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In mathematics, the Dirichlet–Jordan test gives sufficient conditions for a real-valued, periodic function f to be equal to the sum of its Fourier series at a point of continuity. Moreover, the behavior of the Fourier series at points of discontinuity is determined as well (it is the midpoint of the values of the discontinuity). It is one of many conditions for the convergence of Fourier series. These conditions are named after Peter Gustav Lejeune Dirichlet and Camille Jordan.

Dirichlet–Jordan test for Fourier series

The Dirichlet–Jordan test states^[1] that if a periodic function $f(x)$ is of bounded variation on a period, then the Fourier series $S_{n}f(x)$ converges, as $n\to \infty$ , at each point of the domain to Failed to parse (unknown function "\evarpsilon"): {\displaystyle \lim_{\varepsilon\to 0}\frac{f(x+\evarpsilon)+f(x-\varepsilon)}{2}.} In particular, if $f$ is continuous at $x$ , then the Fourier series converges to $f(x)$ . Moreover, if $f$ is continuous everywhere, then the convergence is uniform.

Stated in terms of a periodic function of period 2π, the Fourier series coefficients are defined as $a_{k}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x)e^{-ikx}\,dx,$ and the partial sums of the Fourier series are $S_{n}f(x)=\sum _{k=-n}^{n}a_{k}e^{ikx}$

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

There is also a pointwise version of the test:^[2] if $f$ is a periodic function in $L^{1}$ , and is of bounded variation in a neighborhood of $x$ , then the Fourier series at $x$ converges to the limit as above $\lim _{\varepsilon \to 0}{\frac {f(x+\varepsilon )+f(x-\varepsilon )}{2}}.$

Jordan test for Fourier integrals

For the Fourier transform on the real line, there is a version of the test as well.^[3] Suppose that $f(x)$ is in $L^{1}(-\infty ,\infty )$ and of bounded variation in a neighborhood of the point $x$ . Then ${\frac {1}{\pi }}\lim _{M\to \infty }\int _{0}^{M}du\int _{-\infty }^{\infty }f(t)\cos u(x-t)\,dt=\lim _{\varepsilon \to 0}{\frac {f(x+\varepsilon )+f(x-\varepsilon )}{2}}.$ If $f$ is continuous in an open interval, then the integral on the left-hand side converges uniformly in the interval, and the limit on the right-hand side is $f(x)$ .

Dirichlet conditions in signal processing

In signal processing, the test is often formulated as a pair of conditions of a signal. First, that the signal should be absolutely integrable (that is $L^{1}$ ) which guarantees the existence of the Fourier series, and secondly that the signal should be bounded with only finitely many local extrema and finitely many discontinuities.^[4] Such a signal can be easily shown to be of bounded variation (the converse is not true). Any signal that can be physically produced in a laboratory satisfies these conditions.^[5]

References

^ Antoni Zygmund (1952), Trigonometric series, Cambridge University Press
^ R. E. Edwards (1967), Fourier series: a modern introduction, Springer.
^ E. C. Titchmarsh (1948), Introduction to the theory of Fourier integrals, Oxford Clarendon Press, p. 13.
^ Alan V. Oppenheim; Alan S. Willsky; Syed Hamish Nawab (1997). Signals & Systems. Prentice Hall. p. 198. ISBN 9780136511755.
^ B P Lathi (2000), Signal processing and linear systems, Oxford

External links

"Dirichlet conditions". PlanetMath.

[1] Antoni Zygmund (1952), Trigonometric series, Cambridge University Press

[2] R. E. Edwards (1967), Fourier series: a modern introduction, Springer.

[3] E. C. Titchmarsh (1948), Introduction to the theory of Fourier integrals, Oxford Clarendon Press, p. 13.

[sands-4] Alan V. Oppenheim; Alan S. Willsky; Syed Hamish Nawab (1997). Signals & Systems. Prentice Hall. p. 198. ISBN 9780136511755.

[5] B P Lathi (2000), Signal processing and linear systems, Oxford

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@@ Line 4: / Line 4: @@
 ==Dirichlet–Jordan test for Fourier series==
 The Dirichlet–Jordan test states<ref>{{citation|author=[[Antoni Zygmund]]|title=Trigonometric series|year=1952|publisher=Cambridge University Press}}</ref> that if a periodic function <math>f(x)</math> is of [[bounded variation]] on a period, then the Fourier series <math>S_nf(x)</math> converges, as <math>n\to\infty</math>, at each point of the domain to
-<math display="block">\lim_{\epsilon\to 0}\frac{f(x+\epsilon)+f(x-\epsilon)}{2}.</math>
+<math display="block">\lim_{\varepsilon\to 0}\frac{f(x+\evarpsilon)+f(x-\varepsilon)}{2}.</math>
 In particular, if <math>f</math> is continuous at <math>x</math>, then the Fourier series converges to <math>f(x)</math>. Moreover, if <math>f</math> is continuous everywhere, then the convergence is uniform.
@@ Line 15: / Line 15: @@
 There is also a pointwise version of the test:<ref>{{citation|author=[[R. E. Edwards]]|title=Fourier series: a modern introduction|publisher=Springer|year=1967}}.</ref> if <math>f</math> is a periodic function in <math>L^1</math>, and is of bounded variation in a neighborhood of <math>x</math>, then the Fourier series at <math>x</math> converges to the limit as above
-<math display="block">\lim_{\epsilon\to 0}\frac{f(x+\epsilon)+f(x-\epsilon)}{2}.</math>
+<math display="block">\lim_{\varepsilon\to 0}\frac{f(x+\varepsilon)+f(x-\varepsilon)}{2}.</math>
 == Jordan test for Fourier integrals ==
 For the [[Fourier transform]] on the real line, there is a version of the test as well.<ref>{{citation|author=[[E. C. Titchmarsh]]|title=Introduction to the theory of Fourier integrals|year=1948|page=13|publisher=Oxford Clarendon Press}}.</ref>  Suppose that <math>f(x)</math> is in <math>L^1(-\infty,\infty)</math> and of bounded variation in a neighborhood of the point <math>x</math>.  Then
-<math display="block">\frac1\pi\lim_{M\to\infty}\int_0^{M}du\int_{-\infty}^\infty f(t)\cos u(x-t)\,dt = \lim_{\epsilon\to 0}\frac{f(x+\epsilon)+f(x-\epsilon)}{2}.</math>
+<math display="block">\frac1\pi\lim_{M\to\infty}\int_0^{M}du\int_{-\infty}^\infty f(t)\cos u(x-t)\,dt = \lim_{\varepsilon\to 0}\frac{f(x+\varepsilon)+f(x-\varepsilon)}{2}.</math>
 If <math>f</math> is continuous in an open interval, then the integral on the left-hand side converges uniformly in the interval, and the limit on the right-hand side is <math>f(x)</math>.