Odd and Even Fourier Series Extension of $f(x)=x$ on $[0,\pi]$

Question

I'm confused on finding the odd and even extensions of $f(x) = x$ on $[0,\pi]$. I know the general forms and how to find the co-efficients, but for the sin series, $f(0)$ =/= $f(\pi)$, so then I only have $f(x)$ related to its extension, not equal to it (Thm: Suppose $f(x)$ is a piecewise smooth on the interval $0\le x\le L$. The Fourier sine series of $f(x)$ will be continuous and will converge to $f(x)$ on the interval provided $f(x)$ is continuous on the interval and $f(0)=0=f(L)$ . So, I would have a discontinuity jump, but I'm not sure how to represent it and apply it. Without it, I have the odd extension $f(x) = x \backsim \sum\limits_{n=1}^{\infty}\sin(nx)(-2/n)(\cos(n \pi))$ and the even extension is $f(x) = x=\pi/2+\sum\limits_{n=1}^{\infty} \cos(nx)(2/(\pi n^2))\cos(n\pi)$.

Answer 1

With your hypotheses, and for the function $f(x)=x$, the Fourier series of sines is equal to $f$ everywhere, except at $x=\pi$, where it is $0$ (as you can check explicitly, at this point, simply substituting $x=\pi$ in the series that you already wrote).

More generally, the Fourier series of sines of a function $f$ with the hypotheses that you described ("piecewise smooth", although strictly speaking one should be more rigorous, as Dirichlet deserves) is zero both at $x=0$ and at $x=\pi$, and is equal to $f$ on all other points.