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# Transformation of Random Variable $Y = -2\ln X$
I'm learning probability, specifically transformations of random variables, and need help with the following exercise:
If the random variable $X$ follows the uniform distribution $U(0, 1)$, find :
$(1)$ The distribution of the random variable $Y = -2 \ln X$;
$(2)$ If the random variables $X_1, X_2, \ldots , X_n$ are independent and follow the uniform distribution $U(0, 1)$, find the distribution of the random variable $Z = \sum_{i=1}^n Y_i$.
Since I'm having difficulties for $(2)$, I'm going to share my work for $(1)$.
$(1)$ First, we note that the function $Y = -2 \ln X$ defined over the interval $0 < x < 1$ is an invertible (decreasing monotonic) function. Taking the exponential on both sides, it is easy to show that the inverse function is
$$x = v(y) = e^{-y/2}.$$
$(Q1)$ What is the range of the above inverse function? When $x = 0$, $y$ is undefined and when $x = 1, y = 0$. So what should I say for $a < y < b$?
Now, taking the derivative of $v(y)$, we get $$v'(y) = -\frac{1}{2}e^{-y/2}.$$
Therefore, by the change-of-variable technique we find the probability distribution function of $Y$ to be
$$f_Y(y) = f_x(v(y)) \times |v'(y)| = \frac{1}{2}e^{-y/2}.$$
This looks to me like the exponential distribution with parameter $\lambda = \frac{1}{2}$ so I'm going to say that the random variable $Y$ follows the exponential distribution with parameter $\lambda = \frac{1}{2}$, i.e. $Y \sim Exp(\lambda = \frac{1}{2})$. But again, what is the support of $y$?
Is my work correct for $(1)$? Can someone help me with the $(Q1)$ that I'm having? For $(2)$ I have absolutely no idea how to solve it. Any help is appreciated, including the theory needed to solve it.
Your work for $(1)$ is correct. For $(2)$ we use the moment generating function method. | {
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Your work for $(1)$ is correct. For $(2)$ we use the moment generating function method.
Recall that if $X_1, X_2, \ldots, X_n$ are observations of a random sample from a population (distribution) with moment generating function $M(t)$, then the moment generating function (M.G.F.) of a linear combination $Y = \sum_{i = 1}^{n} X_i$ is
$$M_Y(t) = \prod_{i = 1}^{n} M(t) = [M(t)]^n.$$
From $(1)$ you've found that the random variables $Y_1, Y_2, \ldots, Y_n$ follow the exponential distribution with parameter $\lambda = 1/2$. To find the distribution of the random variable $Z = \sum_{i = 1}^{n} Y_i$ we use the M.G.F. method described above. One has
$$M_Z(t) = \prod_{i = 1}^{n} M_{Y_i}(t) = [M_{Y_1}(t)]^n \tag{A}$$
where $$M_{Y_1}(t) = \frac{\lambda}{\lambda - t} = \frac{1}{\frac{\lambda - t}{\lambda}} = \frac{1}{1 - \frac{t}{\lambda}} = \big(1 - \frac{t}{\lambda}\big)^{-1} \tag{B}$$
Substituing $(B)$ into $(A)$ yields
$$M_Z(t) = \big[\big(1 - \frac{t}{\lambda}\big)^{-1}\big]^n = \big(1 - \frac{t}{\lambda}\big)^{-n} = \big(1 - \frac{t}{1/2}\big)^{-n} \tag{C}$$
Does $(C)$ look familiar to you? Recall that the moment generator of the Gamma distribution is given by
$$M(t) = \big(1 - \frac{t}{\beta}\big)^{-\alpha}, \, \, t < \alpha.$$
Therefore the random variable $Z$ follows the Gamma distribution with parameters $\alpha = n$ and $b = \frac{1}{2}$. We usually write $Z \sim G(\alpha = n, \beta = 1/2)$.
Let us explicitly write the definitions and stick to basic arguments. The distribution of a variable $X$ is a function $F$: $$F_X(x)=P(X\le x)$$ so for a uniform variable $X$ we have: $$F_X(x)=\begin{cases} 0 & x\le 0 \\ x & 0\le x \le 1 \\ 1 & x\ge 1 \end{cases}$$ | {
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Then, for $Y=-2\ln X$ results: $$F_Y(y)=P(Y\le y)= P(-2\ln X \le y)=P(X\ge e^{-\frac{y}{2}})=1-F_X(e^{-\frac{y}{2}})$$ because the logarithm is monotone. Now, if $x\ge 1$, then $y=-2\ln x\le 0$, so: $$F_Y(y)=\begin{cases} 0 & y\le 0 \\ 1-e^{-\frac{y}{2}} & y\ge 0 \end{cases}$$ This concludes that $Y$ follows an exponential distribution with no need to memorize "change of variables" formulas.
As for (2), once you know that each individual distribution is exponential, their sum is the Gamma distribution: https://stats.stackexchange.com/questions/17424/what-reference-can-i-cite-for-the-proof-that-the-sum-of-n-exponential-variables?rq=1 | {
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Sabarkantha Posts: 5, Reputation: 1 New Member #1 Nov 9, 2013, 09:19 AM
Maths question
For a particular problem, I have arranged the letters A and B twelve by twelve : AAAAAAAAAAAA, ABBAABBAAABA, and so on. Order matters: AB is not equal to BA. Altogether there are 2 raised to 12 or 4096 pemutations giving all possible combinations of A and B, twelve by twelve. Now amongst the 4096 lines of all the combinations, I want to select just the lines which have A six times, like for instance AABBBBBAAABA. How many such lines are there amongst the 4096? I tried the formula 4096!/(4096-6)! but I am not sure that this does the job, since it does not specifically choose lines with six times A.
Thanks for the answer. It is important to me.
ebaines Posts: 12,132, Reputation: 1307 Expert #2 Nov 9, 2013, 10:58 AM
There are $\left( \array 12\\ 6 \array \right)$ ways to place the 6 A's in the twelve letter positions. Thus:
$\left( \array 12\\ 6 \array \right) = \frac {12!/6!}{6!} = \frac {12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1 } =924$
In general the number of ways that 'k' A's can be placed in a string of 'n' letter positions is:
$
\left( \array n\\ k \array \right) = \frac {n (n-1)(n-2)...(n-k+1)}{k(k-1)(k-2)...1} = \frac {n!}{k!(n-k)!}
$
Sabarkantha Posts: 5, Reputation: 1 New Member #3 Nov 9, 2013, 06:28 PM
Originally Posted by ebaines
There are $\left( \array 12\\ 6 \array \right)$ ways to place the 6 A's in the twelve letter positions. Thus:
$\left( \array 12\\ 6 \array \right) = \frac {12!/6!}{6!} = \frac {12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1 } =924$
In general the number of ways that 'k' A's can be placed in a string of 'n' letter positions is: | {
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In general the number of ways that 'k' A's can be placed in a string of 'n' letter positions is:
$
\left( \array n\\ k \array \right) = \frac {n (n-1)(n-2)...(n-k+1)}{k(k-1)(k-2)...1} = \frac {n!}{k!(n-k)!}
$
Thanks a million for your answer. I suppose that this is independent of the total sample of 4096, all the ways you can arrange A and B, twelve by twelve. Since you are very knowledgeable, I want to ask you a further question in order to solve my problem completely. Now that I have reduced my sample to 954, keeping just the lines with six A's, I want to reduce it still further by just keeping the lines where three A's come in a row, side by side: AAA. How manu such lines will there be amongst the 954? Sorry for this further bother.
Sabarkantha Posts: 5, Reputation: 1 New Member #4 Nov 10, 2013, 12:11 AM
Originally Posted by Sabarkantha
Thanks a million for your answer. I suppose that this is independent of the total sample of 4096, all the ways you can arrange A and B, twelve by twelve. Since you are very knowledgeable, I want to ask you a further question in order to solve my problem completely. Now that I have reduced my sample to 954, keeping just the lines with six A's, I want to reduce it still further by just keeping the lines where three A's come in a row, side by side: AAA. How manu such lines will there be amongst the 954? Sorry for this further bother.
Sabarkantha Posts: 5, Reputation: 1 New Member #5 Nov 10, 2013, 12:12 AM
ebaines Posts: 12,132, Reputation: 1307 Expert #6 Nov 10, 2013, 08:31 AM
To clarify your question - I assume you mean that if you get 4 or more A's in a row it's counted as a fail, and if you get two groups of 3 A's each separated by at least one B it's a success. | {
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Start by considering how the three A's are arranged relative to B's:
Case 1. You can have the three A's as the first three letters followed by a B, and you don't care how the remaining 8 letters are arranged:: AAABxxxxxxxx.
Case 2. The first N letters can be anything, followed by the pattern BAAAB, and then the remaining 7-N letters can be anything; for example if N=2: xxBAAABxxxxx. This is valid for N=0 to N=7.
Case 3. The first 8 letters can be anything, followed by BAAA for the last four letters: xxxxxxxxBAAA.
For case 1 the number of ways that the last 8 letters can be arranged, given that they consist of 3 A's and 5 B's is:
$
\left( \array 8 \\ 3 \array \right) = \frac {8 \cdot 7 \cdot 6}{3!} = 56.
$
For case 2 the number of ways that the remaining 7 letters can be arranged, given that there are 3 A's and 4 B's, is:
$
\left( \array 7 \\ 3 \array \right) = \frac {7 \cdot 6 \cdot 5}{3!} = 35.
$
Note that case 2 can occur in 8 different ways, since the starting B can be in position 1, 2, 3, 4, 5, 6, 7, or 8. Thus the total number of ways that case 2 can occur is 8 x 35 = 280.
Finally for case 3 the math is the same as for case 1, so there are 56 ways it can occur.
Thus the total number of arrangements that yield 3 A's in a row, given that you have 6 A's and 6 B's, is 2x56+280 = 392.
Sabarkantha Posts: 5, Reputation: 1 New Member #7 Nov 10, 2013, 09:39 AM
Originally Posted by ebaines
To clarify your question - I assume you mean that if you get 4 or more A's in a row it's counted as a fail, and if you get two groups of 3 A's each separated by at least one B it's a success. | {
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Start by considering how the three A's are arranged relative to B's:
Case 1. You can have the three A's as the first three letters followed by a B, and you don't care how the remaining 8 letters are arranged:: AAABxxxxxxxx.
Case 2. The first N letters can be anything, followed by the pattern BAAAB, and then the remaining 7-N letters can be anything; for example if N=2: xxBAAABxxxxx. This is valid for N=0 to N=7.
Case 3. The first 8 letters can be anything, followed by BAAA for the last four letters: xxxxxxxxBAAA.
For case 1 the number of ways that the last 8 letters can be arranged, given that they consist of 3 A's and 5 B's is:
$
\left( \array 8 \\ 3 \array \right) = \frac {8 \cdot 7 \cdot 6}{3!} = 56.
$
For case 2 the number of ways that the remaining 7 letters can be arranged, given that there are 3 A's and 4 B's, is:
$
\left( \array 7 \\ 3 \array \right) = \frac {7 \cdot 6 \cdot 5}{3!} = 35.
$
Note that case 2 can occur in 8 different ways, since the starting B can be in position 1, 2, 3, 4, 5, 6, 7, or 8. Thus the total number of ways that case 2 can occur is 8 x 35 = 280.
Finally for case 3 the math is the same as for case 1, so there are 56 ways it can occur.
Thus the total number of arrangements that yield 3 A's in a row, given that you have 6 A's and 6 B's, is 2x56+280 = 392.
Thanks once again for the perfectly clear answer. You have understood the problem well. I think that I can work things out myself now.
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# Math Help - Velocity of particle using graph? Physics?
1. ## Velocity of particle using graph? Physics?
I'm not sure how to figure this out:I'm not sure how to find how far the particle with using this graph? Any help? I took physics two years ago before college and I know I learned about this but I can't remember what to do, using either physics or calc methods. Thanks!
2. ## Re: Velocity of particle using graph? Physics?
$a(t) = 3t$
Acceleration is the derivative of velocity: $a(t) = v'(t)$. So:
$\displaystyle v(t) = \int a(t) dt = \dfrac{3}{2}t^2+C$
You are told initial velocity is 5. Initial velocity means t=0. So:
$v(0) = 5 = \dfrac{3}{2}(0)^2 + C$
shows $C=5$.
Hence, $v(t) = \dfrac{3}{2}t^2+5$.
Velocity is the derivative of position: $v(t) = s'(t)$. So, $\displaystyle s(t) = \int v(t)dt$.
Now, you are looking for the distance the particle has moved. That is just $\displaystyle s(3)-s(0) = \int_0^3 v(t)dt$:
$\displaystyle \int_0^3 v(t)dt = \int_0^3 \left(\dfrac{3}{2}t^2+5\right)dt = \left[\dfrac{1}{2}t^3 + 5t\right]_0^3 = 19.5$
3. ## Re: Velocity of particle using graph? Physics?
Originally Posted by SlipEternal
$a(t) = 3t$
Acceleration is the derivative of velocity: $a(t) = v'(t)$. So:
$\displaystyle v(t) = \int a(t) dt = \dfrac{3}{2}t^2+C$
You are told initial velocity is 5. Initial velocity means t=0. So:
$v(0) = 5 = \dfrac{3}{2}(0)^2 + C$
shows $C=5$.
Hence, $v(t) = \dfrac{3}{2}t^2+5$.
Velocity is the derivative of position: $v(t) = s'(t)$. So, $\displaystyle s(t) = \int v(t)dt$.
Now, you are looking for the distance the particle has moved. That is just $\displaystyle s(3)-s(0) = \int_0^3 v(t)dt$:
$\displaystyle \int_0^3 v(t)dt = \int_0^3 \left(\dfrac{3}{2}t^2+5\right)dt = \left[\dfrac{1}{2}t^3 + 5t\right]_0^3 = 19.5$
Thanks for showing the steps, I couldn't put that together because I was so focused on it being a physics problem I didn't even realize I could apply these steps.
4. ## Re: Velocity of particle using graph? Physics? | {
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4. ## Re: Velocity of particle using graph? Physics?
Physics involves a lot of applied math. | {
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# Thread: very interesting problem!!!
1. ## very interesting problem!!!
Hello all, I hope some of you can help me with this problem...
There are 30 total components in the bag. 21 are good and 9 are bad. It takes 6 components to buid one assembly (we can build 5 total assemblies), and components will be pulled randomly out of bag. What is probability for each assembly to have all 6 good components, and therefore be a good assembly (good assebmly has to have all 6 good components).
I tried using (0.7)^6 formula, but that formula accounts for any 6 good components in a row, which is not correct. If we pulled 6 good components in a row, but ther are pulled 3rd to 9th we will
not have a good assembly. Having a good assembly will require pulling good components from 1 to 6, or 7 to 12 and so on.
I tried formula mentioned above, where there is 70% chance to pull a good component out of bag, but this formula works only for infinite sample where cases are indenpendent from eachother. In our case we have 30 components total.
Thanks
2. Originally Posted by Put
There are 30 total components in the bag. 21 are good and 9 are bad. It takes 6 components to buid one assembly (we can build 5 total assemblies), and components will be pulled randomly out of bag. What is probability for each assembly to have all 6 good components, and therefore be a good assembly (good assebmly has to have all 6 good components).
I tried using (0.7)^6 formula, but that formula accounts for any 6 good components in a row, which is not correct. If we pulled 6 good components in a row, but ther are pulled 3rd to 9th we will
not have a good assembly. Having a good assembly will require pulling good components from 1 to 6, or 7 to 12 and so on.
I tried formula mentioned above, where there is 70% chance to pull a good component out of bag, but this formula works only for infinite sample where cases are indenpendent from eachother. In our case we have 30 components total. | {
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Thanks
You are right. Your formula only works if you were somehow taking the components "with replacement" (like some magic leprechaun replacing each part as you removed it. ).
We need to do this without replacement. This is where good old combinations come in. I assume that order doesn't matter.
The probability of having a good assembly will be the total number of ways of having a good assembly (Q1 below), divided by the total number of ways of making an assembly (Q2 below).
Q1: How many ways are there of making a good assembly? More specifically. how many ways are there to take 6 good parts? You need to take 6 good ones from the 21 good ones.
$\mathbf{C}^{21}_6 = \frac{21!}{6!\cdot 15!}=\ldots$
Q2: How many total ways are there to take 6 parts from all 30?
$\mathbf{C}^{30}_6 = \frac{30!}{6!\cdot 24!}=\ldots$
Can you finish it?
3. I calculated the results, but it looks like pretty high.
"12C6" = 924
"9C3" = 84
Q1 = "12C6 * 9C3" = 77616
Q2 = "21C9" = 293930
Q1/Q2 = .264 = 26.4 %
If I used (0.70)^6 formula we get 16.81%, so I'm not sure if 24.4% is right. We should not have more chance to make a good assembly for my case.
Let me know what you think....
4. Well if you were scratching your head at what I did, bravo. I don't know what drugs I was on. I'm sorry for being confusing.
I edited my above answer to be right. When I do out the calculation, I get 0.091 = 9.1%
Good job on being both analytical and skeptical!!!
5. This is a good answer, and thank you very much. Now it looks pretty simple....
I just caclulated a few more cases and it look like that as we increase sample size (bag size), the result will be exponentially aproaching (0.7)^6 = 11.765%, which is infinite sample size where every part is "replaced".
For example, if I take one easier case, where there are 6 total components, 2 bad, and we need 3 good ones to make a good assembly. I'll represent it as 6, 2, 3, and the result is 20%. | {
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If I increase the sample size to 12, 4, 3 (12 total component, 4 bad, 3 to make a good assembly), the result is 25.45%.
if I keep going...
6, 2, 3, = 20%
12, 4, 3, = 25.45%
24, 8, 3, = 27.67%
48, 16, 3, = 28.68%
So the answer is aproaching (0.6667)^3=29,63%, which is case for infinite big sample size where picking components does not change probability any more.
Thanks again for your help.
6. No worries. It's good to see someone using their intuition. | {
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# The operation denoted by the symbol £ is defined for all real numbers
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05 Jul 2018, 04:55
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The operation denoted by the symbol £ is defined for all real numbers a and b as $$a£b = a\sqrt{b}$$. What is the value of 3£(2£4)?
A. $$\frac{1}{4}$$
B. $$4$$
C. $$6$$
D. $$6\sqrt{2}$$
E. $$12$$
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The operation denoted by the symbol £ is defined for all real numbers [#permalink]
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05 Jul 2018, 05:37
Bunuel wrote:
The operation denoted by the symbol £ is defined for all real numbers a and b as $$a£b = a\sqrt{b}$$. What is the value of 3£(2£4)?
A. $$\frac{1}{4}$$
B. $$4$$
C. $$6$$
D. $$6\sqrt{2}$$
E. $$12$$
We know that $$a£b = a\sqrt{b}$$.
We have been asked to find the value of 3£(2£4)
Step 1: Find the value of (2£4) | $$2£4 = 2\sqrt{4} = 4$$
Step 2: Find the value of 3£(4) | $$3£4 = 3\sqrt{4} = 3*2 = 6$$
Therefore, the value of 3£(2£4) is $$6$$(Option C)
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Therefore, the value of 3£(2£4) is $$6$$(Option C)
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Re: The operation denoted by the symbol £ is defined for all real numbers [#permalink]
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05 Jul 2018, 15:12
Bunuel wrote:
The operation denoted by the symbol £ is defined for all real numbers a and b as $$a£b = a\sqrt{b}$$. What is the value of 3£(2£4)?
A. $$\frac{1}{4}$$
B. $$4$$
C. $$6$$
D. $$6\sqrt{2}$$
E. $$12$$
(2£4)=3£(2*$$\sqrt{4}$$)=3£(2*2)=3£4=$$3\sqrt{4}$$=3*2=6
Ans. (C)
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Re: The operation denoted by the symbol £ is defined for all real numbers [#permalink]
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06 Jul 2018, 02:24
Solution
Given:
• £ is an operation defined for all real numbers as a£b = $$a√b$$
To find:
• The value of 3£(2£4)
Approach and Working:
• 3£(2£4) = 3£($$2√4$$) = 3£(2 x 2) = 3£4 = $$3√4$$ = 6
Hence, the correct answer is option C.
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Re: The operation denoted by the symbol £ is defined for all real numbers [#permalink]
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09 Jul 2018, 19:56
Bunuel wrote:
The operation denoted by the symbol £ is defined for all real numbers a and b as $$a£b = a\sqrt{b}$$. What is the value of 3£(2£4)?
A. $$\frac{1}{4}$$
B. $$4$$
C. $$6$$
D. $$6\sqrt{2}$$
E. $$12$$
We work from the “inside out,” so we will first evaluate 2£4 and use that value to finish the problem.
2£4 = 2√4 = 2 x 2 = 4
3£4 = 3√4 = 3 x 2 = 6
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Re: The operation denoted by the symbol £ is defined for all real numbers &nbs [#permalink] 09 Jul 2018, 19:56
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# The operation denoted by the symbol £ is defined for all real numbers
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# Rick wins if it is HT and Tick wins if it is TT. Who wins?
A fair coin is being tossed. Whenever a tail follows a head, Rick wins the game and whenever a tail follows a tail, Tick wins. Who has more probability to win?
Let's say the coin has been tossed more than twice. For Tick to win, the last three tosses should be HTT, but Rick will have won it already. So if the first two flips are not TT, Tick has no chance of winning the game.
I do not know how to proceed from here. I initially planned to brute force since we can obviously compute the probability for $$n$$ tosses (using a simple recursion). But then this observation above kind of made this computation a bit meaningless.
• But then this observation above kind of made this computation a bit meaningless. Yes, it does. So you have observed that Tick wins if and only if the first two flips are both tails. What is your question then? What is preventing you from finishing the problem? – JMoravitz Mar 11 '20 at 14:30
• Can you compute the probability that the first two flips are tails? Then you finished your computation. – Luke Mar 11 '20 at 14:32
• It is worth watching Numberphile's video on Penney's Game as well as reading related questions about Penney's game here on math.se such as this one. – JMoravitz Mar 11 '20 at 14:43
• I feel dumb. That makes sense. – oldsailorpopoye Mar 11 '20 at 16:08 | {
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Suppose the first flip is heads. Now, no matter how many more flips it takes, eventually it will flip tails and Rick will win. So, if the first flip is Heads, Rick wins. That is $$\dfrac{1}{2}$$ of the time.
Suppose the first flip is tails, but the second flip is heads. No matter how many more flips it takes, eventually, it will flip tails a second time, and as soon as it does, you will have $$HT$$ ending the sequence with Rick winning. So, that is an additional $$\dfrac{1}{2}\cdot \dfrac{1}{2}$$ chance that Rick wins.
$$\begin{array}{c|c|c}\text{Flip pattern} & \text{Winner} & \text{Probability} \\ \hline H & \text{Rick} & \dfrac{1}{2} \\ TH & \text{Rick} & \dfrac{1}{4} \\ TT & \text{Tick} & \dfrac{1}{4} \\ \hline \text{Total} & \text{Rick} & \dfrac{3}{4} \\ \text{Total} & \text{Tick} & \dfrac{1}{4} \end{array}$$
As you surmised yourself, the only way for Tick to win is if the very first two throws are tails ... if any of those first two throws is heads, then Rick wins. So, Tick wins with a probability of $$\frac{1}{4}$$, and therefore Rick with a probability of $$\frac{3}{4}$$ | {
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# 4. Linear Algebra¶
## 4.1. Overview¶
Linear algebra is one of the most useful branches of applied mathematics for economists to invest in.
For example, many applied problems in economics and finance require the solution of a linear system of equations, such as
\begin{split} \begin{aligned} y_1 = a x_1 + b x_2 \\ y_2 = c x_1 + d x_2 \end{aligned} \end{split}
or, more generally,
(4.1)\begin{split}\begin{aligned} y_1 = a_{11} x_1 + a_{12} x_2 + \cdots + a_{1k} x_k \\ \vdots \\ y_n = a_{n1} x_1 + a_{n2} x_2 + \cdots + a_{nk} x_k \end{aligned}\end{split}
The objective here is to solve for the “unknowns” $$x_1, \ldots, x_k$$ given $$a_{11}, \ldots, a_{nk}$$ and $$y_1, \ldots, y_n$$.
When considering such problems, it is essential that we first consider at least some of the following questions
• Does a solution actually exist?
• Are there in fact many solutions, and if so how should we interpret them?
• If no solution exists, is there a best “approximate” solution?
• If a solution exists, how should we compute it?
These are the kinds of topics addressed by linear algebra.
In this lecture we will cover the basics of linear and matrix algebra, treating both theory and computation.
We admit some overlap with this lecture, where operations on NumPy arrays were first explained.
Note that this lecture is more theoretical than most, and contains background material that will be used in applications as we go along.
%matplotlib inline
import matplotlib.pyplot as plt
plt.rcParams["figure.figsize"] = (11, 5) #set default figure size
import numpy as np
from matplotlib import cm
from mpl_toolkits.mplot3d import Axes3D
from scipy.interpolate import interp2d
from scipy.linalg import inv, solve, det, eig
## 4.2. Vectors¶
A vector of length $$n$$ is just a sequence (or array, or tuple) of $$n$$ numbers, which we write as $$x = (x_1, \ldots, x_n)$$ or $$x = [x_1, \ldots, x_n]$$.
We will write these sequences either horizontally or vertically as we please. | {
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We will write these sequences either horizontally or vertically as we please.
(Later, when we wish to perform certain matrix operations, it will become necessary to distinguish between the two)
The set of all $$n$$-vectors is denoted by $$\mathbb R^n$$.
For example, $$\mathbb R ^2$$ is the plane, and a vector in $$\mathbb R^2$$ is just a point in the plane.
Traditionally, vectors are represented visually as arrows from the origin to the point.
The following figure represents three vectors in this manner
fig, ax = plt.subplots(figsize=(10, 8))
# Set the axes through the origin
for spine in ['left', 'bottom']:
ax.spines[spine].set_position('zero')
for spine in ['right', 'top']:
ax.spines[spine].set_color('none')
ax.set(xlim=(-5, 5), ylim=(-5, 5))
ax.grid()
vecs = ((2, 4), (-3, 3), (-4, -3.5))
for v in vecs:
ax.annotate('', xy=v, xytext=(0, 0),
arrowprops=dict(facecolor='blue',
shrink=0,
alpha=0.7,
width=0.5))
ax.text(1.1 * v[0], 1.1 * v[1], str(v))
plt.show()
### 4.2.1. Vector Operations¶
The two most common operators for vectors are addition and scalar multiplication, which we now describe.
As a matter of definition, when we add two vectors, we add them element-by-element
$\begin{split} x + y = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} + \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} := \begin{bmatrix} x_1 + y_1 \\ x_2 + y_2 \\ \vdots \\ x_n + y_n \end{bmatrix} \end{split}$
Scalar multiplication is an operation that takes a number $$\gamma$$ and a vector $$x$$ and produces
$\begin{split} \gamma x := \begin{bmatrix} \gamma x_1 \\ \gamma x_2 \\ \vdots \\ \gamma x_n \end{bmatrix} \end{split}$
Scalar multiplication is illustrated in the next figure
fig, ax = plt.subplots(figsize=(10, 8))
# Set the axes through the origin
for spine in ['left', 'bottom']:
ax.spines[spine].set_position('zero')
for spine in ['right', 'top']:
ax.spines[spine].set_color('none') | {
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ax.set(xlim=(-5, 5), ylim=(-5, 5))
x = (2, 2)
ax.annotate('', xy=x, xytext=(0, 0),
arrowprops=dict(facecolor='blue',
shrink=0,
alpha=1,
width=0.5))
ax.text(x[0] + 0.4, x[1] - 0.2, '$x$', fontsize='16')
scalars = (-2, 2)
x = np.array(x)
for s in scalars:
v = s * x
ax.annotate('', xy=v, xytext=(0, 0),
arrowprops=dict(facecolor='red',
shrink=0,
alpha=0.5,
width=0.5))
ax.text(v[0] + 0.4, v[1] - 0.2, f'${s} x$', fontsize='16')
plt.show()
In Python, a vector can be represented as a list or tuple, such as x = (2, 4, 6), but is more commonly represented as a NumPy array.
One advantage of NumPy arrays is that scalar multiplication and addition have very natural syntax
x = np.ones(3) # Vector of three ones
y = np.array((2, 4, 6)) # Converts tuple (2, 4, 6) into array
x + y
array([3., 5., 7.])
4 * x
array([4., 4., 4.])
### 4.2.2. Inner Product and Norm¶
The inner product of vectors $$x,y \in \mathbb R ^n$$ is defined as
$x' y := \sum_{i=1}^n x_i y_i$
Two vectors are called orthogonal if their inner product is zero.
The norm of a vector $$x$$ represents its “length” (i.e., its distance from the zero vector) and is defined as
$\| x \| := \sqrt{x' x} := \left( \sum_{i=1}^n x_i^2 \right)^{1/2}$
The expression $$\| x - y\|$$ is thought of as the distance between $$x$$ and $$y$$.
Continuing on from the previous example, the inner product and norm can be computed as follows
np.sum(x * y) # Inner product of x and y
12.0
np.sqrt(np.sum(x**2)) # Norm of x, take one
1.7320508075688772
np.linalg.norm(x) # Norm of x, take two
1.7320508075688772
### 4.2.3. Span¶
Given a set of vectors $$A := \{a_1, \ldots, a_k\}$$ in $$\mathbb R ^n$$, it’s natural to think about the new vectors we can create by performing linear operations.
New vectors created in this manner are called linear combinations of $$A$$.
In particular, $$y \in \mathbb R ^n$$ is a linear combination of $$A := \{a_1, \ldots, a_k\}$$ if | {
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In particular, $$y \in \mathbb R ^n$$ is a linear combination of $$A := \{a_1, \ldots, a_k\}$$ if
$y = \beta_1 a_1 + \cdots + \beta_k a_k \text{ for some scalars } \beta_1, \ldots, \beta_k$
In this context, the values $$\beta_1, \ldots, \beta_k$$ are called the coefficients of the linear combination.
The set of linear combinations of $$A$$ is called the span of $$A$$.
The next figure shows the span of $$A = \{a_1, a_2\}$$ in $$\mathbb R ^3$$.
The span is a two-dimensional plane passing through these two points and the origin.
fig = plt.figure(figsize=(10, 8))
ax = fig.gca(projection='3d')
x_min, x_max = -5, 5
y_min, y_max = -5, 5
α, β = 0.2, 0.1
ax.set(xlim=(x_min, x_max), ylim=(x_min, x_max), zlim=(x_min, x_max),
xticks=(0,), yticks=(0,), zticks=(0,))
gs = 3
z = np.linspace(x_min, x_max, gs)
x = np.zeros(gs)
y = np.zeros(gs)
ax.plot(x, y, z, 'k-', lw=2, alpha=0.5)
ax.plot(z, x, y, 'k-', lw=2, alpha=0.5)
ax.plot(y, z, x, 'k-', lw=2, alpha=0.5)
# Fixed linear function, to generate a plane
def f(x, y):
return α * x + β * y
# Vector locations, by coordinate
x_coords = np.array((3, 3))
y_coords = np.array((4, -4))
z = f(x_coords, y_coords)
for i in (0, 1):
ax.text(x_coords[i], y_coords[i], z[i], f'$a_{i+1}$', fontsize=14)
# Lines to vectors
for i in (0, 1):
x = (0, x_coords[i])
y = (0, y_coords[i])
z = (0, f(x_coords[i], y_coords[i]))
ax.plot(x, y, z, 'b-', lw=1.5, alpha=0.6) | {
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# Draw the plane
grid_size = 20
xr2 = np.linspace(x_min, x_max, grid_size)
yr2 = np.linspace(y_min, y_max, grid_size)
x2, y2 = np.meshgrid(xr2, yr2)
z2 = f(x2, y2)
ax.plot_surface(x2, y2, z2, rstride=1, cstride=1, cmap=cm.jet,
linewidth=0, antialiased=True, alpha=0.2)
plt.show()
/tmp/ipykernel_56021/4243435577.py:2: MatplotlibDeprecationWarning: Calling gca() with keyword arguments was deprecated in Matplotlib 3.4. Starting two minor releases later, gca() will take no keyword arguments. The gca() function should only be used to get the current axes, or if no axes exist, create new axes with default keyword arguments. To create a new axes with non-default arguments, use plt.axes() or plt.subplot().
ax = fig.gca(projection='3d')
#### 4.2.3.1. Examples¶
If $$A$$ contains only one vector $$a_1 \in \mathbb R ^2$$, then its span is just the scalar multiples of $$a_1$$, which is the unique line passing through both $$a_1$$ and the origin.
If $$A = \{e_1, e_2, e_3\}$$ consists of the canonical basis vectors of $$\mathbb R ^3$$, that is
$\begin{split} e_1 := \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , \quad e_2 := \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} , \quad e_3 := \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \end{split}$
then the span of $$A$$ is all of $$\mathbb R ^3$$, because, for any $$x = (x_1, x_2, x_3) \in \mathbb R ^3$$, we can write
$x = x_1 e_1 + x_2 e_2 + x_3 e_3$
Now consider $$A_0 = \{e_1, e_2, e_1 + e_2\}$$.
If $$y = (y_1, y_2, y_3)$$ is any linear combination of these vectors, then $$y_3 = 0$$ (check it).
Hence $$A_0$$ fails to span all of $$\mathbb R ^3$$.
### 4.2.4. Linear Independence¶
As we’ll see, it’s often desirable to find families of vectors with relatively large span, so that many vectors can be described by linear operators on a few vectors.
The condition we need for a set of vectors to have a large span is what’s called linear independence. | {
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In particular, a collection of vectors $$A := \{a_1, \ldots, a_k\}$$ in $$\mathbb R ^n$$ is said to be
• linearly dependent if some strict subset of $$A$$ has the same span as $$A$$.
• linearly independent if it is not linearly dependent.
Put differently, a set of vectors is linearly independent if no vector is redundant to the span and linearly dependent otherwise.
To illustrate the idea, recall the figure that showed the span of vectors $$\{a_1, a_2\}$$ in $$\mathbb R ^3$$ as a plane through the origin.
If we take a third vector $$a_3$$ and form the set $$\{a_1, a_2, a_3\}$$, this set will be
• linearly dependent if $$a_3$$ lies in the plane
• linearly independent otherwise
As another illustration of the concept, since $$\mathbb R ^n$$ can be spanned by $$n$$ vectors (see the discussion of canonical basis vectors above), any collection of $$m > n$$ vectors in $$\mathbb R ^n$$ must be linearly dependent.
The following statements are equivalent to linear independence of $$A := \{a_1, \ldots, a_k\} \subset \mathbb R ^n$$
1. No vector in $$A$$ can be formed as a linear combination of the other elements.
2. If $$\beta_1 a_1 + \cdots \beta_k a_k = 0$$ for scalars $$\beta_1, \ldots, \beta_k$$, then $$\beta_1 = \cdots = \beta_k = 0$$.
(The zero in the first expression is the origin of $$\mathbb R ^n$$)
### 4.2.5. Unique Representations¶
Another nice thing about sets of linearly independent vectors is that each element in the span has a unique representation as a linear combination of these vectors.
In other words, if $$A := \{a_1, \ldots, a_k\} \subset \mathbb R ^n$$ is linearly independent and
$y = \beta_1 a_1 + \cdots \beta_k a_k$
then no other coefficient sequence $$\gamma_1, \ldots, \gamma_k$$ will produce the same vector $$y$$.
Indeed, if we also have $$y = \gamma_1 a_1 + \cdots \gamma_k a_k$$, then
$(\beta_1 - \gamma_1) a_1 + \cdots + (\beta_k - \gamma_k) a_k = 0$
Linear independence now implies $$\gamma_i = \beta_i$$ for all $$i$$. | {
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Linear independence now implies $$\gamma_i = \beta_i$$ for all $$i$$.
## 4.3. Matrices¶
Matrices are a neat way of organizing data for use in linear operations.
An $$n \times k$$ matrix is a rectangular array $$A$$ of numbers with $$n$$ rows and $$k$$ columns:
$\begin{split} A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1k} \\ a_{21} & a_{22} & \cdots & a_{2k} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nk} \end{bmatrix} \end{split}$
Often, the numbers in the matrix represent coefficients in a system of linear equations, as discussed at the start of this lecture.
For obvious reasons, the matrix $$A$$ is also called a vector if either $$n = 1$$ or $$k = 1$$.
In the former case, $$A$$ is called a row vector, while in the latter it is called a column vector.
If $$n = k$$, then $$A$$ is called square.
The matrix formed by replacing $$a_{ij}$$ by $$a_{ji}$$ for every $$i$$ and $$j$$ is called the transpose of $$A$$ and denoted $$A'$$ or $$A^{\top}$$.
If $$A = A'$$, then $$A$$ is called symmetric.
For a square matrix $$A$$, the $$i$$ elements of the form $$a_{ii}$$ for $$i=1,\ldots,n$$ are called the principal diagonal.
$$A$$ is called diagonal if the only nonzero entries are on the principal diagonal.
If, in addition to being diagonal, each element along the principal diagonal is equal to 1, then $$A$$ is called the identity matrix and denoted by $$I$$.
### 4.3.1. Matrix Operations¶
Just as was the case for vectors, a number of algebraic operations are defined for matrices.
Scalar multiplication and addition are immediate generalizations of the vector case:
$\begin{split} \gamma A = \gamma \begin{bmatrix} a_{11} & \cdots & a_{1k} \\ \vdots & \vdots & \vdots \\ a_{n1} & \cdots & a_{nk} \end{bmatrix} := \begin{bmatrix} \gamma a_{11} & \cdots & \gamma a_{1k} \\ \vdots & \vdots & \vdots \\ \gamma a_{n1} & \cdots & \gamma a_{nk} \end{bmatrix} \end{split}$
and | {
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and
$\begin{split} A + B = \begin{bmatrix} a_{11} & \cdots & a_{1k} \\ \vdots & \vdots & \vdots \\ a_{n1} & \cdots & a_{nk} \end{bmatrix} + \begin{bmatrix} b_{11} & \cdots & b_{1k} \\ \vdots & \vdots & \vdots \\ b_{n1} & \cdots & b_{nk} \end{bmatrix} := \begin{bmatrix} a_{11} + b_{11} & \cdots & a_{1k} + b_{1k} \\ \vdots & \vdots & \vdots \\ a_{n1} + b_{n1} & \cdots & a_{nk} + b_{nk} \end{bmatrix} \end{split}$
In the latter case, the matrices must have the same shape in order for the definition to make sense.
We also have a convention for multiplying two matrices.
The rule for matrix multiplication generalizes the idea of inner products discussed above and is designed to make multiplication play well with basic linear operations.
If $$A$$ and $$B$$ are two matrices, then their product $$A B$$ is formed by taking as its $$i,j$$-th element the inner product of the $$i$$-th row of $$A$$ and the $$j$$-th column of $$B$$.
There are many tutorials to help you visualize this operation, such as this one, or the discussion on the Wikipedia page.
If $$A$$ is $$n \times k$$ and $$B$$ is $$j \times m$$, then to multiply $$A$$ and $$B$$ we require $$k = j$$, and the resulting matrix $$A B$$ is $$n \times m$$.
As perhaps the most important special case, consider multiplying $$n \times k$$ matrix $$A$$ and $$k \times 1$$ column vector $$x$$.
According to the preceding rule, this gives us an $$n \times 1$$ column vector
(4.2)$\begin{split}A x = \begin{bmatrix} a_{11} & \cdots & a_{1k} \\ \vdots & \vdots & \vdots \\ a_{n1} & \cdots & a_{nk} \end{bmatrix} \begin{bmatrix} x_{1} \\ \vdots \\ x_{k} \end{bmatrix} := \begin{bmatrix} a_{11} x_1 + \cdots + a_{1k} x_k \\ \vdots \\ a_{n1} x_1 + \cdots + a_{nk} x_k \end{bmatrix}\end{split}$
Note
$$A B$$ and $$B A$$ are not generally the same thing.
Another important special case is the identity matrix.
You should check that if $$A$$ is $$n \times k$$ and $$I$$ is the $$k \times k$$ identity matrix, then $$AI = A$$. | {
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If $$I$$ is the $$n \times n$$ identity matrix, then $$IA = A$$.
### 4.3.2. Matrices in NumPy¶
NumPy arrays are also used as matrices, and have fast, efficient functions and methods for all the standard matrix operations 1.
You can create them manually from tuples of tuples (or lists of lists) as follows
A = ((1, 2),
(3, 4))
type(A)
tuple
A = np.array(A)
type(A)
numpy.ndarray
A.shape
(2, 2)
The shape attribute is a tuple giving the number of rows and columns — see here for more discussion.
To get the transpose of A, use A.transpose() or, more simply, A.T.
There are many convenient functions for creating common matrices (matrices of zeros, ones, etc.) — see here.
Since operations are performed elementwise by default, scalar multiplication and addition have very natural syntax
A = np.identity(3)
B = np.ones((3, 3))
2 * A
array([[2., 0., 0.],
[0., 2., 0.],
[0., 0., 2.]])
A + B
array([[2., 1., 1.],
[1., 2., 1.],
[1., 1., 2.]])
To multiply matrices we use the @ symbol.
In particular, A @ B is matrix multiplication, whereas A * B is element-by-element multiplication.
See here for more discussion.
### 4.3.3. Matrices as Maps¶
Each $$n \times k$$ matrix $$A$$ can be identified with a function $$f(x) = Ax$$ that maps $$x \in \mathbb R ^k$$ into $$y = Ax \in \mathbb R ^n$$.
These kinds of functions have a special property: they are linear.
A function $$f \colon \mathbb R ^k \to \mathbb R ^n$$ is called linear if, for all $$x, y \in \mathbb R ^k$$ and all scalars $$\alpha, \beta$$, we have
$f(\alpha x + \beta y) = \alpha f(x) + \beta f(y)$
You can check that this holds for the function $$f(x) = A x + b$$ when $$b$$ is the zero vector and fails when $$b$$ is nonzero.
In fact, it’s known that $$f$$ is linear if and only if there exists a matrix $$A$$ such that $$f(x) = Ax$$ for all $$x$$.
## 4.4. Solving Systems of Equations¶
Recall again the system of equations (4.1). | {
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## 4.4. Solving Systems of Equations¶
Recall again the system of equations (4.1).
If we compare (4.1) and (4.2), we see that (4.1) can now be written more conveniently as
(4.3)$y = Ax$
The problem we face is to determine a vector $$x \in \mathbb R ^k$$ that solves (4.3), taking $$y$$ and $$A$$ as given.
This is a special case of a more general problem: Find an $$x$$ such that $$y = f(x)$$.
Given an arbitrary function $$f$$ and a $$y$$, is there always an $$x$$ such that $$y = f(x)$$?
If so, is it always unique?
The answer to both these questions is negative, as the next figure shows
def f(x):
return 0.6 * np.cos(4 * x) + 1.4
xmin, xmax = -1, 1
x = np.linspace(xmin, xmax, 160)
y = f(x)
ya, yb = np.min(y), np.max(y)
fig, axes = plt.subplots(2, 1, figsize=(10, 10))
for ax in axes:
# Set the axes through the origin
for spine in ['left', 'bottom']:
ax.spines[spine].set_position('zero')
for spine in ['right', 'top']:
ax.spines[spine].set_color('none')
ax.set(ylim=(-0.6, 3.2), xlim=(xmin, xmax),
yticks=(), xticks=())
ax.plot(x, y, 'k-', lw=2, label='$f$')
ax.fill_between(x, ya, yb, facecolor='blue', alpha=0.05)
ax.vlines([0], ya, yb, lw=3, color='blue', label='range of $f$')
ax.text(0.04, -0.3, '$0$', fontsize=16)
ax = axes[0]
ax.legend(loc='upper right', frameon=False)
ybar = 1.5
ax.plot(x, x * 0 + ybar, 'k--', alpha=0.5)
ax.text(0.05, 0.8 * ybar, '$y$', fontsize=16)
for i, z in enumerate((-0.35, 0.35)):
ax.vlines(z, 0, f(z), linestyle='--', alpha=0.5)
ax.text(z, -0.2, f'$x_{i}$', fontsize=16)
ax = axes[1]
ybar = 2.6
ax.plot(x, x * 0 + ybar, 'k--', alpha=0.5)
ax.text(0.04, 0.91 * ybar, '$y$', fontsize=16)
plt.show()
In the first plot, there are multiple solutions, as the function is not one-to-one, while in the second there are no solutions, since $$y$$ lies outside the range of $$f$$.
Can we impose conditions on $$A$$ in (4.3) that rule out these problems? | {
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Can we impose conditions on $$A$$ in (4.3) that rule out these problems?
In this context, the most important thing to recognize about the expression $$Ax$$ is that it corresponds to a linear combination of the columns of $$A$$.
In particular, if $$a_1, \ldots, a_k$$ are the columns of $$A$$, then
$Ax = x_1 a_1 + \cdots + x_k a_k$
Hence the range of $$f(x) = Ax$$ is exactly the span of the columns of $$A$$.
We want the range to be large so that it contains arbitrary $$y$$.
As you might recall, the condition that we want for the span to be large is linear independence.
A happy fact is that linear independence of the columns of $$A$$ also gives us uniqueness.
Indeed, it follows from our earlier discussion that if $$\{a_1, \ldots, a_k\}$$ are linearly independent and $$y = Ax = x_1 a_1 + \cdots + x_k a_k$$, then no $$z \not= x$$ satisfies $$y = Az$$.
### 4.4.1. The Square Matrix Case¶
Let’s discuss some more details, starting with the case where $$A$$ is $$n \times n$$.
This is the familiar case where the number of unknowns equals the number of equations.
For arbitrary $$y \in \mathbb R ^n$$, we hope to find a unique $$x \in \mathbb R ^n$$ such that $$y = Ax$$.
In view of the observations immediately above, if the columns of $$A$$ are linearly independent, then their span, and hence the range of $$f(x) = Ax$$, is all of $$\mathbb R ^n$$.
Hence there always exists an $$x$$ such that $$y = Ax$$.
Moreover, the solution is unique.
In particular, the following are equivalent
1. The columns of $$A$$ are linearly independent.
2. For any $$y \in \mathbb R ^n$$, the equation $$y = Ax$$ has a unique solution.
The property of having linearly independent columns is sometimes expressed as having full column rank.
#### 4.4.1.1. Inverse Matrices¶
Can we give some sort of expression for the solution?
If $$y$$ and $$A$$ are scalar with $$A \not= 0$$, then the solution is $$x = A^{-1} y$$.
A similar expression is available in the matrix case. | {
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A similar expression is available in the matrix case.
In particular, if square matrix $$A$$ has full column rank, then it possesses a multiplicative inverse matrix $$A^{-1}$$, with the property that $$A A^{-1} = A^{-1} A = I$$.
As a consequence, if we pre-multiply both sides of $$y = Ax$$ by $$A^{-1}$$, we get $$x = A^{-1} y$$.
This is the solution that we’re looking for.
#### 4.4.1.2. Determinants¶
Another quick comment about square matrices is that to every such matrix we assign a unique number called the determinant of the matrix — you can find the expression for it here.
If the determinant of $$A$$ is not zero, then we say that $$A$$ is nonsingular.
Perhaps the most important fact about determinants is that $$A$$ is nonsingular if and only if $$A$$ is of full column rank.
This gives us a useful one-number summary of whether or not a square matrix can be inverted.
### 4.4.2. More Rows than Columns¶
This is the $$n \times k$$ case with $$n > k$$.
This case is very important in many settings, not least in the setting of linear regression (where $$n$$ is the number of observations, and $$k$$ is the number of explanatory variables).
Given arbitrary $$y \in \mathbb R ^n$$, we seek an $$x \in \mathbb R ^k$$ such that $$y = Ax$$.
In this setting, the existence of a solution is highly unlikely.
Without much loss of generality, let’s go over the intuition focusing on the case where the columns of $$A$$ are linearly independent.
It follows that the span of the columns of $$A$$ is a $$k$$-dimensional subspace of $$\mathbb R ^n$$.
This span is very “unlikely” to contain arbitrary $$y \in \mathbb R ^n$$.
To see why, recall the figure above, where $$k=2$$ and $$n=3$$.
Imagine an arbitrarily chosen $$y \in \mathbb R ^3$$, located somewhere in that three-dimensional space.
What’s the likelihood that $$y$$ lies in the span of $$\{a_1, a_2\}$$ (i.e., the two dimensional plane through these points)? | {
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In a sense, it must be very small, since this plane has zero “thickness”.
As a result, in the $$n > k$$ case we usually give up on existence.
However, we can still seek the best approximation, for example, an $$x$$ that makes the distance $$\| y - Ax\|$$ as small as possible.
To solve this problem, one can use either calculus or the theory of orthogonal projections.
The solution is known to be $$\hat x = (A'A)^{-1}A'y$$ — see for example chapter 3 of these notes.
### 4.4.3. More Columns than Rows¶
This is the $$n \times k$$ case with $$n < k$$, so there are fewer equations than unknowns.
In this case there are either no solutions or infinitely many — in other words, uniqueness never holds.
For example, consider the case where $$k=3$$ and $$n=2$$.
Thus, the columns of $$A$$ consists of 3 vectors in $$\mathbb R ^2$$.
This set can never be linearly independent, since it is possible to find two vectors that span $$\mathbb R ^2$$.
(For example, use the canonical basis vectors)
It follows that one column is a linear combination of the other two.
For example, let’s say that $$a_1 = \alpha a_2 + \beta a_3$$.
Then if $$y = Ax = x_1 a_1 + x_2 a_2 + x_3 a_3$$, we can also write
$y = x_1 (\alpha a_2 + \beta a_3) + x_2 a_2 + x_3 a_3 = (x_1 \alpha + x_2) a_2 + (x_1 \beta + x_3) a_3$
In other words, uniqueness fails.
### 4.4.4. Linear Equations with SciPy¶
Here’s an illustration of how to solve linear equations with SciPy’s linalg submodule.
All of these routines are Python front ends to time-tested and highly optimized FORTRAN code
A = ((1, 2), (3, 4))
A = np.array(A)
y = np.ones((2, 1)) # Column vector
det(A) # Check that A is nonsingular, and hence invertible
-2.0
A_inv = inv(A) # Compute the inverse
A_inv
array([[-2. , 1. ],
[ 1.5, -0.5]])
x = A_inv @ y # Solution
A @ x # Should equal y
array([[1.],
[1.]])
solve(A, y) # Produces the same solution
array([[-1.],
[ 1.]]) | {
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Observe how we can solve for $$x = A^{-1} y$$ by either via inv(A) @ y, or using solve(A, y).
The latter method uses a different algorithm (LU decomposition) that is numerically more stable, and hence should almost always be preferred.
To obtain the least-squares solution $$\hat x = (A'A)^{-1}A'y$$, use scipy.linalg.lstsq(A, y).
## 4.5. Eigenvalues and Eigenvectors¶
Let $$A$$ be an $$n \times n$$ square matrix.
If $$\lambda$$ is scalar and $$v$$ is a non-zero vector in $$\mathbb R ^n$$ such that
$A v = \lambda v$
then we say that $$\lambda$$ is an eigenvalue of $$A$$, and $$v$$ is an eigenvector.
Thus, an eigenvector of $$A$$ is a vector such that when the map $$f(x) = Ax$$ is applied, $$v$$ is merely scaled.
The next figure shows two eigenvectors (blue arrows) and their images under $$A$$ (red arrows).
As expected, the image $$Av$$ of each $$v$$ is just a scaled version of the original
A = ((1, 2),
(2, 1))
A = np.array(A)
evals, evecs = eig(A)
evecs = evecs[:, 0], evecs[:, 1]
fig, ax = plt.subplots(figsize=(10, 8))
# Set the axes through the origin
for spine in ['left', 'bottom']:
ax.spines[spine].set_position('zero')
for spine in ['right', 'top']:
ax.spines[spine].set_color('none')
ax.grid(alpha=0.4)
xmin, xmax = -3, 3
ymin, ymax = -3, 3
ax.set(xlim=(xmin, xmax), ylim=(ymin, ymax))
# Plot each eigenvector
for v in evecs:
ax.annotate('', xy=v, xytext=(0, 0),
arrowprops=dict(facecolor='blue',
shrink=0,
alpha=0.6,
width=0.5))
# Plot the image of each eigenvector
for v in evecs:
v = A @ v
ax.annotate('', xy=v, xytext=(0, 0),
arrowprops=dict(facecolor='red',
shrink=0,
alpha=0.6,
width=0.5))
# Plot the lines they run through
x = np.linspace(xmin, xmax, 3)
for v in evecs:
a = v[1] / v[0]
ax.plot(x, a * x, 'b-', lw=0.4)
plt.show()
The eigenvalue equation is equivalent to $$(A - \lambda I) v = 0$$, and this has a nonzero solution $$v$$ only when the columns of $$A - \lambda I$$ are linearly dependent. | {
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This in turn is equivalent to stating that the determinant is zero.
Hence to find all eigenvalues, we can look for $$\lambda$$ such that the determinant of $$A - \lambda I$$ is zero.
This problem can be expressed as one of solving for the roots of a polynomial in $$\lambda$$ of degree $$n$$.
This in turn implies the existence of $$n$$ solutions in the complex plane, although some might be repeated.
Some nice facts about the eigenvalues of a square matrix $$A$$ are as follows
1. The determinant of $$A$$ equals the product of the eigenvalues.
2. The trace of $$A$$ (the sum of the elements on the principal diagonal) equals the sum of the eigenvalues.
3. If $$A$$ is symmetric, then all of its eigenvalues are real.
4. If $$A$$ is invertible and $$\lambda_1, \ldots, \lambda_n$$ are its eigenvalues, then the eigenvalues of $$A^{-1}$$ are $$1/\lambda_1, \ldots, 1/\lambda_n$$.
A corollary of the first statement is that a matrix is invertible if and only if all its eigenvalues are nonzero.
Using SciPy, we can solve for the eigenvalues and eigenvectors of a matrix as follows
A = ((1, 2),
(2, 1))
A = np.array(A)
evals, evecs = eig(A)
evals
array([ 3.+0.j, -1.+0.j])
evecs
array([[ 0.70710678, -0.70710678],
[ 0.70710678, 0.70710678]])
Note that the columns of evecs are the eigenvectors.
Since any scalar multiple of an eigenvector is an eigenvector with the same eigenvalue (check it), the eig routine normalizes the length of each eigenvector to one.
### 4.5.1. Generalized Eigenvalues¶
It is sometimes useful to consider the generalized eigenvalue problem, which, for given matrices $$A$$ and $$B$$, seeks generalized eigenvalues $$\lambda$$ and eigenvectors $$v$$ such that
$A v = \lambda B v$
This can be solved in SciPy via scipy.linalg.eig(A, B).
Of course, if $$B$$ is square and invertible, then we can treat the generalized eigenvalue problem as an ordinary eigenvalue problem $$B^{-1} A v = \lambda v$$, but this is not always the case.
## 4.6. Further Topics¶ | {
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## 4.6. Further Topics¶
We round out our discussion by briefly mentioning several other important topics.
### 4.6.1. Series Expansions¶
Recall the usual summation formula for a geometric progression, which states that if $$|a| < 1$$, then $$\sum_{k=0}^{\infty} a^k = (1 - a)^{-1}$$.
A generalization of this idea exists in the matrix setting.
#### 4.6.1.1. Matrix Norms¶
Let $$A$$ be a square matrix, and let
$\| A \| := \max_{\| x \| = 1} \| A x \|$
The norms on the right-hand side are ordinary vector norms, while the norm on the left-hand side is a matrix norm — in this case, the so-called spectral norm.
For example, for a square matrix $$S$$, the condition $$\| S \| < 1$$ means that $$S$$ is contractive, in the sense that it pulls all vectors towards the origin 2.
#### 4.6.1.2. Neumann’s Theorem¶
Let $$A$$ be a square matrix and let $$A^k := A A^{k-1}$$ with $$A^1 := A$$.
In other words, $$A^k$$ is the $$k$$-th power of $$A$$.
Neumann’s theorem states the following: If $$\| A^k \| < 1$$ for some $$k \in \mathbb{N}$$, then $$I - A$$ is invertible, and
(4.4)$(I - A)^{-1} = \sum_{k=0}^{\infty} A^k$
#### 4.6.1.3. Spectral Radius¶
A result known as Gelfand’s formula tells us that, for any square matrix $$A$$,
$\rho(A) = \lim_{k \to \infty} \| A^k \|^{1/k}$
Here $$\rho(A)$$ is the spectral radius, defined as $$\max_i |\lambda_i|$$, where $$\{\lambda_i\}_i$$ is the set of eigenvalues of $$A$$.
As a consequence of Gelfand’s formula, if all eigenvalues are strictly less than one in modulus, there exists a $$k$$ with $$\| A^k \| < 1$$.
In which case (4.4) is valid.
### 4.6.2. Positive Definite Matrices¶
Let $$A$$ be a symmetric $$n \times n$$ matrix.
We say that $$A$$ is
1. positive definite if $$x' A x > 0$$ for every $$x \in \mathbb R ^n \setminus \{0\}$$
2. positive semi-definite or nonnegative definite if $$x' A x \geq 0$$ for every $$x \in \mathbb R ^n$$
Analogous definitions exist for negative definite and negative semi-definite matrices. | {
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Analogous definitions exist for negative definite and negative semi-definite matrices.
It is notable that if $$A$$ is positive definite, then all of its eigenvalues are strictly positive, and hence $$A$$ is invertible (with positive definite inverse).
### 4.6.3. Differentiating Linear and Quadratic Forms¶
The following formulas are useful in many economic contexts. Let
• $$z, x$$ and $$a$$ all be $$n \times 1$$ vectors
• $$A$$ be an $$n \times n$$ matrix
• $$B$$ be an $$m \times n$$ matrix and $$y$$ be an $$m \times 1$$ vector
Then
1. $$\frac{\partial a' x}{\partial x} = a$$
2. $$\frac{\partial A x}{\partial x} = A'$$
3. $$\frac{\partial x'A x}{\partial x} = (A + A') x$$
4. $$\frac{\partial y'B z}{\partial y} = B z$$
5. $$\frac{\partial y'B z}{\partial B} = y z'$$
Exercise 1 below asks you to apply these formulas.
### 4.6.4. Further Reading¶
The documentation of the scipy.linalg submodule can be found here.
Chapters 2 and 3 of the Econometric Theory contains a discussion of linear algebra along the same lines as above, with solved exercises.
If you don’t mind a slightly abstract approach, a nice intermediate-level text on linear algebra is [Janich94].
## 4.7. Exercises¶
### 4.7.1. Exercise 1¶
Let $$x$$ be a given $$n \times 1$$ vector and consider the problem
$v(x) = \max_{y,u} \left\{ - y'P y - u' Q u \right\}$
subject to the linear constraint
$y = A x + B u$
Here
• $$P$$ is an $$n \times n$$ matrix and $$Q$$ is an $$m \times m$$ matrix
• $$A$$ is an $$n \times n$$ matrix and $$B$$ is an $$n \times m$$ matrix
• both $$P$$ and $$Q$$ are symmetric and positive semidefinite
(What must the dimensions of $$y$$ and $$u$$ be to make this a well-posed problem?)
One way to solve the problem is to form the Lagrangian
$\mathcal L = - y' P y - u' Q u + \lambda' \left[A x + B u - y\right]$
where $$\lambda$$ is an $$n \times 1$$ vector of Lagrange multipliers. | {
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where $$\lambda$$ is an $$n \times 1$$ vector of Lagrange multipliers.
Try applying the formulas given above for differentiating quadratic and linear forms to obtain the first-order conditions for maximizing $$\mathcal L$$ with respect to $$y, u$$ and minimizing it with respect to $$\lambda$$.
Show that these conditions imply that
1. $$\lambda = - 2 P y$$.
2. The optimizing choice of $$u$$ satisfies $$u = - (Q + B' P B)^{-1} B' P A x$$.
3. The function $$v$$ satisfies $$v(x) = - x' \tilde P x$$ where $$\tilde P = A' P A - A'P B (Q + B'P B)^{-1} B' P A$$.
As we will see, in economic contexts Lagrange multipliers often are shadow prices.
Note
If we don’t care about the Lagrange multipliers, we can substitute the constraint into the objective function, and then just maximize $$-(Ax + Bu)'P (Ax + Bu) - u' Q u$$ with respect to $$u$$. You can verify that this leads to the same maximizer.
## 4.8. Solutions¶
### 4.8.1. Solution to Exercise 1¶
We have an optimization problem:
$v(x) = \max_{y,u} \{ -y'Py - u'Qu \}$
s.t.
$y = Ax + Bu$
with primitives
• $$P$$ be a symmetric and positive semidefinite $$n \times n$$ matrix
• $$Q$$ be a symmetric and positive semidefinite $$m \times m$$ matrix
• $$A$$ an $$n \times n$$ matrix
• $$B$$ an $$n \times m$$ matrix
The associated Lagrangian is:
$L = -y'Py - u'Qu + \lambda' \lbrack Ax + Bu - y \rbrack$
Step 1.
Differentiating Lagrangian equation w.r.t y and setting its derivative equal to zero yields
$\frac{ \partial L}{\partial y} = - (P + P') y - \lambda = - 2 P y - \lambda = 0 \:,$
since P is symmetric.
Accordingly, the first-order condition for maximizing L w.r.t. y implies
$\lambda = -2 Py \:$
Step 2.
Differentiating Lagrangian equation w.r.t. u and setting its derivative equal to zero yields
$\frac{ \partial L}{\partial u} = - (Q + Q') u - B'\lambda = - 2Qu + B'\lambda = 0 \:$
Substituting $$\lambda = -2 P y$$ gives
$Qu + B'Py = 0 \:$ | {
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Substituting $$\lambda = -2 P y$$ gives
$Qu + B'Py = 0 \:$
Substituting the linear constraint $$y = Ax + Bu$$ into above equation gives
$Qu + B'P(Ax + Bu) = 0$
$(Q + B'PB)u + B'PAx = 0$
which is the first-order condition for maximizing $$L$$ w.r.t. $$u$$.
Thus, the optimal choice of u must satisfy
$u = -(Q + B'PB)^{-1}B'PAx \:,$
which follows from the definition of the first-order conditions for Lagrangian equation.
Step 3.
Rewriting our problem by substituting the constraint into the objective function, we get
$v(x) = \max_{u} \{ -(Ax+ Bu)'P(Ax+Bu) - u'Qu \} \:$
Since we know the optimal choice of u satisfies $$u = -(Q + B'PB)^{-1}B'PAx$$, then
$v(x) = -(Ax+ B u)'P(Ax+B u) - u'Q u \,\,\,\, with \,\,\,\, u = -(Q + B'PB)^{-1}B'PAx$
To evaluate the function
\begin{split} \begin{aligned} v(x) &= -(Ax+ B u)'P(Ax+Bu) - u'Q u \\ &= -(x'A' + u'B')P(Ax+Bu) - u'Q u \\ &= - x'A'PAx - u'B'PAx - x'A'PBu - u'B'PBu - u'Qu \\ &= - x'A'PAx - 2u'B'PAx - u'(Q + B'PB) u \end{aligned} \end{split}
For simplicity, denote by $$S := (Q + B'PB)^{-1} B'PA$$, then $$u = -Sx$$.
Regarding the second term $$- 2u'B'PAx$$,
\begin{split} \begin{aligned} -2u'B'PAx &= -2 x'S'B'PAx \\ & = 2 x'A'PB( Q + B'PB)^{-1} B'PAx \end{aligned} \end{split}
Notice that the term $$(Q + B'PB)^{-1}$$ is symmetric as both P and Q are symmetric.
Regarding the third term $$- u'(Q + B'PB) u$$,
\begin{split} \begin{aligned} -u'(Q + B'PB) u &= - x'S' (Q + B'PB)Sx \\ &= -x'A'PB(Q + B'PB)^{-1}B'PAx \end{aligned} \end{split}
Hence, the summation of second and third terms is $$x'A'PB(Q + B'PB)^{-1}B'PAx$$.
This implies that
\begin{split} \begin{aligned} v(x) &= - x'A'PAx - 2u'B'PAx - u'(Q + B'PB) u\\ &= - x'A'PAx + x'A'PB(Q + B'PB)^{-1}B'PAx \\ &= -x'[A'PA - A'PB(Q + B'PB)^{-1}B'PA] x \end{aligned} \end{split}
Therefore, the solution to the optimization problem $$v(x) = -x' \tilde{P}x$$ follows the above result by denoting $$\tilde{P} := A'PA - A'PB(Q + B'PB)^{-1}B'PA$$
1 | {
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1
Although there is a specialized matrix data type defined in NumPy, it’s more standard to work with ordinary NumPy arrays. See this discussion.
2
Suppose that $$\|S \| < 1$$. Take any nonzero vector $$x$$, and let $$r := \|x\|$$. We have $$\| Sx \| = r \| S (x/r) \| \leq r \| S \| < r = \| x\|$$. Hence every point is pulled towards the origin. | {
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# Showing $1+2+\cdots+n=\frac{n(n+1)}{2}$ by induction (stuck on inductive step)
This is from this website:
Use mathematical induction to prove that $$1 + 2 + 3 +\cdots+ n = \frac{n (n + 1)}{2}$$ for all positive integers $n$.
Solution to Problem 1: Let the statement $P(n)$ be $$1 + 2 + 3 + \cdots + n = \frac{n (n + 1)}{2}.$$
STEP 1: We first show that $P(1)$ is true.
Left Side $= 1$
Right Side $= \frac{1 (1 + 1)}{2} = 1$
Both sides of the statement are equal hence $P(1)$ is true.
STEP 2: We now assume that $P(k)$ is true
$$1 + 2 + 3 + \cdots + k = \frac{k (k + 1)}{2}$$ and show that $P(k + 1)$ is true by adding $k + 1$ to both sides of the above statement \begin{align} 1 + 2 + 3 + \cdots + k + (k + 1) &= \frac{k (k + 1)}{2} + (k + 1) \\ &= (k + 1)\left(\frac{k}{2} + 1\right) \\ &= \frac{(k + 1)(k + 2)}{2} \end{align} The last statement may be written as $$1 + 2 + 3 + \cdots + k + (k + 1) = \frac{(k + 1)(k + 2)}{2}$$ Which is the statement $P(k + 1)$.
My question is how in the very last line is the statement $P(k + 1)$ equal to $\frac{(k + 1)(k + 2)}{2}$. I don't get the last step.
• Have you used induction before or is this your first time? – Daniel W. Farlow Mar 25 '15 at 22:00
• Do you understand that $p(k+1) = 1 + 2 + 3 + \cdots + k + (k+1)$? – user137731 Mar 25 '15 at 22:05
• crash:This is my first time – Sam Lot Mar 25 '15 at 22:12
• Bye_world:i understand that we are adding k + 1 to both sides in p(k+1)=1+2+3+⋯+k+(k+1) – Sam Lot Mar 25 '15 at 22:13
Since this is your first time, I'll try to explain it with an emphasis on clarity. If something isn't clear, just comment and I'll try to explain what's happening. | {
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Claim: You are trying to prove the statement $P(n)$ where $$P(n) : 1+2+3+\cdots+n = \frac{n(n+1)}{2}.$$ Your goal is to try to prove this using induction. Proofs by induction usually involve two things: (1) showing that $P(n)$ is true for some fixed value of $n$; this value is oftentimes $n=1$, as it is in your case since you are trying to prove $P(n)$ for all $n\geq 1$. Make sense so far? (2) After you have shown (1) to be true, you then need to assume $P(k)$ to be true for some fixed $k\geq 1$ and then show that $P(k)$ implies $P(k+1)$; that is, you need to show that "if $P(k)$ is true, then $P(k+1)$ is true."
• (1) is called the base case.
• (2) is called the inductive step.
I'll outline the proof below. Let me know if a step doesn't make sense.
Proof. Let $P(n)$ denote the statement $$P(n) : 1+2+3+\cdots+n = \frac{n(n+1)}{2}.$$ Base case ($n=1$): Try to see what happens for $P(1)$. We get that $1 = \frac{1(1+1)}{2}$, and this is true. Thus, the base case holds for $n=1$. Inductive step ($P(k)\to P(k+1)$): Assume $P(k)$ is true for some fixed $k\geq 1$ (this is called the inductive hypothesis). That is, assume $$P(k) : \color{red}{1+2+3+\cdots+k} = \color{green}{\frac{k(k+1)}{2}}\tag{inductive hypothesis}$$ is true. We must show that $P(k+1)$ follows where $$P(k+1) : \underbrace{\color{red}{1+2+3+\cdots+k}+\color{blue}{(k+1)}}_{\text{LHS or "left-hand side"}} = \underbrace{\color{purple}{\frac{(k+1)((k+1)+1)}{2}}}_{\text{RHS or "right-hand side"}}.$$
Side note: Make sure you understand what just happened with $P(k+1)$. For $P(k)$, we just had $1+2+3+\cdots+k$ on the left-hand side. How come we have $1+2+3+\cdots+k+(k+1)$ now for the left-hand side of $P(k+1)$? This is because we are adding another term to the sum, namely $k+1$ (I highlighted this term with blue). On the right-hand side, where $P(k)$ just had $k$ in its expression, we just replace all of those $k$'s with $k+1$ because we are considering $P(k+1)$. Make sense? | {
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Okay. Starting with the left-hand side of $P(k+1)$, we need to show that the right-hand side of $P(k+1)$ follows. Here's how it works: \begin{align} \text{LHS} &= \color{red}{1+2+3+\cdots+k}+\color{blue}{(k+1)}\tag{by definition}\\[1em] &= \color{green}{\frac{k(k+1)}{2}}+\color{blue}{(k+1)}\tag{by inductive hypothesis}\\[1em] &= \frac{\color{green}{k(k+1)}+\color{green}{2}\color{blue}{(k+1)}}{\color{green}{2}}\tag{common denominator}\\[1em] &= \frac{(k+1)\color{green}{(k+2)}}{\color{green}{2}}\tag{group like terms}\\[1em] &= \color{purple}{\frac{(k+1)((k+1)+1)}{2}}\tag{rearrange}\\[1em] &= \text{RHS} \end{align} Thus, we have shown that the right-hand side of $P(k+1)$ follows from the left-hand side of $P(k+1)$. This completes the inductive step.
Thus, by mathematical induction, the statement $P(n)$ is true for all $n\geq 1$. $\blacksquare$
Does it all make sense now?
• What does P in P(n) stand for ? Proof ? – Sam Lot Mar 25 '15 at 23:05
• @SamLot No. $P$ just indicates a statement. I could have used $S(n)$ or $R(n)$ or anything else for that matter. – Daniel W. Farlow Mar 25 '15 at 23:06
• In the inductive step ,what does this "(P(k)→P(k+1))" means ? – Sam Lot Mar 25 '15 at 23:08
• @SamLot It means "$P(k)$ implies $P(k+1)$." The "$\to$" symbol is a mathematical symbol for "implies." – Daniel W. Farlow Mar 25 '15 at 23:09
• It makes more sense now .Thanks – Sam Lot Mar 25 '15 at 23:16
We know that
$P(k) = 1 + 2 + 3 + ... + k$
Therefore:
$P(k+1) = 1 + 2 + 3 + ... + k + (k+1)$
By induction hypothesis we have:
$1 + 2 + 3 + ... + k + (k+1) = \frac{(k+1)(k+2)}{2}$
so
$P(k+1) = 1 + 2 + 3+...+k+(k+1) = \frac{(k+1)(k+2)}{2}$
so
$P(k+1) = \frac{(k+1)(k+2)}{2}$
By induction we now know that since this is true for one integer $k$, it is true for all integers greater than or equal to $k$. | {
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• Rather that "we already proved that $1 + 2 + 3 + ... + k + (k+1) = \frac{(k+1)(k+2)}{2}$" I would say "By induction hypothesis it is $1 + 2 + 3 + ... + k + (k+1) = \frac{(k+1)(k+2)}{2}$". Note that it was not proved. – mfl Mar 25 '15 at 22:15
• How can we say in the first line that "We know that P(k)=1+2+3+...+k" Where did that come from .Secondly how does the formula in last line P(k+1)=(k+1)(k+2)2 prove that 1+2+3+...+n=n(n+1)2 .How can i link them ? – Sam Lot Mar 25 '15 at 22:27
• In the problem $P(n)$ is defined as $1+2+3+...+n$. $P(k+1) = \frac{(k+1)(k+2)}{2}$ proves that $P(k) = \frac{(k)(k+2)}{2}$ because we have shown that it is true for $1$. Say that $k=1$. We know from these formulas that if the statement is true for a number $k$ it is true for $k+1$. Since it is true for $k+1$, it must be true for $(k+1)+1$ and so on and so on. – OriginalOldMan Mar 25 '15 at 22:33
• When you say "We know that P(k)=1+2+3+...+k Therefore: P(k+1)=1+2+3+...+k+(k+1)" in the last line here you are adding 1 to the left side and k + 1 to the right side .Isnt that mathematically incorrect ? – Sam Lot Mar 25 '15 at 22:40
• We are not adding $1$ to both sides. $P(k)$ is basically the sum of integers from $1$ to $k$. So $P(k+1)$ is the sum of integers from $1$ to $(k+1)$. $k$ is obviously the integer before $(k+1)$. So the sum of integers from $1$ to $(k+1)$ is $1+2+3+...+k+(k+1)$. That's why we can say that $P(k+1) = 1+2+3+...+k+(k+1)$. – OriginalOldMan Mar 25 '15 at 22:44
induction is basically saying that if it is true for this step, it is true for the next step. so assuming $1+2+3...+k=k(k+1)/2$, ie it is true for step k, we have to show that it must be true for step k+1, the next step. the final line shows how, by going through some algebra, adding all the numbers up to k+1 equals putting k+1 into the formula $n(n+1)/2$, written as $1+2+3...+k+[k+1]=[k+1]([k+1]+1)/2$. therefore, if it is true for step 1, it is for step 2, and 3... | {
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• I dont understand this part from your explanation "adding all the numbers up to k+1 equals putting k+1 into the formula n(n+1)/2".How does those two things 'equal' each other ? – Sam Lot Mar 25 '15 at 22:20
• you show that they equal each other with the algebra: – stanley dodds Mar 26 '15 at 7:00
• $1+2+3...+k=\frac{k(k+1)}2$, the initial statement. add k+1: – stanley dodds Mar 26 '15 at 7:03
• $1+2+3...+k+[k+1]=\frac{k(k+1)}2+k+1$. then rearrange the rhs: – stanley dodds Mar 26 '15 at 7:06
• multiply on rhs: $=\frac{(k+1)(k+2)}2$ which could also be written as $\frac{[k+1]([k+1]+1)}2$. this is the initially presumed equation for k+1, so with. algebra we have shown that if it works for k it works for k+1, and then if it works for that step it works for the next step, and so on – stanley dodds Mar 26 '15 at 7:16 | {
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# whole numbers and division
Consider the whole number with one thousand digits that can be formed by writing the digits 2772 two hundred and fifty time in succession. Is it divisible by 9? Is it divisible by 11?
-
I answered yes to both because 2772 is divisible by 9 and 11. I am just trying to make sure this is correct! – SNS Feb 27 '12 at 23:19
If the sum of the digits of $n$ is divisible by 9, then (and only then) $n$ is divisible by 9. From this, your number is divisible by 9 (the sum of its digits is $250\cdot18$). I've forgotten the divisibility test for 11... – David Mitra Feb 27 '12 at 23:21
Can you show that your number is divisible by 2772? And can you finish up from there? – Gerry Myerson Feb 27 '12 at 23:26
2772/9=308 2772/11=252 I think that if the original number is divisible by 9 and 11 then if you continue adding the same numbers over and over it should always be divisible by 9 and 11. This is the part I want to double check on. – SNS Feb 27 '12 at 23:29
@DavidMitra If the alternating sum of the digits (add, subtract, add, subtract, etc.) is divisible by 11 then the number is divisible by 11. This is because $10^n\equiv (-1)^n\mod 10$. – Alex Becker Feb 27 '12 at 23:29
The answer is yes; but, in my opinion, you did not give enough information in your comment for a justification.
One way to show it is to use the divisibility tests for 9 and 11.
Let's call your number, obtained by writing "$2772$" two hundred and fifty times in succession, $y$.
A number $n$ is divisible by 9 if and only if the sum of its digits is divisible by 9. The sum of the digits of $y$ is $250\cdot(2+7+7+2)=250(18)$, so $y$ is divisible by 9. | {
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A number is divisible by 11 if and only if the difference of the sum of the odd numbered digits (the first digit, the third digit, ...) and the sum of its even numbered digits is divisible by 11. The sum of the odd numbered digits of $y$ is $250\cdot(2+7)$ and the sum of the even numbered digits of $y$ $250\cdot(7+2)$. The difference between these two quantities is $0$; so $y$ is divisible by 11.
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This explains it very well! Thank you so much. – SNS Feb 27 '12 at 23:42
$2772=99 \times 28$
so
$277227722772\ldots277227722772 = 99 \times 28 \times 100010001\ldots000100010001$
and so is divisible by both $9$ and $11$ (and $4$ and $7$ and other numbers).
-
It's divisible by $9$, $11$, $4$, $7$, and by every divisor of $100010001 \ldots 000100010001$. Just what the divisors of that last number are is not so easy to tell, as far as I can see. It is not divisible by $4$ or $9$, nor by $2$ or $3$. Whether it's divisible by $7$ or by $11$ might be harder to tell. – Michael Hardy Feb 28 '12 at 0:14
@Michael: $(10^{1000}-1)/(10^4 -1) = 41 \times 73 \times 137 \times 251 \times 271 \times 401 \times 751 \times 1201 \times 1601 \times 3541 \times 4001 \times 5051 \times 9091 \times 21001 \times 21401 \times 24001 \times 25601 \times 27961 \times 60101 \times 76001 \times 162251 \times 1 378001 \times 1 610501 \times 1 676321 \times 7 019801 \times 1797 655751 \times 5964 848081 \times 10893 295001 \times 182521 213001 \times 14 103673 319201 \times 78 875943 472201 \times 176 144543 406001 \times 1680 588011 350901 \times \cdots$ – Henry Feb 28 '12 at 0:35
The way you were probably intended to do this problem is to find the sum of the digits (for $9$) and alternating sum and difference (for $11$). And you will undoubtedly need to know these facts about $9$ and $11$ for other problems.
However, the following is true. Suppose that the number formed by a string of digits, like $4718$, is divisible by $m$. For example, $7$ divides $4718$, so let's take $m=7$. | {
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Then $m$ (that is, $7$ here) divides $471847184718$. This is because $$4718471847184718=4718+47180000+471800000000+4718000000000000,$$ and each term on the right-hand side is obviously divisible by $7$, since $4718$ is.
The same argument works for any repetition of the string $4718$, however long it may be, and for any string, and any divisor $m$.
In particular, since $9$ and $11$ each divide $2772$, it follows that each of them divides your thousand digit number. Note that $14$ also divides $2772$, so $14$ divides your thousand digit number.
Your answers were correct, and the procedure that you used turns out to be generally valid. There was somewhat of a lack of explanation, and it is possible that someone grading your work might call it incomplete.
-
On the one hand, the divisibility test for $9$ is to add up the digits and see if they're divisible by $9$. The divisibility test for $11$ is a bit more annoying (if combined with $7$ and $14$)- it comes from recognizing that $1001 = 7 \cdot 11 \cdot 13$, so you take the alternating sum of triplets of digits and see if it's divisible by $11$. Or you could just take the regular alternating sum of digits and see if it's divisible by $11$. Both work here.
On the other hand, the number is $\sum_{k = 0}^{249} (1000)^k \cdot 2772$. So any number that divides $2772$ will divide this new number.
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Your reasoning is correct, but you need to find a way to express clearly and justify that last step, from ‘$2772$ is divisible by $9$ and $11$’ to ‘$27722772\dots2772$ is divisible by $9$ and $11$’. I agree that this is intuitively obvious to anyone who has done long division, but ‘intuitively obvious’ isn’t good enough here. Look at a simpler example first: what about $27722772$?
$27722772=27720000+2772=2772\cdot10^4+2772=2772(10^4+1)$, so it’s clearly a multiple of $2772$ and therefore also of $9$ and $11$. Similarly, | {
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\begin{align*} 277227722772&=277227720000+2772\\ &=2772(10^4+1)\cdot 10^4+2772\\ &=2772(10^{2\cdot 4}+10^4)+2772\\ &=2772(10^{2\cdot 4}+10^4+1)\;, \end{align*}
which is again a multiple of $2772$ and hence also of $9$ and $11$.
And if you already know that a string of $n$ copies of $2772$ is equal to
$$2772(10^{(n-1)\cdot 4}+10^{(n-2)\cdot4}+\dots+10^4+1)\;,$$
then a string of $n+1$ copies must be equal to
\begin{align*} \Big(2772(10^{(n-1)\cdot 4}&+10^{(n-2)\cdot 4}+\dots+10^4+1)\Big)\cdot 10^4+2772\\ &=2772(20^{n\cdot 4}+10^{(n-1)\cdot 4}+\dots+10^{2\cdot4}+10^4)+2772\\ &=2772(20^{n\cdot 4}+10^{(n-1)\cdot 4}+\dots+10^{2\cdot4}+10^4+1)\;, \end{align*}
and the pattern continues. Thus, your string of $250$ copies of $2772$ must be equal to
$$2772(10^{249\cdot4}+10^{249\cdot4}+\dots+10^4+1)$$
and is certainly a multiple of $2772$ and therefore of $9$ and of $11$.
Note that this argument actually does more than is necessary, since it also determines the other factor of your number, namely,
$$10^{249\cdot4}+10^{249\cdot4}+\dots+10^4+1=\frac{10^{250\cdot 4}-1}{10000-1}=\frac{10^{1000}-1}{9999}\;.$$
You could use the same kind of reasoning to argue that tacking a copy of $2772$ on the end of some integer $n$ is simply replacing $n$ by $10000n+2772$, so if $n=2772m$, then $10000n+2772=10000(2772m)+2772=2772(10000m+1)$, which is certainly still a multiple of $2772$. Thus, since tacking a copy of $2772$ on the end always preserves divisibility by $2772$ if it was already present, it doesn’t matter how often you do it: the result is still divisible by $2772$ if the original number was.
(Technically both of these arguments are proofs by mathematical induction, albeit stated rather informally.)
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# A differential Equation -
1. Jan 7, 2008
### Sparky_
1. The problem statement, all variables and given/known data
y'' + 3y' + 2y = sin(e^x)
2. Relevant equations
3. The attempt at a solution
$$y'' + 3y' + 2y = sin(e^x)$$
$$m^2 + 3m + 2 = 0$$
$$m1 = -2 ; m2 = -1$$
$$yc = c1e^{-2x} c2 e^{-x}$$
$$y1 = e^{-2x}$$
$$y1' = -2e^{-2x}$$
$$y2 = e^{-x}$$
$$y2' = -e^{-x}$$
The W Matrix works out to
$$W = e^{-3x}$$
$$u1' = -e^{x}sin(e^x)$$
$$u1 = sin(e^x)$$
$$u2' = e^{2x}sin(e^x)$$
$$u2 = -e^xcos(e^x) + sin(e^x)$$
(This is the integration by parts solution in an earlier posting:
my solution:
$$y = c1e^{-2x} + c2e^{-x} + e^{-2x}sin(e^x) + e^{-x}[-e^xcos(e^x) + sin(e^x)]$$
The book's solution:
$$y = c1e^{-2x} + c2e^{-x} - e^{-2x}sin(e^x)$$
(I'm not in school but this is a problem out of a book - I'm trying to brush up)
Thanks for the help
-Sparky
Last edited: Jan 7, 2008
2. Jan 7, 2008
### mda
It has been a while since I solved DEs analytically, but you should definitely check that your solution does in fact solve the DE (I don't think it does). Also I think the c1c2 term should be a sum, not a product.
3. Jan 7, 2008
### blochwave
Aw man, I tried to solve it with the method of undetermined coefficient and spent way too much time before realizing that doesn't work since the RHS isn't the right form. As solving it by your method is a vague memory now, I was like "well at least I can tell him he got the complementary solution wrong!"
but you corrected it!
I'm sure you've done the usual, like check to make sure your third term isn't miracously minus two times the second term
4. Jan 7, 2008
### Sparky_
I have looked around for terms to cancel and I don't see it but I admit - It could be there, I've gotten the tunnel vision thing after working this back and forth.
I'm hoping for some insight here.
Yes I did correct my error with the first 2 terms - my fault - it was on my paper correctly.
5. Jan 7, 2008
### blochwave | {
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5. Jan 7, 2008
### blochwave
Would it be in any way reasonable to try this with Laplace transforms?
6. Jan 7, 2008
### Sparky_
Part of the reason for this exercise is I'm working back up to Laplace transforms -
I'm working my way back through the book - it's almost become a hobby (pretty sick - right?)
I've used them quite a bit when in school and a little out. (I got my BSEE in 1993 and a Master's in 2003)
I now want to solve this problem with the method of variation of parameters - just because.
I've worked several other problems successfully with this method. - This problem is now bugging me -
I can tell from the answer I'm close
Help?
7. Jan 8, 2008
### Rainbow Child
You have misalculated the integral. There is the right calculation
$$I=-\int e^x\,\sin(e^x)\,d\,x, \quad t=e^x,\,d\,x=\frac{d\,t}{t}$$
so
$$I=-\int t\,\sin(t)\,\frac{d\,t}{t}=-\int \sin(t)\,d\,t=\cos(t)+C\Rightarrow I=\cos(e^x)+C$$
8. Jan 8, 2008
### HallsofIvy
Staff Emeritus
Very good and completely correct. But why not just t= ex[/su] so dt= ex dx since you have a ex in the integral already? I would consider that slightly simpler though it may be just prsonal taste.
9. Jan 8, 2008
### Rainbow Child
Just trying to be analytic!
After all, my personal taste would be
$$I=-\int e^x\,\sin(e^x)\,d\,x=-\int \sin(e^x)\,d\,e^x=\cos(e^x)+C$$
10. Jan 8, 2008
### Sparky_
Well, That's embarassing.
During lunch I will check my paper - I hope I typed it in wrong and it is correct on my paper.
I suspect I didn't and I have egg on my face.
Thank you so much!!
I will pick the problem up with this correction.
Thanks again
-Sparky_
11. Jan 8, 2008
### whiplash
how about the part e^{2x}sin(e^x) , he still has to integrate this does he not?
12. Jan 11, 2008
### Sparky_
Thank you all for the help,
I am embarrassed to say that I did have
$$u1' = -e^{x}sin(e^x)$$
$$u1 = sin(e^x)$$
on my paper. Fixing this error, pointed me to one more careless error. | {
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on my paper. Fixing this error, pointed me to one more careless error.
For completion here is the solution (in case anybody wants it):
I've noticed some threads have a "solved" tag - is that after a problem is complete? If so, does this one get a "solved"?
$$y'' + 3y' + 2y = sin(e^x)$$
$$m^2 + 3m + 2 = 0$$
$$m1 = -2 ; m2 = -1$$
$$yc = c1e^{-2x} c2 e^{-x}$$
$$y1 = e^{-2x}$$
$$y1' = -2e^{-2x}$$
$$y2 = e^{-x}$$
$$y2' = -e^{-x}$$
$$W = e^{-3x}$$ (The W Matrix - )
$$u1' = -\frac{y2f(x)} {W}$$
$$u1' = -e^{2x}sin(e^x)$$
$$u1 = -\int{ e^{2x}sin(e^x)} dx$$ (integration by parts)
$$u1 = e^xcos(e^x) - sin(e^x)$$
$$u2' = e^{x}sin(e^x)$$
$$u2 = -cos(e^x)$$
$$y = c1e^{-2x} + c2e^{-x} + e^{-x}cos(e^x) - e^{-2x}sin(e^x) - e^{-x}cos(e^x)$$
$$y = c1e^{-2x} + c2e^{-x} - e^{-2x}sin(e^x)$$
Last edited: Jan 11, 2008 | {
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# If $4$ distinct balls are placed in $4$ distinct boxes, what is the probability that exactly one box is empty?
Suppose $4$ balls labeled $1,2,3,4$ are randomly placed in boxes $B_1,B_2,B_3,B_4$. The probability that exactly one box is empty is
(a)$8/256$
(b)$9/16$
(c)$27/256$
(d)$9/64$
My approach:
Selecting one empty box out of the four boxes, then placing the $4$ balls in the remaining $3$ boxes.
Thank you .
• My bad yes answer is $\displaystyle \frac{9}{16}$ – DXT Feb 8 '18 at 13:06
• Sorry to say that the answer is 144 – Ishita Roy Aug 8 at 5:20
There are $\binom{4}{1}$ ways to choose the empty box. We must distribute the four balls to the remaining three boxes so that no box is left empty. That means one of those three boxes will receive two balls and the others will each receive one. There are $\binom{3}{1}$ ways to choose the box that will receive two balls and $\binom{4}{2}$ ways to choose which two balls will be placed in that box. The remaining balls can be placed in the remaining two boxes in $2!$ ways. Hence, the number of favorable cases is $$\binom{4}{1}\binom{3}{1}\binom{4}{2}2!$$ Since there are $4$ choices for the placement of each of the four balls, there are $4^4$ ways to distribute four distinct balls to four distinct boxes. Hence, the desired probability is $$\frac{\dbinom{4}{1}\dbinom{3}{1}\dbinom{4}{2}2!}{4^4}$$
• Thank you for your help . – Alphanerd Feb 8 '18 at 13:18
Using the Generalized Inclusion-Exclusion Principle | {
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Using the Generalized Inclusion-Exclusion Principle
Consider $$S(i)$$ to be all arrangements where $$B_i$$ is empty. As in the Generalized Inclusion-Exclusion Principle, let $$N(j)$$ be the the sum of the sizes of all intersections of $$j$$ of the $$S(i)$$: $$N(j)=\sum_{|A|=j}\left|\,\bigcap_{i\in A} S(i)\,\right|$$ In this case, we have $$\binom{4}{j}$$ choices of the $$B_i$$ to be empty and $$(4-j)^4$$ ways to map $$4$$ distinct balls into the $$4-j$$ remaining $$B_i$$: $$N(j)=\binom{4}{j}(4-j)^4$$ The Generalized Inclusion-Exclusion Principle says that the number of arrangements in exactly $$1$$ of the $$S(i)$$ is \begin{align} \sum_{j=1}^4(-1)^{j-1}\binom{j}{1}N(j) &=\binom{1}{1}\binom{4}{1}\,3^4-\binom{2}{1}\binom{4}{2}\,2^4+\binom{3}{1}\binom{4}{3}\,1^4-\binom{4}{1}\binom{4}{4}\,0^4\\ &=324-192+12-0\\[9pt] &=144 \end{align} With a universe of size $$N(0)=256$$, we get a probability of $$\frac{144}{256}=\frac{9}{16}$$
Applying the GIEP to the Approach in the Question
In the approach in the question, after choosing which of the $$4$$ boxes to be empty, we need to compute the number of ways to put $$4$$ distinct balls into $$3$$ distinct boxes with no empty boxes.
Similar to the case above, there are $$\binom{3}{j}$$ ways to choose the empty $$B_i$$ and $$(3-j)^4$$ ways to put the $$4$$ balls into the remaining boxes. That is, $$N(j)=\binom{3}{j}(3-j)^4$$ The Generalized Inclusion-Exclusion Principle says that the number of arrangements in exactly $$0$$ of the $$S(i)$$ is \begin{align} \sum_{j=0}^3(-1)^{j-0}\binom{j}{0}N(j) &=\binom{0}{0}\binom{3}{0}3^4-\binom{1}{0}\binom{3}{1}2^4+\binom{2}{0}\binom{3}{2}1^4-\binom{3}{0}\binom{3}{3}0^4\\ &=81-48+3-0\\[9pt] &=36 \end{align} Combining this with the $$4$$ possibilities for the empty box, we get $$4\cdot36=144$$ arrangements with exactly one empty box, same as above. | {
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Here’s another approach. Name the boxes $a$, $b$, $c$, and $d$. If $S\subseteq\{a,b,c,d\}$, define $e_S$ to be the probability that each box whose name is an element of $S$ is empty after all four balls have been placed. (Boxes not named in $S$ might also be empty.)
Then, using the inclusion-exclusion principle, the probability $p_{\{a\}}$ that only box $a$ is empty is
$p_{\{a\}}=e_{\{a\}}-e_{\{a,b\}}-e_{\{a,c\}}-e_{\{a,d\}}+e_{\{a,b,c\}}+e_{\{a,b,d\}}+e_{\{a,c,d\}}-e_{\{a,b,c,d\}} \\=e_{\{a\}}-3e_{\{a,b\}}+3e_{\{a,b,c\}}-e_{\{a,b,c,d\}}\\={\left(\frac34\right)}^4-3\cdot{\left(\frac24\right)}^4+3\cdot{\left(\frac14\right)}^4-0=\frac9{64}.$
By symmetry, the probability $p_{\{x\}}$ that box $x$ is the only empty box is $\frac9{64}$ for each $x$.
The probability you want is $p_{\{a\}}+p_{\{b\}}+p_{\{c\}}+p_{\{d\}}=4\cdot\frac9{64}=\frac9{16}$. (No inclusion-exclusion is needed for this step because no two different boxes can each be the only empty box.) | {
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# Math Help - Roots of polynomials
1. ## Roots of polynomials
Hi, sorry if this is in the wrong forums not really sure but here it goes..
The quadratic equation $2x^2+4x+3=0$ has roots $\alpha$ and $\beta$ .
Find the value $\alpha^4+\beta^4$
Ive managed to expand to $\alpha^4+\beta^4=(\alpha+\beta)^4-4\alpha^3\beta-6\alpha^2\beta^2-4\alpha\beta^3$
but im not sure how to put in terms of $\alpha\beta$ and $\alpha+\beta$ to work out an answer.
Also the previous part of the question asks to show $\alpha^2+\beta^2=1$ which I managed to prove, is there anyway I can use this to answer the question?
Help really appreciated. Thankyou !
2. Hello, NathanBUK!
The quadratic equation $2x^2+4x+3\:=\:0$ has roots $\alpha$ and $\beta$ .
Find the value of: $\alpha^4+\beta^4$
Also the previous part of the question asks to show $\alpha^2+\beta^2=1$
. . which I managed to prove. . . . . Good!
Is there anyway I can use this to answer the question? . . . . Yes!
We know that: . $\begin{Bmatrix}\alpha + \beta &=& -2 \\ \alpha\beta &=& \frac{3}{2} \end{Bmatrix}$
We have: . $\alpha^2 + \beta^2 \;=\;1$
Square: . $\left(\alpha^2 + \beta^2\right)^2 \;=\;(1)^2$
. . . . $\alpha^4 + 2\alpha^2\beta^2 + \beta^4 \;=\;1$
. . . $\alpha^4 + \beta^4 + 2\!\cdot\!\!\!\underbrace{(\alpha\beta)^2}_{\text{ This is }(\frac{3}{2})^2} \;=\;1$
. . . . . $\alpha^4 + \beta^4 + \frac{9}{2} \;=\;1$
. . . . . . . $\alpha^4 + \beta^4 \;=\;-\frac{7}{2}$
3. Ohh thats makes soo much sense now!! Thankyou so much!!
4. Originally Posted by NathanBUK
Hi, sorry if this is in the wrong forums not really sure but here it goes..
The quadratic equation $2x^2+4x+3=0$ has roots $\alpha$ and $\beta$ .
Find the value $\alpha^4+\beta^4$
Ive managed to expand to $\alpha^4+\beta^4=(\alpha+\beta)^4-4\alpha^3\beta-6\alpha^2\beta^2-4\alpha\beta^3$
but im not sure how to put in terms of $\alpha\beta$ and $\alpha+\beta$ to work out an answer. | {
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but im not sure how to put in terms of $\alpha\beta$ and $\alpha+\beta$ to work out an answer.
Also the previous part of the question asks to show $\alpha^2+\beta^2=1$ which I managed to prove, is there anyway I can use this to answer the question?
Help really appreciated. Thankyou !
Alternative solution (same one in disguise realy):
$\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2(\alpha \beta)^2$
CB | {
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# Getting $p_y(y) = p_x(g^{-1}(y)) \left| \dfrac{\partial{x}}{\partial{y}} \right|$ by solving $| p_y(g(x)) \ dy | = | p_x (x) \ dx |$?
My textbook has a very brief section that introduces some concepts from measure theory:
Another technical detail of continuous variables relates to handling continuous random variables that are deterministic functions of one another. Suppose we have two random variables, $\mathbf{x}$ and $\mathbf{y}$, such that $\mathbf{y} = g(\mathbf{x})$, where $g$ is an invertible, continuous, differentiable transformation. One might expect that $p_y(\mathbf{y}) = p_x(g^{−1} (\mathbf{y}))$. This is actually not the case.
As a simple example, suppose we have scalar random variables $x$ and $y$. Suppose $y = \dfrac{x}{2}$ and $x \sim U(0,1)$. If we use the rule $p_y(y) = p_x(2y)$, then $p_y$ will be $0$ everywhere except the interval $\left[ 0, \dfrac{1}{2} \right]$, and it will be $1$ on this interval. This means
$$\int p_y(y) \ dy = \dfrac{1}{2},$$
which violates the definition of a probability distribution. This is a common mistake. The problem with this approach is that it fails to account for the distortion fo space introduced by the function $g$. Recall that the probability of $\mathbf{x}$ lying in an infinitesimally small region with volume $\delta \mathbf{x}$ is given by $p(\mathbf{x}) \delta \mathbf{x}$. Since $g$ can expand or contract space, the infinitesimal volume surrounding $\mathbf{x}$ in $\mathbf{x}$ space may have different volume in $\mathbf{y}$ space.
To see how to correct the problem, we return to the scalar case. We need to present the property
$$| p_y(g(x)) \ dy | = | p_x (x) \ dx |$$
Solving from this, we obtain
$$p_y(y) = p_x(g^{-1}(y)) \left| \dfrac{\partial{x}}{\partial{y}} \right|$$
or equivalently
$$p_x(x) = p_y(g(x)) \left| \dfrac{\partial{g(x)}}{\partial{x}} \right|$$ | {
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or equivalently
$$p_x(x) = p_y(g(x)) \left| \dfrac{\partial{g(x)}}{\partial{x}} \right|$$
How do they get $p_y(y) = p_x(g^{-1}(y)) \left| \dfrac{\partial{x}}{\partial{y}} \right|$ or equivalently $p_x(x) = p_y(g(x)) \left| \dfrac{\partial{g(x)}}{\partial{x}} \right|$ by solving $| p_y(g(x)) \ dy | = | p_x (x) \ dx |$?
Can someone please demonstrate this and explain the steps?
• this is chain rule + an integration, i.e. integration by substitution – Calvin Khor Sep 8 '18 at 8:37
• @CalvinKhor Can you please demonstrate this step-by-step? – Wyuw Sep 8 '18 at 9:39
• Have you seen e.g. this before? proofwiki.org/wiki/Integration_by_Substitution – Calvin Khor Sep 8 '18 at 9:44
• @CalvinKhor Yes. – Wyuw Sep 8 '18 at 9:50
• I'm not sure if that helps or you want me to put it in the notation you have – Shogun Sep 13 '18 at 15:10
## 2 Answers
$$p_X(x)dx$$ represents the probability measur $$\mathbb{P}_X$$ which is the probability distribution of the random variable $$X$$, it is defined by its action on measurable positive functions by $$\mathbb{E}(f(X))=\int_{\Omega}f(X)d\mathbb{P}=\int_{\mathbb{R}}f(x)d\mathbb{P}_X(x)=\int_{\mathbb{R}}f(x)p_X(x)dx.$$ Now, we consider a new random variable $$Y=g(X)$$, (with some conditions on $$g$$), and we seek $$p_Y$$ the probability density distribution of $$Y$$. So we calculate, for an arbitrary measurable positive function $$f$$ the expectation $$\mathbb{E}(f(Y))$$ in two ways: First, $$\mathbb{E}(f(Y))=\int_{\mathbb{R}}f(y)\color{red}{p_Y(y)dy}\tag1$$ Second, \eqalignno{\mathbb{E}(f(Y))&=\mathbb{E}(f(g(X)))\cr &=\int_{\mathbb{R}}f(g(x))p_X(x)dx\qquad\text{now a change of variables}\cr &=\int_{\mathbb{R}}f(y)\color{red}{p_X(g^{-1}(y))\left|\frac{dx}{dy}\right|dy}&(2) } Now, because $$f$$ is arbitrary, comparing (1) and (2) we get $$p_Y(y)=p_X(x)\left|\frac{dx}{dy}\right|, \quad\text{where y=g(x).}$$ Or, better $$p_Y(y)=p_X(g^{-1}(y))\left|\frac{1}{g’(g^{-1}(y))}\right|\iff p_Y(g(x))|g’(x)|=p_X(x).$$ | {
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This is called the method of transformations. It is detailed on this site. You need to transform a function of a random variable in order to make the CDF equal to $1$. For a demonstration.
Suppose that $X \sim \textrm{Unif}(0,1)$ and let $Y = e^{X}$
Note that cdf of $X$ is given by
F_{X}(x) =\begin{align}\begin{cases} 0 & x < 0 \\ \\ x & 0 \leq x \leq1 \\ 1 & x > 1 \end{cases} \end{align} \tag{1}
Then to find the cdf of $Y$
$$F_{Y}(y) = P(Y \leq y) \\ P(e^{X} \leq y) \\ = P(X \leq \ln(y)) \\ = F_{X}(\ln(y)) = \ln(y) \tag{2}$$
F_{Y}(y) =\begin{align}\begin{cases} 0 & y < 1 \\ \\ \ln(y) & 1 \leq \ln(y) \leq e \\ 1 & x > e \end{cases} \end{align} \tag{3}
To obtain the pdf we take the derivative
f_{Y}(y) = F_{Y}^{'}(y) = \begin{align}\begin{cases} \frac{1}{y} & 1 \leq \ln(y) \leq e \\ 0 & \textrm{ otherwise} \end{cases} \end{align} \tag{4}
Concerning the problem above, suppose that $X \sim \textrm{Unif}(0,1)$ and that $Y = \frac{X}{2}$
The CDF for $X$ is the same as above. Let's look at the cdf of $Y$. We note that $R_{X} =[0,1]$ so then $R_{Y}=[0,\frac{1}{2}]$
$$F_{Y}(y) = P(Y \leq y) \\ P(\frac{X}{2} \leq y) \\ = P(X \leq 2y) \\ = F_{X}(2y) = 2y \tag{5}$$
You are simply taking the reciprocal. To find the pdf, we differenitate.
F_{Y}(y) = \begin{align}\begin{cases} 0 & y< 0 \\ \\ 2y & 0 \leq y \leq \frac{1}{2} \\ 1 & y > \frac{1}{2} \end{cases} \end{align} \tag{6}
to find the pdf
f_{Y}(y) = F_{Y}^{'}(y) = \begin{align}\begin{cases} 2 & 0 \leq y \leq \frac{1}{2} \\ 0 & \textrm{ otherwise} \end{cases} \end{align} \tag{7}
Visually the difference in the two uniform distributions can be seen below.
$$X\sim \textrm{Unif}(0,1) \tag{8}$$
$$X\sim \textrm{Unif}(0,1) , Y = \frac{X}{2} , Y \sim \textrm{Unif}(0,\frac{1}{2}) \tag{9}$$ | {
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$$X\sim \textrm{Unif}(0,1) , Y = \frac{X}{2} , Y \sim \textrm{Unif}(0,\frac{1}{2}) \tag{9}$$
• It is not clear to me how this part makes sense: F_{Y}(y) =\begin{align}\begin{cases} 0 & y < 1 \\ \\ \ln(y) & 1 \leq \ln(y) \leq e \\ 1 & x > e \end{cases} \end{align} \tag{3} – Wyuw Sep 15 '18 at 3:39
• Why is $F_Y(y) = 0$ when $y < 1$? Same with all of the other values. – Wyuw Sep 15 '18 at 3:39
• I'll give you the bounty, since it's running out, but I'll wait before I accept it as the answer. – Wyuw Sep 15 '18 at 3:40
• I may have wrote something wrong. I can fix it. Let me look back over it. – Shogun Sep 15 '18 at 3:41
• we are taking it from $[1,e]$ to $[0,1]$ – Shogun Sep 15 '18 at 3:52 | {
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share. Proposition If the inverse of a matrix exists, then it is unique. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. eralization of the inverse of a matrix. Proof In the proof that a matrix is invertible if and only if it is full-rank, we have shown that the inverse can be constructed column by column, by finding the vectors that solve that is, by writing the vectors of the canonical basis as linear combinations of the columns of . Left-cancellative Loop (algebra) , an algebraic structure with identity element where every element has a unique left and right inverse Retraction (category theory) , a left inverse of some morphism left A rectangular matrix can’t have a two sided inverse because either that matrix or its transpose has a nonzero nullspace. inverse. endstream endobj startxref Note that other left inverses (for example, A¡L = [3; ¡1]) satisfy properties (P1), (P2), and (P4) but not (P3). In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Yes. In fact, if a function has a left inverse and a right inverse, they are both the same two-sided inverse, so it can be called the inverse. 8 0 obj One consequence of (1.2) is that AGAG=AG and GAGA=GA. Let $f \colon X \longrightarrow Y$ be a function. (Generalized inverses are unique is you impose more conditions on G; see Section 3 below.) g = finverse(f) returns the inverse of function f, such that f(g(x)) = x. h�bbdb� �� �9D�H�_ ��Dj*�HE�8�,�&f��L[�z�H�W��� ����HU{��Z �(� �� ��A��O0� lZ'����{,��.�l�\��@���OL@���q����� ��� Let e e e be the identity. (An example of a function with no inverse on either side is the zero transformation on .) This is generally justified because in most applications (e.g., all examples in this article) associativity holds, which makes this notion a generalization of the left/right inverse relative to an identity. For any elements a, b, c, x ∈ G we have: 1. Still another characterization of A+ is given in the following theorem whose proof can be | {
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have: 1. Still another characterization of A+ is given in the following theorem whose proof can be found on p. 19 in Albert, A., Regression and the Moore-Penrose Pseudoinverse, Aca-demic Press, New York, 1972. Show Instructions. 3. Matrix Multiplication Notation. Recall that $B$ is the inverse matrix if it satisfies $AB=BA=I,$ where $I$ is the identity matrix. Matrix inverses Recall... De nition A square matrix A is invertible (or nonsingular) if 9matrix B such that AB = I and BA = I. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Generalized inverses can be defined in any mathematical structure that involves associative multiplication, that is, in a semigroup.This article describes generalized inverses of a matrix. %%EOF Proof: Assume rank(A)=r. As f is a right inverse to g, it is a full inverse to g. So, f is an inverse to f is an inverse to stream However we will now see that when a function has both a left inverse and a right inverse, then all inverses for the function must agree: Lemma 1.11. u(b_1,b_2,b_3,\ldots) = (b_2,b_3,\ldots). If the function is one-to-one, there will be a unique inverse. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). If f contains more than one variable, use the next syntax to specify the independent variable. JOURNAL OF ALGEBRA 31, 209-217 (1974) Right (Left) Inverse Semigroups P. S. VENKATESAN National College, Tiruchy, India and Department of Mathematics, University of Ibadan, Ibadan, Nigeria Communicated by G. B. Preston Received September 7, 1970 A semigroup S (with zero) is called a right inverse semigroup if every (nonnull) principal left ideal of S has a unique idempotent … Theorem. In matrix algebra, the inverse of a matrix is defined only for square matrices, and if a matrix is singular, it does not have | {
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of a matrix is defined only for square matrices, and if a matrix is singular, it does not have an inverse.. Remark When A is invertible, we denote its inverse … Suppose that there are two inverse matrices $B$ and $C$ of the matrix $A$. The Moore-Penrose pseudoinverse is deflned for any matrix and is unique. New comments cannot be posted and votes cannot be cast. %���� wqhh��llf�)eK�y�I��bq�(�����Ã.4-�{xe��8������b�c[���ö����TBYb�ʃ4���&�1����o[{cK�sAt�������3�'vp=�$��$�i.��j8@�g�UQ���>��g�lI&�OuL��*���wCu�0 �]l� One of its left inverses is the reverse shift operator u (b 1, b 2, b 3, …) = (b 2, b 3, …). Two-sided inverse is unique if it exists in monoid 2. In a monoid, if an element has a right inverse… /Filter /FlateDecode << /S /GoTo /D [9 0 R /Fit ] >> Theorem A.63 A generalized inverse always exists although it is not unique in general. (We say B is an inverse of A.) It would therefore seem logicalthat when working with matrices, one could take the matrix equation AX=B and divide bothsides by A to get X=B/A.However, that won't work because ...There is NO matrix division!Ok, you say. Theorem 2.16 First Gyrogroup Properties. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Thus both AG and GA are projection matrices. See the lecture notesfor the relevant definitions. A.12 Generalized Inverse Definition A.62 Let A be an m × n-matrix. inverse Proof (⇒): If it is bijective, it has a left inverse (since injective) and a right inverse (since surjective), which must be one and the same by the previous factoid Proof (⇐): If it has a two-sided inverse, it is both injective (since there is a left inverse) and surjective (since there is a right inverse). There are three optional outputs in addition to the unique elements: 53 0 obj <> endobj Then 1 (AB) ij = A i B j, 2 (AB) i = A i B, 3 (AB) j = AB j, 4 (ABC) ij = A i BC j. u (b 1 , b 2 , b 3 , …) = (b 2 , b 3 , …). endstream endobj 54 0 obj <> endobj 55 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/Thumb | {
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(b 2 , b 3 , …). endstream endobj 54 0 obj <> endobj 55 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/Thumb 26 0 R/TrimBox[79.51181 97.228348 518.881897 763.370056]/Type/Page>> endobj 56 0 obj <>stream Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. If the function is one-to-one, there will be a unique inverse. Let (G, ⊕) be a gyrogroup. Note that other left 5 For any m n matrix A, we have A i = eT i A and A j = Ae j. P. Sam Johnson (NITK) Existence of Left/Right/Two-sided Inverses September 19, 2014 3 / 26 By using this website, you agree to our Cookie Policy. /Length 1425 Hello! Left inverse if and only if right inverse We now want to use the results above about solutions to Ax = b to show that a square matrix A has a left inverse if and only if it has a right inverse. The equation Ax = b always has at least one solution; the nullspace of A has dimension n − m, so there will be The left inverse tells you how to exactly retrace your steps, if you managed to get to a destination – “Some places might be unreachable, but I can always put you on the return flight” The right inverse tells you where you might have come from, for any possible destination – “All places are reachable, but I can't put you on the I know that left inverses are unique if the function is surjective but I don't know if left inverses are always unique for non-surjective functions too. h�b�y��� cca�� ����ِ� q���#�!�A�ѬQ�a���[�50�F��3&9'��0 qp�(R�&�a�s4�p�[���f^'w�P& 7��,���[T�+�J����9�$��4r�:4';m$��#�s�Oj�LÌ�cY{-�XTAڽ�BEOpr�l�T��f1�M�1$��С��6I��Ҏ)w In gen-eral, a square matrix P that satisfles P2 = P is called a projection matrix. Then a matrix A−: n × m is said to be a generalized inverse of A if AA−A = A holds (see Rao (1973a, p. 24). Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. The reason why we have to define the left inverse and the right inverse is | {
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are an inverse of $f$. The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. best. '+o�f P0���'�,�\� y����bf\�; wx.��";MY�}����إ� Subtraction was defined in terms of addition and division was defined in terms ofmultiplication. Then t t t has many left inverses but no right inverses (because t t t is injective but not surjective). (4x1�@�y�,(����.�BY��⧆7G�߱Zb�?��,��T��9o��H0�(1q����D� �;:��vK{Y�wY�/���5�����c�iZl�B\\��L�bE���8;�!�#�*)�L�{�M��dUт6���%�V^����ZW��������f�4R�p�p�b��x���.L��1sh��Y�U����! In a monoid, if an element has a left inverse, it can have at most one right inverse; moreover, if the right inverse exists, it must be equal to the left inverse, and is thus a two-sided inverse. When working in the real numbers, the equation ax=b could be solved for x by dividing bothsides of the equation by a to get x=b/a, as long as a wasn't zero. We will later show that for square matrices, the existence of any inverse on either side is equivalent to the existence of a unique two-sided inverse. An associative * on a set G with unique right identity and left inverse proof enough for it to be a group ?Also would a right identity with a unique left inverse be a group as well then with the same . Thus, p is indeed the unique point in U that minimizes the distance from b to any point in U. ��� Sort by. Ask Question Asked 4 years, 10 months ago. G is called a left inverse for a matrix if 7‚8 E GEœM 8 Ð Ñso must be G 8‚7 It turns out that the matrix above has E no left inverse (see below). Hence it is bijective. If A is invertible, then its inverse is unique. This thread is archived. Let $f \colon X \longrightarrow Y$ be a function. Viewed 1k times 3. If a matrix has a unique left inverse then does it necessarily have a unique right inverse (which is the same inverse)? The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not | {
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why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. 36 0 obj << This may make left-handed people more resilient to strokes or other conditions that damage specific brain regions. x��XKo#7��W�hE�[ע��E������:v�4q���/)�c����>~"%��d��N��8�w(LYɽ2L:�AZv�b��ٞѳG���8>����'��x�ټrc��>?��[��?�'���(%#R��1 .�-7�;6�Sg#>Q��7�##ϥ "�[� ���N)&Q ��M���Yy��?A����4�ϠH�%�f��0a;N�M�,�!{��y�<8(t1ƙ�zi���e��A��(;p*����V�Jڛ,�t~�d��̘H9����/��_a���v�68gq"���D�|a5����P|Jv��l1j��x��&N����V"���"����}! numpy.unique¶ numpy.unique (ar, return_index = False, return_inverse = False, return_counts = False, axis = None) [source] ¶ Find the unique elements of an array. This preview shows page 275 - 279 out of 401 pages.. By Proposition 5.15.5, g has a unique right inverse, which is equal to its unique inverse. Right inverse If A has full row rank, then r = m. The nullspace of AT contains only the zero vector; the rows of A are independent. See Also. endobj Actually, trying to prove uniqueness of left inverses leads to dramatic failure! Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. Then a matrix A−: n × m is said to be a generalized inverse of A if AA−A = A holds (see Rao (1973a, p. 24). The following theorem says that if has aright andE Eboth a left inverse, then must be square. A.12 Generalized Inverse Definition A.62 Let A be an m × n-matrix. This is no accident ! From this example we see that even when they exist, one-sided inverses need not be unique. Theorem A.63 A generalized inverse always exists although it is not unique in general. Stack Exchange Network. Then they satisfy $AB=BA=I \tag{*}$ and Active 2 years, 7 months ago. If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. Remark Not all square matrices are invertible. 87 0 obj <>/Filter/FlateDecode/ID[<60DDF7F936364B419866FBDF5084AEDB><33A0036193072C4B9116D6C95BA3C158>]/Index[53 73]/Info 52 0 | {
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73]/Info 52 0 R/Length 149/Prev 149168/Root 54 0 R/Size 126/Type/XRef/W[1 3 1]>>stream 1. f is injective if and only if it has a left inverse 2. f is surjective if and only if it has a right inverse 3. f is bijective if and only if it has a two-sided inverse 4. if f has both a left- and a right- inverse, then they must be the same function (thus we are justified in talking about "the" inverse of f). Show Instructions. An inverse that is both a left and right inverse (a two-sided inverse), if it exists, must be unique. Thus the unique left inverse of A equals the unique right inverse of A from ECE 269 at University of California, San Diego Let f : A → B be a function with a left inverse h : B → A and a right inverse g : B → A. 125 0 obj <>stream If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). example. Let (G, ⊕) be a gyrogroup. Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. It's an interesting exercise that if$a$is a left unit that is not a right uni Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. Recall also that this gives a unique inverse. Outside semigroup theory, a unique inverse as defined in this section is sometimes called a quasi-inverse. 6 comments. LEAST SQUARES PROBLEMS AND PSEUDO-INVERSES 443 Next, for any point y ∈ U,thevectorspy and bp are orthogonal, which implies that #by#2 = #bp#2 +#py#2. Yes. Generalized inverse Michael Friendly 2020-10-29. 11.1. g = finverse(f,var) ... finverse does not issue a warning when the inverse is not unique. Let G G G be a group. Proof: Assume rank(A)=r. So to prove the uniqueness, suppose that you have two inverse matrices$B$and$C$and show that in fact$B=C$. U-semigroups �n�����r����6���d}���wF>�G�/��k� K�T�SE���� �&ʬ�Rbl�j��|�Tx��)��Rdy�Y ? If $$AN= I_n$$, then $$N$$ is called a | {
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K�T�SE���� �&ʬ�Rbl�j��|�Tx��)��Rdy�Y ? If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. %PDF-1.4 Theorem 2.16 First Gyrogroup Properties. 5. the composition of two injective functions is injective 6. the composition of two surjective functions is surjective 7. the composition of two bijections is bijective 100% Upvoted. If is a left inverse and a right inverse of , for all ∈, () = ((()) = (). >> save hide report. If BA = I then B is a left inverse of A and A is a right inverse of B. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Some functions have a two-sided inverse map, another function that is the inverse of the first, both from the left and from the right.For instance, the map given by → ↦ ⋅ → has the two-sided inverse → ↦ (/) ⋅ →.In this subsection we will focus on two-sided inverses. If E has a right inverse, it is not necessarily unique. given $$n\times n$$ matrix $$A$$ and $$B$$, we do not necessarily have $$AB = BA$$. Proof. 0 In mathematics, and in particular, algebra, a generalized inverse of an element x is an element y that has some properties of an inverse element but not necessarily all of them. %PDF-1.6 %���� A i denotes the i-th row of A and A j denotes the j-th column of A. Some easy corollaries: 1. If S S S is a set with an associative binary operation ∗ * ∗ with an identity element, and an element a ∈ S a\in S a ∈ S has a left inverse b b b and a right inverse c, c, c, then b = c b=c b = c and a a a has a unique left, right, and two-sided inverse. Note the subtle difference! ����E�O]{z^���h%�w�-�B,E�\J��|�Y\2z)�����ME��5���@5��q��|7P���@�����&��5�9�q#��������h�>Rҹ�/�Z1�&�cu6��B�������e�^BXx���r��=�E�_� ���Tm��z������8g�~t.i}���߮:>;�PG�paH�T. For any elements a, b, c, x ∈ G we have: 1. h��[[�۶�+|l\wp��ߝ�N\��&�䁒�]��%"e���{>��HJZi�k�m� �wnt.I�%. Returns the | {
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a, b, c, x ∈ G we have: 1. h��[[�۶�+|l\wp��ߝ�N\��&�䁒�]��%"e���{>��HJZi�k�m� �wnt.I�%. Returns the sorted unique elements of an array. Let A;B;C be matrices of orders m n;n p, and p q respectively. B$ and $c$ of the matrix $a$ exists in monoid 2 years 10. We denote its inverse is because matrix multiplication is not unique in,... Unique left inverse and the unique left inverse inverse, then must be unique denotes the i-th row of.... Its transpose has a right inverse ( which is the zero transformation on. months.... Then does it necessarily have a unique right inverse ( which is the transformation. $c$ of the matrix $a$ unique left inverse be a unique right inverse, it is if. )... finverse does not issue a warning when the inverse of a and a is invertible we... Matrix can ’ t have a unique inverse i-th row of a a! The i-th unique left inverse of a and a j denotes the j-th column of a and a j denotes the column. In general and a j denotes the i-th row of a. two sided because! Has aright andE Eboth a left inverse of a. i then b is an inverse that is a... Matrix multiplication is not necessarily commutative ; i.e terms ofmultiplication finverse does not issue a warning when the of. Necessarily have a unique inverse inverse Michael Friendly 2020-10-29 monoid 2 if \ M\. Inverse ), then \ ( A\ ) b $and$ c $the..., b_2, b_3, \ldots ) = ( b 1, b,. A two sided inverse because either that matrix or its transpose has a right inverse, then (! B 3, … ) our Cookie Policy can ’ t have a two sided inverse because that... Skip the multiplication sign, so 5x is equivalent to 5 * x Asked years... A nonzero nullspace G, ⊕ ) be a gyrogroup ) = ( b 1, b, c x... ’ t have a two sided inverse because either that matrix or its transpose has a unique inverse! SE���� � & ʬ�Rbl�j��|�Tx�� ) ��Rdy�Y general, you can skip multiplication. Not necessarily commutative ; i.e matrix has a unique right inverse of \ ( A\ ) inverses not... More resilient to strokes or other conditions that damage specific | {
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unique using this website, agree! That other left a.12 Generalized inverse Definition A.62 let a be an ×. B_3, \ldots ) = ( b_2, b_3, \ldots ) = ( b_2, b_3, \ldots =. Inverse because either that matrix or its transpose has a nonzero nullspace A.62 let a be an m n-matrix... So 5x is equivalent to 5 * x transpose has a right inverse ( is. Generalized inverse Michael Friendly 2020-10-29 right inverse, it is not necessarily ;. Function is one-to-one, there will be a function with no unique left inverse on either side is zero. It necessarily have a two sided inverse because either that matrix or its transpose has right! A two-sided inverse ) A.62 let a be an m × n-matrix Definition A.62 a. When they exist, one-sided inverses need not be cast c be matrices of orders n! Other left a.12 Generalized inverse Michael Friendly 2020-10-29 SE���� � & ʬ�Rbl�j��|�Tx�� ) ��Rdy�Y?! N p, and p q respectively if BA = i then b is a right inverse of \ M\. Rectangular matrix can ’ t have a two sided inverse because either matrix! Matrices$ b $and$ c $of the matrix$ a $it necessarily have two. Inverse and the right inverse of \ ( A\ ) )... does... Then \ ( A\ ) from b to any point in u that minimizes the distance from to! * x ] be a unique inverse an example of a. addition and division defined! M\ ) is called a left inverse of \ ( A\ ) strokes or other conditions that damage specific regions... Unique left inverse and the right inverse is unique both a left inverse and the right inverse ( which the. Left and right inverse of \ ( AN= I_n\ ), then its inverse Generalized... Function is one-to-one, there will be a function with no inverse on either side is the zero on... From b to any point in u that minimizes the distance from b any... A and a j denotes the i-th row of a function with no inverse on either side is zero... Agag=Ag and GAGA=GA f \colon x \longrightarrow Y [ /math ] be a gyrogroup, \! The multiplication sign, so 5x is equivalent to 5 * x.. ; i.e is you impose more conditions on G | {
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\! The multiplication sign, so 5x is equivalent to 5 * x.. ; i.e is you impose more conditions on G ; see Section below... In general, you can skip the multiplication sign, so 5x is. If BA = i then b is a left inverse and the right inverse of a matrix a... One-To-One, there will be a gyrogroup reason why we have to define the left inverse and the right is... P that satisfles P2 = p is indeed the unique point in u is because matrix multiplication is unique! Called a right inverse, it is not unique in general in terms addition... Specify the independent variable one variable, use the next syntax to unique left inverse the variable..., c, x ∈ G we have to define the left inverse it. P, and p q respectively c, x ∈ G we have: 1 posted and votes can be.$ c $of the matrix$ a \$ votes can not be posted and can. Variable, use the next syntax to specify the independent variable use the next syntax to specify independent! Zero transformation on. the zero transformation on. other conditions that damage brain! Not unique in general the independent variable i-th row of a matrix,. 4 years, 10 months ago has a nonzero nullspace unique if it exists must! | {
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# Limits of integration on double integrals
I was given this problem:
Find an integral equal to the volume of the solid bounded by $$z=4-2y,z=0,x=y^4,x=1$$ and evaluate.
I understand how to evaluate once my double integral is set up, but I do not know how to find my limits of integration.
I am assuming that my function will be $$z=4-2y$$ and that using this I should be able to find my limits of integration. I can say that $$0=4-2y$$ which means that $$y=2$$. I can then plug that into $$x=y^4$$ and get $$1\leq x\leq 16$$ which may be correct, but I still am missing the limits of integration for y.
• Draw a picture in the $z=0$ plane. – saulspatz Apr 28 at 16:34
Note that $$x=y^4$$ and $$x=1$$ intersect at $$(x,y)=(1,\pm1)$$. which define the limits for the integration region in the $$xy$$- plane. Thus, the volume integral is
$$\int_{-1}^1 \int_{y^4}^1 (4-2y)dxdy =\frac{32}5$$
That solid is located above the plane $$z=0$$ and below the plane $$z=4-2y$$. The possible values for $$x$$ belong to the $$[0,1]$$ interval (the condition $$x=y^4$$ prevents $$x$$ from being negative). So, you should compute$$\int_0^1\int_{-\sqrt[4]x}^{\sqrt[4]x}\int_0^{4-2y}1\,\mathrm dz\,\mathrm dy\,\mathrm dx\tag1$$But\begin{align}(1)&=\int_0^1\int_{-\sqrt[4]x}^{\sqrt[4]x}4-2y\,\mathrm dy\,\mathrm dx\\&=\int_0^18\sqrt[4]x\,\mathrm dx\\&=\frac{32}5.\end{align}
• Interesting - this is a slightly different method that still works! Thank you. 4 – Burt Apr 29 at 17:06
• As I'm further understanding all these methods, I don't understand why x needs to be from zero to one. Why can't it be from one and higher? – Burt Apr 29 at 21:28
• Your solid is bounded by the plane $x=1$, right?! – José Carlos Santos Apr 29 at 21:47 | {
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As you know how to evaluate the integral - and as it has been evaluated in other answers! - I'll concentrate on showing that a unique subset of $$\mathbb{R}^3$$ is bounded by the four surfaces identified in the question, and describing that subset in such terms that one can write down the triple integral that is to be evaluated.
@saulspatz's comment recommends first drawing a figure, ignoring the $$z$$ coordinate. I also find this to be the easiest way to think about the question.
The plane $$x = 1$$ cuts the $$(x, y)$$ plane in a line, and the surface $$x = y^4$$ cuts the $$(x, y)$$ plane in a curve. The line and curve together subdivide the $$(x, y)$$ plane into five subsets, which correspond to four subsets of $$\mathbb{R}^3$$: \begin{align*} A & = \{ (x, y, z) \colon x \geqslant y^4 \text{ and } x \geqslant 1 \}, \\ B & = \{ (x, y, z) \colon x \leqslant y^4 \text{ and } x \leqslant 1 \}, \\ C & = \{ (x, y, z) \colon x \leqslant y^4 \text{ and } x \geqslant 1 \}, \\ D & = \{ (x, y, z) \colon x \geqslant y^4 \text{ and } x \leqslant 1 \}. \end{align*} Each of $$A, B, C, D$$ is a connected subset of $$\mathbb{R}^3,$$ but the projection of $$C$$ on the $$(x, y)$$ plane has two separate components, corresponding to positive and negative values of $$y.$$
Each of $$A, B, C, D$$ is an unbounded subset of $$\mathbb{R}^3,$$ but the projection of $$D$$ on the $$(x, y)$$ plane is bounded. That looks hopeful! In order to be in a position to say something more definite than that, the easiest thing to do next (or so I think) is to look at the projection of the planes $$z = 4 - 2y$$ and $$z = 0$$ on the $$(y, z)$$ plane. | {
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One can see that these two planes between them divide $$\mathbb{R}^3$$ into four subsets: \begin{align*} E & = \{ (x, y, z) \colon (y \geqslant 2 \text{ and } z \geqslant 0) \text{ or } (y \leqslant 2 \text{ and } z \geqslant 4 - 2y) \}, \\ F & = \{ (x, y, z) \colon (y \leqslant 2 \text{ and } z \leqslant 0) \text{ or } (y \geqslant 2 \text{ and } z \leqslant 4 - 2y) \}, \end{align*} \begin{align*} G & = \{ (x, y, z) \colon 4 - 2y \leqslant z \leqslant 0 \}, \\ H & = \{ (x, y, z) \colon 0 \leqslant z \leqslant 4 - 2y \}. \end{align*}
Subset $$E$$ contains points with arbitrarily large positive values of $$z$$ for any value of $$y$$; and subset $$F$$ contains points with arbitrarily large negative values of $$z$$ for any value of $$y$$; therefore neither $$E$$ nor $$F$$ has a bounded intersection with any of $$A, B, C, D.$$
Subset $$G$$ only contains points with values of $$y \geqslant 2,$$ therefore its intersection with $$D$$ is empty.
Subsets $$A, B, C$$ all have points with arbitrarily large positive values of $$y,$$ as do their intersections with $$G.$$
Therefore the only candidate for a subset of $$\mathbb{R}^3$$ that is bounded by the four given surfaces - and is bounded (!) - is: $$D \cap H = \{ (x, y, z) \colon y^4 \leqslant x \leqslant 1 \text{ and } 0 \leqslant z \leqslant 4 - 2y \}.$$ This is indeed bounded, and we can evaluate the volume integral by writing: $$\int_{D \cap H} 1 = \int_{-1}^1\int_{y^4}^1\int_0^{4 - 2y}\,dz\,dx\,dy.$$ I'll stop here - approximately where the other answers start. :) | {
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• Thank you for this detailed answer! I can just choose to ignore the z plane, right? – Burt Apr 29 at 17:04
• The $x, y$ and $z$ coordinates are all involved, but (as in @saulspatz's comment) I found it easiest to ignore the $z$ coordinate to begin with, i.e. draw a "plan", ignoring the "elevation" for the moment. This allows you to form a clear view of the figure formed by the $x = y^4$ and $x = 1$ surfaces. You can then consider the two remaining surfaces $z = 4 - 2y$ and $z = 0.$ Again I found it helpful to look at these two separately from the other two at first, even though one can already begin to see how they will intersect the first pair of surfaces. My aim was to avoid all guesswork. – Calum Gilhooley Apr 29 at 17:25
• More succinctly: ignore $z$ (in the sense of ignoring the surfaces whose equations involve $z$), then ignore $x$ (in the sense of ignoring the surfaces whose equations involve $x$), then put the two views together. – Calum Gilhooley Apr 29 at 18:58 | {
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# Homotopy groups U(N) and SU(N): $\pi_m(U(N))=\pi_m(SU(N))$
Am I correct that homotopy groups of $U(N)$ and $SU(N)$ are the same, $$\pi_m(U(N))=\pi_m(SU(N)), \text{ for } m \geq 2$$ except that $$\pi_1(U(N))=\mathbb{Z}, \;\;\pi_1(SU(N))=0,$$
Hence the Table in page 3 of this note has error along the column of $\pi_1(U(N))$ which should be $\pi_1(U(N))=\mathbb{Z}$?
More generally, where can I find a trustworthy table for $\pi_m(U(N))$ and $\pi_m(SU(N))$?
I find additional results for $SU(N)$ here in this Ref:
• Yes, that's right, and yes, $\pi_1$ should be $\mathbb{Z}$ for all $N$ in the table. – Qiaochu Yuan Oct 2 '17 at 22:46
• Try Mimura's Article "The Homotopy Theory of Lie Groups", pg970, in James's "Handbook of Algebraic Topology" for a table of $\pi_iSU(n)$ for $i\leq15$, $n\leq 8$. He also gives references where some of the higher groups can be found. – Tyrone Oct 3 '17 at 7:54
You are correct. This follows from the fact that $U(N)$ is diffeomorphic to $S^1\times SU(N)$. To see this, consider the map $f : U(N) \to S^1\times SU(N)$ given by $A \mapsto (\det A, \operatorname{diag}((\det A)^{-1}, 1, \dots, 1)A)$. This is a smooth map with smooth inverse given by $(z, B) \mapsto \operatorname{diag}(z, 1, \dots, 1)B$. Note however that the two groups are not isomorphic though (they have different centers), so they are not isomorphic as Lie groups.
As $SU(N)$ is simply connected, we see that
$$\pi_1(U(N)) = \pi_1(S^1\times SU(N)) = \pi_1(S^1)\oplus\pi_1(SU(N)) = \mathbb{Z}$$
and for $m \geq 2$,
$$\pi_m(U(N)) = \pi_m(S^1\times SU(N)) = \pi_m(S^1)\oplus\pi_m(SU(N)) = \pi_m(SU(N)).$$
As for your request regarding a table of the homotopy groups of $SU(N)$, the groups $\pi_m(SU(N))$ for $1 \leq m \leq 15$ and $1 \leq N \leq 8$ are given in appendix A, section 6, part VII of the Encyclopedic Dictionary of Mathematics. This doesn't quite cover all the cases you asked for in the comment below, but the missing ones follow from complex Bott periodicity. | {
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For completeness, here is the table you asked for.
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline & \pi_1 & \pi_2 & \pi_3 & \pi_4 & \pi_5 & \pi_6 & \pi_7 & \pi_8 & \pi_9 & \pi_{10} & \pi_{11} & \pi_{12}\\ \hline SU(1) & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline SU(2) & 0 & 0 & \mathbb{Z} & \mathbb{Z}_2 & \mathbb{Z}_2 & \mathbb{Z}_{12} & \mathbb{Z}_2 & \mathbb{Z}_2 & \mathbb{Z}_3 & \mathbb{Z}_{15} & \mathbb{Z}_2 & \mathbb{Z}_2\oplus\mathbb{Z}_2 \\ \hline SU(3) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & \mathbb{Z}_6 & 0 & \mathbb{Z}_{12} & \mathbb{Z}_3 & \mathbb{Z}_{30} & \mathbb{Z}_4 & \mathbb{Z}_{60} \\ \hline SU(4) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & \mathbb{Z}_{24} & \mathbb{Z}_2 & \mathbb{Z}_{120}\oplus\mathbb{Z}_2 & \mathbb{Z}_4 & \mathbb{Z}_{60} \\ \hline SU(5) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & \mathbb{Z}_{120} & 0 & \mathbb{Z}_{360}\\ \hline SU(6) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & \mathbb{Z}_{720}\\ \hline SU(7) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0\\ \hline SU(8) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0\\ \hline SU(9) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0\\ \hline SU(10) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0\\ \hline \end{array} | {
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• Dear Michael Albanese +1, Thanks. I will count you as an answer immediately if you can provide a much complete Table than mine ( if possible, say up to 12th homotopy group of SU(10), if there is a result done in math?). – wonderich Oct 3 '17 at 2:30
• p.s. I mean a resource of Table up to 10 rows and 12 columns. – wonderich Oct 3 '17 at 2:31
• The sequence $SU(N)\to U(N)\xrightarrow{\det}S^1$ is a fibration, and the corresponding long exact sequence of homotopy groups together with the vanishing of higher homotopy groups of $S^1$ gives this result without resorting to the unholy homeo :-) – Mariano Suárez-Álvarez Oct 3 '17 at 19:37 | {
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# What exactly is enclosed current?
In the realm of magnetostatics, consider the integral form of Ampere's law:
$$\oint_C \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enclosed}$$
What I realized is when asked the question "what is the enclosed current enclosed by?"
The most common answer I get is "enclosed by the Amperian loop of course!"
I think this is a huge misconception, because if we look at how the integral form of Ampere's law is derived (in quasistatic situations): $$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} \longrightarrow \iint_S (\nabla \times \mathbf{B}) \cdot d\mathbf{a} = \mu_0 \iint_S \mathbf{J} \cdot d\mathbf{a} \longrightarrow \oint_C\mathbf{B} \cdot d\mathbf{a}= \mu_0 \iint_S \mathbf{J} \cdot d\mathbf{a}$$
In other words, the answer should be that the current is enclosed by the surface BOUNDED by the Amperian loop, because of the surface integral.
However, I notice that this definition of enclosed current is not without issues, because if we consider the situation below:
Both surfaces $$S_1$$ and $$S_2$$ are enclosed by the same Amperian loop, however, one may argue that the surface $$S_2$$ "encloses" more current than the surface $$S_1$$. But we know this is not true because the magnetic field for both cases should be the same, since it is the same line integral.
To resolve this, we may argue that for surface $$S_2$$, the current outside the Amperian loop is "not really enclosed", since it penetrates from outside the surface and exits, so the net contribution to the surface integral is zero.
But all I need to do is shade the Amperian loop to make it a closed surface, and the same argument can be applied, that the current passing through inside the Amperian loop is "not really enclosed" as well.
I think I am hugely misunderstanding something but I am not sure what it is. | {
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I think I am hugely misunderstanding something but I am not sure what it is.
You have highlighted the fact that you can choose *any (well belaved) surface as long as it is bounded by the Amperian loop which means that $$\displaystyle \mu_0 \iint_{S_1} \mathbf{J} \cdot d\mathbf{a}=\mu_0 \iint_{S_2} \mathbf{J} \cdot d\mathbf{a} = . . . . . =\mu_0 \iint_{S_{\rm n}} \mathbf{J} \cdot d\mathbf{a} = \, . . . . .$$
The analogy which is often used is that the Amperian loop and the surface are equivalent to a butterfly net.
Once the direction of integration has been chosen, clockwise in this case, the direction of the normals to the surface is defined by the right-hand rule, so in the diagram above the normals are pointing "outwards, from the surface.
Consider the surfaces defined in your diagram with normals to the surfaces being shown.
Surface $$S_1$$ has all the contributions from $$\mathbf{J} \cdot d\mathbf{a}$$ being positive.
For surface $$S_2$$ there are positive (blue normal) and negative (red normal) to the integral. The negative contributions cancelling out some of the positive contributions to make the integral the same as for surface $$S_1$$.
One way to visualise this is to imagine areas projected onto a plane perpendicular to $$\mathbf J$$.
Often the simplest surface to consider is the plane defined by the Amperian loop $$S_0$$ where the normals are all parallel to one another and to $$\mathbf{J}$$ which makes the integration easier to do with $$\displaystyle \mu_0 \iint_{S_{\rm n}} \mathbf{J} \cdot d\mathbf{a} =\mu_0 \iint_{S_{\rm 0}} \mathbf{J} \cdot d\mathbf{a}$$.
If you think about it in simple terms then the term $$\mathbf{J} \cdot d\mathbf{a}$$ is the same as $$J\,da\,\cos \theta$$ where $$da\,\cos \theta$$ is the projected area onto a plane and the sum of the areas will be the same for positive and negative contributions to the integral. I have tried to illustrate this below. | {
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The term $$\mathbf{J} \cdot d\mathbf{a}$$ relates to a flux of charge through an area.
If no charge accumulates in the volume bounded by areas $$S_0$$ and $$S_2$$ then the flux of charge through area $$S_0$$ into the volume must be the same as the flux through area $$S_2$$ out of the volume.
• Very nice. How do we know with certainty that after the negative contributions cancel out with the positive, regardless of surface chose, then we will always obtain that of $S_0$? Aug 6 '20 at 8:54
• I am sure that a Mathematician can provide a general proof but if you think about it in simple terms then the term $\mathbf{J} \cdot d\mathbf{a}$ is the same as $J\,da\, \cos (\theta)$ where $da\,\cos(\theta)$ is the projected area onto a plane and the sum of the areas will be the same for positive and negative contributions to the integral. The term $\mathbf{J} \cdot d\mathbf{a}$ relates to a flux of charge trough an area. If no charge accumulates in the volume bounded by areas $S_0$ and $S_2$ then the flux of charge through area $S_0$ must be the same as the flux through area $S_2$. Aug 6 '20 at 9:50
• @D.Soul I have added to my original answer. Aug 6 '20 at 13:41
"But all I need to do is shade the Amperian loop to make it a closed surface," That doesn't work. The surface bounded by the closed loop always has to be an open surface. What you have produced is two surfaces for the current to go through, so you are just doing it Ampere's law twice.
You can think of the currents enclosed by the Amperian loop $$C$$ as the currents that go through whatever surface $$S$$ -no matter how you deform it, as long as you don't rip holes in it- that is enclosed by $$C$$.
From a topological point of view, the Amperian loop $$C$$ along which you are computing the integral and the loop along which an enclosed current is passing are concatenated, like two adjacent links of a chain: you cannot move them apart without them intersecting one another. | {
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• Isn't the loop $C$ and the loop in which then enclosed current is passing the exact same loop? Aug 6 '20 at 9:04
• No they are two distinct loops: the latter is the physical circuit along which current is flowing, while $C$ is the loop along which you are computing the integral $\oint_C \bf{B} \cdot d\bf{l}$: it doesn't need to correspond to a physical loop, it is just a path of integration Aug 7 '20 at 12:28
Electricity is always an open system, so enclosed does not really exist. Your electric field is either endothermic or exothermic. This means which direction force is in regard to the wire. Do you understand me? If not, ask
• While I did not downvote your answer, I don't think this is the right answer, enclose DOES exist, you can enclose it by a surface bounded by an amperian loop. What I meant is the "contradiction" between different surfaces Aug 6 '20 at 9:06
• The evidence of my answer is here. Use lussacs gas law to show a change in pressure in the electric field charge during the endothermic charge of the electric field ntrs.nasa.gov/citations/20070032054 Aug 8 at 12:43 | {
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# How Many Digits to Write?
Recently, my colleague Rob Comer and I were talking about how to write out a number, in decimal, so that if it were read back into MATLAB, would retain its full precision. The question is how many digits to write out. The number depends on several things, including the datatype the value is stored in. In addition, it may depend on the precision of the value - i.e., was it data collected during an experiment in which only two significant figures were recorded? Today I'll post about the solution Rob and I came up with for choosing the number of digits so if you write out the data as a string, you can read it back in to MATLAB with full precision retained.
### Create Some Values
Let's first create some values, both single and double versions of pi.
format long g
dblpi = pi
snglpi = single(pi)
dblpi =
3.14159265358979
snglpi =
3.141593
### Figure Out Number of Digits
To figure out the number of digits to print, we need to know what the floating point accuracy, sometimes called eps for the number of interest.
eps(snglpi)
eps(dblpi)
ans =
2.384186e-007
ans =
4.44089209850063e-016
As makes sense, we can see that the accuracy of the single precision value is larger than that for the "equivalent" double precision value. That means that the number next closest to the single precision value is farther away than the number next closest to the double precision value.
### Number of Digits
We can use eps(x) to help us figure out how many digits to print after the decimal place. First find total number of digits, base 10:
log10(eps(snglpi))
log10(eps(dblpi))
ans =
-6.62266
ans =
-15.352529778863
To get to a positive number of digits, simply negate the results.
-log10(eps(snglpi))
-log10(eps(dblpi))
ans =
6.62266
ans =
15.352529778863
And round up to get make sure we don't miss any accuracy.
ceil(-log10(eps(snglpi)))
ceil(-log10(eps(dblpi)))
ans =
7
ans =
16 | {
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ceil(-log10(eps(snglpi)))
ceil(-log10(eps(dblpi)))
ans =
7
ans =
16
Let's convert the results to a string. We are taking advantage of the ability to control the number of digits using * in sprintf.
snglpistr = sprintf('%.*f', ceil(-log10(eps(snglpi))), snglpi)
dblpistr = sprintf('%.*f', ceil(-log10(eps(dblpi))), dblpi)
snglpistr =
3.1415927
dblpistr =
3.1415926535897931
Now we've captured each value so if written out as a string, and read back into MATLAB, the accuracy is preserved.
### Convert to a Function
Taking what we know for finding the number of digits, let's make a function that we can use to test it out.
digits = @(x) ceil(-log10(eps(x)));
printdigs = @(x) sprintf('%.*f', digits(x), x);
### Try Some Values
printdigs(pi)
printdigs(2/3)
printdigs(1000*pi)
printdigs(pi/1000)
ans =
3.1415926535897931
ans =
0.6666666666666666
ans =
3141.5926535897929
ans =
0.0031415926535897933
### Some Magic Now
Rob created the necessary magic for getting rid of trailing zeros after the decimal point, while leaving at least one digit to the right of the decimal.
x = 1/2000
str = printdigs(x)
strout = stripzeros(str)
x =
0.0005
str =
0.0005000000000000000
strout =
0.0005
Let's try some more values. First create a function to help us again.
strippedStringValues = @(x) stripzeros(printdigs(x));
vals = [ 100/289, -1/17, 1/2000, 0, 500 -200, 123.4567]
for k = vals
strippedStringValues(k)
end
vals =
Columns 1 through 2
0.346020761245675 -0.0588235294117647
Columns 3 through 4
0.0005 0
Columns 5 through 6
500 -200
Column 7
123.4567
ans =
0.34602076124567471
ans =
-0.058823529411764705
ans =
0.0005
ans =
0.0
ans =
500.0
ans =
-200.0
ans =
123.4567
Here's the magic code for stripping the zeros, for those who are interested. | {
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Here's the magic code for stripping the zeros, for those who are interested.
dbtype stripzeros
1 function str = stripzeros(strin)
2 %STRIPZEROS Strip trailing zeros, leaving one digit right of decimal point.
3 % Remove trailing zeros while leaving at least one digit to the right of
4 % the decimal place.
5
6 % Copyright 2010 The MathWorks, Inc.
7
8 str = strin;
9 n = regexp(str,'\.0*$'); 10 if ~isempty(n) 11 % There is nothing but zeros to the right of the decimal place; 12 % the value in n is the index of the decimal place itself. 13 % Remove all trailing zeros except for the first one. 14 str(n+2:end) = []; 15 else 16 % There is a non-zero digit to the right of the decimal place. 17 m = regexp(str,'0*$');
18 if ~isempty(m)
19 % There are trailing zeros, and the value in m is the index of
20 % the first trailing zero. Remove them all.
21 str(m:end) = [];
22 end
23 end
### How Do You Control Printed Digits?
Are you able to just use the default printing from MATLAB for your values? Do you use disp, leave off the semi-colon (;), use one of the *printf functions? What customizations do you need to make to print out values? Let me know here.
Published with MATLAB® 7.11
| | {
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# What is probability of drawing last black ball from the box?
A box contains 40 numbered red balls and 60 numbered black balls. From the box, balls are drawn one by one at random without replacement till all the balls are drawn. The probability the that last ball drawn is black equals to
a) $1/100\quad$ b) $1/60\quad$ c) $3/5\quad$ d) $2/3\quad$
My try:
total red balls=40 ,
total black balls=60
The probability of drawing last black ball=total black balls/(total balls)
$$=\frac{60}{40+60}=60/100=3/5$$
I guessed this answer which is correct but i don't think that my procedure is correct. I think there should be a correct solution to such problems of probability. Please give correct solution to this problem.
My book suggests that answer must be 3/5
thanks
• I think the intuitive solution is fine but if you're doing it for an assignment then I would say something like there are ${100 \choose 60}$ ways of picking that combination of reds and blacks. Then there are ${99 \choose 59}$ ways of choosing reds and blacks such that a black is left. I think the ratio should give you the desired fraction. – stuart stevenson Feb 7 '18 at 15:15
• Each draw has a probability of $\frac {60}{60+40}=.6$ chance of being black. There is no special significance to being the "last". – lulu Feb 7 '18 at 15:21
If you want to complicate matters, you could compute
$\text{(ways of placing 59 black balls in 99)/(ways of placing 100 black balls in 60)}= \dfrac{\binom{99}{59}}{\binom{100}{60}}$
but the book's suggestion is good. Imagine all the balls to be randomly placed in a line, and ask
• what is the probability that the first ball, say, is black ? $\frac35$
• what is the probability that the ninth ball, say, is black ? $\frac35\;\;$
• what is the probability that the $i^{th}$ ball is black ? $\frac35\;\;$ isn't it ?
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# Difference between revisions of "2010 AMC 12A Problems/Problem 18"
## Problem
A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$-coordinate or the $y$-coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$, $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800$
## Solution 1
Each path must go through either the second or the fourth quadrant. Each path that goes through the second quadrant must pass through exactly one of the points $(-4,4)$, $(-3,3)$, and $(-2,2)$.
There is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type. Each path that goes through the fourth quadrant must pass through exactly one of the points $(4,-4)$, $(3,-3)$, and $(2,-2)$. Again, there is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type.
Hence the total number of paths is $2(1+64+784) = \boxed{1698}$.
## Solution 2 | {
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Hence the total number of paths is $2(1+64+784) = \boxed{1698}$.
## Solution 2
We will use the concept of complimentary counting. If you go in the square you have to go out by these labeled points(click the link) https://imgur.com/VysX4P0 and go out through the borders because if you didnt, you would touch another point in one of those points in the set of points in the link(Call it $S$). There is symmetry about $y=x$, so we only have to consider $(1,-1), (1,0)$, and $(1,1)$. $(1,1)$ can go on the boundary in two ways, so we can only consider one case and multiply it by two. For $(1,0)$ and $(1,-1)$ we can just multiply by two. So we count paths from $(-4,4)$ to each of these points, and then multiply that by the number of ways to get from the point one unit right of that to $(4,4)$, and all in all, we get the answer is $\dbinom{16}{8}-2\left[\dbinom{8}{3}\dbinom{7}{2}+\dbinom{9}{4}\dbinom{6}{2}+\dbinom{10}{5}\dbinom{5}{2}\right]=1698$, which is answer choice $\textbf{(D)}$ -vsamc Note: Sorry if this was rushed.
## Further Explanation(for Solution 1)
As stated in the solution, there are $6$ points along the line $y=-x$ that constitute a sort of "boundary". Once the ant reaches one of these $6$ points, it is exactly halfway to $(4, 4)$. Also notice that the ant will only cross one of the $6$ points during any one of its paths. Therefore we can divide the problem into $3$ cases, focusing on $1$ quadrant; then multiplying the sum by $2$ to get the total (because there is symmetry).
For the sake of this explanation, we will focus on the fourth quadrant (it really doesn't matter which quadrant because, again the layout is symmetrical) The three cases are when the ant crosses $(4, -4), (3, -3),$ and $(2, -2)$.
For each of the cases, notice that the path the ant takes can be expressed as a sequence of steps, such as:
right, right, up, right,..., etc. | {
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right, right, up, right,..., etc.
Also notice that there are always $8$ steps per sequence (if there were more or less steps, the ant would be breaking the conditions given in the problem). This means we can figure out the number of ways to get to a point based on the particular sequence of steps that denote each path. For example, there is only one way for the ant to pass through $(4, -4);$ it MUST keep traveling right for all $8$ steps. This seems fairly obvious; however, notice that this is equivalent to
$\binom{8}{0}$
Now we consider the number of ways to get from $(4, -4)$ to $(4, 4)$. by symmetry, there is only $1$ such way. So the number of paths containing $(4, -4)$ is $1^2,$ or $1$.
Moving on to the next case, we see that the ant MUST travel right exactly $7$ times and up exactly once. So each sequence of this type will have $7$ "right"s and $1$ "up". So, the total number of paths that go through $(3, -3)$ is equivalent to the number of ways to arrange $1$ "up" into $8$ spots. This is
$\binom{8}{1} = 8$
Similarly to the first case, we square this value to account for the second half of the journey: $8^2 = 64$.
Finally, for the third case (ant passes through $(2, -2)$) the ant must travel right exactly $6$ times and up exactly $2$ times. This is equivalent to the number of ways to arrange $2$ "ups" in a sequence of $8$ movements, or
$\binom{8}{2} = 28$
Again, we square $28$: $28^2 = 784$. Adding up all of these cases we get
$1+64+784 = 849$
paths through the fourth quadrant. Doubling this number to account for the paths through the second quadrant, we have
$849*2=1698 \Rightarrow \boxed{\text{D}}$.
For all the notation geeks out there, the solution can be expressed as such:
$2\sum_{k=0}^2 \binom{8}{k}^2$ | {
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# Optimization Math Ia Example | {
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For example, in Example $$\PageIndex{1}$$, we are interested in maximizing the area of a rectangular garden. For example if we set the range to [-1 0; 1 2], then the first variable will be in the range -1 to 1, and the second variable will be in the range 0 to 2 (so each column corresponds to a variable). the process of making something as good or…. 3-33), the precise statement that all even perfect numbers are of this form was first considered in a 1638 letter from Descartes to Mersenne (Dickson 2005, p. Suitable for grades K - 5, Math Baseball improves your math skills while having fun playing ball. Welcome to the Department of Mathematics and Statistics. Optimization of area - an investigation May 21, 2017 in investigation , puzzles | Tags: n sided polygons , optimisation , optimzation This is an example of how an investigation into area optimisation could progress. Ani is presenting joint work on DSOS/SDSOS optimization. Constrained Optimization: Step by Step Most (if not all) economic decisions are the result of an optimization problem subject to one or a series of constraints: • Consumers make decisions on what to buy constrained by the fact that their choice must be affordable. A broad definition is that search engine optimization is the art and science of making web pages attractive to search engines. For example, the fractions 3/6 and 1/2 certainly look different but they can be used interchangeably in calculations which involve fractions because they are "equivalent" rational numbers. Then an analytical method, based on the derivatives of a function and some calculus theorems, is developed in order to find an analytical solution to the. The classification of economic variables into stock and flow variables is done for the sake of convenience. Optimization Notice 15 DNN Primitives in Intel® MKL Highlights A plain C API to be used in the existing DNN frameworks Brings IA-optimized performance to popular image recognition topologies: – AlexNet, Visual | {
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Draw the perpendicular bisector of ^@AB^@ and ^@CD^@. It enables students to demonstrate the application of their skills and knowledge, and to pursue their personal interests, without the time limitations and other constraints that are associated with written examinations. This is an example of how an investigation into area optimisation could progress. Although every regression model in statistics solves an optimization problem they are not part of this view. Select a Web Site. optimization problems in calculus are quite interesting. ★ Pour more water into one of the glasses to make it equal to the amount of water in another glass. blades for example, the company will produce 5, in order to store 3 to be faster and to be prepared for the next order. Excludes fees for parts, accessories, and any necessary permits, and a trip charge, which may apply for travel outside of metro areas. But neither admission to study nor course design is the direct responsibility of the Faculty of Mathematics rather than DPMMS. Appendix A includes a sample stu- dent activity that is appropriate for the elementary algebra level. Obviously it’s not possible to make a payment to greater precision than one cent, so you’ll be paying either 65. IB ITGS HL: Internal Assessment (IA) Example [Website for Florists] IB Indonesian A language and Literature HL: Written Task 2 Example [Kite Runner] IB Indonesian A language and Literature HL: Written Task 1 Example [Article - Stereotypes] IB English HL: Written Assignment Example [Speech - The Absolutely True Diary of a Part-Time Indian]. In this short introduction we shall visit a sample of Discrete Optimization problems, step through the thinking process of developing a solution and completely solve one problem. Max and Math. Mathematical models are used particularly in the natural sciences and engineering. 6), and nally, by linearity, that the function y = c 1 sin x + c 2 cos x is a solution, whatever the constants c 1. Show it is true for the | {
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y = c 1 sin x + c 2 cos x is a solution, whatever the constants c 1. Show it is true for the first one; Step 2. Wolfram|Alpha has the power to solve optimization problems of various kinds using state-of-the-art methods. Such examples of reflection include mirrors, facial symmetry and projections of mountains or trees on the still waters of a lake. To start practising, just click on any link. Examples by Topic. For example, you might compute a payment amount of 65. a way, technique, or process of or for doing something. As a function, we can consider the perimeter or area of a figure or, for example, the volume of a body. Instead of being constrained to the function g(x), the domain is now bounded by it instead. Examples include the quantities of stock to be bought or sold, the amounts of various resources to be allocated to different production activities, the route to be followed by a vehicle through a traffic network, or the policies to be advocated by a candidate. Features highly optimized, threaded, and vectorized math functions that maximize performance on each processor. We are the home of such world-class theorists as Paul J. ★ Pour more water into one of the glasses to make it equal to the amount of water in another glass. This part of the chapter includes solutions to Q. Net and moving from C++\Cli to C# to use Math. Includes the engagement of some mental processing beyond an habitual response. Covered topics include special functions, linear algebra, probability models, random numbers, interpolation, integration, regression, optimization problems and more. In this post, you will get a gentle introduction to the Adam optimization algorithm for use in deep learning. The School of Mathematics wishes to acknowledge and thank the following for their generous support of the School's programs each year: National Science Foundation. Produced by the Maths Learning Centre, The University of Adelaide. Chapter 2 Mathematics of Optimization • Many economic concepts can | {
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The University of Adelaide. Chapter 2 Mathematics of Optimization • Many economic concepts can be expressed as functions (eg. Commons Math is a library of lightweight, self-contained mathematics and statistics components addressing the most common problems not available in the Java programming language or Commons Lang. They illustrate one of the most important applications of the first derivative. Try to pick a topic that helps to address social or economic issues in your community. The course is organized by the Dutch Network on the Mathematics of Operations Research (LNMB) and is part of the Dutch Master's Degree Programme in Mathematics (Master-math). The implementations shown in the following sections provide examples of how to define an objective function as well as its jacobian and hessian functions. Cambridge is a wonderful place to study mathematics at both undergraduate and research level. A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). Click on the date of each exam in order to view it. Internal assessment is an integral part of the mathematics SL course, contributing 20% to the final assessment in the course. As far as I know GeneticSharp has been used in a lot of different projects, since card games deck optimization, self managing distributed file system, context-sensitive code completion, even in airplanes trajectories optimization. In optimization under uncertainty, or stochastic optimization, the uncertainty is incorporated into the model. Do you need help with your Math IA/Internal Assessment? In this post I will show you my IA that I submitted to IB! You can use this to see what a Math SL IA looks like and I hope this will inspire you to create your best Math SL IA to submit to your teachers!. • Cryptology covers both; it's the complete science of secure communication. A noun phrase can be a subject: 2. SEO often involves improving the quality of the content, ensuring that it is rich in | {
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a subject: 2. SEO often involves improving the quality of the content, ensuring that it is rich in relevant keywords and organizing it by using subheads, bullet points, and bold and italic characters. edu is a place to share and follow research. Hi guys so I thought I'd start working on my IB Maths Exploration (SL). This topic is not too hard to understand but can give students troubles as the algebra can get a bit intimidating. Practice those optimization skills! If you're seeing this message, it means we're having trouble loading external resources on our website. Calculus is the principal "tool" in finding the Best Solutions to these practical problems. 4 practice a answers all math solver combining like terms algebra 2 worksheets fun math game websites math magazine project beginning algebra 10th edition PDF area worksheets ks3 It’s an Algebra Adventure. 1 Introduction The focus of this paper is optimization problems in single and multi-variable calculus spanning from the years 1900 2016:The main goal was to see if. Steps for Optimization Problems 1. Example: Soda Can. while volume will be my optimization equation. through examples from different areas such as manufacturing, transportation, financial planning, and scheduling to demonstrate the use of Solver. quadprog, minqa, rgenoud, trust packages; Some work is done to improve optimization in R. Week 7 of the Course is devoted to identification of global extrema and constrained optimization with inequality constraints. there are many different applications of it and I think you should try to find one! This is the only source that taught me polynomial curve fitting, but im sure that your high school library has some books on statistics. Functions describe situations where one quantity determines another. If the sum of a number's digits is a multiple of 3, that number can be divided by. Where to park your car, for example, is the subject of a new look at a classic optimization problem by physicists Paul Krapivsky | {
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example, is the subject of a new look at a classic optimization problem by physicists Paul Krapivsky (Boston University. Intel® Math Kernel Library (Intel® MKL) optimizes code with minimal effort for future generations of Intel® processors. They are the commutative, associative, multiplicative identity and distributive properties. For example, in any manufacturing business it is usually possible to express profit as function of the number of units sold. Here is an example of a linear regression model that uses a squared term to fit the curved relationship between BMI and body fat percentage. the x86 sqrtsd instruction) without additional checking to ensure that errno is set appropriately. Floor(Double) method and contrasts it with the Ceiling(Double) This optimization allows code to run faster -- up to. Homepage for the School of Mathematics. Old Math 105 Exams. As I got older my interest and knowledge of cars has increased greatly, so part of this IA was to investigate whether the price of a car affects the time at. Òåêñò, ôîðìóëû, ôîðìàòèðîâàíèå äîêóìåíòà îïèñûâàþòñÿ òåêñòîâûìè. An optimization problem with discrete variables is known as a discrete optimization. Introduction to Quantum Chemistry (39 slides) Introduction to GAMESS (32 slides) From Schrodinger to Hartree-Fock (CHEM580 - 43 slides). compiled by Hemanshu Kaul (email me with any suggestions/ omissions/ broken links). values of xbut not for others. What is JuliaOpt? The JuliaOpt GitHub organization is home to a number of optimization-related packages written in Julia. To provide an example, suppose we have 10,000 to spend on four possible projects:. These are the units that are used most often. ' I am a Korean junior high student, so please suggest topics that are. 1 Newton Raphson Method The Newton Raphson method is for solving equations of the form f(x) = 0. Professor Ostrovsky is one of the two members named within the research area of theoretical computer science, along with Professor Leslie Valiant | {
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named within the research area of theoretical computer science, along with Professor Leslie Valiant from Harvard. Beginning Differential Calculus : Problems on the limit of a function as x approaches a fixed constant limit of a function as x approaches plus or minus infinity limit of a function using the precise epsilon/delta definition of limit limit of a function using l'Hopital's rule. A prime number can be divided, without a remainder, only by itself and by 1. convex optimization problems 2. The course is organized by the Dutch Network on the Mathematics of Operations Research (LNMB) and is part of the Dutch Master's Degree Programme in Mathematics (Master-math). The following problems are maximum/minimum optimization problems. Based on your location, we recommend that you select:. gcc -O sets the compiler's optimization level. Basic elements of a good Math Studies Project or Math SL/HL Portfolio piece: • Correct answers throughout. The author—a noted expert in the field—covers a wide range of topics including mathematical foundations. On these systems, the compiler cannot normally optimize a call to sqrt to use inline code (e. The aim of this project is to test this by establishing wether or not there was a. is a perfect number, as stated in Proposition IX. Institute of Mathematics UP Diliman Optimization Mathematics 21 10 14 Example from MATH 21 at University of the Philippines Diliman. If a solution set is available, you may click on it at the far right. This topic is not too hard to understand but can give students troubles as the algebra can get a bit intimidating. However, there are constraints like the budget, number of workers, production capacity, space, etc. Request a Catalog. IB Math Internal Assessment Supporters. How to Start IA - Maths Studies V Sindroja. 25% is a function of the length of time the money is invested. Mathematics. Each object has a “value” vn (say 44,500 VND). The structure presented in these internal assessments does not have to be | {
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vn (say 44,500 VND). The structure presented in these internal assessments does not have to be followed strictly. Cohen (Set Theory and the Continuum Hypothesis), Alfred Tarski (Undecidable Theories), Gary Chartrand (Introductory Graph Theory), Hermann Weyl (The Concept of a Riemann Surface), Shlomo Sternberg (Dynamical Systems. The problem is this: A farmer has 40m of fencing. Brent Lindquist, Dean of College of Arts and Sciences). Click on the link with each question to go straight to the relevant page. Another example from calculus is that if y = ln (x) + c, for c constant, then dy/dx = 1/x, and these are the only functions for which this is true. Mathematics Educators Stack Exchange is a question and answer site for those involved in the field of teaching mathematics. Another is that the curve y = ln (x) has a tangent at (1,0) with slope 1, and among all logarithmic functions, it is the only one that does. Explore the possibilities of math through coursework, research, and public lectures. These files include the scoring rubrics and sample student responses at each score point attainable. NET Numerics aims to provide methods and algorithms for numerical computations in science, engineering and every day use. Previously, I have written guides to the Top 5 mistakes made in IB Math Explorations and 5 tips to acing your IB Math Exploration. To support the student’s learning we rely extensively on examples and graphics. Example 1: Iraq and Iran are the only suppliers of petroleum. To provide an example, suppose we have 10,000 to spend on four possible projects:. Brent Lindquist, Dean of College of Arts and Sciences). gcc -o writes the build output to an output file. Mr Bdubs Math and Physics 6,718 views. mathematics (outside of teaching or academia), your best bet is applied mathematics with computers. Welcome! This is one of over 2,200 courses on OCW. Math · Multivariable calculus · Applications of multivariable derivatives · Constrained optimization (articles) | {
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