How do I find the function to which the fourier series
$$g(x) = -\frac{4}{\pi} \sum_{n =1}^\infty \frac{sin((2n-1)x)}{(2n-1)^3}$$
if I know that the function
$$|x| = \frac{\pi}{2} -\frac{4}{\pi} \sum_{n =1}^\infty \frac{\cos((2n-1)x)}{(2n-1)^3}$$
for all $x\in \mathbb{T}^1$, and the convergence is uniform?
From what I understand, since I know that the fourier serie of $|x|$ converges uniformely, then, I can rewrite the equality to
$$|x| - \frac{\pi}{2} =-\frac{4}{\pi} \sum_{n =1}^\infty \frac{sin((2n-1)x)}{(2n-1)^3}$$
Also from uniform convergence, I know that I can exchange limits process, thefore
\begin{align*} \int_{0}^x |t| - \frac{\pi}{2} dt &= \int_{0}^x \left(-\frac{4}{\pi} \sum_{n =1}^\infty \frac{\cos((2n-1)t)}{(2n-1)^3}\right) dt\\ &= -\frac{4}{\pi} \sum_{n =1}^\infty \int_0^x \frac{\cos((2n-1)t)}{(2n-1)^3}dt \\ &= -\frac{4}{\pi} \sum_{n =1}^\infty \frac{\sin((2n-1)x)}{(2n-1)^3} \end{align*} So normally I would juste integrate the absulte value by separating in two cases, i.e $$\int_0^x |t| - \frac{\pi}{2} dt = \frac{x(x-\pi)}{2}$$ if $x \in [0, \pi]$ and $$\int_0^x |t| - \frac{\pi}{2} dt = \frac{-x(x-\pi)}{2}$$ if $x \in (0, \pi)$ But then, do I have two piecewise polynomials on for the function to which the serie
$$-\frac{4}{\pi} \sum_{n =1}^\infty \frac{\sin((2n-1)x)}{(2n-1)^3}$$
converges, i.e for which $x\in \mathbb{T}^1$ is this true?
(I have a problem set on fourier series and it asks to find, knowing that, a piecewiese polynomial such that the serie converges on subintervals of the form (xi, xi+1). However I can't see why it shouldnt be on the whole interval)
Thank you!