operations ? and ⊕ are distributed over each other. Definition 5.4.5 Modular Lattice: A lattice hL, ∧, ∨i is called modular if for all x, y, z ∈ L, x ≤ z =⇒ x ∨ (y ∧ z) = (x ∨ y) ∧ z
(modular equations).
Remark 5.4.6 We have (by modular inequality) if x ≤ z =⇒ x ∨ (y ∧ z) = (x ∨ y) ∧ z holds in any lattice. Therefore, to show that a lattice L is modular, it is enough to show if x ≤ z =⇒ x ∨ (y ∧ z) ≥ (x ∨ y) ∧ z
holds in L.
Theorem 5.4.7 Every chain is a distributive lattice. Proof. Let hL, ≤i be a chain. Let a, b, c ∈ L. Consider the following possible cases: (i) a ≤ b or a ≤ c (ii) a ≥ b and a ≥ c. We shall now show the distributive law a ? (b ⊕ c) = (a ? b) ⊕ (a ? c). In case (i), if a ≤ b or a ≤ c, then we have a ? b = a, a ⊕ a = a, a ? c = c and =⇒
a ≤ b ⊕ c.
(5.16)
a ? (b ⊕ c) = a
Hence,
(a ? b) ⊕ (a ? c) = a ⊕ a = a.
and
(5.17)
From (5.16) and (5.17), we get a ? (b ⊕ c) = (a ? b) ⊕ (a ? c). In case (ii), if a ≥ b and a ≥ c, then we have a ? b = b, a ? c = c and b ⊕ c ≤ a, so that a ? (b ⊕ c) = b ⊕ c (5.18) and
(a ? b) ⊕ (a ? c) = b ⊕ c.
From (5.18) and (5.19), we get a ? (b ⊕ c) = (a ? b) ⊕ (a ? c).
(5.19)
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Discrete Mathematical Structures
Theorem 5.4.8 Let hL, ?, ⊕i be a distributive lattice. For any a, b, c ∈ L, (a ? b = a ? c) ∧ (a ⊕ b = a ⊕ c) =⇒ b = c. Proof. (a ? b) ⊕ c = (a ? c) ⊕ c = c.
(5.20)
(a ? b) ⊕ c = (a ⊕ c) ? (b ⊕ c) = (a ⊕ b) ? (b ⊕ c) = b ⊕ (a ? c) = b ⊕ (a ? b) = b.
(5.21)
From (5.20) and (5.21), we have b = c. Theorem 5.4.9 Every distributive lattice is modular. Proof. Let hL, ≤i be a distributive lattice. For all a, b, c ∈ L, we have a ⊕ (b ? c) = (a ⊕ b) ? (a ⊕ c). Thus, if a ≤ c, then a ⊕ c = c and a ⊕ (b ? c) = (a ⊕ b) ? c. Hence, if a ≤ c, the modular equation is satisfied, and L is modular.
5.4.1
Solved Problems
1. Show that a chain of three or more elements is not complemented. Solution. In a chain, we have that any two elements are comparable. Let 0, x, 1 be any three elements in a chain hL, ≤i with least element 0 and greatest element 1. We have 0 ≤ x ≤ 1. Now, 0 ∧ x = 0 and 0 ∨ x = x. Similarly, x ∧ 1 = x and x ∨ 1 = 1. Therefore, x does not have any complement. Hence, any chain with three or more elements is not complemented. 2. Find all sublattices of hD30 , |i where | is the divisor relation. Solution. The Hasse diagram of hD30 , |i is shown in Figure 5.6.
Lattices and Boolean Algebra
243 30
6
10 5
2
15 3
1
FIGURE 5.6 Hasse diagram of hD30 , |i Therefore, the sublattices are D6 D10 D15 S1 S2
= {1, 2, 3, 6} = {1, 2, 5, 10} = {1, 3, 5, 15} = {5, 10, 15, 30} = {3, 5, 15, 30}, etc. are lattices.
In general, if m|n, then Dm is a sublattice of Dn , and Dkm is also a sublattice of Dn . 3. Show that the lattices given by the diagrams are not distributive. 1
1 a2 b1
a1
b2
b3
a3 0 (i)
0 (ii)
Hasse diagrams of given lattices Solution. In lattice (i), a3 ? (a1 ⊕ a2 ) = a3 ? 1 = a3 = (a3 ? a1 ) ⊕ (a3 ? a2 ) a1 ? (a2 ⊕ a3 ) = 0 = (a1 ? a2 ) ⊕ (a1 ? a3 ) but a2 ? (a1 ⊕ a3 ) = a2 ? 1 = a2 (a2 ? a1 ) ⊕ (a2 ? a3 ) = 0 ⊕ a3 = a3 . Hence, the lattice (i) is not distributive. In lattice (ii),
244
Discrete Mathematical Structures b1 ? (b2 ⊕ b3 ) = b1 while (b1 ? b2 ) ⊕ (b1 ? b3 ) = 0 which shows that the lattice is not distributive. 4. If Dn denotes the lattice of all the divisors of the integer n, draw the Hasse diagrams of D10 , D15 , D32 , and D45 . Solution. The Hasse diagrams are shown below. 45
32 15
10
16 5
2
3
5
4 1
1
9
15
3
5
8
2 1 1
D10
D15
D32
D45
Hasse diagrams of given lattices 5. Prove that in a distributive lattice, the complement of an element is unique. Solution. Let a be an element with two distinct complements b and c. Then a ? b = 0 and a ? c = 0 =⇒ a ? b = a ? c. Also, a ⊕ b = 1 and a ⊕ c = 1 =⇒ a ⊕ b = a ⊕ c. By a theorem, we have b = c. 6. Let L be a complemented, distributive lattice. Then for a, b ∈ L, show the following are equivalent: (i) a ≤ b (ii) a ? b0 = 0 (iii) a0 ⊕ b = 1 (iv) b0 ≤ a0 where 0 denotes corresponding complement. or Show that the following hold in a distributive and complemented lattice L: a ≤ b ⇐⇒ a ? b0 = 0 ⇐⇒ a0 ⊕ b = 1 ⇐⇒ b0 ≤ a0 for a, b ∈ L.
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245
Solution. a ≤ b =⇒ a ⊕ b = b =⇒ (a ⊕ b) ? b0 = 0 since b ? b0 = 0 =⇒ (a ? b0 ) ⊕ (b ? b0 ) = 0 =⇒ a ? b0 = 0 since b ? b0 = 0. Hence (i) =⇒ (ii). a ? b0 = 0 =⇒ (a ? b)0 = 1 =⇒ a0 ⊕ (b0 )0 = 1 =⇒ a0 ⊕ b = 1. Hence (ii) =⇒ (iii). a0 ⊕ b = 1 =⇒ (a0 ⊕ b) ? b0 = b0 =⇒ (a0 ? b0 ) ⊕ (b ? b0 ) = b0 (using distributive law) =⇒ a0 ? b0 = b0 since b ? b0 = 0 =⇒ b0 ≤ a0 . Hence (iii) =⇒ (iv). 7. Let hL, ∧, ∨i be a distributive lattice and a, b, c ∈ L. If a ∧ b = a ∧ c and a ∨ b = a ∨ c, then b = c. or Show that the cancellation laws are valid in a distributive lattice. Solution. Let hL, ∧, ∨i be a distributive lattice and a, b, c ∈ L, such that a ∧ b = a ∧ c and a ∨ b = a ∨ c. Now, (a ∧ b) ∨ c = (a ∨ c) ∧ (b ∨ c) = (a ∨ b) ∧ (b ∨ c) = (b ∨ a) ∧ (b ∨ c) = b ∨ (a ∧ c) = b ∨ (a ∧ b) =b and
(since L is distributive)
(a ∧ b) ∨ c = (a ∧ c) ∨ c = c. Thus, b = (a ∧ b) ∨ c = c, so that a ∧ b = a ∧ c and a ∨ b = a ∨ c =⇒ b = c. That is, the cancellation law is valid in a distributive lattice.
8. Show that the direct product of any two distributive lattices is a distributive lattice.
246
Discrete Mathematical Structures Solution. Let L1 and L2 be two distributive lattices. Let x, y, z ∈ L1 × L2 , the direct product (lattice) of L1 and L2 . Then, x = (a1 , a2 ), y = (b1 , b2 ), and z = (c1 , c2 ) for some a1 , b1 , c1 ∈ L1 and a2 , b2 , c2 ∈ L2 . Now, x ∨ (y ∧ z) = (a1 , a2 ) ∨ ((b1 , b2 ) ∧ (c1 , c2 )) = (a1 , a2 ) ∨ (b1 ∧ c1 , b2 ∧ c2 ) = (a1 ∨ (b1 ∧ c1 ), a2 ∨ (b2 ∧ c2 )) = ((a1 ∨ b1 ) ∧ (a1 ∨ c1 ), (a2 ∨ b2 ) ∧ (a2 ∨ c2 )) (since L1 and L2 are distributive lattices) = ((a1 ∨ b1 ), (a2 ∨ b2 )) ∧ ((a1 ∨ c1 ), (a2 ∨ c2 )) = ((a1 , a2 ) ∨ (b1 , b2 )) ∧ ((a1 , a2 ) ∨ (c1 , c2 )) = (x ∨ y) ∧ (x ∨ z). Hence, for all x, y, z ∈ L1 × L2 , x ∨ (y ∧ z) = (x ∨ z) ∧ (x ∨ z). Therefore, if L1 and L2 are distributive lattices, then the direct product L1 × L2 is also a distributive lattice. 9. Prove that the lattice is modular. 1
a
b
c
0
Hasse diagram of given lattice Solution. The elements a, b, and c are symmetric in the lattice. It is enough to prove for any one of a, b, c. We have the cases a < 1 and 0 < a. Case (i): Let a < 1. Let x1 = a and x3 = 1. Then x1 ∨ (x2 ∧ x3 ) = a ∨ (x1 ∧ 1) = a ∨ x2 and (x1 ∨ x2 ) ∧ x3 = (a ∨ x2 ) ∧ 1 = a ∨ x2 . Hence, x1 ∨ (x2 ∧ x3 ) = (x1 ∨ x2 ) ∧ x3 . Case (ii): Let 0 < a. Let x1 = 0 and x3 = a. Then x1 ∨ (x2 ∧ x3 ) = 0 ∨ (x2 ∧ a) = x2 ∧ a and (x1 ∨ x2 ) ∧ x3 = (0 ∨ x2 ) ∧ a = x2 ∧ a. Hence, x1 ∨ (x2 ∧ x3 ) = (x1 ∨ x2 ) ∧ x3 . Therefore, the above lattice is modular.
Lattices and Boolean Algebra
5.4.2
247
Problems for Practice
1. Find the complements, if they exist, of the elements a, b, c of the lattice, whose Hasse diagram is given below. Can the lattice be complemented? 1
e
d c
a
b
0
2. Give an example for a distributive and complemented lattice. 3. Examine whether the lattice given in the following Hasse diagram is distributive or not. 1
a
c
b
0
4. In a distributive complemented lattice, show that the following are equivalent: (i) (ii) (iii) (iv)
a≤b a ∧ ¯b = 0 a ¯∨b=1 ¯b ≤ a ¯.
5. Let hL, ≤, ∨, ∧i be a distributive lattice and a, b ∈ L if a ∧ b = a ∧ c and a ∨ b = a ∨ c. Then, show that b = c. 6. Define a lattice. Give a suitable example. 7. In a complemented and distributive lattice, prove that the complement of each element is unique. 8. State modular inequality of lattices. 9. Show that cancellation laws are valid in a distributive lattice.
248
Discrete Mathematical Structures
5.5
Boolean Algebra
Definition 5.5.1 Boolean Algebra: A Boolean algebra is a complemented distributive lattice. A Boolean algebra will generally be denoted by hB, ?, ⊕, 0 , 0, 1i, and it satisfies the following properties in which a, b, and c denote any element of the set B. 1. hB, ?, ⊕, 0 , 0, 1i is a lattice and satisfies the following: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)
a?a=a a?b=b?a (a ? b) ? c = a ? (b ? c) a ? (a ⊕ b) = a a⊕a=a a⊕b=b⊕a (a ⊕ b) ⊕ c = a ⊕ (b ⊕ c) a ⊕ (a ? b) = a.
2. hB, ?, ⊕i is a distributive lattice and satisfies the following: (i) (ii) (iii) (iv)
a ? (b ⊕ c) = (a ? b) ⊕ (a ? c) a ⊕ (b ? c) = (a ⊕ b) ? (a ⊕ c) (a ? b) ⊕ (b ? c) ⊕ (c ? a) = (a ⊕ b) ? (b ⊕ c) ? (c ⊕ a) (a ? b) = (a ? c) and (a ⊕ b) = (a ⊕ c) =⇒ b = c.
3. hB, ?, ⊕, 0 , 0, 1i is a bounded lattice and satisfies the following: (i) (ii) (iii) (iv) (v)
0≤a≤1 a?0=0 a?1=a a⊕0=a a ⊕ 1 = 1.
4. hB, ?, ⊕, 0 , 0, 1i is a complemented lattice in which the complement of any element a ∈ B is denoted by a0 ∈ B and satisfies the following: (i) (ii) (iii) (iv) (v) (vi)
a ? a0 = 0 a ⊕ a0 = 1 00 = 1 10 = 0 (a ? b)0 = a0 ⊕ b0 a ⊕ b0 = a0 ? b0 .
Lattices and Boolean Algebra
249
5. There exists a partial ordering ≤ on B such that (i) (ii) (iii) (iv)
a ? b = GLB{a, b} a ⊕ b = LU B{a, b} a ≤ b ⇐⇒ a ? b = a ⇐⇒ a ⊕ b = b a ≤ b ⇐⇒ a ? b0 = 0 ⇐⇒ b0 ≤ a0 ⇐⇒ a0 ⊕ b = 1.
Example 5.5.2 Let A = {a, b, c} and consider the lattice hP (A), ∩, ∪i as shown below. {a,b,c}
{a,b}
{a}
{a,c}
{b}
{b,c}
{c}
Hasse diagram of hP (A), ∩, ∪i Clearly, hP (A), ∩, ∪i is a Boolean algebra. Example 5.5.3 Let B = {0, 1} be a set. The operations ?, ⊕,0 on B are defined in the table below. ? 0 1
Tables showing Operations of ?, ⊕,0 on B 0 1 ⊕ 0 1 x 0 0 0 0 1 0 0 1 1 1 1 1
x0 1 0
Clearly, hB, ?, ⊕,0 , 0, 1i is a Boolean algebra. Definition 5.5.4 Sub-Boolean Algebra: Let hB, ?, ⊕,0 , 0, 1i be a Boolean algebra and S ⊆ B. If S contains the elements 0 and 1 and is closed under the operations ?, ⊕, 0 , then hS, ?, ⊕,0 , 0, 1i is called a sub-Boolean algebra. Remark 5.5.5 A sub-Boolean algebra of a Boolean algebra is itself a Boolean algebra. Remark 5.5.6 A subset of a Boolean algebra can be a Boolean algebra. However, it may not be a sub-Boolean algebra because it may not close with respect to the operations in the Boolean algebra.
250
Discrete Mathematical Structures
Definition 5.5.7 Direct Product of Boolean Algebra: Let hB1 , ?1 , ⊕1 ,0 , 01 , 11 i and hB2 , ?2 , ⊕2 ,00 , 02 , 12 i be any two Boolean algebras. The direct product of the two Boolean algebras is defined to be a Boolean algebra that is given by hB1 × B2 , ?3 , ⊕3 ,000 , 03 , 13 i in which the following operations are defined for any (a1 , b1 ), (a2 , b2 ) ∈ B1 × B2 as (a1 , b1 ) ?3 (a2 , b2 ) = ((a1 ?1 a2 ), (b1 ?2 b2 )) (a1 , b1 ) ⊕3 (a2 , b2 ) = ((a1 ⊕1 a2 ), (b1 ⊕2 b2 )) (a1 , b1 )000 = (a01 , b001 ) 03 = (01 , 02 ) and 13 = (11 , 12 ). Definition 5.5.8 Join-irreducible: Let hL, ?, ⊕i be a lattice. An element a ∈ L is called join-irreducible if it cannot be expressed as the join of two distinct elements of L. In other words, a ∈ L is join-irreducible, if for any a1 , a2 ∈ L, a = a1 ⊕ a2 =⇒ (a = a1 ) ? (a = a2 ). Definition 5.5.9 Boolean Homomorphism: Let hB, ?, ⊕,0 , 0, 1i and hP (A), ∪, ∩,c , α, βi be any two Boolean algebras, where A is a set. Then, a mapping f : B −→ P (A) is called a Boolean homomorphism, if for any a, b ∈ B, f (a ? b) = f (a) ∩ f (b) f (a ⊕ b) = f (a) ∪ f (b) f (a0 ) = [f (a)]c f (0) = α f (1) = β. Remark 5.5.10 The binary operations ? and ⊕ are preserved under Boolean homomorphism. Remark 5.5.11 Let hL, ?, ⊕, ≤i and hS, ∧, ∨, ≤0 i be two Boolean algebras. Then, a mapping g : L =⇒ S is called an order homomorphism, then a ≤ b =⇒ g(a) ≤0 g(b), for all a, b ∈ L. Theorem 5.5.12 In a Boolean algebra, De Morgan’s laws hold. Proof. Let hL, ?, ⊕,− , 0, 1i be a Boolean algebra. Then, L is a complemented and distributive lattice. De Morgan’s laws are ¯ b=a a⊕ ¯ ? ¯b, a ¯? b = a ¯ ⊕ ¯b, for all a ¯, a, b ∈ L. Assume that a, b ∈ L. There exist elements a ¯, ¯b ∈ L such that a⊕a ¯ = 1, a ? a ¯ = 0, b ⊕ ¯b = 1, b ? ¯b = 0.
Lattices and Boolean Algebra
251
¯ b=a (i) Claim: a ⊕ ¯ ? ¯b. (a ⊕ b) ⊕ (¯ a ? ¯b) = [(a ⊕ b) ⊕ a ¯] ? [(a ⊕ b) ⊕ ¯b] = [a ⊕ ? a ⊕ b] ? [a ⊕ b ⊕ ¯b] = [a ⊕ b] ? [a ⊕ a] = 1 ? 1 = 1. ¯ (a ⊕ b) ? (¯ a ? b) = [(a ⊕ b) ? a ¯] ? [(a ⊕ b) ? ¯b] = [(a ? a ¯) ⊕ (b ? a ¯)] ? [(a ? ¯b) ⊕ (b ? ¯b)] = [0 ⊕ (b ? a ¯)] ? [(a ? ¯b) ⊕ 0] = (b ? a ¯) ? (a ? ¯b) = b ? (¯ a ? a) ? b = ¯b ? 0 ? ¯b = 0. Hence, claim (i) is proved. (ii) Claim: a ¯? b = a ¯ ⊕ ¯b. (a ? b) ⊕ (¯ a ⊕ ¯b) = [(a ? b) ⊕ a ¯] ⊕ [(a ? b) ⊕ ¯b] = [(a ⊕ a ¯) ? (b ⊕ a ¯)] ⊕ [(a ⊕ ¯b) ? (b ⊕ ¯b)] = [1 ? (b ⊕ a ¯)] ⊕ [(a ⊕ ¯b) ? 1] = (b ⊕ a ¯) ⊕ (a ⊕ ¯b) = b ⊕ (¯ a ⊕ a) ⊕ ¯b = b ⊕ 1 ⊕ ¯b = b ⊕ ¯b = 1. (a ? b) ? (¯ a ⊕ ¯b) = [(a ? b) ? a ¯] ⊕ [(a ? b) ? ¯b] = (a ? a ¯ ? b) ⊕ (a ? b ? ¯b) = (0 ? b) ⊕ (a ? 0) = 0 ? 0 = 0. Hence, claim (ii) is proved. Therefore, De Morgan’s laws are proved. Theorem 5.5.13 In a Boolean algebra hL, ?, ⊕i, the complement a ¯ of any element a ∈ L is unique. Proof. Let a ∈ L have two complements b, c ∈ L. By definition, we have a ? b = 0, a ⊕ b = 1, Then, we have b=b?1 = b ? (a ⊕ c) = (b ? a) ⊕ (b ? c) = 0 ⊕ (b ? c) =b?c
a ? c = 0,
a ⊕ c = 1.
(5.22)
252
Discrete Mathematical Structures and
c=c?1 = c ? (a ⊕ b) = (c ? a) ⊕ (c ? b) = 0 ⊕ (c ? b) =c?b = b ? c.
(5.23)
From (5.22) and (5.23), we have b = c. Therefore, every element of L has a unique complement.
5.5.1
Solved Problems
1. Show that hP (A), ∪, ∩, ⊆i is a Boolean algebra, where A is any set. Solution. We know that hP (A), ∪, ∩, ⊆i is a lattice. For any X, Y, Z ∈ P (A), X ∩ (Y ∪ Z) = (X ∩ Y ) ∪ (X ∩ Z) X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z). ¯ of A such that Also, for all X ∈ P (A), there exists a subset X ¯ ¯ X ∪ X = A, X ∩ X = { } = φ. Zero element of P (A) is { } = least element. The greatest element of P (A) is A. Therefore, hP (A), ∪, ∩, ⊆i is a Boolean algebra. 2. Show that in any Boolean algebra, (a + b)(a0 + c) = ac + a0 b + bc. Solution. Let hB, +, ·, 0 i be a Boolean algebra. Let a, b, c ∈ B. (a + b)(a0 + c) = (a + b)a0 + (a + b)c = aa0 + ba0 + ac + bc = 0 + a0 b + ac + bc = ac + a0 b + bc. 3. In any Boolean algebra, show that a = b if and only if a¯b + a ¯b = 0. Solution. Let hB, +·, − , 0, 1i be any Boolean algebra. Let a, b ∈ B and a = b. To show that: a¯b + a ¯b = 0. a · ¯b + a ¯ · b = a¯ a+a ¯a = 0 + 0 = 0. Now, let a¯b + a ¯b = 0, for all a, b ∈ B. Then a¯b + a ¯b = 0
Lattices and Boolean Algebra =⇒ =⇒ =⇒ =⇒ =⇒ =⇒ =⇒ 4. Simplify
253 a¯ a + a¯b + a ¯b + b¯b = 0 a(¯ a + ¯b) + b(¯ a + ¯b) = 0 (a + b)(¯ a + ¯b) = 0 ¯ (a + b)ab = 0 ab = a + b GLB{a, b} = LU B{a, b} a = b.
(i) (a ? b)0 ⊕ (a ⊕ b)0 (ii) (a0 ? b0 ? c) ⊕ (a ? b0 ? c) ⊕ (a ? b0 ? c0 ).
Solution. (i)
(ii)
(a ? b)0 ⊕ (a ⊕ b)0 = (a0 ⊕ b0 ) ⊕ (a0 ? b0 ) = (a0 ⊕ b0 ⊕ a0 ) ? (a0 ⊕ b0 ⊕ b0 ) = (a0 ⊕ b0 ) ? (a0 ? b0 ) = a0 ? b0 . (a0 ? b0 ? c) ⊕ (a ? b0 ? c) ⊕ (a ? b0 ? c0 ) = (a0 ⊕ a) ? (b0 ? c) = 1 ? (b0 ? c) = b0 ? c.
5. Let a, b, c be any elements in a Boolean algebra B. Prove that (i) a ? a = a (ii) a ⊕ a = a. Solution. (i) To prove: a ? a = a. Let
a=a?1 = a ? (a ⊕ a0 ) = a ? a ⊕ a ? a0 = (a ? a) ⊕ 0 = a ? a.
(by (by (by (by (by
identity law) complement law) distributive law) complement law) identity law)
(ii) To prove: a ⊕ a = a. Let
a=a⊕0 = a ⊕ (a ? a0 ) = (a ⊕ a) ? (a ⊕ a0 ) = (a ⊕ a) ? 1 = a ⊕ a.
(by (by (by (by (by
identity law) complement law) distributive law) complement law) identity law)
254
Discrete Mathematical Structures 6. Let a, b, c be any elements in a Boolean algebra B. Show that (i) a ⊕ a = 1
(ii) a ? 0 = 0.
Solution. (i) To prove: a ⊕ 1 = 1. Let
a ⊕ 1 = (a ⊕ 1) ? 1 = (a ⊕ 1) ? (a ⊕ a0 ) = a ⊕ (a ? a0 ) = a ⊕ (a0 ? 1) = a ⊕ a0 = 1.
(by (by (by (by (by (by
identity law) complement law) distributive law) commutative law) identity law) complement law)
(ii) To prove: a ? 0 = 0. Let
a ? 0 = (a ? 0) ⊕ 0 = (a ? 0) ⊕ (a ? a0 ) = a ? (0 ⊕ a0 ) = a ? (a0 ⊕ 0) = a ? a0 = 0.
(by (by (by (by (by (by
identity law) complement law) distributive law) commutative law) identity law) complement law)
7. Prove that a ⊕ (a0 ? b) = a ⊕ b. Solution. a ⊕ (a0 ? b) = (a ⊕ a0 ) ? (a ⊕ b) = 1 ? (a ⊕ b) = a ⊕ b. 8. Prove that a ? (a0 ⊕ b) = a ? b. Solution. a ? (a0 ⊕ b) = (a ? a0 ) ⊕ (a ? b) = 0 ⊕ (a ? b) = a ? b. 9. Prove that (a ? b) ⊕ (a ? b0 ) = a. Solution. (a ? b) ⊕ (a ? b0 ) = a ? (b ⊕ b0 ) = a ? 1 = a. 10. In any Boolean algebra, hB, ·, +, 0 , 0, 1i, show that (a + b0 )(b + c0 )(c + a0 ) = (a0 + b)(b0 + c)(c0 + a). Solution. (a + b0 )(b + c0 )(c + a0 ) = (a + b0 + 0)(b + c0 + 0)(c + a0 + 0) = (a + b0 + cc0 )(b + c0 + aa0 )(c + a0 + bb0 ) = (a + b0 + c)(a + b0 + c0 )(b + c0 + a) (b + c0 + a0 )(c + a0 + b)(c + a0 + b0 ) = [(a0 + b + c)(a0 + b + c0 )]
Lattices and Boolean Algebra
255 [(b0 + c + a)(b0 + c + a0 )] [(c0 + a + b)(c0 + a + b0 )] = (a0 + b + cc0 )(b0 + c + aa0 )(c0 + a + bb0 ) = (a0 + b + 0)(b0 + c + 0)(c0 + a + 0) = (a0 + b)(b0 + c)(c0 + a).
11. In any Boolean algebra, hB, ·, +, 0 , 0, 1i, show that a = 0 ⇐⇒ ab0 + a0 b = b. Solution. If a = 0, then it directly follows that ab0 + a0 b = 0 + 1b = 0 + b = b. Suppose
b = ab0 + a0 b.
(5.24)
Therefore, 0 = b0 b = b0 (ab0 + a0 b) = ab0 + 0 = ab0 . Using De Morgan’s law, from (5.24) we obtain b0 (a0 + b)(a + b0 ). Therefore, 0 = ab0 = a(a0 + b)(a + b0 ) = (aa0 + ab)(a + b0 ) = (0 + ab)(a + b0 ) = ab(a + b0 ) = aba + abb0 = ab + 0 = ab. Therefore, Therefore,
5.5.2
0 = ab = ab0 . 0 = ab + ab0 = a(b + b0 ) = a1 = a. Hence, a = 0.
Problems for Practice
1. What values of the Boolean variables x and y satisfy xy = x + y? 2. Show that De Morgan’s laws hold in a Boolean algebra. That is, show that for all x and y, x ∨ y = x ∧ y and x ∧ y = x ∨ y. 3. Does a Boolean algebra contain six elements? Justify your answer. 4. If P (S) is the power set of a non-empty set S, prove that hP (S), ∪, ∩,c , φ, Si is a Boolean algebra. 5. Prove that in a Boolean algebra, (a ∨ b)0 = a0 ∧ b0 . 6. Give an example of a two-element Boolean algebra. 7. Write the Boolean algebra whose Hasse diagram is a chain. 8. Is there a Boolean algebra with five elements? Justify your answer. 9. Show that a lattice homomorphism on a Boolean algebra which preserves 0 and 1 is a Boolean homomorphism.
256
Discrete Mathematical Structures
10. Prove that a lattice with five elements is not a Boolean algebra. 11. Show that in a Boolean algebra, a ⊕ (a0 ? b) = a ⊕ b. 12. Show that in a Boolean algebra, a ? (a0 ⊕ b) = a ? b. 13. Show that in a Boolean algebra, (a ? b) ⊕ (a ? b0 ) = a. 14. Show that in a Boolean algebra, (a ? b ? c) ⊕ (a ? b) = a ? b. 15. Show that in a Boolean algebra, a ≤ b =⇒ a + bc = b(a + c). 16. Simplify the Boolean expression: (a ? c) ⊕ c ⊕ [(b ⊕ b0 ) ? a]. 17. Simplify the Boolean expression: (1 ? a) ⊕ (0 ? a0 ).
Bibliography
G. Balaji, Discrete Mathematics, G. Balaji Publishers, Chennai, India, 2017. R. K. Bisht and H. S. Dhami, Discrete Mathematics, Oxford University Press, New Delhi, India, 2015. Susanna S. Epp, Discrete Mathematics with Applications, Fourth Edition, Brooks/Cole, Cengage Learning, U.S.A., 2011. Rowan Garnier and John Taylor, Discrete Mathematics: Proofs, Structures and Applications, Third Edition, CRC Press, Boca Raton, FL, 2010. Seymour Lipschutz and Marc Lipson, Discrete Mathematics, Third Edition, Schaum’s Outlines, Tata McGraw-Hill Company, New Delhi, India, 2007. Kenneth H. Rosen, Discrete Mathematics and its Applications, Seventh Edition, McGraw-Hill Education, New York, U.S.A., 2007. J. P. Tremblay and R. Manohar, Discrete Mathematical Structures with Applications to Computer Science, Tata McGraw-Hill Publishing Company Limited, New Delhi, India, 2008.
257
Index
A Abelian group, 174, 184 Absorption law, 233 Adjacency matrix, 149 Adjacent edges, 137 Adjacent vertices, 137 Algebraic structure, 174 Algebra, 173 Algebraic system, 173 Associative property, 173
B Biconditional, 4 Binary operation, 173 Bipartite graph, 143 Block, 158 Boolean algebra, 248 Boolean homomorphism, 250 Bounded variable, 23
C Cancellation property, 174 Canonical form, 9 Cayley’s representation theorem, 196 Chain, 231 Characteristic equation, 91 Chinese postman problem, 163 Circuit, 157 Circular path, 157 Closed walk, 157 Closure property, 173 Combination, 80 Combinatorics, 39 Commutative group, 184 Commutative ring, 219 Commutativity, 173
Complemented lattice, 240 Complete bipartite graph, 144 Complete graph, 142 Complete lattice, 240 Component, 158 Conditional statement, 3 Congruence relation, 175 Conjunction, 2 Connected graph, 158 Complement element, 240 Connectivity, 156 Contingency, 6 Contradiction, 6 Cosets, 197 Cycle, 157 Cycle graph, 142 Cyclic group, 193 Cyclic permutation, 209
D Degree of a vertex, 138 Directed graph, 138 Direct product of Boolean algebra, 250 Direct product of lattices, 239 Direct proof, 22 Disconnected graph, 158 Disjoint cycles, 209 Disjunction, 3 Distributive lattice, 241 Distributive properties, 174 Duality law, 7 Dual lattices, 231
E Elementary cycle, 157 Elementary path, 157 259
260 Endomorphism, 196 Eulerian circuit, 161 Eulerian graph, 161 Eulerian path, 161 Eulerian trail, 161 Even permutation, 210 Existence of identity, 173, 183 Existence of inverse, 173, 183 Existential generalization, 28 Existential specification, 28
F Factor group, 199 Field, 218 Free variable, 23 Fundamental theorem, 202
G Generalized pigeonhole principle, 58 Generating function, 103 Graph colouring, 145 Graph isomorphism, 149 Graphs, 135 Group, 183
H Hamiltonian circuit, 163 Hamiltonian path, 163 Handshaking theorem, 139 Hasse diagram, 224 Homogenous recurrence relation, 88 Homomorphism, 195
I Idempotent element, 174 Idempotent law, 232 Incidence matrix, 150 Inclusion-exclusion principle, 117 In-degree of a vertex, 139 Indirect method, 19 Inference theory, 13 Inference theory for predicate calculus, 28
Index Integral domain, 218 Isolated vertex, 137 Isomorphic graphs, 151 Isomorphism, 196
J Join-irreducible, 250
K Kernel of a homormorphism, 195 Konisberg bridge, 168
L Lagrange’s theorem, 198 Lattice, 231 Lattice automorphism, 239 Lattice endomorphism, 239 Lattice homomorphism, 239 Lattice isomorphism, 239 Left coset, 197 Linearly ordered set, 231 Linear recurrence relation, 88 Lower bound, 231
M Mathematical induction, 39 Method of contradiction, 19 Method of contrapositive, 21 Mixed graph, 138 Modular lattice, 241 Monoid, 174 Multigraph, 138
N Negation, 2 Normal form, 9 Normal subgroup, 199
O Odd permutation, 210 Open walk, 157 Order isomorphic partially ordered set, 240 Order of a group, 193 Order-preserving mapping, 240 Out-degree of a vertex, 139
Index
P Parallel edges, 136 Partially ordered set, 223 Partial order relation, 223 Path, 157 Permutation, 70 Permutation function, 208 Permutation group, 185 Permutation with repetition, 71 Pigeonhole principle, 58 Poset, 223 Predicate calculus, 22 Principle of mathematical induction, 39 Proposition, 1 Pseudo graph, 138
Q Quantifier, 23 Quotient group, 199
R Recurrence relation, 87 Regular graph, 142 Right coset, 197 Ring, 217 Ring homomorphism, 218 Rules of inference, 14
S Self loop, 136 Semigroup, 174 Simple graph, 137 Simple path, 157 Special lattices, 240
261 Star graph, 144 Strong induction, 57 Sub-Boolean algebra, 249 Subgraph, 142 Subgroup, 192 Sublattice, 238 Subring, 218 Substitution property, 175 Symmetric group, 185
T Tautology, 6 Terminal vertex, 157 Totally ordered set, 223 Transposition, 210 Trivial proof, 21 Truth table, 2
U Undirected graph, 138 Unicursal graph, 164 Universal generalization, 28 Universal specification, 28 Universe of discourse, 23 Upper bound, 231
V Vacuous proof, 22
W Walk, 156 Well-ordering property, 57 Well-ordered set, 224 Wheel graph, 143
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# Heuristics of the General Relativity
posted in: Methodology, Relativity
The essential principle of coordination in GR is the principle of equivalence, including a negative heuristic. The argument “is not that all reference frames are equivalent, but that the classical coordination of uniform motion in a straight line with the paths of force-free particles cannot be carried out unambiguously or consistently.” [1] The principle of equivalence states that the decomposition of the gravitational motion into a uniform motion and gravitational acceleration cannot be unique, since free fall is not distinguishable locally from uniform motion. However, such a decomposition implies a violation of the general covariance, because it represents an arbitrary choice of a coordinate system[2]. For any coordinate system, if we identify its lines with the geodesic lines, we can construct the gravitational field so that the difference between these geodesics and the actual motions can be differentiated. [3]
Einstein’s special theory of relativity (SR) is built on two fundamental postulates. the postulate of light (the speed of light, in the “rest frame”, is independent of the speed of the source), and the principle of relativity. The latter was explicitly adopted by Einstein as a means of restricting the form of laws, whatever their detailed structure. Thus, we have the difference between a “constructive” theory and a “principle” theory. The general theory of relativity was developed using as a nucleus a principle of symmetry: the principle of general covariance[4]. Initially, Einstein saw the principle of general covariance as an extension of the principle of relativity in classical mechanics, and in SR. For Einstein, the principle of general covariance was a crucial postulate in the development of GR. The freedom of the GR diffeomorphism (the invariance of the form of the laws under transformations of the coordinates depending on the arbitrary functions of space and time) is a “local” spacetime symmetry, as opposed to the “global” spacetime symmetries of the SR (which depend instead on the constant parameters ).
In recent years, there have been numerous debates in physics and philosophy regarding certain types of symmetries that act in the space of theories. Such symmetries are interpreted as achieving an “equivalence” between two theories that are said to be related to a “dual symmetry” (in the case of a “symmetry” in the strict sense of an automorphism, these are called “self-dualities”). Katherine Brading[5] exemplifies with the dualities between quantum field theories (such as generalized magnetic/electrical duality), between string theories (such as T and S dualities) and between physical descriptions that are, such as a quantum field theory and a string theory, as in the case of gauge/gravity dualities[6]. Other examples are the position-momentum duality, the wave-particle duality, or the Kramers-Wannier duality of the two-dimensional Ising model in statistical physics. Dualities are transformations between theories, while symmetry is a mapping between solutions of the same theory. A symmetry can be exact (unconditional validity), approximate (valid under certain conditions) or broken (depending on the object considered and its context). The symmetries functioned normatively, like constraints, in Einstein’s general covariance in establishing the equations of general relativity.
Elie Zahar said that Einstein’s development of relativity was due to his vague metaphysical beliefs, corresponding to some of his own “heuristic prescriptions” that became a specific and powerful tool. Zahar states that Kuhn’s scientific revolution does not apply to Einstein’s case. According to him, two “heuristic devices” led to the discovery of the theory of relativity: the internal requirement of coherence, and the claim that, “since God is no deceiver, there can be no accidents in Nature.” Natural symmetries are fundamental at the ontological level, and the heuristic rule takes precedence over a theory that does not explain symmetries as deeper manifestations. [7]
According to Newton, gravity is not a primary quality like inertia or impenetrability. Therefore, inertia and gravity are independent properties. But Newton states that the inertial mass is equal to the gravitational mass, without explaining the reason for this identity (there is a symmetry that contradicts the independence of the two properties). In Michelson’s experience, by applying the ether as a universal medium, it is undetectable, which is a paradox. Einstein became aware of this paradox. Einstein eliminates the asymmetry between gravity and inertia by proposing that all gravitational fields be inertial. He also had other objections to classical physics: Lorentz’s electromagnetic theory faced a dualism between discrete charged particles governed by Newton’s laws and a continuous field that respected Maxwell’s equations; relativity applies to Lorentz mechanics, but not to electrodynamics; the idea of absolute space (there is a privileged inertial framework), although its elimination does not influence classical mechanics.
Einstein appreciated, on the principle of relativity, its universality and its unifying role for mechanics and electrodynamics, this being the first principle used to develop its general theory of relativity. The second principle is that of light but, epistemologically, Einstein’s second starting point in developing the general theory of relativity was not the principle of light, but the idea that Maxwell’s equations are covariant and express a law of nature. The principle of light results from this idea, as does the principle of relativity, according to Zahar. [8]
Basically, Einstein had the choice of developing general relativity based on Maxwell’s equations or Newton’s laws. But in the dualism between particles and fields, all attempts at mechanical explanation of field behavior failed.
According to Zahar, no “crucial” experiment could have been conceived between Lorentz theory and Einstein’s in 1905. But Minkowski and Planck abandon the classical program for special relativity, contrary to Kuhn’s methodology. Moreover, Einstein was at that time a quasi-stranger, while Lorentz was a recognized authority. And Lorentz’s theory was very clear from that of Einstein, which involved a major overhaul of the notions of space and time. Also, there were no anomalies that Einstein’s theory would have solved better than Lorentz. In addition, Lorentz himself was finally convinced of the new perspective[9]. Whittaker[10] regards Lorentz and Poincaré as the true authors of special relativity, Einstein’s credit being that of developing general relativity. Thus, Lorentz’s etheric program was not defeated by the program of relativity but was practically developed in it. Zahar contradicts it, based on the fact that the two programs have very different heuristics. [11]
In the case of the Copernican revolution, the Platonic program for modeling the phenomenon through circular and spherical movements was initially successful, with each planet on a real physical crystalline sphere in axial rotation. It was later discovered that the distance between the earth and the planets varies, so that additional assumptions were made through eccentricities, epicycles and screens, to explain the new observations. When one tried to determine the motion of the celestial bodies towards the earth due to the uneven movements, there appeared differences between phenomena and mathematical methods that allowed only circular motions with the earth in the center of the universe. Copernicus, although he considered the Sun fixed, did not resolve this difference, still using epicycles. Kepler was the one who abolished the epicycles and found the laws of the elliptical motion of the planets with the Sun in a focus. Lorentz used the Galilean transformations, eliminating the epicycles but giving the etheric frame a privileged status. Just as Copernicus was aware of the idealization of his planetary model, Lorentz later understood that the effective coordinates, not the Galilean ones, are the quantities measured in the moving frame. Einstein gave up the Galilean transformations and identified the actual coordinates measured as the only real ones. Einstein’s heuristics are based on a general requirement of Lorentz covariance for all physical laws, requiring the renunciation of Galilean transformations.
Zahar claims that Lorentz and Einstein used different heuristics in their research programs[12]. The etheric program was practically replaced by a program with greater heuristic power, which is why Planck abandoned Lorentz’s theory in favor of Einstein just before Einstein’s program became progressively empirical. The two theories are similar in terms of the “hard core” (negative heuristics) and can be considered as bifurcated programs. The difference between the positive heuristics was what led to the choice of scientists of Einstein’s program at the beginning of the last century. Lorentz’s positive heuristic consisted in providing the ether with properties that would explain many physical phenomena, including the electromagnetic field and Newtonian mechanics. This approach allowed a rapid development of Lorentz’s program, but by the end of the 19th century heuristics had reached a saturation point. A number of degenerate programmes have emerged as mechanical models to resolve ether anomalies. To explain certain electromagnetic phenomena, Lorentz introduced the postulate of the ether at rest, but subsequent calculations contradicted this hypothesis.
The differences between Lorentz and Einstein’s views were metaphysical: Lorentz believed that the universe respects intelligible laws (there is a propagating environment, an absolute “now”, etc.), while for Einstein the universe is governed by coherent mathematical principles. (covariant laws, etc.) Zahar states that all major scientific revolutions were accompanied by an increase in mathematical coherence accompanied by a (temporary) loss of intelligibility (Newtonian astronomy is more coherent than Ptolemaic, but remote action was not accepted before Newton, then accepted at the end of the 18th century and again rejected after Maxwell). In Lorentz’s research programme, the behavior of the electromagnetic field had come to dictate the properties of the ether, even improbable (for example, resting ether and acting by zero net forces). Basically, Lorentz’s heuristic strategy has reversed: instead to deduce a theory from the ether considered fundamental, he reaches the ether based on the field. Einstein’s heuristics were based on the requirement that all physical laws be Lorentz-covariant (to take the same form regardless of the frame of reference), and classical law to emerge from the new law as a limit case.
In order to obtain a relativistic theory of gravity, Einstein maintained the principle of equivalence, decided to treat all coordinate systems equally and to impose a condition of general covariance on all laws. The empirical success of general relativity through the correct prediction of the behavior of Mercury’s perihelion has proved crucial for the further development of the programme.
Since 1905, the program of relativity has proved to be heuristically superior compared to the classical one. But special relativity has failed to outperform the Lorentz program. Bucherer’s experiment[13] confirmed both hypotheses, and Kaufmann’s experiment[14] denied both. Before the emergence of general relativity, the scientific community spoke of Lorentz-Einstein’s theory, considering them as equivalent from an observer’s point of view. General relativity has succeeded empirically replacing the Lorentz program by successfully explaining the “abnormal precession” of Mercury’s perihelion. This prediction was an empirical progress. In addition, general relativity has been found to be more falsifiable.
Nicholas Maxwell proposes also a method for the unification of two “mutually contradictory principles.” [15] The method proposed by him for establishing the unified theory is as follows: from the two theories, the common elements that do not contradict are chosen, the contradictory elements are removed, and on this basis the new theory is developed. He does not sufficiently exemplify, in my opinion, what would be those common elements in the case of classical mechanics and classical electrodynamics, considered by all scientists as two contradictory theories from which the special theory of relativity was born. Also, Nicholas Maxwell imposes the existence of a “crucial assumption”, whose falsifiability allows the acceptance of the theory as a result of a method of discovery based on empirical purpose. In today’s physics, there are countless examples of unifying theories (such as the M theory proposing the union of all fundamental forces, including gravity) that have not set out to become falsifiable by “crucial assumptions”.
General relativity is the result of Einstein’s unification of Newton’s theory of universal gravity (with the instantaneous action at a distance of the gravity) and the special theory of relativity (with the limitation of any speed to the constant value of the speed of light, c). These two principles contradict each other. So, according to Maxwell, it should be removed from the future unifying theory.
## Notes
[1] Robert Disalle, “Spacetime Theory as Physical Geometry,” Erkenntnis 42, no. 3 (1995): 317–337.
[2] A. Einstein, “The Foundation of the General Theory of Relativity,” in The Principle of Relativity. Dover Books on Physics. June 1, 1952. 240 Pages. 0486600815, p. 109-164, 1952, 114, http://adsabs.harvard.edu/abs/1952prel.book..109E.
[3] Einstein, 142–43.
[4] Katherine Brading, Elena Castellani, and Nicholas Teh, “Symmetry and Symmetry Breaking,” in The Stanford Encyclopedia of Philosophy, ed. Edward N. Zalta, Winter 2017 (Metaphysics Research Lab, Stanford University, 2017), https://plato.stanford.edu/archives/win2017/entries/symmetry-breaking/.
[5] Katherine Brading and Harvey R. Brown, “Symmetries and Noether’s Theorems,” in Symmetries in Physics: Philosophical Reflections, ed. Katherine A. Brading and Elena Castellani (Cambridge University Press, 2003), 89–109.
[6] Brading, Castellani, and Teh, “Symmetry and Symmetry Breaking.”
[7] Elie Zahar, “Why Did Einstein’s Programme Supersede Lorentz’s? (II),” British Journal for the Philosophy of Science 24, no. 3 (1973): 223–262.
[8] Zahar.
[9] Zahar.
[10] Edmund Taylor Whittaker, A History of the Theories of Aether and Electricity (Harper, 1960).
[11] Zahar, “Why Did Einstein’s Programme Supersede Lorentz’s?”
[12] Zahar.
[13] A. H. Bucherer, “Die Experimentelle Bestätigung Des Relativitätsprinzips,” Annalen Der Physik 333 (1909): 513–36, https://doi.org/10.1002/andp.19093330305.
[14] W. Kaufmann, “Über Die Konstitution Des Elektrons,” Annalen Der Physik 324 (1906): 949, https://doi.org/10.1002/andp.19063240303.
[15] Nicholas Maxwell, Karl Popper, Science and Enlightenment (London: UCL Press, 2017).
## Bibliography
• Brading, Katherine, and Harvey R. Brown. “Symmetries and Noether’s Theorems.” In Symmetries in Physics: Philosophical Reflections, edited by Katherine A. Brading and Elena Castellani, 89. Cambridge University Press, 2003.
• Brading, Katherine, Elena Castellani, and Nicholas Teh. “Symmetry and Symmetry Breaking.” In The Stanford Encyclopedia of Philosophy, edited by Edward N. Zalta, Winter 2017. Metaphysics Research Lab, Stanford University, 2017. https://plato.stanford.edu/archives/win2017/entries/symmetry-breaking/.
• Bucherer, A. H. “Die Experimentelle Bestätigung Des Relativitätsprinzips.” Annalen Der Physik 333 (1909): 513–36. https://doi.org/10.1002/andp.19093330305.
• Disalle, Robert. “Spacetime Theory as Physical Geometry.” Erkenntnis 42, no. 3 (1995): 317–337.
• Einstein, A. “The Foundation of the General Theory of Relativity.” In The Principle of Relativity. Dover Books on Physics. June 1, 1952. 240 Pages. 0486600815, p. 109-164, 109–64, 1952. http://adsabs.harvard.edu/abs/1952prel.book..109E.
• Kaufmann, W. “Über Die Konstitution Des Elektrons.” Annalen Der Physik 324 (1906): 487–553. https://doi.org/10.1002/andp.19063240303.
• Maxwell, Nicholas. Karl Popper, Science and Enlightenment. London: UCL Press, 2017.
• Whittaker, Edmund Taylor. A History of the Theories of Aether and Electricity. Harper, 1960.
• Zahar, Elie. “Why Did Einstein’s Programme Supersede Lorentz’s? (II).” British Journal for the Philosophy of Science 24, no. 3 (1973): 223–262.
Nicolae Sfetcu
Email: nicolae@sfetcu.com | 3,696 | 16,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-10 | latest | en | 0.91643 |
https://convertit.com/Go/SmartPages/Measurement/Converter.ASP?From=rod | 1,701,623,527,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00129.warc.gz | 227,039,556 | 3,699 | New Online Book! Handbook of Mathematical Functions (AMS55)
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https://de.mathworks.com/matlabcentral/answers/1856818-complex-ifft-output-for-hermitian-input | 1,675,207,921,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499891.42/warc/CC-MAIN-20230131222253-20230201012253-00159.warc.gz | 213,992,029 | 28,205 | # complex ifft output for hermitian input
4 views (last 30 days)
Md. Al-Imran Abir on 19 Nov 2022
Commented: Md. Al-Imran Abir on 19 Nov 2022
I have some S11 data (complex) from a simulation and I want to find the corresponding time response. The values are for positive frequencies (5-15 GHz). So, I added conjugates at negative frequencies and padded zeros for frequencies in between. So, I expected the output of the MATLAB ifft to give a real response but got complex values with significat imaginary part. I have added my code and the data here. Can you point at what I am doing wrong?
clc
clear
close all
%% Read values from the file
posFreq = s11(:, 1) * 1e9; %positive frequencies
posVal = s11(:, 2);
%% Conjugates at negative frequencies
negFreq = -flip(posFreq);
negVal = conj(flip(posVal));
step = posFreq(2) - posFreq(1);
zeroFreq = (negFreq(end)+step : step : posFreq(1)-step)';
zeroVal = zeros(length(zeroFreq), 1);
%% Merging all frequencies
val = [negVal; zeroVal; posVal];
freq = [negFreq; zeroFreq; posFreq];
%% Plot S11
figure
subplot(2, 1, 1)
plot(freq, real(val), '-.')
title('Real S11')
subplot(2, 1, 2)
plot(freq, imag(val), '-.')
title('Imaginary S11')
%% IFFT
timeResp = ifftshift(ifft(val));
fs = freq(end)-freq(1);
ts = 1/fs;
n = length(freq);
t = (0:n-1)*ts;
%% Plot time response
figure
plot(t, real(timeResp))
title('Real of time resp')
figure
plot(t, imag(timeResp))
title('Imag of time resp')
Paul on 19 Nov 2022
Hi Md.
I didn't inspect the whole code, but did notice that the calcuation of timeResp seemed to have the ifft/ifftshift operations in the wrong order. Is the result below closer to what you expect?
clc
clear
close all
%% Read values from the file
posFreq = s11(:, 1) * 1e9; %positive frequencies
posVal = s11(:, 2);
%% Conjugates at negative frequencies
negFreq = -flip(posFreq);
negVal = conj(flip(posVal));
step = posFreq(2) - posFreq(1);
zeroFreq = (negFreq(end)+step : step : posFreq(1)-step)';
zeroVal = zeros(length(zeroFreq), 1);
%% Merging all frequencies
val = [negVal; zeroVal; posVal];
freq = [negFreq; zeroFreq; posFreq];
%% IFFT
% timeResp = ifftshift(ifft(val));
timeResp = ifft(ifftshift(val));
fs = freq(end)-freq(1);
ts = 1/fs;
n = length(freq);
t = (0:n-1)*ts;
%% Plot time response
figure
plot(t, real(timeResp))
title('Real of time resp')
figure
plot(t, imag(timeResp))
title('Imag of time resp'
Md. Al-Imran Abir on 19 Nov 2022
Thank you.
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Translated by | 786 | 2,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-06 | longest | en | 0.677421 |
https://metanumbers.com/7652 | 1,656,428,640,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103556871.29/warc/CC-MAIN-20220628142305-20220628172305-00419.warc.gz | 447,612,239 | 7,269 | # 7652 (number)
7,652 (seven thousand six hundred fifty-two) is an even four-digits composite number following 7651 and preceding 7653. In scientific notation, it is written as 7.652 × 103. The sum of its digits is 20. It has a total of 3 prime factors and 6 positive divisors. There are 3,824 positive integers (up to 7652) that are relatively prime to 7652.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 20
• Digital Root 2
## Name
Short name 7 thousand 652 seven thousand six hundred fifty-two
## Notation
Scientific notation 7.652 × 103 7.652 × 103
## Prime Factorization of 7652
Prime Factorization 22 × 1913
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 3826 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 7,652 is 22 × 1913. Since it has a total of 3 prime factors, 7,652 is a composite number.
## Divisors of 7652
1, 2, 4, 1913, 3826, 7652
6 divisors
Even divisors 4 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 6 Total number of the positive divisors of n σ(n) 13398 Sum of all the positive divisors of n s(n) 5746 Sum of the proper positive divisors of n A(n) 2233 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 87.4757 Returns the nth root of the product of n divisors H(n) 3.42678 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 7,652 can be divided by 6 positive divisors (out of which 4 are even, and 2 are odd). The sum of these divisors (counting 7,652) is 13,398, the average is 2,233.
## Other Arithmetic Functions (n = 7652)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 3824 Total number of positive integers not greater than n that are coprime to n λ(n) 1912 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 974 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 3,824 positive integers (less than 7,652) that are coprime with 7,652. And there are approximately 974 prime numbers less than or equal to 7,652.
## Divisibility of 7652
m n mod m 2 3 4 5 6 7 8 9 0 2 0 2 2 1 4 2
The number 7,652 is divisible by 2 and 4.
• Arithmetic
• Deficient
• Polite
## Base conversion (7652)
Base System Value
2 Binary 1110111100100
3 Ternary 101111102
4 Quaternary 1313210
5 Quinary 221102
6 Senary 55232
8 Octal 16744
10 Decimal 7652
12 Duodecimal 4518
16 Hexadecimal 1de4
20 Vigesimal j2c
36 Base36 5wk
## Basic calculations (n = 7652)
### Multiplication
n×y
n×2 15304 22956 30608 38260
### Division
n÷y
n÷2 3826 2550.67 1913 1530.4
### Exponentiation
ny
n2 58553104 448048351808 3428465988034816 26234621740442412032
### Nth Root
y√n
2√n 87.4757 19.7057 9.35285 5.98074
## 7652 as geometric shapes
### Circle
Radius = n
Diameter 15304 48078.9 1.8395e+08
### Sphere
Radius = n
Volume 1.87678e+12 7.358e+08 48078.9
### Square
Length = n
Perimeter 30608 5.85531e+07 10821.6
### Cube
Length = n
Surface area 3.51319e+08 4.48048e+11 13253.7
### Equilateral Triangle
Length = n
Perimeter 22956 2.53542e+07 6626.83
### Triangular Pyramid
Length = n
Surface area 1.01417e+08 5.2803e+10 6247.83
## Cryptographic Hash Functions
md5 341cd40532980c4909c8c647f2138c03 2e25e899bdf1e7128f80c5b39161d27839e6140e f71a6b4573d3c51925048c251732914b0badc55ab8c9bd524893c295fda5ae3a 65aef4e1bd5ce6fb2736a43b4102ecd2c5c23b688df1c1db9183ecc08e59a9496a2c2ae2e8d6840b88825e1e103d5618687ce8206b9ceab9cd6dceb0fd96b343 c3c202dde5d84836c83cef0de6cf536407d09cf7 | 1,446 | 4,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2022-27 | latest | en | 0.808609 |
https://www.calcudoku.org/en/2021-11-22/7/3/1 | 1,642,402,633,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300343.4/warc/CC-MAIN-20220117061125-20220117091125-00390.warc.gz | 684,367,183 | 18,327 | 17 Jan 2022
Easy:
4×4
5×5
6×6
Killer
Sudoku
Medium:
4×4
6×6
7×7
Sudoku
Difficult:
4×4
5×5
6×6
7×7
9×9
Extra
Difficult 7 × 7, Rating: ★★★★★ Points: 9 Solved 149 times
abcdefg
1
1^
2187^
1^
8+
2
24×
5-
1 mod
3
20×
3-
3-
4
1-
120×
4-
5
5+
30×
6
1 :
0-
13+
7
5+
5
Note: this puzzle also uses the modulo operator mod
a mod b = r means that when dividing a by b, the remainder is r. For example, 5 mod 2 = 1, and 3 mod 5 = 3.
Look here for a longer explanation.
Note: this puzzle also uses the exponentiation operator ^
Example: 2^3 = 2 x 2 x 2 = 8
Or, with three digits: 2^4^2 = 2^(4^2) = 2^16 = 65536
Note: evaluation is right to left
total
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# Maths Pedagogy Questions for CTET,KVS Exam :16th October 2018(Solutions)
Dear Students!!! There is most general as well as a scoring section in all the competitive entrance examinations in the teaching field i.e “Mathematics”.Because in this section only one thing is work i.e your accuracy and that could be nourished with the daily practice. So, for this, we are providing you the daily quiz for all teaching exams i.e CTET Exam 2018, DSSSB ,KVS,STET Exam.
Q1. Teaching techniques are based on
(a) Projection theory
(b) classical theory of human organisation
(c) cognitive theory of development
(d) modern theory of human organisation
Q2. Exciting examination system provides opportunities for
(a) remedial teaching
(b) enriched instructions
(c) mutual consolation
(d) None of these
Q3. Existing pattern of examination mainly measures the objective of which aspect ?
(a) Cognitive aspect
(b) Affective aspect
(c) Psychomotor aspect
(d) None of these
Q4. The most important characteristics of objective type test is
(a) reliability
(b) discrimination power
(c) objectivity
(d) All of these
Q5. Which of the following is the correct reason to assign homework to the students of upper primary classes ?
(a) To relieve the teacher from teaching some part of syllabus in the class
(b) To make students practice
(c) To ensure that the students do not have too much of leisure time available at home
(d) To deal with the problem of covering all topics and sub topics mentioned in the curriculum
Q6. Arjun solves –7 – 3 = +10. The error is committed as
(a) Arjun has not understood the concept of multiplication of integers
(b) Arjun is careless
(c) Arjun is not clear about the concept of addition of integers
(d) Arjun needs to practice solving problems of similar type
Q7. While solving a problem based on Pythagoras theorem, a teacher draws the following ∆MNO.Pankaj argued that ∆MNO is not drawn correctly. The only way to draw it is Pankaj has the misconception is
(a) he lacks in analytical ability
(b) he has dysgraphia
(c) his teacher must have always drawn the triangle in this particular way
(d) he is weak in geometrical concepts
Q8. Which one of the following functions is not the remedial function of Mathematics teaching?
(a) Determining objectives
(b) Determining content
(c) Arrangement of feedback system
(d) To arrange remedial programme according to diagnosis
Q9. The evaluative work of Mathematics teaching variable is
(a) construction of criterion test
(b) evaluation of behavioural change
(c) Both of the above
(d) None of the above
Q10. In analytic method, we proceed from
(a) concrete to abstract
(b) unknown to known
(c) known to unknown
(d) particular to general
Solutions
S1. Ans.(b)
Sol. Teaching techniques/methods are based on classical theory of human organisation whereas teaching strategies are based on modern theory of human organisation.
S2. Ans.(d)
Sol. Present examination system suffers from a number of imperfections. If focuses only on cognitive learning outcomes and completely ignores the non-cognitive aspect. Hence, we can say that exciting examination system provides opportunities for cognitive learning.
S3. Ans.(a)
Sol. Existing pattern of examination mainly measures the objective of cognitive aspect in which evaluation of a child is done only in one area i.e. academics.
S4. Ans.(d)
Sol. The most important characteristics of objective type test are its reliability, discrimination power and objectivity.
S5. Ans.(b)
Sol. Homework is given to students primarily to inculcate the habit of self study and to practice the lesson taught in school.
S6. Ans.(c)
Sol. The error is committed as Arjun is not clear about the concept of addition of integers.
S7. Ans.(a)
Sol. Pankaj has the misconception that he lacks in analytical ability.
Analytical skill is the ability to visualize, articulate, conceptualise or solve both complex and uncomplicated problems by making decisions that are sensible given the available information.
S8. Ans.(b)
Sol. Determining content is one of the following functions which is not the remedial function of Mathematics teaching.
S9. Ans.(c)
Sol. The evaluative work of Mathematics teaching variable is construction of criterion test and evaluation of behavioural change.
S10. Ans.(b)
Sol. In analytic method, we proceed from unknown to known. This method includes breaking up the unknown problem into simpler parts which can be recombined to find solutions.
You may also like to read: | 1,018 | 4,557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-49 | latest | en | 0.90766 |
https://www.erp-information.com/calculators/profit-margin-calculator | 1,726,888,403,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00707.warc.gz | 698,980,285 | 16,715 | # Profit Margin Calculator Online
Last updated on by Editorial Staff
## How to Use the Profit Margin Calculator?
Our Profit Margin Calculator is designed to be simple and intuitive. Follow these steps to calculate your profit margin:
• Select Your Currency: Choose the currency you are working with from the dropdown menu.
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• Enter Selling Price: Input the selling price of your product. This is how much you sell the product to your customers.
• Calculate: Click the "Calculate" button to see your profit margin displayed as a percentage and an absolute amount.
• Reset if Needed: Use the "Reset" button to clear all fields and perform a new calculation.
## Formula
Profit margin (%) = (Selling Price - Cost Price / Selling Price) * 100
## What is a profit margin?
A profit margin is a financial metric that shows the percentage of sales that have turned into profits. It's calculated by dividing net income by revenue.
## Who Can Use This Calculator?
• Business Owners: Understand the profit margins of products or services to make informed pricing and marketing decisions.
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## FAQs
### Why is knowing your profit margin important?
Knowing your profit margin helps you understand how much profit you make on each sales dollar. It's crucial for making informed business and pricing decisions.
### Can this calculator help with pricing strategies?
Yes, by understanding your profit margins, you can set prices that maximize profitability while remaining competitive in the market.
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Yes. Online retailers and e-commerce businesses can use this calculator to price their products effectively.
## Conclusion
Our Profit Margin Calculator is an essential tool for anyone involved in selling products or services.
It provides clear insights into your profit margins, helping you make informed decisions that drive business success. Whether you're a seasoned business owner or a new entrepreneur, understanding your profit margins is key to long-term profitability and growth.
With this user-friendly tool, calculating profit margins is quick and easy, allowing you to focus on what matters most - growing your business. | 553 | 2,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.88813 |
https://mathoverflow.net/questions/372450/parity-of-the-multiplicative-order-of-2-modulo-p | 1,603,119,669,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107863364.0/warc/CC-MAIN-20201019145901-20201019175901-00572.warc.gz | 408,551,866 | 29,752 | # Parity of the multiplicative order of 2 modulo p
Let $$\operatorname{ord}_p(2)$$ be the order of 2 in the multiplicative group modulo $$p$$. Let $$A$$ be the subset of primes $$p$$ where $$\operatorname{ord}_p(2)$$ is odd, and let $$B$$ be the subset of primes $$p$$ where $$\operatorname{ord}_p(2)$$ is even. Then how large is $$A$$ compared to $$B$$?
• $A/(A+B)$ tends to $7/24$ ? (not proved yet). – Henri Cohen Sep 23 at 21:39
• Seems like an interesting question, and clearly generalizable quite a lot. However, if you're going to ask many questions on this site, it would be a good idea to learn a little bit of TeX formatting. I've fixed the formatting of your question, so if you click on "edit", you'll be able to see what I did to make it more readable. I also changed the title of your question to make it even clearer what you're asking. – Joe Silverman Sep 23 at 21:54
• @HenriCohen how did you determine $A/(A+B)$ to be $7/24$ while also writing "not proved yet"? The proportion of $p \leq 100000$ for which $2 \bmod p$ has odd order is $2797/9591$, which as a continued fraction is $[0,3,2,3,44,9]$, and the truncated continued fraction $[0,3,2,3]$ is $7/24$. I'd be interested to know if you did that or something else. – KConrad Sep 24 at 3:26
• Note: the set $A$ is at OEIS: oeis.org/A014663, while its complement $B$ is oeis.org/A091317. Among the 46 primes below 200, $A$ consists of the 14 primes 7, 23, 31, 47, 71, 73, 79, 89, 103, 127, 151, 167, 191, 199. – YCor Sep 24 at 9:44
• see this answer – René Gy Sep 24 at 15:39
This problem was asked by Sierpinski in 1958 and answered by Hasse in the 1960s.
For each nonzero rational number $$a$$ (take $$a \in \mathbf Z$$ if you wish) and each prime $$\ell$$, let $$S_{a,\ell}$$ be the set of primes $$p$$ not dividing the numerator or denominator of $$a$$ such that $$a \bmod p$$ has multiplicative order divisible by $$\ell$$. When $$a = \pm 1$$, $$S_{a,\ell}$$ is empty except that $$S_{-1,2}$$ is all odd primes. From now on, suppose $$a \not= \pm 1$$.
In Math. Ann. 162 (1965/66), 74–76 (the paper is at https://eudml.org/doc/161322 and on MathSciNet see MR0186653) Hasse treated the case $$\ell \not= 2$$. Let $$e$$ be the largest nonnegative integer such that $$a$$ in $$\mathbf Q$$ is an $$\ell^e$$-th power. (For example, if $$a$$ is squarefree then $$e = 0$$ for every $$\ell$$ not dividing $$a$$.) The density of $$S_{a,\ell}$$ is $$\ell/(\ell^e(\ell^2-1))$$. This is $$\ell/(\ell^2-1)$$ when $$e = 0$$ and $$1/(\ell^2-1)$$ when $$e = 1$$.
In Math. Ann. 166 (1966), 19–23 (the paper is at https://eudml.org/doc/161442 and on MathSciNet see MR0205975) Hasse treated the case $$\ell = 2$$. The general answer in this case is more complicated, as issues involving $$\ell$$-th roots of unity in the ground field (like $$\pm 1$$ in $$\mathbf Q$$ when $$\ell = 2$$) often are. The density of $$S_{a,2}$$ for "typical" $$a$$ is $$1/3$$, such as when $$a \geq 3$$ is squarefree. But $$S_{2,2}$$ has density 17/24, so the set of $$p$$ for which $$2 \bmod p$$ has even order has density $$17/24$$ and the set of $$p$$ for which $$2 \bmod p$$ has odd order has density $$1 - 17/24 = 7/24$$.
For example, there are $$167$$ odd primes up to $$1000$$, $$1228$$ odd primes up to $$10000$$, and $$9591$$ odd primes up to $$100000$$. There are $$117$$ odd primes $$p \leq 1000$$ such that $$2 \bmod p$$ has even order, $$878$$ odd primes $$p \leq 10000$$ such that $$2 \bmod p$$ has even order, and $$6794$$ odd primes $$p \leq 100000$$ such that $$2 \bmod p$$ has even order. The proportion of odd primes up $$1000$$, $$10000$$, and $$100000$$ for which $$2 \bmod p$$ has even order is $$117/167 \approx .700059$$, $$878/1228 \approx .71498$$, and $$6794/9591 \approx .70837$$, while $$17/24 \approx .70833$$.
The math.stackexchange page here treats $$S_{7,2}$$ in some detail and at the end mentions the case of $$S_{2,2}$$.
• Thanks @KConrad for the expression. I was thinking about a combinatorial property of cyclic groups of prime order(called acyclic matching property) and I proved the above mentioned sequence of primes does not hold it. See Proposition 2.3 of core.ac.uk/download/pdf/33123051.pdf Anyway I will like to mention this result in my ongoing research work as a remark, and hence I ask your permission for the same, of course with acknowledgment. – Mohsen Sep 25 at 18:09
• Since the result is due to Hasse, cite his paper when you want to indicate who first showed that the density exists and what its value is. – KConrad Sep 25 at 19:35 | 1,484 | 4,532 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 78, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-45 | longest | en | 0.903905 |
https://www.kaysonseducation.co.in/questions/p-span-sty_1101 | 1,670,428,409,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711200.6/warc/CC-MAIN-20221207153419-20221207183419-00705.warc.gz | 861,944,174 | 12,701 | The normal at an end of a latus rectum of the ellipse passes through an end of the minor axis if : Kaysons Education
# The Normal At An End Of A Latus Rectum Of The Ellipse passes Through An End Of The Minor Axis If
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## Question
### Solution
Correct option is
e4 + e2 = 1
Let an end of a latus rectum be , then the equation of the normal at this end is
If will pass through the end (0, –b) if
Or
Or
#### SIMILAR QUESTIONS
Q1
Consider the two curves C1 : y2 = 4x ; C2 : x2 + y2 – 6x + 1, then
Q2
If F1 = (3, 0), F2 = (–3, 0) and P is any point on the curve 16x2 +25y2 = 400, then PF1 + PF2 equal
Q3
The locus of the points of intersection of the tangents at the extremities of the chords of the ellipse x2 + 2y2 = 6 which touch the ellipse x2 + 4y2 = 4 is
Q4
If an ellipse slides between two perpendicular straight line, then the locus of its center is
Q5
If the tangent at a point on the ellipse meets the auxillary circle in two points, the chords joining them subtends a right angle at the center; then the eccentricity of the ellipse is given by
Q6
The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse. x2 + 9y2 = 9, meets the auxillary circle at the point M, then the area of the triangle with vertices at A, M and the origin is
Q7
The normal at a point P on the ellipse x2 + 4y2 = 16 meets the x-axis at Q, then locus of M intersects the latus rectums of the given ellipse at the points.
Q8
Let a and b be non-zero real numbers. Then the equation (ax2 + by2 + c)(x2 – 5xy + 6y2) = 0 represents
Q9
The pints of intersection of the two ellipse x2 + 2y2 – 6x – 12y + 23 = 0 and 4x2 + 2y2 – 20x – 12y + 35 = 0.
Q10
The tangent at any point P of the hyperbola makes an intercept of length p between the point of contact and the transverse axis of the hyperbola, p1p2 are the lengths of the perpendiculars drawn from the foci on the normal at P, then | 814 | 2,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2022-49 | longest | en | 0.699114 |
https://tutorialhorizon.com/js-algo/negate-a-number-using-addition-operator/ | 1,702,186,517,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101195.85/warc/CC-MAIN-20231210025335-20231210055335-00127.warc.gz | 446,105,759 | 9,570 | Negate a number using addition operator
Problem:
Given a number X, flip it to -X.
O ( X )
Logic :
The negation can be implemented by adding -1, X times. | 42 | 157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-50 | longest | en | 0.737437 |
https://topbinhxqfznm.netlify.app/negus69342ny/formula-for-calculating-annual-percentage-rate-rezy.html | 1,726,502,553,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00593.warc.gz | 530,441,381 | 13,547 | ## Formula for calculating annual percentage rate
This Mortgage APR Calculator takes all of that into account to determine what your APR will be on a home loan. It will also calculate what your monthly payments
Annual percentage rate (APR) is a measure that attempts to calculate what percentage of the principal you’ll pay per period (in this case a year), taking every charge from monthly payments over the course of the loan, upfront fees, etc. into account. APR is the annual rate of interest that is paid on an investment, Annual Percentage Rate (APR) Formula. Even though annual percentage rate (APR) is simple in concept, its calculation might be tricky. Example. Angela, who must choose between two payday loans, each for \$3,000 and 14-days: Loan A Weakness of Annual Percentage Rate. We may quickly conclude that APR Explained: Annual Percentage Rate (APR) is the equivalent interest rate considering all the added costs to a given loan.Naturally, it is a function of the loan amount, the interest rate, the total added cost, and the terms. The APR would equal the interest rate if there is no additional costs to a given loan. How to Calculate an Annual Percentage Growth Rate - Calculating Annual Growth over Multiple Years Get the starting value. Get the final value. Determine the number of years. Calculate the annual growth rate. The term annual percentage rate of charge (APR), corresponding sometimes to a nominal APR and sometimes to an effective APR (EAPR), is the interest rate for a whole year (annualized), rather than just a monthly fee/rate, as applied on a loan, mortgage loan, credit card, etc. It is a finance charge expressed as an annual rate. Now, 2/20 = 0.10, so the APR is 10%. This is a one-year loan at an interest rate of 10% and an APR of 10%. Now suppose you lend me \$20 for a year at 10% interest, but you are also charging me a \$3 fee. And I can pay you the fee at the end of the year. At the end of the year I will owe you 20 + (20 x 10%) + 3 = 20 + 2 + 3 = \$25.
## Free calculator to find out the real APR of a loan, considering all the fees and extra charges. The real APR, or annual percentage rate, considers these costs as well as the To find the actual amount of interest paid, use this formula instead:
6 Jan 2020 Banks use an APR calculation formula to determine how much interest you pay on your outstanding balance. It can be calculated daily or The Annual Percentage Yield APY is accurate as of 3/16/2020. The interest rate and corresponding APY for savings is variable and is set at our discretion. This is a Annual Percentage Rate (APR) is commonly used to calculate the actual interest rates by the lenders. It also signifies the "Actual Interest" borrowers pay in How to calculate credit card APR charges. Understanding how your credit card's Annual Percentage Rate (APR) is calculated and applied to your outstanding Loan APR Calculator. This calculator will help you compute the average combined interest rate you are paying on up to fifteen of your outstanding debts. Calculate the APR (Annual Percentage Rate) of any loan using this calculator by specifying the interest rate, finance charges and the term of the loan.
### Effective Annual Rate Calculator. Below is a screenshot of CFI’s free effective annual rate (EAR) calculator. As you can see in the example above, a nominal interest rate of 8.0% with 12 compounding periods per year equates to an effective annual percentage rate (EAPR) of 8.3%. Download the Free Template
How to Calculate Annual Percentage Rate - Calculating APR for Credit Cards Divide your finance charges by the total balance, then multiply by 1200 to get your APR. Find the current balance on your card using the most recent statement. Find the finance charge on your card using the most recent
### How to Calculate Annual Percentage Rate - Calculating APR for Credit Cards Divide your finance charges by the total balance, then multiply by 1200 to get your APR. Find the current balance on your card using the most recent statement. Find the finance charge on your card using the most recent
For example, let's derive the compound annual growth rate of a company's sales over 10 years: The CAGR of sales for the decade is 5.43%. A more complex situation arises when the measurement period is not in even years. This is a near-certainty when talking about investment returns, compared to annual sales figures. APY can sometimes be called EAPR, effective annual percentage rate, or EAR, effective annual rate. The main difference between these and APR is that the former considers compounded interest while APR doesn't. Because financial institutions want to advertise the most enticing rates possible to their clientele, Calculate the APR (Annual Percentage Rate) of a loan with pre-paid or added finance charges. APR Explained: Annual Percentage Rate (APR) is the equivalent interest rate considering all the added costs to a given loan.Naturally, it is a function of the loan amount, the interest rate, the total added cost, and the terms. The APR would equal the interest rate if there is no additional costs to a given loan. The formula for Effective Annual Rate can be calculated by using the following three steps: Step 1: Firstly, figure out the nominal rate of interest for the given investment Step 2: Next, try to determine the number of compounding periods per year and Step 3: Finally, in the case of Effective Annual Rate Calculator. Below is a screenshot of CFI’s free effective annual rate (EAR) calculator. As you can see in the example above, a nominal interest rate of 8.0% with 12 compounding periods per year equates to an effective annual percentage rate (EAPR) of 8.3%. Download the Free Template Formula Step 1: Calculate the percent change from one period to another using the following formula: Step 2: Calculate the percent growth rate using the following formula:
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This Mortgage APR Calculator takes all of that into account to determine what your APR will be on a home loan. It will also calculate what your monthly payments How to calculate annual percentage yield; Difference between APR and APY; FAQ. APY Calculator is a 12 Feb 2020 How do you calculate the APR using this information? Just add these two steps: Divide the total loan by 100; Multiply the result by the fixed fee for Solving for annual percentage rate(APR). Inputs: interest rate (i). times per year compounded(q) Calculate the annual percentage rate for a loan. *, Loan Amount: \$. *, Interest Rate: Interest Calculation Methodology and Annual Percentage Rate of Charge. Interest calculation methodology; How APRC is calculated in business lending. The If, indeed, you want to calculate the annual percentage rate given the loan amount, the monthly payment and the number of months then you have nearly
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https://users.rust-lang.org/t/need-advice-on-data-layout-for-an-algorithm/86610 | 1,702,094,833,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00028.warc.gz | 639,948,173 | 7,977 | # Need advice on data layout for an algorithm
I have an algorithm that was originally implemented in another language than rust.
This algorithm finds the right order of rendering for isometric objects. I have a goal of implementing this algorithm in rust.
I'll try to be short and not go into too much detail, but basically, in the original implementation, it works like this:
• Every isometric object is a dynamically allocated.
• Every isometric object has some isometric points, which are also dynamically allocated. Object have references to these points, and points have references to their object.
• There is a special container that holds points grouped by their depth. Actually it is a hash map of vectors of points, using depth as key. Every point holds a reference to the vector it is in, and every vector holds a reference to the hash map.
• When an object is added, all of its points are added to the container.
• When an object moves, its points are moved inside a container so that they are in the right place. This is done through the point object and uses references to the vector and to the hash map. Basically removes point from the original vector and adds to the right vector according to the new depth.
• When an object gets removed, all of its points are removed from the container. And here again, references from the point to the vector are very handy.
• To iterate objects in the right order we iterate over all vectors and over all of the points in these vectors. During this iteration, we get objects from points and do some things with them, like storing references to each other if one object is blocked from drawing by another object, but this is mostly outside of the scope of this question.
My previous rust experience is mostly solving Advent of Code puzzles, and this mostly did not involve any complex data structures. At least I've never used `Rc<RefCell<_>>`, or may be used it once and do not remember it. I've read the documentation and I think I understand the concepts.
First of all, I've decided to implement a point container because it is a self-contained thing.
Some pseudo-code, so that I can speak in more detail:
``````struct Point {depth: Depth, owner: PointSubContainer};
struct PointContainer {map: HashMap<Depth, PointSubContainer>}
struct PointSubContainer{depth: Depth, owner: PointContainer}
impl Point {
fn new(depth, container) Point{depth, owner: container.get_or_create_sub_for_depth(depth)}
fn change_depth(depth) {
self.owner.remove(self);
self.depth = depth;
self.owner = self.owner.owner.get_or_create_sub_for_depth(depth);
}
fn dispose() {/*remove from owner*/}
}
``````
Because points are referenced by both objects and containers I had to store them in `Rc`, and because I have to modify them, that became `Rc<RefCell>`. The same thing is with point containers and their sub-containers: they also became `Rc<RefCell>`. I created most of the data structure and a code for adding points. I decided that container will store weak references to points and auto-remove them when point gets dropped. That was not very hard, but I felt like I lose the power of rust because now any reference to anything is actually mutable. But the main problems were when I needed to create an iterator over the container: Return an iterator from struct in RefCell - #7 by romamik.
Here is what I've got at the moment: Rust Playground
It all feels like I'm doing something wrong from the beginning. May I had not to store references to containers in points, but have the `change_depth` method accept the root container as an argument and find actual sub-containers dynamically, that's more expensive than having references to them, but not too much. I will still have to have references to the points from both containers and objects, and that's a problem: I can imagine points being stored and owned by a container, like vector stores it's items for example, and having references to them in objects, but that will make container immutable because there always will be references to it's contents. I can eliminate references from points to objects, and find points dynamically when I need to modify this, but that can be expensive at runtime.
So, what is the rust approach to such algorithms?
I’ve looked at this code, but before I can suggest possible alternative approaches I need to understand the deal with these “objects” you mention in the previous text / description. In particular, in the code I only see one type, “`DepthPoint`”, which I assume to be about points, not objects… is the code simply incomplete? You’re talking about references between points and objects though here:
Assuming the “objects” are not part of the “what I've got at the moment” at all yet, could you elaborate slightly on what references you imagine between points and objects, and what you had in mind when discussing the alternative to “eliminate” some of them with an expensive runtime overhead?
Yes, "Objects" are not part of what I've got at the moment, because I decided to start with PointContainer and it is not ready still.
"Objects" are the main owners of "Points". And they should be easy to find given the object.
The bigger structure will be something like this:
``````struct ObjectContainer {
objects: HashMap<ObjectId, Object>,
points: PointContainer
}
Point {depth,sub_container,object}
Object {points: Vec<Point>}
impl ObjectContainer {
let object = create_object_from_desct(obj_descr); // this will also create points
self.objects[object.id] = object;
}
fn update_object(obj_descr) {
let object = self.object[obj_descr.id];
object.update(obj_descr); // that will modify object points, making them change their position in PointContainer
}
fn iterate() -> Iterator<Object> {
for sub_container in self.points {
for point in sub_container {
yield point.object; // that's an oversimplification but gives an idea
}
}
}
``````
As you can see object has reference to its points and vice versa. I can have only references from objects to points and have points to hold the Ids of the objects instead of the references. That will make me look up objects by Id, when iterating over points. And that will make me find points somehow in the containers when I modify an object.
I would try to avoid `Rc` and `RefCell`.
You don't need points to store pointers into "depth vector" in the hash map -- you can simply look it up in the hash map by the hash map key (depth). Storing pointers seems like a micro-optimization but the benefits aren't clear: look-ups in a hash map are O(1) anyway, and storing pointers requires additional storage and maintenance to keep them in sync. If you really need to optimize the hash map, there are other things you can do, such as changing the hash function etc.
You don't need the "depth vectors" in the "depth hash map" to store pointers back to the hash map. It's unnecessary for hash map values to store pointers to the hash map.
Since depth is an ordered key, you may want to use `BTreeMap` instead of `HashMap`, so that it will keep track of the depth order for you.
Similarly, points shouldn't need to store pointers back to the objects they are in or any Ids of those objects. This relationship is implicit in the object. Instead, I would have your `HashMap` / `BTreeMap` contain some sort of mapping "depth -> (object id, point id)".
I think it is not right stylistically to have methods on `Point` that perform operations on things that the `Point` doesn't own (the object, the hash map, etc). It seems like this is what leads you to a complicated design where a `Point` stores pointers to things that own it. Instead, they should be functions outside of `Point`.
3 Likes
I've tried an approach without Rc and RefCell and it worked very well, the code has reduced in size by a factor of 3 and it was super easy to write. Thanks.
I still have to check the performance, but probably you are right: maintaining all these backlinks plus reference counting in Rc and borrow checking in RefCell will also take time so more map access will not be a problem.
2 Likes
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HOME > > Intersection of Vector Valued Functions
Intersection of Vector Valued Functions
[From: ] [author: ] [Date: 12-06-06] [Hit: ]
the point of intersection is r1(2) = r2(-1) = .I hope this helps!......
Need help finding the intersection of these two vector valued functions: r1(t)= and r2(s)=<4t+6,4t^2,7-t>. Thanks!
-
Given r1(t) = and r2(s) = <4s + 6, 4s^2, 7 - s>:
Equate corresponding entries to obtain
t = 4s + 6
t^2 = 4s^2
t^3 = 7 - s.
The second equation implies that t = 2s or t = -2s.
(i) If t = 2s, then substituting this into the first equation yields
2s = 4s + 6 ==> s = -3 (and t = -6).
Double check in third equation: -216 = 10, which is false (so ignore this!).
-----------
(ii) If t = -2s, then substituting this into the first equation yields
-2s = 4s + 6 ==> s = -1 (and t = 2).
Double check in third equation: 8 = 8, which is true.
So, the point of intersection is r1(2) = r2(-1) = <2, 4, 8>.
I hope this helps!
1
keywords: Vector,Functions,Valued,Intersection,of,Intersection of Vector Valued Functions
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https://math.stackexchange.com/questions/482895/if-two-norms-are-equivalent-on-a-dense-subspace-of-a-normed-space-are-they-equi/482995 | 1,713,791,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818293.64/warc/CC-MAIN-20240422113340-20240422143340-00483.warc.gz | 333,630,917 | 41,936 | # If two norms are equivalent on a dense subspace of a normed space, are they equivalent?
Given a vector space $$V$$ equipped with two norms $$|\cdot|$$ and $$\|\cdot\|$$ which are equivalent on a subspace $$W$$ which is $$\|\cdot\|$$-dense in $$V$$, are the two norms necessarily equivalent?
The statement seems like a relatively straightforward thing to show, but I can't manage it. Having played around with a few proof strategies and not getting anywhere, I'm starting to think that that the statement isn't true, but I'm not yet familiar with many normed spaces and can't think of a counter-example. Any help or hints would be much appreciated.
EDIT: To clear things up, in the book that I took this question from (Linear Analysis by Béla Bollobás) a normed space is defined to be a real or complex vector space, so I think that the intention is for $$V$$ to be over $$\mathbb{R}$$ or $$\mathbb{C}$$.
• Those things are tricky very often. I bet that the statement is false but with unnatural counter-examples. Let me think a bit about it Sep 3, 2013 at 10:34
• This is an interesting similar question which has a negative answer. Sep 3, 2013 at 10:36
• Just to make sure (cf. Mark Bennet's answer), we're dealing with vector spaces over $\mathbb{R}$ or $\mathbb{C}$? Sep 3, 2013 at 11:43
• @DanielFischer having just checked the book that this question came from (Linear Analysis by Bela Bollobas) I'm fairly sure that this is the intention since a normed space is defined to be a vector space over $\mathbb{R}$ or $\mathbb{C}$. Sep 3, 2013 at 11:47
I submit this counterexample which, in my opinion, proves that the statement is false.
In the vein of this MathOverflow post by Gerald Edgar, let $V$ denote the real vector space of all polynomials of one variable and let $$\lVert P\rVert=\max_{x\in[0, 1]} \lvert P(x)\rvert,\qquad \forall P\in V.$$ Moreover, let $$W=\{a_0+a_2x^2+a_4x^4+\dots+a_{2k}x^{2k}\ :\ a_j\in \mathbb{R}\ k\in \mathbb{N}\}.$$ This is a dense subspace of $V$ (cfr. linked post).
Now consider the following linear operator: $$T(x^n)=\begin{cases} x^n & n\ \text{even} \\ nx^n& n\ \text{odd}\end{cases}$$ Its peculiarities are that:
1. $T\equiv I$ (identity) on the dense subspace $W$;
2. $T$ is not bounded.
3. $T$ is $1:1$.
Define $$\lvert P \rvert=\lVert T(P)\rVert.$$ Since $T$ is linear and $1:1$, this defines a norm on $V$. Moreover, this norms agrees with $\lVert\cdot\rVert$ on $W$. Nevertheless, the two norms are not equivalent, because this would imply boundedness of the operator $T$.
• This is interesting because it seems to use no choice at all. But of course, this is because the space $V$ has countable algebraic dimension... By the way, something strange happened: I wanted to upvote your answer, but I was told that this is not possible because I cannot upvote... my own post (!) Sep 3, 2013 at 13:36
• Thanks a lot, this is exactly the kind of thing I was looking for! Sep 3, 2013 at 19:22
$\mathbb Q(\sqrt 2)$ has two norms which are equivalent on $\mathbb Q$ derived from the two embeddings into $\mathbb R$
• Two embeddings? Sep 3, 2013 at 10:50
• @Vishal: You can send $a+b\sqrt 2 \in \mathbb Q(\sqrt 2)$ to $a+b\sqrt 2\in \mathbb R$ or to $a-b\sqrt 2 \in \mathbb R$ Sep 3, 2013 at 10:55
• Oh! I thought you were referring to two embeddings of $\mathbb{Q}$ into $\mathbb{R}$. Sep 3, 2013 at 10:58
• I find this example a bit unsatisfactory because $\mathbb{Q}$ is not a real vector space. I think that the OP intended this when speaking of a "vector space" with "norms". Sep 3, 2013 at 11:01
• @GiuseppeNegro Well I gave an example for the question as asked. And it does show that if there is any chance of proving something like this for a real vector space it will have to involve particular properties of the real numbers in some essential way - you can't just prove it from the properties of norms and vector spaces. Sep 3, 2013 at 11:07
This is not true in general; and in fact this is "never" true.
First, take for example $V=\mathcal C([0,1])$, the space of all continuous functions on $[0,1]$, and let $\Vert\;\Vert=\Vert\,\Vert_\infty$. Let also $W\subset V$ be the space of all polynomial functions on $[0,1]$. Then $W$ is dense in $(V,\Vert\;\Vert)$. Let $E\subset V$ be a linear subspace such that $E\oplus W=V$ (you can find this $E$ with some amount of "choice"). Then $E$ certainly has infinite dimension; so one can find (again with some amount of "choice") a norm $\vert \;\vert_E$ on $E$ which is not equivalent to (the restriction of) $\Vert\;\Vert$ to $E$. Now define the norm $\vert\;\vert$ on $V$ by $\vert e\oplus w\vert=\vert e\vert_E+\Vert w\Vert$ for every $(e,w)\in E\times W$. Then $\vert\;\vert$ are equal on $W$ but not equivalent.
This "construction" works in fact on any (infinite-dimensional) vector space $V$ with any given norm $\Vert\;\Vert$ on $V$. That is, the following holds true: For any infinite-dimensional vector space $V$ and any nor $\Vert\;\Vert$, one can find another norm $\vert \;\vert$ on $V$ such that $\vert\;\vert$ and $\Vert\; \Vert$ coincide on a dense (with respect to $\Vert\;\Vert$) linear subspace $W\subset V$ and yet are not equivalent.
To prove this, all what you need is to find a dense linear subspace $W\subset V$ such that the quotient space $V/W$ is infinite-dimensional; equivalently, for which you can write $V=E\oplus W$ with $\dim E=\infty$.
Now, it is "well known" that you can find a dense set $D\subset W$ (with respect to $\Vert\;\Vert$) which is also linearly independent; se e.g. here: Does there exist a linearly independent and dense subset? Take any sequence $(x_n)\subset D$ such that $\Vert x_n\Vert\to 0$. Then $D\setminus\{ x_n;\; n\in\mathbb N\}$ is again dense in $V$ (wrt $\Vert\;\Vert$). (Indeed, since the set $A=\{ 0\}\cup\{ x_n;\; n\in\mathbb N\}$ is countable, it has empty interior wrt $\Vert\;\Vert$ because nonempty open sets in a normed vector space are uncountable, and moreover $A$ is $\Vert\;\Vert$-closed in $V$; so $V\setminus A$ is dense and open in $V$ wrt $\Vert\;\Vert$, and hence $D\setminus A$ is also $\Vert\;\Vert$-dense in $V$). If you denote by $W$ the linear span of $D\setminus\{ x_n;\; n\in\mathbb N\}$ and by $E$ the linear span of $\{ x_n;\; n\in\mathbb N\}$, then everything is OK.
Note also that the "construction" does not contradict Giuseppe's statement since there is no reason for $\vert\;\vert$ to be complete; but perhaps it could be suitably modified?
• This is a nice post. Actually I think that both the example I have written above and the one by Mark Bennett are, roughly speaking, versions of this general construction that you present here. Sep 3, 2013 at 16:42
• For your example, I completely agree. For the one by Mark Bennet, I cannot say (I would need more details...) Sep 3, 2013 at 17:11
$V$ is complete with respect to $\lvert \cdot \rvert$ (recall that $W$ is dense with respect to the other norm $\lVert\cdot\rVert$).
Proof. By assumption, constants $C_1, C_2$ exist such that the following inequalities hold: $$\tag{1} \begin{array}{ccc} \lvert y\rvert\le C_1\lVert y\rVert, & \lVert y\rVert\le C_2\lvert y\rvert & \forall y\in W. \end{array}$$ So take $x\in V$. By assumption, we can find a sequence $y_n\in W$ converging to $x$ with respect to $\lVert\cdot\lVert$. Because of the first inequality in (1), $y_n$ is Cauchy with respect to both norms and so, using completeness, we see that a vector $y\in V$ exists such that $$\lvert y_n-y\rvert\to 0.$$ Actually, $x=y$ because $$\lvert x-y\rvert\le \lvert x-y_n\rvert+\lvert y_n-y\rvert\le C_1\lVert x-y_n\rVert + \lvert y_n-y\rvert,$$ and the right hand side of this inequality tends to $0$ as $n\to \infty$. Therefore, by letting $n\to \infty$ in the inequalities $$\lvert y_n\rvert\le C_1\lVert y_n\rVert, \qquad \lVert y_n\rVert\le C_2\lvert y_n\rvert$$ we see that the two norms are equivalent on the whole of $V$.
• I had thought about this approach, but I'm not sure that the result $x=y$ follows directly, since if $x\not\in W$ there is no guarantee that $x-y_n \in W$ and so I don't think we can immediately apply $|x-y_n|\leq C_1 ||x-y_n||$... Sep 3, 2013 at 11:41 | 2,493 | 8,140 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-18 | latest | en | 0.928308 |
https://www.statisticshowto.com/polychotomous-variable | 1,590,479,251,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390448.11/warc/CC-MAIN-20200526050333-20200526080333-00019.warc.gz | 816,038,396 | 17,212 | # Polychotomous Variable Definition
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Types of Variables > Polychotomous Variable
## What is a Polychotomous Variable?
A frequency chart showing multiple (polychotomous) variables.
A polychotomous variable is a variable that can have more than two values (a variable with exactly two values is called a binary variable). Polychotomous variables can be ordered, unordered, or sequential:
• Ordered polychotomous variables: variables that have some kind of order, like: “1” if you earn up to \$25,000, “2” if you earn \$25,001-\$50,000 and “3” if you earn over \$50,000.
• Unordered polychotomous variables: variables that don’t have an implied order, like: “1” for male, “2” for female “3” for trans gendered male and “4” for trans gendered female.
• Sequential polychotomous variables: variables with a sequence. For example: “1” for freshmen, “2” for sophomore, “3” for junior and “4” for senior.
Polychotomous variables are usually qualitative variables, but they can be quantitative variables as well. For example, if studying birth weight of children, you could have the categories of heavy smoker/smoker/light smoker or non-smoker. But it may be more useful to code the “number of cigarettes smoked per day during pregnancy” into categories:
1. 0 cigarettes per day.
2. up to 5 cigarettes per day.
3. Between 6 and 20 cigarettes per day.
4. Over 20 cigarettes per day.
## Polychotomous Independent Variable
If your independent variable has more than two categories, then it’s a polychotomous independent variable. For example: race, country of origin, or school attended are all examples of independent variables that could have more than three categories.
## Dichotomous vs. Polychotomous Variables
A dichotomous variable is the same as a binary variable, i.e. it has two possible values. So a dichotomous variable would have two values, a polychotomous variable would have more than two. In statistics, the term binary variable is preferred. The term dichotomous variable is used in psychometrics.
## Polytomous vs. Polychomotous
The two words mean the same thing. There’s some division in statistics about whether the term polytomous or polychotomous should be used. This Penn State professor states that “The word [polychotomous] does not exist!”, although a quick check in the Mirriam-Webster dictionary says that it does: “dividing or marked by division into many parts, branches, or classes”.
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Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free! | 640 | 2,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-24 | latest | en | 0.889046 |
http://math.wallawalla.edu/academics/teacherEd/mathematics/standard04.phtml | 1,632,339,457,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057388.12/warc/CC-MAIN-20210922193630-20210922223630-00666.warc.gz | 44,068,029 | 3,205 | # Washingon State Teacher Certification: Mathematics (5-12)
## Standard 4: Geometry and Measure
Candidates prove and understand geometric theorems and transformations as they apply to congruence, similarity, lines, triangles, trigonometry, circles, and measurements. In addition, the following content should be included; the relationship between geometryc figures and the Cartesian coordinate system.
4.A:
Understand congruence in terms of rigid motions and prove geometric theorems
4.B:
Apply transformations and use similarity and congruence in mathematical situations
4.C:
Perform geometric constructions physically and/or with technology
4.D:
Define trigonometric ratios and solve problems involving right triangles and general triangles
4.E:
Derive the Pythagorean Theorem and apply it to problem solving situations
4.F:
Identify and describe relationships among angles, radii, and chords
4.G:
Derive formulas for arc lengths and areas of sectors of circles
4.H:
Translate between the geometric description and the equation for a conic section
4.I:
Use coordinates to prove geometric theorems algebraically
4.J:
Derive area, surface area, and volume formulas and use them to solve problems
4.K:
Solve real life and mathematical problems involving angle measures and/or polygons
4.L:
Visualize and describe two-dimensional figures and three-dimensional objects as well as the relationships among them
4.M:
Apply geometric concepts to model real world situations
The B.S. degree in Mathematics, with a concentration in Secondary Education, covers these standards in the courses indicated in the table below.
4.A 4.B 4.C 4.D 4.E 4.F 4.G 4.H 4.I 4.J 4.K 4.L 4.M Precalculus Mathematics I (MATH 121)† 1 1 Precalculus Mathematics II (MATH 122)† 1 3 1 1 1 1 2 2 Calculus Life Sciences I (MATH 131), orCalculus I (MATH 181) 2 2 2 3 Data Analysis (MATH 215) Discrete Mathematics (MATH 250) Calculus II (MATH 281) 2 2 2 2 2 3 Calculus III (MATH 282) 2 2 2 3 2 2 3 3 Calculus IV (MATH 283) 2 3 3 2 2 3 Introduction to Linear Algebra (MATH 289) 1 1 1 1 Differential Equations (MATH 312) 2 3 3 Probability and Statistics (MATH 315) Survey of Geometries (MATH 321) 3 3 3 2 3 3 3 3 2 Junior Seminar I (MATH 396) Junior Seminar II (MATH 397) Real Analysis I (MATH 451) 1 Abstract Algebra I (MATH 461) Senior Seminar I (MATH 496) Senior Seminar II (MATH 497) †While not required for the major, these classes are prerequisites for MATH 181 | 665 | 2,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-39 | latest | en | 0.799707 |
http://mathoverflow.net/questions/80463/unreduced-suspension-isomorphism/80465 | 1,469,329,436,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823935.18/warc/CC-MAIN-20160723071023-00204-ip-10-185-27-174.ec2.internal.warc.gz | 163,141,563 | 16,598 | # Unreduced Suspension Isomorphism
Tending to a lecture on homotopy theory, the following question occured to me (is that a correct sentence?):
Given a pointed space $(X,x)$, is the UNREDUCED suspension map $S:\pi_k(X,x) \rightarrow \pi_{k+1}(SX, \ast)$ a group homomorphism?
Here unreduced suspension refers to $SX = X \times D^1 / \sim$, where $\sim$ collapses $X \times \{1\}$ and $X \times \{-1\}$ respectively, and the basepoint $\ast$ is (the one point set with element) $(x,0)$.
The statement is contained in every book on homotopy theory and almost trivial for the REDUCED suspension. For wellpointed spaces this of course surfices to answer my question, but it has resisted several similar 'general nonsense'-arguments in the general case.
-
## 2 Answers
You can reduce to the well pointed case by observing that every space $X$ admits a weak equivalence $X'\to X$ from a well pointed space. Now the composition $$\pi_kX'\to \pi_kX\to \pi_{k+1}SX$$ (composition of your map with an isomorphism) is the same as the composition $$\pi_kX'\to \pi_{k+1}SX'\to \pi_{k+1}SX$$ in which the first map is a homomorphism because $X'$ is well pointed and the second is a homomorphism because it is induced by a map of spaces
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oooh embarassing. Yupp, that does it. You can even find a pointed map that is a (non-pointed) homotopy equivalence, right? $\left(X \times {0} \cup {x} \times I, (x,1)\right) \rightarrow (X,x)$ the projection, correct? – old account Nov 9 '11 at 15:59
That is correct. – Tom Goodwillie Nov 9 '11 at 17:10
On the linguistic question: I would say "While attending a lecture on homotopy theory, the following question occurred to me."
On the mathematical question: I think you need only prove that $S(u+v)=S(u)+S(v)$ when $(X,u,v)$ is the universal example of a based space with two based maps $u,v:S^k\to X$, ie $(X,u,v)=(S^k\vee S^k,i_1,i_2)$. Here $X$ is well-pointed so there is no problem.
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Again on the linguistic part: I'm not attending the lecture, I'm the guy in charge of the exercise session and so on, but I don't know the english word for that, in german (I'm in münster, with the lecture given by christian ausoni) it's Übungsleiter. On the mathematics: oh right, I tried it for the universal example but used a single sphere, not a wedge. Stupid, not realizing there are two inputs... – old account Nov 9 '11 at 13:32
damn, for the argument to go through i seem to need $S$ to be coproduct preserving, which it is not. could you maybe elaborate on that answer? – old account Nov 9 '11 at 14:06
I think Neil's argument is this: consider the universal sum $f: S^k\to S^k \vee S^k$, and think about $Sf: S(S^k)\to S(S^k\vee S^k)$. Using a homeomorphism $S(S^k)=S^{k+1}$, think of this as an element of $\pi_{k+1}S(S^k\vee S^k)$; you want to show this is the same as the element $S(i_1)+S(i_2)$, i.e., that two maps into $S(S^k\vee S^k)$ are homotopic. Use the homotopy equivalence $S(S^k\vee S^k)\to S^{k+1}\vee S^{k+1}$. – Charles Rezk Nov 9 '11 at 15:52
yes, but it is the reduction to the universal case, that i did not fully understand, when going through it this afternoon. For the reduction step on needs a map SX∨SX→S(X∨X) and this map will not be a homotopyequivalence in general. I overlooked however that the argument does not need it. So by now I'm quite convinced this works as well. But i do not know how to accept multiple answers. – old account Nov 9 '11 at 17:38
@Fabian: Maybe the word you are looking for is teaching assistant: en.wikipedia.org/wiki/Teaching_assistant . – Rasmus Bentmann Jan 5 '12 at 18:55 | 1,069 | 3,570 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2016-30 | latest | en | 0.909261 |
http://www.numbersaplenty.com/4531 | 1,566,271,201,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315222.14/warc/CC-MAIN-20190820024110-20190820050110-00114.warc.gz | 302,556,719 | 3,489 | Search a number
4531 = 23197
BaseRepresentation
bin1000110110011
320012211
41012303
5121111
632551
716132
oct10663
96184
104531
11344a
122757
1320a7
141919
151521
hex11b3
4531 has 4 divisors (see below), whose sum is σ = 4752. Its totient is φ = 4312.
The previous prime is 4523. The next prime is 4547. The reversal of 4531 is 1354.
Adding to 4531 its reverse (1354), we get a palindrome (5885).
It can be divided in two parts, 453 and 1, that added together give a palindrome (454).
It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 1354 = 2677.
It is a cyclic number.
It is not a de Polignac number, because 4531 - 23 = 4523 is a prime.
It is a super-2 number, since 2×45312 = 41059922, which contains 22 as substring.
It is an Ulam number.
It is a Duffinian number.
4531 is an undulating number in base 14.
It is a plaindrome in base 11.
It is not an unprimeable number, because it can be changed into a prime (4561) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 76 + ... + 121.
It is an arithmetic number, because the mean of its divisors is an integer number (1188).
4531 is a deficient number, since it is larger than the sum of its proper divisors (221).
4531 is a wasteful number, since it uses less digits than its factorization.
4531 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 220.
The product of its digits is 60, while the sum is 13.
The square root of 4531 is about 67.3127031102. The cubic root of 4531 is about 16.5474605378.
The spelling of 4531 in words is "four thousand, five hundred thirty-one".
Divisors: 1 23 197 4531 | 563 | 1,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2019-35 | latest | en | 0.918272 |
https://www.easycalculation.com/analytical/learn-subwoofer-tuning-frequency.php | 1,580,232,928,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251779833.86/warc/CC-MAIN-20200128153713-20200128183713-00499.warc.gz | 850,978,653 | 6,103 | English
# Learn How to Calculate Subwoofer Port Tuning Frequency - Tutorial
## Calculate Subwoofer Port Tuning Frequency (F)
#### Definition
Tuning of a subwoofer vent tuning box is established mainly by the grouping of port area, port length, and net volume of the subwoofer box.
#### Formula
F = (23562.5 x D2 x N) / (V x (L + KD))
###### Where,
D = vent diameter N = number of vents V = volume of box L = vent length K = end correction factor
##### Example
A subwoofer contains 2 vents of length 5 cm and diameter of 8 cm. The volume of box is 200 cm3 and it produces end correction factor of 4. Calculate the tuning frequency (F) for the subwoofer box?
##### Given,
Vent diameter (D) = 8 cm Vent length (L) = 5cm End correction factor (K) = 4 Volume of box (V) = 200 Number of vents (N) = 2
##### To Find,
Tuning Frequency (F)
##### Solution
Tuning Frequency (F)= (23562.5 x D2 x N) / (V x (L + KD)) = (23562.5 x 82 x 2) / (200 x 0.001 x (5 + (4 x 8))) = (23562.5 x 64 x 2) / (200 x 0.001 x (5 + 32)) = 3016000 / 7.4 = 407567.568 = 638.41 Hz | 343 | 1,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2020-05 | longest | en | 0.758411 |
http://www.tutorsglobe.com/homework-help/managerial-economics/theory-of-supply-74041.aspx | 1,542,402,858,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743184.39/warc/CC-MAIN-20181116194306-20181116215417-00038.warc.gz | 526,162,515 | 11,488 | #### Theory of Supply - Concepts of Supply Curve
Theory of Supply:
Chapter Summary:
A small seller, that can’t affect the market price, maximizes gain by producing at a rate where its marginal cost equivalents the price. (For a small seller, the price equivalents marginal revenue.) In short run, at least one input is fixed and thus can’t be adjusted. The business breaks even whenever total revenue covers variable cost. In long run, the business can adjust all inputs and depart or enter the industry. It breaks even whenever total revenue covers total cost.
The supply curve exhibits the quantity supplied as a function of price, other things equivalent. The consequence of a change in price is symbolized by a movement all along the supply curve to a new quantity. Modifications in other factors like wages and the prices of other inputs are symbolized by shifts of the whole supply curve.
Seller surplus is the difference among revenue from certain production rate and the minimum amount needed to induce the seller to generate that quantity. Elasticity of supply, measure the responsiveness of supply to modifications in underlying factors which affect supply.
The Math Supplement to this illustrates two things mathematically. First, it exhibits that the profit-maximizing (or loss minimizing) production rate exists where marginal revenue equivalents marginal cost. Second, it exhibits that a change in any factor of supply other than the price of item is symbolized by a shift of the whole market supply curve.
Key Concepts:
General Chapter Objectives:
A) Elucidate why short run costs differ with the production rate and what could cause a shift in the short run cost curves.
B) Find out the profit-maximizing (or loss-minimizing) production level in short run.
C) Elucidate beneath what short-run conditions a business will remain in operation and whenever it is best to shut down.
D) Describe why individual seller’s supply curve is that part of its marginal cost curve above minimum average variable cost.
E) Describe why demand for inputs is derived from level of production.
F) Describe why costs differ with the production rate and what could cause a shift in the cost curves.
G) Find out the profit-maximizing (or loss-minimizing) production level in the long run and elucidate the breakeven situation in the long run.
H) Explain how the market supply curve is derived in short and long run and their properties.
I) Explain seller surplus in words and find out it graphically.
J) Elucidate why labor supply curves usually slope upward in terms of work-leisure tradeoff and the income consequence.
K) Explain the elasticity of supply concept, compute price elasticity of supply coefficients and deduce them.
Notes:
1) Individual supply: long run and short run.
a) Whether to carry on in production and how much to generate based on costs and revenues.
b) Time horizon:
i) Short run: time horizon in which a seller can’t adjust at least one input. In short run, the business should work in the constraints of past commitments like employment contracts and investment in equipment and facilities.
ii) Long run: time horizon long adequate for a seller to adjust all inputs (comprising the size of its facility) and production.
2) Costs and production rate: short run.
A) Costs:
i) Based on estimates of expenditures on wages, rent and other supplies required at different production rates.
ii) Fixed cost = cost of inputs which do not change with the production rate. The whole fixed cost is a sunk cost, that is, a cost which has been committed and can’t be avoided.
iii) Variable cost = cost of inputs which change with the production rate. Therefore, variable cost differs with the scale of operations. This is an avoidable cost.
iv) Total cost = sum of fixed cost and variable cost (that is, C = F + V).
(Note: Various expenses might have both a fixed and variable component.)
v) Average fixed cost = fixed cost divided by output level.
vi) Average variable cost = variable cost divided by output level.
vii) Average (total) cost/unit cost = the total sum of average variable and average fixed cost; as well equivalents total cost divided by production rate;
viii) Marginal cost = the change in total cost (that is, variable cost) related with each additional unit generated; symbolized graphically as the slope of total cost curve.
B) Marginal product:
i) Definition: The increase in output occurring from an extra unit of the input.
ii) Reducing marginal product from the variable inputs: the marginal product becomes smaller with each and every raise in the quantity of variable input.
C) Average total cost (or average cost) usually first drops with raise in the production rate and then increases.
i) Whenever the production rate is greater, the fixed costs will be spread over more units, the average fixed cost will be diminishing.
ii) Whenever the production rate is higher:
• At first, the average variable cost drops (since the variable inputs match the fixed input comparatively better);
• Then, the average variable cost increases (as more of the variable inputs are added up in the combination with fixed input, there will be a mismatch, leading to a reducing marginal product from the variable inputs).
• Accordingly, the average variable, average (or total), and marginal cost curves are U-shaped. The curves will modify with the technology being employed.
3) Revenue and production rate: short run.
a) Total revenue = price x sales.
b) Marginal revenue = the modification in total revenue occurring from selling an additional unit; symbolized graphically as the slope of total revenue line.
4) Individual supply: short run.
a) Suppositions:
i) Business goals to maximize gain.
ii) The business is too small relative to the market which it can sell as much as it would like at going market price.
b) Profit-maximizing (or loss-minimizing) quantity to generate.
i) If continuing in production, profit = total revenue – total cost; that is, Profit = R – F – V.
ii) Profit-maximizing production rate is where marginal revenue equivalents marginal cost (where the total revenue line and total cost curve climb at exactly similar rate).
• If the marginal revenue surpasses the marginal cost, profit will be increased by increasing the production.
• Whenever marginal revenue is less than the marginal cost, profit will be increased by decreasing production.
iii) Whenever a business can sell as much as it would like at the market price, marginal revenue equivalents price (that is, it does not have to decrease price to sell more units); and thus, the profit-maximizing rule for such a business becomes the production rate at which price equivalents marginal cost.
c) To decide whether to carry on production at all.
i) If the business shuts down, it should pay the fixed cost, although not the variable cost; profit = -F.
ii) A business must continue in production:
• If the maximum gain from continuing in production is at least as big as the gain from shutting down; that is, if the business breaks even.
• Break even condition: too long as total revenue covers variable cost; or equivalently, too long as average revenue (or price) covers the average variable cost.
• Sunk costs must be ignored in making a present decision.
d) Individual supply curve – short run.
i) The graph exhibiting the quantity which one seller will supply at every possible price. It exhibits the minimum price which the seller will accept for each and every unit of production.
ii) It is similar to that part of marginal cost curve above the minimum point of the average variable cost curve.
iii) For each and every possible price of its output, a business must generate at the rate which balances its marginal cost with the price, given that the price covers the average variable cost.
iv) Since marginal cost increases when production is expanded, a seller must expand production only when it receives a higher price.
v) A change in the price of output usually causes movement all along a supply curve.
e) Demand for inputs:
i) Since the costs of inputs modifications, the marginal cost curve shifts up or down, the profit-maximizing scale of production modifications, the demand of input as well changes.
ii) By varying the cost of a specific input, we can find out the quantity demanded of that input at each and every possible cost and build the seller’s demand for that input.
iii) Since the quantity demanded of the input will be greater at a lower input price, the demand curve will slope downward.
5) Individual supply: long run.
a) Costs:
i) Based on estimates of expenditures on wages, rent and other supplies required at different production rates whenever all inputs are avoidable.
ii) The business might incur certain costs even at production level of zero, example: maintenance of minimum size facility.
iii) Long run average cost curve is lower and consists of a gentler slope than that of short run. In long run, the seller has more flexibility in adjusting inputs to modifications in the production rate, and can therefore produce at a lower cost. In short run, it might not be able to modify one or more inputs.
b) Profit-maximizing quantity to generate. The profit-maximizing rule is to generate where price equivalents long run marginal cost (essentially similar in the long run as for short run).
c) To decide whether to carry on production at all.
i) If business shuts down, it will acquire no costs (that is, as all costs are avoidable in the long run); then profit = 0.
ii) A business must continue in production:
• When the maximum profit from continuing in production is at least as big as the gain from shutting down; that is, when the business breaks even.
• Break even condition: too long as its total revenue covers total cost; or equivalently, too long as average revenue (or price) covers average total cost.
d) Individual supply curve: long run.
i) The long run individual supply curve is similar to that portion of the long run marginal cost curve above the minimum point of the long run average total cost.
6) Market supply: short run.
a) The market supply curve:
i) This is a graph exhibiting the quantity which the market will supply at every possible price of the output.
ii) The consequence of a change in the price of an output is symbolized by a movement all along the supply curve.
b) The market supply curve is a horizontal summation of different individual seller’s supply curves. The market supply curve starts with the seller which has the lowest average variable cost, then blends in sellers with the higher average variable cost.
c) Market supply curve slopes upward. The higher the price of output, each individual seller will wish to generate more, the market as an entire will too produce more.
d) A change in the price of an input will influence an individual seller’s marginal cost at all production levels and shift the whole marginal cost curve. These changes will too shift the market supply curve.
7) Market supply: long run.
a) The market supply curve.
i) This is a graph exhibiting the quantity which the market will supply at every possible price of output.
ii) The consequence of a change in the price of an output is symbolized by a movement all along the supply curve.
b) The long-run market supply curve might be flat. Quantity supplied can expand via replication of existing businesses. The cost of production supplied by new entrants must be similar as that of the existing businesses.
c) The long-run market supply curve might slope upward. The resources accessible to different suppliers might vary in quality. New entrants might not be capable to replicate the resources (particularly in resource-based industries and where place is important) of existing suppliers and will acquire higher costs.
d) The long-run market supply curve slopes more tenderly upward (that is, more elastic) than the short-run market supply curve.
i) The freedom of entry and exit is a key difference among the short and long run: In the long run, every business consists of complete flexibility in deciding on inputs and production.
ii) In long run, when there is a modification in market price, the quantity supplied will adjust in two manners:
• All existing sellers will adjust their quantities supplied all along their individual supply curves; and
• Some sellers might leave or enter the market.
iii) Sellers whose total revenue can’t cover total costs will leave the industry till all remaining sellers break even.
iv) The industry where businesses are gainful (that is, total revenue surpasses total costs) will attract new entrants. This will raise market supply, decrease market price (that is, push down the gain of all sellers):
• Existing sellers will adjust the quantities supplied all along their individual supply curves;
• Some sellers might enter or leave the market till all sellers just break even.
e) A change in price of an input and other factors, than the price of output will cause a shift in the whole market supply curve.
8) Individual seller surplus:
a) A seller’s profit differs with the price of its output.
b) Seller’s surplus = difference between a seller’s revenue from certain production rate and the minimum amount essential to persuade the seller to generate that quantity.
i) It is equivalent to the difference between price and marginal cost, for each and every quantity from zero up to the production level.
ii) In short run, it is equivalent to the difference between total revenue from various production rate and the variable cost, or R – V.
iii) In long run, it is equivalent to the difference among total revenue from certain production rate and total cost.
c) A buyer can remove the seller’s surplus. The buyer must design a bulk order and pay the seller the minimum amount to persuade production at the desired level, that is, the buyer must purchase up the seller’s marginal cost curve.
d) A seller’s profit differs relatively more than the price of its output.
i) A change in price influences the revenue at the initial production rate; and
ii) Induces the seller to amend its production level.
9) Market seller surplus:
a) Market seller surplus = difference between seller’s revenue from certain production rate and the minimum amount essential to induce the seller to generate that quantity.
i) It is equivalent to the sum of individual seller surpluses; and
ii) It is symbolized graphically by the region between the price line and the market supply curve (that is, in both short run and long run).
b) In the long run:
i) If the market supply curve is flat, then there will be no market seller excess;
ii) If the market supply curves slopes upward, then there will be some market seller surplus. Seller surplus ensue to those who own the superior resources.
10) Supply of labor (that is, as an output, as opposed to being an input).
A) The sellers are individual persons.
B) Individual labor supply curve.
i) It exhibits the quantity of labor which the person will supply at each and every possible wage.
ii) Slope of an individual labor supply curve: work-leisure tradeoff.
a) Leisure is a consumer good.
• A person’s time is restricted to 168 hrs per day.
• Increasing marginal cost of labor: the marginal cost of labor starts from a low level for 0 hrs to labor and increases with raising hrs of labor.
• Reducing marginal profit of leisure time: owing to reducing marginal profit, the marginal profit of leisure begins high and declines with rising leisure time.
b) Rising marginal cost of labor: causes the labor supply curve to slope upward. To maximize total benefit, a person selects the quantity of labor where the marginal cost of labor equivalents the wage. Since the marginal cost of labor rises with the quantity labor, when the wage is higher, a bigger quantity of labor will be supplied.
c) The income effect: it causes the labor supply curve to slope downward. For any specified quantity of labor, income will be higher with a greater wage. The higher income will increase the marginal profit from leisure and the marginal cost of work, decreasing the quantity of labor supplied.
d) The total effect on the supply curve based on the balance of the two consequences.
C) Market supply curve of labor: this is the horizontal summary of all individual supply curves of labor.
11) Price elasticity of supply:
a) Elasticity of supply: the responsiveness of supply to modification in an underlying factor (like the price of the item and inputs). There is an elasticity equivalent to every factor which affects supply.
b) Price elasticity of supply:
i) This is the percentage by which, the quantity supplied will modify if the price of the item increases by 1percent; other things equivalent.
ii) Computed by employing the arc or point approach and interpreting the resultant number.
iii) A pure number which does not based on any units of measure.
• Usually the supply curves of businesses selling goods and services slope upward, therefore the price elasticity of supply is a positive number varying from 0 to infinity.
• When the price elasticity is less than 1, then supply is inelastic.
• When the price elasticity is greater than 1, then supply is elastic.
iv) It might vary all along a supply curve.
v) A change in any of the factors, influence supply might also affect the price elasticity.
c) Intuitive factors:
i) Capacity utilization influences individual and market supply elasticity, example: if capacity is tight, the seller might not raise production much even if the price increases substantially, and supply will be comparatively inelastic.
ii) Time beneath consideration. In short run, some inputs might be costly or not possible to modify, the marginal cost of production will be steep and supply inelastic. In long run, the marginal cost of production will slope more tenderly, and an individual and market supply curve will be much elastic.
12) Forecasting quantity supplied. A change in price will influence revenue:
a) Directly; and
b) Via changing the sales or quantity supplied (that is, employing the price elasticity).
Question-Answer:
Mercury Ltd was finalizing plans for a brick factory. Variable and fixed costs would have been similar to those of Jol Inc.'s existing brick factory. The only main difference between the two companies was that Jol had already sunk the fixed cost of its plant, whereas Mercury had not. Then, an industry analyst predicts that, due to expansion of supply, the long-run price of bricks would drop by 20 percent.
Mercury suspended all investment plans. By disparity, Jol announced that it would carry on production. Mercury and Jol are too small relative to the market that each can sell as much as it would like at market price
a) Describe the short run break-even condition.
b) Elucidate the long run break-even condition.
c) Assume that the new long-run price is less than Mercury’s average cost however higher than Jol's average variable cost. Describe why the two companies came to distinct decisions.
Answer:
a) In short run, the business must continue production too long as its revenue covers variable costs. Or, alternatively, too long as price covers the average variable cost.
b) In long run, the business must continue in production so long as its net revenue covers total cost. Or, alternatively, too long as price covers average cost.
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Instructional Focus DocumentMathematical Models with Applications
TITLE : Unit 06: Management of Personal Finances SUGGESTED DURATION : 14 days
#### Unit Overview
Introduction
This unit bundles student expectations that address personal finance and budgeting, including earnings, banking services and accounts, taxes, and credit options. Concepts are incorporated into both mathematical and real-world problem situations. According to the Texas Education Agency, mathematical process standards including application, tools and techniques, communication, representations, relationships, and justifications should be integrated (when applicable) with content knowledge and skills so that students are prepared to use mathematics in everyday life, society, and the workplace.
Prior to this Unit
In Grade 6, students compared the features and costs of a checking account and a debit card offered by different local financial institutions. They distinguished between debit cards and credit cards and explained the importance of a positive credit history. In Grade 7, students were introduced to the components of a personal budget, including income; planned savings, taxes, and expenses. Students also calculated and compared simple interest and compound interest. In Grade 8, students analyzed situations to determine if they represented financially responsible decisions and identified the benefits of financial responsibility and the costs of financial irresponsibility. Students calculated the total cost of repaying a loan, including credit cards and easy access loans, under various rates of interest over different periods of time. Students also contrasted proportional y = kx and non-proportional y = mx + b linear relationships. In Algebra I, students studied linear functions and were introduced to exponential functions.
During this Unit
Students use linear equations and functions to describe proportional and non-proportional linear relationships and apply them to represent real-world problems involving earnings and budgeting. Students explore methods employees are compensated (e.g., hourly pay, base salary plus commissions, straight commission, salary, etc.) and compare the methods of compensations using various representations. Students use IRS Tax tables to determine personal income taxes and amounts to be deducted from regular pay checks. Students calculate payroll deductions including taxes, pensions, and additional employee benefits, to determine net pay. Students analyze their personal budget based on their net earnings. Students explore types of bank services, including checking and savings options, overdraft protection policies, fees and charges, online banking options, and ATM and card availability and policies, etc. Students analyze banking options based on budgetary and banking needs. Students explore and study models of credit options in retail purchasing and compare relative advantages and disadvantages of each option. Students explain option choices based on income and spending scenarios. Throughout the unit, students develop an economic way of thinking, encouraging financial responsibility for both short-term and long-term goals.
Other considerations: Reference the Mathematics COVID-19 Gap Implementation Tool HS MMA
After this Unit
Students will connect the financial literacy concepts to analyze various loan amortizations, investments for future planning, and insurance options, including life insurance, homeowner’s insurance, and auto insurance.
This unit is supporting the development of the Texas College and Career Readiness Standards (TxCCRS): I. Numeric Reasoning B1; II. Algebraic Reasoning D1, D2; V. Statistical Reasoning A1, B4, C2; VII. Problem Solving and Reasoning A1, A2, A3, A4, A5, B1, C1, C2, D1, D2; VIII. Communication and Representation A1, A2, A3, B1, B2, C1, C2, C3; IX. Connections A1, A2, B1, B2, B3.
Research
According to the Connections Standard for Grades 9-12 from the National Council of Teachers of Mathematics (NCTM), “Instructional programs from pre-kindergarten through grade 12 should enable students to:
• recognize and use connections among mathematical ideas;
• understand how mathematical ideas interconnect and build on one another to produce a coherent whole; and
• recognize and apply mathematics in contexts outside of mathematics.
When students can see the connections across different mathematical content areas, they develop a view of mathematics as an integrated whole. As they build on their previous mathematical understandings while learning new concepts, students become increasingly aware of the connections among various mathematical topics. As students' knowledge of mathematics, their ability to use a wide range of mathematical representations, and their access to sophisticated technology and software increase, the connections they make with other academic disciplines, especially the sciences and social sciences, give them greater mathematical power” (NCTM, 2000, p. 354).
National Council of Teachers of Mathematics. (2000). Principles and standards for school mathematics: Connections standard for grades 9-12. Reston, VA: National Council of Teachers of Mathematics, Inc.
Texas Education Agency & Texas Higher Education Coordinating Board. (2009). Texas college and career readiness standards. Retrieved from http://www.thecb.state.tx.us/index.cfm?objectid=E21AB9B0-2633-11E8-BC500050560100A9
Quantitative relationships model problem situations efficiently and can be used to make generalizations, predictions, and critical judgements in everyday life. What patterns exist within different types of quantitative relationships and where are they found in everyday life? Why is the ability to model quantitative relationships in a variety of ways essential to solving problems in everyday life? Financial and economic knowledge leads to informed and rational decisions allowing for effective management of financial resources when planning for a lifetime of financial security. Why is financial stability important in everyday life? What economic and financial knowledge is critical for planning for a lifetime of financial security? How can mapping one’s financial future lead to significant short and long-term benefits? How can current financial and economic factors in everyday life impact daily decisions and future opportunities?
Unit Understandings
and Questions
Overarching Concepts
and Unit Concepts
Performance Assessment(s)
• Linear function models can be used to compare methods of compensation, including deductions, to make predictions and budgetary judgments.
• How can linear functions be used to represent and describe proportional and non-proportional relationships in methods of compensation?
• What are the differences in methods of compensation, and how are these differences illustrated in representative function models?
• How do net earnings and gross earnings compare?
• What type of deductions can be taken from an individual’s gross pay?
• How are deductions represented in linear functions used to model net earnings?
• Why must a budget be built around personal earnings?
• Mathematical Modeling
• Personal Finance
• Finance and budgeting
• Taxes
• Insurance
• Associated Mathematical Processes
• Application
• Problem Solving Model
• Tools and Techniques
• Communication
• Representations
• Relationships
• Justification
Assessment information provided within the TEKS Resource System are examples that may, or may not, be used by your child’s teacher. In accordance with section 26.006 (2) of the Texas Education Code, "A parent is entitled to review each test administered to the parent’s child after the test is administered." For more information regarding assessments administered to your child, please visit with your child’s teacher.
Financial and economic knowledge leads to informed and rational decisions allowing for effective management of financial resources when planning for a lifetime of financial security. Why is financial stability important in everyday life? What economic and financial knowledge is critical for planning for a lifetime of financial security? How can mapping one’s financial future lead to significant short and long-term benefits? How can current financial and economic factors in everyday life impact daily decisions and future opportunities?
Unit Understandings
and Questions
Overarching Concepts
and Unit Concepts
Performance Assessment(s)
• In order to maintain financial stability, consumers must be knowledgeable of the assessment and collection processes used by the various taxing entities for property taxes and personal income taxes to make predictions and budgetary judgments.
• What types of data help make decisions about personal taxes?
• How are the tax amounts determined and collected for personal property?
• How are …
• gross income
• taxable income
… determined, and why is knowing it important?
• What are …
• personal exemptions
• deductions
… and how do they affect the taxes owed?
• How does an employer determine the amount to withhold for income taxes?
• How do “Refund” and “Amount Due” compare on an IRS filing?
• How does understanding assessment and collection processes used by the various taxing entities for property taxes and personal income taxes promote a more secured financial future?
• Mathematical Modeling
• Personal Finance
• Finance and budgeting
• Taxes
• Associated Mathematical Processes
• Application
• Problem Solving Model
• Tools and Techniques
• Communication
• Representations
• Relationships
• Justification
Assessment information provided within the TEKS Resource System are examples that may, or may not, be used by your child’s teacher. In accordance with section 26.006 (2) of the Texas Education Code, "A parent is entitled to review each test administered to the parent’s child after the test is administered." For more information regarding assessments administered to your child, please visit with your child’s teacher.
Financial and economic knowledge leads to informed and rational decisions allowing for effective management of financial resources when planning for a lifetime of financial security. Why is financial stability important in everyday life? What economic and financial knowledge is critical for planning for a lifetime of financial security? How can mapping one’s financial future lead to significant short and long-term benefits? How can current financial and economic factors in everyday life impact daily decisions and future opportunities?
Unit Understandings
and Questions
Overarching Concepts
and Unit Concepts
Performance Assessment(s)
• In order to maintain financial stability, consumers must analyze bank services according to personal needs to make predictions and budgetary judgments.
• What types of data help make decisions about retail banking?
• How are decisions about banking accounts, services, fees, and protection justified?
• What factors about bank accounts and personal finances must be considered when making decisions about personal banking?
• How are the factors prioritized to realize the maximum benefits and minimum costs related to banking services?
• How does understanding of bank services according to personal needs promote a more secured financial future?
• Mathematical Modeling
• Personal Finance
• Finance and budgeting
• Banking
• Associated Mathematical Processes
• Application
• Problem Solving Model
• Tools and Techniques
• Communication
• Representations
• Relationships
• Justification
Assessment information provided within the TEKS Resource System are examples that may, or may not, be used by your child’s teacher. In accordance with section 26.006 (2) of the Texas Education Code, "A parent is entitled to review each test administered to the parent’s child after the test is administered." For more information regarding assessments administered to your child, please visit with your child’s teacher.
Financial and economic knowledge leads to informed and rational decisions allowing for effective management of financial resources when planning for a lifetime of financial security. Why is financial stability important in everyday life? What economic and financial knowledge is critical for planning for a lifetime of financial security? How can mapping one’s financial future lead to significant short and long-term benefits? How can current financial and economic factors in everyday life impact daily decisions and future opportunities?
Unit Understandings
and Questions
Overarching Concepts
and Unit Concepts
Performance Assessment(s)
• In order to maintain financial stability, consumers must analyze bank credit card services according to personal needs and compare advantages and disadvantages to make predictions and budgetary judgments.
• What types of data help make decisions about retail credit options?
• How are decisions about retail credit justified?
• What factors about credit cards and personal finances must be considered when making decisions about personal banking?
• How are the factors prioritized to realize the maximum benefits and minimum costs related to credit cards?
• How is the average daily balance calculated?
• How is monthly interest calculated using the average daily balance?
• How does understanding the advantages and disadvantages of bank credit card services according to personal needs promote a more secured financial future?
• Mathematical Modeling
• Personal Finance
• Finance and budgeting
• Banking
• Credit options
• Associated Mathematical Processes
• Application
• Problem Solving Model
• Tools and Techniques
• Communication
• Representations
• Relationships
• Justification
Assessment information provided within the TEKS Resource System are examples that may, or may not, be used by your child’s teacher. In accordance with section 26.006 (2) of the Texas Education Code, "A parent is entitled to review each test administered to the parent’s child after the test is administered." For more information regarding assessments administered to your child, please visit with your child’s teacher.
#### MISCONCEPTIONS / UNDERDEVELOPED CONCEPTS
Misconceptions:
• Some students may think that if a relationship is linear, it is always proportional rather than that all proportional relationships are linear, but not all linear relationships are proportional.
Underdeveloped Concepts:
• Some students may have weaknesses in performing calculations involving percentages and operations with decimals within algebraic expressions and equations used in solving problems.
#### Unit Vocabulary
• Annual interest rate (APR) – annual percentage rate (APR) applied to the balance on a loan compounded for a set time frame
• Appraised value – stated value of a property as set by a qualified appraiser or person who works for the local tax assessing office
• Automatic Teller Machine (ATM) – a computer terminal that provides 24/7 access for basic banking services (e.g., cash withdrawals and deposits to various checking and savings accounts) from remote locations
• Budget – a monthly or yearly spending and savings plan for an individual, family, business, or organization
• Commission plus base pay – amount of money paid that includes a base amount plus a percentage of goods sold during a pay period, pay = set percentage • amount of goods sold + base pay
• Compensation – payment for goods or services (e.g., wages, salaries, tips, fees, commissions, etc.)
• Constant rate of change – a ratio when the dependent, y-value, changes at a constant rate for each independent, x-value. All linear relationships have a constant rate of change
• Checking account – a bank account that allows customers access to their money through withdrawals and deposits (e.g., writing a check (or using a debit card) to a payee or to obtain cash from the account, making a deposit to the account, etc.)
• Credit card – a card that can be used to borrow money from financial institutions, stores, or other businesses in order to buy products and services on credit
• Debit card – a bankcard issued by a financial institution that is electronically linked to an individual’s checking account for the purpose of making banking transactions, making payments for services, and/or making purchases
• Debit card/ATM fees – fees charged by a bank for a customer's use of an ATM owned by another bank or credit union
• Deductions – the amount(s) subtracted from gross income for expense(s) allowed by the government that reduces the taxpayers’ taxable income
• Deffered payments – a loan, that may or may not change interest, in which the borrower is allowed to start making payments at some specified date in the future
• Gross pay – the total amount of personal income prior to taxes and deductions
• Hourly rate – amount of money paid per hour worked, pay = hourly rate • hours worked
• Itemized deductions – amounts deducted from gross income based on specified allowable expenses
• Layaway plans – retail sales promotion under which a customer deposits a down payment or a fraction of the cost of the merchandise and the merchandise is held on or before a specified date when the customer completes the payment and picks up the merchandise
• Net pay – the income that remains after taxes and other deductions are taken from an individual’s gross income
• Online banking – a banking system that allows an individual to perform banking services using the internet (e.g., monitor accounts, pay their bills, etc.)
• Overdraft protection – a line of credit a banking institution offers to their customers to cover overdrafts, withdrawals from an account that is greater than the balance within the account
• Payday loan – a high-interest, short term loan that is repaid when the borrower receives their next paycheck
• Payroll deductions – a percentage of money that a company withholds from its employees for the federal and state governments as required by law (e. g., income tax, social security contributions, health insurance, unemployment and disability insurances, supplemental retirement, etc.). Some deductions are determined by formulas set by the government, whereas other deductions are voluntary by the employee.
• Processing fees – fees charged by the bank to the customers for processing account activities (e.g., deposits, checks, transfers, loan payments, bill payments, etc.).
• Personal income tax – a tax based on an individual’s income for federal and/or state governments as required by law. The amount of the tax is usually determined by varying rates on different levels of taxable income which are published annually in tax tables by the federal Internal Revenue Service.
• Personal property taxes – annual state imposed tax on different types of personal properties (e.g., vehicle, farming equipment, boat, etc.) paid by a property owner to local government in which the property is located
• Property taxes – annual tax (county, city, school, etc.) based on an appraised value (land, improvements to land, buildings, etc.) paid by a land owner to local government
• Salary – a fixed annual sum that may or may not be dependent on the number of hours worked and usually paid in regular increments, such as monthly
• Slope of a line – the steepness of a line; rate of change in y (vertical) compared to the rate of change in x (horizontal), or , or , denoted as m in y = mx + b.
• Standard deduction – a flat rate amount based on income and number of exemptions the government allows in place of itemizing expenses for deduction
• Straight commission – amount of money paid based on a percentage of the goods sold, pay = set percentage • amount of goods sold
Related Vocabulary:
Credit limits Bank balance Dental insurance Employees benefits Interest Health insurance Linear function Non-proportional linear relationship Proportional linear relationship Retirement savings Social Security payments
Unit Assessment Items System Resources Other Resources
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Unit Assessment Items that have been published by your district may be accessed through Search All Components in the District Resources tab. Assessment items may also be found using the Assessment Center if your district has granted access to that tool.
System Resources may be accessed through Search All Components in the District Resources Tab.
Texas Higher Education Coordinating Board – Texas College and Career Readiness Standards
Texas Education Agency – Mathematics Curriculum
Texas Education Agency – STAAR Mathematics Resources
Texas Education Agency Texas Gateway – Revised Mathematics TEKS: Vertical Alignment Charts
Texas Education Agency Texas Gateway – Mathematics TEKS: Supporting Information
Texas Education Agency Texas Gateway – Interactive Mathematics Glossary
Texas Education Agency Texas Gateway – Resources Aligned to Mathematical Models with Applications Mathematics TEKS
Texas Instruments – Graphing Calculator Tutorials
TAUGHT DIRECTLY TEKS
TEKS intended to be explicitly taught in this unit.
TEKS/SE Legend:
• Knowledge and Skills Statements (TEKS) identified by TEA are in italicized, bolded, black text.
• Student Expectations (TEKS) identified by TEA are in bolded, black text.
• Portions of the Student Expectations (TEKS) that are not included in this unit but are taught in previous or future units are indicated by a strike-through.
Specificity Legend:
• Supporting information / clarifications (specificity) written by TEKS Resource System are in blue text.
• Unit-specific clarifications are in italicized, blue text.
• Information from Texas Education Agency (TEA), Texas College and Career Readiness Standards (TxCCRS), Texas Response to Curriculum Focal Points (TxRCFP) is labeled.
TEKS# SE# TEKS SPECIFICITY
M.1 Mathematical process standards. The student uses mathematical processes to acquire and demonstrate mathematical understanding. The student is expected to:
M.1A Apply mathematics to problems arising in everyday life, society, and the workplace.
Apply
MATHEMATICS TO PROBLEMS ARISING IN EVERYDAY LIFE, SOCIETY, AND THE WORKPLACE
Including, but not limited to:
• Mathematical problem situations within and between disciplines
• Everyday life
• Society
• Workplace
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• VII.D. Problem Solving and Reasoning – Real-world problem solving
• VII.D.1. Interpret results of the mathematical problem in terms of the original real-world situation.
• IX.A. Connections – Connections among the strands of mathematics
• IX.A.1. Connect and use multiple key concepts of mathematics in situations and problems.
• IX.A.2. Connect mathematics to the study of other disciplines.
• IX.B. Connections – Connections of mathematics to nature, real-world situations, and everyday life
• IX.B.1. Use multiple representations to demonstrate links between mathematical and real-world situations.
• IX.B.2. Understand and use appropriate mathematical models in the natural, physical, and social sciences.
• IX.B.3. Know and understand the use of mathematics in a variety of careers and professions.
M.1B Use a problem-solving model that incorporates analyzing given information, formulating a plan or strategy, determining a solution, justifying the solution, and evaluating the problem-solving process and the reasonableness of the solution.
Use
A PROBLEM-SOLVING MODEL THAT INCORPORATES ANALYZING GIVEN INFORMATION, FORMULATING A PLAN OR STRATEGY, DETERMINING A SOLUTION, JUSTIFYING THE SOLUTION, AND EVALUATING THE PROBLEM-SOLVING PROCESS AND THE REASONABLENESS OF THE SOLUTION
Including, but not limited to:
• Problem-solving model
• Analyze given information
• Formulate a plan or strategy
• Determine a solution
• Justify the solution
• Evaluate the problem-solving process and the reasonableness of the solution
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• I.B. Numeric Reasoning – Number sense and number concepts
• I.B.1. Use estimation to check for errors and reasonableness of solutions.
• V.A. Statistical Reasoning – Design a study
• V.A.1. Formulate a statistical question, plan an investigation, and collect data.
• VII.A. Problem Solving and Reasoning – Mathematical problem solving
• VII.A.1. Analyze given information.
• VII.A.2. Formulate a plan or strategy.
• VII.A.3. Determine a solution.
• VII.A.4. Justify the solution.
• VII.A.5. Evaluate the problem-solving process.
• VII.D. Problem Solving and Reasoning – Real-world problem solving
• VII.D.2. Evaluate the problem-solving process.
M.1C Select tools, including real objects, manipulatives, paper and pencil, and technology as appropriate, and techniques, including mental math, estimation, and number sense as appropriate, to solve problems.
Select
TOOLS, INCLUDING REAL OBJECTS, MANIPULATIVES, PAPER AND PENCIL, AND TECHNOLOGY AS APPROPRIATE, AND TECHNIQUES, INCLUDING MENTAL MATH, ESTIMATION, AND NUMBER SENSE AS APPROPRIATE, TO SOLVE PROBLEMS
Including, but not limited to:
• Appropriate selection of tool(s) and techniques to apply in order to solve problems
• Tools
• Real objects
• Manipulatives
• Paper and pencil
• Technology
• Techniques
• Mental math
• Estimation
• Number sense
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• I.B. Numeric Reasoning – Number sense and number concepts
• I.B.1. Use estimation to check for errors and reasonableness of solutions.
• V.C. Statistical Reasoning – Analyze, interpret, and draw conclusions from data
• V.C.2. Analyze relationships between paired data using spreadsheets, graphing calculators, or statistical software.
M.1D Communicate mathematical ideas, reasoning, and their implications using multiple representations, including symbols, diagrams, graphs, and language as appropriate.
Communicate
MATHEMATICAL IDEAS, REASONING, AND THEIR IMPLICATIONS USING MULTIPLE REPRESENTATIONS, INCLUDING SYMBOLS, DIAGRAMS, GRAPHS, AND LANGUAGE AS APPROPRIATE
Including, but not limited to:
• Mathematical ideas, reasoning, and their implications
• Multiple representations, as appropriate
• Symbols
• Diagrams
• Graphs
• Language
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• II.D. Algebraic Reasoning – Representing relationships
• II.D.1. Interpret multiple representations of equations, inequalities, and relationships.
• II.D.2. Convert [among multiple representations of equations, inequalities, and relationships.
• VIII.A. Communication and Representation – Language, terms, and symbols of mathematics
• VIII.A.1. Use mathematical symbols, terminology, and notation to represent given and unknown information in a problem.
• VIII.A.2. Use mathematical language to represent and communicate the mathematical concepts in a problem.
• VIII.A.3. Use mathematical language for reasoning, problem solving, making connections, and generalizing.
• VIII.B. Communication and Representation – Interpretation of mathematical work
• VIII.B.1. Model and interpret mathematical ideas and concepts using multiple representations.
• VIII.B.2. Summarize and interpret mathematical information provided orally, visually, or in written form within the given context.
• VIII.C. Communication and Representation – Presentation and representation of mathematical work
• VIII.C.1. Communicate mathematical ideas, reasoning, and their implications using symbols, diagrams, models, graphs, and words.
• VIII.C.2. Create and use representations to organize, record, and communicate mathematical ideas.
• VIII.C.3. Explain, display, or justify mathematical ideas and arguments using precise mathematical language in written or oral communications.
• IX.B. Connections – Connections of mathematics to nature, real-world situations, and everyday life
• IX.B.1. Use multiple representations to demonstrate links between mathematical and real-world situations.
M.1E Create and use representations to organize, record, and communicate mathematical ideas.
Create, Use
REPRESENTATIONS TO ORGANIZE, RECORD, AND COMMUNICATE MATHEMATICAL IDEAS
Including, but not limited to:
• Representations of mathematical ideas
• Organize
• Record
• Communicate
• Evaluation of the effectiveness of representations to ensure clarity of mathematical ideas being communicated
• Appropriate mathematical vocabulary and phrasing when communicating mathematical ideas
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• VIII.B. Communication and Representation – Interpretation of mathematical work
• VIII.B.1. Model and interpret mathematical ideas and concepts using multiple representations.
• VIII.B.2. Summarize and interpret mathematical information provided orally, visually, or in written form within the given context.
• VIII.C. Communication and Representation – Presentation and representation of mathematical work
• VIII.C.1. Communicate mathematical ideas, reasoning, and their implications using symbols, diagrams, models, graphs, and words.
• VIII.C.2. Create and use representations to organize, record, and communicate mathematical ideas.
M.1F Analyze mathematical relationships to connect and communicate mathematical ideas.
Analyze
MATHEMATICAL RELATIONSHIPS TO CONNECT AND COMMUNICATE MATHEMATICAL IDEAS
Including, but not limited to:
• Mathematical relationships
• Connect and communicate mathematical ideas
• Conjectures and generalizations from sets of examples and non-examples, patterns, etc.
• Current knowledge to new learning
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• VII.A. Problem Solving and Reasoning – Mathematical problem solving
• VII.A.1. Analyze given information.
• VIII.A. Communication and Representation – Language, terms, and symbols of mathematics
• VIII.A.1. Use mathematical symbols, terminology, and notation to represent given and unknown information in a problem.
• VIII.A.2. Use mathematical language to represent and communicate the mathematical concepts in a problem.
• VIII.A.3. Use mathematical language for reasoning, problem solving, making connections, and generalizing.
• VIII.B. Communication and Representation – Interpretation of mathematical work
• VIII.B.1. Model and interpret mathematical ideas and concepts using multiple representations.
• VIII.C. Communication and Representation – Presentation and representation of mathematical work
• VIII.C.1. Communicate mathematical ideas, reasoning, and their implications using symbols, diagrams, models, graphs, and words.
• VIII.C.2. Create and use representations to organize, record, and communicate mathematical ideas.
• VIII.C.3. Explain, display, or justify mathematical ideas and arguments using precise mathematical language in written or oral communications.
• IX.A. Connections – Connections among the strands of mathematics
• IX.A.1. Connect and use multiple key concepts of mathematics in situations and problems.
• IX.A.2. Connect mathematics to the study of other disciplines.
M.1G Display, explain, and justify mathematical ideas and arguments using precise mathematical language in written or oral communication.
Display, Explain, Justify
MATHEMATICAL IDEAS AND ARGUMENTS USING PRECISE MATHEMATICAL LANGUAGE IN WRITTEN OR ORAL COMMUNICATION
Including, but not limited to:
• Mathematical ideas and arguments
• Validation of conclusions
• Displays to make work visible to others
• Diagrams, visual aids, written work, etc.
• Explanations and justifications
• Precise mathematical language in written or oral communication
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• VII.A. Problem Solving and Reasoning – Mathematical problem solving
• VII.A.4. Justify the solution.
• VII.B. Problem Solving and Reasoning – Proportional reasoning
• VII.B.1. Use proportional reasoning to solve problems that require fractions, ratios, percentages, decimals, and proportions in a variety of contexts using multiple representations.
• VII.C. Problem Solving and Reasoning – Logical reasoning
• VII.C.1. Develop and evaluate convincing arguments.
• VIII.A. Communication and Representation – Language, terms, and symbols of mathematics
• VIII.A.3. Use mathematical language for reasoning, problem solving, making connections, and generalizing.
• VIII.B. Communication and Representation – Interpretation of mathematical work
• VIII.B.1. Model and interpret mathematical ideas and concepts using multiple representations.
• VIII.B.2. Summarize and interpret mathematical information provided orally, visually, or in written form within the given context.
• VIII.C. Communication and Representation – Presentation and representation of mathematical work
• VIII.C.3. Explain, display, or justify mathematical ideas and arguments using precise mathematical language in written or oral communications.
M.2 Mathematical modeling in personal finance. The student uses mathematical processes with graphical and numerical techniques to study patterns and analyze data related to personal finance. The student is expected to:
M.2A Use rates and linear functions to solve problems involving personal finance and budgeting, including compensations and deductions.
Use
RATES AND LINEAR FUNCTIONS INVOLVING PERSONAL FINANCE AND BUDGETING
Including, but not limited to:
• Rates of change
• Slope of a line – the steepness of a line; rate of change in y (vertical) compared to change in x (horizontal), or or , denoted as m in y = mx + b
• Constant rate of change – a ratio when the dependent, y-value, changes at a constant rate for each independent, x-value. All linear relationships have a constant rate of change.
• Linear functions
• Linear proportional relationship (direct variation)
• Linear
• Represented by y = kx
• Constant rate of change, k
• Passes through the origin
• Linear non-proportional relationship
• Linear
• Represented by y = mx + b
• Constant slope, m
• y-intercept, b, where b ≠ 0
To Solve
PROBLEMS INVOLVING PERSONAL FINANCE AND BUDGETING, INCLUDING COMPENSATIONS AND DEDUCTIONS
Including, but not limited to:
• Compensations – payment for goods or services (e.g., wages, salaries, tips, fees, commissions, etc.)
• Hourly rate – amount of money paid per hour worked, pay = hourly rate • hours worked
• Proportional relationship (direct variation)
• Linear
• Represented by y = kx
• Constant rate of change, k, is the hourly rate.
• Passes through the origin, zero hours worked is zero pay.
• Commission
• Straight commission – amount of money paid based on a percentage of the goods sold, pay = set percentage • amount of goods sold
• Proportional relationship (direct variation)
• Linear
• Represented by y = kx
• Constant rate of change, k, is the set percentage
• Passes through the origin, zero goods sold is zero pay
• Commission plus base pay – amount of money paid that includes a base amount plus a percentage of goods sold during a pay period, pay = set percentage • amount of goods sold + base pay
• Non-proportional linear relationship
• Represented by slope-intercept form, y = mx + b
• Base pay is the y-intercept, b
• Set percentage is the slope, m
• Salary – a fixed annual sum that may or may not be dependent on the number of hours worked and usually paid in regular increments, such as monthly
• Deductions – the amount(s) subtracted from gross income for expense(s) allowed by the government that reduces the taxpayers’ taxable income
• Gross pay – the total amount of personal income prior to taxes and deductions
• Net pay – the income that remains after taxes and other deductions are taken from an individual’s gross income
• Payroll deductions – a percentage of money that a company withholds from its employees for the federal and state governments as required by law (e. g., income tax, social security contributions, health insurance, unemployment and disability insurances, supplemental retirement, etc.). Some deductions are determined by formulas set by the government, whereas other deductions are voluntary by the employee.
• Income tax withholdings are taxes withheld from an employee’s wages that are paid directly to the government.
• Social security and medical payments are percentages withheld from an employee’s pay that is matched by the employer and deposited into the federal retirement system for that employee’s social security retirement benefits.
• Retirement savings are optional savings plans or accounts requested by the employee to which the employer can make direct deposits from the employee's pay.
• Employee-paid benefits are optional benefits provided through the employer such as individual and/or family health insurance, dental insurance, and life insurance.
• Budget – a monthly or yearly spending and savings plan for an individual, family, business, or organization
• Basics of creating a budget
• Determine amount of money available for spending
• Factors to consider in budgeting spending
• Housing (mortgage, rent, insurance, taxes, maintenance)
• Transportation (auto payment, insurance, maintenance, fuel)
• Clothing and personal needs (clothes, personal hygiene and appearance, health needs)
• Food (groceries and restaurants)
• Savings
• Other (charitable giving, gifts, leisure and entertainment, including travel, etc.)
Note(s):
• Grade 7 identified the components of a personal budget, including income; planned savings for college, retirement, and emergencies; taxes; and fixed and variable expenses, and calculate what percentage each category comprises of the total budget.
• Grade 8 compared and contrasted proportional y = kx and non-proportional y = mx + b linear relationships.
• Algebra I introduced the concept of a function in terms of the linear relationship.
• Various mathematical process standards will be applied to this student expectation as appropriate.
• TxCCRS
• VII.A. Problem Solving and Reasoning – Mathematical problem solving
• VII.A.3. Determine a solution.
• VII.D. Problem Solving and Reasoning – Real-world problem solving
• VII.D.1. Interpret results of the mathematical problem in terms of the original real-world situation.
• VIII.A. Communication and Representation – Language, terms, and symbols of mathematics
• VIII.A.1. Use mathematical symbols, terminology, and notation to represent given and unknown information in a problem.
• VIII.A.2. Use mathematical language to represent and communicate the mathematical concepts in a problem.
• VIII.A.3. Use mathematical language for reasoning, problem solving, making connections, and generalizing.
• VIII.B. Communication and Representation – Interpretation of mathematical work
• VIII.B.2. Summarize and interpret mathematical information provided orally, visually, or in written form within the given context.
• VIII.C. Communication and Representation – Presentation and representation of mathematical work
• VIII.C.1. Communicate mathematical ideas, reasoning, and their implications using symbols, diagrams, models, graphs, and words.
• IX.A. Connections – Connections among the strands of mathematics
• IX.A.1. Connect and use multiple key concepts of mathematics in situations and problems.
• IX.A.2. Connect mathematics to the study of other disciplines.
• IX.B. Connections – Connections of mathematics to nature, real-world situations, and everyday life
• IX.B.1. Use multiple representations to demonstrate links between mathematical and real-world situations.
• IX.B.3. Know and understand the use of mathematics in a variety of careers and professions.
M.2B Solve problems involving personal taxes.
Solve
PROBLEMS INVOLVING PERSONAL TAXES
Including, but not limited to:
• Property taxes – annual tax (county, city, school, etc.) based on an appraised value (land, improvements to land, buildings, etc.) paid by a land owner to local government
• Appraised value – stated value of a property as set by a qualified appraiser or person who works for the local tax assessing office
• Personal property taxes – annual state imposed tax on different types of personal properties (e.g., vehicle, farming equipment, boat, etc.) paid by a property owner to local government in which the property is located
• Personal income tax – a tax based on an individual's income for federal and/or state governments as required by law. The amount of the tax is usually determined by varying rates on different levels of taxable income which are published annually in tax tables by the federal Internal Revenue Service.
• Deductions – the amount(s) subtracted from gross income for expense(s) allowed by the government that reduces the taxpayers’ taxable income
• Standard deduction – a flat rate amount based on income and number of exemptions the government allows in place of itemizing expenses for deduction
• Itemized deductions – amounts deducted from gross income based on specified allowable expenses
Note(s):
• Mathematical Models with Applications examines personal taxes and how taxes are calculated.
• Various mathematical process standards will be applied to this student expectation as appropriate
• TxCCRS
• VII.A. Problem Solving and Reasoning – Mathematical problem solving
• VII.A.3. Determine a solution.
• VII.D. Problem Solving and Reasoning – Real-world problem solving
• VII.D.1. Interpret results of the mathematical problem in terms of the original real-world situation.
• VIII.A. Communication and Representation – Language, terms, and symbols of mathematics
• VIII.A.1. Use mathematical symbols, terminology, and notation to represent given and unknown information in a problem.
• VIII.A.2. Use mathematical language to represent and communicate the mathematical concepts in a problem.
• VIII.A.3. Use mathematical language for reasoning, problem solving, making connections, and generalizing.
• VIII.B. Communication and Representation – Interpretation of mathematical work
• VIII.B.2. Summarize and interpret mathematical information provided orally, visually, or in written form within the given context.
• VIII.C. Communication and Representation – Presentation and representation of mathematical work
• VIII.C.1. Communicate mathematical ideas, reasoning, and their implications using symbols, diagrams, models, graphs, and words.
• IX.A. Connections – Connections among the strands of mathematics
• IX.A.1. Connect and use multiple key concepts of mathematics in situations and problems.
• IX.A.2. Connect mathematics to the study of other disciplines.
• IX.B. Connections – Connections of mathematics to nature, real-world situations, and everyday life
• IX.B.1. Use multiple representations to demonstrate links between mathematical and real-world situations.
• IX.B.3. Know and understand the use of mathematics in a variety of careers and professions.
M.2C Analyze data to make decisions about banking, including options for online banking, checking accounts, overdraft protection, processing fees, and debit card/ATM fees.
Analyze
DATA TO MAKE DECISIONS ABOUT BANKING, INCLUDING OPTIONS FOR ONLINE BANKING, CHECKING ACCOUNTS, OVERDRAFT PROTECTION, PROCESSING FEES, AND DEBIT CARD/ATM FEES
Including, but not limited to:
• Decisions about banking options
• Online banking – a banking system that allows an individual to perform banking services using the internet (e.g., monitor accounts, pay their bills, etc.)
• Checking account – a bank account that allows customers access to their money through withdrawals and deposits (e.g., writing a check (or using a debit card) to a payee or to obtain cash from the account, making a deposit to the account, etc.)
• Automatic Teller Machine (ATM) – a computer terminal that provides 24/7 access for basic banking services (e.g., cash withdrawals and deposits to various checking and saving accounts) from remote locations
• Overdraft protection – a line of credit a banking institution offers to their customers to cover overdrafts, withdrawals from an account that is greater than the balance within the account
• Processing fees – fees charged by the bank to the customers for processing account activities (e.g., deposits, checks, transfers, loan payments, bill payments, etc.).
• Debit card – a bankcard issued by a financial institution that is electronically linked to an individual’s checking account for the purpose of making banking transactions, making payments for services, and/or making purchases
• Debit card/ATM fees – fees charged by a bank for a customer's use of an ATM owned by another bank or credit union
Note(s):
• Grade 6 compared features and costs of a checking account and a debit card offered by different local financial institutions.
• Grade 6 distinguished between debit cards and credit cards.
• Grade 6 balanced a check register that includes deposits, withdrawals, and transfers.
• Mathematical Models with Applications examines the analysis of bank services and accounts.
• Various mathematical process standards will be applied to this student expectation as appropriate.
• TxCCRS
• V.B. Statistical Reasoning – Describe data
• V.B.4. Describe patterns and departure from patterns in the study of data.
• VII.A. Problem Solving and Reasoning – Mathematical problem solving
• VII.A.1. Analyze given information.
• VII.D. Problem Solving and Reasoning – Real-world problem solving
• VII.D.1. Interpret results of the mathematical problem in terms of the original real-world situation.
• VIII.A. Communication and Representation – Language, terms, and symbols of mathematics
• VIII.A.1. Use mathematical symbols, terminology, and notation to represent given and unknown information in a problem.
• VIII.A.2. Use mathematical language to represent and communicate the mathematical concepts in a problem.
• VIII.A.3. Use mathematical language for reasoning, problem solving, making connections, and generalizing.
• VIII.B. Communication and Representation – Interpretation of mathematical work
• VIII.B.2. Summarize and interpret mathematical information provided orally, visually, or in written form within the given context.
• VIII.C. Communication and Representation – Presentation and representation of mathematical work
• VIII.C.1. Communicate mathematical ideas, reasoning, and their implications using symbols, diagrams, models, graphs, and words.
• VIII.C.2. Create and use representations to organize, record, and communicate mathematical ideas.
• IX.A. Connections – Connections among the strands of mathematics
• IX.A.1. Connect and use multiple key concepts of mathematics in situations and problems.
• IX.A.2. Connect mathematics to the study of other disciplines.
• IX.B. Connections – Connections of mathematics to nature, real-world situations, and everyday life
• IX.B.1. Use multiple representations to demonstrate links between mathematical and real-world situations.
• IX.B.3. Know and understand the use of mathematics in a variety of careers and professions.
M.3 Mathematical modeling in personal finance. The student uses mathematical processes with algebraic formulas, graphs, and amortization modeling to solve problems involving credit. The student is expected to:
M.3B Analyze personal credit options in retail purchasing and compare relative advantages and disadvantages of each option.
Analyze
PERSONAL CREDIT OPTIONS IN RETAIL PURCHASING
Including, but not limited to:
• Credit card – a card that can be used to borrow money from financial institutions, stores, or other businesses in order to buy products and services on credit
• Bank credit cards and store specific credit cards
• An annual fee is charged by some credit cards.
• Minimum payment each month is based on the balance.
• Credit limits define the maximum amount that can be charged on an account.
• Limits are determined using various criteria such as credit history, credit score, and/or ability to pay based on income.
• Interest is usually charged on average balance throughout a month if entire balance is not paid each month.
• Rewards programs in the form of airline travel miles, points that can be exchanged for merchandise or services, donations to charities or scholarship funds
• Layaway plans – retail sales promotion under which a customer deposits a down payments or a fraction of the cost of the merchandise, and the merchandise is held until on or before a specified date when the customer completes the payment and picks up the merchandise
• Deferred payments – a loan, that may or may not charge interest, in which the borrower is allowed to start making payments at some specified date in the future
• Payday loan – a high-interest, short term loan that is repaid when the borrower receives their next paycheck
Compare
RELATIVE ADVANTAGES AND DISADVANTAGES OF EACH CREDIT OPTION
Including, but not limited to:
• Interest rates
• Credit card interest can vary according to credit history, balance, and other factors.
• Layaway plans charge a small fee and set the time to pay but do not charge interest.
• Deferred payment interest can vary according to the loan agreement and may or may not apply during the deferment period. The borrow can purchase large items and make payments at a later time when they feel they are more capable of meeting the monthly payment.
• Payday loans are short term, but annual interest rates (APR) is higher than with credit cards.
• Payment options
• Credit cards require a minimum payment each month based on account balance
• Layaway requires regular payments on a schedule in order to get the full amount paid within a set time period, such as 8, 10, or 12 weeks. The longer the period, the higher the fee.
• Deferred payment plans require a monthly payment which begins at some specified date in the future. Interest usually begins to apply when monthly payments begin, but can apply during the deferment period.
• Payday or cash advance loans require full payment by the borrower's next payday or the loan is "rolled over" and more interest at a higher rate is charged. The interest is charged in advance for a specified time.
• Accounting formulas for credit cards (monthly interest rate, annual interest rate, etc.)
• Annual interest rate – annual percentage rate (APR) applied to the balance on a loan compounded for a set time frame
• Periodic (daily) interest rate:
• Average daily balance (ADB) is interest charged on the amount owed at the end of each day
• Most credit card companies charge interest based on average daily balance, which means that the interest rate is effectively slightly higher than the stated APR.
Note(s):
• Grade 8 calculated the total cost of repaying a loan, including credit cards and easy access loans, under various rates of interest and over different periods using an online calculator.
• Grade 8 identified and explained the advantages and disadvantages of different payment methods.
• Algebra I studied the exponential parent function f(x) = abx.
• Mathematical Models with Applications examines loans and interest charges.
• Various mathematical process standards will be applied to this student expectation as appropriate.
• TxCCRS
• VII.A. Problem Solving and Reasoning – Mathematical problem solving
• VII.A.1. Analyze given information.
• VII.A.4. Justify the solution.
• VII.B. Problem Solving and Reasoning – Proportional reasoning
• VII.B.1. Use proportional reasoning to solve problems that require fractions, ratios, percentages, decimals, and proportions in a variety of contexts using multiple representations.
• VII.C. Problem Solving and Reasoning – Logical reasoning
• VII.C.2. Understand attributes and relationships with inductive and deductive reasoning.
• VIII.C. Communication and Representation – Presentation and representation of mathematical work
• VIII.C.3. Explain, display, or justify mathematical ideas and arguments using precise mathematical language in written or oral communications.
• IX.A. Connections – Connections among the strands of mathematics
• IX.A.2. Connect mathematics to the study of other disciplines.
• IX.B. Connections – Connections of mathematics to nature, real-world situations, and everyday life
• IX.B.1. Use multiple representations to demonstrate links between mathematical and real-world situations.
• IX.B.3. Know and understand the use of mathematics in a variety of careers and professions. | 10,417 | 52,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-24 | latest | en | 0.951741 |
https://www.cheenta.com/problem-on-integral-inequality-isi-msqms-b-2015/ | 1,610,809,538,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703506697.14/warc/CC-MAIN-20210116135004-20210116165004-00536.warc.gz | 716,995,539 | 46,510 | Categories
# Problem on Integral Inequality | ISI – MSQMS – B, 2015
Try this problem from ISI MSQMS 2015 which involves the concept of Integral Inequality and real analysis. You can use the sequential hints provided to solve the problem.
Try this problem from ISI-MSQMS 2015 which involves the concept of Integral Inequality.
## INTEGRAL INEQUALITY | ISI 2015 | MSQMS | PART B | PROBLEM 7b
Show that $1<\int_{0}^{1} e^{x^{2}} d x<e$
### Key Concepts
Real Analysis
Inequality
Numbers
But Try the Problem First…
ISI – MSQMS – B, 2015, Problem 7b
“INEQUALITIES: AN APPROACH THROUGH PROBLEMS BY BJ VENKATACHALA”
## Try with Hints
We have to show that ,
$1<\int_{0}^{1} e^{x^{2}} d x<e$
$0< x <1$
It implies, $0 < x^2 <1$
Now with this reduced form of the equation why don’t you give it a try yourself, I am sure you can do it.
Thus, $e^0 < e^{x^2} <e^1$
i.e $1 < e^{x^2} <e$
So you are just one step away from solving your problem, go on………….
Therefore, Integrating the inequality with limits $0$ to $1$ we get, $\int\limits_0^1 \mathrm dx < \int\limits_0^1 e^{x^2} \mathrm dx < \int\limits_0^1e \mathrm dx$
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This site uses Akismet to reduce spam. Learn how your comment data is processed. | 398 | 1,243 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2021-04 | longest | en | 0.791023 |
http://mathhelpforum.com/advanced-math-topics/61337-limit-analysis-question.html | 1,480,891,269,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541426.52/warc/CC-MAIN-20161202170901-00321-ip-10-31-129-80.ec2.internal.warc.gz | 179,293,749 | 10,635 | 1. ## Limit Analysis Question
Hey.
I have played around with this question for my analysis class and I would like some help:
Given: L= lim f1(x), M=lim f2(x)
Show that if f1<=f2 for all x in some interval (a,b), then L<=M
2. It's probably easiest to do this by contradiction. Suppose that L>M and let $\varepsilon = \tfrac12(L-M)$. For x sufficiently close to some limit point c (the limit point isn't specified in the question), it will be true that $|f_1(x)-L|< \varepsilon$ and $|f_2(x)-M|< \varepsilon$. From that, and the fact that $f_1(x)\leqslant f_2(x)$, you should be able to use the triangle inequality to get a contradiction.
3. I am still struggling here. I can't find a way to make the triangle inequality work for this because if I use it on $|f_1(x)-L|< \varepsilon$ or $|f_2(x)-M|< \varepsilon$ I know nothing about the relationship between $|f_1(x)| + |-L|$ and epsilon.
4. Originally Posted by Uriah
I am still struggling here. I can't find a way to make the triangle inequality work for this because if I use it on $|f_1(x)-L|< \varepsilon$ or $|f_2(x)-M|< \varepsilon$ I know nothing about the relationship between $|f_1(x)| + |-L|$ and epsilon.
Okay, $|f_1(x)-L|< \varepsilon$ is equivalent to $L-\varepsilon. Similarly, $M-\varepsilon. From those, you can extract $L-\varepsilon. Can you finish it from there?
5. Duh. I can't believe I didn't look at it like that. Thanks! | 433 | 1,402 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2016-50 | longest | en | 0.880448 |
https://www.physicsforums.com/threads/graphing-angles.858218/ | 1,532,349,635,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596336.96/warc/CC-MAIN-20180723110342-20180723130342-00300.warc.gz | 963,048,840 | 15,146 | # Graphing angles
1. Feb 19, 2016
### tony873004
I once saw a graph of angle vs time. In order to avoid a big shift in the graph when the angle crossed over from 359 degrees to 1 degree, this graph was circular, with time along the radius of the circle.
Is there a name for this type of graph?
2. Feb 19, 2016
### Staff: Mentor
Do you mean something like this?
https://en.wikipedia.org/wiki/Polar_coordinate_system
3. Feb 19, 2016
### tony873004
Yes, except the 3 and 4 in that graph represent the magnitude of the vector. In the graph I'm think of (sorry, can't find a link to it at the moment), the 3 and 4 would represent time. The angle would change over time, so the green and blue lines would not be straight lines, and 1 standard deviation would be shaded around the line.
4. Feb 19, 2016
### Staff: Mentor
I meant, e.g. the graph of the Archimedean spiral on that Wiki-page. Polar coordinates is the closest I know to describe what you could have meant. Sorry, if it doesn't help.
5. Feb 19, 2016
### tony873004
Thanks, Archimedean spiral looks like the term I was searching for.
6. Feb 19, 2016
### Staff: Mentor
Well, it is a special case of your graph, with constant derivative. I don't know if there is some name for the general case.
7. Feb 19, 2016
### tony873004
Thanks, I noticed that after I posted. Archimedean spiral is just the data plotted in the type of graph I was describing.
The reason I'm asking is because I want to make such a graph of orbital elements vs. time, and if there is a convention that is commonly used I want to use that.
For example, would time run from the center to the circumference (that would be my guess), or from the circumference to the center? Is 0 degrees the along the +x axis? I'm also guessing yes. And if there is no convention, then I get to do it my way!
8. Feb 19, 2016
### Staff: Mentor
There are probably thousands of types of graphs out there. The 'right' choice depends only on which information has to be in it and then minimizing on how long it takes to read the information it provides. Whether the time starts or ends in the origin will likely depend on whether your data provide a starting or an ending point of time. Time might as well be along a radius or a varying radius like during the drawing of a spiral.
I guess you would have better chances to get a satisfying answer if you show us an example of how you would draw it and then read the feedbacks. (Sorry if I sounded strange, I'm lacking some English vocabulary.)
9. Feb 20, 2016
### Staff: Mentor
Orbital elements can have other critical points if the orbit changes over time.
Seeing such a plot, I would expect time to increase with increasing radius, and having zero angle to the right is probably a good choice as well. | 690 | 2,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-30 | latest | en | 0.959298 |
https://www.yummymath.com/?s=halloween | 1,632,405,225,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057424.99/warc/CC-MAIN-20210923135058-20210923165058-00050.warc.gz | 1,088,122,291 | 18,514 | # 13 search results for "halloween"
## 14 activities for Halloween
Fourteen activities that will please and intrigue your students as October comes to an end.
## Operation Gratitude – Halloween candy buy back
This is a great idea! Motivate kids to get rid of all of that candy for a good cause. Operation Gratitude and dentists are encouraging healthy eating, motivated giving, and offering a good use for all of that leftover candy.…
## How Much Halloween Candy Will You Collect?
Students consider how to determine the best candy haul prediction. The math ideas of measurement error and percent error are introduced. Students then predict their own Halloween candy collections numbers and compare their predictions to their actual haul using measurement and percent error.
## Costume spending 2020
Use our current data on Halloween spending in America to engage your students in an analysis of the ratios and percents of population, participation, and money spent in honor of October 31st. This activity takes kids through finding parts of wholes as well as ratio, percents and proportion problems. Beware: the tasks in this activity involve some large numbers as well as some challenging fraction/ratio/proportion problems.
## Sadly, Jorge passed away yesterday.
Jorge would have been 15 years old on May 5, 2020. We wondered how old that would have been in human years. Does that 7 times his dog years formula really predict all dog’s aging?
## How long will it take you to trick-or-treat?
It’s Halloween and this is your neighborhood. How long will it take you to trick or treat at every house in your neighborhood? Possible conversation with students: How long will it take you to get from house to house? You…
## Collecting the most candy
Students calculate the volume of candy containers; bags, globe, basket, can, cone, and bucket.
## Ghost Whisperer
As the Halloween season begins, let your students play a spooky game. How does this work? Students can play this game online at Ghost Whisperer Crystal Ball and then Figure out why it works. Construct viable arguments Critique the work of others…
## Make a costume from a sheet
I’m going to make a costume pattern to sell. People can buy my plan and make awesome, long, full, scary, ghost costumes. This scheme will be so simple. Perhaps I’ll get rich! In this activity young students can reason about…
## Holiday candy sales
Which holidays are associated with candy? For which of those holidays is the most candy sold? What percents of annual candy sales does each holiday contribute? Consider using some independent think time, then small group discussion, then whole group discussion as…
## Cheap-otle?
One of my favorite foods is a burrito. One of my favorite places to get a burrito is Chipotle. Their food is delicious and environmentally friendly. For Halloween 2016, Chipotle ran a promotion and charity drive. Chipotle offered \$3 burritos…
## Testimonials
We thought that you might like to see some of the feedback that we receive about YummyMath. On May 22, 2018 Dear Leslie, First, I want you to know how much I have enjoyed using your website to address the…
## The size of chocolates
In this activity students approximate the volume (through finding surface area) of three different pieces of chocolate. The chocolates are shaped like a heart, a bat (for Halloween) and a chocolate turtle irregular shape. | 726 | 3,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-39 | longest | en | 0.935398 |
http://www.jiskha.com/display.cgi?id=1373601694 | 1,462,446,069,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860126502.50/warc/CC-MAIN-20160428161526-00158-ip-10-239-7-51.ec2.internal.warc.gz | 595,878,878 | 3,491 | Thursday
May 5, 2016
# Homework Help: physics
Posted by gina on Friday, July 12, 2013 at 12:01am.
A 9.90-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.372. Determine the kinetic frictional force that acts on the box when the elevator is (a) stationary, (b) accelerating upward with an acceleration whose magnitude is 1.65 m/s2, and (c) accelerating downward with an acceleration whose magnitude is 1.65 m/s2. | 132 | 498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2016-18 | longest | en | 0.91516 |
https://forum.allaboutcircuits.com/threads/simple-inverter-reverse-polarity-protection-cable-project.123785/ | 1,695,767,636,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00412.warc.gz | 286,699,403 | 27,910 | # simple inverter reverse polarity protection cable project
#### edro
Joined May 5, 2016
16
I need to make a robust inverter cable with simple inverter reverse polarity protection. I need help using diodes, fuse application. Note, please, that I am a decent software engineer but flunked my first (basic) electronics class in college in 1972. Either the help you can give is really simple or really thoroughly explained. I have an electronics brown thumb having fried or "popped" more 555 TTL chips than I can remember.
Here are the calculations that I made, probably with a lot of mistakes. Please feel free to correct or even scold me! The inverter is 1000 watts, 2000 peak, so 2000/120v AC is 16 amps, times 12v is 200 watts 12 volts DC. So a 200 watt in-line fuse is important. If I want to make a cable that prevents reverse polarity hookup then I was thinking that I can get put several 3 or 5 amp diodes in parallel then put that (maybe better wording would say "those") in series with the cable. If someone connects reverse polarity I figure that the diodes would create enough resistance to blow the fuse without damaging the inverter. Then only replacing the fuse would allow it all to work again.
The other idea I found would be to use several self healing resettable fuses (PTC) in parallel to add up to 200 amps. I'm unfamiliar with these fuses, never heard of them before this week. I also saw a YouTube of a circuit using a transister circuit but it was over my head. I need something simple that I can build myself!
Do you have any advice to give me on this idea? Do you think this will work? Any better plans that I can implement?
Thank you!
Ed
#### mcgyvr
Joined Oct 15, 2009
5,394
Whats wrong with a keyed plug that only goes in one way?
#### crutschow
Joined Mar 14, 2008
32,874
If the inverter output is 1000W then the input will be more than 1000W due to converter inefficiency.
This means an input current of more than 1000W / 12V = 83A.
You could use fuse in series with a diode in parallel to the input.
If you want to avoid fuses, you could use a MOSFET in series as a low-loss diode.
Here's an article on that.
#### edro
Joined May 5, 2016
16
Great idea, but the cables have large battery clips so a person could possibly put the battery clips on the wrong battery post. Some of the batteries I have, it's difficult to see which is positive or negative because it's black on black lettering, and there's sometimes crud over the lettering, so it's just difficult to see. Even when the posts are well marked it is possible for some moron (such as I am sometimes) to connect it the wrong way before realizing it.
#### edro
Joined May 5, 2016
16
Thank you crutschow. Evident that you know more than I do. I went to the link you provided, the article on how to use a "MOSFET in series as a low-loss diode"... brain freeze looking at the article, no idea. I understand a battery and a diode. The MOSFET article is over my head. Maybe you can help me with a simple schematic.
Here is what I can understand. (graphic below). Can you post the part number for the right size for my inverter? A link to the correct part would be even better. Some websites have so many variations for parts.
I'm not even sure my graphic (Visio 2003) is correct or not.
Ed
Last edited:
#### Dodgydave
Joined Jun 22, 2012
10,946
Your diode is wrong, it needs to go the other way.
Last edited:
#### edro
Joined May 5, 2016
16
If the inverter output is 1000W then the input will be more than 1000W due to converter inefficiency.
This means an input current of more than 1000W / 12V = 83A.
You could use fuse in series with a diode in parallel to the input.
If you want to avoid fuses, you could use a MOSFET in series as a low-loss diode.
Here's an article on that.
ah HA! I looked up "MOSFET reverse polatity schematic" and found this:
[image deleted]
Please help me to see that this will work -- will it will allow current to go to the inverter but if the battery is hooked up incorrectly, it will not allow current to flow? I would still put a fuse in between the battery + and the 'D' pin on the MOSFET. Will reverse polarity blow the fuse or will it just not pass the current?
What size specification MOSFET do I need to protect a 1,000 watt inverter?
And about the fuse, I don't know that a 83 amp ANL fuse is made. Would 100 amp blow before damaging a typical 1000 watt inverter? Also, what about the 2000 surge watts that the inverter allows? Would a 100 amp fuse blow prematurely if 2000 watts load happens for 1 or 2 seconds?
Thank you again!
Last edited:
#### edro
Joined May 5, 2016
16
Your diode is wrong, it needs to go in series with the fuse,...
That makes more sense. Yes. Thank you.
#### crutschow
Joined Mar 14, 2008
32,874
Your diode is wrong, it needs to go in series with the fuse,...
The diode is in the correct position, just the wrong polarity. You want the fuse to blow if there's an reverse connection.
If you put the diode is series then you will have the large forward loss at the high inverter current.
#### crutschow
Joined Mar 14, 2008
32,874
ah HA! I looked up "MOSFET reverse polatity schematic" and found this:
Please help me to see that this will work -- will it will allow current to go to the inverter but if the battery is hooked up incorrectly, it will not allow current to flow? I would still put a fuse in between the battery + and the 'D' pin on the MOSFET. Will reverse polarity blow the fuse or will it just not pass the current?
What size specification MOSFET do I need to protect a 1,000 watt inverter? .................
The MOSFET acts like the series diode. It conducts well in the forward direction (the same direction as the substrate diode) and blocks in the reverse direction.
You want a MOSFET with a low ON resistance (no more than a milliohm) as even that will dissipate over 6W at 80A and will require a heat-sink.
If the ground of the inverter is isolated from the vehicle chassis, then it would be better to use an N-MOSFET and put it in the minus lead, since N-MOSFETs have lower ON resistance for a given size chip as compared to P-MOSFETs.
#### tcmtech
Joined Nov 4, 2013
2,867
Great idea, but the cables have large battery clips so a person could possibly put the battery clips on the wrong battery post. Some of the batteries I have, it's difficult to see which is positive or negative because it's black on black lettering, and there's sometimes crud over the lettering, so it's just difficult to see. Even when the posts are well marked it is possible for some moron (such as I am sometimes) to connect it the wrong way before realizing it.
View attachment 105504
Buy a red and a white paint pen and draw the (+) in red and the (-) in white on the batteries.
If there is enough light to see the battery lugs to connect the clamps to there is enough to distinguish a red (+) from a (-) white.
#### edro
Joined May 5, 2016
16
Buy a red and a white paint pen and draw the (+) in red and the (-) in white on the batteries.
If there is enough light to see the battery lugs to connect the clamps to there is enough to distinguish a red (+) from a (-) white.
Another fine idea but it doesn't address some of the complications that exist - which I neglected to explain:
1. The batteries are not already hooked up to anything thus there are no existing color coded connections. (I look to see which is connected to chassis ground but that's an assumption that the chassis is a negative ground..)
2. The batteries are various, not in my possession, not mine to paint red, white, or any other color indicators.
3. And, (this one is the most relevant): I'm not there. Someone else (less able to distinguish anything other than there are two connections to make), they are the ones that I'm trying to prevent from making a reverse polarity mistake.
Still, thank you for the idea. A fine one barring these complications.
#### edro
Joined May 5, 2016
16
The diode is in the correct position, just the wrong polarity. You want the fuse to blow if there's an reverse connection.
If you put the diode is series then you will have the large forward loss at the high inverter current.
Thank you for clarifying!!! And good for Dodgydave to edit their post!!! I was a little too quick to just take you all at your word 'cause I'm such a newbie..
#### edro
Joined May 5, 2016
16
The MOSFET acts like the series diode. It conducts well in the forward direction (the same direction as the substrate diode) and blocks in the reverse direction.
You want a MOSFET with a low ON resistance (no more than a milliohm) as even that will dissipate over 6W at 80A and will require a heat-sink.
If the ground of the inverter is isolated from the vehicle chassis, then it would be better to use an N-MOSFET and put it in the minus lead, since N-MOSFETs have lower ON resistance for a given size chip as compared to P-MOSFETs.
Wow. I understood your first sentence. But from "You want a MOSFET with a low ON resistance..." and so on it's pretty much greek to me. It isn't that I don't want to learn something here but I'm pretty newbie in electronics.
I can tell you this:
- inverter will be isolated from the vehicle chassis.
- adding a heat-sink won't be a problem.
Here are my questions:
- will this simply not pass the current along if connected with reverse polarity (open circuit) OR
- will this stop the current, creating so much resistance that the battery fuse will blow?
I have only a simple understanding of a transistor, it can pass current or not depending on the input (base? last time I thought about it was over 40 years ago, in 1972)
#### crutschow
Joined Mar 14, 2008
32,874
......................
Here are my questions:
- will this simply not pass the current along if connected with reverse polarity (open circuit) OR
- will this stop the current, creating so much resistance that the battery fuse will blow?
...............
The MOSFET will completely stop any reverse current.
If you stop the current then the fuse won't blow.
Fuses blow when there's too much current (low resistance), not with a high resistance.
MOSFETs fully turn on when the gate voltage is greater than the source (typically +10V Vgs for standard N-MOSFETs and -10V Vgs for standard P-MOSFETs).
The are fully off when the gate voltage equals the source voltage.
Last edited:
#### edro
Joined May 5, 2016
16
The MOSFET will completely stop any reverse current.
If you stop the current then the fuse won't blow.
Fuses blow when there's too much current (low resistance), not with a high resistance.
MOSFETs fully turn on when the gate voltage is greater than the source (typically +10V Vgs for standard N-MOSFETs and -10V Vgs for standard P-MOSFETs).
The are fully off when the gate voltage equals the source voltage.
Got it. I did some "research", really just some YouTube tutorials, and wow! MOSFETs are really cool! So I understand now how they work, how to "hook it up", and all is well. Just one thing now: I cannot understand the details on the data sheet. Would you be willing to tell me the Jameco part number for what I need? And how many I need? --do I put them in parallel to cover the amps that I need to pass through? To refresh, application is a 1,000 watt inverter hooked to a 12 volt battery ... 83 amps, double that for peak.
Ed
#### edro
Joined May 5, 2016
16
Or, do I just need one, and keep it cool, 25 degrees, right? can use a 12v fan, + alum heatsink.
Just need a Jameco part number for right-sized one for me, please.
#### crutschow
Joined Mar 14, 2008
32,874
This IRF2804PBF is the lowest ON resistance (2.3mΩ) MOSFET that Jameco sells.
It will dissipate about 15W @ 80A so would need an appropriate heat sink.
Adding more MOSFETs in parallel would reduce that--
2 in parallel would halve the power and 4 in parallel would dissipate ¼ the total power.
#### edro
Joined May 5, 2016
16
This IRF2804PBF is the lowest ON resistance (2.3mΩ) MOSFET that Jameco sells.
It will dissipate about 15W @ 80A so would need an appropriate heat sink.
Adding more MOSFETs in parallel would reduce that--
2 in parallel would halve the power and 4 in parallel would dissipate ¼ the total power.
Considering the cost, why wouldn't all inverters be built with this built in? Seems pretty plain to me. I'm making 2 of these "idiot-proof" cables (a term my dad used to say a lot, and I think that was because of me, boo hoo). I only have one inverter but I'm looking to purchase a pure sine wave inverter.
Thanks so much crutschow! I'll take a photo when I'm all done and post it here. | 3,188 | 12,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-40 | longest | en | 0.94435 |
https://cs.nyu.edu/courses/fall98/V22.0101-003/hw3.html | 1,516,210,868,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886952.14/warc/CC-MAIN-20180117173312-20180117193312-00081.warc.gz | 648,463,366 | 2,729 | V22.0101 Homework Assignment 3 Fall 1998
A Game of Craps
Section 3; Marateck
Assigned: Thur Oct 8
Due: Thurs Oct 29, 11:59 pm
Midterm Exam: Tues Oct 27
In this assignment you will write a program to play craps and to bet on the outcomes of the games. It is very important that you make sufficient progress on the assignment before the midterm exam. Especially, you should get some practice correctly implementing while and/or repeat loops before the exam.
Craps is a simple dice game. You roll two dice. If you get a 7 or an 11 on this first roll you win; if you get a 2 or a 12 on the first roll you lose; otherwise the result of the first roll is called your point. You then continue to roll the dice until you match the point (in which case you win) or until you roll a 7 (in which case you lose).
You are to write a program that allows the user to place a bet (on either winning or losing) and play a game of craps. It should continue until the person doesn't want to play anymore or until he or she runs out of money. Have the player start with \$100 dollars. Don't allow bets that are less than 0 or more than the player has.
The computer rolls the dice by setting each of the two dice values to a random value between 1 and 6 and adding the two values together.
Here is a sample run.
[Some introductory remarks....]
You have \$100 dollars. How much would you like to bet? \$40
Are you betting on winning(y/n)? y
Press return key to roll the dice.
You rolled a 3 and a 4 for a total of 7.
Congratulations! You won the game and the bet!
You won \$40.
You now have \$140.
Would you like to play again (y/n)? j
That was not a valid response. Please type y or n.
Would you like to play again(y/n)? y
================================================================
You have \$140. How much would you like to bet? \$150
That's more money than you have.
You have \$140. How much would you like to bet? \$130
Are you betting on winning(y/n)? n
You rolled a 3 and a 2 for a total of 5.
Press return key to roll the dice.
You rolled a 3 and a 5 for a total of 8.
Press return key to roll the dice.
You rolled a 1 and a 4 for a total of 5.
You won the game but you lost the bet.
You lost \$130.
You now have \$10.
Would you like to play again(y/n)? n
==================================================================
Bye. You lost \$90.
As the sample shows, you should reject anything other than a ``y" or a ``n" when asking if the user wants to play again. Feel free to make the output nicer as long as it includes the basic information shown here.
The main execution of your program should call the following procedures (which, yes, you need to define):
get_bet
This needs to take three parameters: one for the current amount the that person has, one for the amount of the bet, and a boolean that tells whether the player bet to win. It should tell the player the current amount of money that he or she has and then prompt for the amount of the wager and whether the bet is to win. It should do the error checking outlined above.
play_game
This takes one parameter, a boolean that states whether the player wins. This procedure does the bulk of the work. It should have some procedures of its own, but the design of that is left to you.
calc_money
The purpose of this procedure is to calculate the amount of money that the player has after a game, so it needs four parameters: the amount of money the player has, the amount of the bet, whether the bet was to win, and the result of the game.
disp_message
This procedure takes four parameters: the result of the game, whether the bet was to win, the amount of the bet and the amount of money that the player now has. It prints an appropriate message (``Oh, too bad. You won the game but you bet against yourself. You lost \$25. You now have \$75").
play_again
This also takes one parameter, a boolean that states whether the player wants to play again. It should prompt for and then read the player's answer, doing appropriate error checking.
Be sure not to ask a player who has run out of money if he or she would like to play again.
Design your program so that it fits the description given above, though you may (and should, in some cases) add extra procedures. Your assignment will be graded partially on whether it executes correctly, but programming style will also count considerably: comments, good identifier names, appropriate constants, no non-local variable accesses, a parameter should be declared as var parameters only if it needs to be, and generally readable code.
Sam Marateck
WED Oct 7 13:15:39 EDT 1998 | 1,079 | 4,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-05 | latest | en | 0.937138 |
http://www.docstoc.com/docs/107932581/Listeningtostarsdoc | 1,438,366,426,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988310.3/warc/CC-MAIN-20150728002308-00336-ip-10-236-191-2.ec2.internal.warc.gz | 410,822,922 | 39,822 | # zones: by pB9GwF8
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``` zones:
astronomical/atmospheric
1. moon phase (other weather related)
2. magnetic fields – magnetometer
3. cosmic rays - neutrino detectors
4. solar flares – neutrino detectors
5. thermal radiation (heat) – thermometer, infrared
spectroscope
6. ultraviolet radiation - circular dichroism
spectrometer
7. ionizing radiation (gamma, etc…) - Geiger counters
and scintillation counters
List of questions for Astronomy Contacts:
can we borrow equipment?
are there historical artifacts or charts we could use?
are there any live cosmic listening posts we can
access?
how much light does the ambient light of Kansas City
filter out?
can we calculate this in terms of years – ie stars of a
certain age are visible?
Is there a way to convert the density of the universe
into sound?
Is there a way to listen to ambient microwaves?
Is there a way to take the COBE maps of temp. variation
and turn it into sound?
How to extend the compressed sound so that it plays
longer?
The color of the universe:
http://www.pha.jhu.edu/~kgb/cosspec/
The sound of the bigbang:
http://www.astro.virginia.edu/~dmw8f/sounds/cdromfiles/
index.php
http://www.astro.virginia.edu/~dmw8f/BBA_web/index_fram
es.html
Notes on Astronomical DeepTime:
The Universe in a Nutshell, Stephen Hawking
Einstein’s relativity requires abandoning the idea of universal time.
Isaac Newton provided first mathematical model for space in Principia Mathematica
(1687) In this model space and time were static backgrounds against which events took
place; time was separate from space.
Replaces it with subjectivy time – time/space are conjoined.
“Space/Time is transformed from a passive background in which events take place to an
active participant in the dynamics of the universe.” – p. 21
You cannot curve space without curving time, thus time has a shape although it still has
only one direction.
Confirmed by two clocks placed on two different airplanes circling the Earth in opposite
directions. The higher velocity plane experiences time slower.
Acceleration and gravity work the same and both affect time.
Space/time is curved not flat – high density objects create a well in space/time just as
high velocity objects create a peak.
Gravity is just an effect of the warp of space/time (note this works only in general
relativity, MTheory and Brane theory view gravity differently)
15 billion year lifespan of the universe is predicted by Einstein’s theory (rate of
expansion vs gravitational pull)
Kant refered to the problem of either a finite or infinite universe as an “antimony of pure
reason” in that it seemed to be a logical contradiction.
If the universe was infinite then every point in the night sky would be occupied by a star
and all of their light would have already reached Earth, lighting up the night sky like one
giant sun.
What did god do before creating the universe? “He was preparing Hell for those who pry
too deep?” – St. Augustine.
Deterministic universe as announced by Einstein’s theory does not function on the
quantum level as per discoveries by Max Planck and Werner Heisenberg.
Particles do not have a definite position, only probably histories (phases/wavestates).
We cannot know what time is, only describe what has been found using mathematical
formulas in order to make predictions.
Because the universe is expanding, when we look at the night sky (back in time) we see
regions of higher density. We observe faint residual big-bang microwave radiation from
when the universe was much denser and hotter. This density curves space/time.
Measureument of this microwave spectrum shows that it passed through curved
space/time confirming expansion of universe. P.40
As one goes back in time, the cross sections of our past light cone reach a maximum size
and then contract depending upon the matter density of the universe.
The past is shaped like a pear.
p.74 Expanding Universe.
Hubble observed the Doppler effect at work in light from other galaxies. Galaxies
moving away have elongated frequencies. In the light spectrum they appear as red-shift.
If the galaxies were moving towards us they would be of a higher frequency (blue-shift).
p. 58 Imaginary Time and quantum theory
Imaginary time is time measured in imaginary numbers; it exists at right angles to
conventional space/time.
Imaginary time provides predictive mathematical models.
Because imaginary time exists at right angles to conventional time/space it can be said to
embody, along with C-Space/time, all possibilities.
Points in imaginary time can overlap with regular points of space/time.
Both real time and imaginary time come to a stop at singularities.
What we interpret as reality may be a four dimensional holographic reflections (3Brane)
of a higher dimensional construct.
p. 80 Quantum Uncertainty
Universe unpredictable on quantum scale – doesn’t have a single history, but every
possible history each with its own probability. (Richard Feynman) These possibilities
exist in Imaginary Time.
Real time determines Imaginary Time and vice versa.
Imaginary Time behaves just like another direction in space – it has no end or beginning,
just different directions.
History of the universe in Imaginary Time is shaped like a four-dimensional sphere. It is
a closed shape that has no specific boundary conditions.
The Anthropic principle states that the universe has to be as we see it because we are able
to see it. In other words, it takes 15 billion years for intelligent life to evolve in order to
observe the universe and the universe looks like the context that would support this type
of life.
p.94
```
To top | 1,257 | 5,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2015-32 | longest | en | 0.850148 |
https://wattsupwiththat.com/2018/06/30/dr-hansens-statistics/ | 1,702,334,126,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679518883.99/warc/CC-MAIN-20231211210408-20231212000408-00072.warc.gz | 645,799,270 | 106,126 | # Dr. Hansen’s Statistics
### Guest Post by Willis Eschenbach
OK, this post has ended up having two parts, because as usual, I got side-tractored while looking at the first part. It’s the problem with science, too many interesting trails leading off the main highway …
### Part The First
I wanted to point out an overlooked part of Dr. James Hansen’s 1988 oral testimony to the US Senate. At the time Dr. Hansen was the Director of GISS, the Goddard Institute of Space Studies. He told the Congresspersonages, or whatever the modern politically correct term is for that class of Politicritters, the following:
The observed warming during the past 30 years, which is the period when we have accurate measurements of atmospheric composition, is shown by the heavy black line in this graph. The warming is almost 0.4 degrees Centigrade by 1987 relative to climatology, which is defined as the 30 year mean, 1950 to 1980 and, in fact, the warming is more than 0.4 degrees Centigrade in 1988. The probability of a chance warming of that magnitude is about 1 percent. So, with 99 percent confidence we can state that the warming during this time period is a real warming trend.
SOURCE
Here is his accompanying graphic …
Now, I am either cursed or blessed with what I call a “nose for bad numbers”. It is a curious talent that I ascribe inter alia to using a slide rule when I was growing up. A slide rule has no decimal point. So if an answer from the slide rule is say 3141, you have to estimate the answer in order to decide if it means 314.1, or 3.141, or .003141, or 31,410. After doing this for years, I developed an innate sense about whether a result seems reasonable or not.
So when I saw Hansen’s claim above, I thought “Nope. Bad numbers”. And when I looked deeper … worse numbers.
First thing I did was to see if I could replicate Hansens’ results. Unfortunately, he was using the old GISS temperature record, made before they were as adjusted as they are today. His statement was that “The warming is almost 0.4 degrees Centigrade by 1987. But in the modern GISS data, I found slightly more warming, 0.5°C.
OK, fair enough. So I went and digitized the dataset above so I could use Dr.Hansen’s data, and it turns out that his “almost 0.4 degrees Centigrade” increase by 1987″ is actually 0.32°C. You can see it in the graphic above. Hmmm … Dr. Hansen’s alarmism is unquenchable. Also, note that Dr. Hansen has spliced into the graphic and discussed the 1988 “annual” average even though at the time he only had a few months of 1988 data … bad scientist, no cookies. Comparisons gotta be apples to apples.
Next, his claim is that there is only one chance in a hundred that the 1987 warmth is a random result. That means his 1987 temperature should be 2.6 standard deviations warmer than the 1951-1980 mean. But once again, Dr. Hansen is exaggerating, although this time only slightly—it’s only 2.5 standard deviations away from the mean, not 2.6.
However, that’s not the real problem. In common with most climate-related temperature datasets, the GISS temperature dataset Hansen used has a high “Hurst Exponent”. This means that the GISS temperature dataset will be what has been called “naturally trendy”. In such datasets, large swings are more common than in purely random datasets.
How much more common? Well, we can actually test that. He’s comparing the 30-year “climatology” period 1951-1980 to the year 1987. So what I did was the exact same thing, but starting in different years, e.g. comparing the thirty-year period 1901-1930 to the year 1937, seeing how unusual that result is, and so on.
When we do that for all possible years of the GISS 1988 dataset, we find that being 2.5 standard deviations away from the climatological mean is not uncommon at all, occurring about one year out of fourteen.
And if we do the same analysis on the full GISS dataset up until today, we find it’s even more common. It has occurred in the historical record about one year out of seven. So Hansen’s “one percent chance” that the 1988 temperature was unusual was actually a fourteen percent chance … more alarmist misrepresentation, which is no surprise considering the source.
### Conclusions the First
Regarding the warmth of 1987, which was 2.5 standard deviations warmer than the 30-year climatology average, Hansen claimed that “The probability of a chance warming of that magnitude is about 1 percent.”
In actuality, this kind of warming occurred in the record that he used about once every fourteen years or so … and it occurs in the modern GISS record about once every seven years. So the probability of a chance warming of that magnitude in the GISS temperature record is not one percent, it is between seven and fourteen percent … which means that it is not unusual in any way.
### Part The Second
In the process of researching the first part of this post, I realized why there is so much debate about whether Hansen’s predictions were right or wrong. The problem is that we’re living in what the most imaginative and talented cartoonist yclept “Josh” calls “The Adjustocene”
The problem is that Dr. James Hansen is not only the guy who made the 1988 alarmist predictions. He’s also the guy who has been in charge of the GISS temperature record that he has long been hoping would make his prediction come true.
So … here are the changes between the version of the GISS temperature record that Hansen used in 1988, and the 2018 version of the GISS temperature record.
(GISS 2018 data available here. )
Gotta say, those are some significant changes. In the old GISS record (red), 1920 to 1950 were much warmer than in the new record. As a result, in the old record temperatures cooled pretty radically from about 1940 to 1970 … but in the new record that’s all gone.
And things don’t get any better when we add another modern record to the mix. Here’s the Hadley Center’s HadCRUT global average temperature, shown in blue …
Note that HadCRUT (blue) shows the same drop in temperature 1940-1970 that we see in the 1988 version of the GISS temperature record (red). More to the current point, the post-1988 divergence between the HadCRUT and the GISS record is enough to rule out any possibility of determining whether Hansen was right or wrong. The overall trend in the GISS 2018 data is about 40% larger than the trend in the HadCRUT data, so you can get the answer you wish by simply picking the right dataset.
### Conclusions the Second
Depending on the dataset chosen, someone can show that Dr. Hansen’s predictions either did or did not come true … it’s the perfect Schrodinger’s Cat of predictions.
Finally, as an aside, just what is an “Institute of Space Studies” doing studying the climate? I’ve heard of “mission creep” before, but that’s more than mission creep, that is extra-terrestrial movement. Don’t know if the Goddard folks have noticed, but there is no climate in space … how about if they go back to, you know, studying the myriad of fascinating things that happen in space, and leave studying the climate to less alarmist folk?
Best regards to all,
w.
### Short Version Of My Usual Request:
QUOTE THE EXACT WORDS YOU ARE DISCUSSING.
### Digitized Hansen Data from Figure 1:
```Year, Anom
1880, -0.403
1881, -0.366
1882, -0.427
1883, -0.464
1884, -0.729
1885, -0.541
1886, -0.461
1887, -0.547
1888, -0.388
1889, -0.184
1890, -0.38
1891, -0.438
1892, -0.44
1893, -0.481
1894, -0.382
1895, -0.408
1896, -0.274
1897, -0.177
1898, -0.38
1899, -0.223
1900, -0.025
1901, -0.086
1902, -0.282
1903, -0.357
1904, -0.493
1905, -0.254
1906, -0.175
1907, -0.45
1908, -0.317
1909, -0.334
1910, -0.313
1911, -0.289
1912, -0.316
1913, -0.254
1914, -0.053
1915, -0.009
1916, -0.258
1917, -0.474
1918, -0.363
1919, -0.197
1920, -0.154
1921, -0.079
1922, -0.143
1923, -0.128
1924, -0.119
1925, -0.097
1926, 0.133
1927, -0.006
1928, 0.066
1929, -0.165
1930, -0.002
1931, 0.085
1932, 0.049
1933, -0.158
1934, 0.047
1935, -0.016
1936, 0.055
1937, 0.17
1938, 0.188
1939, 0.052
1940, 0.111
1941, 0.126
1942, 0.094
1943, 0.034
1944, 0.108
1945, -0.027
1946, 0.035
1947, 0.152
1948, 0.034
1949, -0.018
1950, -0.136
1951, 0.02
1952, 0.071
1953, 0.2
1954, -0.028
1955, -0.069
1956, -0.184
1957, 0.094
1958, 0.113
1959, 0.061
1960, 0.006
1961, 0.077
1962, 0.027
1963, 0.022
1964, -0.264
1965, -0.174
1966, -0.09
1967, -0.024
1968, -0.128
1969, 0.028
1970, 0.034
1971, -0.117
1972, -0.077
1973, 0.168
1974, -0.09
1975, -0.039
1976, -0.235
1977, 0.164
1978, 0.1
1979, 0.131
1980, 0.267
1981, 0.359
1982, 0.058
1983, 0.305
1984, 0.096
1985, 0.053
1986, 0.173
1987, 0.325
1988, 0.562 (five months only)```
Article Rating
Inline Feedbacks
Peter D. Tillman
June 30, 2018 11:45 pm
Nice, careful analysis, Willis. Love the Josh cartoon! The Climate Alarmists have been political from the get-go. And very successful at it, for many years. Starting to wear thin, now that it’s more obvious that the gloom ‘n’ doom stuff just ain’t happening…..
Menicholas
July 1, 2018 5:10 pm
Starting to wear thin?
Yeah, and the ocean is getting a little wet.
Sunsettommy
June 30, 2018 11:46 pm
Quoting Willis:
“Finally, as an aside, just what is an “Institute of Space Studies” doing studying the climate? I’ve heard of “mission creep” before, but that’s more than mission creep, that is extra-terrestrial movement. Don’t know if the Goddard folks have noticed, but there is no climate in space … how about if they go back to, you know, studying the myriad of fascinating things that happen in space, and leave studying the climate to less alarmist folk?”
The original mission for GISS was substantially changed in the 1980’s AFTER DR. Hansen became Director for GISS, to greatly allow for increased weather/climate research that was more into global warming issues and less for space studies of other planets and lunar atmospheres.
June 30, 2018 11:59 pm
The interesting thing is not that Hansen’s ‘detection’ science was rubbish — that was known at the time, and everyone, including the authors of the IPCC detection chapter (Wigley, Barnett), publicly said so at the time.
See:
No, what is interesting is how that did not matter. At the very beginning of the climate policy push, already the science did not matter. More important for the rest of us outside the USA was what happened today 30 years ago, when the Toronto conference statement was released, opening with:
Humanity is conducting an unintended, uncontrolled, globally pervasive experiment whose ultimate consequences could be second only to a global nuclear war.
At that meeting John Houghton and John Zillman from the IPCC protested at its outrageous claims. That did not matter. The science never mattered. The funding for science kept coming. The science did not matter. Why? And what does this mean about the role of science in society?
Santa
July 1, 2018 12:26 am
The dictatorship of the proletariat Failed so they try other ways and means. Here they try The dictatorship of Nature.
Curious George
July 1, 2018 8:15 am
Is the proletariat the working class or the welfare class? The working class elected Trump. The welfare class is caught in Pelosi’s safety net.
Santa
July 1, 2018 12:53 am
“Humanity is conducting an unintended, uncontrolled, globally pervasive experiment whose ultimate consequences could be second only to a global nuclear war.” If you want to rule the Humanity, with the Dictatorship of Nature, this is what we all have to believe. Stalin once said “ Ideas are more powerful than guns” I think he is right so be careful what ideas you let into our minds?
July 1, 2018 4:56 am
“Humanity is conducting an unintended, uncontrolled, globally pervasive experiment whose ultimate consequences could be second only to a global nuclear war.”
It’s called ‘Socialism’ and ‘renewable energy;
Hivemind
July 1, 2018 5:23 am
Especially if the people with wacky ideas have guns.
July 1, 2018 4:55 am
To the the post modern New Left, the role of science in society is to support the Revolution to bring about the New World Order.
Politically incorrect conclusions will not get future funding.
RobR
July 2, 2018 2:28 am
Politically incorrect science will not get funding either.
dennisambler
July 1, 2018 6:19 am
“At that meeting John Houghton and John Zillman from the IPCC protested at its outrageous claims” – but subsequently jumped onboard….
Mardler
July 2, 2018 2:36 am
Houghton soon got on board with Hansen.
Nick Stokes
July 2, 2018 2:43 am
“everyone, including the authors of the IPCC detection chapter (Wigley, Barnett), publicly said so at the time”
What did they say? No quotes are visible.
Jeff
July 1, 2018 12:05 am
I would like to see the raw data record chart against the current and Hansen’s 1988 GISS record.
Alley
July 1, 2018 6:23 am
I love comparing the raw oceans and land data to the adjusted. Raw shows more warming.
July 1, 2018 11:56 am
Alley
When government bureaucrats
REDUCE the average temperature
in the past,
and do NOT REDUCE
today’s average temperature
by the same amount,
then the net result
is MORE GLOBAL WARMING
from the past (reduced) to today (not reduced).
Do you understand that, Alley?
Probably one-third of the “warming”
since 1880 was caused by “adjustments”
to raw data.
Alley
July 2, 2018 5:27 am
Richard, when scientists INCREASE the ocean temps of the past, the net result is less warming.
Do you even understand that, “Richard”? Probably 1/3 of the “cooling” since 1880 was caused by “adjustments” to raw data.
Why do you even bother to pretend to know how temperature has been adjusted?
Clyde Spencer
July 2, 2018 9:27 am
Alley,
I have read Karl’s paper. He adjusted the temperatures of the recent, cutting-edge technology of the Argos floats upward, to match the poor quality ship water-intake temperatures so that he could claim there had been no pause in temperature increase. You should get out and read more.
Nick Stokes
July 2, 2018 11:21 am
Clyde,
You should read more carefully. Karl’s paper had nothing to do with Argo floats.
Clyde Spencer
July 2, 2018 8:40 pm
Nick,
From the paper: “… the data are sea surface temperature (SST) observations taken primarily by thousands of commercial ships and drifting surface buoys.”
It is not so much a matter of not reading carefully as it is not remembering the details after three years. So, you are right that the drifting and moored buoys used are not a part of the Argo network.
But, your talent for nit picking to create a diversion from the essence of the claim is what earns you so many negative votes. You do not win an argument or gain respect from red herrings and straw men.
You might have just said “The Argo network was not a part of the floating and moored buoy data that Karl analyzed.” But, that would have implied that my claim was essentially correct with regard to how Karl et al. adjusted the SSTs. Instead, you attempted to make me look like I didn’t know what I was talking about. You have instead, damaged your credibility by making people realize that you will do anything to win an argument, even if truth is not served.
Nick Stokes
July 2, 2018 9:17 pm
“is what earns you so many negative votes”
I’m trying to get ahead of Steven Mosher. My fear is that I’ll be on -999 when WUWT pulls the plug on the scheme 🙁
Wim Röst
July 2, 2018 11:54 pm
Nick Stokes, the content of your remarks most times is very good, worth reading. But the position you take is not always understood. Nevertheless, you keep us sharp and that is why we need people like you.
But it would be good to hear from you when you have doubts on something on ‘the other side’. That would make the discussion more open, more balanced (from your side) and with less emotions (by the readers here). There is enough we don’t know about weather and climate, let’s be open about that. In fact, we don’t know that much. And the search for truth is an interesting one.
By the good information you give and by keeping everyone sharp, you are doing well. But why not ‘a bit more two-sided’?
Nick Stokes
July 3, 2018 12:13 am
Wim,
I generally try to stick to facts rather than opinion – I don’t always succeed, but I try. If the facts are right, they generally don’t need my affirmation – I speak up when I think they are wrong. But sometimes I do need to speak up in favour, and then I do. Recent example here.
Wim Röst
July 3, 2018 1:32 am
Nick, I know you generally stick to facts rather than opinion. But by the choice of facts that you are opposing to, you seem to express an opinion as well. And that opinion seems (!) to be that ‘the other side’ is doing everything well, big mistakes don’t need to be criticized and ‘on this side/site’ everyone is doing wrong and even small errors need to be corrected. Which might not be what you are aiming to. It might be that you are trying to correct errors on both sides (even big ones) but that we don’t see that or hear that, at least not on this side/site.
A personal note from my side. It seems that everyone may make ‘alarm’, but never needs to apologize when he/she was doing completely wrong. I am not talking about you, I just notice that no ‘green person’ or ‘alarmist’ ever makes excuses for the damage he/she makes. I am worried about that. It is good to know that someone is able to be open about main mistakes that have been made in the past or about the position someone was taking in. Here you are mainly correcting mistakes or possible mistakes that are made by people on this site. Which is good, but possibly only half of the work that should be done.
My personal position – I was clear about that here https://wattsupwiththat.com/2015/11/29/ipcc-science-ipcc-government/ – is that the IPCC is not a scientific institute but a government organization. The IPCC is an organization that never accepted the scientific method (to find the truth) but instead used ‘its own rules’ to promote a certain (government) opinion, trying to reach a certain ‘government goal’. Which brought that institute far from the science that we need to understand weather and climate, which was (and is) at the detriment of society itself. After that government trick, all people that were paid by governmental organizations (also universities) needed to speak ‘government (IPCC) language’. At the detriment of science. Now, when the real independence of universities and research institutes is gone, large corrections for the points of view as expressed /or NOT expressed by scientific groups and for their kind of science have to be made. We need (your) corrections for those ‘big mistakes’ as well.
Clyde Spencer
July 3, 2018 7:35 pm
OK, Nick. I’ll have to give you a +point for a sense of humor.
Anthony Banton
July 3, 2018 2:36 am
“I have read Karl’s paper. He adjusted the temperatures of the recent, cutting-edge technology of the Argos floats upward, to match the poor quality ship water-intake temperatures so that he could claim there had been no pause in temperature increase. You should get out and read more.”
Clyde:
Ships have been found to overread by 0.12C as compared to buoys.
Now because buoys are increasing in number vs ships then there has to be a correction made to either the ships or the buoys, else the trend will be falsely altered via that introduced bias of changing relative numbers.
It doesn’t matter to the trend which way you do it. Buoys up or ships down.
Not to do with “cutting-edge” buoy readings being adjusted to claim anything.
Karl found that this had introduced that bias during the “Hiatus” period (an artifact affecting the trend that was not physical).
Given that there are more ships readings than buoys they applied the 0.12C to the buoys
If you don’t correct (because of increasing buoy observations vs ships) the temperature gradually slides from ship temp to buoy temp weighting, which is 0.12°C cooler. That is a falsely introduced bias to the trend, which is what Karl et al 2015 revealed.
Clyde Spencer
July 3, 2018 7:50 pm
Anthony,
I perfectly understand that the offset had to be adjusted. But, only in climatology would someone get away with adjusting high-quality data to agree with inferior quality data! The increasing number of buoys is a weak excuse in these days of high-speed computers. Start the adjustment run on Friday night before going home for the weekend, or borrow a computer from someone taking their annual 2-3 week vacation. It isn’t like you have to hire a temporary bunch of monks to erase parchment and scribe the new numbers!
I does make a difference whether the ship intakes are adjusted down or the buoys are adjusted up. By adjusting the buoy temps up, it reinforces the claim that there was no ‘pause’ and that the oceans are warming more than they probably are.
Nick Stokes
July 3, 2018 8:20 pm
“I does make a difference whether the ship intakes are adjusted down or the buoys are adjusted up.”
It makes absolutely no difference. The adjustments are only done in the context of calculating an anomaly. Suppose instead of adding 0.12C to buoys, you subtract it from ships. Then you add 0.12C to everything. The result is the same as adding to buoys. What does the effect of that last have on anomalies? None at all, because you add the same to the temperature and to the mean that you subtract. So arithmetically there is no difference.
It seems to have some kind of symbolism for some people. I suspect that is mixed with arithmetical misunderstanding. But there is a simple practical reason for adjusting buoys. The records are a lot shorter. If you adjust ships, you have to go back to 1900’s. And that affects a much greater number of places where the data is used and has been written down.
DC Cowboy
Editor
July 1, 2018 6:35 am
I believe that the ‘raw’ data no longer exists for either GISS or HadCRUT.
Alley
July 2, 2018 5:28 am
Nice beliefs. They exist.
Clyde Spencer
July 2, 2018 9:28 am
Can you provide a link to where they can be accessed by the public?
commieBob
July 1, 2018 12:05 am
Since the mid twentieth century warming was nearly the same as the late twentieth century warming, that presents a big problem for the warmistas. link Hansen had to have known that. It should have been kinda obvious to him even without the application of statistical analysis. (Yes, I realize that the century wasn’t over yet when he testified before Congress.)
Terry Dyson
July 1, 2018 12:08 am
Side-tractored? Remember WEillis, autocorrect is your worst enema.
FrankH
July 1, 2018 1:12 am
I like “tractored”, especially when travelling along “trails leading off the main highway”.
July 1, 2018 5:02 am
Never mind Willis, at least it was a ‘Green’ tractor
Duane Johnson
July 1, 2018 8:43 am
Deere me!
ralfellis
July 1, 2018 5:31 am
I think that was deliberate….
R Taylor
July 1, 2018 5:50 am
Anyone who knows “yclept” likely uses “side-tractored” humorously. Thanks for both.
Jeff Alberts
July 1, 2018 8:36 am
Yup. Willis even says “could care less” when he means “couldn’t care less”.
colin smith
July 1, 2018 9:29 am
He was slide-tractored too 🙂
Mary Brown
July 1, 2018 6:10 pm
Yclept is a Haitian rapper
RoHa
July 1, 2018 10:18 pm
Very effective. Suggests being dragged off track, rather than simply wandering off.
Trevor
July 2, 2018 2:40 am
Does this also mean that you have HEARING PROBLEMS from said same tractor ?
Red94ViperRT10
July 1, 2018 11:52 am
“…side tractored…”… That’s a phrase my Dad would have used, and maybe even did. He had me believing for years that disconhookeded was a real word.
HotScot
July 1, 2018 3:19 pm
Red94ViperRT10
“disconhookeded”
You’re claiming it’s not a real word?!
MOD, report this man to his father immediately!
M Courtney
July 1, 2018 12:51 am
When we do that for all possible years of the GISS 1988 dataset, we find that being 2.5 standard deviations away from the climatological mean is not uncommon at all, occurring about one year out of fourteen.
So, twice then?
I agree that 2 is 100% different to 1.
I agree that Hansen was biased in his presentation.
But it wasn’t as implausible in the ’80s as it is now.
We should all beware of hindsight when making judgements.
Lewis P Buckingham
July 1, 2018 2:11 am
So therefore the precautionary principal still applies.
However as warming is not according to the models of climate catastrophe, we
need not take unworkable and uneconomic precautions for something that may be benign or in fact, beneficial.
Neil Jordan
July 1, 2018 10:45 am
Even more worser. L’Hopital’s rule on operations with zero and infinity show that chance of occurence approaching zero x consequence (end of world as we know it) yields zero risk.
July 1, 2018 11:38 pm
The Precautionary Principle: “Do not stand up, lest you fall down!”
Fortunately, early man ignored that one…
BTW Willis – great post!
Best, Allan
Phoenix44
July 1, 2018 4:53 am
But it was, because assuming a small sample of new data is outside normal is dangerous. The problem might easily be that your sample is simply too small or unrepresentative or that there is no normal data set with normal distribution.
30 years seems like an incredibly small sample of data to use for something like climate, particularly using one averaged figure that is likely to be pretty inaccurate and meaningless.
Pierre
July 1, 2018 6:52 am
M Courtney said “But it wasn’t as implausible in the ’80s as it is now.
We should all beware of hindsight when making judgements” I agree, Hansen did take that into account. The fox (Hansen), was watching the hen house, (GISS Data ) and we got the Adjustocene.
Rich Davis
July 1, 2018 10:47 am
Overly charitable of you to say he didn’t look, isn’t it, Wiillis? If he didn’t look, then we need to conclude that he made up the 1% claim out of whole cloth. But it seems far more likely that he was well aware of his deception, just as 0.32 was nearly 0.4, and a few months was sufficient to estimate the whole year.
Rich Davis
July 2, 2018 8:37 am
of course I’m engaging in crazy conspiracy theory here. After all it’s not as if he predicted the most likely day for a heat wave, disabled the air conditioning, or left the windows open prior to the hearings to ensure that the room was stiflingly hot, or something. Shame on me.
Nick Stokes
July 2, 2018 11:19 am
Actually, there is no evidence at all that Hansen did any of that.
Clyde Spencer
July 3, 2018 7:53 pm
Senator Wirth is a figment of someone’s imagination? I don’t imagine that Hansen personally did the dirty deeds, but he was almost certainly aware of them and was therefore a party to them.
Crispin in Waterloo
July 6, 2018 6:47 pm
No one has mentioned that from 1977-1987, according to the anomaly charted values, the globe warmed 0.5C. That’s a rate of 5 C per century. So 30 years later we should already be up more than 2 more due to the baked in warming and all the additional CO2 that was emitted above even Hansen’s worst case scenario. What’s (not) happening here? Global warming seems awfully non-linear.
tty
July 1, 2018 10:30 am
“But it wasn’t as implausible in the ’80s as it is now.”
I disagree. It was perfectly well-known that temperature time series are strongly autocorrelated and that probabilities based on normal distributions (=independent data points) will be way off. Hansen undoubtedly knew this. Modern “climate scientists” know this. Do they care? No.
Red94ViperRT10
July 1, 2018 12:07 pm
Courtney July 1, 2018 12:51 am you say, “…So, twice then? I agree that 2 is 100% different to 1…”. But you’re making a comparison Willis never made. “…[Hansen’s] claim is that there is only one chance in a hundred that the 1987 warmth is a random result…” So the valid comparison is 2 out of 30, or 7%, vs. 1 out of 100, or 1%. Using your strawman statistics, Hansen is 700% wrong!
Richard S Courtney
July 2, 2018 1:55 am
Red94ViperRT10:
You demonstrate that M Courtney was right when he wrote,
“We should all beware of hindsight when making judgements.”
You attribute his post which said that to “Mike Courtney”, but M Courtney was christened Matthew (not Mike), and those who know him call him Matt (not Mike), and he calls me Pater.
Paraphrasing his comment, I say “We should all beware of our unwarranted assumptions when making judgements.”
Richard
jono1066
July 1, 2018 12:54 am
Its raining over here in Devon,
The MET office statistics page was checked yesterday (actually 6.5 hours ago as I didnt switch off till 0130 hrs), “10 % chance of rain through until mid-day”.
today it says the same , 10% chance of rain and jumps to 22% chance of rain at noon. The statistics were absolutely correct, A 10% value was proven right, it is raining, whats even better is the now blanket of yellow crayon from end to end across the whole south west of official thunderstorm warning
The only problem with this is that folks like to think 10% means its not going to happen and that the MET office looks further ahead than a couple of hours.
Gut feel seems to be far better at reality than statistics ever will be !
Steven Fraser
July 1, 2018 12:37 pm
Gee, I thought 10% chance of rain meant ‘will rain in 10% of the area’, or ‘will rain everywhere for 10% of the time period’…
John F. Hultquist
July 1, 2018 2:37 pm
Or there is a 10% chance there will be rain somewhere in the area. In our case (USA) over the last 50 years the areas have gotten larger and, thus, the forecasts more accurate.
Mary Brown
July 1, 2018 6:15 pm
Most statistical models that produce a probability of precipitation forecast calculate the probability that it will rain At a single point in a given time frame
Clyde Spencer
July 2, 2018 9:32 am
As a practical matter, that is not very useful. I want to know if I need an umbrella, not it those in the next county need an umbrella!
July 1, 2018 1:06 am
When looking at the bad things done in climate science, some skeptics can imply that ‘consensus scientists’ are self-serving and dishonest. It’s not like that; ‘they believe in‘ what they do. The problems in climate science are part of a more general problem in modern science. Mission failure. Many scientists, perhaps most, are careerists with no great commitment to a wider truth. For them, ‘truth’ is what they can first get themselves, then the rest of us, to believe. A game played with numbers. I have little doubt in my head that Mann regarded the statistical tricks he used to create the Hockey Stick from 3 tree rings as valid. Andrew Gelman, tells the long tale of how so much got so bad. The result of this mission failure is a replication crisis in science. One big cause of bad science being far too many ‘researcher degrees of freedom‘. This was once about cherry-picking, putting undue emphasis on small data sets, claiming significance where it isn’t, … mostly bad statistics. The climate saga added treating ‘models as data‘, not reporting what the actual model is, etc. Bad modeling. That’s a whole new dimension to the mis-science of ‘researcher degrees of freedom‘. When scientists’ methods allow them so many research degrees of freedom; they must almost always get it wrong.
Santa
July 1, 2018 1:45 am
When things get politicized it usually bring its own logic to. What supports us and our Agenda is logical and everything else is not. First they politicized the environmental movement then leftist politicians let them into scientific areas so they could infect those to.
Edwin
July 1, 2018 9:20 am
Santa, raised in the Conservation movement that begin in and around the turn of 20th Century I was fascinated with the modern environmental movement which really only began in late 1960s. I was intimately involved with the environmental movement. I can assure you that from early days the leaders in the movement planned it to be a political movement, though in early days it was not leftist just liberal. Worse, unlike the Conservation movement which assume humans are part of nature, both part of the problem and the solution, the modern environmental movement quickly made humans the bad guy, something in total opposition to nature. At first it was just evil corporations; today it is all humans, but especially those in Western Democracies that are supported by capitalism.
I do agree that leftist politicians allowed or in some cases forced science into the realm of big P politics. As research funding became more and more dominated by government it became easier and easier to politicize the game. Climate science is certainly not the only field corrupted.
HotScot
July 1, 2018 3:28 pm
Edwin
michael hart
July 1, 2018 6:13 am
Well said, mark Pawelek.
Climate science also suffers from a relative paucity of data, which is why they spend so much time and effort manipulating what they do have. It mitigates against levels of reproducibility seen in other scientific disciplines.
If somebody reports a significant new method of, say, producing pluripotent stem cells, then you can bet the farm that some competitors will be trying to reproduce it within a few weeks. Often quietly, if they have reason to think it suspect. This helps reduce confirmation bias and outright fraud, which can still occur.
Cli-sci is a smaller and more incestuous field where significant real-world data is more difficult and time-consuming to obtain, where ‘interpretaion’ of data can build a career, and where shysters can be retired before they are rumbled.
July 1, 2018 9:53 am
Thank you Michael,
July 1, 2018 11:44 am
Michael Hart
Climate science does not use real data.
– Over 99.999% of Earth’s history
has no real time temperature measurements,
– For surface temperatures since 1880,
most of the Earth’s grids have no data,
so are infilled with wild guesses by government bureaucrats
who have predicted a lot of global warming,
and would like their predictions to come true, and
– For surface measurements since 1880,
a minority of grids do have raw measurements,
but those data are not used to compile
a global average temperature.
What is used are “adjusted” data,
Once raw data have been “adjusted”,
they are no longer real data —
they are government bureaucrat estimates
of what the data would have been,
if measured accurately in the first place.
So, in conclusion, climate science does not use any real data
to compile the global average temperature … which is a
temperature no one actually lives in !
Not that having perfectly accurate temperature data
since 1880, or even for thousands of years,
would help us predict what the future climate will be !
My climate change blog:
http://www.elOnionBloggle.Blogspot.com
HotScot
July 1, 2018 3:30 pm
Richard Greene
Brilliant, thank you for a post a layman can understand.
richard verney
July 1, 2018 1:08 am
Willis
Gotta say, those are some significant changes. In the old GISS record (red), 1920 to 1950 were much warmer than in the new record.
This is particularly concerning since we know from the Climategate emails, that the Team quite blazenly said that they have to get rid of the 1940 warming.
We therefore know that the reduction in the 1940 warming, and the ensuing revisions to past warming was deliberate. These adjustments go even beyond a priori bias, because we know that the Team were aware that the 1920 to 1940 warming episode causes real problems to displacing the null hypothesis, ie., that all variations in the temperature record are of natural origin, and deliberately went about the task of flattening the warming out of the record.
In my opinion, you proved the adjusted temperature reconstruction is wrong, the other day when you posted your interesting article on stacking up volcanos. In my opinion, the reason why you were unable to find the signal to volcanos in your analysis, is because the data sets (and they are not data) have been so bastardised by endless adjustments that they are now no longer a reliable source for real world temperature change.
PS. I am sure that someone who is more familiar with Climategate than I am, can post the Climategate email that I am thinking of.
Nick Stokes
July 1, 2018 2:42 am
If they do, I think you’ll find that they didn’t say that.
Charlie
July 1, 2018 2:48 am
September, 2009. Tom Wigley writes to Phil Jones:
Here are some speculations on correcting sea temperatures to partly
explain the 1940s warming blip.
If you look at the attached plot you will see that the land also shows the
1940s blip (as I’m sure you know). So, if we could reduce the ocean blip
by, say, 0.15 degrees Celsius, then this would be significant for the global
average—but we’d still have to explain the land blip.
I’ve chosen 0.15 degrees Celsius here deliberately. This still leaves an ocean
blip, and I think one needs to have some form of ocean blip to explain the
land blip …
It would be good to remove at least part of the 1940s blip, but we are still
left with “why the blip?”
Nick Stokes
July 1, 2018 3:25 am
He’s trying to explain the blip. If they are still left with “why the blip?” they haven’t got rid of it. And anyway, as he said, it was speculation. No adjustment actually happened as a result. Neither Wigley nor Jones had custody of any SST data.
Wim Röst
July 1, 2018 4:03 am
Tom Wigley: “if we could reduce the ocean blip”
WR: No. Facts are facts, data are data. In Holland professors are dismissed when they manipulate the data to create a certain result. Because of fraud. In Dutch:
https://nos.nl/artikel/554459-overzicht-fraude-in-de-wetenschap.html
https://nl.wikipedia.org/wiki/Diederik_Stapel
A scientist can not discuss ‘reducing the ocean blip’ when there IS an ocean blip. Tom Wigley is talking about reducing “the ocean blip” that was found in the data.
But, as you say, Nick Stokes, “Neither Wigley nor Jones had custody of any SST data”. So they COULD not change the data.
Even the intention to change data is contra-scientific. In Holland: close to effective scientific fraud which is concluded when data are changed to get a certain (desired) result. Resulting in a dismissal.
Nick Stokes
July 1, 2018 4:27 am
In fact, there is no suggestion that raw data would be changed. The question is, what does it mean? There is something odd here – it could be an error. How much error?Scientists are allowed to think about that.
Phoenix44
July 1, 2018 4:58 am
Except that is not remotely what the quote says. So yes, if the quote actually said something along the lines of what you say, perhaps. But it doesn’t. That’s how words work you see.
Nick Stokes
July 1, 2018 11:29 am
What the quote certainly does not say was the original claim:
” the Team quite blazenly said that they have to get rid of the 1940 warming”
In fact they are talking about an odd-looking dip in 1945. And wondering if it is an artefact.
richard verney
July 1, 2018 1:46 pm
Nick,
I have spent about 30 years studying ship’s data and I know that it is anything but robust and not fit for scientific study.
You state:
In fact they are talking about an odd-looking dip in 1945. And wondering if it is an artefact.
They are not talking about a dip in just one year, ie., 1945. It was a dip occurring during the war years, in particular during the convoy era.
During the war, shipping lanes changed, and ships followed closely hurdled close to one another in convoys such that the ocean was being agitated far more than would normally be the case. During the war years, one is simply not measuring like with like and that is before one even considers the inherent problems with bucket measurements. Later, of course, there was the change between bucket measurements and water cooling inlet measurements.
But whilst there may be reasons to consider that post 1940 data is suspect, there is no reason at all to reduce the 1920 to 1940 data. The peak of 1940 should stand.
The issue with the war years data should have been dealt with by setting out a wider margin of error during that period, and again during the change from bucket to inlet manifold.
Joseph Murphy
July 2, 2018 7:19 am
Nick, it is not odd. There is a complete lack of oddness. It is in the wrong direction.
Nick Stokes
July 2, 2018 11:18 am
“wrong direction”
For whom? Folks here are always complaining that scientists are always cooling the past. But the suggestion here is that late 1940’s SST should have been warmer. The 1945 drop was spurious.
Clyde Spencer
July 3, 2018 7:56 pm
” the Team quite blazenly said that they have to get rid of the 1940 warming” That is not the same as saying “I don’t understand that blip.
We need to be able to explain it.”
Wim Röst
July 1, 2018 5:05 am
Nick: “There is something odd here – it could be an error. How much error?Scientists are allowed to think about that.”
WR: perhaps it is subtle, but there is a big difference in searching for errors at random and searching for SPECIFIC errors as you do when you are specifically searching for errors that REDUCE a certain data result: Tom Wigley to Phil Jones “if we could reduce the ocean blip by, say, 0.15 degrees Celsius, “.
The last sentence has the intention to change the data in a specific direction. Isn’t that the intention to scientific fraud?
R. Shearer
July 1, 2018 6:26 am
And they did in fact change the record and they are at it today.
Red94ViperRT10
July 1, 2018 1:11 pm
I need the assistance of those more gifted in WayBack Machine than I…, I became interested in the Global Warming thing because I had to. Back in 2006 I took a consulting job to look for energy conservation measures at a U.S. military installation. As part of the contract we attended the conference each year, held at various places, to discuss various aspects of government energy policy. The conference in 2008 was held in August in Phoenix, AZ, and it seemed to me it was virtually dedicated to Global Warming. A keynote speaker (can’t recall his name, and although the website presenting the “minutes of the meetings” if you will lists keynote speakers, none rings and bell, and there is no indication of the time of day they actually presented, so that’s no help. It could have been Gavin Schmidt himself, but he’s not listed on the program or minutes) the second day began his speech with, “You don’t need to Google, or do an internet search, on Global Warming. Everything you need to know about climate is right here…” and gave the web address for Real Climate. That’s right, Gavin Schmidt’s (supposedly his personal, but done while he is on the clock at his day job, working for the government) blog. The speaker went on to say the time for debate was over, the discussion was closed, and there would BE no further discussion. All of which got my back up, so to speak.
So when I got back to a computer, I did exactly what he told us not to do… I Googled Global Warming. As an auto-complete suggestion, I saw, “Global warming scam…” I clicked on that. What I read was astounding. That’s how I discovered WUWT, it still showed up near the top of the search results in Google back then.
By the end of the week, though, I wanted to know why it was so important to the speaker that I NOT pursue any other information aside from that presented on the Real Climate website. So I went there. I quickly deduced why I was not to visit anywhere else, I could quickly tell all of the website was completely one-sided, and no debate or discussion was allowed. Curiosity, though, compelled me to click on the right-hand tool-bar, where it was organized by the author of the post at the beginning of each thread. I think I picked Gavin Schmidt (80% confidence level *smirk*). Now here is where my memory gets fuzzy, I think I picked the opening post of a random thread (50% confidence level, it might have been a reply to another comment), in which, about half-way down the page the author proclaimed (this is from memory, so may not be exact),
“We know Global Warming is happening, all the models say so, but we’re not seeing it in the records. So clearly, the records must be wrong. (italics mine, if they show up). But we have somebody working on that.”!!! (Exclamations mine)
Imagine that for a second, he’s admitting not only that there is no Global Warming in the data, but also that he has assigned people to set about changing the data!!! That is when I became not just a skeptic, but possibly even a conspiracy theorist. But this is where I need the WayBack Machine drivers, I can’t find that post anymore! It was made sometime in the second half of 2008 (very remotely possibly as late as 1st quarter 2009). And beyond that, I have given everything I can recall that may help uncover it.
End rant.
Robert W. Turner
July 1, 2018 7:04 am
ROFL… Man science is easy, when after collecting data that doesn’t conform to your preconceived notions, you simply invent an “error” and adjust the data.
Nick Stokes
July 1, 2018 10:38 am
No, it doesn’t say they want to change data. The blip is a calculated average derived from data. They think that there is something wrong with that average, and are trying to figure out what might be causing it.
But there is context here. These are not secret plottings. Jones and others had already written a much discussed paper in Nature on exactly this matter. They thought the peculiar discontinuity was a changeover from buckets to engine intakes, and postwar variations in the intake. Those are things scientists should think about. And publishing a paper in Nature is a very public and proper way of doing it. And if an adjustment is made, they will have to very publicly persuade the actual managers of SST indices to do it. SST is actually a combination of different data types, and you have to figure out how to combine them. That is not data in itself, but calculation. Jones et al thought the calculation method needed to be revised.
HotScot
July 1, 2018 3:52 pm
Nick Stokes
Is there any scientific data relative to the heights of people reading mercury thermometers in the 18th/19th/20th Century?
Is there any data relative to the types of paints used on Stephenson screens globally?
Is there any scientific data on tea boys being sent out in the snow to record data when the resident ‘scientist’ was on his day off/on holiday/drunk/couldn’t be bothered?
Is there any data on the types of buckets used in ship readings?
Is there any data on the number of cabin boys sent to chuck a bucket over the side of a ship and not bother?
Is there any data that describes the wholesale falsification of climate data when these few variables are not intimately understood?
I could go on, but a cursory examination of historic recording methods ought to throw up many more conditions that blatantly undermine historic data.
Mann’s hockey stick describes temperature measurement down to tenth’s of a degree, but that’s simply not credible when proxy data going back hundreds of years is involved.
Torture the data and the methods all you want mate, the fact is, the wild claims of Hansen et al have not manifested themselves.
They were wrong then, and scientific observation proves them wrong now.
Nick Stokes
July 1, 2018 5:16 pm
” many more conditions that blatantly undermine historic data”
So what do you suggest? Give up on historic data completely? I can just imagine what people here would say then. Look at the criticism of BoM for not including data before 1910 in their ACORN set.
HotScot
July 2, 2018 1:43 am
Nick Stokes
I don’t think I suggested we give up on historic data. You conjured that up out the blue to score a point.
What I’m suggesting is that it should be treated with extreme caution and not used to bludgeon humanity to death with, as irrefutable fact. Just by way of example, you may recognise it:
“The problems with K/A dating are more profound, especially for the fossil record. A study in Hawaii gave an age for rock of 2.3 ± 0.3 million years. I don’t care about the specific age except that it is relatively recent geologically. What troubles me is the ± 0.3 million years. That is 300,000 years or a full error range of 600,000 years. How much happened in the last 600,000 years? Of course, as you go back in time the error increases. A one-million-year error range is not unusual.”
https://wattsupwiththat.com/2018/07/01/evolutionary-and-global-warming-theory-predictable-responses-with-no-empirical-evidence/
In fact, I could go further. I wonder if numerous scientific studies couldn’t be consigned to the bin because specific dates of historical events, based on paleo records are used, which don’t acknowledge the error ranges.
Clyde Spencer
July 2, 2018 9:36 am
Nick,
You said, “So what do you suggest? Give up on historic data completely?” Isn’t that what has happened with pH data?
DaveS
July 1, 2018 6:25 am
For goodness sake. There is a very clear and explicit order to that last sentence. Asking ‘why the blip’ comes after ‘it would be good to remove at least part of (it)’. Scientists most certainly are not allowed to think like that.
Nick Stokes
July 1, 2018 11:33 am
No, scientists often do think like that. There is something odd in that result that doesn’t look right. What would it take to remove it? It is a way of quantitatively looking for cause. It doesn’t mean you just change things to remove it, and Wigley, in thinking about that land component etc is clearly indicating that in thinking his way through it. But if you know there is a variation of something that could explain it, then you can focus on whether that something is right.
Michael Moon
July 1, 2018 10:06 pm
No Nick, scientists don’t. Real scientists say, “Oh hell, these numbers are no good, let’s get a better instrument and start again!”
And they do not publish until they have better numbers.
“Climate Scientists,” on the other hand…
Nick Stokes
July 2, 2018 2:54 am
“let’s get a better instrument and start again”
Hard to do with historic temperatures.
Robert W. Turner
July 1, 2018 7:01 am
Only in the crazy climate cult are statements like ” So, if we could reduce the ocean blip by, say, 0.15 degrees Celsius”, “I’ve chosen 0.15 degrees Celsius here deliberately.”, “It would be good to remove at least part of the 1940s blip” seen as normal ways to discuss the data.
Red94ViperRT10
July 1, 2018 1:18 pm
There is a big, yuuuge difference between, “That blip looks odd and is not supported by other data (in fact it WAS supported by other data, the “blip” appeared in both the land data and the SST data). Let’s look into it, such as instrument calibration, recording, verification, etc.” and “…It would be good to remove at least part of the 1940s blip…” Do you see the difference there? And Stokes, if you try to tell us those statements are identical, you lose all credibility. And you do have some left, BTW, but you’re pushing it.
Michael Jankowski
July 1, 2018 10:41 am
You are the biggest “denier” on the face of the earth, Nick. Ethically, you’re statistically insignificantly different from zero.
No, there’s no suggestion the “raw data would be changed”…raw data is raw data. It would be adjusted and reported data. Duh. Playing semantic games here as usual?
There’s certainly a suggestion on how to product adjusted data…”speculations on how to correct sea surface temperatures” is just that. They even discuss a “deliberate” selection of a quantity.
As with Hansen’s failed claims about increases in tornadic strength that you tried to spin-away as a fairy tale, the climategate quotes are plain and damning.
hunter
July 1, 2018 8:35 pm
Nick,
They were conspiring to deceive fellow scientists and the politicians naive enogh to trust them.
David Paul Zimmerman
July 1, 2018 6:12 am
Makes sense. Holland has a great need to deal with reality. Ocean level rise needs to be accurate when reclaiming land from the ocean via dikes.
Phoenix44
July 1, 2018 4:57 am
Really, you refuse to read that as a scientists saying he wants to change the data to fit his theory?
Let’s try this then: do we know that plenty of scientists in every field of science have done exactly that? Yes we do.
Does Nick Stokes consistently and constantly claim that climate scientists never do and never would? Yes he does.
Is somebody therefore being a bit stupid?
Robert W. Turner
July 1, 2018 7:06 am
I wouldn’t say stupid, it’s probably a psychological defense mechanism as a result of cultish brainwashing.
Nick Stokes
July 1, 2018 11:49 am
“Really, you refuse to read that as a scientists saying he wants to change the data to fit his theory?”
Yes. Firstly, one simple item that people just don’t want to deal with. He can’t change the data! He is a mid-level academic in a minor English University. The data is held by a large US Federal agency.
But second is this local enthusiasm for picking over stolen emails to glean truth, while ignoring the actual path of science. Jones and others had just published a much-discussed paper in Nature explaining why they thought the blip was due to a change in methods of SST measurement. That put the case for modifying the index squarely and publicly. Musings in a Wigley email are of no significance in comparison.
Richard S Courtney
July 2, 2018 2:26 am
Nick,
You wrongly assert that the Climategate emails were “stolen”.
All the ‘Climategate’ emails (including those from me) were leaked. They were NOT “stolen”.
The emails were of discussions of scientific matters that are not covered by intellectual property rights. Also, those emails do not contain statements that say they are subject to personal or commercial or financial or security or military confidentiality.
Assertions that the Climategate emails were “stolen” are attempts to side-track considerations of their contents.
Richard
Robert Austin
July 1, 2018 9:27 am
Nick Mannsplaining why Wigley and Jones should retain their halos.
Philip Schaeffer
July 1, 2018 7:53 pm
Yeah, well, to quote Richelieu:
“If you give me six lines written by the hand of the most honest of men, I will find something in them which will hang him.”
Fortunately in this case you can look at the actual science they produced. But ain’t nobody got time for that! There’s a witch to be burned.
July 1, 2018 9:55 am
If the land and ocean temperatures show the same ‘blip’, that means the ‘blip’ is real. So one should not get rid of it.
Rich Davis
July 1, 2018 10:54 am
You’re right as usual Nick. Conspiracy to commit murder is no crime, only the actual murder.
Nothing see here. Move on!
Nick Stokes
July 1, 2018 3:49 am
In fact, the “1940’s blip” was a topic of discussion at the time, and it wasn’t at all about “trying to get rid of 1940’s warming”. Quite the contrary. There is an article from the time about it here. There was a sudden dip in temperature in 1945. A lot of people thought it was an artefact of some measurement problem. A paper, of which Phil Jones was a co-author, had recently appeared in Nature, titled “A large discontinuity in the mid-twentieth century in observed global-mean surface temperature”. Their argument was
“We argue that the abrupt temperature drop of 0.3°C in 1945 is the apparent result of uncorrected instrumental biases in the sea surface temperature record.”
They did not think that the 1940’s warmth was wrong, but the SST temperatures immediately following 1945 were too low.
dh-mtl
July 1, 2018 6:00 am
Regarding this abrupt drop in temperature in 1945.
It seems that the drop in temperature between 1940 and 1960 has been either erased from or minimized in the various global temperature record databases. However it still shows up in many local temperature records, such as for the continental U.S. temperature or for the AMO index.
I also noticed in a recent post on WUWT that correlations with the global temperature record broke down over this same period. I have noticed this phenomenon in the past as well. To me, this indicates that corrections made to the global temperature record over the period of 1940 to 1960, which are apparently related to the measurement of sea-surface temperatures, are inappropriate.
Without these corrections, which lower temperatures prior to 1960 by approximately 0.3 – 0.4 C, the whole AGW narative falls apart.
ferd berple
July 1, 2018 7:41 am
There is the problem. Scientists placing their beliefs ahead of the data.
The data says the SST was cold. The scientists can’t explain why so they assume the data is wrong and needs to be adjusted.
They fail to consider that the data is correct and there is some fact they have failed to consider. Or more likely some factor they have not yet discovered.
And once they have adjusted the data the unknown factor will be permanently hidden from discovery. Holding back science for at least a generation.
Jeff Alberts
July 1, 2018 9:34 am
“They fail to consider that the data is correct and there is some fact they have failed to consider. Or more likely some factor they have not yet discovered.”
Or, that temperature is very localized, and doesn’t follow some phantom “global mean”.
Geoff Sherrington
July 1, 2018 8:40 am
Nick,
I was getting email from Phil Jones at the time, some about how land and sea temperatures for NZ matched each other, while those for Australia did not.
Today, after convoluted purposeful adjustments, not only are all matches good, but land and sea are now combined for global.
Can this be done by honest, open science? Geoff.
Michael Jankowski
July 1, 2018 10:46 am
Nick, you are flat-out lying or playing stupid again.
The emails explicitly refer to the “1940s warming blip.”
You’re pretending it was “a sudden dip in temperature in 1945” and that they were trying to explain aberrational post-1945 cooling.
If that were the case, they’d be talking about the 1940s COOLING blip. Do you need someone to explain the difference between warming and cooling to you?
How do you sleep at night?
Nick Stokes
July 1, 2018 12:32 pm
“If that were the case, they’d be talking about the 1940s COOLING blip. Do you need someone to explain the difference between warming and cooling to you?”
Well, here is Willis’ complaint about that from above:
“As a result, in the old record temperatures cooled pretty radically from about 1940 to 1970 … but in the new record that’s all gone.”
So there wasn’t actually a “1940’s warming blip” to remove. Willis is grumbling about the removal of cooling.
The context is clearly the paper that Jones had co-authored in Nature the year before. It begins
“Data sets used to monitor the Earth’s climate indicate that the surface of the Earth warmed from 1910 to 1940, cooled slightly from 1940 to 1970, and then warmed markedly from 1970 onward1 . The weak cooling apparent in the middle part of the century has been interpreted in the context of a variety of physical factors, such as atmosphere–ocean interactions and anthropogenic emissions of sulphate aerosols2 .
Here we call attention to a previously overlooked discontinuity in the record at 1945, which is a prominent feature of the cooling trend in the midtwentieth century. The discontinuity is evident in published versions of the global-mean temperature time series1 , but stands out more clearly after the data are filtered for the effects of internal climate variability. We argue that the abrupt temperature drop of 0.3 6C in 1945 is the apparent result of uncorrected instrumental biases in the sea surface temperature record. Corrections for the discontinuity are expected to alter the character of mid-twentieth century temperature variability but not estimates of the century long trend in global-mean temperatures.”
Michael Moon
July 1, 2018 10:13 pm
They must pay you a lot of money to make this stuff up, your explanations are nuanced and almost convincing, until we think, “Why is this man saying this, for the last ten years?”
You must be very proud of yourself puta
Frank
July 1, 2018 11:17 pm
Nick: My complaint with all of the SST adjustments is that they have been done from the desk. When someone goes out on a boat with a variety of buckets and a way to sample temperature of the water at the engine intake valve under various configurations, and a buoy and measures the difference in temperatures under a variety of conditions, then perhaps I’ll believe there is something scientific involved in adjusting SST records.
Nick Stokes
July 2, 2018 3:09 am
Frank,
Lots of people have tinkered with buckets, on ship and ashore, and engine intakes There is a review here. The more difficult area seems to be working out what mix of methods was actually used, rather than the variation of the methods themselves.
zazove
July 1, 2018 1:30 am
Meantime the soup gets warmer.
Editor
July 1, 2018 1:45 am
zazove
July 1, 2018 4:17 am
Why stop at the surface David? Cherrypick?
Robert W. Turner
July 1, 2018 7:43 am
Put on you geometry cap for a second. We can measure 100% of the sea surface temperature via satellites. It is a 2D geometry or approximately 150,000,000 m^2 and all heat in and out of the ocean must pass through this 2D surface.
Now think of 2,000 m of ocean depth. This is a 3D shape of approximately 1X10^12 cubic meters. A single transect, essentially a 1D line passing through this shape, is an infinitesimal % of the overall volume, and even 2,500 transects adds up to a very small % of this volume actually being measured. Heat can be transferred to or from the atmosphere (must pass through the 2D skin of the ocean) but also to and from the 1,750 m of ocean below where no heat content data is collected.
In other words, the surface temperature is a well constrained measurement with little error, whereas heat content of the upper 2,000 m of the ocean is highly uncertain and meaningless when you consider that heat can be transferred to or from a place where we aren’t even looking.
Edwin
July 1, 2018 9:39 am
Robert, Good explanation!
Since probably before we put missiles on a nuclear submarine the Navy has been “mapping” the “thermal structure” of the oceans. Why? to hide submarines or find out where they might be hiding. I have often wondered how the Navy data would match to other data. My guess is that it certainly would be a better estimate of the thermal capacity of the oceans. Of course such data remains Top Secret or did the last I checked.
Nick Stokes
July 2, 2018 2:34 am
“the surface temperature is a well constrained measurement with little error”
That isn’t a common view here. But satellites aren’t magic. In fact, they aren’t used in SST indices. Wiki notes:
“here are several difficulties with satellite-based absolute SST measurements. First, in infrared remote sensing methodology the radiation emanates from the top “skin” of the ocean, approximately the top 0.01 mm or less, which may not represent the bulk temperature of the upper meter of ocean due primarily to effects of solar surface heating during the daytime, reflected radiation, as well as sensible heat loss and surface evaporation. All these factors make it somewhat difficult to compare satellite data to measurements from buoys or shipboard methods, complicating ground truth efforts.[17] Secondly, the satellite cannot look through clouds, creating a cool bias in satellite-derived SSTs within cloudy areas.[2] However, passive microwave techniques can accurately measure SST and penetrate cloud cover.[13] Within atmospheric sounder channels on weather satellites, which peak just above the ocean’s surface, knowledge of the sea surface temperature is important to their calibration.[2]”
richard verney
July 2, 2018 1:02 am
Pre ARGO, there is all but no measurements down to 2000 metres. Talk about sparse data!
And unfortunately, ARGO was corrupted as soon as it was rolled out, which has forever tainted its data. When first rolled out, it showed cooling. Due to a priori bias, this was thought to be erroneous. The solution was to simply delete from the data set, the buoys that showed most cooling. This was done without returning a random sample of buoys to the laboratory to see whether there was any equipment fault/calibration error.
Wim Röst
July 2, 2018 2:12 am
“More recently, Argo buoys were field calibrated against very accurate CTD (conductivity-temperature-depth) measurements and exhibited average RMS errors of ±0.56 C. [37] This is similar in magnitude to the reported average ±0.58 C buoy-Advanced Microwave Scanning Radiometer (AMSR) satellite SST difference. [38]”
Alley
July 1, 2018 6:41 am
Why start at the last El Nino? You should always have at least 10 years and start-stop at the same ENSO.
Alley
July 1, 2018 8:18 am
Why start at 2015.5?
Sunsettommy
July 1, 2018 1:55 am
Translation: I have no counterpoint against the blog post, therefore try to create a deflection with a pretty chart.
lee
July 1, 2018 2:33 am
And in Zetajoules no less. 😉
zazove
July 1, 2018 4:29 am
No, I just don’t see the point on focussing on Hansen and how accurate or otherwie 30 yo scenarios are. The atmosphere is far more prone to shorter term variations than 2km of seawater so if you want to know what the trend is – that pretty graph is a good place to start. Who know though, maybe its getting colder down deeper and that graph is misleading. Willis?
zazove
July 2, 2018 5:10 am
None the less an indicator far less prone to the vicissitudes of short term fluctuation and as that graph indicates: the warming proceeds apace regardless of the accuracy oc 30yo scenarios.
July 1, 2018 6:14 am
Heat capacity of water = 3993 J/kg/K
Total mass of water on earth = 1.35 10²¹ kg
I don’t know about the chart above, but when I looked at NOAA’s data I saw a 180 ZJ rise in 30 years. 6 ZJ/year.
So, total heat capacity of earth’s surface water: 3993 × 1.35 . 10²¹ = 5390 x 10²¹
Divide that by 6ZJ, to get an average ocean temperature increase of about 0.001 C per year. In 1000 years time, oceans could be 1 C warmer at this rate. Sadly, for all I know the next deep glaciation may have kicked in by then.
Shanghai Dan
July 1, 2018 10:59 am
You stop that! A value of 0.001 deg C is nowhere near as apocalyptic sounding as 60,000,000,000,000,000,000,000 Joules per year! You’ll get people no longer screaming about impending DOOM!
Rich Davis
July 1, 2018 11:10 am
Yes, but if they change the aspect ratio of the chart so that the y axis is much taller and the time base is compressed, it would be much more alarming.
Menicholas
July 1, 2018 10:50 pm
Yes, the data should really be presented in Hiroshima’s of energy.
Hey, where are the error bars on that graph?
Alley
July 1, 2018 6:26 am
No matter what you plot, it’s definitely warmer. They love woodsfortrees, so here it is from 1980:
Sunsettommy
July 1, 2018 11:31 am
Alley, who claims it isn’t warming since 1979?
Also have you noticed that you just smashed the AGW conjecture since the warming rate from 1979 are about the same or cooler than earlier warming rate trends going back to the1800’s.
Dr. Jones BBC interview
http://news.bbc.co.uk/2/hi/8511670.stm
There is no indication that CO2 is driving the undenied warming trend since 1979.
dennisambler
July 1, 2018 6:31 am
Shades of CRU graphical presentation, meaningless. This begs for the before and likely after, so a full cyclical picture can be seen.
Alley
July 1, 2018 6:40 am
Great chart. Thanks.
Alley
July 1, 2018 8:18 am
Great graph.
Jeff Alberts
July 1, 2018 9:35 am
Zazove, do you think that ocean heat content is controlled by atmospheric CO2?
Red94ViperRT10
July 1, 2018 1:26 pm
I see hand-waving, red-herring, and strawman all in one pretty little chart. Because it all comes down to, so what? Of course the temperature of the oceans change, they have been doing it for millions or even billions of years, and undoubtedly will continue to change in the future. Various proxy data indicates it has been both warmer and cooler than anything measured in our very, very, very, very short instrumental temperature record. Did I mention that you have presented a short record? You show <60 years of data (it doesn't even go back as far as the 1940s "blip" than began this thread), and I'm pretty sure there is proof we have had oceans for several billion years, what percentage is that? I am not alarmed. FAIL.
zazove
July 2, 2018 5:20 am
If the graph had been going down I’m sure you’d find it far less disturbing. No comment on David’s 3 year plot?
Peta of Newark
July 1, 2018 1:37 am
The Human Animal cannot pass off untruths – it is a very discomforting thing to even attempt to do.
Even ‘by omission’ because the worst person for keeping a secret is the very person whose secret it is.
Government Types will claim they ‘miss-spoke’ and if you kick up a fuss, will hit you with a new tax.
Sometimes they get it wrong and give themselves away.
Big beautiful shiny diamond was buried within the Martian dust-storm story.
The ‘scientist’ told us that dust storms of the Martian style couldn’t happen on Planet Earth because:
Planet Earth “has plants”
And THAT is ALL you need to know.
Peta of Newark
July 1, 2018 1:44 am
PS You now know why Hansen was so hot & sweaty during his ‘testimony’
Cute huh
MattS
July 1, 2018 1:40 am
It isn’t CO2. ERBE shows that quite clearly. As the earth has got warmer it has emitted more, not less energy.
Lawrence Todd
July 1, 2018 1:44 am
I am happy to learn that the post 1945 era that I grew up in was not as hot as I remember.
Steven Mosher
July 1, 2018 1:48 am
Good job on the Hurst aspect and the other nits.
On the temperature, I believe 1988 would be land only, and not a very complete record as it predates construction of ghcn.
An apples to apples comparison is hard, because station inventories change and there are needed adjustments.
Nevertheless hansen’s achievement is spectacular given the complexity of the system. History will record the broadly correct findings and ignore the nits.
Sunsettommy
July 1, 2018 1:53 am
Dr. Hansen’s original 1988 chart (posted above) stated that it is a GLOBAL Temperature Trend, in fact ALL of the charts posted are Global based data.
Nick Stokes
July 1, 2018 2:40 am
He also said, in the paper with Lebedeff which introduced the Ts index:
“The principal limitation of this data set for global or hemispheric analysis is the incomplete spatial coverage, illustrated in Figure 1 for four dates. Although the number and geographical extent of recording stations on land areas increased strongly between 1870 and 1900, there were still large areas in Africa and South America, and all of Antarctica, without coverage in 1900. Substantial station data for Antarctica begins in the 1950s. Large ocean areas remain without fixed meteorological stations at all times.”
Sunsettommy
July 1, 2018 4:15 am
Uh huh so what!
Nick Stokes
July 1, 2018 4:31 am
A whole lot of other station records were collected later (GHCN), and this filled in gaps and affected the estimated global average a little.
Robert W. Turner
July 1, 2018 7:57 am
http://noconsensus.files.wordpress.com/2010/01/ghcncrucompare21.png
verb
1.
join (something) to something else so as to increase the size, number, or amount.
Robert W. Turner
July 1, 2018 7:49 am
But then he stands in front of Congress and tells them the graph he is showing is “global average temperature”, doesn’t discuss uncertainties and errors, and instead says he is 99% certain. So you’re in agreement that his 1988 testimony to congress was political and not scientific?
Nick Stokes
July 1, 2018 11:08 am
” doesn’t discuss uncertainties and errors, and instead says he is 99% certain”
It isn’t true that he didn’t discuss errors. Here is his Fig 1 showing the uncertainties, with a range similar to the variation we have seen between his version then and later versions. His 99% relates to the question of whether, even allowing for error, the graph is showing a rise that could have happened by chance.
Clyde Spencer
July 1, 2018 12:35 pm
Why do there appear to be fewer “error estimates” than annual mean temperatures, and why are they placed on the lines connecting the data points? Can we add Hansenization to Karlization?
Red94ViperRT10
July 1, 2018 1:53 pm
I don’t see any error estimates on the trendlines, or near the data. I see a footnote sort of thing at the bottom of the chart, indicating “Error estimates (95% confidence)”. I never took a statistics class, and clearly Hansen didn’t either, but we did some discussion of data and error estimates in Mechanical Engineering classes. I don’t think that’s how we were taught to do error estimates. I don’t see his supporting data or supplemental information where he calculated the error, so we can’t tell for sure.
Nick Stokes
July 1, 2018 5:05 pm
This is 1988. You can’t just wave the mouse and get a shaded colored background in the graph. At least, not easily. I think where I was we were still using mechanical plotters with pens. The estimates vary gradually and are given at sample length. Even then it is cluttered.
richard verney
July 2, 2018 1:10 am
Fair enough, but they are not realistic. Who in their right mind would think that the margin of error is only about 0.1 to about 0.18 degC.
Clyde Spencer
July 2, 2018 9:58 am
Nick,
I had an Atari with good graphics and a spreadsheet in 1979, and an Amiga with superior graphics, an improved spreadsheet, and a decent Paint program in 1984. While I didn’t get my first IBM PC until 1991, the first IBM PC came out in 1981 and many, if not most, businesses and government agencies had them by 1988. I would be surprised if NASA didn’t have them in abundance in 1988. I was routinely using a desktop PC at work in 1988, and had access to color printers.
Edwin
July 1, 2018 9:49 am
Nick, SO what Lebedeff was saying is there are large areas of the Earth, especially the oceans, with no temperature data especially historical data. And for portions of the Earth there is probably data that is inaccurate. In other words there is a huge amount of uncertainty in the data being used and that is before adjustment.
Nick Stokes
July 1, 2018 10:58 am
Well, we now have a measure of the uncertainty caused by lack of stations. It is shown in Willis’ discrepancy plot which varies by about 0.1°C. But it doesn’t change the broad conclusion. The old index with limited stations and the new both followed a similar upward path.
Red94ViperRT10
July 1, 2018 1:35 pm
Well, we now have a measure of the uncertainty caused by lack of stations. It is shown in Willis’ discrepancy plot which varies by about 0.1°C.
I don’t think that’s what it shows at all. Comparing inaccurate data to inaccurate data doesn’t prove anything, least of all how inaccurate either (or both) data set(s) may be.
Matthew R Marler
July 1, 2018 10:09 am
Nick Stokes:
He also said, in the paper with Lebedeff which introduced the Ts index:
His Congressional testimony and comments to reporters and other non-peer-reviewed communications were more alarmist than what he wrote in his peer-reviewed publications.
Rich Davis
July 1, 2018 11:39 am
Oh ok, that explains everything then. There was a 1% chance it occurred by random variation except that there was a 33% chance that the data was crap. Is that the explanation?
Michael Jankowski
July 1, 2018 11:46 am
“…The principal limitation of this data set for global or hemispheric analysis is the incomplete spatial coverage…”
Fact: It was STILL global and not just land, so Mosh was wrong.
Fact: There is STILL “incomplete spatial coverage” today.
Fact: You’ve gone out of your way any number of times to show that a very limited number of stations are needed to accurately represent global anomalies, so why are you going out-of-your-way to defend Mosh’s false claim with this “incomplete spatial coverage” point?
richard verney
July 2, 2018 1:14 am
Mosh himself also suggested that one needed only about 50 stations (but he might have placed a caveat that they be spatially evenly distributed).
Red94ViperRT10
July 1, 2018 1:30 pm
Now I read that to mean he didn’t have enough data to make any meaningful projections, and then he went ahead and did it anyway!
Ray
July 2, 2018 12:06 am
“Nevertheless hansen’s achievement is spectacular given the complexity of the system.”
May history please quote you on this, as well?
Nick Stokes
July 1, 2018 1:51 am
Willis,
“So I went and digitized the dataset above so I could use Dr.Hansen’s data”
GISS has a handy history page, which not only graphs the old data, but has numbers for download. There is a file of old TS versions here. It includes a 1987 version, which is what Hansen said he used.
“here are the changes between the version of the GISS temperature record that Hansen used in 1988, and the 2018 version of the GISS temperature record”
It is a very different set of stations. GISS Ts in 1987 is from Hansen and Lebedeff. GHCN did not exist then; they used mainly WWR data. They showed a plot of data available; it has a peak of about 1800 in the years 1950-75, but then a substantial drop-off, down to under 1000 in 1980. So the 2018 version has at least three times the number of stations.
“More to the current point, the post-1988 divergence between the HadCRUT and the GISS record is enough to rule out any possibility of determining whether Hansen was right or wrong.”
This is not like with like. GISS Ts, which Hansen used, is land stations only. HADCRUT is land and SST, but dominated by SST.
richard verney
July 1, 2018 2:26 am
Nick
Always good to see your comments, but I did not see any comments from you on the recent article by Willis on stacking up volcanoes (https://wattsupwiththat.com/2018/06/25/stacking-up-volcanoes/)
If you have time, it would be useful to have your comments on that interesting article, in particular, since you are an expert mathematician, whether you have any issues with the maths behind Willis’ analysis.
Willis found that with respect to the 24 largest volcano eruptions, these did not show up in the BEST and CET temperature reconstructions as producing even a short term cooling.
No one raised issues with the methodology of Willis’s analysis, and Willis concluded that contrary to popular views, volcanos do not lead to cooling. Personally, whilst I consider that Willis’s analysis supports that conclusion, it is nonetheless incorrect, and the real conclusion to be drawn from that analysis is that the temperature reconstructions are not fit for scientific scrutiny, and it is because they are defective that one cannot see the cooling effect of any one of the 24 largest volcano eruptions.
You have alluded to the reason why, namely that the set consists of continuously changing stations such that one is never able to compare like with like, ie., a point similar to your comment: “It is a very different set of stations. GISS Ts in 1987 is from Hansen and Lebedeff. GHCN did not exist then; they used mainly WWR data. And of course, Steven Mosher makes a similar point, namely: An apples to apples comparison is hard, because station inventories change and there are needed adjustments.
As I often suggest to you, what is needed is to isolate say the best 200 stations, least impacted with local environmental change these past 80 or so years, retrofit these (with same LIG thermometers, enclosures painted with the same historic paints), and simply compare modern day unadjusted RAW data collected at each station, using the same TOB as used at each individual station, with the historic RAW data for that station, ie., simply compare each station with itself, and not try and construct some hemispherical or global construct.
Please, if you have time, have a look at Willis’s analysis and please comment on why you consider it failed to reveal the impact of volcanoes
Nick Stokes
July 1, 2018 2:34 am
Richard,
I don’t have a strong opinion on the matter. It seems to me that sometimes (eg Pinatubo) volcanoes seem to have an effect; sometimes it is hard to detect, and there may in fact be none.
richard verney
July 1, 2018 2:53 am
Thanks, Nick.
Wim Röst
July 1, 2018 4:50 am
Nick Stokes: “This is not like with like. GISS Ts, which Hansen used, is land stations only. HADCRUT is land and SST, but dominated by SST.”
WR: You are right that it is not like with like. But Hansen was suggesting that he was telling about the warming of the Earth. And the Earth is dominated by the oceans, also in extension: 71% of the surface is ocean. So HADCRUT is a good choice to compare with the graphic of Hansen as shown above, a graphic which has the title “Global Temperature Trend”. Global. Exactly what HADCRUT is supposed to demonstrate.
Hansen should have said that he was using mostly Land Stations which could give a very different picture than the picture for the Earth as a whole. That would have been correct. It was Hansen who was not correct and preferred to give a biased view on temperature developments on the Earth. Well seen by Willis.
Nick Stokes, another question for which I am interested in your answer: how much might Land temperatures go up without (!) a comparable rise in Ocean temperatures, more than a half degree? If so it has to be explained in which way (!) oceans can rise as fast in temperature: by which mechanism?
Nick Stokes
July 1, 2018 10:54 am
“Hansen should have said that he was using mostly Land Stations”
Hansen was not actually using stations at all for the prediction. He was calculating a global air temperature with a GCM. He compared it with an air temperature measure; the best available at the time. Indices made of combined air temperatures over land and SST for ocean simply did not exist at the time.
It is clear that land temperatures can and do rise faster than SST during warming. I don’t know how far they could get out of line, but it clearly has, by a fraction of a degree. During that 33 year period from 1988 to 2017, HADCRUT rose by 0.53°C, while CRUTEM rose by 0.85°C. HADSST3 rose by 0.42°C.
Wim Röst
July 1, 2018 12:57 pm
Nicke Stokes: “I don’t know how far they could get out of line, but it clearly has, by a fraction of a degree.”
WR: As I have shown here: https://wattsupwiththat.com/2017/01/26/warming-and-the-pause-explained-by-wind-upwelling-and-mixing/
sea surface temperatures can get ‘out of line’ simply by more or less wind over the oceans: more or less mixing of surface waters or more or less cold deep upwelling will result in important changes.
As the graphics of Ole Humlum show, most time a change of temperature of the oceans is followed by a change of the temperature of land – and not the other way round: page 41 from http://www.climate4you.com/Text/Climate4you_May_2018.pdf
If the oceans as a whole are only warming by a 0.001 degree a year (see calculation above of Mark Pawelek), any ‘dangerous warming’ will soon find its limits in the but slowly rising overall temperatures of the oceans. Even if there is a mechanism (less wind) which temporarily could enhance surface temperatures over the oceans. But as soon as wind will be enhanced, cooling of the oceans will be the result.
To continue ‘fear mongering’, a mechanism must be known of how to definitely break the strong relationship between ocean temperatures and land temperatures, something else than ‘urban heat island effects’ that influence land temperature measurements. If ‘climate science’ does not know a mechanism other than ‘temporarily less wind over the oceans’, all climate scientists should say that any warming over decades or over a century must be very limited. Because no other mechanism is known to heat the oceans faster than with a 0.001 degree in a year.
Without such a mechanism, the relatively fast warming over the last century should be near its limits.
Clyde Spencer
July 1, 2018 10:15 am
Largely because of the difference in specific heat of water and land, it is inappropriate to conflate the two temperature sets. But, I’m not really telling you anything that you don’t already know, but willfully choose to pretend isn’t true.
Red95ViperRT10
July 1, 2018 2:20 pm
Stokes This is why you have credibility left. You are great with actual data, and data sources. It’s the conclusions you reach from said data that boggle my mind!
richard verney
July 1, 2018 1:52 am
Willis
I suspect that your own article was written contemporaneously with that of Clyde Spencer’s (https://wattsupwiththat.com/2018/06/30/analysis-of-james-hansens-1988-prediction-of-global-temperatures-for-the-last-30-years/).
In your article, you note, quoting Hansen himself, that:
The warming is almost 0.4 degrees Centigrade by 1987 relative to climatology, which is defined as the 30 year mean, 1950 to 1980 and, in fact, the warming is more than 0.4 degrees Centigrade in 1988. The probability of a chance warming of that magnitude is about 1 percent. So, with 99 percent confidence we can state that the warming during this time period is a real warming trend.
As Clyde points out:
“The warming is almost 0.4 degrees Centigrade [sic] by 1987 relative to … the 30 year mean, 1950 to 1980 … The probability of a chance warming of that magnitude is about 1 percent.” The first graph, above, with the red line, shows that 0.3 °C would be a more accurate estimate. One should be suspicious of such a claim when his own data demonstrated that the temperature had already exceeded that for one season in 1981! Are we to believe that at least two events with a 1% probability occur within 7 years of each other? (my emphasis)
Further to that very point, Hansen ought to have reviewed the rate of warming for the 30 year period between about 1870 to 1900, and 32 years between about 1905 to about 1937. Whilst these are cherry picks it demonstrates that the rate of warming during the modern period is not unusual and is statistically indistinguishable from other periods of warming set out in his own plot.
I consider that the problem is not whether there has been a real warming trend, or a statistically significant warming trend. but whether such a warming trend can be considered unusual, such that some inference can be drawn as to cause, ie., inferring that the warming trend is caused by manmade actions.
I emphasise that the modern rate of warming is not statistically distinguishable from two earlier warming episodes, which earlier warming episodes were not caused by manmade action, and this undermines any inference to be drawn as to the cause of the modern warming period.
Shanghai Dan
July 1, 2018 11:09 am
“Are we to believe that at least two events with a 1% probability occur within 7 years of each other?”
Of course! Every storm from here on out is to be labeled “a once in 100 year event”, because how else can we claim that things are getting So Much Worse and we need to Be Concerned?
Michael Jankowski
July 1, 2018 12:00 pm
Much of that is due to ignorance (reporters), sometimes willful (meteorologists and scientists).
A 100-yr storm event is precipitation that falls in a given amount of time. There is a 100-yr rainfall for a 24-hr storm event. There’s one for 48-hrs. 72-hrs. 15-minutes. 30-minutes. There’s one for ANY storm duration. So there are any number of possible durations for which a 100-yr storm event can occur in a given year. There’s isn’t one true “once in 100 year event.”
Now that’s not true for flood events…there is a “once in 100 year event” for flooding. But that is so out-of-whack for most locations because of changes in land use.
Alasdair
July 1, 2018 2:58 am
For me the concept of “Global Temperature “ suffers from cyclical logic; for it is merely the result of a series of definitions, assumptions and measurement techniques by which the result is obtained. A flaw in any one of these renders the result suspect and any change in the process negates the previous result.
Hence the never ending debates upon its value and behaviour. However it is a useful concept if treated with due respect for its vagueness; but should never ever be used as a basis for prediction. That is a matter for Chaos Theory to deal with.
zazove
July 1, 2018 4:36 am
That is why I prefer Ocean heat content (J) and seaice extent (km2).
esalil
July 1, 2018 3:05 pm
Sea ice volume is more dependent on heat content than sea ice extent.
zazove
July 2, 2018 5:24 am
But extent is easier to measure.
Jeff Alberts
July 1, 2018 9:43 am
“For me the concept of “Global Temperature “ suffers from cyclical logic; for it is merely the result of a series of definitions, assumptions and measurement techniques by which the result is obtained. A flaw in any one of these renders the result suspect and any change in the process negates the previous result.”
It’s not just that. Temperature is an intensive property of the point being measured. Averaging intensive properties from different locations doesn’t give you anything meaningful. Anomalies derived from such are also physically meaningless.
InterZonKomizar
July 1, 2018 3:08 am
Oh No! The truth is worse than we thought. Thank you Minister of Reality. And a quart of hero medals for Komrad Willis.
.
.
Sandy,
Minister of Future
Javier
July 1, 2018 3:36 am
but there is no climate in space
There is climate in most planets and large moons of the solar system, and interplanetary solar wind conditions and cosmic ray rates are usually referred as space weather.
Phoenix44
July 1, 2018 4:46 am
It’s simply nonsensical to use the 1950-1980 to define your data set and then say things outside that dataset are unusual. They might be, but they might not be. What you are trying to discover is whether they are. But you cannot assume they are, then exclude them from the data set and say “look, if I exclude them, they are excluded.”
You have to be able to show that your data set is “normal”, but that is entirely what we are trying to discover. Should the warming be in the normal data set or not?
Geoff Sherrington
July 1, 2018 8:55 am
The 1950 to 1980 base includes the great global climate shift of 1970-75 or so.
Not a good choice, in hindsight. Or then. Another reason to prefer actual temperatures to these silly animal things.
In my State of Victoria, there are old mile markers by the railway lines. Between Melbourne and Bendigo, IIRC, there are measurement errors that were tracked to the use of 3 or 4 different definitions of the length of the standard mile at the time. You need stable, reliable, understood base measurements for best results. Geoff
Clyde Spencer
July 1, 2018 10:28 am
Phoenix44,
I agree completely. Hansen used the standard deviation of the 1950-1980 data set, instead of the standard deviation of the 1958-1988 data set, to arrive at a probability for a measurement in the latter data set. He was comparing peaches and nectarines, inasmuch as there was some overlap. As I explained yesterday in my article, if Hansen wanted to establish the probability of the early-1988 seasonal average occurring as a random variation, he should have de-trended the data set containing the measurement, thereby removing the upward trending bias. He then should have calculated the standard deviation of the residuals. He should have known those things were necessary, but didn’t do them. That leaves two possible conclusions: He is incompetent in statistics, or he willfully lied for political reasons.
July 1, 2018 4:48 am
If course there is climate in space , Willis, never heard of the ‘solar wind’ ?
🙂
milocrabtree
July 1, 2018 5:53 am
Solar wind is what happens when we eat too many sunflower seeds.
John F. Hultquist
July 1, 2018 2:51 pm
We began feeding Black Oil Sunflower seeds to quail. We then noticed they take off faster and more vertically. We had no idea. Thanks for the explanation.
David L. Hagen
July 1, 2018 6:12 am
Thanks Willis for exposing those numerous errors.
Yes accurate Global Temperature must address BOTH Type A (Statistical) AND Type B (Systematic etc.) errors. Yes temperature statistics must include both locational, temporal (HURST) Climate Persistence variations, both non-random.
Systematic temperature errors also need to be addressed, per the BIMP international standard “Guide to the expression of Uncertainty in Measurement” https://www.bipm.org/en/publications/guides/gum.html e.g. as exposed by Anthony Watts Surface Stations Project. http://www.surfacestations.org/
Richard S Courtney
July 2, 2018 11:51 pm
David L Hagen,
You say,
“Yes accurate Global Temperature must address BOTH Type A (Statistical) AND Type B (Systematic etc.) errors. ”
Sadly, at present there cannot be a determination of accurate Global Temperature”. The two basic reasons for this are as follows.
1.
There is no agreed definition of ‘Global Temperature;.
Each group that provides data for ‘Global Temperature’ uses a unique definition of Global Temperature alters the definition it uses almost every month. Effects of this have been presented in this thread as discussions of
(a) whether Hansen’s 1988 data were global (as he said) or merely “land based” (as some here assert)
and
(b) whether it was right to ‘cool the past’ by use of different definitions of Global Temperature (i.e. different data sets collated in different ways and subjected to different data ‘adjustents’). The severity of the changes to Global Temperature GISS has obtained by choosing different definitions of Global Climate is clearly seen at a glance by this
http://jonova.s3.amazonaws.com/graphs/giss/hansen-giss-1940-1980.gif
2.
If there were an agreed definition of Global Temperature then the accuracy of any determination of Global Climate could not be known because there is no possibility of a calibration standard for it.
An assessment of these problems is provided by Annex C of the item at
https://publications.parliament.uk/pa/cm200910/cmselect/cmsctech/memo/climatedata/uc0102.htm
Richard
R Taylor
July 1, 2018 6:39 am
Sceptics should simply acknowledge Hansen’s second monumental achievement, the Hansen Factor. To determine the value of an unknown (e.g. global temperature), one multiplies the value of an economically essential known (e.g. carbon dioxide) by its Hansen Factor. Et voila, the answer to “where should humanity aim.”
Humanity, however, must let history decide whether the Hansen Factor eclipses the Hansenometer, a numerical device that changes the past and divines the future.
Robert W. Turner
July 1, 2018 6:49 am
And it’s even harder to say a prediction is right or wrong when it is about a graph of a highly interpolated average. It’s like arguing over the meaning of an abstract painting. It’s hard to say who is right and who is wrong, and more importantly, who cares?
It’s much easier to test a tangible prediction, like the West Side Highway being underwater this year. Someone check, any kayaks paddling down the highway yet?
Phil.
July 1, 2018 8:17 am
It’s much easier to test a tangible prediction, like the West Side Highway being underwater this year. Someone check, any kayaks paddling down the highway yet?
No one predicted that the West Side Highway would be underwater this year.
There was some flooding in April though.
mkelly
July 1, 2018 9:47 am
Phil Hansen predicted the Westside Highway would be under water.
https://realclimatescience.com/2018/06/hansen-got-it-right/
Jeff Alberts
July 1, 2018 9:53 am
“Phil Hansen predicted…”
Who’s Phil Hansen?
Editor
July 1, 2018 10:06 am
I think he’s one of the Hanson brothers…
Jeff Alberts
July 1, 2018 11:50 am
Since many people here misspell his name “Hanson”, you could be right, David.
Phil.
July 1, 2018 6:35 pm
On the contrary that link agrees with me, I suggest you read it.
Michael Jankowski
July 1, 2018 12:24 pm
April flooding was a temporary event due to the stormwater system failing to handle heavy rainfall. The West Side Highway has had “some flooding” regularly during storm events, including many times before Hansen’s prediction. That’s not what he was talking about.
Hansen’s claim was based on sea level rise and thermal expansion backing-up the Hudson River so that it would cover the West Side Highway.
Phil.
July 1, 2018 7:08 pm
Neither Hansen nor the interviewer said that as far as I’m aware, in fact Hansen described a storm. The West Side Highway does flood periodically, so saying ‘well the WSH will be flooded (again!)’ would be something of a local in-joke. The highway that he mentioned no longer exists in any case.
Richard S Courtney
July 2, 2018 2:54 am
” The highway that he mentioned no longer exists in any case.”
Hmmm.
In that case, Hansen’s prediction was the most wrong that was possible.
Richard
monosodiumg
July 1, 2018 8:10 am
>After doing this for years, I developed an innate sense…
July 1, 2018 8:20 am
I am going to watch and see. I say AGW is over and this year will be the transitional year. The present – next few years will be telling because we have increasing CO2( warmer), and very low solar( cooling). Prior to this time 1850-2005 natural climatic factors favored warming. Only in 2005 did that start to change and lag times have to be factored which brings us to year 2018.
In year 2018 the solar criteria the two solar conditions I have called for in order for solar to exert a more direct influence on the climate are in, which are 10+ years of sub solar activity in general followed by a time of very low average value solar parameters which are equal to or exceed in magnitude and duration of time typical solar minimums between so called normal solar cycles.
The geo magnetic field moderates given solar activity and because it happens to be in sync with solar (both magnetic fields are weakening) this will compound given solar activity.
My theory is very low prolonged solar conditions will result in overall oceanic cooling which is happening for the past year along with a slightly higher albedo the result cooling.
An decrease in UV /Near UV light equates to lower overall surface sea temperatures.
An increase in global cloud coverage, snow coverage, major volcanic activity equates to a slightly higher albedo.
The above tied to very low solar activity due to an increase in GALACTIC COSMIC RAYS ,decrease in EUV ,AP INDEX ,and the SOLAR WIND.
SOLAR IRRADIANCE – decreases by a very slight amount and is not the main reason for the cooling climate just a small part of it.
In ending I say it happens now moving forward and if it does not happen now( the cooling that is) I do not think it is going to happen.
Will find out now moving forward.
Jeff Alberts
July 1, 2018 8:26 am
“Now, I am either cursed or blessed with what I call a “nose for bad numbers”.”
Presenting a single number for “global temperature”, by anyone, is use of bad numbers.
Geoff Sherrington
July 1, 2018 8:28 am
Thanks, Willis.
Australia is similar, with older, plausible, official temperature data sets mismatched with modern ones. We have studied the Government Year Book records and CSIR summaries published before the 1950s, to the stage of using the same stations. These Year Books were like the National Bible for the state of the Nation. No matter how we try for a match, we find those years before 1950 or so being warmer than the modern reconstructions, with the effect of giving Australia global warming 1910 to 2010 of 0.9C versus 0.5C at most by the older data sets. The extra half degree of alleged warming comes mainly from recent official adjustments to pre-1950 data, often for no supportable reason.
In our case as well, there is official reluctance to do a proper comparison and explanation, with various trite explanations then dismissal.
If only it was not serious, as we see our industrial base has crumbled via huge increases in cost of electricity because politicians of all shapes are wedded to Paris agreement CO2 reductions. Geoff.
Nick Stokes
July 1, 2018 4:48 pm
Geoff,
I think you should list any record you have found in Year Books which are not also in the modern records.
Acorn data is of course adjusted. But I’m pretty sure you could calculate the same average using modern unadjusted data. Provided of course that you used the same station set, area weighting etc.
Geoff Sherrington
July 1, 2018 10:15 pm
Nick,
Several of us worked together a few years ago to get these results that Chris Gilham summarised on his web site.
http://www.waclimate.net/year-book-csir.html
It is a pretty thorough analysis with no adjustments or cherry picking by us. There are other documents if you find gaps in this one.
Our introduction is –
“Unadjusted temperatures published by the Weather Bureau in the mid 20th century indicate a warmer Australian climate before 1940 than calculated with RAW or ACORN adjustments and suggest warming from the 1800s to 2014 at approximately half the rate calculated by ACORN since 1910.”
Geoff.
Geoff Sherrington
July 1, 2018 10:37 pm
Nick,
It is all here in great detail.
Our summary intro was “Unadjusted temperatures published by the Weather Bureau in the mid 20th century indicate a warmer Australian climate before 1940 than calculated with RAW or ACORN adjustments and suggest warming from the 1800s to 2014 at approximately half the rate calculated by ACORN since 1910.”
http://www.waclimate.net/year-book-csir.html
Nick Stokes
July 2, 2018 2:48 am
Geoff,
I have looked at the spreadsheet. It is big, with many sheets. But I could not find anywhere that you have calculated averages with old and new data over the same period. It seems to be all calculating old data over an old period, and new data against some period finishing recently. So the differences could well be just warming over time.
I really would like to see evidence that you are finding old data that isn’t already in the current BoM dataset. They can read year-books too. In fact, the year books would have got the data from BoM.
Geoff Sherrington
July 2, 2018 4:37 am
Nick,
It is not as easy as that because of the different ways the data were aggregated, with different start and finish dates, some daily, some monthly, missing values etc.
As a guide to what can be extracted, have a look at the highest recorded temperatures for each site pre-1930s from the CSIR lists, then compare with the hottest day in the last decade or so of our 2014 analysis. The hotter of the pair, by a 2:1 majority, is the pre-1910 data from CSIR. Yet we are repeatedly told that that recent decade had the hottest number of the hottest days in Australia, a piece of propaganda not supported by the CSIR figures.
Geoff.
Nick Stokes
July 2, 2018 10:53 am
Geoff,
“As a guide to what can be extracted, have a look at the highest recorded temperatures for each site pre-1930s from the CSIR lists, then compare with the hottest day in the last decade or so of our 2014 analysis.”
This again mixes up climate and measurement issues. You can’t learn much about measurement by looking at non-matching periods. You need to look at alternate measurements of the same thing. Are those hot days the same in the CSIR and unadjusted BoM? I suspect they are. As I said, the reports you are looking at really came from the BoM anyway. A test would be just one case, one day or month, where CSIR and modern records differ in the same place. Then we could see if there was a reason, or if modern unadjusted records really had been altered. I have not found any cases where they were. Sometimes it is unclear whether the stations are really the same.
I did some of this in looking at Melbourne records here. I looked at hot days from long ago in the current unadjusted record, and compared with old newspaper reports. They always matched. One calibration point I keep in mind is Melbourne, 13 Jan 1939. I have known for 60 years that the temperature was 114.1°F. So I check records to see if they say that. The unadjusted modern records always do (OK, now it is usually 45.6°C).
Pamela Gray
July 1, 2018 8:30 am
For Hansen to describe a modern warming rise a significant event within the directly measured data set may indeed be significant within that tiny time span. But it is a gnat’s ass view of Earth’s history of temperature variations. This kind of science is nothing more or better than humanity’s thoughts of its first experience of a mass flood event. They would say it was significant, and likely the consequence of human related behavior that needed to be changed. Hansen’s legacy will, if future generations care about him at all, be remembered as the silly dance of a witch doctor, with an equally silly group-dance of fans.
SAMURAI
July 1, 2018 8:36 am
HADCRUT4 and GISTEMP are now worthless datasets due to all the data tampering.
Rather than adjusting CAGW’s global warming projections to better reflect the empirical evidence, CAGW advocates decided to adjust the empirical evidence to fit the CAGW projections, which isn’t how science actually works…
UAH6 global satellite data is the only reliable dataset remaining, and CAGW advocates have an impossible task of explaining the huge disparity that diets between UAH6 and GISTEMP:
http://www.woodfortrees.org/plot/uah6/from:1979/plot/uah6/from:1979/trend/plot/esrl-co2/from:1979/normalise/trend/plot/esrl-co2/from:1979/normalise/plot/gistemp-dts/from:1979/plot/gistemp-dts/from:1979/trend/plot/gistemp-dts/from:1979/trend
TDBraun
July 1, 2018 8:40 am
When a scientist makes a huge public prediction,
and then is in charge of the data that will test that prediction,
and then multiple times adjusts that data so that it is closer to his prediction…
… kinda makes you wonder about that data, right?
JimG1
July 1, 2018 8:56 am
Statistical significance only references sample error such as sample size or sample selection techniques,for example, which could cause the sample to not be a true representation of the universe one is attempting to describe and does not include possible instrumental and other error and does not imply causality. End of story
Common sense should tell one that looking at paleo data, even with the limitations caused by the proxies used for that data, there is no climate relationship to cause co2 to raise temperature that is of a causal nature. Some periods of time do show a potential for temperature to increase co2 in a potentially causal manner and that has physical sciences to back it up much more so than the supposed radiative effects of co2 upon temperature. Warmer oceans release co2 and we are a 70% water covered planet.
Bryan A
July 1, 2018 9:21 am
Willis,
In defense of Goddard (something few should attempt)
If…a warming atmosphere expands and forces the envelope further away from the planet…
Then…I could see them following temps for the reason of satellite altitude and potential atmospheric drag imparted on the lowest orbiting satellites.
Not that they could elevate any existing potentially affected satellites to a higher orbit.
At least the Geostationary (GOES) Satellites are far away from this potential.
Alley
July 1, 2018 10:06 am
Since adjustments to global temps (proper, of course) have obviously lessened the warming trend, why do so many people still think that adjustments did the opposite?
Rich Davis
July 1, 2018 12:19 pm
Wow, Alley. As Barbie said, “Math is hard.”
Adjust the temperatures from the 1920s to 1940s down and adjust the recent temperatures up. How does that “lessen” the warming trend?
Clyde Spencer
July 1, 2018 12:23 pm
Because the new temperature data that I have seen removed the warming in the ’30s, increasing the long-term warming trend.
Anthony Banton
July 2, 2018 10:37 am
“Because the new temperature data that I have seen removed the warming in the ’30s, increasing the long-term warming trend.”
No, as Alley said – not on the global index.
There is a decreased long-term trend of GMT’s, via warming the 30’s.
John F. Hultquist
July 1, 2018 10:51 am
Great post. Thanks.
Finally, as an aside, just what is an “Institute of Space Studies” doing studying the climate?
I wonder if it did not start with the issue of “drag on satellites.”
This would be of interest to many groups and could have been contracted out, or done in-house. Here I include this phrase “I really don’t care. Do U?”** — only because, well, just because I do care, but not enough to research the history of mission creep within GISS.
** I think we need to use this phrase as much as possible.
I think I might put it on a personal ID card along with a few other pithy sayings.
Suggestions accepted.
Julian Flood
July 1, 2018 11:22 am
Where’s the blip?
JF
Red94ViperRT10
July 1, 2018 2:35 pm
I don’t think there’s anybody back there!
Eben
July 1, 2018 6:30 pm
according to the best record of satellite measurements the temperature today is exactly the same as 30 years ago when Hansen made his prediction ,
https://goo.gl/wy79AY
zazove
July 2, 2018 5:34 am
Call me when the 13 month mean has spent a year or two below zero.
Sunsettommy
July 2, 2018 7:23 am
Call me when the RATE of warming per decade exceed the MINIMUM .30C/ Decade.
Eben
July 4, 2018 10:03 am
Which part of “temperature today is exactly the same as in 1988” you don’t understand ???
Nylo
July 2, 2018 3:18 am
Willis:
“How much more common? Well, we can actually test that. He’s comparing the 30-year “climatology” period 1951-1980 to the year 1987. So what I did was the exact same thing, but starting in different years, e.g. comparing the thirty-year period 1901-1930 to the year 1937, seeing how unusual that result is, and so on.
When we do that for all possible years of the GISS 1988 dataset, we find that being 2.5 standard deviations away from the climatological mean is not uncommon at all, occurring about one year out of fourteen.
And if we do the same analysis on the full GISS dataset up until today, we find it’s even more common. It has occurred in the historical record about one year out of seven. So Hansen’s “one percent chance” that the 1988 temperature was unusual was actually a fourteen percent chance … more alarmist misrepresentation, which is no surprise considering the source”.
While all of that is true, I don’t think it invalidates Hansen’s “1%” claim, if it is interpreted in the obvious (to me) way that he meant, i.e. that the probabilities would be 1% IF it weren’t because of humans emitting carbon dioxide. He could claim that you can find it in one every seven years in the data set because we have been emitting carbon dioxide throughout the whole data set. A whole different thing is whether Hansen can PROVE or not his claim of the 1%. He can’t. But the data set does not disprove it either.
Nick Stokes
July 2, 2018 4:20 am
I haven’t been able to emulate the result exactly. But what I do find is that occasions where years were above the 1987 level relative to the 30 year period ending 7 years ago, were in the decade leading up to 1987. IOW it was indeed a rare event prior to about 1980.
Clyde Spencer
July 2, 2018 10:05 am
Michael Lorrey
July 2, 2018 2:33 pm
This is why Willis went back to the turn of the century, to show that this 2.5 std’s effect happened 1 in 14 years no matter which time span in the century you chose. Now, the fact that it went from 1 in 14 to 1 in 7 is interesting. Lets look at prior centuries data and chart the range of standard deviations. Use Hansens own proxies.
Nylo
July 2, 2018 11:01 pm
But Willis said 1 in 14 years, and didn’t mention WHEN those years actually happened. I don’t really know if they were at the turn of the century or much later. Do you? In any case, humans were already emitting CO2 at the beginning of the 20th century, even if in much smaller ammounts. Hansen cannot be disproven because he is just talking about a what-if scenario (if we hadn’t emitted CO2) that didn’t happen and for which we don’t have data. Which also means that his claim has no merit at all.
Nick Stokes
July 2, 2018 11:43 pm
Nylo,
I’m not sure if I’m calculating the same thing as Willis. I took each year and compared with the 30-year period ending 7 years earlier. so I compared 1987 with (1951-1980), 1986 with (1950-1979) etc. I got 72 years, of which 7 were t>2.5, so that is 1 in 10. The years were 1973 1977 1979 1980 1981 1983 1987. But I got t= 3.12 for 1987, and only 1980,1981 and 1983 exceeded that. So it looks like the distribution varies in time. But there are plenty of ways I could be doing something different. The 1987 t-values don’t match.
Clyde Spencer
July 3, 2018 9:23 am
Nick and Nylo,
I have a suggestion for you. Create a synthetic time series of noise-free data. A simple progression of {1, 2, 3,…n}, with n at least 20, which is a line with a slope of 1, is adequate. Calculate the standard deviation of the set and the number of standard deviations from the mean of the last data point, n. Now, increase n, say to 30, and recalculate the mean,SD, and t-value of the last number in the series. It should be instructive. What I hope that you would conclude is that both the slope and the length of time time series (n) determine the calculated SD, and hence the ‘probabilities.’ Now, in the real world we have noise. The noise, or annual variance, is what is actually interesting because Hansen’s claim was that random variation alone had a 1% probability of accounting for the early 1988 temperature anomalies. However, to obtain a normal distribution (and be justified in using parametric statistics), one has to de-trend (a good start, but not necessarily sufficient) the anomaly data. From that, one can see which annual anomalies have a large standard deviation and, hence, low probability.
Nylo
July 4, 2018 11:54 pm
Thanks Nick but I think that you didn’t get my point. I say that you cannot evaluate Hansen’s statement by evaluating the existing data set because his claim is that, should there be NO human interference (no CO2 emitted), the probabilities of this happening would be 1%. To evaluate that, you would need a temperature data set which is NOT affected by human interference (CO2), and see if this phenomenom happens in THAT data set. But such a data set does not exist.
Regards.
Andy in Epsom
July 2, 2018 5:05 am
When I was about to click on the link for this article the advertisement beneath was a picture of a bag full of £50 and £20 notes. Why do I feel that was so appropriate?
ResourceGuy
July 2, 2018 5:59 am
Hansen’s worst professional misconduct was in not showing all of the medium and long run ocean temperature cycles that give context for an uninformed general audience. That too is a sure sign of bias that bias spotters first notice. When one focuses on a particular model selection bias, they also act to shut out doubt and ‘possible confusion’ from other factors. Throw in activists and political donations and you have the makings of a landmark policy distortion.
Michael Lorrey
July 2, 2018 11:27 am
Great article, but given that astrophysicists are now talking about “Space Weather” in re solar wind, solar flares, etc, there obviously must be space climate relating to long term changes in such conditions. If only Hansen would stick to solar wind, but of course, there are no solar windmills for him to tilt against quixotically.
ThinkingScientist
July 3, 2018 7 | 28,468 | 114,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-50 | latest | en | 0.966956 |
https://analyticsindiamag.com/deterministic-vs-stochastic-machine-learning/?utm_source=rss&utm_medium=rss&utm_campaign=deterministic-vs-stochastic-machine-learning | 1,653,779,042,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00732.warc.gz | 146,654,051 | 36,394 | # Deterministic vs Stochastic Machine Learning
A deterministic approach is a simple and comprehensible compared to stochastic approach.
In machine learning, deterministic and stochastic methods are utilised in different sectors based on their usefulness. A deterministic process believes that known average rates with no random deviations are applied to huge populations. A stochastic process, on the other hand, defines a collection of time-ordered random variables that reflect the potential sample pathways. In this article, we will be discussing the key differences between their functioning and their applications. The major points to be discussed in this article are outlined below.
1. Deterministic and Stochastic process modelling
2. When could they both be used?
3. How do these approaches work?
4. Different forms of stochastic and deterministic algorithms
5. Benefits and drawbacks of Deterministic and Stochastic
6. Applications of Deterministic and Stochastic algorithms
## Deterministic and Stochastic process modelling
Deterministic modelling produces consistent outcomes for a given set of inputs, regardless of how many times the model is recalculated. The mathematical characteristics are known in this case. None of them is random, and each problem has just one set of specified values as well as one answer or solution. The unknown components in a deterministic model are external to the model. It deals with the definitive outcomes as opposed to random results and doesn’t make allowances for error.
In contrast, stochastic modelling is intrinsically unpredictable, and the unknown components are integrated into the model. The model generates a large number of answers, estimates, and outcomes, much like adding variables to a difficult maths problem to see how they affect the solution. The identical procedure is then done several times in different settings.
Are you looking for a complete repository of Python libraries used in data science, check out here.
## When could they both be used?
A deterministic model is applied where outcomes are precisely determined through a known relationship between states and events where there is no randomness or uncertainty.
For example, If we know that consuming a fixed amount of sugar ‘y’ will increase the fat in one’s body by ‘2x’ times. Then ‘y’ can always be determined exactly when the value of ‘x’ is known.
Similarly, when the relationship between variables is unknown or uncertain then stochastic modelling could be used because it relies on likelihood estimation of the probability of events.
For example, the insurance sector primarily depends on stochastic modelling to forecast how firm balance sheets will appear in the future.
## How do these approaches work?
As deterministic models show the relationship between results and the factors affecting the outcomes. For this kind of model, the relationship between the variables should be known or determined.
Let’s consider building a machine learner that can help an athlete in a 100-metre sprint, the most important factor in the 100-metre sprint is time. The objective of the model would be to minimize the time of the athlete. The two most important factors affecting time are speed and distance.
The distance covered by every athlete is the same, it’s constant for everyone, the only thing that varies is speed. But varying speed could be controlled as the factors affecting speed are known as the position of the body, the flight time, etc. Since we know time is dependent on speed and distance this makes this problem deterministic.
The stochastic aspect of machine learning algorithms is most evident in complicated and nonlinear approaches used to solve classification and regression predictive modelling issues. These methods employ randomization in the process of building a model from the training data, resulting in a different model fitting each time the same algorithm is performed on the same data.
As a result, when tested on a holdout test dataset, the slightly modified models perform differently. Because of this stochastic behaviour, the model’s performance must be described using summary statistics that indicate the model’s mean or predicted performance rather than the model’s performance from any single training session.
Let’s consider a die-rolling problem. You are rolling a die in a casino. If you roll a six or a one, you win the cash prize. Initially, a sample space that includes all possibilities for die roll outcomes will be generated. The probability for any number being rolled is computed which is ‘0.17’. But we are only interested in two numbers, ‘6’ and ‘1’. So the final probability would be 0.33. This is how a stochastic model would work.
Let’s have a look at how a linear regression model can work both as a deterministic as well as a stochastic model in different scenarios.
Deterministic models define a precise link between variables. In the deterministic scenario, linear regression has three components. The dependent variable ‘y’, the independent variable ‘x’ and the intercept ‘c’. There is no room for mistakes in predicting y for a given x. Here is an equation as an example to replicate the above explanation.
F=95C+32
Image source
The above equation would have a graph something like this with all data points in a straight line.
A stochastic model that takes into account random error. There is a deterministic component as well as a random error component. A probabilistic link between y and x is hypothesised in this paradigm. Here is an equation as an example to replicate the above explanation.
y= 1.5x+error
Image source
In the above graph, it could be observed that due to the error component in the linear regression equation there is randomness in the data.
## Different forms of stochastic and deterministic algorithms
### Principal Component Analysis (PCA)
PCA is a deterministic approach as there are no parameters to initialize. PCA finds the line through the centroid with the smallest sum of squared distances between the points given a set of points in n-dimensional space. Identifying the line for which the projections of the points onto that line are as large as feasible is the same thing (as measured by the sum of squared lengths).
Then, subject to the restriction of being orthogonal to the first line, it finds the line through the centroid with the smallest sum of squared distances to the points. The third principle component, the fourth, and so on. Because all of these procedures are simply geometric, the main components are deterministic data functions.
### Weighted nearest neighbours
A weighted nearest neighbours method also could be called a basic KNN is a deterministic method. This technique employs a statistic known as the “Weighing function.” The weight is determined by taking the inverse of the distance. Because the distance between each data point and the query point would be the same in each iteration, the weights would be a deterministic term.
### Poissons Process
The Poisson method is a stochastic process that displays a random number of points or occurrences across time. The number of points in a process that falls between zero and a specific period is characterised as a time-dependent Poisson random variable. The index set of this process is made up of non-negative integers, whereas the state space is made up of natural numbers. This approach is known as the Poisson counting process because it may be thought of as a counting operation.
### Bernoulli Process
The Bernoulli process is a set of randomly distributed random variables, each with a chance of one or zero. This procedure is analogous to continually flipping a coin, with the probability of winning being p and the value being one, and the likelihood of obtaining a tail being zero. As the result is probabilistic that’s the reason this method is a stochastic process.
### Random Walk
The simple random walk is a discrete-time stochastic process using integers as the state space that is based on a Bernoulli process with each Bernoulli variable taking either a positive or a negative value.
## Benefits and drawbacks of Deterministic and Stochastic
Let’s have a look at the benefits and drawbacks of both of these processes.
Benefits
• Deterministic models get the advantage of being simple.
• Deterministic is simpler to grasp and hence may be more suitable for some cases.
• Stochastic models provide a variety of possible outcomes and the relative likelihood of each.
• The Stochastic model uses the commonest approach for getting the outcomes.
Drawbacks
• In the deterministic approach, there are no cumulative probabilities due to which low reserve cases are overoptimistic.
• In the stochastic approach, the model is more complex, also called the black-box approach.
• The biases may be hidden in the stochastic model and it focuses on extremes.
## Applications of Deterministic and Stochastic algorithms
• Deterministic models are used in the analysis of flood risk.
• The deterministic model used in the Turing machine is a machine (automaton) capable of enumerating any arbitrary subset of acceptable alphabet strings; these strings are part of a recursively enumerable set. A Turing machine has an infinitely long tape on which to execute read and write operations.
• Stochastic investing models aim to estimate price changes, returns on assets (ROA), and asset classes (such as bonds and equities) across time. It uses Monte Carlo simulation, which may simulate how a portfolio would perform based on the probability distributions of individual stock returns.
• Stochastic modelling influences the marketing and shifting movement of audience tastes and preferences, as well as the solicitation and scientific appeal of specific motion picture cameos (i.e., opening weekends, word-of-mouth, top-of-mind knowledge among surveyed groups, star name recognition, and other elements of social media outreach and advertising).
## Conclusion
A deterministic approach has a simple and comprehensible structure which could be applied only when the relationship between variables is determined; on the other hand, a stochastic approach has a complex and incomprehensible structure which works on the likelihood of probabilities. With this article, we have understood the difference between the deterministic and stochastic approaches in machine learning.
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Chapter: Problem:
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Chapter: Problem:
• Step 1 of 4
The work done needed to climb a stair against the gravitation pull is equal to the potential energy. Calculate the potential energy for a man to climb the stair using the formula.
Here, is mass of the man, is acceleration due to gravity and is height of the stair.
Substitute for approximate mass of the man, for and for.
• Step 2 of 4
Convert the unit for time from hour to s.
Calculate the energy consumption by a light using the formula.
Here, is power consumption and is time.
Substitute for and for.
• Step 3 of 4
The total energy of the man is equal to the energy consumed by the light that is
• Step 4 of 4
The energy the energy consumed by the man to climb one stair is. So, the number of stairs the man can climb with energy of is,
Therefore, the man can climb stairs.
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# What is (20 - (2010 - 201)) + (2010 - (201 - 20)) ?
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29 Mar 2019, 01:41
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What is $$(20 - (2010 - 201)) + (2010 - (201 - 20))$$ ?
(A) -4020
(B) 0
(C) 40
(D) 401
(E) 4020
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Re: What is (20 - (2010 - 201)) + (2010 - (201 - 20)) ? [#permalink]
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29 Mar 2019, 02:55
By "opening" the parentheses , we quickly notice that all numbers except the two 20s are cancelled out. So answer is C = 40.
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Re: What is (20 - (2010 - 201)) + (2010 - (201 - 20)) ? [#permalink]
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29 Mar 2019, 02:57
Solution
To find:
• The value of the given expression (20 − (2010 − 201)) + (2010 − (201 − 20))
Approach and Working:
We can simplify the given expression in the following manner:
• (20 − (2010 − 201)) + (2010 − (201 − 20))
= (20 – 1809) + (2010 – 181)
= -1789 + 1829
= 40
Hence, the correct answer is option C.
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Re: What is (20 - (2010 - 201)) + (2010 - (201 - 20)) ? [#permalink]
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29 Mar 2019, 03:00
Alternate Solution
To find:
• The value of the given expression (20−(2010−201)) + (2010−(201−20))
Approach and Working:
We can simplify the given expression in the following manner:
• (20 − (2010 − 201)) + (2010 − (201 − 20))
= 20 – 2010 + 201 + 2010 – 201 + 20
= 20 + 20 – 2010 + 2010 + 201 – 201
= 20 + 20 = 40
Hence, the correct answer is option C.
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Re: What is (20 - (2010 - 201)) + (2010 - (201 - 20)) ? [#permalink]
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29 Mar 2019, 03:36
Bunuel wrote:
What is $$(20 - (2010 - 201)) + (2010 - (201 - 20))$$ ?
(A) -4020
(B) 0
(C) 40
(D) 401
(E) 4020
$$(20 - (2010 - 201)) + (2010 - (201 - 20))$$ ==> $$20-2010+201+2010-201+20$$ = 40
Hence C
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Re: What is (20 - (2010 - 201)) + (2010 - (201 - 20)) ? [#permalink]
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29 Mar 2019, 04:01
The expression
20 - 2010 + 201 + 2010 - 201 + 20
= 40
C
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Re: What is (20 - (2010 - 201)) + (2010 - (201 - 20)) ? [#permalink]
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29 Mar 2019, 08:54
Bunuel wrote:
What is $$(20 - (2010 - 201)) + (2010 - (201 - 20))$$ ?
(A) -4020
(B) 0
(C) 40
(D) 401
(E) 4020
open the expression
we get
20-2010+201+2010-201+20
IMO C ; 40
Re: What is (20 - (2010 - 201)) + (2010 - (201 - 20)) ? [#permalink] 29 Mar 2019, 08:54
Display posts from previous: Sort by | 1,488 | 4,089 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-43 | latest | en | 0.809235 |
https://www.powershow.com/view4/6fa6c6-NmQ4Z/Topic_4_-_Image_Mapping_-_I_powerpoint_ppt_presentation | 1,722,979,071,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640508059.30/warc/CC-MAIN-20240806192936-20240806222936-00688.warc.gz | 756,885,396 | 36,828 | # Topic 4 - Image Mapping - I - PowerPoint PPT Presentation
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## Topic 4 - Image Mapping - I
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### Department of Physics and Astronomy DIGITAL IMAGING Course 3624 Topic 4 - Image Mapping - I Professor Bob Warwick – PowerPoint PPT presentation
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Title: Topic 4 - Image Mapping - I
1
Topic 4 - Image Mapping - I
Department of Physics and Astronomy
DIGITAL IMAGING Course 3624
Professor Bob Warwick
2
Typical Image Processing Steps
ORIGINAL IMAGE
PRE-PROCESSING STEPS
ENHANCEMENT RESTORATION
IMPROVED IMAGE
IMAGE ANALYSIS
3
Image Mapping Processes
Image Mapping encompasses a range of enhancement
methods which adjust the way the image data are
displayed (ie how the data are "mapped" onto the
display device).
4
Histograms of a Colour Image
5
The Form of the Image Histogram
The form of the image histogram P(f) provides
useful information on the content/quality of the
image
P(f) ?
P(f) ?
P(f) ?
f ?
f ?
f ?
Good contrast Poor contrast
Saturated?
Image histogram modification techniques aim to
improve the gray level distribution in the
displayed image so as to make as much use as
possible of the rather limited ability of the eye
to discern gray shades.
6
Discriminating between Gray Levels - I
I Intensity of Scene
7
Discriminating between Gray Levels - II
Typically we are able to discern 32 25 gray
levels in any particular image
8
Discriminating between Gray Levels - III
Small squares have different intensity but same
apparent brightness.
Small squares have same intensity but different
apparent brightness.
9
Image Enhancement by Histogram Modification
Original Image
New" image
(Inefficient) Implementation Method Once fout
T(fin) has been defined, we compute a new image
by fin ? fout on a pixel-by-pixel basis 15 20
12 25 30 16 15 22
? 25 32
10
Forms of T(f) A Linear Contrast Stretch
• The parameters of the process f1 f2 might be
determined
• Interactively
• Automatically
11
Example of Contrast Stretching
12
Improved Contrast?
13
Forms of T(f) Increased Gamma
14
Forms of T(f) Decreased Gamma
15
4.2 Image Enhancement by Histogram Matching
The objective is to set up the displayed image so
that its histogram has a specified form.
16
Histogram Equalisation Problem
Note that the result is only a crude
approximation to the target uniform distribution
due to the very coarse digitization of the
input image data
17
Comments on Implementation
Highly Efficient Method Load the look-up table
of the display device with the required
transformation
18
Histogram Equalisation in Action
Original Image
Original Histogram
Final Image
Equalised Histogram
19
Histogram Equalisation in Action
Original Image
Final Image
Equalised Histogram
Original Histogram
20
The General Case
The general formula above can be applied to give
any form for the output image histogram. The
procedure to apply this formula is
Equalization General
f
f
• A practical implementation might involve
• For each fin calculate C1(fin)
• Compute a look-up table of fout versus C2(fout)
• For each fin find the nearest C2 value to C1(fin)
• Determine the fout value the C2 value
• Load the resulting mapping fin ? fout into the
display device look-up table
21
Image Enhancement by Histogram Specification
22
Example Histogram Specification
Image P(f)
f
23
Histogram to be matched taken from a second image
Target P(f)
f
24
Histogram Matching Example
Image CDF
Target CDF
User Comments (0)
About PowerShow.com | 870 | 3,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-33 | latest | en | 0.788898 |
https://cs.stackexchange.com/questions/77636/bvl-balanced-tree | 1,716,625,015,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058789.0/warc/CC-MAIN-20240525065824-20240525095824-00895.warc.gz | 152,120,966 | 40,276 | # BVL Balanced Tree
I have an issue about proving the next problem:
Let's define a BVL tree, which is a binary tree, who satisfied the feature that the difference between the heights of the children of a node, is at most 2.
Meaning|height(node.left) - height(node.right)| <= 2 Prove that BVL tree is a balanced tree.
Now I do know it's some kind of extension of AVL tree. Well, as we know, the defenition of a balanced tree is related to its height, which is O(logn)
However, trying to perform an induction on N for BVL tree, ends with failure..
Can someong turn on the light and give me a hint?
Thanks!
• You can generalize height balance to arbitrary boundaries; as long as it's constant, the height will be logarithmic. The proof is very similar to the one for AVL trees; just go through and adapt. Jul 6, 2017 at 19:00
If you want to show that the height is $h=\mathcal O(\log(n))$ then I would suggest the following:
Define $n_h$ as the minimum vertices in tree with height $h$
Then to get the minimum vertices for given height:
$n_h = n_{h-1} + n_{h-3}\\ n_h \ge 2n_{h-3}$
Now from the inequality you know that:
$n_{h-3} \ge 2*n_{h-6}$
(Just changing index)
Placing it to the former inequality doing it again and again:
$n_h \ge 2n_{h-3} \ge 2 * (2n_{h-6}) \ge 2 * 2 * (2n_{h-9}) \ge ...\ge 2*2*...*2*c\\$
So lets say that you did 'i' iterations, then $n_h \ge 2^i*c$
$\Rightarrow n_h \ge 2^i*c$
where $c$ is the minimum amount of vertices in the base case and $c \ge 1$
The recursion will stop when the index will be ($0/1/2$):
$\Rightarrow h - 3 - 3i = 0\\ \Rightarrow i = \frac{h-3}{3}\\ \Rightarrow n_h \ge 2^\frac{h-3}{3}\\ 3(1 + \log(n_h)) \ge h\\ \Rightarrow h = \mathcal O(\log(n))$
• Welcome to Computer Science! Note that you can use LaTeX here to typeset mathematics in a more readable way. See here for a short introduction. Jul 6, 2017 at 19:00
• Hi Omer, Thanks for the details!. However, I completely lost it, when u added that i to the inequality, can u explain that ? Jul 6, 2017 at 19:33
• Hey, i've edited it to explain more deeply how I got this 'i'. Jul 7, 2017 at 6:42
• Thanks again, it's more clearly now. How did you manage to reach the line h - 3- 3i? Jul 9, 2017 at 17:27 | 691 | 2,221 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-22 | latest | en | 0.914935 |
http://forums.wolfram.com/mathgroup/archive/2011/Oct/msg00617.html | 1,713,307,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817112.71/warc/CC-MAIN-20240416222403-20240417012403-00538.warc.gz | 9,795,804 | 7,699 | DSolve bug for complicated forcing functions in a 2nd order ODE
• To: mathgroup at smc.vnet.net
• Subject: [mg122394] DSolve bug for complicated forcing functions in a 2nd order ODE
• From: Dan Dubin <ddubin at ucsd.edu>
• Date: Thu, 27 Oct 2011 06:31:28 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
```Folks -- reporting a bug in DSolve, I think. The following Module solves
a 2nd order differential equation with periodic forcing, written as a
Fourier series. The number of Fourier coefficients kept in the series
is M. For M less than or equal to 40, the solution is correct, but for M
greater than 40 it starts to go wrong. Try M=41, for instance. The
resulting solution no longer satisfies the ODE. Any ideas what is
happening here?
PS[M_] := Module[{},
T = 1/2;
=CF=89[n_] = 2 Pi n/T;
f[n_] = 1/T Integrate[1 Exp[I =CF=89[n] t], {t, 0, 1/4}];
f[0] = Limit[f[n], n -> 0];
fapprox[t_] = Sum[f[n] Exp[-I =CF=89[n] t], {n, -M, M}];
s = DSolve[{x''[t] + 16 =CF=80^2 x[t] == fapprox[t], x[0] == 0, x'[0] == 0},
x[t], t];
xs[t_] = x[t] /. s[[1]];
xs''[t] + 16 Pi^2 xs[t] - fapprox[t] // Simplify]
Prof. Dan Dubin
Dept of Physics, UCSD La Jolla CA 92093-0319
858-534-4174 fax: 858-534-0173
ddubin at ucsd.edu
```
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https://www.numwords.com/words-to-number/en/4437 | 1,560,878,302,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998808.17/warc/CC-MAIN-20190618163443-20190618185443-00016.warc.gz | 858,743,183 | 1,843 | NumWords.com
# How to write Four thousand four hundred thirty-seven in numbers in English?
We can write Four thousand four hundred thirty-seven equal to 4437 in numbers in English
< Four thousand four hundred thirty-six :||: Four thousand four hundred thirty-eight >
Eight thousand eight hundred seventy-four = 8874 = 4437 × 2
Thirteen thousand three hundred eleven = 13311 = 4437 × 3
Seventeen thousand seven hundred forty-eight = 17748 = 4437 × 4
Twenty-two thousand one hundred eighty-five = 22185 = 4437 × 5
Twenty-six thousand six hundred twenty-two = 26622 = 4437 × 6
Thirty-one thousand fifty-nine = 31059 = 4437 × 7
Thirty-five thousand four hundred ninety-six = 35496 = 4437 × 8
Thirty-nine thousand nine hundred thirty-three = 39933 = 4437 × 9
Forty-four thousand three hundred seventy = 44370 = 4437 × 10
Forty-eight thousand eight hundred seven = 48807 = 4437 × 11
Fifty-three thousand two hundred forty-four = 53244 = 4437 × 12
Fifty-seven thousand six hundred eighty-one = 57681 = 4437 × 13
Sixty-two thousand one hundred eighteen = 62118 = 4437 × 14
Sixty-six thousand five hundred fifty-five = 66555 = 4437 × 15
Seventy thousand nine hundred ninety-two = 70992 = 4437 × 16
Seventy-five thousand four hundred twenty-nine = 75429 = 4437 × 17
Seventy-nine thousand eight hundred sixty-six = 79866 = 4437 × 18
Eighty-four thousand three hundred three = 84303 = 4437 × 19
Eighty-eight thousand seven hundred forty = 88740 = 4437 × 20
Sitemap | 424 | 1,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2019-26 | latest | en | 0.818497 |
https://multi-converter.com/seconds-to-milliseconds | 1,725,741,612,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650920.0/warc/CC-MAIN-20240907193650-20240907223650-00662.warc.gz | 379,730,296 | 6,057 | # Seconds to Milliseconds
Convert s to ms
Change to Milliseconds to Seconds
Share:
## How to convert Seconds to Milliseconds
1 [Seconds] = 1000 [Milliseconds]
[Milliseconds] = [Seconds] * 1000
To convert Seconds to Milliseconds multiply Seconds * 1000.
## Example
95 Seconds to Milliseconds
95 [s] * 1000 = 95000 [ms]
## Conversion table
Seconds Milliseconds
0.01 s10 ms
0.1 s100 ms
1 s1000 ms
2 s2000 ms
3 s3000 ms
4 s4000 ms
5 s5000 ms
10 s10000 ms
15 s15000 ms
50 s50000 ms
100 s100000 ms
500 s500000 ms
1000 s1000000 ms | 179 | 530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-38 | latest | en | 0.408352 |
http://pari.math.u-bordeaux.fr/archives/pari-users-1211/msg00007.html | 1,529,416,523,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863043.35/warc/CC-MAIN-20180619134548-20180619154548-00541.warc.gz | 245,892,138 | 3,340 | Ewan Delanoy on Mon, 12 Nov 2012 18:35:45 +0100
Using Mod inside functions
Hello all,
below is some PARI-GP code showing a function called âcleaner_for_leading_coefficient
â that does not work.
Surprisingly, however, everything goes fine if one executes all the instructions wrapped inside the function step by step, as shown in the code below.
Judging from the error message, the problem is related to my use of âModâ inside the function.
Thanks in advance for any help.
Ewan
GP/PARI CALCULATOR Version 2.5.2 (released)
i386 running darwin (x86-64/GMP-5.0.4 kernel) 64-bit version
compiled: Sep 18 2012, gcc-4.5.4 (MacPorts gcc45 4.5.4_5)
(readline v6.2 enabled, extended help enabled)
Copyright (C) 2000-2011 The PARI Group
PARI/GP is free software, covered by the GNU General Public License, and comes
WITHOUT ANY WARRANTY WHATSOEVER.
Type ? for help, \q to quit.
Type ?12 for how to get moral (and possibly technical) support.
parisize = 8000000, primelimit = 500509
? paul1=(-6566965496*beth^5 + 1069572966280*beth^3 - 23940418649316*beth)*nuu^3 + (3283482748*beth^6 - 703675242270*beth^4 + 26192533760510*beth^2 + 184509608906896)*nuu^2 + (-938137928*beth^7 + 379974579348*beth^5 - 34472816584284*beth^3 + 39850128820580*beth)*nuu + (117267241*beth^8 - 63329096558*beth^6 + 8591936284087*beth^4 - 17222997429586*beth^2 - 15148802119203)
%1 = 117267241*beth^8 - 938137928*nuu*beth^7 + (3283482748*nuu^2 - 63329096558)*beth^6 + (-6566965496*nuu^3 + 379974579348*nuu)*beth^5 + (-703675242270*nuu^2 + 8591936284087)*beth^4 + (1069572966280*nuu^3 - 34472816584284*nuu)*beth^3 + (26192533760510*nuu^2 - 17222997429586)*beth^2 + (-23940418649316*nuu^3 + 39850128820580*nuu)*beth + (184509608906896*nuu^2 - 15148802119203)
? paul2=30020413696*beth^8 - 1801531152512*beth^6 + 27258536763952*beth^4 - 7293106986984*beth^2 + 468152903089
%2 = 30020413696*beth^8 - 1801531152512*beth^6 + 27258536763952*beth^4 - 7293106986984*beth^2 + 468152903089
?
?
{
local(symb,temp,temp2,temp3,agatha);
symb=Mod(agatha,subst(annulator,var_x,agatha));
temp2=subst(temp,var_x,symb);
temp3=1/temp2;
return(lift(temp3));
}
?
*** ^--------------------
*** ...(temp,var_x,symb);temp3=1/temp2;return(lift(t
*** ^--------------------
*** _/_: impossible inverse modulo: Mod(0, 468152903089).
*** Break loop: type 'break' to go back to GP
break>
break> break
? pol=paul1
%4 = 117267241*beth^8 - 938137928*nuu*beth^7 + (3283482748*nuu^2 - 63329096558)*beth^6 + (-6566965496*nuu^3 + 379974579348*nuu)*beth^5 + (-703675242270*nuu^2 + 8591936284087)*beth^4 + (1069572966280*nuu^3 - 34472816584284*nuu)*beth^3 + (26192533760510*nuu^2 - 17222997429586)*beth^2 + (-23940418649316*nuu^3 + 39850128820580*nuu)*beth + (184509608906896*nuu^2 - 15148802119203)
? annulator=paul2
%5 = 30020413696*beth^8 - 1801531152512*beth^6 + 27258536763952*beth^4 - 7293106986984*beth^2 + 468152903089
? var_x=beth
%6 = beth
? var_y=nuu
%7 = nuu
? symb=Mod(agatha,subst(annulator,var_x,agatha))
%8 = Mod(agatha, 30020413696*agatha^8 - 1801531152512*agatha^6 + 27258536763952*agatha^4 - 7293106986984*agatha^2 + 468152903089)
%9 = -6566965496*beth^5 + 1069572966280*beth^3 - 23940418649316*beth
? temp2=subst(temp,var_x,symb)
%10 = Mod(-6566965496*agatha^5 + 1069572966280*agatha^3 - 23940418649316*agatha, 30020413696*agatha^8 - 1801531152512*agatha^6 + 27258536763952*agatha^4 - 7293106986984*agatha^2 + 468152903089)
? temp3=1/temp2
%11 = Mod(20969910094342573625112778762878346210816/7852515084056523941826305327703635391125038356414739097*agatha^7 - 1260104881146110861084127972946836114952704/7852515084056523941826305327703635391125038356414739097*agatha^5 + 19097036931589896718062702970453586827359488/7852515084056523941826305327703635391125038356414739097*agatha^3 - 5124393596497362934274073300552845177336080/7852515084056523941826305327703635391125038356414739097*agatha, 30020413696*agatha^8 - 1801531152512*agatha^6 + 27258536763952*agatha^4 - 7293106986984*agatha^2 + 468152903089)
? second_trial=lift(temp3)
%12 = 20969910094342573625112778762878346210816/7852515084056523941826305327703635391125038356414739097*agatha^7 - 1260104881146110861084127972946836114952704/7852515084056523941826305327703635391125038356414739097*agatha^5 + 19097036931589896718062702970453586827359488/7852515084056523941826305327703635391125038356414739097*agatha^3 - 5124393596497362934274073300552845177336080/7852515084056523941826305327703635391125038356414739097*agatha | 1,770 | 4,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-26 | latest | en | 0.46479 |
https://www.cqmode.com/how-to-calculate-the-paps/ | 1,675,173,986,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499871.68/warc/CC-MAIN-20230131122916-20230131152916-00279.warc.gz | 731,246,864 | 12,906 | # How to calculate the PAPS?
### How to calculate the PAPS?
The PAPS is calculated by the formula PAPS = 4V2 + POD, where V is the peak tricuspid insufficiency flow velocity measured in continuous Doppler, and POD is an estimate of the average right atrial pressure (Figure 1).
### What is normal lung pressure?
The level of lung pressure is an important clinical element for assessing the severity and tolerance of cardiac damage. Usually in a population of healthy subjects, the pressure arterial pulmonary systolic (PAPs) normal is less than 30 mmHg.
### How to measure cardiac output?
Measure by doppler cardiac This is the simplest and most widely used method: debit is measured by multiplying the time-velocity integral (area under the continuous Doppler curve) of the aortic flow by the area of the outflow orifice calculated in two-dimensional imaging.
### What is the relationship between DC FC and ves?
cardiac output is the result of the integrated functioning of the cardiovascular system. VS’is the amount of blood ejected by each ventricle in one minute. He is equal to the product of the volume ejected with each beat (VES) by heart rate.
### How is an Eto?
I’TEE requires the introduction of a specific ultrasound probe into the esophagus through the mouth and pharynx. It is performed in our center under brief general anesthesia requiring a pre-anaesthetic consultation a few days before the examination and that you be well on an empty stomach.
### What are the symptoms of pulmonary hypertension?
From symptoms associated withpulmonary hypertensionlet us mention:
• shortness of breath on exertion – when climbing stairs or during everyday activities.
• severe fatigue.
• dizziness or fainting.
• chest pain.
• swelling in the ankles, legs or abdomen.
### How to measure VES?
the VES = VTIAo X (πD2/4), with the DC which will be equal to this value multiplied by the heart rate during the measure (figure 1).
### What is resting cardiac output?
Her value to rest is about 5 liters per minute, but depends on the size and build of the subject; during intense effort, it can reach 40 liters per minute.
### What is the relationship between heart rate and blood pressure?
the link between a cardiac frequency high and hypertension arterial is found in most studies. He is established that this association is secondary to an overactive nervous system through increased sympathetic tone and decreased parasympathetic tone [10].
### What is normal cardiac output?
normally equal to 4 to 6 l/min.
### Which anesthesia for an Eto?
The examination is carried out under anesthesia generally brief. A pre-consultationanesthesia dedicated must therefore be carried out a few days before the date of the procedure, except in an emergency. You must be on an empty stomach (food, drink, cigarette) within 6 hours of the exam.
### How is a throat ultrasound performed?
How’s it going ? The exam to takes place in a hospital environment, in the presence of a nurse and a cardiologist. You are lying on a table, you wear a blouse. You will be required to gargle with lidocaine gel and be given a spray of numbing spray in your throat. | 681 | 3,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-06 | latest | en | 0.911713 |
http://www.800score.com/forum/viewtopic.php?p=4059& | 1,566,190,481,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314641.41/warc/CC-MAIN-20190819032136-20190819054136-00558.warc.gz | 210,008,679 | 5,429 | It is currently Mon Aug 19, 2019 12:54 am
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Post subject: GRE Arithmetic (Select One)Posted: Tue Sep 14, 2010 2:58 am
Joined: Sun May 30, 2010 3:15 am
Posts: 424
A prize of \$600 is to be distributed among 20 winners, each of whom must be awarded at least \$20. If 2/5 of the prize will be distributed to 3/5 of the winners, what is the greatest possible individual award?
A. \$20
B. \$25
C. \$45
D. \$220
E. \$300
(D) We are told that 2/5 of the prize is distributed to 3/5 of the winners:
2/5 of the prize = 2/5 × \$600 = \$240.
3/5 of the winners = 3/5 × 20 = 12 winners.
Since \$240/12 = \$20, we know that each of these people received the minimum of \$20.
The remaining \$360 will be divided among the 8 remaining winners. For one of these winners to receive the greatest possible award, the other seven winners must receive the minimum \$20. These seven winners receive a total of: 7 × \$20 = \$140.
The eighth winner would receive all the remaining prize money:\$360 – \$140 = \$220.
The correct answer is choice (D).
-------------
In the above question, you calculate a minimum of \$20 to be distributed among the "12" people to whom \$240 would be distributed. Then you claim that that \$20 is also a minimum for the remaining 8 winners. It seems that the minimum amount that each of the remaining 8 winners could receive is 360/8 = \$45.
So for one of these 8 winners to receive the greatest possible award, the remaining 7 winners must receive 7 × \$45 (not \$20) = \$315.
So the eighth winner would receive the remaining of the prize:
\$360 – \$315 = \$45
(\$240 + \$315 + \$45 = \$600)
Please explain if my logic is flawed.
Thank you.
Top
Post subject: Re: math (test 4, question 5): arithmetic, logic.Posted: Tue Sep 14, 2010 3:24 am
Joined: Sun May 30, 2010 2:23 am
Posts: 498
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Jump to: Select a forum ------------------ GMAT GMAT: Quantitative Section (Math) GMAT: Verbal Section GMAT: Integrated Reasoning GMAT: General Questions GRE GRE: Quantitative Reasoning (Math) GRE: Verbal Reasoning GRE: General Questions General questions Other questions | 747 | 2,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-35 | latest | en | 0.948653 |
https://calculator.tutorpace.com/electric-field-calculator-online-tutoring | 1,623,809,656,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621699.22/warc/CC-MAIN-20210616001810-20210616031810-00047.warc.gz | 162,481,410 | 8,464 | #### A PHP Error was encountered
Severity: Warning
Message: count(): Parameter must be an array or an object that implements Countable
Filename: controllers/calculator.php
Line Number: 37
Electric Field Calculator
# Electric Field Calculator
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Electric field is the space around an electric charge because of which when another charge enters this
space, it experiences either its force of attraction or the force of repulsion. Electric field calculator is the
instant online tool which can calculate the electric field around a charge.
Example 1: Given a charge is of 0.25C. If the force produced by the charge is 60N, then calculate
the electric field of the charge.
In order to find the Electric field around the given electron, we can use this formula:
E = F/q
where E = Electric field
F = Force produced by a charge = 60N
q = electric charge = 0.25C
Electric field, E = 60/0.25
Electric field, E = 240N/C
Example 2:If the force produced by an electron is 2.00 * 10-17N, then calculate the electric field
around this electron. (Charge of an electron is 1.60 * 10-19C).
In order to find the Electric field around the given electron, we can use this formula:
E = F/q
where E = Electric field
F = Force produced by a charge = 2.00 * 10-17N
q = electric charge = 1.60 * 10-19C
Electric field, E = 2.00 * 10-17/ (1.60 * 10-19)
Electric field, E = 125N/C | 391 | 1,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-25 | longest | en | 0.88418 |
http://lemon.cs.elte.hu/trac/lemon/changeset/e72bacfea6b78fd9d84df2128824bc027f6e5afa/lemon/lemon/gomory_hu.h | 1,606,264,025,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141177607.13/warc/CC-MAIN-20201124224124-20201125014124-00675.warc.gz | 53,569,562 | 6,227 | # Changeset 592:e72bacfea6b7 in lemon for lemon/gomory_hu.h
Ignore:
Timestamp:
02/25/09 12:10:57 (12 years ago)
Branch:
default
Phase:
public
Message:
Remane GomoryHuTree? to GomoryHu? (#66)
File:
1 moved
Unmodified
Removed
• ## lemon/gomory_hu.h
r591 typename CAP = typename GR::template EdgeMap > class GomoryHuTree { class GomoryHu { public: /// \param graph The graph the algorithm will run on. /// \param capacity The capacity map. GomoryHuTree(const Graph& graph, const Capacity& capacity) GomoryHu(const Graph& graph, const Capacity& capacity) : _graph(graph), _capacity(capacity), _pred(0), _weight(0), _order(0) /// /// Destructor ~GomoryHuTree() { ~GomoryHu() { destroyStructures(); } /// This iterator class lists the nodes of a minimum cut found by /// GomoryHuTree. Before using it, you must allocate a GomoryHuTree class, /// and call its \ref GomoryHuTree::run() "run()" method. /// GomoryHu. Before using it, you must allocate a GomoryHu class, /// and call its \ref GomoryHu::run() "run()" method. /// /// This example counts the nodes in the minimum cut separating \c s from /// \c t. /// \code /// GomoruHuTree gom(g, capacities); /// GomoruHu gom(g, capacities); /// gom.run(); /// int sum=0; /// for(GomoruHuTree::MinCutNodeIt n(gom,s,t);n!=INVALID;++n) ++sum; /// for(GomoruHu::MinCutNodeIt n(gom,s,t);n!=INVALID;++n) ++sum; /// \endcode class MinCutNodeIt /// Constructor /// MinCutNodeIt(GomoryHuTree const &gomory, ///< The GomoryHuTree class. You must call its MinCutNodeIt(GomoryHu const &gomory, ///< The GomoryHu class. You must call its /// run() method /// before initializing this iterator /// This iterator class lists the edges of a minimum cut found by /// GomoryHuTree. Before using it, you must allocate a GomoryHuTree class, /// and call its \ref GomoryHuTree::run() "run()" method. /// GomoryHu. Before using it, you must allocate a GomoryHu class, /// and call its \ref GomoryHu::run() "run()" method. /// /// This example computes the value of the minimum cut separating \c s from /// \c t. /// \code /// GomoruHuTree gom(g, capacities); /// GomoruHu gom(g, capacities); /// gom.run(); /// int value=0; /// for(GomoruHuTree::MinCutEdgeIt e(gom,s,t);e!=INVALID;++e) /// for(GomoruHu::MinCutEdgeIt e(gom,s,t);e!=INVALID;++e) /// value+=capacities[e]; /// \endcode /// the result will be the same as it is returned by /// \ref GomoryHuTree::minCostValue() "gom.minCostValue(s,t)" /// \ref GomoryHu::minCostValue() "gom.minCostValue(s,t)" class MinCutEdgeIt { public: MinCutEdgeIt(GomoryHuTree const &gomory, ///< The GomoryHuTree class. You must call its MinCutEdgeIt(GomoryHu const &gomory, ///< The GomoryHu class. You must call its /// run() method /// before initializing this iterator
Note: See TracChangeset for help on using the changeset viewer. | 843 | 2,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-50 | latest | en | 0.620027 |
https://train.co.uk/tag/calculations/ | 1,544,851,075,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826715.45/warc/CC-MAIN-20181215035757-20181215061757-00451.warc.gz | 775,147,740 | 24,143 | ## Calculations in Excel
Learn the 4 steps that make entering a calculation into an Excel spreadsheet work reliably every time.
Never entered a calculation into an Excel spreadsheet? When you’ve finished this short tutorial, you’ll be able to create formulas in Excel to make your business life just a little bit easier…
The video refers to buttons below – these only appear when viewing from the www.tekbites.com website.
We’re just getting started (this is our first video) so the resources available on the tekbites.com website will be quite limited to begin with – apologies in advance.
## What percentage does a value represent of the total amount?
#### 7 answers out of 12 were correct. What is the percentage score?
This type of calculation is best approached as a fraction.
The partial value, in this case 7, is divided by the total value. Formatting is then used to display the resulting number as a percentage.
In the example below, 7 answers were correct out of a possible 12.
Finally, change the format of the answer so that Excel displays it as a percentage. This can be done using the percentage icon on the toolbar.
It is usually worthwhile increasing the number of decimal places displayed so that a more accurate value is shown.
A score of 7 correct out of 12 questions represents a total score of 58.33%.
Try it yourself using the examples below…
## Decrease a number by a percentage
#### Calculate a discount of 9% on a purchase of 120
This is best approached in two stages.
First, calculate the percentage required using the technique described here – Finding a percentage of a number. Then subtract this amount from the original.
In the example below a 9% discount is applied to a purchase of 120.
If you get a 9% discount on a purchase of 120, you only pay 109.20.
Try it yourself using the examples below…
## Increase a number by a percentage
#### Increase 24,000 by 4%
This is best approached in two stages.
First, calculate the percentage required using the technique described here – Finding a percentage of a number. Then add this amount to the original.
In the example below, an annual salary of 24,000 has been increased by 4%.
A 4% increase on a salary of 24,000 is 24,960.
Try it yourself using the examples below…
## Count using two conditions
#### How many hats in this list are red?
There are several ways in which this problem may be solved. The simplest is first to determine which of the rows in the table contain Red Hats. This is done using the logical function AND.
The AND function needs two or more parameters. In this case there are two questions to be answered. ‘Is the item a hat?’ and ‘Is the item Red?’.
For the first row, these translate into B3 = “HAT” and C3 = “RED”. Once again the text must be enclosed in quotes and the match is not case sensitive.
If all the criteria in the AND function are met, the function returns a TRUE value.
The calculation is then copied for each of the other rows in the table.
Any row containing a Red Hat will have a value of TRUE. To find the number of Red Hats, COUNTIF is used to count all entries with a value of TRUE.
TRUE need not be enclosed in speech marks as it is a logical value and not a text value.
Since the individual TRUE and FALSE values do not need to be seen, the answer can be moved to E14 and column D can be hidden.
Try it yourself using the examples below…
## Using COUNTIF to count specific items in an area
#### How many of the items in this list are red?
The COUNTIF function can be used to count only certain specific entries in a list. It requires two pieces of information.
The area containing the items to be counted and some way to tell which entries are required.
In the example below, which was created in the UK in the dead of winter, the colours of the items to be counted are contained within the range C3 to C12
If only the red items are to be counted, these are identified using the word RED. To indicate that this is literally the word RED, the text must be enclosed within speech marks.
Please note that a text match of this type is not case sensitive.
To find out how many hats are in the list, a similar procedure is followed. The only difference is that the item type is listed in the range B3 to B12. Once again, the text criteria is enclosed in speech marks and it does not matter whether upper or lower case text is used.
And if you want to know how many red hats are in the list, click here
Try it yourself using the examples below…
## Percentage of a number using Excel
#### What is 7% of 2,500?
This is probably the easiest percentage calculation to do. Simply multiply the number by the percentage required.
In the example below, a tax rate of 7% is applied to an original amount of 2,500…
7% of 2,500 is 175.
To find the total amount payable, add the 2,500 and the 175…
Try it yourself using the examples below…
## Count everything in an area
#### How many entries have been completed?
In the example below, an ‘x’ has been used to indicate who has shown an interest in a neighbourhood watch scheme. To find out how many people are interested, the COUNT function might be used.
COUNT, like all functions, has a set of brackets that contain the details of the area containing the items to be counted.
The screen shot below shows the result of using COUNT on the column containing the ‘x’ characters.
The COUNT function only ever counts numbers. Since ‘x’ is a text character, the result comes out as 0.
To count numbers, text and pretty much anything else, use the COUNTA function. You can think of the A as meaning All or Anything.
The screen shot below shows the result of using COUNTA on the same column.
If dates had been used instead of a text character, the COUNT function would have been fine since properly entered dates are numbers in Excel!
Try it yourself using the examples below…
## Counting stuff in Excel
The COUNT function is used in Excel to find out how many items appear in a specific area of the worksheet.
Unfortunately for the unwary, COUNT only works for numbers. To get the most out of this family of functions, you need to know about COUNTA and COUNTIF as well
Try these short tutorials – the easiest one is at the top…
Count everything in an area
(How many entries have been completed?)
Only count specific items in an area
(How many of the items in this list are red?)
Count using two criteria
(How many hats in this list are red?)
## Percentages in Excel
The use of percentages in Excel are not the easiest concept to understand.
The percentages you use with Excel are just the same as the ones you might calculate with a calculator but it helps if you think about them slightly differently when calculating in Excel.
Always enter a percentage in Excel as a number followed by a percent sign. If you work this way, you don’t have to worry about all that multiplying and dividing by 100!
Try these short tutorials – the easier ones are at the top…
## Percentage of a number
(What is 7% of 2,500?)
## Increase a number by a percentage
(Increase 24,000 by 4%)
## Decrease a number by a percentage
(Calculate a discount of 9% on a purchase of 120)
## What percentage does a value represent of the total amount?
(7 answers out of 12 were correct. What is the percentage score?)
## Difference between two numbers as a percentage
(Customer numbers changed from 2,800 to 3,200. What percentage increase is that?)
(Customer numbers changed from 3,500 to 3,400. What percentage decrease is that?) | 1,659 | 7,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-51 | latest | en | 0.898267 |
https://tomhopper.me/2014/07/04/the-most-useful-data-plot-youve-never-used/?shared=email&msg=fail | 1,558,641,479,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257361.12/warc/CC-MAIN-20190523184048-20190523210048-00395.warc.gz | 654,634,643 | 24,690 | # The Most Useful Data Plot You’ve Never Used
Those of us working in industry with Excel are familiar with scatter plots, line graphs, bar charts, pie charts and maybe a couple of other graph types. Some of us have occasionally used the Analysis Pack to create histograms that don’t update when our data changes (though there is a way to make dynamic histograms in Excel; perhaps I’ll cover this in another blog post).
One of the most important steps in data analysis is to just look at the data. What does the data look like? When we have time-dependent data, we can lay it out as a time-series or, better still, as a control chart (a.k.a. “natural process behavior chart”). Sometimes we just want to see how the data looks as a group. Maybe we want to look at the product weight or the cycle time across production shifts.
Unless you have Minitab, R or another good data analysis tool at your disposal, you have probably never used—maybe never heard of—boxplots. That’s unfortunate, because boxplots should be one of the “go-to” tools in your data analysis tool belt. It’s a real oversight that Excel doesn’t provide a good way to create them.
For the purpose of demonstration, let’s start with creating some randomly generated data:
```head(df)
```
```## variable value
## 1 group1 -1.5609
## 2 group1 -0.3708
## 3 group1 1.4242
## 4 group1 1.3375
## 5 group1 0.3007
## 6 group1 1.9717
```
```tail(df)
```
```## variable value
## 395 group1 1.4591
## 396 group1 -1.5895
## 397 group1 -0.4692
## 398 group1 0.1450
## 399 group1 -0.3332
## 400 group1 -2.3644
```
If we don’t have much data, we can just plot the points:
```library(ggplot2)
ggplot(data = df[1:10,]) +
geom_point(aes(x = variable, y = value)) +
coord_flip() +
theme_bw()
```
But if we have lots of data, it becomes hard to see the distribution due to overplotting:
```ggplot(data = df) +
geom_point(aes(x = variable, y = value)) +
coord_flip() +
theme_bw()
```
We can try to fix this by changing some parameters, like adding semi-transparency (alpha blending) and using an open plot symbol, but for the most part this just makes the data points harder to see; the distribution is largely lost:
```ggplot(data = df) +
geom_point(aes(x = variable, y = value), alpha = 0.3, shape = 1) +
coord_flip() +
theme_bw()
```
The natural solution is to use histograms, another “go-to” data analysis tool that Excel doesn’t provide in a convenient way:
```ggplot(data = df) +
geom_histogram(aes(x = value), binwidth = 1) +
theme_bw()
```
But histograms don’t scale well when you want to compare multiple groups; the histograms get too short (or too narrow) to really provide useful information. Here I’ve broken the data into eight groups:
```head(df)
```
```## variable value
## 1 group1 -1.5609
## 2 group1 -0.3708
## 3 group1 1.4242
## 4 group1 1.3375
## 5 group1 0.3007
## 6 group1 1.9717
```
```tail(df)
```
```## variable value
## 395 group8 -0.6384
## 396 group8 -3.0245
## 397 group8 1.5866
## 398 group8 1.9747
## 399 group8 0.2377
## 400 group8 -0.3468
```
```ggplot(data = df) +
geom_histogram(aes(x = value), binwidth = 1) +
facet_grid(variable ~ .) +
theme_bw()
```
Either the histograms need to be taller, making the stack too tall to fit on a page, or we need a better solution.
The solution is the box plot:
```ggplot() +
geom_boxplot(data = df, aes(y = value, x = variable)) +
coord_flip() +
theme_bw()
```
The boxplot provides a nice, compact representation of the distribution of a set of data, and makes it easy to compare across a large number of groups.
There’s a lot of information packed into that graph, so let’s unpack it:
Median
A measure of the central tendency of the data that is a little more robust than the mean (or arithmetic average). Half (50%) of the data falls below this mark. The other half falls above it.
First quartile (25th percentile) hinge
Twenty-five percent (25%) of the data falls below this mark.
Third quartile (75th percentile) hinge
Seventy-five percent (75%) of the data falls below this mark.
Inter-Quartile Range (IQR)
The middle half (50%) of the data falls within this band, drawn between the 25th percentile and 75th percentile hinges.
Lower whisker
The lower whisker connects the first quartile hinge to the lowest data point within 1.5 * IQR of the hinge.
Upper whisker
The upper whisker connects the third quartile hinge to the highest data point within 1.5 * IQR of the hinge.
Outliers
Any data points below 1.5 * IQR of the first quartile hinge, or above 1.5 * IQR of the third quartile hinge, are marked individually as outliers.
We can add additional values to these plots. For instance, it’s sometimes useful to add the mean (average) when the distributions are heavily skewed:
```ggplot(data = df, aes(y = value, x = variable)) +
geom_boxplot() +
stat_summary(fun.y = mean, geom="point", shape = 10, size = 3, colour = "blue") +
coord_flip() +
theme_bw()
```
Graphs created in the R programming language using the ggplot2 and gridExtra packages.
### References
1. R Core Team (2014). R: A language and environment for statistical computing. R Foundation for Statistical
Computing, Vienna, Austria. URL http://www.R-project.org/.
2. H. Wickham. ggplot2: elegant graphics for data analysis. Springer New York, 2009.
3. Baptiste Auguie (2012). gridExtra: functions in Grid graphics. R package version 0.9.1.
http://CRAN.R-project.org/package=gridExtra
## One thought on “The Most Useful Data Plot You’ve Never Used”
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,550 | 5,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2019-22 | latest | en | 0.842647 |
https://fmforums.com/topic/46211-date-and-time-calculations/ | 1,716,243,252,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00490.warc.gz | 230,448,167 | 19,014 | # Date and Time Calculations
This topic is 6457 days old. Please don't post here. Open a new topic instead.
## Recommended Posts
Here is my situation. I've got various files which contain a value which is the date and time in seconds since 1st Jan 1904. Some files contain this as a negative value, others as a positive value.
I have an Excel calc that handles both:
IF(A1<0,DATE(50,1,1)+((2843274496+A1)/86400),(A1/86400))
You simply replace the A1 references to the cell containing the value, and it'll work out if it's neg or pos and then perform the correct calculation on it.
What I'm looking for is a similar thing for Filemaker Pro, as I want to import the files into filemaker rather than use Excel.
From the limited amount I know, I *think* I would need to define three fields in total, one for the neg/pos value, one for the date (it would be a calculation with the format set to date) and one for the time (another calc, this time the format set to time).
Any help on acomplishing this appreciated.
##### Share on other sites
Hi
give a try to this custom function:
http://www.briandunning.com/cf/423
##### Share on other sites
Try
date = Div ( 60052752000 + yourNumber ; 86400 ) + 1 ;
time = Mod ( 60052752000 + yourNumber ; 86400 )
---
A simpler way would be to use the Timestamp() function, but there seems to be a bug:
Timestamp ( Date ( 1 ; 1 ; 1904 ) ; -1 )
should return:
Dec-31-1903 23:59:59
but instead it returns:
Jan-01-1904 0:00:00
It seems to work with SOME negative numbers, but not all.
##### Share on other sites
Hi comment
this calc ( result Timestamp ) seems to work:
Timestamp ( Date ( 1 ; 1 ; 1904 ) ; 0 ) - 1
return:
Dec-31-1903 23:59:59
##### Share on other sites
Indeed, but Timestamp ( Date ( 1 ; 1 ; 1904 ) ; 0 ) is a constant, so it can be pre-computed - that's exactly what I have done above.
Which, BTW, could be further simplified to:
date = Div ( 60052838400 + yourNumber ; 86400 ) ;
time = Mod ( yourNumber ; 86400 )
Edited by Guest
##### Share on other sites
This topic is 6457 days old. Please don't post here. Open a new topic instead.
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## Create an account
Sign up for a new account in our community. It's easy!
Register a new account | 637 | 2,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.887361 |
http://www.elmerfem.org/forum/viewtopic.php?f=8&t=597&p=1963 | 1,571,734,096,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987813307.73/warc/CC-MAIN-20191022081307-20191022104807-00194.warc.gz | 257,998,553 | 7,346 | ## Eigen Modes
Post processing utility for Elmer
arend
Posts: 4
Joined: 06 Aug 2010, 15:40
### Eigen Modes
Hello,
I am looking for how to visualise the eigen modes (eingenvectors) on the Elmer Post.
The solver generated the eigen values but I don't know how to set the eigen mode / associed displacement view in the Elmer Post.
tks,
Arend
Code: Select all
``````...
Solver 1
Equation = "Stress Analysis"
Eigen Analysis = Logical True
Eigen System Values = Integer 5
...
``````
raback
Posts: 3467
Joined: 22 Aug 2009, 11:57
Antispam: Yes
Location: Espoo, Finland
Contact:
### Re: Eigen Modes
Hi Arend
Eigenmodes are treated as timesteps. So you should have five "timesteps" in your .ep file.
-Peter
arend
Posts: 4
Joined: 06 Aug 2010, 15:40
### Re: Eigen Modes
Hi Peter,
Yes, the .ep file shows five #time steps. The question is how to display it. The color mesh variables shows only the displacement x, y, z and abs.
tks
raback
Posts: 3467
Joined: 22 Aug 2009, 11:57
Antispam: Yes
Location: Espoo, Finland
Contact:
### Animating timedependent and harmonic data
Hi,
Here is a copy-paste from some older discussion:
ElmerPost assumes that the node coordinates are defined in the "nodes" matrix
such that nodes=[x y z]. Now if you want to have the displacement/deflection
included in the nodes you should alter it accordingly i.e. x=x0+dx etc. For
that we 1st save the original nodes:
math n0=nodes
a) The simplest case is if you have just one timestep/eigenvalue read:
In the math window of the main window
math nodes=n0+Displacement*scale
and press the phsychedelic button in the upper righthand corner. Here "scale"
is a scaling factor of your choice
b) for many steps
open "Timestep control" panel, write the following in
the "Do after Frame" field:
math nodes=n0+Displacement(0:2,time(\$t))
and press the "Loop" button.
c) for time-harmonic / eigenmode case
You need to create time-harmonic behavior on-the-fly in the main math window
using a small loop. For example over 2*pi:
do i 0 20 { math phi=2*pi*\$i/20; math
nodes=n0+Displacement(0:2,time(0))*sin(phi)*0.02; display}
or if you have a complex function, like in this case a complex pressure,
do i 1 50 { math phi=2*pi*\$i/50; math nodes=n+Displacement*sin(phi)*1.0e-5;
math pres=Pressure.1*cos(phi)-Pressure.2*sin(phi); display; savempg append }
In this case its a good idea to a priori set the scaling and fix it since
otherwise the colorscale will be chanching all the time. For sinusoidal
movement the interval sould be [-max(Displacement_abs),max(Displacement_abs)].
animations
=======
To make an animation using the "savempg" module on the right-hand-corner do
2) Put within the loop the command "savempg append" and loop over the
timesteps (like in the example)
3) Stop
Or on older way, if the modules are not available you could (in Unix/Linux)
take a number of screenshots
\$ xwd -name "ELMER POST GRAPHICS" -out fig\$i.xwd
The possibilities are endless
BR, Peter
arend
Posts: 4
Joined: 06 Aug 2010, 15:40
### Re: Eigen Modes
Peter,
To show a displacement in x,y and z is not the issue.
In my application I selected 5 main modes (Eigen System Values = Integer 5), how to display each one ?
So many tks,
Arend
arend
Posts: 4
Joined: 06 Aug 2010, 15:40
### Re: Eigen Modes
Peter,
I found an example ( ElasticPlateEigenmodes) in the new Elmer Tutorial .pdf document (my last tutorial doc. version did not have this one).
Now is clear the time step procedure.
Thanks and sorry the mess,
Arend | 997 | 3,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-43 | longest | en | 0.77617 |
https://www.raucci.net/2021/09/06/ | 1,632,651,630,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057857.27/warc/CC-MAIN-20210926083818-20210926113818-00079.warc.gz | 970,054,387 | 17,425 | ## Archive for Settembre 6th, 2021
### Python Crash Course…
lunedì, Settembre 6th, 2021
## Libraries
Python is a high-level open-source language. But the Python world is inhabited by many packages or libraries that provide useful things like array operations, plotting functions, and much more. We can import libraries of functions to expand the capabilities of Python in our programs.
OK! We’ll start by importing a few libraries to help us out. First: our favorite library is NumPy, providing a bunch of useful array operations (similar to MATLAB). We will use it a lot! The second library we need is Matplotlib, a 2D plotting library which we will use to plot our results.
import numpy as np
from matplotlib import pyplot as plt
We are importing one library named numpy (that we will call np)and we are importing a module called pyplot (we refer to this library as plt) of a big library called matplotlib. To use a function belonging to one of these libraries, we have to tell Python where to look for it. For that, each function name is written following the library name, with a dot in between. So if we want to use the NumPy function linspace(), which creates an array with equally spaced numbers between a start and end, we call it by writing:
import numpy as np
from matplotlib import pyplot as plt
myarray = np.linspace(0,5,10)
print(myarray)
## Variables
Python doesn’t require explicitly declared variable types like Fortran and other languages.
a =5 #a is an integer
b= 'Five' #b is a string
c = 5.0 #c is a floating point number
print(type(a))
print(type(b))
print(type(c))
we have:
Note: if you divide an integer by an integer that yields a remainder, the result will be converted to a float.
## Whitespace in Python
Python uses indents and whitespace to group statements together. To write a short loop in Fortran, you might use:
b= "Five"
do i = 1, 5
print(b)
end do
Python does not use end do blocks like Fortran, so the same program as above is written in Python as follows:
b= 'Five' #b is a string
for i in range(5):
print(b)
If you have nested for-loops, there is a further indent for the inner loop.
for i in range(3):
for j in range(3):
print(i, j)
print("This statement is within the i-loop, but not the j-loop")
## Slicing Arrays
In NumPy, you can look at portions of arrays in the same way as in Matlab, with a few extra tricks thrown in. Let’s take an array of values from 1 to 5.
vec = np.array([1, 2, 3, 4, 5])
print(vec)
Python uses a zero-based index, so let’s look at the first and last element in the array vec
print(vec)
print(vec[0], vec[4])
it has this output:
[1 2 3 4 5]
1 5
Arrays can also be ‘sliced‘, grabbing a range of values. Let’s look at the first three elements:
vec = np.array([1, 2, 3, 4, 5])
print(vec)
print(vec[0], vec[4])
print(vec[0:3])
Note here, the slice is inclusive on the front end and exclusive on the back, so the above command gives us the values of vec[0], vec[1] and vec[2], but not vec[3].
## Assigning Array Variables
One of the strange little quirks/features in Python that often confuses people comes up when assigning and comparing arrays of values. Here is a quick example. Let’s start by defining a 1-D array called vecx:
vecx = np.linspace(1,5,5)
print(vecx)
The linspace command return evenly spaced numbers over a specified interval. In this case, for vecx we have:
[1. 2. 3. 4. 5.]
OK, so we have an array vecx, with the values 1 through 5. I want to make a copy of that array, called vecy, so I’ll try the following:
vecy = vecx
print(vecy)
and we have:
[1. 2. 3. 4. 5.]
Great. So vecx has the values 1 through 5 and now so does vecy. Now that I have a backup of vecx, I can change its values without worrying about losing data (or so I may think!).
vecx[2]=10
print(vecx)
print(vecy)
we have:
And that’s how things go wrong! When you use a statement like vecx=vecy, rather than copying all the values of vecx
into a new array called vecy, Python just creates an alias (or a pointer) called vecy and tells it to route us to vecx. So if we change a value in vecx then vecy will reflect that change (technically, this is called assignment by reference).
If you want to make a true copy of the array, you have to tell Python to copy every element of vecx into a new array. Let’s call it vecz:
vecx = np.linspace(1,5,5)
print(vecx)
vecy = vecx
vecz=vecx.copy()
vecx[2]=10
print(vecx)
print(vecy)
print(vecz)
## Matrix Multiplication
A matrix is a 2D array, where each element in the array has 2 indices. For example, [[1, 2], [3, 4]] is a matrix, and the index of 1 is (0,0).
We can prove this using Python and Numpy.
import numpy as np
A = [[1, 2], [3, 4]]
print(np.array(A)[0,0])
The simple form of matrix multiplication is called scalar multiplication, multiplying a scalar by a matrix.
Scalar multiplication is generally easy. Each value in the input matrix is multiplied by the scalar, and the output has the same shape as the input matrix.
In Python:
import numpy as np
b=7
A = [[1, 2], [3, 4]]
C=np.dot(b,A)
print(np.array(C))
There are a few things to keep in mind.
1. Order matters now. $$A \times B \neq B \times A$$;
2. Matrices can be multiplied if the number of columns in the 1st equals the number of rows in the 2nd;
3. Multiplication is the dot product of rows and columns. Rows of the 1st matrix with columns of the 2nd.
Let’s replicate the result in Python.
import numpy as np
A = [[1, 2], [3, 4]]
B = [[5, 6], [7, 8]]
C = np.dot(A,B)
print(np.array(C))
## Inner product
The inner product takes two vectors of equal size and returns a single number (scalar). This is calculated by multiplying the corresponding elements in each vector and adding up all of those products. In numpy, vectors are defined as one-dimensional numpy arrays.
To get the inner product, we can use either np.inner() or np.dot(). Both give the same results. The inputs for these functions are two vectors and they should be the same size.
import numpy as np
# Vectors as 1D numpy arrays
a = np.array([1, 2, 3])
b = np.array([4, 5, 6])
print("a= ", a)
print("b= ", b)
print("\ninner:", np.inner(a, b))
print("dot:", np.dot(a, b))
## Transpose
The transpose of a matrix is found by switching its rows with its columns. We can use np.transpose() function or NumPy ndarray.transpose() method or ndarray.T (a special method which does not require parentheses) to get the transpose. All give the same output.
import numpy as np
a = np.array([[1, 2], [3, 4], [5, 6]])
print("a = ")
print(a)
print("\nWith np.transpose(a) function")
print(np.transpose(a))
print("\nWith ndarray.transpose() method")
print(a.transpose())
print("\nWith ndarray.T short form")
print(a.T)
## Trace
The trace is the sum of diagonal elements in a square matrix. There are two methods to calculate the trace. We can simply use the trace() method of an ndarray object or get the diagonal elements first and then get the sum.
import numpy as np
a = np.array([[2, 2, 1],
[1, 3, 1],
[1, 2, 2]])
print("a = ")
print(a)
print("\nTrace:", a.trace())
print("Trace:", sum(a.diagonal()))
## Rank
The rank of a matrix is the dimensions of the vector space spanned (generated) by its columns or rows. In other words, it can be defined as the maximum number of linearly independent column vectors or row vectors.
The rank of a matrix can be found using the matrix_rank() function which comes from the numpy linalg package.
import numpy as np
a = np.arange(1, 10)
a.shape = (3, 3)
print("a = ")
print(a)
rank = np.linalg.matrix_rank(a)
print("\nRank:", rank)
## Determinant
The determinant of a square matrix can be calculated det() function which also comes from the numpy linalg package. If the determinant is 0, that matrix is not invertible. It is known as a singular matrix in algebra terms.
import numpy as np
a = np.array([[2, 2, 1],
[1, 3, 1],
[1, 2, 2]])
print("a = ")
print(a)
det = np.linalg.det(a)
print("\nDeterminant:", np.round(det))
## True inverse
The true inverse of a square matrix can be found using the inv() function of the numpy linalg package. If the determinant of a square matrix is not 0, it has a true inverse.
import numpy as np
a = np.array([[2, 2, 1],
[1, 3, 1],
[1, 2, 2]])
print("a = ")
print(a)
det = np.linalg.det(a)
print("\nDeterminant:", np.round(det))
inv = np.linalg.inv(a)
print("\nInverse of a = ")
print(inv)
## Flatten
Flatten is a simple method to transform a matrix into a one-dimensional numpy array. For this, we can use the flatten() method of an ndarray object.
import numpy as np
a = np.arange(1, 10)
a.shape = (3, 3)
print("a = ")
print(a)
print("\nAfter flattening")
print("------------------")
print(a.flatten())
## Eigenvalues and eigenvectors
Let A be an n x n matrix. A scalar $$\lambda$$ is called an eigenvalue of A if there is a non-zero vector x satisfying the following equation.
$$A x=\lambda x$$
The vector x is called the eigenvector of A corresponding to $$\lambda$$.
In numpy, both eigenvalues and eigenvectors can be calculated simultaneously using the eig() function.
import numpy as np
a = np.array([[2, 2, 1],
[1, 3, 1],
[1, 2, 2]])
print("a = ")
print(a)
w, v = np.linalg.eig(a)
print("\nEigenvalues:")
print(w)
print("\nEigenvectors:")
print(v)
The sum of eigenvalues (1+5+1=7) is equal to the trace (2+3+2=7) of the same matrix! The product of the eigenvalues (1x5x1=5) is equal to the determinant (5) of the same matrix!
### […]
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# 3 Introduction to Linear Programming - Introduction to...
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± ± ± ± ± ± ± ± ± ± ± EXAMPLE 1 Giapetto’s Woodcarving Introduction to Linear Programming Linear programming (LP) is a tool for solving optimization problems. In 1947, George Dantzig de- veloped an efFcient method, the simplex algorithm, for solving linear programming problems (also called LP). Since the development of the simplex algorithm, LP has been used to solve optimiza- tion problems in industries as diverse as banking, education, forestry, petroleum, and trucking. In a survey of ±ortune 500 Frms, 85% of the respondents said they had used linear programming. As a measure of the importance of linear programming in operations research, approximately 70% of this book will be devoted to linear programming and related optimization techniques. In Section 3.1, we begin our study of linear programming by describing the general char- acteristics shared by all linear programming problems. In Sections 3.2 and 3.3, we learn how to solve graphically those linear programming problems that involve only two variables. Solv- ing these simple LPs will give us useful insights for solving more complex LPs. The remainder of the chapter explains how to formulate linear programming models of real-life situations. 3.1 What Is a Linear Programming Problem? In this section, we introduce linear programming and defne important terms that are used to describe linear programming problems. Giapetto’s Woodcarving, Inc., manuFactures two types oF wooden toys: soldiers and trains. A soldier sells For \$27 and uses \$10 worth oF raw materials. Each soldier that is manu- Factured increases Giapetto’s variable labor and overhead costs by \$14. A train sells For \$21 and uses \$9 worth oF raw materials. Each train built increases Giapetto’s variable la- bor and overhead costs by \$10. The manuFacture oF wooden soldiers and trains requires two types oF skilled labor: carpentry and fnishing. A soldier requires 2 hours oF fnishing labor and 1 hour oF carpentry labor. A train requires 1 hour oF fnishing and 1 hour oF car- pentry labor. Each week, Giapetto can obtain all the needed raw material but only 100 fn- ishing hours and 80 carpentry hours. Demand For trains is unlimited, but at most 40 sol- diers are bought each week. Giapetto wants to maximize weekly proft (revenues ± costs). ±ormulate a mathematical model oF Giapetto’s situation that can be used to maximize Gi- apetto’s weekly proft. Solution In developing the Giapetto model, we explore characteristics shared by all linear pro- gramming problems. Decision Variables We begin by defning the relevant decision variables. In any linear programming model, the decision variables should completely describe the decisions to be made (in this case, by Giapetto). Clearly, Giapetto must decide how many soldiers and trains should be manuFactured each week. With this in mind, we defne
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x 1 ± number of soldiers produced each week x 2 ± number of trains produced each week Objective Function In any linear programming problem, the decision maker wants to max-
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### mixed model of artificial and conduction-based cells
Posted: Tue Apr 09, 2013 12:07 pm
I'm trying to make a mixed model, where some multicompartment or single-compartment cells may be arbitrarily replaced by Izhikevich model(39948). Conditions of Izh. model should be the same as for conduction-based cells: the same Exp2Syn synapses, InNp noise current generator and so on. I'm not sure that it is possible, because Exp2Syn need voltage and current from segment. I'm wondering, maybe somebody has solved this problem before? If yes, could you please provide any links?
One more question. Is it possible to manipulate with voltage in module (mod file) mechanism? If it is possible, why can we not present quadratic part of Izh. model in right side as non-specific current and use standard approach for voltage and current?
Thanks,
Ruben
### Re: mixed model of artificial and conduction-based cells
Posted: Tue Apr 09, 2013 10:52 pm
Not aware that anyone has done this yet but definitely what i had in mind when i posted the model so i would be happy to help you out with getting this developed and posted.
It would appear that the reason i couldn't use a cell V for the 'v' state variable in izhikevich formlation is that the izhikevich v (called vv in the mod file) is a quadratic form: vv' = e*vv*vv + f*vv + g - u + I - gsyn*(vv-erev)
Because of this discrepancy it would appear to be necessary to create the formula for gsyn within a variant of the izh.mod file, probably by making gsyn a state variable so that it could provide exponential falloff in the manner of the weight var in Exp2Syn.
bill
### Re: mixed model of artificial and conduction-based cells
Posted: Wed Apr 10, 2013 11:39 am
Thank you, Bill.
Right now I have a question. You use vv, because v is embedded in standard segment. Why can we not use variable v and i in standard segment for Izh. model? We can rewrite his model as v'=(Izh(v)+I)/c and where Izh(v)=e*v*v+f*v+g+u - is the non-specific current, c is equal 1 and u is one more dynamic variable. The problem is that: in this way we have to have access to voltage variable v in a segment to reset it, when it reaches threshold. Is it possible? Maybe it is possible use some c-code insertions to access v for manipulation?
Ruben
### Re: mixed model of artificial and conduction-based cells
Posted: Wed Apr 10, 2013 11:49 am
sounds like a good idea -- give it a try?
### Re: mixed model of artificial and conduction-based cells
Posted: Wed Apr 10, 2013 5:43 pm
Well, here is a first shot.
Module for current, izhcur.mod
Code: Select all
``````TITLE IzhikevichCurrent
COMMENT
It is a dirty trick for implementation of Izhikevich model as a
non-specific current.
This mod file treats Izhikevich model as
v'=(Iizh(v)+i)/cm
where Iizh=e*v^2+f*v+g-u;
u is treated as gating variable
Do not forget setup cm in 1mF
scratched by Ruben A. Tikidji-Hamburyan
ENDCOMMENT
NEURON {
SUFFIX izhcur
NONSPECIFIC_CURRENT i_izh
RANGE a,b,c,d,e,f,g
}
PARAMETER {
a = 0.1 (1)
b = 0.2 (1)
c = -65 (mV)
d = 2 (1)
e = 0.04 (1)
f = 5 (1)
g = 140 (1)
}
UNITS {
(mV) = (millivolt)
(mA) = (milliamp)
}
ASSIGNED {
v (mV)
i_izh (mA/cm2)
}
STATE { u }
BREAKPOINT {
SOLVE states METHOD cnexp
i_izh = -0.000001*(vinfi(v)-u)
if(v>0.0){
printf ("spike! t=%g,v=%g, u=%g\n",t,v,u)
v = c
u = u+d
}
}
INITIAL {
u = 0
}
DERIVATIVE states {
UNITSOFF
u'= a*(b*v-u)
UNITSON
}
FUNCTION vinfi(v (mV)) {
UNITSOFF
vinfi = e*v*v + f*v + g
UNITSON
}
``````
Test example izh.hoc
Code: Select all
``````create soma
access soma
soma{
L=1
diam=10/PI
nseg=1
insert izhcur
cm=0.1
}
objref istim
istim = new IClamp(.5)
istim.del = 200
istim.dur = 500
istim.amp = 0.0001
tstop = 1000
tstep = 0.1
objref g
g = new Graph()
g.size(0,tstop,-80,40)
graphList[0].append(g)
g.addexpr("v(.5)", 1, 1, 0.8, 0.9, 2)
run()
``````
It produces some activity, for example, for regular spiking:
and fast spiking modes
but voltage resetting doesn't work:
Code: Select all
``````
spike! t=206.662,v=0.106934, u=-10.1618
spike! t=206.662,v=0.105934, u=-8.16178
spike! t=206.687,v=0.682722, u=-6.14606
spike! t=206.687,v=0.681722, u=-4.14606
spike! t=206.712,v=1.25796, u=-2.14007
spike! t=206.712,v=1.25696, u=-0.140073
spike! t=206.737,v=1.83267, u=1.8562
spike! t=206.737,v=1.83167, u=3.8562
spike! t=206.762,v=2.40687, u=5.84278
spike! t=206.762,v=2.40587, u=7.84278
spike! t=206.787,v=2.98057, u=9.81969
spike! t=206.787,v=2.97957, u=11.8197
spike! t=206.812,v=3.55378, u=13.787
spike! t=206.812,v=3.55278, u=15.787
spike! t=206.837,v=4.12654, u=17.7446
spike! t=206.837,v=4.12554, u=19.7446
spike! t=206.862,v=4.69885, u=21.6927
spike! t=206.862,v=4.69785, u=23.6927
spike! t=206.887,v=5.27072, u=25.6311
spike! t=206.887,v=5.26972, u=27.6311
spike! t=206.912,v=5.84219, u=29.5601
spike! t=206.912,v=5.84119, u=31.5601
spike! t=206.937,v=6.41325, u=33.4795
spike! t=206.937,v=6.41225, u=35.4795
spike! t=206.962,v=6.98394, u=37.3894
spike! t=206.962,v=6.98294, u=39.3894
spike! t=206.987,v=7.55426, u=41.2898
spike! t=206.987,v=7.55326, u=43.2898
spike! t=207.012,v=8.12424, u=45.1808
spike! t=207.012,v=8.12324, u=47.1808
spike! t=207.037,v=8.69388, u=49.0623
spike! t=207.037,v=8.69288, u=51.0623
spike! t=207.062,v=9.26322, u=52.9345
spike! t=207.062,v=9.26222, u=54.9345
spike! t=207.087,v=9.83226, u=56.7972
spike! t=207.087,v=9.83126, u=58.7972
.......
``````
So, it isn't working model at all! Don't use it, please
If anybody has any idea how to reset voltage, I'll appreciate.
Ruben
### Re: mixed model of artificial and conduction-based cells
Posted: Wed Apr 10, 2013 5:51 pm
as above
you need to use a WATCH statement
looks like you are not doing so
bill
### Re: mixed model of artificial and conduction-based cells
Posted: Wed Apr 10, 2013 6:03 pm
yes Bill, because NET_RECEIVE section can exist only in POINT_PROCESS module.
### Re: mixed model of artificial and conduction-based cells
Posted: Wed Apr 10, 2013 6:16 pm
maybe we can insert two modules, one for current and the other one for threshold (POINT_PROCESS), but how the latter one will reach the u variable in current module, I don't know....
### Re: mixed model of artificial and conduction-based cells
Posted: Wed Apr 10, 2013 7:07 pm
yes, this needs to be a POINT_PROCESS which is what it was before
### Re: mixed model of artificial and conduction-based cells
Posted: Wed Apr 10, 2013 7:32 pm
@ insert two modules -- just 1 module; this is as it was before; start with the orig and edit that
NEURON {
POINT_PROCESS IZH
### Re: mixed model of artificial and conduction-based cells
Posted: Thu Apr 11, 2013 2:36 pm
Bill,
It seems it works. Here alpha version for testing. Please take a look. I didn't check all 21 modes, but what I've checked works perfect. I'll appreciate for any comments and tests.
file izhcur.mod:
Code: Select all
``````TITLE IzhikevichCurrent
COMMENT
This dirty trick implements Izhikevich model as a non-specific current.
This module reats Izhikevich model as
v'=(Iizh(v)+i)/cm
where Iizh=e*v^2+f*v+g-u is a inward current and u is treated as gating variable
Do not forget setup:
cm in 1uF
L in 1um
diam in 10/PI
Here an example, who to use it in hoc file.
objref izh
soma{
L=1
diam=10/PI
nseg=1
izh = new izhcur(0.5)
cm=1
}
Authors: Ruben Tikidji-Hamburyan rtikid at lsuhsc.edu, rth at nisms.krinc.ru
Code partially based on izh.mod file written by William Lytton [billl at neurosim.downstate.edu];
ENDCOMMENT
NEURON {
POINT_PROCESS izhcur
NONSPECIFIC_CURRENT i_izh
RANGE a,b,c,d,e,f,g
}
PARAMETER {
a = 0.01 (1)
b = 0.2 (1)
c = -65 (mV)
d = 2 (1)
e = 0.04 (1)
f = 5 (1)
g = 140 (1)
}
UNITS {
(mV) = (millivolt)
(mA) = (milliamp)
}
ASSIGNED {
v (mV)
i_izh (mA)
}
STATE { u }
BREAKPOINT {
SOLVE states METHOD cnexp
i_izh = -1e-5*(vinfi(v)-u) :minus, because it is inward current
}
INITIAL {
u = -14.0
net_send(0,1) :we have to send first event
}
DERIVATIVE states {
UNITSOFF
u'= a*(b*v-u)
UNITSON
}
FUNCTION vinfi(v (mV)) {
UNITSOFF
vinfi = e*v*v + f*v + g
UNITSON
}
if (flag == 1) {
WATCH (v > 30.0) 2
} else {
net_event(t)
v = c
u = u+d
}
}
``````
Simple example, who to use it:
Code: Select all
``````create soma
access soma
objref izh
soma{
L=1
diam=10/PI
nseg=1
izh = new izhcur(0.5)
cm=1
}
objref istim
istim = new IClamp(.5)
istim.del = 200
istim.dur = 500
istim.amp = 0.0001
tstop = 1000
tstep = 0.1
objref g
g = new Graph()
g.size(0,tstop,-80,40)
graphList[0].append(g)
g.addexpr("v(.5)", 1, 1, 0.8, 0.9, 2)
run()
``````
The results for regular spiking mode:
for fast spiking
Ruben
### Re: mixed model of artificial and conduction-based cells
Posted: Thu Apr 11, 2013 3:15 pm
correction: diam should be 1/PI
### Re: mixed model of artificial and conduction-based cells
Posted: Thu Apr 11, 2013 4:05 pm | 3,217 | 8,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-04 | latest | en | 0.930191 |
https://math.libretexts.org/Courses/Mission_College/Math_1%3A_College_Algebra_(Carr)/07%3A_Conics/7.02%3A_The_Ellipse | 1,726,059,994,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00446.warc.gz | 352,063,638 | 40,633 | # 7.2: The Ellipse
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$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
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$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
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$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
###### Learning Objectives
• Write equations of ellipses in standard form.
• Graph ellipses centered at the origin.
• Graph ellipses not centered at the origin.
• Solve applied problems involving ellipses.
Can you imagine standing at one end of a large room and still being able to hear a whisper from a person standing at the other end? The National Statuary Hall in Washington, D.C., shown in Figure $$\PageIndex{1}$$, is such a room. It is an oval-shaped room called a whispering chamber because the shape makes it possible for sound to travel along the walls. In this section, we will investigate the shape of this room and its real-world applications, including how far apart two people in Statuary Hall can stand and still hear each other whisper.
Figure $$\PageIndex{1}$$: The National Statuary Hall in Washington, D.C. (credit: Greg Palmer, Flickr)
## Writing Equations of Ellipses in Standard Form
A conic section, or conic, is a shape resulting from intersecting a right circular cone with a plane. The angle at which the plane intersects the cone determines the shape, as shown in Figure $$\PageIndex{2}$$.
Figure $$\PageIndex{2}$$
Conic sections can also be described by a set of points in the coordinate plane. Later in this chapter, we will see that the graph of any quadratic equation in two variables is a conic section. The signs of the equations and the coefficients of the variable terms determine the shape. This section focuses on the four variations of the standard form of the equation for the ellipse. An ellipse is the set of all points $$(x,y)$$ in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus(plural: foci).
We can draw an ellipse using a piece of cardboard, two thumbtacks, a pencil, and string. Place the thumbtacks in the cardboard to form the foci of the ellipse. Cut a piece of string longer than the distance between the two thumbtacks (the length of the string represents the constant in the definition). Tack each end of the string to the cardboard, and trace a curve with a pencil held taut against the string. The result is an ellipse. See Figure $$\PageIndex{3}$$.
Figure $$\PageIndex{3}$$
Every ellipse has two axes of symmetry. The longer axis is called the major axis, and the shorter axis is called the minor axis. Each endpoint of the major axis is the vertex of the ellipse (plural: vertices), and each endpoint of the minor axis is a co-vertex of the ellipse. The center of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci (Figure $$\PageIndex{4}$$).
Figure $$\PageIndex{4}$$
In this section, we restrict ellipses to those that are positioned vertically or horizontally in the coordinate plane. That is, the axes will either lie on or be parallel to the $$x$$- and $$y$$-axes. Later in the chapter, we will see ellipses that are rotated in the coordinate plane.
To work with horizontal and vertical ellipses in the coordinate plane, we consider two cases: those that are centered at the origin and those that are centered at a point other than the origin. First we will learn to derive the equations of ellipses, and then we will learn how to write the equations of ellipses in standard form. Later we will use what we learn to draw the graphs.
### Deriving the Equation of an Ellipse Centered at the Origin
To derive the equation of an ellipse centered at the origin, we begin with the foci $$(−c,0)$$ and $$(c,0)$$. The ellipse is the set of all points $$(x,y)$$ such that the sum of the distances from $$(x,y)$$ to the foci is constant, as shown in Figure $$\PageIndex{5}$$.
Figure $$\PageIndex{5}$$
If $$(a,0)$$ is a vertex of the ellipse, the distance from $$(−c,0)$$ to $$(a,0)$$ is $$a−(−c)=a+c$$. The distance from $$(c,0)$$ to $$(a,0)$$ is $$a−c$$. The sum of the distances from the foci to the vertex is
$$(a+c)+(a−c)=2a$$
If $$(x,y)$$ is a point on the ellipse, then we can define the following variables:
• $$d_1=$$ the distance from $$(−c,0)$$ to $$(x,y)$$
• $$d_2=$$ the distance from $$(c,0)$$ to $$(x,y)$$
By the definition of an ellipse, $$d_1+d_2$$ is constant for any point $$(x,y)$$ on the ellipse. We know that the sum of these distances is $$2a$$ for the vertex $$(a,0)$$. It follows that $$d_1+d_2=2a$$ for any point on the ellipse. We will begin the derivation by applying the distance formula. The rest of the derivation is algebraic.
\begin{align*} d_1+d_2&= 2a\\ \sqrt{{(x-(-c))}^2+{(y-0)}^2}+\sqrt{{(x-c)}^2+{(y-0)}^2}&=2a\qquad \text{Distance formula}\\ \sqrt{{(x+c)}^2+y^2}+\sqrt{{(x-c)}^2+y^2}&=2a\qquad \text{Simplify expressions.}\\ \sqrt{{(x+c)}^2+y^2}&=2a-\sqrt{{(x-c)}^2+y^2}\qquad \text{Move radical to opposite side.}\\ {(x+c)}^2+y^2&={\left[2a-\sqrt{{(x-c)}^2+y^2}\right]}^2\qquad \text{Square both sides.}\\ x^2+2cx+c^2+y^2&=4a^2-4a\sqrt{{(x-c)}^2+y^2}+{(x-c)}^2+y^2\qquad \text{Expand the squares.}\\ x^2+2cx+c^2+y^2&=4a^2-4a\sqrt{{(x-c)}^2+y^2}+x^2-2cx+c^2+y^2\qquad \text{Expand remaining squares.}\\ 2cx&=4a^2-4a\sqrt{{(x-c)}^2+y^2}-2cx\qquad \text{Combine like terms.}\\ 4cx-4a^2&=-4a\sqrt{{(x-c)}^2+y^2}\qquad \text{Isolate the radical.}\\ cx-a^2&=-a\sqrt{{(x-c)}^2+y^2}\qquad \text{Divide by 4.}\\ {\left[ cx-a^2\right]}^2&=a^2{\left[ \sqrt{{(x-c)}^2+y^2}\right] }^2\qquad \text{Square both sides.}\\ c^2x^2-2a^2cx+a^4&=a^2(x^2-2cx+c^2+y^2)\qquad \text{Expand the squares.}\\ c^2x^2-2a^2cx+a^4&=a^2x^2-2a^2cx+a^2c^2+a^2y^2\qquad \text{Distribute } a^2\\ a^2x^2-c^2x^2+a^2y^2&=a^4-a^2c^2\qquad \text{Rewrite.}\\ x^2(a^2-c^2)+a^2y^2&=a^2(a^2-c^2)\qquad \text{Factor common terms.}\\ x^2b^2+a^2y^2&=a^2b^2\qquad \text{Set } b^2=a^2-c^2\\ \dfrac{x^2b^2}{a^2b^2}+\dfrac{a^2y^2}{a^2b^2}&=\dfrac{a^2b^2}{a^2b^2}\qquad \text{Divide both sides by } a^2b^2\\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}&=1\qquad \text{Simplify} \end{align*}
Thus, the standard equation of an ellipse is $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$.This equation defines an ellipse centered at the origin. If $$a>b$$,the ellipse is stretched further in the horizontal direction, and if $$b>a$$, the ellipse is stretched further in the vertical direction.
### Writing Equations of Ellipses Centered at the Origin in Standard Form
Standard forms of equations tell us about key features of graphs. Take a moment to recall some of the standard forms of equations we’ve worked with in the past: linear, quadratic, cubic, exponential, logarithmic, and so on. By learning to interpret standard forms of equations, we are bridging the relationship between algebraic and geometric representations of mathematical phenomena.
The key features of the ellipse are its center, vertices, co-vertices, foci, and lengths and positions of the major and minor axes. Just as with other equations, we can identify all of these features just by looking at the standard form of the equation. There are four variations of the standard form of the ellipse. These variations are categorized first by the location of the center (the origin or not the origin), and then by the position (horizontal or vertical). Each is presented along with a description of how the parts of the equation relate to the graph. Interpreting these parts allows us to form a mental picture of the ellipse.
###### STANDARD FORMS OF THE EQUATION OF AN ELLIPSE WITH CENTER $$(0,0)$$
The standard form of the equation of an ellipse with center $$(0,0)$$ and major axis on the $$x$$-axis is
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
where
• $$a>b$$
• the length of the major axis is $$2a$$
• the coordinates of the vertices are $$(\pm a,0)$$
• the length of the minor axis is $$2b$$
• the coordinates of the co-vertices are $$(0,\pm b)$$
• the coordinates of the foci are $$(\pm c,0)$$, where $$c^2=a^2−b^2$$. See Figure $$\PageIndex{6a}$$.
The standard form of the equation of an ellipse with center $$(0,0)$$ and major axis on the $$y$$-axis is
$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$
where
• $$a>b$$
• the length of the major axis is $$2a$$
• the coordinates of the vertices are $$(0,\pm a)$$
• the length of the minor axis is $$2b$$
• the coordinates of the co-vertices are $$(\pm b,0)$$
• the coordinates of the foci are $$(0,\pm c)$$, where $$c^2=a^2−b^2$$. See Figure $$\PageIndex{6b}$$.
Note that the vertices, co-vertices, and foci are related by the equation $$c^2=a^2−b^2$$. When we are given the coordinates of the foci and vertices of an ellipse, we can use this relationship to find the equation of the ellipse in standard form.
Figure $$\PageIndex{6}$$: (a) Horizontal ellipse with center $$(0,0)$$ (b) Vertical ellipse with center $$(0,0)$$
###### How to: Given the vertices and foci of an ellipse centered at the origin, write its equation in standard form
1. Determine whether the major axis lies on the x- or y-axis.
• If the given coordinates of the vertices and foci have the form $$(\pm a,0)$$ and $$(\pm c,0)$$ respectively, then the major axis is the x-axis. Use the standard form $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
• If the given coordinates of the vertices and foci have the form $$(0,\pm a)$$ and $$(\pm c,0)$$,respectively, then the major axis is the y-axis. Use the standard form $$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$$
2. Use the equation $$c^2=a^2−b^2$$, along with the given coordinates of the vertices and foci, to solve for $$b^2$$.
3. Substitute the values for $$a^2$$ and $$b^2$$ into the standard form of the equation determined in Step 1.
###### Example $$\PageIndex{1}$$: Writing the Equation of an Ellipse Centered at the Origin in Standard Form
What is the standard form equation of the ellipse that has vertices $$(\pm 8,0)$$ and foci $$(\pm 5,0)$$?
Solution
The foci are on the $$x$$-axis, so the major axis is the $$x$$-axis. Thus, the equation will have the form $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
The vertices are $$(\pm 8,0)$$,so $$a=8$$ and $$a^2=64$$.
The foci are $$(\pm 5,0)$$,so $$c=5$$ and $$c^2=25$$.
We know that the vertices and foci are related by the equation $$c^2=a^2−b^2$$. Solving for $$b^2$$, we have:
\begin{align*} c^2&=a^2-b^2\\ 25&=64-b^2\qquad \text{Substitute for } c^2 \text{ and } a^2\\ b^2&=39\qquad \text{Solve for } b^2 \end{align*}
Now we need only substitute $$a^2=64$$ and $$b^2=39$$ into the standard form of the equation. The equation of the ellipse is $$\dfrac{x^2}{64}+\dfrac{y^2}{39}=1$$.
###### Exercise $$\PageIndex{1}$$
What is the standard form equation of the ellipse that has vertices $$(0,\pm 4)$$ and foci $$(0,\pm \sqrt{15})$$?
$$x^2+\dfrac{y^2}{16}=1$$
###### Q&A
Can we write the equation of an ellipse centered at the origin given coordinates of just one focus and vertex?
Yes. Ellipses are symmetrical, so the coordinates of the vertices of an ellipse centered around the origin will always have the form $$(\pm a,0)$$ or $$(0, \pm a)$$. Similarly, the coordinates of the foci will always have the form $$(\pm c,0)$$ or $$(0, \pm c)$$. Knowing this, we can use $$a$$ and $$c$$ from the given points, along with the equation $$c^2=a^2−b^2$$, to find $$b^2$$.
### Writing Equations of Ellipses Not Centered at the Origin
Like the graphs of other equations, the graph of an ellipse can be translated. If an ellipse is translated $$h$$ units horizontally and $$k$$ units vertically, the center of the ellipse will be $$(h,k)$$. This translation results in the standard form of the equation we saw previously, with $$x$$ replaced by $$(x−h)$$ and y replaced by $$(y−k)$$.
###### STANDARD FORMS OF THE EQUATION OF AN ELLIPSE WITH CENTER $$(H, K)$$
The standard form of the equation of an ellipse with center $$(h, k)$$ and major axis parallel to the $$x$$-axis is
$\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1$
where
• $$a>b$$
• the length of the major axis is $$2a$$
• the coordinates of the vertices are $$(h\pm a,k)$$
• the length of the minor axis is $$2b$$
• the coordinates of the co-vertices are $$(h,k\pm b)$$
• the coordinates of the foci are $$(h\pm c,k)$$,where $$c^2=a^2−b^2$$. See Figure $$\PageIndex{7a}$$.
The standard form of the equation of an ellipse with center $$(h,k)$$ and major axis parallel to the $$y$$-axis is
$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1$
where
• $$a>b$$
• the length of the major axis is $$2a$$
• the coordinates of the vertices are $$(h,k\pm a)$$
• the length of the minor axis is $$2b$$
• the coordinates of the co-vertices are $$(h\pm b,k)$$
• the coordinates of the foci are $$(h,k\pm c)$$, where $$c^2=a^2−b^2$$. See Figure $$\PageIndex{7b}$$.
Just as with ellipses centered at the origin, ellipses that are centered at a point $$(h,k)$$ have vertices, co-vertices, and foci that are related by the equation $$c^2=a^2−b^2$$. We can use this relationship along with the midpoint and distance formulas to find the equation of the ellipse in standard form when the vertices and foci are given.
Figure $$\PageIndex{7}$$: (a) Horizontal ellipse with center $$(h,k)$$ (b) Vertical ellipse with center $$(h,k)$$
###### How to: Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form
1. Determine whether the major axis is parallel to the $$x$$- or $$y$$-axis.
• If the y-coordinates of the given vertices and foci are the same, then the major axis is parallel to the $$x$$-axis. Use the standard form $$\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1$$
• If the x-coordinates of the given vertices and foci are the same, then the major axis is parallel to the y-axis. Use the standard form $$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1$$
2. Identify the center of the ellipse $$(h,k)$$ using the midpoint formula and the given coordinates for the vertices.
3. Find $$a^2$$ by solving for the length of the major axis, $$2a$$, which is the distance between the given vertices.
4. Find $$c^2$$ using $$h$$ and $$k$$, found in Step 2, along with the given coordinates for the foci.
5. Solve for $$b^2$$ using the equation $$c^2=a^2−b^2$$.
6. Substitute the values for $$h$$, $$k$$, $$a^2$$, and $$b^2$$ into the standard form of the equation determined in Step 1.
###### Example $$\PageIndex{2}$$: Writing the Equation of an Ellipse Centered at a Point Other Than the Origin
What is the standard form equation of the ellipse that has vertices $$(−2,−8)$$ and $$(−2,2)$$ and foci $$(−2,−7)$$ and $$(−2,1)$$?
Solution
The $$x$$-coordinates of the vertices and foci are the same, so the major axis is parallel to the $$y$$-axis. Thus, the equation of the ellipse will have the form
$$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1 \nonumber$$
First, we identify the center, $$(h,k)$$. The center is halfway between the vertices, $$(−2,−8)$$ and $$(−2,2)$$. Applying the midpoint formula, we have:
\begin{align} (h,k) &=\left(\dfrac{−2+(−2)}{2},\dfrac{−8+2}{2}\right) \nonumber \\ &=(−2,−3) \nonumber \end{align} \nonumber
Next, we find $$a^2$$. The length of the major axis, $$2a$$, is bounded by the vertices. We solve for $$a$$ by finding the distance between the y-coordinates of the vertices.
\begin{align} 2a &=2−(−8) \nonumber \\ 2a &=10 \nonumber\\ a&=5 \nonumber \end{align} \nonumber
So $$a^2=25$$.
Now we find $$c^2$$. The foci are given by $$(h,k\pm c)$$. So, $$(h,k−c)=(−2,−7)$$ and $$(h,k+c)=(−2,1)$$. We substitute $$k=−3$$ using either of these points to solve for $$c$$.
\begin{align} k+c &=1 \nonumber \\ −3+c&=1 \nonumber \\ c&=4 \nonumber \end{align} \nonumber
So $$c^2=16$$.
Next, we solve for $$b^2$$ using the equation $$c^2=a^2−b^2$$.
\begin{align} c^2&=a^2−b^2 \nonumber \\ 16&=25−b^2 \nonumber \\ b^2&=9 \nonumber \end{align} \nonumber
Finally, we substitute the values found for $$h$$, $$k$$, $$a^2$$, and $$b^2$$ into the standard form equation for an ellipse:
$\dfrac{{(x+2)}^2}{9}+\dfrac{{(y+3)}^2}{25}=1 \nonumber$
###### Exercise $$\PageIndex{2}$$
What is the standard form equation of the ellipse that has vertices $$(−3,3)$$ and $$(5,3)$$ and foci $$(1−2\sqrt{3},3)$$ and $$(1+2\sqrt{3},3)$$?
$$\dfrac{{(x−1)}^2}{16}+\dfrac{{(y−3)}^2}{4}=1 \nonumber$$
## Graphing Ellipses Centered at the Origin
Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1, a>b$$ for horizontal ellipses
and
$$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1, a>b$$ for vertical ellipses
###### How to: Given the standard form of an equation for an ellipse centered at $$(0, 0)$$, sketch the graph.
1. Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci.
• If the equation is in the form $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$, where $$a>b$$, then
• the major axis is the $$x$$-axis
• the coordinates of the vertices are $$(\pm a,0)$$
• the coordinates of the co-vertices are $$(0,\pm b)$$
• the coordinates of the foci are $$(\pm c,0)$$
• If the equation is in the form $$x^2b^2+y^2a^2=1$$,where $$a>b$$, then
• the major axis is the $$y$$-axis
• the coordinates of the vertices are $$(0,\pm a)$$
• the coordinates of the co-vertices are $$(\pm b,0)$$
• the coordinates of the foci are $$(0,\pm c)$$
2. Solve for $$c$$ using the equation $$c^2=a^2−b^2$$.
3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.
###### Example $$\PageIndex{3}$$: Graphing an Ellipse Centered at the Origin
Graph the ellipse given by the equation, $$\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$$. Identify and label the center, vertices, co-vertices, and foci.
Solution
First, we determine the position of the major axis. Because $$25>9$$,the major axis is on the $$y$$-axis. Therefore, the equation is in the form $$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$$,where $$b^2=9$$ and $$a^2=25$$. It follows that:
• the center of the ellipse is $$(0,0)$$
• the coordinates of the vertices are $$(0,\pm a)=(0,\pm \sqrt{25})=(0,\pm 5)$$
• the coordinates of the co-vertices are $$(\pm b,0)=(\pm 9,0)=(\pm 3,0)$$
• the coordinates of the foci are $$(0,\pm c)$$, where $$c^2=a^2−b^2$$ Solving for $$c$$, we have:
\begin{align} c&=\pm \sqrt{a^2−b^2} \nonumber \\ &=\pm \sqrt{25−9} \nonumber\\ &=\pm \sqrt{16} \nonumber\\ &=\pm 4 \nonumber \end{align} \nonumber
Therefore, the coordinates of the foci are $$(0,\pm 4)$$.
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See Figure $$\PageIndex{8}$$.
Figure $$\PageIndex{8}$$
###### Exercise $$\PageIndex{3}$$
Graph the ellipse given by the equation $$\dfrac{x^2}{36}+\dfrac{y^2}{4}=1$$. Identify and label the center, vertices, co-vertices, and foci.
center: $$(0,0)$$; vertices: $$(\pm 6,0)$$; co-vertices: $$(0,\pm 2)$$; foci: $$(\pm 4\sqrt{2},0)$$
Figure $$\PageIndex{9}$$
###### Example $$\PageIndex{4}$$: Graphing an Ellipse Centered at the Origin from an Equation Not in Standard Form
Graph the ellipse given by the equation $$4x^2+25y^2=100$$. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.
Solution
First, use algebra to rewrite the equation in standard form.
\begin{align} 4x^2+25y^2&=100 \nonumber \\ \dfrac{4x^2}{100}+\dfrac{25y^2}{100}&=\dfrac{100}{100} \nonumber \\ \dfrac{x^2}{25}+\dfrac{y^2}{4}&=1 \nonumber \end{align} \nonumber
Next, we determine the position of the major axis. Because $$25>4$$, the major axis is on the $$x$$-axis. Therefore, the equation is in the form $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$, where $$a^2=25$$ and $$b^2=4$$. It follows that:
• the center of the ellipse is $$(0,0)$$
• the coordinates of the vertices are $$(\pm a,0)=(\pm \sqrt{25},0)=(\pm 5,0)$$
• the coordinates of the co-vertices are $$(0,\pm b)=(0,\pm \sqrt{4})=(0,\pm 2)$$
• the coordinates of the foci are $$(\pm c,0)$$, where $$c^2=a^2−b^2$$. Solving for $$c$$, we have:
\begin{align} c&=\pm \sqrt{a^2−b^2} \nonumber \\ &=\pm \sqrt{25−4} \nonumber \\ &=\pm \sqrt{21} \nonumber \end{align} \nonumber
Therefore the coordinates of the foci are $$(\pm \sqrt{21},0)$$.
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.
Figure $$\PageIndex{10}$$
###### Exercise $$\PageIndex{4}$$
Graph the ellipse given by the equation $$49x^2+16y^2=784$$. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.
Standard form: $$\dfrac{x^2}{16}+\dfrac{y^2}{49}=1$$; center: $$(0,0)$$; vertices: $$(0,\pm 7)$$; co-vertices: $$(\pm 4,0)$$; foci: $$(0,\pm \sqrt{33})$$
Figure $$\PageIndex{11}$$
## Graphing Ellipses Not Centered at the Origin
When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, $$(h,k)$$,we use the standard forms $$\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1$$, $$a>b$$ for horizontal ellipses and $$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1$$, $$a>b$$ for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.
###### How to: Given the standard form of an equation for an ellipse centered at $$(h, k)$$, sketch the graph.
1. Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci.
• If the equation is in the form $$\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1$$, where $$a>b$$, then
• the center is $$(h,k)$$
• the major axis is parallel to the $$x$$-axis
• the coordinates of the vertices are $$(h\pm a,k)$$
• the coordinates of the co-vertices are $$(h,k\pm b)$$
• the coordinates of the foci are $$(h\pm c,k)$$
• If the equation is in the form $$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1$$, where $$a>b$$, then
• the center is $$(h,k)$$
• the major axis is parallel to the $$y$$-axis
• the coordinates of the vertices are $$(h,k\pm a)$$
• the coordinates of the co-vertices are $$(h\pm b,k)$$
• the coordinates of the foci are $$(h,k\pm c)$$
2. Solve for $$c$$ using the equation $$c^2=a^2−b^2$$.
3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.
###### Example $$\PageIndex{5}$$: Graphing an Ellipse Centered at $$(h, k)$$
Graph the ellipse given by the equation, $$\dfrac{{(x+2)}^2}{4}+\dfrac{{(y−5)}^2}{9}=1$$. Identify and label the center, vertices, co-vertices, and foci.
Solution
First, we determine the position of the major axis. Because $$9>4$$, the major axis is parallel to the $$y$$-axis. Therefore, the equation is in the form $$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1$$, where $$b^2=4$$ and $$a^2=9$$. It follows that:
• the center of the ellipse is $$(h,k)=(−2,5)$$
• the coordinates of the vertices are $$(h,k\pm a)=(−2,5\pm \sqrt{9})=(−2,5\pm 3)$$, or $$(−2,2)$$ and $$(−2,8)$$
• the coordinates of the co-vertices are $$(h\pm b,k)=(−2\pm \sqrt{4},5)=(−2\pm 2,5)$$, or $$(−4,5)$$ and $$(0,5)$$
• the coordinates of the foci are $$(h,k\pm c)$$, where $$c^2=a^2−b^2$$. Solving for $$c$$,we have:
\begin{align} c&=\pm \sqrt{a^2−b^2} \nonumber \\[4pt] &=\pm \sqrt{9−4} \nonumber \\[4pt] &=\pm \sqrt{5} \nonumber \end{align} \nonumber
Therefore, the coordinates of the foci are $$(−2,5−\sqrt{5})$$ and $$(−2,5+\sqrt{5})$$.
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.
Figure $$\PageIndex{12}$$
###### Exercise $$\PageIndex{5}$$
Graph the ellipse given by the equation $$\dfrac{{(x−4)}^2}{36}+\dfrac{{(y−2)}^2}{20}=1$$. Identify and label the center, vertices, co-vertices, and foci.
Center: $$(4,2)$$; vertices: $$(−2,2)$$ and $$(10,2)$$; co-vertices: $$(4,2−2\sqrt{5})$$ and $$(4,2+2\sqrt{5})$$; foci: $$(0,2)$$ and $$(8,2)$$
Figure $$\PageIndex{13}$$
###### How to: Given the general form of an equation for an ellipse centered at $$(h, k)$$, express the equation in standard form.
1. Recognize that an ellipse described by an equation in the form $$ax^2+by^2+cx+dy+e=0$$ is in general form.
2. Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation.
3. Factor out the coefficients of the $$x^2$$ and $$y^2$$ terms in preparation for completing the square.
4. Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant, $$m_1{(x−h)}^2+m_2{(y−k)}^2=m_3$$, where $$m_1$$, $$m_2$$,and $$m_3$$ are constants.
5. Divide both sides of the equation by the constant term to express the equation in standard form.
###### Example $$\PageIndex{6}$$: Graphing an Ellipse Centered at $$(h, k)$$ by First Writing It in Standard Form
Graph the ellipse given by the equation $$4x^2+9y^2−40x+36y+100=0$$. Identify and label the center, vertices, co-vertices, and foci.
Solution
We must begin by rewriting the equation in standard form.
$$4x^2+9y^2−40x+36y+100=0$$
Group terms that contain the same variable, and move the constant to the opposite side of the equation.
$$(4x^2−40x)+(9y^2+36y)=−100$$
Factor out the coefficients of the squared terms.
$$4(x^2−10x)+9(y^2+4y)=−100$$
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
$$4(x^2−10x+25)+9(y^2+4y+4)=−100+100+36$$
Rewrite as perfect squares.
$$4{(x−5)}^2+9{(y+2)}^2=36$$
Divide both sides by the constant term to place the equation in standard form.
$$\dfrac{{(x−5)}^2}{9}+\dfrac{{(y+2)}^2}{4}=1$$
Now that the equation is in standard form, we can determine the position of the major axis. Because $$9>4$$, the major axis is parallel to the $$x$$-axis. Therefore, the equation is in the form $$\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1$$, where $$a^2=9$$ and $$b^2=4$$. It follows that:
• the center of the ellipse is $$(h,k)=(5,−2)$$
• the coordinates of the vertices are $$(h\pm a,k)=(5\pm \sqrt{9},−2)=(5\pm 3,−2)$$, or $$(2,−2)$$ and $$(8,−2)$$
• the coordinates of the co-vertices are $$(h,k\pm b)=(5,−2\pm \sqrt{4})=(5,−2\pm 2)$$, or $$(5,−4)$$ and $$(5,0)$$
• the coordinates of the foci are $$(h\pm c,k)$$, where $$c^2=a^2−b^2$$. Solving for $$c$$, we have:
\begin{align*} c&=\pm \sqrt{a^2-b^2}\\ &=\pm \sqrt{9-4}\\ &=\pm \sqrt{5} \end{align*}
Therefore, the coordinates of the foci are $$(5−\sqrt{5},−2)$$ and $$(5+\sqrt{5},−2)$$.
Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse as shown in Figure $$\PageIndex{14}$$.
Figure $$\PageIndex{14}$$
###### Exercise $$\PageIndex{6}$$
Express the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and foci of the ellipse.
$$4x^2+y^2−24x+2y+21=0$$
$$\dfrac{{(x−3)}^2}{4}+\dfrac{{(y+1)}^2}{16}=1$$; center: $$(3,−1)$$; vertices: $$(3,−5)$$ and $$(3,3)$$; co-vertices: $$(1,−1)$$ and $$(5,−1)$$; foci: $$(3,−1−2\sqrt{3})$$ and $$(3,−1+2\sqrt{3})$$
## Solving Applied Problems Involving Ellipses
Many real-world situations can be represented by ellipses, including orbits of planets, satellites, moons and comets, and shapes of boat keels, rudders, and some airplane wings. A medical device called a lithotripter uses elliptical reflectors to break up kidney stones by generating sound waves. Some buildings, called whispering chambers, are designed with elliptical domes so that a person whispering at one focus can easily be heard by someone standing at the other focus. This occurs because of the acoustic properties of an ellipse. When a sound wave originates at one focus of a whispering chamber, the sound wave will be reflected off the elliptical dome and back to the other focus (Figure $$\PageIndex{15}$$). In the whisper chamber at the Museum of Science and Industry in Chicago, two people standing at the foci—about $$43$$ feet apart—can hear each other whisper.
Figure $$\PageIndex{15}$$: Sound waves are reflected between foci in an elliptical room, called a whispering chamber.
###### Example $$\PageIndex{7}$$: Locating the Foci of a Whispering Chamber
The Statuary Hall in the Capitol Building in Washington, D.C. is a whispering chamber. Its dimensions are $$46$$ feet wide by $$96$$ feet long as shown in Figure $$\PageIndex{16}$$.
1. What is the standard form of the equation of the ellipse representing the outline of the room? Hint: assume a horizontal ellipse, and let the center of the room be the point $$(0,0)$$.
2. If two senators standing at the foci of this room can hear each other whisper, how far apart are the senators? Round to the nearest foot.
Figure $$\PageIndex{16}$$
Solution
1. We are assuming a horizontal ellipse with center $$(0,0)$$, so we need to find an equation of the form $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$, where $$a>b$$. We know that the length of the major axis, $$2a$$, is longer than the length of the minor axis, $$2b$$. So the length of the room, 96, is represented by the major axis, and the width of the room, 46, is represented by the minor axis.
Therefore, the equation of the ellipse is (dfrac{x^2}{2304}+\dfrac{y^2}{529}=1\) • Solving for $$a$$, we have $$2a=96$$, so $$a=48$$, and $$a^2=2304$$. • Solving for $$b$$, we have $$2b=46$$, so $$b=23$$, and $$b^2=529$$. 2. To find the distance between the senators, we must find the distance between the foci, $$(\pm c,0)$$, where $$c^2=a^2−b^2$$. Solving for $$c$$,we have: \[\begin{align*} c^2&=a^2-b^2\\ c^2&=2304-529\qquad \text{Substitute using the values found in part } (a)\\ c&=\pm \sqrt{2304-529}\qquad \text{Take the square root of both sides.}\\ c&=\pm \sqrt{1775}\qquad \text{Subtract.}\\ c&\approx \pm 42\qquad \text{Round to the nearest foot.} \end{align*}
The points $$(\pm 42,0)$$ represent the foci. Thus, the distance between the senators is $$2(42)=84$$ feet.
###### Exercise $$\PageIndex{7}$$
Suppose a whispering chamber is $$480$$ feet long and $$320$$ feet wide.
1. What is the standard form of the equation of the ellipse representing the room? Hint: assume a horizontal ellipse, and let the center of the room be the point $$(0,0)$$.
2. If two people are standing at the foci of this room and can hear each other whisper, how far apart are the people? Round to the nearest foot.
$$\dfrac{x^2}{57,600}+\dfrac{y^2}{25,600}=1$$
The people are standing $$358$$ feet apart.
###### Media
Access these online resources for additional instruction and practice with ellipses.
## Key Equations
Horizontal ellipse, center at origin $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$, $$a>b$$ Vertical ellipse, center at origin $$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$$, $$a>b$$ Horizontal ellipse, center $$(h,k)$$ $$\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1$$, $$a>b$$ Vertical ellipse, center $$(h,k)$$ $$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1$$, $$a>b$$
## Key Concepts
• An ellipse is the set of all points $$(x,y)$$ in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci).
• When given the coordinates of the foci and vertices of an ellipse, we can write the equation of the ellipse in standard form. See Example $$\PageIndex{1}$$ and Example $$\PageIndex{2}$$.
• When given an equation for an ellipse centered at the origin in standard form, we can identify its vertices, co-vertices, foci, and the lengths and positions of the major and minor axes in order to graph the ellipse. See Example $$\PageIndex{3}$$ and Example $$\PageIndex{4}$$.
• When given the equation for an ellipse centered at some point other than the origin, we can identify its key features and graph the ellipse. See Example $$\PageIndex{5}$$ and Example $$\PageIndex{6}$$.
• Real-world situations can be modeled using the standard equations of ellipses and then evaluated to find key features, such as lengths of axes and distance between foci. See Example $$\PageIndex{7}$$.
This page titled 7.2: The Ellipse is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. | 12,761 | 36,664 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 8, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-38 | latest | en | 0.193832 |
https://forum.allaboutcircuits.com/threads/mosfet-as-switch-instead-of-a-transistor.162320/ | 1,675,112,304,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499829.29/warc/CC-MAIN-20230130201044-20230130231044-00609.warc.gz | 273,769,012 | 21,161 | # MOSFET as switch instead of a transistor
#### ZwareValk
Joined Aug 14, 2019
3
We were given the circuit in the image below. With the following data:
Vin = 0V or 5V
Vcc = 12V
β = 50
The desired current through the LED is 50mA and the voltage over that LED is 2.7V in that case.
Then we were asked what would happen if the npn-transistor was replaced by an n-channel MOSFET and if a Vin of 5V would still be sufficient. This is the part where I'm having trouble: How can I tell if 5V would have the desired effect without a datasheet or anything given about the MOSFET?
What I know:
With Vds is the voltage over drain-source, Vgs is the voltage over gate-source and Vt is the voltage threshold of the MOSFET.
Of course Vgs is where the 5 Volts would be applied.
• Saturation occurs when Vds >= Vgs - Vt
• ID = K(Vgs - Vt
However, this doesn't seems insufficient to solve this problem.
Where could I start to solve this?
Joined Mar 10, 2018
4,057
You have to have the datasheet for the MOSFET to tell if it is
a logic level MOSFET and whats its Rdson is as well to do
the design.
Regards, Dana.
#### ZwareValk
Joined Aug 14, 2019
3
Thank you, that is what I thought as well.
Good to know I'm not going crazy.
#### crutschow
Joined Mar 14, 2008
31,112
A small nit, but a MOSFET is also a transistor.
#### ZwareValk
Joined Aug 14, 2019
3
A small nit, but a MOSFET is also a transistor.
Woops, that's right.
#### Audioguru
Joined Dec 20, 2007
11,248
You were told that the hFE of the original transistor is 50. But hFE is used when the transistor is a linear amplifier with plenty of C to E voltage. To turn on the LED then the transistor must saturate with a very low C to E voltage when hFE is not used. The datasheet for most little transistors shows that they saturate well when the base current is 1/10th the collector current.
Most Mosfets conduct well when the gate to source voltage is 10V. But some Mosfets are called "logic level" and conduct well when the G to S voltage is 4.5V or 5V. | 575 | 2,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-06 | latest | en | 0.966406 |
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