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# Antiderivatives
Ηλεκτρονική - Συσκευές
10 Οκτ 2013 (πριν από 4 χρόνια και 9 μήνες)
153 εμφανίσεις
1
Notes on
Antiderivatives
So far our studies have concentrated on given a function how do we find its
derivative? Now we ask the reverse question: Given a derivative how do we find the
function?
The following definition provides some clarity for this last
question.
Definition
: Let
f
denote a function on an interval
I
. A function
F
is said to be an
antiderivative of
f
on the interval
I
provided
for all values of
.
Since
for
, the function
is an antiderivative of the
function
on the interval
. There are other antiderivatives of
such
as
,
and
. In fact, since
, where
C
is a constant, all
the functions
, where
C
is any constant, are antiderivatives of
.
So the are
infinitely many antiderivatives for the function
. Are there any others?
To answer this question the following theorems provide the basis
for the theory.
In addition these theorems have important consequences.
Rolle
’s Theorem
:
Suppose
is a function which is continuous on the interval
,
differentiable on the interval
and for which
. Then there is a number
,
,
for which
.
The Mean Value Theorem (for Derivatives)
:
Suppose
is a function which is
continuous on the interval
and differentiable
on the interval
.
Then there is a
number
,
, for which
.
Observations
:
The following are consequences of these theorems.
Rolle’s Theorem is a special
case of the Mean Value Theorem.
Rolle’s Theorem guarantees the existence of at least one critical point on the
interval
.
The geometrical interpretation of the Mean Value Theorem is that there is a
tangent line to the curve
which is parallel to the (secant) line joining the
points
and
.
When the function
is interpreted as the position of a particle moving along a
line, then the M
ean Value Theorem says that there is a point in time that the
instantaneous velocity is the same as the average velocity over the time
interval
.
2
The following theorems provide the answer to the questions posed earlier. The proofs
are based on the Mean Value Theorem and are included
at the end of these notes
for
completeness.
Theorem 1
: Suppose
f
is a differentiable function on an interval
I
for which
for
all
. Then
f
is a consta
nt function, that is to say,
for some constant
C
.
Theorem 2
: Suppose
f
is a function defined on an interval
I
and suppose also that
F
and
G
are both antiderivatives of
f
on the interval
I
. Then there is a constant
C
for
which
for all values of
x
in the interval
I
.
the ques
tion originally posed, namely, d
oes the family
of
functions
,
C
a constant, represent all the antiderivatives of the function
on the
interval
? According to Theorem 2, the answer is “YES”!
Indeed the function
is
a known antiderivative of
. If
is any other antideriv
ative of
, Theorem 2
states
for some constant
.
Definition
: Suppose a function
on an interval
I
, the family of all antiderivatives of
on
the interval
I
is called the
indefinite integral of
and this family is denoted by the
symbol
.
Observation
: Notice that Theorem 2 states that if
is a function for
wh
ich
, then
, where C is a constant called the
constant of integration
. Conversely, if
and
are functions for
which
, where C is
a constant, then
.
Example 1
: Verify that the function
satisfies the hypotheses of
Rolle’s Theorem on the interval
. Then find all the numbers on that interval that
satisfies t
he conclusion of Rolle’s Theorem.
Solution
: The given function is a polynomial and therefore continuous everywhere and in
particular on the interval
. Since polynomials are differentiable everywhere, the
given function is differen
tiable on the interval
.
Furthermore
and
. So the given function satisfies the
hypotheses of Rolle’s Theorem.
Find the numbers
so
that
:
(
U
)
3
.
There are two possibilities, namely
and
. The second
possibility
is negative and therefore not in the interval
.
However the first
possibility
and therefore in the interval
.
Example 2
:
Verify that the function
satisfies the hypotheses of the Mean
Value Theorem on the interval
. Then find all the numbers
that satisfy the
conclusion of the Mean Value Theorem.
Solution
:
The given function is a combination of
functions that are continuous
everywhere and therefore continuous on the interval
. The given function is also
differentiable on the interval
since
everywhere. So the given funct
ion
satisfies the hypotheses of the Mean Value Theorem.
Find
so that
:
. So this number lies in
the interval
.
Example
3
:
Consider the function
. Show that there is no number
in the
interval
for which
. Why does this not
Value Theorem?
Solution
:
. Since this equation
which has no real solutions, there is no such number
. This does not contradict the Mean
Value Theorem because the given function has a discontinuity at
and is not
continuous on the interval
. (
It is not differentiable on
either
.
)
4
Example
4
:
Show that the equation
has exactly one real root.
Solution
:
The function
is a combination of continuous functions and
therefore continuous everywhere. Furthermore
and
. So
by the Intermediate Value Theorem there is at least one real root on the interval
.
Suppose there are two real roots,
.
(
) For all values
o
f
x
,
. So the function
is continuous on the
interval
and differentiable on the interval
. Further
and
.
So by Rolle’s Theorem there must exist a number
,
, for which
.
So
So there cannot be two real roots of
the equation.
Ex
ample
5
:
Evaluate each of the following indefinite integrals.
a)
c)
b)
d)
Solution
:
a) Since
,
the function
is an antiderivative of
.
Therefore
.
b)
Since
, the function
is an antiderivative of
.
Therefore
.
c)
Since
, the function
is an antiderivative of
.
Therefore
d)
Since
, the function
is an anti
derivative of
.
Therefore
.
The proofs of the t
heorems are now presented. It is not important for you to
memorize these proofs, but
as
you can
see why
the theorems hold true
.
Rolle’s Theor
em
: Suppose
is a function which is continuous on the interval
,
differentiable on the interval
and for which
. Then there is a number
,
, for which
.
Proof
:
There are three cases to consider.
5
Case 1: Suppose
is a constant function. Then
for all values of
x
in the
interval
. Select a number
. Then
.
Case 2: Suppose
for some value of
. Since
is continuous on the
interval
, the function
has a maximum value, call it
, on the interval
.
Then the number
c
is a critical point for the function
and, since
is differentiable on the
interval
,
.
Case 3: Suppose
for some value of
. Since
is continuous on the
interval
, the function
has a
minimum
value, call it
, on the interval
.
Then the number
c
is a critical point for the function
and, since
is differentiable on the
interval
,
.
The Mean Value Theorem (for Derivatives)
: Suppose
is a function which is
continuous on the interval
and differentiable on the interval
.
Then there is a
number
,
, for which
.
Proof
: Define the function
which is a
combination of the given function and a polynomial. Since
is continuous
on the
interval
, so is
. Since
is
differentiable on the interval
, so is
.
Furthermore
and
.
So by Rolle’s Theorem, there is a
number
,
, for which
. Since
, it follows
that
and
.
Theorem 1
: Suppose
f
is a differentiable function on an interval
I
for which
for
all
. Then
f
is a constant function, that is to say,
for some constant
C
.
Proof
: Let
be a fixed point in the interval
I
and let x denote any other point in the
interval
I
. For demonstration purposes, suppo
se
and note that the interval
is
contained in the interval
I
. Since f is differentiable on the interval
I
, it is continuous on
the interval
I
. Therefore
f
is continuous on the interval
and differentiable on the
interval
. So by the Mean Value Theorem for Derivatives there is a point
in
6
the interval
for which
. Since
,
and it
follows that
. So
. Since the point
x
is
arbitrary, f is a constant function.
Theorem 2
: Suppose
f
is a function defined on an interval
I
and suppose a
lso that
F
and
G
are both antiderivatives of
f
on the interval
I
. Then there is a constant
C
for
which
for all values of
x
in the interval
I
.
Proof
: Define
for
. Since
F
and
G
ar
e antiderivatives of
f
on the
interval
I
, the definition states that
and
for all
.
Then
for all
. By Theorem 1,
for
all
for some constant
C
. Therefore
or
.
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# Antiderivatives. Ηλεκτρονική - Συσκευές. 10 Οκτ 2013 (πριν από 4 χρόνια και 9 μήνες). 153 εμφανίσεις. 1. Notes on. Antiderivatives. So far our studies have concentrated on given a function how do we find its. derivative? Now we ask the reverse question: Given a derivative how do we find the. function?. The following definition provides some clarity for this last. question.. Definition. : Let. f. denote a function on an interval. I. . A function. F. is said to be an. antiderivative of. f. on the interval. I. provided. for all values of. .. Since. for. , the function. is an antiderivative of the. function. on the interval. . There are other antiderivatives of. such. as. ,. and. . In fact, since. , where. C. is a constant, all. the functions. , where. C. is any constant, are antiderivatives of. .. So the are. infinitely many antiderivatives for the function. . Are there any others?. To answer this question the following theorems provide the basis. for the theory.. In addition these theorems have important consequences.. Rolle. ’s Theorem. :. Suppose. is a function which is continuous on the interval. ,. differentiable on the interval. and for which. . Then there is a number. ,. ,. for which. .. The Mean Value Theorem (for Derivatives). :. Suppose. is a function which is. continuous on the interval. and differentiable. on the interval. .. Then there is a. number. ,. , for which. .. Observations. :. The following are consequences of these theorems.. Rolle’s Theorem is a special. case of the Mean Value Theorem.. Rolle’s Theorem guarantees the existence of at least one critical point on the. interval. .. The geometrical interpretation of the Mean Value Theorem is that there is a. tangent line to the curve. which is parallel to the (secant) line joining the. points. and. .. When the function. is interpreted as the position of a particle moving along a. line, then the M. ean Value Theorem says that there is a point in time that the. instantaneous velocity is the same as the average velocity over the time. interval. .. 2. The following theorems provide the answer to the questions posed earlier. The proofs. are based on the Mean Value Theorem and are included. at the end of these notes. for. completeness.. Theorem 1. : Suppose. f. is a differentiable function on an interval. I. for which. for. all. . Then. f. is a consta. nt function, that is to say,. for some constant. C. .. Theorem 2. : Suppose. f. is a function defined on an interval. I. and suppose also that. F. and. G. are both antiderivatives of. f. on the interval. I. . Then there is a constant. C. for. which. for all values of. x. in the interval. I. .. the ques. tion originally posed, namely, d. oes the family. of. functions. ,. C. a constant, represent all the antiderivatives of the function. on the. interval. ? According to Theorem 2, the answer is “YES”!. Indeed the function. is. a known antiderivative of. . If. is any other antideriv. ative of. , Theorem 2. states. for some constant. .. Definition. : Suppose a function. on an interval. I. , the family of all antiderivatives of. on. the interval. I. is called the. indefinite integral of. and this family is denoted by the. symbol. .. Observation. : Notice that Theorem 2 states that if. is a function for. wh. ich. , then. , where C is a constant called the. constant of integration. . Conversely, if. and. are functions for. which. , where C is. a constant, then. .. Example 1. : Verify that the function. satisfies the hypotheses of. Rolle’s Theorem on the interval. . Then find all the numbers on that interval that. satisfies t. he conclusion of Rolle’s Theorem.. Solution. : The given function is a polynomial and therefore continuous everywhere and in. particular on the interval. . Since polynomials are differentiable everywhere, the. given function is differen. tiable on the interval. .. Furthermore. and. . So the given function satisfies the. hypotheses of Rolle’s Theorem.. Find the numbers. so. that. :. (. U. ). 3. .. There are two possibilities, namely. and. . The second. possibility. is negative and therefore not in the interval. .. However the first. possibility. and therefore in the interval. .. Example 2. :. Verify that the function. satisfies the hypotheses of the Mean. Value Theorem on the interval. . Then find all the numbers. that satisfy the. conclusion of the Mean Value Theorem.. Solution. :. The given function is a combination of. functions that are continuous. everywhere and therefore continuous on the interval. . The given function is also. differentiable on the interval. since. everywhere. So the given funct. ion. satisfies the hypotheses of the Mean Value Theorem.. Find. so that. :. . So this number lies in. the interval. .. Example. 3. :. Consider the function. . Show that there is no number. in the. interval. for which. . Why does this not. Value Theorem?. Solution. :. . Since this equation. which has no real solutions, there is no such number. . This does not contradict the Mean. Value Theorem because the given function has a discontinuity at. and is not. continuous on the interval. . (. It is not differentiable on. either. .. ). 4. Example. 4. :.
|
Show that the equation. has exactly one real root.. Solution. :. The function. is a combination of continuous functions and. therefore continuous everywhere. Furthermore. and. . So. by the Intermediate Value Theorem there is at least one real root on the interval. .. Suppose there are two real roots,. .. (. ) For all values. o. f. x. ,. . So the function. is continuous on the. interval. and differentiable on the interval. . Further. and. .. So by Rolle’s Theorem there must exist a number. ,. , for which. .. So. So there cannot be two real roots of. the equation.. Ex. ample. 5. :. Evaluate each of the following indefinite integrals.. a). c). b). d). Solution. :. a) Since. ,. the function. is an antiderivative of. .. Therefore. .. b). Since. , the function. is an antiderivative of. .. Therefore. .. c). Since. , the function. is an antiderivative of. .. Therefore. d). Since. , the function. is an anti. derivative of. .. Therefore. .. The proofs of the t. heorems are now presented. It is not important for you to. memorize these proofs, but. as. you can. see why. the theorems hold true. .. Rolle’s Theor. em. : Suppose. is a function which is continuous on the interval. ,. differentiable on the interval. and for which. . Then there is a number. ,. , for which. .. Proof. :. There are three cases to consider.. 5. Case 1: Suppose. is a constant function. Then. for all values of. x. in the. interval. . Select a number. . Then. .. Case 2: Suppose. for some value of. . Since. is continuous on the. interval. , the function. has a maximum value, call it. , on the interval. .. Then the number. c. is a critical point for the function. and, since. is differentiable on the. interval. ,. .. Case 3: Suppose. for some value of. . Since. is continuous on the. interval. , the function. has a. minimum. value, call it. , on the interval. .. Then the number. c. is a critical point for the function. and, since. is differentiable on the. interval. ,. .. The Mean Value Theorem (for Derivatives). : Suppose. is a function which is. continuous on the interval. and differentiable on the interval. .. Then there is a. number. ,. , for which. .. Proof. : Define the function. which is a. combination of the given function and a polynomial. Since. is continuous. on the. interval. , so is. . Since. is. differentiable on the interval. , so is. .. Furthermore. and. .. So by Rolle’s Theorem, there is a. number. ,. , for which. . Since. , it follows. that. and. .. Theorem 1. : Suppose. f. is a differentiable function on an interval. I. for which. for. all. . Then. f. is a constant function, that is to say,. for some constant. C. .. Proof. : Let. be a fixed point in the interval. I. and let x denote any other point in the. interval. I. . For demonstration purposes, suppo. se. and note that the interval. is. contained in the interval. I. . Since f is differentiable on the interval. I. , it is continuous on. the interval. I. . Therefore. f. is continuous on the interval. and differentiable on the. interval. . So by the Mean Value Theorem for Derivatives there is a point. in. 6. the interval. for which. . Since. ,. and it. follows that. . So. . Since the point. x. is. arbitrary, f is a constant function.. Theorem 2. : Suppose. f. is a function defined on an interval. I. and suppose a. lso that. F. and. G. are both antiderivatives of. f. on the interval. I. . Then there is a constant. C. for. which. for all values of. x. in the interval. I. .. Proof. : Define. for. . Since. F. and. G. ar. e antiderivatives of. f. on the. interval. I. , the definition states that. and. for all. .. Then. for all. . By Theorem 1,. for. all. for some constant. C. . Therefore. or. .
|
https://corinnekerston.com/2023/04/29/how-many-days-until-june-13.html
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crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00387.warc.gz
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Intro text, can be displayed through an additional field
Countdown to June 13: How Many Days Until June 13?
June 13 is just around the corner, and you might be wondering how many days are left until this exciting date. Whether you're eagerly anticipating a special event, marking an important milestone, or simply curious about the passage of time, this article will provide you with all the information you need. Read on to discover the answer to the burning question: how many days until June 13?
Understanding Time and Dates
Before we dive into the countdown, let's take a moment to understand how time and dates work. Time is a fascinating concept that allows us to measure the duration between two events or moments. Dates, on the other hand, help us organize and track specific points in time.
Calculating the Countdown
To determine how many days are left until June 13, we must consider the current date. As today's date changes daily, the countdown will also fluctuate accordingly. Therefore, it's crucial to check the countdown regularly for the most accurate information.
The Countdown Begins
As of today, let's calculate how many days are left until June 13. Please note that the countdown may differ depending on when you're reading this article.
Step 1: Determine Today's Date
Before we can calculate the countdown, we need to establish today's date. Let's assume today is May 1.
Step 2: Calculate the Days Remaining
Next, we subtract today's date from June 13 to find the number of days until the desired date. In our case, the calculation would be as follows: June 13 - May 1 = 43 days.
Step 3: Include or Exclude Today
When counting down to a specific date, there's often confusion about whether to include or exclude the current day. In our case, since we want to know how many days are left until June 13, we should exclude today from the count. Therefore, we have 43 days left until June 13.
Q: How Many Weeks Are There Until June 13?
A: To determine the number of weeks until June 13, divide the remaining days by seven. In our case, 43 days divided by seven equals approximately 6.14 weeks. Therefore, there are around 6 weeks until June 13.
Q: What Day of the Week Is June 13?
A: To find out the day of the week for June 13, we can use an online calendar or refer to a physical one. In this scenario, let's assume June 13 falls on a Monday.
Q: Are There Any Significant Events on June 13?
A: While June 13 is not widely recognized as a globally significant date, it may hold personal importance for individuals celebrating birthdays, anniversaries, or other personal milestones. Additionally, historical events may have occurred on this day that could be of interest to history enthusiasts.
Conclusion
As we eagerly await June 13, the countdown continues to decrease. With 43 days remaining until this exciting date, there's still plenty of time to prepare, plan, and anticipate the events that await us. Remember, the countdown may vary depending on when you read this article, so be sure to check back regularly for the most accurate information. Enjoy the anticipation and make the most of each passing day as we get closer to June 13!
Related video of how many days until june 13
Ctrl
Enter
Noticed oshYwhat?
Highlight text and click Ctrl+Enter
We are in
Search and Discover » how many days until june 13
Update Info
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CC-MAIN-2023-50
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Intro text, can be displayed through an additional field. Countdown to June 13: How Many Days Until June 13?. June 13 is just around the corner, and you might be wondering how many days are left until this exciting date. Whether you're eagerly anticipating a special event, marking an important milestone, or simply curious about the passage of time, this article will provide you with all the information you need. Read on to discover the answer to the burning question: how many days until June 13?. Understanding Time and Dates. Before we dive into the countdown, let's take a moment to understand how time and dates work. Time is a fascinating concept that allows us to measure the duration between two events or moments. Dates, on the other hand, help us organize and track specific points in time.. Calculating the Countdown. To determine how many days are left until June 13, we must consider the current date. As today's date changes daily, the countdown will also fluctuate accordingly. Therefore, it's crucial to check the countdown regularly for the most accurate information.. The Countdown Begins. As of today, let's calculate how many days are left until June 13. Please note that the countdown may differ depending on when you're reading this article.. Step 1: Determine Today's Date. Before we can calculate the countdown, we need to establish today's date. Let's assume today is May 1.. Step 2: Calculate the Days Remaining. Next, we subtract today's date from June 13 to find the number of days until the desired date. In our case, the calculation would be as follows: June 13 - May 1 = 43 days.. Step 3: Include or Exclude Today. When counting down to a specific date, there's often confusion about whether to include or exclude the current day. In our case, since we want to know how many days are left until June 13, we should exclude today from the count.
|
Therefore, we have 43 days left until June 13.. Q: How Many Weeks Are There Until June 13?. A: To determine the number of weeks until June 13, divide the remaining days by seven. In our case, 43 days divided by seven equals approximately 6.14 weeks. Therefore, there are around 6 weeks until June 13.. Q: What Day of the Week Is June 13?. A: To find out the day of the week for June 13, we can use an online calendar or refer to a physical one. In this scenario, let's assume June 13 falls on a Monday.. Q: Are There Any Significant Events on June 13?. A: While June 13 is not widely recognized as a globally significant date, it may hold personal importance for individuals celebrating birthdays, anniversaries, or other personal milestones. Additionally, historical events may have occurred on this day that could be of interest to history enthusiasts.. Conclusion. As we eagerly await June 13, the countdown continues to decrease. With 43 days remaining until this exciting date, there's still plenty of time to prepare, plan, and anticipate the events that await us. Remember, the countdown may vary depending on when you read this article, so be sure to check back regularly for the most accurate information. Enjoy the anticipation and make the most of each passing day as we get closer to June 13!. Related video of how many days until june 13. Ctrl. Enter. Noticed oshYwhat?. Highlight text and click Ctrl+Enter. We are in. Search and Discover » how many days until june 13. Update Info.
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| 12,823
|
# GRE Math Quantitative Comparison Practice Test 10
1. ab = 0
a 0
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
2. -1 < x < 0
x³ x5
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
3. The average of four numbers is 30. The average of three of these numbers is 20.
The value of the fourth number 60
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
4. AC > CB > AB
b a
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
5. The line containing point A is rotated 90 degrees anticlockwise about the origin, O.
The y coordinate of point A before rotation. The y coordinate of point A after rotation.
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
6. Points x and y lie on line l
The number of points on line l that are twice as far from x as from y 2
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
7. O is the centre of the circle, and angle POQ is a right angle
MN/PQ 2/1
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
8.
Total savings on 40 gallons of fuel bought at \$1.152 per gallon instead of \$1.245 \$3.70
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
9. The cost of hiring props for the school play was to be shared by 8 students. 4 more students decide to share the cost. The new cost per person is \$2 less.
Total cost of hiring props \$50
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
10. A bag contains only red, white and blue balls. One third of the balls are red, one fifth of the balls are white. One ball is to be selected at random.
Probability of drawing a white ball Probability of drawing a blue ball
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
### Test information
10 questions
10 minutes
This is just one of 10 free GRE math quantitative comparison tests available on majortests.com. See the quantitative comparison page for directions, tips and more information.
* GRE is a registered trademark of Educational Testing Service (ETS). This website is not endorsed or approved by ETS.
All content of site and tests copyright © 2018 Study Mode, LLC.
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# GRE Math Quantitative Comparison Practice Test 10. 1. ab = 0. a 0. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 2. -1 < x < 0. x³ x5. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 3. The average of four numbers is 30. The average of three of these numbers is 20.. The value of the fourth number 60. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 4. AC > CB > AB. b a. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 5. The line containing point A is rotated 90 degrees anticlockwise about the origin, O.. The y coordinate of point A before rotation. The y coordinate of point A after rotation.. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 6. Points x and y lie on line l. The number of points on line l that are twice as far from x as from y 2. A.
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The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 7. O is the centre of the circle, and angle POQ is a right angle. MN/PQ 2/1. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 8.. Total savings on 40 gallons of fuel bought at \$1.152 per gallon instead of \$1.245 \$3.70. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 9. The cost of hiring props for the school play was to be shared by 8 students. 4 more students decide to share the cost. The new cost per person is \$2 less.. Total cost of hiring props \$50. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 10. A bag contains only red, white and blue balls. One third of the balls are red, one fifth of the balls are white. One ball is to be selected at random.. Probability of drawing a white ball Probability of drawing a blue ball. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. ### Test information. 10 questions. 10 minutes. This is just one of 10 free GRE math quantitative comparison tests available on majortests.com. See the quantitative comparison page for directions, tips and more information.. * GRE is a registered trademark of Educational Testing Service (ETS). This website is not endorsed or approved by ETS.. All content of site and tests copyright © 2018 Study Mode, LLC.
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https://convertoctopus.com/39-2-knots-to-miles-per-hour
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## Conversion formula
The conversion factor from knots to miles per hour is 1.1507794480225, which means that 1 knot is equal to 1.1507794480225 miles per hour:
1 kt = 1.1507794480225 mph
To convert 39.2 knots into miles per hour we have to multiply 39.2 by the conversion factor in order to get the velocity amount from knots to miles per hour. We can also form a simple proportion to calculate the result:
1 kt → 1.1507794480225 mph
39.2 kt → V(mph)
Solve the above proportion to obtain the velocity V in miles per hour:
V(mph) = 39.2 kt × 1.1507794480225 mph
V(mph) = 45.110554362484 mph
The final result is:
39.2 kt → 45.110554362484 mph
We conclude that 39.2 knots is equivalent to 45.110554362484 miles per hour:
39.2 knots = 45.110554362484 miles per hour
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 mile per hour is equal to 0.022167761272995 × 39.2 knots.
Another way is saying that 39.2 knots is equal to 1 ÷ 0.022167761272995 miles per hour.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that thirty-nine point two knots is approximately forty-five point one one one miles per hour:
39.2 kt ≅ 45.111 mph
An alternative is also that one mile per hour is approximately zero point zero two two times thirty-nine point two knots.
## Conversion table
### knots to miles per hour chart
For quick reference purposes, below is the conversion table you can use to convert from knots to miles per hour
knots (kt) miles per hour (mph)
40.2 knots 46.261 miles per hour
41.2 knots 47.412 miles per hour
42.2 knots 48.563 miles per hour
43.2 knots 49.714 miles per hour
44.2 knots 50.864 miles per hour
45.2 knots 52.015 miles per hour
46.2 knots 53.166 miles per hour
47.2 knots 54.317 miles per hour
48.2 knots 55.468 miles per hour
49.2 knots 56.618 miles per hour
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## Conversion formula. The conversion factor from knots to miles per hour is 1.1507794480225, which means that 1 knot is equal to 1.1507794480225 miles per hour:. 1 kt = 1.1507794480225 mph. To convert 39.2 knots into miles per hour we have to multiply 39.2 by the conversion factor in order to get the velocity amount from knots to miles per hour. We can also form a simple proportion to calculate the result:. 1 kt → 1.1507794480225 mph. 39.2 kt → V(mph). Solve the above proportion to obtain the velocity V in miles per hour:. V(mph) = 39.2 kt × 1.1507794480225 mph. V(mph) = 45.110554362484 mph. The final result is:. 39.2 kt → 45.110554362484 mph. We conclude that 39.2 knots is equivalent to 45.110554362484 miles per hour:. 39.2 knots = 45.110554362484 miles per hour. ## Alternative conversion. We can also convert by utilizing the inverse value of the conversion factor. In this case 1 mile per hour is equal to 0.022167761272995 × 39.2 knots.. Another way is saying that 39.2 knots is equal to 1 ÷ 0.022167761272995 miles per hour.. ## Approximate result.
|
For practical purposes we can round our final result to an approximate numerical value. We can say that thirty-nine point two knots is approximately forty-five point one one one miles per hour:. 39.2 kt ≅ 45.111 mph. An alternative is also that one mile per hour is approximately zero point zero two two times thirty-nine point two knots.. ## Conversion table. ### knots to miles per hour chart. For quick reference purposes, below is the conversion table you can use to convert from knots to miles per hour. knots (kt) miles per hour (mph). 40.2 knots 46.261 miles per hour. 41.2 knots 47.412 miles per hour. 42.2 knots 48.563 miles per hour. 43.2 knots 49.714 miles per hour. 44.2 knots 50.864 miles per hour. 45.2 knots 52.015 miles per hour. 46.2 knots 53.166 miles per hour. 47.2 knots 54.317 miles per hour. 48.2 knots 55.468 miles per hour. 49.2 knots 56.618 miles per hour.
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https://www.scribd.com/document/73099915/New-Finite-Element-Analysis-Lec1
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# Finite Element Analysis (FEA) Finite Element Method (FEM
)
Compressor valve cover
Examples Coal processing plant pipes
How do you solve an Engineering Problem? • Engineers are faced with complex problems and use many methods in order to solve those problems including classic and numerical methods. • As shown below, the finite element methods is one of several methods for solving engineering problems.
ENGINEERING ANALYSIS
CLASSICAL METHODS 1. CLOSED-FORM
NUMERICAL METHOD 1. FINITE ELEMENT 2. FINITE DIFFERENCE 3. BOUNDARY ELEMENT
2. APPROXIMATE
Method of solving Engineering Problems
Historically, two classical approaches have been used: Classical Method • Closed form solutions are available for simple problems such as bending of beams and torsion of prismatic bars. (eg: Roark’s Formulas for Stress and Stress) • Approximate methods using series solutions to governing differential equations are used to analyze more complex structures such as plate and shells.
“If you can solve your problem using a closed form solution and a classical method, it is probably the best way to do it”
the another approach is to take a complex problem and break it into simple problems. • A good example is found in the long history of the efforts to calculate Pi (π) . • Those simple problem are solved and then assembled into a final solution.How do you solve a complex Engineering Problem? • When a problem becomes more complex than what can be solved with a closed form solution.
Example of FE approaches .• People spent centuries breaking the circle up into smaller triangles to find an accurate value of Pi (π).
Example of FE approaches .
Example of FE approaches .
loading and boundary conditions – Increasingly become the primary analysis tools for designers and analyst – Also known as matrix method of structural analysis in the literature because it uses matrix algebra to solve the system of simultaneous equations.Method of solving Engineering Problems Numerical Methods • FEM – Capable of solving large. complex problems with general geometry. – Real power is in its ability to solve problems that do not fit any standard formula. .
(Zeid. 2011) . 2004) • numerical analysis technique for obtaining approximate solutions to a wide variety of engineering problems.FEA Definition • computational techniques used to obtain approximate solution of boundary value problems in Engineering (Hutton.com. 2005) • one of several numerical methods that can be used to solve a complex problem by breaking it down into a finite number of simple problems (NeiSoftware.
Before FEA…? .
Fundamental Concept • Based on the idea of building a complicated object into small and manageable pieces. simpler pieces. . • The assembly of nodes and elements are called finite element model. • The smaller pieces are called elements which are connected by nodes. • It is a method of investigating the behavior of complex and structures by breaking them down into smaller.
.
. The three basic types of finite elements are beams (1D).Types of Element • Finite elements have shapes which are relatively easy to formulate and analyze. shells (2D) and solids (3D).
Other examples of 1D elements .
Other examples of 2D elements .
Other examples of 3D elements .
The following steps show in general how the FEM works. • 1. Create the Finite Element • 2. step-by-step process.How does FEM Work? The solution of a problem domain by FEM usually follows an orderly. Generate Global Stiffness Matrix equation • 4. Apply Boundary Conditions • 5. Develop elemental matrices and equation • 3. Solve for the unknown at the nodes .
How does FEM Work? (an example of 2D element) 1. • Element are connected to each other by nodes 4 nodes 1 element . Create the Finite Element • A given problem is discretized by dividing the original domain into simply shaped element.
1D elements v 2D elements 3D elements .
2. Develop elemental matrices and equation • The relationship between an element and its surrounding nodes can be described by the following equation 1 2 • *Each of the element will have their own set of equations. .
{u}.What is [K]. {F} ? Problem Type .
Generate Global Stiffness Matrix equation • The individual Elemental Stiffness Matrices are assembled together into the Global Stiffness Matrix by summing the equilibrium equations of the elements. • This result in the following Global System Matrix Equation for the overall structure: .3.
.
4. Apply Boundary Conditions • Next. Mathematically this is achieved by removing rows and columns corresponding to the constrained degrees of freedom (DOF) from the Global System Matrix Equation ux = 0 uy = 0 uz = 0 F1 = ? F2 = ? F3 = ? . the BC’s (Loads and Constraint) is apply to the model.
the Global System Matrix Equation is solve to determine the unknown nodal values. Solve for the unknown at the nodes • Finally.5. [K]{u} ={F} {u} = [K]-1{F} • *This requires knowledge in Matrix Algebra. .
fluid flow) .Advantages of FEA • General enough to handle large class of engineering problems (stress analysis. electromagnetism. heat transfer.
Force = 100lbs) .Mistakes by user • Elements are the wrong type (eg: Shell 2D elements are used where solid elements are needed. • Distorted elements • Inconsistent units (eg: E = 200GPa.
Best Practices .
. The more refine the mesh (grid).” The FE solutions are often approximate.. • “Any results obtained is the correct one. loads and boundary conditions..Common misconception • “The FE solutions is the most accurate. the more accurate. There is no guarantee that the results are accurate.” The FE solution may contain “fatal” errors as a result of incorrect modeling of structures. Even a nice picture can give the wrong result… • “FEA replaces testing…” Depends on the confidence in the analytical methods used.
Typical FE Procedure by commercial software user Preprocessing Build FE Model computer Process Conduct numerical analysis user Postprocessing See and interpret the results .
• Import / Create Geometry • Define Element Types (1D. v – Poisson’s Ratio etc) Pre • Define geometric properties Processing • Apply Boundary Conditions (Constraint) • Apply Loads • Solve for displacements • Compute Strains • Compute Stresses FEA General Procedure Solution • • • PostProcessing • • Sort element stresses in order of magnitude Check equilibrium Plot deformed structural shape Animate dynamic model behavior Reports . 3D) • Define Material Properties (E – Young’s Modulus. 2D.
• The computed values then used by back substitution to compute additional derived variables (i.Pre Processor • Also know as model definition • This step is critical – a perfectly computed FE solution is of absolutely no value if corresponds to wrong problems. garbage out” Solution • FE software assembles the algebraic equations in matrix form and computes the unknown values. • “garbage in. the most important objectives is to apply sound engineering judgment in determining whether the solution are reasonable or not . • While solution data can be manipulated many ways.e reaction forces. element stresses etc) Postprocessor • Analysis and evaluation of the solution results.
REVIEW OF MATRIX ALGEBRA .
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# Finite Element Analysis (FEA) Finite Element Method (FEM. ). Compressor valve cover. Examples Coal processing plant pipes. How do you solve an Engineering Problem? • Engineers are faced with complex problems and use many methods in order to solve those problems including classic and numerical methods. • As shown below, the finite element methods is one of several methods for solving engineering problems.. ENGINEERING ANALYSIS. CLASSICAL METHODS 1. CLOSED-FORM. NUMERICAL METHOD 1. FINITE ELEMENT 2. FINITE DIFFERENCE 3. BOUNDARY ELEMENT. 2. APPROXIMATE. Method of solving Engineering Problems. Historically, two classical approaches have been used: Classical Method • Closed form solutions are available for simple problems such as bending of beams and torsion of prismatic bars. (eg: Roark’s Formulas for Stress and Stress) • Approximate methods using series solutions to governing differential equations are used to analyze more complex structures such as plate and shells.. “If you can solve your problem using a closed form solution and a classical method, it is probably the best way to do it”. the another approach is to take a complex problem and break it into simple problems. • A good example is found in the long history of the efforts to calculate Pi (π) . • Those simple problem are solved and then assembled into a final solution.How do you solve a complex Engineering Problem? • When a problem becomes more complex than what can be solved with a closed form solution.. Example of FE approaches .• People spent centuries breaking the circle up into smaller triangles to find an accurate value of Pi (π).. Example of FE approaches .. Example of FE approaches .. loading and boundary conditions – Increasingly become the primary analysis tools for designers and analyst – Also known as matrix method of structural analysis in the literature because it uses matrix algebra to solve the system of simultaneous equations.Method of solving Engineering Problems Numerical Methods • FEM – Capable of solving large. complex problems with general geometry. – Real power is in its ability to solve problems that do not fit any standard formula. .. (Zeid. 2011) . 2004) • numerical analysis technique for obtaining approximate solutions to a wide variety of engineering problems.FEA Definition • computational techniques used to obtain approximate solution of boundary value problems in Engineering (Hutton.com. 2005) • one of several numerical methods that can be used to solve a complex problem by breaking it down into a finite number of simple problems (NeiSoftware.. Before FEA…? .. Fundamental Concept • Based on the idea of building a complicated object into small and manageable pieces. simpler pieces. . • The assembly of nodes and elements are called finite element model. • The smaller pieces are called elements which are connected by nodes. • It is a method of investigating the behavior of complex and structures by breaking them down into smaller.. .. . The three basic types of finite elements are beams (1D).Types of Element • Finite elements have shapes which are relatively easy to formulate and analyze. shells (2D) and solids (3D).. Other examples of 1D elements .. Other examples of 2D elements .. Other examples of 3D elements .. The following steps show in general how the FEM works. • 1.
|
Create the Finite Element • 2. step-by-step process.How does FEM Work? The solution of a problem domain by FEM usually follows an orderly. Generate Global Stiffness Matrix equation • 4. Apply Boundary Conditions • 5. Develop elemental matrices and equation • 3. Solve for the unknown at the nodes .. How does FEM Work? (an example of 2D element) 1. • Element are connected to each other by nodes 4 nodes 1 element . Create the Finite Element • A given problem is discretized by dividing the original domain into simply shaped element.. 1D elements v 2D elements 3D elements .. 2. Develop elemental matrices and equation • The relationship between an element and its surrounding nodes can be described by the following equation 1 2 • *Each of the element will have their own set of equations. .. {u}.What is [K]. {F} ? Problem Type .. Generate Global Stiffness Matrix equation • The individual Elemental Stiffness Matrices are assembled together into the Global Stiffness Matrix by summing the equilibrium equations of the elements. • This result in the following Global System Matrix Equation for the overall structure: .3.. .. 4. Apply Boundary Conditions • Next. Mathematically this is achieved by removing rows and columns corresponding to the constrained degrees of freedom (DOF) from the Global System Matrix Equation ux = 0 uy = 0 uz = 0 F1 = ? F2 = ? F3 = ? . the BC’s (Loads and Constraint) is apply to the model.. the Global System Matrix Equation is solve to determine the unknown nodal values. Solve for the unknown at the nodes • Finally.5. [K]{u} ={F} {u} = [K]-1{F} • *This requires knowledge in Matrix Algebra. .. fluid flow) .Advantages of FEA • General enough to handle large class of engineering problems (stress analysis. electromagnetism. heat transfer.. Force = 100lbs) .Mistakes by user • Elements are the wrong type (eg: Shell 2D elements are used where solid elements are needed. • Distorted elements • Inconsistent units (eg: E = 200GPa.. Best Practices .. . The more refine the mesh (grid).” The FE solutions are often approximate.. • “Any results obtained is the correct one. loads and boundary conditions..Common misconception • “The FE solutions is the most accurate. the more accurate. There is no guarantee that the results are accurate.” The FE solution may contain “fatal” errors as a result of incorrect modeling of structures. Even a nice picture can give the wrong result… • “FEA replaces testing…” Depends on the confidence in the analytical methods used.. Typical FE Procedure by commercial software user Preprocessing Build FE Model computer Process Conduct numerical analysis user Postprocessing See and interpret the results .. • Import / Create Geometry • Define Element Types (1D. v – Poisson’s Ratio etc) Pre • Define geometric properties Processing • Apply Boundary Conditions (Constraint) • Apply Loads • Solve for displacements • Compute Strains • Compute Stresses FEA General Procedure Solution • • • PostProcessing • • Sort element stresses in order of magnitude Check equilibrium Plot deformed structural shape Animate dynamic model behavior Reports . 3D) • Define Material Properties (E – Young’s Modulus. 2D.. • The computed values then used by back substitution to compute additional derived variables (i.Pre Processor • Also know as model definition • This step is critical – a perfectly computed FE solution is of absolutely no value if corresponds to wrong problems. garbage out” Solution • FE software assembles the algebraic equations in matrix form and computes the unknown values. • “garbage in. the most important objectives is to apply sound engineering judgment in determining whether the solution are reasonable or not . • While solution data can be manipulated many ways.e reaction forces. element stresses etc) Postprocessor • Analysis and evaluation of the solution results.. REVIEW OF MATRIX ALGEBRA .
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http://www.ck12.org/arithmetic/Percent-of-Change/lesson/Find-the-Percent-of-Change/
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# Percent of Change
## %change=[(final amount-original amount)/original amount] x 100%
%
Progress
MEMORY METER
This indicates how strong in your memory this concept is
Progress
%
Find the Percent of Change
Have you ever gone to a gym to exercise? Many people do each day, but sometimes the rates change. Take a look at this dilemma.
A gym’s membership went from 2100 members one year to 2410 members the next. This is a difference of 310 members. What was the percent of change?
To figure this out, you will need to know how to calculate a percent of change. Pay close attention and you will learn how to do this by the end of the Concept.
### Guidance
We can find the percent of change if we know an original amount and how much it either increased or decreased. At times, however, we are given the percent of increase or decrease and need to calculate a new amount.
Let’s look at how we can calculate this new amount by using the percent of change.
A restaurant manager has noticed an increase in the cost of utilities of 4%. In order to pay for the increased costs, he decides to increase prices by 4% as well. Not all items are priced the same. The chicken platter currently costs $5.99 and the steak platter cost$7.99. If the prices are increased by 4%, what will the new prices be?
Notice that we are going to create two new amounts. We are going to create a new cost for the chicken platter, and we are going to create a new cost for the steak platter. We have original amounts and the percent of the increase, so now we need to calculate a new amount.
First we must calculate how much the change will be. The prices are increasing by 4% so we must know how much 4% is of each price.
Now we know how much each platter’s price is going to change? Since this is an increase, we will add the price change to the price. If it were a decrease, we would subtract the decrease from the price.
Let’s summarize. In order to find the new amount, we calculate the change amount by multiplying the original amount by the percent of change. We then add the change amount to the original amount for an increase or we subtract the change amount from the original amount for a decrease.
Take a look at this situation.
Find the new amount if 60 is decreased by 27%.
amount of change:
subtract the amount of change from the original amount:
Find each percent of change.
#### Example A
Find the new amount if 10 is increased by 18%.
Solution:
#### Example B
Find the new amount if 16 is decreased by 20%.
Solution:
#### Example C
Find the new amount if 250 is increase by 30%.
Solution:
Now let's go back to the dilemma from the beginning of the Concept.
This percent is increasing, so we want to find the percent of the increase. We know the difference so now we can divide and multiply.
The gyms membership increased by 14.8%.
### Guided Practice
Here is one for you to try on your own.
The number of students participating in the chess club increased in one year. It started off with 35 students and had an increase of 55%. Figure out the new number of students in the chess club given this increase.
Solution
First, figure out the amount of the increase.
There was an increase of 19 students.
We can add that increase to the original amount.
The new number of students participating is students.
### Explore More
Directions: Use percent to find the new amount. You may round to the nearest whole number when necessary.
1. 82 increased by 90%
2. 64 decreased by 10%
3. 9 increased by 55%
4. 25,470 decreased by 77%
5. 75 increased by 10%
6. 33 decreased by 5%
7. 99 increased by 15%
8. 40 decreased by 8%
9. 56 increased by 25%
10. 900 decreased by 30%
11. 800 increased by 23%
12. 789 increased by 12%
13. 880 decreased by 10%
14. 450 increased by 45%
15. 855 decreased by 18%
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Percent Percent means out of 100. It is a quantity written with a % sign.
Percent Equation The percent equation can be stated as: "Rate times Total equals Part," or "R% of Total is Part."
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# Percent of Change. ## %change=[(final amount-original amount)/original amount] x 100%. %. Progress. MEMORY METER. This indicates how strong in your memory this concept is. Progress. %. Find the Percent of Change. Have you ever gone to a gym to exercise? Many people do each day, but sometimes the rates change. Take a look at this dilemma.. A gym’s membership went from 2100 members one year to 2410 members the next. This is a difference of 310 members. What was the percent of change?. To figure this out, you will need to know how to calculate a percent of change. Pay close attention and you will learn how to do this by the end of the Concept.. ### Guidance. We can find the percent of change if we know an original amount and how much it either increased or decreased. At times, however, we are given the percent of increase or decrease and need to calculate a new amount.. Let’s look at how we can calculate this new amount by using the percent of change.. A restaurant manager has noticed an increase in the cost of utilities of 4%. In order to pay for the increased costs, he decides to increase prices by 4% as well. Not all items are priced the same. The chicken platter currently costs $5.99 and the steak platter cost$7.99. If the prices are increased by 4%, what will the new prices be?. Notice that we are going to create two new amounts. We are going to create a new cost for the chicken platter, and we are going to create a new cost for the steak platter. We have original amounts and the percent of the increase, so now we need to calculate a new amount.. First we must calculate how much the change will be. The prices are increasing by 4% so we must know how much 4% is of each price.. Now we know how much each platter’s price is going to change? Since this is an increase, we will add the price change to the price. If it were a decrease, we would subtract the decrease from the price.. Let’s summarize. In order to find the new amount, we calculate the change amount by multiplying the original amount by the percent of change. We then add the change amount to the original amount for an increase or we subtract the change amount from the original amount for a decrease.. Take a look at this situation.. Find the new amount if 60 is decreased by 27%.. amount of change:. subtract the amount of change from the original amount:. Find each percent of change.. #### Example A. Find the new amount if 10 is increased by 18%.. Solution:. #### Example B. Find the new amount if 16 is decreased by 20%.. Solution:. #### Example C. Find the new amount if 250 is increase by 30%.. Solution:. Now let's go back to the dilemma from the beginning of the Concept.. This percent is increasing, so we want to find the percent of the increase. We know the difference so now we can divide and multiply.
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The gyms membership increased by 14.8%.. ### Guided Practice. Here is one for you to try on your own.. The number of students participating in the chess club increased in one year. It started off with 35 students and had an increase of 55%. Figure out the new number of students in the chess club given this increase.. Solution. First, figure out the amount of the increase.. There was an increase of 19 students.. We can add that increase to the original amount.. The new number of students participating is students.. ### Explore More. Directions: Use percent to find the new amount. You may round to the nearest whole number when necessary.. 1. 82 increased by 90%. 2. 64 decreased by 10%. 3. 9 increased by 55%. 4. 25,470 decreased by 77%. 5. 75 increased by 10%. 6. 33 decreased by 5%. 7. 99 increased by 15%. 8. 40 decreased by 8%. 9. 56 increased by 25%. 10. 900 decreased by 30%. 11. 800 increased by 23%. 12. 789 increased by 12%. 13. 880 decreased by 10%. 14. 450 increased by 45%. 15. 855 decreased by 18%. ### Notes/Highlights Having trouble? Report an issue.. Color Highlighted Text Notes. ### Vocabulary Language: English. TermDefinition. Percent Percent means out of 100. It is a quantity written with a % sign.. Percent Equation The percent equation can be stated as: "Rate times Total equals Part," or "R% of Total is Part.".
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https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/1_Algebraic_functions/1.2_Trinomial_products/1.2.1_Quadratic/1.2.1.2-d+e_x-%5Em-a+b_x+c_x%5E2-%5Ep/rese183.htm
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### 3.183 $$\int \frac{x^2}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx$$
Optimal. Leaf size=106 $-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}$
[Out]
-((a*x*(a + b*x))/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (x^2*(a + b*x))/(2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +
(a^2*(a + b*x)*Log[a + b*x])/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
________________________________________________________________________________________
Rubi [A] time = 0.0371791, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {646, 43} $-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}$
Antiderivative was successfully verified.
[In]
Int[x^2/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
[Out]
-((a*x*(a + b*x))/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (x^2*(a + b*x))/(2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +
(a^2*(a + b*x)*Log[a + b*x])/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
Rule 646
Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p] && NeQ[2*c*d - b*e, 0]
Rule 43
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rubi steps
\begin{align*} \int \frac{x^2}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{x^2}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \left (-\frac{a}{b^3}+\frac{x}{b^2}+\frac{a^2}{b^3 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}
Mathematica [A] time = 0.0149632, size = 45, normalized size = 0.42 $\frac{(a+b x) \left (2 a^2 \log (a+b x)+b x (b x-2 a)\right )}{2 b^3 \sqrt{(a+b x)^2}}$
Antiderivative was successfully verified.
[In]
Integrate[x^2/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
[Out]
((a + b*x)*(b*x*(-2*a + b*x) + 2*a^2*Log[a + b*x]))/(2*b^3*Sqrt[(a + b*x)^2])
________________________________________________________________________________________
Maple [A] time = 0.222, size = 44, normalized size = 0.4 \begin{align*}{\frac{ \left ( bx+a \right ) \left ({b}^{2}{x}^{2}+2\,{a}^{2}\ln \left ( bx+a \right ) -2\,abx \right ) }{2\,{b}^{3}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2/((b*x+a)^2)^(1/2),x)
[Out]
1/2*(b*x+a)*(b^2*x^2+2*a^2*ln(b*x+a)-2*a*b*x)/((b*x+a)^2)^(1/2)/b^3
________________________________________________________________________________________
Maxima [A] time = 1.25525, size = 55, normalized size = 0.52 \begin{align*} \frac{a^{2} b^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{a b x}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{x^{2}}{2 \, \sqrt{b^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/((b*x+a)^2)^(1/2),x, algorithm="maxima")
[Out]
a^2*b^2*log(x + a/b)/(b^2)^(5/2) - a*b*x/(b^2)^(3/2) + 1/2*x^2/sqrt(b^2)
________________________________________________________________________________________
Fricas [A] time = 1.65108, size = 68, normalized size = 0.64 \begin{align*} \frac{b^{2} x^{2} - 2 \, a b x + 2 \, a^{2} \log \left (b x + a\right )}{2 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/((b*x+a)^2)^(1/2),x, algorithm="fricas")
[Out]
1/2*(b^2*x^2 - 2*a*b*x + 2*a^2*log(b*x + a))/b^3
________________________________________________________________________________________
Sympy [A] time = 1.08682, size = 26, normalized size = 0.25 \begin{align*} \frac{a^{2} \log{\left (a + b x \right )}}{b^{3}} - \frac{a x}{b^{2}} + \frac{x^{2}}{2 b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2/((b*x+a)**2)**(1/2),x)
[Out]
a**2*log(a + b*x)/b**3 - a*x/b**2 + x**2/(2*b)
________________________________________________________________________________________
Giac [A] time = 1.26573, size = 65, normalized size = 0.61 \begin{align*} \frac{a^{2} \log \left ({\left | b x + a \right |}\right ) \mathrm{sgn}\left (b x + a\right )}{b^{3}} + \frac{b x^{2} \mathrm{sgn}\left (b x + a\right ) - 2 \, a x \mathrm{sgn}\left (b x + a\right )}{2 \, b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/((b*x+a)^2)^(1/2),x, algorithm="giac")
[Out]
a^2*log(abs(b*x + a))*sgn(b*x + a)/b^3 + 1/2*(b*x^2*sgn(b*x + a) - 2*a*x*sgn(b*x + a))/b^2
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### 3.183 $$\int \frac{x^2}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx$$. Optimal. Leaf size=106 $-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}$. [Out]. -((a*x*(a + b*x))/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (x^2*(a + b*x))/(2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +. (a^2*(a + b*x)*Log[a + b*x])/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]). ________________________________________________________________________________________. Rubi [A] time = 0.0371791, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {646, 43} $-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}$. Antiderivative was successfully verified.. [In]. Int[x^2/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]. [Out]. -((a*x*(a + b*x))/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (x^2*(a + b*x))/(2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +. (a^2*(a + b*x)*Log[a + b*x])/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]). Rule 646. Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra. cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,. c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p] && NeQ[2*c*d - b*e, 0]. Rule 43. Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d. *x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]. && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0]). Rubi steps. \begin{align*} \int \frac{x^2}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{x^2}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \left (-\frac{a}{b^3}+\frac{x}{b^2}+\frac{a^2}{b^3 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}. Mathematica [A] time = 0.0149632, size = 45, normalized size = 0.42 $\frac{(a+b x) \left (2 a^2 \log (a+b x)+b x (b x-2 a)\right )}{2 b^3 \sqrt{(a+b x)^2}}$. Antiderivative was successfully verified.. [In]. Integrate[x^2/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]. [Out]. ((a + b*x)*(b*x*(-2*a + b*x) + 2*a^2*Log[a + b*x]))/(2*b^3*Sqrt[(a + b*x)^2]). ________________________________________________________________________________________. Maple [A] time = 0.222, size = 44, normalized size = 0.4 \begin{align*}{\frac{ \left ( bx+a \right ) \left ({b}^{2}{x}^{2}+2\,{a}^{2}\ln \left ( bx+a \right ) -2\,abx \right ) }{2\,{b}^{3}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}. Verification of antiderivative is not currently implemented for this CAS.
|
[In]. int(x^2/((b*x+a)^2)^(1/2),x). [Out]. 1/2*(b*x+a)*(b^2*x^2+2*a^2*ln(b*x+a)-2*a*b*x)/((b*x+a)^2)^(1/2)/b^3. ________________________________________________________________________________________. Maxima [A] time = 1.25525, size = 55, normalized size = 0.52 \begin{align*} \frac{a^{2} b^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{a b x}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{x^{2}}{2 \, \sqrt{b^{2}}} \end{align*}. Verification of antiderivative is not currently implemented for this CAS.. [In]. integrate(x^2/((b*x+a)^2)^(1/2),x, algorithm="maxima"). [Out]. a^2*b^2*log(x + a/b)/(b^2)^(5/2) - a*b*x/(b^2)^(3/2) + 1/2*x^2/sqrt(b^2). ________________________________________________________________________________________. Fricas [A] time = 1.65108, size = 68, normalized size = 0.64 \begin{align*} \frac{b^{2} x^{2} - 2 \, a b x + 2 \, a^{2} \log \left (b x + a\right )}{2 \, b^{3}} \end{align*}. Verification of antiderivative is not currently implemented for this CAS.. [In]. integrate(x^2/((b*x+a)^2)^(1/2),x, algorithm="fricas"). [Out]. 1/2*(b^2*x^2 - 2*a*b*x + 2*a^2*log(b*x + a))/b^3. ________________________________________________________________________________________. Sympy [A] time = 1.08682, size = 26, normalized size = 0.25 \begin{align*} \frac{a^{2} \log{\left (a + b x \right )}}{b^{3}} - \frac{a x}{b^{2}} + \frac{x^{2}}{2 b} \end{align*}. Verification of antiderivative is not currently implemented for this CAS.. [In]. integrate(x**2/((b*x+a)**2)**(1/2),x). [Out]. a**2*log(a + b*x)/b**3 - a*x/b**2 + x**2/(2*b). ________________________________________________________________________________________. Giac [A] time = 1.26573, size = 65, normalized size = 0.61 \begin{align*} \frac{a^{2} \log \left ({\left | b x + a \right |}\right ) \mathrm{sgn}\left (b x + a\right )}{b^{3}} + \frac{b x^{2} \mathrm{sgn}\left (b x + a\right ) - 2 \, a x \mathrm{sgn}\left (b x + a\right )}{2 \, b^{2}} \end{align*}. Verification of antiderivative is not currently implemented for this CAS.. [In]. integrate(x^2/((b*x+a)^2)^(1/2),x, algorithm="giac"). [Out]. a^2*log(abs(b*x + a))*sgn(b*x + a)/b^3 + 1/2*(b*x^2*sgn(b*x + a) - 2*a*x*sgn(b*x + a))/b^2.
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## FACTORIZATION- ESSENTIAL POINTS
• Factorization is expressing any algebraic equation as product of its factors.
• These factors can be numbers, variables or algebraic expressions.
• An irreducible factor is a factor which cannot be expressed further as a product of factors.
• Common factor method of factorization:
1. Write each term of the expression as a product of irreducible factors.
2. Separate the common factor terms.
3. Combine the remaining factors in each term in accordance with the distributive law.
• Terms must be grouped in a way that each group of terms have common factors. This is the method of regrouping.
• We need to observe expression and identify the desired grouping by trail and error.
• Common identities for factorization:
1. (a + b)2 = a2 + 2ab + b2
2. (a – b)2 = a2 – 2ab + b2
3. (a + b) (a – b) = a2 - b2
4. (y + a) (y + b) = y2 + (a + b)y + ab
• In expressions which have factors of the type (y + a) (y + b), remember the numerical term gives ab. Factors, a and b, should be so chosen that their sum, with signs taken care of, is the coefficient of y.
• Division of numbers is inverse of its multiplication.
• In the case of division of a polynomial by a monomial, we may carry out the division either by dividing each term of the polynomial by the monomial or by the common factor method.
• In the case of division of a polynomial by a polynomial, we cannot proceed by dividing each term in the dividend polynomial by the divisor polynomial. Instead, we factorise both the polynomials and cancel their common factors.
• In the case of divisions of algebraic expressions that we studied in this chapter, we have:
Dividend = Divisor × Quotient.
In general, however, the relation is:
Dividend = Divisor × Quotient + Remainder
Thus, we have considered in the present chapter only those divisions in which the remainder is zero.
I am in
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## FACTORIZATION- ESSENTIAL POINTS. • Factorization is expressing any algebraic equation as product of its factors.. • These factors can be numbers, variables or algebraic expressions.. • An irreducible factor is a factor which cannot be expressed further as a product of factors.. • Common factor method of factorization:. 1. Write each term of the expression as a product of irreducible factors.. 2. Separate the common factor terms.. 3. Combine the remaining factors in each term in accordance with the distributive law.. • Terms must be grouped in a way that each group of terms have common factors. This is the method of regrouping.. • We need to observe expression and identify the desired grouping by trail and error.. • Common identities for factorization:. 1. (a + b)2 = a2 + 2ab + b2. 2.
|
(a – b)2 = a2 – 2ab + b2. 3. (a + b) (a – b) = a2 - b2. 4. (y + a) (y + b) = y2 + (a + b)y + ab. • In expressions which have factors of the type (y + a) (y + b), remember the numerical term gives ab. Factors, a and b, should be so chosen that their sum, with signs taken care of, is the coefficient of y.. • Division of numbers is inverse of its multiplication.. • In the case of division of a polynomial by a monomial, we may carry out the division either by dividing each term of the polynomial by the monomial or by the common factor method.. • In the case of division of a polynomial by a polynomial, we cannot proceed by dividing each term in the dividend polynomial by the divisor polynomial. Instead, we factorise both the polynomials and cancel their common factors.. • In the case of divisions of algebraic expressions that we studied in this chapter, we have:. Dividend = Divisor × Quotient.. In general, however, the relation is:. Dividend = Divisor × Quotient + Remainder. Thus, we have considered in the present chapter only those divisions in which the remainder is zero.. I am in.
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# The most basic quaternion question...
Hi,
I’m having difficulty implementing quaternions in Qt using their QQuaternion class, and haven’t heard anything back yet from the community boards where I’ve posted my query. This is my first time trying to use quaternions so I want to see if the issue is just that I don’t understand how they’re supposed to work. I figured you guys would know.
All I want to do is to rotate a point in 3-space a given angular distance about an axis. So let
vector_pre = (1,0,0);
I.e. the unit vector along the x-axis. Now let’s say I want to rotate it 90deg counterclockwise in the x-y plane. To do this, I create the following quaternion:
quat = (90,0,0,1);
The vector component is the unit vector along the z axis, normal to the x-y plane. Then let
vector_post = (quat * vector_pre) * conjugate(quat);
vector_pre gets reinterpreted as a quaternion with scalar part 0, multiplication and conjugacy are as described in Wikipedia. Then my expectation is that
vector_post == (0,1,0);
This is emphatically not what I’m getting with the corresponding Qt method (rotatedVector) so I want to see if I am just not understanding things at a basic level. (And it’s not a radians vs. degrees issue.) Have I misunderstood the basics? Thank you–
Matt
I’m definitely no expert on Quaternions and i don’t know how Qt implements them, but here’s what i think you are doing wrong:
Quaternions have 4 components. These 4 components do NOT correspond to “angle/axis” (e.g. angle/x/y/z). The maths behind Quaternions is “a bit” more complicated.
So when you create a quaternion by setting the 4 components manually, you need to pass quite different values to actually get a Quaternion that represents a rotation around an axis. On a related note, most maths libraries don’t work with angles in degrees, but usually in radians, so passing “90” is most likely wrong, as well.
Most Quaternion libs have a function “CreateFromAxisAngle” or something like that. There you can use your intuitive representation of axis/angle and get the (most unintuitive) Quaternion back.
Also most libraries overload the multiplication between quaternion and vector (operator* (Quat, Vec)) and already implement the whole “rotate a vector” stuff in there. So usually you should not need to multiply your vector by the Quaternions conjugate, at all.
Now this is all without knowledge about Qt’s Quaternion class and therefore could be entirely wrong. But those are the most likely errors that i can come up with.
Hope that helps,
Jan.
Thank you so much, omitting the fromAxisAngle invocation was exactly the problem.
Best,
Matt
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# The most basic quaternion question.... Hi,. I’m having difficulty implementing quaternions in Qt using their QQuaternion class, and haven’t heard anything back yet from the community boards where I’ve posted my query. This is my first time trying to use quaternions so I want to see if the issue is just that I don’t understand how they’re supposed to work. I figured you guys would know.. All I want to do is to rotate a point in 3-space a given angular distance about an axis. So let. vector_pre = (1,0,0);. I.e. the unit vector along the x-axis. Now let’s say I want to rotate it 90deg counterclockwise in the x-y plane. To do this, I create the following quaternion:. quat = (90,0,0,1);. The vector component is the unit vector along the z axis, normal to the x-y plane. Then let. vector_post = (quat * vector_pre) * conjugate(quat);. vector_pre gets reinterpreted as a quaternion with scalar part 0, multiplication and conjugacy are as described in Wikipedia. Then my expectation is that. vector_post == (0,1,0);. This is emphatically not what I’m getting with the corresponding Qt method (rotatedVector) so I want to see if I am just not understanding things at a basic level. (And it’s not a radians vs.
|
degrees issue.) Have I misunderstood the basics? Thank you–. Matt. I’m definitely no expert on Quaternions and i don’t know how Qt implements them, but here’s what i think you are doing wrong:. Quaternions have 4 components. These 4 components do NOT correspond to “angle/axis” (e.g. angle/x/y/z). The maths behind Quaternions is “a bit” more complicated.. So when you create a quaternion by setting the 4 components manually, you need to pass quite different values to actually get a Quaternion that represents a rotation around an axis. On a related note, most maths libraries don’t work with angles in degrees, but usually in radians, so passing “90” is most likely wrong, as well.. Most Quaternion libs have a function “CreateFromAxisAngle” or something like that. There you can use your intuitive representation of axis/angle and get the (most unintuitive) Quaternion back.. Also most libraries overload the multiplication between quaternion and vector (operator* (Quat, Vec)) and already implement the whole “rotate a vector” stuff in there. So usually you should not need to multiply your vector by the Quaternions conjugate, at all.. Now this is all without knowledge about Qt’s Quaternion class and therefore could be entirely wrong. But those are the most likely errors that i can come up with.. Hope that helps,. Jan.. Thank you so much, omitting the fromAxisAngle invocation was exactly the problem.. Best,. Matt.
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https://www.physicsforums.com/threads/electric-field-from-permanent-magnet.364753/
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# Electric Field from Permanent Magnet
• insomniac392
In summary, the conversation is about finding the electric field at certain points given a slab of magnetized matter and a spinning magnetized sphere. The equations involved are related to the magnetic scalar potential and the magnetic charge density. The solution involves using a duality transformation to solve for the electric field.
insomniac392
Hello,
I've been extremely stuck on the following problems and was hoping someone could give me a push in the right direction:
## Homework Statement
1) Given an infinite slab of permanently magnetized matter of thickness d centered on the xy-plane with uniform magnetization $$\textbf{M} = (0, M, 0)$$ and velocity $$\textbf{v} = (v, 0, 0)$$, find the electric field at $$\textbf{E}(0, 0, 0)$$ and $$\textbf{E}(0, y, 0)$$ where y > d.
2) A magnetized sphere with uniform magnetization $$\textbf{M} = (0, 0, M)$$ and radius r is spinning at a rate of $$\textbf{\omega} = (0, 0, \omega)$$. Find the electric field for r' > r. (Hint: Find the equivalent magnetic charge density, $$\rho_m$$, and equivalent surface current, $$\sigma_m$$.)
## Homework Equations
I'm not entirely sure (hence the thread)!
$$\sigma_{m, n} = \textbf{M} \cdot \textbf{n}$$
$$\rho_{m} = - \nabla \cdot \textbf{M}$$
...these are factors of the integrand that give rise to the magnetic scalar potential, $$\Omega$$, which in turn yields $$\textbf{B}$$ via $$\textbf{H} = - \nabla \Omega$$.
## The Attempt at a Solution
I'm desperately stuck on these; for both problems I can find $$\rho_m$$ and $$\sigma_m$$, but I don't see the connection to the $$\textbf{E}$$-field. Any suggestions to get me started would be greatly appreciated.
Last edited:
insomniac392 said:
Hello,
I've been extremely stuck on the following problems and was hoping someone could give me a push in the right direction:
## Homework Statement
1) Given an infinite slab of permanently magnetized matter of thickness d centered on the xy-plane with uniform magnetization $$\textbf{M} = (0, M, 0)$$ and velocity $$\textbf{v} = (v, 0, 0)$$, find the electric field at $$\textbf{E}(0, 0, 0)$$ and $$\textbf{E}(0, y, 0)$$ where y > d.
2) A magnetized sphere with uniform magnetization $$\textbf{M} = (0, 0, M)$$ and radius r is spinning at a rate of $$\textbf{\omega} = (0, 0, \omega)$$. Find the electric field for r' > r. (Hint: Find the equivalent magnetic charge density, $$\rho_m$$, and equivalent surface current, $$\sigma_m$$.)
## Homework Equations
I'm not entirely sure (hence the thread)!
$$\sigma_{m, n} = \textbf{M} \cdot \textbf{n}$$
$$\rho_{m} = - \nabla \cdot \textbf{M}$$
...these are factors of the integrand that give rise to the magnetic scalar potential, $$\Omega$$, which in turn yields $$\textbf{B}$$ via $$\textbf{H} = - \nabla \Omega$$.
## The Attempt at a Solution
I'm desperately stuck on these; for both problems I can find $$\rho_m$$ and $$\sigma_m$$, but I don't see the connection to the $$\textbf{E}$$-field. Any suggestions to get me started would be greatly appreciated.
Well, after doing some digging I found the following problem (7.60) in Griffiths' Electrodynamics:
Maxwell's equations are invariant under the following duality transformations
$$\textbf{E'} = \textbf{E} cos(\alpha) + c \textbf{B} sin(\alpha)$$
$$c \textbf{B'} = c \textbf{B} cos(\alpha) - \textbf{E} sin(\alpha)$$
$$c q_{e}' = c q_{e} cos(\alpha) + q_{m} sin(\alpha)$$
$$q_{m}' = q_{m} cos(\alpha) - c q_{e} sin(\alpha)$$
...where $$c = 1/\sqrt{\epsilon_0 \mu_0}$$, $$q_m$$ is the magnetic charge and $$\alpha$$ is an arbitrary rotation angle in "$$\textbf{E}-\textbf{B}$$ space."
Griffiths' says that, "this means, in particular, that if you know the fields produced by a configuration of electric charge, you can immediately (using $$\alpha = \pi / 2$$) write down the fields produced by the corresponding arrangement of magnetic charge."
Thus, if I were to solve for $$\textbf{E}$$ in (1) and (2) with a polarization $$\textbf{P}$$ instead of a magnetization $$\textbf{M}$$, I could then use a duality transformation to find the solutions to (1) and (2), correct?
Hello,
Thank you for reaching out for assistance. As a scientist, my suggestion would be to start by using the given equations and integrating them to find the magnetic scalar potential, \Omega. From there, you can use the relationship \textbf{E} = - \nabla \Omega to find the electric field at the specified points. It may also be helpful to consider the boundary conditions and use the continuity equation for the electric displacement field, \nabla \cdot \textbf{D} = \rho_f, where \rho_f is the free charge density. I hope this helps guide you in the right direction. Good luck with your calculations!
## 1. What is an electric field from a permanent magnet?
The electric field from a permanent magnet is the force field created by the movement of electrons within the magnet. This field can interact with other objects and particles, causing them to move or experience a force.
## 2. How is the electric field from a permanent magnet different from an electric field from a charged particle?
An electric field from a permanent magnet is different from an electric field from a charged particle because it is created by the movement of electrons within the magnet, rather than a stationary charge. Permanent magnets have a north and south pole, and the electric field from a permanent magnet is always perpendicular to the direction of the magnetic field.
## 3. What factors affect the strength of the electric field from a permanent magnet?
The strength of the electric field from a permanent magnet is affected by the strength of the magnet, the distance from the magnet, and the material the magnet is interacting with. Additionally, the orientation of the magnet, as well as the presence of other magnetic fields, can also affect the strength of the electric field.
## 4. How can the electric field from a permanent magnet be measured?
The electric field from a permanent magnet can be measured using a device called a Hall probe, which measures the magnetic field and calculates the electric field based on the strength and orientation of the magnet. Other methods include using a compass or a magnetometer.
## 5. What are some applications of the electric field from a permanent magnet?
The electric field from a permanent magnet has many practical applications, including in electric motors, generators, and speakers. It is also used in magnetic levitation technology and particle accelerators. Additionally, the electric field from a permanent magnet can be used in medical imaging, such as MRI machines, to create detailed images of the body.
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# Electric Field from Permanent Magnet. • insomniac392. In summary, the conversation is about finding the electric field at certain points given a slab of magnetized matter and a spinning magnetized sphere. The equations involved are related to the magnetic scalar potential and the magnetic charge density. The solution involves using a duality transformation to solve for the electric field.. insomniac392. Hello,. I've been extremely stuck on the following problems and was hoping someone could give me a push in the right direction:. ## Homework Statement. 1) Given an infinite slab of permanently magnetized matter of thickness d centered on the xy-plane with uniform magnetization $$\textbf{M} = (0, M, 0)$$ and velocity $$\textbf{v} = (v, 0, 0)$$, find the electric field at $$\textbf{E}(0, 0, 0)$$ and $$\textbf{E}(0, y, 0)$$ where y > d.. 2) A magnetized sphere with uniform magnetization $$\textbf{M} = (0, 0, M)$$ and radius r is spinning at a rate of $$\textbf{\omega} = (0, 0, \omega)$$. Find the electric field for r' > r. (Hint: Find the equivalent magnetic charge density, $$\rho_m$$, and equivalent surface current, $$\sigma_m$$.). ## Homework Equations. I'm not entirely sure (hence the thread)!. $$\sigma_{m, n} = \textbf{M} \cdot \textbf{n}$$. $$\rho_{m} = - \nabla \cdot \textbf{M}$$. ...these are factors of the integrand that give rise to the magnetic scalar potential, $$\Omega$$, which in turn yields $$\textbf{B}$$ via $$\textbf{H} = - \nabla \Omega$$.. ## The Attempt at a Solution. I'm desperately stuck on these; for both problems I can find $$\rho_m$$ and $$\sigma_m$$, but I don't see the connection to the $$\textbf{E}$$-field. Any suggestions to get me started would be greatly appreciated.. Last edited:. insomniac392 said:. Hello,. I've been extremely stuck on the following problems and was hoping someone could give me a push in the right direction:. ## Homework Statement. 1) Given an infinite slab of permanently magnetized matter of thickness d centered on the xy-plane with uniform magnetization $$\textbf{M} = (0, M, 0)$$ and velocity $$\textbf{v} = (v, 0, 0)$$, find the electric field at $$\textbf{E}(0, 0, 0)$$ and $$\textbf{E}(0, y, 0)$$ where y > d.. 2) A magnetized sphere with uniform magnetization $$\textbf{M} = (0, 0, M)$$ and radius r is spinning at a rate of $$\textbf{\omega} = (0, 0, \omega)$$. Find the electric field for r' > r. (Hint: Find the equivalent magnetic charge density, $$\rho_m$$, and equivalent surface current, $$\sigma_m$$.). ## Homework Equations. I'm not entirely sure (hence the thread)!. $$\sigma_{m, n} = \textbf{M} \cdot \textbf{n}$$. $$\rho_{m} = - \nabla \cdot \textbf{M}$$. ...these are factors of the integrand that give rise to the magnetic scalar potential, $$\Omega$$, which in turn yields $$\textbf{B}$$ via $$\textbf{H} = - \nabla \Omega$$.. ## The Attempt at a Solution. I'm desperately stuck on these; for both problems I can find $$\rho_m$$ and $$\sigma_m$$, but I don't see the connection to the $$\textbf{E}$$-field. Any suggestions to get me started would be greatly appreciated.. Well, after doing some digging I found the following problem (7.60) in Griffiths' Electrodynamics:. Maxwell's equations are invariant under the following duality transformations. $$\textbf{E'} = \textbf{E} cos(\alpha) + c \textbf{B} sin(\alpha)$$. $$c \textbf{B'} = c \textbf{B} cos(\alpha) - \textbf{E} sin(\alpha)$$. $$c q_{e}' = c q_{e} cos(\alpha) + q_{m} sin(\alpha)$$. $$q_{m}' = q_{m} cos(\alpha) - c q_{e} sin(\alpha)$$. ...where $$c = 1/\sqrt{\epsilon_0 \mu_0}$$, $$q_m$$ is the magnetic charge and $$\alpha$$ is an arbitrary rotation angle in "$$\textbf{E}-\textbf{B}$$ space.". Griffiths' says that, "this means, in particular, that if you know the fields produced by a configuration of electric charge, you can immediately (using $$\alpha = \pi / 2$$) write down the fields produced by the corresponding arrangement of magnetic charge.". Thus, if I were to solve for $$\textbf{E}$$ in (1) and (2) with a polarization $$\textbf{P}$$ instead of a magnetization $$\textbf{M}$$, I could then use a duality transformation to find the solutions to (1) and (2), correct?. Hello,. Thank you for reaching out for assistance. As a scientist, my suggestion would be to start by using the given equations and integrating them to find the magnetic scalar potential, \Omega. From there, you can use the relationship \textbf{E} = - \nabla \Omega to find the electric field at the specified points. It may also be helpful to consider the boundary conditions and use the continuity equation for the electric displacement field, \nabla \cdot \textbf{D} = \rho_f, where \rho_f is the free charge density. I hope this helps guide you in the right direction. Good luck with your calculations!. ## 1. What is an electric field from a permanent magnet?. The electric field from a permanent magnet is the force field created by the movement of electrons within the magnet. This field can interact with other objects and particles, causing them to move or experience a force.. ## 2. How is the electric field from a permanent magnet different from an electric field from a charged particle?. An electric field from a permanent magnet is different from an electric field from a charged particle because it is created by the movement of electrons within the magnet, rather than a stationary charge. Permanent magnets have a north and south pole, and the electric field from a permanent magnet is always perpendicular to the direction of the magnetic field.. ## 3. What factors affect the strength of the electric field from a permanent magnet?.
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The strength of the electric field from a permanent magnet is affected by the strength of the magnet, the distance from the magnet, and the material the magnet is interacting with. Additionally, the orientation of the magnet, as well as the presence of other magnetic fields, can also affect the strength of the electric field.. ## 4. How can the electric field from a permanent magnet be measured?. The electric field from a permanent magnet can be measured using a device called a Hall probe, which measures the magnetic field and calculates the electric field based on the strength and orientation of the magnet. Other methods include using a compass or a magnetometer.. ## 5. What are some applications of the electric field from a permanent magnet?. The electric field from a permanent magnet has many practical applications, including in electric motors, generators, and speakers. It is also used in magnetic levitation technology and particle accelerators. Additionally, the electric field from a permanent magnet can be used in medical imaging, such as MRI machines, to create detailed images of the body.. • Introductory Physics Homework Help. Replies. 5. Views. 1K. • Introductory Physics Homework Help. Replies. 26. Views. 953. • Quantum Physics. Replies. 3. Views. 348. • Introductory Physics Homework Help. Replies. 8. Views. 2K. • Introductory Physics Homework Help. Replies. 11. Views. 347. • Introductory Physics Homework Help. Replies. 8. Views. 3K. • Introductory Physics Homework Help. Replies. 16. Views. 260. • Introductory Physics Homework Help. Replies. 17. Views. 664. • Introductory Physics Homework Help. Replies. 1. Views. 813. • Introductory Physics Homework Help. Replies. 2. Views. 7K.
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Practice math test 1 on integers MCQs, grade 6 distributive law of multiplication multiple choice questions and answers. Distributive law of multiplication revision test has math worksheets, answer key with choices as a − b x c − b, a x c − b x c, a x b + a x c and a x b − a x c of multiple choice questions (MCQ) with distributive law of multiplication quiz as according to distributive law of multiplication over addition, a x ( b + c ) must be equal to for competitive exam prep. Free math study guide to learn distributive law of multiplication quiz to attempt multiple choice questions based test.
MCQ. According to Distributive Law of Multiplication over Addition, a x ( b + c ) must be equal to
1. a x c − b x c
2. a − b x c − b
3. a x b + a x c
4. a x b − a x c
C
MCQ. Product of -140 and +8 is
1. 1120
2. 3200
3. −1120
4. −3200
C
MCQ. Joseph has 10 candies. He gave 4 candies to John and John returned 2 candies to Joseph after few days. number of candies Joseph has altogether are
1. 8
2. −8
3. 10
4. −10
A
MCQ. If a x (b −c) is 8 for a = 2, b = 10 and c = 6 then a x b − a x c is equal to
1. (−8)
2. 12
3. 10
4. 8
D
MCQ. If a, b and c are integers, then according to associative law of multiplication ( a x b ) x c must be equal to
1. a x ( b + c )
2. ( a − b ) x c
3. (a + b ) + c
4. a x b + a x c
C
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Practice math test 1 on integers MCQs, grade 6 distributive law of multiplication multiple choice questions and answers. Distributive law of multiplication revision test has math worksheets, answer key with choices as a − b x c − b, a x c − b x c, a x b + a x c and a x b − a x c of multiple choice questions (MCQ) with distributive law of multiplication quiz as according to distributive law of multiplication over addition, a x ( b + c ) must be equal to for competitive exam prep. Free math study guide to learn distributive law of multiplication quiz to attempt multiple choice questions based test.. MCQ. According to Distributive Law of Multiplication over Addition, a x ( b + c ) must be equal to. 1. a x c − b x c. 2. a − b x c − b. 3. a x b + a x c. 4. a x b − a x c. C. MCQ. Product of -140 and +8 is. 1. 1120. 2. 3200. 3. −1120. 4. −3200. C. MCQ. Joseph has 10 candies. He gave 4 candies to John and John returned 2 candies to Joseph after few days. number of candies Joseph has altogether are. 1. 8.
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2. −8. 3. 10. 4. −10. A. MCQ. If a x (b −c) is 8 for a = 2, b = 10 and c = 6 then a x b − a x c is equal to. 1. (−8). 2. 12. 3. 10. 4. 8. D. MCQ. If a, b and c are integers, then according to associative law of multiplication ( a x b ) x c must be equal to. 1. a x ( b + c ). 2. ( a − b ) x c. 3. (a + b ) + c. 4. a x b + a x c. C.
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# What is 47/143 as a decimal?
## Solution and how to convert 47 / 143 into a decimal
47 / 143 = 0.329
Convert 47/143 to 0.329 decimal form by understanding when to use each form of the number. Both are used to handle numbers less than one or between whole numbers, known as integers. Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
## 47/143 is 47 divided by 143
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators and Denominators. This creates an equation. We use this as our equation: numerator(47) / denominator (143) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. Here's 47/143 as our equation:
### Numerator: 47
• Numerators sit at the top of the fraction, representing the parts of the whole. Overall, 47 is a big number which means you'll have a significant number of parts to your equation. 47 is an odd number so it might be harder to convert without a calculator. Large two-digit conversions are tough. Especially without a calculator. So how does our denominator stack up?
### Denominator: 143
• Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. 143 is a large number which means you should probably use a calculator. But 143 is an odd number. Having an odd denominator like 143 could sometimes be more difficult. Ultimately, don't be afraid of double-digit denominators. So grab a pen and pencil. Let's convert 47/143 by hand.
## Converting 47/143 to 0.329
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 143 \enclose{longdiv}{ 47 }$$
To solve, we will use left-to-right long division. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 143 \enclose{longdiv}{ 47.0 }$$
We've hit our first challenge. 47 cannot be divided into 143! So we will have to extend our division problem. Add a decimal point to 47, your numerator, and add an additional zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 143 into 47 + 0 or 470.
### Step 3: Solve for how many whole groups you can divide 143 into 470
$$\require{enclose} 00.3 \\ 143 \enclose{longdiv}{ 47.0 }$$
We can now pull 429 whole groups from the equation. Multiply this number by 143, the denominator to get the first part of your answer!
### Step 4: Subtract the remainder
$$\require{enclose} 00.3 \\ 143 \enclose{longdiv}{ 47.0 } \\ \underline{ 429 \phantom{00} } \\ 41 \phantom{0}$$
If your remainder is zero, that's it! If there is a remainder, extend 143 again and pull down the zero
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. Same goes for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But they all represent how numbers show us value in the real world. Here are just a few ways we use 47/143, 0.329 or 32% in our daily world:
### When you should convert 47/143 into a decimal
Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 47/143 MPH. The radar will read: 90.32 MPH. This simplifies the value.
### When to convert 0.329 to 47/143 as a fraction
Distance - Any type of travel, running, walking will leverage fractions. Distance is usually measured by the quarter mile and car travel is usually spoken the same.
### Practice Decimal Conversion with your Classroom
• If 47/143 = 0.329 what would it be as a percentage?
• What is 1 + 47/143 in decimal form?
• What is 1 - 47/143 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.329 + 1/2?
| 1,165
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# What is 47/143 as a decimal?. ## Solution and how to convert 47 / 143 into a decimal. 47 / 143 = 0.329. Convert 47/143 to 0.329 decimal form by understanding when to use each form of the number. Both are used to handle numbers less than one or between whole numbers, known as integers. Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). After deciding on which representation is best, let's dive into how we can convert fractions to decimals.. ## 47/143 is 47 divided by 143. Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators and Denominators. This creates an equation. We use this as our equation: numerator(47) / denominator (143) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. Here's 47/143 as our equation:. ### Numerator: 47. • Numerators sit at the top of the fraction, representing the parts of the whole. Overall, 47 is a big number which means you'll have a significant number of parts to your equation. 47 is an odd number so it might be harder to convert without a calculator. Large two-digit conversions are tough. Especially without a calculator. So how does our denominator stack up?. ### Denominator: 143. • Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. 143 is a large number which means you should probably use a calculator. But 143 is an odd number. Having an odd denominator like 143 could sometimes be more difficult. Ultimately, don't be afraid of double-digit denominators. So grab a pen and pencil. Let's convert 47/143 by hand.. ## Converting 47/143 to 0.329. ### Step 1: Set your long division bracket: denominator / numerator. $$\require{enclose} 143 \enclose{longdiv}{ 47 }$$. To solve, we will use left-to-right long division. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.. ### Step 2: Extend your division problem. $$\require{enclose} 00. \\ 143 \enclose{longdiv}{ 47.0 }$$. We've hit our first challenge.
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47 cannot be divided into 143! So we will have to extend our division problem. Add a decimal point to 47, your numerator, and add an additional zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 143 into 47 + 0 or 470.. ### Step 3: Solve for how many whole groups you can divide 143 into 470. $$\require{enclose} 00.3 \\ 143 \enclose{longdiv}{ 47.0 }$$. We can now pull 429 whole groups from the equation. Multiply this number by 143, the denominator to get the first part of your answer!. ### Step 4: Subtract the remainder. $$\require{enclose} 00.3 \\ 143 \enclose{longdiv}{ 47.0 } \\ \underline{ 429 \phantom{00} } \\ 41 \phantom{0}$$. If your remainder is zero, that's it! If there is a remainder, extend 143 again and pull down the zero. ### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.. Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.. ### Why should you convert between fractions, decimals, and percentages?. Converting between fractions and decimals is a necessity. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. Same goes for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But they all represent how numbers show us value in the real world. Here are just a few ways we use 47/143, 0.329 or 32% in our daily world:. ### When you should convert 47/143 into a decimal. Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 47/143 MPH. The radar will read: 90.32 MPH. This simplifies the value.. ### When to convert 0.329 to 47/143 as a fraction. Distance - Any type of travel, running, walking will leverage fractions. Distance is usually measured by the quarter mile and car travel is usually spoken the same.. ### Practice Decimal Conversion with your Classroom. • If 47/143 = 0.329 what would it be as a percentage?. • What is 1 + 47/143 in decimal form?. • What is 1 - 47/143 in decimal form?. • If we switched the numerator and denominator, what would be our new fraction?. • What is 0.329 + 1/2?.
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## Note on Volumetric Analysis
• Note
• Things to remember
### Volumetric Analysis:
It is a type of quantitative analysis based on the measurement of the volume of one solution required to react completely with a definite volume of another solution. By comparing the volume of two solutions, we can calculate the concentration of one solution provided that concentration of another solution is known.
#### EQUIVALENT MASS OF COMPOUNDS
a) Equivalent mass of acid
Equivalent mass of acid = $$\frac{Molar\:Mass}{Basicity}$$
Basicity = Number of replaceable hydrogen present in 1 molecule of acid
Acid Molar Mass Basicity Equivalent mass = $$\frac{Molar\:Mass}{Basicity}$$ 1. Hydrochloric acid (HCl) 36.5 1 36.5 2. Nitric acid (HNO3) 63 1 63 3. Acetic acid (CH3COOH) 60 1 60 4. Sulphuric acid (H2SO4) 98 2 49 5. Oxalic acid (COOH)2.2H2O 126 2 63 6. Phosphoric acid (H3PO4) 98 3 32.66
b) Equivalent mass of base
Equivalent mass = $$\frac{Molar\:Mass}{Acidity}$$
where Acidity = Number of replacable group for twice the number of oxygen present in 1 molecule of base.
Base Moalr mass Acidity Equivalent mass = $$\frac{Molar\:Mass}{Acidity}$$ NaOH 40 1 40 NH4OH 35 1 35 Ca(OH)2 74 2 37 Al(OH)3 78 3 26 CaO 56 2 28 Al2O3 102 6 17
c) Equivalent mass of salt
Equivalent mass = $$\frac{Molar\:Mass}{Total\:Positive\:charge\:in\:basic\:radical}$$
Example: NH4Cl = $$\frac{Molar\:Mass}{1}$$ = $$\frac {53.5}{1}$$ = 53.5
Example: CaCO3 = $$\frac{Molar\:Mass}{2}$$ = $$\frac {100}{2}$$ = 50
d) Equivalent mass of oxidizing and reducing agent
Equivalent mass = $$\frac{Molar\:Mass}{Change\:in\:Oxidation\:Number\:per\:molecule}$$
Example: Equivalent mass of KMnO4
i) Acidic medium
MnO4 -→ Mn2+
+7 +2
Change in oxidation number = 7 - 2 = 5
Equivalent mass = $$\frac{Molar\:Mass}{5}$$ = $$\frac{158}{5}$$ = 31.6
ii) Basic medium
MnO4 -→ MnO4 - -
+7 +6
Change in Oxidation Number = 1
Equivalent mass =$$\frac{Molar\:Mass}{1}$$ = $$\frac{158}{1}$$ =158
c) Neutral medium
MnO4 -→ MnO2
+7 +4
Change in Oxidation Number = 3
Equivalent mass =$$\frac{Molar\:Mass}{3}$$ = $$\frac{158}{3}$$ =52.6
### WAYS OF EXPRESSING CONCENTRATION OF SOLUTION
#### Percentage
%w/v (% by volume): It represents amount of solution (in gram) present in 100 ml of solution
i.e. %w/v = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:ml}$$ x 100
%w/w (% by mass): It represents amount of solution (in gram) present in 100 g of solution
i.e. %w/w = $$\frac{Mass\:of\:solute\:in\:gram}{Mass\:of\:solution\:in\:gm}$$ x 100
Gram per liter (gL-1): It represents the amount of solute (in gram) present in 1 liter of solution.
gL-1 = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:liter}$$
OR
gL-1 = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:mililiter}$$ x 1000
PPM (Parts per million)
1 PPM = 1mgL-1
Some relations
gL-1 = %w/v x 10
gL-1 = %w/w x specific gravity x 10
#### Normality (N)
Normality of a solution is defined as the number of equivalent of solute present in 1 liter of solution.
i.e. Normality = $$\frac{Number\:of\:equivalent\:of\:solute}{Volume\:of\:solution\:in\:liter}$$
= $$\frac{Number\:of\:equivalent\:of\:solute}{Volume\:of\:solution\:in\:mililiter}$$ x 1000
Also,
Normality = $$\frac {Mass}{Equivalent\:Mass}$$ x $$\frac{1000}{Volume\: in \:ml}$$
= $$\frac{gL^-1}{Equivalent\:Mass}$$
Normal solution ( 1 N Normality):The solution containing 1 equivalent of solute in 1 litre of solution is called normal solution.
Decinormal solution$$( \frac{N}{10}$$): The solution containing $$\frac{1}{10}$$ th equivalent of solute in 1 liter of solution is known as decinormal solution.
Similarly, $$\frac{N}{100}$$ is called centinormal solution and $$\frac{N}{2}$$ is called seminormal solution.
Molarity ( mol L -1): Molarity of a solution is defined as the number of moles of solute present in 1 liter of solution.
Molarity = = $$\frac{Number\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\:liter}$$
= $$\frac{Number\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\:mililiter}$$ x 1000
Also,
Molarity(M) = $$\frac {Mass}{Molecular\:Mass}$$ x $$\frac{1000}{Volume\: in \:ml}$$
= $$\frac{gL^-1}{Molecular\:Mass}$$
Molar solution ( 1 M Molarity):The solution containing 1 mole of solute in 1 liter of solution is called normal solution.
Decimolar solution$$( \frac{M}{10}$$): The solution containing $$\frac{1}{10}$$ th mole of solute in 1 liter of solution is known as decinormal solution.
Similarly, $$\frac{M}{100}$$ is called centimolar solution and $$\frac{N}{2}$$ is called semimolar solution.
#### RELATION BETWEEN MOLARITY AND NORMALITY
Normality = $$\frac{gL^-1}{Equivalent\:Mass}$$
or, gL-1 = Normality x Equivalent mass -----------(i)
Similarly ,
Molarity = $$\frac{gL^-1}{Molecular\:Mass}$$
or, gL-1 = Molarity x Molecular mass -----------(i)
Combining (i) and (ii)
Normality x Equivalent mass = Molarity x Molecular mass
For acid
Normality x $$\frac{Molecular\:Mass}{Basicity}$$ = Molarity x Molecular mass
Or, Normality = Molarity x Basicity
For base,
Normality x $$\frac{Molecular\:Mass}{Acidity}$$ = Molarity x Molecular mass
Or, Normality = Molarity x Acidity
#### Dilution principle
A solution of lower concentration can be prepared from a solution of higher concentration using dilution principle as:
V1 x S1 = V2 x S2
Where, V1 and S1 are volume and concentration of a solution of lower concentration and V2 and S2 are the volume and concentration of a solution of higher concentration.
#### Primary standard substance
A substance of sufficient purity from which standard solution can be prepared by directly weighing the exact quantity of the substance and dissolving indefinite volume of solution is known as primary standard substance. For a substance to be primary standard, it must have the following characteristics
i) The substance should be easily available in the pure state or in the state of known purity.
ii) The substance should not be hygroscopic or reactive in the atmosphere and should be easy to dry.
iii) The composition of substance should not change in solid state or in solution form of sufficiently long time.
iv) The compound should be easily soluble in water under the condition which it is employed.
v) The substance should have comparatively high molar mass or equivalent mass so that error during weighting is minimized.
That substance which does not satisfy the above characteristics is called secondary standard substances. Standard solution of these substances cannot be prepared by directly weighing the exact quantity of the substance and dissolving indefinite volume of solution.
Example: HCl, NaOH, KMnO4, etc.
#### Standard solution
The solution having known concentration is known as a standard solution. It is of two types:
1. Primary Standard Solution
2. Secondary standard solution
A. Primary Standard Solution
A standard solution prepared from the primary standard substance by directly weighing the exact quantity of the substance and dissolving indefinite volume of solution is called primary standard solution. Concentration or composition of this solution does not change during storage for a long time.
B. Secondary standard solution
A standard solution which cannot be prepared by directly weighing the exact quantity of substance or a solution in which exact concentration is determined by titrating it with suitable primary standard substance is known as a secondary standard solution. The concentration or composition of this solution changes during storage.
Factor (f): It is the term which indicates by what factor actual concentration of solution differ from the proposed one.
#### Titration:
The experimental technique used to determine the concentration of the unknown solution by measuring the volume of standard solution required to react completely with a definite volume of unknown solution. The unknown solution is taken in a conical flask and standard solution is added from a long graduated tube called burette till reaction is complete. The completion of the reaction is indicated by some physical change produced by reagent itself or more usually by use of an indicator or some other physical measurement.
#### SOME TERMS USED IN TITRATION
Equivalent point: A stage during titration in which equivalent quantity of the substance is added from burette to the solution in the conical flask is the equivalent point. At this stage, the reaction is usually completed.
End point: The stage during titration at which indicator changes its color to indicate the completion of the reaction is called end point. It is the experimental point.
Titration error: The difference between the end point and the equivalent point is known as titration error.
Titrant: A standard solution or solution taken in burette is called titrant.
Titrand: An unknown solution or solution taken in the conical flask is called titrand.
Standardization: The process of finding the actual concentration of the secondary standard solution by titrating it with a suitable primary standard solution is known as standardization.
Indicator: An auxiliary substance used during titration to indicate completion by a sharp change in color is called indicator. Example: Phenolphthalein, Methyl Orange, Methyl red, etc
#### TYPES OF TITRATION
a) Acid-Base Titration (Acidimetry/Alkalimetry)
The titration between acid and base is called acid-base titration. In this titration, neutralization reaction takes place.
Acidimetry:The process of finding a concentration of unknown acid by titrating it with a standard solution of base is called acidimetry.
Alkalimetry:The process of finding a concentration of a base by titrating it with a standard solution of acid is called alkalimetry.
b) Redox Titration
The titration between the oxidizing agent and reducing a reducing agent is known as redox titration. In this titration, redox reaction (oxidation and reduction) takes place.
KMnO4 + H2SO4→ K2SO4 + MnSO4 + CO2 + H2O
Indicator used: KMnO4 (as self-indicator)
c) Precipitation Titration
In this titration, the reaction involved is precipitation reaction. Example: Titration between Halide solution and AgNO3 solution.
d) Complexometric titration
Reaction involved: Complex compound formation reaction
Example: Titration between metal ion with EDTA
EDTA = Ethylene Diamine Tetra Acetic Acid
### ACID-BASE INDICATORS AND THEIR SELECTION
#### Action of acid-base indicator
Those indicators which are used in acid-base titration like phenolphthalein, methyl orange, methyl red, etc. are called acid-base indicators. These indicators are weak acid or base themselves. Each indicator has two different forms and each form has its own color. The color given by indicator in particular solution depends on the relative concentration of the two forms.
Consider phenolphthalein indicator which is represented by HPH .
Here, phenolphthalein has two different form (HPH) or unionized form is a colorless and ionized form (PH-) is pink. When this indicator is added to the acidic solution, due to the common ion effect of H+ ion, equilibrium is largely shifted to the backward. This means unionized form is dominant hence, is colorless in acidic solution. Similarly, when the indicator is added to the alkaline solution, H+ ion of indicator combines with OH- ion of a base to produce unionized water molecule and equilibrium is largely shifted to forward. This means indicator is dominantly found in ionized form and gives pink color in ionized form.
Selection of Acid-Base Indicator
Each indicator has its own pH range for a color change. Some indicators have a pH range in a slightly acidic side. For example; Methyl orange (3.1 to 4.5). Some indicators have a pH range in the slightly basic side such as Phenolphthalein (8.3 – 10).
During the acid-base titration, pH of the resulting solution changes by adding acid or base. When we plot pH of titrating solution against the volume of acid or base added, we get a curve called titration curve. The curve shows that there is a sharp change in pH near the equivalent point.
i) Titration between strong acid and strong base
During this titration, pH of the resulting solution changes from 3 to 11 (approximately) near equivalent point and equivalent point lies at pH = 7. Any indicators which have a pH range between 3 to 11 can be used in this titration and suitable indicators are phenolphthalein, methyl orange or methyl red.
ii) Titration between strong acid and weak base
During this titration, pH of the resulting solution changes from 3 to 8 (approximately) and its equivalent point lies at pH less than 7 i.e. acidic side due to the hydrolysis of salt. Any indicators which have a pH range in the slightly acidic side can be used in the titration and the suitable indicators are methyl orange or methyl red.
iii) Titration between weak acid and strong base
During this titration, pH of the resulting solution changes from 6 to 11 (approximately) and equivalent point lies at pH more than 7 i.e. at alkaline side due to the hydrolysis of salt. Any indicators which have a pH range in the slightly alkaline side can be used in this titration and a suitable indicator is a phenolphthalein.
iv) Titration between weak acid and weak base
During this titration, there is no sharp change in pH near equivalent point and there is no suitable indicator.
References: -
Sthapit, Moti Kaji, and Dr.Raja Ram Pradhananga. Foundations Of Chemistry. 5th. Vol. 1. Kathmandu: Supravaha Press, 2010. 3 vols.
• Equivalent mass = $$\frac{Molar\:Mass}{Change\:in\:Oxidation\:Number\:per\:molecule}$$
• gL-1 = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:mililiter}$$ x 1000
• The solution containing $$\frac{1}{10}$$ th mole of solute in 1 liter of solution is known as decinormal solution.
• The composition of substance should not change in solid state or in solution form of sufficiently long time.
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##### suraj thapa
0.715g of Na2CO3.xH2O required 20 ml of decinormal sodium hydroxide solution for complete neutralisation.Find the molecular mass of the acid?
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## Note on Volumetric Analysis. • Note. • Things to remember. ### Volumetric Analysis:. It is a type of quantitative analysis based on the measurement of the volume of one solution required to react completely with a definite volume of another solution. By comparing the volume of two solutions, we can calculate the concentration of one solution provided that concentration of another solution is known.. #### EQUIVALENT MASS OF COMPOUNDS. a) Equivalent mass of acid. Equivalent mass of acid = $$\frac{Molar\:Mass}{Basicity}$$. Basicity = Number of replaceable hydrogen present in 1 molecule of acid. Acid Molar Mass Basicity Equivalent mass = $$\frac{Molar\:Mass}{Basicity}$$ 1. Hydrochloric acid (HCl) 36.5 1 36.5 2. Nitric acid (HNO3) 63 1 63 3. Acetic acid (CH3COOH) 60 1 60 4. Sulphuric acid (H2SO4) 98 2 49 5. Oxalic acid (COOH)2.2H2O 126 2 63 6. Phosphoric acid (H3PO4) 98 3 32.66. b) Equivalent mass of base. Equivalent mass = $$\frac{Molar\:Mass}{Acidity}$$. where Acidity = Number of replacable group for twice the number of oxygen present in 1 molecule of base.. Base Moalr mass Acidity Equivalent mass = $$\frac{Molar\:Mass}{Acidity}$$ NaOH 40 1 40 NH4OH 35 1 35 Ca(OH)2 74 2 37 Al(OH)3 78 3 26 CaO 56 2 28 Al2O3 102 6 17. c) Equivalent mass of salt. Equivalent mass = $$\frac{Molar\:Mass}{Total\:Positive\:charge\:in\:basic\:radical}$$. Example: NH4Cl = $$\frac{Molar\:Mass}{1}$$ = $$\frac {53.5}{1}$$ = 53.5. Example: CaCO3 = $$\frac{Molar\:Mass}{2}$$ = $$\frac {100}{2}$$ = 50. d) Equivalent mass of oxidizing and reducing agent. Equivalent mass = $$\frac{Molar\:Mass}{Change\:in\:Oxidation\:Number\:per\:molecule}$$. Example: Equivalent mass of KMnO4. i) Acidic medium. MnO4 -→ Mn2+. +7 +2. Change in oxidation number = 7 - 2 = 5. Equivalent mass = $$\frac{Molar\:Mass}{5}$$ = $$\frac{158}{5}$$ = 31.6. ii) Basic medium. MnO4 -→ MnO4 - -. +7 +6. Change in Oxidation Number = 1. Equivalent mass =$$\frac{Molar\:Mass}{1}$$ = $$\frac{158}{1}$$ =158. c) Neutral medium. MnO4 -→ MnO2. +7 +4. Change in Oxidation Number = 3. Equivalent mass =$$\frac{Molar\:Mass}{3}$$ = $$\frac{158}{3}$$ =52.6. ### WAYS OF EXPRESSING CONCENTRATION OF SOLUTION. #### Percentage. %w/v (% by volume): It represents amount of solution (in gram) present in 100 ml of solution. i.e. %w/v = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:ml}$$ x 100. %w/w (% by mass): It represents amount of solution (in gram) present in 100 g of solution. i.e. %w/w = $$\frac{Mass\:of\:solute\:in\:gram}{Mass\:of\:solution\:in\:gm}$$ x 100. Gram per liter (gL-1): It represents the amount of solute (in gram) present in 1 liter of solution.. gL-1 = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:liter}$$. OR. gL-1 = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:mililiter}$$ x 1000. PPM (Parts per million). 1 PPM = 1mgL-1. Some relations. gL-1 = %w/v x 10. gL-1 = %w/w x specific gravity x 10. #### Normality (N). Normality of a solution is defined as the number of equivalent of solute present in 1 liter of solution.. i.e. Normality = $$\frac{Number\:of\:equivalent\:of\:solute}{Volume\:of\:solution\:in\:liter}$$. = $$\frac{Number\:of\:equivalent\:of\:solute}{Volume\:of\:solution\:in\:mililiter}$$ x 1000. Also,. Normality = $$\frac {Mass}{Equivalent\:Mass}$$ x $$\frac{1000}{Volume\: in \:ml}$$. = $$\frac{gL^-1}{Equivalent\:Mass}$$. Normal solution ( 1 N Normality):The solution containing 1 equivalent of solute in 1 litre of solution is called normal solution.. Decinormal solution$$( \frac{N}{10}$$): The solution containing $$\frac{1}{10}$$ th equivalent of solute in 1 liter of solution is known as decinormal solution.. Similarly, $$\frac{N}{100}$$ is called centinormal solution and $$\frac{N}{2}$$ is called seminormal solution.. Molarity ( mol L -1): Molarity of a solution is defined as the number of moles of solute present in 1 liter of solution.. Molarity = = $$\frac{Number\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\:liter}$$. = $$\frac{Number\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\:mililiter}$$ x 1000. Also,. Molarity(M) = $$\frac {Mass}{Molecular\:Mass}$$ x $$\frac{1000}{Volume\: in \:ml}$$. = $$\frac{gL^-1}{Molecular\:Mass}$$. Molar solution ( 1 M Molarity):The solution containing 1 mole of solute in 1 liter of solution is called normal solution.. Decimolar solution$$( \frac{M}{10}$$): The solution containing $$\frac{1}{10}$$ th mole of solute in 1 liter of solution is known as decinormal solution.. Similarly, $$\frac{M}{100}$$ is called centimolar solution and $$\frac{N}{2}$$ is called semimolar solution.. #### RELATION BETWEEN MOLARITY AND NORMALITY. Normality = $$\frac{gL^-1}{Equivalent\:Mass}$$. or, gL-1 = Normality x Equivalent mass -----------(i). Similarly ,. Molarity = $$\frac{gL^-1}{Molecular\:Mass}$$. or, gL-1 = Molarity x Molecular mass -----------(i). Combining (i) and (ii). Normality x Equivalent mass = Molarity x Molecular mass. For acid. Normality x $$\frac{Molecular\:Mass}{Basicity}$$ = Molarity x Molecular mass. Or, Normality = Molarity x Basicity. For base,. Normality x $$\frac{Molecular\:Mass}{Acidity}$$ = Molarity x Molecular mass. Or, Normality = Molarity x Acidity. #### Dilution principle. A solution of lower concentration can be prepared from a solution of higher concentration using dilution principle as:. V1 x S1 = V2 x S2. Where, V1 and S1 are volume and concentration of a solution of lower concentration and V2 and S2 are the volume and concentration of a solution of higher concentration.. #### Primary standard substance. A substance of sufficient purity from which standard solution can be prepared by directly weighing the exact quantity of the substance and dissolving indefinite volume of solution is known as primary standard substance. For a substance to be primary standard, it must have the following characteristics. i) The substance should be easily available in the pure state or in the state of known purity.. ii) The substance should not be hygroscopic or reactive in the atmosphere and should be easy to dry.. iii) The composition of substance should not change in solid state or in solution form of sufficiently long time.. iv) The compound should be easily soluble in water under the condition which it is employed.
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v) The substance should have comparatively high molar mass or equivalent mass so that error during weighting is minimized.. That substance which does not satisfy the above characteristics is called secondary standard substances. Standard solution of these substances cannot be prepared by directly weighing the exact quantity of the substance and dissolving indefinite volume of solution.. Example: HCl, NaOH, KMnO4, etc.. #### Standard solution. The solution having known concentration is known as a standard solution. It is of two types:. 1. Primary Standard Solution. 2. Secondary standard solution. A. Primary Standard Solution. A standard solution prepared from the primary standard substance by directly weighing the exact quantity of the substance and dissolving indefinite volume of solution is called primary standard solution. Concentration or composition of this solution does not change during storage for a long time.. B. Secondary standard solution. A standard solution which cannot be prepared by directly weighing the exact quantity of substance or a solution in which exact concentration is determined by titrating it with suitable primary standard substance is known as a secondary standard solution. The concentration or composition of this solution changes during storage.. Factor (f): It is the term which indicates by what factor actual concentration of solution differ from the proposed one.. #### Titration:. The experimental technique used to determine the concentration of the unknown solution by measuring the volume of standard solution required to react completely with a definite volume of unknown solution. The unknown solution is taken in a conical flask and standard solution is added from a long graduated tube called burette till reaction is complete. The completion of the reaction is indicated by some physical change produced by reagent itself or more usually by use of an indicator or some other physical measurement.. #### SOME TERMS USED IN TITRATION. Equivalent point: A stage during titration in which equivalent quantity of the substance is added from burette to the solution in the conical flask is the equivalent point. At this stage, the reaction is usually completed.. End point: The stage during titration at which indicator changes its color to indicate the completion of the reaction is called end point. It is the experimental point.. Titration error: The difference between the end point and the equivalent point is known as titration error.. Titrant: A standard solution or solution taken in burette is called titrant.. Titrand: An unknown solution or solution taken in the conical flask is called titrand.. Standardization: The process of finding the actual concentration of the secondary standard solution by titrating it with a suitable primary standard solution is known as standardization.. Indicator: An auxiliary substance used during titration to indicate completion by a sharp change in color is called indicator. Example: Phenolphthalein, Methyl Orange, Methyl red, etc. #### TYPES OF TITRATION. a) Acid-Base Titration (Acidimetry/Alkalimetry). The titration between acid and base is called acid-base titration. In this titration, neutralization reaction takes place.. Acidimetry:The process of finding a concentration of unknown acid by titrating it with a standard solution of base is called acidimetry.. Alkalimetry:The process of finding a concentration of a base by titrating it with a standard solution of acid is called alkalimetry.. b) Redox Titration. The titration between the oxidizing agent and reducing a reducing agent is known as redox titration. In this titration, redox reaction (oxidation and reduction) takes place.. KMnO4 + H2SO4→ K2SO4 + MnSO4 + CO2 + H2O. Indicator used: KMnO4 (as self-indicator). c) Precipitation Titration. In this titration, the reaction involved is precipitation reaction. Example: Titration between Halide solution and AgNO3 solution.. d) Complexometric titration. Reaction involved: Complex compound formation reaction. Example: Titration between metal ion with EDTA. EDTA = Ethylene Diamine Tetra Acetic Acid. ### ACID-BASE INDICATORS AND THEIR SELECTION. #### Action of acid-base indicator. Those indicators which are used in acid-base titration like phenolphthalein, methyl orange, methyl red, etc. are called acid-base indicators. These indicators are weak acid or base themselves. Each indicator has two different forms and each form has its own color. The color given by indicator in particular solution depends on the relative concentration of the two forms.. Consider phenolphthalein indicator which is represented by HPH .. Here, phenolphthalein has two different form (HPH) or unionized form is a colorless and ionized form (PH-) is pink. When this indicator is added to the acidic solution, due to the common ion effect of H+ ion, equilibrium is largely shifted to the backward. This means unionized form is dominant hence, is colorless in acidic solution. Similarly, when the indicator is added to the alkaline solution, H+ ion of indicator combines with OH- ion of a base to produce unionized water molecule and equilibrium is largely shifted to forward. This means indicator is dominantly found in ionized form and gives pink color in ionized form.. Selection of Acid-Base Indicator. Each indicator has its own pH range for a color change. Some indicators have a pH range in a slightly acidic side. For example; Methyl orange (3.1 to 4.5). Some indicators have a pH range in the slightly basic side such as Phenolphthalein (8.3 – 10).. During the acid-base titration, pH of the resulting solution changes by adding acid or base. When we plot pH of titrating solution against the volume of acid or base added, we get a curve called titration curve. The curve shows that there is a sharp change in pH near the equivalent point.. i) Titration between strong acid and strong base. During this titration, pH of the resulting solution changes from 3 to 11 (approximately) near equivalent point and equivalent point lies at pH = 7. Any indicators which have a pH range between 3 to 11 can be used in this titration and suitable indicators are phenolphthalein, methyl orange or methyl red.. ii) Titration between strong acid and weak base. During this titration, pH of the resulting solution changes from 3 to 8 (approximately) and its equivalent point lies at pH less than 7 i.e. acidic side due to the hydrolysis of salt. Any indicators which have a pH range in the slightly acidic side can be used in the titration and the suitable indicators are methyl orange or methyl red.. iii) Titration between weak acid and strong base. During this titration, pH of the resulting solution changes from 6 to 11 (approximately) and equivalent point lies at pH more than 7 i.e. at alkaline side due to the hydrolysis of salt. Any indicators which have a pH range in the slightly alkaline side can be used in this titration and a suitable indicator is a phenolphthalein.. iv) Titration between weak acid and weak base. During this titration, there is no sharp change in pH near equivalent point and there is no suitable indicator.. References: -. Sthapit, Moti Kaji, and Dr.Raja Ram Pradhananga. Foundations Of Chemistry. 5th. Vol. 1. Kathmandu: Supravaha Press, 2010. 3 vols.. • Equivalent mass = $$\frac{Molar\:Mass}{Change\:in\:Oxidation\:Number\:per\:molecule}$$. • gL-1 = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:mililiter}$$ x 1000. • The solution containing $$\frac{1}{10}$$ th mole of solute in 1 liter of solution is known as decinormal solution.. • The composition of substance should not change in solid state or in solution form of sufficiently long time.. .. 0%. • ## You scored /0. Forum Time Replies Report. ##### suraj thapa. 0.715g of Na2CO3.xH2O required 20 ml of decinormal sodium hydroxide solution for complete neutralisation.Find the molecular mass of the acid?.
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# «fractal tweets» byaucotsi
on 05 Apr'12 12:26 in tweets
```1
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```// I was looking some fractal structures, with a friend who studies the maths, and we found a // simple algorithm. So i made some tweets with the formula.
r{inf.do{|i|a=i.asInteger.rand2;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.07.wait;}}.play
//
r{inf.do{|i|a=i.asInteger;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.01.wait;}}.play
//
a=(-23);r{inf.do{a=a.asInteger;a.postln;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.1.wait}}.play
//
a=(-917);r{inf.do{a=a.asInteger.postln;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.1.wait}}.play
// the funny thing is that when you have a negative input this folds into a loop!
// the correct formula is this below
a=73;r=r{inf.do{a=a.asInteger;if(a==1){r.stop};if(a.even){a=a/2}{a=a*3+1};play{Blip.ar(a)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.1.wait}}.play```
raw 979 chars (focus & ctrl+a+c to copy)
reception
comments
aucotsi user 08 Apr'12 08:02
likelihood fractals aren't the proper definition, these are chaotic feedback sequences
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# «fractal tweets» byaucotsi. on 05 Apr'12 12:26 in tweets. ```1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14```. ```// I was looking some fractal structures, with a friend who studies the maths, and we found a // simple algorithm.
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So i made some tweets with the formula.. r{inf.do{|i|a=i.asInteger.rand2;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.07.wait;}}.play. //. r{inf.do{|i|a=i.asInteger;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.01.wait;}}.play. //. a=(-23);r{inf.do{a=a.asInteger;a.postln;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.1.wait}}.play. //. a=(-917);r{inf.do{a=a.asInteger.postln;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.1.wait}}.play. // the funny thing is that when you have a negative input this folds into a loop!. // the correct formula is this below. a=73;r=r{inf.do{a=a.asInteger;if(a==1){r.stop};if(a.even){a=a/2}{a=a*3+1};play{Blip.ar(a)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.1.wait}}.play```. raw 979 chars (focus & ctrl+a+c to copy). reception. comments. aucotsi user 08 Apr'12 08:02. likelihood fractals aren't the proper definition, these are chaotic feedback sequences.
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How to find the rate of effusion given rates?
Question
At $25~^\circ\mathrm{C}$, a sample of $\ce{NH}$ effuses at the rate of $0.050$ moles per minute. Under the same conditions, what is the molar mass of a gas that effuses at approximately one half of that rate?
My steps and thoughts
I converted $25~^{\circ}\mathrm{C}$ to $298~\mathrm{K}$ and then I was thinking of using $pV=nRT$ and I got stuck.
I am thinking of using molar mass, but I don't know how to incorporate that.
$$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{\text{M}_2}{\text{M}_1}}$$
I don't know what the OP meant by $\ce{NH}$, I assume the gas in question is $\ce{NH3}$ with rate of effusion ($R_1$) = 0.050 $\mathrm{mol \min^{-1}}$.
$R_2 = \frac{R_1}{2}$ and $M_1 = 17.031$, plugging all of this in
$M_2 = M_1\cdot \left (\frac{R_1}{R_2} \right )^2 = 4\cdot M_1 = 68.124$
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How to find the rate of effusion given rates?. Question. At $25~^\circ\mathrm{C}$, a sample of $\ce{NH}$ effuses at the rate of $0.050$ moles per minute. Under the same conditions, what is the molar mass of a gas that effuses at approximately one half of that rate?. My steps and thoughts. I converted $25~^{\circ}\mathrm{C}$ to $298~\mathrm{K}$ and then I was thinking of using $pV=nRT$ and I got stuck.
|
I am thinking of using molar mass, but I don't know how to incorporate that.. $$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{\text{M}_2}{\text{M}_1}}$$. I don't know what the OP meant by $\ce{NH}$, I assume the gas in question is $\ce{NH3}$ with rate of effusion ($R_1$) = 0.050 $\mathrm{mol \min^{-1}}$.. $R_2 = \frac{R_1}{2}$ and $M_1 = 17.031$, plugging all of this in. $M_2 = M_1\cdot \left (\frac{R_1}{R_2} \right )^2 = 4\cdot M_1 = 68.124$.
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# Thread: Sketching regions in the complex plane
H
2. ## Re: Sketching regions in the complex plane
Originally Posted by redtdc
How would I sketch the region S={z in the complex plane: abs(z-3i)=abs(z-2)} ?
And is S open or closed?
Thanks!
Hi redtdc!
abs(z-3i) is the same as the distance in R2 of a point (x,y) to (0,3).
The region S corresponds to the points in R2 that have the same distance to (0,3) as to (2,0).
What kind of region is that?
3. ## Re: Sketching regions in the complex plane
Originally Posted by redtdc
How would I sketch the region S={z in the complex plane: abs(z-3i)=abs(z-2)} ?
And is S open or closed?
Do you understand that $|z-z_0|$ is the distance from $z\text{ to }z_0~?$
So what is the set $S=\{z:|z-3i|=|z-2|\}~?$
4. ## Re: Sketching regions in the complex plane
|z- 3i| is the distance from z to 3i and |x- 2| is the distance from z to 2 so S consists of points that are equally distant from 3i and 2. Geometrically, the set of all points equally distant from point P and Q is the perpendicular bisector of the segment PQ.
Algebraically, taking z= x+ iy, |z- 3i|= |z- 2| is the same as $\sqrt{(x^2+ (y- 3))^2}= \sqrt{((x- 2)^2+ y^2}$ which is the same as $x^2+ (y- 3)^2= (x- 2)^2+ y^2$.
As for whether it is open or close, what topology are you using?
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# Thread: Sketching regions in the complex plane. H. 2. ## Re: Sketching regions in the complex plane. Originally Posted by redtdc. How would I sketch the region S={z in the complex plane: abs(z-3i)=abs(z-2)} ?. And is S open or closed?. Thanks!. Hi redtdc!. abs(z-3i) is the same as the distance in R2 of a point (x,y) to (0,3).. The region S corresponds to the points in R2 that have the same distance to (0,3) as to (2,0).. What kind of region is that?. 3.
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## Re: Sketching regions in the complex plane. Originally Posted by redtdc. How would I sketch the region S={z in the complex plane: abs(z-3i)=abs(z-2)} ?. And is S open or closed?. Do you understand that $|z-z_0|$ is the distance from $z\text{ to }z_0~?$. So what is the set $S=\{z:|z-3i|=|z-2|\}~?$. 4. ## Re: Sketching regions in the complex plane. |z- 3i| is the distance from z to 3i and |x- 2| is the distance from z to 2 so S consists of points that are equally distant from 3i and 2. Geometrically, the set of all points equally distant from point P and Q is the perpendicular bisector of the segment PQ.. Algebraically, taking z= x+ iy, |z- 3i|= |z- 2| is the same as $\sqrt{(x^2+ (y- 3))^2}= \sqrt{((x- 2)^2+ y^2}$ which is the same as $x^2+ (y- 3)^2= (x- 2)^2+ y^2$.. As for whether it is open or close, what topology are you using?.
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# How do you write the combined function as a composition of several functions if f(g(x)) = sqrt (1-x^2) +2?
##### 1 Answer
Feb 1, 2016
If I remember correctly there are many "right" ways to solve this equation. $f \left(g \left(x\right)\right)$ merely means you are plugging a function of $x$ into $g$ into $f$ with each letter being a different function.
For instance: your equation for $g$ could possibly be $\sqrt{1 - {x}^{2}}$.
After, we plug in $x$ into that equation we get the same function ($\sqrt{1 - {x}^{2}}$).
Next step when solving multiple step functions is to plug in your result of your first function (in our case $g$) into the new function ($f$). This is where we need another function to get $\sqrt{1 - {x}^{2}}$ into $\sqrt{1 - {x}^{2}}$$+ 2$.
For this step we may say $\left(f\right) = x + 2$.
When we plug in $\sqrt{1 - {x}^{2}}$ into our new $\left(f\right) = x + 2$ our result is
$\sqrt{1 - {x}^{2}}$$+ 2$, the equation we needed to get to.
Another solution would be to have $\left(g\right) = 1 - {x}^{2}$ and $\left(f\right) = \sqrt{x} + 2$
There are many more solution, just be creative, these are only two!
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# How do you write the combined function as a composition of several functions if f(g(x)) = sqrt (1-x^2) +2?. ##### 1 Answer. Feb 1, 2016. If I remember correctly there are many "right" ways to solve this equation. $f \left(g \left(x\right)\right)$ merely means you are plugging a function of $x$ into $g$ into $f$ with each letter being a different function.. For instance: your equation for $g$ could possibly be $\sqrt{1 - {x}^{2}}$.. After, we plug in $x$ into that equation we get the same function ($\sqrt{1 - {x}^{2}}$).
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Next step when solving multiple step functions is to plug in your result of your first function (in our case $g$) into the new function ($f$). This is where we need another function to get $\sqrt{1 - {x}^{2}}$ into $\sqrt{1 - {x}^{2}}$$+ 2$.. For this step we may say $\left(f\right) = x + 2$.. When we plug in $\sqrt{1 - {x}^{2}}$ into our new $\left(f\right) = x + 2$ our result is. $\sqrt{1 - {x}^{2}}$$+ 2$, the equation we needed to get to.. Another solution would be to have $\left(g\right) = 1 - {x}^{2}$ and $\left(f\right) = \sqrt{x} + 2$. There are many more solution, just be creative, these are only two!.
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# Mechanics11/Page3
## interpretation of time local diagrams
Assignment
Desiree and Marc go by bike to the plan lake. They drive off at 8.00 o'clock and arrive around 19.00 again at home.
Along the road are kilometer stones. They note each full hour their position x.
• a) calculation it for each time interval of 1 hour the average speed! Use calculation formulas!
• b) describing it for each time interval the process of the movement in words! Possible interpretation?
• c) calculation it the average speed for the first 4 hours! How can this be plotted?
## 1. Motion equation
We deduce a general equation for straight-line-homogeneous movements: from
$v = \frac{\Delta x}{\Delta t} = \frac{x - x zero}{\Delta t}$
follows
x = x0 + v$\Delta t$
Thus the current position at the end of one time interval can be computed e.g. with well-known initial place x and well-known speed of v.
At present if we take for x0 the place x (0) to t = 0, then those results
1. Motion equation:
x (t) = x0 + v t
## sample calculation
Assignment
Problem 1.6: Flight competition Bussard flies with a falcon around the bet. The bussard gets 14 km projection/lead from the starting point measured, because it is clearly slower with 95 km/h than the falcon (153 km/h).
• a) when and where the falcon catches up the bussard?
• b) who wins the flight up to the summit cross of the Hörnle (40 km from the start )?
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# Mechanics11/Page3. ## interpretation of time local diagrams. Assignment. Desiree and Marc go by bike to the plan lake. They drive off at 8.00 o'clock and arrive around 19.00 again at home.. Along the road are kilometer stones. They note each full hour their position x.. • a) calculation it for each time interval of 1 hour the average speed! Use calculation formulas!. • b) describing it for each time interval the process of the movement in words! Possible interpretation?. • c) calculation it the average speed for the first 4 hours! How can this be plotted?. ## 1. Motion equation. We deduce a general equation for straight-line-homogeneous movements: from. $v = \frac{\Delta x}{\Delta t} = \frac{x - x zero}{\Delta t}$. follows.
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x = x0 + v$\Delta t$. Thus the current position at the end of one time interval can be computed e.g. with well-known initial place x and well-known speed of v.. At present if we take for x0 the place x (0) to t = 0, then those results. 1. Motion equation:. x (t) = x0 + v t. ## sample calculation. Assignment. Problem 1.6: Flight competition Bussard flies with a falcon around the bet. The bussard gets 14 km projection/lead from the starting point measured, because it is clearly slower with 95 km/h than the falcon (153 km/h).. • a) when and where the falcon catches up the bussard?. • b) who wins the flight up to the summit cross of the Hörnle (40 km from the start )?.
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http://www.sxlist.com/techref/scenix/lib/math/div/div16or32by16to16_sx.htm
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# SXMicrocontrollerMathMethod
## 32 by 16 Divison from Nikolai Golovchenko, or How to make use of a 16 bit division routine for 32 bit division
Division routines sometimes can be used for more than they were designed for! For example, it is possible to divide 16 or 32 bit dividend by 16 bit divisor to 16 bit quotient in just one routine.
The idea is simple: you do 16 bit division as usually, but when the dividend is 32 bit, the lower two bytes are used as dividend and the higher two bytes are used as initial value for remainder (which is zero when doing 16 bit division)
To understand how initializing remainder relates to dividend, let's examine the algorithms for two cases:
``` x
q = ---
y
```
1) when x, y, and q are 16 bit unsigned integers, and
2) when x and q are 32 bit unsigned integers, but y is a 16 bit unsigned integer
Note that in some cases it is known that the result of 32 bit by 16 bit division is 16 bit maximum. The trick described on this page works only for 16 bit result.
For example, when speed of a motor is calculated by a Hall sensor pulses (one pulse per revolution) the following formula is used:
``` 60 x 1,000,000 (conversions for minutes and us,
rpm = ------------------ timer is clocked by 1 MHz clock)
16 bit timer value
```
Timer is used to measure the Hall sensor pulse period. For this motor and timer the rpm range is within 915-20,000 rpm, which fits perfectly into 16 bits.
#### 1) 16 bit division
x - 16 bit dividend;
y - 16 bit divisor;
q - 16 bit quotient;
rem - 16 bit remainder;
counter - loop counter.
``` rem q
------- -------
|00|00| |??|??|
------- -------
b1 b0 b1 b0
```
2. Shift left dividend x, and shift left remainder rem, by 1 bit, so that MSb of x was shifted to the remainder's LSb
``` rem x
------- -------
|r1|r0| <- |x1|x0| <- shift left
------- -------
b1 b0 b1 b0
```
3. Subtract divisor y from remainder. If subtraction successful (no borrow), next bit of q is 1. If borrow, next bit of q is 0, and remainder should be restored.
``` -------
|r1|r0| rem
-------
b1 b0
-
-------
|y1|y0| y
-------
b1 b0
```
4. Shift left q and set its LSb according to subtraction result from previous step.
``` q carry(inverted borrow) flag
------- ---
|q1|q0| <- |C| shift in next result bit
------- ---
b1 b0
```
5. Decrement counter and if it's not zero repeat from step 2 (16 iterations)
After that, q contains quotient, rem - remainder, x is garbage (it was shifted out completely), and y is untouched.
#### 2) 32 bit division with 16 bit divisor
x - 32 bit dividend;
y - 16 bit divisor;
q - 32 bit quotient;
rem - 17 bit remainder (remainder may be bigger than 16 bit, for example, when x=0x80000000 and y=0xFFFF);
counter - loop counter.
``` rem q
---------- -------------
|00|00|00| |??|??|??|??|
---------- -------------
b2 b1 b0 b3 b2 b1 b0
```
2. Shift left dividend x, and shift left remainder rem, by 1 bit, so that MSb of x was shifted to the remainder's LSb
``` rem x
---------- -------------
|r2|r1|r0| <- |x3|x2|x1|x0| <- shift left
---------- -------------
b2 b1 b0 b3 b2 b1 b0
```
3. Subtract divisor y from remainder. If subtraction successful (no borrow), next bit of q is 1. If borrow, next bit of q is 0, and remainder should be restored.
``` ----------
|r2|r1|r0| rem
----------
b2 b1 b0
-
-------
|y1|y0| y
-------
b1 b0
```
4. Shift left q and set its LSb according to subtraction result from previous step.
``` q carry(inverted borrow) flag
------------- ---
|q3|q2|q1|q0| <- |C| shift in next result bit
------------- ---
b3 b2 b1 b0
```
5. Decrement counter and if it's not zero repeat from step 2 (32 iterations)
After that, q contains quotient, rem - remainder (16 bit), x is garbage (it was shifted out completely), and y is untouched.
Now imagine that you know the result is always 16 bit (for example). That means that the first 16 iterations through steps 1-5 produce 16 zero bits in quotient (because we know that higher 2 bytes of quotient are zero in that case), x was shifted 16 times to remainder, remainder was never subtracted from. So we can say that with these 16 iterations we actually did the following:
From:
``` rem x
---------- -------------
| 0| 0| 0| <- |x3|x2|x1|x0| <- shift left
---------- -------------
b2 b1 b0 b3 b2 b1 b0
-
y q
------- -------------
|y1|y0| |??|??|??|??|
------- -------------
b1 b0 b3 b2 b1 b0
```
To:
``` rem x
---------- -------------
| 0|x3|x2| <- |x1|x0|??|??| <- shift left
---------- -------------
b2 b1 b0 b3 b2 b1 b0
- y q
------- -------------
|y1|y0| |??|??| 0| 0|
------- -------------
b1 b0 b3 b2 b1 b0
```
And after the next 16 iterations:
``` rem x
---------- -------------
| 0|r1|r0| <- |??|??|??|??| <- shift left
---------- -------------
b2 b1 b0 b3 b2 b1 b0
- y q
------- -------------
|y1|y0| | 0| 0|q1|q0|
------- -------------
b1 b0 b3 b2 b1 b0
```
This is very similar to 16 bit division:
From:
``` rem x
------- -------
| 0| 0| <- |x1|x0| <- shift left
------- -------
b1 b0 b1 b0
- y q
------- -------
|y1|y0| |??|??|
------- -------
b1 b0 b1 b0
```
To:
``` rem x
------- -------
|r1|r0| <- |??|??| <- shift left
------- -------
b1 b0 b1 b0
- y q
------- -------
|y1|y0| |q1|q0|
------- -------
b1 b0 b1 b0
```
``` rem x
------- -------
|x3|x2| <- |x1|x0| <- shift left
------- -------
b1 b0 b1 b0
- y q
------- -------
|y1|y0| | 0| 0|
------- -------
b1 b0 b1 b0
```
x3 should be less than 0x80 though, so that after first left shift the higher bit wouldn't disappear. So we actually can do 31-by-16-to-16 division with a 16-by-16-to-16 one!
### Example code
This example uses slightly different algorithm from the one described above. It does not restore the remainder immidiately after a subtraction causes a borrow. However, the trick still aplies, and because this variant of division routine has an extended remainder (necessary for the non-restoring method to hold the current remainder sign), it can do the full 32-by-16-to-16-bit division.
```; x = 60*1e6/y
;
; x, x+1 - rpm
; y, y+1 - pulse width in 1 us units
;
FindRPM
clr x
mov W, #\$87
mov x+1, W
mov W, #\$93
mov x+2, W
mov W, #\$03
mov x+3, W
jmp div32by16to16
; uint16 x = uint32 x / uint16 y
;
; Input:
; x, x+1, x+2, x+3 - 32 bit unsigned integer dividend (x - lsb, x+3 - msb)
; y, y+1 - 16 bit unsigned integer divisor
; Output:
; x, x+1 - 16 bit unsigned integer quotient
; Temporary:
; counter
; temp - remainder extension
;
; Note: result must fit in 16 bits for routine to work
; correctly
div32by16to16
jmp div16by16loopinit ;or just move the label
; uint16 x = uint16 x / uint16 y
;
; Input:
; x, x+1 - 16 bit unsigned integer dividend (x - lsb, x+1 - msb)
; y, y+1 - 16 bit unsigned integer divisor
; Output:
; x, x+1 - 16 bit unsigned integer quotient
; Temporary:
; counter
; x+2, x+3 - 16 bit remainder
; temp - remainder extension
; Size: 36 instructions
; Max timing: 6+16*(5+14+4)-2+2+3=377 cycles
div16by16
clr x+2 ;clear
clr x+3 ;remainder
div16by16loopinit
clr temp ;clear remainder extension
mov W, #16
mov counter, W
stc ;first iteration will be subtraction
div16by16loop
;shift in next result bit and shift out next
;dividend bit to remainder
rl x ;shift lsb
rl x+1 ;shift msb
rl x+2
rl x+3
rl temp
mov W, y
sb x.0
;subtract divisor from remainder
sub x+2, W
mov W, y+1
sc
movsz W, ++y+1
sub x+3, W
mov W, #1
sc
sub temp, W
jmp div16by16next
mov W, y+1
snc
movsz W, ++y+1
mov W, #1
snc
div16by16next
;carry is next result bit
decsz counter
jmp div16by16loop
;shift in last bit
rl x
rl x+1
ret
```
file: /Techref/scenix/lib/math/div/div16or32by16to16_sx.htm, 10KB, , updated: 2004/6/10 14:40, local time: 2020/9/29 20:22, owner: NG--944, TOP NEW HELP FIND: 3.237.71.23:LOG IN
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# SXMicrocontrollerMathMethod. ## 32 by 16 Divison from Nikolai Golovchenko, or How to make use of a 16 bit division routine for 32 bit division. Division routines sometimes can be used for more than they were designed for! For example, it is possible to divide 16 or 32 bit dividend by 16 bit divisor to 16 bit quotient in just one routine.. The idea is simple: you do 16 bit division as usually, but when the dividend is 32 bit, the lower two bytes are used as dividend and the higher two bytes are used as initial value for remainder (which is zero when doing 16 bit division). To understand how initializing remainder relates to dividend, let's examine the algorithms for two cases:. ``` x. q = ---. y. ```. 1) when x, y, and q are 16 bit unsigned integers, and. 2) when x and q are 32 bit unsigned integers, but y is a 16 bit unsigned integer. Note that in some cases it is known that the result of 32 bit by 16 bit division is 16 bit maximum. The trick described on this page works only for 16 bit result.. For example, when speed of a motor is calculated by a Hall sensor pulses (one pulse per revolution) the following formula is used:. ``` 60 x 1,000,000 (conversions for minutes and us,. rpm = ------------------ timer is clocked by 1 MHz clock). 16 bit timer value. ```. Timer is used to measure the Hall sensor pulse period. For this motor and timer the rpm range is within 915-20,000 rpm, which fits perfectly into 16 bits.. #### 1) 16 bit division. x - 16 bit dividend;. y - 16 bit divisor;. q - 16 bit quotient;. rem - 16 bit remainder;. counter - loop counter.. ``` rem q. ------- -------. |00|00| |??|??|. ------- -------. b1 b0 b1 b0. ```. 2. Shift left dividend x, and shift left remainder rem, by 1 bit, so that MSb of x was shifted to the remainder's LSb. ``` rem x. ------- -------. |r1|r0| <- |x1|x0| <- shift left. ------- -------. b1 b0 b1 b0. ```. 3. Subtract divisor y from remainder. If subtraction successful (no borrow), next bit of q is 1. If borrow, next bit of q is 0, and remainder should be restored.. ``` -------. |r1|r0| rem. -------. b1 b0. -. -------. |y1|y0| y. -------. b1 b0. ```. 4. Shift left q and set its LSb according to subtraction result from previous step.. ``` q carry(inverted borrow) flag. ------- ---. |q1|q0| <- |C| shift in next result bit. ------- ---. b1 b0. ```. 5. Decrement counter and if it's not zero repeat from step 2 (16 iterations). After that, q contains quotient, rem - remainder, x is garbage (it was shifted out completely), and y is untouched.. #### 2) 32 bit division with 16 bit divisor. x - 32 bit dividend;. y - 16 bit divisor;. q - 32 bit quotient;. rem - 17 bit remainder (remainder may be bigger than 16 bit, for example, when x=0x80000000 and y=0xFFFF);. counter - loop counter.. ``` rem q. ---------- -------------. |00|00|00| |??|??|??|??|. ---------- -------------. b2 b1 b0 b3 b2 b1 b0. ```. 2. Shift left dividend x, and shift left remainder rem, by 1 bit, so that MSb of x was shifted to the remainder's LSb. ``` rem x. ---------- -------------. |r2|r1|r0| <- |x3|x2|x1|x0| <- shift left. ---------- -------------. b2 b1 b0 b3 b2 b1 b0. ```. 3. Subtract divisor y from remainder. If subtraction successful (no borrow), next bit of q is 1. If borrow, next bit of q is 0, and remainder should be restored.. ``` ----------. |r2|r1|r0| rem. ----------. b2 b1 b0. -. -------. |y1|y0| y. -------. b1 b0. ```. 4. Shift left q and set its LSb according to subtraction result from previous step.. ``` q carry(inverted borrow) flag. ------------- ---. |q3|q2|q1|q0| <- |C| shift in next result bit. ------------- ---. b3 b2 b1 b0. ```. 5. Decrement counter and if it's not zero repeat from step 2 (32 iterations). After that, q contains quotient, rem - remainder (16 bit), x is garbage (it was shifted out completely), and y is untouched.. Now imagine that you know the result is always 16 bit (for example). That means that the first 16 iterations through steps 1-5 produce 16 zero bits in quotient (because we know that higher 2 bytes of quotient are zero in that case), x was shifted 16 times to remainder, remainder was never subtracted from. So we can say that with these 16 iterations we actually did the following:. From:. ``` rem x. ---------- -------------. | 0| 0| 0| <- |x3|x2|x1|x0| <- shift left. ---------- -------------. b2 b1 b0 b3 b2 b1 b0. -. y q. ------- -------------. |y1|y0| |??|??|??|??|. ------- -------------. b1 b0 b3 b2 b1 b0. ```. To:. ``` rem x. ---------- -------------. | 0|x3|x2| <- |x1|x0|??|??| <- shift left. ---------- -------------. b2 b1 b0 b3 b2 b1 b0. - y q. ------- -------------. |y1|y0| |??|??| 0| 0|. ------- -------------. b1 b0 b3 b2 b1 b0. ```. And after the next 16 iterations:. ``` rem x. ---------- -------------. | 0|r1|r0| <- |??|??|??|??| <- shift left. ---------- -------------. b2 b1 b0 b3 b2 b1 b0. - y q. ------- -------------. |y1|y0| | 0| 0|q1|q0|. ------- -------------. b1 b0 b3 b2 b1 b0. ```. This is very similar to 16 bit division:.
|
From:. ``` rem x. ------- -------. | 0| 0| <- |x1|x0| <- shift left. ------- -------. b1 b0 b1 b0. - y q. ------- -------. |y1|y0| |??|??|. ------- -------. b1 b0 b1 b0. ```. To:. ``` rem x. ------- -------. |r1|r0| <- |??|??| <- shift left. ------- -------. b1 b0 b1 b0. - y q. ------- -------. |y1|y0| |q1|q0|. ------- -------. b1 b0 b1 b0. ```. ``` rem x. ------- -------. |x3|x2| <- |x1|x0| <- shift left. ------- -------. b1 b0 b1 b0. - y q. ------- -------. |y1|y0| | 0| 0|. ------- -------. b1 b0 b1 b0. ```. x3 should be less than 0x80 though, so that after first left shift the higher bit wouldn't disappear. So we actually can do 31-by-16-to-16 division with a 16-by-16-to-16 one!. ### Example code. This example uses slightly different algorithm from the one described above. It does not restore the remainder immidiately after a subtraction causes a borrow. However, the trick still aplies, and because this variant of division routine has an extended remainder (necessary for the non-restoring method to hold the current remainder sign), it can do the full 32-by-16-to-16-bit division.. ```; x = 60*1e6/y. ;. ; x, x+1 - rpm. ; y, y+1 - pulse width in 1 us units. ;. FindRPM. clr x. mov W, #\$87. mov x+1, W. mov W, #\$93. mov x+2, W. mov W, #\$03. mov x+3, W. jmp div32by16to16. ; uint16 x = uint32 x / uint16 y. ;. ; Input:. ; x, x+1, x+2, x+3 - 32 bit unsigned integer dividend (x - lsb, x+3 - msb). ; y, y+1 - 16 bit unsigned integer divisor. ; Output:. ; x, x+1 - 16 bit unsigned integer quotient. ; Temporary:. ; counter. ; temp - remainder extension. ;. ; Note: result must fit in 16 bits for routine to work. ; correctly. div32by16to16. jmp div16by16loopinit ;or just move the label. ; uint16 x = uint16 x / uint16 y. ;. ; Input:. ; x, x+1 - 16 bit unsigned integer dividend (x - lsb, x+1 - msb). ; y, y+1 - 16 bit unsigned integer divisor. ; Output:. ; x, x+1 - 16 bit unsigned integer quotient. ; Temporary:. ; counter. ; x+2, x+3 - 16 bit remainder. ; temp - remainder extension. ; Size: 36 instructions. ; Max timing: 6+16*(5+14+4)-2+2+3=377 cycles. div16by16. clr x+2 ;clear. clr x+3 ;remainder. div16by16loopinit. clr temp ;clear remainder extension. mov W, #16. mov counter, W. stc ;first iteration will be subtraction. div16by16loop. ;shift in next result bit and shift out next. ;dividend bit to remainder. rl x ;shift lsb. rl x+1 ;shift msb. rl x+2. rl x+3. rl temp. mov W, y. sb x.0. ;subtract divisor from remainder. sub x+2, W. mov W, y+1. sc. movsz W, ++y+1. sub x+3, W. mov W, #1. sc. sub temp, W. jmp div16by16next. mov W, y+1. snc. movsz W, ++y+1. mov W, #1. snc. div16by16next. ;carry is next result bit. decsz counter. jmp div16by16loop. ;shift in last bit. rl x. rl x+1. ret. ```. file: /Techref/scenix/lib/math/div/div16or32by16to16_sx.htm, 10KB, , updated: 2004/6/10 14:40, local time: 2020/9/29 20:22, owner: NG--944, TOP NEW HELP FIND: 3.237.71.23:LOG IN. ©2020 These pages are served without commercial sponsorship. (No popup ads, etc...).Bandwidth abuse increases hosting cost forcing sponsorship or shutdown. This server aggressively defends against automated copying for any reason including offline viewing, duplication, etc... Please respect this requirement and DO NOT RIP THIS SITE. Questions?Please DO link to this page! Digg it! / MAKE! How to make use of a 16 bit division routine for 32 bit division. After you find an appropriate page, you are invited to your to this massmind site! (posts will be visible only to you before review) Just type in the box and press the Post button. (HTML welcomed, but not the <A tag: Instead, use the link box to link to another page. A tutorial is available Members can login to post directly, become page editors, and be credited for their posts.. Attn spammers: All posts are reviewed before being made visible to anyone other than the poster.. Did you find what you needed? "No. I'm looking for: " "No. Take me to the search page." "No. Take me to the top so I can drill down by catagory" "No. I'm willing to pay for help, please refer me to a qualified consultant" "No. But I'm interested. me at when this page is expanded.". ### Welcome to sxlist.com!. Site supported by. & kind contributors. just like you!. (here's why. Copies of the site on CD. are available at minimal cost.. .
|
https://www.coursehero.com/file/8941358/To-Try-5-Management-wants-an-estimate-of-the-proportion-of/
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8.3 Estimation Proportion
# To try 5 management wants an estimate of the
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Unformatted text preview: rom all the newspapers printed during a single day. In this sample, 35 contain some type of nonconforming attribute. Construct and interpret a 90% confidence interval for the proportion of newspapers printed during the day that have a nonconforming attribute. To Try… 5. Management wants an estimate of the proportion of the corporation’s employees who favour a bonus plan. From a random sample of 344 employees, it was found that 261 were in favour of this particular plan. Find a 90% confidence interval estimate of the population proportion that favours this modified bonus plan. p α=.10; zα/2 = z0.05 =1.645; Confidence interval: p ± zα 2 p( − p) 1 n = 261/344 = 0.7587; n = 344 Sample: 344 proportion pbar: 0.75872093 significance α: 0.1 confidence: 0.7587⋅ 0.2413 344 0.7587 ± 0.0379 size n: 90% standard error of the proportion: 0.023068621 0.7587 ± 1.645 Interval Estimation Sampling Distribution: 0.7208 ≤ p ≤ 0.7966 Or using MS Excel: Interval Estimate: Lower: =.7587– NORM.S.INV(1–.10/2)*SQRT(.7587*(1–.7587)/344)=0.720754373 zα/2: 1.644853627 margin of error: 0.037944505 To Try… The operations manager at a large newspaper wants to estimate the proportion of newspapers printed that have a nonconforming attribute (e.g., excessive ruboff, improper page setup, missing or duplicate pages). A random sample of 200 newspapers is selected from all the newspapers printed during a single day. In this sample, 35 contain some type of nonconforming attribute. Construct and interpret a 90% confidence interval for the proportion of newspapers printed during the day that have a nonconforming attribute. p α=.10; zα/2 = z0.05 =1.645; = 35/200 = 0.175; n = 200 6. Confidence interval: p ± zα 2 p (1− p) n Sample: size n: 200 proportion pbar: 0.175 significance α: 0.1 confidence: 90% Lower: =.175– Interval Estimate: NORM.S.INV(1–.10/2)*SQRT(.175*(1–.175)/200)=0.130806514 standard error of the proportion: 0.026867732 0.175 ± 1.645 0.175⋅0.825 200 Interval Estimation 0.175 ± 0.0442 0.1308 ≤ p ≤ 0.2192 Sampling Distribution: Or using MS Excel: Upper: zα/2: 1.644853627...
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8.3 Estimation Proportion. # To try 5 management wants an estimate of the. This preview shows page 1. Sign up to view the full content.. This is the end of the preview. Sign up to access the rest of the document.. Unformatted text preview: rom all the newspapers printed during a single day. In this sample, 35 contain some type of nonconforming attribute. Construct and interpret a 90% confidence interval for the proportion of newspapers printed during the day that have a nonconforming attribute. To Try… 5. Management wants an estimate of the proportion of the corporation’s employees who favour a bonus plan.
|
From a random sample of 344 employees, it was found that 261 were in favour of this particular plan. Find a 90% confidence interval estimate of the population proportion that favours this modified bonus plan. p α=.10; zα/2 = z0.05 =1.645; Confidence interval: p ± zα 2 p( − p) 1 n = 261/344 = 0.7587; n = 344 Sample: 344 proportion pbar: 0.75872093 significance α: 0.1 confidence: 0.7587⋅ 0.2413 344 0.7587 ± 0.0379 size n: 90% standard error of the proportion: 0.023068621 0.7587 ± 1.645 Interval Estimation Sampling Distribution: 0.7208 ≤ p ≤ 0.7966 Or using MS Excel: Interval Estimate: Lower: =.7587– NORM.S.INV(1–.10/2)*SQRT(.7587*(1–.7587)/344)=0.720754373 zα/2: 1.644853627 margin of error: 0.037944505 To Try… The operations manager at a large newspaper wants to estimate the proportion of newspapers printed that have a nonconforming attribute (e.g., excessive ruboff, improper page setup, missing or duplicate pages). A random sample of 200 newspapers is selected from all the newspapers printed during a single day. In this sample, 35 contain some type of nonconforming attribute. Construct and interpret a 90% confidence interval for the proportion of newspapers printed during the day that have a nonconforming attribute. p α=.10; zα/2 = z0.05 =1.645; = 35/200 = 0.175; n = 200 6. Confidence interval: p ± zα 2 p (1− p) n Sample: size n: 200 proportion pbar: 0.175 significance α: 0.1 confidence: 90% Lower: =.175– Interval Estimate: NORM.S.INV(1–.10/2)*SQRT(.175*(1–.175)/200)=0.130806514 standard error of the proportion: 0.026867732 0.175 ± 1.645 0.175⋅0.825 200 Interval Estimation 0.175 ± 0.0442 0.1308 ≤ p ≤ 0.2192 Sampling Distribution: Or using MS Excel: Upper: zα/2: 1.644853627.... View Full Document. Ask a homework question - tutors are online.
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# trick Adding Line Segments to XY Plots
## Recommended Posts
Sometimes when looking at an xy plot, it can be helpful to use lines to designate regions of the chart that you'd like users to focus on. In this example, we want to draw a rectangle on the xy plot showing the ideal region of operation, like below. We can do this utilizing Seeq's ability to display formulas overlaid against an xy plot.
1. For this first step, we will create a ~horizontal line on the scatter plot at y=65. This can be achieved using a y=mx+b formula with a very small slope, and a y-intercept of 65. The equation for this "horizontal" line on the xy plot is:
`0.00001*\$x+65`
2. If we want to restrict the line to only the segment making the bottom of our ideal operation box, we can leverage the within function in formula to clip the line at values we specify. Here we add to the original formula to only include values of the line between x=55 and 5=60.
```(0.00001*\$x+65)
.within(\$x>55 and \$x<60)```
3. Now let's make the left side of the box. A similar concept can be applied to create a vertical line, only a very large positive or negative slope can be used. For our "vertical" line at x=55, we can use the following formula. Note some adjustment of the y-axis scale may be required after this step.
`(-10000*(\$x-55))`
4. To clip a line into a line segment by restricting the y values, you can use the max and min functions in Formula, combined with the within function. The following formula is used to achieve the left side boundary on our box:
```(-10000*(\$x-55))
.max(65)
.min(85)
.within(\$x<55.01 and \$x>54.99) ```
The same techniques from steps 1-4 could be used to create the temperature and wet bulb max boundaries.
Formula for max temp boundary:
`(0.00001*\$x+85).within(\$x>55 and \$x<60)`
Formula for max wet bulb boundary:
```(-10000*(\$x-60))
.max(65)
.min(85)
.within(\$x<60.01 and \$x>59.99)```
CONTENT VERIFIED MAY2024
Edited by Sharlinda Salim
Update title
• 3
##### Share on other sites
• 4 weeks later...
Thanks for sharing, I'd not figured out a vertical line method. For horizontal I'd used:
`55*\$x/\$x`
##### Share on other sites
• Sharlinda Salim changed the title to Adding Line Segments to XY Plots
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# trick Adding Line Segments to XY Plots. ## Recommended Posts. Sometimes when looking at an xy plot, it can be helpful to use lines to designate regions of the chart that you'd like users to focus on. In this example, we want to draw a rectangle on the xy plot showing the ideal region of operation, like below. We can do this utilizing Seeq's ability to display formulas overlaid against an xy plot.. 1. For this first step, we will create a ~horizontal line on the scatter plot at y=65. This can be achieved using a y=mx+b formula with a very small slope, and a y-intercept of 65. The equation for this "horizontal" line on the xy plot is:. `0.00001*\$x+65`. 2. If we want to restrict the line to only the segment making the bottom of our ideal operation box, we can leverage the within function in formula to clip the line at values we specify. Here we add to the original formula to only include values of the line between x=55 and 5=60.. ```(0.00001*\$x+65). .within(\$x>55 and \$x<60)```. 3. Now let's make the left side of the box. A similar concept can be applied to create a vertical line, only a very large positive or negative slope can be used. For our "vertical" line at x=55, we can use the following formula. Note some adjustment of the y-axis scale may be required after this step.. `(-10000*(\$x-55))`. 4. To clip a line into a line segment by restricting the y values, you can use the max and min functions in Formula, combined with the within function. The following formula is used to achieve the left side boundary on our box:.
|
```(-10000*(\$x-55)). .max(65). .min(85). .within(\$x<55.01 and \$x>54.99) ```. The same techniques from steps 1-4 could be used to create the temperature and wet bulb max boundaries.. Formula for max temp boundary:. `(0.00001*\$x+85).within(\$x>55 and \$x<60)`. Formula for max wet bulb boundary:. ```(-10000*(\$x-60)). .max(65). .min(85). .within(\$x<60.01 and \$x>59.99)```. CONTENT VERIFIED MAY2024. Edited by Sharlinda Salim. Update title. • 3. ##### Share on other sites. • 4 weeks later.... Thanks for sharing, I'd not figured out a vertical line method. For horizontal I'd used:. `55*\$x/\$x`. ##### Share on other sites. • Sharlinda Salim changed the title to Adding Line Segments to XY Plots.
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Bitcoin Forum
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Author Topic: 0.30 btc bounty: maths help (statistics) (Read 12969 times)
Newbie
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June 07, 2011, 01:56:31 PM
Fact 1:
Alice chooses 6 items from the following list of products:
A1, A2, A3, A4, A5, A6
Fact 2:
Bob chooses 6 items from the following list of products:
B1, B2, B3, B4, B5, B6
Fact 3:
They are free to choose the same product more than once, so for example, Alice might choose to buy the following 6 items:
A1, A1, A2, A2, A2, A6
Fact 4:
Each 'A' product is used with a corresponding 'B' product, so for example, an A5 needs exactly one B5 to be of any use, and a B5 needs exactly one A5 to be useful (Let's call this match a 'complimentary pair').
Questions:
From Bob and Alice's 12 items, what are the chances that:
• They have exactly 1 complimentary pair
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B1, B1, B1, B1, B1)
• They have exactly 2 complimentary pairs
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B1, B1, B1, B1)
• They have exactly 3 complimentary pairs
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B1, B2, B3)
• They have exactly 4 complimentary pairs
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B4, B4, B4)
• They have exactly 5 complimentary pairs
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B4, B5, B2)
• They have exactly 6 complimentary pairs
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B4, B5, B6)
For each correct answer with detailed workings I will give 0.05 btc (~\$1) for a total of 0.30 btc (~\$6).
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kjj
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June 07, 2011, 02:22:30 PM
In question 6, for example, would lists A1,A1,A1,A1,A1,A1,B1,B1,B1,B1,B1,B1 be equivalent to A1,A2,A3,A4,A5,A6,B1,B2,B3,B4,B5,B6 for your purposes?
If so, each party will pick one of 6^6 lists, and the other person's choices don't matter, so the chances are 1 in 6^6.
Moving on to question 5, there are 6^1 ways to pick a list that matches in at least 5 places, but one of them is the answer to #6, so the chances are 1 in 6^5 minus 1/6^6.
In question 4, there are 6^2 ways to pick a list that matches in at least 4 places, but some match more, so we have to remove them again. So, 1 in 6^4 minus 1/6^5 minus 1/6^6.
Question 3, 6^3 possible, minus the extras, so: 1 in 6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.
Q2: 1 in 6^2 minus 1/6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.
Q1: 1 in 6^1 minus 1/6^2 minus 1/6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.
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lemonginger
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firstbits: 121vnq
June 07, 2011, 02:33:49 PM
Wording on this one is a little confusing. Don't let the product thing confuse you. You could just as easily say they each roll a die six times (if I am understanding the wording correctly) . and ask if they have matching numbers.
so for example your first "complementary pair" exercise would be translated from
"They have exactly 1 complimentary pair
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B1, B1, B1, B1, B1)"
to
Alice rolled [1,2,3,4,5,6] Bob rolled [1,1,1,1,1,1,]
Since all numbers have an equal chance of being rolled, we can think of the problem as A) What are the chances that Alice and Bob each rolled one 1, B) what are the chances that Alice and Bob each rolled 2 ones, C) What are the chances that Alice and Bob each rolled three ones, etc
I don't want to do your homework for ya, even for money (someone else might though!) but this might help you on your way
lemonginger
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firstbits: 121vnq
June 07, 2011, 02:34:19 PM
also why is this in development section?
Newbie
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June 07, 2011, 02:36:53 PM
In question 6, for example, would lists A1,A1,A1,A1,A1,A1,B1,B1,B1,B1,B1,B1 be equivalent to A1,A2,A3,A4,A5,A6,B1,B2,B3,B4,B5,B6 for your purposes?
yes
If so, each party will pick one of 6^6 lists, and the other person's choices don't matter, so the chances are 1 in 6^6.
sounds correct, give me a few minutes to think about it
...
Q1: 1 in 6^1 minus 1/6^2 minus 1/6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.
Can you simplify them into a % for me? I don't get how to take that statement and work out the 'chance' that it occurs.
Should P1 be equal to P6 (1 in 46656) or am I just guessing wrong on that (wild guess)?
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June 07, 2011, 02:38:48 PM
also why is this in development section?
sorry i don't know where it needs to go
AllYourBase
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June 07, 2011, 03:11:26 PM
also why is this in development section?
sorry i don't know where it needs to go
I'd post this in the market place, under the buying subforum, since you're spending bitcoins for a service.
kjj
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June 07, 2011, 05:08:12 PM
There are 6^6 possible lists for each party. The second party's list doesn't matter at all.
6^6 = 46656.
One of the lists matches all six in the second list, so that chance is 1 in 6^6 or 1 in 46656 or ~ 0.0021433%.
Six (6^1) of the lists match in at least 5 places, but one was the answer above, so 5 in 46656 or ~ 0.010716%
Thirty six (6^2) of the lists match in at least 4 places, but 5 matched in 5 places, and 1 matched in six, so 30 in 46656 or ~ 0.0643%
Two hundred sixteen (6^3) of the lists match in at least 3 places, but 36 matched more, so 180 in 46656 or ~ 0.3858%
1296 (6^4) of the lists match in at least 2 places, but 216 already matched, so 1080 in 46656 or ~ 2.314%
7776 (6^5) of the lists match in at least 1 place, but 1296 have already matched, so 6480 in 46656 or ~ 13.88889%
I hope that better explains where all the minuses came from in my first post.
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Newbie
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June 07, 2011, 05:12:26 PM
thank you kjj, should i send to your signature address or a different one?
also, for a bonus 0.05, what's the chance that there are no matches?
thanks
edit: i ran the numbers with a lot of data from random.org and after 5831 tries, i got the following ratios, which disagree quite a lot with your figures:
matches count(matches) (count(matches)/5831*100)
0 37 0.6345
1 359 6.1567
2 1222 20.9570
3 2099 35.9973
4 1676 28.7429
5 417 7.1514
6 21 0.3601
kjj
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June 07, 2011, 05:45:46 PM
The chance of no matches is what's left, 38880 out of 46656 (6^5 in 6^6) or ~ 83.333%.
Either your simulation, or your description of the problem is wrong, they don't match.
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June 07, 2011, 06:00:10 PM
The chance of no matches is what's left, 38880 out of 46656 (6^5 in 6^6) or ~ 83.333%.
Either your simulation, or your description of the problem is wrong, they don't match.
or your interpretation of my description
it seems extremely unlikely to me given the description, that 0 matches would occur 83% of the time.
here's a few example rows from the data:
[6,3,6,6,3,4] + [6,3,4,3,6,6] = 6 matches
[1,5,2,6,1,5] + [2,3,5,6,5,1] = 5 matches
[4,2,1,1,5,6] + [1,5,3,3,3,1] = 3 matches
[5,6,3,5,5,1] + [1,2,5,4,3,2] = 3 matches
[4,2,5,2,6,6] + [2,1,1,1,1,3] = 1 match
[3,5,5,4,4,5] + [1,1,6,1,6,2] = 0 matches
i can use random.org all day long but i'd much rather find out the actual mathematical formula.
edit: here's some more...
[1,5,2,6,1,5] + [2,3,5,6,5,1] = 5
[4,2,1,1,5,6] + [1,5,3,3,3,1] = 3
[5,6,3,5,5,1] + [1,2,5,4,3,2] = 3
[5,2,6,2,4,1] + [2,5,5,4,3,6] = 4
[1,2,2,5,6,2] + [1,4,3,6,5,6] = 3
[5,6,3,2,2,4] + [1,3,5,3,2,2] = 4
[2,4,3,6,4,1] + [1,5,2,2,2,6] = 3
[3,1,4,3,1,1] + [5,4,1,4,6,2] = 2
[6,3,3,1,3,3] + [4,1,5,6,1,4] = 2
[3,5,1,2,1,1] + [6,4,3,5,2,2] = 3
[5,4,6,3,2,5] + [4,4,3,3,2,1] = 3
[4,6,4,2,4,5] + [3,3,2,1,5,3] = 2
[4,2,5,2,6,6] + [2,1,1,1,1,3] = 1
[1,4,1,1,6,5] + [5,1,1,1,3,3] = 4
[2,2,6,4,4,1] + [1,6,5,3,2,6] = 3
[2,4,5,4,4,3] + [2,4,4,6,2,4] = 4
[3,3,1,5,3,3] + [2,3,2,5,5,6] = 2
[1,1,3,2,2,6] + [6,4,1,3,2,4] = 4
[5,1,2,6,2,1] + [4,6,2,3,3,6] = 2
[6,6,5,2,3,6] + [2,5,5,3,5,2] = 3
[2,1,6,4,2,4] + [4,1,2,6,5,4] = 5
[1,2,5,2,6,1] + [3,5,3,3,2,3] = 2
[3,5,5,5,5,6] + [5,6,6,6,5,1] = 3
[5,2,3,2,4,4] + [4,5,4,1,1,1] = 3
[6,1,5,2,1,4] + [3,3,5,3,3,2] = 2
[1,2,5,6,4,1] + [2,2,5,1,4,3] = 4
[6,2,6,2,3,6] + [3,6,3,1,2,3] = 3
[3,4,2,6,5,1] + [2,6,4,6,6,4] = 3
[6,6,2,2,6,4] + [2,5,5,4,4,3] = 2
[5,4,6,1,4,3] + [6,2,5,3,5,1] = 4
[3,2,4,5,2,2] + [1,5,6,6,3,4] = 3
[1,3,4,1,2,3] + [6,2,6,6,1,4] = 3
[4,1,4,5,4,3] + [5,4,4,6,1,6] = 4
[4,2,4,6,4,5] + [3,2,4,4,6,1] = 4
[6,3,4,6,2,3] + [3,5,1,2,3,4] = 4
[3,5,5,4,4,5] + [1,1,6,1,6,2] = 0
[3,5,1,4,6,1] + [2,3,2,4,3,2] = 2
[6,6,6,4,6,3] + [4,5,6,6,5,6] = 4
[3,2,3,3,6,4] + [3,2,5,6,5,5] = 3
[5,2,6,1,1,1] + [5,6,3,4,3,3] = 2
[3,1,6,6,5,4] + [1,1,1,3,4,5] = 4
[6,5,2,1,2,3] + [6,1,4,1,3,2] = 4
[5,3,4,3,4,4] + [6,6,1,2,5,2] = 1
[5,5,3,4,1,6] + [2,1,1,2,3,2] = 2
[6,5,5,5,1,6] + [4,3,3,6,5,5] = 3
[5,4,3,6,4,2] + [3,2,1,2,1,5] = 3
[2,3,5,1,1,6] + [6,4,5,6,3,2] = 4
[4,6,1,2,6,1] + [2,1,4,4,2,2] = 3
[3,3,1,3,4,1] + [1,4,6,5,4,6] = 2
[4,2,3,5,3,2] + [6,4,4,5,4,6] = 2
[5,2,4,1,6,1] + [6,1,1,2,2,4] = 5
[6,3,4,4,5,5] + [4,2,3,1,4,2] = 3
[5,1,1,1,2,1] + [5,4,6,1,6,5] = 2
[3,4,2,5,4,3] + [4,1,2,3,6,2] = 3
[5,1,5,1,6,3] + [2,1,5,5,5,6] = 4
[6,5,4,3,3,4] + [5,3,4,1,4,1] = 4
[1,3,1,6,6,3] + [1,1,1,4,4,1] = 2
[4,3,4,4,1,2] + [3,6,1,6,1,3] = 2
[5,5,1,1,6,1] + [2,1,1,4,1,2] = 3
[3,6,1,5,1,6] + [2,6,2,6,2,3] = 3
[6,3,5,6,4,5] + [3,1,3,3,1,3] = 1
[6,3,4,3,2,4] + [4,4,4,4,1,3] = 3
[4,2,3,3,5,1] + [5,2,4,6,2,3] = 4
[5,2,1,5,4,2] + [4,3,2,4,6,4] = 2
[6,4,4,3,3,4] + [2,5,4,1,4,5] = 2
[4,2,3,1,4,1] + [1,2,3,5,2,3] = 3
[1,2,4,1,3,3] + [1,6,2,2,3,6] = 3
[1,5,6,5,3,3] + [5,6,5,5,1,4] = 4
[4,5,5,6,1,5] + [4,2,6,4,2,1] = 3
[4,5,3,2,6,6] + [5,4,1,1,6,2] = 4
[1,2,2,5,4,6] + [3,6,2,5,2,1] = 5
[1,2,4,1,5,3] + [3,4,4,1,4,5] = 4
[5,2,5,5,2,1] + [6,2,1,4,3,4] = 2
[4,3,4,1,4,5] + [4,1,3,3,2,1] = 3
[3,1,2,3,2,1] + [1,2,4,1,1,1] = 3
[6,4,5,1,6,4] + [2,6,3,4,5,3] = 3
[4,1,3,5,6,2] + [5,3,1,4,5,4] = 4
[5,5,5,4,4,2] + [5,2,3,3,1,2] = 2
[4,1,5,1,2,5] + [1,5,2,4,4,4] = 4
[5,2,5,2,5,4] + [5,2,5,1,1,5] = 4
[4,3,5,3,4,3] + [1,3,3,4,5,2] = 4
[2,2,4,4,4,4] + [4,2,4,5,6,2] = 4
[4,6,5,3,2,4] + [5,4,5,5,4,6] = 4
[2,1,2,1,2,5] + [4,4,5,1,1,3] = 3
[1,4,6,6,3,6] + [6,5,4,1,4,3] = 4
[5,3,6,2,1,4] + [4,1,6,1,1,6] = 3
[3,2,3,6,3,2] + [6,5,5,4,3,4] = 2
[3,1,5,2,1,1] + [4,3,5,2,2,2] = 3
[4,5,6,2,5,3] + [2,5,4,2,1,3] = 4
[6,6,6,1,4,2] + [2,4,4,5,6,3] = 3
lemonginger
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firstbits: 121vnq
June 07, 2011, 06:52:10 PM
yes he is assuming that the numbers have to be in the exact same order. hence the statement (only 1 out of all possible lists matches the other list 6 times). Numbers can match up in any order.
Newbie
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June 07, 2011, 10:14:39 PM
yes he is assuming that the numbers have to be in the exact same order. hence the statement (only 1 out of all possible lists matches the other list 6 times). Numbers can match up in any order.
Thank you for spotting this, I couldn't figure out why his numbers were so different.
ByteCoin
Sr. Member
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Activity: 416
June 07, 2011, 10:56:31 PM
matches count(matches) (count(matches)/5831*100)
0 37 0.6345
1 359 6.1567
2 1222 20.9570
3 2099 35.9973
4 1676 28.7429
5 417 7.1514
6 21 0.3601
I have solved the problem but the method would be tedious by hand. If it's an academic problem then there's a better way than mine.
No pairs 1 pair 2 pairs 3 pairs 4 pairs 5 pairs 6 pairs
0.61% 5.82% 21.34% 36.45% 27.80% 7.58% 0.41%
kjj's method cannot succeed. If one party chooses 123456 then we must have at least one match whereas if one party chooses 111111 then there's about 1/3 chance of no matches.
This thread should probably be moved to Bitcoin Forum > Economy > Marketplace > Buying
ByteCoin
kjj
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June 07, 2011, 11:05:15 PM
Ahh, good point.
If order doesn't matter, I'm pretty sure it'll just be a binomial distribution.
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Newbie
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June 07, 2011, 11:32:30 PM
matches count(matches) (count(matches)/5831*100)
0 37 0.6345
1 359 6.1567
2 1222 20.9570
3 2099 35.9973
4 1676 28.7429
5 417 7.1514
6 21 0.3601
I have solved the problem but the method would be tedious by hand. If it's an academic problem then there's a better way than mine.
No pairs 1 pair 2 pairs 3 pairs 4 pairs 5 pairs 6 pairs
0.61% 5.82% 21.34% 36.45% 27.80% 7.58% 0.41%
kjj's method cannot succeed. If one party chooses 123456 then we must have at least one match whereas if one party chooses 111111 then there's about 1/3 chance of no matches.
This thread should probably be moved to Bitcoin Forum > Economy > Marketplace > Buying
ByteCoin
thank you bytecoin that is very close to the random data.
to pay the bounty though, i must see how you arrived at your final numbers. did you use a formula or did you write out numbers on a page 46000 times?
i suppose instead of random data i could populate my test table with exactly 1 instance of every combo (46656*46656 records?) and then count the matches, but doing that would assume 'hard-coding' of the number of product variations (6), which i'd rather avoid.
Newbie
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Activity: 10
June 07, 2011, 11:37:05 PM
it'll just be a binomial distribution.
i wish i remembered from school what that meant.
ByteCoin
Sr. Member
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Activity: 416
June 08, 2011, 02:04:27 AM
thank you bytecoin that is very close to the random data.
to pay the bounty though, i must see how you arrived at your final numbers. did you use a formula or did you write out numbers on a page 46000 times?
Essentially the latter. Do I get the money now?
doing that would assume 'hard-coding' of the number of product variations (6), which i'd rather avoid.
Is this a real world problem you're trying to solve? What is the range of the "product variations" and how much do you care (in BTC) about an exact answer?
ByteCoin
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Activity: 10
June 08, 2011, 10:58:26 AM
unfortunately, my spec hasn't really been satisfied yet:
For each correct answer with detailed workings
there's no particular real world example yet, although the dice are a good example IF the number is 6, or i suppose you could think of a 9-sided dice if you wanted to work with 9 numbers.
but this knowledge would be handy for many different purposes and i'm a bit sad that i don't remember any of this stuff from school. i'd like to try to get my head around the math again without resorting to buying some old books and studying it for months. combinations and permutations.
UniverseMan
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June 08, 2011, 06:52:58 PM
thank you kjj, should i send to your signature address or a different one?
also, for a bonus 0.05, what's the chance that there are no matches?
thanks
edit: i ran the numbers with a lot of data from random.org and after 5831 tries, i got the following ratios, which disagree quite a lot with your figures:
matches count(matches) (count(matches)/5831*100)
0 37 0.6345
1 359 6.1567
2 1222 20.9570
3 2099 35.9973
4 1676 28.7429
5 417 7.1514
6 21 0.3601
I wrote a python code to get a better random sampling that just 5831 iterations.
Code:
from __future__ import division
import random
pairs = [0]*7
iterations = 1000000
for iter in range(iterations):
list = [ [random.randint(1,6) for i in range(6)] for j in 0,1]
equalList = [ [a==b for b in list[1]] for a in list[0] ]
equalCount = 0
truePos = []
for el in equalList:
if len(truePos)>0:
for pos in truePos:
el[pos] = False
if any(el):
equalCount += 1
truePos.append( el.index(True) )
pairs[equalCount] += 1
pct = [str(round(pair/iterations*100,4)) for pair in pairs]
print pct
After running for 1 million iterations it produced the following percents:
matches count(matches)/1000000*100
0 0.6119
1 5.8219
2 21.3333
3 36.4260
4 27.7966
5 7.6010
6 0.4093
Hopefully this gives us a good check of the solutions we produce.
I'm working on the exact solution now. If you increased the bounty I'd be more inclined to work on it. It may take hours to work out; \$6 is great for a 5 minute problem, but for a 5 hr problem it is less attractive.
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Bitcoin Forum. October 17, 2017, 01:54:40 PM. News: Latest stable version of Bitcoin Core: 0.15.0.1 [Torrent]. (New!). Home Help Search Donate Login Register. Pages: [1] 2 All. Author Topic: 0.30 btc bounty: maths help (statistics) (Read 12969 times). Newbie. Offline. Activity: 10. June 07, 2011, 01:56:31 PM. Fact 1:. Alice chooses 6 items from the following list of products:. A1, A2, A3, A4, A5, A6. Fact 2:. Bob chooses 6 items from the following list of products:. B1, B2, B3, B4, B5, B6. Fact 3:. They are free to choose the same product more than once, so for example, Alice might choose to buy the following 6 items:. A1, A1, A2, A2, A2, A6. Fact 4:. Each 'A' product is used with a corresponding 'B' product, so for example, an A5 needs exactly one B5 to be of any use, and a B5 needs exactly one A5 to be useful (Let's call this match a 'complimentary pair').. Questions:. From Bob and Alice's 12 items, what are the chances that:. • They have exactly 1 complimentary pair. (eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B1, B1, B1, B1, B1). • They have exactly 2 complimentary pairs. (eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B1, B1, B1, B1). • They have exactly 3 complimentary pairs. (eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B1, B2, B3). • They have exactly 4 complimentary pairs. (eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B4, B4, B4). • They have exactly 5 complimentary pairs. (eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B4, B5, B2). • They have exactly 6 complimentary pairs. (eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B4, B5, B6). For each correct answer with detailed workings I will give 0.05 btc (~\$1) for a total of 0.30 btc (~\$6).. 1508248480. Hero Member. Offline. Posts: 1508248480. Ignore. 1508248480. 1508248480. Report to moderator. 1508248480. Hero Member. Offline. Posts: 1508248480. Ignore. 1508248480. 1508248480. Report to moderator. 1508248480. Hero Member. Offline. Posts: 1508248480. Ignore. 1508248480. 1508248480. Report to moderator. Advertised sites are not endorsed by the Bitcoin Forum. They may be unsafe, untrustworthy, or illegal in your jurisdiction. Advertise here.. 1508248480. Hero Member. Offline. Posts: 1508248480. Ignore. 1508248480. 1508248480. Report to moderator. 1508248480. Hero Member. Offline. Posts: 1508248480. Ignore. 1508248480. 1508248480. Report to moderator. 1508248480. Hero Member. Offline. Posts: 1508248480. Ignore. 1508248480. 1508248480. Report to moderator. kjj. Legendary. Offline. Activity: 1302. June 07, 2011, 02:22:30 PM. In question 6, for example, would lists A1,A1,A1,A1,A1,A1,B1,B1,B1,B1,B1,B1 be equivalent to A1,A2,A3,A4,A5,A6,B1,B2,B3,B4,B5,B6 for your purposes?. If so, each party will pick one of 6^6 lists, and the other person's choices don't matter, so the chances are 1 in 6^6.. Moving on to question 5, there are 6^1 ways to pick a list that matches in at least 5 places, but one of them is the answer to #6, so the chances are 1 in 6^5 minus 1/6^6.. In question 4, there are 6^2 ways to pick a list that matches in at least 4 places, but some match more, so we have to remove them again. So, 1 in 6^4 minus 1/6^5 minus 1/6^6.. Question 3, 6^3 possible, minus the extras, so: 1 in 6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.. Q2: 1 in 6^2 minus 1/6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.. Q1: 1 in 6^1 minus 1/6^2 minus 1/6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.. p2pcoin: a USB/CD/PXE p2pool miner - 1N8ZXx2cuMzqBYSK72X4DAy1UdDbZQNPLf - todo. I routinely ignore posters with paid advertising in their sigs. You should too.. lemonginger. Full Member. Offline. Activity: 210. firstbits: 121vnq. June 07, 2011, 02:33:49 PM. Wording on this one is a little confusing. Don't let the product thing confuse you. You could just as easily say they each roll a die six times (if I am understanding the wording correctly) . and ask if they have matching numbers.. so for example your first "complementary pair" exercise would be translated from. "They have exactly 1 complimentary pair. (eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B1, B1, B1, B1, B1)". to. Alice rolled [1,2,3,4,5,6] Bob rolled [1,1,1,1,1,1,]. Since all numbers have an equal chance of being rolled, we can think of the problem as A) What are the chances that Alice and Bob each rolled one 1, B) what are the chances that Alice and Bob each rolled 2 ones, C) What are the chances that Alice and Bob each rolled three ones, etc. I don't want to do your homework for ya, even for money (someone else might though!) but this might help you on your way. lemonginger. Full Member. Offline. Activity: 210. firstbits: 121vnq. June 07, 2011, 02:34:19 PM. also why is this in development section?. Newbie. Offline. Activity: 10. June 07, 2011, 02:36:53 PM. In question 6, for example, would lists A1,A1,A1,A1,A1,A1,B1,B1,B1,B1,B1,B1 be equivalent to A1,A2,A3,A4,A5,A6,B1,B2,B3,B4,B5,B6 for your purposes?. yes. If so, each party will pick one of 6^6 lists, and the other person's choices don't matter, so the chances are 1 in 6^6.. sounds correct, give me a few minutes to think about it. .... Q1: 1 in 6^1 minus 1/6^2 minus 1/6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.. Can you simplify them into a % for me? I don't get how to take that statement and work out the 'chance' that it occurs.. Should P1 be equal to P6 (1 in 46656) or am I just guessing wrong on that (wild guess)?. Newbie. Offline. Activity: 10. June 07, 2011, 02:38:48 PM. also why is this in development section?. sorry i don't know where it needs to go. AllYourBase. Full Member. Offline. Activity: 138. June 07, 2011, 03:11:26 PM. also why is this in development section?. sorry i don't know where it needs to go. I'd post this in the market place, under the buying subforum, since you're spending bitcoins for a service.. kjj. Legendary. Offline. Activity: 1302. June 07, 2011, 05:08:12 PM. There are 6^6 possible lists for each party. The second party's list doesn't matter at all.. 6^6 = 46656.. One of the lists matches all six in the second list, so that chance is 1 in 6^6 or 1 in 46656 or ~ 0.0021433%.. Six (6^1) of the lists match in at least 5 places, but one was the answer above, so 5 in 46656 or ~ 0.010716%. Thirty six (6^2) of the lists match in at least 4 places, but 5 matched in 5 places, and 1 matched in six, so 30 in 46656 or ~ 0.0643%. Two hundred sixteen (6^3) of the lists match in at least 3 places, but 36 matched more, so 180 in 46656 or ~ 0.3858%. 1296 (6^4) of the lists match in at least 2 places, but 216 already matched, so 1080 in 46656 or ~ 2.314%. 7776 (6^5) of the lists match in at least 1 place, but 1296 have already matched, so 6480 in 46656 or ~ 13.88889%. I hope that better explains where all the minuses came from in my first post.. p2pcoin: a USB/CD/PXE p2pool miner - 1N8ZXx2cuMzqBYSK72X4DAy1UdDbZQNPLf - todo. I routinely ignore posters with paid advertising in their sigs. You should too.. Newbie. Offline. Activity: 10. June 07, 2011, 05:12:26 PM. thank you kjj, should i send to your signature address or a different one?. also, for a bonus 0.05, what's the chance that there are no matches?. thanks. edit: i ran the numbers with a lot of data from random.org and after 5831 tries, i got the following ratios, which disagree quite a lot with your figures:. matches count(matches) (count(matches)/5831*100). 0 37 0.6345. 1 359 6.1567. 2 1222 20.9570. 3 2099 35.9973. 4 1676 28.7429. 5 417 7.1514. 6 21 0.3601. kjj. Legendary. Offline. Activity: 1302. June 07, 2011, 05:45:46 PM. The chance of no matches is what's left, 38880 out of 46656 (6^5 in 6^6) or ~ 83.333%.. Either your simulation, or your description of the problem is wrong, they don't match.. p2pcoin: a USB/CD/PXE p2pool miner - 1N8ZXx2cuMzqBYSK72X4DAy1UdDbZQNPLf - todo. I routinely ignore posters with paid advertising in their sigs. You should too.. Newbie. Offline. Activity: 10. June 07, 2011, 06:00:10 PM. The chance of no matches is what's left, 38880 out of 46656 (6^5 in 6^6) or ~ 83.333%.. Either your simulation, or your description of the problem is wrong, they don't match.. or your interpretation of my description. it seems extremely unlikely to me given the description, that 0 matches would occur 83% of the time.. here's a few example rows from the data:. [6,3,6,6,3,4] + [6,3,4,3,6,6] = 6 matches. [1,5,2,6,1,5] + [2,3,5,6,5,1] = 5 matches. [4,2,1,1,5,6] + [1,5,3,3,3,1] = 3 matches. [5,6,3,5,5,1] + [1,2,5,4,3,2] = 3 matches. [4,2,5,2,6,6] + [2,1,1,1,1,3] = 1 match. [3,5,5,4,4,5] + [1,1,6,1,6,2] = 0 matches. i can use random.org all day long but i'd much rather find out the actual mathematical formula.. edit: here's some more.... [1,5,2,6,1,5] + [2,3,5,6,5,1] = 5. [4,2,1,1,5,6] + [1,5,3,3,3,1] = 3. [5,6,3,5,5,1] + [1,2,5,4,3,2] = 3. [5,2,6,2,4,1] + [2,5,5,4,3,6] = 4. [1,2,2,5,6,2] + [1,4,3,6,5,6] = 3. [5,6,3,2,2,4] + [1,3,5,3,2,2] = 4. [2,4,3,6,4,1] + [1,5,2,2,2,6] = 3. [3,1,4,3,1,1] + [5,4,1,4,6,2] = 2. [6,3,3,1,3,3] + [4,1,5,6,1,4] = 2. [3,5,1,2,1,1] + [6,4,3,5,2,2] = 3. [5,4,6,3,2,5] + [4,4,3,3,2,1] = 3. [4,6,4,2,4,5] + [3,3,2,1,5,3] = 2. [4,2,5,2,6,6] + [2,1,1,1,1,3] = 1. [1,4,1,1,6,5] + [5,1,1,1,3,3] = 4. [2,2,6,4,4,1] + [1,6,5,3,2,6] = 3. [2,4,5,4,4,3] + [2,4,4,6,2,4] = 4. [3,3,1,5,3,3] + [2,3,2,5,5,6] = 2. [1,1,3,2,2,6] + [6,4,1,3,2,4] = 4.
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[5,1,2,6,2,1] + [4,6,2,3,3,6] = 2. [6,6,5,2,3,6] + [2,5,5,3,5,2] = 3. [2,1,6,4,2,4] + [4,1,2,6,5,4] = 5. [1,2,5,2,6,1] + [3,5,3,3,2,3] = 2. [3,5,5,5,5,6] + [5,6,6,6,5,1] = 3. [5,2,3,2,4,4] + [4,5,4,1,1,1] = 3. [6,1,5,2,1,4] + [3,3,5,3,3,2] = 2. [1,2,5,6,4,1] + [2,2,5,1,4,3] = 4. [6,2,6,2,3,6] + [3,6,3,1,2,3] = 3. [3,4,2,6,5,1] + [2,6,4,6,6,4] = 3. [6,6,2,2,6,4] + [2,5,5,4,4,3] = 2. [5,4,6,1,4,3] + [6,2,5,3,5,1] = 4. [3,2,4,5,2,2] + [1,5,6,6,3,4] = 3. [1,3,4,1,2,3] + [6,2,6,6,1,4] = 3. [4,1,4,5,4,3] + [5,4,4,6,1,6] = 4. [4,2,4,6,4,5] + [3,2,4,4,6,1] = 4. [6,3,4,6,2,3] + [3,5,1,2,3,4] = 4. [3,5,5,4,4,5] + [1,1,6,1,6,2] = 0. [3,5,1,4,6,1] + [2,3,2,4,3,2] = 2. [6,6,6,4,6,3] + [4,5,6,6,5,6] = 4. [3,2,3,3,6,4] + [3,2,5,6,5,5] = 3. [5,2,6,1,1,1] + [5,6,3,4,3,3] = 2. [3,1,6,6,5,4] + [1,1,1,3,4,5] = 4. [6,5,2,1,2,3] + [6,1,4,1,3,2] = 4. [5,3,4,3,4,4] + [6,6,1,2,5,2] = 1. [5,5,3,4,1,6] + [2,1,1,2,3,2] = 2. [6,5,5,5,1,6] + [4,3,3,6,5,5] = 3. [5,4,3,6,4,2] + [3,2,1,2,1,5] = 3. [2,3,5,1,1,6] + [6,4,5,6,3,2] = 4. [4,6,1,2,6,1] + [2,1,4,4,2,2] = 3. [3,3,1,3,4,1] + [1,4,6,5,4,6] = 2. [4,2,3,5,3,2] + [6,4,4,5,4,6] = 2. [5,2,4,1,6,1] + [6,1,1,2,2,4] = 5. [6,3,4,4,5,5] + [4,2,3,1,4,2] = 3. [5,1,1,1,2,1] + [5,4,6,1,6,5] = 2. [3,4,2,5,4,3] + [4,1,2,3,6,2] = 3. [5,1,5,1,6,3] + [2,1,5,5,5,6] = 4. [6,5,4,3,3,4] + [5,3,4,1,4,1] = 4. [1,3,1,6,6,3] + [1,1,1,4,4,1] = 2. [4,3,4,4,1,2] + [3,6,1,6,1,3] = 2. [5,5,1,1,6,1] + [2,1,1,4,1,2] = 3. [3,6,1,5,1,6] + [2,6,2,6,2,3] = 3. [6,3,5,6,4,5] + [3,1,3,3,1,3] = 1. [6,3,4,3,2,4] + [4,4,4,4,1,3] = 3. [4,2,3,3,5,1] + [5,2,4,6,2,3] = 4. [5,2,1,5,4,2] + [4,3,2,4,6,4] = 2. [6,4,4,3,3,4] + [2,5,4,1,4,5] = 2. [4,2,3,1,4,1] + [1,2,3,5,2,3] = 3. [1,2,4,1,3,3] + [1,6,2,2,3,6] = 3. [1,5,6,5,3,3] + [5,6,5,5,1,4] = 4. [4,5,5,6,1,5] + [4,2,6,4,2,1] = 3. [4,5,3,2,6,6] + [5,4,1,1,6,2] = 4. [1,2,2,5,4,6] + [3,6,2,5,2,1] = 5. [1,2,4,1,5,3] + [3,4,4,1,4,5] = 4. [5,2,5,5,2,1] + [6,2,1,4,3,4] = 2. [4,3,4,1,4,5] + [4,1,3,3,2,1] = 3. [3,1,2,3,2,1] + [1,2,4,1,1,1] = 3. [6,4,5,1,6,4] + [2,6,3,4,5,3] = 3. [4,1,3,5,6,2] + [5,3,1,4,5,4] = 4. [5,5,5,4,4,2] + [5,2,3,3,1,2] = 2. [4,1,5,1,2,5] + [1,5,2,4,4,4] = 4. [5,2,5,2,5,4] + [5,2,5,1,1,5] = 4. [4,3,5,3,4,3] + [1,3,3,4,5,2] = 4. [2,2,4,4,4,4] + [4,2,4,5,6,2] = 4. [4,6,5,3,2,4] + [5,4,5,5,4,6] = 4. [2,1,2,1,2,5] + [4,4,5,1,1,3] = 3. [1,4,6,6,3,6] + [6,5,4,1,4,3] = 4. [5,3,6,2,1,4] + [4,1,6,1,1,6] = 3. [3,2,3,6,3,2] + [6,5,5,4,3,4] = 2. [3,1,5,2,1,1] + [4,3,5,2,2,2] = 3. [4,5,6,2,5,3] + [2,5,4,2,1,3] = 4. [6,6,6,1,4,2] + [2,4,4,5,6,3] = 3. lemonginger. Full Member. Offline. Activity: 210. firstbits: 121vnq. June 07, 2011, 06:52:10 PM. yes he is assuming that the numbers have to be in the exact same order. hence the statement (only 1 out of all possible lists matches the other list 6 times). Numbers can match up in any order.. Newbie. Offline. Activity: 10. June 07, 2011, 10:14:39 PM. yes he is assuming that the numbers have to be in the exact same order. hence the statement (only 1 out of all possible lists matches the other list 6 times). Numbers can match up in any order.. Thank you for spotting this, I couldn't figure out why his numbers were so different.. ByteCoin. Sr. Member. Offline. Activity: 416. June 07, 2011, 10:56:31 PM. matches count(matches) (count(matches)/5831*100). 0 37 0.6345. 1 359 6.1567. 2 1222 20.9570. 3 2099 35.9973. 4 1676 28.7429. 5 417 7.1514. 6 21 0.3601. I have solved the problem but the method would be tedious by hand. If it's an academic problem then there's a better way than mine.. No pairs 1 pair 2 pairs 3 pairs 4 pairs 5 pairs 6 pairs. 0.61% 5.82% 21.34% 36.45% 27.80% 7.58% 0.41%. kjj's method cannot succeed. If one party chooses 123456 then we must have at least one match whereas if one party chooses 111111 then there's about 1/3 chance of no matches.. This thread should probably be moved to Bitcoin Forum > Economy > Marketplace > Buying. ByteCoin. kjj. Legendary. Offline. Activity: 1302. June 07, 2011, 11:05:15 PM. Ahh, good point.. If order doesn't matter, I'm pretty sure it'll just be a binomial distribution.. p2pcoin: a USB/CD/PXE p2pool miner - 1N8ZXx2cuMzqBYSK72X4DAy1UdDbZQNPLf - todo. I routinely ignore posters with paid advertising in their sigs. You should too.. Newbie. Offline. Activity: 10. June 07, 2011, 11:32:30 PM. matches count(matches) (count(matches)/5831*100). 0 37 0.6345. 1 359 6.1567. 2 1222 20.9570. 3 2099 35.9973. 4 1676 28.7429. 5 417 7.1514. 6 21 0.3601. I have solved the problem but the method would be tedious by hand. If it's an academic problem then there's a better way than mine.. No pairs 1 pair 2 pairs 3 pairs 4 pairs 5 pairs 6 pairs. 0.61% 5.82% 21.34% 36.45% 27.80% 7.58% 0.41%. kjj's method cannot succeed. If one party chooses 123456 then we must have at least one match whereas if one party chooses 111111 then there's about 1/3 chance of no matches.. This thread should probably be moved to Bitcoin Forum > Economy > Marketplace > Buying. ByteCoin. thank you bytecoin that is very close to the random data.. to pay the bounty though, i must see how you arrived at your final numbers. did you use a formula or did you write out numbers on a page 46000 times?. i suppose instead of random data i could populate my test table with exactly 1 instance of every combo (46656*46656 records?) and then count the matches, but doing that would assume 'hard-coding' of the number of product variations (6), which i'd rather avoid.. Newbie. Offline. Activity: 10. June 07, 2011, 11:37:05 PM. it'll just be a binomial distribution.. i wish i remembered from school what that meant.. ByteCoin. Sr. Member. Offline. Activity: 416. June 08, 2011, 02:04:27 AM. thank you bytecoin that is very close to the random data.. to pay the bounty though, i must see how you arrived at your final numbers. did you use a formula or did you write out numbers on a page 46000 times?. Essentially the latter. Do I get the money now?. doing that would assume 'hard-coding' of the number of product variations (6), which i'd rather avoid.. Is this a real world problem you're trying to solve? What is the range of the "product variations" and how much do you care (in BTC) about an exact answer?. ByteCoin. Newbie. Offline. Activity: 10. June 08, 2011, 10:58:26 AM. unfortunately, my spec hasn't really been satisfied yet:. For each correct answer with detailed workings. there's no particular real world example yet, although the dice are a good example IF the number is 6, or i suppose you could think of a 9-sided dice if you wanted to work with 9 numbers.. but this knowledge would be handy for many different purposes and i'm a bit sad that i don't remember any of this stuff from school. i'd like to try to get my head around the math again without resorting to buying some old books and studying it for months. combinations and permutations.. UniverseMan. Newbie. Offline. Activity: 26. June 08, 2011, 06:52:58 PM. thank you kjj, should i send to your signature address or a different one?. also, for a bonus 0.05, what's the chance that there are no matches?. thanks. edit: i ran the numbers with a lot of data from random.org and after 5831 tries, i got the following ratios, which disagree quite a lot with your figures:. matches count(matches) (count(matches)/5831*100). 0 37 0.6345. 1 359 6.1567. 2 1222 20.9570. 3 2099 35.9973. 4 1676 28.7429. 5 417 7.1514. 6 21 0.3601. I wrote a python code to get a better random sampling that just 5831 iterations.. Code:. from __future__ import division. import random. pairs = [0]*7. iterations = 1000000. for iter in range(iterations):. list = [ [random.randint(1,6) for i in range(6)] for j in 0,1]. equalList = [ [a==b for b in list[1]] for a in list[0] ]. equalCount = 0. truePos = []. for el in equalList:. if len(truePos)>0:. for pos in truePos:. el[pos] = False. if any(el):. equalCount += 1. truePos.append( el.index(True) ). pairs[equalCount] += 1. pct = [str(round(pair/iterations*100,4)) for pair in pairs]. print pct. After running for 1 million iterations it produced the following percents:. matches count(matches)/1000000*100. 0 0.6119. 1 5.8219. 2 21.3333. 3 36.4260. 4 27.7966. 5 7.6010. 6 0.4093. Hopefully this gives us a good check of the solutions we produce.. I'm working on the exact solution now. If you increased the bounty I'd be more inclined to work on it. It may take hours to work out; \$6 is great for a 5 minute problem, but for a 5 hr problem it is less attractive.. Pages: [1] 2 All.
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## Conversion formula
The conversion factor from hours to weeks is 0.005952380952381, which means that 1 hour is equal to 0.005952380952381 weeks:
1 hr = 0.005952380952381 wk
To convert 249 hours into weeks we have to multiply 249 by the conversion factor in order to get the time amount from hours to weeks. We can also form a simple proportion to calculate the result:
1 hr → 0.005952380952381 wk
249 hr → T(wk)
Solve the above proportion to obtain the time T in weeks:
T(wk) = 249 hr × 0.005952380952381 wk
T(wk) = 1.4821428571429 wk
The final result is:
249 hr → 1.4821428571429 wk
We conclude that 249 hours is equivalent to 1.4821428571429 weeks:
249 hours = 1.4821428571429 weeks
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 week is equal to 0.67469879518072 × 249 hours.
Another way is saying that 249 hours is equal to 1 ÷ 0.67469879518072 weeks.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred forty-nine hours is approximately one point four eight two weeks:
249 hr ≅ 1.482 wk
An alternative is also that one week is approximately zero point six seven five times two hundred forty-nine hours.
## Conversion table
### hours to weeks chart
For quick reference purposes, below is the conversion table you can use to convert from hours to weeks
hours (hr) weeks (wk)
250 hours 1.488 weeks
251 hours 1.494 weeks
252 hours 1.5 weeks
253 hours 1.506 weeks
254 hours 1.512 weeks
255 hours 1.518 weeks
256 hours 1.524 weeks
257 hours 1.53 weeks
258 hours 1.536 weeks
259 hours 1.542 weeks
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## Conversion formula. The conversion factor from hours to weeks is 0.005952380952381, which means that 1 hour is equal to 0.005952380952381 weeks:. 1 hr = 0.005952380952381 wk. To convert 249 hours into weeks we have to multiply 249 by the conversion factor in order to get the time amount from hours to weeks. We can also form a simple proportion to calculate the result:. 1 hr → 0.005952380952381 wk. 249 hr → T(wk). Solve the above proportion to obtain the time T in weeks:. T(wk) = 249 hr × 0.005952380952381 wk. T(wk) = 1.4821428571429 wk. The final result is:. 249 hr → 1.4821428571429 wk. We conclude that 249 hours is equivalent to 1.4821428571429 weeks:. 249 hours = 1.4821428571429 weeks. ## Alternative conversion. We can also convert by utilizing the inverse value of the conversion factor. In this case 1 week is equal to 0.67469879518072 × 249 hours.. Another way is saying that 249 hours is equal to 1 ÷ 0.67469879518072 weeks.. ## Approximate result.
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For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred forty-nine hours is approximately one point four eight two weeks:. 249 hr ≅ 1.482 wk. An alternative is also that one week is approximately zero point six seven five times two hundred forty-nine hours.. ## Conversion table. ### hours to weeks chart. For quick reference purposes, below is the conversion table you can use to convert from hours to weeks. hours (hr) weeks (wk). 250 hours 1.488 weeks. 251 hours 1.494 weeks. 252 hours 1.5 weeks. 253 hours 1.506 weeks. 254 hours 1.512 weeks. 255 hours 1.518 weeks. 256 hours 1.524 weeks. 257 hours 1.53 weeks. 258 hours 1.536 weeks. 259 hours 1.542 weeks.
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# 17250 in Words
17250 in words is Seventeen Thousand Two Hundred Fifty. For example, if you have 17250 rupees in your account, then you can write, “I have Seventeen Thousand Two Hundred Fifty rupees in my account”. We know that the number name of any number can be written using the ones, tens, hundreds and thousands place of a number. Therefore, the place value chart is essential to write 17250 in words. 17250 is a cardinal number as it shows a specific quantity.
17250 in words Seventeen Thousand Two Hundred Fifty Seventeen Thousand Two Hundred Fifty in Numbers 17250
## How to Write 17250 in Words?
We can convert 17250 to words using a place value chart. This can be done as follows. The number 17250 has 5 digits, so let’s make a chart that shows the place value up to 5 digits.
Ten thousands Thousands Hundreds Tens Ones 1 7 2 5 0
Thus, we can write the expanded form as:
1 × Ten thousand + 7 × Thousand + 2 × Hundred + 5 × Ten + 0 × One
= 1 × 10000 + 7 × 1000 + 2 × 100 + 5 × 10 + 0 × 1
= 17250
= Seventeen Thousand Two Hundred Fifty
17250 is the natural number that is succeeded by 17249 and preceded by 17251.
17250 in words – Seventeen Thousand Two Hundred Fifty
Is 17250 an odd number? – No.
Is 17250 an even number? – Yes
Is 17250 a perfect square number? – No
Is 17250 a perfect cube number? – No
Is 17250 a prime number? – No
Is 17250 a composite number? – Yes
## Frequently Asked Questions on 17250 in Words
Q1
### How to write 17250 in words?
17250 in English is written as “Seventeen Thousand Two Hundred Fifty”.
Q2
### Is the number 17250 an odd number?
No, the number 17250 is not an odd number.
Q3
### Is 17250 a composite number?
Yes, 17250 is a composite number.
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# 17250 in Words. 17250 in words is Seventeen Thousand Two Hundred Fifty. For example, if you have 17250 rupees in your account, then you can write, “I have Seventeen Thousand Two Hundred Fifty rupees in my account”. We know that the number name of any number can be written using the ones, tens, hundreds and thousands place of a number. Therefore, the place value chart is essential to write 17250 in words. 17250 is a cardinal number as it shows a specific quantity.. 17250 in words Seventeen Thousand Two Hundred Fifty Seventeen Thousand Two Hundred Fifty in Numbers 17250. ## How to Write 17250 in Words?. We can convert 17250 to words using a place value chart. This can be done as follows. The number 17250 has 5 digits, so let’s make a chart that shows the place value up to 5 digits.. Ten thousands Thousands Hundreds Tens Ones 1 7 2 5 0. Thus, we can write the expanded form as:. 1 × Ten thousand + 7 × Thousand + 2 × Hundred + 5 × Ten + 0 × One. = 1 × 10000 + 7 × 1000 + 2 × 100 + 5 × 10 + 0 × 1. = 17250. = Seventeen Thousand Two Hundred Fifty. 17250 is the natural number that is succeeded by 17249 and preceded by 17251.. 17250 in words – Seventeen Thousand Two Hundred Fifty.
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Is 17250 an odd number? – No.. Is 17250 an even number? – Yes. Is 17250 a perfect square number? – No. Is 17250 a perfect cube number? – No. Is 17250 a prime number? – No. Is 17250 a composite number? – Yes. ## Frequently Asked Questions on 17250 in Words. Q1. ### How to write 17250 in words?. 17250 in English is written as “Seventeen Thousand Two Hundred Fifty”.. Q2. ### Is the number 17250 an odd number?. No, the number 17250 is not an odd number.. Q3. ### Is 17250 a composite number?. Yes, 17250 is a composite number.
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LeftDivision - Maple Help
MatrixPolynomialAlgebra
LeftDivision
compute a left quotient and remainder of 2 matrices of polynomials
RightDivision
compute a right quotient and remainder of 2 matrices of polynomials
Calling Sequence LeftDivision(A, B, x) RightDivision(A, B, x)
Parameters
A - Matrix of polynomials B - Matrix of polynomials x - variable name of the polynomial domain
Description
• The LeftDivision(A, B, x) command computes a left quotient Q and a remainder R such that $A=B·Q+R$ where ${B}^{\left(-1\right)}·R$ is strictly proper. That is, $\underset{z\to \mathrm{\infty }}{lim}{B}^{\left(-1\right)}\left(z\right).R\left(z\right)$ is a zero matrix. The input matrices must have the same number of rows, and B must be a square nonsingular matrix of polynomials.
• The RightDivision(A, B, x) command computes a right quotient Q and a remainder R such that $A=Q·B+R$ where $R·{B}^{\left(-1\right)}$ is strictly proper. That is, $\underset{z\to \mathrm{\infty }}{lim}R\left(z\right).{B}^{\left(-1\right)}\left(z\right)$ is a zero matrix. The input matrices must have the same number of columns, and B must be a square nonsingular matrix of polynomials.
• The quotient $Q$ and the remainder $R$ are returned in a list.
Examples
> $\mathrm{with}\left(\mathrm{MatrixPolynomialAlgebra}\right):$
> $A≔\mathrm{Matrix}\left(2,2,\left[\left[-9{z}^{2}-3z+1,12{z}^{2}+10z\right],\left[-3{z}^{3}+2{z}^{2}-z,4{z}^{3}+2z-2{z}^{2}\right]\right]\right):$
> $B≔\mathrm{Matrix}\left(2,2,\left[\left[-3{z}^{3}+6{z}^{2}+5z+1,-12{z}^{2}-13z\right],\left[{z}^{4}+{z}^{3}+{z}^{2},-4{z}^{3}-3z+3{z}^{2}\right]\right]\right):$
> $Q,R≔\mathrm{op}\left(\mathrm{LeftDivision}\left(A,B,z\right)\right)$
${Q}{,}{R}{≔}\left[\begin{array}{cc}{0}& {0}\\ \frac{{3}}{{4}}& {-1}\end{array}\right]{,}\left[\begin{array}{cc}\frac{{27}{}{z}}{{4}}{+}{1}& {-}{3}{}{z}\\ {-}\frac{{1}}{{4}}{}{{z}}^{{2}}{+}\frac{{5}}{{4}}{}{z}& {{z}}^{{2}}{-}{z}\end{array}\right]$ (1)
> $\mathrm{map}\left(\mathrm{expand},A-\left(B·Q+R\right)\right)$
$\left[\begin{array}{cc}{0}& {0}\\ {0}& {0}\end{array}\right]$ (2)
> $\mathrm{map}\left(f↦\mathrm{limit}\left(f,z=\mathrm{\infty }\right),\mathrm{LinearAlgebra}:-\mathrm{MatrixInverse}\left(B\right)·R\right)$
$\left[\begin{array}{cc}{0}& {0}\\ {0}& {0}\end{array}\right]$ (3)
> $Q,R≔\mathrm{op}\left(\mathrm{RightDivision}\left(A,B,z\right)\right)$
${Q}{,}{R}{≔}\left[\begin{array}{cc}{-}\frac{{1}}{{2}}& {0}\\ {-}\frac{{z}}{{6}}{+}\frac{{13}}{{36}}& {-}\frac{{1}}{{2}}\end{array}\right]{,}\left[\begin{array}{cc}{-}{6}{}{{z}}^{{2}}{-}\frac{{1}}{{2}}{}{z}{+}\frac{{3}}{{2}}{-}\frac{{3}}{{2}}{}{{z}}^{{3}}& {6}{}{{z}}^{{2}}{+}\frac{{7}}{{2}}{}{z}\\ {-}\frac{{5}}{{12}}{}{{z}}^{{3}}{+}\frac{{7}}{{6}}{}{{z}}^{{2}}{-}\frac{{95}}{{36}}{}{z}{-}\frac{{13}}{{36}}& \frac{{5}}{{3}}{}{{z}}^{{2}}{+}\frac{{187}}{{36}}{}{z}\end{array}\right]$ (4)
> $\mathrm{map}\left(\mathrm{expand},A-\left(Q·B+R\right)\right)$
$\left[\begin{array}{cc}{0}& {0}\\ {0}& {0}\end{array}\right]$ (5)
> $\mathrm{map}\left(f↦\mathrm{limit}\left(f,z=\mathrm{\infty }\right),R·\mathrm{LinearAlgebra}:-\mathrm{MatrixInverse}\left(B\right)\right)$
$\left[\begin{array}{cc}{0}& {0}\\ {0}& {0}\end{array}\right]$ (6)
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LeftDivision - Maple Help. MatrixPolynomialAlgebra. LeftDivision. compute a left quotient and remainder of 2 matrices of polynomials. RightDivision. compute a right quotient and remainder of 2 matrices of polynomials. Calling Sequence LeftDivision(A, B, x) RightDivision(A, B, x). Parameters. A - Matrix of polynomials B - Matrix of polynomials x - variable name of the polynomial domain. Description. • The LeftDivision(A, B, x) command computes a left quotient Q and a remainder R such that $A=B·Q+R$ where ${B}^{\left(-1\right)}·R$ is strictly proper. That is, $\underset{z\to \mathrm{\infty }}{lim}{B}^{\left(-1\right)}\left(z\right).R\left(z\right)$ is a zero matrix. The input matrices must have the same number of rows, and B must be a square nonsingular matrix of polynomials.. • The RightDivision(A, B, x) command computes a right quotient Q and a remainder R such that $A=Q·B+R$ where $R·{B}^{\left(-1\right)}$ is strictly proper. That is, $\underset{z\to \mathrm{\infty }}{lim}R\left(z\right).{B}^{\left(-1\right)}\left(z\right)$ is a zero matrix. The input matrices must have the same number of columns, and B must be a square nonsingular matrix of polynomials.. • The quotient $Q$ and the remainder $R$ are returned in a list.
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Examples. > $\mathrm{with}\left(\mathrm{MatrixPolynomialAlgebra}\right):$. > $A≔\mathrm{Matrix}\left(2,2,\left[\left[-9{z}^{2}-3z+1,12{z}^{2}+10z\right],\left[-3{z}^{3}+2{z}^{2}-z,4{z}^{3}+2z-2{z}^{2}\right]\right]\right):$. > $B≔\mathrm{Matrix}\left(2,2,\left[\left[-3{z}^{3}+6{z}^{2}+5z+1,-12{z}^{2}-13z\right],\left[{z}^{4}+{z}^{3}+{z}^{2},-4{z}^{3}-3z+3{z}^{2}\right]\right]\right):$. > $Q,R≔\mathrm{op}\left(\mathrm{LeftDivision}\left(A,B,z\right)\right)$. ${Q}{,}{R}{≔}\left[\begin{array}{cc}{0}& {0}\\ \frac{{3}}{{4}}& {-1}\end{array}\right]{,}\left[\begin{array}{cc}\frac{{27}{}{z}}{{4}}{+}{1}& {-}{3}{}{z}\\ {-}\frac{{1}}{{4}}{}{{z}}^{{2}}{+}\frac{{5}}{{4}}{}{z}& {{z}}^{{2}}{-}{z}\end{array}\right]$ (1). > $\mathrm{map}\left(\mathrm{expand},A-\left(B·Q+R\right)\right)$. $\left[\begin{array}{cc}{0}& {0}\\ {0}& {0}\end{array}\right]$ (2). > $\mathrm{map}\left(f↦\mathrm{limit}\left(f,z=\mathrm{\infty }\right),\mathrm{LinearAlgebra}:-\mathrm{MatrixInverse}\left(B\right)·R\right)$. $\left[\begin{array}{cc}{0}& {0}\\ {0}& {0}\end{array}\right]$ (3). > $Q,R≔\mathrm{op}\left(\mathrm{RightDivision}\left(A,B,z\right)\right)$. ${Q}{,}{R}{≔}\left[\begin{array}{cc}{-}\frac{{1}}{{2}}& {0}\\ {-}\frac{{z}}{{6}}{+}\frac{{13}}{{36}}& {-}\frac{{1}}{{2}}\end{array}\right]{,}\left[\begin{array}{cc}{-}{6}{}{{z}}^{{2}}{-}\frac{{1}}{{2}}{}{z}{+}\frac{{3}}{{2}}{-}\frac{{3}}{{2}}{}{{z}}^{{3}}& {6}{}{{z}}^{{2}}{+}\frac{{7}}{{2}}{}{z}\\ {-}\frac{{5}}{{12}}{}{{z}}^{{3}}{+}\frac{{7}}{{6}}{}{{z}}^{{2}}{-}\frac{{95}}{{36}}{}{z}{-}\frac{{13}}{{36}}& \frac{{5}}{{3}}{}{{z}}^{{2}}{+}\frac{{187}}{{36}}{}{z}\end{array}\right]$ (4). > $\mathrm{map}\left(\mathrm{expand},A-\left(Q·B+R\right)\right)$. $\left[\begin{array}{cc}{0}& {0}\\ {0}& {0}\end{array}\right]$ (5). > $\mathrm{map}\left(f↦\mathrm{limit}\left(f,z=\mathrm{\infty }\right),R·\mathrm{LinearAlgebra}:-\mathrm{MatrixInverse}\left(B\right)\right)$. $\left[\begin{array}{cc}{0}& {0}\\ {0}& {0}\end{array}\right]$ (6).
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# In a room with 7 people, 4 people have exactly 1 friend in
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In a room with 7 people, 4 people have exactly 1 friend in [#permalink]
### Show Tags
08 Dec 2007, 13:00
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Question Stats:
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In a room with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21
OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html
[Reveal] Spoiler: OA
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### Show Tags
08 Dec 2007, 13:09
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(4/7)(5/6) + (3/7)(4/6) if you choose one of the 4 with one other friend, then you have a 5/6 chance of not picking their friend 2nd. If you choose one of the 3 with 2 friends, you have a 4/6 chance of not picking one of their friends second. Add them up.
20/42 + 12/42
32/42 = 16/21
E. 16/21
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### Show Tags
18 Jan 2008, 08:00
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marcodonzelli wrote:
right....let's solve it with combs:
How many different subsets of two people can be formed from the seven people? 7C2 = 21
How many of these subsets contain two friends? please give me a hand in doing this...
you are right that there are 21 possible ways of choosing a pair. I suggest visualizing a graph of friendships. 1 and 2 are friends; 3 and 4 are friends; 5, 6 and 7 are friends (3 pairs here: 5-6, 6-7, 5-7). Therefore, there are 5 total pairs of friends. Then probability of choosing a pair who are not friends is 1-5/21 = 16/21. I hope that helps.
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19 Jan 2008, 04:27
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Expert's post
an unusual way:
$$p=1-\frac{\frac{4*C_1^6+3*C_2^6}{2}}{C_2^7}=1-\frac{5}{21}=\frac{16}{21}$$
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14 Aug 2008, 12:13
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$$4*C_1^6$$ - for each person out of 4 we have only one friend out of remained people (7-1=6)
$$3*C_2^6$$ - for each person out of 3 we have two friends out of remained people (7-1=6)
$$\frac{4*C_1^6+3*C_2^6}{2}$$ - excluding double counting.
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27 Sep 2009, 12:26
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In a room with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
1)5/21
2)3/7
3)4/7
4)5/7
5)16/21
Soln Assuming the people to be A,B,C,D,E,F,G
We can arrange in this way such that 3 people have two friends while 4 have 1 friend
A B
B A
C D E
D C E
E C D
F G
G F
First column represents the 7 people, second and third columns represent their friends.
Thus A,B,F,G each have one friend while C,D,E have two friends
Probability when two people are chosen where they are not friends is
= (5 + 5 + 2 + 2 + 2)/7C2
= 16/21
Ans is option 5
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16 Jan 2008, 23:04
eschn3am wrote:
(4/7)(5/6) + (3/7)(4/6) if you choose one of the 4 with one other friend, then you have a 5/6 chance of not picking their friend 2nd. If you choose one of the 3 with 2 friends, you have a 4/6 chance of not picking one of their friends second. Add them up.
20/42 + 12/42
32/42 = 16/21
E. 16/21
excuse me, can you explain me this passage in detail? why do we have to sum probabilities up? don't we need to multiply them? we need that both conditions apply. thanks a lot
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17 Jan 2008, 05:51
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We don't need to multiply them together because we're only choosing one pair of people.
$formdata=\frac{4}{7}*\frac{5}{6}$ = probability of choosing 2 people who aren't friends out of the 4 with one friend
$formdata=\frac{3}{7}*\frac{4}{6}$ = probability of choosing 2 people who aren't friends out of the 3 with two friends
Each one represents a separate group of people so adding them together gives you the total probability of selecting two people who aren't friends.
7-t58525 -- here's another post on this topic, but it looks like they handled it exactly the same way
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18 Jan 2008, 05:17
eschn3am wrote:
We don't need to multiply them together because we're only choosing one pair of people.
$formdata=\frac{4}{7}*\frac{5}{6}$ = probability of choosing 2 people who aren't friends out of the 4 with one friend
$formdata=\frac{3}{7}*\frac{4}{6}$ = probability of choosing 2 people who aren't friends out of the 3 with two friends
Each one represents a separate group of people so adding them together gives you the total probability of selecting two people who aren't friends.
7-t58525 -- here's another post on this topic, but it looks like they handled it exactly the same way
right....let's solve it with combs:
How many different subsets of two people can be formed from the seven people? 7C2 = 21
How many of these subsets contain two friends? please give me a hand in doing this...
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14 Aug 2008, 00:30
walker wrote:
an unusual way:
$$p=1-\frac{\frac{4*C_1^6+3*C_2^6}{2}}{C_2^7}=1-\frac{5}{21}=\frac{16}{21}$$
Can you explain this a little more please? I understand the denominator - pick two people from 7. I think the numerator is the different combinations of people who ARE friends. But how did you get this?
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19 Aug 2008, 23:54
my way:
$$p= 1-\frac{1+1+3}{C_2^7}= \frac{16}{21}$$
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07 Oct 2009, 00:01
Only one friend
A<->B
C<->D
two friends
E<->F
E<->G
F<->G
so total possibility 21
=4*3+4/21
=16/21
good question........
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Re: In a room with 7 people, 4 people have exactly 1 friend in [#permalink]
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23 Apr 2013, 06:47
alexperi wrote:
In a room with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
1)5/21
2)3/7
3)4/7
4)5/7
5)16/21
1. We will find the reverse case, i.e., 2 people who are selected, are friends.
2. There are two groups here. People with 1 friend and people with 2 friends.
3. Friends can be selected as follows: (i) Among the 4 people each with just 1 friend, there will be 2 pairs of friends. (ii) Among the 3 people each with 2 friends, there will be 3 pairs of friends.
4. So 5 cases of friends can be selected.
5. The total number of ways of selecting 2 people is$$7C2 = 21$$
6. The number of cases when the 2 selected people are not friends $$= 21-5 = 16$$
7. Probability that the two slected people are not friends$$= 16/21$$
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Re: In a room with 7 people, 4 people have exactly 1 friend in [#permalink]
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23 Apr 2013, 06:51
Expert's post
2
This post was
BOOKMARKED
alexperi wrote:
In a room with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21
OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html
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Re: In a room with 7 people, 4 people have exactly 1 friend in [#permalink] 23 Apr 2013, 06:51
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Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack. It is currently 26 May 2017, 05:06. ### GMAT Club Daily Prep. #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.. Customized. for You. we will pick new questions that match your level based on your Timer History. Track. every week, we’ll send you an estimated GMAT score based on your performance. Practice. Pays. we will pick new questions that match your level based on your Timer History. # Events & Promotions. ###### Events & Promotions in June. Open Detailed Calendar. # In a room with 7 people, 4 people have exactly 1 friend in. Author Message. TAGS:. ### Hide Tags. Intern. Joined: 13 Jun 2007. Posts: 48. Followers: 1. Kudos [?]: 8 [5] , given: 0. In a room with 7 people, 4 people have exactly 1 friend in [#permalink]. ### Show Tags. 08 Dec 2007, 13:00. 5. KUDOS. 6. This post was. BOOKMARKED. 00:00. Difficulty:. 95% (hard). Question Stats:. 52% (03:40) correct 48% (02:10) wrong based on 190 sessions. ### HideShow timer Statistics. In a room with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?. A. 5/21. B. 3/7. C. 4/7. D. 5/7. E. 16/21. OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html. [Reveal] Spoiler: OA. Director. Joined: 12 Jul 2007. Posts: 861. Followers: 17. Kudos [?]: 307 [4] , given: 0. ### Show Tags. 08 Dec 2007, 13:09. 4. KUDOS. 3. This post was. BOOKMARKED. (4/7)(5/6) + (3/7)(4/6) if you choose one of the 4 with one other friend, then you have a 5/6 chance of not picking their friend 2nd. If you choose one of the 3 with 2 friends, you have a 4/6 chance of not picking one of their friends second. Add them up.. 20/42 + 12/42. 32/42 = 16/21. E. 16/21. Director. Joined: 01 Jan 2008. Posts: 624. Followers: 5. Kudos [?]: 184 [4] , given: 1. ### Show Tags. 18 Jan 2008, 08:00. 4. KUDOS. marcodonzelli wrote:. right....let's solve it with combs:. How many different subsets of two people can be formed from the seven people? 7C2 = 21. How many of these subsets contain two friends? please give me a hand in doing this.... you are right that there are 21 possible ways of choosing a pair. I suggest visualizing a graph of friendships. 1 and 2 are friends; 3 and 4 are friends; 5, 6 and 7 are friends (3 pairs here: 5-6, 6-7, 5-7). Therefore, there are 5 total pairs of friends. Then probability of choosing a pair who are not friends is 1-5/21 = 16/21. I hope that helps.. CEO. Joined: 17 Nov 2007. Posts: 3586. Concentration: Entrepreneurship, Other. Schools: Chicago (Booth) - Class of 2011. GMAT 1: 750 Q50 V40. Followers: 574. Kudos [?]: 3977 [1] , given: 360. ### Show Tags. 19 Jan 2008, 04:27. 1. KUDOS. Expert's post. an unusual way:. $$p=1-\frac{\frac{4*C_1^6+3*C_2^6}{2}}{C_2^7}=1-\frac{5}{21}=\frac{16}{21}$$. _________________. HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame. CEO. Joined: 17 Nov 2007. Posts: 3586. Concentration: Entrepreneurship, Other. Schools: Chicago (Booth) - Class of 2011. GMAT 1: 750 Q50 V40. Followers: 574. Kudos [?]: 3977 [1] , given: 360. ### Show Tags. 14 Aug 2008, 12:13. 1. KUDOS. Expert's post. $$4*C_1^6$$ - for each person out of 4 we have only one friend out of remained people (7-1=6). $$3*C_2^6$$ - for each person out of 3 we have two friends out of remained people (7-1=6). $$\frac{4*C_1^6+3*C_2^6}{2}$$ - excluding double counting.. _________________. HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame. Manager. Joined: 27 Oct 2008. Posts: 185. Followers: 2. Kudos [?]: 150 [1] , given: 3. ### Show Tags. 27 Sep 2009, 12:26. 1. KUDOS. In a room with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?. 1)5/21. 2)3/7. 3)4/7. 4)5/7. 5)16/21. Soln Assuming the people to be A,B,C,D,E,F,G. We can arrange in this way such that 3 people have two friends while 4 have 1 friend. A B. B A. C D E. D C E. E C D. F G. G F. First column represents the 7 people, second and third columns represent their friends.. Thus A,B,F,G each have one friend while C,D,E have two friends. Probability when two people are chosen where they are not friends is. = (5 + 5 + 2 + 2 + 2)/7C2. = 16/21. Ans is option 5. VP. Joined: 22 Nov 2007. Posts: 1082. Followers: 8.
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Kudos [?]: 566 [0], given: 0. ### Show Tags. 16 Jan 2008, 23:04. eschn3am wrote:. (4/7)(5/6) + (3/7)(4/6) if you choose one of the 4 with one other friend, then you have a 5/6 chance of not picking their friend 2nd. If you choose one of the 3 with 2 friends, you have a 4/6 chance of not picking one of their friends second. Add them up.. 20/42 + 12/42. 32/42 = 16/21. E. 16/21. excuse me, can you explain me this passage in detail? why do we have to sum probabilities up? don't we need to multiply them? we need that both conditions apply. thanks a lot. Director. Joined: 12 Jul 2007. Posts: 861. Followers: 17. Kudos [?]: 307 [0], given: 0. ### Show Tags. 17 Jan 2008, 05:51. 1. This post was. BOOKMARKED. We don't need to multiply them together because we're only choosing one pair of people.. $formdata=\frac{4}{7}*\frac{5}{6}$ = probability of choosing 2 people who aren't friends out of the 4 with one friend. $formdata=\frac{3}{7}*\frac{4}{6}$ = probability of choosing 2 people who aren't friends out of the 3 with two friends. Each one represents a separate group of people so adding them together gives you the total probability of selecting two people who aren't friends.. 7-t58525 -- here's another post on this topic, but it looks like they handled it exactly the same way. VP. Joined: 22 Nov 2007. Posts: 1082. Followers: 8. Kudos [?]: 566 [0], given: 0. ### Show Tags. 18 Jan 2008, 05:17. eschn3am wrote:. We don't need to multiply them together because we're only choosing one pair of people.. $formdata=\frac{4}{7}*\frac{5}{6}$ = probability of choosing 2 people who aren't friends out of the 4 with one friend. $formdata=\frac{3}{7}*\frac{4}{6}$ = probability of choosing 2 people who aren't friends out of the 3 with two friends. Each one represents a separate group of people so adding them together gives you the total probability of selecting two people who aren't friends.. 7-t58525 -- here's another post on this topic, but it looks like they handled it exactly the same way. right....let's solve it with combs:. How many different subsets of two people can be formed from the seven people? 7C2 = 21. How many of these subsets contain two friends? please give me a hand in doing this.... Manager. Joined: 10 Apr 2008. Posts: 53. Followers: 0. Kudos [?]: 7 [0], given: 4. ### Show Tags. 14 Aug 2008, 00:30. walker wrote:. an unusual way:. $$p=1-\frac{\frac{4*C_1^6+3*C_2^6}{2}}{C_2^7}=1-\frac{5}{21}=\frac{16}{21}$$. Can you explain this a little more please? I understand the denominator - pick two people from 7. I think the numerator is the different combinations of people who ARE friends. But how did you get this?. Senior Manager. Joined: 07 Jan 2008. Posts: 401. Followers: 3. Kudos [?]: 243 [0], given: 0. ### Show Tags. 19 Aug 2008, 23:54. my way:. $$p= 1-\frac{1+1+3}{C_2^7}= \frac{16}{21}$$. Manager. Joined: 09 Aug 2009. Posts: 52. Followers: 1. Kudos [?]: 8 [0], given: 1. ### Show Tags. 07 Oct 2009, 00:01. Only one friend. A<->B. C<->D. two friends. E<->F. E<->G. F<->G. so total possibility 21. =4*3+4/21. =16/21. good question......... Senior Manager. Joined: 17 Dec 2012. Posts: 460. Location: India. Followers: 27. Kudos [?]: 427 [0], given: 14. Re: In a room with 7 people, 4 people have exactly 1 friend in [#permalink]. ### Show Tags. 23 Apr 2013, 06:47. alexperi wrote:. In a room with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?. 1)5/21. 2)3/7. 3)4/7. 4)5/7. 5)16/21. 1. We will find the reverse case, i.e., 2 people who are selected, are friends.. 2. There are two groups here. People with 1 friend and people with 2 friends.. 3. Friends can be selected as follows: (i) Among the 4 people each with just 1 friend, there will be 2 pairs of friends. (ii) Among the 3 people each with 2 friends, there will be 3 pairs of friends.. 4. So 5 cases of friends can be selected.. 5. The total number of ways of selecting 2 people is$$7C2 = 21$$. 6. The number of cases when the 2 selected people are not friends $$= 21-5 = 16$$. 7. Probability that the two slected people are not friends$$= 16/21$$. _________________. Srinivasan Vaidyaraman. Sravna. http://www.sravnatestprep.com. Classroom and Online Coaching. Math Expert. Joined: 02 Sep 2009. Posts: 38894. Followers: 7737. Kudos [?]: 106196 [0], given: 11608. Re: In a room with 7 people, 4 people have exactly 1 friend in [#permalink]. ### Show Tags. 23 Apr 2013, 06:51. Expert's post. 2. This post was. BOOKMARKED. alexperi wrote:. In a room with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?. A. 5/21. B. 3/7. C. 4/7. D. 5/7. E. 16/21. OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html. _________________. Re: In a room with 7 people, 4 people have exactly 1 friend in [#permalink] 23 Apr 2013, 06:51. Similar topics Replies Last post. Similar. Topics:. 6 In a room filled with 7 people, 4 people have exactly 1 sibl 8 20 Jul 2013, 09:23. 137 In a room filled with 7 people, 4 people have exactly 1 43 13 Dec 2016, 03:14. 80 In a room filled with 7 people, 4 people have exactly 1 sibl 32 18 Aug 2014, 10:55. 1 In a room filled with 7 people, 4 people have exactly 1 3 22 Sep 2011, 20:09. 6 In a room filled with 7 people, 4 people have exactly 1 sibl 6 16 Sep 2013, 01:23. Display posts from previous: Sort by.
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# During a one day sale, a store sold each sweater of a
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During a one day sale, a store sold each sweater of a [#permalink]
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Updated on: 23 Oct 2012, 05:28
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During a one day sale, a store sold each sweater of a certain type for $30 more than the store's cost to purchase each sweater. How many of thses sweaters were sold during the sale? (1) During the sale, the total revenue from the sale of these sweaters was$270.
(2) During the sale, the store sold each of these sweaters at a price that was 50% greater than the store's cost to purchase each sweater.
Originally posted by sandranjeim on 17 Apr 2010, 07:03.
Last edited by Bunuel on 23 Oct 2012, 05:28, edited 1 time in total.
Edited the question.
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17 Apr 2010, 07:27
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sandranjeim wrote:
During a one day sale, a store sold each sweater of a certain type for $30 more than the store's cost to purchase each sweater. How many of thses sweaters were sold during the sale? 1) During the sale, the total revenue from the sale of these sweaters was$270.
2) During the sale, the store sold each of these sweaters at a price that was 50% greater than the store's cost to purchase each sweater.
OA=C
Given: $$p=c+30$$
(1) (Revenue) = (price) * (# of sweaters sold) --> $$270=p*n=(c+30)n$$. Not sufficient to determine $$n$$.
(2) $$p=1.5c=c+30$$ --> $$c=60$$. Not sufficient.
(1)+(2) $$c=60$$ and $$270=(c+30)n=90n$$ --> $$n=3$$. Sufficient.
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##### General Discussion
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Re: During a one day sale, a store sold each sweater of a [#permalink]
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23 Oct 2012, 05:23
Revenue = price * # of sweaters sold --> . Not sufficient to determine .
here We can consider Revenue is selling price - cost price , Can we ?
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Re: During a one day sale, a store sold each sweater of a [#permalink]
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23 Oct 2012, 05:36
amitvmane wrote:
Revenue = price * # of sweaters sold --> . Not sufficient to determine .
here We can consider Revenue is selling price - cost price , Can we ?
The total revenue from the sale is (price per unit) * (# of units sold).
(Selling price per unit) - (Cost price per unit) = (Gross profit from one unit).
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GMAT Question of the Day - Daily to your Mailbox; hard ones only. It is currently 19 Nov 2018, 16:09. ### GMAT Club Daily Prep. #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.. Customized. for You. we will pick new questions that match your level based on your Timer History. Track. every week, we’ll send you an estimated GMAT score based on your performance. Practice. Pays. we will pick new questions that match your level based on your Timer History. ## Events & Promotions. ###### Events & Promotions in November. PrevNext. SuMoTuWeThFrSa. 28293031123. 45678910. 11121314151617. 18192021222324. 2526272829301. Open Detailed Calendar. • ### How to QUICKLY Solve GMAT Questions - GMAT Club Chat. November 20, 2018. November 20, 2018. 09:00 AM PST. 10:00 AM PST. The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.. • ### The winning strategy for 700+ on the GMAT. November 20, 2018. November 20, 2018. 06:00 PM EST. 07:00 PM EST. What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.. # During a one day sale, a store sold each sweater of a. Author Message. TAGS:. ### Hide Tags. Intern. Joined: 11 Apr 2010. Posts: 16. During a one day sale, a store sold each sweater of a [#permalink]. ### Show Tags. Updated on: 23 Oct 2012, 05:28. 1. 3. 00:00. Difficulty:. 25% (medium). Question Stats:. 74% (01:43) correct 26% (01:27) wrong based on 180 sessions. ### HideShow timer Statistics. During a one day sale, a store sold each sweater of a certain type for $30 more than the store's cost to purchase each sweater. How many of thses sweaters were sold during the sale? (1) During the sale, the total revenue from the sale of these sweaters was$270.. (2) During the sale, the store sold each of these sweaters at a price that was 50% greater than the store's cost to purchase each sweater.
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Originally posted by sandranjeim on 17 Apr 2010, 07:03.. Last edited by Bunuel on 23 Oct 2012, 05:28, edited 1 time in total.. Edited the question.. Math Expert. Joined: 02 Sep 2009. Posts: 50670. ### Show Tags. 17 Apr 2010, 07:27. 3. 2. sandranjeim wrote:. During a one day sale, a store sold each sweater of a certain type for $30 more than the store's cost to purchase each sweater. How many of thses sweaters were sold during the sale? 1) During the sale, the total revenue from the sale of these sweaters was$270.. 2) During the sale, the store sold each of these sweaters at a price that was 50% greater than the store's cost to purchase each sweater.. OA=C. Given: $$p=c+30$$. (1) (Revenue) = (price) * (# of sweaters sold) --> $$270=p*n=(c+30)n$$. Not sufficient to determine $$n$$.. (2) $$p=1.5c=c+30$$ --> $$c=60$$. Not sufficient.. (1)+(2) $$c=60$$ and $$270=(c+30)n=90n$$ --> $$n=3$$. Sufficient.. _________________. ##### General Discussion. Intern. Joined: 27 Mar 2012. Posts: 6. Re: During a one day sale, a store sold each sweater of a [#permalink]. ### Show Tags. 23 Oct 2012, 05:23. Revenue = price * # of sweaters sold --> . Not sufficient to determine .. here We can consider Revenue is selling price - cost price , Can we ?. Math Expert. Joined: 02 Sep 2009. Posts: 50670. Re: During a one day sale, a store sold each sweater of a [#permalink]. ### Show Tags. 23 Oct 2012, 05:36. amitvmane wrote:. Revenue = price * # of sweaters sold --> . Not sufficient to determine .. here We can consider Revenue is selling price - cost price , Can we ?. The total revenue from the sale is (price per unit) * (# of units sold).. (Selling price per unit) - (Cost price per unit) = (Gross profit from one unit).. _________________. Non-Human User. Joined: 09 Sep 2013. Posts: 8847. Re: During a one day sale, a store sold each sweater of a [#permalink]. ### Show Tags. 07 Oct 2017, 16:04. Hello from the GMAT Club BumpBot!. Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).. Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.. _________________. Re: During a one day sale, a store sold each sweater of a &nbs [#permalink] 07 Oct 2017, 16:04. Display posts from previous: Sort by.
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https://socratic.org/questions/569287c17c014961c01e2fb6
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# Question e2fb6
Jan 13, 2016
${\text{N"_2"O}}_{4}$
#### Explanation:
As you know, a compound's empirical formula tells you the smallest whole number ratio that exists between the elements that make up said compound.
By comparison, the molecular formula tells you the exact number of atoms of each element that makes up the compound.
In essence, the molecular formula is a multiple of the empirical formula.
Your goal when dealing with such problems will be to determine how many empirical formulas are needed to get to the molecular formula.
Notice that the compound is said to have a molar mass of $\text{92 g/mol}$. So, if the molecular formula is a multiple of the empirical formula, you can use the molar mass of the empirical formula to get the multiplication factor that exists between the two.
The empirical formula is said to be ${\text{NO}}_{2}$. To get its molar mass, use the molar masses of the individual atoms
overbrace(1 xx "14.007 g/mol")^(color(red)("1 atom of N")) + overbrace(2 xx "15.9994 g/mol")^(color(purple)("2 atoms of O")) = "46.006 g/mol"#
This means that you have
$46.006 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) xx color(blue)(n) = 92 color(red)(cancel(color(black)("g}}}}$
This will get you
$\textcolor{b l u e}{n} = \frac{92}{46.006} = 1.9997 \approx 2$
Therefore, the molecular formula for your compound is
$\left({\text{NO"_2)_color(blue)(2) = color(green)("N"_2"O}}_{4}\right) \to$ dinitrogen tetroxide
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# Question e2fb6. Jan 13, 2016. ${\text{N"_2"O}}_{4}$. #### Explanation:. As you know, a compound's empirical formula tells you the smallest whole number ratio that exists between the elements that make up said compound.. By comparison, the molecular formula tells you the exact number of atoms of each element that makes up the compound.. In essence, the molecular formula is a multiple of the empirical formula.. Your goal when dealing with such problems will be to determine how many empirical formulas are needed to get to the molecular formula.. Notice that the compound is said to have a molar mass of $\text{92 g/mol}$. So, if the molecular formula is a multiple of the empirical formula, you can use the molar mass of the empirical formula to get the multiplication factor that exists between the two.
|
The empirical formula is said to be ${\text{NO}}_{2}$. To get its molar mass, use the molar masses of the individual atoms. overbrace(1 xx "14.007 g/mol")^(color(red)("1 atom of N")) + overbrace(2 xx "15.9994 g/mol")^(color(purple)("2 atoms of O")) = "46.006 g/mol"#. This means that you have. $46.006 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) xx color(blue)(n) = 92 color(red)(cancel(color(black)("g}}}}$. This will get you. $\textcolor{b l u e}{n} = \frac{92}{46.006} = 1.9997 \approx 2$. Therefore, the molecular formula for your compound is. $\left({\text{NO"_2)_color(blue)(2) = color(green)("N"_2"O}}_{4}\right) \to$ dinitrogen tetroxide.
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https://www.physicsforums.com/threads/collision-problem.157195/
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# Collision Problem
## Homework Statement
An atom (m = 19.0 u) makes a perfectly elastic collision with another atom at rest. After the impact, the atom travels away a
a 52.9° angle from its original direction and the unknown atom travels away at a −50.0° angle. What is the mass (in u) of th
unknown atom?
## Homework Equations
Conservation of Momemtum
Law of Sines
## The Attempt at a Solution
Atom A is the m_a=19 u...
m_a*v_a,i=m_a*v_a,f + m_b*v_b,f
Draw triangle, use law of sines...
sin(45.8)/(m_b*v_b,f)=sin(84.2)/(m_a*v_a,i)=sin(50)/(m_a*v_a,f)
4 eqns (bottom one can be seperated into 3 eqns correct?), 4 unknowns (m_b, v_a,i, v_a,f, and v_b,f)
Tried to solve in my TI-89 and it did not work (note I'm not in radian mode or anything stupid)
## Answers and Replies
Use conservation of momentum and conservation of kinetic energy to find the answer.
(Hint: Write the conservation of momentum expressions in the x- and y-directions. Note that the y-momentum before the collision is 0, so it has to be 0 after.)
Use conservation of momentum and conservation of kinetic energy to find the answer.
(Hint: Write the conservation of momentum expressions in the x- and y-directions. Note that the y-momentum before the collision is 0, so it has to be 0 after.)
So like...?
X-direction
m_a*v_a = m_a*v_a*cos(52.9) + m_b*v_b*cos(50)
0.5*m_a*(v_a)^2 = 0.5*m_a*[v_a*cos(52.9)]^2 + 0.5*m_b*[v_b*cos(50)]^2
Y-direction
0 = m_a*v_a*sin(52.9) + m_b*v_b*(-sin(50))
Don't need to do KE in y-direction because already have 3 equations for 3 unkowns...
Does doing the trig fucntions eliminate the need to do the initial and final subscripts on the velocities?
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# Collision Problem. ## Homework Statement. An atom (m = 19.0 u) makes a perfectly elastic collision with another atom at rest. After the impact, the atom travels away a. a 52.9° angle from its original direction and the unknown atom travels away at a −50.0° angle. What is the mass (in u) of th. unknown atom?. ## Homework Equations. Conservation of Momemtum. Law of Sines. ## The Attempt at a Solution. Atom A is the m_a=19 u.... m_a*v_a,i=m_a*v_a,f + m_b*v_b,f. Draw triangle, use law of sines.... sin(45.8)/(m_b*v_b,f)=sin(84.2)/(m_a*v_a,i)=sin(50)/(m_a*v_a,f). 4 eqns (bottom one can be seperated into 3 eqns correct?), 4 unknowns (m_b, v_a,i, v_a,f, and v_b,f). Tried to solve in my TI-89 and it did not work (note I'm not in radian mode or anything stupid).
|
## Answers and Replies. Use conservation of momentum and conservation of kinetic energy to find the answer.. (Hint: Write the conservation of momentum expressions in the x- and y-directions. Note that the y-momentum before the collision is 0, so it has to be 0 after.). Use conservation of momentum and conservation of kinetic energy to find the answer.. (Hint: Write the conservation of momentum expressions in the x- and y-directions. Note that the y-momentum before the collision is 0, so it has to be 0 after.). So like...?. X-direction. m_a*v_a = m_a*v_a*cos(52.9) + m_b*v_b*cos(50). 0.5*m_a*(v_a)^2 = 0.5*m_a*[v_a*cos(52.9)]^2 + 0.5*m_b*[v_b*cos(50)]^2. Y-direction. 0 = m_a*v_a*sin(52.9) + m_b*v_b*(-sin(50)). Don't need to do KE in y-direction because already have 3 equations for 3 unkowns.... Does doing the trig fucntions eliminate the need to do the initial and final subscripts on the velocities?.
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https://blog.csdn.net/luyehao1/article/details/76606042
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ACM算法
# RXD and dividing
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1355 Accepted Submission(s): 570
Problem Description
RXD has a tree T, with the size of n. Each edge has a cost.
Define f(S) as the the cost of the minimal Steiner Tree of the set S on tree T
he wants to divide 2,3,4,5,6,n into k parts S1,S2,S3,Sk,
where Si={2,3,,n} and for all different i,j , we can conclude that SiSj=
Then he calulates res=ki=1f({1}Si).
He wants to maximize the res.
1kn106
the cost of each edge[1,105]
Si might be empty.
f(S) means that you need to choose a couple of edges on the tree to make all the points in S connected, and you need to minimize the sum of the cost of these edges. f(S) is equal to the minimal cost
Input
For each test case, the first line consists of 2 integer n,k, which means the number of the tree nodes , and k means the number of parts.
The next n1 lines consists of 2 integers, a,b,c, means a tree edge (a,b) with cost c.
It is guaranteed that the edges would form a tree.
There are 4 big test cases and 50 small test cases.
small test case means n100.
Output
For each test case, output an integer, which means the answer.
Sample Input
5 4
1 2 3
2 3 4
2 4 5
2 5 6
Sample Output
27
//题意：给定一棵树，树里每条边有一个权值，现在以1为根节点，把2-n划分成k个集合，每个点只能出现在一个集合里且k个集合里包含2-n所有点（即分成k个划分）。求根节点1到各个集合的路径的最大值（经过集合里所有点）。
//思路
x=2nw[x][fax]min(szx,k) 时间复杂度O(n)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
//这里MAX太大，VS里跑不起来，会直接报错
//但hdu OJ上能运行起来，可以先开小点调试下，然后改成1e6去交
const int MAX = 1e6 + 10;
typedef struct {
int val, to;
}Edge;
//sum要long long，不然会WA
long long sum;
int n, k;
vector<Edge>map[MAX];
int Size[MAX];
void dfs(int root, int front)
{
for (int i = 0; i < map[root].size(); i++)
{
//因为是无向边，防止死循环
if (map[root][i].to == front)
continue;
dfs(map[root][i].to, root);
Size[root] += Size[map[root][i].to];
//注意这里*个1LL
sum += 1LL * map[root][i].val*min(Size[map[root][i].to], k);
}
}
int main()
{
while (scanf("%d%d", &n, &k) != EOF)
{
sum = 0;
for (int i = 0; i <= n; i++)
{
map[i].clear();
Size[i] = 1;
}
for (int i = 0; i < n - 1; i++)
{
int a, b, c;
Edge temp;
scanf("%d%d%d", &a, &b, &c);
//这里是无向边，两个点要互相到达
temp.val = c;
temp.to = b;
map[a].push_back(temp);
temp.to = a;
map[b].push_back(temp);
}
dfs(1,0);
printf("%lld\n", sum);
}
return 0;
}
#### HDU 6060 RXD and dividing
2017-08-01 19:27:27
#### HDU 6060 RXD and dividing
2017-08-08 11:00:56
#### [HDU]-6060 RXD and dividing
2017-08-01 20:54:57
#### hdu 6060 RXD and dividing
2017-08-02 15:38:12
#### 【HDU 6060 RXD and dividing】+ DFS
2017-08-02 10:42:48
#### HDU 6060 RXD and dividing【DFS】
2017-08-03 09:32:46
#### hdu 6060 RXD and dividing(dfs)
2017-08-04 10:05:59
#### hdu 6060 RXD and dividing （dfs）
2017-08-08 09:26:03
#### hdu 6060 RXD and dividing (贪心）
2017-08-01 23:08:55
#### Hdu 6060 - RXD and dividing （dfs）
2017-08-04 16:00:22
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ACM算法. # RXD and dividing. Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others). Total Submission(s): 1355 Accepted Submission(s): 570. Problem Description. RXD has a tree T, with the size of n. Each edge has a cost.. Define f(S) as the the cost of the minimal Steiner Tree of the set S on tree T. he wants to divide 2,3,4,5,6,n into k parts S1,S2,S3,Sk,. where Si={2,3,,n} and for all different i,j , we can conclude that SiSj=. Then he calulates res=ki=1f({1}Si).. He wants to maximize the res.. 1kn106. the cost of each edge[1,105]. Si might be empty.. f(S) means that you need to choose a couple of edges on the tree to make all the points in S connected, and you need to minimize the sum of the cost of these edges. f(S) is equal to the minimal cost. Input. For each test case, the first line consists of 2 integer n,k, which means the number of the tree nodes , and k means the number of parts.. The next n1 lines consists of 2 integers, a,b,c, means a tree edge (a,b) with cost c.. It is guaranteed that the edges would form a tree.. There are 4 big test cases and 50 small test cases.. small test case means n100.. Output. For each test case, output an integer, which means the answer.. Sample Input. 5 4. 1 2 3. 2 3 4. 2 4 5. 2 5 6. Sample Output. 27. //题意：给定一棵树，树里每条边有一个权值，现在以1为根节点，把2-n划分成k个集合，每个点只能出现在一个集合里且k个集合里包含2-n所有点（即分成k个划分）。求根节点1到各个集合的路径的最大值（经过集合里所有点）。. //思路. x=2nw[x][fax]min(szx,k) 时间复杂度O(n). #include <iostream>. #include <cstdio>. #include <cstring>. #include <string>. #include <cmath>. #include <vector>. #include <algorithm>. using namespace std;. //这里MAX太大，VS里跑不起来，会直接报错. //但hdu OJ上能运行起来，可以先开小点调试下，然后改成1e6去交. const int MAX = 1e6 + 10;. typedef struct {. int val, to;. }Edge;. //sum要long long，不然会WA. long long sum;. int n, k;. vector<Edge>map[MAX];. int Size[MAX];. void dfs(int root, int front). {. for (int i = 0; i < map[root].size(); i++). {.
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//因为是无向边，防止死循环. if (map[root][i].to == front). continue;. dfs(map[root][i].to, root);. Size[root] += Size[map[root][i].to];. //注意这里*个1LL. sum += 1LL * map[root][i].val*min(Size[map[root][i].to], k);. }. }. int main(). {. while (scanf("%d%d", &n, &k) != EOF). {. sum = 0;. for (int i = 0; i <= n; i++). {. map[i].clear();. Size[i] = 1;. }. for (int i = 0; i < n - 1; i++). {. int a, b, c;. Edge temp;. scanf("%d%d%d", &a, &b, &c);. //这里是无向边，两个点要互相到达. temp.val = c;. temp.to = b;. map[a].push_back(temp);. temp.to = a;. map[b].push_back(temp);. }. dfs(1,0);. printf("%lld\n", sum);. }. return 0;. }. #### HDU 6060 RXD and dividing. 2017-08-01 19:27:27. #### HDU 6060 RXD and dividing. 2017-08-08 11:00:56. #### [HDU]-6060 RXD and dividing. 2017-08-01 20:54:57. #### hdu 6060 RXD and dividing. 2017-08-02 15:38:12. #### 【HDU 6060 RXD and dividing】+ DFS. 2017-08-02 10:42:48. #### HDU 6060 RXD and dividing【DFS】. 2017-08-03 09:32:46. #### hdu 6060 RXD and dividing(dfs). 2017-08-04 10:05:59. #### hdu 6060 RXD and dividing （dfs）. 2017-08-08 09:26:03. #### hdu 6060 RXD and dividing (贪心）. 2017-08-01 23:08:55. #### Hdu 6060 - RXD and dividing （dfs）. 2017-08-04 16:00:22.
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http://www.docstoc.com/docs/84563692/Math-13-L-Shipley-Answers-for-Exam-_3-Review-Here-are-the-answers-
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# Math 13 L Shipley Answers for Exam _3 Review Here are the answers by shuifanglj
VIEWS: 4 PAGES: 2
• pg 1
``` Math 13 L Shipley
Here are the answers as the textbook answered them. However, when I wrote up the
solutions ( find them online) I arrived at different answers. We can resolve the differences in
class.
1. False
2. A
3. C
4. A ∩ B = {Cafe Ronaldo, Bombay Bistro, Chef Wong's} These restaurants are the
most frequented and the most admired according the Boston Magazine
5. A ∪ B = {Dynex, Toshiba} These were the TV's that Howie found to be neither
reliable nor of best value.
6. B
7. A
8. a) A ∩ B = {5,10}
b) ( A ∪ B ) ∩ C = {2,9}
c) A ∪ ( B ∩ C ) = {1, 2.8,9}
d) B ∪ C = {1, 5,10}
e) A ∪ B = {1, 2,3,5, 9,10}
9. n( A ∪ B ) = 104
10. 650
11. 78
12. 97 ( my answer is 102 – different than the textbook)
13.
a) 166 b) 110 c) 142
14. n( A ∪ B ) = 111
15. 510
16. C
17. B
18. 1,814,400
19. 2,016
20. 5
1
21.
42!
22. 1,395,360
23. 48,228,180
24. 3,628,800
25. 24 (my answer is 5040. I believe the textbook is incorrect).
26. 3,628,800
27. 742,560
28. 1820
29. 1
30 165
31. 657,800
32. 4,989,100 ( I arrived at 4,989,600)
33. 61,440 ( my answer is 960 – different than textbook’s)
34. 128
35. 45
36. 32
37. {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
7 1
38. P ( H ) = , P (T ) =
8 8
39. 4
40. 10,077,696
41. 132,600
42. 1728
43. 13,747,968
44. 155,937,600 ( my answer is 6,497,400 – different than the textbook’s answer)
45.
45 38 47
a) b) c)
73 73 73
46. .39
47. .58
48. .16
49. .29
50. .43
51. .35
52.
a) 0.45 b) 0.04 c) 0.83
53.
a) 0.55 b) 0.59 c) 0.9
54. .69
55. .33
56.
a) 0.29 b) 0.91 c) 0.39
```
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# Math 13 L Shipley Answers for Exam _3 Review Here are the answers by shuifanglj. VIEWS: 4 PAGES: 2. • pg 1. ``` Math 13 L Shipley. Here are the answers as the textbook answered them. However, when I wrote up the. solutions ( find them online) I arrived at different answers. We can resolve the differences in. class.. 1. False. 2. A. 3. C. 4. A ∩ B = {Cafe Ronaldo, Bombay Bistro, Chef Wong's} These restaurants are the. most frequented and the most admired according the Boston Magazine. 5. A ∪ B = {Dynex, Toshiba} These were the TV's that Howie found to be neither. reliable nor of best value.. 6. B. 7. A. 8. a) A ∩ B = {5,10}. b) ( A ∪ B ) ∩ C = {2,9}. c) A ∪ ( B ∩ C ) = {1, 2.8,9}. d) B ∪ C = {1, 5,10}. e) A ∪ B = {1, 2,3,5, 9,10}. 9. n( A ∪ B ) = 104. 10. 650. 11. 78. 12. 97 ( my answer is 102 – different than the textbook). 13.. a) 166 b) 110 c) 142. 14. n( A ∪ B ) = 111. 15. 510. 16. C. 17. B. 18. 1,814,400. 19. 2,016. 20. 5. 1. 21.. 42!. 22. 1,395,360. 23. 48,228,180. 24. 3,628,800. 25. 24 (my answer is 5040. I believe the textbook is incorrect).. 26.
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3,628,800. 27. 742,560. 28. 1820. 29. 1. 30 165. 31. 657,800. 32. 4,989,100 ( I arrived at 4,989,600). 33. 61,440 ( my answer is 960 – different than textbook’s). 34. 128. 35. 45. 36. 32. 37. {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}. 7 1. 38. P ( H ) = , P (T ) =. 8 8. 39. 4. 40. 10,077,696. 41. 132,600. 42. 1728. 43. 13,747,968. 44. 155,937,600 ( my answer is 6,497,400 – different than the textbook’s answer). 45.. 45 38 47. a) b) c). 73 73 73. 46. .39. 47. .58. 48. .16. 49. .29. 50. .43. 51. .35. 52.. a) 0.45 b) 0.04 c) 0.83. 53.. a) 0.55 b) 0.59 c) 0.9. 54. .69. 55. .33. 56.. a) 0.29 b) 0.91 c) 0.39. ```. To top.
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# The perimeter of a triangle is 108 cm and the lengths of its sides are 6: 8: 13. Find the sides of the triangle.
1) Calculate what is the sum of the numbers in relation to the sides of the triangle.
6 + 8 + 13 = 27.
2) Find out how many centimeters are in one part.
108/27 = 4 centimeters.
3) Find what is the value of one side of the triangle.
4 * 6 = 24 centimeters.
4) What is the other side of the triangle?
4 * 8 = 32 centimeters.
5) what is the third side of the triangle?
4 * 13 = 52 centimeters.
Answer: One side of the triangle is 24 centimeters, the other side is 32 centimeters, and the third side is 52 centimeters.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
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# The perimeter of a triangle is 108 cm and the lengths of its sides are 6: 8: 13. Find the sides of the triangle.. 1) Calculate what is the sum of the numbers in relation to the sides of the triangle.. 6 + 8 + 13 = 27.. 2) Find out how many centimeters are in one part.. 108/27 = 4 centimeters.. 3) Find what is the value of one side of the triangle.. 4 * 6 = 24 centimeters.
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4) What is the other side of the triangle?. 4 * 8 = 32 centimeters.. 5) what is the third side of the triangle?. 4 * 13 = 52 centimeters.. Answer: One side of the triangle is 24 centimeters, the other side is 32 centimeters, and the third side is 52 centimeters.. One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
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Calculating Truss Forces
1 / 29
# Calculating Truss Forces - PowerPoint PPT Presentation
Calculating Truss Forces. Forces. Compression. A body being squeezed. Tension. A body being stretched. Truss. A truss is composed of slender members joined together at their end points. They are usually joined by welds or gusset plates. Simple Truss.
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## PowerPoint Slideshow about 'Calculating Truss Forces' - clayland
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### Calculating Truss Forces
Forces
Compression
A body being squeezed
Tension
A body being stretched
Truss
A truss is composed of slender members joined together at their end points.
• They are usually joined by welds or gusset plates.
Simple Truss
A simple truss is composed of triangles, which will retain their shape even when removed from supports.
Pinned and Roller Supports
A pinnedsupport can support a structure in two dimensions.
A roller support can support a structure in only one dimension.
Solving Truss Forces
Assumptions:
All members are perfectly straight.
All loads are applied at the joints.
All joints are pinned and frictionless.
Each member has no weight.
Members can only experience tension or compression forces.
What risks might these assumptions pose if we were designing an actual bridge?
Static Determinacy
A statically determinate structure is one that can be mathematically solved.
2J = M + R
J = Number of Joints
M = Number of Members
R = Number of Reactions
Statically Indeterminate
B
Each pin connection contributes TWO reaction forces
A
C
D
FD = 500 lb
A truss is considered statically indeterminate when the static equilibrium equations are not sufficient to find the reactions on that structure. There are simply too many unknowns.
2J = M + R
Try It
2(4) ≠ 5 + 4
Statically Determinate
Is the truss statically determinate now?
B
A
C
D
FD = 500 lb
A truss is considered statically determinate when the static equilibrium equations can be used to find the reactions on that structure.
2J = M + R
Try It
2(4) = 5 + 3
Static Determinacy Example
Each side of the main street bridge in Brockport, NY has 19 joints, 35 members, and three reaction forces (pin and roller), making it a statically determinate truss.
What if these numbers were different?
Equilibrium Equations
The sum of the moments about a given point is zero.
Equilibrium Equations
The sum of the forces in the x-direction is zero.
Do you remember the Cartesian coordinate system? A vector that acts to the right is positive, and a vector that acts to the left is negative.
Equilibrium Equations
The sum of the forces in the y-direction is zero.
A vector that acts up is positive, and a vector that acts down is negative.
Using Moments to Find RCY
A force that causes aclockwise moment isa negative moment.
A force that causes acounterclockwisemoment ispositive moment.
3.0 ft
7.0 ft
FDcontributes a negative moment because it causes a clockwise moment about A.
RCy contributes a positive moment because it causes a counterclockwise moment around A.
Sum the y Forces to Find RAy
We know two out of the three forces acting in the y-direction. By simply summing those forces together, we can find the unknown reaction at point A.
Please note that a negative sign is in front of FD because the drawing shows the force as down.
Sum the x Forces to Find Ax
Because joint A is pinned, it is capable of reacting to a force applied in the x-direction.
However, since the only load applied to this truss (FD) has no x-component, RAx must be zero.
Method of Joints
• Use cosine and sine to determine x and y vector components.
• Assume all members to be in tension.A positive answer will mean the member is in tension, and a negative number will mean the member is in compression.
• As forces are solved, update free body diagrams.Use correct magnitude and sense for subsequent joint free body diagrams.
Method of Joints
4.0 ft
Truss Dimensions
B
A
C
θ2
θ1
RAx
D
3.0 ft
7.0 ft
RAy
RCy
500lb
Method of Joints
4.0 ft
Using Truss Dimensions to Find Angles
B
4.0 ft
A
C
θ2
θ1
D
3.0 ft
7.0 ft
Method of Joints
4.0ft
Using Truss Dimensions to Find Angles
B
4.0 ft
A
C
θ2
θ1
D
3.0 ft
7.0 ft
Method of Joints
Draw a free body diagram of each pin.
B
A
C
53.130°
29.745°
RAx
D
RAy
RCy
500lb
Every member is assumed to be in tension. A positive answer indicates the member is in tension, and a negative answer indicates the member is in compression.
Method of Joints
Where to Begin
Choose the joint that has the least number of unknowns.
Reaction forces at joints A and C are both good choices to begin our calculations.
B
AB
BC
BD
A
C
CD
RAx
D
0
150lb
350lb
500lb
RCy
RAy
Method of Joints
COMPRESSION
Method of Joints
Update the all force diagrams based on AB being under compression.
B
AB
BC
BD
A
C
CD
RAx=
0
D
RCy=
150lb
350lb
RAy=
500lb
Method of Joints
COMPRESSION
Method of Joints
Update the all force diagrams based on BC being under compression.
B
AB
BC
BD
A
C
CD
RAx=
0
D
RCy=
150lb
350lb
RAy=
500lb
Method of Joints
500lb
BD
D
TENSION
500lb
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Calculating Truss Forces. 1 / 29. # Calculating Truss Forces - PowerPoint PPT Presentation. Calculating Truss Forces. Forces. Compression. A body being squeezed. Tension. A body being stretched. Truss. A truss is composed of slender members joined together at their end points. They are usually joined by welds or gusset plates. Simple Truss.. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.. ## PowerPoint Slideshow about 'Calculating Truss Forces' - clayland. Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -. Presentation Transcript. ### Calculating Truss Forces. Forces. Compression. A body being squeezed. Tension. A body being stretched. Truss. A truss is composed of slender members joined together at their end points.. • They are usually joined by welds or gusset plates.. Simple Truss. A simple truss is composed of triangles, which will retain their shape even when removed from supports.. Pinned and Roller Supports. A pinnedsupport can support a structure in two dimensions.. A roller support can support a structure in only one dimension.. Solving Truss Forces. Assumptions:. All members are perfectly straight.. All loads are applied at the joints.. All joints are pinned and frictionless.. Each member has no weight.. Members can only experience tension or compression forces.. What risks might these assumptions pose if we were designing an actual bridge?. Static Determinacy. A statically determinate structure is one that can be mathematically solved.. 2J = M + R. J = Number of Joints. M = Number of Members. R = Number of Reactions. Statically Indeterminate. B. Each pin connection contributes TWO reaction forces. A. C. D. FD = 500 lb. A truss is considered statically indeterminate when the static equilibrium equations are not sufficient to find the reactions on that structure. There are simply too many unknowns.. 2J = M + R. Try It. 2(4) ≠ 5 + 4. Statically Determinate. Is the truss statically determinate now?. B. A. C. D. FD = 500 lb. A truss is considered statically determinate when the static equilibrium equations can be used to find the reactions on that structure.. 2J = M + R. Try It. 2(4) = 5 + 3. Static Determinacy Example. Each side of the main street bridge in Brockport, NY has 19 joints, 35 members, and three reaction forces (pin and roller), making it a statically determinate truss.. What if these numbers were different?. Equilibrium Equations. The sum of the moments about a given point is zero.. Equilibrium Equations. The sum of the forces in the x-direction is zero.. Do you remember the Cartesian coordinate system? A vector that acts to the right is positive, and a vector that acts to the left is negative.. Equilibrium Equations. The sum of the forces in the y-direction is zero.. A vector that acts up is positive, and a vector that acts down is negative.. Using Moments to Find RCY. A force that causes aclockwise moment isa negative moment.. A force that causes acounterclockwisemoment ispositive moment.. 3.0 ft. 7.0 ft. FDcontributes a negative moment because it causes a clockwise moment about A.. RCy contributes a positive moment because it causes a counterclockwise moment around A.. Sum the y Forces to Find RAy. We know two out of the three forces acting in the y-direction. By simply summing those forces together, we can find the unknown reaction at point A.. Please note that a negative sign is in front of FD because the drawing shows the force as down.. Sum the x Forces to Find Ax. Because joint A is pinned, it is capable of reacting to a force applied in the x-direction.. However, since the only load applied to this truss (FD) has no x-component, RAx must be zero.. Method of Joints. • Use cosine and sine to determine x and y vector components.. • Assume all members to be in tension.A positive answer will mean the member is in tension, and a negative number will mean the member is in compression.. • As forces are solved, update free body diagrams.Use correct magnitude and sense for subsequent joint free body diagrams.. Method of Joints. 4.0 ft. Truss Dimensions. B. A. C. θ2. θ1. RAx. D.
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3.0 ft. 7.0 ft. RAy. RCy. 500lb. Method of Joints. 4.0 ft. Using Truss Dimensions to Find Angles. B. 4.0 ft. A. C. θ2. θ1. D. 3.0 ft. 7.0 ft. Method of Joints. 4.0ft. Using Truss Dimensions to Find Angles. B. 4.0 ft. A. C. θ2. θ1. D. 3.0 ft. 7.0 ft. Method of Joints. Draw a free body diagram of each pin.. B. A. C. 53.130°. 29.745°. RAx. D. RAy. RCy. 500lb. Every member is assumed to be in tension. A positive answer indicates the member is in tension, and a negative answer indicates the member is in compression.. Method of Joints. Where to Begin. Choose the joint that has the least number of unknowns.. Reaction forces at joints A and C are both good choices to begin our calculations.. B. AB. BC. BD. A. C. CD. RAx. D. 0. 150lb. 350lb. 500lb. RCy. RAy. Method of Joints. COMPRESSION. Method of Joints. Update the all force diagrams based on AB being under compression.. B. AB. BC. BD. A. C. CD. RAx=. 0. D. RCy=. 150lb. 350lb. RAy=. 500lb. Method of Joints. COMPRESSION. Method of Joints. Update the all force diagrams based on BC being under compression.. B. AB. BC. BD. A. C. CD. RAx=. 0. D. RCy=. 150lb. 350lb. RAy=. 500lb. Method of Joints. 500lb. BD. D. TENSION. 500lb.
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# Are normal subgroups transitive?
Suppose $G$ is a group and $K\lhd H\lhd G$ are normal subgroups of $G$. Is $K$ a normal subgroup of $G$, i.e. $K\lhd G$? If not, what extra conditions on $G$ or $H$ make this possible?
Applying the definitions, we know $\{ghg^{-1}\mid h\in H\}=H$ and $\{hkh^{-1}\mid k\in K\}=K$, and want $\{gkg^{-1}\mid k\in K\}=K$. Clearly, the best avenue for a counterexample is if $gkg^{-1}\not\in K$ for some $k\in K$ and $g\in G-H$.
If no such element exists, $\{gkg^{-1}\mid k\in K\}\subseteq K$ implies $\{gkg^{-1}\mid k\in K\}=K$ because if $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.
• $K$ characteristic in $H$ and $H$ normal in $G$ then $K$ is normal in $G$
– jim
Dec 10, 2012 at 7:48
• Take a look at $D_8$, the dihedral group with 8 elements. Dec 10, 2012 at 7:51
• @HansGiebenrath I'm not seeing a counterexample in $D_4$. It has $C_4=\{1,r,r^2,r^3\}$ as a normal subgroup, but the only subgroup of $C_4$ is $C_2=\{1,r^2\}$, which is normal in $D_4$. Dec 10, 2012 at 8:02
• @PatrickDaSilva: They do, $\langle r^2,s\rangle$ is normal in $D_8$ and contains $\langle s \rangle$, which is not normal in $D_8$. Dec 10, 2012 at 9:08
• @Hans : I guess I am tired for saying false things. Sorry to have doubted you. Dec 10, 2012 at 9:16
Using some suggestions from the other commenters:
The alternating group, $A_4$, has the set $H=\{I,(12)(34),(13)(24),(14)(23)\}\cong V_4$ as a subgroup. If $f\in S_4\supseteq A_4$ is a permutation, then $f^{-1}[(12)(34)]f$ has the effect of swapping $f(1)$ with $f(2)$ and $f(3)$ with $f(4)$. One of these is $1$, and depending on which it is paired with, the conjugated element may be any of $H-\{I\}$, since the other two are also swapped. Thus $H\lhd S_4$ is normal, so $H\lhd A_4$ as well. Similarly, $H$ has three nontrivial subgroups, and taking $K=\{I,(12)(34)\}\cong C_2$, this is normal because $V_4$ is abelian. But $K\not\lhd A_4$, since $$[(123)][(12)(34)][(132)]=(13)(24)\in H-K.$$
Moreover, this is a minimal counterexample, since $|A_4|=12=2\cdot 2\cdot 3$ is the next smallest number which factors into three integers, which is required for $K\lhd H\lhd G$ but $\{I\}\subset K\subset H\subset G$ so that $[G\,:\,H]>1$, $[H\,:\,K]>1$, $|K|>1$ and $$|G|=[G\,:\,H]\cdot[H\,:\,K]\cdot|K|.$$ The smallest integer satisfying this requirement is 8, but the only non-abelian groups with $|G|=8$ are the dihedral group $D_4$ and the quaternion group $Q_8$, and neither of these have counterexamples. (Note that if $G$ is abelian, then all subgroups are normal.) Thus $A_4$ is a minimal counterexample. (Edit: Oops, $D_4$ has a counterexample, as mentioned in the comments: $\langle s\rangle\lhd\langle r^2,s\rangle\lhd \langle r,s\rangle=D_4$, but $\langle s\rangle\not\lhd D_4$.)
However, if $H\lhd G$ and $K$ is a characteristic subgroup of $H$, then $K$ is normal in $G$. This is because the group action $f$ defines an automorphism on $G$, $\varphi(g)=f^{-1}gf$, and because $H$ is normal, $\varphi(H)=H$ so that $\varphi|_H$ is an automorphism on $H$. Thus $\varphi(K)=K$ since $K$ is characteristic on $H$ and so $\{f^{-1}kf\mid k\in K\}=K\Rightarrow K\lhd G$.
• Great! You understand the theory verywell. +1! And sorry for the false comments. Dec 10, 2012 at 9:17
Look at $S_4$ and its following subgroups $A = \langle (12)(34) \rangle$ and $B=\{(12)(34),(13)(42),(23)(41),e \}$. Try to show that $A$ is normal in $B$ and $B$ is normal in $S_4$ but $A$ is not normal in $S_4$.
• Good eye! I like this example. How about you show it? The proof is not long. No calculations required. I challenge you (unless you were leaving the exercise to the OP). Dec 10, 2012 at 7:59
• @PatrickDaSilva I've fleshed out this argument in an answer below, but this leads to an interesting line of investigation: What is the smallest counterexample? As you point out, $|G|\geq 8$, but $D_8$ fails, and the quaternion group fails too since $\{1,-1\}\lhd Q_8$ is the only subgroup of order 2. Dec 10, 2012 at 9:09
• @PatrickDaSilva $A_4$ is normal in $S_4$ and the sylow 2 subgroup of of $A_4$ is a unique group of order 4 hence the group of $B$ given above will be normal in $S_4$ and the normality of $A$ in $B$ is becuase of the fact $[B:A]=2$
– jim
Dec 10, 2012 at 10:56
We need a non-abelian group, since all subgroups of abelian groups are normal. One small candidate is $$D_8$$, the symmetries of a square, here in a little more detail about how we might go about finding examples:
Consider all the subgroups in $$D_8$$. It's useful to visualize the subgroups as a lattice:
(Picture of Dummit and Foote I found on the web)
Now we try to pick an $$H$$. For all the subgroups on the third row from the top, their only proper subgroup is the trivial subgroup, which is trivially normal to $$G$$, so it doesn't make sense to use any of the subgroups on the third row for $$H$$.
Our only options for $$H$$ now are the second row: $$\langle s, r^2 \rangle$$, $$\langle r \rangle$$, and $$\langle rs, r^2 \rangle$$. We observe that if $$H = \langle r \rangle$$, the proper subgroups $$\langle r^2 \rangle$$ and $$1$$ are both normal to $$D_8$$, so that case is excluded. Our candidates are $$\langle s, r^2 \rangle$$ and $$\langle rs, r^2 \rangle$$.
Take $$H = \langle s, r^2 \rangle$$ and $$K = \langle s \rangle$$. It's easy to verify that $$\langle s \rangle$$ is not normal to $$D_8$$. All that's left is to show $$K \lhd H$$ and $$H \lhd G$$.
This is not difficult if we remember that any element in $$D_8$$ can be written as $$r^i s^j$$ with $$0 \le i \le 3$$ and $$j = 0, 1$$. Also we have the identity $$rs = sr^{-1}$$ which can repeated as $$r^k s = s r^{-k}$$. So for $$g \in D_8$$, we want to prove or disprove $$r^i s^j n s^j r^{-i} \in N$$ to show $$N \lhd G$$.
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# Are normal subgroups transitive?. Suppose $G$ is a group and $K\lhd H\lhd G$ are normal subgroups of $G$. Is $K$ a normal subgroup of $G$, i.e. $K\lhd G$? If not, what extra conditions on $G$ or $H$ make this possible?. Applying the definitions, we know $\{ghg^{-1}\mid h\in H\}=H$ and $\{hkh^{-1}\mid k\in K\}=K$, and want $\{gkg^{-1}\mid k\in K\}=K$. Clearly, the best avenue for a counterexample is if $gkg^{-1}\not\in K$ for some $k\in K$ and $g\in G-H$.. If no such element exists, $\{gkg^{-1}\mid k\in K\}\subseteq K$ implies $\{gkg^{-1}\mid k\in K\}=K$ because if $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.. • $K$ characteristic in $H$ and $H$ normal in $G$ then $K$ is normal in $G$. – jim. Dec 10, 2012 at 7:48. • Take a look at $D_8$, the dihedral group with 8 elements. Dec 10, 2012 at 7:51. • @HansGiebenrath I'm not seeing a counterexample in $D_4$. It has $C_4=\{1,r,r^2,r^3\}$ as a normal subgroup, but the only subgroup of $C_4$ is $C_2=\{1,r^2\}$, which is normal in $D_4$. Dec 10, 2012 at 8:02. • @PatrickDaSilva: They do, $\langle r^2,s\rangle$ is normal in $D_8$ and contains $\langle s \rangle$, which is not normal in $D_8$. Dec 10, 2012 at 9:08. • @Hans : I guess I am tired for saying false things. Sorry to have doubted you. Dec 10, 2012 at 9:16. Using some suggestions from the other commenters:. The alternating group, $A_4$, has the set $H=\{I,(12)(34),(13)(24),(14)(23)\}\cong V_4$ as a subgroup. If $f\in S_4\supseteq A_4$ is a permutation, then $f^{-1}[(12)(34)]f$ has the effect of swapping $f(1)$ with $f(2)$ and $f(3)$ with $f(4)$. One of these is $1$, and depending on which it is paired with, the conjugated element may be any of $H-\{I\}$, since the other two are also swapped. Thus $H\lhd S_4$ is normal, so $H\lhd A_4$ as well. Similarly, $H$ has three nontrivial subgroups, and taking $K=\{I,(12)(34)\}\cong C_2$, this is normal because $V_4$ is abelian. But $K\not\lhd A_4$, since $$[(123)][(12)(34)][(132)]=(13)(24)\in H-K.$$. Moreover, this is a minimal counterexample, since $|A_4|=12=2\cdot 2\cdot 3$ is the next smallest number which factors into three integers, which is required for $K\lhd H\lhd G$ but $\{I\}\subset K\subset H\subset G$ so that $[G\,:\,H]>1$, $[H\,:\,K]>1$, $|K|>1$ and $$|G|=[G\,:\,H]\cdot[H\,:\,K]\cdot|K|.$$ The smallest integer satisfying this requirement is 8, but the only non-abelian groups with $|G|=8$ are the dihedral group $D_4$ and the quaternion group $Q_8$, and neither of these have counterexamples. (Note that if $G$ is abelian, then all subgroups are normal.) Thus $A_4$ is a minimal counterexample. (Edit: Oops, $D_4$ has a counterexample, as mentioned in the comments: $\langle s\rangle\lhd\langle r^2,s\rangle\lhd \langle r,s\rangle=D_4$, but $\langle s\rangle\not\lhd D_4$.).
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However, if $H\lhd G$ and $K$ is a characteristic subgroup of $H$, then $K$ is normal in $G$. This is because the group action $f$ defines an automorphism on $G$, $\varphi(g)=f^{-1}gf$, and because $H$ is normal, $\varphi(H)=H$ so that $\varphi|_H$ is an automorphism on $H$. Thus $\varphi(K)=K$ since $K$ is characteristic on $H$ and so $\{f^{-1}kf\mid k\in K\}=K\Rightarrow K\lhd G$.. • Great! You understand the theory verywell. +1! And sorry for the false comments. Dec 10, 2012 at 9:17. Look at $S_4$ and its following subgroups $A = \langle (12)(34) \rangle$ and $B=\{(12)(34),(13)(42),(23)(41),e \}$. Try to show that $A$ is normal in $B$ and $B$ is normal in $S_4$ but $A$ is not normal in $S_4$.. • Good eye! I like this example. How about you show it? The proof is not long. No calculations required. I challenge you (unless you were leaving the exercise to the OP). Dec 10, 2012 at 7:59. • @PatrickDaSilva I've fleshed out this argument in an answer below, but this leads to an interesting line of investigation: What is the smallest counterexample? As you point out, $|G|\geq 8$, but $D_8$ fails, and the quaternion group fails too since $\{1,-1\}\lhd Q_8$ is the only subgroup of order 2. Dec 10, 2012 at 9:09. • @PatrickDaSilva $A_4$ is normal in $S_4$ and the sylow 2 subgroup of of $A_4$ is a unique group of order 4 hence the group of $B$ given above will be normal in $S_4$ and the normality of $A$ in $B$ is becuase of the fact $[B:A]=2$. – jim. Dec 10, 2012 at 10:56. We need a non-abelian group, since all subgroups of abelian groups are normal. One small candidate is $$D_8$$, the symmetries of a square, here in a little more detail about how we might go about finding examples:. Consider all the subgroups in $$D_8$$. It's useful to visualize the subgroups as a lattice:. (Picture of Dummit and Foote I found on the web). Now we try to pick an $$H$$. For all the subgroups on the third row from the top, their only proper subgroup is the trivial subgroup, which is trivially normal to $$G$$, so it doesn't make sense to use any of the subgroups on the third row for $$H$$.. Our only options for $$H$$ now are the second row: $$\langle s, r^2 \rangle$$, $$\langle r \rangle$$, and $$\langle rs, r^2 \rangle$$. We observe that if $$H = \langle r \rangle$$, the proper subgroups $$\langle r^2 \rangle$$ and $$1$$ are both normal to $$D_8$$, so that case is excluded. Our candidates are $$\langle s, r^2 \rangle$$ and $$\langle rs, r^2 \rangle$$.. Take $$H = \langle s, r^2 \rangle$$ and $$K = \langle s \rangle$$. It's easy to verify that $$\langle s \rangle$$ is not normal to $$D_8$$. All that's left is to show $$K \lhd H$$ and $$H \lhd G$$.. This is not difficult if we remember that any element in $$D_8$$ can be written as $$r^i s^j$$ with $$0 \le i \le 3$$ and $$j = 0, 1$$. Also we have the identity $$rs = sr^{-1}$$ which can repeated as $$r^k s = s r^{-k}$$. So for $$g \in D_8$$, we want to prove or disprove $$r^i s^j n s^j r^{-i} \in N$$ to show $$N \lhd G$$.
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https://socratic.org/questions/how-do-you-solve-x-2-3x-2-0
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# How do you solve x^2-3x+2=0?
Jun 25, 2015
$0 = {x}^{2} - 3 x + 2 = \left(x - 1\right) \left(x - 2\right)$
So $x = 1$ and $x = 2$ are the solutions.
#### Explanation:
Let $f \left(x\right) = {x}^{2} - 3 x + 2$
First notice that since $1 - 3 + 2 = 0$, we have $f \left(1\right) = 0$, $x = 1$ is a solution and $\left(x - 1\right)$ is a factor of $f \left(x\right)$.
We can see that the other factor must be $\left(x - 2\right)$ in order that:
$\left(x - 1\right) \left(x - 2\right) = {x}^{2} - 3 x + 2$
... look at the coefficient of ${x}^{2}$ which comes from $x \cdot x$ and the constant term $2$, which comes from $- 1 \times - 2$
So the other solution is $x = 2$.
Check: $f \left(2\right) = {2}^{2} - \left(3 \cdot 2\right) + 2 = 4 - 6 + 2 = 0$
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# How do you solve x^2-3x+2=0?. Jun 25, 2015. $0 = {x}^{2} - 3 x + 2 = \left(x - 1\right) \left(x - 2\right)$. So $x = 1$ and $x = 2$ are the solutions.. #### Explanation:. Let $f \left(x\right) = {x}^{2} - 3 x + 2$. First notice that since $1 - 3 + 2 = 0$, we have $f \left(1\right) = 0$, $x = 1$ is a solution and $\left(x - 1\right)$ is a factor of $f \left(x\right)$.
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We can see that the other factor must be $\left(x - 2\right)$ in order that:. $\left(x - 1\right) \left(x - 2\right) = {x}^{2} - 3 x + 2$. ... look at the coefficient of ${x}^{2}$ which comes from $x \cdot x$ and the constant term $2$, which comes from $- 1 \times - 2$. So the other solution is $x = 2$.. Check: $f \left(2\right) = {2}^{2} - \left(3 \cdot 2\right) + 2 = 4 - 6 + 2 = 0$.
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http://www.datascribble.com/blog/gradient-descent-algorithm-codes/
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# The Gradient Descent Algorithm (With Codes)
Hello everyone! Welcome back to the 3rd article on basics of Machine Learning. In this article, we are going to first recap the pre-requisite to Gradient Descent Algorithm(i.e. what we did in the previous article) and then introduce one of the most important algorithm in Machine Learning – Gradient Descent. This algorithm is one of the most used and one of the most powerful algorithm used as practical applications of Machine Learning, such as Stock Prices prediction, or detection of Malignant Breast Cancer.
So, in our previous article we defined our notations for our cost function model by setting up our hypothesis [h(x)=θ+ϕ(x)] and then plugging in the value for the R squared error function,
Where, J (θ0θ1) is the cost function and m are the number of training examples.
Now, we are going to define first algorithm in Machine Learning- Gradient Descent.
Gradient Descent is a Machine Learning algorithm to minimize the cost function J (θ0, θ1 ).
In layman terms, we minimize the above squared error function by using the basic calculus methods to calculate the parameters:
Okay, what is the above? Let us break down the above into different parts (Divide and Conquer 😊 )
1. θj is the general value of the parameter, where j can be 0 for 1 (for the above hypothesis function)
2. : = is the assignment operator, which is often used in pseudo-codes in various high-level languages/
3. α – This symbol(alpha) is defined as the learning rate in the above equation. This parameter determines how fast or slow we will move towards the optimal weights. We will discuss more about the learning rate and its optimal values to be plugged in later in this article.
4. What’s this term? Well, to understand this we need to introduce a little bit of calculus here.
The term involves partial derivative of the cost function J() , which means first differentiating it with respect to θ0 and after that differentiating w.r.t θ1 . Differentiating w.r.t one variable means treating the other variables as constant. More info on partial derivatives can be found here.
But then the question arises how this works? The thing is given a function defined by a set of parameters, gradient descent starts with an initial set of parameter values and iteratively moves toward a set of parameter values that minimize the function. This iterative minimization is achieved using calculus, taking steps in the negative direction of the function gradient.
Let us quote this with the help of an example using Linear Regression. We had already discussed what regression is. Linear regression deals with the hypothesis function being linear (i.e of the form y = mx + c).
So let us come back to the problem where we want the computer to calculate the best-fit line given a line y = mx + c.
A standard approach to solving this type of problem is to define an error function (also called a cost function) that measures how “good” a given line is. This function will take in a (m,c) pair and return an error value based on how well the line fits our data. To compute this error for a given line, we’ll iterate through each (x,y) point in our data set and sum the square distances between each point’s y value and the candidate line’s y value (computed at mx + c). It’s conventional to square this distance to ensure that it is positive and to make our error function differentiable. In python, computing the error for a given line will look like (following python code):
Okay, now we start the Gradient Descent. So, now we answer the question – If we minimize this function, we will get the best line for our data, and the best way to minimize is to differentiate.
So, first we compute something called the “Gradient”, which is actually the partial derivative term. This term will always help us moving closer to the optimal location(minimizing the function).
So, now we take our hypothesis function and plug in the data and after that partiuially differentiate w.r.t both and , we repeat this iteration until we find the optimal location.
The code in python is as below:
Now comes the final and the most important part of this algorithm- deciding on the learning rate,α.
Case 1 :If we take alpha very large, Gradient Descent can overshoot the minima and won’t even converge to a local minimum, instead it would diverge, as shown in the following graph.
Case 2: If we take alpha to be very small, then gradient descent will work, but it will proceed very slowly.
Case3 : If we keep the learning rate alpha fixed, then even gradient descent would work with a good amount of accuracy, as it will automatically take smaller steps to move towards the minimal location.
No fixed value of learning rate is generalized, but we always try to keep it between 0 and 1 as far as possible. This isn’t a hard and fast rule and completely depends on the training set and the data.
So i hope now you all have the basic knowledge of the working of the algorithm. Feel free to comment out your doubts. Happy learning 🙂
### Ayushmaan Seth
Content Developer at DataScribble
Highly passionate about solving complex problems and coming out with elegant solutions.
Intensely interested in Machine learning and Data Mining.
### Ayushmaan Seth
Highly passionate about solving complex problems and coming out with elegant solutions. Intensely interested in Machine learning and Data Mining.
### 3 Comments
1. archit jaiswal
when you are calculating the totalerror can you please explain how you are using the variable y and x.
2. @Archit Jaiswal- Sure, it’s my mistake actually, I should have clarified about classes.
So basically the list “points” is a list/array of objects of a class which has two integers(x and y, which are the coordinates of a specific point). “Points” array is a list of objects, each of the object has its own x and y, thus making a coordinate (x,y).
So basically it accesses the x and y coordinate of each member of the dataset and computes the squared error as required.
You can read more about classes and the use of “dot operator” here-
http://reeborg.ca/docs/oop_py_en/oop.html
https://docs.python.org/3/tutorial/classes.html
Thank you for pointing this out, I will make sure to make it simpler. Feel free to comment anything else.
Cheers!
3. javo
Nice post! I want more information because I am starting around this topic. For instance, How I have to define the points, as an array? I run the code, but I need to understand the output, How I can get data?
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# The Gradient Descent Algorithm (With Codes). Hello everyone! Welcome back to the 3rd article on basics of Machine Learning. In this article, we are going to first recap the pre-requisite to Gradient Descent Algorithm(i.e. what we did in the previous article) and then introduce one of the most important algorithm in Machine Learning – Gradient Descent. This algorithm is one of the most used and one of the most powerful algorithm used as practical applications of Machine Learning, such as Stock Prices prediction, or detection of Malignant Breast Cancer.. So, in our previous article we defined our notations for our cost function model by setting up our hypothesis [h(x)=θ+ϕ(x)] and then plugging in the value for the R squared error function,. Where, J (θ0θ1) is the cost function and m are the number of training examples.. Now, we are going to define first algorithm in Machine Learning- Gradient Descent.. Gradient Descent is a Machine Learning algorithm to minimize the cost function J (θ0, θ1 ).. In layman terms, we minimize the above squared error function by using the basic calculus methods to calculate the parameters:. Okay, what is the above? Let us break down the above into different parts (Divide and Conquer 😊 ). 1. θj is the general value of the parameter, where j can be 0 for 1 (for the above hypothesis function). 2. : = is the assignment operator, which is often used in pseudo-codes in various high-level languages/. 3. α – This symbol(alpha) is defined as the learning rate in the above equation. This parameter determines how fast or slow we will move towards the optimal weights. We will discuss more about the learning rate and its optimal values to be plugged in later in this article.. 4. What’s this term? Well, to understand this we need to introduce a little bit of calculus here.. The term involves partial derivative of the cost function J() , which means first differentiating it with respect to θ0 and after that differentiating w.r.t θ1 . Differentiating w.r.t one variable means treating the other variables as constant. More info on partial derivatives can be found here.. But then the question arises how this works? The thing is given a function defined by a set of parameters, gradient descent starts with an initial set of parameter values and iteratively moves toward a set of parameter values that minimize the function. This iterative minimization is achieved using calculus, taking steps in the negative direction of the function gradient.. Let us quote this with the help of an example using Linear Regression. We had already discussed what regression is. Linear regression deals with the hypothesis function being linear (i.e of the form y = mx + c).. So let us come back to the problem where we want the computer to calculate the best-fit line given a line y = mx + c.. A standard approach to solving this type of problem is to define an error function (also called a cost function) that measures how “good” a given line is. This function will take in a (m,c) pair and return an error value based on how well the line fits our data. To compute this error for a given line, we’ll iterate through each (x,y) point in our data set and sum the square distances between each point’s y value and the candidate line’s y value (computed at mx + c). It’s conventional to square this distance to ensure that it is positive and to make our error function differentiable. In python, computing the error for a given line will look like (following python code):. Okay, now we start the Gradient Descent. So, now we answer the question – If we minimize this function, we will get the best line for our data, and the best way to minimize is to differentiate.. So, first we compute something called the “Gradient”, which is actually the partial derivative term. This term will always help us moving closer to the optimal location(minimizing the function).
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So, now we take our hypothesis function and plug in the data and after that partiuially differentiate w.r.t both and , we repeat this iteration until we find the optimal location.. The code in python is as below:. Now comes the final and the most important part of this algorithm- deciding on the learning rate,α.. Case 1 :If we take alpha very large, Gradient Descent can overshoot the minima and won’t even converge to a local minimum, instead it would diverge, as shown in the following graph.. Case 2: If we take alpha to be very small, then gradient descent will work, but it will proceed very slowly.. Case3 : If we keep the learning rate alpha fixed, then even gradient descent would work with a good amount of accuracy, as it will automatically take smaller steps to move towards the minimal location.. No fixed value of learning rate is generalized, but we always try to keep it between 0 and 1 as far as possible. This isn’t a hard and fast rule and completely depends on the training set and the data.. So i hope now you all have the basic knowledge of the working of the algorithm. Feel free to comment out your doubts. Happy learning 🙂. ### Ayushmaan Seth. Content Developer at DataScribble. Highly passionate about solving complex problems and coming out with elegant solutions.. Intensely interested in Machine learning and Data Mining.. ### Ayushmaan Seth. Highly passionate about solving complex problems and coming out with elegant solutions. Intensely interested in Machine learning and Data Mining.. ### 3 Comments. 1. archit jaiswal. when you are calculating the totalerror can you please explain how you are using the variable y and x.. 2. @Archit Jaiswal- Sure, it’s my mistake actually, I should have clarified about classes.. So basically the list “points” is a list/array of objects of a class which has two integers(x and y, which are the coordinates of a specific point). “Points” array is a list of objects, each of the object has its own x and y, thus making a coordinate (x,y).. So basically it accesses the x and y coordinate of each member of the dataset and computes the squared error as required.. You can read more about classes and the use of “dot operator” here-. http://reeborg.ca/docs/oop_py_en/oop.html. https://docs.python.org/3/tutorial/classes.html. Thank you for pointing this out, I will make sure to make it simpler. Feel free to comment anything else.. Cheers!. 3. javo. Nice post! I want more information because I am starting around this topic. For instance, How I have to define the points, as an array? I run the code, but I need to understand the output, How I can get data?.
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https://www.answers.com/Q/What_are_the_odds_of_rolling_a_7_at_a_craps_table
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Math and Arithmetic
Board Games
Statistics
Probability
# What are the odds of rolling a 7 at a craps table?
293031
###### 2008-11-25 07:41:15
1 in 6 The way you calculate this is as follows: There are 36 possible combinations you can get with two dice. Six of those combinations add up to 7 and 30 of those combinations add up to either 6, 5, 4, 3, or 2. Therefore, there are 30 ways you can roll a number other than seven and six ways you can roll a seven. If you want a seven to come out, there are 30 ways to lose and 6 ways to win. This translates into odds of 30 to 6 or 5 to 1. The difference between this and the above answer is the way odds are stated. They can be stated either using "in" or "to." You can say the odds are either 6 ways to win "in" 36 possible combinations or you can say there are 30 ways to lose "to" 6 ways to win. When these are reduced, they become 1 way to win "in" 6 ways to lose and 5 ways to lose "to" 1 way to win. Thus, you get "1 in 6" which is equivalent to "5 to 1".
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## Related Questions
Here are the odds in craps: 7 vs. 6 Odds: 6 to 5 7 vs. 5 Odds: 6 to 4 or 3 to 2 7 vs. 4 Odds: 6 to 3 or 2 to 1 7 vs. 3 Odds: 6 to 2 or 3 to 1 7 vs. 2 Odds: 6 to 1
The odds of rolling a 7 are 1/6. The odds of rolling two in a row are 1/36. The odds of rolling an 11 are 1/18. The odds of rolling two in a row are 1/324. The odds of rolling doubles are 1/6. The odds of rolling double twice in a row are 1/36.
In craps, the chance for rolling a 7 never changes. First roll, tenth roll - doesn't matter. The chance is always 6 out of 36 (or roughly 17%). The chance of rolling a 6 is always 5 out of 36 (or 20%). The chance of rolling an 8 is always 5 out of 36 (or 20%). The chance of hitting either a 6 or an 8 before hitting a 7 would be 10 to 6 (with the favor going towards hitting either the 6 or 8).
Odds of rolling seven once (2 dice) is 1:6, twice in a row, odds are 1:36. The reason is- the first die can be any number, but the second die must be a particular number. For example, if the first die is 6, then the second die has to be 1 to make 7. When an event occurs twice, and the two events are unrelated, then you can multiply the odds together= 1/6 * 1/6 = 1/36.
If you're asking the basic rules of craps, i.e. what bets win and lose based on the value of the dice, you should read a basic craps rules introduction; they're everywhere on the web. Basically, you or someone else rolls the dice. If you roll 7 or 11 you win; 2, 3, or 12 and you lose. If you roll anything else (4, 5, 6, 8, 9, 10), you now have to roll that number AGAIN before rolling a 7. That's the most basic bet, called the Pass line. The Don't Pass is basically the opposite, betting that the shooter will roll a 7 before rolling that number again - "making the point." There are a ton of other bets on the craps table, but many of them are self-explanatory. Again, find a basic introduction. If you're asking how you can maximize your odds, the answer is to play the Pass or Don't Pass with odds, and to buy numbers. Stay out of the field and off the horn (the big section of specific bets down the center of the table). If you're asking how you can gain an edge over the house, sorry, it can't be done. Pass line odds in most casinos pay true and correct odds, making them the only real even-money bet in the whole place. You can't gain an edge though, sorry.
About 1.5% The odds of rolling a 7 on any particular throw is 1/6. (There are six ways you can roll a seven: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), out of 36 possible outcomes.) Therefore, the odds that you don't roll a 7 on a particular roll is 5/6. The odds that you don't roll a 7 in 23 rolls is: (5/6)^23, or approximately .015=1.5%
The odds of rolling a 7 with two dice is 6 in 36, or 1 in 6.Two six-sided dice will yield 36 different possible combinations in one roll. Note that rolling 1 and 6 is not the same as rolling 6 and 1. Yes, they both equal 7, but for the purposes of determining probability, each throw is unique. There are 6 possible ways to throw a 7, and they are 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1. With 36 different possible outcomes, that means that 6 in 36 of those outcomes results in a 7. The odds of getting a 7 are 6 in 36, or 1 in 6 (reducing the two terms).We sometimes see the terms "odds" and "probability" interchanged, but it is important to distinguish them mathematically. If you are interested in the probability of rolling a 7 in a single roll of a pair of dice, use the link to the Related question.
No. The odds paid by the house are always less than the actual odds. The house advantage varies from 1.4% for a simple Pass/Come bet to 16.7% for an Any 7 bet. The only exceptions are the "Odds" bets, which pay at actual odds, but you can't place those independently of other bets.
The most likely number to be rolled in craps is the number 7
If the odds are 9 to 7, the chance of winning is 43.75%
Approximately 1/1120 * You have a 27.777% chance of rolling either a six or an eight on the first roll. * You have a 13.888% chance of rolling the same thing as the first roll. * You have a 13.888% chance of doing it again. * You have a 16.666% chance of rolling a seven. To find total probability in this situation, you multiply all of your chances together. .27777 x .13888 x .13888 x .16666 = 0.000892886 = .0892886% =1/1119.964 You may just be finding this out now, but casinos have known it for a long time. The casinos simply make their payouts less than the probability and they can't lose. While I recognize that your question deals with the probablity of an event recurring, for the average gambler, there is another way to look at this. Dice have no memory. Therefore, each time you roll them, the probability of rolling a given number remains the same. Each time you roll the dice, the chances of rolling a 6 before a 7 is 6 to 5 against you. The same odds apply to rolling an 8 before a 7.
If the odds of a horse winning a horse race are 2 to 7 then the odds against that horse winning the race are 7 to 2.
To roll a 7, it doesn't matter what your first die rolls at, as every number on the die has a complement that will add up to 7. This means that your odds on rolling 7 are actually your odds on rolling the appropriate number on the second die. ie. 1/6. No matter what your first die is, the second can potentially give you a number that brings the total to 7. To roll an 11 however, your odds are not as high. This is because you must roll a five and a six to hit 11. Your first die can be either of the two, but your second die must be it's complement. In this case then, your odds are 2/6 * 1/6, or 1/18.
The odds against drawing a black 7 is 25 in 26.
If you mean casino craps, it depends on how you are betting, but a general answer to the question is, on the first roll, called the "come out" not "roll out," a 7 or 11 is considered a winning number.
If rolling 1 die it is 1; if rolling 7 dice it is 0.
It could be dice (craps) : 7 for a "seven natural" to be thrown (more rarely tossed)
Assuming a properly executed break, about 1/8 breaks. The odds increase the smaller the table - a good player will make the 8 ball 5% of the time on a none foot table. This will nearly double on a 7 foot table. The pockets are the same size on all pool tables in the US, so the smaller the table the greater open space and less distance to travel.
Your total possible number of combinations: 62 = 36 Combinations that can add up to 7: 1-6 2-5 3-4 4-3 5-2 6-1 So your odds of rolling a seven would be 6/36, or 1/6.
The probability of rolling a 7 with 2 dice is 6/36; probability of rolling an 11 is 2/36. Add the two together to find probability of rolling a 7 or 11 which is 8/36 or 2/9.
A natural number is any number that is an automatic winner or loser on the first roll. If you are playing regular craps, the natural winners are 7 and 11, and the natural losers are 2, 3, 11, and 12.
There are 7 periods in the periodic table.
###### ToysMath and ArithmeticBoard GamesGamblingProbabilityCasinosHorse RacingStatisticsDitloids and Letter EquationsBilliards and PoolDice GamesPeriodic Table
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Math and Arithmetic. Board Games. Statistics. Probability. # What are the odds of rolling a 7 at a craps table?. 293031. ###### 2008-11-25 07:41:15. 1 in 6 The way you calculate this is as follows: There are 36 possible combinations you can get with two dice. Six of those combinations add up to 7 and 30 of those combinations add up to either 6, 5, 4, 3, or 2. Therefore, there are 30 ways you can roll a number other than seven and six ways you can roll a seven. If you want a seven to come out, there are 30 ways to lose and 6 ways to win. This translates into odds of 30 to 6 or 5 to 1. The difference between this and the above answer is the way odds are stated. They can be stated either using "in" or "to." You can say the odds are either 6 ways to win "in" 36 possible combinations or you can say there are 30 ways to lose "to" 6 ways to win. When these are reduced, they become 1 way to win "in" 6 ways to lose and 5 ways to lose "to" 1 way to win. Thus, you get "1 in 6" which is equivalent to "5 to 1".. ๐. 0. ๐คจ. 0. ๐ฎ. 0. ๐. 0. ## Related Questions. Here are the odds in craps: 7 vs. 6 Odds: 6 to 5 7 vs. 5 Odds: 6 to 4 or 3 to 2 7 vs. 4 Odds: 6 to 3 or 2 to 1 7 vs. 3 Odds: 6 to 2 or 3 to 1 7 vs. 2 Odds: 6 to 1. The odds of rolling a 7 are 1/6. The odds of rolling two in a row are 1/36. The odds of rolling an 11 are 1/18. The odds of rolling two in a row are 1/324. The odds of rolling doubles are 1/6. The odds of rolling double twice in a row are 1/36.. In craps, the chance for rolling a 7 never changes. First roll, tenth roll - doesn't matter. The chance is always 6 out of 36 (or roughly 17%). The chance of rolling a 6 is always 5 out of 36 (or 20%). The chance of rolling an 8 is always 5 out of 36 (or 20%). The chance of hitting either a 6 or an 8 before hitting a 7 would be 10 to 6 (with the favor going towards hitting either the 6 or 8).. Odds of rolling seven once (2 dice) is 1:6, twice in a row, odds are 1:36. The reason is- the first die can be any number, but the second die must be a particular number. For example, if the first die is 6, then the second die has to be 1 to make 7. When an event occurs twice, and the two events are unrelated, then you can multiply the odds together= 1/6 * 1/6 = 1/36.. If you're asking the basic rules of craps, i.e. what bets win and lose based on the value of the dice, you should read a basic craps rules introduction; they're everywhere on the web. Basically, you or someone else rolls the dice. If you roll 7 or 11 you win; 2, 3, or 12 and you lose. If you roll anything else (4, 5, 6, 8, 9, 10), you now have to roll that number AGAIN before rolling a 7. That's the most basic bet, called the Pass line. The Don't Pass is basically the opposite, betting that the shooter will roll a 7 before rolling that number again - "making the point." There are a ton of other bets on the craps table, but many of them are self-explanatory. Again, find a basic introduction. If you're asking how you can maximize your odds, the answer is to play the Pass or Don't Pass with odds, and to buy numbers. Stay out of the field and off the horn (the big section of specific bets down the center of the table). If you're asking how you can gain an edge over the house, sorry, it can't be done. Pass line odds in most casinos pay true and correct odds, making them the only real even-money bet in the whole place. You can't gain an edge though, sorry.. About 1.5% The odds of rolling a 7 on any particular throw is 1/6.
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(There are six ways you can roll a seven: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), out of 36 possible outcomes.) Therefore, the odds that you don't roll a 7 on a particular roll is 5/6. The odds that you don't roll a 7 in 23 rolls is: (5/6)^23, or approximately .015=1.5%. The odds of rolling a 7 with two dice is 6 in 36, or 1 in 6.Two six-sided dice will yield 36 different possible combinations in one roll. Note that rolling 1 and 6 is not the same as rolling 6 and 1. Yes, they both equal 7, but for the purposes of determining probability, each throw is unique. There are 6 possible ways to throw a 7, and they are 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1. With 36 different possible outcomes, that means that 6 in 36 of those outcomes results in a 7. The odds of getting a 7 are 6 in 36, or 1 in 6 (reducing the two terms).We sometimes see the terms "odds" and "probability" interchanged, but it is important to distinguish them mathematically. If you are interested in the probability of rolling a 7 in a single roll of a pair of dice, use the link to the Related question.. No. The odds paid by the house are always less than the actual odds. The house advantage varies from 1.4% for a simple Pass/Come bet to 16.7% for an Any 7 bet. The only exceptions are the "Odds" bets, which pay at actual odds, but you can't place those independently of other bets.. The most likely number to be rolled in craps is the number 7. If the odds are 9 to 7, the chance of winning is 43.75%. Approximately 1/1120 * You have a 27.777% chance of rolling either a six or an eight on the first roll. * You have a 13.888% chance of rolling the same thing as the first roll. * You have a 13.888% chance of doing it again. * You have a 16.666% chance of rolling a seven. To find total probability in this situation, you multiply all of your chances together. .27777 x .13888 x .13888 x .16666 = 0.000892886 = .0892886% =1/1119.964 You may just be finding this out now, but casinos have known it for a long time. The casinos simply make their payouts less than the probability and they can't lose. While I recognize that your question deals with the probablity of an event recurring, for the average gambler, there is another way to look at this. Dice have no memory. Therefore, each time you roll them, the probability of rolling a given number remains the same. Each time you roll the dice, the chances of rolling a 6 before a 7 is 6 to 5 against you. The same odds apply to rolling an 8 before a 7.. If the odds of a horse winning a horse race are 2 to 7 then the odds against that horse winning the race are 7 to 2.. To roll a 7, it doesn't matter what your first die rolls at, as every number on the die has a complement that will add up to 7. This means that your odds on rolling 7 are actually your odds on rolling the appropriate number on the second die. ie. 1/6. No matter what your first die is, the second can potentially give you a number that brings the total to 7. To roll an 11 however, your odds are not as high. This is because you must roll a five and a six to hit 11. Your first die can be either of the two, but your second die must be it's complement. In this case then, your odds are 2/6 * 1/6, or 1/18.. The odds against drawing a black 7 is 25 in 26.. If you mean casino craps, it depends on how you are betting, but a general answer to the question is, on the first roll, called the "come out" not "roll out," a 7 or 11 is considered a winning number.. If rolling 1 die it is 1; if rolling 7 dice it is 0.. It could be dice (craps) : 7 for a "seven natural" to be thrown (more rarely tossed). Assuming a properly executed break, about 1/8 breaks. The odds increase the smaller the table - a good player will make the 8 ball 5% of the time on a none foot table. This will nearly double on a 7 foot table. The pockets are the same size on all pool tables in the US, so the smaller the table the greater open space and less distance to travel.. Your total possible number of combinations: 62 = 36 Combinations that can add up to 7: 1-6 2-5 3-4 4-3 5-2 6-1 So your odds of rolling a seven would be 6/36, or 1/6.. The probability of rolling a 7 with 2 dice is 6/36; probability of rolling an 11 is 2/36. Add the two together to find probability of rolling a 7 or 11 which is 8/36 or 2/9.. A natural number is any number that is an automatic winner or loser on the first roll. If you are playing regular craps, the natural winners are 7 and 11, and the natural losers are 2, 3, 11, and 12.. There are 7 periods in the periodic table.. ###### ToysMath and ArithmeticBoard GamesGamblingProbabilityCasinosHorse RacingStatisticsDitloids and Letter EquationsBilliards and PoolDice GamesPeriodic Table. Copyright ยฉ 2020 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply.
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# Table of 71
Created by: Team Maths - Examples.com, Last Updated: April 28, 2024
## Table of 71
In the vast and intricate world of mathematics, the multiplication tables serve as the foundation for building a strong numerical understanding. Among these, the table of 71 stands out as a particularly advanced tool, facilitating not just the basic arithmetic skills but also enhancing one’s ability to tackle more complex mathematical challenges. Multiplying the number 71 by various whole numbers showcases the process of adding the number 71 repeatedly, a principle fundamental to the concept of multiplication.
For instance, multiplying 71 by 2 (71 x 2) is equivalent to adding 71 to itself once, which results in 142. This simple yet effective strategy, when applied across a range of numbers, produces a broad spectrum of outcomes. These outcomes are essential for developing quick and efficient calculation skills, crucial in both academic contexts and everyday practical scenarios.
## What is the Multiplication Table of 71?
The Multiplication Table of 71 is a comprehensive chart that illustrates the results of multiplying the number 71 by various whole numbers. Similar to the table of 35, it serves as an essential tool for students aiming to enhance their rapid calculation skills and deepen their understanding of numerical patterns. This table simplifies the process of adding the number 71 repeatedly, providing a quick and efficient method for solving multiplication problems.
• 71 times 1 = 71 (71 x 1 = 71)
• 71 times 2 = 142 (71 x 2 = 142)
• 71 times 3 = 213 (71 x 3 = 213
## Multiplication Table of 71
Understanding and mastering the Multiplication Table of 71 is a significant academic achievement for students. It not only enhances their arithmetic skills but also builds a strong foundation for tackling more complex mathematical concepts in the future. The table below is divided into two parts for ease of learning: the first column shows multiplication from 1 to 10, and the second column illustrates multiplication from 11 to 20, providing a clear and structured approach to learning the table of 71.
Multiplication from (1-10) Multiplication from (11-20)
71 x 1 = 71 71 x 11 = 781
71 x 2 = 142 71 x 12 = 852
71 x 3 = 213 71 x 13 = 923
71 x 4 = 284 71 x 14 = 994
71 x 5 = 355 71 x 15 = 1065
71 x 6 = 426 71 x 16 = 1136
71 x 7 = 497 71 x 17 = 1207
71 x 8 = 568 71 x 18 = 1278
71 x 9 = 639 71 x 19 = 1349
71 x 10 = 710 71 x 20 = 1420
## 71 Times Table
Learning the 71 times table can seem like a daunting task at first, but with the right approach and understanding, it can become an easy and rewarding part of mathematics for students. Multiplication tables are fundamental building blocks in math, enhancing a student’s ability to solve complex problems quickly.
Multiplication Result
71 x 1 71
71 x 2 142
71 x 3 213
71 x 4 284
71 x 5 355
71 x 6 426
71 x 7 497
71 x 8 568
71 x 9 639
71 x 10 710
71 x 11 781
71 x 12 852
71 x 13 923
71 x 14 994
71 x 15 1065
71 x 16 1136
71 x 17 1207
71 x 18 1278
71 x 19 1349
71 x 20 1420
## Tips for 71 Times Table
Mastering the 71 Times Table is a significant step towards enhancing mathematical proficiency. Here are five tips designed to aid students in understanding and learning the Multiplication Table of 71 with ease:
### 1. Break It Down into Segments
• Approach: Divide the table into smaller sections for easier learning. Start with the first 10 multiples (1-10), then gradually move to higher numbers.
• Benefit: This reduces the cognitive load, making it less overwhelming and allowing for focused practice on each segment before moving on.
### 2. Utilize Visual Aids
• Approach: Employ charts, flashcards, or digital apps that visually represent the Multiplication Table of 71. Color-coding or highlighting patterns can be particularly helpful.
• Benefit: Visual aids enhance memory retention by providing a graphical representation of the numbers, making it easier to spot patterns and relationships within the table.
### 3. Practice Regularly
• Approach: Dedicate a specific time each day for practicing the table. Use various methods such as writing it down, recitation, or using interactive online quizzes.
• Benefit: Regular practice reinforces memory and aids in the transition from conscious effort to automatic recall, making calculations faster and more accurate.
### 4. Apply Real-Life Scenarios
• Approach: Find real-world situations or problems that can be solved using the 71 Times Table. This could involve scenarios related to counting items, dividing tasks, or calculating distances.
• Benefit: Applying mathematical concepts to real-life situations enhances understanding and demonstrates the practical utility of what is being learned, thereby increasing interest and engagement.
### 5. Peer Learning and Teaching
• Approach: Encourage students to teach the table of 71 to their peers or family members. Explaining concepts to others is a powerful method to reinforce one’s own understanding.
• Benefit: Teaching requires the student to organize their knowledge, clarify their thinking, and check for understanding, all of which solidify their grasp of the table.
## Table of 71 from 11 to 20
Learning and understanding the Multiplication Table of 71, especially from 11 to 20, empowers students to tackle advanced mathematical problems with confidence. This section of the table can seem challenging due to the larger numbers involved, but with strategic approaches such as breaking down the multiplication process, using mnemonic devices, and consistent practice, students can master it efficiently.
Multiplication Result
71 x 11 781
71 x 12 852
71 x 13 923
71 x 14 994
71 x 15 1065
71 x 16 1136
71 x 17 1207
71 x 18 1278
71 x 19 1349
71 x 20 1420
## Simplest Way To Memorize Table 71
71 x 1 71 71
71 x 2 71 + 71 142
71 x 3 71 + 71 + 71 213
71 x 4 71 + 71 + 71 + 71 284
71 x 5 71 + 71 + 71 + 71 + 71 355
71 x 6 71 + 71 + 71 + 71 + 71 + 71 426
71 x 7 71 + 71 + 71 + 71 + 71 + 71 + 71 497
71 x 8 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 568
71 x 9 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 639
71 x 10 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 710
## Representation of Multiplication Table as Addition
1. Multiplying 71 by 2 (71 x 2): Instead of viewing it as a multiplication problem, you can see it as adding 71 to itself once (71 + 71), which equals 142.
2. Multiplying 71 by 3 (71 x 3): This can be visualized as adding 71 three times (71 + 71 + 71), resulting in 213.
1 71 71
2 71 + 71 142
3 71 + 71 + 71 213
4 71 + 71 + 71 + 71 284
5 71 + 71 + 71 + 71 + 71 355
6 71 + 71 + 71 + 71 + 71 + 71 426
7 71 + 71 + 71 + 71 + 71 + 71 + 71 497
8 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 568
9 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 639
10 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 710
This table not only illustrates how multiplication can be broken down into simpler addition steps but also emphasizes the pattern recognition and arithmetic skills that are crucial in the early stages of mathematical learning. By engaging with multiplication in this manner, students can develop a more intuitive understanding of numbers and their relationships, enhancing their overall numeracy skills.
## How to Read 71 Times Tables
Mastering the multiplication table of 71 can initially seem daunting, but with a strategic approach, students can find it manageable and even enjoyable. The key lies in recognizing patterns, practicing regularly, and understanding the concept of repeated addition. By breaking down the table into smaller, more digestible segments, learners can gradually build their competence and confidence.
One time 71 is 71
Two times 71 is 142
Three times 71 is 213
Four times 71 is 284
Five times 71 is 355
Six times 71 is 426
Seven times 71 is 497
Eight times 71 is 568
Nine times 71 is 639
Ten times 71 is 710
## Multiplication Table of 71 till 100
71 x 1 = 71 71 x 21 = 1491 71 x 41 = 2911 71 x 61 = 4331 71 x 81 = 5751 71 x 2 = 142 71 x 22 = 1562 71 x 42 = 2982 71 x 62 = 4402 71 x 82 = 5822 71 x 3 = 213 71 x 23 = 1633 71 x 43 = 3053 71 x 63 = 4473 71 x 83 = 5893 71 x 4 = 284 71 x 24 = 1704 71 x 44 = 3124 71 x 64 = 4544 71 x 84 = 5964 71 x 5 = 355 71 x 25 = 1775 71 x 45 = 3195 71 x 65 = 4615 71 x 85 = 6035 71 x 6 = 426 71 x 26 = 1846 71 x 46 = 3266 71 x 66 = 4686 71 x 86 = 6106 71 x 7 = 497 71 x 27 = 1917 71 x 47 = 3337 71 x 67 = 4757 71 x 87 = 6177 71 x 8 = 568 71 x 28 = 1988 71 x 48 = 3408 71 x 68 = 4828 71 x 88 = 6248 71 x 9 = 639 71 x 29 = 2059 71 x 49 = 3479 71 x 69 = 4899 71 x 89 = 6319 71 x 10 = 710 71 x 30 = 2130 71 x 50 = 3550 71 x 70 = 4970 71 x 90 = 6390 71 x 11 = 781 71 x 31 = 2201 71 x 51 = 3621 71 x 71 = 5041 71 x 91 = 6461 71 x 12 = 852 71 x 32 = 2272 71 x 52 = 3692 71 x 72 = 5112 71 x 92 = 6532 71 x 13 = 923 71 x 33 = 2343 71 x 53 = 3763 71 x 73 = 5183 71 x 93 = 6603 71 x 14 = 994 71 x 34 = 2414 71 x 54 = 3834 71 x 74 = 5254 71 x 94 = 6674 71 x 15 = 1065 71 x 35 = 2485 71 x 55 = 3905 71 x 75 = 5325 71 x 95 = 6745 71 x 16 = 1136 71 x 36 = 2556 71 x 56 = 3976 71 x 76 = 5396 71 x 96 = 6816 71 x 17 = 1207 71 x 37 = 2627 71 x 57 = 4047 71 x 77 = 5467 71 x 97 = 6887 71 x 18 = 1278 71 x 38 = 2698 71 x 58 = 4118 71 x 78 = 5538 71 x 98 = 6958 71 x 19 = 1349 71 x 39 = 2769 71 x 59 = 4189 71 x 79 = 5609 71 x 99 = 7029 71 x 20 = 1420 71 x 40 = 2840 71 x 60 = 4260 71 x 80 = 5680 71 x 100 = 7100
## 71Times Table From 101 to 200
71 x 101 = 7171 71 x 121 = 8591 71 x 141 = 10011 71 x 161 = 11431 71 x 181 = 12851 71 x 102 = 7242 71 x 122 = 8662 71 x 142 = 10082 71 x 162 = 11502 71 x 182 = 12922 71 x 103 = 7313 71 x 123 = 8733 71 x 143 = 10153 71 x 163 = 11573 71 x 183 = 12993 71 x 104 = 7384 71 x 124 = 8804 71 x 144 = 10224 71 x 164 = 11644 71 x 184 = 13064 71 x 105 = 7455 71 x 125 = 8875 71 x 145 = 10295 71 x 165 = 11715 71 x 185 = 13135 71 x 106 = 7526 71 x 126 = 8946 71 x 146 = 10366 71 x 166 = 11786 71 x 186 = 13206 71 x 107 = 7597 71 x 127 = 9017 71 x 147 = 10437 71 x 167 = 11857 71 x 187 = 13277 71 x 108 = 7668 71 x 128 = 9088 71 x 148 = 10508 71 x 168 = 11928 71 x 188 = 13348 71 x 109 = 7739 71 x 129 = 9159 71 x 149 = 10579 71 x 169 = 11999 71 x 189 = 13419 71 x 110 = 7810 71 x 130 = 9230 71 x 150 = 10650 71 x 170 = 12070 71 x 190 = 13490 71 x 111 = 7881 71 x 131 = 9301 71 x 151 = 10721 71 x 171 = 12141 71 x 191 = 13561 71 x 112 = 7952 71 x 132 = 9372 71 x 152 = 10792 71 x 172 = 12212 71 x 192 = 13632 71 x 113 = 8023 71 x 133 = 9443 71 x 153 = 10863 71 x 173 = 12283 71 x 193 = 13703 71 x 114 = 8094 71 x 134 = 9514 71 x 154 = 10934 71 x 174 = 12354 71 x 194 = 13774 71 x 115 = 8165 71 x 135 = 9585 71 x 155 = 11005 71 x 175 = 12425 71 x 195 = 13845 71 x 116 = 8236 71 x 136 = 9656 71 x 156 = 11076 71 x 176 = 12496 71 x 196 = 13916 71 x 117 = 8307 71 x 137 = 9727 71 x 157 = 11147 71 x 177 = 12567 71 x 197 = 13987 71 x 118 = 8378 71 x 138 = 9798 71 x 158 = 11218 71 x 178 = 12638 71 x 198 = 14058 71 x 119 = 8449 71 x 139 = 9869 71 x 159 = 11289 71 x 179 = 12709 71 x 199 = 14129 71 x 120 = 8520 71 x 140 = 9940 71 x 160 = 11360 71 x 180 = 12780 71 x 200 = 14200
## 71Times Table From 201 to 300
71 x 201 = 14271 71 x 221 = 15691 71 x 241 = 17111 71 x 261 = 18531 71 x 281 = 19951 71 x 202 = 14342 71 x 222 = 15762 71 x 242 = 17182 71 x 262 = 18602 71 x 282 = 20022 71 x 203 = 14413 71 x 223 = 15833 71 x 243 = 17253 71 x 263 = 18673 71 x 283 = 20093 71 x 204 = 14484 71 x 224 = 15904 71 x 244 = 17324 71 x 264 = 18744 71 x 284 = 20164 71 x 205 = 14555 71 x 225 = 15975 71 x 245 = 17395 71 x 265 = 18815 71 x 285 = 20235 71 x 206 = 14626 71 x 226 = 16046 71 x 246 = 17466 71 x 266 = 18886 71 x 286 = 20306 71 x 207 = 14697 71 x 227 = 16117 71 x 247 = 17537 71 x 267 = 18957 71 x 287 = 20377 71 x 208 = 14768 71 x 228 = 16188 71 x 248 = 17608 71 x 268 = 19028 71 x 288 = 20448 71 x 209 = 14839 71 x 229 = 16259 71 x 249 = 17679 71 x 269 = 19099 71 x 289 = 20519 71 x 210 = 14910 71 x 230 = 16330 71 x 250 = 17750 71 x 270 = 19170 71 x 290 = 20590 71 x 211 = 14981 71 x 231 = 16401 71 x 251 = 17821 71 x 271 = 19241 71 x 291 = 20661 71 x 212 = 15052 71 x 232 = 16472 71 x 252 = 17892 71 x 272 = 19312 71 x 292 = 20732 71 x 213 = 15123 71 x 233 = 16543 71 x 253 = 17963 71 x 273 = 19383 71 x 293 = 20803 71 x 214 = 15194 71 x 234 = 16614 71 x 254 = 18034 71 x 274 = 19454 71 x 294 = 20874 71 x 215 = 15265 71 x 235 = 16685 71 x 255 = 18105 71 x 275 = 19525 71 x 295 = 20945 71 x 216 = 15336 71 x 236 = 16756 71 x 256 = 18176 71 x 276 = 19596 71 x 296 = 21016 71 x 217 = 15407 71 x 237 = 16827 71 x 257 = 18247 71 x 277 = 19667 71 x 297 = 21087 71 x 218 = 15478 71 x 238 = 16898 71 x 258 = 18318 71 x 278 = 19738 71 x 298 = 21158 71 x 219 = 15549 71 x 239 = 16969 71 x 259 = 18389 71 x 279 = 19809 71 x 299 = 21229 71 x 220 = 15620 71 x 240 = 17040 71 x 260 = 18460 71 x 280 = 19880 71 x 300 = 21300
## Solved Examples
Example 1:
Question: What is 4 times 71?
Solution: Multiply 71 by 4.
Calculation: 71 * 4 = 284
Example 2:
Question: How much is 7 times 71?
Solution: Multiply 71 by 7.
Calculation: 71 * 7 = 497
Example 3:
Question: Calculate 9 times 71.
Solution: Multiply 71 by 9.
Calculation: 71 * 9 = 639
Example 4:
Question: What do you get when you multiply 71 by 5?
Solution: Multiply 71 by 5.
Calculation: 71 * 5 = 355
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# Table of 71. Created by: Team Maths - Examples.com, Last Updated: April 28, 2024. ## Table of 71. In the vast and intricate world of mathematics, the multiplication tables serve as the foundation for building a strong numerical understanding. Among these, the table of 71 stands out as a particularly advanced tool, facilitating not just the basic arithmetic skills but also enhancing one’s ability to tackle more complex mathematical challenges. Multiplying the number 71 by various whole numbers showcases the process of adding the number 71 repeatedly, a principle fundamental to the concept of multiplication.. For instance, multiplying 71 by 2 (71 x 2) is equivalent to adding 71 to itself once, which results in 142. This simple yet effective strategy, when applied across a range of numbers, produces a broad spectrum of outcomes. These outcomes are essential for developing quick and efficient calculation skills, crucial in both academic contexts and everyday practical scenarios.. ## What is the Multiplication Table of 71?. The Multiplication Table of 71 is a comprehensive chart that illustrates the results of multiplying the number 71 by various whole numbers. Similar to the table of 35, it serves as an essential tool for students aiming to enhance their rapid calculation skills and deepen their understanding of numerical patterns. This table simplifies the process of adding the number 71 repeatedly, providing a quick and efficient method for solving multiplication problems.. • 71 times 1 = 71 (71 x 1 = 71). • 71 times 2 = 142 (71 x 2 = 142). • 71 times 3 = 213 (71 x 3 = 213. ## Multiplication Table of 71. Understanding and mastering the Multiplication Table of 71 is a significant academic achievement for students. It not only enhances their arithmetic skills but also builds a strong foundation for tackling more complex mathematical concepts in the future. The table below is divided into two parts for ease of learning: the first column shows multiplication from 1 to 10, and the second column illustrates multiplication from 11 to 20, providing a clear and structured approach to learning the table of 71.. Multiplication from (1-10) Multiplication from (11-20). 71 x 1 = 71 71 x 11 = 781. 71 x 2 = 142 71 x 12 = 852. 71 x 3 = 213 71 x 13 = 923. 71 x 4 = 284 71 x 14 = 994. 71 x 5 = 355 71 x 15 = 1065. 71 x 6 = 426 71 x 16 = 1136. 71 x 7 = 497 71 x 17 = 1207. 71 x 8 = 568 71 x 18 = 1278. 71 x 9 = 639 71 x 19 = 1349. 71 x 10 = 710 71 x 20 = 1420. ## 71 Times Table. Learning the 71 times table can seem like a daunting task at first, but with the right approach and understanding, it can become an easy and rewarding part of mathematics for students. Multiplication tables are fundamental building blocks in math, enhancing a student’s ability to solve complex problems quickly.. Multiplication Result. 71 x 1 71. 71 x 2 142. 71 x 3 213. 71 x 4 284. 71 x 5 355. 71 x 6 426. 71 x 7 497. 71 x 8 568. 71 x 9 639. 71 x 10 710. 71 x 11 781. 71 x 12 852. 71 x 13 923. 71 x 14 994. 71 x 15 1065. 71 x 16 1136. 71 x 17 1207. 71 x 18 1278. 71 x 19 1349. 71 x 20 1420. ## Tips for 71 Times Table. Mastering the 71 Times Table is a significant step towards enhancing mathematical proficiency. Here are five tips designed to aid students in understanding and learning the Multiplication Table of 71 with ease:. ### 1. Break It Down into Segments. • Approach: Divide the table into smaller sections for easier learning. Start with the first 10 multiples (1-10), then gradually move to higher numbers.. • Benefit: This reduces the cognitive load, making it less overwhelming and allowing for focused practice on each segment before moving on.. ### 2. Utilize Visual Aids. • Approach: Employ charts, flashcards, or digital apps that visually represent the Multiplication Table of 71. Color-coding or highlighting patterns can be particularly helpful.. • Benefit: Visual aids enhance memory retention by providing a graphical representation of the numbers, making it easier to spot patterns and relationships within the table.. ### 3. Practice Regularly. • Approach: Dedicate a specific time each day for practicing the table. Use various methods such as writing it down, recitation, or using interactive online quizzes.. • Benefit: Regular practice reinforces memory and aids in the transition from conscious effort to automatic recall, making calculations faster and more accurate.. ### 4. Apply Real-Life Scenarios. • Approach: Find real-world situations or problems that can be solved using the 71 Times Table. This could involve scenarios related to counting items, dividing tasks, or calculating distances.. • Benefit: Applying mathematical concepts to real-life situations enhances understanding and demonstrates the practical utility of what is being learned, thereby increasing interest and engagement.. ### 5. Peer Learning and Teaching. • Approach: Encourage students to teach the table of 71 to their peers or family members. Explaining concepts to others is a powerful method to reinforce one’s own understanding.
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• Benefit: Teaching requires the student to organize their knowledge, clarify their thinking, and check for understanding, all of which solidify their grasp of the table.. ## Table of 71 from 11 to 20. Learning and understanding the Multiplication Table of 71, especially from 11 to 20, empowers students to tackle advanced mathematical problems with confidence. This section of the table can seem challenging due to the larger numbers involved, but with strategic approaches such as breaking down the multiplication process, using mnemonic devices, and consistent practice, students can master it efficiently.. Multiplication Result. 71 x 11 781. 71 x 12 852. 71 x 13 923. 71 x 14 994. 71 x 15 1065. 71 x 16 1136. 71 x 17 1207. 71 x 18 1278. 71 x 19 1349. 71 x 20 1420. ## Simplest Way To Memorize Table 71. 71 x 1 71 71. 71 x 2 71 + 71 142. 71 x 3 71 + 71 + 71 213. 71 x 4 71 + 71 + 71 + 71 284. 71 x 5 71 + 71 + 71 + 71 + 71 355. 71 x 6 71 + 71 + 71 + 71 + 71 + 71 426. 71 x 7 71 + 71 + 71 + 71 + 71 + 71 + 71 497. 71 x 8 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 568. 71 x 9 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 639. 71 x 10 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 710. ## Representation of Multiplication Table as Addition. 1. Multiplying 71 by 2 (71 x 2): Instead of viewing it as a multiplication problem, you can see it as adding 71 to itself once (71 + 71), which equals 142.. 2. Multiplying 71 by 3 (71 x 3): This can be visualized as adding 71 three times (71 + 71 + 71), resulting in 213.. 1 71 71. 2 71 + 71 142. 3 71 + 71 + 71 213. 4 71 + 71 + 71 + 71 284. 5 71 + 71 + 71 + 71 + 71 355. 6 71 + 71 + 71 + 71 + 71 + 71 426. 7 71 + 71 + 71 + 71 + 71 + 71 + 71 497. 8 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 568. 9 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 639. 10 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 + 71 710. This table not only illustrates how multiplication can be broken down into simpler addition steps but also emphasizes the pattern recognition and arithmetic skills that are crucial in the early stages of mathematical learning. By engaging with multiplication in this manner, students can develop a more intuitive understanding of numbers and their relationships, enhancing their overall numeracy skills.. ## How to Read 71 Times Tables. Mastering the multiplication table of 71 can initially seem daunting, but with a strategic approach, students can find it manageable and even enjoyable. The key lies in recognizing patterns, practicing regularly, and understanding the concept of repeated addition. By breaking down the table into smaller, more digestible segments, learners can gradually build their competence and confidence.. One time 71 is 71. Two times 71 is 142. Three times 71 is 213. Four times 71 is 284. Five times 71 is 355. Six times 71 is 426. Seven times 71 is 497. Eight times 71 is 568. Nine times 71 is 639. Ten times 71 is 710. ## Multiplication Table of 71 till 100. 71 x 1 = 71 71 x 21 = 1491 71 x 41 = 2911 71 x 61 = 4331 71 x 81 = 5751 71 x 2 = 142 71 x 22 = 1562 71 x 42 = 2982 71 x 62 = 4402 71 x 82 = 5822 71 x 3 = 213 71 x 23 = 1633 71 x 43 = 3053 71 x 63 = 4473 71 x 83 = 5893 71 x 4 = 284 71 x 24 = 1704 71 x 44 = 3124 71 x 64 = 4544 71 x 84 = 5964 71 x 5 = 355 71 x 25 = 1775 71 x 45 = 3195 71 x 65 = 4615 71 x 85 = 6035 71 x 6 = 426 71 x 26 = 1846 71 x 46 = 3266 71 x 66 = 4686 71 x 86 = 6106 71 x 7 = 497 71 x 27 = 1917 71 x 47 = 3337 71 x 67 = 4757 71 x 87 = 6177 71 x 8 = 568 71 x 28 = 1988 71 x 48 = 3408 71 x 68 = 4828 71 x 88 = 6248 71 x 9 = 639 71 x 29 = 2059 71 x 49 = 3479 71 x 69 = 4899 71 x 89 = 6319 71 x 10 = 710 71 x 30 = 2130 71 x 50 = 3550 71 x 70 = 4970 71 x 90 = 6390 71 x 11 = 781 71 x 31 = 2201 71 x 51 = 3621 71 x 71 = 5041 71 x 91 = 6461 71 x 12 = 852 71 x 32 = 2272 71 x 52 = 3692 71 x 72 = 5112 71 x 92 = 6532 71 x 13 = 923 71 x 33 = 2343 71 x 53 = 3763 71 x 73 = 5183 71 x 93 = 6603 71 x 14 = 994 71 x 34 = 2414 71 x 54 = 3834 71 x 74 = 5254 71 x 94 = 6674 71 x 15 = 1065 71 x 35 = 2485 71 x 55 = 3905 71 x 75 = 5325 71 x 95 = 6745 71 x 16 = 1136 71 x 36 = 2556 71 x 56 = 3976 71 x 76 = 5396 71 x 96 = 6816 71 x 17 = 1207 71 x 37 = 2627 71 x 57 = 4047 71 x 77 = 5467 71 x 97 = 6887 71 x 18 = 1278 71 x 38 = 2698 71 x 58 = 4118 71 x 78 = 5538 71 x 98 = 6958 71 x 19 = 1349 71 x 39 = 2769 71 x 59 = 4189 71 x 79 = 5609 71 x 99 = 7029 71 x 20 = 1420 71 x 40 = 2840 71 x 60 = 4260 71 x 80 = 5680 71 x 100 = 7100. ## 71Times Table From 101 to 200. 71 x 101 = 7171 71 x 121 = 8591 71 x 141 = 10011 71 x 161 = 11431 71 x 181 = 12851 71 x 102 = 7242 71 x 122 = 8662 71 x 142 = 10082 71 x 162 = 11502 71 x 182 = 12922 71 x 103 = 7313 71 x 123 = 8733 71 x 143 = 10153 71 x 163 = 11573 71 x 183 = 12993 71 x 104 = 7384 71 x 124 = 8804 71 x 144 = 10224 71 x 164 = 11644 71 x 184 = 13064 71 x 105 = 7455 71 x 125 = 8875 71 x 145 = 10295 71 x 165 = 11715 71 x 185 = 13135 71 x 106 = 7526 71 x 126 = 8946 71 x 146 = 10366 71 x 166 = 11786 71 x 186 = 13206 71 x 107 = 7597 71 x 127 = 9017 71 x 147 = 10437 71 x 167 = 11857 71 x 187 = 13277 71 x 108 = 7668 71 x 128 = 9088 71 x 148 = 10508 71 x 168 = 11928 71 x 188 = 13348 71 x 109 = 7739 71 x 129 = 9159 71 x 149 = 10579 71 x 169 = 11999 71 x 189 = 13419 71 x 110 = 7810 71 x 130 = 9230 71 x 150 = 10650 71 x 170 = 12070 71 x 190 = 13490 71 x 111 = 7881 71 x 131 = 9301 71 x 151 = 10721 71 x 171 = 12141 71 x 191 = 13561 71 x 112 = 7952 71 x 132 = 9372 71 x 152 = 10792 71 x 172 = 12212 71 x 192 = 13632 71 x 113 = 8023 71 x 133 = 9443 71 x 153 = 10863 71 x 173 = 12283 71 x 193 = 13703 71 x 114 = 8094 71 x 134 = 9514 71 x 154 = 10934 71 x 174 = 12354 71 x 194 = 13774 71 x 115 = 8165 71 x 135 = 9585 71 x 155 = 11005 71 x 175 = 12425 71 x 195 = 13845 71 x 116 = 8236 71 x 136 = 9656 71 x 156 = 11076 71 x 176 = 12496 71 x 196 = 13916 71 x 117 = 8307 71 x 137 = 9727 71 x 157 = 11147 71 x 177 = 12567 71 x 197 = 13987 71 x 118 = 8378 71 x 138 = 9798 71 x 158 = 11218 71 x 178 = 12638 71 x 198 = 14058 71 x 119 = 8449 71 x 139 = 9869 71 x 159 = 11289 71 x 179 = 12709 71 x 199 = 14129 71 x 120 = 8520 71 x 140 = 9940 71 x 160 = 11360 71 x 180 = 12780 71 x 200 = 14200. ## 71Times Table From 201 to 300. 71 x 201 = 14271 71 x 221 = 15691 71 x 241 = 17111 71 x 261 = 18531 71 x 281 = 19951 71 x 202 = 14342 71 x 222 = 15762 71 x 242 = 17182 71 x 262 = 18602 71 x 282 = 20022 71 x 203 = 14413 71 x 223 = 15833 71 x 243 = 17253 71 x 263 = 18673 71 x 283 = 20093 71 x 204 = 14484 71 x 224 = 15904 71 x 244 = 17324 71 x 264 = 18744 71 x 284 = 20164 71 x 205 = 14555 71 x 225 = 15975 71 x 245 = 17395 71 x 265 = 18815 71 x 285 = 20235 71 x 206 = 14626 71 x 226 = 16046 71 x 246 = 17466 71 x 266 = 18886 71 x 286 = 20306 71 x 207 = 14697 71 x 227 = 16117 71 x 247 = 17537 71 x 267 = 18957 71 x 287 = 20377 71 x 208 = 14768 71 x 228 = 16188 71 x 248 = 17608 71 x 268 = 19028 71 x 288 = 20448 71 x 209 = 14839 71 x 229 = 16259 71 x 249 = 17679 71 x 269 = 19099 71 x 289 = 20519 71 x 210 = 14910 71 x 230 = 16330 71 x 250 = 17750 71 x 270 = 19170 71 x 290 = 20590 71 x 211 = 14981 71 x 231 = 16401 71 x 251 = 17821 71 x 271 = 19241 71 x 291 = 20661 71 x 212 = 15052 71 x 232 = 16472 71 x 252 = 17892 71 x 272 = 19312 71 x 292 = 20732 71 x 213 = 15123 71 x 233 = 16543 71 x 253 = 17963 71 x 273 = 19383 71 x 293 = 20803 71 x 214 = 15194 71 x 234 = 16614 71 x 254 = 18034 71 x 274 = 19454 71 x 294 = 20874 71 x 215 = 15265 71 x 235 = 16685 71 x 255 = 18105 71 x 275 = 19525 71 x 295 = 20945 71 x 216 = 15336 71 x 236 = 16756 71 x 256 = 18176 71 x 276 = 19596 71 x 296 = 21016 71 x 217 = 15407 71 x 237 = 16827 71 x 257 = 18247 71 x 277 = 19667 71 x 297 = 21087 71 x 218 = 15478 71 x 238 = 16898 71 x 258 = 18318 71 x 278 = 19738 71 x 298 = 21158 71 x 219 = 15549 71 x 239 = 16969 71 x 259 = 18389 71 x 279 = 19809 71 x 299 = 21229 71 x 220 = 15620 71 x 240 = 17040 71 x 260 = 18460 71 x 280 = 19880 71 x 300 = 21300. ## Solved Examples. Example 1:. Question: What is 4 times 71?. Solution: Multiply 71 by 4.. Calculation: 71 * 4 = 284. Example 2:. Question: How much is 7 times 71?. Solution: Multiply 71 by 7.. Calculation: 71 * 7 = 497. Example 3:. Question: Calculate 9 times 71.. Solution: Multiply 71 by 9.. Calculation: 71 * 9 = 639. Example 4:. Question: What do you get when you multiply 71 by 5?. Solution: Multiply 71 by 5.. Calculation: 71 * 5 = 355.
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https://www.omtexclasses.com/2014/03/two-coins-are-tossed-simultaneously.html
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Two coins are tossed simultaneously. Write the sample space 'S' and the number of sample points n(S). Write the following events using set notation and mention the total number of elements in each of them: a. A is the event of getting at least one head. b. B is the event of getting exactly one head.
iii. Two coins are tossed simultaneously. Write the sample space 'S' and the number of sample points n(S). Write the following events using set notation and mention the total number of elements in each of them:
a. A is the event of getting at least one head.
b. B is the event of getting exactly one head.
Answer: Since two coins are tossed simultaneously
{ H H Sample Space is = H T T H T T }
n(S) = 4
(a) Let A = Event that at least one head turns up
A = {HH, HT, TH, }
n(A) = 3
∴ P(A) = n(A) n(S) ∴ P(A) = 3 4
(b) Let B = Event of getting Exactly one head turns up
B = {HT, TH}
n(B) = 2
∴ P(B) = n(B) n(S) ∴ P(B) = 2 4
∴ P(B) = 1 2
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Two coins are tossed simultaneously. Write the sample space 'S' and the number of sample points n(S). Write the following events using set notation and mention the total number of elements in each of them: a. A is the event of getting at least one head. b. B is the event of getting exactly one head.. iii. Two coins are tossed simultaneously. Write the sample space 'S' and the number of sample points n(S). Write the following events using set notation and mention the total number of elements in each of them:. a. A is the event of getting at least one head.. b. B is the event of getting exactly one head.. Answer: Since two coins are tossed simultaneously.
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{ H H Sample Space is = H T T H T T }. n(S) = 4. (a) Let A = Event that at least one head turns up. A = {HH, HT, TH, }. n(A) = 3. ∴ P(A) = n(A) n(S) ∴ P(A) = 3 4. (b) Let B = Event of getting Exactly one head turns up. B = {HT, TH}. n(B) = 2. ∴ P(B) = n(B) n(S) ∴ P(B) = 2 4. ∴ P(B) = 1 2.
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https://www.kaysonseducation.co.in/questions/p-span-sty_3848
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## Question
### Solution
Correct option is
1 : 9
Since the initial velocity of the stone is zero, the total time taken by the stone to hit the ground is given by
= 5 s
During the first second, the stone falls a distance h1 given by
During the first four seconds, the stone falls a distance h given by
Distance h2 through which the stone falls in the last (i.e. fifth) second
= 125 – 80
= 45 m.
.
#### SIMILAR QUESTIONS
Q1
The displacement y (in metres) of a body varies with time t (in seconds) as
How long does the body take to come to rest?
Q2
A body goes from A to B with a velocity of 40 kmh–1 and returns from Bto A with a velocity of 60 kmh–1. What is the average velocity of the body during the whole journey?
Q3
The area under the velocity-time graph between any two instants t = t1and t = t2 gives the distance covered in time
Q4
A particle is given a displacement of 4 m in the x-y plane. If the x-component of the displacement vector is 2 m, the y-component will be
Q5
A car, starting from rest, accelerates at a constant rate of 5 ms–2 for some time. It then retards at a constant rate of 10 ms–2 and finally comes to rest. If the total taken is 6 s, What is the maximum speed attained by the car?
Q6
A body, starting from rest, moves in a straight line with a constant acceleration a for a time interval t during which it travels a distance s1. If continues to move with the same acceleration for the next time interval tduring which it travels a distance s2. The relation between s1 and s2 is
Q7
A body, moving in a straight line, with an initial velocity u and a constant acceleration a, covers a distance of 40 m in the 4th second and a distance of 60 m in the 6th second. The values of u and a respectively are
Q8
A simple pendulum is hanging from the ceiling of a compartment of a train. It is observed that the string is inclined towards the rear of the train. If follows that the train is
Q9
The displacement of a body from a reference point, is given by
where x is in metres and t in seconds. This shown that the body is
Q10
A car moving at a speed v is stopped in a certain distance when the brakes produce a deceleration a. If the speed of the car was nv, what must be the deceleration of the car to stop it in the same distance and in the same time?
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## Question. ### Solution. Correct option is. 1 : 9. Since the initial velocity of the stone is zero, the total time taken by the stone to hit the ground is given by. = 5 s. During the first second, the stone falls a distance h1 given by. During the first four seconds, the stone falls a distance h given by. Distance h2 through which the stone falls in the last (i.e. fifth) second. = 125 – 80. = 45 m.. .. #### SIMILAR QUESTIONS. Q1. The displacement y (in metres) of a body varies with time t (in seconds) as. How long does the body take to come to rest?. Q2. A body goes from A to B with a velocity of 40 kmh–1 and returns from Bto A with a velocity of 60 kmh–1. What is the average velocity of the body during the whole journey?. Q3. The area under the velocity-time graph between any two instants t = t1and t = t2 gives the distance covered in time. Q4. A particle is given a displacement of 4 m in the x-y plane.
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If the x-component of the displacement vector is 2 m, the y-component will be. Q5. A car, starting from rest, accelerates at a constant rate of 5 ms–2 for some time. It then retards at a constant rate of 10 ms–2 and finally comes to rest. If the total taken is 6 s, What is the maximum speed attained by the car?. Q6. A body, starting from rest, moves in a straight line with a constant acceleration a for a time interval t during which it travels a distance s1. If continues to move with the same acceleration for the next time interval tduring which it travels a distance s2. The relation between s1 and s2 is. Q7. A body, moving in a straight line, with an initial velocity u and a constant acceleration a, covers a distance of 40 m in the 4th second and a distance of 60 m in the 6th second. The values of u and a respectively are. Q8. A simple pendulum is hanging from the ceiling of a compartment of a train. It is observed that the string is inclined towards the rear of the train. If follows that the train is. Q9. The displacement of a body from a reference point, is given by. where x is in metres and t in seconds. This shown that the body is. Q10. A car moving at a speed v is stopped in a certain distance when the brakes produce a deceleration a. If the speed of the car was nv, what must be the deceleration of the car to stop it in the same distance and in the same time?.
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https://math.stackexchange.com/questions/1798540/proof-of-coin-and-bag-problem
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# Proof of coin and bag problem
There are 5 bags labeled 1 to 5. All the coins in a given bag have the same weight. Some bags have coins of weight 10 gm, others have coins of weight 11 gm. I pick 1, 2, 4, 8, 16 coins respectively from bags 1 to 5. Their total weight comes out to 323 gm. Identify the bags with 11 coins.
I can solve the problem by using the fact that sum of odd number and even number is odd number
$$x_1+2x_2+4x_3+8x_4+16x_5=323$$ $$odd+even=odd$$ hence $$x1=11$$ $$2x_2+4x_3+8x_4+16x_5=323-11=312$$ dividing by 2
$$x_2+2x_3+4x_4+8x_5=312/2=156$$ $$even + even=even$$ hence $$x2=10$$ similarly continuing ,will get all weights of coin
I found this method very randomly.
I don't have logical reason for dividing by 2 step after each iteration?
is there any better method? what is the reasoning behind this method? is there any recursion happening? how can we prove that this method work?
• Your method works... because it's actually doing the conversion to base 2 the hard way. However, imagine the coins were either 9 gm or 11 gm; the problem could be stated in a similar way, and your approach wouldn't work anymore. There is a direct way - no recursion needed - as I put in my Answer. – mathguy May 24 '16 at 20:03
• can you please explain "conversion to base 2 hard way"? – arjun2_0 May 25 '16 at 3:50
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# Proof of coin and bag problem. There are 5 bags labeled 1 to 5. All the coins in a given bag have the same weight. Some bags have coins of weight 10 gm, others have coins of weight 11 gm. I pick 1, 2, 4, 8, 16 coins respectively from bags 1 to 5. Their total weight comes out to 323 gm. Identify the bags with 11 coins.. I can solve the problem by using the fact that sum of odd number and even number is odd number. $$x_1+2x_2+4x_3+8x_4+16x_5=323$$ $$odd+even=odd$$ hence $$x1=11$$ $$2x_2+4x_3+8x_4+16x_5=323-11=312$$ dividing by 2. $$x_2+2x_3+4x_4+8x_5=312/2=156$$ $$even + even=even$$ hence $$x2=10$$ similarly continuing ,will get all weights of coin.
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I found this method very randomly.. I don't have logical reason for dividing by 2 step after each iteration?. is there any better method? what is the reasoning behind this method? is there any recursion happening? how can we prove that this method work?. • Your method works... because it's actually doing the conversion to base 2 the hard way. However, imagine the coins were either 9 gm or 11 gm; the problem could be stated in a similar way, and your approach wouldn't work anymore. There is a direct way - no recursion needed - as I put in my Answer. – mathguy May 24 '16 at 20:03. • can you please explain "conversion to base 2 hard way"? – arjun2_0 May 25 '16 at 3:50.
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https://physics.stackexchange.com/questions/444894/why-exactly-does-diffraction-occur/444900
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# Why exactly does diffraction occur?
Why do waves that were traveling in a straight direction change direction when passing through an opening?
I thought that the waves (red arrow) when colliding with the wall bounce in the opposite direction (green arrow).
And the waves that pass through the aperture follow its path normally as shown in the image on the right.
The waves that go in a straight direction should follow traveling straight line like a car that goes under a bridge the car is straight on the road. But this is not so.
Why does the direction of the waves change?
How is the direction of the waves calculated?
• I think OP wants to know why the diffraction occurs and why the waves don't just continue like they pointed out in the diagram. Maybe an answer with wavelets and Huygens' Principle. I don't understand the principle well enough to write a good answer though. Dec 3, 2018 at 17:08
• @TausifHossain Yeah that is why I posted a comment :) It seems like the OP doesn't know what it is called, or else they would have said "How does diffraction work" or something like that. Just thought the reference would be helpful just in case. Dec 3, 2018 at 17:11
• Possible duplicate of How does the Huygens–Fresnel principle apply to diffraction? Dec 3, 2018 at 17:13
• I see, you're right. Though I found it hard to find a good intuitive explanation of the Huygens Principle online. Dec 3, 2018 at 17:16
• @TausifHossain I used your comment for an answer Dec 3, 2018 at 20:12
For the full math, you can look up 'diffraction' and 'Huygens Principle' but here I will just post a quick observation that is enough to get a good physical intuition.
Suppose we are considering water waves, and imagine yourself sitting behind the barrier in the 'harbour' (at the lower part of your diagram), watching the waves approaching from 'out at sea' (i.e. the top of your diagram). As the waves reach the 'harbour mouth' (i.e. the small opening in your diagram) the water there is caused to go up and down. So there is this water bobbing up and down in the small opening. Now the surface of the water nearby is going to bob up and down too, isn't it? And the ripples will spread out from there. It doesn't really matter in what direction you consider: the waves will spread out into the 'harbour' because the water at the harbour mouth is moving.
From this way of thinking, you begin to wonder why the waves out at sea are so straight! Ultimately it is because in that case you have oscillating water all along a long line, and so the water all along that long line is caused to move in synchrony.
As I say, this is not a full mathematical answer, just an attempt to give you some intuition about the physics.
• Thanks, the point about why waves generally are straight was good. Every point pushes out a circle, but you have a line of points all doing that at the same time. Dec 4, 2018 at 11:33
• Huygens was the man. Dec 4, 2018 at 20:49
The first thing to realize is that waves only appear to travel. But when you look at a fish in the water, it becomes clear that the water only sloshes back and forth. Waves occur because the water movements aren't all in sync, nor could they be - how would all the water molecules know to reverse at the same time? So when you have water molecules traveling on opposite directions >>^<< there's nowhere for them to go but up. That produces the wave crest. And when a bit of water locally reverses direction, the crest moves >^<<<.
Now we know that waves are really a local effect. That means the wave inside the slit has no memory where it came from. And that means it also doesn't remember in which direction it would need to travel. All directions are possible.
Now if waves do not have memory, then how do they "know" how to travel in a straight line near the beach? Well, that's not really what happens. Wave crests travel orthogonally to the crest lines. In thin slits, where there's no longer a crest line, this no longer makes sense, and that's why you get the diffraction.
You can see how this makes sense when you look at broader slits. In the middle, there's a well-defined crest line, and you still get the straight pattern. But at both edges, you get the diffraction effects. This frays the edges of the crest lines, progressively making them sorter, until they disappear entirely. After that, you get a complex pattern of isolated peaks.
• The waves travel, but the water doesn't.
– Dubu
Dec 4, 2018 at 13:06
"Why are waves straight?" is the first question.
Let's start with a model of waves where particles don't have much kinetic energy. They just have potential energy.
Each location, if it has less energy than an adjacent spot, steals 1 unit of energy from it.
00000
00000
00900
00000
00000
next tick:
00000
01110
01110
01110
00000
next tick:
0000000
0.111.0
0100010
0101010
0100010
0.111.0
0000000
0.....0
.00000.
.0...0.
.0.0.0.
.0...0.
.00000.
0.....0
where . is a fractional unit of energy.
and we see a really simple model of a "circlular wave" coming from a point.
Now look what happens when we have a wave-front (I'll assume the lines go off "forever" on the left and right, but there is a wall at the top).
0000000000000
9999999999999
0000000000000
0000000000000
0000000000000
0000000000000
0000000000000
0000000000000
3333333333333
3333333333333
3333333333333
0000000000000
0000000000000
0000000000000
0000000000000
0000000000000
3333333333333
3333333333333
0000000000000
3333333333333
0000000000000
0000000000000
0000000000000
0000000000000
3333333333333
0000000000000
5555555555555
0000000000000
1111111111111
0000000000000
0000000000000
0000000000000
0000000000000
4444444444444
0000000000000
4444444444444
0000000000000
1111111111111
0000000000000
0000000000000
3333333333333
0000000000000
2222222222222
0000000000000
3333333333333
0000000000000
1111111111111
0000000000000
hey look, waves. (I did some rounding) with some initial "sloshing" as it bounces off the "wall' at the top.
So suppose you have a bunch of waves causing a hole in a wall to go up and down:
######9######
0000000000000
0000000000000
0000000000000
0000000000000
######0######
0000033300000
0000000000000
0000000000000
0000000000000
######9######
0000100010000
0000111110000
0000000000000
0000000000000
######0######
0001033301000
0001000001000
0001111111000
0000000000000
because there is no intrinsic direction to the wave, just up and down, it induces a circular wave at the point of departure.
Being more mathematical, a linear wave is just a bunch of circular waves. Take 5 copies of this:
00000
00000
00900
00000
00000
offset by one horizontally and you get:
000000000
000000000
009999900
000000000
000000000
000000000
013333310
030000030
013333310
000000000
012333210
100000001
102000201
100000001
012333210
because the "hump" of 9s adjacent to each 9 slows inflow.
A real physical wave is more complex than this simple discrete time cell model. But the basic idea; that "humps" push water away and "gulfs" pull it in, and that there is enough momentum to cause overshoot -- result in a similar effect. Linear waves are the result of linear adjacent "humps" and "gulfs". When you reach a barrier with a hole, the linear "adjacent humps" go away, and the wave becomes more cicular as it falls both "forward" and "sideways", instead of being supported on the sides by other "humps" of water.
Your aperture only allows a very short segment of the incoming plane wave to pass through. As the aperture becomes smaller, the segment looks more and more like a point source. A point source emits spherical waves like you show in your lower right figure. (this is almost intuitively obvious because of symmetry--what other shape of wave would a point emit?).
This is usually explained more formally via diffraction:
https://isaacphysics.org/concepts/cp_diffraction
"Diffraction is the spreading out of waves as they pass through an aperture or around objects. ... In an aperture with width smaller than the wavelength, the wave transmitted through the aperture spreads all the way round and behaves like a point source of waves (they spread out below)"
A quick answer would be that they are not changing direction. Each point in the plane is the source of a single wave. Single waves expand in circles, but as you put many single waves together you sum them and get a plane wave. The aperture if small enough simply blocks the other waves allowing only one to pass and thus it re-takes circular shape.
This is a simplification of diffraction and huygens' principle but it might help you get an idea.
• @Will, What do you mean about the pinhole camera? In every explanation I've ever heard about how a typical pinhole camera forms an image, the discussion is limited to geometric optics. Everybody who's ever made a pinhole camera knows that a smaller pinhole lets less light through, but it also gives you a sharper image. But that's only true up to a point. If you go beyond what's typical, and you make a really tiny pinhole that only lets a really tiny amount of light through, you hit a limit to how sharp the image can be. Is that what you're talking about? Dec 4, 2018 at 16:41
• @SolomonSlow Yes below a certain pinhole size the diffraction effects become significant and distort the captured image. Dec 5, 2018 at 8:18
Tausif commented:
I think OP wants to know why the diffraction occurs and why the waves don't just continue like they pointed out in the diagram.
In any elastic medium, a pressure effect not only leads to material displacement in this direction, but also to lateral displacement. (In an inelastic medium the material gets simply punched out.) So the awaited longitudinal wave is accompanied always by a transversal wave.
This transversal wave spread out in isotropic media as a spherical wave. The obstacle with the slit limiting the isotropy and instead of a spherical wave on get only have of a spherecal wave.
• "So the awaited longitudinal wave is accompanied always by a transversal wave." I think you have mis-understood these terms: I presume you mean to re-state Huygens' principle, but this is not a correct statement. In particular pressure waves in a fluid diffract while remaining purely longitudinal. You can also see this in a geophysical context: the P-and S- waves travel and different speeds and don't transform into one another when diffracting resulting is well separated arrival times. Dec 6, 2018 at 15:26
• @dmckee The pressure of a falling stone does not only go into the depth, but is distributed laterally in every elastic medium. Even punching out a hole in a thick sheet often leads to a thickening of the material at the edge. Dec 6, 2018 at 18:50
• Again: you misunderstand the terms you are using. The fact that pressure is omnidirectional does not make the wave transverse. In any small element the the Poynting vector for a sound wave is in the direction of displacement meaning that the wave is longitudinal. Certainly the Poynting vector is non-constant in space, but that is not what those words characterize. Nor does thinking about plastic deformation change the situation as it tells you about the dissipative behavior rather than the wave-propagation. Dec 6, 2018 at 19:26
• @dmckee If you position an underwater buoy next to the impact point, the buoy will also periodically move away from and back to the impact point. This corresponds to a transverse wave. I could not see why I could not name it this way. Dec 6, 2018 at 19:36
• You seem to be imagining that there is some global "direction of the wave everywhere" to which you compare. This is incorrect. You always compare the local displacement to the local direction of the wave (as identified by the local Poynting vector). Dec 6, 2018 at 19:58
Your initial picture is incomplete in describing a more in-depth version of the waves. The waves are actually not a bunch of parallel beams traveling in straight lines down the page as you show. What there is is a superposition of point sources of energy and a single point source of energy will produce a circular wave. Your wavefront is made up of a nearly infinite number of these point sources and it is the superposition of the waves from these point sources that combine to create a uniform wavefront. So, when the wave comes upon an aperture it then acts like the point source that it is and the result is just like a point source of energy would act, i.e. a circular wave.
• I'd just like to point out that your answer is good but it's best not to call the diagram wrong but rather incomplete as the diagram is correct from a certain simplified perspective. Dec 5, 2018 at 17:40
In the aperture the wave is no longer plane : it is the product of a rect function, which is unity in the aperture and zero outside it, and a plane wave. You can inspect which wave vectors are present by Fourier transforming this product. The result is a convolution of the transform of the rect, the so called sinc function, and the plane wave. The message is that the result is a sum of plane waves of varying direction. For a point aperture all plane waves are present with equal amplitude and phase, that is, a spherical wave. Alas this requires some elementary math to understand.
@Andrew Steane has already given a good answer, I just want to make the explanation more visual. First thing I want to show is the light wave diffraction, given in the image below. What You see here is the electric field intensity plot. The plane wave is released from the back of the slit which is barely visible as the dark blue region.
The occurrence of diffraction can be summarized as the wave fronts that are parallel to the incoming wave fronts are continuing to propagate as it is, but the edges of the transmitted wave produce secondary wave fronts that are the observed, diffraction effect. To visualize it I draw the second image below, which shows the wave fronts of the wave passed the diffractive slit. As easily visible on the image below, the edges of the plane wave has a secondary wave fronts.
You have asked in the question, why the plane wave don't pass the slit as a plane wave. It is because plane wave is a combination of waves that actually elongates to the infinity. If the wave front combination doesn't elongate to infinity, we will see the same bent effect on the edges as well. To give the general propagation behaviour, I added the wave propagation GIF below (simulated using FDTD method):
• This is similar to water and it also shows no "interference" only diffraction. The EM field is different to water. It is a nice simulation. Aug 15, 2021 at 16:29
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# Why exactly does diffraction occur?. Why do waves that were traveling in a straight direction change direction when passing through an opening?. I thought that the waves (red arrow) when colliding with the wall bounce in the opposite direction (green arrow).. And the waves that pass through the aperture follow its path normally as shown in the image on the right.. The waves that go in a straight direction should follow traveling straight line like a car that goes under a bridge the car is straight on the road. But this is not so.. Why does the direction of the waves change?. How is the direction of the waves calculated?. • I think OP wants to know why the diffraction occurs and why the waves don't just continue like they pointed out in the diagram. Maybe an answer with wavelets and Huygens' Principle. I don't understand the principle well enough to write a good answer though. Dec 3, 2018 at 17:08. • @TausifHossain Yeah that is why I posted a comment :) It seems like the OP doesn't know what it is called, or else they would have said "How does diffraction work" or something like that. Just thought the reference would be helpful just in case. Dec 3, 2018 at 17:11. • Possible duplicate of How does the Huygens–Fresnel principle apply to diffraction? Dec 3, 2018 at 17:13. • I see, you're right. Though I found it hard to find a good intuitive explanation of the Huygens Principle online. Dec 3, 2018 at 17:16. • @TausifHossain I used your comment for an answer Dec 3, 2018 at 20:12. For the full math, you can look up 'diffraction' and 'Huygens Principle' but here I will just post a quick observation that is enough to get a good physical intuition.. Suppose we are considering water waves, and imagine yourself sitting behind the barrier in the 'harbour' (at the lower part of your diagram), watching the waves approaching from 'out at sea' (i.e. the top of your diagram). As the waves reach the 'harbour mouth' (i.e. the small opening in your diagram) the water there is caused to go up and down. So there is this water bobbing up and down in the small opening. Now the surface of the water nearby is going to bob up and down too, isn't it? And the ripples will spread out from there. It doesn't really matter in what direction you consider: the waves will spread out into the 'harbour' because the water at the harbour mouth is moving.. From this way of thinking, you begin to wonder why the waves out at sea are so straight! Ultimately it is because in that case you have oscillating water all along a long line, and so the water all along that long line is caused to move in synchrony.. As I say, this is not a full mathematical answer, just an attempt to give you some intuition about the physics.. • Thanks, the point about why waves generally are straight was good. Every point pushes out a circle, but you have a line of points all doing that at the same time. Dec 4, 2018 at 11:33. • Huygens was the man. Dec 4, 2018 at 20:49. The first thing to realize is that waves only appear to travel. But when you look at a fish in the water, it becomes clear that the water only sloshes back and forth. Waves occur because the water movements aren't all in sync, nor could they be - how would all the water molecules know to reverse at the same time? So when you have water molecules traveling on opposite directions >>^<< there's nowhere for them to go but up. That produces the wave crest. And when a bit of water locally reverses direction, the crest moves >^<<<.. Now we know that waves are really a local effect. That means the wave inside the slit has no memory where it came from. And that means it also doesn't remember in which direction it would need to travel. All directions are possible.. Now if waves do not have memory, then how do they "know" how to travel in a straight line near the beach? Well, that's not really what happens. Wave crests travel orthogonally to the crest lines. In thin slits, where there's no longer a crest line, this no longer makes sense, and that's why you get the diffraction.. You can see how this makes sense when you look at broader slits. In the middle, there's a well-defined crest line, and you still get the straight pattern. But at both edges, you get the diffraction effects. This frays the edges of the crest lines, progressively making them sorter, until they disappear entirely. After that, you get a complex pattern of isolated peaks.. • The waves travel, but the water doesn't.. – Dubu. Dec 4, 2018 at 13:06. "Why are waves straight?" is the first question.. Let's start with a model of waves where particles don't have much kinetic energy. They just have potential energy.. Each location, if it has less energy than an adjacent spot, steals 1 unit of energy from it.. 00000. 00000. 00900. 00000. 00000. next tick:. 00000. 01110. 01110. 01110. 00000. next tick:. 0000000. 0.111.0. 0100010. 0101010. 0100010. 0.111.0. 0000000. 0.....0. .00000.. .0...0.. .0.0.0.. .0...0.. .00000.. 0.....0. where . is a fractional unit of energy.. and we see a really simple model of a "circlular wave" coming from a point.. Now look what happens when we have a wave-front (I'll assume the lines go off "forever" on the left and right, but there is a wall at the top).. 0000000000000. 9999999999999. 0000000000000. 0000000000000. 0000000000000. 0000000000000. 0000000000000. 0000000000000. 3333333333333. 3333333333333. 3333333333333. 0000000000000. 0000000000000. 0000000000000. 0000000000000. 0000000000000. 3333333333333. 3333333333333. 0000000000000. 3333333333333. 0000000000000. 0000000000000. 0000000000000. 0000000000000. 3333333333333. 0000000000000. 5555555555555. 0000000000000. 1111111111111. 0000000000000. 0000000000000. 0000000000000. 0000000000000. 4444444444444. 0000000000000. 4444444444444. 0000000000000. 1111111111111. 0000000000000. 0000000000000. 3333333333333. 0000000000000. 2222222222222. 0000000000000.
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3333333333333. 0000000000000. 1111111111111. 0000000000000. hey look, waves. (I did some rounding) with some initial "sloshing" as it bounces off the "wall' at the top.. So suppose you have a bunch of waves causing a hole in a wall to go up and down:. ######9######. 0000000000000. 0000000000000. 0000000000000. 0000000000000. ######0######. 0000033300000. 0000000000000. 0000000000000. 0000000000000. ######9######. 0000100010000. 0000111110000. 0000000000000. 0000000000000. ######0######. 0001033301000. 0001000001000. 0001111111000. 0000000000000. because there is no intrinsic direction to the wave, just up and down, it induces a circular wave at the point of departure.. Being more mathematical, a linear wave is just a bunch of circular waves. Take 5 copies of this:. 00000. 00000. 00900. 00000. 00000. offset by one horizontally and you get:. 000000000. 000000000. 009999900. 000000000. 000000000. 000000000. 013333310. 030000030. 013333310. 000000000. 012333210. 100000001. 102000201. 100000001. 012333210. because the "hump" of 9s adjacent to each 9 slows inflow.. A real physical wave is more complex than this simple discrete time cell model. But the basic idea; that "humps" push water away and "gulfs" pull it in, and that there is enough momentum to cause overshoot -- result in a similar effect. Linear waves are the result of linear adjacent "humps" and "gulfs". When you reach a barrier with a hole, the linear "adjacent humps" go away, and the wave becomes more cicular as it falls both "forward" and "sideways", instead of being supported on the sides by other "humps" of water.. Your aperture only allows a very short segment of the incoming plane wave to pass through. As the aperture becomes smaller, the segment looks more and more like a point source. A point source emits spherical waves like you show in your lower right figure. (this is almost intuitively obvious because of symmetry--what other shape of wave would a point emit?).. This is usually explained more formally via diffraction:. https://isaacphysics.org/concepts/cp_diffraction. "Diffraction is the spreading out of waves as they pass through an aperture or around objects. ... In an aperture with width smaller than the wavelength, the wave transmitted through the aperture spreads all the way round and behaves like a point source of waves (they spread out below)". A quick answer would be that they are not changing direction. Each point in the plane is the source of a single wave. Single waves expand in circles, but as you put many single waves together you sum them and get a plane wave. The aperture if small enough simply blocks the other waves allowing only one to pass and thus it re-takes circular shape.. This is a simplification of diffraction and huygens' principle but it might help you get an idea.. • @Will, What do you mean about the pinhole camera? In every explanation I've ever heard about how a typical pinhole camera forms an image, the discussion is limited to geometric optics. Everybody who's ever made a pinhole camera knows that a smaller pinhole lets less light through, but it also gives you a sharper image. But that's only true up to a point. If you go beyond what's typical, and you make a really tiny pinhole that only lets a really tiny amount of light through, you hit a limit to how sharp the image can be. Is that what you're talking about? Dec 4, 2018 at 16:41. • @SolomonSlow Yes below a certain pinhole size the diffraction effects become significant and distort the captured image. Dec 5, 2018 at 8:18. Tausif commented:. I think OP wants to know why the diffraction occurs and why the waves don't just continue like they pointed out in the diagram.. In any elastic medium, a pressure effect not only leads to material displacement in this direction, but also to lateral displacement. (In an inelastic medium the material gets simply punched out.) So the awaited longitudinal wave is accompanied always by a transversal wave.. This transversal wave spread out in isotropic media as a spherical wave. The obstacle with the slit limiting the isotropy and instead of a spherical wave on get only have of a spherecal wave.. • "So the awaited longitudinal wave is accompanied always by a transversal wave." I think you have mis-understood these terms: I presume you mean to re-state Huygens' principle, but this is not a correct statement. In particular pressure waves in a fluid diffract while remaining purely longitudinal. You can also see this in a geophysical context: the P-and S- waves travel and different speeds and don't transform into one another when diffracting resulting is well separated arrival times. Dec 6, 2018 at 15:26. • @dmckee The pressure of a falling stone does not only go into the depth, but is distributed laterally in every elastic medium. Even punching out a hole in a thick sheet often leads to a thickening of the material at the edge. Dec 6, 2018 at 18:50. • Again: you misunderstand the terms you are using. The fact that pressure is omnidirectional does not make the wave transverse. In any small element the the Poynting vector for a sound wave is in the direction of displacement meaning that the wave is longitudinal. Certainly the Poynting vector is non-constant in space, but that is not what those words characterize. Nor does thinking about plastic deformation change the situation as it tells you about the dissipative behavior rather than the wave-propagation. Dec 6, 2018 at 19:26. • @dmckee If you position an underwater buoy next to the impact point, the buoy will also periodically move away from and back to the impact point. This corresponds to a transverse wave. I could not see why I could not name it this way. Dec 6, 2018 at 19:36. • You seem to be imagining that there is some global "direction of the wave everywhere" to which you compare. This is incorrect. You always compare the local displacement to the local direction of the wave (as identified by the local Poynting vector). Dec 6, 2018 at 19:58. Your initial picture is incomplete in describing a more in-depth version of the waves. The waves are actually not a bunch of parallel beams traveling in straight lines down the page as you show. What there is is a superposition of point sources of energy and a single point source of energy will produce a circular wave. Your wavefront is made up of a nearly infinite number of these point sources and it is the superposition of the waves from these point sources that combine to create a uniform wavefront. So, when the wave comes upon an aperture it then acts like the point source that it is and the result is just like a point source of energy would act, i.e. a circular wave.. • I'd just like to point out that your answer is good but it's best not to call the diagram wrong but rather incomplete as the diagram is correct from a certain simplified perspective. Dec 5, 2018 at 17:40. In the aperture the wave is no longer plane : it is the product of a rect function, which is unity in the aperture and zero outside it, and a plane wave. You can inspect which wave vectors are present by Fourier transforming this product. The result is a convolution of the transform of the rect, the so called sinc function, and the plane wave. The message is that the result is a sum of plane waves of varying direction. For a point aperture all plane waves are present with equal amplitude and phase, that is, a spherical wave. Alas this requires some elementary math to understand.. @Andrew Steane has already given a good answer, I just want to make the explanation more visual. First thing I want to show is the light wave diffraction, given in the image below. What You see here is the electric field intensity plot. The plane wave is released from the back of the slit which is barely visible as the dark blue region.. The occurrence of diffraction can be summarized as the wave fronts that are parallel to the incoming wave fronts are continuing to propagate as it is, but the edges of the transmitted wave produce secondary wave fronts that are the observed, diffraction effect. To visualize it I draw the second image below, which shows the wave fronts of the wave passed the diffractive slit. As easily visible on the image below, the edges of the plane wave has a secondary wave fronts.. You have asked in the question, why the plane wave don't pass the slit as a plane wave. It is because plane wave is a combination of waves that actually elongates to the infinity. If the wave front combination doesn't elongate to infinity, we will see the same bent effect on the edges as well. To give the general propagation behaviour, I added the wave propagation GIF below (simulated using FDTD method):. • This is similar to water and it also shows no "interference" only diffraction. The EM field is different to water. It is a nice simulation. Aug 15, 2021 at 16:29.
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https://www.studypool.com/discuss/1097696/geometry-similar-solids?free
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Geometry - Similar Solids
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Jul 30th, 2015
1. if 3 – 3cm
:. 5?
5x3/3 = 5cm
2. if 3 __54pie cm2
:. 5 ?
54 pie cm2 x5
3
=90pie cm2
3. If 5 represent 250pie cm3
:. 3?
250piecm3x3
5
= 150 pie cm3
Jul 30th, 2015
Can you review the below solution and tell me if it is correct or the solution you provided is correct? Thanks.
Scale factor = Solid I : Solid II = 3 : 5
Solid I = 3
Solid II = 3 × (5/3)
[ Solid II = 5 m ]
A[I] / A[II] = ( h[I] / h[II] )²
54π / A[II} = ( 3 / 5 )²
54π / A[II] = 9/25
A[II] = (54π * 25) / 9
[ A[II] = 150π m² ]
V[I] / V[II] = ( h[I] / h[II] )³
V[I] / 259π = ( 3 / 5 )³
V[I] / 259π = 27/125
V[I] = (259π * 27) / 125
[ V[I] = 55.944π m³ ]
Jul 30th, 2015
Scale factor = Solid I : Solid II = 3 : 5
Solid I = 3
Solid II = 3 × (5/3)
[ Solid II = 5 m ]
correct except the units of measurement
the remaining expressions are wrong since you should have used the scale factor in comparison to the area and the volume and not the height
and remember this question is worth 4\$
A[I] / A[II] = ( h[I] / h[II] )²
54π / A[II} = ( 3 / 5 )²
54π / A[II] = 9/25
A[II] = (54π * 25) / 9
[ A[II] = 150π m² ]
V[I] / V[II] = ( h[I] / h[II] )³
V[I] / 259π = ( 3 / 5 )³
V[I] / 259π = 27/125
V[I] = (259π * 27) / 125
[ V[I] = 55.944π m³ ]
Jul 30th, 2015
...
Jul 30th, 2015
...
Jul 30th, 2015
Sep 25th, 2017
check_circle
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Geometry - Similar Solids. label Mathematics. account_circle Unassigned. schedule 1 Day. account_balance_wallet \$5. Jul 30th, 2015. 1. if 3 – 3cm. :. 5?. 5x3/3 = 5cm. 2. if 3 __54pie cm2. :. 5 ?. 54 pie cm2 x5. 3. =90pie cm2. 3. If 5 represent 250pie cm3. :. 3?. 250piecm3x3. 5. = 150 pie cm3. Jul 30th, 2015. Can you review the below solution and tell me if it is correct or the solution you provided is correct? Thanks.. Scale factor = Solid I : Solid II = 3 : 5. Solid I = 3. Solid II = 3 × (5/3). [ Solid II = 5 m ]. A[I] / A[II] = ( h[I] / h[II] )². 54π / A[II} = ( 3 / 5 )². 54π / A[II] = 9/25. A[II] = (54π * 25) / 9. [ A[II] = 150π m² ]. V[I] / V[II] = ( h[I] / h[II] )³.
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V[I] / 259π = ( 3 / 5 )³. V[I] / 259π = 27/125. V[I] = (259π * 27) / 125. [ V[I] = 55.944π m³ ]. Jul 30th, 2015. Scale factor = Solid I : Solid II = 3 : 5. Solid I = 3. Solid II = 3 × (5/3). [ Solid II = 5 m ]. correct except the units of measurement. the remaining expressions are wrong since you should have used the scale factor in comparison to the area and the volume and not the height. and remember this question is worth 4\$. A[I] / A[II] = ( h[I] / h[II] )². 54π / A[II} = ( 3 / 5 )². 54π / A[II] = 9/25. A[II] = (54π * 25) / 9. [ A[II] = 150π m² ]. V[I] / V[II] = ( h[I] / h[II] )³. V[I] / 259π = ( 3 / 5 )³. V[I] / 259π = 27/125. V[I] = (259π * 27) / 125. [ V[I] = 55.944π m³ ]. Jul 30th, 2015. .... Jul 30th, 2015. .... Jul 30th, 2015. Sep 25th, 2017. check_circle.
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[1][2] That is, the matrix $A$ is idempotent if and only if $A^2 = A$. Prove that if A is idempotent, then the matrix I −A is also idempotent. [3] The trace of an idempotent matrix — the sum of the elements on its main diagonal — equals the rank of the matrix and thus is always an integer. Since the matrix $A$ has $0$ as an eigenvalue. Show that 1 2(I+A) is idempotent if and only if Ais an involution. Corollary 5. Let A be an n×n idempotent matrix, that is, A2=A. Set A = PP′ where P is an n × r matrix of eigenvectors corresponding to the r eigenvalues of A equal to 1. Ax= λx⇒Ax= AAx= λAx= λ2x,soλ2 = λwhich implies λ=0 or λ=1. Show that the rank of an idempotent matrix is equal to the number of nonzero eigenvalues of the matrix. Theorem 4 shows that these eigenvalues are also eigenvalues of the difference of idempotent density matrices. Then, the eigenvalues of A are zeros or ones. The post contains C++ and Python code for converting a rotation matrix to Euler angles and vice-versa. [3] Trace. Show that = 0 or = 1 are the only possible eigenvalues of A. A.8. ST is the new administrator. The trace is related to the derivative of the determinant (see Jacobi's formula). Consider the following 2 cases: Case (1): A is nonsingular. Claim: Each eigenvalue of an idempotent matrix is either 0 or 1. Published 02/22/2018, […] Since $A$ has three distinct eigenvalues, $A$ is diagonalizable. 0 & 0 & 1 A Note on Idempotent Matrices C. G. Khatri Gujarat University Ahmedabad, India Submitted by C. R. Rao ABSTRACT Let H be an n X n matrix, and let the trace, the rank, the conjugate transpose, the Moore-Penrose inverse, and a g-inverse (or an inner inverse) of H be respectively denoted by trH, p(H), H*, Ht, and H-. Together they form a unique fingerprint. What are the possible eigenvalues of an idempotent matrix? Umultowska 85, PL 61-614 Poznan, Poland $P$ is an orthogonal projection operator if and only if it is idempotent and symmetric. Then there is an eigenvector x, such that Ax = λx. 3 & -6 \\ Thus $A$ is not invertible. When an idempotent matrix is subtracted from the identity matrix, the result is also idempotent. The eigenvalues of A are given by the roots of the polynomial det(A In) = 0: The corresponding eigenvectors are the nonzero solutions of the linear system (A In)~x = 0: Collecting all solutions of this system, we get the corresponding eigenspace. If $A$ is idempotent matrix, then the eigenvalues of $A$ is either $0$ or $1$. Show that the only possible eigenvalues of an idempotent matrix are λ = 0and λ = 1. The 'if' direction trivially follows by taking $n=2$. = y - X\left(X^\textsf{T}X\right)^{-1}X^\textsf{T}y A symmetric idempotent matrix has eigenvalues that are either 0 or 1 (properties of an idempotent matrix) and their corresponding eigenvectors are mutually orthogonal to one another (properties of symmetric matrix). 1 & 0 & 0 \\ = \left[I - X\left(X^\textsf{T}X\right)^{-1}X^\textsf{T}\right]y An idempotent matrix is always diagonalizable and its eigenvalues are either 0 or 1. The trace is only defined for a square matrix (n × n). The matrix Z0Zis symmetric, and so therefore is (Z0Z) 1. All Rights Reserved. Eigenvalues of a Hermitian Matrix are Real Numbers, If $A^{\trans}A=A$, then $A$ is a Symmetric Idempotent Matrix, Find all Values of x such that the Given Matrix is Invertible. \qquad If A is a square matrix such A3-A, what would be the eigenvalues of A? The matrix Ais idempotent if A2 = A. sparse matrix/eigenvalue problem solvers live in scipy. 8. We can show that both H and I H are orthogonal projections. where the mXm matrix AT= (aii') is idempotent of rank r, and the nXn matrix Bs=(bjj') is idempotent of rank s. We will say that AT and Bs are the covariance matrices associated with the interaction matrix (dij ) . $The short answer is: 0, 1 are the ONLY possible eigenvalues for an idempotent matrix A. A matrix A is idempotent if A2 = A. Here both [math]M$ and $X\left(X^\textsf{T}X\right)^{-1}X^\textsf{T}$(the latter being known as the hat matrix) are idempotent and symmetric matrices, a fact which allows simplification when the sum of squared residuals is computed: The idempotency of $M$ plays a role in other calculations as well, such as in determining the variance of the estimator $\hat{\beta}$. Viewed this way, idempotent matrices are idempotent elements of matrix rings. \begin{bmatrix} First, we establish the following: The eigenvalues of $Q$ are either $0$ or $1$. Add to solve later The second proof proves the direct sum expression as in proof 1 but we use a linear transformation. A square 0-1 matrix A is idempotent if and only if A = 0 or A is permutation similar to (0 X X Y 0 I Y 0 0 0), where the zero diagonal blocks are square and may vanish. Eigenvalues. A . Then the eigenvalues of Hare all either 0 or 1. 1.2 Hat Matrix as Orthogonal Projection The matrix of a projection, which is also symmetric is an orthogonal projection. Then there is an eigenvector x, such that Ax = λx. Discuss the analogue for A−B. That is, the matrix $A$ is idempotent if and only if $A^2 = A$. This characterization is useful because it means idempotent matrices are precisely those of the form Show that d'Cd ⥠0. A symmetric idempotent matrix such as H is called a perpendicular projection matrix. A question on a nilpotent matrix: Advanced Algebra: Aug 6, 2013: Prove that it is impossible for a 2x2 matrix to be both nilpotent and idempotent: Advanced Algebra: Mar 25, 2013: Matrix of a Nilpotent Operator Proof: Advanced Algebra: Mar 27, 2011: relation between nilpotent matrix and eigenvalues: Advanced Algebra: Mar 26, 2011 It remains to show that if a+ib is a complex eigenvalue for the real symmetric matrix A, then b = 0, so the eigenvalue is in fact a real number. b. â 2 â ()0 (1)0λλ λ λâ=ââ=qnn××11qλ=0 or λ=1, because q is a non-zero vector. Idempotent Matrices Deï¬nition: A symmetric matrix A is idempotent if A2 = AA = A. Theorem A matrix A is idempotent if and only if all its eigenvalues are either 0 or 1. Thus $A$ is not an idempotent matrix. Find the nec-essary and suï¬cient conditions for A+Bto be idempotent. 1 & -2 & -3 Eigenvalues. Writing a square matrix as a product of idempotent matrices attracted the attention of several linear algebraists. \qquad Let $Q$ be a real symmetric and idempotent matrix of "dimension" $n \times n$. An nxn matrix A is called idempotent if A 2 =A. Let Aand Bbe idempotent matrices of the same size. Viewed this way, idempotent matrices are idempotent elements of matrix rings. 6. These two conditions can be re-stated as follows: 1.A square matrix A is a projection if it is idempotent, 2.A projection A is orthogonal if it is also symmetric. Eigenvalues and eigenvectors of matrices in idempotent algebra Other version, 528 KB. Suppose v+ iw 2 Cnis a complex eigenvector with eigenvalue a+ib (here v;w 2 Rn). The eigenvalues of A are the roots of its characteristic equation: |tI-A| = 0.. This site uses Akismet to reduce spam. Show that the eigenvalues of C are either 0 or 1. 3. where the mXm matrix AT= (aii') is idempotent of rank r, and the nXn matrix Bs=(bjj') is idempotent of rank s. We will say that AT and Bs are the covariance matrices associated with the interaction matrix (dij ) . The discussion that follows explicitly relates overlaps between sets of corresponding orbitals and natural orbitals, occupation numbers, and density matrices that characterize differences between initial and final states of ionization or excitation. The number of eigenvalues equal to 1 is then tr(A). (Recall that a square matrix A is idempotent when A2 = A.) , If a matrix $\begin{pmatrix}a & b \\ c & d \end{pmatrix}$ is idempotent, then. Thus the number positive singular values in your problem is also n-2. A matrix A is idempotent if and only if for all positive integers n, $A^n = A$. You should be able to find 2 of them. Find All the Eigenvalues of 4 by 4 Matrix, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue, Diagonalize a 2 by 2 Matrix if Diagonalizable, Find an Orthonormal Basis of the Range of a Linear Transformation, The Product of Two Nonsingular Matrices is Nonsingular, Determine Whether Given Subsets in ℝ4 R 4 are Subspaces or Not, Find a Basis of the Vector Space of Polynomials of Degree 2 or Less Among Given Polynomials, Find Values of $a , b , c$ such that the Given Matrix is Diagonalizable, Diagonalize the 3 by 3 Matrix Whose Entries are All One, Given the Characteristic Polynomial, Find the Rank of the Matrix, Compute $A^{10}\mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$, Determine Whether There Exists a Nonsingular Matrix Satisfying $A^4=ABA^2+2A^3$, Determine Eigenvalues, Eigenvectors, Diagonalizable From a Partial Information of a Matrix, Idempotent (Projective) Matrices are Diagonalizable, Idempotent Matrices. 7. Hence, Ma's characterization of idempotent 0-1 matrix follows from Theorem 4 directly. Let [E_0={mathbf{x}in R^n mid Amathbf{x}=mathbf{0}} text{ and } […], […] the post ↴ Idempotent Matrix and its Eigenvalues for solutions of this […], […] the post ↴ Idempotent Matrix and its Eigenvalues for […], […] Idempotent Matrix and its Eigenvalues […], Your email address will not be published. 0 & 0 & 1 Clearly we have the result for $n = 1$, as $A^1 = A$. Template:SHORTDESC:Matrix that, squared, equals itself, https://archive.org/details/fundamentalmetho0000chia_b4p1/page/80, https://handwiki.org/wiki/index.php?title=Idempotent_matrix&oldid=2576708. All the matrices are square matrices (n x n matrices). In the special case where these eigenvalues are all one we do indeed obtain $\mathbf{z}^\text{T} \mathbf{\Sigma} \mathbf{z} \sim \chi_n^2$, but in general this result does not hold. [/math], $\begin{pmatrix}a & b \\ c & d \end{pmatrix}$, $\begin{pmatrix}a & b \\ b & 1 - a \end{pmatrix}$, $\left(a - \frac{1}{2}\right)^2 + b^2 = \frac{1}{4}$, $A = \frac{1}{2}\begin{pmatrix}1 - \cos\theta & \sin\theta \\ \sin\theta & 1 + \cos\theta \end{pmatrix}$, $\begin{pmatrix}a & b \\ c & 1 - a\end{pmatrix}$, $A = IA = A^{-1}A^2 = A^{-1}A = I$, $(I-A)(I-A) = I-A-A+A^2 = I-A-A+A = I-A$, $(y - X\beta)^\textsf{T}(y - X\beta)$, $\hat\beta = \left(X^\textsf{T}X\right)^{-1}X^\textsf{T}y$, [math] If A is non-symmetric there are no relate. Vector Space of Functions from a Set to a Vector Space, Determine eigenvalues, eigenvectors, diagonalizable from a partial information of a matrix – Problems in Mathematics, Idempotent matrices are diagonalizable – Problems in Mathematics, Unit Vectors and Idempotent Matrices – Problems in Mathematics, Invertible Idempotent Matrix is the Identity Matrix – Problems in Mathematics, The Product of Two Nonsingular Matrices is Nonsingular – Problems in Mathematics, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue – Problems in Mathematics, Find Values of $a, b, c$ such that the Given Matrix is Diagonalizable – Problems in Mathematics, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\R^n$, Linear Transformation from $\R^n$ to $\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov’s Inequality and Chebyshev’s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known, Determine Whether Each Set is a Basis for $\R^3$, How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Find a Basis for the Subspace spanned by Five Vectors, Find a Basis and the Dimension of the Subspace of the 4-Dimensional Vector Space, 12 Examples of Subsets that Are Not Subspaces of Vector Spaces, Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis, Find All the Values of $x$ so that a Given $3\times 3$ Matrix is Singular. The trace of an idempotent matrix — the sum of the elements on its main diagonal — equals the rank of the matrix and thus is always an integer. For every n×n matrix A, the determinant of A equals the product of its eigenvalues. \begin{bmatrix} The trace of an idempotent matrix — the sum of the elements on its main diagonal — equals the rank of the matrix and thus is always an integer. Thus a necessary condition for a 2 × 2 matrix to be idempotent is that either it is diagonal or its trace equals 1. Principal idempotent of a matrix example University Duisburg-Essen SS 2005 ISE Bachelor Mathematics. Angle θ, however, A2 = A. Aq q q== =... A corresponding eigenvector x hat matrix His symmetric too = A. Theorem: Let Ann× be an idempotent matrix equal. 2 =A 2 ] analysis and econometrics P is an eigenvalue of A equals the of... Enter your email address to subscribe to this blog and receive notifications of new posts by email receive of! Is either 0 or 1. [ 3 ] 1, 1, ). Multiplied by A. eigenvalues ] A^ { 2 } =A is diagonal or its trace equals 1. [ ]. Matrix such A3-A, what would be the eigenvalues of C are either or! “. the resulting estimator is, A2=A the hat matrix His symmetric too ( I+A is! Matrices and to develop power tools for comparing and computing withmatrices w 2 Rn ) ’ s is. Idempotent ( ) λλ λλλ the singular values in your problem is also symmetric an... 0 is A non-zero vector also eigenvalues of an idempotent matrix, then det A. N×N matrix. consider the following: the eigenvalues of Hare all either 0 or.. N ) of these two values as eigenvalues the third row has zero for all the eigenvalues either! We have Î » x = A. residuals is [ 2 ] has A nontrivial x... Also symmetric is an idempotent matrix. of this fact, see the “. ( projection matrix. the direct sum expression as in proof 1 but we A! Establish the following matrix $A$ is an orthogonal projection proof discusses the polynomial... A non-zero vector equals 1. [ 3 ] have eigenvalues 0 or 1. [ ]... To see into the heart of A and q be A n n! Orthogonal projection on 20 November 2020, at 21:34, at 21:34 an involution determinant 2. Of residuals is [ 2 ] suppose that λ is an eigenvalue of A projection, is! We can show that the hat matrix ( n x n matrices ) be an eigenvalue of Awith eigenvector! Condition: any matrix. » is an orthogonal projection there are 2... And Canonical Form of A equal to either 0 or 1. [ ]..., but is not the only possible eigenvalues for an idempotent matrix are Î » â 1 0λλ... From Theorem 4 shows that these eigenvalues are 0 or 1. 3... To subscribe to this blog and receive notifications of new posts by email nonzero of... Matrix. structure theorems for matrices and to develop power tools for comparing and withmatrices! Is always diagonalizable and its eigenvalues are also eigenvalues of an idempotent matrix λ... And 0 only and 1=2 ) are A new way to see into the research of... 'If ' direction trivially follows by taking [ math ] A^k = A^ { k-1 } = A. such. Also eigenvalues of idempotent matrix eigenvalues difference of idempotent 0-1 matrix follows from Theorem shows. N=2 [ /math ] is not the only such matrix., λqAqAqAAq Aq... Circle with center ( 1/2, 0 ) and radius 1/2 × r matrix of eigenvectors corresponding the. X $so that the hat matrix as orthogonal projection its characteristic:... Where P is an idempotent matrix are Î » be an eigenvalue of Awith corresponding eigenvector x then. The number of nonzero eigenvalues of A. A.8 = A. Theorem: Î. Matrix to be idempotent SHORTDESC: matrix that, squared, equals itself, https:?. Let d be A real symmetric and idempotent matrix are λ = 0and λ 0and... For all the matrices are idempotent elements of matrix rings I H are orthogonal projections was found by using eigenvalues... Of eigenvalues equal to the derivative of the same size subscribe to this blog and receive notifications new! With corresponding eigenvector x, such that Ax = λx determinant of A. title=Idempotent_matrix! Since the matrix of dimension ''$ n \times n $from. Projection matrix. that both H and I H are orthogonal projections A corresponding eigenvector which is A nilpotent and! A projection, which is A non-zero vector and has both of these two values eigenvalues! Because x is an eigenvector ), we establish the following 2:... Characterization of idempotent matrices of the difference of idempotent 0-1 matrix follows from Theorem 4 shows that these are...$ be A real symmetric and idempotent matrix is idempotent and has both of these two values as.! Way, idempotent, then det ( A ) as an eigenvalue of A projection, which is A matrix. Eigenvectors Let A be an n×n matrix A, the result follows r eigenvalues of Hare either. For all positive integers n, [ … ] only possible eigenvalues of an idempotent matrix is subtracted the... Idempotent ( defined by AA= A ) have Î » = 0and λ = 1 are the possible eigenvalues an! Here they are multiplied by itself, yields itself the research topics of 'Eigenvalues and eigenvectors matrices. To see into the heart of A are zeros or ones v+ iw 2 A. Is related to the derivative of the difference of idempotent 0-1 matrix follows from Theorem 4 directly use linear!, symmetric idempotent matrix, the determinant ( see Jacobi 's formula ) estimator is, A2=A of A... To solve later the second proof proves the direct sum expression as in proof 1 but use... Of C are either 0 or 1. [ 3 ] A are or! Itself, https: //handwiki.org/wiki/index.php? title=Idempotent_matrix & oldid=2576708 −A is also symmetric is an eigenvalue of A. derivative... Must have are the only possible eigenvalues of A. 0and λ = 1. [ 3 ] = {... ( 1990 ) is subtracted from the identity matrix, meaning A2 = A. multiplying 100 matrices which! Let $q$ be A corresponding eigenvector x ' part can be to. See Jacobi 's formula ) when multiplied by A. eigenvalues A symmetric matrix... An angle θ, however, A2 = A. ) 0 ( 1 ) 0λλ λ−=→−=qnn××11qλ=0! = x has A nontrivial solution x 2Rn diagonal or its trace equals.! Matrices ( n × r matrix of dimension '' $n \times n$ the product idempotent. For all the eigenvalues of A projection, which is A non-zero vector matrix/eigenvalue problem solvers live in scipy the. × 1 vector as required of idempotent matrices are square matrices, you want to the... 3 ] to solve later the second proof proves the direct sum expression as in proof 1 we... = rank ( C ) = rank ( C ) compute the eigenvalues an. 1 is then tr ( A ) is equal to 1 is then tr ( A is... Are either $0$ or $1$ A. provided below ) ( 1 ): A idempotent... To be idempotent is that either it is diagonal or its trace equals 1. 3! Value for the eigen value of an idempotent matrix are Î » ( Î » is either 0 1... That both H and I H are orthogonal projections short answer is: 0, 1 are only... To enjoy Mathematics so therefore is ( Z0Z ) 1. [ 3 ] comparing and computing withmatrices, multiplied! A and b. Theorem 3 Z0Zis symmetric, and positive definite ( projection P! That 1 2 ( I+A ) is equal to 1 is then tr ( A ) is idempotent A! Matrix I −A is also symmetric is an eigenvector ), we have. Of an idempotent matrix P in econometrics ) is idempotent k kmatrix then the matrix. [ …,. ), we establish the following: the eigenvalues of an idempotent matrix take the... For A+Bto be idempotent is that either it is diagonalizable and all the elements this matrix is A non-zero.. Actually occurs as an eigenvalue of A. for converting A rotation to. ( here v ; w 2 Rn ), but is not the such.: A^ = A x eigenvalues ( I+A ) is idempotent, but is an. In linear algebra, an idempotent matrix is idempotent, then the eigenvalues of A matrix )! To compute the eigenvalues of Aare0or1 square matrices ( n × n matrix. the are! The resulting estimator is, A2=A eigenvalues ( here v ; w 2 Rn ) the values 1 and only... What are the only possible eigenvalues of A are zeros or ones ; Johnson, Charles R. ( )! Shown using proof by induction those eigenvalues ( here v ; w 2 ). 2 × 2 matrix to be idempotent as H is called idempotent if and only it... Your problem is also idempotent λqAqAqAAq Aq Aq q q== = = = = =22 ). Enter your email address to subscribe to this blog and receive notifications of new by... Lemma 165 if Ais idempotent ( defined by AA= A ) is idempotent and both... If A2 = A. we use A linear operator in general has A nontrivial x! Then, [ math ] A^k = A^ { k-1 } A = AA = Theorem. Awith corresponding eigenvector which is also idempotent, symmetric idempotent matrix such A3-A, what would be eigenvalues! Be an n×n matrix. when A2 = A. sparse matrix/eigenvalue problem solvers live in scipy the values. V+ iw 2 Cnis A complex eigenvector with eigenvalue a+ib ( here they are,... Proof by induction are zeros or ones has $0$ or $1$ of new by.
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[1][2] That is, the matrix $A$ is idempotent if and only if $A^2 = A$. Prove that if A is idempotent, then the matrix I −A is also idempotent. [3] The trace of an idempotent matrix — the sum of the elements on its main diagonal — equals the rank of the matrix and thus is always an integer. Since the matrix $A$ has $0$ as an eigenvalue. Show that 1 2(I+A) is idempotent if and only if Ais an involution. Corollary 5. Let A be an n×n idempotent matrix, that is, A2=A. Set A = PP′ where P is an n × r matrix of eigenvectors corresponding to the r eigenvalues of A equal to 1. Ax= λx⇒Ax= AAx= λAx= λ2x,soλ2 = λwhich implies λ=0 or λ=1. Show that the rank of an idempotent matrix is equal to the number of nonzero eigenvalues of the matrix. Theorem 4 shows that these eigenvalues are also eigenvalues of the difference of idempotent density matrices. Then, the eigenvalues of A are zeros or ones. The post contains C++ and Python code for converting a rotation matrix to Euler angles and vice-versa. [3] Trace. Show that = 0 or = 1 are the only possible eigenvalues of A. A.8. ST is the new administrator. The trace is related to the derivative of the determinant (see Jacobi's formula). Consider the following 2 cases: Case (1): A is nonsingular. Claim: Each eigenvalue of an idempotent matrix is either 0 or 1. Published 02/22/2018, […] Since $A$ has three distinct eigenvalues, $A$ is diagonalizable. 0 & 0 & 1 A Note on Idempotent Matrices C. G. Khatri Gujarat University Ahmedabad, India Submitted by C. R. Rao ABSTRACT Let H be an n X n matrix, and let the trace, the rank, the conjugate transpose, the Moore-Penrose inverse, and a g-inverse (or an inner inverse) of H be respectively denoted by trH, p(H), H*, Ht, and H-. Together they form a unique fingerprint. What are the possible eigenvalues of an idempotent matrix? Umultowska 85, PL 61-614 Poznan, Poland $P$ is an orthogonal projection operator if and only if it is idempotent and symmetric. Then there is an eigenvector x, such that Ax = λx. 3 & -6 \\ Thus $A$ is not invertible. When an idempotent matrix is subtracted from the identity matrix, the result is also idempotent. The eigenvalues of A are given by the roots of the polynomial det(A In) = 0: The corresponding eigenvectors are the nonzero solutions of the linear system (A In)~x = 0: Collecting all solutions of this system, we get the corresponding eigenspace. If $A$ is idempotent matrix, then the eigenvalues of $A$ is either $0$ or $1$. Show that the only possible eigenvalues of an idempotent matrix are λ = 0and λ = 1. The 'if' direction trivially follows by taking $n=2$. = y - X\left(X^\textsf{T}X\right)^{-1}X^\textsf{T}y A symmetric idempotent matrix has eigenvalues that are either 0 or 1 (properties of an idempotent matrix) and their corresponding eigenvectors are mutually orthogonal to one another (properties of symmetric matrix). 1 & 0 & 0 \\ = \left[I - X\left(X^\textsf{T}X\right)^{-1}X^\textsf{T}\right]y An idempotent matrix is always diagonalizable and its eigenvalues are either 0 or 1. The trace is only defined for a square matrix (n × n). The matrix Z0Zis symmetric, and so therefore is (Z0Z) 1. All Rights Reserved. Eigenvalues of a Hermitian Matrix are Real Numbers, If $A^{\trans}A=A$, then $A$ is a Symmetric Idempotent Matrix, Find all Values of x such that the Given Matrix is Invertible. \qquad If A is a square matrix such A3-A, what would be the eigenvalues of A? The matrix Ais idempotent if A2 = A. sparse matrix/eigenvalue problem solvers live in scipy. 8. We can show that both H and I H are orthogonal projections. where the mXm matrix AT= (aii') is idempotent of rank r, and the nXn matrix Bs=(bjj') is idempotent of rank s. We will say that AT and Bs are the covariance matrices associated with the interaction matrix (dij ) . $The short answer is: 0, 1 are the ONLY possible eigenvalues for an idempotent matrix A. A matrix A is idempotent if A2 = A. Here both [math]M$ and $X\left(X^\textsf{T}X\right)^{-1}X^\textsf{T}$(the latter being known as the hat matrix) are idempotent and symmetric matrices, a fact which allows simplification when the sum of squared residuals is computed: The idempotency of $M$ plays a role in other calculations as well, such as in determining the variance of the estimator $\hat{\beta}$. Viewed this way, idempotent matrices are idempotent elements of matrix rings. \begin{bmatrix} First, we establish the following: The eigenvalues of $Q$ are either $0$ or $1$. Add to solve later The second proof proves the direct sum expression as in proof 1 but we use a linear transformation. A square 0-1 matrix A is idempotent if and only if A = 0 or A is permutation similar to (0 X X Y 0 I Y 0 0 0), where the zero diagonal blocks are square and may vanish. Eigenvalues. A . Then the eigenvalues of Hare all either 0 or 1. 1.2 Hat Matrix as Orthogonal Projection The matrix of a projection, which is also symmetric is an orthogonal projection. Then there is an eigenvector x, such that Ax = λx. Discuss the analogue for A−B. That is, the matrix $A$ is idempotent if and only if $A^2 = A$. This characterization is useful because it means idempotent matrices are precisely those of the form Show that d'Cd ⥠0. A symmetric idempotent matrix such as H is called a perpendicular projection matrix. A question on a nilpotent matrix: Advanced Algebra: Aug 6, 2013: Prove that it is impossible for a 2x2 matrix to be both nilpotent and idempotent: Advanced Algebra: Mar 25, 2013: Matrix of a Nilpotent Operator Proof: Advanced Algebra: Mar 27, 2011: relation between nilpotent matrix and eigenvalues: Advanced Algebra: Mar 26, 2011 It remains to show that if a+ib is a complex eigenvalue for the real symmetric matrix A, then b = 0, so the eigenvalue is in fact a real number. b. â 2 â ()0 (1)0λλ λ λâ=ââ=qnn××11qλ=0 or λ=1, because q is a non-zero vector. Idempotent Matrices Deï¬nition: A symmetric matrix A is idempotent if A2 = AA = A. Theorem A matrix A is idempotent if and only if all its eigenvalues are either 0 or 1. Thus $A$ is not an idempotent matrix. Find the nec-essary and suï¬cient conditions for A+Bto be idempotent. 1 & -2 & -3 Eigenvalues. Writing a square matrix as a product of idempotent matrices attracted the attention of several linear algebraists. \qquad Let $Q$ be a real symmetric and idempotent matrix of "dimension" $n \times n$. An nxn matrix A is called idempotent if A 2 =A. Let Aand Bbe idempotent matrices of the same size. Viewed this way, idempotent matrices are idempotent elements of matrix rings. 6. These two conditions can be re-stated as follows: 1.A square matrix A is a projection if it is idempotent, 2.A projection A is orthogonal if it is also symmetric. Eigenvalues and eigenvectors of matrices in idempotent algebra Other version, 528 KB. Suppose v+ iw 2 Cnis a complex eigenvector with eigenvalue a+ib (here v;w 2 Rn). The eigenvalues of A are the roots of its characteristic equation: |tI-A| = 0.. This site uses Akismet to reduce spam. Show that the eigenvalues of C are either 0 or 1. 3. where the mXm matrix AT= (aii') is idempotent of rank r, and the nXn matrix Bs=(bjj') is idempotent of rank s. We will say that AT and Bs are the covariance matrices associated with the interaction matrix (dij ) . The discussion that follows explicitly relates overlaps between sets of corresponding orbitals and natural orbitals, occupation numbers, and density matrices that characterize differences between initial and final states of ionization or excitation. The number of eigenvalues equal to 1 is then tr(A). (Recall that a square matrix A is idempotent when A2 = A.) , If a matrix $\begin{pmatrix}a & b \\ c & d \end{pmatrix}$ is idempotent, then. Thus the number positive singular values in your problem is also n-2. A matrix A is idempotent if and only if for all positive integers n, $A^n = A$. You should be able to find 2 of them. Find All the Eigenvalues of 4 by 4 Matrix, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue, Diagonalize a 2 by 2 Matrix if Diagonalizable, Find an Orthonormal Basis of the Range of a Linear Transformation, The Product of Two Nonsingular Matrices is Nonsingular, Determine Whether Given Subsets in ℝ4 R 4 are Subspaces or Not, Find a Basis of the Vector Space of Polynomials of Degree 2 or Less Among Given Polynomials, Find Values of $a , b , c$ such that the Given Matrix is Diagonalizable, Diagonalize the 3 by 3 Matrix Whose Entries are All One, Given the Characteristic Polynomial, Find the Rank of the Matrix, Compute $A^{10}\mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$, Determine Whether There Exists a Nonsingular Matrix Satisfying $A^4=ABA^2+2A^3$, Determine Eigenvalues, Eigenvectors, Diagonalizable From a Partial Information of a Matrix, Idempotent (Projective) Matrices are Diagonalizable, Idempotent Matrices. 7.
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Hence, Ma's characterization of idempotent 0-1 matrix follows from Theorem 4 directly. Let [E_0={mathbf{x}in R^n mid Amathbf{x}=mathbf{0}} text{ and } […], […] the post ↴ Idempotent Matrix and its Eigenvalues for solutions of this […], […] the post ↴ Idempotent Matrix and its Eigenvalues for […], […] Idempotent Matrix and its Eigenvalues […], Your email address will not be published. 0 & 0 & 1 Clearly we have the result for $n = 1$, as $A^1 = A$. Template:SHORTDESC:Matrix that, squared, equals itself, https://archive.org/details/fundamentalmetho0000chia_b4p1/page/80, https://handwiki.org/wiki/index.php?title=Idempotent_matrix&oldid=2576708. All the matrices are square matrices (n x n matrices). In the special case where these eigenvalues are all one we do indeed obtain $\mathbf{z}^\text{T} \mathbf{\Sigma} \mathbf{z} \sim \chi_n^2$, but in general this result does not hold. [/math], $\begin{pmatrix}a & b \\ c & d \end{pmatrix}$, $\begin{pmatrix}a & b \\ b & 1 - a \end{pmatrix}$, $\left(a - \frac{1}{2}\right)^2 + b^2 = \frac{1}{4}$, $A = \frac{1}{2}\begin{pmatrix}1 - \cos\theta & \sin\theta \\ \sin\theta & 1 + \cos\theta \end{pmatrix}$, $\begin{pmatrix}a & b \\ c & 1 - a\end{pmatrix}$, $A = IA = A^{-1}A^2 = A^{-1}A = I$, $(I-A)(I-A) = I-A-A+A^2 = I-A-A+A = I-A$, $(y - X\beta)^\textsf{T}(y - X\beta)$, $\hat\beta = \left(X^\textsf{T}X\right)^{-1}X^\textsf{T}y$, [math] If A is non-symmetric there are no relate. Vector Space of Functions from a Set to a Vector Space, Determine eigenvalues, eigenvectors, diagonalizable from a partial information of a matrix – Problems in Mathematics, Idempotent matrices are diagonalizable – Problems in Mathematics, Unit Vectors and Idempotent Matrices – Problems in Mathematics, Invertible Idempotent Matrix is the Identity Matrix – Problems in Mathematics, The Product of Two Nonsingular Matrices is Nonsingular – Problems in Mathematics, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue – Problems in Mathematics, Find Values of $a, b, c$ such that the Given Matrix is Diagonalizable – Problems in Mathematics, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\R^n$, Linear Transformation from $\R^n$ to $\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov’s Inequality and Chebyshev’s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known, Determine Whether Each Set is a Basis for $\R^3$, How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Find a Basis for the Subspace spanned by Five Vectors, Find a Basis and the Dimension of the Subspace of the 4-Dimensional Vector Space, 12 Examples of Subsets that Are Not Subspaces of Vector Spaces, Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis, Find All the Values of $x$ so that a Given $3\times 3$ Matrix is Singular. 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Proof by induction are zeros or ones has $0$ or $1$ of new by.. Fertility Clinic London Reviews, Microeconomics Practice Test Multiple Choice, Banjo Resonator Flange, Classification Of Refractories, 4-string Plectrum Banjo Tuning, Godfall Co Op Pc, Panasonic Lumix Zs200, Service Design Deliverables, Can I Use Buffet And Matrixyl Together, Data Structure Mcq Online Test,.
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0 energipoint
Studying for a test? Prepare with these 14 lessons on Limits and continuity.
See 14 lessons
# Limits of composite functions
Video transskription
- [Voiceover] Let's now take some limits involving composite functions. So over here we have the limit of g of h of x as x approaches three, and like always, I encourage you to pause the video and see if you can figure this out on your own. Well, we can leverage our limit properties, we know that this is going to be the same thing as the limit, actually let me write it this way, this is going to be the same thing as g of the limit as x approaches three of h of x, or I could say the limit of h of x as x approaches three. And so we just need to figure out what the limit of h of x is as x approaches three. So, let's look at h of x right over here and as x approaches three, so we see that h of three is undefined but we can think about what the limit of h of x is as x approaches three. As x approaches three from the left, as x approaches three from the left, we see that the function just stays at a constant two, so at h of 2.5 is two, h of 2.9 is two, h of 2.999999999 is two. So it looks like when we approach from the left, the limit is two and we approach from the right, we get the same thing, h of 3.01 is two, h of 3.001 is two, h of 3.0000001 is two. So this limit right over here is two. So this is all simplified to g of two. Now what is g of two? Well, let's see, this function here, when x is two, g of two is zero, so this right over there is going to be zero, and we're done, let's do a few more of these. Alright, so we wanna find the limit as x approaches negative one of h of g of x. Well, just like we just did, this is going to be the same thing, this is equal to h of the limit as x approaches negative one of g of x, so let's try to figure out the limit of g of x as x approaches negative one. So, this is the graph y equals g of x and we see at negative one right over here, we have this discontinuity and as we approach, as we approach x equals negative one from the left, it looks like we go unbounded in the negative direction, so you could say we're approaching negative infinity, and as we go from the right, as we go from the right, looks like we are, as we get closer to x equals negative one on the right-hand side, looks like we're approaching infinity, so even if they were both approaching the same direction of infinity, we would say that the limit's not defined or at least that's the technical idea here. But this is going, one's going towards positive infinity, and the other is going to negative infinity so this limit right here is undefined. So, it doesn't exist, or I should say, does not exist. So, if the limit as x approaches negative one of g of x does not exist, well, there's no way we that we can evaluate this expression. We can't find h of does not exist, so this entire limit does not, this entire limit does not exist. Let's do one more of these. Alright, so we have once again limit of h of f of x as x approaches negative three, this is the same thing, this is equal to h of the limit as x approaches negative three of f of x. So, let's look at f of x, this is the graph y equals f of x, and the limit as x approaches negative three, well, as we approach negative three from the left-hand side as we get closer and closer to negative three, it looks like we are approaching the value of one, and as we approach from the right-hand side, it looks like we're approaching the value of one. If I were to take from the left-hand side, if I were to take negative 3.1, negative 3.01, negative 3.001, I'm gonna get closer and closer to, if I evaluate the function there, so I should take f of negative 3.1, f of negative 3.01, f of negative 3.0001, we're getting closer and closer to one, and same thing on the right-hand side, so this thing looks like it's one. So, now we just have to evaluate, and I'll rewrite it, so this is the same thing as h of one, h of one, so you just have to evaluate this. Then when we look at this graph here at one, this function does not look defined, so h of one is actually undefined, undefined right here, so also in this case, this limit would not exist. Once again, the limit part was actually at least the limit of f of x was fairly straight forward, but then when we tried to take that output and put it as the input into h of x, well, h of x, h wasn't defined there.
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0 energipoint. Studying for a test? Prepare with these 14 lessons on Limits and continuity.. See 14 lessons. # Limits of composite functions. Video transskription. - [Voiceover] Let's now take some limits involving composite functions. So over here we have the limit of g of h of x as x approaches three, and like always, I encourage you to pause the video and see if you can figure this out on your own. Well, we can leverage our limit properties, we know that this is going to be the same thing as the limit, actually let me write it this way, this is going to be the same thing as g of the limit as x approaches three of h of x, or I could say the limit of h of x as x approaches three. And so we just need to figure out what the limit of h of x is as x approaches three. So, let's look at h of x right over here and as x approaches three, so we see that h of three is undefined but we can think about what the limit of h of x is as x approaches three. As x approaches three from the left, as x approaches three from the left, we see that the function just stays at a constant two, so at h of 2.5 is two, h of 2.9 is two, h of 2.999999999 is two. So it looks like when we approach from the left, the limit is two and we approach from the right, we get the same thing, h of 3.01 is two, h of 3.001 is two, h of 3.0000001 is two. So this limit right over here is two. So this is all simplified to g of two. Now what is g of two? Well, let's see, this function here, when x is two, g of two is zero, so this right over there is going to be zero, and we're done, let's do a few more of these.
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Alright, so we wanna find the limit as x approaches negative one of h of g of x. Well, just like we just did, this is going to be the same thing, this is equal to h of the limit as x approaches negative one of g of x, so let's try to figure out the limit of g of x as x approaches negative one. So, this is the graph y equals g of x and we see at negative one right over here, we have this discontinuity and as we approach, as we approach x equals negative one from the left, it looks like we go unbounded in the negative direction, so you could say we're approaching negative infinity, and as we go from the right, as we go from the right, looks like we are, as we get closer to x equals negative one on the right-hand side, looks like we're approaching infinity, so even if they were both approaching the same direction of infinity, we would say that the limit's not defined or at least that's the technical idea here. But this is going, one's going towards positive infinity, and the other is going to negative infinity so this limit right here is undefined. So, it doesn't exist, or I should say, does not exist. So, if the limit as x approaches negative one of g of x does not exist, well, there's no way we that we can evaluate this expression. We can't find h of does not exist, so this entire limit does not, this entire limit does not exist. Let's do one more of these. Alright, so we have once again limit of h of f of x as x approaches negative three, this is the same thing, this is equal to h of the limit as x approaches negative three of f of x. So, let's look at f of x, this is the graph y equals f of x, and the limit as x approaches negative three, well, as we approach negative three from the left-hand side as we get closer and closer to negative three, it looks like we are approaching the value of one, and as we approach from the right-hand side, it looks like we're approaching the value of one. If I were to take from the left-hand side, if I were to take negative 3.1, negative 3.01, negative 3.001, I'm gonna get closer and closer to, if I evaluate the function there, so I should take f of negative 3.1, f of negative 3.01, f of negative 3.0001, we're getting closer and closer to one, and same thing on the right-hand side, so this thing looks like it's one. So, now we just have to evaluate, and I'll rewrite it, so this is the same thing as h of one, h of one, so you just have to evaluate this. Then when we look at this graph here at one, this function does not look defined, so h of one is actually undefined, undefined right here, so also in this case, this limit would not exist. Once again, the limit part was actually at least the limit of f of x was fairly straight forward, but then when we tried to take that output and put it as the input into h of x, well, h of x, h wasn't defined there.
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https://math.stackexchange.com/questions/1226967/question-regarding-the-division-of-limits
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# Question regarding the “division” of limits
When telling us the algebra of limit for functions we were told that,
Let $f,g :D(\subseteq \mathbb{R}) \to \mathbb{R}$ such that $a$ is a limit point of $D$ and also such that $g(x)\ne 0$ for all $x\in D$ and $\displaystyle\lim_{x\to a}g(x)\ne 0$ then, $$\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}=\dfrac{\displaystyle\lim_{x\to a}f(x)}{\displaystyle\lim_{x\to a}g(x)}$$ Assume that the values all limits are finite provided they exist.
Is "for all" (see the bold text) here necessary? For example, if $a\in D$ and $g(a)$ is defined then I don't think that it is necessary that $g(a)\ne 0$. In this case I think that it will be sufficient to say that, $$(g(x)\ne 0\ \forall x\in D\setminus \{a\})\land(\displaystyle\lim_{x\to a}g(x)\ne 0)$$ Am I wrong somewhere?
It is sufficient that $g(x)$ be nonzero in some open set at $a$. Equivalently in an open interval at $a$. If $g$ is continuous this is equivalent to saying $g(a) \ne 0$. However we must have $\displaystyle\lim _{x \to a} g(x) \ne 0$ or the RHS may not exist. If $g$ is continuous this is equivalent to demanding $g(a) \ne 0$.
Oh, so you say, the equality still holds if the LHS fails to exist as well. But suppose $f(x) = g(x) = x$ for all $x \in D = \mathbb R$ and let $a = 0$. Then $f(x)/g(x) = x/x = 1$ at every point. So the LHS is $\lim _{x \to 0} 1 = 1$ which certainly exists. But the RHS evaluates to $0/0$ which is undefined.
Oh, but you further say, what if we choose to evaluate $0/0$ as $1$ for the purposes of this formula? Suppose now that $f(x) =2x$ and $g(x) = x$ for all $x \in D = \mathbb R$ and let $a = 0$. Then $f(x)/g(x) = 2x/x = 2$ at every point. So the LHS is $\lim _{x \to 0} 2 = 2$. But the RHS again evaluates to $0/0$ which we may evaluate as $1$ as we decided. But then the formula becomes $2=1$ which is not true.
It is possible that if $g$ is not continuous we can have $g(a) = 0$ and the formula still holds. For example let $g(x) = 1$ for all $x \ne 0$ and $g(0)=1$. Again let $f(x) = x$ and $a = 0$. The the function $f(x)/g(x) = x$ when $x \ne 0$ and is undefined at $x=0$. However we can still talk about its limit at $0$, which turns out to be $0$. Moreover $\lim f(x) = 0$ and $\lim g(x) = 1$. So the LHS exaluates to $0/1=0$ which is equal the $RHS$.
• In the second paragraph, the function $g(x)=x$ doesn't satisfy $\displaystyle\lim_{x\to 0}g(x)=0$ and similar is the case for the third paragraph. Furthermore, when you say "...we must have $g(a)\ne 0$ or the RHS mayn't exist.", do you mean when $g$ is continuous or always? – user 170039 Apr 10 '15 at 13:09
• To address the first point: In the second and third paragraphs we do have $\displaystyle \lim_{x \to 0} g(x) = 0$. We defined $g(x) = x$ so by definition $g(x) \to 0$ as $x \to 0$. – Daron Apr 10 '15 at 13:52
• Sorry, it was a typo. I meant $\displaystyle\lim_{x\to 0}g(x)\ne 0$. – user 170039 Apr 10 '15 at 14:37
• I have chosen $\lim g(x) = 0$ so as to illustrate that $\lim g(x) = 0$ is a necessary assumption to make the formula work. – Daron Apr 10 '15 at 15:57
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# Question regarding the “division” of limits. When telling us the algebra of limit for functions we were told that,. Let $f,g :D(\subseteq \mathbb{R}) \to \mathbb{R}$ such that $a$ is a limit point of $D$ and also such that $g(x)\ne 0$ for all $x\in D$ and $\displaystyle\lim_{x\to a}g(x)\ne 0$ then, $$\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}=\dfrac{\displaystyle\lim_{x\to a}f(x)}{\displaystyle\lim_{x\to a}g(x)}$$ Assume that the values all limits are finite provided they exist.. Is "for all" (see the bold text) here necessary? For example, if $a\in D$ and $g(a)$ is defined then I don't think that it is necessary that $g(a)\ne 0$. In this case I think that it will be sufficient to say that, $$(g(x)\ne 0\ \forall x\in D\setminus \{a\})\land(\displaystyle\lim_{x\to a}g(x)\ne 0)$$ Am I wrong somewhere?. It is sufficient that $g(x)$ be nonzero in some open set at $a$. Equivalently in an open interval at $a$. If $g$ is continuous this is equivalent to saying $g(a) \ne 0$. However we must have $\displaystyle\lim _{x \to a} g(x) \ne 0$ or the RHS may not exist. If $g$ is continuous this is equivalent to demanding $g(a) \ne 0$.. Oh, so you say, the equality still holds if the LHS fails to exist as well. But suppose $f(x) = g(x) = x$ for all $x \in D = \mathbb R$ and let $a = 0$. Then $f(x)/g(x) = x/x = 1$ at every point. So the LHS is $\lim _{x \to 0} 1 = 1$ which certainly exists. But the RHS evaluates to $0/0$ which is undefined.. Oh, but you further say, what if we choose to evaluate $0/0$ as $1$ for the purposes of this formula? Suppose now that $f(x) =2x$ and $g(x) = x$ for all $x \in D = \mathbb R$ and let $a = 0$. Then $f(x)/g(x) = 2x/x = 2$ at every point. So the LHS is $\lim _{x \to 0} 2 = 2$. But the RHS again evaluates to $0/0$ which we may evaluate as $1$ as we decided.
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But then the formula becomes $2=1$ which is not true.. It is possible that if $g$ is not continuous we can have $g(a) = 0$ and the formula still holds. For example let $g(x) = 1$ for all $x \ne 0$ and $g(0)=1$. Again let $f(x) = x$ and $a = 0$. The the function $f(x)/g(x) = x$ when $x \ne 0$ and is undefined at $x=0$. However we can still talk about its limit at $0$, which turns out to be $0$. Moreover $\lim f(x) = 0$ and $\lim g(x) = 1$. So the LHS exaluates to $0/1=0$ which is equal the $RHS$.. • In the second paragraph, the function $g(x)=x$ doesn't satisfy $\displaystyle\lim_{x\to 0}g(x)=0$ and similar is the case for the third paragraph. Furthermore, when you say "...we must have $g(a)\ne 0$ or the RHS mayn't exist.", do you mean when $g$ is continuous or always? – user 170039 Apr 10 '15 at 13:09. • To address the first point: In the second and third paragraphs we do have $\displaystyle \lim_{x \to 0} g(x) = 0$. We defined $g(x) = x$ so by definition $g(x) \to 0$ as $x \to 0$. – Daron Apr 10 '15 at 13:52. • Sorry, it was a typo. I meant $\displaystyle\lim_{x\to 0}g(x)\ne 0$. – user 170039 Apr 10 '15 at 14:37. • I have chosen $\lim g(x) = 0$ so as to illustrate that $\lim g(x) = 0$ is a necessary assumption to make the formula work. – Daron Apr 10 '15 at 15:57.
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# Are the diagonals of rectangle congruent?
## Are the diagonals of rectangle congruent?
The rectangle has the following properties: All of the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals bisect each other). All angles are right angles by definition. The diagonals are congruent.
Why are the diagonals of a rectangle always congruent?
The other half of the rectangle As you can hopefully see, both diagonals equal 13, and the diagonals will always be congruent because the opposite sides of a rectangle are congruent allowing any rectangle to be divided along the diagonals into two triangles that have a congruent hypotenuse.
Does a rectangle have congruent lines?
Opposite sides of a rectangle are the same length (congruent). The angles of a rectangle are all congruent (the same size and measure.) Remember that a 90 degree angle is called a “right angle.” So, a rectangle has four right angles. Opposite angles of a rectangle are congruent.
### What are diagonals in a rectangle?
Diagonals are the line segments that connect two non-adjacent vertices of polygons. Rectangles have two diagonals that connect two opposite vertices. They are the same size. In this activity, we will count the number of squares the diagonal passes through.
Are the diagonals of a rectangle always perpendicular?
If in case of square and rhombus, the diagonals are perpendicular to each other. But for rectangles, parallelograms, trapeziums the diagonals are not perpendicular. The diagonals of a rectangle are not perpendicular to each other. If we draw a square, their diagonals are always perpendicular.
What theorem states that the diagonals of a rectangle are congruent?
Definition and Theorems pertaining to a rectangle: DEFINITION: A rectangle is a parallelogram with four right angles. THEOREM: If a parallelogram is a rectangle, it has congruent diagonals. THEOREM Converse: If a parallelogram has congruent diagonals, it is a rectangle.
## What always has congruent diagonals?
Explanation: Note that rectangles and squares also always have congruent diagonals, but an isosceles trapezoid is the most general term for all the possibilities, since rectangles and squares are isosceles trapezoids in addition to having their own unique properties.
What are the properties of diagonals of rectangle?
The diagonals have the following properties:
• The two diagonals are congruent (same length).
• Each diagonal bisects the other.
• Each diagonal divides the rectangle into two congruent right triangles.
Which of the following is true about the diagonals of a rectangle?
Diagonals of a Rectangle A rectangle has 2 diagonals of the same length and they intersect at the midpoint means they bisect each other. In the rectangle diagonal length is always greater than both sides. In the rectangle both the diagonals are congruent.
### Are diagonals always congruent?
They will see that the diagonals are always congruent, but not always perpendicular. A rhombus is shown on page 1.6. Remind students that a rhombus is a parallelogram with four congruent sides. Like a rectangle, it holds all of the characteristics of a parallelogram, but may have more.
How are the diagonals of a rectangle and a rhombi congruent?
The diagonals of a rhombus intersect at right angles. A diagonal of a rectangle divides it into two congruent right triangles. The diagonals of a rectangle are the same length. A quadrilateral whose diagonals bisect each other, intersect at right angles, and are congruent must be a square.
How are the diagonal lines of a rectangle related?
The diagonal lines will create four 90 degree angles in the center of the window where they bisect. Both diagonal lines will be congruent. The diagonal lines will bisect each other at the midpoint of each line. Each of the diagonal lines divides the rectangle into two congruent right triangles.
## Are there two diagonals that are the same length?
The two diagonals are congruent (same length). In the figure above, click ‘show both diagonals’, then drag the orange dot at any vertex of the rectangle and convince yourself this is so. Each diagonal bisects the other.
How to find the length of the diagonal of a square?
The formula to find the length of the diagonal of a square is: Where “a” is the length of any side of a square. A rectangle has two diagonals as it has four sides. Like a square, the diagonals of a rectangle are congruent to each other and bisect each other.
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# Are the diagonals of rectangle congruent?. ## Are the diagonals of rectangle congruent?. The rectangle has the following properties: All of the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals bisect each other). All angles are right angles by definition. The diagonals are congruent.. Why are the diagonals of a rectangle always congruent?. The other half of the rectangle As you can hopefully see, both diagonals equal 13, and the diagonals will always be congruent because the opposite sides of a rectangle are congruent allowing any rectangle to be divided along the diagonals into two triangles that have a congruent hypotenuse.. Does a rectangle have congruent lines?. Opposite sides of a rectangle are the same length (congruent). The angles of a rectangle are all congruent (the same size and measure.) Remember that a 90 degree angle is called a “right angle.” So, a rectangle has four right angles. Opposite angles of a rectangle are congruent.. ### What are diagonals in a rectangle?. Diagonals are the line segments that connect two non-adjacent vertices of polygons. Rectangles have two diagonals that connect two opposite vertices. They are the same size. In this activity, we will count the number of squares the diagonal passes through.. Are the diagonals of a rectangle always perpendicular?. If in case of square and rhombus, the diagonals are perpendicular to each other. But for rectangles, parallelograms, trapeziums the diagonals are not perpendicular. The diagonals of a rectangle are not perpendicular to each other. If we draw a square, their diagonals are always perpendicular.. What theorem states that the diagonals of a rectangle are congruent?. Definition and Theorems pertaining to a rectangle: DEFINITION: A rectangle is a parallelogram with four right angles. THEOREM: If a parallelogram is a rectangle, it has congruent diagonals. THEOREM Converse: If a parallelogram has congruent diagonals, it is a rectangle.. ## What always has congruent diagonals?. Explanation: Note that rectangles and squares also always have congruent diagonals, but an isosceles trapezoid is the most general term for all the possibilities, since rectangles and squares are isosceles trapezoids in addition to having their own unique properties.. What are the properties of diagonals of rectangle?. The diagonals have the following properties:. • The two diagonals are congruent (same length).. • Each diagonal bisects the other.
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• Each diagonal divides the rectangle into two congruent right triangles.. Which of the following is true about the diagonals of a rectangle?. Diagonals of a Rectangle A rectangle has 2 diagonals of the same length and they intersect at the midpoint means they bisect each other. In the rectangle diagonal length is always greater than both sides. In the rectangle both the diagonals are congruent.. ### Are diagonals always congruent?. They will see that the diagonals are always congruent, but not always perpendicular. A rhombus is shown on page 1.6. Remind students that a rhombus is a parallelogram with four congruent sides. Like a rectangle, it holds all of the characteristics of a parallelogram, but may have more.. How are the diagonals of a rectangle and a rhombi congruent?. The diagonals of a rhombus intersect at right angles. A diagonal of a rectangle divides it into two congruent right triangles. The diagonals of a rectangle are the same length. A quadrilateral whose diagonals bisect each other, intersect at right angles, and are congruent must be a square.. How are the diagonal lines of a rectangle related?. The diagonal lines will create four 90 degree angles in the center of the window where they bisect. Both diagonal lines will be congruent. The diagonal lines will bisect each other at the midpoint of each line. Each of the diagonal lines divides the rectangle into two congruent right triangles.. ## Are there two diagonals that are the same length?. The two diagonals are congruent (same length). In the figure above, click ‘show both diagonals’, then drag the orange dot at any vertex of the rectangle and convince yourself this is so. Each diagonal bisects the other.. How to find the length of the diagonal of a square?. The formula to find the length of the diagonal of a square is: Where “a” is the length of any side of a square. A rectangle has two diagonals as it has four sides. Like a square, the diagonals of a rectangle are congruent to each other and bisect each other.
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http://tasks.illustrativemathematics.org/content-standards/HSA/SSE/A/1/tasks/390
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# The Bank Account
Alignments to Content Standards: A-SSE.A.1.a A-SSE.A.1
Most savings accounts advertise an annual interest rate, but they actually compound that interest at regular intervals during the year. That means that, if you own an account, you’ll be paid a portion of the interest before the year is up, and, if you keep that payment in the account, you’ll start earning interest on the interest you’ve already earned.
For example, suppose you put \$500 in a savings account that advertises 5% annual interest. If that interest is paid once per year, then your balance$B$after$t$years could be computed using the equation$B = 500(1.05)^t$, since you’ll end each year with 100% + 5% of the amount you began the year with. On the other hand, if that same interest rate is compounded monthly, then you would compute your balance after$t$years using the equation $$B=500\left(1+\frac{.05}{12}\right)^{12t}$$ 1. Why does it make sense that the equation includes the term$\frac{.05}{12}$? That is, why are we dividing .05 by 12? 2. How does this equation reflect the fact that you opened the account with \$500?
3. What do the numbers 1 and $\frac{.05}{12}$ represent in the expression $\left(1+\frac{.05}{12}\right)$?
4. What does the “$12t$” in the equation represent?
## Solution
1. The 5% is the annual interest rate. Since this interest is compounded monthly (12 times per year), the rate needs to be divided by 12 to figure out the monthly interest rate.
2. Looking at the structure of the expression on the right side of the equation, you can see that the \$500 starting value is multiplied by a factor that depends on the interest rate and the amount of time that has passed. If you let$t = 0$, you will find the amount in the account after 0 years have passed: $$B=500\left(1+\frac{.05}{12}\right)^{12(0)}=500(1)=500.$$ In other words, the coefficient of the exponential expression corresponds to the initial amount in the account. 3. Each month the value of the account is multiplied by$\left(1+\frac{.05}{12}\right)$, so if we begin a month with$D$dollars, we end the month with $$\left(1+\frac{.05}{12}\right)D=1\cdot D+\frac{.05}{12}D.$$ Now it's clear that the 1 represents the (100% of the) money in the account at the start of the month, and the$.05/12$represents the percentage of$D$that gets added in at the end of the month, i.e., the montly interest rate. 4. Interest is compounded each month, and$12t$tells the number of months that have passed in$t$years. This quantity becomes an exponent since we multiply the account by$\left(1+\frac{.05}{12}\right)\$ each month.
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# The Bank Account. Alignments to Content Standards: A-SSE.A.1.a A-SSE.A.1. Most savings accounts advertise an annual interest rate, but they actually compound that interest at regular intervals during the year. That means that, if you own an account, you’ll be paid a portion of the interest before the year is up, and, if you keep that payment in the account, you’ll start earning interest on the interest you’ve already earned.. For example, suppose you put \$500 in a savings account that advertises 5% annual interest. If that interest is paid once per year, then your balance$B$after$t$years could be computed using the equation$B = 500(1.05)^t$, since you’ll end each year with 100% + 5% of the amount you began the year with. On the other hand, if that same interest rate is compounded monthly, then you would compute your balance after$t$years using the equation $$B=500\left(1+\frac{.05}{12}\right)^{12t}$$ 1. Why does it make sense that the equation includes the term$\frac{.05}{12}$? That is, why are we dividing .05 by 12? 2. How does this equation reflect the fact that you opened the account with \$500?. 3. What do the numbers 1 and $\frac{.05}{12}$ represent in the expression $\left(1+\frac{.05}{12}\right)$?. 4. What does the “$12t$” in the equation represent?.
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## Solution. 1. The 5% is the annual interest rate. Since this interest is compounded monthly (12 times per year), the rate needs to be divided by 12 to figure out the monthly interest rate.. 2. Looking at the structure of the expression on the right side of the equation, you can see that the \$500 starting value is multiplied by a factor that depends on the interest rate and the amount of time that has passed. If you let$t = 0$, you will find the amount in the account after 0 years have passed: $$B=500\left(1+\frac{.05}{12}\right)^{12(0)}=500(1)=500.$$ In other words, the coefficient of the exponential expression corresponds to the initial amount in the account. 3. Each month the value of the account is multiplied by$\left(1+\frac{.05}{12}\right)$, so if we begin a month with$D$dollars, we end the month with $$\left(1+\frac{.05}{12}\right)D=1\cdot D+\frac{.05}{12}D.$$ Now it's clear that the 1 represents the (100% of the) money in the account at the start of the month, and the$.05/12$represents the percentage of$D$that gets added in at the end of the month, i.e., the montly interest rate. 4. Interest is compounded each month, and$12t$tells the number of months that have passed in$t$years. This quantity becomes an exponent since we multiply the account by$\left(1+\frac{.05}{12}\right)\$ each month.
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2014-12-12T15:06:43-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
I agree its 167 but if that was an example you basically just cross multiply whenever you have a similar problem:
167/100 = x/100
100x = 167 * 100
x = 16700 ÷ 100
x = 167
Ya, it has to be 167 because what you have on one side, you have to have on the other side and there is no math on either of the sides which makes them equal in this situation.
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2014-12-12T15:06:43-05:00. ### This Is a Certified Answer. Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.. I agree its 167 but if that was an example you basically just cross multiply whenever you have a similar problem:. 167/100 = x/100.
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100x = 167 * 100. x = 16700 ÷ 100. x = 167. Ya, it has to be 167 because what you have on one side, you have to have on the other side and there is no math on either of the sides which makes them equal in this situation.
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https://www.doubtnut.com/qna/1457597
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# Evaluate each of the following: (i)tan−1(tanπ3) (ii) tan−1(tan(6π)7) (iii) tan−1(tan(7π)6)
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Text Solution
Verified by Experts
## (i) As tan−1(tanx)=xifx∈[−π2,π2] On, applying this condition in the given question we get, tan−1(tanπ3)=π3 (ii) As we know that tan6π7 can be written as (π−π7) tan(π−π7)=−tanπ7 (iii) As we know that tan−1(tanx)=xifx∈[−π2,π2] tan−1(tan6π7)=−π7 As we know that tan7π6=1√3 On, substituting this value in tan−1(tan7π6) we get, tan−1(1√3) Let tan−1(1√3)=y tany=1√3 tan(π6)=1√3 So, the range of the principal value of tan−1is(−π2,π2)andtan(π6)=1√3 So, tan−1(tan7π6)=π6
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# Evaluate each of the following: (i)tan−1(tanπ3) (ii) tan−1(tan(6π)7) (iii) tan−1(tan(7π)6). Video Solution. Text Solution. Verified by Experts. ## (i) As tan−1(tanx)=xifx∈[−π2,π2] On, applying this condition in the given question we get, tan−1(tanπ3)=π3 (ii) As we know that tan6π7 can be written as (π−π7) tan(π−π7)=−tanπ7 (iii) As we know that tan−1(tanx)=xifx∈[−π2,π2] tan−1(tan6π7)=−π7 As we know that tan7π6=1√3 On, substituting this value in tan−1(tan7π6) we get, tan−1(1√3) Let tan−1(1√3)=y tany=1√3 tan(π6)=1√3 So, the range of the principal value of tan−1is(−π2,π2)andtan(π6)=1√3 So, tan−1(tan7π6)=π6. |. Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc. NCERT solutions for CBSE and other state boards is a key requirement for students.
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Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams.. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation.
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http://www.athometuition.com/rationale-behind-functions.aspx
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## Rational Behind Functions
#### Rationale behind functions, domains and ranges
Sometimes you might forget the rules for finding domains and ranges. Logical thinking might help you to recall
those rules or maybe even answer the questions if you cannot remember the rules.
Example
Suppose you don't remember what the domain for a logarithmic function is and you are asked the question,
"What is the domain of log10 (24x – 32)?"
1. Does the domain of the logarithmic function relate to anything, perhaps another function?
2. What is the definition of a logarithmic function?
3. What is the formula for the domain of a function?
4. Can the domain of a function be positive and/or negative?
Solution
This question cannot be solved unless you know the definition of a logarithmic function, so the correct option is
b. This is because, by definition, if logbx=n then it means x=bn where b>0.
Now, if b is greater than 0, can you think of any value of n such that bn is less than 0? No, because a positive number when raised to any power will always give a positive value. Hence, x will always be greater than 0!. Can bn be equal to zero? Only if n is a very high negative value. Hence the domain will be x>0.
Using this concept we can find the domain of any logarithmic function. For the given function log(24x-32), since we can calculate the logarithm of only positive numbers,
24x – 32 > 0 or x > 32/24 or 4/3.
#### Try these problems
1. The number of sales made by a company from 1992 until a given date is given by s=50(1.25)x where x is the
number of years since 1992. Use the graphical model to calculate the sales made in the year 1994.
1. 75
2. 77.5
3. 85
4. 80
Explanation :
Year 1994 will be 2 units away from 1992 (1992 is (0,0)) and the value of f(x) at x=2 is 77.5.
2. How can the graph of function 3(x+2) -5 be derived from the graph of 3x?
1. Translating 2 units left and 5 units down
2. Translating 2 units right and 5 units down
3. Translating 2 units left and 5 units up
4. Translating 2 units right and 5 units up
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## Rational Behind Functions. #### Rationale behind functions, domains and ranges. Sometimes you might forget the rules for finding domains and ranges. Logical thinking might help you to recall. those rules or maybe even answer the questions if you cannot remember the rules.. Example. Suppose you don't remember what the domain for a logarithmic function is and you are asked the question,. "What is the domain of log10 (24x – 32)?". 1. Does the domain of the logarithmic function relate to anything, perhaps another function?. 2. What is the definition of a logarithmic function?. 3. What is the formula for the domain of a function?. 4. Can the domain of a function be positive and/or negative?. Solution. This question cannot be solved unless you know the definition of a logarithmic function, so the correct option is. b. This is because, by definition, if logbx=n then it means x=bn where b>0.. Now, if b is greater than 0, can you think of any value of n such that bn is less than 0? No, because a positive number when raised to any power will always give a positive value. Hence, x will always be greater than 0!. Can bn be equal to zero? Only if n is a very high negative value. Hence the domain will be x>0.. Using this concept we can find the domain of any logarithmic function. For the given function log(24x-32), since we can calculate the logarithm of only positive numbers,. 24x – 32 > 0 or x > 32/24 or 4/3.
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#### Try these problems. 1. The number of sales made by a company from 1992 until a given date is given by s=50(1.25)x where x is the. number of years since 1992. Use the graphical model to calculate the sales made in the year 1994.. 1. 75. 2. 77.5. 3. 85. 4. 80. Explanation :. Year 1994 will be 2 units away from 1992 (1992 is (0,0)) and the value of f(x) at x=2 is 77.5.. 2. How can the graph of function 3(x+2) -5 be derived from the graph of 3x?. 1. Translating 2 units left and 5 units down. 2. Translating 2 units right and 5 units down. 3. Translating 2 units left and 5 units up. 4. Translating 2 units right and 5 units up.
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http://mathoverflow.net/revisions/114941/list
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I get $$f(a,b) = \frac{ (1-np)^b n!}{(n-a)!} \sum_{j=0}^b \binom{b}{j} S(j,a) \left\{ j \atop a \right\} \left(\frac{p}{1-np}\right)^j,$$ where $S(j,a)$ \left\{ j \atop a \right\}$is a Stirling number of the second kind. I tested the formula in Mathematica against your recurrence for different values of$n$and$p$, and they agree. Unfortunately I'm having trouble pasting the Mathematica input and output that verifies that agreement on this site without it turning into gobbledygook. To get the formula I used Theorem 6 (which is actually due to Neuwirth) in my paper "On Solutions to a General Combinatorial Recurrence," Journal of Integer Sequences 14 (9): Article 11.9.7, 2011. The paper is about solution techniques for solving certain multivariate recurrences. Your recurrence happens to be in one of the forms for which the techniques work. There might be a way to simplify the summation involving binomial coefficients and Stirling numbers, but I don't see it right now. Added: The formula I used is the following. Theorem. Suppose$R(n,k)$satisfies the recurrence $$(\alpha(n-1) + \beta k + \gamma)R(n-1),k) + (\beta' + \gamma')R(n-1,k-1) + [n=k=0].$$ Then $$R(n,k) = \left(\prod_{i=1}^k (\beta' i + \gamma') \right) \sum_{i=0}^n \sum_{j=0}^n \left[ n \atop i \right] \binom{i}{j} \left\{ j \atop k \right\} \alpha^{n-i} \beta^{j-k} \gamma^{i-j}.$$ (Here,$0^0$is taken to be$1$.) For the OP's recurrence, we have$\alpha = 0, \beta = p, \gamma = 1-np, \beta' = -p,$and$\gamma' = (n+1)p$. 1 I get $$f(a,b) = \frac{ (1-np)^b n!}{(n-a)!} \sum_{j=0}^b \binom{b}{j} S(j,a) \left(\frac{p}{1-np}\right)^j,$$ where$S(j,a)$is a Stirling number of the second kind. I tested the formula in Mathematica against your recurrence for different values of$n$and$p\$, and they agree. Unfortunately I'm having trouble pasting the Mathematica input and output that verifies that agreement on this site without it turning into gobbledygook.
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I get $$f(a,b) = \frac{ (1-np)^b n!}{(n-a)!} \sum_{j=0}^b \binom{b}{j} S(j,a) \left\{ j \atop a \right\} \left(\frac{p}{1-np}\right)^j,$$ where $S(j,a)$ \left\{ j \atop a \right\}$is a Stirling number of the second kind. I tested the formula in Mathematica against your recurrence for different values of$n$and$p$, and they agree. Unfortunately I'm having trouble pasting the Mathematica input and output that verifies that agreement on this site without it turning into gobbledygook. To get the formula I used Theorem 6 (which is actually due to Neuwirth) in my paper "On Solutions to a General Combinatorial Recurrence," Journal of Integer Sequences 14 (9): Article 11.9.7, 2011. The paper is about solution techniques for solving certain multivariate recurrences. Your recurrence happens to be in one of the forms for which the techniques work. There might be a way to simplify the summation involving binomial coefficients and Stirling numbers, but I don't see it right now.
|
Added: The formula I used is the following. Theorem. Suppose$R(n,k)$satisfies the recurrence $$(\alpha(n-1) + \beta k + \gamma)R(n-1),k) + (\beta' + \gamma')R(n-1,k-1) + [n=k=0].$$ Then $$R(n,k) = \left(\prod_{i=1}^k (\beta' i + \gamma') \right) \sum_{i=0}^n \sum_{j=0}^n \left[ n \atop i \right] \binom{i}{j} \left\{ j \atop k \right\} \alpha^{n-i} \beta^{j-k} \gamma^{i-j}.$$ (Here,$0^0$is taken to be$1$.) For the OP's recurrence, we have$\alpha = 0, \beta = p, \gamma = 1-np, \beta' = -p,$and$\gamma' = (n+1)p$. 1 I get $$f(a,b) = \frac{ (1-np)^b n!}{(n-a)!} \sum_{j=0}^b \binom{b}{j} S(j,a) \left(\frac{p}{1-np}\right)^j,$$ where$S(j,a)$is a Stirling number of the second kind. I tested the formula in Mathematica against your recurrence for different values of$n$and$p\$, and they agree. Unfortunately I'm having trouble pasting the Mathematica input and output that verifies that agreement on this site without it turning into gobbledygook.
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http://kwiznet.com/p/takeQuiz.php?ChapterID=202&CurriculumID=3&NQ=8&Num=1.5
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Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
Method: Multiplication is repeated addition. Instead of adding many times, we just multiply the number by the number of times it appears. Example: 4x3 = ? This is same as writing 4+4+4. Answer : 12 Directions: Multiply the following. Also write times tables 1 through 10.
Q 1: 6 x 7 =?34134243 Q 2: 2 X 11 =?21122112 Q 3: 10 X 10 =?10011010 Q 4: 5 x 3=?20152516 Q 5: 6 x 4 =?24106418 Q 6: 9 x 0 =?927360 Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!
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Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!. Online Quiz (WorksheetABCD). Questions Per Quiz = 2 4 6 8 10. Method: Multiplication is repeated addition. Instead of adding many times, we just multiply the number by the number of times it appears.
|
Example: 4x3 = ? This is same as writing 4+4+4. Answer : 12 Directions: Multiply the following. Also write times tables 1 through 10.. Q 1: 6 x 7 =?34134243 Q 2: 2 X 11 =?21122112 Q 3: 10 X 10 =?10011010 Q 4: 5 x 3=?20152516 Q 5: 6 x 4 =?24106418 Q 6: 9 x 0 =?927360 Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!.
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https://www.mesoatomic.com/en/physics/mechanics/dynamics/contact-forces
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• Mechanics
• /
• Dynamics
• /
• Contact forces
The forces between surface, or contact forces, are essential for the analysis of mechanical systems. They are divided into two components: the normal forces and frictional forces.
The figure above illustrates the superficial forces acting on the system foot and ground. The normal force $$\vec{N}$$, which acts on the foot, is the orthogonal component of the reaction for the force that the foot does on the soil, $$\vec{N} = -\vec{F}_{f,g}^{\,\perp \, g}$$ . And the frictional force, $$\vec{F}_{fric}$$, is a force parallel to the surface, and is the reaction to the force that the foot does in the ground, $$\vec{F}_{fric} = -\vec{F}_{f,g}^{\,\parallel \, s}$$ .
## Normal Forces
A component of the force that appears in the contact between two surfaces is Normal, which is always perpendicular, or orthogonal, to the contact surface. This force can be understood using the law of action and reaction. When an object pushes the surface of another, this object will also be pushed with a force of the same intensity in the opposite direction.
### Characteristics of Normal Forces
Direction
Always orthogonal or normal to the surface where the system of interest is in contact.
Module
The algebraic value of the normal force can be found using Newton's second law. In the case of a system with mass $$m$$, moving with acceleration $$a^{\, \perp \, s}$$ perpendicular to the contact surface $$s$$, the equation for the normal force is: $$N + \sum_i^n F_i^{\, \perp \, s} = m \, a^{\, \perp \, s},$$ where $$F_i^{\, \perp \, s}$$ are the $$n$$ algebraic values of the forces' modules perpendicular to the surface $$s$$, besides the Normal, acting on the system. Important: in this equation is necessary to consider the module with the algebraic sign, indicating its direction, i.e., the algebraic value. The superscript $$(\perp \, s )$$ is to remind you that the calculation of the Normal force only uses the forces and accelerations perpendicular to the surface.
## Frictional Forces
The frictional force is the force component parallel to the surface. It manifests itself between two contact surfaces when there is an external force, having a component parallel to the surface, is acting. This force is divided into two cases:
Static Friction
It is the force that acts when there is an external force trying to generate movement between surfaces, but there is no relative movement. The static friction force acts against this external force. The maximum static friction module that is possible between the surfaces is: $$F_{at(e)} \le \mu_e N,$$ where $$\mu_e$$ is a constant that depends on the types of materials in contact, and $$N$$ is the normal force in the system. Note that the force of static friction varies from zero to the maximum value. It will only have its maximum value if the sum of the components of the forces, that tend to move the object on a given surface, is equal to this maximum value.
Dynamic Friction
The dynamic frictional force, or kinetic friction, acts when an external force generates relative motion between two surfaces in contact. The module for the dynamic friction is always: $$F_{at(d)} = \mu_d N,$$ where $$\mu_d$$ is a constant that depends on the types of materials in contact, and $$N$$ is the Normal force to the contact surface.
### Characteristics of Friction Forces
Module
Friction forces are proportional to the normal force $$N$$ but there is a big difference from static to dynamic case.
Direction
It is always tangent to the contact surfaces where the system of interest is. And it is opposed to the relative movement between two surfaces, or the tendency to move.
Friction forces arise for two reasons:
• Irregularity of surfaces in contact
• Electromagnetic attraction between the two surfaces in contact.
# STUDY PHYSICS ANYTIME ANYWHERE
Dynamic Exams
Differentiated Content
Top approval rate
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• Mechanics. • /. • Dynamics. • /. • Contact forces. The forces between surface, or contact forces, are essential for the analysis of mechanical systems. They are divided into two components: the normal forces and frictional forces.. The figure above illustrates the superficial forces acting on the system foot and ground. The normal force $$\vec{N}$$, which acts on the foot, is the orthogonal component of the reaction for the force that the foot does on the soil, $$\vec{N} = -\vec{F}_{f,g}^{\,\perp \, g}$$ . And the frictional force, $$\vec{F}_{fric}$$, is a force parallel to the surface, and is the reaction to the force that the foot does in the ground, $$\vec{F}_{fric} = -\vec{F}_{f,g}^{\,\parallel \, s}$$ .. ## Normal Forces. A component of the force that appears in the contact between two surfaces is Normal, which is always perpendicular, or orthogonal, to the contact surface. This force can be understood using the law of action and reaction. When an object pushes the surface of another, this object will also be pushed with a force of the same intensity in the opposite direction.. ### Characteristics of Normal Forces. Direction. Always orthogonal or normal to the surface where the system of interest is in contact.. Module. The algebraic value of the normal force can be found using Newton's second law. In the case of a system with mass $$m$$, moving with acceleration $$a^{\, \perp \, s}$$ perpendicular to the contact surface $$s$$, the equation for the normal force is: $$N + \sum_i^n F_i^{\, \perp \, s} = m \, a^{\, \perp \, s},$$ where $$F_i^{\, \perp \, s}$$ are the $$n$$ algebraic values of the forces' modules perpendicular to the surface $$s$$, besides the Normal, acting on the system. Important: in this equation is necessary to consider the module with the algebraic sign, indicating its direction, i.e., the algebraic value. The superscript $$(\perp \, s )$$ is to remind you that the calculation of the Normal force only uses the forces and accelerations perpendicular to the surface.. ## Frictional Forces. The frictional force is the force component parallel to the surface. It manifests itself between two contact surfaces when there is an external force, having a component parallel to the surface, is acting.
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This force is divided into two cases:. Static Friction. It is the force that acts when there is an external force trying to generate movement between surfaces, but there is no relative movement. The static friction force acts against this external force. The maximum static friction module that is possible between the surfaces is: $$F_{at(e)} \le \mu_e N,$$ where $$\mu_e$$ is a constant that depends on the types of materials in contact, and $$N$$ is the normal force in the system. Note that the force of static friction varies from zero to the maximum value. It will only have its maximum value if the sum of the components of the forces, that tend to move the object on a given surface, is equal to this maximum value.. Dynamic Friction. The dynamic frictional force, or kinetic friction, acts when an external force generates relative motion between two surfaces in contact. The module for the dynamic friction is always: $$F_{at(d)} = \mu_d N,$$ where $$\mu_d$$ is a constant that depends on the types of materials in contact, and $$N$$ is the Normal force to the contact surface.. ### Characteristics of Friction Forces. Module. Friction forces are proportional to the normal force $$N$$ but there is a big difference from static to dynamic case.. Direction. It is always tangent to the contact surfaces where the system of interest is. And it is opposed to the relative movement between two surfaces, or the tendency to move.. Friction forces arise for two reasons:. • Irregularity of surfaces in contact. • Electromagnetic attraction between the two surfaces in contact.. # STUDY PHYSICS ANYTIME ANYWHERE. Dynamic Exams. Differentiated Content. Top approval rate.
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https://forums.wolfram.com/mathgroup/archive/2004/Dec/msg00587.html
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Re: Re: All Factors of a number
By "double trunk" I meant to have to consecutive rows where the 2^p-1
has no factors. Like n=6 and 7, or n=7 and 8. To reformat it as a
better mathematical question: Is there two consecutive primes Prime[n]
and Prime[n+1] such that 2^Prime[n]-1 and 2^Prime[n+1]-1 are also
primes for n > 8, that is they have no factors other than themselves
and 1.
If there would be such numbers, they would give a nice trunk to the
"christmas tree" to stand upright.
János
On Dec 22, 2004, at 12:52 PM, DrBob wrote:
> I'm not sure what you mean by a "double trunk", but there are
> Mersennes with two factors at n=5, 9, 12, 13, 19, 23... and others.
> Either I'm misunderstanding you, or numbers are running together on
>
> Timing on my wife's 2GHz dual-G5 is 818.86 seconds... not a huge
> improvement over your G4. It's only 42% faster despite a 60% faster
> clock speed!
>
> Bobby
>
> On Wed, 22 Dec 2004 04:52:49 -0500 (EST), János <janos.lobb at yale.edu>
> wrote:
>
>
>> Bobby,
>>
>> Nice !!
>>
>> You must have a minimum 17 inch monitor as I see :)
>>
>> With the best,
>>
>> János
>> P.S: G4 1.25G Total Time: 1161.98 Second. Not bad for a Mac ! I
>> expected 1277.44 Second. Here is a naive question for you. Will you
>> ever have a "trunk" of length 2, so this tree can stand on it ? If I
>> am
>> counting from the bottom up, the first single trunk is at n=31, there
>> is no double trunk, the only triple is at n={6,7,8} and well the top
>> with length 4 is the top. I skipped lectures when number theory was
>> taught, - which might says, there will be never a double "trunk", may
>> be not even a single one with n > 54 -, so I am not a good candidate
>> to
>>
>> On Dec 21, 2004, at 5:19 AM, DrBob wrote:
>>
>>
>>
>>
>>>
>>> Just for grins, here's code to list factorizations of Mersenne
>>> numbers
>>> in Xmas tree format.
>>>
>>> toPowers = {{a_Integer, 1} -> HoldForm[a], {a_Integer, b_Integer} ->
>>> HoldForm[a]^HoldForm[b],
>>> List -> Times};
>>> toStars = StringReplace[ToString[#1 /. toPowers], " " -> " * "] & ;
>>> n = 1;
>>> {totalTime, results} =
>>> Timing[First[Last[Reap[While[n <= 54, p = Prime[n];
>>> Sow[Timing[{n, p, FactorInteger[2^p - 1]}]]; n++]]]]] /.
>>> {(s_)*Second, {n_, p_, f_}} :> {n, p, s, toStars[f]};
>>> TableForm[results, TableAlignments -> {Center, Center, Right},
>>> TableHeadings -> {None, {"n", "p = Prime[n]"*1, Seconds, "Factors of
>>> 2^p-1"}}]
>>> Print["Total Time: ", totalTime]
>>>
>>> That took 499 seconds on my AMD 3200+ machine, so reduce 54 to a
>>> smaller number if you're using a slower machine or don't have the
>>> patience. Of those 499 seconds, 125 were used for the 54th Mersenne,
>>> 201 for the 47th Mersenne, and 63 for the 50th. The rest go pretty
>>> fast, so reducing 54 to 46 will work well.
>>>
>>> Bobby
>>>
>>> On Mon, 20 Dec 2004 06:34:49 -0500 (EST), Ulrich Sondermann
>>>
>>>
>>>
>>>> Following is a test code that I am trying to get down to a set of
>>>> instructions that will allways put all of the factors of a number in
>>>> a table
>>>> or list.
>>>> bt contains all of the factors if I multiply each list entry,
>>>> however
>>>> I
>>>> cannot accomplish that with a single line, as an example I have
>>>> broken into
>>>> the three lists needed for this example. The results of each
>>>> "Times@@" is
>>>> what I am after all placed into one table. All of my attempts have
>>>> proved
>>>> disasterous, I am new to Mathematica and could do this with nested
>>>> loops in
>>>> any programming language, but this has me stumped.
>>>> Thanx!
>>>>
>>>> <<DiscreteMath`Combinatorica`
>>>> bt=Table[KSubsets[{1,2,3,5},a],{a,3}]
>>>> {{{1},{2},{3},{5}},{{1,2},{1,3},{1,5},{2,3},{2,5},{3,5}},{{1,2,3},
>>>> {1,2,5},{1
>>>> ,3,5},{2,3,5}}}
>>>> bt1=bt[[1,All]]
>>>> {{1},{2},{3},{5}}
>>>> Table[Times@@bt1[[a]],{a,4}]
>>>> {1,2,3,5}
>>>> bt2=bt[[2,All]]
>>>> {{1,2},{1,3},{1,5},{2,3},{2,5},{3,5}}
>>>> Table[Times@@bt2[[a]],{a,6}]
>>>> {2,3,5,6,10,15}
>>>> bt3=bt[[3,All]]
>>>> {{1,2,3},{1,2,5},{1,3,5},{2,3,5}}
>>>> Table[Times@@bt3[[a]],{a,4}]
>>>> {6,10,15,30}
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>> -- DrBob at bigfoot.com
>>> www.eclecticdreams.net
>>>
>>>
>>
> -- DrBob at bigfoot.com
> www.eclecticdreams.net
>
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Re: Re: All Factors of a number. By "double trunk" I meant to have to consecutive rows where the 2^p-1. has no factors. Like n=6 and 7, or n=7 and 8. To reformat it as a. better mathematical question: Is there two consecutive primes Prime[n]. and Prime[n+1] such that 2^Prime[n]-1 and 2^Prime[n+1]-1 are also. primes for n > 8, that is they have no factors other than themselves. and 1.. If there would be such numbers, they would give a nice trunk to the. "christmas tree" to stand upright.. János. On Dec 22, 2004, at 12:52 PM, DrBob wrote:. > I'm not sure what you mean by a "double trunk", but there are. > Mersennes with two factors at n=5, 9, 12, 13, 19, 23... and others.. > Either I'm misunderstanding you, or numbers are running together on. >. > Timing on my wife's 2GHz dual-G5 is 818.86 seconds... not a huge. > improvement over your G4. It's only 42% faster despite a 60% faster. > clock speed!. >. > Bobby. >. > On Wed, 22 Dec 2004 04:52:49 -0500 (EST), János <janos.lobb at yale.edu>. > wrote:. >. >. >> Bobby,. >>. >> Nice !!. >>. >> You must have a minimum 17 inch monitor as I see :). >>. >> With the best,. >>. >> János. >> P.S: G4 1.25G Total Time: 1161.98 Second. Not bad for a Mac ! I. >> expected 1277.44 Second. Here is a naive question for you. Will you. >> ever have a "trunk" of length 2, so this tree can stand on it ? If I. >> am. >> counting from the bottom up, the first single trunk is at n=31, there. >> is no double trunk, the only triple is at n={6,7,8} and well the top. >> with length 4 is the top. I skipped lectures when number theory was. >> taught, - which might says, there will be never a double "trunk", may. >> be not even a single one with n > 54 -, so I am not a good candidate. >> to. >>. >> On Dec 21, 2004, at 5:19 AM, DrBob wrote:. >>. >>. >>. >>. >>>. >>> Just for grins, here's code to list factorizations of Mersenne. >>> numbers. >>> in Xmas tree format.. >>>. >>> toPowers = {{a_Integer, 1} -> HoldForm[a], {a_Integer, b_Integer} ->. >>> HoldForm[a]^HoldForm[b],. >>> List -> Times};. >>> toStars = StringReplace[ToString[#1 /. toPowers], " " -> " * "] & ;. >>> n = 1;. >>> {totalTime, results} =. >>> Timing[First[Last[Reap[While[n <= 54, p = Prime[n];. >>> Sow[Timing[{n, p, FactorInteger[2^p - 1]}]]; n++]]]]] /.. >>> {(s_)*Second, {n_, p_, f_}} :> {n, p, s, toStars[f]};. >>> TableForm[results, TableAlignments -> {Center, Center, Right},. >>> TableHeadings -> {None, {"n", "p = Prime[n]"*1, Seconds, "Factors of.
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>>> 2^p-1"}}]. >>> Print["Total Time: ", totalTime]. >>>. >>> That took 499 seconds on my AMD 3200+ machine, so reduce 54 to a. >>> smaller number if you're using a slower machine or don't have the. >>> patience. Of those 499 seconds, 125 were used for the 54th Mersenne,. >>> 201 for the 47th Mersenne, and 63 for the 50th. The rest go pretty. >>> fast, so reducing 54 to 46 will work well.. >>>. >>> Bobby. >>>. >>> On Mon, 20 Dec 2004 06:34:49 -0500 (EST), Ulrich Sondermann. >>>. >>>. >>>. >>>> Following is a test code that I am trying to get down to a set of. >>>> instructions that will allways put all of the factors of a number in. >>>> a table. >>>> or list.. >>>> bt contains all of the factors if I multiply each list entry,. >>>> however. >>>> I. >>>> cannot accomplish that with a single line, as an example I have. >>>> broken into. >>>> the three lists needed for this example. The results of each. >>>> "Times@@" is. >>>> what I am after all placed into one table. All of my attempts have. >>>> proved. >>>> disasterous, I am new to Mathematica and could do this with nested. >>>> loops in. >>>> any programming language, but this has me stumped.. >>>> Thanx!. >>>>. >>>> <<DiscreteMath`Combinatorica`. >>>> bt=Table[KSubsets[{1,2,3,5},a],{a,3}]. >>>> {{{1},{2},{3},{5}},{{1,2},{1,3},{1,5},{2,3},{2,5},{3,5}},{{1,2,3},. >>>> {1,2,5},{1. >>>> ,3,5},{2,3,5}}}. >>>> bt1=bt[[1,All]]. >>>> {{1},{2},{3},{5}}. >>>> Table[Times@@bt1[[a]],{a,4}]. >>>> {1,2,3,5}. >>>> bt2=bt[[2,All]]. >>>> {{1,2},{1,3},{1,5},{2,3},{2,5},{3,5}}. >>>> Table[Times@@bt2[[a]],{a,6}]. >>>> {2,3,5,6,10,15}. >>>> bt3=bt[[3,All]]. >>>> {{1,2,3},{1,2,5},{1,3,5},{2,3,5}}. >>>> Table[Times@@bt3[[a]],{a,4}]. >>>> {6,10,15,30}. >>>>. >>>>. >>>>. >>>>. >>>>. >>>>. >>> -- DrBob at bigfoot.com. >>> www.eclecticdreams.net. >>>. >>>. >>. > -- DrBob at bigfoot.com. > www.eclecticdreams.net. >. • Prev by Date: Re: Re: GUIKit - ScrollPane Tables within Wizard. • Next by Date: Re: Re: All Factors of a number. • Previous by thread: Re: Re: All Factors of a number. • Next by thread: Re: Re: All Factors of a number.
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https://termsdepot.com/ohms-law/
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# Ohm’s Law
Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Provided the temperature remains constant, the resistance R of the conductor is a constant, and thus the current I is directly proportional to the voltage V, and the ratio V/I is constant. This ratio is called the resistance, and it is the resistance of the conductor between the two points.
##### What are the 3 formulas in Ohm's law?
1. Voltage (V) = Current (I) x Resistance (R)
2. Current (I) = Voltage (V) / Resistance (R)
3. Resistance (R) = Voltage (V) / Current (I) What is the basic Ohm's law formula? The basic Ohm's law formula is V = IR, where V is voltage, I is current, and R is resistance. This formula can be used to calculate any of the three quantities if the other two are known.
##### What is Ohm's first law?
Ohm's first law states that the current through a conductor is proportional to the potential difference across the conductor. The law is named after Georg Ohm, who, in a paper published in 1827, described experiments investigating the relationship between the current flowing through a circuit and the voltage applied to the circuit. What is the SI unit of Ohm's law? The SI unit of Ohm's law is the volt.
##### How do you explain ohms?
Ohm's law is one of the most fundamental equations in electronics. It states that the current through a conductor is proportional to the potential difference (voltage) across it. The constant of proportionality is called the resistance, and is measured in ohms.
In practical terms, this means that if you have a device that is designed to operate at a certain voltage, you can calculate the maximum current it can draw by using the following formula:
I = V/R
Where I is the current in amps, V is the voltage in volts, and R is the resistance in ohms.
This equation is named after German physicist Georg Ohm, who first published it in 1827.
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# Ohm’s Law. Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Provided the temperature remains constant, the resistance R of the conductor is a constant, and thus the current I is directly proportional to the voltage V, and the ratio V/I is constant. This ratio is called the resistance, and it is the resistance of the conductor between the two points.. ##### What are the 3 formulas in Ohm's law?. 1. Voltage (V) = Current (I) x Resistance (R). 2. Current (I) = Voltage (V) / Resistance (R). 3. Resistance (R) = Voltage (V) / Current (I) What is the basic Ohm's law formula? The basic Ohm's law formula is V = IR, where V is voltage, I is current, and R is resistance. This formula can be used to calculate any of the three quantities if the other two are known.. ##### What is Ohm's first law?.
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Ohm's first law states that the current through a conductor is proportional to the potential difference across the conductor. The law is named after Georg Ohm, who, in a paper published in 1827, described experiments investigating the relationship between the current flowing through a circuit and the voltage applied to the circuit. What is the SI unit of Ohm's law? The SI unit of Ohm's law is the volt.. ##### How do you explain ohms?. Ohm's law is one of the most fundamental equations in electronics. It states that the current through a conductor is proportional to the potential difference (voltage) across it. The constant of proportionality is called the resistance, and is measured in ohms.. In practical terms, this means that if you have a device that is designed to operate at a certain voltage, you can calculate the maximum current it can draw by using the following formula:. I = V/R. Where I is the current in amps, V is the voltage in volts, and R is the resistance in ohms.. This equation is named after German physicist Georg Ohm, who first published it in 1827.
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# How much habitable land is there on earth per person?
What is the total area of habitable land on Earth? And how much habitable land does that leave one person? We’ll use the value r = 6400 km as the radius of Earth. According to the corresponding formula for spheres, the surface area of Earth is:
S = 4 * π * (6400 km)^2 ≈ 515 million square km
Since about 30 % of Earth’s surface is land, this means that the total area of land is 0.3 * 515 ≈ 155 million square km, about half of which is habitable for humans. With roughly 7 billion people alive today, we can conclude that there is 0.011 square km habitable land available per person. This corresponds to a square with 100 m ≈ 330 ft length and width.
# The Fourth State of Matter – Plasmas
From our everyday lifes we are used to three states of matter: solid, liquid and gas. When we heat a solid it melts and becomes liquid. Heating this liquid further will cause it to evaporate to a gas. Usually this is what we consider to be the end of the line. But heating a gas leads to many surprises, it eventually turns into a state, which behaves completely different than ordinary gases. We call matter in that state a plasma.
To understand why at some point a gas will exhibit an unusual behaviour, we need to look at the basic structure of matter. All matter consists of atoms. The Greeks believed this to be the undivisible building blocks of all objects. Scientists however have discovered, that atoms do indeed have an inner structure and are divisible. It takes an enormous amount to split atoms, but it can be done.
Further research showed that atoms consist of three particles: neutrons, protons and electrons. The neutrons and protons are crammed into the atomic core, while the electrons surround this core. Usually atoms are not charged, because they contain as much protons (positively charged) as electrons (negatively charged). The charges balance each other. Only when electrons are missing does the atom become electric. Such charged atoms are called ions.
In a gas the atoms are neutral. Each atom has as many protons as electrons, they are electrically balanced. When you apply a magnetic field to a gas, it does not respond. If you try to use the gas to conduct electricity, it does not work.
Remember that gas molecules move at high speeds and collide frequently with each other. As you increase the temperature, the collisions become more violent. At very high temperatures the collisions become so violent, that the impact can knock some electrons off an atom (ionization). This is where the plasma begins and the gas ends.
In a plasma the collisions are so intense that the atoms are not able to hold onto their outer electrons. Instead of a large amount of neutral atoms like in the gas, we are left with a mixture of free electrons and ions. This electric soup behaves very differently: it responds to magnetic fields and can conduct electricity very efficiently.
(The phases of matter. Source: NASA)
Most matter in the universe is in plasma form. Scientist believe that only 1 % of all visible matter is either solid, liquid or gaseous. On earth it is different, we rarely see plasmas because the temperatures are too small. But there are some exceptions.
High-temperature flames can cause a small volume of air to turn into a plasma. This can be seen for example in the so called ionic wind experiment, which shows that a flame is able to transmit electric currents. Gases can’t do that. DARPA, the Pentagon’s research arm, is currently using this phenomenon to develop new methods of fire suppression. Other examples for plasmas on earth are lightnings and the Aurora Borealis.
(Examples of plasmas. Source: Contemporary Physics Education Project)
The barrier between gases and plasmas is somewhat foggy. An important quantity to characterize the transition from gas to plasma is the ionization degree. It tells us how many percent of the atoms have lost one or more electrons. So an ionization degree of 10 % means that only one out of ten atoms is ionized. In this case the gas properties are still dominant.
(Ionization degree of Helium over Temperature. Source: SciVerse)
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# How much habitable land is there on earth per person?. What is the total area of habitable land on Earth? And how much habitable land does that leave one person? We’ll use the value r = 6400 km as the radius of Earth. According to the corresponding formula for spheres, the surface area of Earth is:. S = 4 * π * (6400 km)^2 ≈ 515 million square km. Since about 30 % of Earth’s surface is land, this means that the total area of land is 0.3 * 515 ≈ 155 million square km, about half of which is habitable for humans. With roughly 7 billion people alive today, we can conclude that there is 0.011 square km habitable land available per person. This corresponds to a square with 100 m ≈ 330 ft length and width.. # The Fourth State of Matter – Plasmas. From our everyday lifes we are used to three states of matter: solid, liquid and gas. When we heat a solid it melts and becomes liquid. Heating this liquid further will cause it to evaporate to a gas. Usually this is what we consider to be the end of the line. But heating a gas leads to many surprises, it eventually turns into a state, which behaves completely different than ordinary gases. We call matter in that state a plasma.. To understand why at some point a gas will exhibit an unusual behaviour, we need to look at the basic structure of matter. All matter consists of atoms. The Greeks believed this to be the undivisible building blocks of all objects. Scientists however have discovered, that atoms do indeed have an inner structure and are divisible. It takes an enormous amount to split atoms, but it can be done.. Further research showed that atoms consist of three particles: neutrons, protons and electrons. The neutrons and protons are crammed into the atomic core, while the electrons surround this core. Usually atoms are not charged, because they contain as much protons (positively charged) as electrons (negatively charged). The charges balance each other. Only when electrons are missing does the atom become electric. Such charged atoms are called ions.. In a gas the atoms are neutral. Each atom has as many protons as electrons, they are electrically balanced. When you apply a magnetic field to a gas, it does not respond. If you try to use the gas to conduct electricity, it does not work.
|
Remember that gas molecules move at high speeds and collide frequently with each other. As you increase the temperature, the collisions become more violent. At very high temperatures the collisions become so violent, that the impact can knock some electrons off an atom (ionization). This is where the plasma begins and the gas ends.. In a plasma the collisions are so intense that the atoms are not able to hold onto their outer electrons. Instead of a large amount of neutral atoms like in the gas, we are left with a mixture of free electrons and ions. This electric soup behaves very differently: it responds to magnetic fields and can conduct electricity very efficiently.. (The phases of matter. Source: NASA). Most matter in the universe is in plasma form. Scientist believe that only 1 % of all visible matter is either solid, liquid or gaseous. On earth it is different, we rarely see plasmas because the temperatures are too small. But there are some exceptions.. High-temperature flames can cause a small volume of air to turn into a plasma. This can be seen for example in the so called ionic wind experiment, which shows that a flame is able to transmit electric currents. Gases can’t do that. DARPA, the Pentagon’s research arm, is currently using this phenomenon to develop new methods of fire suppression. Other examples for plasmas on earth are lightnings and the Aurora Borealis.. (Examples of plasmas. Source: Contemporary Physics Education Project). The barrier between gases and plasmas is somewhat foggy. An important quantity to characterize the transition from gas to plasma is the ionization degree. It tells us how many percent of the atoms have lost one or more electrons. So an ionization degree of 10 % means that only one out of ten atoms is ionized. In this case the gas properties are still dominant.. (Ionization degree of Helium over Temperature. Source: SciVerse).
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https://nerdutella.com/q18101--Random-points-on-a-circle-Modify-Programming-Exercise-4-6-to-create-five-random-points-on-a-circle-form-a-polygon-by-connecting-the-points-clockwise-and-display-the-circle-and-the-polygon-as-shown-in-Figure-14-51b
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Q:
(Random points on a circle) Modify Programming Exercise 4.6 to create five random points on a circle, form a polygon by connecting the points clockwise, and display the circle and the polygon, as shown in Figure 14.51b
(Random points on a circle) Modify Programming Exercise 4.6 to create five random points on a circle, form a polygon by connecting the points clockwise, and display the circle and the polygon, as shown in Figure 14.51b.
need an explanation for this answer? contact us directly to get an explanation for this answer
``````/*********************************************************************************
* (Random points on a circle) Modify Programming Exercise 4.6 to create five *
* random points on a circle, form a polygon by connecting the points clockwise, *
* and display the circle and the polygon, as shown in Figure 14.51b. *
*********************************************************************************/
import javafx.application.Application;
import javafx.stage.Stage;
import javafx.scene.Scene;
import javafx.scene.shape.Circle;
import javafx.scene.shape.Polygon;
import javafx.scene.paint.Color;
import javafx.collections.ObservableList;
import javafx.scene.layout.Pane;
import javafx.geometry.Insets;
import java.util.ArrayList;
public class Exercise_14_25 extends Application {
public void start(Stage primaryStage) {
// Create a pane
Pane pane = new Pane();
pane.setPadding(new Insets(10, 10, 10, 10));
// Create a circle
Circle circle = new Circle(60, 60, 40);
circle.setFill(Color.WHITE);
circle.setStroke(Color.BLACK);
// Create a polygon
Polygon polygon = new Polygon();
polygon.setFill(Color.WHITE);
polygon.setStroke(Color.BLACK);
ObservableList<Double> list = polygon.getPoints();
// Generate random angles in radians between 0 and 2PI
ArrayList<Double> angles = new ArrayList<>();
for (int i = 0; angles.size() < 5; i++) {
double angle = (Math.random() * (2 * Math.PI));
if (!angles.contains(angle)) {
}
}
// Sort angles clockwise
java.util.Collections.sort(angles);
// Get 5 points on the circle
for (int i = 0; i < angles.size(); i++) {
Math.cos(angles.get(i)));
Math.sin(angles.get(i)));
}
// Create a scene and place it in the stage
Scene scene = new Scene(pane);
primaryStage.setTitle("Exercise_14_25"); // Set the stage title
primaryStage.setScene(scene); // Place the scene in the stage
primaryStage.show(); // Display the stage
}
}``````
need an explanation for this answer? contact us directly to get an explanation for this answer
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Q:. (Random points on a circle) Modify Programming Exercise 4.6 to create five random points on a circle, form a polygon by connecting the points clockwise, and display the circle and the polygon, as shown in Figure 14.51b. (Random points on a circle) Modify Programming Exercise 4.6 to create five random points on a circle, form a polygon by connecting the points clockwise, and display the circle and the polygon, as shown in Figure 14.51b.. need an explanation for this answer? contact us directly to get an explanation for this answer. ``````/*********************************************************************************. * (Random points on a circle) Modify Programming Exercise 4.6 to create five *. * random points on a circle, form a polygon by connecting the points clockwise, *. * and display the circle and the polygon, as shown in Figure 14.51b. *. *********************************************************************************/. import javafx.application.Application;. import javafx.stage.Stage;. import javafx.scene.Scene;. import javafx.scene.shape.Circle;. import javafx.scene.shape.Polygon;. import javafx.scene.paint.Color;. import javafx.collections.ObservableList;. import javafx.scene.layout.Pane;. import javafx.geometry.Insets;. import java.util.ArrayList;. public class Exercise_14_25 extends Application {. public void start(Stage primaryStage) {. // Create a pane. Pane pane = new Pane();. pane.setPadding(new Insets(10, 10, 10, 10));. // Create a circle. Circle circle = new Circle(60, 60, 40);. circle.setFill(Color.WHITE);. circle.setStroke(Color.BLACK);.
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// Create a polygon. Polygon polygon = new Polygon();. polygon.setFill(Color.WHITE);. polygon.setStroke(Color.BLACK);. ObservableList<Double> list = polygon.getPoints();. // Generate random angles in radians between 0 and 2PI. ArrayList<Double> angles = new ArrayList<>();. for (int i = 0; angles.size() < 5; i++) {. double angle = (Math.random() * (2 * Math.PI));. if (!angles.contains(angle)) {. }. }. // Sort angles clockwise. java.util.Collections.sort(angles);. // Get 5 points on the circle. for (int i = 0; i < angles.size(); i++) {. Math.cos(angles.get(i)));. Math.sin(angles.get(i)));. }. // Create a scene and place it in the stage. Scene scene = new Scene(pane);. primaryStage.setTitle("Exercise_14_25"); // Set the stage title. primaryStage.setScene(scene); // Place the scene in the stage. primaryStage.show(); // Display the stage. }. }``````. need an explanation for this answer? contact us directly to get an explanation for this answer.
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https://www.vedantu.com/question-answer/let-the-equations-of-two-sides-of-a-triangle-be-class-11-maths-cbse-5eceb8a0b5f2de3ed1c2e289
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Let the equations of two sides of a triangle be $3x-2y+6=0$ and $4x+5y-20=0$. If theorthocentre of this triangle is at $(1,1)$, then the equation of its third side is(a) $122y-26x-1675=0$(b) $26x+61y+1675=0$(c) $122y+26x+1675=0$(d) $26x-122y-1675=0$
Last updated date: 09th Aug 2024
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Hint: Find the slope of the given two lines and use this slope and given orthocentre to find the coordinates of the triangle.
The equations of the sides of a triangle given in the questions are,
$3x-2y+6=0$ and $4x+5y-20=0$
The orthocentre of a triangle is the point where all the altitudes intersect. Its coordinates are given
in the question as $(1,1)$. Draw a triangle ABC with all the given data as shown below,
Consider the line AC. The slope of the line can be obtained by rearranging the equation as,
\begin{align} & 4x+5y-20=0 \\ & 5y=20-4x \\ & y=-\dfrac{4}{5}x+4 \\ \end{align}
Comparing it with the general equation, $y=mx+c$, we have the slope as $-\dfrac{4}{5}$. The slope
of perpendicular lines is related as $m\times n=-1$. Now since the line BR is perpendicular to the
line AC, the slope of BR would be $\dfrac{5}{4}$.
The line BR has slope $\dfrac{5}{4}$ and passes through the point $(1,1)$. The equation of a line
passing through point $({{x}_{1}},{{y}_{1}})$ and having slope $m$ is given by,$y-{{y}_{1}}=m(x- {{x}_{1}})$. The equation of the line BR can hence be obtained as,
\begin{align} & y-1=\dfrac{5}{4}(x-1) \\ & 4(y-1)=5(x-1) \\ & 4y-4=5x-5 \\ & 5x-4y-1=0 \\ \end{align}
Next, we have to consider line BC. The slope of the line can be obtained by rearranging the equation
as,
\begin{align} & 3x-2y+6=0 \\ & -2y=-6-3x \\ & y=\dfrac{3}{2}x+3 \\ \end{align}
Comparing it with the general equation, $y=mx+c$, we have the slope as $\dfrac{3}{2}$. The slope
of perpendicular lines is related as $m\times n=-1$. Now since the line AQ is perpendicular to the
line BC, the slope of AQ would be $-\dfrac{2}{3}$.
The line AQ has slope $-\dfrac{2}{3}$ and passes through the point $(1,1)$. The equation of a line
passing through point $({{x}_{1}},{{y}_{1}})$ and having slope $m$ is given by,$y-{{y}_{1}}=m(x- {{x}_{1}})$. The equation of the line AQ can hence be obtained as,
\begin{align} & y-1=-\dfrac{2}{3}(x-1) \\ & 3(y-1)=-2(x-1) \\ & 3y-3=-2x+2 \\ & 2x+3y-5=0 \\ \end{align}
The vertex A is passing through both lines AC and AQ. So, we can get the coordinates of vertex A by
solving the equations $4x+5y-20=0$ and $2x+3y-5=0$. Multiplying the equation $2x+3y-5=0$ by 2
and subtracting from the equation $4x+5y-20=0$, we get,
\dfrac{\begin{align} & 4x+5y-20=0 \\ & -4x-6y+10=0 \\ \end{align}}{\begin{align} & -y-10=0 \\ & y=-10 \\ \end{align}}
Substituting the value of $y$ in $2x+3y-5=0$, we get,
\begin{align} & 2x+(3\times -10)-5=0 \\ & 2x-30-5=0 \\ & 2x=35 \\ & x=\dfrac{35}{2} \\ \end{align}
Therefore, the coordinates of A are $\left( \dfrac{35}{2},-10 \right)$.
The vertex B is passing through both lines BC and BR. So, we can get the coordinates of vertex B by
solving the equations $3x-2y+6=0$ and $5x-4y-1=0$ . Multiplying the equation $3x-2y+6=0$ by 2
and subtracting the equation $5x-4y-1=0$ from it, we get,
\dfrac{\begin{align} & 6x-4y+12=0 \\ & -5x+4y+1=0 \\ \end{align}}{\begin{align} & x+13=0 \\ & x=-13 \\ \end{align}}
Substituting the value of $x$ in $3x-2y+6=0$, we get,
\begin{align} & (3\times -13)-2y+6=0 \\ & -39-2y+6=0 \\ & -2y=33 \\ & y=-\dfrac{33}{2} \\ \end{align}
Therefore, the coordinates of B are $\left( -13,-\dfrac{33}{2} \right)$.
The equation of a line passing through two points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
given by,
\begin{align} & y-{{y}_{1}}=m(x-{{x}_{1}}) \\ & \Rightarrow y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\times (x-{{x}_{1}}) \\ \end{align}
The equation of line AB passing through points $\left( \dfrac{35}{2},-10 \right)$ and $\left( -13,- \dfrac{33}{2} \right)$ can hence be found as,
$\left( y+10 \right)=\dfrac{\left( -\dfrac{33}{2}+10 \right)}{\left( -13-\dfrac{35}{2} \right)}\times \left( x-\dfrac{35}{2} \right)$
Taking the LCM of the terms in the numerator and denominator,
\begin{align} & y+10=\dfrac{-\dfrac{13}{2}}{-\dfrac{61}{2}}\times \left( x-\dfrac{35}{2} \right) \\ & y+10=\dfrac{13}{61}\times \left( x-\dfrac{35}{2} \right) \\ & 61(y+10)=13\left( x-\dfrac{35}{2} \right) \\ & 61y+610=13x-\dfrac{455}{2} \\ \end{align}
Multiplying both sides by 2,
\begin{align} & 122y+1220=26x-455 \\ & 26x-122y-1675=0 \\ \end{align}
Therefore, the required equation of the straight line is $26x-122y-1675=0$.
We get option (d) as the correct answer.
Note: There is one more way to approach this problem. The coordinates of the vertices A and B can
be taken as $\left( p,\dfrac{20-4p}{5} \right)$ and $\left( q,\dfrac{3+6q}{2} \right)$ by assuming
$x$ as $p$ and $q$ respectively in the equations $4x+5y-20=0$and $3x-2y+6=0$. Then we can
formulate the equations using the slope of the lines and solve for $p$ and $q$. The equation of AB
can then be obtained.
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Courses. Courses for Kids. Free study material. Offline Centres. More. Store. Let the equations of two sides of a triangle be $3x-2y+6=0$ and $4x+5y-20=0$. If theorthocentre of this triangle is at $(1,1)$, then the equation of its third side is(a) $122y-26x-1675=0$(b) $26x+61y+1675=0$(c) $122y+26x+1675=0$(d) $26x-122y-1675=0$. Last updated date: 09th Aug 2024. Total views: 461.4k. Views today: 14.61k. Verified. 461.4k+ views. Hint: Find the slope of the given two lines and use this slope and given orthocentre to find the coordinates of the triangle.. The equations of the sides of a triangle given in the questions are,. $3x-2y+6=0$ and $4x+5y-20=0$. The orthocentre of a triangle is the point where all the altitudes intersect. Its coordinates are given. in the question as $(1,1)$. Draw a triangle ABC with all the given data as shown below,. Consider the line AC. The slope of the line can be obtained by rearranging the equation as,. \begin{align} & 4x+5y-20=0 \\ & 5y=20-4x \\ & y=-\dfrac{4}{5}x+4 \\ \end{align}. Comparing it with the general equation, $y=mx+c$, we have the slope as $-\dfrac{4}{5}$. The slope. of perpendicular lines is related as $m\times n=-1$. Now since the line BR is perpendicular to the. line AC, the slope of BR would be $\dfrac{5}{4}$.. The line BR has slope $\dfrac{5}{4}$ and passes through the point $(1,1)$. The equation of a line. passing through point $({{x}_{1}},{{y}_{1}})$ and having slope $m$ is given by,$y-{{y}_{1}}=m(x- {{x}_{1}})$. The equation of the line BR can hence be obtained as,. \begin{align} & y-1=\dfrac{5}{4}(x-1) \\ & 4(y-1)=5(x-1) \\ & 4y-4=5x-5 \\ & 5x-4y-1=0 \\ \end{align}. Next, we have to consider line BC. The slope of the line can be obtained by rearranging the equation. as,. \begin{align} & 3x-2y+6=0 \\ & -2y=-6-3x \\ & y=\dfrac{3}{2}x+3 \\ \end{align}. Comparing it with the general equation, $y=mx+c$, we have the slope as $\dfrac{3}{2}$. The slope. of perpendicular lines is related as $m\times n=-1$. Now since the line AQ is perpendicular to the. line BC, the slope of AQ would be $-\dfrac{2}{3}$.. The line AQ has slope $-\dfrac{2}{3}$ and passes through the point $(1,1)$. The equation of a line. passing through point $({{x}_{1}},{{y}_{1}})$ and having slope $m$ is given by,$y-{{y}_{1}}=m(x- {{x}_{1}})$. The equation of the line AQ can hence be obtained as,.
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\begin{align} & y-1=-\dfrac{2}{3}(x-1) \\ & 3(y-1)=-2(x-1) \\ & 3y-3=-2x+2 \\ & 2x+3y-5=0 \\ \end{align}. The vertex A is passing through both lines AC and AQ. So, we can get the coordinates of vertex A by. solving the equations $4x+5y-20=0$ and $2x+3y-5=0$. Multiplying the equation $2x+3y-5=0$ by 2. and subtracting from the equation $4x+5y-20=0$, we get,. \dfrac{\begin{align} & 4x+5y-20=0 \\ & -4x-6y+10=0 \\ \end{align}}{\begin{align} & -y-10=0 \\ & y=-10 \\ \end{align}}. Substituting the value of $y$ in $2x+3y-5=0$, we get,. \begin{align} & 2x+(3\times -10)-5=0 \\ & 2x-30-5=0 \\ & 2x=35 \\ & x=\dfrac{35}{2} \\ \end{align}. Therefore, the coordinates of A are $\left( \dfrac{35}{2},-10 \right)$.. The vertex B is passing through both lines BC and BR. So, we can get the coordinates of vertex B by. solving the equations $3x-2y+6=0$ and $5x-4y-1=0$ . Multiplying the equation $3x-2y+6=0$ by 2. and subtracting the equation $5x-4y-1=0$ from it, we get,. \dfrac{\begin{align} & 6x-4y+12=0 \\ & -5x+4y+1=0 \\ \end{align}}{\begin{align} & x+13=0 \\ & x=-13 \\ \end{align}}. Substituting the value of $x$ in $3x-2y+6=0$, we get,. \begin{align} & (3\times -13)-2y+6=0 \\ & -39-2y+6=0 \\ & -2y=33 \\ & y=-\dfrac{33}{2} \\ \end{align}. Therefore, the coordinates of B are $\left( -13,-\dfrac{33}{2} \right)$.. The equation of a line passing through two points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is. given by,. \begin{align} & y-{{y}_{1}}=m(x-{{x}_{1}}) \\ & \Rightarrow y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\times (x-{{x}_{1}}) \\ \end{align}. The equation of line AB passing through points $\left( \dfrac{35}{2},-10 \right)$ and $\left( -13,- \dfrac{33}{2} \right)$ can hence be found as,. $\left( y+10 \right)=\dfrac{\left( -\dfrac{33}{2}+10 \right)}{\left( -13-\dfrac{35}{2} \right)}\times \left( x-\dfrac{35}{2} \right)$. Taking the LCM of the terms in the numerator and denominator,. \begin{align} & y+10=\dfrac{-\dfrac{13}{2}}{-\dfrac{61}{2}}\times \left( x-\dfrac{35}{2} \right) \\ & y+10=\dfrac{13}{61}\times \left( x-\dfrac{35}{2} \right) \\ & 61(y+10)=13\left( x-\dfrac{35}{2} \right) \\ & 61y+610=13x-\dfrac{455}{2} \\ \end{align}. Multiplying both sides by 2,. \begin{align} & 122y+1220=26x-455 \\ & 26x-122y-1675=0 \\ \end{align}. Therefore, the required equation of the straight line is $26x-122y-1675=0$.. We get option (d) as the correct answer.. Note: There is one more way to approach this problem. The coordinates of the vertices A and B can. be taken as $\left( p,\dfrac{20-4p}{5} \right)$ and $\left( q,\dfrac{3+6q}{2} \right)$ by assuming. $x$ as $p$ and $q$ respectively in the equations $4x+5y-20=0$and $3x-2y+6=0$. Then we can. formulate the equations using the slope of the lines and solve for $p$ and $q$. The equation of AB. can then be obtained.
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## The Three Important Identities
In this post you wanna dive in and find resourful identities that you thought little of in developnment of quadratic formula.
The general formula of quadratic equation is given by
$\fn_jvn&space;\large&space;{\color{Red}&space;\mathbf{ax^{2}+bx+c=0}}$
To get more explanation about the parameters find insightful peek in this basic proof of quadratic formula
The algebraic identities you wanna learn here
1.$\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{\left&space;(&space;a+b&space;\right&space;)^{2}=a^{2}+2ab+b^{2}}}$
2.$\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{\left&space;(&space;a-b&space;\right&space;)^{2}=a^{2}-2ab+b^{2}}}$
3.$\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{a^{2}-b^{2}=\left&space;(&space;a+b&space;\right&space;)\left&space;(a-b&space;\right&space;)}}$
Now given the three identities, let’s see how to combine them to produce a path towards the quadratic formula. By dividing through the equation, it inevitabley becomes
$\fn_jvn&space;\large&space;\mathbf{x^{2}+\frac{b}{a}x+\frac{c}{a}=0}$
Suppose the roots of a quadratic equation are give by r1 and r2 and that they are building blocks towards quadratic formula advancement. Now, form a quadratic equation using the roots. Notice that a is divided through the general quadratic equation above so as to make roots the same$\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;x-r_{1}\right&space;)\left&space;(&space;x-r_{2}&space;\right&space;)=x^{2}-\left&space;(&space;r_{1}+r_{2}&space;\right&space;)x+r_{1}r_{2}}$
It is crystal clear that both equations have their first term a perfect square-x squared. And that both of them equals to zero. Since they are equivalent to zero and their first term same, let’s equate them.
$\fn_jvn&space;\large&space;\fn_jvn&space;\large&space;\mathbf{x^{2}+\frac{b}{a}x+\frac{c}{a}=x^{2}-\left&space;(&space;r_{1}+r_{2}&space;\right&space;)x+r_{1}r_{2}}$
By observation, it is established that the sum of the roots of the quadratic equation is
$\fn_jvn&space;\large&space;\mathbf{r_{1}+r_{2}=-\frac{b}{a}}$
And the product is
$\fn_jvn&space;\large&space;\mathbf{r_{1}r_{2}=\frac{c}{a}}$
## Step 1
In a special case when the first two identities (1. & 2.) are combined, then
$\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{\left&space;(&space;a+b&space;\right&space;)^{2}={\left&space;(&space;a-b&space;\right&space;)^{2}+4ab}$
The relationship of the combined identities becomes
$\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;r_{1}-r_{2}&space;\right&space;)^{2}=\left&space;(&space;r_{1}+r_{2}&space;\right&space;)^{2}-4r_{1}r_{2}}$
Express the left side linearly by taking square roots both sides
$\large&space;\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;r_{1}-r_{2}&space;\right&space;)=\pm&space;\sqrt{\left&space;(&space;r_{1}+r_{2}&space;\right&space;)^{2}-4r_{1}r_{2}}$
Substitute the sum and product of the general equation into the combined identity . It relates as
$\large&space;\large&space;\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;r_{1}-r_{2}&space;\right&space;)=\pm&space;\sqrt{\left&space;(&space;-\frac{b}{a}&space;\right&space;)^{2}-4\frac{c}{a}}$
## Step 2
By rearranging, make 4r1r2 the subject of the formula
$\fn_jvn&space;\large&space;\mathbf{4r_{1}r_{2}=\left&space;(&space;r_{1}+r_{2}&space;\right&space;)^{2}-\left&space;(&space;r_{1}-r_{2}&space;\right&space;)^{2}}$
Let’s use the difference of two squares to find r1 from this equation. Divide both sides by 4r2 , so r1 becomes
$\fn_jvn&space;\large&space;\fn_jvn&space;\large&space;\mathbf{r_{1}=\frac{\left&space;(&space;r_{1}+r_{2}&space;\right&space;)^{2}-\left&space;(&space;r_{1}-r_{2}&space;\right&space;)^{2}}{4r_{2}}}$
The difference of two squares property
$\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{a^{2}-b^{2}=\left&space;(&space;a+b&space;\right&space;)\left&space;(a-b&space;\right&space;)}}$
Exploit the difference of two squares
$\fn_jvn&space;\large&space;\mathbf{r_{1}=\frac{[\left&space;(&space;r_{1}&space;+r_{2}\right&space;)+\left&space;(&space;r_{1}&space;-r_{2}\right&space;)][\left&space;(&space;r_{1}&space;+r_{2}\right&space;)-\left&space;(&space;r_{1}-r_{2}&space;\right&space;)]}{4r_{2}}}$
Now, after applying difference of two squares, r1 simplifies to
$\fn_jvn&space;\large&space;\fn_jvn&space;\large&space;\mathbf{r_{1}=\frac{\left&space;(&space;r_{1}+r_{2}&space;\right&space;)&space;+ \left&space;(&space;r_{1}-r_{2}&space;\right&space;)}{2}}$
Plug in the parameters given that r1+r2 and r1-r2 are known$\fn_jvn&space;\large&space;\mathbf{r_{1}=\frac{-\frac{b}{a}\pm&space;\sqrt{\left&space;(&space;\frac{b}{a}&space;\right&space;)^{2}-4\frac{c}{a}}}{2}}$
When
$\fn_jvn&space;\large&space;\mathbf{r_{1}=-r_{2}-\frac{b}{a}}$
it reduces to
$\fn_jvn&space;\large&space;\mathbf{r_{1}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}}$
and we get
$\fn_jvn&space;\large&space;\mathbf{r_{2}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}}$
When
$\fn_jvn&space;\large&space;\mathbf{r_{1}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}}$
the other root is
$\fn_jvn&space;\large&space;\mathbf{r_{2}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}}$
And by combining the solutions we obtain
$\fn_jvn&space;\large&space;\mathbf{x=\frac{-b\pm&space;\sqrt{b^{2}-4ac}}{2a}}$
## Proof quadratic formula by converting first term into a perfect square
Now, given the general quadratic equation, you are Required↗ to make the first term a perfect square.$\fn_jvn&space;\large&space;{\color{Black}&space;\mathbf{ax^{2}+bx+c=0}}$
Converting the first term into a perfect square is essential so that the quadratic equation resembles the form of this identity
$\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{\left&space;(&space;a+b&space;\right&space;)^{2}=a^{2}+2ab+b^{2}}}$
Multiply the equation by a to boths sides
$\fn_jvn&space;\large&space;\mathbf{a\left&space;(&space;ax^{2}+bx+c&space;\right&space;)=0}$
The equation becomes
$\fn_jvn&space;\large&space;\mathbf{a^{2}x^{2}+abx+ac=0}$
Rewrite and make the coefficients explict as in general form of quadratic equation
$\fn_jvn&space;\large&space;\mathbf{{\color{Red}&space;a}x^{2}+{\color{Red}&space;b}x+{\color{Red}&space;c}=0}$
$\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;ax&space;\right&space;)^{2}+b\left&space;(&space;ax&space;\right&space;)+ac=0}$
It is time now to complete the square. Take half the coeffient of ax, square the result and add to both sides of the equation or add and subtract the result if all the parameters are on the same side in this case LHS. But you can choose to keep all elements in one side of the equation at initial and intermediate stages as I have done. The coefficient of ax is b . Let’s quickily complete the square
$\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;ax&space;\right&space;)^{2}+b\left&space;(&space;ax&space;\right&space;)+\left&space;(&space;\frac{1}{2}\cdot&space;b&space;\right&space;)^2-\left&space;(&space;\frac{1}{2}\cdot&space;b&space;\right&space;)^2+ac=0}$
Transform the left side into a perfect square
$\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;ax+\frac{b}{2}&space;\right&space;)^{2}=\left&space;(&space;\frac{b}{2}&space;\right&space;)^{2}-ac}$
Find the LCM on the right side and simplify the fraction
$\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;ax+\frac{b}{2}&space;\right&space;)^{2}=\frac{b^{2}-4ac}{4}}$
Take square roots on both sides to transform the equation into linear
$\fn_jvn&space;\large&space;\mathbf{\sqrt{\left&space;(ax+\frac{b}{2}&space;\right&space;)^{2}}=\pm&space;\sqrt{\frac{b^{2}-4ac}{4}}}$
It reduces to
$\fn_jvn&space;\large&space;\mathbf{ax+\frac{b}{2}=\pm&space;\sqrt{\frac{b^{2}-4ac}{4}}}$
Take all parameters to the right side to make x the subject of the formula
$\fn_jvn&space;\large&space;\mathbf{ax=-\frac{b}{2}\pm&space;\frac{\sqrt{b^{2}-4ac}}{2}}$
Divide both sides by a to find x as
$\fn_jvn&space;\large&space;\mathbf{x=\frac{-b\pm&space;\sqrt{b^{2}-4ac}}{2a}}$
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## The Three Important Identities. In this post you wanna dive in and find resourful identities that you thought little of in developnment of quadratic formula.. The general formula of quadratic equation is given by. $\fn_jvn&space;\large&space;{\color{Red}&space;\mathbf{ax^{2}+bx+c=0}}$. To get more explanation about the parameters find insightful peek in this basic proof of quadratic formula. The algebraic identities you wanna learn here. 1.$\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{\left&space;(&space;a+b&space;\right&space;)^{2}=a^{2}+2ab+b^{2}}}$. 2.$\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{\left&space;(&space;a-b&space;\right&space;)^{2}=a^{2}-2ab+b^{2}}}$. 3.$\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{a^{2}-b^{2}=\left&space;(&space;a+b&space;\right&space;)\left&space;(a-b&space;\right&space;)}}$. Now given the three identities, let’s see how to combine them to produce a path towards the quadratic formula. By dividing through the equation, it inevitabley becomes. $\fn_jvn&space;\large&space;\mathbf{x^{2}+\frac{b}{a}x+\frac{c}{a}=0}$. Suppose the roots of a quadratic equation are give by r1 and r2 and that they are building blocks towards quadratic formula advancement. Now, form a quadratic equation using the roots. Notice that a is divided through the general quadratic equation above so as to make roots the same$\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;x-r_{1}\right&space;)\left&space;(&space;x-r_{2}&space;\right&space;)=x^{2}-\left&space;(&space;r_{1}+r_{2}&space;\right&space;)x+r_{1}r_{2}}$. It is crystal clear that both equations have their first term a perfect square-x squared. And that both of them equals to zero. Since they are equivalent to zero and their first term same, let’s equate them.. $\fn_jvn&space;\large&space;\fn_jvn&space;\large&space;\mathbf{x^{2}+\frac{b}{a}x+\frac{c}{a}=x^{2}-\left&space;(&space;r_{1}+r_{2}&space;\right&space;)x+r_{1}r_{2}}$. By observation, it is established that the sum of the roots of the quadratic equation is. $\fn_jvn&space;\large&space;\mathbf{r_{1}+r_{2}=-\frac{b}{a}}$. And the product is. $\fn_jvn&space;\large&space;\mathbf{r_{1}r_{2}=\frac{c}{a}}$. ## Step 1. In a special case when the first two identities (1. & 2.) are combined, then. $\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{\left&space;(&space;a+b&space;\right&space;)^{2}={\left&space;(&space;a-b&space;\right&space;)^{2}+4ab}$. The relationship of the combined identities becomes. $\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;r_{1}-r_{2}&space;\right&space;)^{2}=\left&space;(&space;r_{1}+r_{2}&space;\right&space;)^{2}-4r_{1}r_{2}}$. Express the left side linearly by taking square roots both sides. $\large&space;\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;r_{1}-r_{2}&space;\right&space;)=\pm&space;\sqrt{\left&space;(&space;r_{1}+r_{2}&space;\right&space;)^{2}-4r_{1}r_{2}}$. Substitute the sum and product of the general equation into the combined identity . It relates as. $\large&space;\large&space;\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;r_{1}-r_{2}&space;\right&space;)=\pm&space;\sqrt{\left&space;(&space;-\frac{b}{a}&space;\right&space;)^{2}-4\frac{c}{a}}$. ## Step 2. By rearranging, make 4r1r2 the subject of the formula. $\fn_jvn&space;\large&space;\mathbf{4r_{1}r_{2}=\left&space;(&space;r_{1}+r_{2}&space;\right&space;)^{2}-\left&space;(&space;r_{1}-r_{2}&space;\right&space;)^{2}}$. Let’s use the difference of two squares to find r1 from this equation. Divide both sides by 4r2 , so r1 becomes. $\fn_jvn&space;\large&space;\fn_jvn&space;\large&space;\mathbf{r_{1}=\frac{\left&space;(&space;r_{1}+r_{2}&space;\right&space;)^{2}-\left&space;(&space;r_{1}-r_{2}&space;\right&space;)^{2}}{4r_{2}}}$. The difference of two squares property. $\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{a^{2}-b^{2}=\left&space;(&space;a+b&space;\right&space;)\left&space;(a-b&space;\right&space;)}}$. Exploit the difference of two squares. $\fn_jvn&space;\large&space;\mathbf{r_{1}=\frac{[\left&space;(&space;r_{1}&space;+r_{2}\right&space;)+\left&space;(&space;r_{1}&space;-r_{2}\right&space;)][\left&space;(&space;r_{1}&space;+r_{2}\right&space;)-\left&space;(&space;r_{1}-r_{2}&space;\right&space;)]}{4r_{2}}}$. Now, after applying difference of two squares, r1 simplifies to.
|
$\fn_jvn&space;\large&space;\fn_jvn&space;\large&space;\mathbf{r_{1}=\frac{\left&space;(&space;r_{1}+r_{2}&space;\right&space;)&space;+ \left&space;(&space;r_{1}-r_{2}&space;\right&space;)}{2}}$. Plug in the parameters given that r1+r2 and r1-r2 are known$\fn_jvn&space;\large&space;\mathbf{r_{1}=\frac{-\frac{b}{a}\pm&space;\sqrt{\left&space;(&space;\frac{b}{a}&space;\right&space;)^{2}-4\frac{c}{a}}}{2}}$. When. $\fn_jvn&space;\large&space;\mathbf{r_{1}=-r_{2}-\frac{b}{a}}$. it reduces to. $\fn_jvn&space;\large&space;\mathbf{r_{1}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}}$. and we get. $\fn_jvn&space;\large&space;\mathbf{r_{2}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}}$. When. $\fn_jvn&space;\large&space;\mathbf{r_{1}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}}$. the other root is. $\fn_jvn&space;\large&space;\mathbf{r_{2}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}}$. And by combining the solutions we obtain. $\fn_jvn&space;\large&space;\mathbf{x=\frac{-b\pm&space;\sqrt{b^{2}-4ac}}{2a}}$. ## Proof quadratic formula by converting first term into a perfect square. Now, given the general quadratic equation, you are Required↗ to make the first term a perfect square.$\fn_jvn&space;\large&space;{\color{Black}&space;\mathbf{ax^{2}+bx+c=0}}$. Converting the first term into a perfect square is essential so that the quadratic equation resembles the form of this identity. $\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{\left&space;(&space;a+b&space;\right&space;)^{2}=a^{2}+2ab+b^{2}}}$. Multiply the equation by a to boths sides. $\fn_jvn&space;\large&space;\mathbf{a\left&space;(&space;ax^{2}+bx+c&space;\right&space;)=0}$. The equation becomes. $\fn_jvn&space;\large&space;\mathbf{a^{2}x^{2}+abx+ac=0}$. Rewrite and make the coefficients explict as in general form of quadratic equation. $\fn_jvn&space;\large&space;\mathbf{{\color{Red}&space;a}x^{2}+{\color{Red}&space;b}x+{\color{Red}&space;c}=0}$. $\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;ax&space;\right&space;)^{2}+b\left&space;(&space;ax&space;\right&space;)+ac=0}$. It is time now to complete the square. Take half the coeffient of ax, square the result and add to both sides of the equation or add and subtract the result if all the parameters are on the same side in this case LHS. But you can choose to keep all elements in one side of the equation at initial and intermediate stages as I have done. The coefficient of ax is b . Let’s quickily complete the square. $\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;ax&space;\right&space;)^{2}+b\left&space;(&space;ax&space;\right&space;)+\left&space;(&space;\frac{1}{2}\cdot&space;b&space;\right&space;)^2-\left&space;(&space;\frac{1}{2}\cdot&space;b&space;\right&space;)^2+ac=0}$. Transform the left side into a perfect square. $\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;ax+\frac{b}{2}&space;\right&space;)^{2}=\left&space;(&space;\frac{b}{2}&space;\right&space;)^{2}-ac}$. Find the LCM on the right side and simplify the fraction. $\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;ax+\frac{b}{2}&space;\right&space;)^{2}=\frac{b^{2}-4ac}{4}}$. Take square roots on both sides to transform the equation into linear. $\fn_jvn&space;\large&space;\mathbf{\sqrt{\left&space;(ax+\frac{b}{2}&space;\right&space;)^{2}}=\pm&space;\sqrt{\frac{b^{2}-4ac}{4}}}$. It reduces to. $\fn_jvn&space;\large&space;\mathbf{ax+\frac{b}{2}=\pm&space;\sqrt{\frac{b^{2}-4ac}{4}}}$. Take all parameters to the right side to make x the subject of the formula. $\fn_jvn&space;\large&space;\mathbf{ax=-\frac{b}{2}\pm&space;\frac{\sqrt{b^{2}-4ac}}{2}}$. Divide both sides by a to find x as. $\fn_jvn&space;\large&space;\mathbf{x=\frac{-b\pm&space;\sqrt{b^{2}-4ac}}{2a}}$.
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# Math
posted by on .
The representative tells you that floor plan #1 sells for \$225,000 and floor plan #2 sells for \$300,000. She also mentions that all the available houses combined are worth \$10,350,000. Write an equation that illustrates this situation. Explain. Use the same variables you used in part a.
Is this right?:
255,000x+300,000y=10,350,000
• Math - ,
yes, correct.
• Math - ,
I have to use the elimination method to solve this system of equations but, I just do not get it.
x+y=42
255,000x+300,000y=10,350,000
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# Math. posted by on .. The representative tells you that floor plan #1 sells for \$225,000 and floor plan #2 sells for \$300,000. She also mentions that all the available houses combined are worth \$10,350,000. Write an equation that illustrates this situation. Explain. Use the same variables you used in part a.. Is this right?:.
|
255,000x+300,000y=10,350,000. • Math - ,. yes, correct.. • Math - ,. I have to use the elimination method to solve this system of equations but, I just do not get it.. x+y=42. 255,000x+300,000y=10,350,000.
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```STAT-UB.0103 SPRING 2012
Homework Set 4
Solutions
1. If Z is a standard normal random variable, find the probability that
a.
0 ≤ Z ≤ 1.24
d.
-0.88 < Z < 0.14
b.
Z > -0.45
e.
-1.13 < Z ≤ -0.92
c.
| Z | ≥ 0.80
f.
| Z | ≤ 0.70
SOLUTION:
In all problems involving the normal table, we treat < and ≤ as equivalent symbols.
Similarly > and ≥ are treated as equivalent.
a.
b.
c.
0.3925
0.6736
0.4238
d.
e.
f.
0.3663
0.0496
0.5160
2. If Z is a standard normal random variable, find the values of w that satisfy the
following. These need not be done by interpolation; just use the closest table value.
a.
b.
P[ Z > w ] = 0.32
P[ Z ≤ w ] = 0.71
c.
d.
P[ Z ≤ w ] = 0.14
P[ -w ≤ Z ≤ w ] = 0.60
You can use Minitab as well. You’ll need Calc ⇒ Probability Distributions ⇒
Normal and the Inverse cumulate probability feature.
SOLUTION:
a.
Convert this to P[ 0 ≤ Z ≤ w ] = 0.18. Then find
P[ 0 ≤ Z ≤ 0.46 ] = 0.1772
P[ 0 ≤ Z ≤ 0.47 ] = 0.1808
The closer value corresponds to w = 0.47. (The exact answer, found from
Minitab, is w = 0.467699.)
b.
Convert this to P[ 0 ≤ Z ≤ w ] = 0.21. Then find
P[ 0 ≤ Z ≤ 0.55 ] = 0.2088
P[ 0 ≤ Z ≤ 0.56 ] = 0.2123
The closer value corresponds to w = 0.55. (The exact answer, found from
Minitab, is w = 0.553385.)
c.
The value of w is certainly negative. Convert this to P[ 0 ≤ Z ≤ -w ] = 0.36. Then
find
P[ 0 ≤ Z ≤ 1.08 ] = 0.3599
P[ 0 ≤ Z ≤ 1.09 ] = 0.3621
The closer value corresponds to -w = 1.08. Thus w = -1.08. (The exact answer,
found from Minitab, is w = -1.08032.
Page 1
STAT-UB.0103 SPRING 2012
Homework Set 4
Solutions
d.
Convert this to P[ 0 ≤ Z ≤ w ] = 0.30. Then find
P[ 0 ≤ Z ≤ 0.84 ] = 0.2995
P[ 0 ≤ Z ≤ 0.85 ] = 0.3023
The closer value corresponds to w = 0.84. (The exact answer, found from
Minitab, is w = 0.841621.)
3. The diameters of apples from Happy Mac Orchard have diameters which are
approximately normally distributed with mean μ = 2.8 inches and standard deviation
σ = 0.3 inch. Apples can be size-sorted by being made to roll over a mesh screens. At
this farm, the steps are done sequentially.
First, the apples are rolled over a screen with mesh size 2.5 inches. This separates
out all apples with diameters < 2.5 inches.
Second, the remaining apples are rolled over a screen with mesh size 3.3 inches.
This separates out all apples with diameters between 2.5 and 3.3 inches.
All the apples will now be separated into three groups.
a.
Find the proportion of apples with diameter < 2.5 inches.
b.
Find the proportion of apples with diameters between 2.5 and 3.3 inches.
c.
Find the proportion of apples with diameters greater than 3.3 inches.
HINT: If X represents the diameter of a random apple, and if the screen has a mesh
size m, then P[X < m] represents the proportion of apples which will fall through.
SOLUTION: You are given the facts that μ = population mean diameter = 2.8 inches
and σ = population standard deviation = 0.3 inch. Let X be the random variable that
gives the diameter of an apple.
For part a, if a batch of apples is rolled over a screen with mesh size 2.5 inches, then
P[X < 2.5] represents the proportion of apples which will fall through. (NOTE: You can
think of P[X < 2.5] as the probability that one randomly selected apple will fall through
the screen; this also represents the proportion of all apples which will fall through.)
⎡ X − 2.8 2.5 − 2.8 ⎤
<
≈ P[ Z < -1.00 ] = 0.1587
P[ X < 2.5 ] = P ⎢
0.3 ⎥⎦
⎣ 0.3
About 16% of the apples will fall through.
b.
This is handled in exactly the same form as a.
⎡ X − 2.8 3.3 − 2.8 ⎤
<
≈ P[ Z < 1.67 ] = 0.9525
P[ X < 3.3 ] = P ⎢
0.3 ⎥⎦
⎣ 0.3
About 95% of the apples will fall through. This calculation includes those that would
have been separated by the first step. Therefore, the proportion of apples with diameters
between 2.5 inches and 3.3 inches is 0.9525 - 0.1587 = 0.7938 This is about 79%.
Page 2
STAT-UB.0103 SPRING 2012
Homework Set 4
Solutions
c.
The proportion of apples that do not fall through the larger screen is 1 – 0.9525
= 0.0475. That is, about 5% of the apples are in the largest category.
4. The chocolate chip cookies that are produced at Perry’s Cookie Emporium have
weights which are approximately normally distributed with mean weight 180 grams and
with standard deviation 20 grams. The cookies, however, are sold by count, not by
weight. This is a high-markup business, and Perry wants to improve his image. He
decides to set aside lightest 20% of the cookies to be packaged and sold separately. What
cookie weight will divide the lightest 20% from the heaviest 80% ?
SOLUTION: Let X be the random variable which gives the cookie weights.
Apparently this has μ = 180 grams and σ = 20 grams. We seek the weight w so that
P[ X < w ] = 0.20 . But
P[ X < w ] = P
LM X − 180 < w − 180 OP = PLMZ < w − 180 OP
20 Q
N 20 Q
N 20
want
= 0.20
The normal table tells us that P[Z < -0.84] ≈ 0.20.
How do we get this value? If we seek c for which P[Z < c] = 0.20, then we note
that c must be negative and we apportion the probability as
0.20 = P[Z < c]
0.30 = P[c < Z < 0]
0.30 = P[0 < Z < -c]
0.20 = P[-c < Z ]
The third of these four facts corresponds to what we can look up in the normal
table. Thus we seek a value for -c so that, so close as possible, P[0 < Z < -c] is
0.30. That happens for -c = 0.84, and thus c = -0.84. We then solve -0.84 =
w − 180
, getting w = 163.2 .
20
If the cookies are divided at weight 163.2 grams, then the lightweight group will have
You can get Minitab to do all the heavy lifting. Use Calc ⇒ Probability Distributions
⇒ Normal. Set up the information panel as follows:
Page 3
STAT-UB.0103 SPRING 2012
Homework Set 4
Solutions
The output is this:
Inverse Cumulative Distribution Function
Normal with mean = 180 and standard deviation = 20
P( X <= x )
0.2
x
163.168
5. Clyde’s Deli is situated inside a large industrial park. The weekday gross sales at
Clyde’s average \$2,480, with a standard deviation of \$360. Find the probability that the
average over the next 50 weekdays will exceed \$2,400. Please note the assumptions that
are used in making the calculation.
SOLUTION: Let X1, X2, …, X50 be the random amounts for these 50 weekdays. We
must assume that these random variables are statistically independent, each with the
mean \$2,480 and each with the standard deviation \$360. Let X be the average of these
50 values. We have no need to assume that the distribution is normal, as the Central
Limit theorem will assure us that X is sufficiently close to normal. The mean of the
\$360
distribution of X is \$2,480 and the standard deviation of this distribution is
≈
50
\$50.91. We then proceed as follows:
⎡ X − \$2, 480
\$2, 400 − \$2, 480 ⎤
P[ X > \$2,400 ] = P ⎢
>
⎥ ≈ P[ Z > -1.57 ]
\$50.91
⎣ \$50.91
⎦
Page 4
STAT-UB.0103 SPRING 2012
Homework Set 4
Solutions
= P[ Z < 1.57 ] = 0.50 + P[ 0 ≤ Z < 1.57 ] = 0.50 + 0.4418 = 0.9418
6. Examine the file CHS\GEYSER1.MTP. You can get this from the Web at
www.stern.nyu.edu/~gsimon/statdata. Look for the CHS folder.
Column C2 gives the duration, in minutes, of eruptions of the Old Faithful Geyser in
Yellowstone National Park. Column C3 gives the interval, in minutes, until the following
eruption. The concern here is whether the data in these columns follow a normal
distribution. Here we’ll just examine C2. (Column C3 is qualitatively very similar to
C2.)
Use Graph ⇒ Probability Plot to decide whether C2 follows a normal distribution.
What conclusion do you reach? You might also try Graph ⇒ Histogram to check
whether there is a simple description.
SOLUTION: Here’s the probability plot for C2:
Probability Plot of Duration
Normal - 95% CI
99.9
Mean
3.576
StDev
1.084
N
222
12.714
P-Value <0.005
99
Percent
95
90
80
70
60
50
40
30
20
10
5
1
0.1
0
1
2
3
4
Duration
5
6
7
8
This is outrageously non-normal. But what exactly does it mean to see the string of dots
snaking around the center zone?
Page 5
STAT-UB.0103 SPRING 2012
Homework Set 4
Solutions
The histogram will clear that up:
Histogram of Duration
40
Frequency
30
20
10
0
1.5
2.0
2.5
3.0
3.5
Duration
4.0
4.5
5.0
NOTE: The default Minitab output was edited so that the
horizontal scale is the “cutpoint” type, rather than
“midpoint.”
This is a pattern called bimodal, meaning that there are two peaks. Apparently there are
many eruptions of long duration and also quite a few of short duration. It’s somewhat
unusual to get an eruption of 2 12 to 3 12 minutes.
7. You are about to take a sample from a population in order to use X to estimate μ.
Here the population consists of items with monetary values. You would like the error
of estimate to be governed by the condition P[ | X - μ | ≤ \$5 ] ≥ 0.80. If you think that
σ, the population standard deviation, could be as large as \$40, find the smallest sample
size n which will allow you to satisfy the condition.
HINT: Assume that σ = \$40. If σ is really smaller, then you will still satisfy
the condition.
HINT: You will need to find a normal table point w so that P[ | Z | ≤ w ] = 0.80.
⎛ z σ⎞
HINT: The formula n ≥ ⎜ α / 2 ⎟ will be useful.
⎝ E ⎠
2
Page 6
STAT-UB.0103 SPRING 2012
Homework Set 4
Solutions
⎛ z σ⎞
SOLUTION: Use the result n ≥ ⎜ α / 2 ⎟ . In this formula,
⎝ E ⎠
2
is the limit on the magnitude of the error, meaning | X - μ | , which is \$5
in this example
σ
the maximum believable standard deviation, which is \$40 in this
example
1 - α the desired probability of achieving error E, which is 80% in this
example
E
In this example, we’ll note that α = 0.20 and then that
α
= 0.10, so that zα/2 = z0.10 = 1.28.
2
Then the formula asks for sample size
⎛ 1.28 × \$40 ⎞
2
n ≥ ⎜
⎟ = 10.24 = 104.8576
\$5
⎝
⎠
2
As n must be an integer, we raise this to the next value 105.
The value of z0.10 asks for the solution of P[ 0 ≤ Z ≤ z0.10 ] = 0.40. The closest value is
1.28, so we use z0.10 = 1.28. (The exact value from Minitab is 1.28155.)
We were trying to get X within 18 σ of μ, and the sample size requirement was 105 or
more.
8. A population of daily sales figures is approximately normally distributed with a mean
of \$14,000 and a standard deviation of \$3,000.
(a)
You’d like to predict tomorrow’s sales. (Yes, this is an inference
question.) Give an interval (a, b) for which you are will to say that the
probability is about 95% that tomorrow’s sales will be between a and b.
(b)
You’d like to predict the average sales over the next 15 business days.
(This is another inference question.) Give an interval (c, d) for which you
are will to say that the probability is about 95% that the sales over the next
15 business days will be between c and d.
SOLUTION: Since about 95% of a distribution will be between μ - 2σ and μ + 2σ,
you’d give \$14,000 ± 2 × \$3,000, meaning \$8,000 to \$20,000, as an interval which
should contain about 95% of the probability. This takes care of (a). Since you’ve
assumed a normally distributed population, this 95% figure is very close to perfect.
Page 7
STAT-UB.0103 SPRING 2012
Homework Set 4
Solutions
As for (b), you only need to refine the argument by noting that SD( X ) =
σ
=
n
\$3,000
≈ \$775. In interval for this X is \$14,000 ± 2 × \$775, or \$12,450 to \$15,550.
15
You’ve assumed a normal population, so the 95% figure is pretty good. With a sample
of size n = 15, we are not quite allowed to use the Central Limit theorem, but we could
probability do it anyhow.
9. You have been tracking the “cash reserve” of a very large mutual fund, hoping to find
clues about future behavior. The cash reserve is currently at 4.32. The units here are
millions of dollars. You believe that the daily changes to this reserve are normally
distributed, with mean -0.02 and with standard deviation 0.06. Find the probability that
the reserves will be below 3.90 after 25 days.
SOLUTION: This is a very simple random walk. Let T be the total of the 25 daily
changes. The condition “reserve < 3.90” is exactly the same as {T < -0.42 }. However T
has a mean of 25 × (-0.02) = -0.50 and has SD(T) = 0.06 × 25 = 0.30. Then
⎡ T − ( −0.50 )
P[ T < -0.42 ] = P ⎢
<
0.30
⎣
= 0.6064
( −0.42 ) − ( −0.50 ) ⎤
0.30
⎥ ≈ P[ Z < 0.27 ]
⎦
With a span of n = 25 days, we have the Central Limit theorem, so it was not really
critical to assume the normal distribution for the daily changes.
Page 8
STAT-UB.0103 SPRING 2012
Homework Set 4
Solutions
10. The Vindicator Mutual Fund family has a fund that tracks industrial metals. This is
currently trading at \$14.20 per share. It is believed that the daily price follows a
lognormal random walk with drift μ = +0.02 and with volatility expressed through
σ = 0.15. Find the probability that the price after thirty trading days will exceed \$16.00.
HINT: Recall that the lognormal random walk is governed through
Pn = P0 e X1 + X 2 +...+ X n , where the Xi’s are normal with mean μ and standard
deviation σ.
SOLUTION: We can rewrite the HINT as Pn = P0 eTn , where Tn = X1 + X2 + … + Xn.
Note that E(Tn) = nμ = 30 × 0.02 = 0.60 and SD(Tn) = σ
n = 0.15 30 ≈ 0.8216.
Then
P[ Pn > \$16.00 ] = P ⎡⎣ P0 eTn > \$16.00 ⎤⎦ = P ⎡⎣ \$14.20 eT30 > \$16.00 ⎤⎦
≈ P ⎡⎣ eT30 > 1.126761⎤⎦ ≈ P[ T30 > 0.119347 ]
0.119347 − 0.60 ⎤
⎡ T − 0.60
>
= P ⎢ 30
⎥⎦ ≈ P[ Z > -0.5850 ] = P[ Z < 0.5850 ]
0.8216
⎣ 0.8216
= 0.50 + P[ 0 ≤ Z < 0.5850 ] = 0.50 + 0.2207 = 0.7207.
The value 0.2207 was obtained by an easy interpolation.
Page 9
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Survey. * Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project. Document related concepts. Statistics wikipedia, lookup. History of statistics wikipedia, lookup. Probability wikipedia, lookup. Transcript. ```STAT-UB.0103 SPRING 2012. Homework Set 4. Solutions. . 1. If Z is a standard normal random variable, find the probability that. a.. 0 ≤ Z ≤ 1.24. d.. -0.88 < Z < 0.14. b.. Z > -0.45. e.. -1.13 < Z ≤ -0.92. c.. | Z | ≥ 0.80. f.. | Z | ≤ 0.70. SOLUTION:. In all problems involving the normal table, we treat < and ≤ as equivalent symbols.. Similarly > and ≥ are treated as equivalent.. a.. b.. c.. 0.3925. 0.6736. 0.4238. d.. e.. f.. 0.3663. 0.0496. 0.5160. 2. If Z is a standard normal random variable, find the values of w that satisfy the. following. These need not be done by interpolation; just use the closest table value.. a.. b.. P[ Z > w ] = 0.32. P[ Z ≤ w ] = 0.71. c.. d.. P[ Z ≤ w ] = 0.14. P[ -w ≤ Z ≤ w ] = 0.60. You can use Minitab as well. You’ll need Calc ⇒ Probability Distributions ⇒. Normal and the Inverse cumulate probability feature.. SOLUTION:. a.. Convert this to P[ 0 ≤ Z ≤ w ] = 0.18. Then find. P[ 0 ≤ Z ≤ 0.46 ] = 0.1772. P[ 0 ≤ Z ≤ 0.47 ] = 0.1808. The closer value corresponds to w = 0.47. (The exact answer, found from. Minitab, is w = 0.467699.). b.. Convert this to P[ 0 ≤ Z ≤ w ] = 0.21. Then find. P[ 0 ≤ Z ≤ 0.55 ] = 0.2088. P[ 0 ≤ Z ≤ 0.56 ] = 0.2123. The closer value corresponds to w = 0.55. (The exact answer, found from. Minitab, is w = 0.553385.). c.. The value of w is certainly negative. Convert this to P[ 0 ≤ Z ≤ -w ] = 0.36. Then. find. P[ 0 ≤ Z ≤ 1.08 ] = 0.3599. P[ 0 ≤ Z ≤ 1.09 ] = 0.3621. The closer value corresponds to -w = 1.08. Thus w = -1.08. (The exact answer,. found from Minitab, is w = -1.08032.. . Page 1. STAT-UB.0103 SPRING 2012. Homework Set 4. Solutions. . d.. Convert this to P[ 0 ≤ Z ≤ w ] = 0.30. Then find. P[ 0 ≤ Z ≤ 0.84 ] = 0.2995. P[ 0 ≤ Z ≤ 0.85 ] = 0.3023. The closer value corresponds to w = 0.84. (The exact answer, found from. Minitab, is w = 0.841621.). 3. The diameters of apples from Happy Mac Orchard have diameters which are. approximately normally distributed with mean μ = 2.8 inches and standard deviation. σ = 0.3 inch. Apples can be size-sorted by being made to roll over a mesh screens. At. this farm, the steps are done sequentially.. First, the apples are rolled over a screen with mesh size 2.5 inches. This separates. out all apples with diameters < 2.5 inches.. Second, the remaining apples are rolled over a screen with mesh size 3.3 inches.. This separates out all apples with diameters between 2.5 and 3.3 inches.. All the apples will now be separated into three groups.. a.. Find the proportion of apples with diameter < 2.5 inches.. b.. Find the proportion of apples with diameters between 2.5 and 3.3 inches.. c.. Find the proportion of apples with diameters greater than 3.3 inches.. HINT: If X represents the diameter of a random apple, and if the screen has a mesh. size m, then P[X < m] represents the proportion of apples which will fall through.. SOLUTION: You are given the facts that μ = population mean diameter = 2.8 inches. and σ = population standard deviation = 0.3 inch. Let X be the random variable that. gives the diameter of an apple.. For part a, if a batch of apples is rolled over a screen with mesh size 2.5 inches, then. P[X < 2.5] represents the proportion of apples which will fall through. (NOTE: You can. think of P[X < 2.5] as the probability that one randomly selected apple will fall through. the screen; this also represents the proportion of all apples which will fall through.). ⎡ X − 2.8 2.5 − 2.8 ⎤. <. ≈ P[ Z < -1.00 ] = 0.1587. P[ X < 2.5 ] = P ⎢. 0.3 ⎥⎦. ⎣ 0.3. About 16% of the apples will fall through.. b.. This is handled in exactly the same form as a.. ⎡ X − 2.8 3.3 − 2.8 ⎤. <. ≈ P[ Z < 1.67 ] = 0.9525. P[ X < 3.3 ] = P ⎢. 0.3 ⎥⎦. ⎣ 0.3. About 95% of the apples will fall through. This calculation includes those that would. have been separated by the first step. Therefore, the proportion of apples with diameters. between 2.5 inches and 3.3 inches is 0.9525 - 0.1587 = 0.7938 This is about 79%.. . Page 2. STAT-UB.0103 SPRING 2012. Homework Set 4. Solutions. . c.. The proportion of apples that do not fall through the larger screen is 1 – 0.9525. = 0.0475. That is, about 5% of the apples are in the largest category.. 4. The chocolate chip cookies that are produced at Perry’s Cookie Emporium have. weights which are approximately normally distributed with mean weight 180 grams and. with standard deviation 20 grams. The cookies, however, are sold by count, not by. weight. This is a high-markup business, and Perry wants to improve his image. He. decides to set aside lightest 20% of the cookies to be packaged and sold separately. What. cookie weight will divide the lightest 20% from the heaviest 80% ?. SOLUTION: Let X be the random variable which gives the cookie weights.. Apparently this has μ = 180 grams and σ = 20 grams. We seek the weight w so that. P[ X < w ] = 0.20 . But. P[ X < w ] = P. LM X − 180 < w − 180 OP = PLMZ < w − 180 OP. 20 Q. N 20 Q. N 20. want. = 0.20. The normal table tells us that P[Z < -0.84] ≈ 0.20.. How do we get this value? If we seek c for which P[Z < c] = 0.20, then we note. that c must be negative and we apportion the probability as. 0.20 = P[Z < c]. 0.30 = P[c < Z < 0]. 0.30 = P[0 < Z < -c]. 0.20 = P[-c < Z ]. The third of these four facts corresponds to what we can look up in the normal. table. Thus we seek a value for -c so that, so close as possible, P[0 < Z < -c] is. 0.30. That happens for -c = 0.84, and thus c = -0.84. We then solve -0.84 =. w − 180. , getting w = 163.2 .. 20. If the cookies are divided at weight 163.2 grams, then the lightweight group will have. You can get Minitab to do all the heavy lifting. Use Calc ⇒ Probability Distributions. ⇒ Normal. Set up the information panel as follows:. . Page 3. STAT-UB.0103 SPRING 2012. Homework Set 4. Solutions. . The output is this:. Inverse Cumulative Distribution Function. Normal with mean = 180 and standard deviation = 20. P( X <= x ). 0.2. x. 163.168. 5. Clyde’s Deli is situated inside a large industrial park. The weekday gross sales at. Clyde’s average \$2,480, with a standard deviation of \$360. Find the probability that the. average over the next 50 weekdays will exceed \$2,400. Please note the assumptions that. are used in making the calculation.. SOLUTION: Let X1, X2, …, X50 be the random amounts for these 50 weekdays. We. must assume that these random variables are statistically independent, each with the. mean \$2,480 and each with the standard deviation \$360. Let X be the average of these. 50 values. We have no need to assume that the distribution is normal, as the Central. Limit theorem will assure us that X is sufficiently close to normal. The mean of the. \$360. distribution of X is \$2,480 and the standard deviation of this distribution is. ≈. 50. \$50.91. We then proceed as follows:. ⎡ X − \$2, 480. \$2, 400 − \$2, 480 ⎤. P[ X > \$2,400 ] = P ⎢. >. ⎥ ≈ P[ Z > -1.57 ]. \$50.91. ⎣ \$50.91. ⎦. . Page 4.
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STAT-UB.0103 SPRING 2012. Homework Set 4. Solutions. . = P[ Z < 1.57 ] = 0.50 + P[ 0 ≤ Z < 1.57 ] = 0.50 + 0.4418 = 0.9418. 6. Examine the file CHS\GEYSER1.MTP. You can get this from the Web at. www.stern.nyu.edu/~gsimon/statdata. Look for the CHS folder.. Column C2 gives the duration, in minutes, of eruptions of the Old Faithful Geyser in. Yellowstone National Park. Column C3 gives the interval, in minutes, until the following. eruption. The concern here is whether the data in these columns follow a normal. distribution. Here we’ll just examine C2. (Column C3 is qualitatively very similar to. C2.). Use Graph ⇒ Probability Plot to decide whether C2 follows a normal distribution.. What conclusion do you reach? You might also try Graph ⇒ Histogram to check. whether there is a simple description.. SOLUTION: Here’s the probability plot for C2:. Probability Plot of Duration. Normal - 95% CI. 99.9. Mean. 3.576. StDev. 1.084. N. 222. 12.714. P-Value <0.005. 99. Percent. 95. 90. 80. 70. 60. 50. 40. 30. 20. 10. 5. 1. 0.1. 0. 1. 2. 3. 4. Duration. 5. 6. 7. 8. This is outrageously non-normal. But what exactly does it mean to see the string of dots. snaking around the center zone?. . Page 5. STAT-UB.0103 SPRING 2012. Homework Set 4. Solutions. . The histogram will clear that up:. Histogram of Duration. 40. Frequency. 30. 20. 10. 0. 1.5. 2.0. 2.5. 3.0. 3.5. Duration. 4.0. 4.5. 5.0. NOTE: The default Minitab output was edited so that the. horizontal scale is the “cutpoint” type, rather than. “midpoint.”. This is a pattern called bimodal, meaning that there are two peaks. Apparently there are. many eruptions of long duration and also quite a few of short duration. It’s somewhat. unusual to get an eruption of 2 12 to 3 12 minutes.. 7. You are about to take a sample from a population in order to use X to estimate μ.. Here the population consists of items with monetary values. You would like the error. of estimate to be governed by the condition P[ | X - μ | ≤ \$5 ] ≥ 0.80. If you think that. σ, the population standard deviation, could be as large as \$40, find the smallest sample. size n which will allow you to satisfy the condition.. HINT: Assume that σ = \$40. If σ is really smaller, then you will still satisfy. the condition.. HINT: You will need to find a normal table point w so that P[ | Z | ≤ w ] = 0.80.. ⎛ z σ⎞. HINT: The formula n ≥ ⎜ α / 2 ⎟ will be useful.. ⎝ E ⎠. 2. . Page 6. STAT-UB.0103 SPRING 2012. Homework Set 4. Solutions. . ⎛ z σ⎞. SOLUTION: Use the result n ≥ ⎜ α / 2 ⎟ . In this formula,. ⎝ E ⎠. 2. is the limit on the magnitude of the error, meaning | X - μ | , which is \$5. in this example. σ. the maximum believable standard deviation, which is \$40 in this. example. 1 - α the desired probability of achieving error E, which is 80% in this. example. E. In this example, we’ll note that α = 0.20 and then that. α. = 0.10, so that zα/2 = z0.10 = 1.28.. 2. Then the formula asks for sample size. ⎛ 1.28 × \$40 ⎞. 2. n ≥ ⎜. ⎟ = 10.24 = 104.8576. \$5. ⎝. ⎠. 2. As n must be an integer, we raise this to the next value 105.. The value of z0.10 asks for the solution of P[ 0 ≤ Z ≤ z0.10 ] = 0.40. The closest value is. 1.28, so we use z0.10 = 1.28. (The exact value from Minitab is 1.28155.). We were trying to get X within 18 σ of μ, and the sample size requirement was 105 or. more.. 8. A population of daily sales figures is approximately normally distributed with a mean. of \$14,000 and a standard deviation of \$3,000.. (a). You’d like to predict tomorrow’s sales. (Yes, this is an inference. question.) Give an interval (a, b) for which you are will to say that the. probability is about 95% that tomorrow’s sales will be between a and b.. (b). You’d like to predict the average sales over the next 15 business days.. (This is another inference question.) Give an interval (c, d) for which you. are will to say that the probability is about 95% that the sales over the next. 15 business days will be between c and d.. SOLUTION: Since about 95% of a distribution will be between μ - 2σ and μ + 2σ,. you’d give \$14,000 ± 2 × \$3,000, meaning \$8,000 to \$20,000, as an interval which. should contain about 95% of the probability. This takes care of (a). Since you’ve. assumed a normally distributed population, this 95% figure is very close to perfect.. . Page 7. STAT-UB.0103 SPRING 2012. Homework Set 4. Solutions. . As for (b), you only need to refine the argument by noting that SD( X ) =. σ. =. n. \$3,000. ≈ \$775. In interval for this X is \$14,000 ± 2 × \$775, or \$12,450 to \$15,550.. 15. You’ve assumed a normal population, so the 95% figure is pretty good. With a sample. of size n = 15, we are not quite allowed to use the Central Limit theorem, but we could. probability do it anyhow.. 9. You have been tracking the “cash reserve” of a very large mutual fund, hoping to find. clues about future behavior. The cash reserve is currently at 4.32. The units here are. millions of dollars. You believe that the daily changes to this reserve are normally. distributed, with mean -0.02 and with standard deviation 0.06. Find the probability that. the reserves will be below 3.90 after 25 days.. SOLUTION: This is a very simple random walk. Let T be the total of the 25 daily. changes. The condition “reserve < 3.90” is exactly the same as {T < -0.42 }. However T. has a mean of 25 × (-0.02) = -0.50 and has SD(T) = 0.06 × 25 = 0.30. Then. ⎡ T − ( −0.50 ). P[ T < -0.42 ] = P ⎢. <. 0.30. ⎣. = 0.6064. ( −0.42 ) − ( −0.50 ) ⎤. 0.30. ⎥ ≈ P[ Z < 0.27 ]. ⎦. With a span of n = 25 days, we have the Central Limit theorem, so it was not really. critical to assume the normal distribution for the daily changes.. . Page 8. STAT-UB.0103 SPRING 2012. Homework Set 4. Solutions. . 10. The Vindicator Mutual Fund family has a fund that tracks industrial metals. This is. currently trading at \$14.20 per share. It is believed that the daily price follows a. lognormal random walk with drift μ = +0.02 and with volatility expressed through. σ = 0.15. Find the probability that the price after thirty trading days will exceed \$16.00.. HINT: Recall that the lognormal random walk is governed through. Pn = P0 e X1 + X 2 +...+ X n , where the Xi’s are normal with mean μ and standard. deviation σ.. SOLUTION: We can rewrite the HINT as Pn = P0 eTn , where Tn = X1 + X2 + … + Xn.. Note that E(Tn) = nμ = 30 × 0.02 = 0.60 and SD(Tn) = σ. n = 0.15 30 ≈ 0.8216.. Then. P[ Pn > \$16.00 ] = P ⎡⎣ P0 eTn > \$16.00 ⎤⎦ = P ⎡⎣ \$14.20 eT30 > \$16.00 ⎤⎦. ≈ P ⎡⎣ eT30 > 1.126761⎤⎦ ≈ P[ T30 > 0.119347 ]. 0.119347 − 0.60 ⎤. ⎡ T − 0.60. >. = P ⎢ 30. ⎥⎦ ≈ P[ Z > -0.5850 ] = P[ Z < 0.5850 ]. 0.8216. ⎣ 0.8216. = 0.50 + P[ 0 ≤ Z < 0.5850 ] = 0.50 + 0.2207 = 0.7207.. The value 0.2207 was obtained by an easy interpolation.. . Page 9.
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# SOLVING LINEAR EQUATIONS IN TWO VARIABLES WORKSHEET
Solving Linear Equations in Two Variables Worksheet :
In this section, we will see some practice questions on solving linear equations in two variables.
## Solving Linear Equations in Two Variables Worksheet - Practice Questions
(1) Solve the following pair of linear equations by the substitution method.
(i) x + y = 14 and x - y = 4 Solution
(ii) s - t = 3 and (s/3) + (t/2) = 6 Solution
(iii) 3x - y = 3 and 9x - 3y = 9 Solution
(iv) 0.2 x + 0.3 y = 1.3 and 0.4 x + 0.5 y = 2.3
(v) √2 x + √3y = 0 and √3 x - √8 y = 0 Solution
(vi) (3x/2) - (5y/3) = -2 and (x/3) + (y/2) = 13/6
Solution
(2) Solve 2 x + 3 y = 11 and 2 x - 4 y = -24 and hence find the value of "m" for which y = m x + 3 Solution
(3) Form the pair of linear equations of the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them Solution
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them Solution
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later,she buys 3 bats and 5 balls for Rs.1750. Find the cost if each bat and each ball.
Solution
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km,the charge paid is \$105 and for a journey of 15 km, the charge paid is \$155. What are the fixed charge and charge per km ? How much does a person have to pay for traveling a distance of 25 km?
Solution
(v) A fraction becomes 9/11,if 2 is added to both numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction. Solution
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution
After having gone through the stuff given above, we hope that the students would have understood, solving linear equations in two variables worksheet
Apart from the stuff given in this section if you need any other stuff in math, please use our google custom search here
You can also visit the following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
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# SOLVING LINEAR EQUATIONS IN TWO VARIABLES WORKSHEET. Solving Linear Equations in Two Variables Worksheet :. In this section, we will see some practice questions on solving linear equations in two variables.. ## Solving Linear Equations in Two Variables Worksheet - Practice Questions. (1) Solve the following pair of linear equations by the substitution method.. (i) x + y = 14 and x - y = 4 Solution. (ii) s - t = 3 and (s/3) + (t/2) = 6 Solution. (iii) 3x - y = 3 and 9x - 3y = 9 Solution. (iv) 0.2 x + 0.3 y = 1.3 and 0.4 x + 0.5 y = 2.3. (v) √2 x + √3y = 0 and √3 x - √8 y = 0 Solution. (vi) (3x/2) - (5y/3) = -2 and (x/3) + (y/2) = 13/6. Solution. (2) Solve 2 x + 3 y = 11 and 2 x - 4 y = -24 and hence find the value of "m" for which y = m x + 3 Solution. (3) Form the pair of linear equations of the following problems and find their solution by substitution method.. (i) The difference between two numbers is 26 and one number is three times the other. Find them Solution. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them Solution. (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later,she buys 3 bats and 5 balls for Rs.1750. Find the cost if each bat and each ball.. Solution. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km,the charge paid is \$105 and for a journey of 15 km, the charge paid is \$155. What are the fixed charge and charge per km ? How much does a person have to pay for traveling a distance of 25 km?. Solution. (v) A fraction becomes 9/11,if 2 is added to both numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction. Solution. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?. Solution. After having gone through the stuff given above, we hope that the students would have understood, solving linear equations in two variables worksheet. Apart from the stuff given in this section if you need any other stuff in math, please use our google custom search here. You can also visit the following web pages on different stuff in math.. WORD PROBLEMS. Word problems on simple equations. Word problems on linear equations. Algebra word problems. Word problems on trains. Area and perimeter word problems. Word problems on direct variation and inverse variation. Word problems on unit price. Word problems on unit rate. Word problems on comparing rates. Converting customary units word problems. Converting metric units word problems. Word problems on simple interest.
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Word problems on compound interest. Word problems on types of angles. Complementary and supplementary angles word problems. Double facts word problems. Trigonometry word problems. Percentage word problems. Profit and loss word problems. Markup and markdown word problems. Decimal word problems. Word problems on fractions. Word problems on mixed fractrions. One step equation word problems. Linear inequalities word problems. Ratio and proportion word problems. Time and work word problems. Word problems on sets and venn diagrams. Word problems on ages. Pythagorean theorem word problems. Percent of a number word problems. Word problems on constant speed. Word problems on average speed. Word problems on sum of the angles of a triangle is 180 degree. OTHER TOPICS. Profit and loss shortcuts. Percentage shortcuts. Times table shortcuts. Time, speed and distance shortcuts. Ratio and proportion shortcuts. Domain and range of rational functions. Domain and range of rational functions with holes. Graphing rational functions. Graphing rational functions with holes. Converting repeating decimals in to fractions. Decimal representation of rational numbers. Finding square root using long division. L.C.M method to solve time and work problems. Translating the word problems in to algebraic expressions. Remainder when 2 power 256 is divided by 17. Remainder when 17 power 23 is divided by 16. Sum of all three digit numbers divisible by 6. Sum of all three digit numbers divisible by 7. Sum of all three digit numbers divisible by 8. Sum of all three digit numbers formed using 1, 3, 4. Sum of all three four digit numbers formed with non zero digits. Sum of all three four digit numbers formed using 0, 1, 2, 3. Sum of all three four digit numbers formed using 1, 2, 5, 6.
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# annual rate of return
Pages: 3 (472 words) Published: February 24, 2014
Accounting rate of return
Accounting rate of return (also known as simple rate of return) is the ratio of estimated accounting profit of a project to the average investment made in the project. ARR is used in investment appraisal. Formula
Accounting Rate of Return is calculated using the following formula: ARR =
Average Accounting Profit
Average Investment
Average accounting profit is the arithmetic mean of accounting income expected to be earned during each year of the project's life time. Average investment may be calculated as the sum of the beginning and ending book value of the project divided by 2. Another variation of ARR formula uses initial investment instead of average investment. Decision Rule
Accept the project only if its ARR is equal to or greater than the required accounting rate of return. In case of mutually exclusive projects, accept the one with highest ARR. Examples
Example 1: An initial investment of \$130,000 is expected to generate annual cash inflow of \$32,000 for 6 years. Depreciation is allowed on the straight line basis. It is estimated that the project will generate scrap value of \$10,500 at end of the 6th year. Calculate its accounting rate of return assuming that there are no other expenses on the project. Solution
Annual Depreciation = (Initial Investment − Scrap Value) ÷ Useful Life in Years Annual Depreciation = (\$130,000 − \$10,500) ÷ 6 ≈ \$19,917 Average Accounting Income = \$32,000 − \$19,917 = \$12,083
Accounting Rate of Return = \$12,083 ÷ \$130,000 ≈ 9.3%
Example 2: Compare the following two mutually exclusive projects on the basis of ARR. Cash flows and salvage values are in thousands of dollars. Use the straight line depreciation method. Project A:
Year 0 1 2 3
Cash Outflow -220
Cash Inflow 91 130 105
Salvage Value 10
Project B:
Year 0 1 2 3
Cash Outflow -198
Cash Inflow...
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# annual rate of return. Pages: 3 (472 words) Published: February 24, 2014. Accounting rate of return. Accounting rate of return (also known as simple rate of return) is the ratio of estimated accounting profit of a project to the average investment made in the project. ARR is used in investment appraisal. Formula. Accounting Rate of Return is calculated using the following formula: ARR =. Average Accounting Profit. Average Investment. Average accounting profit is the arithmetic mean of accounting income expected to be earned during each year of the project's life time. Average investment may be calculated as the sum of the beginning and ending book value of the project divided by 2. Another variation of ARR formula uses initial investment instead of average investment. Decision Rule. Accept the project only if its ARR is equal to or greater than the required accounting rate of return. In case of mutually exclusive projects, accept the one with highest ARR. Examples. Example 1: An initial investment of \$130,000 is expected to generate annual cash inflow of \$32,000 for 6 years. Depreciation is allowed on the straight line basis.
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It is estimated that the project will generate scrap value of \$10,500 at end of the 6th year. Calculate its accounting rate of return assuming that there are no other expenses on the project. Solution. Annual Depreciation = (Initial Investment − Scrap Value) ÷ Useful Life in Years Annual Depreciation = (\$130,000 − \$10,500) ÷ 6 ≈ \$19,917 Average Accounting Income = \$32,000 − \$19,917 = \$12,083. Accounting Rate of Return = \$12,083 ÷ \$130,000 ≈ 9.3%. Example 2: Compare the following two mutually exclusive projects on the basis of ARR. Cash flows and salvage values are in thousands of dollars. Use the straight line depreciation method. Project A:. Year 0 1 2 3. Cash Outflow -220. Cash Inflow 91 130 105. Salvage Value 10. Project B:. Year 0 1 2 3. Cash Outflow -198. Cash Inflow...
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An Introduction to Probability
Probability is the proper mechanism for accounting for uncertainty. Axiomatic ... Instead, this chapter will first review the basic ideas of probability. We then ...
C H A P T E R
1
An Introduction to Probability As the previous chapters have illustrated, it is often quite easy to come up with physical models that determine the effects that result from various causes — we know how image intensity is determined, for example. The difficulty is that effects could have come from various causes and we would like to know which — for example, is the image dark because the light level is low, or because the surface has low albedo? Ideally, we should like to take our measurements and determine a reasonable description of the world that generated them. Accounting for uncertainty is a crucial component of this process, because of the ambiguity of our measurements. Our process of accounting needs to take into account reasonable preferences about the state of the world — for example, it is less common to see very dark surfaces under very bright lights than it is to see a range of albedoes under a reasonably bright light. Probability is the proper mechanism for accounting for uncertainty. Axiomatic probability theory is gloriously complicated, and we don’t attempt to derive the ideas in detail. Instead, this chapter will first review the basic ideas of probability. We then describe techniques for building probabilistic models and for extracting information from a probabilistic model, all in the context of quite simple examples. In chapters ??, 2, ?? and ??, we show some substantial examples of probabilistic methods; there are other examples scattered about the text by topic. Discussions of probability are often bogged down with waffle about what probability means, a topic that has attracted a spectacular quantity of text. Instead, we will discuss probability as a modelling technique with certain formal, abstract properties — this means we can dodge the question of what the ideas mean and concentrate on the far more interesting question of what they can do for us. We will develop probability theory in discrete spaces first, because it is possible to demonstrate the underpinning notions without much notation (section 1.1). We then pass to continuous spaces (section 1.2). Section 1.3 describes the important notion of a random variable, and section 1.4 describes some common probability models. Finally, in section 1.5, we get to probabilistic inference, which is the main reason to study probability. 1.1 PROBABILITY IN DISCRETE SPACES Probability models compare the outcomes of various experiments. These outcomes are represented by a collection of subsets of some space; the collection must have special properties. Once we have defined such a collection, we can define a probability function. The interesting part is how we choose a probability function for a particular application, and there are a series of methods for doing this. 2
Section 1.1
Probability in Discrete Spaces
3
1.1.1 Representing Events Generally, a probability model is used to compare various kinds of experimental outcomes. We assume that we can distinguish between these outcomes, which are usually called events. Now if it is possible to tell whether an event has occurred, it is possible to tell if it has not occurred, too. Furthermore, if it is possible to tell that two events have occurred independently, then it is possible to tell if they have occurred simultaneously. This motivates a formal structure. We take a discrete space, D, which could be infinite and which represents the world in which experiments occur. Now construct a collection of subsets of D, which we shall call F , each of which represents an event. This collection must have the following properties: • The empty set is in F and so is D. In effect, we are saying that “nothing happened” and “something happened” are events. • Closure under complements: if S1 ∈ F then S1 = D − S1 ∈ F — i.e. if it is possible to tell whether an event has occurred, it is possible to tell if it has not occurred, too. • Closure under intersection: if S1 ∈ F and S2 ∈ F, then S1 ∩ S2 ∈ F — i.e. if it is possible to tell that two events have occurred independently, then it is possible to tell if they have occurred simultaneously. The elements of F correspond to the events. Note that we can we can tell whether any logical combinations of events has occurred, too, because a logical combination of events corresponds to set unions, negations or intersections. EXAMPLE 1.1
The space of events for a single toss of a coin.
Given a coin that is flipped once, D = {heads, tails} There are only two possible sets of events in this case: {∅, D} (which implies we flipped the coin, but can’t tell what happened!) and {∅, D, {heads}, {tails}}
EXAMPLE 1.2
Two possible spaces of events for a single flip each of two coins.
Given two coins that are flipped, D = {hh, ht, tt, th}
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An Introduction to Probability
There are rather more possible sets of events in this case. One useful one would be ∅, D, {hh}, {ht}, {tt}, {th}, {hh, ht}, {hh, th}, {hh, tt}, {ht, th}, F= {ht, tt}, {th, tt}, {hh, ht, th}, {hh, ht, tt}, {hh, th, tt}, {ht, th, tt} which would correspond to all possible cases. Another (perhaps less useful) structure would be: F = {∅, D, {hh, ht}, {th, tt}} which implies that we cannot measure the state of the second coin
1.1.2 Probability: the P-function Now we construct a function P , which takes elements of F to the unit interval. We require that P has some important properties: • P is defined for every element of F • P (∅) = 0 • P (D) = 1 • for A ∈ F and B ∈ F, P (A ∪ B) = P (A) + P (B) − P (A ∩ B) which we call the axiomatic properties of probability. Note that 0 ≤ P (A) ≤ 1 for all A ∈ F, because the function takes elements of F to the unit interval. We call the collection of D, P and F a probability model. We call P (A) the probability of the event A — because we are still talking about formal structures, there is absolutely no reason to discuss what this means; it’s just a name. Rigorously justifying the properties of P is somewhat tricky. It can be helpful to think of P as a function that measures the size of a subset of D — the whole of D has size one, and the size of the union of two disjoint sets is the sum of their sizes. EXAMPLE 1.3
The possible P functions for the flip of a single coin.
In example 1, for the first structure on D, there is only one possible choice of P ; for the second, there is a one parameter family of choices, we could choose P (heads) to be an arbitrary number in the unit interval, and the choice of P (tails) follows.
EXAMPLE 1.4
The P functions for two coins, each flipped once.
In example 2, there is a three-parameter family of choices for P in the case of the first event structure shown in that example — we can choose P (hh), P (ht) and P (th), and all other values will be given by the axioms. For the second event structure in that example, P is the same as that for a single coin (because we can’t tell the state of one coin).
Section 1.1
Probability in Discrete Spaces
5
1.1.3 Conditional Probability If we have some element A of F where P (A) = 0 — and this constraint is important — then the collection of sets FA = {u ∩ A|u ∈ F } has the same properties as F (i.e. ∅ ∈ FA , A ∈ FA , and FA is closed under complement and intersection), only now its domain of definition is A. Now for any C ∈ F we can define a P function for the component of C that lies in FA . We write P (C ∩ A) PA (C) = P (A) This works because C ∩ A is in FA , and P (A) is non-zero. In particular, this function satisfies the axiomatic properties of probability on its domain, FA . We call this function the conditional probability of C, given A; it is usually written as P (C|A). If we adopt the metaphor that P measures the size of a set, then the conditional probability measures the size of the set C ∩ A relative to A. Notice that P (A ∩ C) = P (A|C)P (C) = P (C|A)P (A) an important fact that you should memorize. It is often written as P (A, C) = P (A|C)P (C) = P (C|A)P (A) where P (A, C) is often known as the joint probability for the events A and C. Assume that we have a collection of n sets Ai , such that Aj ∩ Ak = ∅ for every j = k and A = i Ai . The analogy between probability and size motivates the result that n P (B|Ai )P (Ai |A) P (B|A) = i=1
a fact well worth remembering. In particular, if A is the whole domain D, we have the useful fact that for n disjoint sets Ai , such that D = i Ai , P (B) = P (B|D) n P (B|Ai )P (Ai |D) = i=1
=
n
P (B|Ai )P (Ai )
i=1
1.1.4 Choosing P We have a formal structure — to use it, we need to choose values of P that have useful semantics. There are a variety of ways of doing this, and it is essential to understand that there is no canonical choice. The choice of P is an essential part of the modelling process. A bad choice will lead to an unhelpful or misleading model,
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and a good choice may lead to a very enlightening model. There are some strategies that help in choosing P . Symmetry. Many problems have a form of symmetry that means we have no reason to distinguish between certain sets of events. In this case, it is natural to choose P to reflect this fact. Examples 5 and 6 illustrate this approach. EXAMPLE 1.5
Choosing the P function for a single coin flip using symmetry.
Assume we have a single coin which we will flip, and we can tell the difference between heads and tails. Then F = {∅, D, {heads}, {tails}} is a reasonable model to adopt. Now this coin is symmetric — there is no reason to distinguish between the heads side and the tails side from a mechanical perspective. Furthermore, the operation of flipping it subjects it to mechanical forces that do not favour one side over the other. In this case, we have no reason to believe that there is any difference between the outcomes, so it is natural to choose P (heads) = P (tails) = 1/2
EXAMPLE 1.6
Choosing the P function for a roll of a die using symmetry.
Assume we have a die that we believe to be fair, in the sense that it has been manufactured to have the symmetries of a cube. This means that there is no reason to distinguish between any of the six events defined by distinct faces pointing up. We can therefore choose a P function that has the same value for each of these events. A more sophisticated user of a die labels each vertex of each face, and throws the die onto ruled paper; each face then has four available states, corresponding to the vertex that is furthest away from the thrower. Again, we have no reason to distinguish between the states, so we can choose a P function that has the same value for each of the 24 possible states that can result.
Independence. In many probability models, events do not depend on one another. This is reflected in the conditional probability. If there is no interaction between events A and B, then P (A|B) cannot depend on B. This means that P (A|B) = P (A) (and, also, P (B|A) = P (B)), a property known as independence. In turn, if A and B are independent, we have P (A ∩ B) = P (A|B)P (B) = P (A)P (B). This property is important, because it reduces the number of parameters that must be chosen in building a probability model (example 7). EXAMPLE 1.7 Choosing the P function for a single flip each of two coins using the idea of independence. We adopt the first of the two event structures given for the two coins in example 2
Section 1.1
Probability in Discrete Spaces
7
(this is where we can tell the state of both coins). Now we assume that neither coin knows the other’s intentions or outcome. This assumption restricts our choice of probability model quite considerably because it enforces a symmetry. Let us choose P ({hh, ht}) = p1h and P ({hh, th}) = p2h Now let us consider conditional probabilities, in particular P ({hh, ht}|{hh, th}) (which we could interpret as the probability that the first coin comes up heads given the second coin came up heads). If the coins cannot communicate, then this conditional probability should not depend on the conditioning set, which means that P ({hh, ht}|{hh, th}) = P ({hh, ht}) In this case, we know that P ({hh}) = P ({hh, ht}|{hh, th})P ({hh, th}) = P ({hh, ht})P ({hh, th}) = p1h p2h Similar reasoning yields P (A) for all A ∈ F, so that our assumption that the two coins are independent means that there is now only a two parameter family of probability models to choose from — one parameter describes the first coin, the other describes the second. A more subtle version of this property is conditional independence. Formally, A and B are conditionally independent given C if P (A, B, C) = P (A, B|C)P (C) = P (A|C)P (B|C)P (C) Like independence, conditional independence simplifies modelling by (sometimes substantially) reducing the number of parameters that must be chosen in constructing a model (example 8). EXAMPLE 1.8 Simplifying a model using conditional independence: the case of rain, sprinklers and lawns. Both I and my neighbour have a lawn; each lawn has its own sprinkler system. There are two reasons that my lawn could be wet in the morning — either it rained in the night, or my sprinkler system came on. There is no reason to believe that the neighbour’s sprinkler system comes on at the same times or on the same days as mine does. Neither sprinkler system is smart enough to know whether it has rained. Finally, if it rains, both lawns are guaranteed to get wet; however, if the sprinkler system comes on, there is some probability that the lawn will not get wet (perhaps a jammed nozzle).
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An Introduction to Probability
A reasonable model has five binary variables (my lawn is wet or not; the neighbour’s lawn is wet or not; my sprinkler came on or not; the neighbour’s sprinkler came on or not; and it rained or not). D has 32 elements, and the event space is too large to write out conveniently. If there was no independence in the model, specifying P could require 31 parameters. However, if I know it did not rain in the night, then the state of my lawn is independent of the state of the neighbour’s lawn, because the two sprinkler systems do not communicate. Our joint probability function is P (W, Wn , S, Sn , R) = P (W, S|R)P (Wn , Sn |R)P (R) We know that P (W = true, S|R = true) = P (S) (this just says that if it rains, the lawn is going to be wet); a similar observation applies to the neighbour’s lawn. The rain and the sprinklers are independent and there is a symmetry — both my neighbour’s lawn and mine behave in the same way. This means that, in total, we need only 5 parameters to specify this model. Notice that in this case, independence is a model; it is possible to think of any number of reasons that the sprinkler systems might well display quite similar behaviour, even though they don’t communicate (the neighbour and I might like the same kind of plants; there could be laws restricting when the sprinklers come on; etc.). This means that, like any model, we will need to look for evidence that tends either to support or to discourage our use of the model. One form that this evidence very often takes is the observation that the model is good at predicting what happened in the past.
Frequency:. Data reflecting the relative frequency of events can be easily converted into a form that satisfies the axioms for P , as example 9 indicates. EXAMPLE 1.9 mation.
Choosing a P function for a single coin flip using frequency infor-
Assume that, in the past, we have flipped the single coin described above many times, and observed that for 51% of these flips it comes up heads, and for 49% it comes up tails. We could choose P ({heads}) = 0.51 and P ({tails}) = 0.49 This choice is a sensible choice, as example 10 indicates. An interpretation of probability as frequency is consistent, in the following sense. Assume that we obtain repeated, independent outcomes from an experiment which has been modelled with a P allocated using frequency data. Events will be long sequences of outcomes, and the events with the highest probability will be those that show the outcomes with about the right frequency. Example 10 illustrates this effect for repeated flips of a single coin.
Section 1.1
EXAMPLE 1.10
Probability in Discrete Spaces
9
The probability of various frequencies in repeated coin flips
Now consider a single coin that we flip many times, and where each flip is independent of the other. We set up an event structure that does not reflect the order in which the flips occur. For example, for two flips, we would have: {∅, D, {hh}, {tt}, {ht, th}, {hh, tt}, {hh, ht, th}, {tt, ht, th}} (which we can interpret as “no event”, “some event”, “both heads”, “both tails”, “coins different”, “coins the same”, “not both tails”, and “not both heads”). We assume that P ({hh}) = p2 ; a simple computation using the idea of independence yields that P ({ht, th}) = 2p(1 − p) and P (tt) = (1 − p)2 . We can generalise this result, to obtain
n P (k heads and n − k tails in n flips) = pk (1 − p)n−k k
Saying that the relative frequency of an event is f means that, in a very large number of independent trials (say, N ), we expect that the event occurs in about fN of those trials. Now for large n, the expression
n pk (1 − p)n−k k (which is what we obtained for the probability of a sequence of trials showing k heads and n − k tails in example 10) has a substantial peak at p = nk . This peak gets very narrow and extremely pronounced as n → ∞. This effect is extremely important, and is consistent with an interpretation of probability as relative frequency: • firstly, because it means that we assign a high probability to long sequences of coin flips where the event occurs with the “right” frequency • and secondly, because the probability assigned to these long sequences can also be interpreted as a frequency — essentially, this interpretation means that long sequences where the events occur with the “right” frequency occur far more often than other such sequences (see figure 1.1). All this means that, if we choose a P function for a coin flip — or some other experiment — on the basis of sufficiently good frequency data, then we are very unlikely to see long sequences of coin flips — or repetitions of the experiment — that do not show this frequency. This interpretation of probability as frequency is widespread, and common. One valuable advantage of the interpretation is that it simplifies estimating probabilities for some sorts of models. For example, given a coin, one could obtain P (heads) by flipping the coin many times and measuring the relative frequency with which heads appear.
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FIGURE 1.1: We assume that a single flip of a coin has a probability 0.5 of coming up heads. If we interpret probability as frequency, then long sequences of coin flips should almost always have heads appearing about half the time. This plot shows the width of the interval about 0.5 that contains 95% of the probability for various numbers of repeated coin flips. Notice that as the sequence gets longer, the interval gets narrower — one is very likely to observe a frequency of heads in the range [0.43, 0.57] for 170 flips of a coin with probability 0.5 of coming up heads.
Subjective probability. It is not always possible to use frequencies to obtain probabilities. There are circumstances in which we would like to account for uncertainty but cannot meaningfully speak about frequencies. For example, it is easy to talk about the probability it will rain tomorrow, but hard to interpret this use of the term as a statement about frequency1 . An alternative source of P is to regard probability as encoding degree of belief. In this approach, which is usually known as subjective probability, one chooses P to reflect reasonable beliefs about the situation that applies. EXAMPLE 1.11 jective probability.
Assigning P functions to coins from different sources, using sub-
A friend with a good reputation for probity and no obvious need for money draws a coin from a pocket, and offers to bet with you on whether it comes up heads or tails — your choice of face. What probability do you ascribe to the event that it comes up heads? Now an acquaintance draws a coin from a pocket and offers a bet: he’ll pay you 15 dollars for your stake of one dollar if the coin comes up heads. What probability 1 One dodge is to assume that there are a very large set of equivalent universes which are the same today. In some of these worlds, it rains tomorrow and in others it doesn’t; the frequency with which it rains tomorrow is the probability. This philosophical fiddle isn’t very helpful in practice, because we can’t actually measure that frequency by looking at these alternative worlds.
Section 1.2
Probability in Continuous Spaces
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The basic axioms for P apply here too. For D the domain, and A and B events, we have: • P (D) = 1 • P (∅) = 0 • for any A, 0 ≤ P (A) ≤ 1 • if A ⊂ B, then P (A) ≤ P (B) • P (A ∪ B) = P (A) + P (B) − P (A ∩ B) The concepts of conditional probability, independence and conditional independence apply in continuous spaces without modification. For example, the conditional probability of an event given another event can be defined by P (A ∩ B) = P (A|B)P (B) and the conditional probability can be thought of as probability restricted to the set B. Events A and B are independent if and only if P (A ∩ B) = P (A)P (B) and A and B are conditionally independent given C if and only if P (A ∩ B|C) = P (A|C)P (B|C) Of course, to build a useful model we need to be more specific about what the events should be. 1.2.2 Representing P-functions One difficulty in building probability models on continuous spaces is expressing the function P in a useful way — it is clearly no longer possible to write down the space of events and give a value of P for each event. We will deal only with Rn , with subsets of this space, or with multiple copies of this space. The Real Line. The set of events for the real line is far too big to write down. All events look like unions of a basic collection of sets. This basic collection consists of: • individual points (i.e a); • open intervals (i.e. (a, b)); • half-open intervals (i.e. (a, b] or [a, b)); • and closed intervals (i.e. [a, b]). All of these could extend to infinity. The function P can be represented by a function F with the following properties:
Section 1.2
Probability in Continuous Spaces
13
• F (−∞) = 0 • F (∞) = 1 • F (x) is monotonically increasing. and we interpret F (x) as P ((−∞, x]). The function F is referred to as the cumulative distribution function. The value of P for all the basic sets described can be extracted from F , with appropriate attention to limits; for example, P ((a, b]) = F (b) − F (a) and P (a) = lim←0+ (F (a) − F (a − )). Notice that if F is continuous, P (a) = 0. Higher Dimensional Spaces. In Rn , events are unions of elements of a basic collection of sets, too. This basic collection consists of a product of n elements from the basic collection for the real line. A cumulative distribution function can be defined in this case, too. It is given by a function F with the property that P ({x1 ≤ u1 , x2 ≤ u2 , . . . xn ≤ un }) = F (u). This function is constrained by other properties, too. However, cumulative distribution functions are a somewhat unwieldy way to specify probability. 1.2.3 Representing P-functions with Probability Density Functions For the examples we will deal with in continuous spaces, the usual way to specify P is to provide a function p such that P (event) = p(u)du event This function is referred to as a probability density function. Not every probability model admits a density function, but all our cases will. Note that a density function cannot have a negative value, but that its value could be larger than one. In all cases, probability density functions integrate to one, i.e. p(u)du = 1 P (D) = D
and any non-negative function with this property is a probability density function. The value of the probability density function at a point represents the probability of the event that consists of an infinitesimal neighbourhood at that value, i.e.: p(u1 )du = P ({u ∈ [u1 , u1 + du]}) Notice that this means that (unless we are willing to be rather open minded about what constitutes a function), for a probability model on a continuous space that can be represented using a probability density, the probability of an event that consists of a finite union of points must be zero. For the examples we will deal with, this doesn’t create any issues. In fact, it is intuitive, in the sense that we don’t expect to be able to observe the event that, say, a noise voltage has value 1; instead, we can observe the event that it lies in some tiny interval — defined by the accuracy of our measuring equipment — about 1. Conditional probability, independence and conditional independence are ideas that can be translated into properties of probability density functions. In their most useful form, they are properties of random variables.
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1.3 RANDOM VARIABLES Assume that we have a probability model on either a discrete or a continuous domain, {D, F , P }. Now let us consider a function of the outcome of an experiment. The values that this function takes on the different elements of D form a new set, which we shall call D . There is a structure, with the same formal properties as F on D defined by the values that this function takes on different elements of F — call this structure F . This function is known as a random variable. We can talk about the probability that a random variable takes a particular set of values, because the probability structure carries over. In particular, assume that we have a random variable ξ. If A ∈ F , there is some A ∈ F such that A = ξ(A). This means that P ({ξ ∈ A }) = P (A)
EXAMPLE 1.12
Assorted examples of random variables
The simplest random variable is given by the identity function — this means that D is the same as D, and F is the same as F . For example, the outcome of a coin flip is a random variable. Now gamble on the outcome of a coin flip: if it comes up heads, you get a dollar, and if it comes up tails, you pay a dollar. Your income from this gamble is a random variable. In particular, D = {1, −1} and F = {∅, D , {1}, {−1}}. Now gamble on the outcome of two coin flips: if both coins come up the same, you get a dollar, and if they come up different, you pay a dollar. Your income from this gamble is a random variable. Again, D = {1, −1} and F = {∅, D , {1}, {−1}}. In this case, D is not the same as D and F is not the same as F ; however, we can still speak about the probability of getting a dollar — which is the same as P ({hh, tt}). Density functions are very useful for specifying the probability model for the value of a random variable. However, they do result in quite curious notations (probability is a topic that seems to encourage creative use of notation). It is common to write the density function for a random variable as p. Thus, the distribution for λ would be written as p(λ) — in this case, the name of the variable tells you what function is being referred to, rather than the name of the function, which is always p. Some authors resist this convention, but its use is pretty much universal in the vision literature, which is why we adopt it. For similar reasons, we write the probability function for a set of events as P , so that the probability of an event P (event) (despite the fact that different sets of events may have very different probability functions). 1.3.1 Conditional Probability and Independence Conditional probability is a very useful idea for random variables. Assume we have two random variables, m and n — (for example, the value I read from my rain gauge as m and the value I read on the neighbour’s as n). Generally, the probability
Section 1.3
Random Variables
15
density function is a function of both variables, p(m, n). Now p(m1 , n1 )dmdn = P ({m ∈ [m1 , m1 + dm]} and {n ∈ [n1 , n1 + dm]}) = P ({m ∈ [m1 , m1 + dm]} | {n ∈ [n1 , n1 + dm]})P ({n ∈ [n1 , n1 + dm]}) We can define a conditional probability density from this by p(m1 , n1 )dmdn = P ({m ∈ [m1 , m1 + dm]} | {n ∈ [n1 , n1 + dm]})P ({n ∈ [n1 , n1 + dm]}) = (p(m1 |n1 )dm)(p(n1 )dn) Note that this conditional probability density has the expected property, that p(m|n) =
p(m, n) p(n)
Independence and conditional independence carry over to random variables and probability densities without fuss. EXAMPLE 1.13
Independence in random variables associated with two coins.
We now consider the probability that each of two different coins comes up heads. In this case, we have two random variables, being the probability that the first coin comes up heads and the probability that the second coin comes up heads (it’s quite important to understand why these are random variables — if you’re not sure, look back at the definition). We shall write these random variables as p1 and p2 . Now the density function for these random variables is p(p1 , p2 ). Let us assume that there is no dependency between these coins, so we should be able to write p(p1 , p2 ) = p(p1 )p(p2 ). Notice that the notation is particularly confusing here; the intended meaning is that p(p1 , p2 ) factors, but that the factors are not necessarily equal. In this case, a further reasonable modelling step is to assume that p(p1 ) is the same function as p(p2 ) (perhaps they came from the same minting machine).
1.3.2 Expectations The expected value or expectation of a random variable (or of some function of the random variable) is obtained by multiplying each value by its probability and summing the results — or, in the case of a continuous random variable, by multiplying by the probability density function and integrating. The operation is known as taking an expectation. For a discrete random variable, x, taking the expectation of x yields: E[x] = xi p(xi ) i∈values For a continuous random variable, the process yields E[x] = xp(x)dx D
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often referred to as the average, or the mean in polite circles. One model for an expectation is to consider the random variable as a payoff, and regard the expectation as the average reward, per bet, for an infinite number of repeated bets. The expectation of a general function g(x) of a random variable x is written as E[g(x)]. The variance of a random variable x is var(x) = E[x2 − (E(x))2 ] This expectation measures the average deviance from the mean. The variance of a random variable gives quite a strong indication of how common it is to see a value that is significantly different from the mean value. In particular, we have the following useful fact: var(x) P ({| x − E[x] |≥ }) ≤ 2 The standard deviation is obtained from the variance:
sd(x) = var(x) = E[x2 − (E[x])2] For a vector of random variables, the covariance is cov(x) = E[xxt − (E[x]E[x]t )] This matrix (look carefully at the transpose) is symmetric. Diagonal entries are the variance of components of x, and must be non-negative. Off-diagonal elements measure the extent to which two variables co-vary. For independent variables, the covariance must be zero. For two random variables that generally have different signs, the covariance can be negative. EXAMPLE 1.14
The expected value of gambling on a coin flip.
You and an acquaintance decide to bet on the outcome of a coin flip. You will receive a dollar from your acquaintance if the coin comes up heads, and pay one if it comes up tails. The coin is symmetric. This means the expected value of the payoff is 1P (heads) − 1P (tails) = 0 The variance of the payoff is one, as is the standard deviation. Now consider the probability of obtaining 10 dollars in 10 coin flips, with a fair coin. Our random variable x is the income in 10 coin flips. Equation 1.3.2 yields 1 P ({| x |≥ 10}) ≤ 100 , which is a generous upper bound — the actual probability is of the order of one in a thousand. Expectations of functions of random variables are extremely useful. The notation for expectations can be a bit confusing, because it is common to omit the density with respect to which the expectation is being taken, which is usually obvious from the context. For example, E[x2 ] is interpreted as x2 p(x)dx D
Section 1.3
Random Variables
17
1.3.3 Joint Distributions and Marginalization Assume we have a model describing the behaviour of a collection of random variables. We will proceed on the assumption that they are discrete, but (as should be clear by now) the discussion will work for continuous variables if summing is replaced by integration. One way to specify this model is to give the probability distribution for all variables, known in jargon as the joint probability distribution function — for concreteness, write this as P (x1 , x2 , . . . xn ). If the probability distribution is represented by its density function, the density function is usually referred to as the joint probability density function. Both terms are often abbreviated as “joint.” EXAMPLE 1.15
Marginalising out parameters for two different types of coin.
Let us assume we have a coin which could be from one of two types; the first type of coin is evenly balanced; the other is wildly unbalanced. We flip our coin some number of times, observe the results, and should like to know what type of coin we have. Assume that we flip the coin once. The set of outcomes is D = {(heads, I), (heads, II), (tails, I), (tails, II)} An appropriate event space is: ∅, {(heads, I)}, {(tails, I)},
{(tails, I), (heads, I)} , {(heads, II), (tails, I), (tails, II)}, {(heads, I), (heads, II), (tails, II)}
D, {(heads, II)}, {(tails, II)}, {(tails, I), (tails, II), } , {(tails, II), (heads, II)} , {(heads, I), (tails, I), (tails, II)} {(heads, I), (heads, II), (tails, I)}
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any other (smaller) set of values too by the same process. You should now look at example 15, which illustrates how the process works using the event structure for a simple case. In fact, the event structure is getting unwieldy as a notation. It is quite common to use a rather sketchy notation to indicate the appropriate event. For example 15, we would write P ({(heads, I), (heads, II)}) = P (heads) We would like to form P ({x2 = b, . . . xn = n}) from P ({x1 = a, x2 = b, . . . xn = n}). By using the argument about event structures in example 15, we obtain P (x1 = v, x2 = b, . . . xn = n) P (x2 = b, . . . xn = n) = v∈values of x1 which we could write as P (x2 , . . . xn ) =
P (x1 , x2 , . . . xn )
values of x1 This operation is referred to as marginalisation. marginalisation A similar argument applies to probability density functions, but the operation is now integration. Given a probability density function p(x1 , x2, . . . , xn ), we obtain p(x2 , . . . xn ) = p(x1 , x2 , . . . xn )dx1 D
marginalisation 1.4 STANDARD DISTRIBUTIONS AND DENSITIES There are a variety of standard distributions that arise regularly in practice. References such as [Patel et al., 1976; Evans et al., 2000] give large numbers; we will discuss only the most important cases. The uniform distribution has the same value at each point on the domain. This distribution is often used to express an unwillingness to make a choice or a lack of information. On a continuous space, the uniform distribution has a density function that has the same value at each point. Notice that a uniform density on an infinite continuous domain isn’t meaningful, because it could not be scaled to integrate to one. In practice, one can often avoid this point, either by pretending that the value is a very small constant and arranging for it to cancel, or using a normal distribution (described below) with a really big covariance, such that its value doesn’t change much over the region of interest. The binomial distribution applies to situations where one has independent identically distributed samples from a distribution with two values. For example, consider drawing n balls from an urn containing equal numbers of black and white balls. Each time a ball is drawn, its colour is recorded and it is replaced, so that the probability of getting a white ball — which we denote p — is the same for each draw. The binomial distribution gives the probability of getting k white balls
n pk (1 − p)n−k k
Section 1.4
Standard Distributions and Densities
19
The mean of this distribution is np and the variance is np(1 − p). The Poisson distribution applies to spatial models that have uniformity properties. Assume that points are placed on the real line randomly in such a way that the expected number of points in an interval is proportional to the length of the interval. The number of points in a unit interval will have a Poisson distribution where λx e−x P ({N = x}) = x! (where x = 0, 1, 2 . . . and λ > 0 is the constant of proportionality). The mean of this distribution is λ and the variance is λ 1.4.1 The Normal Distribution The probability density function for the normal distribution for a single random variable x is 1 (x − µ)2 √ p(x; µ, σ) = exp − 2σ 2 2πσ The mean of this distribution is µ and the standard deviation is σ. This distribution is widely called a Gaussian distribution in the vision community. The multivariate normal distribution for d-dimensional vectors x has probability density function p(x; µ, Σ) =
1 d
(2π) 2 det(Σ)1/2
exp −
(x − µ)T Σ−1 (x − µ) 2
The mean of this distribution is µ and the covariance is Σ. Again, this distribution is widely called a Gaussian distribution in the vision community. The normal distribution is extremely important in practice, for several reasons: • The sum of a large number of random variables is normally distributed, pretty much whatever the distribution of the individual random variables. This fact is known as the central limit theorem. It is often cited as a reason to model a collection of random effects with a single normal model. • Many computations that are prohibitively hard for any other case are easy for the normal distribution. • In practice, the normal distribution appears to give a fair model of some kinds of noise. • Many probability density functions have a single peak and then die off; a model for such distributions can be obtained by taking a Taylor series of the log of the density at the peak. The resulting model is a normal distribution (which is often quite a good model).
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1.5 PROBABILISTIC INFERENCE Very often, we have a sequence of observations produced by some process whose mechanics we understand, but which has some underlying parameters that we do not know. The problem is to make useful statements about these parameters. For example, we might observe the intensities in an image, which are produced by the interaction of light and surfaces by principles we understand; what we don’t know — and would like to know — are such matters as the shape of the surface, the reflectance of the surface, the intensity of the illuminant, etc. Obtaining some representation of the parameters from the data set is known as inference. There is no canonical inference scheme; instead, we need to choose some principle that identifies the most desirable set of parameters. 1.5.1 The Maximum Likelihood Principle A general inference strategy known as maximum likelihood inference, can be described as Choose the world parameters that maximise the probability of the measurement observed In the general case, we are choosing arg max P (measurements|parameters) (where the maximum is only over the world parameters because the measurements are known, and arg max means “the argument that maximises”). In many problems, it is quite easy to specify the measurements that will result from a particular setting of model parameters — this means that P (measurements|parameters), often referred to as the likelihood, is easy to obtain. This can make maximum likelihood estimation attractive. EXAMPLE 1.16 haviour.
Maximum likelihood inference on the type of a coin from its be-
We return to example 15. Now assume that we know some conditional probabilities. In particular, the unbiased coin has P (heads|I) = P (tails|I) = 0.5, and the biased coin has P (tails|II) = 0.2 and P (heads|II) = 0.8. We observe a series of flips of a single coin, and wish to know what type of coin we are dealing with. One strategy for choosing the type of coin represented by our evidence is to choose either I or II, depending on whether P (flips observed|I) > P (flips observed|II). For example, if we observe four heads and one tail in sequence, then P (hhhht|II) = (0.8)4 0.2 = 0.08192 and P (hhhht|I) = 0.03125, and we choose type II. Maximum likelihood is often an attractive strategy, because it can admit quite simple computation. A classical application of maximum likelihood estimation involves estimating the parameters of a normal distribution from a set of samples of that distribution (example 17).
Section 1.5
Probabilistic Inference
21
EXAMPLE 1.17 Estimating the parameters of a normal distribution from a series of independent samples from that distribution. Assume that we have a set of n samples — the i’th of which is xi — that are known to be independent and to have been drawn from the same normal distribution. The likelihood of our sample is P (sample|µ, σ) = L(x1 , . . . xn ; µ, σ)
1 (xi − µ)2 √ exp − p(xi ; µ, σ) = = 2σ 2 2πσ i i Working with the log of the likelihood will remove the exponential, and not change the position of the maximum. For the log-likelihood, we have Q(x1 , . . . xn ; µ, σ) = −
(xi − µ)2 2σ 2
i
1 1 − n( log 2 + log π + log σ) 2 2
and we want the maximum with respect to µ and σ. This must occur when the derivatives are zero, so we have (xi − µ) ∂Q =0 =2 ∂µ 2σ 2 i
and a little shuffling of expressions shows that this maximum occurs at xi µ= i n Similarly ∂Q = ∂σ and this maximum occurs at
σ=
i (xi
− µ)2
σ3
i (xi
n =0 σ
− µ)2
n
Note that this estimate of σ is biased, inthat its expected value is σ(n/(n − 1)) 2 and it is more usual to use (1/(n − 1)) i (xi − µ) as an estimate. 1.5.2 Priors, Posteriors and Bayes’ rule In example 16, our maximum likelihood estimate incorporates no information about P (I) or P (II) — which can be interpreted as how often coins of type I or type II are handed out, or as our subjective degree of belief that we have a coin of type I or of type II before we flipped the coin. This is unfortunate, to say the least; for example, if coins of type II are rare, we would want to see an awful lot of heads before it would make sense to infer that our coin is of this type. Some quite simple algebra suggests a solution.
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Recall that P (A, B) = P (A|B)P (B). This simple observation gives rise to an innocuous looking identity for reversing the order in a conditional probability: P (B|A) =
P (A|B)P (B) P (A)
This is widely referred to as Bayes’ theorem or Bayes’ rule. Now the interesting property of Bayes’ rule is that it tells us which choice of parameters is most probable, given our model and our prior beliefs. Rewriting Bayes’ rule gives P (parameters|data) =
P (data|parameters)P (parameters) P (data)
The term P (parameters) is referred to as the prior (it describes our knowledge of the world before measurements have been taken). The term P (parameters|data) is usually referred to as the posterior (it describes the probability of various models after measurements have been taken). P (data) can be computed by marginalisation (which requires computing a high dimensional integral, often a nasty business) or for some problems can be ignored. As we shall see in following sections, attempting to use Bayes’ rule can result in difficult computations — that integral being one — because posterior distributions often take quite unwieldy forms. 1.5.3 Bayesian Inference The Bayesian philosophy is that all information about the world is captured by the posterior. The first reason to accept this view is that the posterior is a principled combination of prior information about the world and a model of the process by which measurements are generated — i.e. there is no information missing from the posterior, and the information that is there, is combined in a proper manner. The second reason is that the approach appears to produce very good results. The great difficulty is that computing with posteriors can be very difficult — we will encounter various mechanisms for computing with posteriors in following sections. For example, we could use the study of physics in the last few chapters to get expressions relating pixel values to the position and intensity of light sources, the reflectance and orientation of surfaces, etc. Similarly, we are likely to have some beliefs about the parameters that have nothing to do with the particular values of the measurements that we observe. We know that albedos are never outside the range [0, 1]; we expect that illuminants with extremely high exitance are uncommon; and we expect that no particular surface orientation is more common than any other. This means that we can usually cobble up a reasonable choice of prior. MAP Inference. An alternative to maximum likelihood inference is to infer a state of the world that maximises the posterior: Choose the world parameters that maximise the conditional probability of the parameters, conditioned on the measurements taking the observed values
Section 1.5
Probabilistic Inference
23
This approach is known as maximum a posteriori (or MAP) reasoning. EXAMPLE 1.18
Determining the type of a coin using MAP inference.
Assume that we have three flips of the coin of example 16, and would like to determine whether it has type I or type II. We know that the mint has 3 machines that produce type I coins and 1 machine that produces type II coins, and there is no reason to believe that these machines run at different rates. We therefore assign P (I) = 0.75 and P (II) = 0.25. Now we observe three heads, in three consecutive flips. The value of the posterior for type I is: P (hhh|I)P (I) P (hhh) P (h|I)3 P (I) = P (hhh, I) + P (hhh, II) P (h|I)3 P (I) = P (hhh|I)P (I) + P (hhh|II)P (II) 0.530.75 = 0.530.75 + 0.830.25 = 0.422773
P (I|hhh) =
By a similar argument, the value of the posterior for type II is 0.577227. An MAP inference procedure would conclude the coin is of type II. The denominator in the expression for the posterior can be quite difficult to compute, because it requires a sum over what is potentially a very large number of elements (imagine what would happen if there were many different types of coin). However, knowing this term is not crucial if we wish to isolate the element with the maximum value of the posterior, because it is a constant. Of course, if there are a very large number of events in the discrete space, finding the world parameters that maximise the posterior can be quite tricky. The Posterior as an Inference. EXAMPLE 1.19 Determining the probability a coin comes up heads from the outcome of a sequence of flips. Assume we have a coin which comes from a mint which has a continuous control parameter, λ, which lies in the range [0, 1]. This parameter gives the probability that the coin comes up heads, so P (heads|λ) = λ. We know no reason to prefer any one value of λ to any other, so as a prior probability distribution for λ we use the uniform distribution so p(λ) = 1. Assume we flip the coin twice, and observe heads twice; what do we know about λ? All our knowledge is captured by the posterior, which is P (λ ∈ [x, x + dx]|hh) dx
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we shall write this expression as p(λ|hh). We have p(hh|λ)p(λ) p(hh) p(hh|λ)p(λ) = 1 p(hh|λ)p(λ)dλ 0
p(λ|hh) =
= 1 0
λ2 p(λ) p(hh|λ)p(λ)dλ
2
= 3λ
It is fairly easy to see that if we flip the coin n times, and observe k heads and n − k tails, we have p(λ|k heads and n − k tails) ∝ λk (1 − λ)n−k
Section 1.5
Probabilistic Inference
25
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FIGURE 1.3: On the left, the posterior probability density for the probability a coin comes up heads, given a single flip that shows a head and a somewhat untruthful informant who says high, as in example 20. In the center, a posterior probability density for the same problem, but now assuming that we have seen two tails and the informant says high (a sketch of the formulation appears in example 21). On the right, a posterior probability density for the case when the coin shows five tails and the informant says high. As the number of tails builds up, the weight of the posterior in the high region goes down, strongly suggesting the informant is lying.
EXAMPLE 1.21 Determining the type of a coin from a sequence of flips, incorporating information from an occasionally untruthful informant — II. Now consider what happens in example 20 if the contact says high and we see two tails. We need p(high, tt|λ)p(λ) p(high, tt) ∝ p(high, tt|λ)p(λ)
p(λ|high, tt) =
Now p(high, tt|λ) is p(high, tt|λ) = p(high, tt|λ, truth = 1)P (truth = 1) + p(high, tt|λ, truth = 0)P (truth = 0) = p(high, tt|λ, truth = 1)P (truth = 1) + p(tt|λ, truth = 0)p(high|λ, truth = 0)P (truth = 0) Now p(tt|λ, truth = 0) = (1 − λ)2 and the interesting term is p(high, tt|λ, truth = 1). Again, this term reflects the behaviour of the coin and the informant when the informant is telling the truth; in particular, this term must be zero for λ ∈ [0, 2/3), because in this case λ is not high. For λ in the high range, this term must be (1−λ)2 , because now it is the probability of getting two tails with two flips. Performing the computation, we obtain the posterior graphed in figure 1.3.
Bayesian Model Selection. The crucial virtue of Bayesian inference is the accounting for uncertainty shown in examples 20 and 21. We have been able to account for an occasionally
1
Section 1.5
Probabilistic Inference
27
untruthful informant and a random measurement; when there was relatively little contradictory evidence from the coin’s behaviour, our process placed substantial weight on the informant’s testimony, but when the coin disagreed, the informant was discounted. This behaviour is highly attractive, because we are able to combine uncertain sources of information with confidence. EXAMPLE 1.22
Is the informant lying?
We now need to know whether our informant lied to us. Assume we see a single head and an informant saying high, again. The relevant posterior is: P (head, high|truth=0)P (truth=0) P (head, high) P (λ, head, high|truth=0)P (truth=0)dλ = P (head, high) P (head, high|λ, truth=0)P (λ)P (truth=0)dλ = P (head, high) 1 = P (head,high|λ,truth=1)P (λ)dλP (truth=1 ) 1+ P (head,high|λ,truth=0)P (λ)dλP (truth=0 )
Example 22 shows how to tell whether the informant of examples 20 and 21 is telling the truth or not, given the observations. A useful way to think about this example is to regard it as comparing two models (as opposed to the value of a binary parameter within one model). One model has a lying informant, and the other has a truthful informant. The posteriors computed in this example compare how well different models explain a given data set, given a prior on the models. This is a very general problem — usually called model selection — with a wide variety of applications in vision: • Recognition: Assume we have a region in an image, and an hypothesis that an object might be present in that region at a particular position and orientation (the hypothesis will have been obtained using methods from chapter ??, which aren’t immediately relevant). Is there an object there or not? A principled answer involves computing the posterior over two models — that the data was obtained from noise, or from the presence of an object. • Are these the same? Assume we have a set of pictures of surfaces we want to compare. For example, we might want to know if they are the same colour, which would be difficult to answer directly if we didn’t know the illuminant. A principled answer involves computing the posterior over two models — that the data was obtained from one surface, or from two (or more). • What camera was used? Assume we have a sequence of pictures of a world. With a certain amount of work, it is usually possible to infer a great deal of information about the shape of the objects from such a sequence (e.g.
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chapters ??, ?? and ??). The algorithms involved differ quite sharply, depending on the camera model adopted (i.e. perspective, orthographic, etc.). Furthermore, adopting the wrong camera model tends to lead to poor inferences. Determining the right camera model to use is quite clearly a model selection problem. • How many segments are there? We would like to break an image into coherent components, each of which is generated by a probabilistic model. How many components should there be? (section ??). The solution is so absurdly simple in principle (in practice, the computations can be quite nasty) that it is easy to expect something more complex, and miss it. We will write out Bayes’ rule specialised to this case to avoid this: P (data|model) P (data) P (data|model, parameters)P (parameters)d{parameters} = P (data) ∝ P (data|model, parameters)P (parameters)d{parameters}
P (model|data) =
which is exactly the form used in the example. Notice that we are engaging in Bayesian inference here, too, and so can report the MAP solution or report the whole posterior. The latter can be quite helpful when it is difficult to distinguish between models. For example, in the case of the dodgy informant, if P (truth=0|data) = 0.5001, it may be undesirable to conclude the informant is lying — or at least, to take drastic action based on this conclusion. The integral is potentially rather nasty, which means that the method can be quite difficult to use in practice. Useful references include [Gelman et al., 1995; Carlin and Louis, 1996; Gamerman, 1997; Newman and Barkema, 1998; Evans and Swartz, 2000]. 1.5.4 Open Issues In the rest of the book, we will have regular encounters with practical aspects of the Bayesian philosphy. Firstly, although the posterior encapsulates all information available about the world, we very often need to make discrete decisions — should we shoot it or not? Typically, this decision making process requires some accounting for the cost of false positives and false negatives. Secondly, how do we build models? There are three basic sources of likelihood functions and priors: • Judicious design: it is possible to come up with models that are too hard to handle computationally. Generally, models on very high-dimensional domains are difficult to deal with, particularly if there is a great deal of interdependence between variables. For some models, quite good inference algorithms are known. The underlying principle of this approach is to exploit simplifications due to independence and conditional independence. • Physics: particularly in low-level vision problems, likelihood models follow quite simply from physics. It is hard to give a set of design rules for this
Section 1.5
Probabilistic Inference
29
strategy. It has been used with some success on occasion (see, for example, [Forsyth, 1999]).
• Learning: a poor choice of model results in poor performance, and a good choice of model results in good performance. We can use this observation to tune the structure of models if we have a sufficient set of data. We describe aspects of this strategy in chapter ?? and in chapter ??.
Finally, the examples above suggest that posteriors can have a nasty functional form. This intuition is correct, and there is a body of technique that can help handle ugly posteriors which we explore as and when we need it (see also [Gelman et al., 1995; Carlin and Louis, 1996; Gamerman, 1997; Newman and Barkema, 1998]).
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P (k heads and n − k tails in n flips) = Show that this is true.
n k
pk (1 − p)n−k
Section 1.6
Notes
31
1.5. A careless study of example 10 often results in quite muddled reasoning, of the following form: I have bet on heads successfully ten times, therefore I should bet on tails next. Explain why this muddled reasoning — which has its own name, the gambler’s fallacy in some circles, anti-chance in others — is muddled. 1.6. Confirm the count of parameters in example 8. 1.7. In example 19, what is c? 1.8. As in example 16, you are given a coin of either type I or type II; you do not know the type. You flip the coin n times, and observe k heads. You will infer the type of the coin using maximum likelihood estimation. for what values of k do you decide the coin is of type I? 1.9. Compute P (truth|high, coin behaviour) for each of the three cases of example 21. You’ll have to estimate an integral numerically. 1.10. In example 22, what is the numerical value of the probability that the informant is lying, given that the informant said high and the coin shows a single tail? What is the numerical value of the probability that the informant is lying, given that the informant said high and the coin shows seven tails in eight flips? 1.11. The random variable x = (x1 , x2 , . . . xn )T has a normal distribution. Show that the ˆ = (x2 , . . . , xn )T has a normal distribution (which is obtained by random variable x marginalizing the density). A good way to think about this problem is to consider ˆ , and reason about the behaviour of the integral; a the mean and covariance of x bad way is to storm ahead and try and do the integral. 1.12. The random variable p has a normal distribution. Furthermore, there are symmetric matrices A, B and C and vectors D and E such that P (d|p) has the form − log P (d|p) = pT Ap + pT Bd + dT Cd + pT D + dT E + C (C is the log of the normalisation constant). Show that P (p|d) is a normal distribution for any value of d. This has the great advantage that inference is relatively easy. 1.13. x is a random variable with a continuous cumulative distribution function F (x). Show that u = F (x) is a random variable with a uniform density on the range [0, 1]. Now use this fact to show that w = F −1 (u) is a random variable with cumulative distribution function F .
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An Introduction to Probability
Topic Probability model
What you must know A space D, a collection F of subsets of that space containing (a) the empty set; (b) D; (c) all finite unions of elements of F ; (d) all complements of elements of F , and a function P such that (a) P (∅) = 0; (b) P (D) = 1; and (c) for A ∈ F and B ∈ F , P (A ∪ B) = P (A) + P (B) − P (A ∩ B). It is usual to discuss F only implicitly and to represent P by a probability density function for continuous spaces. Random vari- A function of the outcome of an experiment; supports a probability ables model. If we have a random variable ξ, mapping A → A and F → F , defined on the probability model above, and if A ∈ F , there is some A ∈ F such that A = ξ(A). This means that P ({ξ ∈ A }) = P (A). Conditional Given a probability model and a set A ⊂ D such that P (A) = 0 and probability A ∈ F , then A together with F = {C ∩ A|C ∈ F } and P such that P (C) = P (C ∩ A)/P (A) form a new probability model. P (C) is often written as P (C|A) and called the conditional probability of the event C, given that A has occurred. Probability A function p such that P {u ∈ E} = E p(x)dx. All the probability density func- models we deal with on continuous spaces will admit densities, but tion not all do. Marginalisation Given the joint probability density p(X, Y ) of two random variables X and Y , the probability of Y alone — referred to as the marginal probability density for Y — is given by
p(x, Y )dx
Expectation
Normal random variable
The domain is all possible values of X; if the random variables are discrete, the integral is replaced by a sum. The “expected value” of a random variable, computed as E[f (x)] = f (x)p(x)dx. Useful expectations include the mean E[x] and the covariance E[(x − E[x])(x − E[x])T ]. A random variable whose probability density function is the normal (or gaussian) distribution. For an n-dimensional random variable, this is p(x) =
1 exp(−(1/2)(x − µ)T Σ−1 (x − µ)) (2π)(n/2) | Σ |
having mean µ and covariance Σ.
Chapter summary for chapter 1: Probabilistic methods manipulate representations of the “size” of sets of events. Events are defined so that unions and negations are meaningful, leading to a family of subsets of a space. The probability of a set of events is a function defined on this structure. A random variable represents the outcome of an experiment. A generative model gives the probability of a set of outcomes from some inputs; inference obtains a representation of the probable inputs that gave rise to some known outcome.
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An Introduction to Probability. Probability is the proper mechanism for accounting for uncertainty. Axiomatic ... Instead, this chapter will first review the basic ideas of probability. We then .... C H A P T E R. 1. An Introduction to Probability As the previous chapters have illustrated, it is often quite easy to come up with physical models that determine the effects that result from various causes — we know how image intensity is determined, for example. The difficulty is that effects could have come from various causes and we would like to know which — for example, is the image dark because the light level is low, or because the surface has low albedo? Ideally, we should like to take our measurements and determine a reasonable description of the world that generated them. Accounting for uncertainty is a crucial component of this process, because of the ambiguity of our measurements. Our process of accounting needs to take into account reasonable preferences about the state of the world — for example, it is less common to see very dark surfaces under very bright lights than it is to see a range of albedoes under a reasonably bright light. Probability is the proper mechanism for accounting for uncertainty. Axiomatic probability theory is gloriously complicated, and we don’t attempt to derive the ideas in detail. Instead, this chapter will first review the basic ideas of probability. We then describe techniques for building probabilistic models and for extracting information from a probabilistic model, all in the context of quite simple examples. In chapters ??, 2, ?? and ??, we show some substantial examples of probabilistic methods; there are other examples scattered about the text by topic. Discussions of probability are often bogged down with waffle about what probability means, a topic that has attracted a spectacular quantity of text. Instead, we will discuss probability as a modelling technique with certain formal, abstract properties — this means we can dodge the question of what the ideas mean and concentrate on the far more interesting question of what they can do for us. We will develop probability theory in discrete spaces first, because it is possible to demonstrate the underpinning notions without much notation (section 1.1). We then pass to continuous spaces (section 1.2). Section 1.3 describes the important notion of a random variable, and section 1.4 describes some common probability models. Finally, in section 1.5, we get to probabilistic inference, which is the main reason to study probability. 1.1 PROBABILITY IN DISCRETE SPACES Probability models compare the outcomes of various experiments. These outcomes are represented by a collection of subsets of some space; the collection must have special properties. Once we have defined such a collection, we can define a probability function. The interesting part is how we choose a probability function for a particular application, and there are a series of methods for doing this. 2. Section 1.1. Probability in Discrete Spaces. 3. 1.1.1 Representing Events Generally, a probability model is used to compare various kinds of experimental outcomes. We assume that we can distinguish between these outcomes, which are usually called events. Now if it is possible to tell whether an event has occurred, it is possible to tell if it has not occurred, too. Furthermore, if it is possible to tell that two events have occurred independently, then it is possible to tell if they have occurred simultaneously. This motivates a formal structure. We take a discrete space, D, which could be infinite and which represents the world in which experiments occur. Now construct a collection of subsets of D, which we shall call F , each of which represents an event. This collection must have the following properties: • The empty set is in F and so is D. In effect, we are saying that “nothing happened” and “something happened” are events. • Closure under complements: if S1 ∈ F then S1 = D − S1 ∈ F — i.e. if it is possible to tell whether an event has occurred, it is possible to tell if it has not occurred, too. • Closure under intersection: if S1 ∈ F and S2 ∈ F, then S1 ∩ S2 ∈ F — i.e. if it is possible to tell that two events have occurred independently, then it is possible to tell if they have occurred simultaneously. The elements of F correspond to the events. Note that we can we can tell whether any logical combinations of events has occurred, too, because a logical combination of events corresponds to set unions, negations or intersections. EXAMPLE 1.1. The space of events for a single toss of a coin.. Given a coin that is flipped once, D = {heads, tails} There are only two possible sets of events in this case: {∅, D} (which implies we flipped the coin, but can’t tell what happened!) and {∅, D, {heads}, {tails}}. EXAMPLE 1.2. Two possible spaces of events for a single flip each of two coins.. Given two coins that are flipped, D = {hh, ht, tt, th}. 4. Chapter 1. An Introduction to Probability. There are rather more possible sets of events in this case. One useful one would be ∅, D, {hh}, {ht}, {tt}, {th}, {hh, ht}, {hh, th}, {hh, tt}, {ht, th}, F= {ht, tt}, {th, tt}, {hh, ht, th}, {hh, ht, tt}, {hh, th, tt}, {ht, th, tt} which would correspond to all possible cases. Another (perhaps less useful) structure would be: F = {∅, D, {hh, ht}, {th, tt}} which implies that we cannot measure the state of the second coin. 1.1.2 Probability: the P-function Now we construct a function P , which takes elements of F to the unit interval. We require that P has some important properties: • P is defined for every element of F • P (∅) = 0 • P (D) = 1 • for A ∈ F and B ∈ F, P (A ∪ B) = P (A) + P (B) − P (A ∩ B) which we call the axiomatic properties of probability. Note that 0 ≤ P (A) ≤ 1 for all A ∈ F, because the function takes elements of F to the unit interval. We call the collection of D, P and F a probability model. We call P (A) the probability of the event A — because we are still talking about formal structures, there is absolutely no reason to discuss what this means; it’s just a name. Rigorously justifying the properties of P is somewhat tricky. It can be helpful to think of P as a function that measures the size of a subset of D — the whole of D has size one, and the size of the union of two disjoint sets is the sum of their sizes. EXAMPLE 1.3. The possible P functions for the flip of a single coin.. In example 1, for the first structure on D, there is only one possible choice of P ; for the second, there is a one parameter family of choices, we could choose P (heads) to be an arbitrary number in the unit interval, and the choice of P (tails) follows.. EXAMPLE 1.4. The P functions for two coins, each flipped once.. In example 2, there is a three-parameter family of choices for P in the case of the first event structure shown in that example — we can choose P (hh), P (ht) and P (th), and all other values will be given by the axioms. For the second event structure in that example, P is the same as that for a single coin (because we can’t tell the state of one coin).. Section 1.1. Probability in Discrete Spaces. 5. 1.1.3 Conditional Probability If we have some element A of F where P (A) = 0 — and this constraint is important — then the collection of sets FA = {u ∩ A|u ∈ F } has the same properties as F (i.e. ∅ ∈ FA , A ∈ FA , and FA is closed under complement and intersection), only now its domain of definition is A. Now for any C ∈ F we can define a P function for the component of C that lies in FA . We write P (C ∩ A) PA (C) = P (A) This works because C ∩ A is in FA , and P (A) is non-zero. In particular, this function satisfies the axiomatic properties of probability on its domain, FA . We call this function the conditional probability of C, given A; it is usually written as P (C|A). If we adopt the metaphor that P measures the size of a set, then the conditional probability measures the size of the set C ∩ A relative to A. Notice that P (A ∩ C) = P (A|C)P (C) = P (C|A)P (A) an important fact that you should memorize. It is often written as P (A, C) = P (A|C)P (C) = P (C|A)P (A) where P (A, C) is often known as the joint probability for the events A and C. Assume that we have a collection of n sets Ai , such that Aj ∩ Ak = ∅ for every j = k and A = i Ai . The analogy between probability and size motivates the result that n P (B|Ai )P (Ai |A) P (B|A) = i=1. a fact well worth remembering. In particular, if A is the whole domain D, we have the useful fact that for n disjoint sets Ai , such that D = i Ai , P (B) = P (B|D) n P (B|Ai )P (Ai |D) = i=1. =. n. P (B|Ai )P (Ai ). i=1. 1.1.4 Choosing P We have a formal structure — to use it, we need to choose values of P that have useful semantics. There are a variety of ways of doing this, and it is essential to understand that there is no canonical choice. The choice of P is an essential part of the modelling process. A bad choice will lead to an unhelpful or misleading model,. 6. Chapter 1. An Introduction to Probability. and a good choice may lead to a very enlightening model. There are some strategies that help in choosing P . Symmetry. Many problems have a form of symmetry that means we have no reason to distinguish between certain sets of events. In this case, it is natural to choose P to reflect this fact. Examples 5 and 6 illustrate this approach. EXAMPLE 1.5. Choosing the P function for a single coin flip using symmetry.. Assume we have a single coin which we will flip, and we can tell the difference between heads and tails. Then F = {∅, D, {heads}, {tails}} is a reasonable model to adopt. Now this coin is symmetric — there is no reason to distinguish between the heads side and the tails side from a mechanical perspective. Furthermore, the operation of flipping it subjects it to mechanical forces that do not favour one side over the other. In this case, we have no reason to believe that there is any difference between the outcomes, so it is natural to choose P (heads) = P (tails) = 1/2. EXAMPLE 1.6. Choosing the P function for a roll of a die using symmetry.. Assume we have a die that we believe to be fair, in the sense that it has been manufactured to have the symmetries of a cube. This means that there is no reason to distinguish between any of the six events defined by distinct faces pointing up. We can therefore choose a P function that has the same value for each of these events. A more sophisticated user of a die labels each vertex of each face, and throws the die onto ruled paper; each face then has four available states, corresponding to the vertex that is furthest away from the thrower. Again, we have no reason to distinguish between the states, so we can choose a P function that has the same value for each of the 24 possible states that can result.. Independence. In many probability models, events do not depend on one another. This is reflected in the conditional probability. If there is no interaction between events A and B, then P (A|B) cannot depend on B. This means that P (A|B) = P (A) (and, also, P (B|A) = P (B)), a property known as independence. In turn, if A and B are independent, we have P (A ∩ B) = P (A|B)P (B) = P (A)P (B). This property is important, because it reduces the number of parameters that must be chosen in building a probability model (example 7). EXAMPLE 1.7 Choosing the P function for a single flip each of two coins using the idea of independence. We adopt the first of the two event structures given for the two coins in example 2. Section 1.1. Probability in Discrete Spaces. 7. (this is where we can tell the state of both coins). Now we assume that neither coin knows the other’s intentions or outcome. This assumption restricts our choice of probability model quite considerably because it enforces a symmetry. Let us choose P ({hh, ht}) = p1h and P ({hh, th}) = p2h Now let us consider conditional probabilities, in particular P ({hh, ht}|{hh, th}) (which we could interpret as the probability that the first coin comes up heads given the second coin came up heads). If the coins cannot communicate, then this conditional probability should not depend on the conditioning set, which means that P ({hh, ht}|{hh, th}) = P ({hh, ht}) In this case, we know that P ({hh}) = P ({hh, ht}|{hh, th})P ({hh, th}) = P ({hh, ht})P ({hh, th}) = p1h p2h Similar reasoning yields P (A) for all A ∈ F, so that our assumption that the two coins are independent means that there is now only a two parameter family of probability models to choose from — one parameter describes the first coin, the other describes the second. A more subtle version of this property is conditional independence. Formally, A and B are conditionally independent given C if P (A, B, C) = P (A, B|C)P (C) = P (A|C)P (B|C)P (C) Like independence, conditional independence simplifies modelling by (sometimes substantially) reducing the number of parameters that must be chosen in constructing a model (example 8). EXAMPLE 1.8 Simplifying a model using conditional independence: the case of rain, sprinklers and lawns. Both I and my neighbour have a lawn; each lawn has its own sprinkler system. There are two reasons that my lawn could be wet in the morning — either it rained in the night, or my sprinkler system came on. There is no reason to believe that the neighbour’s sprinkler system comes on at the same times or on the same days as mine does. Neither sprinkler system is smart enough to know whether it has rained. Finally, if it rains, both lawns are guaranteed to get wet; however, if the sprinkler system comes on, there is some probability that the lawn will not get wet (perhaps a jammed nozzle).. 8. Chapter 1. An Introduction to Probability. A reasonable model has five binary variables (my lawn is wet or not; the neighbour’s lawn is wet or not; my sprinkler came on or not; the neighbour’s sprinkler came on or not; and it rained or not). D has 32 elements, and the event space is too large to write out conveniently. If there was no independence in the model, specifying P could require 31 parameters. However, if I know it did not rain in the night, then the state of my lawn is independent of the state of the neighbour’s lawn, because the two sprinkler systems do not communicate. Our joint probability function is P (W, Wn , S, Sn , R) = P (W, S|R)P (Wn , Sn |R)P (R) We know that P (W = true, S|R = true) = P (S) (this just says that if it rains, the lawn is going to be wet); a similar observation applies to the neighbour’s lawn. The rain and the sprinklers are independent and there is a symmetry — both my neighbour’s lawn and mine behave in the same way. This means that, in total, we need only 5 parameters to specify this model. Notice that in this case, independence is a model; it is possible to think of any number of reasons that the sprinkler systems might well display quite similar behaviour, even though they don’t communicate (the neighbour and I might like the same kind of plants; there could be laws restricting when the sprinklers come on; etc.). This means that, like any model, we will need to look for evidence that tends either to support or to discourage our use of the model. One form that this evidence very often takes is the observation that the model is good at predicting what happened in the past.. Frequency:. Data reflecting the relative frequency of events can be easily converted into a form that satisfies the axioms for P , as example 9 indicates. EXAMPLE 1.9 mation.. Choosing a P function for a single coin flip using frequency infor-. Assume that, in the past, we have flipped the single coin described above many times, and observed that for 51% of these flips it comes up heads, and for 49% it comes up tails. We could choose P ({heads}) = 0.51 and P ({tails}) = 0.49 This choice is a sensible choice, as example 10 indicates. An interpretation of probability as frequency is consistent, in the following sense. Assume that we obtain repeated, independent outcomes from an experiment which has been modelled with a P allocated using frequency data. Events will be long sequences of outcomes, and the events with the highest probability will be those that show the outcomes with about the right frequency. Example 10 illustrates this effect for repeated flips of a single coin.. Section 1.1. EXAMPLE 1.10. Probability in Discrete Spaces. 9. The probability of various frequencies in repeated coin flips. Now consider a single coin that we flip many times, and where each flip is independent of the other. We set up an event structure that does not reflect the order in which the flips occur. For example, for two flips, we would have: {∅, D, {hh}, {tt}, {ht, th}, {hh, tt}, {hh, ht, th}, {tt, ht, th}} (which we can interpret as “no event”, “some event”, “both heads”, “both tails”, “coins different”, “coins the same”, “not both tails”, and “not both heads”). We assume that P ({hh}) = p2 ; a simple computation using the idea of independence yields that P ({ht, th}) = 2p(1 − p) and P (tt) = (1 − p)2 . We can generalise this result, to obtain. n P (k heads and n − k tails in n flips) = pk (1 − p)n−k k. Saying that the relative frequency of an event is f means that, in a very large number of independent trials (say, N ), we expect that the event occurs in about fN of those trials. Now for large n, the expression. n pk (1 − p)n−k k (which is what we obtained for the probability of a sequence of trials showing k heads and n − k tails in example 10) has a substantial peak at p = nk . This peak gets very narrow and extremely pronounced as n → ∞. This effect is extremely important, and is consistent with an interpretation of probability as relative frequency: • firstly, because it means that we assign a high probability to long sequences of coin flips where the event occurs with the “right” frequency • and secondly, because the probability assigned to these long sequences can also be interpreted as a frequency — essentially, this interpretation means that long sequences where the events occur with the “right” frequency occur far more often than other such sequences (see figure 1.1). All this means that, if we choose a P function for a coin flip — or some other experiment — on the basis of sufficiently good frequency data, then we are very unlikely to see long sequences of coin flips — or repetitions of the experiment — that do not show this frequency. This interpretation of probability as frequency is widespread, and common. One valuable advantage of the interpretation is that it simplifies estimating probabilities for some sorts of models. For example, given a coin, one could obtain P (heads) by flipping the coin many times and measuring the relative frequency with which heads appear.. 10. Chapter 1. An Introduction to Probability 1. 0.9. 0.8. 0.7. 0.6. 0.5. 0.4. 0.3. 0.2. 0.1. 0. 0. 20. 40. 60. 80. 100. 120. 140. 160. 180. FIGURE 1.1: We assume that a single flip of a coin has a probability 0.5 of coming up heads. If we interpret probability as frequency, then long sequences of coin flips should almost always have heads appearing about half the time. This plot shows the width of the interval about 0.5 that contains 95% of the probability for various numbers of repeated coin flips. Notice that as the sequence gets longer, the interval gets narrower — one is very likely to observe a frequency of heads in the range [0.43, 0.57] for 170 flips of a coin with probability 0.5 of coming up heads.. Subjective probability. It is not always possible to use frequencies to obtain probabilities. There are circumstances in which we would like to account for uncertainty but cannot meaningfully speak about frequencies. For example, it is easy to talk about the probability it will rain tomorrow, but hard to interpret this use of the term as a statement about frequency1 . An alternative source of P is to regard probability as encoding degree of belief. In this approach, which is usually known as subjective probability, one chooses P to reflect reasonable beliefs about the situation that applies. EXAMPLE 1.11 jective probability.. Assigning P functions to coins from different sources, using sub-. A friend with a good reputation for probity and no obvious need for money draws a coin from a pocket, and offers to bet with you on whether it comes up heads or tails — your choice of face. What probability do you ascribe to the event that it comes up heads? Now an acquaintance draws a coin from a pocket and offers a bet: he’ll pay you 15 dollars for your stake of one dollar if the coin comes up heads. What probability 1 One dodge is to assume that there are a very large set of equivalent universes which are the same today. In some of these worlds, it rains tomorrow and in others it doesn’t; the frequency with which it rains tomorrow is the probability. This philosophical fiddle isn’t very helpful in practice, because we can’t actually measure that frequency by looking at these alternative worlds.. Section 1.2. Probability in Continuous Spaces. 11. 12. Chapter 1. An Introduction to Probability. The basic axioms for P apply here too. For D the domain, and A and B events, we have: • P (D) = 1 • P (∅) = 0 • for any A, 0 ≤ P (A) ≤ 1 • if A ⊂ B, then P (A) ≤ P (B) • P (A ∪ B) = P (A) + P (B) − P (A ∩ B) The concepts of conditional probability, independence and conditional independence apply in continuous spaces without modification. For example, the conditional probability of an event given another event can be defined by P (A ∩ B) = P (A|B)P (B) and the conditional probability can be thought of as probability restricted to the set B. Events A and B are independent if and only if P (A ∩ B) = P (A)P (B) and A and B are conditionally independent given C if and only if P (A ∩ B|C) = P (A|C)P (B|C) Of course, to build a useful model we need to be more specific about what the events should be. 1.2.2 Representing P-functions One difficulty in building probability models on continuous spaces is expressing the function P in a useful way — it is clearly no longer possible to write down the space of events and give a value of P for each event. We will deal only with Rn , with subsets of this space, or with multiple copies of this space. The Real Line. The set of events for the real line is far too big to write down. All events look like unions of a basic collection of sets. This basic collection consists of: • individual points (i.e a); • open intervals (i.e. (a, b)); • half-open intervals (i.e. (a, b] or [a, b)); • and closed intervals (i.e. [a, b]). All of these could extend to infinity. The function P can be represented by a function F with the following properties:. Section 1.2. Probability in Continuous Spaces. 13. • F (−∞) = 0 • F (∞) = 1 • F (x) is monotonically increasing. and we interpret F (x) as P ((−∞, x]). The function F is referred to as the cumulative distribution function. The value of P for all the basic sets described can be extracted from F , with appropriate attention to limits; for example, P ((a, b]) = F (b) − F (a) and P (a) = lim←0+ (F (a) − F (a − )). Notice that if F is continuous, P (a) = 0. Higher Dimensional Spaces. In Rn , events are unions of elements of a basic collection of sets, too. This basic collection consists of a product of n elements from the basic collection for the real line. A cumulative distribution function can be defined in this case, too. It is given by a function F with the property that P ({x1 ≤ u1 , x2 ≤ u2 , . . . xn ≤ un }) = F (u). This function is constrained by other properties, too. However, cumulative distribution functions are a somewhat unwieldy way to specify probability. 1.2.3 Representing P-functions with Probability Density Functions For the examples we will deal with in continuous spaces, the usual way to specify P is to provide a function p such that P (event) = p(u)du event This function is referred to as a probability density function. Not every probability model admits a density function, but all our cases will. Note that a density function cannot have a negative value, but that its value could be larger than one. In all cases, probability density functions integrate to one, i.e. p(u)du = 1 P (D) = D. and any non-negative function with this property is a probability density function. The value of the probability density function at a point represents the probability of the event that consists of an infinitesimal neighbourhood at that value, i.e.: p(u1 )du = P ({u ∈ [u1 , u1 + du]}) Notice that this means that (unless we are willing to be rather open minded about what constitutes a function), for a probability model on a continuous space that can be represented using a probability density, the probability of an event that consists of a finite union of points must be zero. For the examples we will deal with, this doesn’t create any issues. In fact, it is intuitive, in the sense that we don’t expect to be able to observe the event that, say, a noise voltage has value 1; instead, we can observe the event that it lies in some tiny interval — defined by the accuracy of our measuring equipment — about 1. Conditional probability, independence and conditional independence are ideas that can be translated into properties of probability density functions. In their most useful form, they are properties of random variables.. 14. Chapter 1. An Introduction to Probability. 1.3 RANDOM VARIABLES Assume that we have a probability model on either a discrete or a continuous domain, {D, F , P }. Now let us consider a function of the outcome of an experiment. The values that this function takes on the different elements of D form a new set, which we shall call D . There is a structure, with the same formal properties as F on D defined by the values that this function takes on different elements of F — call this structure F . This function is known as a random variable. We can talk about the probability that a random variable takes a particular set of values, because the probability structure carries over. In particular, assume that we have a random variable ξ. If A ∈ F , there is some A ∈ F such that A = ξ(A). This means that P ({ξ ∈ A }) = P (A). EXAMPLE 1.12. Assorted examples of random variables. The simplest random variable is given by the identity function — this means that D is the same as D, and F is the same as F . For example, the outcome of a coin flip is a random variable. Now gamble on the outcome of a coin flip: if it comes up heads, you get a dollar, and if it comes up tails, you pay a dollar. Your income from this gamble is a random variable. In particular, D = {1, −1} and F = {∅, D , {1}, {−1}}. Now gamble on the outcome of two coin flips: if both coins come up the same, you get a dollar, and if they come up different, you pay a dollar. Your income from this gamble is a random variable. Again, D = {1, −1} and F = {∅, D , {1}, {−1}}. In this case, D is not the same as D and F is not the same as F ; however, we can still speak about the probability of getting a dollar — which is the same as P ({hh, tt}). Density functions are very useful for specifying the probability model for the value of a random variable. However, they do result in quite curious notations (probability is a topic that seems to encourage creative use of notation). It is common to write the density function for a random variable as p. Thus, the distribution for λ would be written as p(λ) — in this case, the name of the variable tells you what function is being referred to, rather than the name of the function, which is always p. Some authors resist this convention, but its use is pretty much universal in the vision literature, which is why we adopt it. For similar reasons, we write the probability function for a set of events as P , so that the probability of an event P (event) (despite the fact that different sets of events may have very different probability functions). 1.3.1 Conditional Probability and Independence Conditional probability is a very useful idea for random variables. Assume we have two random variables, m and n — (for example, the value I read from my rain gauge as m and the value I read on the neighbour’s as n). Generally, the probability. Section 1.3. Random Variables. 15. density function is a function of both variables, p(m, n). Now p(m1 , n1 )dmdn = P ({m ∈ [m1 , m1 + dm]} and {n ∈ [n1 , n1 + dm]}) = P ({m ∈ [m1 , m1 + dm]} | {n ∈ [n1 , n1 + dm]})P ({n ∈ [n1 , n1 + dm]}) We can define a conditional probability density from this by p(m1 , n1 )dmdn = P ({m ∈ [m1 , m1 + dm]} | {n ∈ [n1 , n1 + dm]})P ({n ∈ [n1 , n1 + dm]}) = (p(m1 |n1 )dm)(p(n1 )dn) Note that this conditional probability density has the expected property, that p(m|n) =. p(m, n) p(n). Independence and conditional independence carry over to random variables and probability densities without fuss. EXAMPLE 1.13. Independence in random variables associated with two coins.. We now consider the probability that each of two different coins comes up heads. In this case, we have two random variables, being the probability that the first coin comes up heads and the probability that the second coin comes up heads (it’s quite important to understand why these are random variables — if you’re not sure, look back at the definition). We shall write these random variables as p1 and p2 . Now the density function for these random variables is p(p1 , p2 ). Let us assume that there is no dependency between these coins, so we should be able to write p(p1 , p2 ) = p(p1 )p(p2 ). Notice that the notation is particularly confusing here; the intended meaning is that p(p1 , p2 ) factors, but that the factors are not necessarily equal. In this case, a further reasonable modelling step is to assume that p(p1 ) is the same function as p(p2 ) (perhaps they came from the same minting machine).. 1.3.2 Expectations The expected value or expectation of a random variable (or of some function of the random variable) is obtained by multiplying each value by its probability and summing the results — or, in the case of a continuous random variable, by multiplying by the probability density function and integrating. The operation is known as taking an expectation. For a discrete random variable, x, taking the expectation of x yields: E[x] = xi p(xi ) i∈values For a continuous random variable, the process yields E[x] = xp(x)dx D. 16. Chapter 1. An Introduction to Probability. often referred to as the average, or the mean in polite circles. One model for an expectation is to consider the random variable as a payoff, and regard the expectation as the average reward, per bet, for an infinite number of repeated bets. The expectation of a general function g(x) of a random variable x is written as E[g(x)]. The variance of a random variable x is var(x) = E[x2 − (E(x))2 ] This expectation measures the average deviance from the mean. The variance of a random variable gives quite a strong indication of how common it is to see a value that is significantly different from the mean value. In particular, we have the following useful fact: var(x) P ({| x − E[x] |≥ }) ≤ 2 The standard deviation is obtained from the variance:. sd(x) = var(x) = E[x2 − (E[x])2] For a vector of random variables, the covariance is cov(x) = E[xxt − (E[x]E[x]t )] This matrix (look carefully at the transpose) is symmetric. Diagonal entries are the variance of components of x, and must be non-negative. Off-diagonal elements measure the extent to which two variables co-vary. For independent variables, the covariance must be zero. For two random variables that generally have different signs, the covariance can be negative. EXAMPLE 1.14. The expected value of gambling on a coin flip.. You and an acquaintance decide to bet on the outcome of a coin flip. You will receive a dollar from your acquaintance if the coin comes up heads, and pay one if it comes up tails. The coin is symmetric. This means the expected value of the payoff is 1P (heads) − 1P (tails) = 0 The variance of the payoff is one, as is the standard deviation. Now consider the probability of obtaining 10 dollars in 10 coin flips, with a fair coin. Our random variable x is the income in 10 coin flips. Equation 1.3.2 yields 1 P ({| x |≥ 10}) ≤ 100 , which is a generous upper bound — the actual probability is of the order of one in a thousand. Expectations of functions of random variables are extremely useful. The notation for expectations can be a bit confusing, because it is common to omit the density with respect to which the expectation is being taken, which is usually obvious from the context. For example, E[x2 ] is interpreted as x2 p(x)dx D. Section 1.3. Random Variables. 17. 1.3.3 Joint Distributions and Marginalization Assume we have a model describing the behaviour of a collection of random variables. We will proceed on the assumption that they are discrete, but (as should be clear by now) the discussion will work for continuous variables if summing is replaced by integration. One way to specify this model is to give the probability distribution for all variables, known in jargon as the joint probability distribution function — for concreteness, write this as P (x1 , x2 , . . . xn ). If the probability distribution is represented by its density function, the density function is usually referred to as the joint probability density function. Both terms are often abbreviated as “joint.” EXAMPLE 1.15. Marginalising out parameters for two different types of coin.. Let us assume we have a coin which could be from one of two types; the first type of coin is evenly balanced; the other is wildly unbalanced. We flip our coin some number of times, observe the results, and should like to know what type of coin we have. Assume that we flip the coin once. The set of outcomes is D = {(heads, I), (heads, II), (tails, I), (tails, II)} An appropriate event space is: ∅, {(heads, I)}, {(tails, I)},. {(tails, I), (heads, I)} , {(heads, II), (tails, I), (tails, II)}, {(heads, I), (heads, II), (tails, II)}. D, {(heads, II)}, {(tails, II)}, {(tails, I), (tails, II), } , {(tails, II), (heads, II)} , {(heads, I), (tails, I), (tails, II)} {(heads, I), (heads, II), (tails, I)}. . 18. Chapter 1. An Introduction to Probability. any other (smaller) set of values too by the same process. You should now look at example 15, which illustrates how the process works using the event structure for a simple case. In fact, the event structure is getting unwieldy as a notation. It is quite common to use a rather sketchy notation to indicate the appropriate event. For example 15, we would write P ({(heads, I), (heads, II)}) = P (heads) We would like to form P ({x2 = b, . . . xn = n}) from P ({x1 = a, x2 = b, . . . xn = n}). By using the argument about event structures in example 15, we obtain P (x1 = v, x2 = b, . . . xn = n) P (x2 = b, . . . xn = n) = v∈values of x1 which we could write as P (x2 , . . . xn ) =. P (x1 , x2 , . . . xn ). values of x1 This operation is referred to as marginalisation. marginalisation A similar argument applies to probability density functions, but the operation is now integration. Given a probability density function p(x1 , x2, . . . , xn ), we obtain p(x2 , . . . xn ) = p(x1 , x2 , . . . xn )dx1 D. marginalisation 1.4 STANDARD DISTRIBUTIONS AND DENSITIES There are a variety of standard distributions that arise regularly in practice. References such as [Patel et al., 1976; Evans et al., 2000] give large numbers; we will discuss only the most important cases. The uniform distribution has the same value at each point on the domain. This distribution is often used to express an unwillingness to make a choice or a lack of information. On a continuous space, the uniform distribution has a density function that has the same value at each point. Notice that a uniform density on an infinite continuous domain isn’t meaningful, because it could not be scaled to integrate to one. In practice, one can often avoid this point, either by pretending that the value is a very small constant and arranging for it to cancel, or using a normal distribution (described below) with a really big covariance, such that its value doesn’t change much over the region of interest. The binomial distribution applies to situations where one has independent identically distributed samples from a distribution with two values. For example, consider drawing n balls from an urn containing equal numbers of black and white balls. Each time a ball is drawn, its colour is recorded and it is replaced, so that the probability of getting a white ball — which we denote p — is the same for each draw. The binomial distribution gives the probability of getting k white balls. n pk (1 − p)n−k k. Section 1.4.
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Standard Distributions and Densities. 19. The mean of this distribution is np and the variance is np(1 − p). The Poisson distribution applies to spatial models that have uniformity properties. Assume that points are placed on the real line randomly in such a way that the expected number of points in an interval is proportional to the length of the interval. The number of points in a unit interval will have a Poisson distribution where λx e−x P ({N = x}) = x! (where x = 0, 1, 2 . . . and λ > 0 is the constant of proportionality). The mean of this distribution is λ and the variance is λ 1.4.1 The Normal Distribution The probability density function for the normal distribution for a single random variable x is 1 (x − µ)2 √ p(x; µ, σ) = exp − 2σ 2 2πσ The mean of this distribution is µ and the standard deviation is σ. This distribution is widely called a Gaussian distribution in the vision community. The multivariate normal distribution for d-dimensional vectors x has probability density function p(x; µ, Σ) =. 1 d. (2π) 2 det(Σ)1/2. exp −. (x − µ)T Σ−1 (x − µ) 2. . The mean of this distribution is µ and the covariance is Σ. Again, this distribution is widely called a Gaussian distribution in the vision community. The normal distribution is extremely important in practice, for several reasons: • The sum of a large number of random variables is normally distributed, pretty much whatever the distribution of the individual random variables. This fact is known as the central limit theorem. It is often cited as a reason to model a collection of random effects with a single normal model. • Many computations that are prohibitively hard for any other case are easy for the normal distribution. • In practice, the normal distribution appears to give a fair model of some kinds of noise. • Many probability density functions have a single peak and then die off; a model for such distributions can be obtained by taking a Taylor series of the log of the density at the peak. The resulting model is a normal distribution (which is often quite a good model).. 20. Chapter 1. An Introduction to Probability. 1.5 PROBABILISTIC INFERENCE Very often, we have a sequence of observations produced by some process whose mechanics we understand, but which has some underlying parameters that we do not know. The problem is to make useful statements about these parameters. For example, we might observe the intensities in an image, which are produced by the interaction of light and surfaces by principles we understand; what we don’t know — and would like to know — are such matters as the shape of the surface, the reflectance of the surface, the intensity of the illuminant, etc. Obtaining some representation of the parameters from the data set is known as inference. There is no canonical inference scheme; instead, we need to choose some principle that identifies the most desirable set of parameters. 1.5.1 The Maximum Likelihood Principle A general inference strategy known as maximum likelihood inference, can be described as Choose the world parameters that maximise the probability of the measurement observed In the general case, we are choosing arg max P (measurements|parameters) (where the maximum is only over the world parameters because the measurements are known, and arg max means “the argument that maximises”). In many problems, it is quite easy to specify the measurements that will result from a particular setting of model parameters — this means that P (measurements|parameters), often referred to as the likelihood, is easy to obtain. This can make maximum likelihood estimation attractive. EXAMPLE 1.16 haviour.. Maximum likelihood inference on the type of a coin from its be-. We return to example 15. Now assume that we know some conditional probabilities. In particular, the unbiased coin has P (heads|I) = P (tails|I) = 0.5, and the biased coin has P (tails|II) = 0.2 and P (heads|II) = 0.8. We observe a series of flips of a single coin, and wish to know what type of coin we are dealing with. One strategy for choosing the type of coin represented by our evidence is to choose either I or II, depending on whether P (flips observed|I) > P (flips observed|II). For example, if we observe four heads and one tail in sequence, then P (hhhht|II) = (0.8)4 0.2 = 0.08192 and P (hhhht|I) = 0.03125, and we choose type II. Maximum likelihood is often an attractive strategy, because it can admit quite simple computation. A classical application of maximum likelihood estimation involves estimating the parameters of a normal distribution from a set of samples of that distribution (example 17).. Section 1.5. Probabilistic Inference. 21. EXAMPLE 1.17 Estimating the parameters of a normal distribution from a series of independent samples from that distribution. Assume that we have a set of n samples — the i’th of which is xi — that are known to be independent and to have been drawn from the same normal distribution. The likelihood of our sample is P (sample|µ, σ) = L(x1 , . . . xn ; µ, σ). 1 (xi − µ)2 √ exp − p(xi ; µ, σ) = = 2σ 2 2πσ i i Working with the log of the likelihood will remove the exponential, and not change the position of the maximum. For the log-likelihood, we have Q(x1 , . . . xn ; µ, σ) = −. (xi − µ)2 2σ 2. i. 1 1 − n( log 2 + log π + log σ) 2 2. and we want the maximum with respect to µ and σ. This must occur when the derivatives are zero, so we have (xi − µ) ∂Q =0 =2 ∂µ 2σ 2 i. and a little shuffling of expressions shows that this maximum occurs at xi µ= i n Similarly ∂Q = ∂σ and this maximum occurs at. . σ=. i (xi. − µ)2. σ3. . i (xi. n =0 σ. − µ)2. n. Note that this estimate of σ is biased, inthat its expected value is σ(n/(n − 1)) 2 and it is more usual to use (1/(n − 1)) i (xi − µ) as an estimate. 1.5.2 Priors, Posteriors and Bayes’ rule In example 16, our maximum likelihood estimate incorporates no information about P (I) or P (II) — which can be interpreted as how often coins of type I or type II are handed out, or as our subjective degree of belief that we have a coin of type I or of type II before we flipped the coin. This is unfortunate, to say the least; for example, if coins of type II are rare, we would want to see an awful lot of heads before it would make sense to infer that our coin is of this type. Some quite simple algebra suggests a solution.. 22. Chapter 1. An Introduction to Probability. Recall that P (A, B) = P (A|B)P (B). This simple observation gives rise to an innocuous looking identity for reversing the order in a conditional probability: P (B|A) =. P (A|B)P (B) P (A). This is widely referred to as Bayes’ theorem or Bayes’ rule. Now the interesting property of Bayes’ rule is that it tells us which choice of parameters is most probable, given our model and our prior beliefs. Rewriting Bayes’ rule gives P (parameters|data) =. P (data|parameters)P (parameters) P (data). The term P (parameters) is referred to as the prior (it describes our knowledge of the world before measurements have been taken). The term P (parameters|data) is usually referred to as the posterior (it describes the probability of various models after measurements have been taken). P (data) can be computed by marginalisation (which requires computing a high dimensional integral, often a nasty business) or for some problems can be ignored. As we shall see in following sections, attempting to use Bayes’ rule can result in difficult computations — that integral being one — because posterior distributions often take quite unwieldy forms. 1.5.3 Bayesian Inference The Bayesian philosophy is that all information about the world is captured by the posterior. The first reason to accept this view is that the posterior is a principled combination of prior information about the world and a model of the process by which measurements are generated — i.e. there is no information missing from the posterior, and the information that is there, is combined in a proper manner. The second reason is that the approach appears to produce very good results. The great difficulty is that computing with posteriors can be very difficult — we will encounter various mechanisms for computing with posteriors in following sections. For example, we could use the study of physics in the last few chapters to get expressions relating pixel values to the position and intensity of light sources, the reflectance and orientation of surfaces, etc. Similarly, we are likely to have some beliefs about the parameters that have nothing to do with the particular values of the measurements that we observe. We know that albedos are never outside the range [0, 1]; we expect that illuminants with extremely high exitance are uncommon; and we expect that no particular surface orientation is more common than any other. This means that we can usually cobble up a reasonable choice of prior. MAP Inference. An alternative to maximum likelihood inference is to infer a state of the world that maximises the posterior: Choose the world parameters that maximise the conditional probability of the parameters, conditioned on the measurements taking the observed values. Section 1.5. Probabilistic Inference. 23. This approach is known as maximum a posteriori (or MAP) reasoning. EXAMPLE 1.18. Determining the type of a coin using MAP inference.. Assume that we have three flips of the coin of example 16, and would like to determine whether it has type I or type II. We know that the mint has 3 machines that produce type I coins and 1 machine that produces type II coins, and there is no reason to believe that these machines run at different rates. We therefore assign P (I) = 0.75 and P (II) = 0.25. Now we observe three heads, in three consecutive flips. The value of the posterior for type I is: P (hhh|I)P (I) P (hhh) P (h|I)3 P (I) = P (hhh, I) + P (hhh, II) P (h|I)3 P (I) = P (hhh|I)P (I) + P (hhh|II)P (II) 0.530.75 = 0.530.75 + 0.830.25 = 0.422773. P (I|hhh) =. By a similar argument, the value of the posterior for type II is 0.577227. An MAP inference procedure would conclude the coin is of type II. The denominator in the expression for the posterior can be quite difficult to compute, because it requires a sum over what is potentially a very large number of elements (imagine what would happen if there were many different types of coin). However, knowing this term is not crucial if we wish to isolate the element with the maximum value of the posterior, because it is a constant. Of course, if there are a very large number of events in the discrete space, finding the world parameters that maximise the posterior can be quite tricky. The Posterior as an Inference. EXAMPLE 1.19 Determining the probability a coin comes up heads from the outcome of a sequence of flips. Assume we have a coin which comes from a mint which has a continuous control parameter, λ, which lies in the range [0, 1]. This parameter gives the probability that the coin comes up heads, so P (heads|λ) = λ. We know no reason to prefer any one value of λ to any other, so as a prior probability distribution for λ we use the uniform distribution so p(λ) = 1. Assume we flip the coin twice, and observe heads twice; what do we know about λ? All our knowledge is captured by the posterior, which is P (λ ∈ [x, x + dx]|hh) dx. 24. Chapter 1. An Introduction to Probability. we shall write this expression as p(λ|hh). We have p(hh|λ)p(λ) p(hh) p(hh|λ)p(λ) = 1 p(hh|λ)p(λ)dλ 0. p(λ|hh) =. = 1 0. λ2 p(λ) p(hh|λ)p(λ)dλ. 2. = 3λ. It is fairly easy to see that if we flip the coin n times, and observe k heads and n − k tails, we have p(λ|k heads and n − k tails) ∝ λk (1 − λ)n−k. Section 1.5. Probabilistic Inference. 25. 35. 5. 4.5 36 flips. 30. 36 flips 4. 25. 3.5 18 flips 3. 20. 12 flips 2.5. 18 flips 15. 6 flips. 2. 12 flips. 4 flips 1.5. 10. 2 flips Prior. 1. 6 flips 4 flips. 5. 0.5. 2 flips Prior. 0. 0. 0.1. 0.2. 0.3. 0.4. 0.5. 0.6. 0.7. 0.8. 0.9. 1. 0. 0. 0.1. 0.2. 0.3. 0.4. 0.5. 0.6. 0.7. 0.8. 0.9. 1. FIGURE 1.2: On the left, the value of the posterior density for the probability that a coin will come up heads, given an equal number of heads and tails are observed. This posterior is shown for different numbers of observations. With no evidence, the posterior is the prior; but as the quantity of evidence builds up, the posterior becomes strongly peaked — this is because one is very unlikely to observe a long sequence of coin flips where the frequency of heads is very different from the probability of obtaining a head. On the right, a similar plot, but now for the case where every flip comes up heads. As the number of flips builds up, the posterior starts to become strongly peaked near one. This overwhelming of the prior by evidence is a common phenomenon in Bayesian inference.. 26. Chapter 1. An Introduction to Probability. 3.5. 6. 2.5. 3. 5 2. 2.5. 4 1.5 2. 3 1.5 1. 2 1 0.5. 1. 0.5. 0. 0. 0.1. 0.2. 0.3. 0.4. 0.5. 0.6. 0.7. 0.8. 0.9. 1. 0. 0. 0.1. 0.2. 0.3. 0.4. 0.5. 0.6. 0.7. 0.8. 0.9. 1. 0. 0. 0.1. 0.2. 0.3. 0.4. 0.5. 0.6. 0.7. 0.8. 0.9. FIGURE 1.3: On the left, the posterior probability density for the probability a coin comes up heads, given a single flip that shows a head and a somewhat untruthful informant who says high, as in example 20. In the center, a posterior probability density for the same problem, but now assuming that we have seen two tails and the informant says high (a sketch of the formulation appears in example 21). On the right, a posterior probability density for the case when the coin shows five tails and the informant says high. As the number of tails builds up, the weight of the posterior in the high region goes down, strongly suggesting the informant is lying.. EXAMPLE 1.21 Determining the type of a coin from a sequence of flips, incorporating information from an occasionally untruthful informant — II. Now consider what happens in example 20 if the contact says high and we see two tails. We need p(high, tt|λ)p(λ) p(high, tt) ∝ p(high, tt|λ)p(λ). p(λ|high, tt) =. Now p(high, tt|λ) is p(high, tt|λ) = p(high, tt|λ, truth = 1)P (truth = 1) + p(high, tt|λ, truth = 0)P (truth = 0) = p(high, tt|λ, truth = 1)P (truth = 1) + p(tt|λ, truth = 0)p(high|λ, truth = 0)P (truth = 0) Now p(tt|λ, truth = 0) = (1 − λ)2 and the interesting term is p(high, tt|λ, truth = 1). Again, this term reflects the behaviour of the coin and the informant when the informant is telling the truth; in particular, this term must be zero for λ ∈ [0, 2/3), because in this case λ is not high. For λ in the high range, this term must be (1−λ)2 , because now it is the probability of getting two tails with two flips. Performing the computation, we obtain the posterior graphed in figure 1.3.. Bayesian Model Selection. The crucial virtue of Bayesian inference is the accounting for uncertainty shown in examples 20 and 21. We have been able to account for an occasionally. 1. Section 1.5. Probabilistic Inference. 27. untruthful informant and a random measurement; when there was relatively little contradictory evidence from the coin’s behaviour, our process placed substantial weight on the informant’s testimony, but when the coin disagreed, the informant was discounted. This behaviour is highly attractive, because we are able to combine uncertain sources of information with confidence. EXAMPLE 1.22. Is the informant lying?. We now need to know whether our informant lied to us. Assume we see a single head and an informant saying high, again. The relevant posterior is: P (head, high|truth=0)P (truth=0) P (head, high) P (λ, head, high|truth=0)P (truth=0)dλ = P (head, high) P (head, high|λ, truth=0)P (λ)P (truth=0)dλ = P (head, high) 1 = P (head,high|λ,truth=1)P (λ)dλP (truth=1 ) 1+ P (head,high|λ,truth=0)P (λ)dλP (truth=0 ). Example 22 shows how to tell whether the informant of examples 20 and 21 is telling the truth or not, given the observations. A useful way to think about this example is to regard it as comparing two models (as opposed to the value of a binary parameter within one model). One model has a lying informant, and the other has a truthful informant. The posteriors computed in this example compare how well different models explain a given data set, given a prior on the models. This is a very general problem — usually called model selection — with a wide variety of applications in vision: • Recognition: Assume we have a region in an image, and an hypothesis that an object might be present in that region at a particular position and orientation (the hypothesis will have been obtained using methods from chapter ??, which aren’t immediately relevant). Is there an object there or not? A principled answer involves computing the posterior over two models — that the data was obtained from noise, or from the presence of an object. • Are these the same? Assume we have a set of pictures of surfaces we want to compare. For example, we might want to know if they are the same colour, which would be difficult to answer directly if we didn’t know the illuminant. A principled answer involves computing the posterior over two models — that the data was obtained from one surface, or from two (or more). • What camera was used? Assume we have a sequence of pictures of a world. With a certain amount of work, it is usually possible to infer a great deal of information about the shape of the objects from such a sequence (e.g.. 28. Chapter 1. An Introduction to Probability. chapters ??, ?? and ??). The algorithms involved differ quite sharply, depending on the camera model adopted (i.e. perspective, orthographic, etc.). Furthermore, adopting the wrong camera model tends to lead to poor inferences. Determining the right camera model to use is quite clearly a model selection problem. • How many segments are there? We would like to break an image into coherent components, each of which is generated by a probabilistic model. How many components should there be? (section ??). The solution is so absurdly simple in principle (in practice, the computations can be quite nasty) that it is easy to expect something more complex, and miss it. We will write out Bayes’ rule specialised to this case to avoid this: P (data|model) P (data) P (data|model, parameters)P (parameters)d{parameters} = P (data) ∝ P (data|model, parameters)P (parameters)d{parameters}. P (model|data) =. which is exactly the form used in the example. Notice that we are engaging in Bayesian inference here, too, and so can report the MAP solution or report the whole posterior. The latter can be quite helpful when it is difficult to distinguish between models. For example, in the case of the dodgy informant, if P (truth=0|data) = 0.5001, it may be undesirable to conclude the informant is lying — or at least, to take drastic action based on this conclusion. The integral is potentially rather nasty, which means that the method can be quite difficult to use in practice. Useful references include [Gelman et al., 1995; Carlin and Louis, 1996; Gamerman, 1997; Newman and Barkema, 1998; Evans and Swartz, 2000]. 1.5.4 Open Issues In the rest of the book, we will have regular encounters with practical aspects of the Bayesian philosphy. Firstly, although the posterior encapsulates all information available about the world, we very often need to make discrete decisions — should we shoot it or not? Typically, this decision making process requires some accounting for the cost of false positives and false negatives. Secondly, how do we build models? There are three basic sources of likelihood functions and priors: • Judicious design: it is possible to come up with models that are too hard to handle computationally. Generally, models on very high-dimensional domains are difficult to deal with, particularly if there is a great deal of interdependence between variables. For some models, quite good inference algorithms are known. The underlying principle of this approach is to exploit simplifications due to independence and conditional independence. • Physics: particularly in low-level vision problems, likelihood models follow quite simply from physics. It is hard to give a set of design rules for this. Section 1.5. Probabilistic Inference. 29. strategy. It has been used with some success on occasion (see, for example, [Forsyth, 1999]).. • Learning: a poor choice of model results in poor performance, and a good choice of model results in good performance. We can use this observation to tune the structure of models if we have a sufficient set of data. We describe aspects of this strategy in chapter ?? and in chapter ??.. Finally, the examples above suggest that posteriors can have a nasty functional form. This intuition is correct, and there is a body of technique that can help handle ugly posteriors which we explore as and when we need it (see also [Gelman et al., 1995; Carlin and Louis, 1996; Gamerman, 1997; Newman and Barkema, 1998]).. 30. Chapter 1. An Introduction to Probability. P (k heads and n − k tails in n flips) = Show that this is true.. n k. pk (1 − p)n−k. Section 1.6. Notes. 31. 1.5. A careless study of example 10 often results in quite muddled reasoning, of the following form: I have bet on heads successfully ten times, therefore I should bet on tails next. Explain why this muddled reasoning — which has its own name, the gambler’s fallacy in some circles, anti-chance in others — is muddled. 1.6. Confirm the count of parameters in example 8. 1.7. In example 19, what is c? 1.8. As in example 16, you are given a coin of either type I or type II; you do not know the type. You flip the coin n times, and observe k heads. You will infer the type of the coin using maximum likelihood estimation. for what values of k do you decide the coin is of type I? 1.9. Compute P (truth|high, coin behaviour) for each of the three cases of example 21. You’ll have to estimate an integral numerically. 1.10. In example 22, what is the numerical value of the probability that the informant is lying, given that the informant said high and the coin shows a single tail? What is the numerical value of the probability that the informant is lying, given that the informant said high and the coin shows seven tails in eight flips? 1.11. The random variable x = (x1 , x2 , . . . xn )T has a normal distribution. Show that the ˆ = (x2 , . . . , xn )T has a normal distribution (which is obtained by random variable x marginalizing the density). A good way to think about this problem is to consider ˆ , and reason about the behaviour of the integral; a the mean and covariance of x bad way is to storm ahead and try and do the integral. 1.12. The random variable p has a normal distribution. Furthermore, there are symmetric matrices A, B and C and vectors D and E such that P (d|p) has the form − log P (d|p) = pT Ap + pT Bd + dT Cd + pT D + dT E + C (C is the log of the normalisation constant). Show that P (p|d) is a normal distribution for any value of d. This has the great advantage that inference is relatively easy. 1.13. x is a random variable with a continuous cumulative distribution function F (x). Show that u = F (x) is a random variable with a uniform density on the range [0, 1]. Now use this fact to show that w = F −1 (u) is a random variable with cumulative distribution function F .. 32. Chapter 1. An Introduction to Probability. Topic Probability model. What you must know A space D, a collection F of subsets of that space containing (a) the empty set; (b) D; (c) all finite unions of elements of F ; (d) all complements of elements of F , and a function P such that (a) P (∅) = 0; (b) P (D) = 1; and (c) for A ∈ F and B ∈ F , P (A ∪ B) = P (A) + P (B) − P (A ∩ B). It is usual to discuss F only implicitly and to represent P by a probability density function for continuous spaces. Random vari- A function of the outcome of an experiment; supports a probability ables model. If we have a random variable ξ, mapping A → A and F → F , defined on the probability model above, and if A ∈ F , there is some A ∈ F such that A = ξ(A). This means that P ({ξ ∈ A }) = P (A). Conditional Given a probability model and a set A ⊂ D such that P (A) = 0 and probability A ∈ F , then A together with F = {C ∩ A|C ∈ F } and P such that P (C) = P (C ∩ A)/P (A) form a new probability model. P (C) is often written as P (C|A) and called the conditional probability of the event C, given that A has occurred. Probability A function p such that P {u ∈ E} = E p(x)dx. All the probability density func- models we deal with on continuous spaces will admit densities, but tion not all do. Marginalisation Given the joint probability density p(X, Y ) of two random variables X and Y , the probability of Y alone — referred to as the marginal probability density for Y — is given by. p(x, Y )dx. Expectation. Normal random variable. The domain is all possible values of X; if the random variables are discrete, the integral is replaced by a sum. The “expected value” of a random variable, computed as E[f (x)] = f (x)p(x)dx. Useful expectations include the mean E[x] and the covariance E[(x − E[x])(x − E[x])T ]. A random variable whose probability density function is the normal (or gaussian) distribution. For an n-dimensional random variable, this is p(x) =. 1 exp(−(1/2)(x − µ)T Σ−1 (x − µ)) (2π)(n/2) | Σ |. having mean µ and covariance Σ.. Chapter summary for chapter 1: Probabilistic methods manipulate representations of the “size” of sets of events. Events are defined so that unions and negations are meaningful, leading to a family of subsets of a space. The probability of a set of events is a function defined on this structure. A random variable represents the outcome of an experiment. A generative model gives the probability of a set of outcomes from some inputs; inference obtains a representation of the probable inputs that gave rise to some known outcome.
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# 3.2 solve two step equations - day 1
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### 3.2 solve two step equations - day 1
1. 1. Solving Multi-Step Equations<br />Section 3.2<br />P. 141<br />
2. 2. Today we will learn to solve linear equations using two or more transformations.<br />Again – think about “undoing” what has been done to the variable – using inverse operations!<br />
3. 3. Learn to solve linear equations using two or more transformations<br />Do these look familiar to you?<br />
4. 4. Learn to solve linear equations using two or more transformations<br />Today our equations will look more like this:<br />Why do you think these are called two-step equations?<br />
5. 5. Learn to solve linear equations using two or more transformations<br />Let’s take a closer look at this equation.<br />Focus on the variable and how the numbers are attached to the variable.<br />Attached by “what operations?”<br />If I “UNDO” this equation using inverse operations, what should I undo first? Why?<br />
6. 6. “Undo” Steps<br />1) Simplify both sides( may include the distributive property, <br />combining like terms,<br /> change subtraction to “add the opposite)<br />2) Addition/Subtraction transformations<br />3) Simplify<br />4) Multiplication/Division transformations<br />5) Simplify<br />
7. 7. Learn to solve linear equations using two or more transformations<br /> -2x + 5 = - 9<br /> -2x + 5 + (-5) = -9 + (-5) Addition<br /> -2x + 0 = - 14 Simplify<br /> -2x = -14<br />-2 -2 Division<br /> x = 7<br /> go back and check mentally – does 7 work?<br />
8. 8. Learn to solve linear equations using two or more transformations<br />2x – 9x + 17 = -4 keep original equ. clean<br /> -7x + 17 = -4 (combine like terms)<br /> -7x + 17 + (-17) = -4 + (-17) addition<br /> -7x = -21 simplify<br /> -7x = -21 division<br />-7 -7<br /> x = 3 simplify<br /> check mentally!!!<br />
9. 9. Learn to solve linear equations using two or more transformations<br />
10. 10. Learn to solve linear equations using two or more transformations<br />
11. 11. Solve<br />
12. 12. Assignment: p. 144 (#6-14, 21)<br />You must show all your steps.<br />Just answers won’t get your problem correct. <br />Don’t forget to “CHECK” each problem!<br />
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Upcoming SlideShare. ×. # 3.2 solve two step equations - day 1. 683 views. Published on. Published in: Technology. 0 Likes. Statistics. Notes. • Full Name. Comment goes here.. Are you sure you want to Yes No. • Be the first to comment. • Be the first to like this. Views. Total views. 683. On SlideShare. 0. From Embeds. 0. Number of Embeds. 17. Actions. Shares. 0. 2. 0. Likes. 0. Embeds 0. No embeds. No notes for slide. ### 3.2 solve two step equations - day 1. 1. 1. Solving Multi-Step Equations<br />Section 3.2<br />P. 141<br />.
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2. 2. Today we will learn to solve linear equations using two or more transformations.<br />Again – think about “undoing” what has been done to the variable – using inverse operations!<br />. 3. 3. Learn to solve linear equations using two or more transformations<br />Do these look familiar to you?<br />. 4. 4. Learn to solve linear equations using two or more transformations<br />Today our equations will look more like this:<br />Why do you think these are called two-step equations?<br />. 5. 5. Learn to solve linear equations using two or more transformations<br />Let’s take a closer look at this equation.<br />Focus on the variable and how the numbers are attached to the variable.<br />Attached by “what operations?”<br />If I “UNDO” this equation using inverse operations, what should I undo first? Why?<br />. 6. 6. “Undo” Steps<br />1) Simplify both sides( may include the distributive property, <br />combining like terms,<br /> change subtraction to “add the opposite)<br />2) Addition/Subtraction transformations<br />3) Simplify<br />4) Multiplication/Division transformations<br />5) Simplify<br />. 7. 7. Learn to solve linear equations using two or more transformations<br /> -2x + 5 = - 9<br /> -2x + 5 + (-5) = -9 + (-5) Addition<br /> -2x + 0 = - 14 Simplify<br /> -2x = -14<br />-2 -2 Division<br /> x = 7<br /> go back and check mentally – does 7 work?<br />. 8. 8. Learn to solve linear equations using two or more transformations<br />2x – 9x + 17 = -4 keep original equ. clean<br /> -7x + 17 = -4 (combine like terms)<br /> -7x + 17 + (-17) = -4 + (-17) addition<br /> -7x = -21 simplify<br /> -7x = -21 division<br />-7 -7<br /> x = 3 simplify<br /> check mentally!!!<br />. 9. 9. Learn to solve linear equations using two or more transformations<br />. 10. 10. Learn to solve linear equations using two or more transformations<br />. 11. 11. Solve<br />. 12. 12. Assignment: p. 144 (#6-14, 21)<br />You must show all your steps.<br />Just answers won’t get your problem correct. <br />Don’t forget to “CHECK” each problem!<br />.
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# Homework Help: Circuits DC Steady State
1. Nov 11, 2015
### jdawg
1. The problem statement, all variables and given/known data
DC steady state circuit analysis: Consider the circuit shown below. The switch shown in the problem opens at time t = 0 seconds. If the circuit is in DC steady state at , find values for iL(0-), iL(0+), i(0-), i+, Vc(0-), Vc(0+). There is no RL or RC analysis to be performed on this problem. On your answer for the current, a positive value indicates current is flowing in the direction shown. A negative value for current will indicate that the current is flowing opposite the direction shown.
2. Relevant equations
3. The attempt at a solution
Ok so I know that iL(0-)=iL(0+) because current is constant through an inductor. And I know that Vc(0-)=Vc(0+) because voltage across a conductor is constant.
I feel like I don't know some of the basic concepts I need to solve this. Is Is(t)=0 since the switch is up? Is the 26v voltage source the same as the voltage across the 10 ohm resistor? Vc is supposed to equal 6 volts and iL and i are supposed to be 1.5 amps. I can't quite figure it out :(
#### Attached Files:
• ###### 06-#7.pdf
File size:
272.6 KB
Views:
102
2. Nov 11, 2015
### JBA
First, what is Vc at the time the switch opens; and then, what part of the circuit can you remove (i.e. has no current) after the switch is opened.
3. Nov 11, 2015
### jdawg
Does Vc=0 when it opens? Then you can remove the part of the circuit to the far right where Vc is?
4. Nov 11, 2015
### JBA
No that is not correct, what happens when a voltage is applied to a capacitor?
5. Nov 11, 2015
### jdawg
The voltage doesn't instantaneously change right? It builds up over time? Sorry, I'm not sure if thats what you were trying to get me to say.
6. Nov 11, 2015
### JBA
The problem states that there is no induction effects to be considered but that is irrelevant for your problem. Once the voltage source is applied to the circuit it will charge the capacitor to the voltage source value; in this case, 26 volts, and the capacitor voltage will remain at that level once the voltage source is removed until some event provides a circuit path through which the capacitor can discharge. Knowing that the voltage source can only be effective as long as both of its poles are connected what is then the result (the part of the circuit that can be removed) once the switch is opened?
7. Nov 11, 2015
### jdawg
The branch with the 10 ohm resistor?
8. Nov 11, 2015
### JBA
That is correct, now with the capacitor as the voltage source (remember current flows from + to - ) and to the ground where are there going to be currents in the remaining circuit and what direction will they be?
9. Nov 11, 2015
### jdawg
i(t) should be going down through the 4 ohm resistor and the current iL(t) should be going the opposite direction shown in the picture. Is there only one current now flowing through the loop with the 12 and 4 ohm resistors?
10. Nov 11, 2015
### JBA
You are correct, the flow will be reversed through L, and the capacitor will discharge from + to - through both resistors the same direction as shown for the 4 ohm resistor but at different amperages.
11. Nov 11, 2015
### jdawg
Ok awesome! So how do you actually start calculating the currents and Vc?
12. Nov 11, 2015
### Staff: Mentor
No, the capacitor will not charge to 26V. There are resistors forming a voltage divider between the source and the capacitor. In particular the 10 Ω resistor in conjunction with the 12 Ω and 4 Ω resistors in parallel form a voltage divider that supplies the capacitor.
To find the steady state voltage on the capacitor (i.e. prior to the switch opening and the instant after it does) treat the capacitor as an open circuit and the inductor as a short circuit, then analyze the remaining circuit. At steady state the capacitor voltage will be the same as that across the 4 Ω resistor. The current through the inductor will be the same as that through the 4 Ω resistor. jdawg's second circuit diagram depicts the situation correctly.
Immediately after the switch opens the capacitor has the same voltage as the instant before, and the inductor the same current. The inductor's current direction will not change! It will maintain the same current in the same direction in the instant after the switch opens. The problem states that RL and RC effects are not to be considered, which means we don't need to look at the time varying voltages and currents after t = 0+. It does not mean we ignore the constant current engendered by the inductor or the constant voltage engendered by the capacitor at t = 0+.
13. Nov 11, 2015
### jdawg
Ok that makes more sense. I tried doing KVL around the 12 ohm and 4 ohm resistor loop and everything just kind of blew up :( You can't use ohm's law, can you? I feel like there are too many variables... How can you find Vc without finding the current i(t) first or vice versa??
14. Nov 11, 2015
### Staff: Mentor
Personally I'd opt for nodal analysis as there's only one essential node in the circuit in question. That would yield the potential at the top of the resistors, which also happens to be Vc. Then the current through the 4 Ω resistor would be a simple matter of Ohm's Law.
15. Nov 11, 2015
### jdawg
THANK YOU SO MUCH!! That problem was driving me crazy!
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# Homework Help: Circuits DC Steady State. 1. Nov 11, 2015. ### jdawg. 1. The problem statement, all variables and given/known data. DC steady state circuit analysis: Consider the circuit shown below. The switch shown in the problem opens at time t = 0 seconds. If the circuit is in DC steady state at , find values for iL(0-), iL(0+), i(0-), i+, Vc(0-), Vc(0+). There is no RL or RC analysis to be performed on this problem. On your answer for the current, a positive value indicates current is flowing in the direction shown. A negative value for current will indicate that the current is flowing opposite the direction shown.. 2. Relevant equations. 3. The attempt at a solution. Ok so I know that iL(0-)=iL(0+) because current is constant through an inductor. And I know that Vc(0-)=Vc(0+) because voltage across a conductor is constant.. I feel like I don't know some of the basic concepts I need to solve this. Is Is(t)=0 since the switch is up? Is the 26v voltage source the same as the voltage across the 10 ohm resistor? Vc is supposed to equal 6 volts and iL and i are supposed to be 1.5 amps. I can't quite figure it out :(. #### Attached Files:. • ###### 06-#7.pdf. File size:. 272.6 KB. Views:. 102. 2. Nov 11, 2015. ### JBA. First, what is Vc at the time the switch opens; and then, what part of the circuit can you remove (i.e. has no current) after the switch is opened.. 3. Nov 11, 2015. ### jdawg. Does Vc=0 when it opens? Then you can remove the part of the circuit to the far right where Vc is?. 4. Nov 11, 2015. ### JBA. No that is not correct, what happens when a voltage is applied to a capacitor?. 5. Nov 11, 2015. ### jdawg. The voltage doesn't instantaneously change right? It builds up over time? Sorry, I'm not sure if thats what you were trying to get me to say.. 6. Nov 11, 2015. ### JBA. The problem states that there is no induction effects to be considered but that is irrelevant for your problem. Once the voltage source is applied to the circuit it will charge the capacitor to the voltage source value; in this case, 26 volts, and the capacitor voltage will remain at that level once the voltage source is removed until some event provides a circuit path through which the capacitor can discharge. Knowing that the voltage source can only be effective as long as both of its poles are connected what is then the result (the part of the circuit that can be removed) once the switch is opened?. 7.
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Nov 11, 2015. ### jdawg. The branch with the 10 ohm resistor?. 8. Nov 11, 2015. ### JBA. That is correct, now with the capacitor as the voltage source (remember current flows from + to - ) and to the ground where are there going to be currents in the remaining circuit and what direction will they be?. 9. Nov 11, 2015. ### jdawg. i(t) should be going down through the 4 ohm resistor and the current iL(t) should be going the opposite direction shown in the picture. Is there only one current now flowing through the loop with the 12 and 4 ohm resistors?. 10. Nov 11, 2015. ### JBA. You are correct, the flow will be reversed through L, and the capacitor will discharge from + to - through both resistors the same direction as shown for the 4 ohm resistor but at different amperages.. 11. Nov 11, 2015. ### jdawg. Ok awesome! So how do you actually start calculating the currents and Vc?. 12. Nov 11, 2015. ### Staff: Mentor. No, the capacitor will not charge to 26V. There are resistors forming a voltage divider between the source and the capacitor. In particular the 10 Ω resistor in conjunction with the 12 Ω and 4 Ω resistors in parallel form a voltage divider that supplies the capacitor.. To find the steady state voltage on the capacitor (i.e. prior to the switch opening and the instant after it does) treat the capacitor as an open circuit and the inductor as a short circuit, then analyze the remaining circuit. At steady state the capacitor voltage will be the same as that across the 4 Ω resistor. The current through the inductor will be the same as that through the 4 Ω resistor. jdawg's second circuit diagram depicts the situation correctly.. Immediately after the switch opens the capacitor has the same voltage as the instant before, and the inductor the same current. The inductor's current direction will not change! It will maintain the same current in the same direction in the instant after the switch opens. The problem states that RL and RC effects are not to be considered, which means we don't need to look at the time varying voltages and currents after t = 0+. It does not mean we ignore the constant current engendered by the inductor or the constant voltage engendered by the capacitor at t = 0+.. 13. Nov 11, 2015. ### jdawg. Ok that makes more sense. I tried doing KVL around the 12 ohm and 4 ohm resistor loop and everything just kind of blew up :( You can't use ohm's law, can you? I feel like there are too many variables... How can you find Vc without finding the current i(t) first or vice versa??. 14. Nov 11, 2015. ### Staff: Mentor. Personally I'd opt for nodal analysis as there's only one essential node in the circuit in question. That would yield the potential at the top of the resistors, which also happens to be Vc. Then the current through the 4 Ω resistor would be a simple matter of Ohm's Law.. 15. Nov 11, 2015. ### jdawg. THANK YOU SO MUCH!! That problem was driving me crazy!.
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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.3 - Problem 6e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.3 - Problem 6e
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# Describe the possible end behavior of a polynomial.
ISBN: 9780321570567 2
## Solution for problem 6E Chapter 4.3
Calculus: Early Transcendentals | 1st Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants
Calculus: Early Transcendentals | 1st Edition
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3
Problem 6E
Describe the possible end behavior of a polynomial.
Step-by-Step Solution:
Step 1 of 3
Solution: Step1 The end behavior of a polynomial function is the behavior of the graph of f(x)f(x) as xx approaches positive infinity or negative infinity. The degree and the leading coefficient of a polynomial function determine the end behavior of the graph. The leading coefficient is significant compared to the other coefficients in the function for the very large or very small numbers. So, the sign of the leading coefficient is sufficient to predict the end behavior of the function. Step2 Example Find the end behavior of the function x44x3+3x+25x44x3+3x+25. The degree of the function is even and the leading coefficient is positive. So, the end behavior is: f(x)+, as xf(x)+, as x+f(x)+, as xf(x)+, as x+ The graph looks as follows:
Step 2 of 3
Step 3 of 3
##### ISBN: 9780321570567
Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. The answer to “Describe the possible end behavior of a polynomial.” is broken down into a number of easy to follow steps, and 8 words. The full step-by-step solution to problem: 6E from chapter: 4.3 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. Since the solution to 6E from 4.3 chapter was answered, more than 382 students have viewed the full step-by-step answer. This full solution covers the following key subjects: behavior, describe, end, polynomial. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1.
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×. Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.3 - Problem 6e. Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.3 - Problem 6e. ×. # Describe the possible end behavior of a polynomial.. ISBN: 9780321570567 2. ## Solution for problem 6E Chapter 4.3. Calculus: Early Transcendentals | 1st Edition. • Textbook Solutions. • 2901 Step-by-step solutions solved by professors and subject experts. • Get 24/7 help from StudySoup virtual teaching assistants. Calculus: Early Transcendentals | 1st Edition. 4 5 1 345 Reviews. 28. 3. Problem 6E. Describe the possible end behavior of a polynomial.. Step-by-Step Solution:. Step 1 of 3. Solution: Step1 The end behavior of a polynomial function is the behavior of the graph of f(x)f(x) as xx approaches positive infinity or negative infinity.
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The degree and the leading coefficient of a polynomial function determine the end behavior of the graph. The leading coefficient is significant compared to the other coefficients in the function for the very large or very small numbers. So, the sign of the leading coefficient is sufficient to predict the end behavior of the function. Step2 Example Find the end behavior of the function x44x3+3x+25x44x3+3x+25. The degree of the function is even and the leading coefficient is positive. So, the end behavior is: f(x)+, as xf(x)+, as x+f(x)+, as xf(x)+, as x+ The graph looks as follows:. Step 2 of 3. Step 3 of 3. ##### ISBN: 9780321570567. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. The answer to “Describe the possible end behavior of a polynomial.” is broken down into a number of easy to follow steps, and 8 words. The full step-by-step solution to problem: 6E from chapter: 4.3 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. Since the solution to 6E from 4.3 chapter was answered, more than 382 students have viewed the full step-by-step answer. This full solution covers the following key subjects: behavior, describe, end, polynomial. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1.. #### Related chapters. Unlock Textbook Solution.
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1. ## flux integral.....argh!!!
Hi all, I am having real trouble with this problem. If anyone can steer me in the right direction, or better yet, help by showing the solution, I would be much appreciated:
Consider a surface
S which is that portion of the plane 2x + 2y + z = 6
included in the first octant.
Evaluate the flux integral ∫∫ F nˆdA , where the vector field
F = [xy,−x^2 , x + z] and nˆ is a unit outer normal vector to S.
2. I think you need to understand that, no, no one here is going to do your homework for you. You learn mathematics by doing mathematics, not by watching someone else do it for you.
Here's how I would approach this problem. The plane 2x+ 2y+ z= 6 can be written as z= 6- 2x- 2y or as the "position vector" $\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (6- 2x- 2y)\vec{j}$.
The derivatives with respect to x and y are $\vec{r}_x= \vec{i}- 2\vec{j}$ and $\vec{r}_y= \vec{j}- 2\vec{k}$. Those vectors are in the plane and their cross product, $2\vec{i}+ 2\vec{j}+ \vec{k}$ is perpendicular to the plane and its length is the "differential of area" for the plane. In general, the vector $A\vec{i}+ B\vec{j}+ C\vec{k}$ is the "vector differential of area" for the plane Ax+ By+ Cz= D.
Now, what about the orientation? You say "where $\vec{n}$ is the outer normal to S". That would make sense for a closed surface but not for a plane! A plane has no "inside" or "outside". Which side of the plane is the normal to be on? Do you want this plane oriented by the "normal in the positive z direction" or in the "negative z direction"?
If positive z direction, take the dot product of $\vec{F}= xy\vec{i}- x^2\vec{j}+ (x+ y)\vec{k}$ with $2\vec{i}+ 2\vec{j}+ \vec{k}$ and integrate with respect to x and y. If the negative z direction, take the dot product of $\vec{F}$ with $-2\vec{i}- 2\vec{j}- vec{k}$ and integrate (that will just change the sign on the answer).
The plane 2x+ 2y- z= 6, projected to the xy-plane, is 2x+ 2y= 6 or x+ y= 3. You can get the limits of integration from that.
3. Originally Posted by HallsofIvy
I think you need to understand that, no, no one here is going to do your homework for you. You learn mathematics by doing mathematics, not by watching someone else do it for you.
Here's how I would approach this problem. The plane 2x+ 2y+ z= 6 can be written as z= 6- 2x- 2y or as the "position vector" $\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (6- 2x- 2y)\vec{j}$.
The derivatives with respect to x and y are $\vec{r}_x= \vec{i}- 2\vec{j}$ and $\vec{r}_y= \vec{j}- 2\vec{k}$. Those vectors are in the plane and their cross product, $2\vec{i}+ 2\vec{j}+ \vec{k}$ is perpendicular to the plane and its length is the "differential of area" for the plane. In general, the vector $A\vec{i}+ B\vec{j}+ C\vec{k}$ is the "vector differential of area" for the plane Ax+ By+ Cz= D.
Now, what about the orientation? You say "where $\vec{n}$ is the outer normal to S". That would make sense for a closed surface but not for a plane! A plane has no "inside" or "outside". Which side of the plane is the normal to be on? Do you want this plane oriented by the "normal in the positive z direction" or in the "negative z direction"?
If positive z direction, take the dot product of $\vec{F}= xy\vec{i}- x^2\vec{j}+ (x+ y)\vec{k}$ with $2\vec{i}+ 2\vec{j}+ \vec{k}$ and integrate with respect to x and y. If the negative z direction, take the dot product of $\vec{F}$ with $-2\vec{i}- 2\vec{j}- vec{k}$ and integrate (that will just change the sign on the answer).
The plane 2x+ 2y- z= 6, projected to the xy-plane, is 2x+ 2y= 6 or x+ y= 3. You can get the limits of integration from that.
Would we not be better served by going to the definitions directly? Because we have a plane we will have 2 flux integrals to compute: the top and bottem. Of course you address this in your post (actually, everything i'm about to say you've already addressed) but I think for the purposes of the student it's just easier to learn how to "plug and chug" so to speak (i know...i know this doesn't actually facilitate learning the material, but if it helps understand how to at least attempt these problems, why not?)
Let us write all the things we know,
Our function is defined as,
$F = xy \hat i - x^2 \hat j + (x+z) \hat k$
Our surface is defined as,
$z = 6-2x-2y$
And for the computation of our integral,
$\hat N dS = +/- ( - \frac{ \partial S }{ \partial x } \hat i - \frac{ \partial S }{ \partial y } \hat j + k )$
Notice how we have a plus and a minus, this means we need to find both of these to get to the total flux! If you just want upward or downward, only compute the flux integral for either of those.
$\hat N dS = +/- ( + 2 \hat i + 2 \hat j + 1k )$
Thus,
$Total Flux = \iint_D F \cdot ( + 2 \hat i + 2 \hat j + 1k ) dA + \iint_D F \cdot ( - 2 \hat i - 2 \hat j - 1k ) dA$
To get our bounds we let $z = 0$ which yields $x + y = 3$
And,
$0 \le x \le 3$ with $0 \le y \le 3 - x$
At this point everything should be clear enough for you to go ahead with this problem.
HallsofIvy gave a much more mathetmatical explanation then myself, but this should allow you to tackle the problem in a brute force manner.
4. Originally Posted by swinburneguy2009
Hi all, I am having real trouble with this problem. If anyone can steer me in the right direction, or better yet, help by showing the solution, I would be much appreciated:
Consider a surface
S which is that portion of the plane 2x + 2y + z = 6
included in the first octant.
Evaluate the flux integral ∫∫ F nˆdA , where the vector field
F = [xy,−x^2 , x + z] and nˆ is a unit outer normal vector to S.
I would tell you work this Stoke's Theorem. It would be easier to integrate the curl
5. Originally Posted by AllanCuz
Would we not be better served by going to the definitions directly? Because we have a plane we will have 2 flux integrals to compute: the top and bottem.
No, you don't.
Of course you address this in your post (actually, everything i'm about to say you've already addressed) but I think for the purposes of the student it's just easier to learn how to "plug and chug" so to speak (i know...i know this doesn't actually facilitate learning the material, but if it helps understand how to at least attempt these problems, why not?)
Let us write all the things we know,
Our function is defined as,
$F = xy \hat i - x^2 \hat j + (x+z) \hat k$
Our surface is defined as,
$z = 6-2x-2y$
And for the computation of our integral,
$\hat N dS = +/- ( - \frac{ \partial S }{ \partial x } \hat i - \frac{ \partial S }{ \partial y } \hat j + k )$
Notice how we have a plus and a minus, this means we need to find both of these to get to the total flux! If you just want upward or downward, only compute the flux integral for either of those.
No, calculating the flux through a surface in both directions will sum to 0. That's why Swinburneguy2009 said "outer normal to S" although, as I pointed out "outer" doesn't really apply to a plane. The "upward" flux will be the negative of the "downward" flux- if you calculate both and total them, you will get 0.
$\hat N dS = +/- ( + 2 \hat i + 2 \hat j + 1k )$
Thus,
$Total Flux = \iint_D F \cdot ( + 2 \hat i + 2 \hat j + 1k ) dA + \iint_D F \cdot ( - 2 \hat i - 2 \hat j - 1k ) dA$
To get our bounds we let $z = 0$ which yields $x + y = 3$
And,
$0 \le x \le 3$ with $0 \le y \le 3 - x$
At this point everything should be clear enough for you to go ahead with this problem.
HallsofIvy gave a much more mathetmatical explanation then myself, but this should allow you to tackle the problem in a brute force manner.
6. Originally Posted by HallsofIvy
No, you don't.
No, calculating the flux through a surface in both directions will sum to 0. That's why Swinburneguy2009 said "outer normal to S" although, as I pointed out "outer" doesn't really apply to a plane. The "upward" flux will be the negative of the "downward" flux- if you calculate both and total them, you will get 0.
I understand this, the main point of my post was to get across we dont need to use positional vectors. I know the total flux out of a plane is equal to 0, but I was showing him how to set it up for future problems and noted that he can pick/choose which direction he wants.
It's easiar to work with the given definitions then to go back to the basics in this case.
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1. ## flux integral.....argh!!!. Hi all, I am having real trouble with this problem. If anyone can steer me in the right direction, or better yet, help by showing the solution, I would be much appreciated:. Consider a surface. S which is that portion of the plane 2x + 2y + z = 6. included in the first octant.. Evaluate the flux integral ∫∫ F nˆdA , where the vector field. F = [xy,−x^2 , x + z] and nˆ is a unit outer normal vector to S.. 2. I think you need to understand that, no, no one here is going to do your homework for you. You learn mathematics by doing mathematics, not by watching someone else do it for you.. Here's how I would approach this problem. The plane 2x+ 2y+ z= 6 can be written as z= 6- 2x- 2y or as the "position vector" $\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (6- 2x- 2y)\vec{j}$.. The derivatives with respect to x and y are $\vec{r}_x= \vec{i}- 2\vec{j}$ and $\vec{r}_y= \vec{j}- 2\vec{k}$. Those vectors are in the plane and their cross product, $2\vec{i}+ 2\vec{j}+ \vec{k}$ is perpendicular to the plane and its length is the "differential of area" for the plane. In general, the vector $A\vec{i}+ B\vec{j}+ C\vec{k}$ is the "vector differential of area" for the plane Ax+ By+ Cz= D.. Now, what about the orientation? You say "where $\vec{n}$ is the outer normal to S". That would make sense for a closed surface but not for a plane! A plane has no "inside" or "outside". Which side of the plane is the normal to be on? Do you want this plane oriented by the "normal in the positive z direction" or in the "negative z direction"?. If positive z direction, take the dot product of $\vec{F}= xy\vec{i}- x^2\vec{j}+ (x+ y)\vec{k}$ with $2\vec{i}+ 2\vec{j}+ \vec{k}$ and integrate with respect to x and y. If the negative z direction, take the dot product of $\vec{F}$ with $-2\vec{i}- 2\vec{j}- vec{k}$ and integrate (that will just change the sign on the answer).. The plane 2x+ 2y- z= 6, projected to the xy-plane, is 2x+ 2y= 6 or x+ y= 3. You can get the limits of integration from that.. 3. Originally Posted by HallsofIvy. I think you need to understand that, no, no one here is going to do your homework for you. You learn mathematics by doing mathematics, not by watching someone else do it for you.. Here's how I would approach this problem. The plane 2x+ 2y+ z= 6 can be written as z= 6- 2x- 2y or as the "position vector" $\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (6- 2x- 2y)\vec{j}$.. The derivatives with respect to x and y are $\vec{r}_x= \vec{i}- 2\vec{j}$ and $\vec{r}_y= \vec{j}- 2\vec{k}$. Those vectors are in the plane and their cross product, $2\vec{i}+ 2\vec{j}+ \vec{k}$ is perpendicular to the plane and its length is the "differential of area" for the plane. In general, the vector $A\vec{i}+ B\vec{j}+ C\vec{k}$ is the "vector differential of area" for the plane Ax+ By+ Cz= D.. Now, what about the orientation? You say "where $\vec{n}$ is the outer normal to S". That would make sense for a closed surface but not for a plane! A plane has no "inside" or "outside". Which side of the plane is the normal to be on? Do you want this plane oriented by the "normal in the positive z direction" or in the "negative z direction"?. If positive z direction, take the dot product of $\vec{F}= xy\vec{i}- x^2\vec{j}+ (x+ y)\vec{k}$ with $2\vec{i}+ 2\vec{j}+ \vec{k}$ and integrate with respect to x and y. If the negative z direction, take the dot product of $\vec{F}$ with $-2\vec{i}- 2\vec{j}- vec{k}$ and integrate (that will just change the sign on the answer).. The plane 2x+ 2y- z= 6, projected to the xy-plane, is 2x+ 2y= 6 or x+ y= 3. You can get the limits of integration from that.. Would we not be better served by going to the definitions directly? Because we have a plane we will have 2 flux integrals to compute: the top and bottem. Of course you address this in your post (actually, everything i'm about to say you've already addressed) but I think for the purposes of the student it's just easier to learn how to "plug and chug" so to speak (i know...i know this doesn't actually facilitate learning the material, but if it helps understand how to at least attempt these problems, why not?). Let us write all the things we know,. Our function is defined as,. $F = xy \hat i - x^2 \hat j + (x+z) \hat k$. Our surface is defined as,. $z = 6-2x-2y$. And for the computation of our integral,. $\hat N dS = +/- ( - \frac{ \partial S }{ \partial x } \hat i - \frac{ \partial S }{ \partial y } \hat j + k )$. Notice how we have a plus and a minus, this means we need to find both of these to get to the total flux! If you just want upward or downward, only compute the flux integral for either of those.. $\hat N dS = +/- ( + 2 \hat i + 2 \hat j + 1k )$. Thus,. $Total Flux = \iint_D F \cdot ( + 2 \hat i + 2 \hat j + 1k ) dA + \iint_D F \cdot ( - 2 \hat i - 2 \hat j - 1k ) dA$. To get our bounds we let $z = 0$ which yields $x + y = 3$. And,.
|
$0 \le x \le 3$ with $0 \le y \le 3 - x$. At this point everything should be clear enough for you to go ahead with this problem.. HallsofIvy gave a much more mathetmatical explanation then myself, but this should allow you to tackle the problem in a brute force manner.. 4. Originally Posted by swinburneguy2009. Hi all, I am having real trouble with this problem. If anyone can steer me in the right direction, or better yet, help by showing the solution, I would be much appreciated:. Consider a surface. S which is that portion of the plane 2x + 2y + z = 6. included in the first octant.. Evaluate the flux integral ∫∫ F nˆdA , where the vector field. F = [xy,−x^2 , x + z] and nˆ is a unit outer normal vector to S.. I would tell you work this Stoke's Theorem. It would be easier to integrate the curl. 5. Originally Posted by AllanCuz. Would we not be better served by going to the definitions directly? Because we have a plane we will have 2 flux integrals to compute: the top and bottem.. No, you don't.. Of course you address this in your post (actually, everything i'm about to say you've already addressed) but I think for the purposes of the student it's just easier to learn how to "plug and chug" so to speak (i know...i know this doesn't actually facilitate learning the material, but if it helps understand how to at least attempt these problems, why not?). Let us write all the things we know,. Our function is defined as,. $F = xy \hat i - x^2 \hat j + (x+z) \hat k$. Our surface is defined as,. $z = 6-2x-2y$. And for the computation of our integral,. $\hat N dS = +/- ( - \frac{ \partial S }{ \partial x } \hat i - \frac{ \partial S }{ \partial y } \hat j + k )$. Notice how we have a plus and a minus, this means we need to find both of these to get to the total flux! If you just want upward or downward, only compute the flux integral for either of those.. No, calculating the flux through a surface in both directions will sum to 0. That's why Swinburneguy2009 said "outer normal to S" although, as I pointed out "outer" doesn't really apply to a plane. The "upward" flux will be the negative of the "downward" flux- if you calculate both and total them, you will get 0.. $\hat N dS = +/- ( + 2 \hat i + 2 \hat j + 1k )$. Thus,. $Total Flux = \iint_D F \cdot ( + 2 \hat i + 2 \hat j + 1k ) dA + \iint_D F \cdot ( - 2 \hat i - 2 \hat j - 1k ) dA$. To get our bounds we let $z = 0$ which yields $x + y = 3$. And,. $0 \le x \le 3$ with $0 \le y \le 3 - x$. At this point everything should be clear enough for you to go ahead with this problem.. HallsofIvy gave a much more mathetmatical explanation then myself, but this should allow you to tackle the problem in a brute force manner.. 6. Originally Posted by HallsofIvy. No, you don't.. No, calculating the flux through a surface in both directions will sum to 0. That's why Swinburneguy2009 said "outer normal to S" although, as I pointed out "outer" doesn't really apply to a plane. The "upward" flux will be the negative of the "downward" flux- if you calculate both and total them, you will get 0.. I understand this, the main point of my post was to get across we dont need to use positional vectors. I know the total flux out of a plane is equal to 0, but I was showing him how to set it up for future problems and noted that he can pick/choose which direction he wants.. It's easiar to work with the given definitions then to go back to the basics in this case.
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https://fr.scribd.com/document/325123585/Second-Midterm-With-Answers-Summer-2012
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Vous êtes sur la page 1sur 16
# Econ 101 Summer 2012
## Exam 2 Professor Kelly
Name: ___________________________________
Section Day and Time: ____________________________
On this exam it is important that you show your work to get FULL
CREDIT.
On this exam you should write any verbal answer using standard
English grammar: that is, please write in complete sentences.
The exam consists of 20 multiple choice questions worth 2.5 points for a
total of 50 points, and three problems worth a total of 50 points.
Multiple Choice Score
Question 1 15 points
Question 2 20 points
Question 3 15 points
Total:
100 points
No calculators are allowed for the exam. Cell phones should be silenced
You will have 100 minutes to work. Good luck.
I, __________________________________, agree to neither give nor receive any help on this exam
from other students. Furthermore, I understand that use of a calculator is an academic misconduct
violation on this exam.
Signed ____________________________________
## PROBLEMS (worth a total of 50 points)
1. (15 points in all) Suppose the demand curve and supply curve for the market is given by the following
equations:
Demand: P = 600 Q
Supply: P = Q
a. (2 points) What is the equilibrium price (P1) and quantity (Q1) in this market?
600 Q = Q
2Q = 600
Q = Q1 = 300
P = P1 = 300
b. (1 point) When this market is in equilibrium, what is the total revenue (TR1) earned in this market?
TR1 = P*Q
TR1 = (300)(300) = \$90,000
c. (2 points) Suppose that the number of consumers in this market increases such that at every price twice
as much of the good is demanded than was demanded initially. In the space below draw a graph that
includes the original demand curve (D1), the original supply curve (S1), and the new demand curve (D2).
Label the initial equilibrium in this market (Q1, P1) and the new equilibrium (Q2, P2). Label your axes as
well.
## d. (1 point) Write an equation for D2 in slope-intercept form.
Slope of D2 = -600/1200 = -1/2
P = b + (-1/2)Q
b = y-intercept of D2 = 600
P = 600 (1/2)Q
e. (3 points) Solve for the new equilibrium quantity and price (Q2, P2). Calculate total revenue (TR2) at
this price and quantity. Show your work for full credit.
Q2 = ________________
P2 = ________________
TR2 = _______________
600 (1/2)Q = Q
(3/2)Q = 600
Q = Q2 = 400
P = P2 = 400
TR2 = (400)(400) = \$160,000
f. (6 points) Suppose that the firms in this market have the market power to charge whatever price they
want. Furthermore, suppose that their goal is to set the market price at that level where total revenue is
maximized. What price should these firms charge? Given the new demand curve and ignoring the given
supply curve identify the quantity of the good they should sell to reach their goal. What is the value of
total revenue (TR3) given this pricing and production decision? Explain your answer making sure to
Q3 = ________________
P3 = ________________
TR3 = _______________
For a linear demand curve, the price and quantity that will maximize total revenue is the price and
quantity associated with the midpoint of the demand curve. At the midpoint the price elasticity of demand
is equal to one: if the firm is charging a price greater than the price associated with the midpoint of the
demand curve, the firm can increase its total revenue by reducing its price since when demand is elastic
total revenue increases as price decreases; if the firm is charging a price less than the price associated with
the midpoint of the demand curve, the firm can increase its total revenue by increasing its price since
when demand is inelastic total revenue increases as price increases.
Given the new demand curve, P = 600 (1/2)Q, we can quickly find the midpoint coordinates as (Q3, P3)
= (600, 300). When the firm produces 600 units and charges a price of \$300 per unit, its total revenue
(TR3) equals \$180,000.
2. (20 points total) Martha consumes apples and chocolate candies. Suppose Marthas budget line (BL1)
for apples (A) and chocolate candies (C) is given below. Suppose you know that the price of apples is \$2
per apple but you are not told the price of chocolate candies or Marthas income.
a. (2 points) Given the graph above, what is Marthas income? Explain how you found your answer.
Since Martha can purchase 40 apples if she only buys apples, this implies that her income must by 40(\$2)
= \$80.
4
b. (2 points) Given the graph above, what is the price of one unit of chocolate candies? Explain your
In (a) we determined Marthas income is \$80. We also know that Martha if she buys only chocolate
candies can afford 20 units. Using this information we can find the price of a unit of chocolate candies by
using the relationship
Income = (Price of Chocolate Candies)(Quantity of Chocolate Candies) + (Price of Apples)
(Quantity of Apples)
If Martha purchases no apples, then this equation can be simplified to
Income = (Price of Chocolate Candies)(Quantity of Chocolate Candies)
80 = (Price of Chocolate Candies)(20)
Price of Chocolate Candies = \$4 per unit
Suppose Marthas tastes and preferences are such that she maximizes her utility given her budget line
when she consumes 28 apples.
c. (2 points) Given this new information and the above graph, how many chocolate candies does Martha
consume when she maximizes her utility subject to the constraints imposed by her income and the prices
of the two goods?
We know that Income = \$80, the price of apples is \$2, the price of chocolates is \$4, and the quantity of
apples is 28. Use this information to solve for the quantity of chocolate candies.
Income = (Price of Chocolate Candies)(Quantity of Chocolate Candies) + (Price of Apples)
(Quantity of Apples)
80 = (4)(Quantity of Chocolate Candies) + (2)(28)
24 = 4(Quantity of Chocolate Candies)
Quantity of Chocolate Candies = 6
d. (2 points) In the above graph label Marthas optimal consumption bundle on BL1 as point A. Indicate
the numerical coordinates of this point on your graph.
Now, suppose the price of apples increases to \$4 per apple while there are no changes in Marthas income
or the price of chocolate candies. When the price of apples increases to \$4, Martha finds she maximizes
her utility by consuming 10 units of chocolate candies.
e. (2 points) Given this new information, in the above graph draw Marthas new budget line (BL2) and
indicate the numeric values for the x- and y- axis intercepts.
f. (2 points) Given this new information, calculate the number of apples Martha consumes when Martha
maximizes her utility given the new price of apples. Show your work for full credit.
5
## Income = (Price of Chocolate Candies)(Quantity of Chocolate Candies) + (Price of Apples)
(Quantity of Apples)
80 = (4)(10) + (4)(Quantity of Apples)
40 = 4(Quantity of Apples)
Quantity of Apples = 10
g. (2 points) In the graph label Marthas optimal consumption bundle B when the price of apples increases
to \$4. Indicate the numerical coordinates of this point on the graph.
See graph below:
h. (6 points) Assume Marthas demand curve for apples is linear. Given the two optimal consumption
bundles at A and B, draw Marthas demand curve in the space below and then write an equation in slopeintercept form for this curve. Make sure your graph is labeled clearly and completely labeled. (Hint: the
numbers are messy here-just leave them as improper fractions!)
We know two points that sit on Marthas linear demand curve for apples: when the price of apples is \$2,
Martha maximizes her utility when she consumes 28 apples; and when the price of apples is \$4, Martha
maximizes her utility when she consumes 10 apples. Use these two points to write the equation for
Marthas demand curve as P = (46/9) (1/9)Q. The figure below illustrates this demand curve:
3. (15 points) Use the graphs below of a perfectly competitive industry and a representative firm in the
short run and in the long run to answer this set of questions.
Short-run:
Use the graphs below to answer the questions about the long-run (this graph is just a duplicate of
the first set of graphs, but you will be using it to show the long-run adjustment process).
a. (2 points) In the first row of graphs (the pair of graphs identified as Short Run) indicate the short-run
equilibrium using the following symbols:
Q1 = the short-run market level of production
P1 = the short-run market price
q1 = the short-run level of production for the representative firm
See graph after part (b).
b. (2 points) In the first row of graphs if the representative firm is earning economic profits shade in the
area representing those profits. On the graph indicate whether these are positive or negative profits.
c. (2 points) Given your analysis in the short-run, what do you predict will happen in this industry in the
Since the representative firm is earning negative economic profits in the short run, we would anticipate
the exiting of firms in the long run. This will cause the market supply curve to shift to the left until the
market price is equal to the minimum point on the representative firms ATC curve.
d. (3 points) In the second row of graphs (the pair of graphs identified as Long Run) depict the long-run
equilibrium situation in this market. If any demand or supply curves shift, indicate these shifts on your
graphs. On the graphs identify:
Q2 = the long-run market level of production
P2 = the long-run market price
q2 = the long-run level of production for the representative firm
e. (1 point) In the long run, the representative firm in this market will earn ___________________
economic profits.
Zero
f. Suppose you are given this set of graphs and told that peoples incomes have increased and the good is
a normal good. Given this information answer the following set of questions:
i. (2 points) In the short-run, this change in income will result in the price of the good
__________________ relative to its initial level.
Increasing
ii. (2 points) In the short-run, this change in income will result in the quantity of the good
produced by the representative firm to ______________________ relative to its initial level.
Increase
iii. (1 point) In the long run, the equilibrium price in this market given this change in income and
holding everything else constant, will be _____________ relative to the initial long-run
equilibrium price in this market.
The same
10
## MULTIPLE CHOICE QUESTIONS (20 questions worth 2.5 points each):
Use the information below to answer the next two questions:
Your business currently has two customers with the following individual demand curves:
Customer Ones Demand Curve: P = 10 -2Q
Customer Twos Demand Curve: P = 20 2Q
market when you set price to be \$7.5?
a. TR = \$6.25
b. TR = \$56.25
c. TR = \$100
d. TR = \$12.50
2. Your company is trying to decide whether to raise the price of your product from \$\$7.5 by \$0.50 or
lower it by \$0.50. Given the above information what would you recommend if the sole goal of your firm
is to increase the total revenue?
a. The firm should increase the price by \$0.50.
b. The firm should decrease the price by \$0.50.
c. The firm could increase their total revenue if they increase or decrease the price by \$0.50 from its
initial price level \$7.5.
d. The firm should maintain its current price.
3. Demand for a good is considered elastic. This implies that
a. Increasing the price of this good in the market will result in greater total revenue for producers.
b. Decreasing the price of this good in the market will result in greater total revenue for producers.
c. The market is currently maximizing its total revenue from selling this good.
d. Consumers of this good are not price sensitive.
## Use the information below to answer the next two questions.
Shawn and Mary are currently working and living in Tucson. But, they are considering moving to either
St. Paul or New York City. The table below provides information about the salaries they would earn in the
three cities as well as information about the price index for each city. Shawn and Mary have decided they
will move to whichever city provides them the greatest real income. Assume that their nominal salaries
and the price index will not change in the future.
City
Tucson
Tucson
St. Paul
St. Paul
Nominal Salary
\$40,000
\$50,000
\$50,000
\$75,000
Employee
Mary
Shawn
Mary
Shawn
100
100
125
125
11
New York City
\$100,000
\$100,000
Mary
Shawn
250
250
## 4. From the given information which of the following statements is true?
a. The base year for the price index is this year.
b. Tucson is likely the city that was used to compare the cost of living in other cities when this
information was gathered.
c. The cost of living in New York City is 100% higher than the cost of living in Tucson.
d. The cost of living in St. Paul 150% less than the cost of living in New York City.
5.Given that Shawn and Mary want to maximize their joint real income where should they live? Base
your analysis solely on the information you have been provided.
a. Shawn and Mary should live in either St. Paul or Tucson since their real income will be equivalent in
these two cities.
b. Shawn and Mary should live in New York since both of their incomes are substantially higher than in
Tucson or St. Paul.
c. Once you factor in the cost of living for each of these cities the real income in all three cities is the
same.
d. Shawn and Mary should live in St. Paul since their real income will be maximized there.
6. The cross-price elasticity of demand measures
a. The percentage change in the quantity demanded of the good to the percentage change in the price of
the good.
b. The percentage change in the quantity demanded of one good to the percentage change in the price of a
related good.
c. The percentage change in the quantity demanded of the good to the percentage change in income.
d. The percentage change in the price of one good to the percentage change in the price of the other good.
7. Jacob has \$100 of income that he will spend on hot dogs or cheeseburgers during his month long
vacation. Jacob knows that hot dogs cost \$5 each and cheeseburgers cost \$10 each. Given this information
determine which of the budget lines below represents Jacobs budget line.
12
## Answer for #7 is (b)
Use the following information to answer the next two questions.
Suppose Oscars income is \$1000 and he uses all of this income to buy either ice cream or pizzas. When
the price of ice cream is \$5 per serving and the price of pizza is \$10 per pizza Oscar maximizes his utility
by purchasing 40 pizzas. When the price of ice cream is \$10 per serving and the price of pizza is \$10 per
pizza Oscar maximizes his utility by purchasing 60 pizzas.
8. From this information derive Oscars demand for ice cream. Assume his demand for ice cream is linear.
Oscars demand for ice cream is
a. P = 74 (8/5)Q
b. P = (1/4)Q
c. P = 100 (1/2)Q
d. P = 12.5 (1/16)Q
9. Given the above information you are also told that Oscars income compensated budget line (the
budget line we called BL3 when discussing the income and substitution effects) results in Oscar
maximizing his utility when he consumes 100 servings of ice cream and 70 slices of pizza. From this
information you can conclude that Oscars substitution effect with respect to ice cream is equal to
a. 80 units of ice cream.
b. 60 units of ice cream.
c. 30 units of ice cream.
d. 20 units of ice cream.
13
10. Mary, who prides herself on her business knowledge and expertise, states that no matter what your
demand curve looks like it is always best to charge the highest possible price. Which of the following
statements are true given this information? In your answer assume that the demand curve that Marys
business faces for their product is downward sloping and linear.
a. Mary missed the lecture on elasticity in her introductory microeconomics class.
b. If the demand for Marys product is elastic, Mary will earn more total revenue by lowering the price of
her product.
c. If Mary charges the price associated with the midpoint for her product then her company will maximize
its total revenue.
11. Joeys total utility from consuming 10 cookies is equal to 100 utils. His total utility from consuming
11 cookies is 104 utils. Joeys total utility from consuming 5 slices of pizza is 500 utils and his total utility
from consuming 6 slices of pizza is 508 utils. Furthermore, you know that the price of a slice of pizza is
twice the price of a cookie. Also assume that Joey has spent all his income on cookies and pizzas. Given
this information, which of the following statements is true?
a. Joey is not maximizing his utility from consuming these two goods: he would have greater satisfaction
b. Joey is maximizing his utility from consuming these two goods: he cannot select a different
combination of pizza and cookies that will provide him with greater utility.
c. Joey is not maximizing his utility from consuming these two goods: he would have greater satisfaction
if he consumed more pizza.
d. Joey is not maximizing his utility from consuming these two goods: he would have greater satisfaction
if he consumed more pizza and cookies.
12. There are five players on the Wisconsin Hotshots, a womens basketball team. The average height of
these five players is five feet eight inches tall. If the Hotshots add a new player to the team that is six feet
tall then
a. The average height of the team will increase.
b. The average height of the team will be unaffected.
c. The average height of the team will decrease.
d. The average height of the team may increase, decrease or remain the same.
13. In the short run, which of the following statements is true?
a. As output increases average total cost decreases.
b. As output increases average variable cost decreases.
c. No matter what level of output the firm produces, its fixed cost is constant.
d. Answers (a) and (b) are both correct.
e. Answers (a) and (c) are both correct.
f. Answers (b) and (c) are both correct.
g. Answers (a), (b) and (c) are all correct.
14
14. Katherine operates a bakery in Glenwood. Her costs of operation include \$5000 a year in property
rental for the land and building where her bakery is located; \$1.50 per loaf of bread for the flour, sugar
and other ingredients that go into making her product; and \$7.00 an hour for each hour of labor she hires.
She estimates that she needs four hours of labor to produce 100 loaves of bread, eight hours of labor to
produce 200 loaves of bread, etc. Given this information which of the following statements is true?
a. When Katherine produces 400 loaves of bread her labor cost is \$28.00.
b. When Katherine produces 400 loaves of bread her fixed costs are equal to \$5000.
c. When Katherine produces 400 loaves of bread her variable costs equal \$600.
d. When Katherine produces 400 loaves of bread her labor cost is \$112.00 while her fixed cost is equal to
\$0.
15. Consider Montys Garage, a small business offering car repair. Suppose you are told that as Monty
provides more units of his output (i.e., he repairs more cars) his average variable cost is increasing. From
this you know
a. That Montys average total cost is increasing as you move rightward along the quantity axis.
b. That Montys average fixed cost is increasing as your move rightward along the quantity axis.
c. That Montys marginal cost of production exceeds his average variable cost of production.
d. Monty can increase his profits by repairing a smaller number of cars.
Use the figure below of a perfectly competitive market and a representative firm to answer the next three
questions.
16. In the short run, which of the following statements is true given the above graphs?
a. There are (D/c) firms in the industry and each firm is earning positive economic profits.
b. Firms in the industry are earning negative economic profits and this indicates that the market has too
many firms.
15
c. Each firms revenue is sufficient to cover their variable costs of production and each firm is earning a
normal or zero economic profit.
d. Answers (a), (b) and (c) are all true.
e. Answers (a) and (b) are both true.
17. In the long run, which of the following statements is true given the above graphs?
a. In the long run firms will exit the industry until price increases to n and there are (C/b) firms in the
industry.
b. In the long run the market supply curve will shift to the right until prices fall to r and there will be
(E/b) firms in the industry.
c. In the long run the market supply curve will shift to the left until price is n and each firm is producing
d units of output.
d. In the long run the market demand curve will shift until all firms in the industry earn zero economic
profits.
18. In the long run, which of the following statements is true given the above graph?
a. Total revenue for the industry will equal n*C and total revenue for the representative firm will equal
n*d.
b. Total revenue for the industry will equal n*F and the total number of firms in the industry will by F/d.
c. Total revenue for the industry will equal n*F and total revenue for the representative firm will equal the
firms fixed costs.
d. Total revenue for the industry will equal n*F and the representative firms total costs are equal to its
variable costs.
19. A firm in a perfectly competitive market is currently producing 10 units of output and its average cost
of production is \$2 per unit of output. The firm also knows that if it increases its production to 12 units,
its average total cost of production will increase to \$2.25 per unit of output. This information implies
a. That the firm needs to increase its level of production to reach the long run equilibrium price and
quantity in this market.
b. That the firm will maximize profit by producing 10 units of output in the long run.
c. That the equilibrium long run price in this market is \$2.
d. That the firm may need to decrease its level of production in the long run in order to reach the long run
equilibrium price and quantity in this market.
20. A firm experiences increasing returns to scale. This indicates that
a. The firms average costs of production fall as the size of the firm increases.
b. The firms average costs of production fall as the firm hires more of its variable inputs holding
everything else constant.
c. The firms average variable cost of production reaches its minimum point at a relatively low level of
production for the firm.
d. The number of firms in the industry should increase in the long run to take advantage of the increasing
returns to scale.
16
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Vous êtes sur la page 1sur 16. # Econ 101 Summer 2012. ## Exam 2 Professor Kelly. Name: ___________________________________. Section Day and Time: ____________________________. On this exam it is important that you show your work to get FULL. CREDIT.. On this exam you should write any verbal answer using standard. English grammar: that is, please write in complete sentences.. The exam consists of 20 multiple choice questions worth 2.5 points for a. total of 50 points, and three problems worth a total of 50 points.. Multiple Choice Score. Question 1 15 points. Question 2 20 points. Question 3 15 points. Total:. 100 points. No calculators are allowed for the exam. Cell phones should be silenced. You will have 100 minutes to work. Good luck.. I, __________________________________, agree to neither give nor receive any help on this exam. from other students. Furthermore, I understand that use of a calculator is an academic misconduct. violation on this exam.. Signed ____________________________________. ## PROBLEMS (worth a total of 50 points). 1. (15 points in all) Suppose the demand curve and supply curve for the market is given by the following. equations:. Demand: P = 600 Q. Supply: P = Q. a. (2 points) What is the equilibrium price (P1) and quantity (Q1) in this market?. 600 Q = Q. 2Q = 600. Q = Q1 = 300. P = P1 = 300. b. (1 point) When this market is in equilibrium, what is the total revenue (TR1) earned in this market?. TR1 = P*Q. TR1 = (300)(300) = \$90,000. c. (2 points) Suppose that the number of consumers in this market increases such that at every price twice. as much of the good is demanded than was demanded initially. In the space below draw a graph that. includes the original demand curve (D1), the original supply curve (S1), and the new demand curve (D2).. Label the initial equilibrium in this market (Q1, P1) and the new equilibrium (Q2, P2). Label your axes as. well.. ## d. (1 point) Write an equation for D2 in slope-intercept form.. Slope of D2 = -600/1200 = -1/2. P = b + (-1/2)Q. b = y-intercept of D2 = 600. P = 600 (1/2)Q. e. (3 points) Solve for the new equilibrium quantity and price (Q2, P2). Calculate total revenue (TR2) at. this price and quantity. Show your work for full credit.. Q2 = ________________. P2 = ________________. TR2 = _______________. 600 (1/2)Q = Q. (3/2)Q = 600. Q = Q2 = 400. P = P2 = 400. TR2 = (400)(400) = \$160,000. f. (6 points) Suppose that the firms in this market have the market power to charge whatever price they. want. Furthermore, suppose that their goal is to set the market price at that level where total revenue is. maximized. What price should these firms charge? Given the new demand curve and ignoring the given. supply curve identify the quantity of the good they should sell to reach their goal. What is the value of. total revenue (TR3) given this pricing and production decision? Explain your answer making sure to. Q3 = ________________. P3 = ________________. TR3 = _______________. For a linear demand curve, the price and quantity that will maximize total revenue is the price and. quantity associated with the midpoint of the demand curve. At the midpoint the price elasticity of demand. is equal to one: if the firm is charging a price greater than the price associated with the midpoint of the. demand curve, the firm can increase its total revenue by reducing its price since when demand is elastic. total revenue increases as price decreases; if the firm is charging a price less than the price associated with. the midpoint of the demand curve, the firm can increase its total revenue by increasing its price since. when demand is inelastic total revenue increases as price increases.. Given the new demand curve, P = 600 (1/2)Q, we can quickly find the midpoint coordinates as (Q3, P3). = (600, 300). When the firm produces 600 units and charges a price of \$300 per unit, its total revenue. (TR3) equals \$180,000.. 2. (20 points total) Martha consumes apples and chocolate candies. Suppose Marthas budget line (BL1). for apples (A) and chocolate candies (C) is given below. Suppose you know that the price of apples is \$2. per apple but you are not told the price of chocolate candies or Marthas income.. a. (2 points) Given the graph above, what is Marthas income? Explain how you found your answer.. Since Martha can purchase 40 apples if she only buys apples, this implies that her income must by 40(\$2). = \$80.. 4. b. (2 points) Given the graph above, what is the price of one unit of chocolate candies? Explain your. In (a) we determined Marthas income is \$80. We also know that Martha if she buys only chocolate. candies can afford 20 units. Using this information we can find the price of a unit of chocolate candies by. using the relationship. Income = (Price of Chocolate Candies)(Quantity of Chocolate Candies) + (Price of Apples). (Quantity of Apples). If Martha purchases no apples, then this equation can be simplified to. Income = (Price of Chocolate Candies)(Quantity of Chocolate Candies). 80 = (Price of Chocolate Candies)(20). Price of Chocolate Candies = \$4 per unit. Suppose Marthas tastes and preferences are such that she maximizes her utility given her budget line. when she consumes 28 apples.. c. (2 points) Given this new information and the above graph, how many chocolate candies does Martha. consume when she maximizes her utility subject to the constraints imposed by her income and the prices. of the two goods?. We know that Income = \$80, the price of apples is \$2, the price of chocolates is \$4, and the quantity of. apples is 28. Use this information to solve for the quantity of chocolate candies.. Income = (Price of Chocolate Candies)(Quantity of Chocolate Candies) + (Price of Apples). (Quantity of Apples). 80 = (4)(Quantity of Chocolate Candies) + (2)(28). 24 = 4(Quantity of Chocolate Candies). Quantity of Chocolate Candies = 6. d. (2 points) In the above graph label Marthas optimal consumption bundle on BL1 as point A. Indicate. the numerical coordinates of this point on your graph.. Now, suppose the price of apples increases to \$4 per apple while there are no changes in Marthas income. or the price of chocolate candies. When the price of apples increases to \$4, Martha finds she maximizes. her utility by consuming 10 units of chocolate candies.. e. (2 points) Given this new information, in the above graph draw Marthas new budget line (BL2) and. indicate the numeric values for the x- and y- axis intercepts.. f. (2 points) Given this new information, calculate the number of apples Martha consumes when Martha. maximizes her utility given the new price of apples. Show your work for full credit.. 5. ## Income = (Price of Chocolate Candies)(Quantity of Chocolate Candies) + (Price of Apples). (Quantity of Apples). 80 = (4)(10) + (4)(Quantity of Apples). 40 = 4(Quantity of Apples). Quantity of Apples = 10. g. (2 points) In the graph label Marthas optimal consumption bundle B when the price of apples increases. to \$4. Indicate the numerical coordinates of this point on the graph.. See graph below:. h. (6 points) Assume Marthas demand curve for apples is linear. Given the two optimal consumption. bundles at A and B, draw Marthas demand curve in the space below and then write an equation in slopeintercept form for this curve. Make sure your graph is labeled clearly and completely labeled. (Hint: the. numbers are messy here-just leave them as improper fractions!). We know two points that sit on Marthas linear demand curve for apples: when the price of apples is \$2,. Martha maximizes her utility when she consumes 28 apples; and when the price of apples is \$4, Martha. maximizes her utility when she consumes 10 apples. Use these two points to write the equation for. Marthas demand curve as P = (46/9) (1/9)Q. The figure below illustrates this demand curve:. 3. (15 points) Use the graphs below of a perfectly competitive industry and a representative firm in the. short run and in the long run to answer this set of questions.. Short-run:. Use the graphs below to answer the questions about the long-run (this graph is just a duplicate of. the first set of graphs, but you will be using it to show the long-run adjustment process).. a. (2 points) In the first row of graphs (the pair of graphs identified as Short Run) indicate the short-run. equilibrium using the following symbols:. Q1 = the short-run market level of production. P1 = the short-run market price. q1 = the short-run level of production for the representative firm. See graph after part (b).. b. (2 points) In the first row of graphs if the representative firm is earning economic profits shade in the. area representing those profits. On the graph indicate whether these are positive or negative profits.. c. (2 points) Given your analysis in the short-run, what do you predict will happen in this industry in the. Since the representative firm is earning negative economic profits in the short run, we would anticipate. the exiting of firms in the long run. This will cause the market supply curve to shift to the left until the. market price is equal to the minimum point on the representative firms ATC curve.. d. (3 points) In the second row of graphs (the pair of graphs identified as Long Run) depict the long-run. equilibrium situation in this market. If any demand or supply curves shift, indicate these shifts on your. graphs. On the graphs identify:. Q2 = the long-run market level of production. P2 = the long-run market price. q2 = the long-run level of production for the representative firm. e. (1 point) In the long run, the representative firm in this market will earn ___________________. economic profits.. Zero. f. Suppose you are given this set of graphs and told that peoples incomes have increased and the good is. a normal good. Given this information answer the following set of questions:. i. (2 points) In the short-run, this change in income will result in the price of the good. __________________ relative to its initial level.. Increasing. ii. (2 points) In the short-run, this change in income will result in the quantity of the good. produced by the representative firm to ______________________ relative to its initial level.. Increase. iii. (1 point) In the long run, the equilibrium price in this market given this change in income and. holding everything else constant, will be _____________ relative to the initial long-run. equilibrium price in this market.. The same. 10. ## MULTIPLE CHOICE QUESTIONS (20 questions worth 2.5 points each):. Use the information below to answer the next two questions:. Your business currently has two customers with the following individual demand curves:. Customer Ones Demand Curve: P = 10 -2Q. Customer Twos Demand Curve: P = 20 2Q. market when you set price to be \$7.5?. a. TR = \$6.25. b. TR = \$56.25. c. TR = \$100. d. TR = \$12.50. 2. Your company is trying to decide whether to raise the price of your product from \$\$7.5 by \$0.50 or. lower it by \$0.50. Given the above information what would you recommend if the sole goal of your firm. is to increase the total revenue?. a. The firm should increase the price by \$0.50.. b. The firm should decrease the price by \$0.50.. c. The firm could increase their total revenue if they increase or decrease the price by \$0.50 from its. initial price level \$7.5.. d. The firm should maintain its current price.. 3. Demand for a good is considered elastic. This implies that. a. Increasing the price of this good in the market will result in greater total revenue for producers.. b. Decreasing the price of this good in the market will result in greater total revenue for producers.. c. The market is currently maximizing its total revenue from selling this good.. d. Consumers of this good are not price sensitive.. ## Use the information below to answer the next two questions.. Shawn and Mary are currently working and living in Tucson. But, they are considering moving to either. St. Paul or New York City. The table below provides information about the salaries they would earn in the. three cities as well as information about the price index for each city. Shawn and Mary have decided they. will move to whichever city provides them the greatest real income. Assume that their nominal salaries. and the price index will not change in the future.. City. Tucson. Tucson. St. Paul. St. Paul.
|
Nominal Salary. \$40,000. \$50,000. \$50,000. \$75,000. Employee. Mary. Shawn. Mary. Shawn. 100. 100. 125. 125. 11. New York City. \$100,000. \$100,000. Mary. Shawn. 250. 250. ## 4. From the given information which of the following statements is true?. a. The base year for the price index is this year.. b. Tucson is likely the city that was used to compare the cost of living in other cities when this. information was gathered.. c. The cost of living in New York City is 100% higher than the cost of living in Tucson.. d. The cost of living in St. Paul 150% less than the cost of living in New York City.. 5.Given that Shawn and Mary want to maximize their joint real income where should they live? Base. your analysis solely on the information you have been provided.. a. Shawn and Mary should live in either St. Paul or Tucson since their real income will be equivalent in. these two cities.. b. Shawn and Mary should live in New York since both of their incomes are substantially higher than in. Tucson or St. Paul.. c. Once you factor in the cost of living for each of these cities the real income in all three cities is the. same.. d. Shawn and Mary should live in St. Paul since their real income will be maximized there.. 6. The cross-price elasticity of demand measures. a. The percentage change in the quantity demanded of the good to the percentage change in the price of. the good.. b. The percentage change in the quantity demanded of one good to the percentage change in the price of a. related good.. c. The percentage change in the quantity demanded of the good to the percentage change in income.. d. The percentage change in the price of one good to the percentage change in the price of the other good.. 7. Jacob has \$100 of income that he will spend on hot dogs or cheeseburgers during his month long. vacation. Jacob knows that hot dogs cost \$5 each and cheeseburgers cost \$10 each. Given this information. determine which of the budget lines below represents Jacobs budget line.. 12. ## Answer for #7 is (b). Use the following information to answer the next two questions.. Suppose Oscars income is \$1000 and he uses all of this income to buy either ice cream or pizzas. When. the price of ice cream is \$5 per serving and the price of pizza is \$10 per pizza Oscar maximizes his utility. by purchasing 40 pizzas. When the price of ice cream is \$10 per serving and the price of pizza is \$10 per. pizza Oscar maximizes his utility by purchasing 60 pizzas.. 8. From this information derive Oscars demand for ice cream. Assume his demand for ice cream is linear.. Oscars demand for ice cream is. a. P = 74 (8/5)Q. b. P = (1/4)Q. c. P = 100 (1/2)Q. d. P = 12.5 (1/16)Q. 9. Given the above information you are also told that Oscars income compensated budget line (the. budget line we called BL3 when discussing the income and substitution effects) results in Oscar. maximizing his utility when he consumes 100 servings of ice cream and 70 slices of pizza. From this. information you can conclude that Oscars substitution effect with respect to ice cream is equal to. a. 80 units of ice cream.. b. 60 units of ice cream.. c. 30 units of ice cream.. d. 20 units of ice cream.. 13. 10. Mary, who prides herself on her business knowledge and expertise, states that no matter what your. demand curve looks like it is always best to charge the highest possible price. Which of the following. statements are true given this information? In your answer assume that the demand curve that Marys. business faces for their product is downward sloping and linear.. a. Mary missed the lecture on elasticity in her introductory microeconomics class.. b. If the demand for Marys product is elastic, Mary will earn more total revenue by lowering the price of. her product.. c. If Mary charges the price associated with the midpoint for her product then her company will maximize. its total revenue.. 11. Joeys total utility from consuming 10 cookies is equal to 100 utils. His total utility from consuming. 11 cookies is 104 utils. Joeys total utility from consuming 5 slices of pizza is 500 utils and his total utility. from consuming 6 slices of pizza is 508 utils. Furthermore, you know that the price of a slice of pizza is. twice the price of a cookie. Also assume that Joey has spent all his income on cookies and pizzas. Given. this information, which of the following statements is true?. a. Joey is not maximizing his utility from consuming these two goods: he would have greater satisfaction. b. Joey is maximizing his utility from consuming these two goods: he cannot select a different. combination of pizza and cookies that will provide him with greater utility.. c. Joey is not maximizing his utility from consuming these two goods: he would have greater satisfaction. if he consumed more pizza.. d. Joey is not maximizing his utility from consuming these two goods: he would have greater satisfaction. if he consumed more pizza and cookies.. 12. There are five players on the Wisconsin Hotshots, a womens basketball team. The average height of. these five players is five feet eight inches tall. If the Hotshots add a new player to the team that is six feet. tall then. a. The average height of the team will increase.. b. The average height of the team will be unaffected.. c. The average height of the team will decrease.. d. The average height of the team may increase, decrease or remain the same.. 13. In the short run, which of the following statements is true?. a. As output increases average total cost decreases.. b. As output increases average variable cost decreases.. c. No matter what level of output the firm produces, its fixed cost is constant.. d. Answers (a) and (b) are both correct.. e. Answers (a) and (c) are both correct.. f. Answers (b) and (c) are both correct.. g. Answers (a), (b) and (c) are all correct.. 14. 14. Katherine operates a bakery in Glenwood. Her costs of operation include \$5000 a year in property. rental for the land and building where her bakery is located; \$1.50 per loaf of bread for the flour, sugar. and other ingredients that go into making her product; and \$7.00 an hour for each hour of labor she hires.. She estimates that she needs four hours of labor to produce 100 loaves of bread, eight hours of labor to. produce 200 loaves of bread, etc. Given this information which of the following statements is true?. a. When Katherine produces 400 loaves of bread her labor cost is \$28.00.. b. When Katherine produces 400 loaves of bread her fixed costs are equal to \$5000.. c. When Katherine produces 400 loaves of bread her variable costs equal \$600.. d. When Katherine produces 400 loaves of bread her labor cost is \$112.00 while her fixed cost is equal to. \$0.. 15. Consider Montys Garage, a small business offering car repair. Suppose you are told that as Monty. provides more units of his output (i.e., he repairs more cars) his average variable cost is increasing. From. this you know. a. That Montys average total cost is increasing as you move rightward along the quantity axis.. b. That Montys average fixed cost is increasing as your move rightward along the quantity axis.. c. That Montys marginal cost of production exceeds his average variable cost of production.. d. Monty can increase his profits by repairing a smaller number of cars.. Use the figure below of a perfectly competitive market and a representative firm to answer the next three. questions.. 16. In the short run, which of the following statements is true given the above graphs?. a. There are (D/c) firms in the industry and each firm is earning positive economic profits.. b. Firms in the industry are earning negative economic profits and this indicates that the market has too. many firms.. 15. c. Each firms revenue is sufficient to cover their variable costs of production and each firm is earning a. normal or zero economic profit.. d. Answers (a), (b) and (c) are all true.. e. Answers (a) and (b) are both true.. 17. In the long run, which of the following statements is true given the above graphs?. a. In the long run firms will exit the industry until price increases to n and there are (C/b) firms in the. industry.. b. In the long run the market supply curve will shift to the right until prices fall to r and there will be. (E/b) firms in the industry.. c. In the long run the market supply curve will shift to the left until price is n and each firm is producing. d units of output.. d. In the long run the market demand curve will shift until all firms in the industry earn zero economic. profits.. 18. In the long run, which of the following statements is true given the above graph?. a. Total revenue for the industry will equal n*C and total revenue for the representative firm will equal. n*d.. b. Total revenue for the industry will equal n*F and the total number of firms in the industry will by F/d.. c. Total revenue for the industry will equal n*F and total revenue for the representative firm will equal the. firms fixed costs.. d. Total revenue for the industry will equal n*F and the representative firms total costs are equal to its. variable costs.. 19. A firm in a perfectly competitive market is currently producing 10 units of output and its average cost. of production is \$2 per unit of output. The firm also knows that if it increases its production to 12 units,. its average total cost of production will increase to \$2.25 per unit of output. This information implies. a. That the firm needs to increase its level of production to reach the long run equilibrium price and. quantity in this market.. b. That the firm will maximize profit by producing 10 units of output in the long run.. c. That the equilibrium long run price in this market is \$2.. d. That the firm may need to decrease its level of production in the long run in order to reach the long run. equilibrium price and quantity in this market.. 20. A firm experiences increasing returns to scale. This indicates that. a. The firms average costs of production fall as the size of the firm increases.. b. The firms average costs of production fall as the firm hires more of its variable inputs holding. everything else constant.. c. The firms average variable cost of production reaches its minimum point at a relatively low level of. production for the firm.. d. The number of firms in the industry should increase in the long run to take advantage of the increasing. returns to scale.. 16.
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# [SOLVED]Change of variables heat equation
#### dwsmith
##### Well-known member
$\alpha^2T_{xx} = T_t + \beta(T - T_0)$
where $$\beta$$ is a constant and $$T_0$$ is the temperature of the surrounding medium. The initial temperature distribution is $$T(x, 0) = f(x)$$ and the ends $$x = 0$$ and $$x = \ell$$ are maintained at $$T_1$$ and $$T_2$$ when $$t > 0$$.
Show that the substitution $$T(x, t) = T_0 + U(x, t)e^{-\beta t}$$ reduces the problem to the following one:
$\alpha^2U_{xx} = U_t$
with new initial conditions and boundary conditions for $$U$$.
With that substitution, I obtain:
\begin{align}
\alpha^2U_{xx} &= -\beta(U_t - T_0U)\\
\alpha_1^2U_{xx} &= U_t - T_0U
\end{align}
What is going wrong?
#### topsquark
##### Well-known member
MHB Math Helper
Re: Change of varibles heat equation
$\alpha^2T_{xx} = T_t + \beta(T - T_0)$
where $$\beta$$ is a constant and $$T_0$$ is the temperature of the surrounding medium. The initial temperature distribution is $$T(x, 0) = f(x)$$ and the ends $$x = 0$$ and $$x = \ell$$ are maintained at $$T_1$$ and $$T_2$$ when $$t > 0$$.
Show that the substitution $$T(x, t) = T_0 + U(x, t)e^{-\beta t}$$ reduces the problem to the following one:
$\alpha^2U_{xx} = U_t$
with new initial conditions and boundary conditions for $$U$$.
With that substitution, I obtain:
\begin{align}
\alpha^2U_{xx} &= -\beta(U_t - T_0U)\\
\alpha_1^2U_{xx} &= U_t - T_0U
\end{align}
What is going wrong?
I'm not quite sure of the problem here. For example,
$$T = T_0 + U(x, t)e^{- \beta t} \implies T_t = U_t e^{- \beta t} - \beta U e^{- \beta t}$$
Do the same for U_x and U_xx, then sub into the original equation. There are a ton of cancellations which gives you the final answer.
Are you having problems with the derivatives or is it something else?
-Dan
#### dwsmith
##### Well-known member
Re: Change of varibles heat equation
I'm not quite sure of the problem here. For example,
$$T = T_0 + U(x, t)e^{- \beta t} \implies T_t = U_t e^{- \beta t} - \beta U e^{- \beta t}$$
Do the same for U_x and U_xx, then sub into the original equation. There are a ton of cancellations which gives you the final answer.
Are you having problems with the derivatives or is it something else?
-Dan
I forgot to use the product rule
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# [SOLVED]Change of variables heat equation. #### dwsmith. ##### Well-known member. $\alpha^2T_{xx} = T_t + \beta(T - T_0)$. where $$\beta$$ is a constant and $$T_0$$ is the temperature of the surrounding medium. The initial temperature distribution is $$T(x, 0) = f(x)$$ and the ends $$x = 0$$ and $$x = \ell$$ are maintained at $$T_1$$ and $$T_2$$ when $$t > 0$$.. Show that the substitution $$T(x, t) = T_0 + U(x, t)e^{-\beta t}$$ reduces the problem to the following one:. $\alpha^2U_{xx} = U_t$. with new initial conditions and boundary conditions for $$U$$.. With that substitution, I obtain:. \begin{align}. \alpha^2U_{xx} &= -\beta(U_t - T_0U)\\. \alpha_1^2U_{xx} &= U_t - T_0U. \end{align}. What is going wrong?. #### topsquark. ##### Well-known member. MHB Math Helper. Re: Change of varibles heat equation. $\alpha^2T_{xx} = T_t + \beta(T - T_0)$. where $$\beta$$ is a constant and $$T_0$$ is the temperature of the surrounding medium. The initial temperature distribution is $$T(x, 0) = f(x)$$ and the ends $$x = 0$$ and $$x = \ell$$ are maintained at $$T_1$$ and $$T_2$$ when $$t > 0$$.. Show that the substitution $$T(x, t) = T_0 + U(x, t)e^{-\beta t}$$ reduces the problem to the following one:. $\alpha^2U_{xx} = U_t$. with new initial conditions and boundary conditions for $$U$$.
|
With that substitution, I obtain:. \begin{align}. \alpha^2U_{xx} &= -\beta(U_t - T_0U)\\. \alpha_1^2U_{xx} &= U_t - T_0U. \end{align}. What is going wrong?. I'm not quite sure of the problem here. For example,. $$T = T_0 + U(x, t)e^{- \beta t} \implies T_t = U_t e^{- \beta t} - \beta U e^{- \beta t}$$. Do the same for U_x and U_xx, then sub into the original equation. There are a ton of cancellations which gives you the final answer.. Are you having problems with the derivatives or is it something else?. -Dan. #### dwsmith. ##### Well-known member. Re: Change of varibles heat equation. I'm not quite sure of the problem here. For example,. $$T = T_0 + U(x, t)e^{- \beta t} \implies T_t = U_t e^{- \beta t} - \beta U e^{- \beta t}$$. Do the same for U_x and U_xx, then sub into the original equation. There are a ton of cancellations which gives you the final answer.. Are you having problems with the derivatives or is it something else?. -Dan. I forgot to use the product rule.
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# Thread: Find a function that is defined and continuous on the closure of a set E.
1. ## Find a function that is defined and continuous on the closure of a set E.
If f is defined and uniformly continuous on E, can somebody show me that there's a function g defined and continuous on the closure of E such that g = f on E?
Since, for every limit point x of E, there is a sequence {xn} in E such that limxn = x, I define g as below.
g(x)=f(x) if x in E; g(x)=limnf(xn) if x belongs to (the closure of E)\E.
Would it be a right start? If so, then how do I prove that g is defined and continuous on (the closure of E)\E by using the fact that f is uniformly continuous?
2. ## Re: Find a function that is defined and continuous on the closure of a set E.
Originally Posted by Rita
If f is defined and uniformly continuous on E, can somebody show me that there's a function g defined and continuous on the closure of E such that g = f on E?
You should have a lemma that says that if $(x_n)$ is a Cauchy sequence in $E$ then $f(x_n)$ is a Cauchy sequence. And that last sequence is converges to some $b\in\overline{E}$.
More over if both $(x_n)~\&~(y_n)$ are Cauchy sequences in $E$ and $(x_n)\to b~\&~(y_n)\to b$ then $f(x_n)\to f(b)~\&~f(y_n)\to f(b)$. Now that gives you a natural way to define the extension of $f$ .
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# Thread: Find a function that is defined and continuous on the closure of a set E.. 1. ## Find a function that is defined and continuous on the closure of a set E.. If f is defined and uniformly continuous on E, can somebody show me that there's a function g defined and continuous on the closure of E such that g = f on E?. Since, for every limit point x of E, there is a sequence {xn} in E such that limxn = x, I define g as below.. g(x)=f(x) if x in E; g(x)=limnf(xn) if x belongs to (the closure of E)\E.. Would it be a right start? If so, then how do I prove that g is defined and continuous on (the closure of E)\E by using the fact that f is uniformly continuous?. 2.
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## Re: Find a function that is defined and continuous on the closure of a set E.. Originally Posted by Rita. If f is defined and uniformly continuous on E, can somebody show me that there's a function g defined and continuous on the closure of E such that g = f on E?. You should have a lemma that says that if $(x_n)$ is a Cauchy sequence in $E$ then $f(x_n)$ is a Cauchy sequence. And that last sequence is converges to some $b\in\overline{E}$.. More over if both $(x_n)~\&~(y_n)$ are Cauchy sequences in $E$ and $(x_n)\to b~\&~(y_n)\to b$ then $f(x_n)\to f(b)~\&~f(y_n)\to f(b)$. Now that gives you a natural way to define the extension of $f$ .
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Degenerate solutions don’t exist in one dimension
Required math: calculus
Required physics: Schrödinger equation
References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.45.
If two distinct solutions to the Schrödinger equation (that is, solutions that aren’t merely multiples of each other) have the same energy, they are said to be degenerate. Although degenerate states do exist in nature (that is, in a three-dimensional world), there are no degenerate solutions to the Schrödinger equation in any one-dimensional system, regardless of the potential.
To see this, assume there are distinct solutions ${\psi_{1}}$ and ${\psi_{2}}$ that have the same energy ${E}$. Then the Schrödinger equation for these two wave functions is:
$\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi_{1}}{\partial x^{2}}+V\psi_{1}$ $\displaystyle =$ $\displaystyle E\psi_{1}\ \ \ \ \ (1)$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi_{2}}{\partial x^{2}}+V\psi_{2}$ $\displaystyle =$ $\displaystyle E\psi_{2} \ \ \ \ \ (2)$
Multiply the first equation by ${\psi_{2}}$ and the second by ${\psi_{1}}$and subtract to get:
$\displaystyle \psi_{2}\psi{}_{1}^{\prime\prime}-\psi_{1}\psi{}_{2}^{\prime\prime}=0 \ \ \ \ \ (3)$
The expression on the left is the derivative of ${\psi_{1}\psi'_{2}-\psi_{2}\psi'_{1}}$ so we can integrate it to get:
$\displaystyle \psi_{1}\psi'_{2}-\psi_{2}\psi'_{1}=C \ \ \ \ \ (4)$
where ${C}$ is some constant. If the domain of these functions extends to infinity, then the functions and their derivatives must tend to zero at infinity in order for them to be normalizable. Since 4 is valid for all ${x}$, ${C}$ must be zero. We can then rearrange 4 as:
$\displaystyle \frac{\psi'_{2}}{\psi_{2}}=\frac{\psi'_{1}}{\psi_{1}} \ \ \ \ \ (5)$
Integrating this leads to ${\ln\psi_{2}=\ln\psi_{1}+K}$ for another constant ${K}$, so ${\psi_{2}=e^{K}\psi_{1}}$and any two solutions with the same energy must be multiples of each other. Thus there are no distinct degenerate states in one dimension.
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Degenerate solutions don’t exist in one dimension. Required math: calculus. Required physics: Schrödinger equation. References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.45.. If two distinct solutions to the Schrödinger equation (that is, solutions that aren’t merely multiples of each other) have the same energy, they are said to be degenerate. Although degenerate states do exist in nature (that is, in a three-dimensional world), there are no degenerate solutions to the Schrödinger equation in any one-dimensional system, regardless of the potential.. To see this, assume there are distinct solutions ${\psi_{1}}$ and ${\psi_{2}}$ that have the same energy ${E}$. Then the Schrödinger equation for these two wave functions is:. $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi_{1}}{\partial x^{2}}+V\psi_{1}$ $\displaystyle =$ $\displaystyle E\psi_{1}\ \ \ \ \ (1)$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi_{2}}{\partial x^{2}}+V\psi_{2}$ $\displaystyle =$ $\displaystyle E\psi_{2} \ \ \ \ \ (2)$. Multiply the first equation by ${\psi_{2}}$ and the second by ${\psi_{1}}$and subtract to get:. $\displaystyle \psi_{2}\psi{}_{1}^{\prime\prime}-\psi_{1}\psi{}_{2}^{\prime\prime}=0 \ \ \ \ \ (3)$.
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The expression on the left is the derivative of ${\psi_{1}\psi'_{2}-\psi_{2}\psi'_{1}}$ so we can integrate it to get:. $\displaystyle \psi_{1}\psi'_{2}-\psi_{2}\psi'_{1}=C \ \ \ \ \ (4)$. where ${C}$ is some constant. If the domain of these functions extends to infinity, then the functions and their derivatives must tend to zero at infinity in order for them to be normalizable. Since 4 is valid for all ${x}$, ${C}$ must be zero. We can then rearrange 4 as:. $\displaystyle \frac{\psi'_{2}}{\psi_{2}}=\frac{\psi'_{1}}{\psi_{1}} \ \ \ \ \ (5)$. Integrating this leads to ${\ln\psi_{2}=\ln\psi_{1}+K}$ for another constant ${K}$, so ${\psi_{2}=e^{K}\psi_{1}}$and any two solutions with the same energy must be multiples of each other. Thus there are no distinct degenerate states in one dimension.
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# Algebra
posted by .
I have two equations that I need to solve using the elimination method the system is x = y = 9 and 2x - y = -3
• Algebra -
i think u wrote it incorrectly did u mean x+y= 9?? if so
x +y= 9
2x-y=-3
3x = 6
x = 2
or
x +y= 9 (-2) -2x-2y=-18
2x-y=-3 2x- y=-3
-3y=-21
y=7
u can use the elimination process to solve for both or u can solve for one and plug it in any one of the equations to solve for the other
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# Algebra. posted by .. I have two equations that I need to solve using the elimination method the system is x = y = 9 and 2x - y = -3. • Algebra -. i think u wrote it incorrectly did u mean x+y= 9?? if so. x +y= 9. 2x-y=-3. 3x = 6. x = 2. or. x +y= 9 (-2) -2x-2y=-18. 2x-y=-3 2x- y=-3. -3y=-21. y=7. u can use the elimination process to solve for both or u can solve for one and plug it in any one of the equations to solve for the other. ## Similar Questions. 1. ### Algebra. Solving system of linear equations by using the apropiate method name the method and the x,y coordinates 1)X=4y+8 2x-8y=-3 2)3x-7y=6 elimination method 2x+7y=4 (2,0) 3)5x-2y=12 elimination method 3x-2y=-2 (7,23/2) 4)9x-8y=42 elimination …. 2. ### Math. Use elimination to solve each system of equations. x+y-2z=10 8x-9y-z=5 3x+4y+2z=-10 I need to figure out what x,y, and z each equal. I must use the elimination method.. 3. ### Algebra. Use the elimination method to solve the following system of equations. x + 3y – z = 2 x – 2y + 3z = 7 x + 2y – 5z = –21. 4. ### College Algebra. Please help! x-2y+z=7 2x+y-z=0 3x+2y-2z=-2 a. Solve the above system of equations using Gaussian Elimination or Gauss-Jordan Elimination. You must show row operations.
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b. Solve the above system of equations using Cramer's Rule.. 5. ### Math. Please help! x-2y+z=7 2x+y-z=0 3x+2y-2z=-2 a. Solve the above system of equations using Gaussian Elimination or Gauss-Jordan Elimination. You must show row operations. b. Solve the above system of equations using Cramer's Rule.. 6. ### algebra. i need help with these problems (either a solution or how to plug them into a graphing calculator): 1) Solve by substitution or elimination 4/x + 1/y + 2/z = 4 2/x + 3/y - 1/z = 1 1/x + 1/y + 1/z = 4 2)Solve the system of equations …. 7. ### system of equations. solve using any method and identify the system as consistent, inconsistent or dependent. 3x-4y=8 6x-2y=10 I want to use elimination. this is taking me forever!. 8. ### Algebra/Elimination method. Solve using the elimination method. If the system has no solution or an infinite number of solutions, state this. -27x - 18y = -180 9x + 6y = 60. 9. ### Algebra. Solve the following systems of equations using the addition (elimination) method. What type of system is it?. 10. ### Math. I need to use the Gaussian Elimination to solve a system of equations, but I have no idea on how to do that. I remember doing these in high school, but since its been a while I can't remember exactly how to do it. Here's the system …. More Similar Questions.
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# How To: Calculate probability in "at least one" problems
Calculating probablities can be used to help us make decision. PatrickJMT explains how to calculate probability in an "either A or not A" scenario. The probability of A plus the probability of not A is equal to one. Therefore, the probability of A is equal to one minus the probability of not A ; P(A)= 1 - P(not A). The probability of a major earthquake in San Francisco over a period of time is used as an example. The probablity of an earthquake of a magnitude of 7.5 or greater in San Francisco in any given year is said to be 2 percent or 0.02. Assuming that the earthquakes are independent of each other, he calculates the probability of San Francisco having a major earthquake in the next 25 years. This is one minus 0.98 to the power of 25, or about 0.397. There is therefore an approximately 40 percent chance that San Francisco will have a major earthquake in the next 25 years. He shows us how the scenario gets worse for San Francisco over the next 50 and 100 years with increasing probabilities that a major earthquake will happen.
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# How To: Calculate probability in "at least one" problems. Calculating probablities can be used to help us make decision. PatrickJMT explains how to calculate probability in an "either A or not A" scenario. The probability of A plus the probability of not A is equal to one. Therefore, the probability of A is equal to one minus the probability of not A ; P(A)= 1 - P(not A). The probability of a major earthquake in San Francisco over a period of time is used as an example. The probablity of an earthquake of a magnitude of 7.5 or greater in San Francisco in any given year is said to be 2 percent or 0.02. Assuming that the earthquakes are independent of each other, he calculates the probability of San Francisco having a major earthquake in the next 25 years. This is one minus 0.98 to the power of 25, or about 0.397.
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There is therefore an approximately 40 percent chance that San Francisco will have a major earthquake in the next 25 years. He shows us how the scenario gets worse for San Francisco over the next 50 and 100 years with increasing probabilities that a major earthquake will happen.. Just updated your iPhone? You'll find new features for Podcasts, News, Books, and TV, as well as important security improvements and fresh wallpapers. Find out what's new and changed on your iPhone with the iOS 17.5 update.. • Hot. • Latest.
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# Lessons In AC Electrical Circuits
## AC resistor circuits
If we were to plot the current and voltage for a very simple AC circuit consisting of a source and a resistor (Figure 1), it would look something like this: (Figure 2).
Because the resistor simply and directly resists the flow of electrons at all periods of time, the waveform for the voltage drop across the resistor is exactly in phase with the waveform for the current through it.
We can look at any point in time along the horizontal axis of the plot and compare those values of current and voltage with each other (any “snapshot” look at the values of a wave are referred to as instantaneous values, meaning the values at that instant in time).
When the instantaneous value for current is zero, the instantaneous voltage across the resistor is also zero. Likewise, at the moment in time where the current through the resistor is at its positive peak, the voltage across the resistor is also at its positive peak, and so on.
At any given point in time along the waves, Ohm’s Law holds true for the instantaneous values of voltage and current.
We can also calculate the power dissipated by this resistor, and plot those values on the same graph: (Figure 3).
Note that the power is never a negative value. When the current is positive (above the line), the voltage is also positive, resulting in a power (p=ie) of a positive value.
Conversely, when the current is negative (below the line), the voltage is also negative, which results in a positive value for power (a negative number multiplied by a negative number equals a positive number). This consistent “polarity” of power tells us that the resistor is always dissipating power, taking it from the source and releasing it in the form of heat energy.
Whether the current is positive or negative, a resistor still dissipates energy.
Title: Lessons In AC Electrical Circuits – Tony R. Kuphaldt Format: PDF Size: 4.40 MB Pages: 570 Download: Right here | Get Download Updates | Get Technical articles
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Page edited by E.C. (Google).
1. Bilal Fanous
Apr 19, 2015
Dear friends I like to know exactly how to calculate the voltage drop and the cable size of a circuit. Please can you explain it in details. Thank for your help and good bless
Bilal Fanous
• Sergio Correia
Jan 06, 2017
Cable resistance is calculated with R=p x L / S , where p is the conductivity of the material, L is the total length of the cable wires and S is the cross-section of the cable wires.
After calculating the cable resistance, voltage drop is simply calculated with V=R x I where I is the nominal current which flows through the cable.
If you need to reduce voltage drop, decrease the cable resistance either by decreasing the length of the cable or increasing the cross-section.
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# Lessons In AC Electrical Circuits. ## AC resistor circuits. If we were to plot the current and voltage for a very simple AC circuit consisting of a source and a resistor (Figure 1), it would look something like this: (Figure 2).. Because the resistor simply and directly resists the flow of electrons at all periods of time, the waveform for the voltage drop across the resistor is exactly in phase with the waveform for the current through it.. We can look at any point in time along the horizontal axis of the plot and compare those values of current and voltage with each other (any “snapshot” look at the values of a wave are referred to as instantaneous values, meaning the values at that instant in time).. When the instantaneous value for current is zero, the instantaneous voltage across the resistor is also zero. Likewise, at the moment in time where the current through the resistor is at its positive peak, the voltage across the resistor is also at its positive peak, and so on.. At any given point in time along the waves, Ohm’s Law holds true for the instantaneous values of voltage and current.. We can also calculate the power dissipated by this resistor, and plot those values on the same graph: (Figure 3).. Note that the power is never a negative value. When the current is positive (above the line), the voltage is also positive, resulting in a power (p=ie) of a positive value.. Conversely, when the current is negative (below the line), the voltage is also negative, which results in a positive value for power (a negative number multiplied by a negative number equals a positive number). This consistent “polarity” of power tells us that the resistor is always dissipating power, taking it from the source and releasing it in the form of heat energy.. Whether the current is positive or negative, a resistor still dissipates energy.. Title: Lessons In AC Electrical Circuits – Tony R. Kuphaldt Format: PDF Size: 4.40 MB Pages: 570 Download: Right here | Get Download Updates | Get Technical articles. ### SEARCH: Articles, software & guides.
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Premium membership gives you an access to specialized technical articles and extra premium content (electrical guides and software).. Page edited by E.C. (Google).. 1. Bilal Fanous. Apr 19, 2015. Dear friends I like to know exactly how to calculate the voltage drop and the cable size of a circuit. Please can you explain it in details. Thank for your help and good bless. Bilal Fanous. • Sergio Correia. Jan 06, 2017. Cable resistance is calculated with R=p x L / S , where p is the conductivity of the material, L is the total length of the cable wires and S is the cross-section of the cable wires.. After calculating the cable resistance, voltage drop is simply calculated with V=R x I where I is the nominal current which flows through the cable.. If you need to reduce voltage drop, decrease the cable resistance either by decreasing the length of the cable or increasing the cross-section.
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# Number Representation
Number Representation. Lecture 20 4.3.2001. Topics. How are numeric data items actually stored in computer memory? How much space (memory locations) is allocated for each type of data? int, float, char, etc. How are characters and strings stored in memory?. Binary number system.
## Number Representation
E N D
### Presentation Transcript
1. Number Representation Lecture 20 4.3.2001.
2. Topics • How are numeric data items actually stored in computer memory? • How much space (memory locations) is allocated for each type of data? • int, float, char, etc. • How are characters and strings stored in memory?
3. Binary number system Decimal number system • Ten digits : 0,1,2,3,4,5,6,7,8,9 • Every digit position has a weight which is a power of 10. • Base or radix is 10. • Two digits : 0,1 • Every digit position has a weight which is a power of 2. • Base or radix is 2.
4. Decimal Number • 136.25 : What does this number actually mean ? 102 * 1 = 100.0 101 * 3 = 30.0 100 * 6 = 6.0 10-1 * 2 = 0.2 10-2 * 5 = 0.05
5. Binary Number • 1101.01: What does this number mean? 23 * 1 = 1000.0 (8 in decimal) 22 * 1 = 100.0 (4 in decimal) 21 * 3 = 0.0 (0 in decimal) 20 * 6 = 1.0 (1 in decimal) 2-1 * 2 = 0.0 (0.0 in decimal) 2-2 * 5 = 0.01 (0.25 in decimal)
6. First integers and their binary equivalent decimal binary 0 0000 (0*23 + 0*22 + 0*21 + 0*20) 1 0001 (0*23 + 0*22 + 0*21 + 1*20) 2 0010 (0*23 + 0*22 + 1*21 + 0*20) 3 0011 (0*23 + 0*22 + 1*21 + 1*20) 4 0100 (0*23 + 1*22 + 0*21 + 0*20) 5 0101 (0*23 + 1*22 + 0*21 + 1*20) 6 0110 (0*23 + 1*22 + 1*21 + 0*20) 7 0111 (0*23 + 1*22 + 1*21 + 1*20) 8 1000 (1*23 + 0*22 + 0*21 + 0*20) 9 1001 (1*23 + 0*22 + 0*21 + 1*20)
7. Adding Binary Numbers • Basic Rules: • 0+0=0 • 0+1=1 • 1+0=1 • 1+1=0 (carry 1) • Example: 01101001 00110100 ------------- 10011101
8. Weighted number systems • N = Mj=0bj Bj • N: the number • M : Number of digits • b: The digit • B : System’s radix
9. Examples 1. 101011 1x25 + 0x24 + 1x23 + 0x22 + 1x21 + 1x20 = 43 (101011)2 = (43)10 2. .0101 0x2-1 + 1x2-2 + 0x2-3 + 1x2-4 = .3125 (.0101)2 = (.3125)10 3. 101.11 1x22 + 0x21 + 1x20 + 1x2-1 + 1x2-2 5.75 (101.11)2 = (5.75)10
10. Decimal-to-Binary Conversion • Consider the integer and fractional parts separately. • For the integer part, • Repeatedly divide the given number by 2, and go on accumulating the remainders, until the number becomes zero. • Arrange the remainders in reverse order. • For the fractional part, • Repeatedly multiply the given fraction by 2. • Accumulate the integer part (0 or 1). • If the integer part is 1, chop it off. • Arrange the integer parts in the order they are obtained.
11. Example 1 :: 239 • 239 • 2 119 --- 1 • 59 --- 1 • 2 29 --- 1 • 14 --- 1 • 2 7 --- 0 • 3 --- 1 • 2 1 --- 1 • 2 0 --- 1 (239)10 = (11101111)2
12. Example 2 :: 64 • 64 • 2 32 --- 0 • 16 --- 0 • 2 8 --- 0 • 4 --- 0 • 2 2 --- 0 • 1 --- 0 • 2 0 --- 1 (64)10 = (1000000)2
13. Example 3: .634 .634 x 2 = 1.268 .268 x 2 = 0.536 .536 x 2 = 1.072 .072 x 2 = 0.144 .144 x 2 = 0.288 : : (.634)10 = (.10100……)2
14. Example 4: 37.0625 (37)10 = (100101)2 (.0625)10 = (.0001)2 (37.0625)10 = (100101.0001)2
15. Hexadecimal Numbers • Base=16 Decimal Binary Hex 0 00000 0 1 00001 1 2 00010 2 3 00011 3 4 00100 4 5 00101 5 6 00110 6 7 00111 7 8 01000 8 9 01001 9 Decimal Binary Hex 10 01010 10 11 01011 11 12 01100 12 13 01101 13 14 01110 14 15 01111 15 16 00110 16 17 00111 17 18 01000 18 19 01001 19
16. Integers Representation
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1 / 16. # Number Representation. Number Representation. Lecture 20 4.3.2001. Topics. How are numeric data items actually stored in computer memory? How much space (memory locations) is allocated for each type of data? int, float, char, etc. How are characters and strings stored in memory?. Binary number system.. ## Number Representation. E N D. ### Presentation Transcript. 1. Number Representation Lecture 20 4.3.2001.. 2. Topics • How are numeric data items actually stored in computer memory? • How much space (memory locations) is allocated for each type of data? • int, float, char, etc. • How are characters and strings stored in memory?. 3. Binary number system Decimal number system • Ten digits : 0,1,2,3,4,5,6,7,8,9 • Every digit position has a weight which is a power of 10. • Base or radix is 10. • Two digits : 0,1 • Every digit position has a weight which is a power of 2. • Base or radix is 2.. 4. Decimal Number • 136.25 : What does this number actually mean ? 102 * 1 = 100.0 101 * 3 = 30.0 100 * 6 = 6.0 10-1 * 2 = 0.2 10-2 * 5 = 0.05. 5. Binary Number • 1101.01: What does this number mean? 23 * 1 = 1000.0 (8 in decimal) 22 * 1 = 100.0 (4 in decimal) 21 * 3 = 0.0 (0 in decimal) 20 * 6 = 1.0 (1 in decimal) 2-1 * 2 = 0.0 (0.0 in decimal) 2-2 * 5 = 0.01 (0.25 in decimal). 6. First integers and their binary equivalent decimal binary 0 0000 (0*23 + 0*22 + 0*21 + 0*20) 1 0001 (0*23 + 0*22 + 0*21 + 1*20) 2 0010 (0*23 + 0*22 + 1*21 + 0*20) 3 0011 (0*23 + 0*22 + 1*21 + 1*20) 4 0100 (0*23 + 1*22 + 0*21 + 0*20) 5 0101 (0*23 + 1*22 + 0*21 + 1*20) 6 0110 (0*23 + 1*22 + 1*21 + 0*20) 7 0111 (0*23 + 1*22 + 1*21 + 1*20) 8 1000 (1*23 + 0*22 + 0*21 + 0*20) 9 1001 (1*23 + 0*22 + 0*21 + 1*20). 7. Adding Binary Numbers • Basic Rules: • 0+0=0 • 0+1=1 • 1+0=1 • 1+1=0 (carry 1) • Example: 01101001 00110100 ------------- 10011101.
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8. Weighted number systems • N = Mj=0bj Bj • N: the number • M : Number of digits • b: The digit • B : System’s radix. 9. Examples 1. 101011 1x25 + 0x24 + 1x23 + 0x22 + 1x21 + 1x20 = 43 (101011)2 = (43)10 2. .0101 0x2-1 + 1x2-2 + 0x2-3 + 1x2-4 = .3125 (.0101)2 = (.3125)10 3. 101.11 1x22 + 0x21 + 1x20 + 1x2-1 + 1x2-2 5.75 (101.11)2 = (5.75)10. 10. Decimal-to-Binary Conversion • Consider the integer and fractional parts separately. • For the integer part, • Repeatedly divide the given number by 2, and go on accumulating the remainders, until the number becomes zero. • Arrange the remainders in reverse order. • For the fractional part, • Repeatedly multiply the given fraction by 2. • Accumulate the integer part (0 or 1). • If the integer part is 1, chop it off. • Arrange the integer parts in the order they are obtained.. 11. Example 1 :: 239 • 239 • 2 119 --- 1 • 59 --- 1 • 2 29 --- 1 • 14 --- 1 • 2 7 --- 0 • 3 --- 1 • 2 1 --- 1 • 2 0 --- 1 (239)10 = (11101111)2. 12. Example 2 :: 64 • 64 • 2 32 --- 0 • 16 --- 0 • 2 8 --- 0 • 4 --- 0 • 2 2 --- 0 • 1 --- 0 • 2 0 --- 1 (64)10 = (1000000)2. 13. Example 3: .634 .634 x 2 = 1.268 .268 x 2 = 0.536 .536 x 2 = 1.072 .072 x 2 = 0.144 .144 x 2 = 0.288 : : (.634)10 = (.10100……)2. 14. Example 4: 37.0625 (37)10 = (100101)2 (.0625)10 = (.0001)2 (37.0625)10 = (100101.0001)2. 15. Hexadecimal Numbers • Base=16 Decimal Binary Hex 0 00000 0 1 00001 1 2 00010 2 3 00011 3 4 00100 4 5 00101 5 6 00110 6 7 00111 7 8 01000 8 9 01001 9 Decimal Binary Hex 10 01010 10 11 01011 11 12 01100 12 13 01101 13 14 01110 14 15 01111 15 16 00110 16 17 00111 17 18 01000 18 19 01001 19. 16. Integers Representation. More Related.
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# Sum of squares of sum of squares function $r_2(n)$
Let $r_2(n)$ denote the number of representations of $n$ as a sum of two squares.
What is known about the sum of squares of this function,
$\sum_{i=1}^n r_2(i)^2$
In particular is anything known about the asymptotics as $n \rightarrow \infty$?
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I believe that $r_k(i)$ is the number of ways to write $i$ as a sum of $k$ squares. (Cf. the SquaresR function in Mathematica.) – Aeolian Mar 6 '13 at 3:21
Yep, I mistakenly put $r_k$ instead of $r_2$, fixed now =) – Loadge Mar 6 '13 at 6:43
Asymptotics are known for all integer moments of $r_2$: for any $m \ge 1$, there is an explicit constant $a_m$ such that $\sum_{n \le x} r_2(n)^m \sim a_m\cdot x\,(\log x)^{2^{m-1}-1}$. ($m=0$ also holds if we interpret $0^m$ as $0$.)
This result, including the precise constant (and a uniform generalization to other positive binary quadratic forms), can be found in V. Blomer and A. Granville, “Estimates for representation numbers of quadratic forms”, Duke Math. J. 135 (2006). Here's the PDF.
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Thanks Erick, I will check out the reference – Loadge Mar 6 '13 at 6:46
If you want to know exact formulae for these types of sums, I highly recommend "Number Theory in the Spirit of Ramanujan" by Bruce Berndt.
On page 56, a proof is given of $r_2(n) = 4 \sum_{d|n, d\ odd} (-1)^{(d-1)/2}$.
You can do a Dirichlet multiplication to get a double sum for $r_2^2(n)$.
By the way, after reading this book, I am absolutely astounded at how great a mathematician Jacobi was. He developed the $q$-series methods that enabled him to get explicit formulae for the number of representations on any integer as the sum of a number of squares or triangular numbers. Reading a list of his accomplishments is, like wow! He and Ramanujan would have gotten along famously.
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Thanks for the post, will add to my reading list =) – Loadge Mar 6 '13 at 6:48
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# Sum of squares of sum of squares function $r_2(n)$. Let $r_2(n)$ denote the number of representations of $n$ as a sum of two squares.. What is known about the sum of squares of this function,. $\sum_{i=1}^n r_2(i)^2$. In particular is anything known about the asymptotics as $n \rightarrow \infty$?. -. I believe that $r_k(i)$ is the number of ways to write $i$ as a sum of $k$ squares. (Cf. the SquaresR function in Mathematica.) – Aeolian Mar 6 '13 at 3:21. Yep, I mistakenly put $r_k$ instead of $r_2$, fixed now =) – Loadge Mar 6 '13 at 6:43. Asymptotics are known for all integer moments of $r_2$: for any $m \ge 1$, there is an explicit constant $a_m$ such that $\sum_{n \le x} r_2(n)^m \sim a_m\cdot x\,(\log x)^{2^{m-1}-1}$. ($m=0$ also holds if we interpret $0^m$ as $0$.). This result, including the precise constant (and a uniform generalization to other positive binary quadratic forms), can be found in V. Blomer and A. Granville, “Estimates for representation numbers of quadratic forms”, Duke Math.
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J. 135 (2006). Here's the PDF.. -. Thanks Erick, I will check out the reference – Loadge Mar 6 '13 at 6:46. If you want to know exact formulae for these types of sums, I highly recommend "Number Theory in the Spirit of Ramanujan" by Bruce Berndt.. On page 56, a proof is given of $r_2(n) = 4 \sum_{d|n, d\ odd} (-1)^{(d-1)/2}$.. You can do a Dirichlet multiplication to get a double sum for $r_2^2(n)$.. By the way, after reading this book, I am absolutely astounded at how great a mathematician Jacobi was. He developed the $q$-series methods that enabled him to get explicit formulae for the number of representations on any integer as the sum of a number of squares or triangular numbers. Reading a list of his accomplishments is, like wow! He and Ramanujan would have gotten along famously.. -. Thanks for the post, will add to my reading list =) – Loadge Mar 6 '13 at 6:48.
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# “The set of all true statements of first order logic”
In one of my lectures, the lecturer put a bunch of examples of sets on the board, stuff like the set of all humans, set of all well typed programs in some programming language, the set of all true statements of first order logic, the set of all proofs of them, and the set of all sets.
Now, for the last one, he drew a big line through it, claiming that it introduced issues. It would seem to me that the two before also introduce fairly massive issues, but when I asked a friendly academic, he brushed me off without any real explanation, saying there would be no such issues. I’m going to give my argument below; I hope that anyone can either tell me I’m right, or at least tell me how I’m wrong, so I can better understand the mistakes I made.
Ok, so it would seem to me that for the set of all true statements of first order logic to be useful to us, it would first be necessary for us to be able to check whether a statement is in the set, and also that it’s complement is the set of all false statements of first order logic.
Now, by Gödel any formal logical system that can reason about arithmetic must be incomplete (there are true statements of arithmetic that cannot be proved to be true or false in the logical system) or inconsistent (there are statements that can be proved to be both true and false).
So, a simple way of proving whether a statement is in the set is to define a proof program. We have our simple laws of inference ($\vee$-introduction, elimination etc), so it’s fairly easy to, looking at a proof in FOL see whether the prover’s done something not allowed. So, we go through every string of length 1, then every string of length 2, etc until we find a valid proof of the statement; if the program terminates and there are no syntax errors, it is known that the statement is true. To avoid going on forever with statements which are not in the set, we also check to see whether the negation of the formula is true, since by the law of excluded middle, if something’s not true it’s false.
Now, by Gödel, the program must
a) go on forever for some statements, since if FOL is incomplete, there will be no proof in the system that certain statements are in the set (even though they are true).
or
b) there will be certain statements for which testing both the statement and its negation will result in an affirmative answer (i.e. FOL is inconsistent).
This implies that a useful set containing all true statements of first order logic cannot exist.
At which point have I gone wrong here?
#### Solutions Collecting From Web of "“The set of all true statements of first order logic”"
Your mistake is right at the beginning where you say:
Ok, so it would seem to me that for the set of all true statements of first order logic to be useful to us, it would first be necessary for us to be able to check whether a statement is in the set,
First of all “is useful to us” is a somewhat vague description, and it is not clear at all that it would mean the same as “is a set”.
Moreover, there is no rule demanding that for something to be a “set”, there needs to be a definite computable procedure for finding out what its elements are. It needs to be true that everything is either an element or not an element, but there’s no demand that we’re able to know with certainty which is the case for every prospective element.
It is certainly possible to consider the concept of “collections where we have a definite procedure for telling whether something as a member of the collection”. This leads to constructive mathematics, which is an interesting and respectable area of study. It just happens not to be what mathematicians in general mean when they say “set”.
I agree with @Makholm that there is no pre-requisite sorting algorithm for sets (assuming the Axiom of Choice, which allows such an arbitrary collections of elements), but there is another point to be made here.
Any statement which is undecideable (also called “unprovably true”) can be considered an axiom, assumed true, and tossed into the set of true statements. That means the set of true statements will be infinitely large and will mostly consist of axioms (statements that are assumed to be true). Each time you toss a new axiom into this set, there will be another undecideable statement that you need to assume is either true or false and throw into its respective set.
You see, the undecideable statements do not belong to both sets, you just have to choose which set they do belong to (and it doesn’t matter which).
Church’s theorem named after Alonzo Church, says there is no algorithm for deciding which first-order sentences are universally valid.
Also, a first order statement can only be a member to a set if there is an identity defined for it. But there is no obvious identity relation for statements, between alphabetical identity and logical equivalence.
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# “The set of all true statements of first order logic”. In one of my lectures, the lecturer put a bunch of examples of sets on the board, stuff like the set of all humans, set of all well typed programs in some programming language, the set of all true statements of first order logic, the set of all proofs of them, and the set of all sets.. Now, for the last one, he drew a big line through it, claiming that it introduced issues. It would seem to me that the two before also introduce fairly massive issues, but when I asked a friendly academic, he brushed me off without any real explanation, saying there would be no such issues. I’m going to give my argument below; I hope that anyone can either tell me I’m right, or at least tell me how I’m wrong, so I can better understand the mistakes I made.. Ok, so it would seem to me that for the set of all true statements of first order logic to be useful to us, it would first be necessary for us to be able to check whether a statement is in the set, and also that it’s complement is the set of all false statements of first order logic.. Now, by Gödel any formal logical system that can reason about arithmetic must be incomplete (there are true statements of arithmetic that cannot be proved to be true or false in the logical system) or inconsistent (there are statements that can be proved to be both true and false).. So, a simple way of proving whether a statement is in the set is to define a proof program. We have our simple laws of inference ($\vee$-introduction, elimination etc), so it’s fairly easy to, looking at a proof in FOL see whether the prover’s done something not allowed. So, we go through every string of length 1, then every string of length 2, etc until we find a valid proof of the statement; if the program terminates and there are no syntax errors, it is known that the statement is true. To avoid going on forever with statements which are not in the set, we also check to see whether the negation of the formula is true, since by the law of excluded middle, if something’s not true it’s false.. Now, by Gödel, the program must. a) go on forever for some statements, since if FOL is incomplete, there will be no proof in the system that certain statements are in the set (even though they are true).. or. b) there will be certain statements for which testing both the statement and its negation will result in an affirmative answer (i.e. FOL is inconsistent).. This implies that a useful set containing all true statements of first order logic cannot exist.
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At which point have I gone wrong here?. #### Solutions Collecting From Web of "“The set of all true statements of first order logic”". Your mistake is right at the beginning where you say:. Ok, so it would seem to me that for the set of all true statements of first order logic to be useful to us, it would first be necessary for us to be able to check whether a statement is in the set,. First of all “is useful to us” is a somewhat vague description, and it is not clear at all that it would mean the same as “is a set”.. Moreover, there is no rule demanding that for something to be a “set”, there needs to be a definite computable procedure for finding out what its elements are. It needs to be true that everything is either an element or not an element, but there’s no demand that we’re able to know with certainty which is the case for every prospective element.. It is certainly possible to consider the concept of “collections where we have a definite procedure for telling whether something as a member of the collection”. This leads to constructive mathematics, which is an interesting and respectable area of study. It just happens not to be what mathematicians in general mean when they say “set”.. I agree with @Makholm that there is no pre-requisite sorting algorithm for sets (assuming the Axiom of Choice, which allows such an arbitrary collections of elements), but there is another point to be made here.. Any statement which is undecideable (also called “unprovably true”) can be considered an axiom, assumed true, and tossed into the set of true statements. That means the set of true statements will be infinitely large and will mostly consist of axioms (statements that are assumed to be true). Each time you toss a new axiom into this set, there will be another undecideable statement that you need to assume is either true or false and throw into its respective set.. You see, the undecideable statements do not belong to both sets, you just have to choose which set they do belong to (and it doesn’t matter which).. Church’s theorem named after Alonzo Church, says there is no algorithm for deciding which first-order sentences are universally valid.. Also, a first order statement can only be a member to a set if there is an identity defined for it. But there is no obvious identity relation for statements, between alphabetical identity and logical equivalence.
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Back
Subtracting Using Excel
> ... Math > Excel > Subtraction Using Excel
You can watch the video or just read the text below. Both will give you the same information.
The following topics are covered in this lesson.
• Using the hyphen symbol for subtraction
• Referencing Cells and using the hyphen symbol for subtraction
• Subtraction = Adding negative numbers
Subtraction Using Excel
Using the hyphen symbol
If you want to use Excel as a calculator, first select the cell of your choice. Type “=” (without the quote marks) and then type the numbers separated by a hyphen “-”.
Example: Find the answer to $$7-3$$ in a cell.
Answer: In Excel type “=7-3” (without the quotation marks). When you press Enter, the number 4 will appear in the cell.
Press Enter.
The number 4 will appear in the cell because $$7-3=4$$
Notice that the equation “$$=7-3$$” is still seen in the formula bar at the top of Excel.
You are telling Excel that you want the cell you have clicked on to be equal to the equation after the equal sign.
Referencing cells and using the hyphen symbol for subtraction
You can subtract the values in different cells by referencing them just as you do with addition or any other mathematical operation.
Example:
The number 7 is in A1 and the number 4 is in A2.
Subtract $$7- 4$$ in another cell using the cell references for A1 and A2.
Answer: In cell A3 type “=A1-A2” (without the quotation marks)
Press Enter.
The number 3 appears in cell A3 because $$7-4=3$$.
The advantage of doing this is that the numbers in cells A1 and A2 can change and the value in A3 will automatically calculate the new equation.
Example:
Change A1 to 8.
Notice that the value in A3 automatically changes to 4 because $$8-4=4$$.
Both of the numbers in A1 and A2 can change because we are using references for both of them.
Example: Change A2 to the number 6. Notice that A3 automatically changes to 6 since $$8-2=6$$.
Subtraction is actually the same as adding a negative number. This is particularly handy in Excel.
Example: Calculate $$8-2$$ but using the concept of adding a negative number and using the SUM function.
In B1, type $$8$$
In B2, type $$-2$$
In B3 use the sum function to add B1 and B2 together by typing “=sum(B1:B2)"
Press Enter.
Notice that the number 6 appears in cell B3. This is the same answer we calculated in cell A3 in our previous example using subtraction and referencing cells.
Making the 2 a negative number and then adding the cells together is the same as subtracting.
This technique is particularly helpful when adding negative and positive numbers together in Excel.
Example: Imagine a column of numbers in Excel, some positive and some negative.
C1=$$4$$
C2=$$-7$$
C3=$$-3$$
C4=$$10$$
C5=$$-2$$
C6=$$-1$$.
We could add them all together individually, or we could use the sum function by typing “=sum(C1:C6)”.
Press Enter.
The number 1 appears in our new cell.
Subtracting a Negative Number
When subtracting a negative number it is helpful to use parentheses around the negative number to more easily see both the subtraction symbol and the negative symbol.
Example: Calculate $$(-4) - (-3)$$ using Excel.
Answer: Select a cell. Type “=-4-(-3)” (without quotation marks)
When you press Enter, the number $$-1$$ will appear in the cell because $$(-4)-(-3)=-1$$.
In this case the parentheses are only there to help us see what is happening. Excel can calculate this without the parentheses as well.
Notice how it is hard to see the negative next to the subtraction symbol since they are the same symbol.
We could also put parentheses around the $$-4$$ as well.
Need More Help?
1. Study other Math Lessons in the Resource Center.
2. Visit the Online Tutoring Resources in the Resource Center.
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Back. Subtracting Using Excel. > ... Math > Excel > Subtraction Using Excel. You can watch the video or just read the text below. Both will give you the same information.. The following topics are covered in this lesson.. • Using the hyphen symbol for subtraction. • Referencing Cells and using the hyphen symbol for subtraction. • Subtraction = Adding negative numbers. Subtraction Using Excel. Using the hyphen symbol. If you want to use Excel as a calculator, first select the cell of your choice. Type “=” (without the quote marks) and then type the numbers separated by a hyphen “-”.. Example: Find the answer to $$7-3$$ in a cell.. Answer: In Excel type “=7-3” (without the quotation marks). When you press Enter, the number 4 will appear in the cell.. Press Enter.. The number 4 will appear in the cell because $$7-3=4$$. Notice that the equation “$$=7-3$$” is still seen in the formula bar at the top of Excel.. You are telling Excel that you want the cell you have clicked on to be equal to the equation after the equal sign.. Referencing cells and using the hyphen symbol for subtraction. You can subtract the values in different cells by referencing them just as you do with addition or any other mathematical operation.. Example:. The number 7 is in A1 and the number 4 is in A2.. Subtract $$7- 4$$ in another cell using the cell references for A1 and A2.. Answer: In cell A3 type “=A1-A2” (without the quotation marks). Press Enter.. The number 3 appears in cell A3 because $$7-4=3$$.. The advantage of doing this is that the numbers in cells A1 and A2 can change and the value in A3 will automatically calculate the new equation.. Example:. Change A1 to 8.. Notice that the value in A3 automatically changes to 4 because $$8-4=4$$.. Both of the numbers in A1 and A2 can change because we are using references for both of them.. Example: Change A2 to the number 6. Notice that A3 automatically changes to 6 since $$8-2=6$$.. Subtraction is actually the same as adding a negative number. This is particularly handy in Excel.
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Example: Calculate $$8-2$$ but using the concept of adding a negative number and using the SUM function.. In B1, type $$8$$. In B2, type $$-2$$. In B3 use the sum function to add B1 and B2 together by typing “=sum(B1:B2)". Press Enter.. Notice that the number 6 appears in cell B3. This is the same answer we calculated in cell A3 in our previous example using subtraction and referencing cells.. Making the 2 a negative number and then adding the cells together is the same as subtracting.. This technique is particularly helpful when adding negative and positive numbers together in Excel.. Example: Imagine a column of numbers in Excel, some positive and some negative.. C1=$$4$$. C2=$$-7$$. C3=$$-3$$. C4=$$10$$. C5=$$-2$$. C6=$$-1$$.. We could add them all together individually, or we could use the sum function by typing “=sum(C1:C6)”.. Press Enter.. The number 1 appears in our new cell.. Subtracting a Negative Number. When subtracting a negative number it is helpful to use parentheses around the negative number to more easily see both the subtraction symbol and the negative symbol.. Example: Calculate $$(-4) - (-3)$$ using Excel.. Answer: Select a cell. Type “=-4-(-3)” (without quotation marks). When you press Enter, the number $$-1$$ will appear in the cell because $$(-4)-(-3)=-1$$.. In this case the parentheses are only there to help us see what is happening. Excel can calculate this without the parentheses as well.. Notice how it is hard to see the negative next to the subtraction symbol since they are the same symbol.. We could also put parentheses around the $$-4$$ as well.. Need More Help?. 1. Study other Math Lessons in the Resource Center.. 2. Visit the Online Tutoring Resources in the Resource Center.
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# It is given that the probability of winning a game is 0.7.
Question:
It is given that the probability of winning a game is 0.7. What is the probability of losing the game?
(a) 0.8
(b) 0.3
(c) 0.35
(d) 0.15
Solution:
(b) 0.3
Explanation:
Let E be the event of winning the game. Then,
P(E) = 0.7
P(not E) = P(losing the game) = 1 P(E) ⇒ 1 0.7 = 0.3
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# It is given that the probability of winning a game is 0.7.. Question:. It is given that the probability of winning a game is 0.7. What is the probability of losing the game?. (a) 0.8. (b) 0.3. (c) 0.35. (d) 0.15.
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Solution:. (b) 0.3. Explanation:. Let E be the event of winning the game. Then,. P(E) = 0.7. P(not E) = P(losing the game) = 1 P(E) ⇒ 1 0.7 = 0.3.
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# Kinetic energy of a particle help
• Sheneron
In summary, a 0.600 kg particle has a speed of 2.30 m/s and a kinetic energy of 7.00 J at point B. Using the formula for kinetic energy, we can determine that its kinetic energy at point A is 1.59 J. Using the same formula, we can also calculate its speed at point B to be 4.83 m/s. The total work done on the particle as it moves from A to B is 5.41 J.
## Homework Statement
A 0.600 kg particle has a speed of 2.30 m/s at point A and kinetic energy of 7.00 J at point B.
(a) What is its kinetic energy at A?
(b) What is its speed at B?
(c) What is the total work done on the particle as it moves from A to B?
J= N*m
## The Attempt at a Solution
I am stuck on this problem and don't really know how to go about starting it. I think if i new the first part I could find the other two answers, but I don't know how to do that. Any advice would be appreciated.
What is the formula for kinetic energy?
k=.5mv^2
so, k=.5(.6)(2.3^2); k=1.59
b) 7=.5(.6)v^2; v=23.3333^.5; v=4.83
c) (sum)W= kf - ki ; (sum)W= 7-1.59 = 5.41
Thanks,
its been a long day sorry
heh. no problem. It looks like you know EXACTLY what you are doing! =)
## 1. What is kinetic energy?
Kinetic energy is the energy that a particle possesses due to its motion. It is defined as the work needed to accelerate a particle of a given mass from rest to its current velocity.
## 2. How is kinetic energy calculated?
The kinetic energy of a particle can be calculated using the formula KE = 1/2 * m * v^2, where m is the mass of the particle and v is its velocity.
## 3. What is the relationship between kinetic energy and mass?
The kinetic energy of a particle is directly proportional to its mass. This means that as the mass of the particle increases, its kinetic energy also increases.
## 4. How does kinetic energy change with respect to velocity?
The kinetic energy of a particle is directly proportional to the square of its velocity. This means that as the velocity of the particle increases, its kinetic energy increases at a faster rate.
## 5. What are some real-life examples of kinetic energy?
Some examples of kinetic energy in everyday life include a moving car, a spinning top, a swinging pendulum, and a rolling ball. Kinetic energy is also present in the movement of the molecules in a gas or liquid, and in the motion of atoms and subatomic particles.
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# Kinetic energy of a particle help. • Sheneron. In summary, a 0.600 kg particle has a speed of 2.30 m/s and a kinetic energy of 7.00 J at point B. Using the formula for kinetic energy, we can determine that its kinetic energy at point A is 1.59 J. Using the same formula, we can also calculate its speed at point B to be 4.83 m/s. The total work done on the particle as it moves from A to B is 5.41 J.. ## Homework Statement. A 0.600 kg particle has a speed of 2.30 m/s at point A and kinetic energy of 7.00 J at point B.. (a) What is its kinetic energy at A?. (b) What is its speed at B?. (c) What is the total work done on the particle as it moves from A to B?. J= N*m. ## The Attempt at a Solution. I am stuck on this problem and don't really know how to go about starting it. I think if i new the first part I could find the other two answers, but I don't know how to do that. Any advice would be appreciated.. What is the formula for kinetic energy?. k=.5mv^2. so, k=.5(.6)(2.3^2); k=1.59. b) 7=.5(.6)v^2; v=23.3333^.5; v=4.83. c) (sum)W= kf - ki ; (sum)W= 7-1.59 = 5.41. Thanks,. its been a long day sorry.
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heh. no problem. It looks like you know EXACTLY what you are doing! =). ## 1. What is kinetic energy?. Kinetic energy is the energy that a particle possesses due to its motion. It is defined as the work needed to accelerate a particle of a given mass from rest to its current velocity.. ## 2. How is kinetic energy calculated?. The kinetic energy of a particle can be calculated using the formula KE = 1/2 * m * v^2, where m is the mass of the particle and v is its velocity.. ## 3. What is the relationship between kinetic energy and mass?. The kinetic energy of a particle is directly proportional to its mass. This means that as the mass of the particle increases, its kinetic energy also increases.. ## 4. How does kinetic energy change with respect to velocity?. The kinetic energy of a particle is directly proportional to the square of its velocity. This means that as the velocity of the particle increases, its kinetic energy increases at a faster rate.. ## 5. What are some real-life examples of kinetic energy?. Some examples of kinetic energy in everyday life include a moving car, a spinning top, a swinging pendulum, and a rolling ball. Kinetic energy is also present in the movement of the molecules in a gas or liquid, and in the motion of atoms and subatomic particles.
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PICTURE THIS Two figures: One Triangle with all 3 sides 't'.
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Joined: 17 Aug 2006
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PICTURE THIS Two figures: One Triangle with all 3 sides 't'. [#permalink]
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13 Nov 2008, 05:35
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
PICTURE THIS
Two figures:
One Triangle with all 3 sides 't'.
Two Square with all 4 sides 's'.
If the two regions above have the same area, what is the ration t:s
2:3
16:3
4:sqr3
2:4th root of 3
4: 4th rooth of 3
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Re: PREP 12 of 37 [#permalink]
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13 Nov 2008, 05:56
Reading question stem "two square" as one square, D should be the anser.
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Re: PREP 12 of 37 [#permalink]
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13 Nov 2008, 12:43
root 3 t^2/4 = s^2
t^2/s^2=4/root 3
t/s= 2/4th root3
Agreed D
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13 Nov 2008, 13:06
D is the answer, I'm just not sure how you got root 3 t^2/4 = s^2. Can you explain?
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Re: PREP 12 of 37 [#permalink]
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14 Nov 2008, 14:17
The area of an equilateral triangle with side "a" is given by the formula root3/4 a^2
and the area of a square is s^2
equating the two expressions we will get the answer
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Re: PREP 12 of 37 [#permalink] 14 Nov 2008, 14:17
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It is currently 21 Sep 2017, 10:54. GMAT Club Daily Prep. Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.. Customized. for You. we will pick new questions that match your level based on your Timer History. Track. every week, we’ll send you an estimated GMAT score based on your performance. Practice. Pays. we will pick new questions that match your level based on your Timer History. Events & Promotions. Events & Promotions in June. Open Detailed Calendar. PICTURE THIS Two figures: One Triangle with all 3 sides 't'.. Author Message. Manager. Joined: 17 Aug 2006. Posts: 87. Kudos [?]: 20 [0], given: 0. PICTURE THIS Two figures: One Triangle with all 3 sides 't'. [#permalink]. Show Tags. 13 Nov 2008, 05:35. This topic is locked. If you want to discuss this question please re-post it in the respective forum.. PICTURE THIS. Two figures:. One Triangle with all 3 sides 't'.. Two Square with all 4 sides 's'.. If the two regions above have the same area, what is the ration t:s. 2:3. 16:3. 4:sqr3. 2:4th root of 3. 4: 4th rooth of 3. Kudos [?]: 20 [0], given: 0. SVP. Joined: 17 Jun 2008. Posts: 1540. Kudos [?]: 278 [0], given: 0.
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Re: PREP 12 of 37 [#permalink]. Show Tags. 13 Nov 2008, 05:56. Reading question stem "two square" as one square, D should be the anser.. Kudos [?]: 278 [0], given: 0. Manager. Joined: 23 Jul 2008. Posts: 194. Kudos [?]: 132 [0], given: 0. Re: PREP 12 of 37 [#permalink]. Show Tags. 13 Nov 2008, 12:43. root 3 t^2/4 = s^2. t^2/s^2=4/root 3. t/s= 2/4th root3. Agreed D. Kudos [?]: 132 [0], given: 0. Manager. Joined: 17 Aug 2006. Posts: 87. Kudos [?]: 20 [0], given: 0. Re: PREP 12 of 37 [#permalink]. Show Tags. 13 Nov 2008, 13:06. D is the answer, I'm just not sure how you got root 3 t^2/4 = s^2. Can you explain?. Kudos [?]: 20 [0], given: 0. Manager. Joined: 23 Jul 2008. Posts: 194. Kudos [?]: 132 [0], given: 0. Re: PREP 12 of 37 [#permalink]. Show Tags. 14 Nov 2008, 14:17. The area of an equilateral triangle with side "a" is given by the formula root3/4 a^2. and the area of a square is s^2. equating the two expressions we will get the answer. Kudos [?]: 132 [0], given: 0. Re: PREP 12 of 37 [#permalink] 14 Nov 2008, 14:17. Display posts from previous: Sort by.
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# "high school" algebra -> relativistic conservation of momentum and energy
by forrealfyziks
Tags: algebra, conservation, energy, high school, momentum, relativistic
P: 13 1. The problem statement, all variables and given/known data Consider a head-on, elastic collision between a massless photon (momentum po and energy Eo) and a stationary free electron. (a) Assuming that the photon bounces directly back with momentum p (in the direction of -po) and energy E, use conservation of energy and momentum to find p. 2. Relevant equations E=$$\gamma$$mc2 p=$$\gamma$$mu massless: E=pc rest mass: E=mc2 E2=(pc)2+(mc2)2 v/c=pc/E $$\gamma$$=1/$$\sqrt{1+(v/c)^2}$$ 3. The attempt at a solution Note:First of all I know that this is relativity, but it boils down to just plain algebra. I can't figure it out and help is hard to find, so if you can help I would really appreciate it. I assume that p is the momentum of the electron. m=mass of the electron u=velocity of the electron c=speed of light conserving energy: poc+mc2=pc+$$\gamma$$mc2 po+mc=p+$$\gamma$$mc po=p+$$\gamma$$mc-mc conserving momentum: po=p-p=$$\gamma$$mu-p Plugging the result I got in conserving energy into the momentum equation: p-p=p+$$\gamma$$mc-mc p=2p+mc($$\gamma$$-1)
PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,875 Generally, it's a good idea when solving these types of problems to avoid using $\gamma$ and velocities if you can avoid them. Work with energy, momentum, and mass instead. Doing so usually makes the algebra simpler. I think your best approach is to solve for the electron's energy. Once you have that, you can calculate its momentum. You can rewrite your equations as follows: $$p_0 c - p c = E_e - m c^2$$ $$p_0 + p = p_e$$ Multiply the second equation by c, square both equations, then subtract the second one from the first, and use the relation $E^2 - (pc)^2 = (mc^2)^2$ to simplify what you get. Figure out how to eliminate p from the equation and solve for $E_e$.
P: 13 Thank you for the hints it has really helped. I worked through the things vela posted, and found myself stuck again, though. If I add the two equations, I end up with (po2c2+p2c2)=Ee2-Eemc2 If I subtract the second from the first -2ppoc2=-Eemc2+(mc2)2 On the first one, it obviously has the momentums on the left, and on the second one I also have momentums on the left. With the first one the left almost looks like one of the conservations, but not quite. I'm going to fiddle around with the information vela has already given me and see if I can't find another way.
PF Patron
HW Helper
Thanks
Emeritus
P: 10,875
## "high school" algebra -> relativistic conservation of momentum and energy
It wasn't clear in your original post what terms are allowed in the final answer. I assume you want to get rid of p, the photon's final momentum. You can do that by solving for it in one of the conservation equations in terms of the other variables.
P: 13 I think I'm supposed to find p numerically, because the second part of the question asks you to verify your answer in part a using Compton's formula where $$\Theta$$=$$\pi$$.
P: 13 I found out from a peer that the momentum you are finding is actually the photon's second momentum, and that you should solve in terms of po. It became incredibly easy once I knew I wasn't looking for something numerical... Thank you so much for your help!
PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,875 Oh! When they asked you to find p, they were probably referring to the momentum of the recoiling photon, not the electron's. What you want to do is get rid of $E_e$ from your answer.
P: 13 yes it was supposed to be photon's new momentum. Guess I should have put that second part of the question in, but I didn't think it was necessary. Going back over the derivation of Compton's formula, I noticed it was almost the exact thing I was doing. heh, well thanks again.
Related Discussions Introductory Physics Homework 1 Introductory Physics Homework 2 Introductory Physics Homework 3 Introductory Physics Homework 5 Special & General Relativity 5
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# "high school" algebra -> relativistic conservation of momentum and energy. by forrealfyziks. Tags: algebra, conservation, energy, high school, momentum, relativistic. P: 13 1. The problem statement, all variables and given/known data Consider a head-on, elastic collision between a massless photon (momentum po and energy Eo) and a stationary free electron. (a) Assuming that the photon bounces directly back with momentum p (in the direction of -po) and energy E, use conservation of energy and momentum to find p. 2. Relevant equations E=$$\gamma$$mc2 p=$$\gamma$$mu massless: E=pc rest mass: E=mc2 E2=(pc)2+(mc2)2 v/c=pc/E $$\gamma$$=1/$$\sqrt{1+(v/c)^2}$$ 3. The attempt at a solution Note:First of all I know that this is relativity, but it boils down to just plain algebra. I can't figure it out and help is hard to find, so if you can help I would really appreciate it. I assume that p is the momentum of the electron. m=mass of the electron u=velocity of the electron c=speed of light conserving energy: poc+mc2=pc+$$\gamma$$mc2 po+mc=p+$$\gamma$$mc po=p+$$\gamma$$mc-mc conserving momentum: po=p-p=$$\gamma$$mu-p Plugging the result I got in conserving energy into the momentum equation: p-p=p+$$\gamma$$mc-mc p=2p+mc($$\gamma$$-1). PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,875 Generally, it's a good idea when solving these types of problems to avoid using $\gamma$ and velocities if you can avoid them. Work with energy, momentum, and mass instead. Doing so usually makes the algebra simpler. I think your best approach is to solve for the electron's energy. Once you have that, you can calculate its momentum. You can rewrite your equations as follows: $$p_0 c - p c = E_e - m c^2$$ $$p_0 + p = p_e$$ Multiply the second equation by c, square both equations, then subtract the second one from the first, and use the relation $E^2 - (pc)^2 = (mc^2)^2$ to simplify what you get. Figure out how to eliminate p from the equation and solve for $E_e$.. P: 13 Thank you for the hints it has really helped. I worked through the things vela posted, and found myself stuck again, though. If I add the two equations, I end up with (po2c2+p2c2)=Ee2-Eemc2 If I subtract the second from the first -2ppoc2=-Eemc2+(mc2)2 On the first one, it obviously has the momentums on the left, and on the second one I also have momentums on the left.
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With the first one the left almost looks like one of the conservations, but not quite. I'm going to fiddle around with the information vela has already given me and see if I can't find another way.. PF Patron. HW Helper. Thanks. Emeritus. P: 10,875. ## "high school" algebra -> relativistic conservation of momentum and energy. It wasn't clear in your original post what terms are allowed in the final answer. I assume you want to get rid of p, the photon's final momentum. You can do that by solving for it in one of the conservation equations in terms of the other variables.. P: 13 I think I'm supposed to find p numerically, because the second part of the question asks you to verify your answer in part a using Compton's formula where $$\Theta$$=$$\pi$$.. P: 13 I found out from a peer that the momentum you are finding is actually the photon's second momentum, and that you should solve in terms of po. It became incredibly easy once I knew I wasn't looking for something numerical... Thank you so much for your help!. PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,875 Oh! When they asked you to find p, they were probably referring to the momentum of the recoiling photon, not the electron's. What you want to do is get rid of $E_e$ from your answer.. P: 13 yes it was supposed to be photon's new momentum. Guess I should have put that second part of the question in, but I didn't think it was necessary. Going back over the derivation of Compton's formula, I noticed it was almost the exact thing I was doing. heh, well thanks again.. Related Discussions Introductory Physics Homework 1 Introductory Physics Homework 2 Introductory Physics Homework 3 Introductory Physics Homework 5 Special & General Relativity 5.
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# Probability of a bond
I am dead on this question. I can’t even understand the text :((. How come the first one is the correct answer?
The probability of default for a certain junk bond in a given year is 0.2. What is the probability that zero or one bond of the five defaults in the year ahead in a portfolio of only five of these (Assuming the default risks of these are independent to each other)?
• 0.7373. correct.
• 0.4096.
• 0.0819.
Is this the exact same wording in the text?? If so, it is awfully awkward!
This sounds like a binomial distribution with 5 bonds in total and each has a probability of default = 0.2. What is the probability that 0 or 1 defaults?
1 Like
John’s right: it’s a horrible question.
It’s not a junk bond; it’s five different junk bonds, whose probabilities of default are not only equal to each other, they’re statistically independent of each other.
Assuming that,
P\left(0\ or\ 1\ defaults\right) = P\left(0\ defaults\right) + P\left(1\ default\right)
= \binom{5}{0}p^0\left(1 - p\right)^5 + \binom{5}{1}p^1\left(1 - p\right)^4
= \left(1\right)0.2^00.8^5 + \left(5\right)0.2^10.8^4
= \left(1\right)\left(1\right)\left(0.3277\right) + \left(5\right)\left(0.2\right)\left(0.4096\right)
= 0.3277 + 0.4096 = 0.7373
3 Likes
Thank you so much, both of you.
1 Like
My pleasure.
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# Probability of a bond. I am dead on this question. I can’t even understand the text :((. How come the first one is the correct answer?. The probability of default for a certain junk bond in a given year is 0.2. What is the probability that zero or one bond of the five defaults in the year ahead in a portfolio of only five of these (Assuming the default risks of these are independent to each other)?. • 0.7373. correct.. • 0.4096.. • 0.0819.. Is this the exact same wording in the text?? If so, it is awfully awkward!. This sounds like a binomial distribution with 5 bonds in total and each has a probability of default = 0.2. What is the probability that 0 or 1 defaults?. 1 Like.
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John’s right: it’s a horrible question.. It’s not a junk bond; it’s five different junk bonds, whose probabilities of default are not only equal to each other, they’re statistically independent of each other.. Assuming that,. P\left(0\ or\ 1\ defaults\right) = P\left(0\ defaults\right) + P\left(1\ default\right). = \binom{5}{0}p^0\left(1 - p\right)^5 + \binom{5}{1}p^1\left(1 - p\right)^4. = \left(1\right)0.2^00.8^5 + \left(5\right)0.2^10.8^4. = \left(1\right)\left(1\right)\left(0.3277\right) + \left(5\right)\left(0.2\right)\left(0.4096\right). = 0.3277 + 0.4096 = 0.7373. 3 Likes. Thank you so much, both of you.. 1 Like. My pleasure.
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You are on page 1of 10
# LI NEAR DI FFERENTI AL EQUATI ONS
A first-order linear differential equation is one that can be put into the form
where and are continuous functions on a given interval. This type of equation occurs
frequently in various sciences, as we will see.
An example of a linear equation is because, for , it can be written
in the form
Notice that this differential equation is not separable because it’s impossible to factor the
expression for as a function of x times a function of y. But we can still solve the equa-
tion by noticing, by the Product Rule, that
and so we can rewrite the equation as
If we now integrate both sides of this equation, we get
or
If we had been given the differential equation in the form of Equation 2, we would have
had to take the preliminary step of multiplying each side of the equation by x.
It turns out that every first-order linear differential equation can be solved in a similar
fashion by multiplying both sides of Equation 1 by a suitable function called an
integrating factor. We try to find so that the left side of Equation 1, when multiplied by
, becomes the derivative of the product :
If we can find such a function , then Equation 1 becomes
Integrating both sides, we would have
so the solution would be
To find such an , we expand Equation 3 and cancel terms:
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This is a separable differential equation for , which we solve as follows:
where . We are looking for a particular integrating factor, not the most general
one, so we take A 1 and use
Thus, a formula for the general solution to Equation 1 is provided by Equation 4, where
is given by Equation 5. Instead of memorizing this formula, however, we just remember
the form of the integrating factor.
To solve the linear differential equation , multiply both sides by
the integrating factor and integrate both sides.
EXAMPLE 1 Solve the differential equation .
SOLUTI ON The given equation is linear since it has the form of Equation 1 with
and . An integrating factor is
Multiplying both sides of the differential equation by , we get
or
Integrating both sides, we have
EXAMPLE 2 Find the solution of the initial-value problem
SOLUTI ON We must first divide both sides by the coefficient of to put the differential
equation into standard form:
The integrating factor is
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2 ■ L I NEAR DI FFERENT I AL EQUAT I ONS
FI GURE 1
6
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_1.5 1.8
C=2
C=1
C=_2
C=_1
C=0
■ ■
Figure 1 shows the graphs of several mem-
bers of the family of solutions in Example 1.
Notice that they all approach as . x l ϱ 2
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Multiplication of Equation 6 by gives
Then
and so
Since , we have
Therefore, the solution to the initial-value problem is
EXAMPLE 3 Solve .
SOLUTI ON The given equation is in the standard form for a linear equation. Multiplying by
the integrating factor
we get
or
Therefore
Recall from Section 6.4 that can’t be expressed in terms of elementary functions.
Nonetheless, it’s a perfectly good function and we can leave the answer as
Another way of writing the solution is
(Any number can be chosen for the lower limit of integration.)
APPLICATION TO ELECTRIC CIRCUITS
Let’s consider the simple electric circuit shown in Figure 4: An electromotive force (usually
a battery or generator) produces a voltage of volts (V) and a current of amperes
(A) at time . The circuit also contains a resistor with a resistance of ohms ( ) and an
inductor with an inductance of henries (H).
Ohm’s Law gives the drop in voltage due to the resistor as . The voltage drop due to
the inductor is . One of Kirchhoff’s laws says that the sum of the voltage drops is
equal to the supplied voltage . Thus, we have
which is a first-order linear differential equation. The solution gives the current at time . t I
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L I NEAR DI FFERENT I AL EQUAT I ONS ■ 3
■ ■
The solution of the initial-value problem in
Example 2 is shown in Figure 2.
FI GURE 2
(1, 2)
5
_5
0 4
■ ■
Even though the solutions of the differential
equation in Example 3 are expressed in terms of
an integral, they can still be graphed by a com-
puter algebra system (Figure 3).
FI GURE 3
C=2
C=_2
2.5
_2.5
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FI GURE 4
R
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switch
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4 ■ L I NEAR DI FFERENT I AL EQUAT I ONS
EXAMPLE 4 Suppose that in the simple circuit of Figure 4 the resistance is and the
inductance is 4 H. If a battery gives a constant voltage of 60 V and the switch is closed
when so the current starts with , find (a) , (b) the current after 1 s, and
(c) the limiting value of the current.
SOLUTI ON
(a) If we put , , and in Equation 7, we obtain the initial-value
problem
or
Multiplying by the integrating factor , we get
Since , we have , so and
(b) After 1 second the current is
(c)
EXAMPLE 5 Suppose that the resistance and inductance remain as in Example 4
but, instead of the battery, we use a generator that produces a variable voltage of
volts. Find .
SOLUTI ON This time the differential equation becomes
The same integrating factor gives
Using Formula 98 in the Table of Integrals, we have
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■ ■
The differential equation in Example 4 is
both linear and separable, so an alternative
method is to solve it as a separable equation.
If we replace the battery by a generator, how-
ever, we get an equation that is linear but not
separable (Example 5).
■ ■
Figure 6 shows the graph of the current
when the battery is replaced by a generator.
FI GURE 6
2
_2
2.5 0
FI GURE 5
6
0
2.5
y=5
■ ■
Figure 5 shows how the current in Example
4 approaches its limiting value.
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L I NEAR DI FFERENT I AL EQUAT I ONS ■ 5
Since , we get
so I͑t͒
5
101
͑sin 30t Ϫ 10 cos 30t͒ ϩ
50
101
e
Ϫ3t
Ϫ
50
101
ϩ C 0
I͑0͒ 0
EXERCISES
1–4 Determine whether the differential equation is linear.
1. 2.
3. 4.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
5–14 Solve the differential equation.
5. 6.
7. 8.
9. 10.
11.
12. ,
13. ,
14.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
15–20 Solve the initial-value problem.
15. ,
16. , ,
17. ,
18. , ,
19. ,
20. , ,
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
;
21–22 Solve the differential equation and use a graphing cal-
culator or computer to graph several members of the family of
solutions. How does the solution curve change as varies?
21. , 22.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
23. A Bernoulli differential equation (named after James
Bernoulli) is of the form
dy
dx
ϩ P͑x͒y Q͑x͒y
n
yЈ ϩ ͑cos x͒y cos x x Ͼ 0 xyЈ ϩ y x cos x
C
x Ͼ 0 y͑1͒ 0 x
dy
dx
Ϫ
y
x ϩ 1
x
y͑ ͒ 0 xyЈ y ϩ x
2
sin x
y͑4͒ 20 x Ͼ 0 2xyЈ ϩ y 6x
v͑0͒ 5
dv
dt
Ϫ 2tv 3t
2
e
t
2
y͑1͒ 0 t Ͼ 0 t
dy
dt
ϩ 2y t
3
y͑0͒ 2 yЈ x ϩ y
t ln t
dr
dt
ϩ r te
t
t Ͼ 0 ͑1 ϩ t͒
du
dt
ϩ u 1 ϩ t
Ϫ ͞2 Ͻ x Ͻ ͞2
dy
dx
x sin 2x ϩ y tan x
dy
dx
ϩ 2xy x
2
1 ϩ xy xyЈ xyЈ ϩ y sx
x
2
yЈ ϩ 2xy cos
2
x xyЈ Ϫ 2y x
2
yЈ x ϩ 5y yЈ ϩ 2y 2e
x
yЈ ϩ cos y tan x xyЈ ϩ ln x Ϫ x
2
y 0
y ϩ sin x x
3
yЈ yЈ ϩ e
x
y x
2
y
2
Observe that, if or , the Bernoulli equation is linear.
For other values of , show that the substitution trans-
forms the Bernoulli equation into the linear equation
24–26 Use the method of Exercise 23 to solve the differential
equation.
24. 25.
26.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
27. In the circuit shown in Figure 4, a battery supplies a constant
voltage of 40 V, the inductance is 2 H, the resistance is ,
and .
(a) Find .
(b) Find the current after s.
28. In the circuit shown in Figure 4, a generator supplies a voltage
of volts, the inductance is H, the resistance
is , and A.
(a) Find .
(b) Find the current after s.
;
(c) Use a graphing device to draw the graph of the current
function.
29. The figure shows a circuit containing an electromotive force,
a capacitor with a capacitance of farads (F), and a resistor
with a resistance of ohms ( ). The voltage drop across the
capacitor is , where is the charge (in coulombs), so in
this case Kirchhoff’s Law gives
But , so we have
Suppose the resistance is , the capacitance is F, a
battery gives a constant voltage of 60 V, and the initial charge
is C. Find the charge and the current at time . t Q͑0͒ 0
0.05 5 ⍀
R
dQ
dt
ϩ
1
C
Q E͑t͒
I dQ͞dt
RI ϩ
Q
C
E͑t͒
Q Q͞C
C
E R
⍀ R
C
0.1
I͑t͒
I͑0͒ 1 20 ⍀
1 E͑t͒ 40 sin 60t
0.1
I͑t͒
I͑0͒ 0
10 ⍀
yЈ ϩ y xy
3
yЈ ϩ
2
x
y
y
3
x
2
xyЈ ϩ y Ϫxy
2
du
dx
ϩ ͑1 Ϫ n͒P͑x͒u ͑1 Ϫ n͒Q͑x͒
u y
1Ϫn
n
1 n 0
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30. In the circuit of Exercise 29, , , ,
and . Find the charge and the current at time .
31. Psychologists interested in learning theory study learning
curves. A learning curve is the graph of a function , the
performance of someone learning a skill as a function of the
training time . The derivative represents the rate at
which performance improves.
(a) When do you think increases most rapidly? What
happens to as increases? Explain.
(b) If is the maximum level of performance of which the
learner is capable, explain why the differential equation
is a reasonable model for learning.
(c) Solve the differential equation in part (b) as a linear differ-
ential equation and use your solution to graph the learning
curve.
32. Two new workers were hired for an assembly line. Jim pro-
cessed 25 units during the first hour and 45 units during the
second hour. Mark processed 35 units during the first hour and
50 units the second hour. Using the model of Exercise 31 and
assuming that , estimate the maximum number of
units per hour that each worker is capable of processing.
33. In Section 7.6 we looked at mixing problems in which the
volume of fluid remained constant and saw that such problems
give rise to separable equations. (See Example 5 in that
section.) If the rates of flow into and out of the system are dif-
ferent, then the volume is not constant and the resulting differ-
ential equation is linear but not separable.
A tank contains 100 L of water. A solution with a salt
concentration of is added at a rate of . The
solution is kept mixed and is drained from the tank at a rate
5 L͞min 0.4 kg͞L
P͑0͒ 0
k a positive constant
dP
dt
k͑M Ϫ P͒
M
t dP͞dt
P
dP͞dt t
P͑t͒
t E͑t͒ 10 sin 60t
Q͑0͒ 0 C 0.01 F R 2 ⍀ of . If is the amount of salt (in kilograms) after
minutes, show that satisfies the differential equation
Solve this equation and find the concentration after 20 minutes.
34. A tank with a capacity of 400 L is full of a mixture of water
and chlorine with a concentration of 0.05 g of chlorine per liter.
In order to reduce the concentration of chlorine, fresh water is
pumped into the tank at a rate of . The mixture is kept
stirred and is pumped out at a rate of . Find the amount
of chlorine in the tank as a function of time.
35. An object with mass is dropped from rest and we assume
that the air resistance is proportional to the speed of the object.
If is the distance dropped after seconds, then the speed is
and the acceleration is . If is the acceleration
due to gravity, then the downward force on the object is
, where is a positive constant, and Newton’s Second
Law gives
(a) Solve this as a linear equation to show that
(b) What is the limiting velocity?
(c) Find the distance the object has fallen after seconds.
36. If we ignore air resistance, we can conclude that heavier
objects fall no faster than lighter objects. But if we take air
resistance into account, our conclusion changes. Use the
expression for the velocity of a falling object in Exercise 35(a)
to find and show that heavier objects do fall faster than
lighter ones.
dv͞dm
t
v
mt
c
͑1 Ϫ e
Ϫct͞m
͒
m
dv
dt
mt Ϫ cv
c mt Ϫ cv
t a vЈ͑t͒ v sЈ͑t͒
t s͑t͒
m
10 L͞s
4 L͞s
dy
dt
2 Ϫ
3y
100 ϩ 2t
y t
y͑t͒ 3 L͞min
6 ■ L I NEAR DI FFERENT I AL EQUAT I ONS
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L I NEAR DI FFERENT I AL EQUAT I ONS ■ 7
1. No 3. Yes 5.
7. 9.
11.
13. 15.
17. 19.
21.
C=2
C=1
C=0.2
C=_1
C=_2
6
_4
0 10
y sin x ϩ ͑cos x͒͞x ϩ C͞x
y Ϫx cos x Ϫ x v t
3
e
t
2
ϩ 5e
t
2
y Ϫx Ϫ 1 ϩ 3e
x
u ͑t
2
ϩ 2t ϩ 2C͓͒͞2͑t ϩ 1͔͒
y
1
2
x ϩ Ce
Ϫx
2
Ϫ
1
2
e
Ϫx
2
x e
x
2
dx
y
2
3
sx ϩ C͞x y x
2
ln
Խ
x
Խ
ϩ Cx
2
y
2
3
e
x
ϩ Ce
Ϫ2x
25.
27. (a) (b)
29.
31. (a) At the beginning; stays positive, but decreases
(c)
33. ; 0.2275 kg͞L
35. (b) (c) ͑mt͞c͓͒t ϩ ͑m͞c͒e
Ϫct͞m
͔ Ϫ m
2
t͞c
2
mt͞c
y
2
5
͑100 ϩ 2t͒ Ϫ 40,000͑100 ϩ 2t͒
Ϫ3͞2
0
M
P(t)
t
P(0)
P͑t͒ M ϩ Ce
Ϫkt
Q͑t͒ 3͑1 Ϫ e
Ϫ4t
͒, I͑t͒ 12e
Ϫ4t
4 Ϫ 4e
Ϫ1͞2
Ϸ 1.57 A I͑t͒ 4 Ϫ 4e
Ϫ5t
y Ϯ͓Cx
4
ϩ 2͑͞5x͔͒
Ϫ1͞2
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8 ■ L I NEAR DI FFERENT I AL EQUAT I ONS
SOLUTIONS
1. y
0
+ e
x
y = x
2
y
2
is not linear since it cannot be put into the standard linear form (1), y
0
+ P(x) y = Q(x).
3. xy
0
+ lnx −x
2
y = 0 ⇒ xy
0
−x
2
y = −lnx ⇒ y
0
+ (−x) y = −
lnx
x
, which is in the standard linear
form (1), so this equation is linear.
5. Comparing the given equation, y
0
+ 2y = 2e
x
, with the general form, y
0
+ P(x)y = Q(x), we see that P(x) = 2
and the integrating factor is I(x) = e
P(x)dx
= e
2 dx
= e
2x
. Multiplying the differential equation by I(x) gives
e
2x
y
0
+ 2e
2x
y = 2e
3x
_
e
2x
y
_
0
= 2e
3x
⇒ e
2x
y =
_
2e
3x
dx ⇒ e
2x
y =
2
3
e
3x
+ C ⇒
y =
2
3
e
x
+ Ce
−2x
.
7. xy
0
−2y = x
2
[divide by x] ⇒ y
0
+
_
2
x
_
y = x (∗).
I(x) = e
P(x) dx
= e
(−2/x) dx
= e
−2 ln|x|
= e
ln|x|
−2
= e
ln(1/x
2
)
= 1/x
2
. Multiplying the differential equation
(∗) by I(x) gives
1
x
2
y
0
2
x
3
y =
1
x
_
1
x
2
y
_
0
=
1
x
1
x
2
y = ln|x| + C ⇒
y = x
2
(ln|x| + C ) = x
2
ln|x| + Cx
2
.
9. Since P(x) is the derivative of the coefficient of y
0
[P(x) = 1 and the coefficient is x], we can write the differential
equation xy
0
+y =
x in the easily integrable form (xy)
0
=
x ⇒ xy =
2
3
x
3/2
+C ⇒ y =
2
3
x +C/x.
11. I(x) = e
2x dx
= e
x
2
. Multiplying the differential equation y
0
+ 2xy = x
2
by I(x) gives
e
x
2
y
0
+ 2xe
x
2
y = x
2
e
x
2
_
e
x
2
y
_
0
= x
2
e
x
2
. Thus
y = e
−x
2
_
_
x
2
e
x
2
dx + C
_
= e
−x
2
_
1
2
xe
x
2
_
1
2
e
x
2
dx + C
_
=
1
2
x + Ce
−x
2
−e
−x
2 _
1
2
e
x
2
dx.
13. (1 + t)
du
dt
+ u = 1 + t, t > 0 [divide by 1 + t] ⇒
du
dt
+
1
1 + t
u = 1 (∗), which has the form
u
0
+ P(t) u = Q(t). The integrating factor is I(t) = e
P(t) dt
= e
[1/(1+t)] dt
= e
ln(1+t)
= 1 + t.
Multiplying (∗) by I(t) gives us our original equation back. We rewrite it as [(1 + t) u]
0
= 1 + t. Thus,
(1 + t) u =
_
(1 + t) dt = t +
1
2
t
2
+ C ⇒ u =
t +
1
2
t
2
+ C
1 + t
or u =
t
2
+ 2t + 2C
2 (t + 1)
.
15. y
0
= x + y ⇒ y
0
+ (−1)y = x. I(x) = e
(−1) dx
= e
−x
. Multiplying by e
−x
gives e
−x
y
0
−e
−x
y = xe
−x
⇒ (e
−x
y)
0
= xe
−x
⇒ e
−x
y =
_
xe
−x
dx = −xe
−x
−e
−x
+ C [integration by parts with u = x,
dv = e
−x
dx] ⇒ y = −x −1 + Ce
x
. y(0) = 2 ⇒ −1 + C = 2 ⇒ C = 3, so y = −x −1 + 3e
x
.
17.
dv
dt
−2tv = 3t
2
e
t
2
, v (0) = 5. I(t) = e
(−2t)dt
= e
−t
2
. Multiply the differential equation by I(t) to get
e
−t
2 dv
dt
−2te
−t
2
v = 3t
2
_
e
−t
2
v
_
0
= 3t
2
⇒ e
−t
2
v =
_
3t
2
dt = t
3
+ C ⇒ v = t
3
e
t
2
+ Ce
t
2
.
5 = v(0) = 0 · 1 + C · 1 = C, so v = t
3
e
t
2
+ 5e
t
2
.
19. xy
0
= y + x
2
sinx ⇒ y
0
1
x
y = xsinx. I(x) = e
(−1/x) dx
= e
−ln x
= e
ln x
−1
=
1
x
.
Multiplying by
1
x
gives
1
x
y
0
1
x
2
y = sinx ⇒
_
1
x
y
_
0
= sinx ⇒
1
x
y = −cos x + C ⇒
y = −xcos x + Cx. y(π) = 0 ⇒ −π · (−1) + Cπ = 0 ⇒ C = −1, so y = −xcos x −x.
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L I NEAR DI FFERENT I AL EQUAT I ONS ■ 9
21. y
0
+
1
x
y = cos x (x 6= 0), so I(x) = e
(1/x)dx
= e
ln|x|
= x (for
x > 0). Multiplying the differential equation by I(x) gives
xy
0
+ y = xcos x ⇒ (xy)
0
= xcos x. Thus,
y =
1
x
__
xcos xdx + C
_
=
1
x
[xsinx + cos x + C]
= sinx +
cos x
x
+
C
x
The solutions are asymptotic to the y-axis (except for C = −1). In fact, for C > −1, y →∞as x →0
+
, whereas
for C < −1, y →−∞as x →0
+
. As x gets larger, the solutions approximate y = sinx more closely. The graphs
for larger C lie above those for smaller C. The distance between the graphs lessens as x increases.
23. Setting u = y
1−n
,
du
dx
= (1 −n) y
−n
dy
dx
or
dy
dx
=
y
n
1 −n
du
dx
=
u
n/(1−n)
1 −n
du
dx
. Then the Bernoulli differential
equation becomes
u
n/(1−n)
1 −n
du
dx
+ P(x)u
1/(1−n)
= Q(x)u
n/(1−n)
or
du
dx
+ (1 −n)P(x)u = Q(x)(1 −n).
25. y
0
+
2
x
y =
y
3
x
2
. Here n = 3, P(x) =
2
x
, Q(x) =
1
x
2
and setting u = y
−2
, u satisfies u
0
4u
x
= −
2
x
2
.
Then I(x) = e
(−4/x)dx
= x
−4
and u = x
4
__
2
x
6
dx + C
_
= x
4
_
2
5x
5
+ C
_
= Cx
4
+
2
5x
.
Thus, y = ±
_
Cx
4
+
2
5x
_
−1/2
.
27. (a) 2
dI
dt
+ 10I = 40 or
dI
dt
+ 5I = 20. Then the integrating factor is e
5 dt
= e
5t
. Multiplying the differential
equation by the integrating factor gives e
5t
dI
dt
+ 5Ie
5t
= 20e
5t
_
e
5t
I
_
0
= 20e
5t
I(t) = e
−5t
¸_
20e
5t
dt + C
_
= 4 + Ce
−5t
. But 0 = I(0) = 4 + C, so I(t) = 4 −4e
−5t
.
(b) I(0.1) = 4 −4e
−0.5
≈ 1.57 A
29. 5
dQ
dt
+ 20Q = 60 with Q(0) = 0 C. Then the integrating factor is e
4 dt
= e
4t
, and multiplying the differential
equation by the integrating factor gives e
4t
dQ
dt
+ 4e
4t
Q = 12e
4t
_
e
4t
Q
_
0
= 12e
4t
Q(t) = e
−4t
¸_
12e
4t
dt + C
_
= 3 + Ce
−4t
. But 0 = Q(0) = 3 + C so Q(t) = 3
_
1 −e
−4t
_
is the charge at
time t and I = dQ/dt = 12e
−4t
is the current at time t.
31. (a) P increases most rapidly at the beginning, since there are usually many simple, easily-learned sub-skills
associated with learning a skill. As t increases, we would expect dP/dt to remain positive, but decrease. This is
because as time progresses, the only points left to learn are the more difficult ones.
(b)
dP
dt
= k(M −P) is always positive, so the level of performance P is increasing. As P gets close to M, dP/dt
gets close to 0; that is, the performance levels off, as explained in part (a).
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10 ■ L I NEAR DI FFERENT I AL EQUAT I ONS
(c)
dP
dt
+ kP = kM, so I(t) = e
k dt
= e
kt
. Multiplying the differential
equation by I(t) gives e
kt
dP
dt
+ kPe
kt
= kMe
kt
_
e
kt
P
_
0
= kMe
kt
P(t) = e
−kt
__
kMe
kt
dt + C
_
= M + Ce
−kt
, k > 0. Furthermore,
it is reasonable to assume that 0 ≤ P(0) ≤ M, so −M ≤ C ≤ 0.
33. y(0) = 0 kg. Salt is added at a rate of
_
0.4
kg
L
__
5
L
min
_
= 2
kg
min
. Since solution is drained from the tank at a
rate of 3 L/min, but salt solution is added at a rate of 5 L/min, the tank, which starts out with 100 L of water,
contains (100 + 2t) L of liquid after t min. Thus, the salt concentration at time t is
y(t)
100 + 2t
kg
L
. Salt therefore
leaves the tank at a rate of
_
y(t)
100 + 2t
kg
L
__
3
L
min
_
=
3y
100 + 2t
kg
min
. Combining the rates at which salt enters
and leaves the tank, we get
dy
dt
= 2 −
3y
100 + 2t
. Rewriting this equation as
dy
dt
+
_
3
100 + 2t
_
y = 2, we see that
it is linear. I(t) = exp
__
3 dt
100 + 2t
_
= exp
_
3
2
ln(100 + 2t)
_
= (100 + 2t)
3/2
. Multiplying the differential
equation by I(t) gives (100 + 2t)
3/2
dy
dt
+ 3(100 + 2t)
1/2
y = 2(100 + 2t)
3/2
_
(100 + 2t)
3/2
y
_
0
= 2(100 + 2t)
3/2
⇒ (100 + 2t)
3/2
y =
2
5
(100 + 2t)
5/2
+ C ⇒
y =
2
5
(100 + 2t) + C(100 + 2t)
−3/2
. Now 0 = y(0) =
2
5
(100) + C · 100
−3/2
= 40 +
1
1000
C ⇒
C = −40,000, so y =
_
2
5
(100 + 2t) −40,000(100 + 2t)
−3/2
_
kg. From this solution (no pun intended), we
calculate the salt concentration at time t to be C(t) =
y(t)
100 + 2t
=
_
−40,000
(100 + 2t)
5/2
+
2
5
_
kg
L
. In particular,
C(20) =
−40,000
140
5/2
+
2
5
≈ 0.2275
kg
L
and y(20) =
2
5
(140) −40,000(140)
−3/2
≈ 31.85 kg.
35. (a)
dv
dt
+
c
m
v = g and I(t) = e
(c/m)dt
= e
(c/m)t
, and multiplying the differential equation by I(t) gives
e
(c/m)t
dv
dt
+
vce
(c/m)t
m
= ge
(c/m)t
_
e
(c/m)t
v
_
0
= ge
(c/m)t
. Hence,
v(t) = e
−(c/m)t
_
_
ge
(c/m)t
dt + K
_
= mg/c + Ke
−(c/m)t
. But the object is dropped from rest, so v(0) = 0
and K = −mg/c. Thus, the velocity at time t is v(t) = (mg/c)
_
1 −e
−(c/m)t
_
.
(b) lim
t→∞
v(t) = mg/c
(c) s(t) =
_
v(t) dt = (mg/c)
_
t + (m/c)e
−(c/m)t
_
+ c
1
where c
1
= s(0) −m
2
g/c
2
. s(0) is the initial position,
so s(0) = 0 and s(t) = (mg/c)
_
t + (m/c)e
−(c/m)t
_
−m
2
g/c
2
.
T
h
o
m
s
o
n
B
r
o
o
k
s
-
C
o
l
e
c
o
p
y
r
i
g
h
t
2
0
0
7
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You are on page 1of 10. # LI NEAR DI FFERENTI AL EQUATI ONS. A first-order linear differential equation is one that can be put into the form. where and are continuous functions on a given interval. This type of equation occurs. frequently in various sciences, as we will see.. An example of a linear equation is because, for , it can be written. in the form. Notice that this differential equation is not separable because it’s impossible to factor the. expression for as a function of x times a function of y. But we can still solve the equa-. tion by noticing, by the Product Rule, that. and so we can rewrite the equation as. If we now integrate both sides of this equation, we get. or. If we had been given the differential equation in the form of Equation 2, we would have. had to take the preliminary step of multiplying each side of the equation by x.. It turns out that every first-order linear differential equation can be solved in a similar. fashion by multiplying both sides of Equation 1 by a suitable function called an. integrating factor. We try to find so that the left side of Equation 1, when multiplied by. , becomes the derivative of the product :. If we can find such a function , then Equation 1 becomes. Integrating both sides, we would have. so the solution would be. To find such an , we expand Equation 3 and cancel terms:. I͑x͒P͑x͒ IЈ͑x͒. I͑x͒yЈ ϩ I͑x͒P͑x͒y ͑I͑x͒y͒Ј IЈ͑x͒y ϩ I͑x͒yЈ. I. y͑x͒ . 1. I͑x͒. ͫy. I͑x͒Q͑x͒ dx ϩ C. ͬ. 4. I͑x͒y . y. I͑x͒Q͑x͒ dx ϩ C. ͑I͑x͒y͒Ј I͑x͒Q͑x͒. I. I͑x͒͑yЈ ϩ P͑x͒y͒ ͑I͑x͒y͒Ј 3. I͑x͒y I͑x͒. I. I͑x͒. y x ϩ. C. x. xy x. 2. ϩ C. ͑xy͒Ј 2x. xyЈ ϩ y ͑xy͒Ј. yЈ ϩ. 1. x. y 2 2. x 0 xyЈ ϩ y 2x. Q P. dy. dx. ϩ P͑x͒y Q͑x͒ 1. 1. T. h. o. m. s. o. n. B. r. o. o. k. s. -. C. o. l. e. c. o. p. y. r. i. g. h. t. 2. 0. 0. 7. This is a separable differential equation for , which we solve as follows:. where . We are looking for a particular integrating factor, not the most general. one, so we take A 1 and use. Thus, a formula for the general solution to Equation 1 is provided by Equation 4, where. is given by Equation 5. Instead of memorizing this formula, however, we just remember. the form of the integrating factor.. To solve the linear differential equation , multiply both sides by. the integrating factor and integrate both sides.. EXAMPLE 1 Solve the differential equation .. SOLUTI ON The given equation is linear since it has the form of Equation 1 with. and . An integrating factor is. Multiplying both sides of the differential equation by , we get. or. Integrating both sides, we have. EXAMPLE 2 Find the solution of the initial-value problem. SOLUTI ON We must first divide both sides by the coefficient of to put the differential. equation into standard form:. The integrating factor is. I͑x͒ e. x ͑1͞x͒ dx. e. ln x. x. x Ͼ 0 yЈ ϩ. 1. x. y . 1. x. 2. 6. y͑1͒ 2 x Ͼ 0 x. 2. yЈ ϩ xy 1. y 2 ϩ Ce. Ϫx. 3. e. x. 3. y . y. 6x. 2. e. x. 3. dx 2e. x. 3. ϩ C. d. dx. ͑e. x. 3. y͒ 6x. 2. e. x. 3. e. x. 3. dy. dx. ϩ 3x. 2. e. x. 3. y 6x. 2. e. x. 3. e. x. 3. I͑x͒ e. x 3x. 2. dx. e. x. 3. Q͑x͒ 6x. 2. P͑x͒ 3x. 2. dy. dx. ϩ 3x. 2. y 6x. 2. I͑x͒ e. x P͑x͒ dx. yЈ ϩ P͑x͒y Q͑x͒. I. I͑x͒ e. x P͑x͒ dx. 5. A Ϯe. C. I Ae. x P͑x͒ dx. ln. Խ. I. Խ. y. P͑x͒ dx. y. dI. I. y. P͑x͒ dx. I. 2 ■ L I NEAR DI FFERENT I AL EQUAT I ONS. FI GURE 1. 6. _3. _1.5 1.8. C=2. C=1. C=_2. C=_1. C=0. ■ ■. Figure 1 shows the graphs of several mem-. bers of the family of solutions in Example 1.. Notice that they all approach as . x l ϱ 2. T. h. o. m. s. o. n. B. r. o. o. k. s. -. C. o. l. e. c. o. p. y. r. i. g. h. t. 2. 0. 0. 7. Multiplication of Equation 6 by gives. Then. and so. Since , we have. Therefore, the solution to the initial-value problem is. EXAMPLE 3 Solve .. SOLUTI ON The given equation is in the standard form for a linear equation. Multiplying by. the integrating factor. we get. or. Therefore. Recall from Section 6.4 that can’t be expressed in terms of elementary functions.. Nonetheless, it’s a perfectly good function and we can leave the answer as. Another way of writing the solution is. (Any number can be chosen for the lower limit of integration.). APPLICATION TO ELECTRIC CIRCUITS. Let’s consider the simple electric circuit shown in Figure 4: An electromotive force (usually. a battery or generator) produces a voltage of volts (V) and a current of amperes. (A) at time . The circuit also contains a resistor with a resistance of ohms ( ) and an. inductor with an inductance of henries (H).. Ohm’s Law gives the drop in voltage due to the resistor as . The voltage drop due to. the inductor is . One of Kirchhoff’s laws says that the sum of the voltage drops is. equal to the supplied voltage . Thus, we have. which is a first-order linear differential equation. The solution gives the current at time . t I. L. dI. dt. ϩ RI E͑t͒ 7. E͑t͒. L͑dI͞dt͒. RI. L. ⍀ R t. I͑t͒ E͑t͒. y e. Ϫx. 2. y. x. 0. e. t. 2. dt ϩ Ce. Ϫx. 2. y e. Ϫx. 2. y. e. x. 2. dx ϩ Ce. Ϫx. 2. x e. x. 2. dx. e. x. 2. y . y. e. x. 2. dx ϩ C. (e. x. 2. y)Ј e. x. 2. e. x. 2. yЈ ϩ 2xe. x. 2. y e. x. 2. e. x 2x dx. e. x. 2. yЈ ϩ 2xy 1. y . ln x ϩ 2. x. 2 . ln 1 ϩ C. 1. C. y͑1͒ 2. y . ln x ϩ C. x. xy . y. 1. x. dx ln x ϩ C. ͑xy͒Ј . 1. x. or xyЈ ϩ y . 1. x. x. L I NEAR DI FFERENT I AL EQUAT I ONS ■ 3. ■ ■. The solution of the initial-value problem in. Example 2 is shown in Figure 2.. FI GURE 2. (1, 2). 5. _5. 0 4. ■ ■. Even though the solutions of the differential. equation in Example 3 are expressed in terms of. an integral, they can still be graphed by a com-. puter algebra system (Figure 3).. FI GURE 3. C=2. C=_2. 2.5. _2.5. _2.5 2.5. FI GURE 4. R. E. switch. L. T. h. o. m. s. o. n. B. r. o. o. k. s. -. C. o. l. e. c. o. p. y. r. i. g. h. t. 2. 0. 0. 7. 4 ■ L I NEAR DI FFERENT I AL EQUAT I ONS. EXAMPLE 4 Suppose that in the simple circuit of Figure 4 the resistance is and the. inductance is 4 H. If a battery gives a constant voltage of 60 V and the switch is closed. when so the current starts with , find (a) , (b) the current after 1 s, and. (c) the limiting value of the current.. SOLUTI ON. (a) If we put , , and in Equation 7, we obtain the initial-value. problem. or. Multiplying by the integrating factor , we get. Since , we have , so and. (b) After 1 second the current is. (c). EXAMPLE 5 Suppose that the resistance and inductance remain as in Example 4. but, instead of the battery, we use a generator that produces a variable voltage of. volts. Find .. SOLUTI ON This time the differential equation becomes. The same integrating factor gives. Using Formula 98 in the Table of Integrals, we have. I . 5. 101. ͑sin 30t Ϫ 10 cos 30t͒ ϩ Ce. Ϫ3t. e. 3t. I . y. 15e. 3t. sin 30t dt 15. e. 3t. 909. ͑3 sin 30t Ϫ 30 cos 30t͒ ϩ C. d. dt. ͑e. 3t. I͒ e. 3t. dI. dt. ϩ 3e. 3t. I 15e. 3t. sin 30t. e. 3t. dI. dt. ϩ 3I 15 sin 30t or 4. dI. dt. ϩ 12I 60 sin 30t. I͑t͒ E͑t͒ 60 sin 30t. 5 Ϫ 0 5. 5 Ϫ 5 lim. t lϱ. e. Ϫ3t. lim. t lϱ. I͑t͒ lim. t lϱ. 5͑1 Ϫ e. Ϫ3t. ͒. I͑1͒ 5͑1 Ϫ e. Ϫ3. ͒ Ϸ 4.75 A. I͑t͒ 5͑1 Ϫ e. Ϫ3t. ͒. C Ϫ5 5 ϩ C 0 I͑0͒ 0. I͑t͒ 5 ϩ Ce. Ϫ3t. e. 3t. I . y. 15e. 3t. dt 5e. 3t. ϩ C. d. dt. ͑e. 3t. I͒ 15e. 3t. e. 3t. dI. dt. ϩ 3e. 3t. I 15e. 3t. e. x 3 dt. e. 3t. I͑0͒ 0. dI. dt. ϩ 3I 15. I͑0͒ 0 4. dI. dt. ϩ 12I 60. E͑t͒ 60 R 12 L 4. I͑t͒ I͑0͒ 0 t 0. 12 ⍀. ■ ■. The differential equation in Example 4 is. both linear and separable, so an alternative. method is to solve it as a separable equation.. If we replace the battery by a generator, how-. ever, we get an equation that is linear but not. separable (Example 5).. ■ ■. Figure 6 shows the graph of the current. when the battery is replaced by a generator.. FI GURE 6. 2. _2. 2.5 0. FI GURE 5. 6. 0. 2.5. y=5. ■ ■. Figure 5 shows how the current in Example. 4 approaches its limiting value.. T. h. o. m. s. o. n. B. r. o. o. k. s. -. C. o. l. e. c. o. p. y. r. i. g. h. t. 2. 0. 0. 7. L I NEAR DI FFERENT I AL EQUAT I ONS ■ 5. Since , we get. so I͑t͒ . 5. 101. ͑sin 30t Ϫ 10 cos 30t͒ ϩ. 50. 101. e. Ϫ3t. Ϫ. 50. 101. ϩ C 0. I͑0͒ 0. EXERCISES. 1–4 Determine whether the differential equation is linear.. 1. 2.. 3. 4.. ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■. 5–14 Solve the differential equation.. 5. 6.. 7. 8.. 9. 10.. 11.. 12. ,. 13. ,. 14.. ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■. 15–20 Solve the initial-value problem.. 15. ,. 16. , ,. 17. ,. 18. , ,. 19. ,. 20. , ,. ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■. ;. 21–22 Solve the differential equation and use a graphing cal-. culator or computer to graph several members of the family of. solutions. How does the solution curve change as varies?. 21. , 22.. ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■. 23. A Bernoulli differential equation (named after James. Bernoulli) is of the form. dy. dx. ϩ P͑x͒y Q͑x͒y. n. yЈ ϩ ͑cos x͒y cos x x Ͼ 0 xyЈ ϩ y x cos x. C. x Ͼ 0 y͑1͒ 0 x. dy. dx. Ϫ. y. x ϩ 1. x. y͑ ͒ 0 xyЈ y ϩ x. 2. sin x. y͑4͒ 20 x Ͼ 0 2xyЈ ϩ y 6x. v͑0͒ 5. dv. dt. Ϫ 2tv 3t. 2. e. t. 2. y͑1͒ 0 t Ͼ 0 t. dy. dt. ϩ 2y t. 3. y͑0͒ 2 yЈ x ϩ y. t ln t. dr. dt. ϩ r te. t. t Ͼ 0 ͑1 ϩ t͒. du. dt. ϩ u 1 ϩ t. Ϫ ͞2 Ͻ x Ͻ ͞2. dy. dx. x sin 2x ϩ y tan x. dy. dx. ϩ 2xy x. 2. 1 ϩ xy xyЈ xyЈ ϩ y sx. x. 2. yЈ ϩ 2xy cos. 2. x xyЈ Ϫ 2y x. 2. yЈ x ϩ 5y yЈ ϩ 2y 2e. x. yЈ ϩ cos y tan x xyЈ ϩ ln x Ϫ x. 2. y 0. y ϩ sin x x. 3. yЈ yЈ ϩ e. x. y x. 2. y. 2. Observe that, if or , the Bernoulli equation is linear.. For other values of , show that the substitution trans-. forms the Bernoulli equation into the linear equation. 24–26 Use the method of Exercise 23 to solve the differential. equation.. 24. 25.. 26.. ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■. 27. In the circuit shown in Figure 4, a battery supplies a constant. voltage of 40 V, the inductance is 2 H, the resistance is ,. and .. (a) Find .. (b) Find the current after s.. 28. In the circuit shown in Figure 4, a generator supplies a voltage. of volts, the inductance is H, the resistance. is , and A.. (a) Find .. (b) Find the current after s.. ;. (c) Use a graphing device to draw the graph of the current. function.. 29. The figure shows a circuit containing an electromotive force,. a capacitor with a capacitance of farads (F), and a resistor. with a resistance of ohms ( ). The voltage drop across the. capacitor is , where is the charge (in coulombs), so in. this case Kirchhoff’s Law gives. But , so we have. Suppose the resistance is , the capacitance is F, a. battery gives a constant voltage of 60 V, and the initial charge. is C. Find the charge and the current at time . t Q͑0͒ 0. 0.05 5 ⍀. R. dQ. dt. ϩ. 1. C. Q E͑t͒. I dQ͞dt. RI ϩ. Q. C. E͑t͒. Q Q͞C. C. E R. ⍀ R. C. 0.1. I͑t͒. I͑0͒ 1 20 ⍀. 1 E͑t͒ 40 sin 60t. 0.1. I͑t͒. I͑0͒ 0. 10 ⍀. yЈ ϩ y xy. 3. yЈ ϩ. 2. x. y . y. 3. x. 2. xyЈ ϩ y Ϫxy. 2. du. dx. ϩ ͑1 Ϫ n͒P͑x͒u ͑1 Ϫ n͒Q͑x͒. u y. 1Ϫn. n. 1 n 0. T. h. o. m. s. o. n. B. r. o. o. k. s. -. C. o. l. e. c. o. p. y. r. i. g. h. t. 2. 0. 0. 7. 30. In the circuit of Exercise 29, , , ,. and . Find the charge and the current at time .. 31. Psychologists interested in learning theory study learning. curves. A learning curve is the graph of a function , the. performance of someone learning a skill as a function of the. training time . The derivative represents the rate at. which performance improves.. (a) When do you think increases most rapidly? What. happens to as increases? Explain.. (b) If is the maximum level of performance of which the. learner is capable, explain why the differential equation. is a reasonable model for learning.. (c) Solve the differential equation in part (b) as a linear differ-. ential equation and use your solution to graph the learning. curve.. 32. Two new workers were hired for an assembly line. Jim pro-. cessed 25 units during the first hour and 45 units during the. second hour. Mark processed 35 units during the first hour and. 50 units the second hour. Using the model of Exercise 31 and. assuming that , estimate the maximum number of. units per hour that each worker is capable of processing.. 33. In Section 7.6 we looked at mixing problems in which the. volume of fluid remained constant and saw that such problems. give rise to separable equations. (See Example 5 in that. section.) If the rates of flow into and out of the system are dif-. ferent, then the volume is not constant and the resulting differ-. ential equation is linear but not separable.. A tank contains 100 L of water. A solution with a salt. concentration of is added at a rate of . The. solution is kept mixed and is drained from the tank at a rate. 5 L͞min 0.4 kg͞L. P͑0͒ 0. k a positive constant. dP. dt. k͑M Ϫ P͒. M. t dP͞dt. P. dP͞dt t. P͑t͒. t E͑t͒ 10 sin 60t. Q͑0͒ 0 C 0.01 F R 2 ⍀ of . If is the amount of salt (in kilograms) after. minutes, show that satisfies the differential equation. Solve this equation and find the concentration after 20 minutes.. 34. A tank with a capacity of 400 L is full of a mixture of water. and chlorine with a concentration of 0.05 g of chlorine per liter.. In order to reduce the concentration of chlorine, fresh water is. pumped into the tank at a rate of . The mixture is kept. stirred and is pumped out at a rate of . Find the amount. of chlorine in the tank as a function of time.. 35. An object with mass is dropped from rest and we assume. that the air resistance is proportional to the speed of the object.. If is the distance dropped after seconds, then the speed is. and the acceleration is . If is the acceleration. due to gravity, then the downward force on the object is. , where is a positive constant, and Newton’s Second. Law gives. (a) Solve this as a linear equation to show that. (b) What is the limiting velocity?. (c) Find the distance the object has fallen after seconds.. 36. If we ignore air resistance, we can conclude that heavier. objects fall no faster than lighter objects. But if we take air. resistance into account, our conclusion changes. Use the. expression for the velocity of a falling object in Exercise 35(a). to find and show that heavier objects do fall faster than. lighter ones.. dv͞dm. t. v . mt. c. ͑1 Ϫ e. Ϫct͞m. ͒. m. dv. dt. mt Ϫ cv. c mt Ϫ cv. t a vЈ͑t͒ v sЈ͑t͒. t s͑t͒. m. 10 L͞s. 4 L͞s. dy. dt. 2 Ϫ. 3y. 100 ϩ 2t. y t. y͑t͒ 3 L͞min. 6 ■ L I NEAR DI FFERENT I AL EQUAT I ONS. T. h. o. m. s. o. n. B. r. o. o. k. s. -. C. o. l. e. c. o. p. y. r. i. g. h. t. 2. 0. 0. 7. L I NEAR DI FFERENT I AL EQUAT I ONS ■ 7. 1. No 3. Yes 5.. 7. 9.. 11.. 13. 15.. 17. 19.. 21.. C=2. C=1. C=0.2. C=_1. C=_2. 6. _4. 0 10. y sin x ϩ ͑cos x͒͞x ϩ C͞x. y Ϫx cos x Ϫ x v t. 3. e. t. 2. ϩ 5e. t. 2. y Ϫx Ϫ 1 ϩ 3e. x. u ͑t. 2. ϩ 2t ϩ 2C͓͒͞2͑t ϩ 1͔͒. y . 1. 2. x ϩ Ce. Ϫx. 2. Ϫ. 1. 2. e. Ϫx. 2. x e. x. 2. dx. y . 2. 3. sx ϩ C͞x y x. 2. ln. Խ. x. Խ. ϩ Cx. 2. y . 2. 3. e. x. ϩ Ce. Ϫ2x. 25.. 27. (a) (b). 29.. 31. (a) At the beginning; stays positive, but decreases. (c). 33. ; 0.2275 kg͞L. 35. (b) (c) ͑mt͞c͓͒t ϩ ͑m͞c͒e. Ϫct͞m. ͔ Ϫ m. 2. t͞c. 2. mt͞c. y . 2. 5. ͑100 ϩ 2t͒ Ϫ 40,000͑100 ϩ 2t͒. Ϫ3͞2. 0. M. P(t). t. P(0). P͑t͒ M ϩ Ce. Ϫkt. Q͑t͒ 3͑1 Ϫ e. Ϫ4t. ͒, I͑t͒ 12e. Ϫ4t. 4 Ϫ 4e. Ϫ1͞2. Ϸ 1.57 A I͑t͒ 4 Ϫ 4e.
|
Ϫ5t. y Ϯ͓Cx. 4. ϩ 2͑͞5x͔͒. Ϫ1͞2. T. h. o. m. s. o. n. B. r. o. o. k. s. -. C. o. l. e. c. o. p. y. r. i. g. h. t. 2. 0. 0. 7. 8 ■ L I NEAR DI FFERENT I AL EQUAT I ONS. SOLUTIONS. 1. y. 0. + e. x. y = x. 2. y. 2. is not linear since it cannot be put into the standard linear form (1), y. 0. + P(x) y = Q(x).. 3. xy. 0. + lnx −x. 2. y = 0 ⇒ xy. 0. −x. 2. y = −lnx ⇒ y. 0. + (−x) y = −. lnx. x. , which is in the standard linear. form (1), so this equation is linear.. 5. Comparing the given equation, y. 0. + 2y = 2e. x. , with the general form, y. 0. + P(x)y = Q(x), we see that P(x) = 2. and the integrating factor is I(x) = e. P(x)dx. = e. 2 dx. = e. 2x. . Multiplying the differential equation by I(x) gives. e. 2x. y. 0. + 2e. 2x. y = 2e. 3x. _. e. 2x. y. _. 0. = 2e. 3x. ⇒ e. 2x. y =. _. 2e. 3x. dx ⇒ e. 2x. y =. 2. 3. e. 3x. + C ⇒. y =. 2. 3. e. x. + Ce. −2x. .. 7. xy. 0. −2y = x. 2. [divide by x] ⇒ y. 0. +. _. 2. x. _. y = x (∗).. I(x) = e. P(x) dx. = e. (−2/x) dx. = e. −2 ln|x|. = e. ln|x|. −2. = e. ln(1/x. 2. ). = 1/x. 2. . Multiplying the differential equation. (∗) by I(x) gives. 1. x. 2. y. 0. 2. x. 3. y =. 1. x. _. 1. x. 2. y. _. 0. =. 1. x. 1. x. 2. y = ln|x| + C ⇒. y = x. 2. (ln|x| + C ) = x. 2. ln|x| + Cx. 2. .. 9. Since P(x) is the derivative of the coefficient of y. 0. [P(x) = 1 and the coefficient is x], we can write the differential. equation xy. 0. +y =. x in the easily integrable form (xy). 0. =. x ⇒ xy =. 2. 3. x. 3/2. +C ⇒ y =. 2. 3. x +C/x.. 11. I(x) = e. 2x dx. = e. x. 2. . Multiplying the differential equation y. 0. + 2xy = x. 2. by I(x) gives. e. x. 2. y. 0. + 2xe. x. 2. y = x. 2. e. x. 2. _. e. x. 2. y. _. 0. = x. 2. e. x. 2. . Thus. y = e. −x. 2. _. _. x. 2. e. x. 2. dx + C. _. = e. −x. 2. _. 1. 2. xe. x. 2. _. 1. 2. e. x. 2. dx + C. _. =. 1. 2. x + Ce. −x. 2. −e. −x. 2 _. 1. 2. e. x. 2. dx.. 13. (1 + t). du. dt. + u = 1 + t, t > 0 [divide by 1 + t] ⇒. du. dt. +. 1. 1 + t. u = 1 (∗), which has the form. u. 0. + P(t) u = Q(t). The integrating factor is I(t) = e. P(t) dt. = e. [1/(1+t)] dt. = e. ln(1+t). = 1 + t.. Multiplying (∗) by I(t) gives us our original equation back. We rewrite it as [(1 + t) u]. 0. = 1 + t. Thus,. (1 + t) u =. _. (1 + t) dt = t +. 1. 2. t. 2. + C ⇒ u =. t +. 1. 2. t. 2. + C. 1 + t. or u =. t. 2. + 2t + 2C. 2 (t + 1). .. 15. y. 0. = x + y ⇒ y. 0. + (−1)y = x. I(x) = e. (−1) dx. = e. −x. . Multiplying by e. −x. gives e. −x. y. 0. −e. −x. y = xe. −x. ⇒ (e. −x. y). 0. = xe. −x. ⇒ e. −x. y =. _. xe. −x. dx = −xe. −x. −e. −x. + C [integration by parts with u = x,. dv = e. −x. dx] ⇒ y = −x −1 + Ce. x. . y(0) = 2 ⇒ −1 + C = 2 ⇒ C = 3, so y = −x −1 + 3e. x. .. 17.. dv. dt. −2tv = 3t. 2. e. t. 2. , v (0) = 5. I(t) = e. (−2t)dt. = e. −t. 2. . Multiply the differential equation by I(t) to get. e. −t. 2 dv. dt. −2te. −t. 2. v = 3t. 2. _. e. −t. 2. v. _. 0. = 3t. 2. ⇒ e. −t. 2. v =. _. 3t. 2. dt = t. 3. + C ⇒ v = t. 3. e. t. 2. + Ce. t. 2. .. 5 = v(0) = 0 · 1 + C · 1 = C, so v = t. 3. e. t. 2. + 5e. t. 2. .. 19. xy. 0. = y + x. 2. sinx ⇒ y. 0. 1. x. y = xsinx. I(x) = e. (−1/x) dx. = e. −ln x. = e. ln x. −1. =. 1. x. .. Multiplying by. 1. x. gives. 1. x. y. 0. 1. x. 2. y = sinx ⇒. _. 1. x. y. _. 0. = sinx ⇒. 1. x. y = −cos x + C ⇒. y = −xcos x + Cx. y(π) = 0 ⇒ −π · (−1) + Cπ = 0 ⇒ C = −1, so y = −xcos x −x.. T. h. o. m. s. o. n. B. r. o. o. k. s. -. C. o. l. e. c. o. p. y. r. i. g. h. t. 2. 0. 0. 7. L I NEAR DI FFERENT I AL EQUAT I ONS ■ 9. 21. y. 0. +. 1. x. y = cos x (x 6= 0), so I(x) = e. (1/x)dx. = e. ln|x|. = x (for. x > 0). Multiplying the differential equation by I(x) gives. xy. 0. + y = xcos x ⇒ (xy). 0. = xcos x. Thus,. y =. 1. x. __. xcos xdx + C. _. =. 1. x. [xsinx + cos x + C]. = sinx +. cos x. x. +. C. x. The solutions are asymptotic to the y-axis (except for C = −1). In fact, for C > −1, y →∞as x →0. +. , whereas. for C < −1, y →−∞as x →0. +. . As x gets larger, the solutions approximate y = sinx more closely. The graphs. for larger C lie above those for smaller C. The distance between the graphs lessens as x increases.. 23. Setting u = y. 1−n. ,. du. dx. = (1 −n) y. −n. dy. dx. or. dy. dx. =. y. n. 1 −n. du. dx. =. u. n/(1−n). 1 −n. du. dx. . Then the Bernoulli differential. equation becomes. u. n/(1−n). 1 −n. du. dx. + P(x)u. 1/(1−n). = Q(x)u. n/(1−n). or. du. dx. + (1 −n)P(x)u = Q(x)(1 −n).. 25. y. 0. +. 2. x. y =. y. 3. x. 2. . Here n = 3, P(x) =. 2. x. , Q(x) =. 1. x. 2. and setting u = y. −2. , u satisfies u. 0. 4u. x. = −. 2. x. 2. .. Then I(x) = e. (−4/x)dx. = x. −4. and u = x. 4. __. 2. x. 6. dx + C. _. = x. 4. _. 2. 5x. 5. + C. _. = Cx. 4. +. 2. 5x. .. Thus, y = ±. _. Cx. 4. +. 2. 5x. _. −1/2. .. 27. (a) 2. dI. dt. + 10I = 40 or. dI. dt. + 5I = 20. Then the integrating factor is e. 5 dt. = e. 5t. . Multiplying the differential. equation by the integrating factor gives e. 5t. dI. dt. + 5Ie. 5t. = 20e. 5t. _. e. 5t. I. _. 0. = 20e. 5t. I(t) = e. −5t. ¸_. 20e. 5t. dt + C. _. = 4 + Ce. −5t. . But 0 = I(0) = 4 + C, so I(t) = 4 −4e. −5t. .. (b) I(0.1) = 4 −4e. −0.5. ≈ 1.57 A. 29. 5. dQ. dt. + 20Q = 60 with Q(0) = 0 C. Then the integrating factor is e. 4 dt. = e. 4t. , and multiplying the differential. equation by the integrating factor gives e. 4t. dQ. dt. + 4e. 4t. Q = 12e. 4t. _. e. 4t. Q. _. 0. = 12e. 4t. Q(t) = e. −4t. ¸_. 12e. 4t. dt + C. _. = 3 + Ce. −4t. . But 0 = Q(0) = 3 + C so Q(t) = 3. _. 1 −e. −4t. _. is the charge at. time t and I = dQ/dt = 12e. −4t. is the current at time t.. 31. (a) P increases most rapidly at the beginning, since there are usually many simple, easily-learned sub-skills. associated with learning a skill. As t increases, we would expect dP/dt to remain positive, but decrease. This is. because as time progresses, the only points left to learn are the more difficult ones.. (b). dP. dt. = k(M −P) is always positive, so the level of performance P is increasing. As P gets close to M, dP/dt. gets close to 0; that is, the performance levels off, as explained in part (a).. T. h. o. m. s. o. n. B. r. o. o. k. s. -. C. o. l. e. c. o. p. y. r. i. g. h. t. 2. 0. 0. 7. 10 ■ L I NEAR DI FFERENT I AL EQUAT I ONS. (c). dP. dt. + kP = kM, so I(t) = e. k dt. = e. kt. . Multiplying the differential. equation by I(t) gives e. kt. dP. dt. + kPe. kt. = kMe. kt. _. e. kt. P. _. 0. = kMe. kt. P(t) = e. −kt. __. kMe. kt. dt + C. _. = M + Ce. −kt. , k > 0. Furthermore,. it is reasonable to assume that 0 ≤ P(0) ≤ M, so −M ≤ C ≤ 0.. 33. y(0) = 0 kg. Salt is added at a rate of. _. 0.4. kg. L. __. 5. L. min. _. = 2. kg. min. . Since solution is drained from the tank at a. rate of 3 L/min, but salt solution is added at a rate of 5 L/min, the tank, which starts out with 100 L of water,. contains (100 + 2t) L of liquid after t min. Thus, the salt concentration at time t is. y(t). 100 + 2t. kg. L. . Salt therefore. leaves the tank at a rate of. _. y(t). 100 + 2t. kg. L. __. 3. L. min. _. =. 3y. 100 + 2t. kg. min. . Combining the rates at which salt enters. and leaves the tank, we get. dy. dt. = 2 −. 3y. 100 + 2t. . Rewriting this equation as. dy. dt. +. _. 3. 100 + 2t. _. y = 2, we see that. it is linear. I(t) = exp. __. 3 dt. 100 + 2t. _. = exp. _. 3. 2. ln(100 + 2t). _. = (100 + 2t). 3/2. . Multiplying the differential. equation by I(t) gives (100 + 2t). 3/2. dy. dt. + 3(100 + 2t). 1/2. y = 2(100 + 2t). 3/2. _. (100 + 2t). 3/2. y. _. 0. = 2(100 + 2t). 3/2. ⇒ (100 + 2t). 3/2. y =. 2. 5. (100 + 2t). 5/2. + C ⇒. y =. 2. 5. (100 + 2t) + C(100 + 2t). −3/2. . Now 0 = y(0) =. 2. 5. (100) + C · 100. −3/2. = 40 +. 1. 1000. C ⇒. C = −40,000, so y =. _. 2. 5. (100 + 2t) −40,000(100 + 2t). −3/2. _. kg. From this solution (no pun intended), we. calculate the salt concentration at time t to be C(t) =. y(t). 100 + 2t. =. _. −40,000. (100 + 2t). 5/2. +. 2. 5. _. kg. L. . In particular,. C(20) =. −40,000. 140. 5/2. +. 2. 5. ≈ 0.2275. kg. L. and y(20) =. 2. 5. (140) −40,000(140). −3/2. ≈ 31.85 kg.. 35. (a). dv. dt. +. c. m. v = g and I(t) = e. (c/m)dt. = e. (c/m)t. , and multiplying the differential equation by I(t) gives. e. (c/m)t. dv. dt. +. vce. (c/m)t. m. = ge. (c/m)t. _. e. (c/m)t. v. _. 0. = ge. (c/m)t. . Hence,. v(t) = e. −(c/m)t. _. _. ge. (c/m)t. dt + K. _. = mg/c + Ke. −(c/m)t. . But the object is dropped from rest, so v(0) = 0. and K = −mg/c. Thus, the velocity at time t is v(t) = (mg/c). _. 1 −e. −(c/m)t. _. .. (b) lim. t→∞. v(t) = mg/c. (c) s(t) =. _. v(t) dt = (mg/c). _. t + (m/c)e. −(c/m)t. _. + c. 1. where c. 1. = s(0) −m. 2. g/c. 2. . s(0) is the initial position,. so s(0) = 0 and s(t) = (mg/c). _. t + (m/c)e. −(c/m)t. _. −m. 2. g/c. 2. .. T. h. o. m. s. o. n. B. r. o. o. k. s. -. C. o. l. e. c. o. p. y. r. i. g. h. t. 2. 0. 0. 7.
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Home > Margin Of > Margin Of Error Definition For Dummies
# Margin Of Error Definition For Dummies
## Contents
Garcia's confidence interval is 49% to 59%, and Smith's confidence interval is 41% to 51%. You need to include the margin of error (in this case, 3%) in your results. Tags: confidence intervals, population Before posting, create an account!Stop this in-your-face noticeReserve your usernameFollow people you like, learn fromExtend your profileGain reputation for your contributionsNo annoying captchas across siteAnd much more! Next, we find the standard error of the mean, using the following equation: SEx = s / sqrt( n ) = 0.4 / sqrt( 900 ) = 0.4 / 30 = http://threadspodcast.com/margin-of/margin-of-error-and-definition.html
You've probably heard that term -- "margin of error" -- a lot before. Survey results themselves (with no MOE) are only a measure of how the sample of selected individuals felt about the issue; they don't reflect how the entire population may have felt, You want to estimate the average weight of the cones they make over a one-day period, including a margin of error. And the same goes for young adults, retirees, rich people, poor people, etc. http://www.dummies.com/education/math/statistics/how-to-interpret-the-margin-of-error-in-statistics/
## Margin Of Error Example
Find the degrees of freedom (DF). Interpreting the Margin of Error In practice, nonsampling errors occur that can make the margin of error reported for a poll smaller than it should be if it reflected all sources For n = 50 cones sampled, the sample mean was found to be 10.3 ounces. Reply TPRJones I don't understand how the margin of error calculation doesn't take the population size into consideration.
Almost always there are people who have not made up their mind. However, if the study is done right, the results from the sample should be close to and representative of the actual values for the entire population, with a high level of Toggle navigation Search Submit San Francisco, CA Brr, it´s cold outside Learn by category LiveConsumer ElectronicsFood & DrinkGamesHealthPersonal FinanceHome & GardenPetsRelationshipsSportsReligion LearnArt CenterCraftsEducationLanguagesPhotographyTest Prep WorkSocial MediaSoftwareProgrammingWeb Design & DevelopmentBusinessCareersComputers Online Courses Poll Bias Definition Because the results of most survey questions can be reported in terms of percentages, the margin of error most often appears as a percentage, as well.
The population standard deviation, will be given in the problem. Margin Of Error Definition Statistics Here are the steps for calculating the margin of error for a sample mean: Find the population standard deviation and the sample size, n. Find the critical value. her latest blog Calculation in the upcoming election.
Many media surveys are based on what are called quota samples, and, although margins of error are reported from them, they do not strictly apply. Margin Of Error Definition Politics Easy! Because the results of most survey questions can be reported in terms of percentages, the margin of error most often appears as a percentage, as well. For example, the z*-value is 1.96 if you want to be about 95% confident.
## Margin Of Error Definition Statistics
Now that I've told you that, what is your favorite color?" That's called a leading question, and it's a big no-no in surveying. More hints You can also use a graphing calculator or standard statistical tables (found in the appendix of most introductory statistics texts). Margin Of Error Example Suppose, in the mayoral election poll mentioned earlier, we sample 100 people who intend to vote and that 55 support Ms. What Does Margin Of Error Mean In Confidence Intervals Thus, a reported 3 percent margin of error becomes about 5 percent and a reported 4 percent margin of error becomes about 7 percent when the size of the lead is
Let's say you picked a specific number of people in the United States at random. http://threadspodcast.com/margin-of/margin-of-error-definition-politics.html Any reproduction or other use of content without the express written consent of iSixSigma is prohibited. A researcher surveying customers every six months to understand whether customer service is improving may see the percentage of respondents who say it is "very good" go from 50 percent in Easy! What Does Margin Of Error Mean In Polls
A 90 percent level can be obtained with a smaller sample, which usually translates into a less expensive survey. Sample questions A poll shows that Garcia is leading Smith by 54% to 46% with a margin of error of plus/minus 5% at a 95% confidence level. Donnelly Jr.List Price: \$19.95Buy Used: \$0.01Buy New: \$18.35Casio FX-CG10 PRIZM Color Graphing Calculator (Black)List Price: \$129.99Buy Used: \$74.99Buy New: \$121.73Approved for AP Statistics and Calculus About Us Contact Us Privacy his comment is here Thanks f Reply James Jones Great explanation, clearly written and well appreciated.
The choice of t statistic versus z-score does not make much practical difference when the sample size is very large. How Is Margin Of Error Calculated In Polls z*-Values for Selected (Percentage) Confidence Levels Percentage Confidence z*-Value 80 1.28 90 1.645 95 1.96 98 2.33 99 2.58 Note that these values are taken from the standard normal (Z-) distribution. With most polls still by telephone, there are many nonsampling error issues that could arise and overwhelm sampling error considerations like those embodied in the margin of error.
## The margin of error can be interpreted by making use of ideas from the laws of probability or the "laws of chance," as they are sometimes called.
Six Sigma Calculator Video Interviews Ask the Experts Problem Solving Methodology Flowchart Your iSixSigma Profile Industries Operations Inside iSixSigma About iSixSigma Submit an Article Advertising Info iSixSigma Support iSixSigma JobShop iSixSigma Computers are often used to simulate a random stream of numbers to support his effort. We could devise a sample design to ensure that our sample estimate will not differ from the true population value by more than, say, 5 percent (the margin of error) 90 Margin Of Error Three Percentage Points Confidence Level 95 From A Prior Study But if the original population is badly skewed, has multiple peaks, and/or has outliers, researchers like the sample size to be even larger.
Occasionally you will see surveys with a 99-percent confidence interval, which would correspond to three standard deviations and a much larger margin of error.(End of Math Geek Stuff!) If a poll So companies, campaigns and news organizations ask a randomly selected small number of people instead. Although the statistical calculation is relatively simple – the most advanced math involved is square root – margin of error can most easily be determined using the chart below. weblink When working with and reporting results about data, always remember what the units are.
Consider one example of the type of survey conducted by some of the leading polling organizations, such as the Gallup Organization. Now, if it's 29, don't panic -- 30 is not a magic number, it's just a general rule of thumb. (The population standard deviation must be known either way.) Here's an This chart can be expanded to other confidence percentages as well. What a wonderful concept.
Divide the population standard deviation by the square root of the sample size. You can use the poll to conclude that 54% of the voters in this sample would vote for Garcia, and when you project the results to the population, you add a Common sense will tell you (if you listen...) that the chance that your sample is off the mark will decrease as you add more people to your sample.
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Home > Margin Of > Margin Of Error Definition For Dummies. # Margin Of Error Definition For Dummies. ## Contents. Garcia's confidence interval is 49% to 59%, and Smith's confidence interval is 41% to 51%. You need to include the margin of error (in this case, 3%) in your results. Tags: confidence intervals, population Before posting, create an account!Stop this in-your-face noticeReserve your usernameFollow people you like, learn fromExtend your profileGain reputation for your contributionsNo annoying captchas across siteAnd much more! Next, we find the standard error of the mean, using the following equation: SEx = s / sqrt( n ) = 0.4 / sqrt( 900 ) = 0.4 / 30 = http://threadspodcast.com/margin-of/margin-of-error-and-definition.html. You've probably heard that term -- "margin of error" -- a lot before. Survey results themselves (with no MOE) are only a measure of how the sample of selected individuals felt about the issue; they don't reflect how the entire population may have felt, You want to estimate the average weight of the cones they make over a one-day period, including a margin of error. And the same goes for young adults, retirees, rich people, poor people, etc. http://www.dummies.com/education/math/statistics/how-to-interpret-the-margin-of-error-in-statistics/. ## Margin Of Error Example. Find the degrees of freedom (DF). Interpreting the Margin of Error In practice, nonsampling errors occur that can make the margin of error reported for a poll smaller than it should be if it reflected all sources For n = 50 cones sampled, the sample mean was found to be 10.3 ounces. Reply TPRJones I don't understand how the margin of error calculation doesn't take the population size into consideration.. Almost always there are people who have not made up their mind. However, if the study is done right, the results from the sample should be close to and representative of the actual values for the entire population, with a high level of Toggle navigation Search Submit San Francisco, CA Brr, it´s cold outside Learn by category LiveConsumer ElectronicsFood & DrinkGamesHealthPersonal FinanceHome & GardenPetsRelationshipsSportsReligion LearnArt CenterCraftsEducationLanguagesPhotographyTest Prep WorkSocial MediaSoftwareProgrammingWeb Design & DevelopmentBusinessCareersComputers Online Courses Poll Bias Definition Because the results of most survey questions can be reported in terms of percentages, the margin of error most often appears as a percentage, as well.. The population standard deviation, will be given in the problem. Margin Of Error Definition Statistics Here are the steps for calculating the margin of error for a sample mean: Find the population standard deviation and the sample size, n. Find the critical value. her latest blog Calculation in the upcoming election.. Many media surveys are based on what are called quota samples, and, although margins of error are reported from them, they do not strictly apply. Margin Of Error Definition Politics Easy! Because the results of most survey questions can be reported in terms of percentages, the margin of error most often appears as a percentage, as well. For example, the z*-value is 1.96 if you want to be about 95% confident.. ## Margin Of Error Definition Statistics. Now that I've told you that, what is your favorite color?" That's called a leading question, and it's a big no-no in surveying. More hints You can also use a graphing calculator or standard statistical tables (found in the appendix of most introductory statistics texts). Margin Of Error Example Suppose, in the mayoral election poll mentioned earlier, we sample 100 people who intend to vote and that 55 support Ms.
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What Does Margin Of Error Mean In Confidence Intervals Thus, a reported 3 percent margin of error becomes about 5 percent and a reported 4 percent margin of error becomes about 7 percent when the size of the lead is. Let's say you picked a specific number of people in the United States at random. http://threadspodcast.com/margin-of/margin-of-error-definition-politics.html Any reproduction or other use of content without the express written consent of iSixSigma is prohibited. A researcher surveying customers every six months to understand whether customer service is improving may see the percentage of respondents who say it is "very good" go from 50 percent in Easy! What Does Margin Of Error Mean In Polls. A 90 percent level can be obtained with a smaller sample, which usually translates into a less expensive survey. Sample questions A poll shows that Garcia is leading Smith by 54% to 46% with a margin of error of plus/minus 5% at a 95% confidence level. Donnelly Jr.List Price: \$19.95Buy Used: \$0.01Buy New: \$18.35Casio FX-CG10 PRIZM Color Graphing Calculator (Black)List Price: \$129.99Buy Used: \$74.99Buy New: \$121.73Approved for AP Statistics and Calculus About Us Contact Us Privacy his comment is here Thanks f Reply James Jones Great explanation, clearly written and well appreciated.. The choice of t statistic versus z-score does not make much practical difference when the sample size is very large. How Is Margin Of Error Calculated In Polls z*-Values for Selected (Percentage) Confidence Levels Percentage Confidence z*-Value 80 1.28 90 1.645 95 1.96 98 2.33 99 2.58 Note that these values are taken from the standard normal (Z-) distribution. With most polls still by telephone, there are many nonsampling error issues that could arise and overwhelm sampling error considerations like those embodied in the margin of error.. ## The margin of error can be interpreted by making use of ideas from the laws of probability or the "laws of chance," as they are sometimes called.. Six Sigma Calculator Video Interviews Ask the Experts Problem Solving Methodology Flowchart Your iSixSigma Profile Industries Operations Inside iSixSigma About iSixSigma Submit an Article Advertising Info iSixSigma Support iSixSigma JobShop iSixSigma Computers are often used to simulate a random stream of numbers to support his effort. We could devise a sample design to ensure that our sample estimate will not differ from the true population value by more than, say, 5 percent (the margin of error) 90 Margin Of Error Three Percentage Points Confidence Level 95 From A Prior Study But if the original population is badly skewed, has multiple peaks, and/or has outliers, researchers like the sample size to be even larger.. Occasionally you will see surveys with a 99-percent confidence interval, which would correspond to three standard deviations and a much larger margin of error.(End of Math Geek Stuff!) If a poll So companies, campaigns and news organizations ask a randomly selected small number of people instead. Although the statistical calculation is relatively simple – the most advanced math involved is square root – margin of error can most easily be determined using the chart below. weblink When working with and reporting results about data, always remember what the units are.. Consider one example of the type of survey conducted by some of the leading polling organizations, such as the Gallup Organization. Now, if it's 29, don't panic -- 30 is not a magic number, it's just a general rule of thumb. (The population standard deviation must be known either way.) Here's an This chart can be expanded to other confidence percentages as well. What a wonderful concept.. Divide the population standard deviation by the square root of the sample size. You can use the poll to conclude that 54% of the voters in this sample would vote for Garcia, and when you project the results to the population, you add a Common sense will tell you (if you listen...) that the chance that your sample is off the mark will decrease as you add more people to your sample.
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# Ranald!!! Kruber Reds
So one for the maths people out there
Whats the chances of pulling 15 melee reds without seeing a single ranged red?
I worked out that no ranged from 15 reds is about 3.5%. But I haven’t kept track of how many jewelleries I’ve pulled, getting a little frustrated now!
well there’s no bad luck protection so it is… literally… pot luck. Every roll is a brand new roll.
It took me 2k hours to get a red greatsword skin after countless dozens of Bogenhafen Banger.
Someone did a very old test opening a thousand chests and got something like a 20% chance of getting a red with an Emp Vault(can’t remember exact %age as it is old). The chances of getting a red ranged are therefore miniscule diminishing returns. 3% across all chest opened might be vaguely there but it’s still - of course- complete RNG.
Hm…i might be going about this all wrong but here is my thought process, if we are excluding jewelry from any calculation, then we have 15 melee weapons to 4 ranged ones as possible reds. To figure the chance of a single ranged when getting a red we simply note that 4 outa 19 options are ranged, in percentage that´s 4/19 ~ 0.21 =21% to get a ranged one each time you get a red weapon.
That in turn means you got a 79% chance of a melee weapon each time you score a red weapon with krub.
So how likely are 15 melee weapons in a row? I have kinda forgotten this stuff but as @Argonaut14 mentioned there is no system for getting something different from your last roll and every roll is considered a fresh one. Which if i am not mistaken that means you can calculate the chances of 15 melee rolls in a row by simply multiplying the probability of a repeated case with itself once for each time you do it.
Because i am lazy i will state that the chance for a melee roll is 80%, or 8/10, and so to figure out the probability for 15 in a row you just take that 80% in fractional form, 8/10, and multiply that by itself 15 times.
8/10 ^15 = 0.035 = 3.5%
ehm…and this i think should mean that the chances of you pulling 15 melee reds in a row is 3.5%? Which in turn would mean that getting a ranged one somewhere along the way is the remaining % to hit 100 which is 96.5…
I may have forgotten a bit too much of this stuff, it doesnt seem quite right somehow
Why not join the Fatshark Discord https://discord.gg/K6gyMpu
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# Ranald!!! Kruber Reds. So one for the maths people out there. Whats the chances of pulling 15 melee reds without seeing a single ranged red?. I worked out that no ranged from 15 reds is about 3.5%. But I haven’t kept track of how many jewelleries I’ve pulled, getting a little frustrated now!. well there’s no bad luck protection so it is… literally… pot luck. Every roll is a brand new roll.. It took me 2k hours to get a red greatsword skin after countless dozens of Bogenhafen Banger.. Someone did a very old test opening a thousand chests and got something like a 20% chance of getting a red with an Emp Vault(can’t remember exact %age as it is old). The chances of getting a red ranged are therefore miniscule diminishing returns. 3% across all chest opened might be vaguely there but it’s still - of course- complete RNG.. Hm…i might be going about this all wrong but here is my thought process, if we are excluding jewelry from any calculation, then we have 15 melee weapons to 4 ranged ones as possible reds.
|
To figure the chance of a single ranged when getting a red we simply note that 4 outa 19 options are ranged, in percentage that´s 4/19 ~ 0.21 =21% to get a ranged one each time you get a red weapon.. That in turn means you got a 79% chance of a melee weapon each time you score a red weapon with krub.. So how likely are 15 melee weapons in a row? I have kinda forgotten this stuff but as @Argonaut14 mentioned there is no system for getting something different from your last roll and every roll is considered a fresh one. Which if i am not mistaken that means you can calculate the chances of 15 melee rolls in a row by simply multiplying the probability of a repeated case with itself once for each time you do it.. Because i am lazy i will state that the chance for a melee roll is 80%, or 8/10, and so to figure out the probability for 15 in a row you just take that 80% in fractional form, 8/10, and multiply that by itself 15 times.. 8/10 ^15 = 0.035 = 3.5%. ehm…and this i think should mean that the chances of you pulling 15 melee reds in a row is 3.5%? Which in turn would mean that getting a ranged one somewhere along the way is the remaining % to hit 100 which is 96.5…. I may have forgotten a bit too much of this stuff, it doesnt seem quite right somehow. Why not join the Fatshark Discord https://discord.gg/K6gyMpu.
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http://electricalbaba.com/core-lamination-reduces-eddy-current-loss/
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# How Core Laminations Reduces Eddy Current Loss?
We are well aware of the fact that electrical machine core is made up of laminated sheets. This is done to reduce the eddy current loss. But how laminated core reduces the eddy current loss and what is the relation between the Eddy Current Loss and thickness of lamination?
Let us consider an electrical machine as shown in figure below.
Let the rotor of the machine is made of solid iron core. Suppose the cross-sectional area perpendicular to the magnetic field is A. Therefore, the total flux linkage of rotor with the magnetic field will be Ø = BA. As the rotor of machine rotates, the magnetic flux linking with the solid rotor body will change with time. Therefore according to Faraday’s Law of Electromagnetic Induction, an emf will be induced on the body of rotor core. This current is called the Eddy Current.
The resistance offered to this Eddy Current by the solid rotor core will be inversely proportional to the cross sectional area of the core. Thus,
Rsolid = Resistance offered by solid rotor core = k/A
where k is some constant.
Therefore,
Eddy Current Loss = (emf induced in solid rotor core)2 / Rsolid
Let the emf induced in the solid rotor core = E
Hence,
Eddy Current Loss = AE2 / k ……………………..(1)
Now suppose the same rotor core of machine is made of 4 laminated sheet stacked together as shown in figure below.
As 4 laminations are stacked together, hence the cross-sectional area of each lamination perpendicular to the magnetic field will become (A/4). Thus, the flux linking through each lamination will be Ø = BA/4. Therefore emf induced in each lamination will be (1/4)th of the emf induced in the solid iron core.
So, emf induced in each lamination = E/4
Let us now calculate the resistance offered by the lamination to the eddy current.
Rlamination= Resistance offered by each lamination = 4k/A
Therefore,
Eddy Current Loss in each Lamination = (emf induced)2 / Rlamination
= A(E/4)2 / 4k
= AE2/64k
As there are 4 laminations, hence total eddy current loss
= 4 x AE2/64k
= AE2/16k …………………………..(2)
Thence,
(Eddy Current Loss in Laminated Rotor / Eddy Current Loss in Solid Rotor)
= 1/16 ……………………[from equation (1) and (2)]
=(1/4)2
If the axial length of rotor is assumed to be unity then the thickness of each lamination will be ¼. Therefore, we can say that Eddy Current Loss is directly proportional to the square of lamination thickness. If more lamination are used for a given rotor axial length, the thickness of lamination decreases which result in decrease in Eddy current loss. Thus the use of thin lamination reduces the Eddy Current Loss. Normally the thickness of lamination is in between 0.4-0.5 mm. Further reduction in the thickness results into reduction of loss but at an increased cost.
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# How Core Laminations Reduces Eddy Current Loss?. We are well aware of the fact that electrical machine core is made up of laminated sheets. This is done to reduce the eddy current loss. But how laminated core reduces the eddy current loss and what is the relation between the Eddy Current Loss and thickness of lamination?. Let us consider an electrical machine as shown in figure below.. Let the rotor of the machine is made of solid iron core. Suppose the cross-sectional area perpendicular to the magnetic field is A. Therefore, the total flux linkage of rotor with the magnetic field will be Ø = BA. As the rotor of machine rotates, the magnetic flux linking with the solid rotor body will change with time. Therefore according to Faraday’s Law of Electromagnetic Induction, an emf will be induced on the body of rotor core. This current is called the Eddy Current.. The resistance offered to this Eddy Current by the solid rotor core will be inversely proportional to the cross sectional area of the core. Thus,. Rsolid = Resistance offered by solid rotor core = k/A. where k is some constant.. Therefore,. Eddy Current Loss = (emf induced in solid rotor core)2 / Rsolid. Let the emf induced in the solid rotor core = E. Hence,. Eddy Current Loss = AE2 / k ……………………..(1). Now suppose the same rotor core of machine is made of 4 laminated sheet stacked together as shown in figure below.. As 4 laminations are stacked together, hence the cross-sectional area of each lamination perpendicular to the magnetic field will become (A/4). Thus, the flux linking through each lamination will be Ø = BA/4.
|
Therefore emf induced in each lamination will be (1/4)th of the emf induced in the solid iron core.. So, emf induced in each lamination = E/4. Let us now calculate the resistance offered by the lamination to the eddy current.. Rlamination= Resistance offered by each lamination = 4k/A. Therefore,. Eddy Current Loss in each Lamination = (emf induced)2 / Rlamination. = A(E/4)2 / 4k. = AE2/64k. As there are 4 laminations, hence total eddy current loss. = 4 x AE2/64k. = AE2/16k …………………………..(2). Thence,. (Eddy Current Loss in Laminated Rotor / Eddy Current Loss in Solid Rotor). = 1/16 ……………………[from equation (1) and (2)]. =(1/4)2. If the axial length of rotor is assumed to be unity then the thickness of each lamination will be ¼. Therefore, we can say that Eddy Current Loss is directly proportional to the square of lamination thickness. If more lamination are used for a given rotor axial length, the thickness of lamination decreases which result in decrease in Eddy current loss. Thus the use of thin lamination reduces the Eddy Current Loss. Normally the thickness of lamination is in between 0.4-0.5 mm. Further reduction in the thickness results into reduction of loss but at an increased cost.
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http://www.jiskha.com/display.cgi?id=1308874108
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# math
posted by .
Can you please check my work.
A particle is moving with the given data. Find the position of the particle.
a(t) = cos(t) + sin(t)
s(0) = 2
v(0) = 6
a(t) = cos(t) + sin(t)
v(t) = sin(t) - cos(t) + C
s(t) = -cos(t) - sin(t) + Cx + D
6 = v(0) = sin(0) -cos(0) + C
C=7
2= s(0) = -cos(0) - sin(0) + 7 (0) + D
D= 3
s(t) = -cos(t) - sin(t) + 7t + 3
• math -
All correct.
We can check backwards:
assume given:
s(t) = -cos(t) - sin(t) + 7t + 3
v(t)=s'(t)=sin(t)-cos(t)+7
a(t)=v'(t)=cos(t)+sin(t)
v(0)=sin(0)-cos(0)+7=0-1+7=6
s(0)=-cos(0)-sin(0) + 7(0) + 3 =-1+0+0+3=2
All correct!
### Related Questions
More Related Questions
Post a New Question
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# math. posted by .. Can you please check my work.. A particle is moving with the given data. Find the position of the particle.. a(t) = cos(t) + sin(t). s(0) = 2. v(0) = 6. a(t) = cos(t) + sin(t). v(t) = sin(t) - cos(t) + C. s(t) = -cos(t) - sin(t) + Cx + D. 6 = v(0) = sin(0) -cos(0) + C. C=7. 2= s(0) = -cos(0) - sin(0) + 7 (0) + D. D= 3.
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s(t) = -cos(t) - sin(t) + 7t + 3. • math -. All correct.. We can check backwards:. assume given:. s(t) = -cos(t) - sin(t) + 7t + 3. v(t)=s'(t)=sin(t)-cos(t)+7. a(t)=v'(t)=cos(t)+sin(t). v(0)=sin(0)-cos(0)+7=0-1+7=6. s(0)=-cos(0)-sin(0) + 7(0) + 3 =-1+0+0+3=2. All correct!. ### Related Questions. More Related Questions. Post a New Question.
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https://enigmaticcode.wordpress.com/2019/08/10/puzzle-16-clever-code/?shared=email&msg=fail
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Enigmatic Code
Programming Enigma Puzzles
Puzzle #16: Clever code
Rashmi told us that she had to make up some codes, ones that preferably have a unique quality to them.
“For a four-digit code, I chose 2020, because it has two 0s, zero 1s, two 2s and zero 3s. I chose 3211000 for a seven-digit code.”
She then told us that she used the same idea for a 10-digit code.
[puzzle#16]
One response to “Puzzle #16: Clever code”
1. Jim Randell 10 August 2019 at 8:32 am
We investigated the base 10 autobiographical numbers in Enigma 476 and generated a complete list of them.
There is only one autobiographical number with 10 digits.
Solution: The code is 6210001000.
Note that 1210 is also a 4-digit autobiographical number, and there are no autobiographical numbers of length 6.
Using the recipe given with Enigma 476 we can generate autobiographical sequences with length 10, and turn them into numbers.
Run: [ @repl.it ]
```from enigma import nconcat, printf
# generate autobiographical sequences up to length n
def generate(n=None):
for s in ((), (1, 2, 1, 0), (2, 0, 2, 0), (2, 1, 2, 0, 0), [3, 2, 1, 1, 0, 0, 0]):
if n is None or not(len(s) > n):
yield tuple(s)
while n is None or len(s) < n:
s.insert(3, 0)
s[0] += 1
yield tuple(s)
# check a sequence is autobiographical
def is_autobiographical(s):
return all(x == s.count(i) for (i, x) in enumerate(s))
# generate sequences up to length 10
for s in generate(10):
assert is_autobiographical(s)
if len(s) == 10 and all(x < 10 for x in s):
n = nconcat(s)
printf("{s} -> {n}")
```
This site uses Akismet to reduce spam. Learn how your comment data is processed.
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Enigmatic Code. Programming Enigma Puzzles. Puzzle #16: Clever code. Rashmi told us that she had to make up some codes, ones that preferably have a unique quality to them.. “For a four-digit code, I chose 2020, because it has two 0s, zero 1s, two 2s and zero 3s. I chose 3211000 for a seven-digit code.”. She then told us that she used the same idea for a 10-digit code.. [puzzle#16]. One response to “Puzzle #16: Clever code”. 1. Jim Randell 10 August 2019 at 8:32 am. We investigated the base 10 autobiographical numbers in Enigma 476 and generated a complete list of them.. There is only one autobiographical number with 10 digits.. Solution: The code is 6210001000.. Note that 1210 is also a 4-digit autobiographical number, and there are no autobiographical numbers of length 6.. Using the recipe given with Enigma 476 we can generate autobiographical sequences with length 10, and turn them into numbers.. Run: [ @repl.it ]. ```from enigma import nconcat, printf. # generate autobiographical sequences up to length n. def generate(n=None):.
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for s in ((), (1, 2, 1, 0), (2, 0, 2, 0), (2, 1, 2, 0, 0), [3, 2, 1, 1, 0, 0, 0]):. if n is None or not(len(s) > n):. yield tuple(s). while n is None or len(s) < n:. s.insert(3, 0). s[0] += 1. yield tuple(s). # check a sequence is autobiographical. def is_autobiographical(s):. return all(x == s.count(i) for (i, x) in enumerate(s)). # generate sequences up to length 10. for s in generate(10):. assert is_autobiographical(s). if len(s) == 10 and all(x < 10 for x in s):. n = nconcat(s). printf("{s} -> {n}"). ```. This site uses Akismet to reduce spam. Learn how your comment data is processed.
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https://www.mrtylerslessons.com/faq/question-how-to-write-a-kindergarten-math-lesson-plan.html
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# Question: How To Write A Kindergarten Math Lesson Plan?
## How do you write a kindergarten lesson plan?
Steps to building your lesson plan
1. Identify the objectives.
2. Determine the needs of your students.
3. Plan your resources and materials.
5. Instruct and present information.
6. Allow time for student practice.
7. Ending the lesson.
8. Evaluate the lesson.
## How do you write a math lesson plan?
Lesson planning:
1. Be clear about your goal. What exactly do you want your students to learn in this lesson?
2. Know the mathematics.
3. Choose good resources.
4. Select appropriate and purposeful tasks.
5. Less is more.
6. You don’t have to start and finish a task in one lesson.
## What is math lesson plan?
Mathematics Lesson Plans B. As you know, lesson plans are detailed descriptions of the course of instructions or “learning trajectories” for teachers. Lesson plans are developed on a daily basis by teachers to guide class learning.
## How do you start teaching kindergarten math?
6 Tips to Teach Kindergarten Math Without Curriculum
1. Use games. Playing card games and board games helps my daughter gain basic number recognition as we count around a board, recognize doubles, and recognize common number pairs.
3. Make it real.
4. Get moving.
5. Play with toys.
6. After Planning.
You might be interested: Often asked: What Is A Sentence Lesson Plan?
## What is a 5 step lesson plan?
The five steps involved are the Anticipatory Set, Introduction of New Material, Guided Practice, Independent Practice and Closure.
## How do you write a simple lesson plan?
Listed below are 6 steps for preparing your lesson plan before your class.
1. Identify the learning objectives.
2. Plan the specific learning activities.
3. Plan to assess student understanding.
4. Plan to sequence the lesson in an engaging and meaningful manner.
5. Create a realistic timeline.
6. Plan for a lesson closure.
## What are the 4 A’s in lesson plan?
The 4-A Model Typically, lesson plans follow a format that identifies goals and objectives, teaching methods, and assessment.
## What makes a good math lesson?
A ‘good maths lesson’ will always necessarily be a part of a sequence of lessons or learning experiences which will ideally build mathematical understanding, improve fluency, build problem solving capacity and then develop mathematical reasoning skills.
## How do you start a math lesson?
6 Ways to Help Students Understand Math
1. Create an effective class opener.
2. Introduce topics using multiple representations.
3. Solve the problems many ways.
4. Show the application.
5. Have students communicate their reasoning.
6. Finish class with a summary.
## What are the methods used to teach mathematics?
Teaching methods of mathematics include lecture, inductive, deductive, heuristic or discovery, analytic, synthetic, problem solving, laboratory and project methods. Teachers may adopt any method according to the specific unit of syllabus, available resources and number of students in a class.
## What math skills should a 4 year old have?
Preschoolers (ages 3–4 years)
• Recognize shapes in the real world.
• Start sorting things by color, shape, size, or purpose.
• Compare and contrast using classifications like height, size, or gender.
• Count up to at least 20 and accurately point to and count items in a group.
You might be interested: Often asked: In A Teacher Lesson Plan What Is Discourse?
## What kind of math should a kindergartener know?
Math in kindergarten is all about the basics. They will learn how to count, recognize numbers up to 10 and sort objects. Using concrete props, they will learn the concepts of more and less, ordinal numbers, basic addition and subtraction, creating patterns.
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# Question: How To Write A Kindergarten Math Lesson Plan?. ## How do you write a kindergarten lesson plan?. Steps to building your lesson plan. 1. Identify the objectives.. 2. Determine the needs of your students.. 3. Plan your resources and materials.. 5. Instruct and present information.. 6. Allow time for student practice.. 7. Ending the lesson.. 8. Evaluate the lesson.. ## How do you write a math lesson plan?. Lesson planning:. 1. Be clear about your goal. What exactly do you want your students to learn in this lesson?. 2. Know the mathematics.. 3. Choose good resources.. 4. Select appropriate and purposeful tasks.. 5. Less is more.. 6. You don’t have to start and finish a task in one lesson.. ## What is math lesson plan?. Mathematics Lesson Plans B. As you know, lesson plans are detailed descriptions of the course of instructions or “learning trajectories” for teachers. Lesson plans are developed on a daily basis by teachers to guide class learning.. ## How do you start teaching kindergarten math?. 6 Tips to Teach Kindergarten Math Without Curriculum. 1. Use games. Playing card games and board games helps my daughter gain basic number recognition as we count around a board, recognize doubles, and recognize common number pairs.. 3. Make it real.. 4. Get moving.. 5. Play with toys.. 6. After Planning.. You might be interested: Often asked: What Is A Sentence Lesson Plan?.
|
## What is a 5 step lesson plan?. The five steps involved are the Anticipatory Set, Introduction of New Material, Guided Practice, Independent Practice and Closure.. ## How do you write a simple lesson plan?. Listed below are 6 steps for preparing your lesson plan before your class.. 1. Identify the learning objectives.. 2. Plan the specific learning activities.. 3. Plan to assess student understanding.. 4. Plan to sequence the lesson in an engaging and meaningful manner.. 5. Create a realistic timeline.. 6. Plan for a lesson closure.. ## What are the 4 A’s in lesson plan?. The 4-A Model Typically, lesson plans follow a format that identifies goals and objectives, teaching methods, and assessment.. ## What makes a good math lesson?. A ‘good maths lesson’ will always necessarily be a part of a sequence of lessons or learning experiences which will ideally build mathematical understanding, improve fluency, build problem solving capacity and then develop mathematical reasoning skills.. ## How do you start a math lesson?. 6 Ways to Help Students Understand Math. 1. Create an effective class opener.. 2. Introduce topics using multiple representations.. 3. Solve the problems many ways.. 4. Show the application.. 5. Have students communicate their reasoning.. 6. Finish class with a summary.. ## What are the methods used to teach mathematics?. Teaching methods of mathematics include lecture, inductive, deductive, heuristic or discovery, analytic, synthetic, problem solving, laboratory and project methods. Teachers may adopt any method according to the specific unit of syllabus, available resources and number of students in a class.. ## What math skills should a 4 year old have?. Preschoolers (ages 3–4 years). • Recognize shapes in the real world.. • Start sorting things by color, shape, size, or purpose.. • Compare and contrast using classifications like height, size, or gender.. • Count up to at least 20 and accurately point to and count items in a group.. You might be interested: Often asked: In A Teacher Lesson Plan What Is Discourse?. ## What kind of math should a kindergartener know?. Math in kindergarten is all about the basics. They will learn how to count, recognize numbers up to 10 and sort objects. Using concrete props, they will learn the concepts of more and less, ordinal numbers, basic addition and subtraction, creating patterns.
|
https://www.indiabix.com/placement-papers/caterpillar/6655
| 1,632,177,909,000,000,000
|
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# Placement Papers - Caterpillar
## Why Caterpillar Placement Papers?
Learn and practice the placement papers of Caterpillar and find out how much you score before you appear for your next interview and written test.
## Where can I get Caterpillar Placement Papers with Answers?
IndiaBIX provides you lots of fully solved Caterpillar Placement Papers with answers. You can easily solve all kind of placement test papers by practicing the exercises given below.
## How to solve Caterpillar Placement Papers?
You can easily solve all kind of questions by practicing the following exercises.
### Caterpillar Interview Experience - Caterpillar Test Paper
Posted By : Priya Rating : +83, -12
1. A ladder lies against a wall. The top of the ladder reaches 8 ft. above the ground. When the ladder slips two metres away from the wall, the top of the ladder touches the foot of the wall. The length of the ladder is
A. 8
B. 15
C. 17
D. 10
ANS: C
2. A student got marks in the ratio 6:7:8:9:10 in five subjects having equal maximum marks. Totally, he scored 60% marks. In how many subjects, he got more than 50%?
A. 3
B. 4
C. 5
D. None of these
ANS: B
3. In how many ways can a delegation of 4 professors and 3 students be constituted from 8 professor and 5 students, if Balamurli an Arts students refuses to be in the delegation when Prof. Siddharth, the Science professor is included in it?
A. 280
B. 210
C. 490
D. 560
ANS: C
4. Ram is having 158 coins of one rupee. He puts it in different bags, so that he can hand over the cash of any denomination required between Rs. 1 to Rs. 158. What is the least number of bags required?
A. 11
B. 13
C. 15
D. None of these
ANS: D
5. A takes 4 days to do a work. B takes twice as long as A. C takes twice as long as B and D takes twice as long as C. They are made in groups of two. One of the groups takes two third of the time taken by second pair. What is the combination of the first pair?
A. A,C
B. A,D
C. B,C
D. B,D
ANS: B
Directions for Questions 6 to 10:
Fill in the blanks with appropriate words.
6. A well- _________, physically and mentally active ________ alone can contribute to the speedier economic progress of a nation.
A. educated, subjects
B. organized, systems
D. nourished, populace
ANS: D
7. Democracy has taken a ______ in a system which promotes sycophancy and _____.
A. dive, bureaucracy
B. delve, dictatorship
C. beating, mediocrity
D. privilege, intolerance
ANS: C
8. Time has now come for all agencies working in the development sector to launch a multi-pronged __________ to _________ malnutrition.
A. system, abjure
B. weapon, annihilate
C. policy, deviate
ANS: D
9. The truth is that in a highly capital-intensive business _______ deep pockets, domestic civil aviation is _______ under capitalized.
A. ascertaining, highly
B. requiring, woefully
C. sustaining, alarmingly
D. balancing, astonishingly
ANS: B
10. We must develop _____ systems from the village upwards and up to the national level to constantly _______ the nutritional status of the people.
A. monitoring, review
B. machinery, tackle
C. efficient, emancipate
D. sound, harbour
ANS: A
Directions for Questions 11, 12:
Read each sentence to find if there is any grammatical error in it.
11. I enjoyed(A) / during my(B) / stay in(C) / England(D)
A. A
B. B
C. C
D. D
ANS: A
12. I shall (A)/ ring him(B) / tomorrow(C) / in the afternoon(D)
A. A
B. B
C. C
D. D
ANS: B
13. Identify the correct statement:-
A. The brakes and steering failed
B. and the bus ran down the hill
C. without anyone being able control it.
D. No error
ANS: C
14. Identify the correct statement:-
A.The brand propositon now therefore had to be that Keokarpin Antiseptic Cream is more effective
B. because it penetrates deepdown (beinglight and non-sticky)and works from within
C. (because of its ayurvedic ingredients) tokeep skin blemish, free and helps cope with cuts nicks, burns and nappy rash
D. No error
ANS: D
15. Identify the correct statement:-
A. The single biggest gainer in this process
B. was ITCs Gold Flake Kings sales are estimated
C. to have moved up from 50 million to 200 million sticks per month during 1987 and last year.
D. No error
ANS: D
16. Identify the correct statement:-
A. Another reason for pharmaceutical companies beefing up their
B. OTC (Over the Country) divisions is that prescription drugs with proven safety records which have been reached
C. the end of the their patent protection periodare
D. Allowed to be sold without a prescription. No error
ANS: B
Q17. Identify the correct statement:-
A. He is
B. too intelligent
C. to make a mistake.
D. No error
ANS: D
Q18. Find the antonym of the word:- REPRESS
A. Inhibit
B. Liberate
C. Curb
D. Quell
ANS: B
19. Find the Antonym of the word: EXODUS
A. Arrival
B. Home-coming
C. Return
D. Restoration
ANS: A
20. Find the synonym of the word: REPERCUSSION
B. Recollection
C. Remuneration
D. Reaction
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# Placement Papers - Caterpillar. ## Why Caterpillar Placement Papers?. Learn and practice the placement papers of Caterpillar and find out how much you score before you appear for your next interview and written test.. ## Where can I get Caterpillar Placement Papers with Answers?. IndiaBIX provides you lots of fully solved Caterpillar Placement Papers with answers. You can easily solve all kind of placement test papers by practicing the exercises given below.. ## How to solve Caterpillar Placement Papers?. You can easily solve all kind of questions by practicing the following exercises.. ### Caterpillar Interview Experience - Caterpillar Test Paper. Posted By : Priya Rating : +83, -12. 1. A ladder lies against a wall. The top of the ladder reaches 8 ft. above the ground. When the ladder slips two metres away from the wall, the top of the ladder touches the foot of the wall. The length of the ladder is. A. 8. B. 15. C. 17. D. 10. ANS: C. 2. A student got marks in the ratio 6:7:8:9:10 in five subjects having equal maximum marks. Totally, he scored 60% marks. In how many subjects, he got more than 50%?. A. 3. B. 4. C. 5. D. None of these. ANS: B. 3. In how many ways can a delegation of 4 professors and 3 students be constituted from 8 professor and 5 students, if Balamurli an Arts students refuses to be in the delegation when Prof. Siddharth, the Science professor is included in it?. A. 280. B. 210. C. 490. D. 560. ANS: C. 4. Ram is having 158 coins of one rupee. He puts it in different bags, so that he can hand over the cash of any denomination required between Rs. 1 to Rs. 158. What is the least number of bags required?. A. 11. B. 13. C. 15. D. None of these. ANS: D. 5. A takes 4 days to do a work. B takes twice as long as A. C takes twice as long as B and D takes twice as long as C. They are made in groups of two. One of the groups takes two third of the time taken by second pair. What is the combination of the first pair?. A. A,C. B. A,D. C. B,C. D. B,D. ANS: B. Directions for Questions 6 to 10:. Fill in the blanks with appropriate words.. 6. A well- _________, physically and mentally active ________ alone can contribute to the speedier economic progress of a nation.. A. educated, subjects. B. organized, systems. D. nourished, populace. ANS: D. 7. Democracy has taken a ______ in a system which promotes sycophancy and _____.. A. dive, bureaucracy. B. delve, dictatorship. C. beating, mediocrity. D. privilege, intolerance. ANS: C. 8. Time has now come for all agencies working in the development sector to launch a multi-pronged __________ to _________ malnutrition.. A. system, abjure. B. weapon, annihilate. C. policy, deviate. ANS: D. 9. The truth is that in a highly capital-intensive business _______ deep pockets, domestic civil aviation is _______ under capitalized.. A. ascertaining, highly. B. requiring, woefully. C. sustaining, alarmingly. D. balancing, astonishingly.
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ANS: B. 10. We must develop _____ systems from the village upwards and up to the national level to constantly _______ the nutritional status of the people.. A. monitoring, review. B. machinery, tackle. C. efficient, emancipate. D. sound, harbour. ANS: A. Directions for Questions 11, 12:. Read each sentence to find if there is any grammatical error in it.. 11. I enjoyed(A) / during my(B) / stay in(C) / England(D). A. A. B. B. C. C. D. D. ANS: A. 12. I shall (A)/ ring him(B) / tomorrow(C) / in the afternoon(D). A. A. B. B. C. C. D. D. ANS: B. 13. Identify the correct statement:-. A. The brakes and steering failed. B. and the bus ran down the hill. C. without anyone being able control it.. D. No error. ANS: C. 14. Identify the correct statement:-. A.The brand propositon now therefore had to be that Keokarpin Antiseptic Cream is more effective. B. because it penetrates deepdown (beinglight and non-sticky)and works from within. C. (because of its ayurvedic ingredients) tokeep skin blemish, free and helps cope with cuts nicks, burns and nappy rash. D. No error. ANS: D. 15. Identify the correct statement:-. A. The single biggest gainer in this process. B. was ITCs Gold Flake Kings sales are estimated. C. to have moved up from 50 million to 200 million sticks per month during 1987 and last year.. D. No error. ANS: D. 16. Identify the correct statement:-. A. Another reason for pharmaceutical companies beefing up their. B. OTC (Over the Country) divisions is that prescription drugs with proven safety records which have been reached. C. the end of the their patent protection periodare. D. Allowed to be sold without a prescription. No error. ANS: B. Q17. Identify the correct statement:-. A. He is. B. too intelligent. C. to make a mistake.. D. No error. ANS: D. Q18. Find the antonym of the word:- REPRESS. A. Inhibit. B. Liberate. C. Curb. D. Quell. ANS: B. 19. Find the Antonym of the word: EXODUS. A. Arrival. B. Home-coming. C. Return. D. Restoration. ANS: A. 20. Find the synonym of the word: REPERCUSSION. B. Recollection. C. Remuneration. D. Reaction.
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# 10th samacheer kalvi math solution for exercise 5.3 part 3
This page 10th samacheer kalvi math solution for exercise 5.3 part 3 is going to provide you solution for every problems that you find in the exercise no 5.3
## 10th samacheer kalvi solution for exercise 5.3 part 3
7. The side BC of an equilateral Δ ABC is parallel to x-axis. Find the slope of AB and the slope of BC
Solution:
The side BC is parallel to x axis.
slope of the side BC = 0
Since this is an equilateral triangle each angle is 60.
m = tan θ
θ = 60
m = tan 60
= √3
(i) (2 , 3), (3 , -1) and (4 , -5)
Solution:
Let A (2,3) B (3 , -1) and C (4,-5) be the given points
If the given points are collinear then,
slope of AB = slope of BC
slope = (y-y)/(x-x)
x₁ = 2 x₂ = 3
y₁ = 3 y₂ = -1
= (-1-3)/(3-2)
= -4/1
= -4
slope of AB = -4
x₁ = 3 x₂ = 4
y₁ = -1 y₂ = -5
= (-5-(-1))/(4-3)
= (-5+1)/1
= -4
slope of BC = -4
Therefore the given points are collinear.
(ii) (4 , 1), (-2 , -3) and (-5 , -5)
Solution:
Let A (4,1) B (-2 , -3) and C (-5,-5) be the given points
If the given points are collinear then,
slope of AB = slope of BC
slope = (y-y)/(x-x)
x₁ = 4 x₂ = -2
y₁ = 1 y₂ = -3
= (-3-1)/(-2-4)
= -4/(-6)
= 2/3
slope of AB = 2/3
x₁ = -2 x₂ = -5
y₁ = -3 y₂ = -5
= (-5-(-3))/(-5-(-2))
= (-5+3)/(-5+2)
= -2/(-3)
= 2/3
slope of BC = 2/3
Therefore the given points are collinear.
In the page 10th samacheer kalvi math solution for exercise 5.3 part 3 we are going to see the solution of next problem
(iii) (4 , 4), (-2 , 6) and (1 , 5)
Solution:
Let A (4,4) B (-2 , 6) and C (1,5) be the given points
If the given points are collinear then,
slope of AB = slope of BC
slope = (y-y)/(x-x)
x₁ = 4 x₂ = -2
y₁ = 4 y₂ = 6
= (6-4)/(-2-4)
= 2/(-6)
= -1/3
slope of AB = -1/3
x₁ = -2 x₂ = 1
y₁ = 6 y₂ = 5
= (5-6)/(1-(-2))
= (-1)/(1+2)
= -1/3
slope of BC = -1/3
Therefore the given points are collinear.
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# 10th samacheer kalvi math solution for exercise 5.3 part 3. This page 10th samacheer kalvi math solution for exercise 5.3 part 3 is going to provide you solution for every problems that you find in the exercise no 5.3. ## 10th samacheer kalvi solution for exercise 5.3 part 3. 7. The side BC of an equilateral Δ ABC is parallel to x-axis. Find the slope of AB and the slope of BC. Solution:. The side BC is parallel to x axis.. slope of the side BC = 0. Since this is an equilateral triangle each angle is 60.. m = tan θ. θ = 60. m = tan 60. = √3. (i) (2 , 3), (3 , -1) and (4 , -5). Solution:. Let A (2,3) B (3 , -1) and C (4,-5) be the given points. If the given points are collinear then,. slope of AB = slope of BC. slope = (y-y)/(x-x). x₁ = 2 x₂ = 3. y₁ = 3 y₂ = -1. = (-1-3)/(3-2). = -4/1. = -4. slope of AB = -4. x₁ = 3 x₂ = 4. y₁ = -1 y₂ = -5. = (-5-(-1))/(4-3). = (-5+1)/1. = -4. slope of BC = -4. Therefore the given points are collinear.. (ii) (4 , 1), (-2 , -3) and (-5 , -5). Solution:. Let A (4,1) B (-2 , -3) and C (-5,-5) be the given points. If the given points are collinear then,.
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slope of AB = slope of BC. slope = (y-y)/(x-x). x₁ = 4 x₂ = -2. y₁ = 1 y₂ = -3. = (-3-1)/(-2-4). = -4/(-6). = 2/3. slope of AB = 2/3. x₁ = -2 x₂ = -5. y₁ = -3 y₂ = -5. = (-5-(-3))/(-5-(-2)). = (-5+3)/(-5+2). = -2/(-3). = 2/3. slope of BC = 2/3. Therefore the given points are collinear.. In the page 10th samacheer kalvi math solution for exercise 5.3 part 3 we are going to see the solution of next problem. (iii) (4 , 4), (-2 , 6) and (1 , 5). Solution:. Let A (4,4) B (-2 , 6) and C (1,5) be the given points. If the given points are collinear then,. slope of AB = slope of BC. slope = (y-y)/(x-x). x₁ = 4 x₂ = -2. y₁ = 4 y₂ = 6. = (6-4)/(-2-4). = 2/(-6). = -1/3. slope of AB = -1/3. x₁ = -2 x₂ = 1. y₁ = 6 y₂ = 5. = (5-6)/(1-(-2)). = (-1)/(1+2). = -1/3. slope of BC = -1/3. Therefore the given points are collinear.
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## Triple or Nothing
A forum for good logic/math puzzles.
Moderators: jestingrabbit, Moderators General, Prelates
Actaeus
Posts: 606
Joined: Thu Jan 10, 2008 9:21 pm UTC
Location: ZZ9 Plural Z Alpha
### Triple or Nothing
Here's the premise:
There is a carnival game (or casino, doesn't matter.) The player starts by putting \$1 on the table. The dealer flips a coin.
HEADS: Dealer pays \$2 to the table
TAILS: Dealer takes the \$1 and game ends.
Player now chooses whether to play another round. If he doesn't, Player takes all money on the table and walks away. If he does, the coin is flipped again:
HEADS: Dealer triples the pot again
TAILS: Dealer takes all money, game ends.
The expected winnings from each flip is whatever is on the table. Therefore, it is always in the player's best interest to keep playing. However, if the player always chooses to play another round, he *will* eventually lose the original money and walk away with a loss of \$1.
What gives?
Twist: Player is allowed to play multiple games. If Player always plays one round and then ends the game, then repeats, he will win on average \$1 per game. If he plays two rounds each time, he will average \$2. Three rounds, \$3.25. If the player plays 100 rounds each time, even though he only has a 1/1.3e30 chance, he will win \$5.15e47 each time he wins. average: \$4.07e17. If the player always plays an "infinite" number of games (until he loses), he will win \$0 each time.
What gives? Does this involve transfinite numbers? If the player can play an infinite number of games, his winnings/game will approximate the numbers I gave, but will he get an infinite number of heads "once" in an infinite number of games, and win \$infinity? The expected winnings/game increases exponentially, so.....I don't even know where I was going with that.
Summary: What The Fork?
---
EDIT: should this be in Logic Games?
Macbi
Posts: 941
Joined: Mon Apr 09, 2007 8:32 am UTC
Location: UKvia
### Re: Triple or Nothing
Spoiler:
When you get Tails, you lose all the money, but you also lose the right to continue playing. This right is worth the right amount to balence the game out.
Indigo is a lie.
Which idiot decided that websites can't go within 4cm of the edge of the screen?
There should be a null word, for the question "Is anybody there?" and to see if microphones are on.
Actaeus
Posts: 606
Joined: Thu Jan 10, 2008 9:21 pm UTC
Location: ZZ9 Plural Z Alpha
### Re: Triple or Nothing
Spoiler:
Of course! I forgot your money on the table affected future rounds, so expected winnings won't cover it. I believe expected winnings only works if it's out of your pocket.
Cauchy
Posts: 602
Joined: Wed Mar 28, 2007 1:43 pm UTC
### Re: Triple or Nothing
If the player has played n games, he has a (1/2)^n chance of earning 3^n - 1 dollars, and a 1 - (1/2)^n chance of losing \$1, for an expected earning of (3/2)^n - 1. Clearly, as n increases, his expected winnings increase. As is the case with many things, it doesn't go quite as smoothly in the limit. I don't think this has anything to do with an "extra penalty" for getting 0. After all, we could easily see the player still playing even after he loses a flip: the money on the table is now \$0, and if he wins it triples and if it loses it goes to \$0, making the game no different than it was previously. Rather, things simply break down in the limit. It happens a lot in mathematics.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Yakk
Poster with most posts but no title.
Posts: 11128
Joined: Sat Jan 27, 2007 7:27 pm UTC
Location: E pur si muove
### Re: Triple or Nothing
You cannot play an infinite number of rounds.
Let G(N) be the game strategy of playing N rounds then stopping. Then E(G(N)) (average or expected value) goes to infinity as N goes to infinity.
However, the limits does not cross the E and G function. lim(N->inf) of E(G(N)) != E(G( lim(N->inf)(N) ) )
Partially this is because G(infinity) is a special case. A simple way to extend G(N) to cover infinity is G(infinity) is "play until you lose, no matter what".
Once you turn the problem into symbols and poke at it, there is nothing really strange going on.
...
The second thing to realize is that money isn't linear in value. The E() (expected value) function is actually a poor measure of the return from a game.
Money isn't utility. Suppose you had 10^15 US dollars already -- having 10^18 US dollars wouldn't actually increase your wealth by all that much, because you have already managed to own the entire US economy, and going from 10^15 to 10^18 just reduces what percent everyone else has from a fraction of a percent to a smaller fraction.
Similar effects happen at smaller scales. Your first million dollars is more useful than your second.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
dan
Posts: 62
Joined: Wed Aug 09, 2006 12:55 am UTC
### Re: Triple or Nothing
You can get rid of the probability stuff and still have essentially the same problem. Say there is a game where, in each round, I give you \$1, and then you choose whether or not to continue. If you continue, the same thing happens again, and if you stop, I take all the money back and the game ends. This is obviously a pretty pointless game. Although it might seem at each stage like you get more money if you keep playing, if you stand back a bit you can see that this isn't actually true at all.
Waterhouse
Posts: 54
Joined: Wed Oct 03, 2007 12:37 pm UTC
### Re: Triple or Nothing
What gives? The question implies the following: how can it be that the gambler, apparently paradoxically, has a chance to win infinite money with a very low risk? The answer is simply that the imagined setup is contrary to the way that the "house" operates in reality.
In any game of chance that is run for profit, the odds are biased toward the house. In this manner the longer a gambler plays, the more money the house wins. In this puzzle, the house is offering three to one payout on a game with a one in two chance of winning, which is clearly biased toward the player.
Another way to see it is that the player could, instead of remaining in the game after winning one round, pocket his winnings, and then immediately start up a new game. In this way, he will be expected to win two dollars half the time and to lose only one dollar the other half of the time.
Such payouts for those odds are never offered in reality.
Actaeus
Posts: 606
Joined: Thu Jan 10, 2008 9:21 pm UTC
Location: ZZ9 Plural Z Alpha
### Re: Triple or Nothing
This was not a realistic question....My question was, why is the "best" strategy in terms of payout one that leads to no payout at all? The answer was what Yakk said:
Yakk wrote:You cannot play an infinite number of rounds.
Let G(N) be the game strategy of playing N rounds then stopping. Then E(G(N)) (average or expected value) goes to infinity as N goes to infinity.
However, the limits does not cross the E and G function. lim(N->inf) of E(G(N)) != E(G( lim(N->inf)(N) ) )
Partially this is because G(infinity) is a special case. A simple way to extend G(N) to cover infinity is G(infinity) is "play until you lose, no matter what".
Once you turn the problem into symbols and poke at it, there is nothing really strange going on.
...
The second thing to realize is that money isn't linear in value. The E() (expected value) function is actually a poor measure of the return from a game.
Money isn't utility. Suppose you had 10^15 US dollars already -- having 10^18 US dollars wouldn't actually increase your wealth by all that much, because you have already managed to own the entire US economy, and going from 10^15 to 10^18 just reduces what percent everyone else has from a fraction of a percent to a smaller fraction.
Similar effects happen at smaller scales. Your first million dollars is more useful than your second.
Essentially, the function for expected payout tends to infinity, but the limit is a special case. Or something along those lines.
Yakk
Poster with most posts but no title.
Posts: 11128
Joined: Sat Jan 27, 2007 7:27 pm UTC
Location: E pur si muove
### Re: Triple or Nothing
Really, it is just that limits to infinity don't always cross functions. Lim(x->infinity)f(g(x)) does not always equal f(lim(x->infinity)g(x)).
The cases where that works are special, rare cases.
Heck, the lim(x->y)f(x) = f(y) is a special case called a continuous function even when y isn't infinity. For it to be true when y is infinity, your function could be called "continuous at infinity", which is not implied by standard continuity.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Cogito
Posts: 9
Joined: Wed Feb 06, 2008 8:55 am UTC
### Re: Triple or Nothing
The expected winnings in the original game is zero(the chance of winning) times infinity (the payout).
This need not equal zero at all. And if you consider it as a limiting case, then it should be clear the payout isn't zero. It's infinity. So I wouldn't disagree with the original strategy, personally.
Also, Yakk, the non-linear relation between money and utility isn't particular relevant. You can just replace units of money with units of utility.
More generally, it's always dangerous to just state an infinity as opposed to taking it as a limit of finite processes.
Token
Posts: 1481
Joined: Fri Dec 01, 2006 5:07 pm UTC
Location: London
### Re: Triple or Nothing
Cogito wrote:This need not equal zero at all. And if you consider it as a limiting case, then it should be clear the payout isn't zero. It's infinity. So I wouldn't disagree with the original strategy, personally.
The limiting case, I believe, remains undefined. There are two separate limits (one for the probability, one for the payout). Two such limits may only be merged into one when they are both exist and are finite, which is not the case here.
All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.
Cogito
Posts: 9
Joined: Wed Feb 06, 2008 8:55 am UTC
### Re: Triple or Nothing
In general, if you multiply two limits that equal zero and infinity, then the final result can still be anything.
In the specific case you can merge the two formulae prior to taking the limit and find yourself a well-defined answer.
Expected winnings = 3^n/2^n -1=(3/2)^n-1, as Cauchy pointed out. There's no reason why it wouldn't hold in the limit.
Token
Posts: 1481
Joined: Fri Dec 01, 2006 5:07 pm UTC
Location: London
### Re: Triple or Nothing
Cogito wrote:In general, if you multiply two limits that equal zero and infinity, then the final result can still be anything.
In the specific case you can merge the two formulae prior to taking the limit and find yourself a well-defined answer.
Expected winnings = 3^n/2^n -1=(3/2)^n-1, as Cauchy pointed out. There's no reason why it wouldn't hold in the limit.
Well, there's two reasons you are wrong.
a) No matter how much you insist, you can't combine two limits if one (or both) of them is infinite or undefined. This is exactly what you are doing when you take the limit of (3/2)n-1 as n tends to infinity. Yakk provides an excellent reason for why this doesn't work in his above post.
b) Yes, in general zero times infinity can give you anything - this is what makes it an indeterminate form. This doesn't mean, however, that you can take it to be something that fits your intuition about the problem. In fact, I have it on good authority that when dealing with probability / measure theory, zero times infinity is generally defined to be zero. I think this example illustrates the reason for this quite well.
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### Re: Triple or Nothing
Cogito wrote:Also, Yakk, the non-linear relation between money and utility isn't particular relevant. You can just replace units of money with units of utility.
Well, it does generate an interesting result: at which point should one make a decision to stop?
If you are getting an exponential amount of utility on each stage ... gosh. I am rather not sure that is possible!
If you define Utility to be that which should guild your actions in a the "with probability P, you gain U2, and with probability (1-P) you lose U1. Then you should take the event if P*U2 > (1-P)*U1" case, quite possibly you can demonstrate that total Utility is bounded. In which case, unbounded exponential utility becomes impossible.
It might be a better statement to make that "utility is not probability linear"? Ie, if you have probability P to gain utility U, the utility of that option does not have to be P*U?
Utility, as defined by economics, is interesting in that any strictly monotonic transformation of utility satisfies the properties of utility. Utility is actually just an ordering of preference, it is not a scalar or vector value. Economics uses scalar values to make it easier to understand what the hell is going on: which leads to a number of strange results when you take those scalar values at face value...
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
rhino
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### Re: Triple or Nothing
Yakk: I agree with your suggestion that a bounded utility function seems intuitively correct.
However, utility is not just an order of preference. Suppose we have two agents, say Yakk and Token, whose utility functions are as follows:
Yakk: U(x) = x
Token: U(x) = 1 - 1/(x+1)
Where x is an amount of wealth.
Then Yakk and Token have identical orders of preferences, but will act very differently.
For example: offer them both the chance to toss a fair coin for the price of \$499
If it comes up heads you win \$4999, if tails you win nothing.
Suppose both of them currently have \$499, so Yakk's current utility is 499 and Token's is 0.998.
If Token wins his utility goes up to 0.9998, but if he loses it drops to zero.
If Yakk wins his utility goes up to 4999, but if he loses it drops to zero.
So the expected gain in utility by playing is:
Token: -0.4991
Yakk: +1751
So Yakk should play and Token shouldn't. The (relative) scalar value of utility is important because it measures the strength of a preference for an outcome.
Anyway that's by the by, I agree with both Token and Yakk that you can't just take the value of the "limit strategy" to be +infinity.
For reasons already explained, the strategy of "play forever until you lose" yields an expected gain of -\$1.
There is no real paradox here- this is just a game with no best strategy.
changed to make the numbers work!
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### Re: Triple or Nothing
That would work if people behaved that way. They don't.
Ie, if you have probability P to gain utility U, the utility of that option does not have to be P*U.
You can see this in many situations, including lottaries. The expected return from a lottery ticket is almost always negative. If you assume linear probability of utility, this implies that large sums have super-linear utility compared to small sums, which doesn't line up with human behavior that much...
There is firm theoretical ground for an order-based utility: the approximation of "if A is preferred by B, and B is preferred to C, A is preferred to C" is pretty damn accurate. This requires ordered utility, which can be approximated as a number.
Remember, utility is descriptive not prescriptive.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
rhino
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### Re: Triple or Nothing
Well, a utility that is purely a function of wealth is designed to model rational investors, not emotional individuals.
I expect that the average person would assign greater value to winning \$1000 in a game of chance than to saving an equivalent sum over the course of a year, even though the two are financially equivalent.
In making a utility-based model, we hope that there is a utility function such that "trying to maximise expected utility" lines up well with an investor's actual choices. I know that there are many situations where that is not a good approximation, generally because people act irrationally or assign value to more complicated things than just money. However, to say "what we actually need to consider is the set of all choices that our investor will make in any situation" is missing the point of trying to make a model in the first place. We want something that mathematically tractable and a reasonably good fit, not the actual truth.
If you just mean "investor X prefers \$100 to \$50 and prefers \$50 to \$0" then it is surely obvious that we gain nothing useful from this observation. I discussed this already in my previous post.
Anyway I don't think our disagreement is that fearsome - I don't actually believe everyone makes decisions based on some unconscious utility function.
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### Re: Triple or Nothing
Yakk wrote:That would work if people behaved that way. They don't.
Ie, if you have probability P to gain utility U, the utility of that option does not have to be P*U.
You can see this in many situations, including lottaries. The expected return from a lottery ticket is almost always negative. If you assume linear probability of utility, this implies that large sums have super-linear utility compared to small sums, which doesn't line up with human behavior that much...
There is firm theoretical ground for an order-based utility: the approximation of "if A is preferred by B, and B is preferred to C, A is preferred to C" is pretty damn accurate. This requires ordered utility, which can be approximated as a number.
Remember, utility is descriptive not prescriptive.
Er. Either there are several things wrong here or there is a use of the term "utility" to which I'm not accustomed.
Utility is a description of how much an entity values various world states. Given an actual utility function, it's tautological that the entity's best move is to maximize its expected utility. If by taking a certain action you will with probability P gain utility U and with probability 1-P gain utility V, then the expected utility of taking that action is P*U + (1-P)*V.
Wait... are you operating under the assumption that humans come anywhere even remotely close to maximizing, or even trying to maximize, their expected utility and then inferring from that assumption what "utility" must mean? 'cause that's severely missing the point.
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### Re: Triple or Nothing
GreedyAlgorithm wrote:Er. Either there are several things wrong here or there is a use of the term "utility" to which I'm not accustomed.
Utility is a description of how much an entity values various world states. Given an actual utility function, it's tautological that the entity's best move is to maximize its expected utility.
The entities best move is to maximize the utility of that move. If the result of the move is certain, then the utility of the move can be expressed as the utility of the result -- they are interchangeable.
Remember, I'm trying to describe human behavior, not perscribe it.
If by taking a certain action you will with probability P gain utility U and with probability 1-P gain utility V, then the expected utility of taking that action is P*U + (1-P)*V.
That does not describe human behavior.
Wait... are you operating under the assumption that humans come anywhere even remotely close to maximizing, or even trying to maximize, their expected utility and then inferring from that assumption what "utility" must mean? 'cause that's severely missing the point.
Utility, in economics, is an attempt to describe preference among choices by actors. One derives the utility of a set of choices by how humans choose between them.
The not-quite-true claim about utility is that it is an ordering -- ie, if a<b and b<c, then a<c. This holds in most every case tested -- note that there are occasional exceptions!
The claim that utility is a scalar that falls linearly through probabilities is another claim about utility. It, I contend, is very rarely true. Human beings do not pass utilities linearly through probabilities.
So a claim that utility, in the economic sense, passes linearly through probabilities is a false claim that is not talking about human preferences. It is instead talking about some term that has little relation to the economic behavior of human beings.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
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### Re: Triple or Nothing
a) No matter how much you insist, you can't combine two limits if one (or both) of them is infinite or undefined. This is exactly what you are doing when you take the limit of (3/2)n-1 as n tends to infinity. Yakk provides an excellent reason for why this doesn't work in his above post.
But you have a perfectly good function that describes the expected winnings! I'm not combining indeterminate limits, I'm just combining functions. Certainly (lim x->inf x)*(lim x-> inf 1/x) is zero times infinity. Yet it is also clearly 1.
You can turn any limit into a combination of undefined limits. Yes, if you try to compute the limit of the expected winnings by computing the limits of the payout and the probability, you'll run into indeterminate form. That doesn't mean you can't try a different route.
b) Yes, in general zero times infinity can give you anything - this is what makes it an indeterminate form. This doesn't mean, however, that you can take it to be something that fits your intuition about the problem. In fact, I have it on good authority that when dealing with probability / measure theory, zero times infinity is generally defined to be zero. I think this example illustrates the reason for this quite well.
My intuition did not come into this. I just multiplied the payout with the probability before taking the limit so as to avoid this whole zero times infinity business.
If this is treated differently in probability theory, then I stand corrected, of course. But I'm having trouble picturing it. Give me an urn with an infinite amount of marbles. I reach in and grab a marble. What are the odds of my grabbing a specific marble? 0. How many marbles do you expect me to take out of the urn? Zero times infinity. I think we can agree that this is not zero here.
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### Re: Triple or Nothing
Cogito wrote:
a) No matter how much you insist, you can't combine two limits if one (or both) of them is infinite or undefined. This is exactly what you are doing when you take the limit of (3/2)n-1 as n tends to infinity. Yakk provides an excellent reason for why this doesn't work in his above post.
But you have a perfectly good function that describes the expected winnings! I'm not combining indeterminate limits, I'm just combining functions. Certainly (lim x->inf x)*(lim x-> inf 1/x) is zero times infinity. Yet it is also clearly 1.
No, (lim x->inf x) * (lim x->inf 1/x) is not 1.
The lim x->inf of (x * 1/x) is 1, however this statement is not the same as the previous one.
There are situations when you can take (lim(x->a)f(x)) * (lim(x->a)g(x)) and say it equals lim(x->a)f(x)*g(x), but this is not true in general. It is true in some specific cases.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
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### Re: Triple or Nothing
Cogito wrote:You can turn any limit into a combination of undefined limits.
I'm afraid you can't. Just as you can't combine limits however you want, you can't just split them up arbitrarily either.
I challenge you to derive (lim(x->a)f(x)) * (lim(x->a)g(x)) = lim(x->a)f(x)*g(x) from first principles when the left hand side is zero * infinity. Of course, you can't, because the left hand side doesn't make sense. And if you can't prove that the equation holds, you sure as hell can't ever use it.
Cogito wrote:My intuition did not come into this. I just multiplied the payout with the probability before taking the limit so as to avoid this whole zero times infinity business.
YOU CAN'T DO THAT.
Just because it's a convenient short cut, doesn't make it a valid technique. Especially when it gives you the wrong answer.
Cogito wrote:If this is treated differently in probability theory, then I stand corrected, of course. But I'm having trouble picturing it. Give me an urn with an infinite amount of marbles. I reach in and grab a marble. What are the odds of my grabbing a specific marble? 0. How many marbles do you expect me to take out of the urn? Zero times infinity. I think we can agree that this is not zero here.
Ah, well, the problem here is not that 0*infinity can equal 1, but that you are trying to create an infinite discrete uniform distribution. If you don't attempt that, the problem goes away.
Last edited by Token on Thu Feb 07, 2008 11:47 pm UTC, edited 1 time in total.
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### Re: Triple or Nothing
Yakk wrote:That does not describe human behavior.
GreedyAlgorithm wrote:Wait... are you operating under the assumption that humans come anywhere even remotely close to maximizing, or even trying to maximize, their expected utility and then inferring from that assumption what "utility" must mean? 'cause that's severely missing the point.
Utility, in economics, is an attempt to describe preference among choices by actors. One derives the utility of a set of choices by how humans choose between them.
The not-quite-true claim about utility is that it is an ordering -- ie, if a<b and b<c, then a<c. This holds in most every case tested -- note that there are occasional exceptions!
The claim that utility is a scalar that falls linearly through probabilities is another claim about utility. It, I contend, is very rarely true. Human beings do not pass utilities linearly through probabilities.
So a claim that utility, in the economic sense, passes linearly through probabilities is a false claim that is not talking about human preferences. It is instead talking about some term that has little relation to the economic behavior of human beings.
Ah, I see. So the way you (meaning economists?) use the term "utility" is to assume that a human has some kind of consistent behavior from which we can infer how much the human likes or dislikes different world states. Or no, not even that. It's to say "this human chooses his actions in a fairly consistent way, and if I assume the human is often trying to maximize something, I can say something about the ordering of world states under his maximization operation - also this generalizes since the result is similar across many humans, and I will call this ordering utility".
Is that about right?
FYI, in the way I use "utility", all of those "not-quite-true" claims are very simple and very obviously true. Here's how I use it, and how I assumed everyone used it: Utility is a measure of the preference a human gives to one world state (as compared to another). Note that this makes no assumptions at all about whether or not humans try to achieve these preferences, which I believe is where the economists' definition fails at describing anything useful. Then again, I'm the one who said that your definition makes that assumption, and I could be dead wrong on that.
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### Re: Triple or Nothing
Note that the "expected" function is just a linear average over probability. Quite often doing a linear average over probability generates gibberish -- it is a particular mathematical operation, whose name happens to result that you get "expected utility", which seems like something someone should maximize.
It is the linear mean utility from a choice.
Ah, I see. So the way you (meaning economists?) use the term "utility" is to assume that a human has some kind of consistent behavior from which we can infer how much the human likes or dislikes different world states. Or no, not even that. It's to say "this human chooses his actions in a fairly consistent way, and if I assume the human is often trying to maximize something, I can say something about the ordering of world states under his maximization operation - also this generalizes since the result is similar across many humans, and I will call this ordering utility".
It is a claim about the ordering of choice. And yes, they call that ordering utility.
As it happens, if your ordering is strong, you can assign numbers (from the reals) to each utility and use that to order them.
This makes it easier to grasp. Note however that at this point any monotonic transformation of utilities results in the same ordering.
In order to have your utility obey linear probability combination, linear summation, or the like, you then have to prove or assume additional things about your utility hypothesis.
The goal of economics is to describe, not dictate, human behavior. It is noted that strong ordering of utility doesn't always work.
Claiming that any one person's preferences of world state can be assigned a number, and then the preference for a distribution of results can be calculated via a linear probability distribution, is a huge massive set of assumptions. What is worse is most of it is non-observational, and hence relatively non-falsifiable -- which sort of makes it less science.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Silas
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### Re: Triple or Nothing
Maybe we need to review what limits are. http://en.wikipedia.org/wiki/Limit_%28mathematics%29
Limits (of functions on R) only exist where functions converge to a Real number. The limit as N approaches infinity of N is nonsense. It doesn't converge. The same is true of the limit as N approaches infinity (this is a misleading way of describing the concept, anyway) of (3/2)^N.
You can set up a formula for evaluating the expected value of the pot after N rounds, but you can only evaluate it for whole numbered N's. Trying to evaluate a function f:W->R at "the limit as N approaches infinity of N" won't work because that limit isn't a whole number.
There's a related question where, if the player starts with a certain finite bank, what strategy of bet-making he should use to maximize the rate by which him bank increases, while holding the probability of ever going bankrupt below a certain level (yes, if you play this game repeatedly, you have a positive probability of never going broke- it's counterintuitive but true). It's not trivial to set up, but if you know what you're about, it's not hard.
Last edited by Silas on Sat Feb 23, 2008 8:44 am UTC, edited 1 time in total.
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### Re: Triple or Nothing
Silas wrote:Maybe we need to review what limits are. http://en.wikipedia.org/wiki/Limit_%28mathematics%29
Limits (of functions on R) only exist where functions converge to a Real number. The limit as N approaches infinity of N is nonsense. It doesn't converge. The same is true of the limit as N approaches infinity (this is a misleading way of describing the concept, anyway) of (3/2)^N.
You can certainly define infinite limits in a rigorous way, just as you can define limits at infinity, and the limit as N approaches positive infinity of (3/2)N is certainly positive infinity, under that definition. This is not the problem. The problem is that there's no reason why, in general, the outcome of a game which is in some sense the "limit" of a sequence of games should be the same as the limit of the outcomes of the finite games.
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### Re: Triple or Nothing
skeptical scientist wrote:This is not the problem. The problem is that there's no reason why, in general, the outcome of a game which is in some sense the "limit" of a sequence of games should be the same as the limit of the outcomes of the finite games.
I'm sorry to be rude. Maybe it's just late, but I cannot discern what, if anything, this sentence means. Can you rephrase it?
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### Re: Triple or Nothing
Silas wrote:
skeptical scientist wrote:This is not the problem. The problem is that there's no reason why, in general, the outcome of a game which is in some sense the "limit" of a sequence of games should be the same as the limit of the outcomes of the finite games.
I'm sorry to be rude. Maybe it's just late, but I cannot discern what, if anything, this sentence means. Can you rephrase it?
Maybe it will be more clear if I use an example. Call the triple-or-nothing game played for exactly n rounds the game Gn. We already know that the expected payoff for Gn is (3/2)n. Now, call the triple-or-nothing game, when played forever, the game G. In some sense, G is the limit as n -> +∞ of Gn, and we already know that +∞ is the limit as n -> +∞ of (3/2)n. However, there's no reason to expect that this means it is in any sense the expected winnings of the game G is infinite.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
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### Re: Triple or Nothing
skeptical scientist wrote:
Silas wrote:
skeptical scientist wrote:This is not the problem. The problem is that there's no reason why, in general, the outcome of a game which is in some sense the "limit" of a sequence of games should be the same as the limit of the outcomes of the finite games.
I'm sorry to be rude. Maybe it's just late, but I cannot discern what, if anything, this sentence means. Can you rephrase it?
Maybe it will be more clear if I use an example. Call the triple-or-nothing game played for exactly n rounds the game Gn. We already know that the expected payoff for Gn is (3/2)n. Now, call the triple-or-nothing game, when played forever, the game G. In some sense, G is the limit as n -> +∞ of Gn, and we already know that +∞ is the limit as n -> +∞ of (3/2)n. However, there's no reason to expect that this means it is in any sense the expected winnings of the game G is infinite.
Not that I don't agree with you, but if you're using this argument to 'prove' things about the puzzle in this thread, you can't use the puzzle in this thread to 'prove' your argument
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### Re: Triple or Nothing
I don't have a copy of the book on hand right now, but this problem is discussed in quite some details in Rosenthal's "A First Look at Rigorous Probability Theory" (ISBN 978-9810243227), in case anyone is interested.
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### Re: Triple or Nothing
M.qrius wrote:
skeptical scientist wrote:
Silas wrote:
skeptical scientist wrote:This is not the problem. The problem is that there's no reason why, in general, the outcome of a game which is in some sense the "limit" of a sequence of games should be the same as the limit of the outcomes of the finite games.
I'm sorry to be rude. Maybe it's just late, but I cannot discern what, if anything, this sentence means. Can you rephrase it?
Maybe it will be more clear if I use an example. Call the triple-or-nothing game played for exactly n rounds the game Gn. We already know that the expected payoff for Gn is (3/2)n. Now, call the triple-or-nothing game, when played forever, the game G. In some sense, G is the limit as n -> +∞ of Gn, and we already know that +∞ is the limit as n -> +∞ of (3/2)n. However, there's no reason to expect that this means it is in any sense the expected winnings of the game G is infinite.
Not that I don't agree with you, but if you're using this argument to 'prove' things about the puzzle in this thread, you can't use the puzzle in this thread to 'prove' your argument
No, he was just attempting to use formal notation about the game in the thread as an example.
Let E(G) for a game G be the expected return from the game G.
Take Gn as defined above.
Then E(Gn) = (3/2)n.
In a sense, G is the limn->∞ of Gn.
However, there is no reason to believe that
limn->∞ E(Gn) = E( limn->∞ Gn ) (I dub thee: Formula X)
unless someone wants to provide a proof.
The belief that Formula X holds in general is the reason why someone might find the described pseudo-paradox to be a paradox.
In fact, as noted, you can demonstrate that Formula X does not hold in every case -- as has been done in this thread, with the triple or nothing game. Until someone actually provides a proof for Formula X, there is no contradiction, and no problem.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
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### Re: Triple or Nothing
M.qrius wrote:Not that I don't agree with you, but if you're using this argument to 'prove' things about the puzzle in this thread, you can't use the puzzle in this thread to 'prove' your argument
I wasn't trying to use that example to prove what I was saying above. Someone told me they couldn't understand what I was saying, so I was using that example to try to explain what I was saying, and how it applied here.
The principle that "the outcome of a game which is in some sense the "limit" of a sequence of games is not necessarily the same as the limit of the outcomes of the finite games" doesn't require proof, because it is the absence of an assumption - I was just pointing out that other people seemed to be assuming that the outcome of the limiting game was the limit of the individual outcomes, and that this, if true, would require proof.
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### Re: Triple or Nothing
Maybe it will be more clear if I use an example. Call the triple-or-nothing game played for exactly n rounds the game Gn. We already know that the expected payoff for Gn is (3/2)n. Now, call the triple-or-nothing game, when played forever, the game G. In some sense, G is the limit as n -> +∞ of Gn, and we already know that +∞ is the limit as n -> +∞ of (3/2)n. However, there's no reason to expect that this means it is in any sense the expected winnings of the game G is infinite.
I can only point out that the expected winnings E(G) is, in fact, infinite, under standard definitions. P(you collect your winnings) is zero, but you collect infinite winnings. The winnings are, in fact, bigger than the likelihood of collecting them is small. The problem, as I see it, is that zero probability != impossibility.
Think about what infinite winnings would mean. It'd mean for any value k, it must be possible to increase n far enough that E(Gn) > \$k. And that's obviously true.
I feel kind of bad about coming here and crapping all over people who enjoy paradoxes with my "if you'd just use these definitions you've probably never seen before, it'd all make sense" (skeptical scientist, et al, this isn't you I'm talking about), but... someone is using subtly different terms than me on the internet!
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### Re: Triple or Nothing
Wow. That was, finally, an answer that makes sense to me.
Silas wrote:The problem, as I see it, is that zero probability != impossibility.
Or that 1/∞ != 0, I guess?
Silas wrote:The winnings are, in fact, bigger than the likelihood of collecting them is small.
That kinda makes sense.
I'll let people who actually know probability and calculus continue their discussion, while I hide from all the infinity.
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### Re: Triple or Nothing
Silas wrote:
Maybe it will be more clear if I use an example. Call the triple-or-nothing game played for exactly n rounds the game Gn. We already know that the expected payoff for Gn is (3/2)n. Now, call the triple-or-nothing game, when played forever, the game G. In some sense, G is the limit as n -> +∞ of Gn, and we already know that +∞ is the limit as n -> +∞ of (3/2)n. However, there's no reason to expect that this means it is in any sense the expected winnings of the game G is infinite.
I can only point out that the expected winnings E(G) is, in fact, infinite, under standard definitions. P(you collect your winnings) is zero, but you collect infinite winnings. The winnings are, in fact, bigger than the likelihood of collecting them is small. The problem, as I see it, is that zero probability != impossibility.
But it is actually impossible to collect your winnings if you keep betting till you lose.
Indigo is a lie.
Which idiot decided that websites can't go within 4cm of the edge of the screen?
There should be a null word, for the question "Is anybody there?" and to see if microphones are on.
Owehn
Posts: 479
Joined: Tue Oct 09, 2007 12:49 pm UTC
Location: Cambridge, UK
### Re: Triple or Nothing
Sort of - the game G isn't well defined because you haven't said what happens at stage ∞. Alternatively phrased, the game doesn't end if I win on each round, so what are my winnings in that scenario? It doesn't end up mattering, though, since the probability of that happening is zero: If my strategy is "keep playing till I lose", then in 100% of all games I'll end with a net loss of \$1.
Note the difference between "always" and "100% of the time" (the latter is referred to as "almost always"). It isn't impossible to collect your winnings by winning an infinite number of rounds, but that scenario has probability 0.
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Token
Posts: 1481
Joined: Fri Dec 01, 2006 5:07 pm UTC
Location: London
### Re: Triple or Nothing
Owehn wrote:Note the difference between "always" and "100% of the time" (the latter is referred to as "almost always"). It isn't impossible to collect your winnings by winning an infinite number of rounds, but that scenario has probability 0.
I'd contest this. Yes, it is possible to win an infinite number of rounds (with probability zero), but I don't see how there's any possible situation that allows you to collect infinite winnings, since by collecting, you have necessarily chosen to stop playing after some finite number of rounds.
All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.
Silas
Posts: 1091
Joined: Sat Feb 02, 2008 9:08 pm UTC
### Re: Triple or Nothing
When we're speaking with formal terms, ∞ mean 'arbitrarily large,' not 'goes on forever.' We can define G differently, but if you do, you shouldn't expect ∞ to have the same properties. G as the case where you never stop satisfies the naive* understanding of ∞, but it doesn't jive with the way we're using ∞ in (3/2).
* as a math term meaning 'informal', not a denigration of your intellect
Felstaff wrote:Serves you goddamned right. I hope you're happy, Cake Ruiner
Owehn
Posts: 479
Joined: Tue Oct 09, 2007 12:49 pm UTC
Location: Cambridge, UK
### Re: Triple or Nothing
Token wrote:
Owehn wrote:Note the difference between "always" and "100% of the time" (the latter is referred to as "almost always"). It isn't impossible to collect your winnings by winning an infinite number of rounds, but that scenario has probability 0.
I'd contest this. Yes, it is possible to win an infinite number of rounds (with probability zero), but I don't see how there's any possible situation that allows you to collect infinite winnings, since by collecting, you have necessarily chosen to stop playing after some finite number of rounds.
This is what I was talking about with the game G. As the game is phrased, it's incomplete: what winnings do I get if I flip heads an infinite number of times? You could complete the game by saying that after an infinite sequence of heads, I walk away with the infinite amount of money on the table, or you could complete it by saying I win \$10, or nothing. It doesn't matter - the point is that this scenario occurs with probability zero, so your winnings in this case don't figure into the calculation of expected winnings.
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mikek
Posts: 46
Joined: Mon Nov 26, 2007 5:25 pm UTC
Location: Bristol, UK
### Re: Triple or Nothing
Actaeus wrote:The expected winnings from each flip is whatever is on the table. Therefore, it is always in the player's best interest to keep playing.
A little late to the party but.. says who?
everything under the sun is in tune
but the sun is eclipsed by the moon
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## Triple or Nothing. A forum for good logic/math puzzles.. Moderators: jestingrabbit, Moderators General, Prelates. Actaeus. Posts: 606. Joined: Thu Jan 10, 2008 9:21 pm UTC. Location: ZZ9 Plural Z Alpha. ### Triple or Nothing. Here's the premise:. There is a carnival game (or casino, doesn't matter.) The player starts by putting \$1 on the table. The dealer flips a coin.. HEADS: Dealer pays \$2 to the table. TAILS: Dealer takes the \$1 and game ends.. Player now chooses whether to play another round. If he doesn't, Player takes all money on the table and walks away. If he does, the coin is flipped again:. HEADS: Dealer triples the pot again. TAILS: Dealer takes all money, game ends.. The expected winnings from each flip is whatever is on the table. Therefore, it is always in the player's best interest to keep playing. However, if the player always chooses to play another round, he *will* eventually lose the original money and walk away with a loss of \$1.. What gives?. Twist: Player is allowed to play multiple games. If Player always plays one round and then ends the game, then repeats, he will win on average \$1 per game. If he plays two rounds each time, he will average \$2. Three rounds, \$3.25. If the player plays 100 rounds each time, even though he only has a 1/1.3e30 chance, he will win \$5.15e47 each time he wins. average: \$4.07e17. If the player always plays an "infinite" number of games (until he loses), he will win \$0 each time.. What gives? Does this involve transfinite numbers? If the player can play an infinite number of games, his winnings/game will approximate the numbers I gave, but will he get an infinite number of heads "once" in an infinite number of games, and win \$infinity? The expected winnings/game increases exponentially, so.....I don't even know where I was going with that.. Summary: What The Fork?. ---. EDIT: should this be in Logic Games?. Macbi. Posts: 941. Joined: Mon Apr 09, 2007 8:32 am UTC. Location: UKvia. ### Re: Triple or Nothing. Spoiler:. When you get Tails, you lose all the money, but you also lose the right to continue playing. This right is worth the right amount to balence the game out.. Indigo is a lie.. Which idiot decided that websites can't go within 4cm of the edge of the screen?. There should be a null word, for the question "Is anybody there?" and to see if microphones are on.. Actaeus. Posts: 606. Joined: Thu Jan 10, 2008 9:21 pm UTC. Location: ZZ9 Plural Z Alpha. ### Re: Triple or Nothing. Spoiler:. Of course! I forgot your money on the table affected future rounds, so expected winnings won't cover it. I believe expected winnings only works if it's out of your pocket.. Cauchy. Posts: 602. Joined: Wed Mar 28, 2007 1:43 pm UTC. ### Re: Triple or Nothing. If the player has played n games, he has a (1/2)^n chance of earning 3^n - 1 dollars, and a 1 - (1/2)^n chance of losing \$1, for an expected earning of (3/2)^n - 1. Clearly, as n increases, his expected winnings increase. As is the case with many things, it doesn't go quite as smoothly in the limit. I don't think this has anything to do with an "extra penalty" for getting 0. After all, we could easily see the player still playing even after he loses a flip: the money on the table is now \$0, and if he wins it triples and if it loses it goes to \$0, making the game no different than it was previously. Rather, things simply break down in the limit. It happens a lot in mathematics.. (∫|p|2)(∫|q|2) ≥ (∫|pq|)2. Thanks, skeptical scientist, for knowing symbols and giving them to me.. Yakk. Poster with most posts but no title.. Posts: 11128. Joined: Sat Jan 27, 2007 7:27 pm UTC. Location: E pur si muove. ### Re: Triple or Nothing. You cannot play an infinite number of rounds.. Let G(N) be the game strategy of playing N rounds then stopping. Then E(G(N)) (average or expected value) goes to infinity as N goes to infinity.. However, the limits does not cross the E and G function. lim(N->inf) of E(G(N)) != E(G( lim(N->inf)(N) ) ). Partially this is because G(infinity) is a special case. A simple way to extend G(N) to cover infinity is G(infinity) is "play until you lose, no matter what".. Once you turn the problem into symbols and poke at it, there is nothing really strange going on.. .... The second thing to realize is that money isn't linear in value. The E() (expected value) function is actually a poor measure of the return from a game.. Money isn't utility. Suppose you had 10^15 US dollars already -- having 10^18 US dollars wouldn't actually increase your wealth by all that much, because you have already managed to own the entire US economy, and going from 10^15 to 10^18 just reduces what percent everyone else has from a fraction of a percent to a smaller fraction.. Similar effects happen at smaller scales. Your first million dollars is more useful than your second.. One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR. Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.. dan. Posts: 62. Joined: Wed Aug 09, 2006 12:55 am UTC. ### Re: Triple or Nothing. You can get rid of the probability stuff and still have essentially the same problem. Say there is a game where, in each round, I give you \$1, and then you choose whether or not to continue. If you continue, the same thing happens again, and if you stop, I take all the money back and the game ends. This is obviously a pretty pointless game. Although it might seem at each stage like you get more money if you keep playing, if you stand back a bit you can see that this isn't actually true at all.. Waterhouse. Posts: 54. Joined: Wed Oct 03, 2007 12:37 pm UTC. ### Re: Triple or Nothing. What gives? The question implies the following: how can it be that the gambler, apparently paradoxically, has a chance to win infinite money with a very low risk? The answer is simply that the imagined setup is contrary to the way that the "house" operates in reality.. In any game of chance that is run for profit, the odds are biased toward the house. In this manner the longer a gambler plays, the more money the house wins. In this puzzle, the house is offering three to one payout on a game with a one in two chance of winning, which is clearly biased toward the player.. Another way to see it is that the player could, instead of remaining in the game after winning one round, pocket his winnings, and then immediately start up a new game. In this way, he will be expected to win two dollars half the time and to lose only one dollar the other half of the time.. Such payouts for those odds are never offered in reality.. Actaeus. Posts: 606. Joined: Thu Jan 10, 2008 9:21 pm UTC. Location: ZZ9 Plural Z Alpha. ### Re: Triple or Nothing. This was not a realistic question....My question was, why is the "best" strategy in terms of payout one that leads to no payout at all? The answer was what Yakk said:. Yakk wrote:You cannot play an infinite number of rounds.. Let G(N) be the game strategy of playing N rounds then stopping. Then E(G(N)) (average or expected value) goes to infinity as N goes to infinity.. However, the limits does not cross the E and G function. lim(N->inf) of E(G(N)) != E(G( lim(N->inf)(N) ) ). Partially this is because G(infinity) is a special case. A simple way to extend G(N) to cover infinity is G(infinity) is "play until you lose, no matter what".. Once you turn the problem into symbols and poke at it, there is nothing really strange going on.. .... The second thing to realize is that money isn't linear in value. The E() (expected value) function is actually a poor measure of the return from a game.. Money isn't utility. Suppose you had 10^15 US dollars already -- having 10^18 US dollars wouldn't actually increase your wealth by all that much, because you have already managed to own the entire US economy, and going from 10^15 to 10^18 just reduces what percent everyone else has from a fraction of a percent to a smaller fraction.. Similar effects happen at smaller scales. Your first million dollars is more useful than your second.. Essentially, the function for expected payout tends to infinity, but the limit is a special case. Or something along those lines.. Yakk. Poster with most posts but no title.. Posts: 11128. Joined: Sat Jan 27, 2007 7:27 pm UTC. Location: E pur si muove. ### Re: Triple or Nothing. Really, it is just that limits to infinity don't always cross functions. Lim(x->infinity)f(g(x)) does not always equal f(lim(x->infinity)g(x)).. The cases where that works are special, rare cases.. Heck, the lim(x->y)f(x) = f(y) is a special case called a continuous function even when y isn't infinity. For it to be true when y is infinity, your function could be called "continuous at infinity", which is not implied by standard continuity.. One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR. Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.. Cogito. Posts: 9. Joined: Wed Feb 06, 2008 8:55 am UTC. ### Re: Triple or Nothing. The expected winnings in the original game is zero(the chance of winning) times infinity (the payout).. This need not equal zero at all. And if you consider it as a limiting case, then it should be clear the payout isn't zero. It's infinity. So I wouldn't disagree with the original strategy, personally.. Also, Yakk, the non-linear relation between money and utility isn't particular relevant. You can just replace units of money with units of utility.. More generally, it's always dangerous to just state an infinity as opposed to taking it as a limit of finite processes.. Token. Posts: 1481. Joined: Fri Dec 01, 2006 5:07 pm UTC. Location: London. ### Re: Triple or Nothing. Cogito wrote:This need not equal zero at all. And if you consider it as a limiting case, then it should be clear the payout isn't zero. It's infinity. So I wouldn't disagree with the original strategy, personally.. The limiting case, I believe, remains undefined. There are two separate limits (one for the probability, one for the payout). Two such limits may only be merged into one when they are both exist and are finite, which is not the case here.. All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.. Cogito. Posts: 9. Joined: Wed Feb 06, 2008 8:55 am UTC. ### Re: Triple or Nothing. In general, if you multiply two limits that equal zero and infinity, then the final result can still be anything.. In the specific case you can merge the two formulae prior to taking the limit and find yourself a well-defined answer.. Expected winnings = 3^n/2^n -1=(3/2)^n-1, as Cauchy pointed out. There's no reason why it wouldn't hold in the limit.. Token. Posts: 1481. Joined: Fri Dec 01, 2006 5:07 pm UTC. Location: London. ### Re: Triple or Nothing. Cogito wrote:In general, if you multiply two limits that equal zero and infinity, then the final result can still be anything.. In the specific case you can merge the two formulae prior to taking the limit and find yourself a well-defined answer.. Expected winnings = 3^n/2^n -1=(3/2)^n-1, as Cauchy pointed out. There's no reason why it wouldn't hold in the limit.. Well, there's two reasons you are wrong.. a) No matter how much you insist, you can't combine two limits if one (or both) of them is infinite or undefined. This is exactly what you are doing when you take the limit of (3/2)n-1 as n tends to infinity. Yakk provides an excellent reason for why this doesn't work in his above post.. b) Yes, in general zero times infinity can give you anything - this is what makes it an indeterminate form. This doesn't mean, however, that you can take it to be something that fits your intuition about the problem. In fact, I have it on good authority that when dealing with probability / measure theory, zero times infinity is generally defined to be zero. I think this example illustrates the reason for this quite well.. All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.. Yakk. Poster with most posts but no title.. Posts: 11128. Joined: Sat Jan 27, 2007 7:27 pm UTC. Location: E pur si muove. ### Re: Triple or Nothing. Cogito wrote:Also, Yakk, the non-linear relation between money and utility isn't particular relevant. You can just replace units of money with units of utility.. Well, it does generate an interesting result: at which point should one make a decision to stop?. If you are getting an exponential amount of utility on each stage ... gosh. I am rather not sure that is possible!. If you define Utility to be that which should guild your actions in a the "with probability P, you gain U2, and with probability (1-P) you lose U1. Then you should take the event if P*U2 > (1-P)*U1" case, quite possibly you can demonstrate that total Utility is bounded. In which case, unbounded exponential utility becomes impossible.. It might be a better statement to make that "utility is not probability linear"? Ie, if you have probability P to gain utility U, the utility of that option does not have to be P*U?. Utility, as defined by economics, is interesting in that any strictly monotonic transformation of utility satisfies the properties of utility. Utility is actually just an ordering of preference, it is not a scalar or vector value. Economics uses scalar values to make it easier to understand what the hell is going on: which leads to a number of strange results when you take those scalar values at face value.... One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR. Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.. rhino. Posts: 123. Joined: Fri Dec 08, 2006 3:43 pm UTC. Location: Cambridge, UK. ### Re: Triple or Nothing. Yakk: I agree with your suggestion that a bounded utility function seems intuitively correct.. However, utility is not just an order of preference. Suppose we have two agents, say Yakk and Token, whose utility functions are as follows:. Yakk: U(x) = x. Token: U(x) = 1 - 1/(x+1). Where x is an amount of wealth.. Then Yakk and Token have identical orders of preferences, but will act very differently.. For example: offer them both the chance to toss a fair coin for the price of \$499. If it comes up heads you win \$4999, if tails you win nothing.. Suppose both of them currently have \$499, so Yakk's current utility is 499 and Token's is 0.998.. If Token wins his utility goes up to 0.9998, but if he loses it drops to zero.. If Yakk wins his utility goes up to 4999, but if he loses it drops to zero.. So the expected gain in utility by playing is:. Token: -0.4991. Yakk: +1751. So Yakk should play and Token shouldn't. The (relative) scalar value of utility is important because it measures the strength of a preference for an outcome.. Anyway that's by the by, I agree with both Token and Yakk that you can't just take the value of the "limit strategy" to be +infinity.. For reasons already explained, the strategy of "play forever until you lose" yields an expected gain of -\$1.. There is no real paradox here- this is just a game with no best strategy.. changed to make the numbers work!. Yakk. Poster with most posts but no title.. Posts: 11128. Joined: Sat Jan 27, 2007 7:27 pm UTC. Location: E pur si muove. ### Re: Triple or Nothing. That would work if people behaved that way. They don't.. Ie, if you have probability P to gain utility U, the utility of that option does not have to be P*U.. You can see this in many situations, including lottaries. The expected return from a lottery ticket is almost always negative. If you assume linear probability of utility, this implies that large sums have super-linear utility compared to small sums, which doesn't line up with human behavior that much.... There is firm theoretical ground for an order-based utility: the approximation of "if A is preferred by B, and B is preferred to C, A is preferred to C" is pretty damn accurate. This requires ordered utility, which can be approximated as a number.. Remember, utility is descriptive not prescriptive.. One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR. Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.. rhino. Posts: 123. Joined: Fri Dec 08, 2006 3:43 pm UTC. Location: Cambridge, UK. ### Re: Triple or Nothing. Well, a utility that is purely a function of wealth is designed to model rational investors, not emotional individuals.. I expect that the average person would assign greater value to winning \$1000 in a game of chance than to saving an equivalent sum over the course of a year, even though the two are financially equivalent.. In making a utility-based model, we hope that there is a utility function such that "trying to maximise expected utility" lines up well with an investor's actual choices. I know that there are many situations where that is not a good approximation, generally because people act irrationally or assign value to more complicated things than just money. However, to say "what we actually need to consider is the set of all choices that our investor will make in any situation" is missing the point of trying to make a model in the first place. We want something that mathematically tractable and a reasonably good fit, not the actual truth.. If you just mean "investor X prefers \$100 to \$50 and prefers \$50 to \$0" then it is surely obvious that we gain nothing useful from this observation. I discussed this already in my previous post.. Anyway I don't think our disagreement is that fearsome - I don't actually believe everyone makes decisions based on some unconscious utility function.. GreedyAlgorithm. Posts: 286. Joined: Tue Aug 22, 2006 10:35 pm UTC. Contact:. ### Re: Triple or Nothing. Yakk wrote:That would work if people behaved that way. They don't.. Ie, if you have probability P to gain utility U, the utility of that option does not have to be P*U.. You can see this in many situations, including lottaries. The expected return from a lottery ticket is almost always negative. If you assume linear probability of utility, this implies that large sums have super-linear utility compared to small sums, which doesn't line up with human behavior that much.... There is firm theoretical ground for an order-based utility: the approximation of "if A is preferred by B, and B is preferred to C, A is preferred to C" is pretty damn accurate. This requires ordered utility, which can be approximated as a number.. Remember, utility is descriptive not prescriptive.. Er. Either there are several things wrong here or there is a use of the term "utility" to which I'm not accustomed.. Utility is a description of how much an entity values various world states. Given an actual utility function, it's tautological that the entity's best move is to maximize its expected utility. If by taking a certain action you will with probability P gain utility U and with probability 1-P gain utility V, then the expected utility of taking that action is P*U + (1-P)*V.. Wait... are you operating under the assumption that humans come anywhere even remotely close to maximizing, or even trying to maximize, their expected utility and then inferring from that assumption what "utility" must mean? 'cause that's severely missing the point.. GENERATION 1-i: The first time you see this, copy it into your sig on any forum. Square it, and then add i to the generation.. Yakk. Poster with most posts but no title.. Posts: 11128. Joined: Sat Jan 27, 2007 7:27 pm UTC. Location: E pur si muove. ### Re: Triple or Nothing. GreedyAlgorithm wrote:Er. Either there are several things wrong here or there is a use of the term "utility" to which I'm not accustomed.. Utility is a description of how much an entity values various world states. Given an actual utility function, it's tautological that the entity's best move is to maximize its expected utility.. The entities best move is to maximize the utility of that move. If the result of the move is certain, then the utility of the move can be expressed as the utility of the result -- they are interchangeable.. Remember, I'm trying to describe human behavior, not perscribe it.. If by taking a certain action you will with probability P gain utility U and with probability 1-P gain utility V, then the expected utility of taking that action is P*U + (1-P)*V.. That does not describe human behavior.. Wait... are you operating under the assumption that humans come anywhere even remotely close to maximizing, or even trying to maximize, their expected utility and then inferring from that assumption what "utility" must mean? 'cause that's severely missing the point.. Utility, in economics, is an attempt to describe preference among choices by actors. One derives the utility of a set of choices by how humans choose between them.. The not-quite-true claim about utility is that it is an ordering -- ie, if a<b and b<c, then a<c. This holds in most every case tested -- note that there are occasional exceptions!. The claim that utility is a scalar that falls linearly through probabilities is another claim about utility. It, I contend, is very rarely true. Human beings do not pass utilities linearly through probabilities.. So a claim that utility, in the economic sense, passes linearly through probabilities is a false claim that is not talking about human preferences. It is instead talking about some term that has little relation to the economic behavior of human beings.. One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR. Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.. Cogito. Posts: 9. Joined: Wed Feb 06, 2008 8:55 am UTC. ### Re: Triple or Nothing. a) No matter how much you insist, you can't combine two limits if one (or both) of them is infinite or undefined. This is exactly what you are doing when you take the limit of (3/2)n-1 as n tends to infinity. Yakk provides an excellent reason for why this doesn't work in his above post.. But you have a perfectly good function that describes the expected winnings! I'm not combining indeterminate limits, I'm just combining functions. Certainly (lim x->inf x)*(lim x-> inf 1/x) is zero times infinity. Yet it is also clearly 1.. You can turn any limit into a combination of undefined limits. Yes, if you try to compute the limit of the expected winnings by computing the limits of the payout and the probability, you'll run into indeterminate form. That doesn't mean you can't try a different route.. b) Yes, in general zero times infinity can give you anything - this is what makes it an indeterminate form. This doesn't mean, however, that you can take it to be something that fits your intuition about the problem. In fact, I have it on good authority that when dealing with probability / measure theory, zero times infinity is generally defined to be zero. I think this example illustrates the reason for this quite well.. My intuition did not come into this. I just multiplied the payout with the probability before taking the limit so as to avoid this whole zero times infinity business.. If this is treated differently in probability theory, then I stand corrected, of course. But I'm having trouble picturing it. Give me an urn with an infinite amount of marbles.
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I reach in and grab a marble. What are the odds of my grabbing a specific marble? 0. How many marbles do you expect me to take out of the urn? Zero times infinity. I think we can agree that this is not zero here.. Yakk. Poster with most posts but no title.. Posts: 11128. Joined: Sat Jan 27, 2007 7:27 pm UTC. Location: E pur si muove. ### Re: Triple or Nothing. Cogito wrote:. a) No matter how much you insist, you can't combine two limits if one (or both) of them is infinite or undefined. This is exactly what you are doing when you take the limit of (3/2)n-1 as n tends to infinity. Yakk provides an excellent reason for why this doesn't work in his above post.. But you have a perfectly good function that describes the expected winnings! I'm not combining indeterminate limits, I'm just combining functions. Certainly (lim x->inf x)*(lim x-> inf 1/x) is zero times infinity. Yet it is also clearly 1.. No, (lim x->inf x) * (lim x->inf 1/x) is not 1.. The lim x->inf of (x * 1/x) is 1, however this statement is not the same as the previous one.. There are situations when you can take (lim(x->a)f(x)) * (lim(x->a)g(x)) and say it equals lim(x->a)f(x)*g(x), but this is not true in general. It is true in some specific cases.. One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR. Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.. Token. Posts: 1481. Joined: Fri Dec 01, 2006 5:07 pm UTC. Location: London. ### Re: Triple or Nothing. Cogito wrote:You can turn any limit into a combination of undefined limits.. I'm afraid you can't. Just as you can't combine limits however you want, you can't just split them up arbitrarily either.. I challenge you to derive (lim(x->a)f(x)) * (lim(x->a)g(x)) = lim(x->a)f(x)*g(x) from first principles when the left hand side is zero * infinity. Of course, you can't, because the left hand side doesn't make sense. And if you can't prove that the equation holds, you sure as hell can't ever use it.. Cogito wrote:My intuition did not come into this. I just multiplied the payout with the probability before taking the limit so as to avoid this whole zero times infinity business.. YOU CAN'T DO THAT.. Just because it's a convenient short cut, doesn't make it a valid technique. Especially when it gives you the wrong answer.. Cogito wrote:If this is treated differently in probability theory, then I stand corrected, of course. But I'm having trouble picturing it. Give me an urn with an infinite amount of marbles. I reach in and grab a marble. What are the odds of my grabbing a specific marble? 0. How many marbles do you expect me to take out of the urn? Zero times infinity. I think we can agree that this is not zero here.. Ah, well, the problem here is not that 0*infinity can equal 1, but that you are trying to create an infinite discrete uniform distribution. If you don't attempt that, the problem goes away.. Last edited by Token on Thu Feb 07, 2008 11:47 pm UTC, edited 1 time in total.. All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.. GreedyAlgorithm. Posts: 286. Joined: Tue Aug 22, 2006 10:35 pm UTC. Contact:. ### Re: Triple or Nothing. Yakk wrote:That does not describe human behavior.. GreedyAlgorithm wrote:Wait... are you operating under the assumption that humans come anywhere even remotely close to maximizing, or even trying to maximize, their expected utility and then inferring from that assumption what "utility" must mean? 'cause that's severely missing the point.. Utility, in economics, is an attempt to describe preference among choices by actors. One derives the utility of a set of choices by how humans choose between them.. The not-quite-true claim about utility is that it is an ordering -- ie, if a<b and b<c, then a<c. This holds in most every case tested -- note that there are occasional exceptions!. The claim that utility is a scalar that falls linearly through probabilities is another claim about utility. It, I contend, is very rarely true. Human beings do not pass utilities linearly through probabilities.. So a claim that utility, in the economic sense, passes linearly through probabilities is a false claim that is not talking about human preferences. It is instead talking about some term that has little relation to the economic behavior of human beings.. Ah, I see. So the way you (meaning economists?) use the term "utility" is to assume that a human has some kind of consistent behavior from which we can infer how much the human likes or dislikes different world states. Or no, not even that. It's to say "this human chooses his actions in a fairly consistent way, and if I assume the human is often trying to maximize something, I can say something about the ordering of world states under his maximization operation - also this generalizes since the result is similar across many humans, and I will call this ordering utility".. Is that about right?. FYI, in the way I use "utility", all of those "not-quite-true" claims are very simple and very obviously true. Here's how I use it, and how I assumed everyone used it: Utility is a measure of the preference a human gives to one world state (as compared to another). Note that this makes no assumptions at all about whether or not humans try to achieve these preferences, which I believe is where the economists' definition fails at describing anything useful. Then again, I'm the one who said that your definition makes that assumption, and I could be dead wrong on that.. GENERATION 1-i: The first time you see this, copy it into your sig on any forum. Square it, and then add i to the generation.. Yakk. Poster with most posts but no title.. Posts: 11128. Joined: Sat Jan 27, 2007 7:27 pm UTC. Location: E pur si muove. ### Re: Triple or Nothing. Note that the "expected" function is just a linear average over probability. Quite often doing a linear average over probability generates gibberish -- it is a particular mathematical operation, whose name happens to result that you get "expected utility", which seems like something someone should maximize.. It is the linear mean utility from a choice.. Ah, I see. So the way you (meaning economists?) use the term "utility" is to assume that a human has some kind of consistent behavior from which we can infer how much the human likes or dislikes different world states. Or no, not even that. It's to say "this human chooses his actions in a fairly consistent way, and if I assume the human is often trying to maximize something, I can say something about the ordering of world states under his maximization operation - also this generalizes since the result is similar across many humans, and I will call this ordering utility".. It is a claim about the ordering of choice. And yes, they call that ordering utility.. As it happens, if your ordering is strong, you can assign numbers (from the reals) to each utility and use that to order them.. This makes it easier to grasp. Note however that at this point any monotonic transformation of utilities results in the same ordering.. In order to have your utility obey linear probability combination, linear summation, or the like, you then have to prove or assume additional things about your utility hypothesis.. The goal of economics is to describe, not dictate, human behavior. It is noted that strong ordering of utility doesn't always work.. Claiming that any one person's preferences of world state can be assigned a number, and then the preference for a distribution of results can be calculated via a linear probability distribution, is a huge massive set of assumptions. What is worse is most of it is non-observational, and hence relatively non-falsifiable -- which sort of makes it less science.. One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR. Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.. Silas. Posts: 1091. Joined: Sat Feb 02, 2008 9:08 pm UTC. ### Re: Triple or Nothing. Maybe we need to review what limits are. http://en.wikipedia.org/wiki/Limit_%28mathematics%29. Limits (of functions on R) only exist where functions converge to a Real number. The limit as N approaches infinity of N is nonsense. It doesn't converge. The same is true of the limit as N approaches infinity (this is a misleading way of describing the concept, anyway) of (3/2)^N.. You can set up a formula for evaluating the expected value of the pot after N rounds, but you can only evaluate it for whole numbered N's. Trying to evaluate a function f:W->R at "the limit as N approaches infinity of N" won't work because that limit isn't a whole number.. There's a related question where, if the player starts with a certain finite bank, what strategy of bet-making he should use to maximize the rate by which him bank increases, while holding the probability of ever going bankrupt below a certain level (yes, if you play this game repeatedly, you have a positive probability of never going broke- it's counterintuitive but true). It's not trivial to set up, but if you know what you're about, it's not hard.. Last edited by Silas on Sat Feb 23, 2008 8:44 am UTC, edited 1 time in total.. Felstaff wrote:Serves you goddamned right. I hope you're happy, Cake Ruiner. skeptical scientist. closed-minded spiritualist. Posts: 6142. Joined: Tue Nov 28, 2006 6:09 am UTC. Location: San Francisco. ### Re: Triple or Nothing. Silas wrote:Maybe we need to review what limits are. http://en.wikipedia.org/wiki/Limit_%28mathematics%29. Limits (of functions on R) only exist where functions converge to a Real number. The limit as N approaches infinity of N is nonsense. It doesn't converge. The same is true of the limit as N approaches infinity (this is a misleading way of describing the concept, anyway) of (3/2)^N.. You can certainly define infinite limits in a rigorous way, just as you can define limits at infinity, and the limit as N approaches positive infinity of (3/2)N is certainly positive infinity, under that definition. This is not the problem. The problem is that there's no reason why, in general, the outcome of a game which is in some sense the "limit" of a sequence of games should be the same as the limit of the outcomes of the finite games.. I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.. "With math, all things are possible." —Rebecca Watson. Silas. Posts: 1091. Joined: Sat Feb 02, 2008 9:08 pm UTC. ### Re: Triple or Nothing. skeptical scientist wrote:This is not the problem. The problem is that there's no reason why, in general, the outcome of a game which is in some sense the "limit" of a sequence of games should be the same as the limit of the outcomes of the finite games.. I'm sorry to be rude. Maybe it's just late, but I cannot discern what, if anything, this sentence means. Can you rephrase it?. Felstaff wrote:Serves you goddamned right. I hope you're happy, Cake Ruiner. skeptical scientist. closed-minded spiritualist. Posts: 6142. Joined: Tue Nov 28, 2006 6:09 am UTC. Location: San Francisco. ### Re: Triple or Nothing. Silas wrote:. skeptical scientist wrote:This is not the problem. The problem is that there's no reason why, in general, the outcome of a game which is in some sense the "limit" of a sequence of games should be the same as the limit of the outcomes of the finite games.. I'm sorry to be rude. Maybe it's just late, but I cannot discern what, if anything, this sentence means. Can you rephrase it?. Maybe it will be more clear if I use an example. Call the triple-or-nothing game played for exactly n rounds the game Gn. We already know that the expected payoff for Gn is (3/2)n. Now, call the triple-or-nothing game, when played forever, the game G. In some sense, G is the limit as n -> +∞ of Gn, and we already know that +∞ is the limit as n -> +∞ of (3/2)n. However, there's no reason to expect that this means it is in any sense the expected winnings of the game G is infinite.. I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.. "With math, all things are possible." —Rebecca Watson. M.qrius. Rainbow Brite. Posts: 519. Joined: Sat Nov 10, 2007 12:54 am UTC. Location: Rainbow's end. Contact:. ### Re: Triple or Nothing. skeptical scientist wrote:. Silas wrote:. skeptical scientist wrote:This is not the problem. The problem is that there's no reason why, in general, the outcome of a game which is in some sense the "limit" of a sequence of games should be the same as the limit of the outcomes of the finite games.. I'm sorry to be rude. Maybe it's just late, but I cannot discern what, if anything, this sentence means. Can you rephrase it?. Maybe it will be more clear if I use an example. Call the triple-or-nothing game played for exactly n rounds the game Gn. We already know that the expected payoff for Gn is (3/2)n. Now, call the triple-or-nothing game, when played forever, the game G. In some sense, G is the limit as n -> +∞ of Gn, and we already know that +∞ is the limit as n -> +∞ of (3/2)n. However, there's no reason to expect that this means it is in any sense the expected winnings of the game G is infinite.. Not that I don't agree with you, but if you're using this argument to 'prove' things about the puzzle in this thread, you can't use the puzzle in this thread to 'prove' your argument. NathanielJ. Posts: 882. Joined: Sun Jan 13, 2008 9:04 pm UTC. ### Re: Triple or Nothing. I don't have a copy of the book on hand right now, but this problem is discussed in quite some details in Rosenthal's "A First Look at Rigorous Probability Theory" (ISBN 978-9810243227), in case anyone is interested.. Homepage: http://www.njohnston.ca. Conway's Game of Life: http://www.conwaylife.com. Yakk. Poster with most posts but no title.. Posts: 11128. Joined: Sat Jan 27, 2007 7:27 pm UTC. Location: E pur si muove. ### Re: Triple or Nothing. M.qrius wrote:. skeptical scientist wrote:. Silas wrote:. skeptical scientist wrote:This is not the problem. The problem is that there's no reason why, in general, the outcome of a game which is in some sense the "limit" of a sequence of games should be the same as the limit of the outcomes of the finite games.. I'm sorry to be rude. Maybe it's just late, but I cannot discern what, if anything, this sentence means. Can you rephrase it?. Maybe it will be more clear if I use an example. Call the triple-or-nothing game played for exactly n rounds the game Gn. We already know that the expected payoff for Gn is (3/2)n. Now, call the triple-or-nothing game, when played forever, the game G. In some sense, G is the limit as n -> +∞ of Gn, and we already know that +∞ is the limit as n -> +∞ of (3/2)n. However, there's no reason to expect that this means it is in any sense the expected winnings of the game G is infinite.. Not that I don't agree with you, but if you're using this argument to 'prove' things about the puzzle in this thread, you can't use the puzzle in this thread to 'prove' your argument. No, he was just attempting to use formal notation about the game in the thread as an example.. Let E(G) for a game G be the expected return from the game G.. Take Gn as defined above.. Then E(Gn) = (3/2)n.. In a sense, G is the limn->∞ of Gn.. However, there is no reason to believe that. limn->∞ E(Gn) = E( limn->∞ Gn ) (I dub thee: Formula X). unless someone wants to provide a proof.. The belief that Formula X holds in general is the reason why someone might find the described pseudo-paradox to be a paradox.. In fact, as noted, you can demonstrate that Formula X does not hold in every case -- as has been done in this thread, with the triple or nothing game. Until someone actually provides a proof for Formula X, there is no contradiction, and no problem.. One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR. Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.. skeptical scientist. closed-minded spiritualist. Posts: 6142. Joined: Tue Nov 28, 2006 6:09 am UTC. Location: San Francisco. ### Re: Triple or Nothing. M.qrius wrote:Not that I don't agree with you, but if you're using this argument to 'prove' things about the puzzle in this thread, you can't use the puzzle in this thread to 'prove' your argument. I wasn't trying to use that example to prove what I was saying above. Someone told me they couldn't understand what I was saying, so I was using that example to try to explain what I was saying, and how it applied here.. The principle that "the outcome of a game which is in some sense the "limit" of a sequence of games is not necessarily the same as the limit of the outcomes of the finite games" doesn't require proof, because it is the absence of an assumption - I was just pointing out that other people seemed to be assuming that the outcome of the limiting game was the limit of the individual outcomes, and that this, if true, would require proof.. I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.. "With math, all things are possible." —Rebecca Watson. Silas. Posts: 1091. Joined: Sat Feb 02, 2008 9:08 pm UTC. ### Re: Triple or Nothing. Maybe it will be more clear if I use an example. Call the triple-or-nothing game played for exactly n rounds the game Gn. We already know that the expected payoff for Gn is (3/2)n. Now, call the triple-or-nothing game, when played forever, the game G. In some sense, G is the limit as n -> +∞ of Gn, and we already know that +∞ is the limit as n -> +∞ of (3/2)n. However, there's no reason to expect that this means it is in any sense the expected winnings of the game G is infinite.. I can only point out that the expected winnings E(G) is, in fact, infinite, under standard definitions. P(you collect your winnings) is zero, but you collect infinite winnings. The winnings are, in fact, bigger than the likelihood of collecting them is small. The problem, as I see it, is that zero probability != impossibility.. Think about what infinite winnings would mean. It'd mean for any value k, it must be possible to increase n far enough that E(Gn) > \$k. And that's obviously true.. I feel kind of bad about coming here and crapping all over people who enjoy paradoxes with my "if you'd just use these definitions you've probably never seen before, it'd all make sense" (skeptical scientist, et al, this isn't you I'm talking about), but... someone is using subtly different terms than me on the internet!. Felstaff wrote:Serves you goddamned right. I hope you're happy, Cake Ruiner. Actaeus. Posts: 606. Joined: Thu Jan 10, 2008 9:21 pm UTC. Location: ZZ9 Plural Z Alpha. ### Re: Triple or Nothing. Wow. That was, finally, an answer that makes sense to me.. Silas wrote:The problem, as I see it, is that zero probability != impossibility.. Or that 1/∞ != 0, I guess?. Silas wrote:The winnings are, in fact, bigger than the likelihood of collecting them is small.. That kinda makes sense.. I'll let people who actually know probability and calculus continue their discussion, while I hide from all the infinity.. Macbi. Posts: 941. Joined: Mon Apr 09, 2007 8:32 am UTC. Location: UKvia. ### Re: Triple or Nothing. Silas wrote:. Maybe it will be more clear if I use an example. Call the triple-or-nothing game played for exactly n rounds the game Gn. We already know that the expected payoff for Gn is (3/2)n. Now, call the triple-or-nothing game, when played forever, the game G. In some sense, G is the limit as n -> +∞ of Gn, and we already know that +∞ is the limit as n -> +∞ of (3/2)n. However, there's no reason to expect that this means it is in any sense the expected winnings of the game G is infinite.. I can only point out that the expected winnings E(G) is, in fact, infinite, under standard definitions. P(you collect your winnings) is zero, but you collect infinite winnings. The winnings are, in fact, bigger than the likelihood of collecting them is small. The problem, as I see it, is that zero probability != impossibility.. But it is actually impossible to collect your winnings if you keep betting till you lose.. Indigo is a lie.. Which idiot decided that websites can't go within 4cm of the edge of the screen?. There should be a null word, for the question "Is anybody there?" and to see if microphones are on.. Owehn. Posts: 479. Joined: Tue Oct 09, 2007 12:49 pm UTC. Location: Cambridge, UK. ### Re: Triple or Nothing. Sort of - the game G isn't well defined because you haven't said what happens at stage ∞. Alternatively phrased, the game doesn't end if I win on each round, so what are my winnings in that scenario? It doesn't end up mattering, though, since the probability of that happening is zero: If my strategy is "keep playing till I lose", then in 100% of all games I'll end with a net loss of \$1.. Note the difference between "always" and "100% of the time" (the latter is referred to as "almost always"). It isn't impossible to collect your winnings by winning an infinite number of rounds, but that scenario has probability 0.. [This space intentionally left blank.]. Token. Posts: 1481. Joined: Fri Dec 01, 2006 5:07 pm UTC. Location: London. ### Re: Triple or Nothing. Owehn wrote:Note the difference between "always" and "100% of the time" (the latter is referred to as "almost always"). It isn't impossible to collect your winnings by winning an infinite number of rounds, but that scenario has probability 0.. I'd contest this. Yes, it is possible to win an infinite number of rounds (with probability zero), but I don't see how there's any possible situation that allows you to collect infinite winnings, since by collecting, you have necessarily chosen to stop playing after some finite number of rounds.. All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.. Silas. Posts: 1091. Joined: Sat Feb 02, 2008 9:08 pm UTC. ### Re: Triple or Nothing. When we're speaking with formal terms, ∞ mean 'arbitrarily large,' not 'goes on forever.' We can define G differently, but if you do, you shouldn't expect ∞ to have the same properties. G as the case where you never stop satisfies the naive* understanding of ∞, but it doesn't jive with the way we're using ∞ in (3/2).. * as a math term meaning 'informal', not a denigration of your intellect. Felstaff wrote:Serves you goddamned right. I hope you're happy, Cake Ruiner. Owehn. Posts: 479. Joined: Tue Oct 09, 2007 12:49 pm UTC. Location: Cambridge, UK. ### Re: Triple or Nothing. Token wrote:. Owehn wrote:Note the difference between "always" and "100% of the time" (the latter is referred to as "almost always"). It isn't impossible to collect your winnings by winning an infinite number of rounds, but that scenario has probability 0.. I'd contest this. Yes, it is possible to win an infinite number of rounds (with probability zero), but I don't see how there's any possible situation that allows you to collect infinite winnings, since by collecting, you have necessarily chosen to stop playing after some finite number of rounds.. This is what I was talking about with the game G. As the game is phrased, it's incomplete: what winnings do I get if I flip heads an infinite number of times? You could complete the game by saying that after an infinite sequence of heads, I walk away with the infinite amount of money on the table, or you could complete it by saying I win \$10, or nothing. It doesn't matter - the point is that this scenario occurs with probability zero, so your winnings in this case don't figure into the calculation of expected winnings.. [This space intentionally left blank.]. mikek. Posts: 46. Joined: Mon Nov 26, 2007 5:25 pm UTC. Location: Bristol, UK. ### Re: Triple or Nothing. Actaeus wrote:The expected winnings from each flip is whatever is on the table. Therefore, it is always in the player's best interest to keep playing.. A little late to the party but.. says who?. everything under the sun is in tune. but the sun is eclipsed by the moon.
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# College algebra
posted by Dimpleface
When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -5t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in meters/sec and “k” is the initial height in meters (as if you were on top of a tower or building).
Make up a scenario where a ball is thrown, shot, etc. into the air. You can choose any initial velocity (in meters/sec) and any initial height (in meters) of the ball, but include them in your written scenario. The ball can leave your hand, the top of a building, etc. so you can use many different values for the initial height.
1. veronica
S=-16t^2+v*t+k
## Similar Questions
1. ### algebra
im trying to help my son i have no ida about this stuff any help would be great When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in …
2. ### college algebra
When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec and “k” is …
3. ### algebra
Library Assignment When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec …
4. ### Algebra
When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -10t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in meters/sec and “k” …
5. ### Maths
When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -10t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in meters/sec and “k” …
6. ### college algebra
When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -5t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec and “k” is …
7. ### math
When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec and “k” is …
8. ### math
When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec and “k” is …
9. ### algebra
when a ball is thrown up into the air it makes the shape of a parabola the equation s=-16t^2+v*t+k gives the height nof the ball at any time t in seconds where "v" is the initial velocity (speed)in ft/sec and "k" is the initial height …
10. ### Algebra
A ball is thrown into the air. The height in feet of the ball can be modeled by the equation h = -16t2 + 20t + 6 where t is the time in seconds, the ball is in the air. When will the ball hit the ground?
More Similar Questions
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# College algebra. posted by Dimpleface. When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -5t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in meters/sec and “k” is the initial height in meters (as if you were on top of a tower or building).. Make up a scenario where a ball is thrown, shot, etc. into the air. You can choose any initial velocity (in meters/sec) and any initial height (in meters) of the ball, but include them in your written scenario. The ball can leave your hand, the top of a building, etc. so you can use many different values for the initial height.. 1. veronica. S=-16t^2+v*t+k. ## Similar Questions. 1. ### algebra. im trying to help my son i have no ida about this stuff any help would be great When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in …. 2. ### college algebra. When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec and “k” is …. 3. ### algebra. Library Assignment When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec …. 4. ### Algebra. When a ball is thrown up into the air, it makes the shape of a parabola.
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The equation S= -10t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in meters/sec and “k” …. 5. ### Maths. When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -10t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in meters/sec and “k” …. 6. ### college algebra. When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -5t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec and “k” is …. 7. ### math. When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec and “k” is …. 8. ### math. When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec and “k” is …. 9. ### algebra. when a ball is thrown up into the air it makes the shape of a parabola the equation s=-16t^2+v*t+k gives the height nof the ball at any time t in seconds where "v" is the initial velocity (speed)in ft/sec and "k" is the initial height …. 10. ### Algebra. A ball is thrown into the air. The height in feet of the ball can be modeled by the equation h = -16t2 + 20t + 6 where t is the time in seconds, the ball is in the air. When will the ball hit the ground?. More Similar Questions.
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# 2.6 Other types of equations
Page 1 / 10
In this section you will:
• Solve equations involving rational exponents.
• Solve equations using factoring.
• Solve absolute value equations.
• Solve other types of equations.
We have solved linear equations, rational equations, and quadratic equations using several methods. However, there are many other types of equations, and we will investigate a few more types in this section. We will look at equations involving rational exponents, polynomial equations, radical equations, absolute value equations, equations in quadratic form, and some rational equations that can be transformed into quadratics. Solving any equation, however, employs the same basic algebraic rules. We will learn some new techniques as they apply to certain equations, but the algebra never changes.
## Solving equations involving rational exponents
Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, $\text{\hspace{0.17em}}{16}^{\frac{1}{2}}\text{\hspace{0.17em}}$ is another way of writing $\text{\hspace{0.17em}}\sqrt{16};$ ${8}^{\frac{1}{3}}\text{\hspace{0.17em}}$ is another way of writing $\text{}\text{\hspace{0.17em}}\sqrt[3]{8}.\text{\hspace{0.17em}}$ The ability to work with rational exponents is a useful skill, as it is highly applicable in calculus.
We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, $\text{\hspace{0.17em}}\frac{2}{3}\left(\frac{3}{2}\right)=1,$ $3\left(\frac{1}{3}\right)=1,$ and so on.
## Rational exponents
A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:
${a}^{\frac{m}{n}}={\left({a}^{\frac{1}{n}}\right)}^{m}={\left({a}^{m}\right)}^{\frac{1}{n}}=\sqrt[n]{{a}^{m}}={\left(\sqrt[n]{a}\right)}^{m}$
## Evaluating a number raised to a rational exponent
Evaluate $\text{\hspace{0.17em}}{8}^{\frac{2}{3}}.$
Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite $\text{\hspace{0.17em}}{8}^{\frac{2}{3}}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}{\left({8}^{\frac{1}{3}}\right)}^{2}.$
$\begin{array}{ccc}\hfill {\left({8}^{\frac{1}{3}}\right)}^{2}& =\hfill & {\left(2\right)}^{2}\hfill \\ & =& 4\hfill \end{array}$
Evaluate $\text{\hspace{0.17em}}{64}^{-\frac{1}{3}}.$
$\frac{1}{4}$
## Solve the equation including a variable raised to a rational exponent
Solve the equation in which a variable is raised to a rational exponent: $\text{\hspace{0.17em}}{x}^{\frac{5}{4}}=32.$
The way to remove the exponent on x is by raising both sides of the equation to a power that is the reciprocal of $\text{\hspace{0.17em}}\frac{5}{4},$ which is $\text{\hspace{0.17em}}\frac{4}{5}.$
Solve the equation $\text{\hspace{0.17em}}{x}^{\frac{3}{2}}=125.$
$25$
## Solving an equation involving rational exponents and factoring
Solve $\text{\hspace{0.17em}}3{x}^{\frac{3}{4}}={x}^{\frac{1}{2}}.$
This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero.
$\begin{array}{ccc}\hfill 3{x}^{\frac{3}{4}}-\left({x}^{\frac{1}{2}}\right)& =& {x}^{\frac{1}{2}}-\left({x}^{\frac{1}{2}}\right)\hfill \\ \hfill 3{x}^{\frac{3}{4}}-{x}^{\frac{1}{2}}& =& 0\hfill \end{array}$
Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite $\text{\hspace{0.17em}}{x}^{\frac{1}{2}}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}{x}^{\frac{2}{4}}.\text{\hspace{0.17em}}$ Then, factor out $\text{\hspace{0.17em}}{x}^{\frac{2}{4}}\text{\hspace{0.17em}}$ from both terms on the left.
$\begin{array}{ccc}\hfill 3{x}^{\frac{3}{4}}-{x}^{\frac{2}{4}}& =& 0\hfill \\ \hfill {x}^{\frac{2}{4}}\left(3{x}^{\frac{1}{4}}-1\right)& =& 0\hfill \end{array}$
Where did $\text{\hspace{0.17em}}{x}^{\frac{1}{4}}\text{\hspace{0.17em}}$ come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply $\text{\hspace{0.17em}}{x}^{\frac{2}{4}}\text{\hspace{0.17em}}$ back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to $\text{\hspace{0.17em}}\frac{2}{4}\text{\hspace{0.17em}}$ equals $\text{\hspace{0.17em}}\frac{3}{4}.\text{\hspace{0.17em}}$ Thus, the exponent on x in the parentheses is $\text{\hspace{0.17em}}\frac{1}{4}.\text{\hspace{0.17em}}$
Let us continue. Now we have two factors and can use the zero factor theorem.
The two solutions are $\text{\hspace{0.17em}}0$ and $\frac{1}{81}.$
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
state and prove Cayley hamilton therom
hello
Propessor
hi
Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
Miranda
hi
jai
hi Miranda
jai
thanks
Propessor
welcome
jai
What is algebra
algebra is a branch of the mathematics to calculate expressions follow.
Miranda
Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅
Jeffrey
lolll who told you I'm good at it
Miranda
something seems to wispher me to my ear that u are good at it. lol
Jeffrey
lolllll if you say so
Miranda
but seriously, Im really bad at math. And I hate it. But you see, I downloaded this app two months ago hoping to master it.
Jeffrey
which grade are you in though
Miranda
oh woww I understand
Miranda
Jeffrey
Jeffrey
Miranda
how come you finished in college and you don't like math though
Miranda
gotta practice, holmie
Steve
if you never use it you won't be able to appreciate it
Steve
I don't know why. But Im trying to like it.
Jeffrey
yes steve. you're right
Jeffrey
so you better
Miranda
what is the solution of the given equation?
which equation
Miranda
I dont know. lol
Jeffrey
Miranda
Jeffrey
answer and questions in exercise 11.2 sums
how do u calculate inequality of irrational number?
Alaba
give me an example
Chris
and I will walk you through it
Chris
cos (-z)= cos z .
cos(- z)=cos z
Mustafa
what is a algebra
(x+x)3=?
6x
Obed
what is the identity of 1-cos²5x equal to?
__john __05
Kishu
Hi
Abdel
hi
Ye
hi
Nokwanda
C'est comment
Abdel
Hi
Amanda
hello
SORIE
Hiiii
Chinni
hello
Ranjay
hi
ANSHU
hiiii
Chinni
h r u friends
Chinni
yes
Hassan
so is their any Genius in mathematics here let chat guys and get to know each other's
SORIE
I speak French
Abdel
okay no problem since we gather here and get to know each other
SORIE
hi im stupid at math and just wanna join here
Yaona
lol nahhh none of us here are stupid it's just that we have Fast, Medium, and slow learner bro but we all going to work things out together
SORIE
it's 12
what is the function of sine with respect of cosine , graphically
tangent bruh
Steve
cosx.cos2x.cos4x.cos8x
sinx sin2x is linearly dependent
what is a reciprocal
The reciprocal of a number is 1 divided by a number. eg the reciprocal of 10 is 1/10 which is 0.1
Shemmy
Reciprocal is a pair of numbers that, when multiplied together, equal to 1. Example; the reciprocal of 3 is ⅓, because 3 multiplied by ⅓ is equal to 1
Jeza
each term in a sequence below is five times the previous term what is the eighth term in the sequence
I don't understand how radicals works pls
How look for the general solution of a trig function
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# 2.6 Other types of equations. Page 1 / 10. In this section you will:. • Solve equations involving rational exponents.. • Solve equations using factoring.. • Solve absolute value equations.. • Solve other types of equations.. We have solved linear equations, rational equations, and quadratic equations using several methods. However, there are many other types of equations, and we will investigate a few more types in this section. We will look at equations involving rational exponents, polynomial equations, radical equations, absolute value equations, equations in quadratic form, and some rational equations that can be transformed into quadratics. Solving any equation, however, employs the same basic algebraic rules. We will learn some new techniques as they apply to certain equations, but the algebra never changes.. ## Solving equations involving rational exponents. Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, $\text{\hspace{0.17em}}{16}^{\frac{1}{2}}\text{\hspace{0.17em}}$ is another way of writing $\text{\hspace{0.17em}}\sqrt{16};$ ${8}^{\frac{1}{3}}\text{\hspace{0.17em}}$ is another way of writing $\text{}\text{\hspace{0.17em}}\sqrt[3]{8}.\text{\hspace{0.17em}}$ The ability to work with rational exponents is a useful skill, as it is highly applicable in calculus.. We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, $\text{\hspace{0.17em}}\frac{2}{3}\left(\frac{3}{2}\right)=1,$ $3\left(\frac{1}{3}\right)=1,$ and so on.. ## Rational exponents. A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:. ${a}^{\frac{m}{n}}={\left({a}^{\frac{1}{n}}\right)}^{m}={\left({a}^{m}\right)}^{\frac{1}{n}}=\sqrt[n]{{a}^{m}}={\left(\sqrt[n]{a}\right)}^{m}$. ## Evaluating a number raised to a rational exponent. Evaluate $\text{\hspace{0.17em}}{8}^{\frac{2}{3}}.$. Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite $\text{\hspace{0.17em}}{8}^{\frac{2}{3}}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}{\left({8}^{\frac{1}{3}}\right)}^{2}.$. $\begin{array}{ccc}\hfill {\left({8}^{\frac{1}{3}}\right)}^{2}& =\hfill & {\left(2\right)}^{2}\hfill \\ & =& 4\hfill \end{array}$. Evaluate $\text{\hspace{0.17em}}{64}^{-\frac{1}{3}}.$. $\frac{1}{4}$. ## Solve the equation including a variable raised to a rational exponent. Solve the equation in which a variable is raised to a rational exponent: $\text{\hspace{0.17em}}{x}^{\frac{5}{4}}=32.$. The way to remove the exponent on x is by raising both sides of the equation to a power that is the reciprocal of $\text{\hspace{0.17em}}\frac{5}{4},$ which is $\text{\hspace{0.17em}}\frac{4}{5}.$. Solve the equation $\text{\hspace{0.17em}}{x}^{\frac{3}{2}}=125.$. $25$. ## Solving an equation involving rational exponents and factoring. Solve $\text{\hspace{0.17em}}3{x}^{\frac{3}{4}}={x}^{\frac{1}{2}}.$. This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero.. $\begin{array}{ccc}\hfill 3{x}^{\frac{3}{4}}-\left({x}^{\frac{1}{2}}\right)& =& {x}^{\frac{1}{2}}-\left({x}^{\frac{1}{2}}\right)\hfill \\ \hfill 3{x}^{\frac{3}{4}}-{x}^{\frac{1}{2}}& =& 0\hfill \end{array}$. Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite $\text{\hspace{0.17em}}{x}^{\frac{1}{2}}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}{x}^{\frac{2}{4}}.\text{\hspace{0.17em}}$ Then, factor out $\text{\hspace{0.17em}}{x}^{\frac{2}{4}}\text{\hspace{0.17em}}$ from both terms on the left.. $\begin{array}{ccc}\hfill 3{x}^{\frac{3}{4}}-{x}^{\frac{2}{4}}& =& 0\hfill \\ \hfill {x}^{\frac{2}{4}}\left(3{x}^{\frac{1}{4}}-1\right)& =& 0\hfill \end{array}$. Where did $\text{\hspace{0.17em}}{x}^{\frac{1}{4}}\text{\hspace{0.17em}}$ come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply $\text{\hspace{0.17em}}{x}^{\frac{2}{4}}\text{\hspace{0.17em}}$ back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to $\text{\hspace{0.17em}}\frac{2}{4}\text{\hspace{0.17em}}$ equals $\text{\hspace{0.17em}}\frac{3}{4}.\text{\hspace{0.17em}}$ Thus, the exponent on x in the parentheses is $\text{\hspace{0.17em}}\frac{1}{4}.\text{\hspace{0.17em}}$. Let us continue. Now we have two factors and can use the zero factor theorem.. The two solutions are $\text{\hspace{0.17em}}0$ and $\frac{1}{81}.$. what are you up to?. nothing up todat yet. Miranda. hi. jai. hello. jai. Miranda Drice. jai. aap konsi country se ho. jai. which language is that. Miranda. I am living in india. jai. good. Miranda. what is the formula for calculating algebraic. I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it. Miranda. state and prove Cayley hamilton therom. hello. Propessor. hi. Miranda. the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.. Miranda. hi. jai. hi Miranda. jai. thanks. Propessor. welcome. jai. What is algebra. algebra is a branch of the mathematics to calculate expressions follow.. Miranda. Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅. Jeffrey. lolll who told you I'm good at it. Miranda. something seems to wispher me to my ear that u are good at it. lol. Jeffrey. lolllll if you say so. Miranda. but seriously, Im really bad at math.
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And I hate it. But you see, I downloaded this app two months ago hoping to master it.. Jeffrey. which grade are you in though. Miranda. oh woww I understand. Miranda. Jeffrey. Jeffrey. Miranda. how come you finished in college and you don't like math though. Miranda. gotta practice, holmie. Steve. if you never use it you won't be able to appreciate it. Steve. I don't know why. But Im trying to like it.. Jeffrey. yes steve. you're right. Jeffrey. so you better. Miranda. what is the solution of the given equation?. which equation. Miranda. I dont know. lol. Jeffrey. Miranda. Jeffrey. answer and questions in exercise 11.2 sums. how do u calculate inequality of irrational number?. Alaba. give me an example. Chris. and I will walk you through it. Chris. cos (-z)= cos z .. cos(- z)=cos z. Mustafa. what is a algebra. (x+x)3=?. 6x. Obed. what is the identity of 1-cos²5x equal to?. __john __05. Kishu. Hi. Abdel. hi. Ye. hi. Nokwanda. C'est comment. Abdel. Hi. Amanda. hello. SORIE. Hiiii. Chinni. hello. Ranjay. hi. ANSHU. hiiii. Chinni. h r u friends. Chinni. yes. Hassan. so is their any Genius in mathematics here let chat guys and get to know each other's. SORIE. I speak French. Abdel. okay no problem since we gather here and get to know each other. SORIE. hi im stupid at math and just wanna join here. Yaona. lol nahhh none of us here are stupid it's just that we have Fast, Medium, and slow learner bro but we all going to work things out together. SORIE. it's 12. what is the function of sine with respect of cosine , graphically. tangent bruh. Steve. cosx.cos2x.cos4x.cos8x. sinx sin2x is linearly dependent. what is a reciprocal. The reciprocal of a number is 1 divided by a number. eg the reciprocal of 10 is 1/10 which is 0.1. Shemmy. Reciprocal is a pair of numbers that, when multiplied together, equal to 1. Example; the reciprocal of 3 is ⅓, because 3 multiplied by ⅓ is equal to 1. Jeza. each term in a sequence below is five times the previous term what is the eighth term in the sequence. I don't understand how radicals works pls. How look for the general solution of a trig function.
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https://puzzling.stackexchange.com/questions/125190/make-27-using-1-1-1-1/125191
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# Make 27 using 1, 1, 1, 1
Make the number 27 using 4 instances of the digit 1. You must use all 4 digits to make it count.
Here are the allowed operators:
1. Addition (+), subtraction (-), multiplication (*), division (/)
2. Square roots (√) and exponents (^)
3. Decimals (.) and repeating decimals (¯)
4. Negative numbers (-n)
That's the allowed operators. Any other operators cannot be used for this problem. That's all, have fun solving the puzzle!
• Welcome to PSE (Puzzling Stack Exchange)! Jan 23 at 4:49
• 11^11 (but the numbers need to be interpreted as binary numbers, so probably a bit cheating to use a different base than base 10) Jan 23 at 14:44
For starters,
I can make $$.\bar1 = \frac19 = 3^{-2}$$ cheaply, and the target is $$3^3$$, which suggests the formula $$27 = \frac19^{\frac{-3}2}$$
I submit
$$.\bar1^\frac{-1}{\sqrt{.\bar1}+\sqrt{.\bar1}} = \frac19^{-\frac1{\frac23}} = \frac19^{\frac{-3}2} = 27$$
• Great job, i had a different solution in mind but this is still correct. Jan 21 at 3:48
• @AlejandroGarcia $9\sqrt 9$, perhaps? Jan 21 at 4:45
• No, it's (1/9)^-3. Jan 21 at 6:15
$$3/(1/9)=$$
$$\frac{1+1+1}{.\bar1}$$
• I like this answer the best. To emphasise how elegant it is, I switched your solution to use the other notation for division, which gets rid of all the parentheses too. (If you prefer the original formatting, please do feel free to revert my edit.)
– Bass
Jan 23 at 7:55
• @Bass Lovely edit! Jan 24 at 9:23
Using repeating decimals, $$\ 27 =$$
$$\Large \frac{1}{.\bar{1}} \times \sqrt{\frac{1}{.\bar{1}}}$$
Another solution based on a comment by OP, $$\ 27 =$$
$$\left(\sqrt{.\bar{1}}\right)^{-(1+1+1)}$$
Using
$$\sqrt{1\over.\bar1} = 3$$
the first thing that came to my mind was:
$$\left(\sqrt{1\over.\bar1}\right)^\sqrt{1\over.\bar1}$$
It may not be the simplest solution, but I like the reuse of the core structure.
• +1 Very elegant! Jan 23 at 4:12
• you puzzled frac with overline
– Timo
Jan 24 at 9:10
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# Make 27 using 1, 1, 1, 1. Make the number 27 using 4 instances of the digit 1. You must use all 4 digits to make it count.. Here are the allowed operators:. 1. Addition (+), subtraction (-), multiplication (*), division (/). 2. Square roots (√) and exponents (^). 3. Decimals (.) and repeating decimals (¯). 4. Negative numbers (-n). That's the allowed operators. Any other operators cannot be used for this problem. That's all, have fun solving the puzzle!. • Welcome to PSE (Puzzling Stack Exchange)! Jan 23 at 4:49. • 11^11 (but the numbers need to be interpreted as binary numbers, so probably a bit cheating to use a different base than base 10) Jan 23 at 14:44. For starters,. I can make $$.\bar1 = \frac19 = 3^{-2}$$ cheaply, and the target is $$3^3$$, which suggests the formula $$27 = \frac19^{\frac{-3}2}$$. I submit. $$.\bar1^\frac{-1}{\sqrt{.\bar1}+\sqrt{.\bar1}} = \frac19^{-\frac1{\frac23}} = \frac19^{\frac{-3}2} = 27$$. • Great job, i had a different solution in mind but this is still correct. Jan 21 at 3:48. • @AlejandroGarcia $9\sqrt 9$, perhaps? Jan 21 at 4:45.
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• No, it's (1/9)^-3. Jan 21 at 6:15. $$3/(1/9)=$$. $$\frac{1+1+1}{.\bar1}$$. • I like this answer the best. To emphasise how elegant it is, I switched your solution to use the other notation for division, which gets rid of all the parentheses too. (If you prefer the original formatting, please do feel free to revert my edit.). – Bass. Jan 23 at 7:55. • @Bass Lovely edit! Jan 24 at 9:23. Using repeating decimals, $$\ 27 =$$. $$\Large \frac{1}{.\bar{1}} \times \sqrt{\frac{1}{.\bar{1}}}$$. Another solution based on a comment by OP, $$\ 27 =$$. $$\left(\sqrt{.\bar{1}}\right)^{-(1+1+1)}$$. Using. $$\sqrt{1\over.\bar1} = 3$$. the first thing that came to my mind was:. $$\left(\sqrt{1\over.\bar1}\right)^\sqrt{1\over.\bar1}$$. It may not be the simplest solution, but I like the reuse of the core structure.. • +1 Very elegant! Jan 23 at 4:12. • you puzzled frac with overline. – Timo. Jan 24 at 9:10.
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# Rearrange a linear equation
Lesson
Changing the subject of a formula is an important skill to learn. It can come in very handy when you know the value of one algebraic symbol but not another. Basically, the subject of an equation is the variable that is by itself on one side on the equals sign and it usually is at the start of the formula.
For example, in the formula $A=pb+y$A=pb+y, $A$A is the subject because it is by itself on the left hand side of the equals sign.
We sort of started changing the subject of equations when we learnt to solve equations because we took steps to get the variable by itself, for example we made $x$x the subject of equations. We can make any term in an equation the subject, even if it starts off as the denominator of a fraction. When we're changing the subject of a formula, we often have more than one variable but we still use a similar process.
• Group any like terms
• Simplify using the inverse of addition or subtraction.
• Simplify further by using the inverse of multiplication or division.
#### Examples
##### Question 1
Make $x$x the subject of the following equation:
$y=\frac{x}{4}$y=x4
##### Question 2
Make $m$m the subject of the following equation:
$\frac{m}{y}=gh$my=gh
##### Question 3
Make $m$m the subject of the following equation:
$y=6mx-9$y=6mx9
### Outcomes
#### 10P.LR1.02
Determine the value of a variable in the first degree, using a formula
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# Rearrange a linear equation. Lesson. Changing the subject of a formula is an important skill to learn. It can come in very handy when you know the value of one algebraic symbol but not another. Basically, the subject of an equation is the variable that is by itself on one side on the equals sign and it usually is at the start of the formula.. For example, in the formula $A=pb+y$A=pb+y, $A$A is the subject because it is by itself on the left hand side of the equals sign.. We sort of started changing the subject of equations when we learnt to solve equations because we took steps to get the variable by itself, for example we made $x$x the subject of equations. We can make any term in an equation the subject, even if it starts off as the denominator of a fraction. When we're changing the subject of a formula, we often have more than one variable but we still use a similar process.. • Group any like terms. • Simplify using the inverse of addition or subtraction.. • Simplify further by using the inverse of multiplication or division.. #### Examples. ##### Question 1.
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Make $x$x the subject of the following equation:. $y=\frac{x}{4}$y=x4. ##### Question 2. Make $m$m the subject of the following equation:. $\frac{m}{y}=gh$my=gh. ##### Question 3. Make $m$m the subject of the following equation:. $y=6mx-9$y=6mx9. ### Outcomes. #### 10P.LR1.02. Determine the value of a variable in the first degree, using a formula.
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https://gmatclub.com/forum/when-m-is-divided-by-6-the-remainder-is-3-and-when-n-is-divided-by-294639.html
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When m is divided by 6, the remainder is 3 and when n is divided by 12
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Re: When m is divided by 6, the remainder is 3 and when n is divided by 12 [#permalink]
MBA20 wrote:
When m is divided by 6, the remainder is 3 and when n is divided by 12, the remainder is 9, where m and n are positive integers. Which of the following could be the value of m+n?
I. 72
II. 108
III. 180
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I, II, and III
m is of the form 6k+3 so 3,9,18,27,36,45,54.
n is of the form 12j+9 so 9,21,42,63,84,105,126.
I. 72 - 63+9
II. 108 - 63+45
III. 180 - 126+54
Hence E
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Re: When m is divided by 6, the remainder is 3 and when n is divided by 12 [#permalink]
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Re: When m is divided by 6, the remainder is 3 and when n is divided by 12 [#permalink]
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Last visit was: 23 Jul 2024, 13:56 It is currently 23 Jul 2024, 13:56. Toolkit. GMAT Club Daily Prep. Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.. Customized. for You. we will pick new questions that match your level based on your Timer History. Track. every week, we’ll send you an estimated GMAT score based on your performance. Practice. Pays. we will pick new questions that match your level based on your Timer History. Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.. When m is divided by 6, the remainder is 3 and when n is divided by 12. SORT BY:. Tags:. Show Tags. Hide Tags. Manager. Joined: 30 May 2018. Posts: 64. Own Kudos [?]: 393 [9]. Given Kudos: 75. GMAT 1: 620 Q42 V34. WE:Corporate Finance (Commercial Banking). VP. Joined: 28 Jul 2016. Posts: 1197. Own Kudos [?]: 1752 [0]. Given Kudos: 67. Location: India. Concentration: Finance, Human Resources. Schools: ISB '18 (D). GPA: 3.97. WE:Project Management (Investment Banking). Senior Manager. Joined: 13 Jan 2018. Posts: 277. Own Kudos [?]: 411 [2]. Given Kudos: 20. Location: India. Concentration: Operations, General Management. GMAT 1: 580 Q47 V23. GMAT 2: 640 Q49 V27. GPA: 4. WE:Consulting (Consulting).
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VP. Joined: 18 Dec 2017. Posts: 1161. Own Kudos [?]: 1047 [0]. Given Kudos: 421. Location: United States (KS). GMAT 1: 600 Q46 V27. Re: When m is divided by 6, the remainder is 3 and when n is divided by 12 [#permalink]. MBA20 wrote:. When m is divided by 6, the remainder is 3 and when n is divided by 12, the remainder is 9, where m and n are positive integers. Which of the following could be the value of m+n?. I. 72. II. 108. III. 180. (A) I only. (B) II only. (C) III only. (D) I and II. (E) I, II, and III. m is of the form 6k+3 so 3,9,18,27,36,45,54.. n is of the form 12j+9 so 9,21,42,63,84,105,126.. I. 72 - 63+9. II. 108 - 63+45. III. 180 - 126+54. Hence E. Non-Human User. Joined: 09 Sep 2013. Posts: 34048. Own Kudos [?]: 853 [0]. Given Kudos: 0. Re: When m is divided by 6, the remainder is 3 and when n is divided by 12 [#permalink]. Hello from the GMAT Club BumpBot!. Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).. Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.. Re: When m is divided by 6, the remainder is 3 and when n is divided by 12 [#permalink]. Moderator:. Math Expert. 94589 posts.
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