Dirichlet kernel

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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, it is the direct reason that the Fourier series of a continuous function may diverge. See convergence of Fourier series for more.

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 is just the sequence of partial sums of this series, can be thought of as an approximate identity. Abstractly speaking it is not however an approximate identity of positive elements (hence the failures mentioned above).

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

$\sum_{k=0}^n a r^k=a\frac{1-r^{n+1}}{1-r}.$

In particular, we have

$\sum_{k=-n}^n r^k=r^{-n}\cdot\frac{1-r^{2n+1}}{1-r}.$

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.

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Categories: Mathematical analysis | Fourier series | Approximation theory

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