Difference between revisions of "HWK11MA527Fall2010" - Rhea

Revision as of 07:00, 6 November 2010

Homework 11 collaboration area

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}{2L}\int_{-L}^L f(x)e^{-inx}\,dx= $

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

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

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

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

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

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?

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?)

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.

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