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:

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

This needs considerable tedious hard slog to complete it. In particular: We need a result that the Fourier series over an interval is the restriction of the resulting periodic function. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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:

\(\ds \map T x\)

\(\sim\)

\(\ds \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \paren {2 n + 1} x\)

\(\ds \)

\(=\)

\(\ds \frac \pi 2 - \frac 4 \pi \paren {\cos x + \frac 1 {3^2} \cos 3 x + \frac 1 {5^2} 5 x + \dotsb}\)