Remember the heat flow equation? We obtained that it's solution could be expressed in the form: 
$$\sum_{n=1}^\infty c_{n}\sin\left( \frac{n\pi x}{L} \right)\quad\text{for}\quad0\leq x\leq L$$
But what is $c_{n}$? They are the coefficients of a fourier transform. We want to develop a way to compute them.
Let's derive how to compute the coefficients of a fourier transform. (feel free to skip to the end)
$f(x)=\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$ where L is length of the rod

This is a Fourier series, it's a more general form of what we have above:
$f(x)=\frac{a_{0}}{2}+\sum_{n=1}^\infty\left( a_{n}\cos\left( \frac{n\pi x}{L} \right) + b_{n}\sin\left( \frac{n\pi x}{L}\right) \right)$
$x \in [-L,L]$
almost everywhere piecewise continuous (?)
has a lot of benefits over taylor series. $f(x)$ doesn't have to be infinitely differentiable (analytic)
f(x) can even have jump discontinuities
lets assume the equation is true when $x \in [-L,L]$
$\int _{-L} ^L f(x) \, dx=\int _{-L}^L \frac{a_{0}}{2} \, dx+\int _{-L}^L (\text{put summation here}) \, dx$
$\int _{-L}^L \cos\left( \frac{n\pi x}{L} \right) \, dx=\frac{L}{n\pi}\sin\left( \frac{n\pi x}{L} \right)|_{-L}^L=0$
same for $\int _{-L}^L \sin\left( \frac{n\pi x}{L} \right)\, dx=0$ (it equals 0)
so
$\int _{-L} ^L f(x) \, dx=\int _{-L}^L \frac{a_{0}}{2} \, dx+\int _{-L}^L0 \, dx$
$\int _{-L} ^L f(x) \, dx=\int _{-L}^L \frac{a_{0}}{2} \, dx+\int _{-L}^L0 \, dx$
$a_{0}=\frac{1}{L}\int _{L}^{2?a_{0}L} f(x) \, dx$
$\int _{-L}^L f(x)\cos\left( \frac{m\pi x}{L} \right)\, dx=\frac{a_{0}}{2}\cancelto{ 0 }{ \int _{-L}^L \cos\left( \frac{m\pi x}{L} \right) \, dx }+\sum_{n=1}^\infty\left( a_{n}\int _{-L}^L\cos\left( \frac{n\pi x}{L} \right)\cos\left( \frac{m\pi x}{L} \right) \, dx \right)+b_{n}\int _{-L}^L \sin\left( \frac{n\pi x}{L} \right)\cos\left( \frac{m\pi x}{L} \right) \, dx$
use trig identities (will be provided on exam):
$\cos(\alpha)\cos(\beta)=\frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta))$
$\sin(\alpha)\cos(\beta)=\frac{1}{2}(\sin(\alpha+\beta)+\sin(\alpha-\beta))$
$\sin(\alpha)\sin(\beta)=\frac{1}{2}(\cos(\alpha-\beta)-\cos(\alpha+\beta))$
$\int _{-L}^L \cos \frac{n\pi x}{L}\cos \frac{m\pi x}{L}\, dx=\frac{1}{2}(\int _{-L}^L \left( \cos \frac{(n-m)\pi x}{L} \right) \, dx+\cancelto{ 0 }{ \frac{\cos(n+m)\pi x}{L} }dx$
$= \begin{cases}0 & n\ne m \\L & n=m\end{cases}$

$\int _{-L}^L \sin \frac{n\pi x}{L}\cos \frac{m\pi}{L} \, dx=0;$
$\int _{-L}^L \sin \frac{n\pi x}{L}\cos \frac{m\pi}{L} \, dx=\begin{cases}0 & n\ne m \\L & n=m\end{cases}$

going back,
$$a_{m}=\frac{1}{L}\int _{-L}^L f(x)\cos \frac{m\pi x}{L} \, dx \quad\text{valid for all }m=0,1,2,\dots$$
$$b_{m}=\frac{1}{L}\int _{-L}^L f(x)\sin \frac{m\pi x}{L} \, dx=b_{m} \quad m=1,2,\dots$$
now we know how to compute the coefficients for Fourier series!

properties:
for functions $f$, $g$, If $\int _{-L}^Lf(x)g(x) \, dx=\begin{cases}0 & f\ne g \\L & f=g \end{cases}$
then $f, g$ are orthogonal
the forier expantion is called an ortho normal expansion, taylor is not ortho normal.
#end of lec 28
#start of lec 29
last lecture we derived how to find the coefficients in a fourier series.
$f(x)=\frac{a_{0}}{2}+\sum_{n=1}^\infty\left( a_{n}\cos\left( \frac{n\pi x}{L} \right) + b_{n}\sin\left( \frac{n\pi x}{L}\right) \right)$
$x \in [-L,L]$
### Theorem:
If $f$ and $f'$ are piecewise continuous on $[-L,L]$, then the fourier series converges to:
$\frac{1}{2}(f(x^-)+f(x^+))$ for all $x \in (-L,L)$
Basically meaning, the fourier series converges.
At $x=\pm L$ the fourier series converges to $\frac{1}{2}(f(-L^+)+f(L^-))$
![[Partial differential equations (lec 26-27) 2023-11-22 13.15.26.excalidraw]]
### Theorem:
If f(x) is continuous on $(-\infty,\infty)$ and $2L$ periodic and if $f'$ is continuous, then the taylor series converges to $f(x)$ everywhere
![[Partial differential equations (lec 26-27) 2023-11-22 13.14.05.excalidraw]]
#ex lets compute the fourier transform of:
$f(x)=\begin{cases}1, & -\pi\leq x\leq 0 \\x, & 0<x\leq \pi\end{cases}$
$L$ here is $\pi$ clearly.
lets find the coefficients $a_{n}$ and $b_{n}$
$a_{n}=\frac{1}{\pi}\left( \int _{-\pi}^0 \cos\left( \frac{n\pi x}{\pi} \right)\, dx +\int _{0}^\pi x\cos(nx)\, dx\right)$
using integration by parts for second term:
$=\frac{1}{\pi}\left( \frac{1}{n}\cancelto{ 0 }{ \sin(nx) }|_{-\pi}^0 +\frac{1}{n}\left( x\cancelto{ 0 }{ \sin(nx) }|_{0}^\pi-\int _{0}^\pi \sin(nx) \, dx \right)\right)$
$a_{n}=\frac{1}{n^2\pi}(\cos(nx)|_{0}^\pi)=\frac{1}{n^2\pi}((-1)^n-1) \quad n=0,1,2,\dots$
now lets find $b_{n}$
$b_{n}=\frac{1}{\pi}\left( \int _{-\pi}^0\sin(nx) \, dx+\int _{0}^\pi x\sin(nx) \, dx \right)$
$=\frac{1}{\pi}\left( \frac{-1}{n}\cos(nx)|_{-\pi}^0-\frac{1}{n}\left( x\cos(nx)|_{0}^\pi-\int _{0}^\pi \cos(nx)\, dx \right) \right)$
$b_{n}=\frac{1}{n\pi}((-1)^n(1-\pi)-1) \quad n=1,2,\dots$



we find that 
$a_{2n}=0$
$a_{2k-1}=-\frac{2}{n^2\pi}$ for $k=1,2,\dots$
what about when $n=0$?
$a_{0}=\frac{1}{\pi}\left( \int _{-\pi}^0 \, dx+\int _{0}^\pi x \, dx \right)$
$=\frac{1}{\pi}\left( x|_{-\pi}^0+\frac{x^2}{2}|_{0}^\pi \right)$
$=\frac{1}{\pi}\left( 0+\pi+\frac{\pi^2}{2} \right)=\frac{\pi}{2}+1$

#ex lets compute the fourier transform of $f(x)=x$ from $-\pi\leq x\leq \pi$
we have to take a windowed form of $f$ to make this possible, $L=\pi$
at the left and right edge of the interval, the fourier series is equal to 0. (from the previous theorem)
find the coefficients:
$a_{n}=\frac{1}{\pi}\int _{-\pi}^\pi x\cos(nx) \, dx=0$
Why is it zero? because the integrand is an odd function. (odd times even is odd.) and because we are integrating from $-\pi$ to $\pi$ (symmetric interval)
definition of odd: $f(x)=-f(-x)$
definition of even: $f(x)=f(-x)$
odd times even is odd.
odd times odd is even.
even times even is even.
huge exam time saving technique.
find $b_{n}$
$b_{n}=\frac{1}{\pi}\int _{-\pi}^\pi x\sin(nx) \, dx=\frac{2}{\pi}\int _{0}^\pi x\sin(nx) \, dx$
using integration by parts:
$b_{n}=\frac{2}{n}(-1)^{n+1}$
another take away: if $f$ is odd, the cos terms are 0
if $f$ is even, the sin terms are 0.

if f is only defined between 0 and L:
you can create an odd extension: $\bar{f}=\begin{cases}f(x)  & 0\leq x\leq L \\-f(-x), & -L\leq x<0 &  & \end{cases}$
this will contain only sin terms
you also have a choise to extend it as an even function, symmetrically across the y axis.
$\bar{f}=\begin{cases}f(x)  & 0\leq x\leq L \\f(-x), & -L\leq x<0 &  & \end{cases}$
this will contain only cos terms.
#end of lec 29
#start of lec 30
from last lecture: 
$f(x)$ is defined on $[0,L]$
odd extention:
$\bar{f}(x)=\begin{cases}f(x), & 0\leq x\leq L \\-f(-x,) & -L\leq x<0\end{cases}$
and the $a$ coeffecients (cos terms) are zero.
not only that, but the b terms are:
$\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right); \qquad b_{n}=\frac{1}{L}\int _{-L}^L\bar{f}(x) \sin\left( \frac{n\pi x}{L} \right) \, dx$
$$b_{n}=\frac{2}{L}\int _{0}^L f(x)\sin\left( \frac{n\pi x}{L} \right)\, dx$$
"How about that, this is called a foureir sin series."
For even extension:
$\bar{f}(x)=\begin{cases}f(x), & 0\leq x\leq L \\f(-x,) & -L\leq x<0\end{cases}$
and the $b$ coeffecients (sin terms) are zero.
not only that but the a terms are:
$\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}\cos\left( \frac{n\pi x}{L} \right)$
$a_{n}=\frac{2}{L}\int _{0}^L f(x)\cos\left( \frac{n\pi x}{L} \right)\, dx$

remember that $\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$ was the expansion of the eigen value function from the heat equation?
then $\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}\cos\left( \frac{n\pi x}{L} \right)$ is also an expansion of some related eigen value function.

#ex Fourier sine series for $f(x)=x^2$ from $0\leq x\leq \pi$
well that means we want the odd extension:
![[Lec 30 2023-11-24 13.15.17.excalidraw]]
the $\cos()$ ($a_{n}$) terms are zero.
the b terms are:
$b_{n}=\frac{2}{\pi}\int _{0}^\pi x^2\sin(nx) \, dx$
$=-\frac{2}{\pi n}\left[ x^2\cos(nx)|_{0}^\pi-2\int _{0}^\pi x\cos(nx)\, dx \right]$
$=-\frac{2}{n\pi}\left[ \pi^2(-1)^n-\frac{2}{n}\left( x\cancelto{ 0 }{ \sin(nx) }|_{0}^\pi-\int _{0}^\pi \sin(nx)\, dx \right) \right]$
$b_{n}=-\frac{2}{n\pi}\left[ \pi^2(-1)^n-\frac{2}{n^2}\cos(nx)|_{0}^\pi \right]=\frac{2\pi}{n}(-1)^{n+1}+\frac{4}{n^3\pi}((-1)^n-1)$
for $n=1,2,3,\dots$ note no $n=0$ so no divison by zero problems here.

#ex fourier cosine series of $f(x)=\sin(x)$ for $0\leq x\leq \pi$
![[Lec 30 2023-11-24 13.23.08.excalidraw]]
$a_{n}=\frac{2}{\pi}\int _{0}^\pi \sin(x)\cos(nx)\, dx$ for $n=0,1,2,\dots$
use trig identity: (by the way the identites will be provided in the exam.)

$=\frac{2}{\pi} \frac{1}{2}\int _{0}^\pi (\sin(1-n)x+\sin(n+1)x)\, dx$
integrating gives you:
$a_{n}=-\frac{1}{\pi} \frac{1}{n+1} (-1)^{n+1}+\frac{1}{\pi} \frac{1}{n+1}+\frac{1}{\pi} \frac{1}{n-1}(-1)^{n-1}-\frac{1}{\pi} \frac{1}{n-1}$
what about when n=0 or n=1?
$a_{0}=\frac{4}{\pi}=\frac{2}{\pi}\int _{0}^\pi \sin(x) \, dx$
$a_{1}=\frac{2}{\pi}\int _{0}^\pi \sin(x)\cos(x) \, dx=\frac{1}{\pi}\int _{0}^\pi \sin(2x)\, dx=0$
"0 is a very very special number it took humanity many numbers of years to invent 0" referring to when dividing by 0.
additionally we know that the terms cancel when:
$a_{2k-1}=0$ for $k=1,2,\dots$
$a_{2k}=\frac{2}{\pi} \frac{1}{2k+1}+\frac{2}{\pi} \frac{1}{2k-1}$ for $k=1,2,\dots$
then:
$$\bar{f}(x)=\frac{2}{\pi}+\frac{2}{\pi}\sum_{k=1}^\infty\left( \frac{1}{2k+1}+\frac{1}{2k-1} \right)\cos(2kx)$$

We have prepared ourselves now, now we start solving PDE's. He's encouraging us to attend the lectures in these last two weeks. He's making it sound like PDE's are hard.