Courtesy of Matthew Jenkins

Fourier Series

<aside> 💡 Fourier Series:

Any periodic function can be written as an infinite sum of cosines and sines with angular frequencies that are integer multiples of the fundemental frequency $\frac{2\pi}{L}$

The Fourier Series of a periodic function $f(x)$ with period/wavelength $L$ is defined as:

$$ f(x) = \frac{a_0}2 + \sum_{n=1}^\infin a_n\cos\frac{2\pi nx}L + \sum_{n=1}^\infin b_n\sin\frac{2\pi nx}L $$

where

$$ \begin{align*} a_n &= \frac2L\int_{x_0}^{x_0+L} f(x)\ \cos\frac{2\pi n x}L \ \ dx \\ \ \\ b_n &= \frac2L\int_{x_0}^{x_0+L} f(x)\ \sin\frac{2\pi n x}L \ \ dx \end{align*} $$

The $a_n$ equation above is defined for every $n$ except $n=0$, so to solve for $a_0$ we can set $n=0$ and solve before integrating, giving us:

$$ a_0 = \frac2L\int_{x_0}^{x_0+L} f(x)\ dx $$

…because $\cos(2\pi n)$ is always $0$. If $a_n=0$, $a_0=0$.

Notes:

• The definition of periodic is $f(x)=f(x\pm nL)$ where $L=\text{period/wavelength}$
• $\frac{a_0}2=$ average of $f$
• If our periodic function is even $b_n=0~~~\forall n$
  • $f(x)=f(-x)$
• If our periodic function is odd $a_n=0~~\forall n$
  • $f(x)=-f(-x)$

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Proof of Concept

Proof/Explanation:

Gibbs’ Phenomenon:

Let’s say we approximate our periodic function with a fourier series approximation with N terms:

$$ f_N(x) = \frac{a_0}2 + \sum_{n=1}^N a_n\cos\frac{2\pi nx}L + \sum_{n=1}^\infin b_n\sin\frac{2\pi nx}L $$

This and all its derivatives are continuous, meaning it approximates $f(x)$ well except at a discontinuity, where it approximates a jump with a smooth function. As a result, there is always a overshoot at the discontinuity resulting in Gibbs’ Peaks.

The peaks overshoot the jump by about 9% for any function with a discontinuity as $N\rarr\infin$.

Complex Fourier Series

Instead of using $\sin\frac{2\pi nx}L$ and $\cos\frac{2\pi nx}L$, we can use $e^{i2\pi nx/L}$:

<aside> 💡

Complex Fourier Series:

The Complex Fourier Series of a function $f(x)$ is defined as:

$$ f(x)=\sum_{n=-\infin}^\infin d_n e^{i2\pi nx/L} $$

Note: The above sum is over all $n$, not just positive.

$d_n$ is given by the following:

$$ d_n=\frac1L \int_{x_0}^{x_0+L} f(x) e^{-i2\pi nx/L} $$

</aside>

By replacing $e^{i2\pi nx/L}$ with $\cos\frac{2\pi nx}L + i\sin\frac{2\pi nx}L$, we can solve for $d_n$ and find that:

$$ \begin{matrix*} a_n = d_n+d_{-n} &&&& b_n = i(d_n-d_{-n})

\end{matrix*} $$

Fourier Transforms Theory

Consider a function that vanishes at $x=\pm\infin$ (pictured above):