Derivation of Fourier Series over General Range

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Theorem

Let $\alpha \in \R$ be a real number.

Let $\lambda \in \R_{>0}$ be a strictly positive real number.

Let $f: \R \to \R$ be a function such that $\ds \int_{\mathop \to \alpha}^{\mathop \to \alpha + 2 \lambda} \map f x \rd x$ converges absolutely.

Let:

$\ds f \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$


The Fourier coefficients for $f$ are calculated by:

\(\ds a_n\) \(=\) \(\ds \dfrac 1 \lambda \int_{\mathop \to \alpha}^{\mathop \to \alpha + 2 \lambda} \map f x \cos \frac {n \pi x} \lambda \rd x\)
\(\ds b_n\) \(=\) \(\ds \dfrac 1 \lambda \int_{\mathop \to \alpha}^{\mathop \to \alpha + 2 \lambda} \map f x \sin \frac {n \pi x} \lambda \rd x\)


Proof

By definition of Fourier series over the range of integration $\openint \alpha {\alpha + 2 \pi}$:

$(1): \quad \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$

where:

\(\ds a_n\) \(=\) \(\ds \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x\)
\(\ds b_n\) \(=\) \(\ds \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \sin n x \rd x\)


Let $\xi = \dfrac {\pi x} \lambda$.

Then let:

$\map \phi \xi \equiv \map f x$


We have that $\ds \int_{\mathop \to \alpha}^{\mathop \to \alpha + 2 \lambda} \map f x \rd x$ converges absolutely.

Thus we are ensured of the existence of:

the limit from the right of $f$ at $\alpha$
the limit from the left of $f$ at $\alpha + 2 \lambda$

Thus $f$ is defined and bounded on $\openint \alpha {\alpha + 2 \lambda}$.


Then:

\(\ds \map f {\alpha^+}\) \(=\) \(\ds \map \phi {\paren {\frac {\pi \alpha} \lambda}^+}\) Definition of Limit from Above
\(\ds \map f {\paren {\alpha + 2 \lambda}^-}\) \(=\) \(\ds \map \phi {\paren {\dfrac {\pi \alpha + \pi 2 \lambda} \lambda}^-}\)
\(\ds \) \(=\) \(\ds \map \phi {\paren {\frac {\pi \alpha} \lambda + 2 \pi}^-}\)

Setting $\beta = \dfrac {\pi \alpha} \lambda$, this allows us:

\(\ds \map f {\alpha^+}\) \(=\) \(\ds \map \phi {\beta^+}\)
\(\ds \map f {\paren {\alpha + 2 \lambda}^-}\) \(=\) \(\ds \map \phi {\paren {\beta + 2 \pi}^-}\)

Thus we have that $\phi$ is defined and bounded on $\openint \beta {\beta + 2 \pi}$.


Then:

$\map \phi \xi \sim \ds \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \xi + b_n \sin \xi}$

where:

\(\ds a_n\) \(=\) \(\ds \dfrac 1 \pi \int_{\mathop \to \pi}^{\mathop \to \alpha + 2 \lambda} \map \phi \xi \cos \xi \rd \xi\)
\(\ds b_n\) \(=\) \(\ds \dfrac 1 \pi \int_{\mathop \to \pi}^{\mathop \to \alpha + 2 \lambda} \map \phi \xi \sin \xi \rd \xi\)


We have that:

$\dfrac {\d \xi} {\d x} = \dfrac \pi \lambda$

and so:

\(\ds a_n\) \(=\) \(\ds \dfrac 1 \lambda \int_{\mathop \to \beta}^{\mathop \to \beta + 2 \lambda} \map f x \cos \frac {n \pi x} \lambda \rd x\)
\(\ds b_n\) \(=\) \(\ds \dfrac 1 \lambda \int_{\mathop \to \beta}^{\mathop \to \beta + 2 \lambda} \map f x \sin \frac {n \pi x} \lambda \rd x\)

and so:

$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$

$\blacksquare$


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