Find the Fourier series expansion of
$$ f(x) = \left\{\begin{aligned} & -1, && -1 < x < 0\\ & \sin(\pi x), && 0 < x < 1 \end{aligned} \right.$$
and find the value of convergence when $x = 1$.
My Work
$a_n = \dfrac{1}{L}\int^L_{-L}f(x) \cos\left( \dfrac{n \pi x}{L} \right) dx$
$b_n = \dfrac{1}{L}\int^L_{-L}f(x) \sin\left( \dfrac{n \pi x}{L} \right) dx$
$a_n$
$\therefore a_n = \dfrac{1}{1} \left( \int^0_{-1} (1)\cos\left( \dfrac{n \pi x}{1} \right) dx + \int^1_0 \sin(\pi x) \cos\left( \dfrac{n \pi x}{1} \right) \right) dx$
$= \dfrac{1}{n \pi}\int^0_{-n\pi} \cos(u) du + \int^1_0 \sin(\pi x) \cos(n \pi x) dx$
We know that $\sin(u)\cos(v) = \dfrac{1}{2}\left[ \sin(u + v) + \sin(u - v) \right]$
$\therefore \sin(\pi x)\cos(n \pi t) = \dfrac{1}{2} \left[ \sin(\pi x + n \pi x) + \sin( \pi x - n \pi x) \right]$
$= \dfrac{1}{n \pi} \int^0_{-n\pi} \cos(u) du + \int^1_0 \dfrac{1}{2} \left[ \sin(\pi x + n \pi x) + \sin(\pi x - n \pi x) \right] dx$
$= \dfrac{1}{n \pi} \int^0_{-n \pi} \cos(u) du + \dfrac{1}{2} \int^1_0 \sin(\pi x + n \pi x) dx + \dfrac{1}{2} \int^1_0 \sin(\pi x - n \pi x) dx$
$= \dfrac{1}{n \pi} \int^0_{-n \pi} \cos(u) du + \dfrac{1}{2}\left( \dfrac{1}{\pi + n \pi} \right) \int_0^{\pi + n \pi} \sin(u) du + \dfrac{1}{2} \left( \dfrac{1}{\pi - n \pi} \right) \int^{\pi - n \pi}_0 \sin(u) du$
$= \dfrac{1}{n \pi} \left[ \sin(u) \right]^0_{-n \pi} + \dfrac{1}{2}\left( \dfrac{1}{\pi + n \pi} \right) \left[ -\cos(u) \right]^{\pi + n \pi}_0 + \dfrac{1}{2} \left( \dfrac{1}{\pi - n \pi} \right) [-\cos(u)]^{\pi - n \pi}_0$
We can see that $\left[ \sin(u) \right]^0_{-n \pi} = 0 \forall n \in \mathbb{N}$.
$= \dfrac{1}{2} \dfrac{1}{\pi + n \pi}(-\cos(\pi + n \pi)) + \dfrac{1}{2} \left( \dfrac{1}{\pi + n \pi} \right)(1) + \dfrac{1}{2} \left( \dfrac{1}{\pi - n \pi} \right)(-\cos(\pi - n \pi)) + \dfrac{1}{2} \left( \dfrac{1}{\pi - n \pi} \right)(1)$
$= \dfrac{1}{2} \left( \dfrac{1}{\pi + n \pi} \right) \left( -cos(\pi + n \pi) + 1 \right) + \dfrac{1}{2} \left( \dfrac{1}{\pi - n \pi} \right) \left( -cos(\pi - n \pi) + 1 \right) = a_n$
$b_n$
$b_n = \dfrac{1}{1} \left[ \int^0_{-1} (1)\sin\left( \dfrac{n \pi x}{1} \right) dx + \int^1_0 \sin(\pi x) \cdot \sin \left( \dfrac{n \pi x}{1} \right) dx \right]$
$= \int^0_{-1} \sin(\pi n x) dx + \int^1_0 \sin(\pi x) \cdot \sin(n \pi x) dx$
$= \dfrac{1}{n \pi} \int^0_{-n \pi} \sin(u) du + \int^1_0 \sin(\pi x) \cdot \sin(n \pi x) dx$
$\sin(u)\sin(v) = \dfrac{1}{2} \left[ \cos(u - v) - \cos(u + v) \right]$
$\therefore \sin(\pi x) \sin( \pi n x) = \dfrac{1}{2} \left[ \cos(\pi x - \pi n x) - \cos(\pi x + \pi n x) \right]$
$= \dfrac{1}{n \pi} \int^0_{-n \pi} \sin(u) du + \int^1_0 \dfrac{1}{2} \left[ \cos(\pi x - \pi n x) - \cos(\pi x + \pi n x) \right] dx$
$= \dfrac{1}{n \pi} \int^0_{-n \pi} \sin(u) du + \dfrac{1}{2} \int^1_0 \cos(\pi x - \pi n x) dx - \dfrac{1}{2} \int^1_0 \cos(\pi x + \pi n x) dx$
$= \dfrac{1}{n \pi}[-\cos(u)]^0_{-n \pi} + \dfrac{1}{2} \left( \dfrac{1}{\pi - \pi n} \right)[\sin(u)]^{\pi - \pi n}_0 - \dfrac{1}{2} \left( \dfrac{1}{\pi + \pi n} \right)[\sin(u)]^{\pi + \pi n}_0$
We can see that $[\sin(u)]^{\pi - \pi n}_0 = 0$ and $[\sin(u)]^{\pi + \pi n}_0 = 0 \forall n \in \mathbb{N}$.
$= \dfrac{1}{n \pi} (-1 + \cos(-\pi n)) = b_n$
$a_0$
$a_n = \dfrac{1}{2} \left( \dfrac{1}{\pi + n \pi} \right) \left( -cos(\pi + n \pi) + 1 \right) + \dfrac{1}{2} \left( \dfrac{1}{\pi - n \pi} \right) \left( -cos(\pi - n \pi) + 1 \right)$
$\therefore a_0 = \dfrac{2}{\pi} $
Calculating the Fourier expansion
$\therefore f(x) = \dfrac{1}{\pi} + \sum^{\infty}_{n = 1} \left[ \dfrac{1}{2} \left( \dfrac{1}{\pi + n \pi} \right) \left( -\cos(\pi + n \pi) + 1 \right) + \dfrac{1}{2} \left( \dfrac{1}{\pi - n \pi}\right) \left( -\cos(\pi - n \pi) + 1 \right) \right] \cdot \cos(\pi n x) + \left[ \dfrac{1}{n \pi} (-1 + \cos(- \pi n)) \right] \cdot \sin(\pi n x)$
We know that $\cos(\pi + n \pi) = (-1)^{n + 1}$, $\cos(\pi - n \pi) = (-1)^{n + 1}$, and $\cos(-\pi n) = (-1)^n$.
$\therefore f(x) = \dfrac{1}{\pi} + \dfrac{1}{2} \sum^{\infty}_{n = 2, even} \left[ \left( \dfrac{1}{\pi + n \pi} \right) ( (-1)^n + 1) + \left( \dfrac{1}{\pi - n \pi}\right)( (-1)^n + 1) \right] \cdot \cos(\pi n x) + \sum^{\infty}_{n = 1, odd} \dfrac{1}{n \pi} (-1 + (-1)^n ) \cdot \sin(\pi n x)$
I suspect that I have made some significant errors in my calculations. I would greatly appreciate it if people could please take the time to review my work and provide feedback.