Fourier Series over General Range from Specific

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Theorem

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

Let $f: \R \to \R$ be a function such that $\ds \int_a^b \map f x \rd x$ converges absolutely.


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

$\ds \frac {A_0} 2 + \sum_{m \mathop = 1}^\infty \paren {A_m \cos \frac {2 m \pi \paren {x - a} } {b - a} + B_m \sin \frac {2 m \pi \paren {x - a} } {b - a} }$


where:

\(\ds A_m\) \(=\) \(\ds \dfrac 2 {b - a} \int_a^b \map f x \cos \frac {2 m \pi \paren {x - a} } {b - a} \rd x\)
\(\ds B_m\) \(=\) \(\ds \dfrac 2 {b - a} \int_a^b \map f x \sin \frac {2 m \pi \paren {x - a} } {b - a} \rd x\)


Proof

Consider the Fourier series:

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

Let $\xi = \dfrac {2 \pi \paren {x - a} } {b - a}$.

Then:

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

Then:

\(\ds x\) \(=\) \(\ds a\)
\(\ds \leadsto \ \ \) \(\ds \xi\) \(=\) \(\ds \dfrac {2 \pi \paren {a - a} } {b - a}\)
\(\ds \) \(=\) \(\ds 0\)


and:

\(\ds x\) \(=\) \(\ds b\)
\(\ds \leadsto \ \ \) \(\ds \xi\) \(=\) \(\ds \dfrac {2 \pi \paren {b - a} } {b - a}\)
\(\ds \) \(=\) \(\ds 2 \pi\)


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

$(2): \quad \map S \xi = \dfrac {A_0} 2 + \ds \sum_{m \mathop = 1}^\infty \paren {A_m \cos m \xi + B_m \sin m \xi}$

which is a Fourier series for a real function $\map g \xi$ such that $\ds \int_0^{2 \pi} \map g \xi \rd x$ converges absolutely.


The Fourier coefficients of $\map g \xi$ are given by:

\(\ds A_n\) \(=\) \(\ds \dfrac 1 \pi \int_0^{2 \pi} \map g \xi \cos n \xi \rd \xi\)
\(\ds B_n\) \(=\) \(\ds \dfrac 1 \pi \int_0^{2 \pi} \map g \xi \sin n \xi \rd \xi\)


Substituting $x$ for $\xi$:

\(\ds A_n\) \(=\) \(\ds \dfrac 1 \pi \dfrac {2 \pi} {b - a} \int_a^b \map g x \cos n \dfrac {2 \pi \paren {x - a} } {b - a} \rd x\)
\(\ds \) \(=\) \(\ds \dfrac 1 {b - a} \int_a^b \map g x \cos \dfrac {2 n \pi \paren {x - a} } {b - a} \rd x\)

and:

\(\ds B_n\) \(=\) \(\ds \dfrac 1 \pi \dfrac {2 \pi} {b - a} \int_a^b \map g x \sin n \dfrac {2 \pi \paren {x - a} } {b - a} \rd x\)
\(\ds \) \(=\) \(\ds \dfrac 1 {b - a} \int_a^b \map g x \sin \dfrac {2 n \pi \paren {x - a} } {b - a} \rd x\)


Thus while $A_n$ and $B_n$ are exactly the Fourier coefficients of $\map g \xi$, they are also exactly the Fourier coefficients of $\map S x$.

Thus $\map S x$ is the Fourier series for $\map f x$.

$\blacksquare$


Sources

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