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# Dirichlet kernel

The Dirichlet kernel, named after Johann Peter Gustav Lejeune Dirichlet,

$D_n(x)=\sum_{k=-n}^n e^{ikx}=1+2\sum_{k=1}^n\cos(kx)=\frac{\sin\left(\left(n+\frac{1}{2}\right)x\right)}{\sin(x/2)},$ is 2π times the nth-degree Fourier series approximation to a "function" with period 2π given by
$\delta_p(x)=\sum_{k=-\infty}^\infty\delta(x-2\pi k)$
where δ is the Dirac delta function, which is not really a function, in the sense of mapping one set into another, but is rather a "generalized function", also called a "distribution". In other words, the Fourier series representation of this "function" is
$\delta_p(x)=\frac{1}{2\pi}\sum_{k=-\infty}^\infty e^{ikx}=\frac{1}{2\pi}\left(1+2\sum_{k=1}^\infty\cos(kx)\right).$
This "periodic delta function" is the identity element for the convolution defined on functions of period 2π by
$(f*g)(x)=\frac{1}{2\pi}\int_{-\pi}^\pi f(y)g(x-y)\,dy.$
In other words, we have
$f*\delta_p=\delta_p*f=f$
for every function f of period 2π. The convolution of Dn(x) with any function f of period 2π is the nth-degree Fourier series approximation to f, i.e., we have
$(D_n*f)(x)=(f*D_n)(x)=\frac{1}{2\pi}\int_{-\pi}^\pi f(y)D_n(x-y)\,dy=\sum_{k=-n}^n \hat{f}(k)e^{ikx},$
where
$\hat{f}(k)=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-ikx}\,dx$
is the kth Fourier coefficient of f.

The trigonometric identity displayed at the top of this article may be established as follows. First recall that the sum of a finite geometric series is

$\sum_{k=0}^n a r^k=a\frac{1-r^{n+1}}{1-r}.$
The first term is a; the common ratio by which each term is multiplied to get the next is r; the number of terms is n + 1. In particular, we have
$\sum_{k=-n}^n r^k=r^{-n}\cdot\frac{1-r^{2n+1}}{1-r}.$
The expression to the left of "=" should make us expect the sum to be a symmetric function of r and 1/r, but the expression to the right of "=" is perhaps less-than-obviously symmetric in those two quantities. The remedy is to multiply both the numerator and the denominator by r-1/2, getting
$\frac{r^{-n-1/2}}{r^{-1/2}}\cdot\frac{1-r^{2n+1}}{1-r}=\frac{r^{-n-1/2}-r^{n+1/2}}{r^{-1/2}-r^{1/2}}.$
In case r = eix we have
$\sum_{k=-n}^n e^{ikx}=\frac{e^{-(n+1/2)ix}-e^{(n+1/2)ix}}{e^{-ix/2}-e^{ix/2}}=\frac{-2i\sin((n+1/2)x)}{-2i\sin(x/2)}$
and then "-2i" cancels.

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