Homework 9 collaboration area
I was wondering why the eigenvalues are v^2 for positive eigenvalues and -v^2 for negative eigenvalues as seen in example 1 of 11.5 and in Dr. Bell's lecture example?
- Michael
Hello all,
I am unsure of where to even begin on # 5 pg. 503. Any thoughts?
Mac
From Eun Young :
you need to show that $ \int_0^{\pi} P_n(\cos \theta) P_m(\cos \theta) \sin \theta d\theta = 0 $ if $ m \neq n $
We already know that $ \int_{-1}^1 P_n(x)P_m(x) dx = 0 $ if $ m \neq n $.
So, use Substitution in the first equation so that you can have the 2nd equation then you're done.
From Jake Eppehimer:
I am not sure how to do number 6 on p. 494. I'm clueless, and there's no answer in the back to verify if I'm doing anything right. Any tips?
From Shawn Whitman: The method of undetermined coefficients for second order nonhomogeneous linear ODEs works well for this problem. See pages 81-84 and use the sum rule. Two of the constants will go to zero. Two others will result in 1/(omega^2-alpha^2) and 1/(omega^2-beta^2); thus the given constraints.
From Mnestero:
I concur with Shawn regarding problem 6. I have a question about the even extension in problem 29. I am getting that the fourier series is 2/pi-4/pi(1/3 cos(2x) + 1/(3*5) cos (4x) + 1/(5*7) *cos(6x)... The answer in the book has odd numbers instead. Their answer doesn't make sense to me. Any thoughts?
From Shawn Whitman: My answer is also different than the one in the back of the book. an is 0 for odd n and -4/pi((n^2)-1) for even n. Note that the (n^2)-1 in the denominator is not a problem for n=1 if you start the summation from 2 instead of 1. This is okay since the odd n’s are zero. ((n^2)-1), for even n, yields 3, 15, 35 or 3*1, 3*5, 5*7 as the book suggests.
From Jayling:
I think the book answer is for Question 28 not 29. If you plot the function it looks like the sawtooth function. Anyway for the Fourier Sine Function did you guys get bn=0 for all n? Which implies that the Fourier Sine Function is equal to zero for all x.
Remark from Steve Bell: Did you get
$ \int_0^\pi \sin(x)\cos(nx)\,dx=-\frac{1+\cos(n\pi)}{n^2-1}? $
Notice that n^2-1=(n-1)(n+1). That might explain where those odd numbers come from when n is even.
Ah yes, I see what you mean. Yes, the answer in the back of the book is wrong for p. 490: 29 (a). The non-zero terms in the cosine series happen for even values of n.
The answer is correct for part (b). It is very much like the fact that x is equal to it's own Taylor series about x=0. Sin(x) is a Fourier Series and the Fourier Series expansion of a function is UNIQUE.
From Jake Eppehimer:
I got that. I have the coefficients as shown in the back, but applied to cos(2x), cos(4x), ... as compared to cos(x), cos(4x) ... just as Mnestero said. Any time n is odd, the numerator is zero. Am I supposed to start the n count at 2, as Shawn said, since n = 1 is undefined?
From Steve Bell: Jake, the n=1 term is the integral
$ \int_0^\pi \sin(x)\cos(x)\,dx $
and that is a very easy integral to compute. You are right that the formula for the general term for n>1 doesn't make sense for n=1.
From Jake Eppehimer:
Thank you! The reference page from 81-84 was very handy because I wasn't sure what to do since there were two sine functions added. I got what you got. And I am glad I'm not the only one having that exact issue on problem 29... So it's likely a book error? Jayling, I got that too, but according to the lecture, the odd expansion of sin(x) is simply sin(x), and it can be seen by graphing it. So that's what I'm going with.
From Ryan Buck:
The odd expansion of sin(x) can be shown using the half range expansion (though intuitively you already know it's sin(x) by the plot). If you do the bn integral using the trig identities Dr. Bell shows below, you get two terms; one with (1-n) as a denominator, on with (1+n) as a denominator. We can see that with the 0 to pi limits of integration that for all n>1 terms will be zero, but n=1 poses an issue. Go back to the original bn integral and solve it with n=1 only. Solving for b1 and using the odd function formula will pop out sin(x) as expected.
From Mike Passalacqua: Problem 6 on p. 494...Having trouble getting the initial Fourier series. I'm taking sin(at) as periodic from -pi to pi, so that coefficients 'a0' and 'an' are both zero. I understand we can find 'bn' for each sine function separately, then add them together to get the total coefficient?
Remark from Steve Bell: Mike, for this problem, you need to solve the second order ODE the way you would have when you were a sophomore. You don't need Fourier series to do it. As they say above, solve the homogeneous problem and then use the Method of Undetermined Coefficients (plus the superposition principal) to get a particular solution. (By the way, sin(at) is not 2pi periodic. It is 2pi/a periodic.)
from Jayling:
This may be a bit out there but r(t) already looks like a fourier series to me, so why bother calculating the coefficients? Also another curve ball has anyone tried to solve Problem 6 by taking the Laplace Transform both sides rather than using Undetermined Coefficients?
from Jake Eppehimer:
I think Jayling is right. I think it already is a Fourier series. And for solving with Laplace, wouldn't you need the initial conditions? Or just assume 0?
from LuoShibo:
Are we allowed to sketch the graph by using matlab and print it? Some graph is too difficult for us to sketch by hand.
from Jake Eppehimer:
The only ones you have to graph are #24 and #29 on Lesson 26. They're very easy to graph, assuming I did it right. But for more complicated ones, we were told that we can graph electronically. On the test, he said he will not ask us to graph anything complicated - he just wants us to demonstrate that we understand the concept.
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From Steve Bell: Here are some handy trig identities for these chapters:
$ \sin\alpha\sin\beta=\frac12(\cos(\alpha-\beta)-\cos(\alpha+\beta)) $
$ \cos\alpha\cos\beta=\frac12(\cos(\alpha-\beta)+\cos(\alpha+\beta)) $
$ \sin\alpha\cos\beta=\frac12(\sin(\alpha-\beta)+\sin(\alpha+\beta)) $
Is 11.4 #11 really supposed to refer to example 1 in 11.1? If so I don't see how that example can be used. Fink7
Based on what you told me, couldn't you say x = pi/2 thus f(x) = k and go through the same steps you did for the problem you helped me with? Hzillmer
Nevermind. I thought I could get away with squaring things but that came back to bite me. Anyone else have an idea? Hzillmer
You can directly pull the Fourier coefficients and f(x) from example 1 in 11.1, then plug those into Parseval's Identity equation (equation #8 in 11.4). You can then split the integral up to account for the two parts of f(x) and go from there. Certain things should cancel themselves out nicely as you go through it... - Mike P
I can't seem to figure out 11.2 #20. I tried setting up Parseval's Identity but can not get anything out of it. Thanks for any help Hzillmer
Fink7 For that one, i let x=1 and therefore f(x) = 1. Plug these into your answer from #11, rearrange so there is the desired pi^2/6 on one side, and the converging series is found on the other side of the equality.
