How to Find B_n in a Fourier Series Problem on the Interval -L < x < L?

[meteorologist1]

Hi, I need help on the following problem on Fourier series:

Let phi(x)=1 for 0<x<pi. Expand
$$1 = \sum\limits_{n = 0}^\infty B_n cos[(n+ \frac{1}{2})x]$$
a) Find B_n.
b) Let -2pi < x < 2pi. For which such x does this series converge? For each such x, what is the sum of the series?
c) Apply Parseval's equality to this series. Use it to calculate the sum
1 + 1/(3^2) + 1/(5^2) + 1/(7^2) + ...

I know the formula for B_n for a function's Fourier series on the interval -L < x < L, so in this question I need to do some kind of odd or even extension for parts a and b, but I don't know how. Please help. Thanks.

 

[meteorologist1]

Ok, actually for part a, I'm pretty sure I should do an even extension of the function at x=0 so that it runs from -pi to pi. And I can then determine the B_n's.

But for part b, it looks like I need to somehow extend to -2pi to 2pi. I looked in books, and does an odd extension at x=pi so that function is -1 from pi to 2i. But I don't understand why. And how would I find the x's for the series to converge?

 

[tacman]

To find B_n in this Fourier series problem on the interval -L < x < L, we can use the following formula:

B_n = \frac{1}{L} \int_{-L}^{L} f(x) cos(n \pi x/L) dx

In this case, L = pi and f(x) = 1 for 0 < x < pi. Therefore, we have:

B_n = \frac{1}{\pi} \int_{0}^{\pi} cos(n \pi x/\pi) dx

Simplifying this integral, we get:

B_n = \frac{1}{\pi} \left[ \frac{sin(n \pi x/\pi)}{n} \right]_{0}^{\pi}

B_n = \frac{1}{\pi} \left[ \frac{sin(n \pi)}{n} - \frac{sin(0)}{n} \right]

B_n = \frac{2}{n \pi} (-1)^n

This gives us the formula for B_n in terms of n.

For part b, we need to consider the convergence of the series on the interval -2pi < x < 2pi. Since the series is a Fourier series, it will converge to the function f(x) = 1 for 0 < x < pi and f(x) = 0 for pi < x < 2pi. Therefore, the series will converge for all values of x in this interval except for x = pi.

To find the sum of the series for each value of x, we can simply plug in the value of x into the series. For example, for x = 0, the sum is:

1 = \sum\limits_{n = 0}^\infty B_n cos[(n+ \frac{1}{2})x]

= B_0 cos(\frac{1}{2}x) + B_1 cos(\frac{3}{2}x) + B_2 cos(\frac{5}{2}x) + ...

= 1 \cdot cos(0) + \frac{2}{1 \pi} (-1)^1 \cdot cos(\frac{3}{2} \cdot 0) + \frac{2}{2 \pi} (-1)^2 \cdot cos(\frac{5}{2} \