The problem asks us to prove the formulas for the Fourier coefficients $a_n$ and $b_n$ given the Fourier series expansion of a function $f(x)$ defined on the interval $(-L, L)$ with period $2L$. The Fourier series is given by $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right)$. We need to prove that $a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx$ and $b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx$.

1. Problem Description

The problem asks us to prove the formulas for the Fourier coefficients $a_n$ and $b_n$ given the Fourier series expansion of a function $f(x)$ defined on the interval $(-L, L)$ with period $2L$. The Fourier series is given by

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

We need to prove that

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

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

2. Solution Steps

(i) Derivation of $a_n$:

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

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

We can interchange the integral and the summation (under appropriate conditions). Then, we get

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

Using the orthogonality properties of sine and cosine functions:

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

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

$\int_{-L}^{L} \cos\left(\frac{m\pi x}{L}\right) dx = 0$ for $m \neq 0$.

Therefore, when $m \neq 0$,

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

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

Replacing $m$ with $n$, we have

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

(ii) Derivation of $b_n$:

Similarly, to find $b_n$, we multiply both sides of the Fourier series equation by $\sin\left(\frac{m\pi x}{L}\right)$ and integrate from $-L$ to $L$:

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

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

Using the orthogonality properties:

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

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

$\int_{-L}^{L} \sin\left(\frac{m\pi x}{L}\right) dx = 0$ for all $m$.

Therefore,

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

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

Replacing $m$ with $n$, we have

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

3. Final Answer

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

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