Dirichlet's Kernel Representation of the Partial Sums of a Fourier Series
Recall from the Dirichlet's Kernel page that Dirichlet's kernel is the collection of functions $D_n$ where for each $n \in \mathbb{N}$:
(1)We also looked at some properties of the functions in Dirichlet's kernel. We saw that for each $n \in \mathbb{N}$, $D_n$ is even, $2\pi$-periodic, and $\displaystyle{\int_0^{2\pi} D_n(t) \: dt = \pi}$. We will now look at a nice theorem which tells us that any function $f \in L([0, 2\pi])$ that is $2\pi$-periodic can have the partial sums of the Fourier series generated by $f$ represented in terms of an integral involving a function in Dirichlet's kernel.
Theorem 1: Let $f \in L([0, 2\pi])$ be a $2\pi$-periodic function. For each $n \in \mathbb{N}$ let $\displaystyle{s_n(x) = \frac{a_0}{2} + \sum_{k=1}^{n} (a_k \cos kx + b_k \sin kx)}$ denote the partial sum of the Fourier series generated by $f$. Then for each $n \in \mathbb{N}$, $\displaystyle{s_n(x) = \frac{2}{\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} D_n(t) \: dt}$. |
Recall that if $f \in L([0, 2\pi])$ then the Fourier series "generated" by $f$ is the Fourier series obtained with respect to the trigonometric system $\displaystyle{\left \{ \frac{1}{\sqrt{2\pi}}, \frac{\cos x}{\sqrt{\pi}}, \frac{\sin x}{\sqrt{\pi}}, \frac{\cos 2x}{\sqrt{\pi}}, \frac{\sin 2x}{\sqrt{\pi}}, ... \right \}}$ and $\displaystyle{f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx)}$ where the $a_n$'s and $b_n$'s are the Fourier coefficients of the Fourier series given by $\displaystyle{a_n = \frac{1}{\pi} \int_0^{2\pi} f(t) \cos nt \: dt}$ and $\displaystyle{b_n = \frac{1}{\pi} \int_0^{2\pi} f(t) \sin nt \: dt}$. Be sure to review the content on the The Fourier Series of Functions Relative to an Orthonormal System page as a refresher!
- Proof: We know that for all $n \in \mathbb{N}$ that $\displaystyle{s_n(x) = \frac{a_0}{2} + \sum_{k=1}^{n} (a_k \cos kx + b_k \sin kx)}$. Compacting this and we get:
- We now use the following trigonometric identity:
- Therefore:
- Notice that $f$ and $D_n$ are both $2\pi$ periodic. Therefore we have that:
- Let $u = t - x$. Then $du =dt$ and thus:
- Since $D_n$ is an even function and $[-\pi, \pi]$ is centered at the origin we have that:
- Replacing the dummy variable $u$ with $t$ yields: