In the following problem I am trying to extend the function $f(x) = x $ defined on the interval $(0,\pi)$ into the interval $(-\pi,0)$ as a even function. Then I need to find the Fourier series of this function.
So I believe I have extending the function onto the interval $(-\pi,0)$ correctly below.
$$f(x) = -x, (-\pi,0)$$
I am having a little trouble understanding the question. I believe I need to find the Fourier series of this function which is now $-x$. Since the function is now an even function the Fourier series should just consist of the terms $a_0$ and $a_n$ since $b_n$ has sin attached to it making it a odd function and therefore making it $0$
However I am a little confused what formula now to use,
should I use the following formulas? and if so what is $L$? once I get these matters figured out I can proceed myself with the calculations, thanks!
$$f(x) = \frac{a_0}{2} + \sum^{\infty}_{n=0} a_n cos(nx)$$
$$a_0 = \frac{2}{L} \int^L_0f(x)dx$$
$$a_n = \frac{2}{L}\int^L_0f(x)cos(\frac{n\pi x}{L})dx$$