Kernels in Fourier Series
This is an introduction to Fourier Series. I am trying to understand Dirichlet’s Kernel taught in class.
Disclaimer: Some of the content below are inspired by my Honours Analysis Notes and Introductory Waves Notes!
Start Here if you don’t know Fourier Series
A fourier series decomposes a function into trigonometric functions.
By adding more and more trigonometric functions together, the summation itself looks more and more like the original function. Look at the graph below, does it look like a sawtooth over time?
The equation above is:
\[1+\frac{4}{\pi}\sum_{n=1}^{m}\frac{\sin\left(\pi nx\right)}{n}\]Where we increased \(m\) over time. For \(m=1000\), we can get the below:
Fourier Series doesn’t work for all functions, there are certain restrictions.
Fourier Series is surprisingly helpful and concise in solving some neat mathematical problems, like Basel Problem:
To understand how and why fourier series work, we can study its convergence (pointwise convergence, \(L^2\) convergence, etc). However, this is slightly too difficult without knowing Bessel’s inequality/Parseval’s Theorem
Start Here if you know Fourier Series
The section below is partly inspired by this amazing video:
A Natural Introduction
Dirichlet’s Kernel is intimately related to Fourier Series, by design.
Below we take \(f(x)\) to be periodic with period \(2π\), continuous, finite, and have continuous and finite first derivative. Now consider the fourier series approximation to the \(N\)th term:
In case you aren’t familiar with Fourier Series, here is a brief reminder:
Within the interval \(−L \leq x \leq L\), we can approximate the function \(f(x)\) as:
\[\begin{aligned} f(x) & =\frac{1}{2} a_0+\sum_{n=1}^{\infty} a_n \cos \left(\frac{n \pi x}{L}\right)+\sum_{n=1}^{\infty} b_n \sin \left(\frac{n \pi x}{L}\right) \\ & =\frac{1}{2} a_0+\sum_{n=1}^{\infty}\left[a_n \cos \left(\frac{n \pi x}{L}\right)+b_n \sin \left(\frac{n \pi x}{L}\right)\right] \end{aligned}\]Where we have:
\[\begin{aligned} a_m & =\frac{1}{L} \int_{-L}^L d x \cos \left(\frac{m \pi x}{L}\right) f(x) \\ \text { Similarly, } \quad b_m & =\frac{1}{L} \int_{-L}^L d x \sin \left(\frac{m \pi x}{L}\right) f(x) . \end{aligned}\]However, considering our limit: \(-\pi \leq L \leq \pi\), we can deduce the below:
\[\begin{aligned} \frac{a_0}{2}=\frac{1}{2 \pi} \int_{-\pi}^\pi f\left(x\right) dx \quad \\ a_n=\frac{1}{ \pi} \int_{-\pi}^\pi f\left(x\right) \operatorname{cos}\left(nx\right) d x \\ b_n=\frac{1}{ \pi} \int_{-\pi}^\pi f\left(x\right) \operatorname{sin}\left(nx\right) d x \end{aligned}\]Using the information above, we can deduce (with a bit of algebraic manipulation):
\[\begin{aligned} f_N(x) & =\frac{a_0}{2}+\sum_{n=1}^N\left(a_n \cos n x+b_n \sin n x\right) \\ & =\int_{-\pi}^\pi[1+2 \sum_{n=1}^N (\cos n y \cos n x+\sin n y \sin n x)] \frac{f(y)}{2 \pi} d y \\ &=\int_{-\pi}^\pi[1+2 \sum_{n=1}^N (\cos n(x-y))] \frac{f(y)}{2 \pi} d y \\ & =\int_{-\pi}^\pi K_N(x-y) f(y) \frac{d y}{2 \pi} \end{aligned}\]Hence (after switching variables) we can naturally define Dirichlet’s Kernel as below:
\[K_N(\theta)=1+2 \sum_{n=1}^N \cos n \theta,\]A Better Form
Using Euler’s Formula \(e^{i \theta}=\cos \theta+i \sin \theta\), we can deduce:
\[K_N(\theta)=\sum_{k=-N}^N e^{i k \theta}=\left(1+2 \sum_{k=1}^N \cos (k \theta)\right)=\frac{\sin ((N+1 / 2) \theta)}{\sin (\theta / 2)}\]- Hint: \(e^{-i \theta}=\cos \theta-i \sin \theta\), then cancel out terms
We can use Desmos to see how it looks like:
Fourier Series Estimation
Going back to \(f_N(x)\), which is our fourier approximate for our (specifically defined) function, we can shift the variables around:
\[\begin{align} f_N(x) &=\int_{-\pi}^\pi K_N(y-x) f(y) \frac{d y}{2 \pi}\\ &=\int_{-\pi-x}^{\pi-x} K_N(\theta) f(x+\theta) \frac{d \theta}{2 \pi} \end{align}\]Note that our function has period \(2\pi\), hence we can imagine shifting the function and getting the same result:
\[\begin{align} \int_{-\pi}^\pi K_N(\theta) f(x+\theta) \frac{d \theta}{2 \pi} \end{align}\]For illustration, consider \(f(x)=\frac{1+ \sin x}{2}\). Notice the area between the vertical moving line never changes.
I hope the above shows you some intuition regarding Dirichlet’s Kernel.
Fejér kernel
Fejér kernel is defined as:
\[K_N(x)=\frac{1}{N+1} \sum_{n=0}^N D_n(x) .\]I will state without proof that:
\[\sum_{n=0}^N D_n(x)=\frac{1-\cos (2 \pi(N+1) x)}{2 \sin (\pi x)^2}\]Approximation of Unity?
Below we define an approximation of unity (under specific conditions where I won’t show here):
\[\sup _{x \in \mathbb{R}}\left|f * k_n(x)-f(x)\right| \rightarrow 0\]1) \(k_n(x) \geq 0\) for all \(x \in \mathbb{R}\)
2) \(\int_{-1 / 2}^{1 / 2} k_n(t) d t=1\)
3) For all \(1 / 2 \geq \delta>0\) we have:
\(\int_{-\delta}^\delta k_n(t) d t \rightarrow 1\) \(\text { as } n \rightarrow \infty\)
The reader can attempt to show that Fejér kernel satisfies all (1), (2), (3) but Dirichlet’s Kernel fails to satisfy (1). Hence Fejér kernel is an approximation of unity!
SideNote: Fejer kernel gives the “Cesàro summation” of Fourier series, it is just the average of Dirichlet’s Kernel!
Fejér kernel, having the “smoothening” factor (by taking average), is more commonly used as it handles discontinuity better, and has less oscillation (almost by definition, since it is the average of Dirihchlet’s Kernel)
So in conclusion, how it works is that both kernels convolve with \(f(x)\) to make an approximation to the \(f(x)\), but Fejér kernel does so more smoothly, as any abrupt discontinity won’t affect too much, due to its averaging nature!
