The problem states that if a function $f(x)$ is defined in an open interval $(-L, L)$ and has period $2L$, its Fourier series expansion is given by $f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos(\frac{n\pi x}{L}) + b_n \sin(\frac{n\pi x}{L}))$. We are asked to prove the formulas for the Fourier coefficients $a_n$ and $b_n$: (i) $a_n = \frac{1}{L}\int_{-L}^{L} f(x) \cos(\frac{n\pi x}{L}) dx$ (ii) $b_n = \frac{1}{L}\int_{-L}^{L} f(x) \sin(\frac{n\pi x}{L}) dx$

1. Problem Description

The problem states that if a function $f(x)$ is defined in an open interval $(-L, L)$ and has period $2L$, its Fourier series expansion is given by

$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos(\frac{n\pi x}{L}) + b_n \sin(\frac{n\pi x}{L}))$.

We are asked to prove the formulas for the Fourier coefficients $a_n$ and $b_n$:

(i) $a_n = \frac{1}{L}\int_{-L}^{L} f(x) \cos(\frac{n\pi x}{L}) dx$

(ii) $b_n = \frac{1}{L}\int_{-L}^{L} f(x) \sin(\frac{n\pi x}{L}) dx$

2. Solution Steps

To find $a_n$, we multiply both sides of the Fourier series equation by $\cos(\frac{m\pi x}{L})$, where $m$ is an integer, and integrate from $-L$ to $L$:

$\int_{-L}^{L} f(x) \cos(\frac{m\pi x}{L}) dx = \int_{-L}^{L} [\frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos(\frac{n\pi x}{L}) + b_n \sin(\frac{n\pi x}{L}))] \cos(\frac{m\pi x}{L}) dx$

We can interchange the integral and summation:

$\int_{-L}^{L} f(x) \cos(\frac{m\pi x}{L}) dx = \int_{-L}^{L} \frac{a_0}{2} \cos(\frac{m\pi x}{L}) dx + \sum_{n=1}^\infty [\int_{-L}^{L} a_n \cos(\frac{n\pi x}{L}) \cos(\frac{m\pi x}{L}) dx + \int_{-L}^{L} b_n \sin(\frac{n\pi x}{L}) \cos(\frac{m\pi x}{L}) dx]$

We use the following orthogonality properties:

$\int_{-L}^{L} \cos(\frac{n\pi x}{L}) \cos(\frac{m\pi x}{L}) dx = \begin{cases} 0, & n \neq m \\ L, & n = m, m \neq 0 \\ 2L, & n = m = 0 \end{cases}$

$\int_{-L}^{L} \sin(\frac{n\pi x}{L}) \cos(\frac{m\pi x}{L}) dx = 0$ for all $n, m$

$\int_{-L}^{L} \cos(\frac{m\pi x}{L}) dx = \begin{cases} 0, & m \neq 0 \\ 2L, & m = 0 \end{cases}$

If $m \neq 0$, then $\int_{-L}^{L} \frac{a_0}{2} \cos(\frac{m\pi x}{L}) dx = 0$. Then the first term in the summation is $a_m L$ when $n=m$, and 0 otherwise. The second term in the summation is always

0. Thus:

$\int_{-L}^{L} f(x) \cos(\frac{m\pi x}{L}) dx = a_m L$

$a_m = \frac{1}{L} \int_{-L}^{L} f(x) \cos(\frac{m\pi x}{L}) dx$

Replacing $m$ with $n$, we get the desired result for $a_n$.

To find $b_n$, we multiply both sides of the Fourier series equation by $\sin(\frac{m\pi x}{L})$, where $m$ is an integer, and integrate from $-L$ to $L$:

$\int_{-L}^{L} f(x) \sin(\frac{m\pi x}{L}) dx = \int_{-L}^{L} [\frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos(\frac{n\pi x}{L}) + b_n \sin(\frac{n\pi x}{L}))] \sin(\frac{m\pi x}{L}) dx$

Interchange the integral and summation:

$\int_{-L}^{L} f(x) \sin(\frac{m\pi x}{L}) dx = \int_{-L}^{L} \frac{a_0}{2} \sin(\frac{m\pi x}{L}) dx + \sum_{n=1}^\infty [\int_{-L}^{L} a_n \cos(\frac{n\pi x}{L}) \sin(\frac{m\pi x}{L}) dx + \int_{-L}^{L} b_n \sin(\frac{n\pi x}{L}) \sin(\frac{m\pi x}{L}) dx]$

We use the following orthogonality properties:

$\int_{-L}^{L} \sin(\frac{n\pi x}{L}) \sin(\frac{m\pi x}{L}) dx = \begin{cases} 0, & n \neq m \\ L, & n = m \end{cases}$

$\int_{-L}^{L} \cos(\frac{n\pi x}{L}) \sin(\frac{m\pi x}{L}) dx = 0$

$\int_{-L}^{L} \sin(\frac{m\pi x}{L}) dx = 0$

Then $\int_{-L}^{L} \frac{a_0}{2} \sin(\frac{m\pi x}{L}) dx = 0$. The first term in the summation is always

0. The second term in the summation is $b_m L$ when $n=m$, and 0 otherwise. Thus:

$\int_{-L}^{L} f(x) \sin(\frac{m\pi x}{L}) dx = b_m L$

$b_m = \frac{1}{L} \int_{-L}^{L} f(x) \sin(\frac{m\pi x}{L}) dx$

Replacing $m$ with $n$, we get the desired result for $b_n$.

3. Final Answer

(i) $a_n = \frac{1}{L}\int_{-L}^{L} f(x) \cos(\frac{n\pi x}{L}) dx$

(ii) $b_n = \frac{1}{L}\int_{-L}^{L} f(x) \sin(\frac{n\pi x}{L}) dx$