Difference between revisions of "HWK11MA527Fall2010" - Rhea

Revision as of 18:17, 9 November 2010

Homework 11 collaboration area

Question: p. 499, #10: It says to convert it to real form, but when I use Euler's formula, I'm getting that there is still both a complex part and a real part for the Fourier series. Am I just supposed to write the real part, or am I doing this problem incorrectly? Thank you!

Answer: If you do it right, the imaginary part should be zero. Look for cancellation when you change cos(-A) to cos(A) and sin(-A) to -sin(A).

Question: p. 499 #10: I'm also left with an imaginary answer. Starting from:

$ f(x)=i \sum_{n=-\infty}^\infty \frac{(-1)^n}{n} e^{inx}  $

Use Euler's formula to get:

$ ie^{inx}=icos(nx)+i^2sin(nx) $

and add that to the sum for -n:

$ ie^{-inx}=icos(nx)-i^2sin(nx) $ The sum is:

$ 2icos(nx) $

So it looks like its the real parts (the i^2 terms) that cancel. What is the mistake here?

Question: I'm having trouble getting HWK 11, Page 499, Problem 3 started.

Answer: You will need to use Euler's identity

$ e^{i\theta}=\cos\theta+i\sin\theta $

and separate the definitions of the complex coefficients into real and imaginary parts. For example,

$ c_n=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}\,dx= $

$ =\frac{1}{2\pi}\int_{-\pi}^\pi f(x)(\cos(-nx)+i\sin(-nx))\,dx= $

$ =\frac{1}{2\pi}\int_{-\pi}^\pi f(x)(\cos(nx)-i\sin(nx))\,dx= $

$ =\frac{1}{2\pi}(\int_{-\pi}^\pi f(x)(\cos(nx)\,dx - i\int_{-\pi}^\pi f(x)\sin(nx)\,dx)= $

$ =\frac{1}{2}(a_n-ib_n). $

Do the same thing for $ c_{-n} $ and combine.

Question: Page 501 #3: (Example 1, really), What is $ C_n $? Is it the same $ C_n $ from the complex fourier series equation? If so, why have we discarded the negative n terms? Example 1 makes some really big algebraic leaps that I'm having trouble following. Can someone explain it more clearly?

Answer: It is a different $ C_n $:

$ C_n = \sqrt{A_n^2+B_n^2} $

where A_n and B_n are the coefficients gotten from doing the Method of Undetermined Coefficients with RHS (4/(n^2pi^2))cos(nt). See page 4 of

Bell's Lecture on Lesson 30

for the algebra. (Response): Thanks, because I wrote that lecture covered sections 11.5 and 11.6, I wasn't looking far forward enough in my notes. I see we did the whole thing out in class in my later pages.

Question: Page 506, Prob 15, if:

$ 2a_o=\frac{2\pi^4}{9} $

and

$ (a_n)^2=\frac{(4\pi^2)(cos)^2(nx)}{9} $

I dont understand where the $ \frac{\pi^4}{4} $ comes from? Can anyone point out what I am doing wrong?

Answer: Hmmm. Problem 15 on page 506 tells you to use Parseval's Identity applied to the function from problem 21 on page 485. The back of the book says that the Fourier Series for f(x)=x^2 between -pi and pi is

$ \frac{\pi^2}{3}-4\cos x + \frac{4}{2^2}\cos 2x-\frac{4}{3^2}\cos 3x + \cdots $

so it seems that

$ 2a_0^2=2\left(\frac{\pi^2}{3}\right)^2 $

and

$ a_n^2=\left(\frac{4}{n^2}\right)^2, $

and since all the b_n are zero, Parseval's Identity says

$ 2a_0^2+\sum_{n=1}^\infty a_n^2=\frac{1}{\pi}\int_{-\pi}^\pi|f(x)|^2\,dx. $

Question: Problem 1 on page 512: The answer key states that

$ A=\frac{1}{(1+w^2)} $

but shouldn't it be

$ A=\frac{2}{(1+w^2)} $

What am I missing here? Also, what are we supposed to do about the pi/2 and 0 terms? I see that they won't affect this problem being instantaneous and zero, respectively, but in general, what would we do if we had an evaluate-able expression for one interval, and then another nonzero expression for a following interval?

Answer: Since f(x)=0 for x<0, you don't multiply by 2 when changing limits from -inf..inf to 0..inf. That is only for even functions. I am with you though on the pi/2 part. Not really sure why that just drops out. (Response:) Thanks, yes, after thinking about it and re-reading I figured out that we are breaking up the integrals (the answer to my hypothetical), but I also don't know what the deal is with the 0-interval pi/2 term.

Answer from Bell: The pi/2 value at zero is there because a Fourier Integral would converge to the value at the middle of a jump. Hence, if that integral is supposed to equal the given function, it would have to be pi/2 at zero. (You don't have to do anything about it.)

Questions: prob 11 on page 512:

Should I still use equation 10 to compute A(w) or should I use equation 12 to compute B(w) since f(x) is odd.

When I find A or B, what should the integral range be? (0 to pi?)

Answer: The function f is only defined for positive x. The Fourier Cosine Integral was cooked up by extending f to the negative real axis in such a way to make it an even function. That made the B(w) integral turn out to be zero.

Hence, you only need to calculate the A(w) integral in the form

A(w)= (2/pi) integral from 0 to infinity ...

Since f(x) is zero after pi, your integral would only really go from 0 to pi.

Question: Problem 1 on page 512:

I'm a little confused as to what this problem is looking for. It seems obvious that the integral is of form (5) with A(w) = 1 / (1+w^2) and B(w) = w/(1+w^2). Are we supposed to just compute the Fourier integral of pi*exp(-x) or is there something else that I'm missing?

Question: Problem 7 on page 502:

I am not sure how to start this problem. How can we find fourier series of r(t)? Also how to proceed on this question?

Does anyone know how to work prob 9 on page 499? I can seem to get the 'i' or get rid of the 1/2pi either. my Cn i keep getting is: $ (-cosn*pi)/(pi*n^2) $

Answer: What I did was first use the integral definition of Cn from -pi to pi. Reduce this until you end up with something having the general form like Constant*i*[e^(ix) + e^(-ix)]. Then plug this back into the formula for f(x) and reduce a little further. Maybe you are changing to sin and cos too early? That was my very last step...

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