Dirichlet–Jordan test

Not to be confused with Dirichlet boundary condition.

In mathematics, the Dirichlet conditions are sufficient condition for a real-valued, periodic function f(x) to be equal the sum of its Fourier series at each point where f is continuous. Moreover, the behavior at points of discontinuity is determined as well. These conditions are named after Johann Peter Gustav Lejeune Dirichlet.

The conditions are:

• f(x) must have a finite number of extrema in any given interval
• f(x) must have a finite number of discontinuities in any given interval
• f(x) must be absolutely integrable over a period.

Dirichlet's Theorem for 1-Dimensional Fourier Series

We state Dirichlet's theorem assuming f is a periodic function of period 2π with Fourier series expansion

$f(x)\sim \sum _{n=-\infty }^{\infty }a_{n}e^{inx}$,

where

$a_{n}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x)e^{-inx}\,dx.$

The analogous statement holds irrespective of what the period of f is, or which version of the Fourier expansion is chosen.

Dirichlet's theorem: If f satisfies Dirichlet conditions, then for all x, we have that the series obtained by plugging x into the Fourier series is convergent, and is given by

$\sum _{n=-\infty }^{\infty }a_{n}e^{inx}={\frac {1}{2}}(f(x+)+f(x-))$,

where the notation

$f(x+)=\lim _{y\to x^{+}}f(y)$

$f(x-)=\lim _{y\to x^{-}}f(y)$

denotes the right/left limits of f.

A function satisfying Dirichlet's conditions must have right and left limits at each point of discontinuity, or else the function would need to oscillate at that point, violating the condition on maxima/minima. Note that at any point where f is continuous,

$\displaystyle f(x+)=f(x-)=f(x)$

so

${\frac {1}{2}}(f(x+)+f(x-))=f(x)$.

Thus Dirichlet's theorem says in particular that the Fourier series for f converges and is equal to f wherever f is continuous.

External links

• "Dirichlet conditions". PlanetMath.