Dirichlet kernel

The Dirichlet kernel,

$D_{n}(x)=\sum _{k=-n}^{n}e^{ikx}=1+2\sum _{k=1}^{n}\cos(kx)={\frac {\sin \left(\left(n+{\frac {1}{2}}\right)x\right)}{\sin(x/2)}},$

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 D_n(x) with any function f of period 2π is the nth-degree Fourier series approximation to f, i.e., we have

$(D_{n}*f)(x)=(f*D_{n})(x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(y)D_{n}(x-y)\,dy=\sum _{k=-n}^{n}{\hat {f}}(k)e^{ikx},$

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

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

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.