I am trying to prove that for the fourier series expansion:
$$f(x) = \frac{a_0}{2} + \sum_{r=1}^\infty \biggl[a_r \cos(\frac{2\pi r x}{L})+b_r \sin(\frac{2 \pi r x}{L}) \biggl]$$
I am trying to derive:
$$a_r = \frac{2}{L} \int_{x_0}^{x_0+L}f(x) \cos(\frac{2\pi r x}{L})dx \tag{1}$$
I multiplied $\cos(\frac{2\pi p x}{L})$ and integrated over one full period in x and hence i get:
$$\int_{x_0}^{x_0+L}f(x) \cos(\frac{2\pi p x}{L})dx = \frac{a_0}{2}\int_{x_0}^{x_0+L}\cos(\frac{2\pi p x}{L})dx \\ + \sum_{r=1}^{\infty}a_r \int_{x_0}^{x_0+L}\cos(\frac{2\pi r x}{L})\cos(\frac{2\pi p x}{L}) dx \\ + \sum_{r=1}^{\infty}b_r \int_{x_0}^{x_0+L}\sin(\frac{2\pi r x}{L})\cos(\frac{2\pi p x}{L}) dx $$
Now using orthogonality for the conditions: $p\neq0$ and $p = r$
I obtain:
$$\int_{x_0}^{x_0+L}f(x) \cos(\frac{2\pi p x}{L})dx = \frac{a_0}{2}\int_{x_0}^{x_0+L}\cos(\frac{2\pi p x}{L})dx + \sum_{r=1}^{\infty} a_r \times \frac{L}{2} + 0 $$
Im not very sure how to continue from here to get $(1)$. Any hints?
I think the reason is more algebraic ! $$\left\{\frac{1}{\sqrt 2}, \cos\left(\frac{2\pi nx}{T}\right),\cos\left(\frac{2\pi nx}{T}\right)\right\}_{n\in\mathbb N}$$ is an orthonormal basis of the $T-$periodic function in $L^2(\mathbb R)$. Now, if a sequence $(f_n)$ converges to $f$ in $L^2(0,T)$ and to $g$ for a.e. $x\in (0,T)$, then $f=g$ a.e. in $(0,T)$. The claim follow.