Fourier Series for |x|: Convergence & Answers

[BearY]

Homework Statement

Find trigonometric Fourier series for ##f(x)=|x|##, ##x∈[−\pi, \pi]##, state points where ##F(x)## fail to converge to ##f(x)##.

Homework Equations

##F(x) = \frac{a_0}{2}+\sum\limits_{n=1}^\infty a_ncos(\frac{n\pi x}{L})+b_nsin(\frac{n\pi x}{L})##
##a_n=\frac{1}{L}\int_{-L}^{L}f(x)cos(\frac{n\pi x}{L})dx##
##b_n=\frac{1}{L}\int_{-L}^{L}f(x)sin(\frac{n\pi x}{L})dx##

The Attempt at a Solution

##f(x)cos(nx)## is an even function##a_n=\frac{2}{\pi}\int_{0}^{\pi}xcos(nx)dx##
##f(x)sin(nx)## is an odd function, so ##b_n = 0##
$$a_0 =\frac{2}{\pi}\int_{0}^{\pi}xdx = \pi$$
$$a_n=\frac{2}{\pi}(\frac{cos(n\pi)-1}{n^2})$$
$$F(x) = \frac{\pi}{2}+\sum\limits_{n=1}^\infty \frac{2}{\pi}(\frac{cos(n\pi)-1}{n^2})cos(nx)$$

My question is, the answer is
$$F(x) = \frac{\pi}{2}-\frac{4}{\pi}\sum\limits_{n=1}^\infty \frac{1}{(2n-1)^2}cos((2n-1)x)$$
I can see that the summation in my ##F(x)## has even terms equal to 0 and hence the ##(2n-1)##. But I am not sure how to get there. Or maybe my answer is wrong.

For completion of answer##F(x)## converge to ##f(x)## everywhere in the domain.

 

[I like Serena]

Hi BearY! :)

Split the summation in even n and odd n.
Note that ##\cos(2k\pi)=1##, which cancels, while ##\cos((2k-1)\pi)=-1##.