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: 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 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 the
[http://www.math.purdue.edu/~bell/MA527/Lectures/lec11-03.pdf |Bell's Lecture on Lesson 30]
for the algebra.
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. $
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.
