# Fourier Series/Triangle Wave

## Theorem

Triangle Wave and $9$th Approximation

Let $\map T x$ be the triangle wave defined on the real numbers $\R$ as:

$\forall x \in \R: \map T x = \begin {cases} \size x & : x \in \closedint {-l} l \\ \map T {x + 2 l} & : x < -l \\ \map T {x - 2 l} & : x > +l \end {cases}$

where:

$l$ is a given real constant
$\size x$ denotes the absolute value of $x$.

Then its Fourier series can be expressed as:

 $\ds \map T x$ $\sim$ $\ds \frac l 2 - \frac {4 l} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} l$ $\ds$ $=$ $\ds \frac l 2 - \frac {4 l} {\pi^2} \paren {\cos \dfrac {\pi x} l + \frac 1 {3^2} \cos \dfrac {3 \pi x} l + \frac 1 {5^2} \cos \dfrac {5 \pi x} l + \dotsb}$

## Proof

Let $\map f x: \openint {-l} l \to \R$ denote the absolute value function on the open interval $\openint {-l} l$:

$\map f x = \size x = \begin{cases} x & : x > 0 \\ 0 & : x = 0 \\ -x & : x < 0 \end{cases}$

From Fourier Series for Absolute Value Function over Symmetric Range, $\map f x$ can immediately be expressed as:

$\ds \map f x \sim \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda$

## Special Cases

### Unit Half Interval

Let $\map T x$ be the triangle wave defined on the real numbers $\R$ as:

$\forall x \in \R: \map T x = \begin {cases} \size x & : x \in \closedint {-1} 1 \\ \map T {x + 2} & : x < -1 \\ \map T {x - 2} & : x > +1 \end {cases}$

where $\size x$ denotes the absolute value of $x$.

Then its Fourier series can be expressed as:

 $\ds \map T x$ $\sim$ $\ds \frac 1 2 - \frac 4 {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \paren {2 n + 1} \pi x$ $\ds$ $=$ $\ds \frac 1 2 - \frac 4 {\pi^2} \paren {\cos \pi x + \frac 1 {3^2} \cos 3 \pi x + \frac 1 {5^2} 5 \pi x + \dotsb}$

### Half Interval $\pi$

Let $\map T x$ be the triangle wave defined on the real numbers $\R$ as:

$\forall x \in \R: \map T x = \begin {cases} \size x & : x \in \closedint {-\pi} \pi \\ \map T {x + 2 \pi} & : x < -\pi \\ \map T {x - 2 \pi} & : x > +\pi \end {cases}$

where $\size x$ denotes the absolute value of $x$.

Then its Fourier series can be expressed as:

 $\ds \map T x$ $\sim$ $\ds \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \paren {2 n + 1} x$ $\ds$ $=$ $\ds \frac \pi 2 - \frac 4 \pi \paren {\cos x + \frac 1 {3^2} \cos 3 x + \frac 1 {5^2} 5 x + \dotsb}$