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Dirichlet kernel

In mathematical analysis, the Dirichlet kernel is the collection of functions

$D_n(x)=\sum_{k=-n}^n e^{ikx}=1+2\sum_{k=1}^n\cos(kx)=\frac{\sin\left(\left(n +\frac{1}{2}\right)x\right)}{\sin(x/2)}.$

It is named after Johann Peter Gustav Lejeune Dirichlet.

The importance of the Dirichlet kernel comes from its relation to Fourier series. The convolution of D_n(x) with any function f of period 2π is the nth-degree Fourier series approximation to f, i.e., we have

$(D_n*f)(x)=\frac{1}{2\pi}\int_{-\pi}^\pi f(y)D_n(x-y)\,dy=\sum_{k=-n}^n \hat{f}(k)e^{ikx},$

where

$\hat{f}(k)=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-ikx}\,dx$

is the kth Fourier coefficient of f. This implies that in order to study convergence of Fourier series it is enough to study properties of the Dirichlet kernel. Of particular importance is the fact that the L¹ norm of D_n diverges to infinity as $n\to\infty$ . This fact is behind many divergence phenomena. For example, the it is the direct reason that the Fourier series of a continuous function may diverge. See convergence of Fourier series for more.

Relation to the delta function

Take the periodic 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", and multiply by 2π. We get the identity element for convolution on functions of period 2π. In other words, we have

$f*(2\pi \delta)=f \,$

for every function f of period 2π. The Fourier series representation of this "function" is

$2\pi \delta(x)\sim\sum_{k=-\infty}^\infty e^{ikx}=\left(1 +2\sum_{k=1}^\infty\cos(kx)\right).$

Therefore the Dirichlet kernel, which are just the partial sums of this series, can be thought of as an approximate identity.

Proof of the trigonometric identity

The trigonometric identity

$\sum_{k=-n}^n e^{ikx} =\frac{\sin\left(\left(n+\frac{1}{2}\right)x\right)}{\sin(x/2)}$

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 = e^ix 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.

Categories: Mathematical analysis

Last updated: 06-02-2005 00:45:50

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09-23-2007 01:00:40
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