Fourier's Theorem

== Theorem ==
Let $\alpha \in \R$ be a real number.
Let $\map f x$ be a real function which is defined and bounded on the interval $\openint \alpha {\alpha + 2 \pi}$.
Let $f$ satisfy the Dirichlet conditions on $\openint \alpha {\alpha + 2 \pi}$:
Outside the interval $\openint \alpha {\alpha + 2 \pi}$, let $f$ be periodic and defined such that:
:$\map f x = \map f {x + 2 \pi}$
Let $f$ be defined by the Fourier series:
:$(1): \quad \ds \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$
such that:
:$\ds a_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x$
:$\ds b_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \sin n x \rd x$
Then for all $a \in \R$, $(1)$ converges to the sum:
:$\ds \frac 1 2 \paren {\lim_{x \mathop \to a^+} \map f x + \lim_{x \mathop \to a^-} \map f x}$
where the $\lim$ symbols denote the limit from the right and the limit from the left.
=== Integral Form ===
This theorem can often be seen presented in the form:
== Proof ==
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
=== Main Theorem ===
Let $\map {S_N} x$ denote the first $N$ terms of the Fourier series:
:$(2): \quad \map {S_N} x = \ds \frac {a_0} 2 + \sum_{n \mathop = 1}^N \paren {a_n \cos n x + b_n \sin n x}$
where:
:$(3): \quad \ds a_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x$
:$(4): \quad \ds b_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \sin n x \rd x$
Substituting from $(3)$ and $(4)$ into $(2)$ and applying Integral of Integrable Function is Additive:
:$\map {S_N} x = \ds \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f u \paren {\frac 1 2 + \sum_{n \mathop = 1}^N \paren {\cos n x \cos n u + \sin n x \sin n u} } \rd u$
Now we have:
Hence:
:$\ds \map {S_N} x = \int_\alpha^{\alpha + 2 \pi} \map \psi u \frac {\map \sin {\paren {N + \frac 1 2} \paren {u - x} } } {2 \, \map \sin {\frac 1 2 \paren {u - x} } } \rd u$
where:
:$\map \psi u = \dfrac 1 \pi \map f u \dfrac {\frac 1 2 \paren {u - x} } {\sin \frac 1 2 \paren {u - x} }$
We have that $\map f u$ satisfies the Dirichlet conditions on $\openint \alpha {\alpha + 2 \pi}$.
Hence $f$ is piecewise smooth on $\openint \alpha {\alpha + 2 \pi}$.
That is, $f$ has right-hand derivative and left-hand derivative at all $x$ in $\openint \alpha {\alpha + 2 \pi}$.
Thus at the point $u = x$, $f$ has right-hand derivative and left-hand derivative, and so does $\map \psi u$.
So by Fourier's Theorem: Lemma 3:
:$\ds \lim_{n \mathop \to N} \map {S_N} x = \frac \pi 2 \paren {\map \psi {x^+} + \map \psi {x^-} }$
Now:
:$\map \psi {x^+} = \ds \frac 1 \pi \map f {x^+} \lim_{u \mathop \to x} \dfrac {\frac 1 2 \paren {u - x} } {\sin \frac 1 2 \paren {u - x} } = \frac 1 \pi \map f {x^+}$
and:
:$\map \psi {x^-} = \ds \frac 1 \pi \map f {x^-} \lim_{u \mathop \to x} \dfrac {\frac 1 2 \paren {u - x} } {\sin \frac 1 2 \paren {u - x} } = \frac 1 \pi \map f {x^-}$
and so:
:$\ds \lim_{n \mathop \to N} \map {S_N} x = \frac 1 2 \paren {\lim_{x \mathop \to a^+} \map f x + \lim_{x \mathop \to a^-} \map f x}$
== Also known as ==
* Fourier's Theorem is also known as Dirichlet's Theorem for 1-Dimensional Fourier Series.
== Sources ==
* : Chapter One: $\S 2$. Fourier Series
* : Chapter Two: $\S 3$. A Fourier Theorem: Theorem $1$
Category:Fourier Series
Category:Piecewise Continuous Functions
Category:Fourier's Theorem