# Fourier Series over General Range from Specific

## Theorem

Let $a, b \in \R$ be real numbers.

Let $f: \R \to \R$ be a function such that $\displaystyle \int_a^b f \left({x}\right) \, \mathrm d x$ converges absolutely.

Then $f$ can be expressed by a Fourier series of the form:

$\displaystyle \frac {A_0} 2 + \sum_{m \mathop = 1}^\infty \left({A_m \cos \frac {2 m \pi \left({x - a}\right)} {b - a} + B_m \sin \frac {2 m \pi \left({x - a}\right)} {b - a} }\right)$

where:

 $\displaystyle A_m$ $=$ $\displaystyle \dfrac 2 {b - a} \int_a^b f \left({x}\right) \cos \frac {2 m \pi \left({x - a}\right)} {b - a} \, \mathrm d x$ $\displaystyle B_m$ $=$ $\displaystyle \dfrac 2 {b - a} \int_a^b f \left({x}\right) \sin \frac {2 m \pi \left({x - a}\right)} {b - a} \, \mathrm d x$

## Proof

Consider the Fourier series:

$(1): \quad \displaystyle S \left({x}\right) = \frac {A_0} 2 + \sum_{m \mathop = 1}^\infty \left({A_m \cos \frac {2 m \pi \left({x - a}\right)} {b - a} + B_m \sin \frac {2 m \pi \left({x - a}\right)} {b - a} }\right)$

Let $\xi = \dfrac {2 \pi \left({x - a}\right)} {b - a}$.

Then:

$\dfrac {\mathrm d \xi} {\mathrm d x} = \dfrac {2 \pi} {b - a}$

Then:

 $\displaystyle x$ $=$ $\displaystyle a$ $\displaystyle \leadsto \ \$ $\displaystyle \xi$ $=$ $\displaystyle \dfrac {2 \pi \left({a - a}\right)} {b - a}$ $\displaystyle$ $=$ $\displaystyle 0$

and:

 $\displaystyle x$ $=$ $\displaystyle b$ $\displaystyle \leadsto \ \$ $\displaystyle \xi$ $=$ $\displaystyle \dfrac {2 \pi \left({b - a}\right)} {b - a}$ $\displaystyle$ $=$ $\displaystyle 2 \pi$

Substituting $\xi$ for $x$ in $(1)$:

$(2): \quad \displaystyle S \left({\xi}\right) = \frac {A_0} 2 + \sum_{m \mathop = 1}^\infty \left({A_m \cos m \xi + B_m \sin m \xi}\right)$

which is a Fourier series for a real function $g \left({\xi}\right)$ such that $\displaystyle \int_0^{2 \pi} g \left({\xi}\right) \, \mathrm d x$ converges absolutely.

The Fourier coefficients of $g \left({\xi}\right)$ are given by:

 $\displaystyle A_n$ $=$ $\displaystyle \dfrac 1 \pi \int_0^{2 \pi} g \left({\xi}\right) \cos n \xi \, \mathrm d \xi$ $\displaystyle B_n$ $=$ $\displaystyle \dfrac 1 \pi \int_0^{2 \pi} g \left({\xi}\right) \sin n \xi \, \mathrm d \xi$

Substituting $x$ for $\xi$:

 $\displaystyle A_n$ $=$ $\displaystyle \dfrac 1 \pi \dfrac {2 \pi} {b - a} \int_a^b g \left({x}\right) \cos n \dfrac {2 \pi \left({x - a}\right)} {b - a} \, \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 {b - a} \int_a^b g \left({x}\right) \cos \dfrac {2 n \pi \left({x - a}\right)} {b - a} \, \mathrm d x$

and:

 $\displaystyle B_n$ $=$ $\displaystyle \dfrac 1 \pi \dfrac {2 \pi} {b - a} \int_a^b g \left({x}\right) \sin n \dfrac {2 \pi \left({x - a}\right)} {b - a} \, \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 {b - a} \int_a^b g \left({x}\right) \sin \dfrac {2 n \pi \left({x - a}\right)} {b - a} \, \mathrm d x$

Thus while $A_n$ and $B_n$ are exactly the Fourier coefficients of $g \left({\xi}\right)$, they are also exactly the Fourier coefficients of $S \left({x}\right)$.

Thus $S \left({x}\right)$ is the Fourier series for $f \left({x}\right)$.

$\blacksquare$