---
title: "SU-ENGR76 APR232024"
source: https://www.jemoka.com/posts/kbhsu_engr76_apr232024/
date: 2024-04-23
---

Fourier Series components form a basis Recall the definition of a basis, and in particular what an orthonormal basis is. In particular, recall that writing a vector as a linear combination of orthonormal basis is a thing you can do very easily.
Recall Importantly, the Fourier Series is defined as:
\begin{equation} f(x) = a_0 + \sum_{k=1}^{\infty} \qty( a_{k} \cos(k \omega x) + b_{k} \sin(k \omega x)) \end{equation}
where \(\omega = \frac{2\pi}{L}\), and
\begin{equation} a_0 = \frac{\langle f, 1 \rangle}{ \langle 1,1 \rangle} = \frac{1}{L} \int_{0}^{L} f(x) \dd{x} \end{equation}
\begin{equation} a_{k} = \frac{\langle f, \cos (k\omega x) \rangle}{\langle \cos (k\omega x), \cos (k\omega x) \rangle} = \frac{2}{L} \int_{0}^{L} f(x) \cos (k\omega x) \dd{x} \end{equation}
\begin{equation} b_{k} = \frac{\langle f, \sin (k\omega x) \rangle}{\langle \sin (k\omega x), \sin (k\omega x) \rangle} = \frac{2}{L} \int_{0}^{L} f(x) \sin (k\omega x) \dd{x} \end{equation}
Check: sinusoids forms an orthonormal basis The functions:
\begin{align} \sin \qty(\omega t), \cos \qty(\omega t), \sin \qty(\omega 2t), \cos \qty(\omega 2t), \dots \end{align}
can be viewed as forming an orthonormal basis by defining an inner product of two \(L\) periodic functions as:
\begin{align} \langle g,h \rangle = \frac{2}{L} \int_{0}^{L} g(x)h(x) \dd{x} \end{align}
this makes our basis above orthonormal.
Verifying that:
\begin{align} \langle \sin \qty(2\pi k_{i} \omega t), \sin \qty(k_{j} \omega t) \rangle &amp;= \frac{2}{L} \int_{0}^{L} \sin \qty(\omega k_{i} t) \sin \qty(\omega k_{j} t) \dd{t} \\ &amp;= \frac{2}{L} \int_{0}^{L} \frac{1}{2} \qty(\cos \qty(\omega t \qty(k_{i}-k_{j})) - \cos \qty(\omega t \qty(k_{i}+k_{j}))) \dd{t} \end{align}
if \(k_{i} \neq k_{j}\), then we will be integrating an at-least \(L\) periodic (i.e. because \(\omega = \frac{2\pi}{L}\)) function in each term of the integral above by \(L\) periods. This makes both terms \(0\).
if \(k_{i} = k_{j}\), then the first term would be \(\cos \qty(0) = 1\), and the second term would still be \(0\). This gives (after normalizing), \(1\).
This means that the sinusoids form an orthonormal basis. By substituting in the relevant double-angle formulas, we can obtain a similar results that checks for the orthonormality of this system. Key insight is that if \(g = \sin\qty(\dots), h= \cos \qty(\dots)\), the angle product formulas would give a difference in \(\sin\), whereby \(\sin(0) = 0\), so the whole thing evaluates to \(0\).
Observation and because we have checked this function as being orthonormal, and we take by faith that the Fourier Series spans, then we can figure our coefficients as needed. This essentially allow us to compute each of the coefficients.
Overall cosine series collapses for odd/even functions this can be attributed to the fact that, say you had an even function that you are trying to take the cosine series of, we&rsquo;ll get intergrals of the shape:
\begin{equation} \int y(t) \sin \qty(k\omega t) \dd{t} \end{equation}
which is a multiplication of an even function to an odd function&mdash;making an odd function (\(g(-x) * f(-x) = -g(x) * f(x) = -(g(x) * f(x)\)). The integral of this, then would give \(0\).
