#start of lec 22
# Power series
#powseries
A power series is defined by:
$$\sum_{n=0}^\infty a_{n}(x-x_{0})^n=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots$$
Where $x_{0}$ is a given point of expansion.
It is convergent if and only if:
$$\sum_{n=0} ^ \infty a_{n}(x-x_{0})^n<\infty \text{ at a given x}$$
If $\sum_{n=0}^\infty \mid a_{n}(x-x_{0})^n\mid$ is convergent
 $\implies\sum_{n=0}^\infty  a_{n}(x-x_{0})^n$ is absolutely convergent
Just because something is convergent doesn't mean it is absolutely convergent. think of the alternating harmonic series. It is convergent but absolutely diverges.
However, if a series is absolutely convergent, then it's definitely convergent as well.

Theorem: With each $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ we can associate a radius of convergence $\rho$ where $0\leq \rho\leq \infty$.
The series is absolutely convergent
for all $x$ such that $\mid x-x_{0}\mid<\rho$, and divergent for all $x$ where $\mid x-x_{0}\mid>\rho$
"Who keeps stealing the whiteboard erases? (jokingly) It's a useless object, anyways"
![[Drawing 2023-10-30 13.12.57.excalidraw.png]]
how can we find $\rho$?
Definition of ratio test: If $\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=L$
then the radius of convergence $\rho$ is: $\rho=\frac{1}{L}$

## Examples:
#ex #powseries
Is this infinite series convergent? divergent? and where so?
$$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}(x-1)^n$$
Determine the convergent set.
Use ratio test:
$\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=\lim_{ n \to \infty } \frac{2^{-(n+1)}}{n+2} \frac{n+1}{2^{-n}}=\lim_{ n \to \infty }\frac{n+1}{2(n+2)}=\frac{1}{2}\implies \rho=2$
It's convergent 2 units away from $x_{0}$
So it's convergent on $-1<x<3$, divergent on $\mid x-1\mid>2$
But what about on the points $-1$ and $3$? Ratio test tells us nothing for these points.
Plug in $x=-1$
$\sum_{n=0}^{\infty} \frac{2^{-n}}{n+1}(-2)^n=\sum_{n=0}^\infty \frac{(-1)^n}{n+1}<\infty$ <- That is the alternating harmonic series, it is convergent.
plug in $x=3$:
$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}2^n=\sum_{n=0}^\infty \frac{1}{n+1}>\infty$ <- harmonic series, this diverges.
so the power series is convergent on 
$$x=[-1,3)$$and diverges otherwise.
</br>
## Theorems:
Assume that $\sum_{n=0}^\infty  a_{n}(x-x_{0})^n$ and $\sum_{n=0}^\infty  b_{n}(x-x_{0})^n$ are convergent with $\rho>0$
Then: 
1.) $\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=0}^{\infty}b_{n}(x-x_{0})^n=\sum_{n=0}^\infty(a_{n}+b_{n})(x-x_{0})^n$
That has a radius of convergence of at least $\rho$.
</br>
2.) $\left( \sum_{n=0}^\infty a_{n}(x-x_{0})^n \right)\left( \sum_{n=0}^\infty b_{n}(x-x_{0})^n \right)=\sum_{n=0}^\infty c_{n}(x-x_{0})^n$ (called the Cauchy product)
Where $c_n=\sum_{k=0}^n a_{k}b_{n-k}$
Here's a demonstration that shows why $c_{n}$ equals the expression above:
$=(a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots)(b_{0}+b_{1}(x-x_{0})+b_{2}(x-x_{0})^2+\dots)$
$=a_{0}b_{0}+(a_{0}b_{1}+a_{1}b_{0})(x-x_{0})+(a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{0})(x-x_{0})^2+\dots$
</br>
3.) If $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ is convergent with $\rho>0$
ie: it's convergent when $\mid x-x_{0}\mid<\rho$
Then we can differentiate this infinite sum and get:
$\implies y'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
$y''(x)=\sum_{n=2}^\infty a_{n}n(n-1)(x-x_{0})^{n-2}$
</br>
Theorem: If $y(x)$ is infinitely many times differentiable on some interval $\mid x-x_{0}\mid<\rho$
then: $y(x)=\sum_{n=0}^\infty \frac{y^{(n)}(x_{0})}{n!}(x-x_{0})^n$ (Taylor series)
"believe me, Taylor series is the most important theorem in engineering."
"I mean engineering is all about approximations, do you know how your calculator computes [...]? Taylor series!"
"Applied mathematics is all about approximating and then measuring how good your approximation is, it's what engineering is all about." -Prof (loosy quotes, can't keep up with how enthusiastic he is!)
</br>
Definition: If $y(x)$ can be represented with a power series on $\mid x-x_{0}\mid$ then $y(x)$ is an analytic function on $(x_{0}-\rho,x_{0}+\rho)$
btw analytic functions are very important in complex calculus.
</br>
### Shifting the index (theorem)
$f(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n<\infty$
$f'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
consider the equation $f'+f=0$
$f(x)+f'(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
let $n-1=k$
$=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{k=0}^\infty a_{k+1}(k+1)(x-x_{0})^{k}$
$=\sum_{n=0}^\infty(a_{n}+a_{n+1}(n+1))(x-x_{0})^n=0$
Last theorem fo' da day:
If $\sum_{n=0}^\infty a_{n}(x-x_{0})^n=0$ for all x$\in(x_{0}-\rho,x_{0}+\rho)$ where $\rho>0$
$\implies a_{n}=0$, $n=0,1,2,\dots$
This means in the above example $a_{n}+a_{n+1}(n+1)=0$
This is called a recursive relation and it will come in handy when solving differential equations.
#end of lec 22 #start of lec 23
Mid terms are almost done being marked!

## Solving DE using series
Let's start using power series to start solving DE!
#ex 
$$y'-2xy=0 \qquad x_{0}=0$$
Note this is separable and linear, so we can already solve this. This time we do it with power series
$y$ should be an analytic function (meaning, infinitely many times differentiable)
therefore we should expect we can represent $y$ as a power series:
$y(x)=\sum_{n=0}^\infty a_{n}x^n$
$y'(x)=\sum_{n=1}^\infty a_{n}nx^{n-1}$
plug these into the equation:
$\sum_{n=1}^\infty a_{n}nx^{n-1}-\sum_{n=0}^\infty 2a_{n}x^{n+1}=0$
if the entire interval is zero, we should expect all the coefficients to equal 0
we need to combine the summations.
shift the index! (so that the exponents on the x are the same)
$k=n-1,\ k=n+1$
$\sum_{k=0}^\infty a_{k+1}(k+1)x^{k}-\sum_{k=1}^\infty 2a_{k-1}x^k=0$


$a_{1}+\sum_{k=1}^\infty (\underbrace{ a_{k+1}(k+1)-2a_{k-1} }_{ =0 })x^k=0$
The whole series equals zero, (due to the theorem from last lecture.)
so $a_{1}=0$ is the first observation
second observation:
$a_{k+1}=\frac{2}{k+1}a_{k-1}$ where $k=1,2,3,\dots$ This is called a recursive relation. (if we know one index we can produce some other index recursively)
from these observations:
$a_{1}, a_{3}, a_{5}, \dots=0$
$a_{2k+1}=0, k=0,1,2,\dots$
this means half of our power series disappears!
what happens with the other half?
$a_{2}$ is related to $a_{0}$ from the above formula
$a_{2}=\frac{2}{2}a_{0}$ ($k=1$)
$a_{4}=\frac{2}{3+1}a_{2}=\frac{a_{0}}{2}$ ($k=2$)
$a_{6}=\frac{2}{5+1} \frac{a_{0}}{2}=\frac{a_{0}}{6}$ ($k=3$)
$a_{8}=\frac{2}{7+1} \frac{a_{0}}{6}=\frac{a_{0}}{24}$ ($k=4$)
$a_{9}=\frac{2}{9+1} \frac{a_{0}}{24}=\frac{a_{0}}{120}$ ($k=5$)
(note that the $k$ value here is used differently than the $k$ above.)
you might start noticing a factorial-y pattern:
$a_{2k}=\frac{1}{k!}a_{0}$ where $k=0,1,2,\dots$
$a_{0}$ is an arbitrary coefficient! We should expect to get one just like we did when solving with previous techniques.
$y(x)=a_{0}\sum_{k=0} ^\infty \frac{1}{k!}x^{2k}=a_{0}\sum_{k=0} ^\infty \frac{1}{k!}(x^2)^k$
Does this look like something from math 101?
Yes! it looks like the Taylor series of $e^{x^2}$
so: 
$$y(x)=a_{0}e^{x^2}$$
"if we are correct--the same is not true in general in real life--but in mathematics if we are correct we should end up with the same solution had we solved with another method." -Prof
#ex 
$$z''-x^2z'-xz=0 \qquad \text{about } x_{0}=0$$
using regular methods will be problematic,
if you use laplace transform you will have problems as well.
>"you try the simplest thing you know, if you know anything :D" (referring to answering a question about how do we know what method to use?)

lets use power series:
assume solution is analytic:
$z(x)=\sum_{n=0}^\infty a_{n}x^n$
$z'(x)=\sum_{n=1}^\infty a_{n}nx^{n-1}$
$z''(x)=\sum_{n=2}^\infty a_{n}n(n-1)x^{n-2}$
plug in:
$\underset{ n-2=k }{ \sum_{n=2}^\infty a_{n}n(n-1)x^{n-2} }-\underset{ n+1=k }{ \sum_{n=1}^\infty a_{n} nx^{n+1} }-\underset{ n+1=k }{ \sum_{n=0}^\infty a_{n}x^{n+1} }=0$

shift the index to equalize the powers.
>loud clash of clans log in sound, class giggles, "whats so funny?" :D "im not a dictator" something about you are not forced to sit through and watch the lecture if you don't like to, "I dont think everybody should like me."

$\sum_{k=0}^\infty a_{k+2}(k+2)(k+1)x^k-\sum_{k=2}^\infty a_{k-1}(k-1)x^k-\sum_{k=1}^\infty a_{k-1}x^k=0$
just as in the previous example, we take out the first terms so that each index starts at the same number.
$2a_{2}+6a_{3}x+\sum_{k=2}^\infty a_{k+2}(k+2)(k+1)x^k-\sum_{k=2}^\infty a_{k-1}(k-1)x^k-a_{0}x-\sum_{k=2}^\infty a_{k-1}x^k=0$
$6a_{2}+(6a_{3}-a_{0})x+\sum_{k=2}^\infty (a_{k+2}(k+1)(k+2)-a_{k-1}(k-\cancel{ 1 })-\cancel{ a_{k-1} })x^k=0$

$\underbrace{ 6a_{2} }_{ =0 }+\underbrace{ (6a_{3}-a_{0})x }_{ =0 }+\sum_{k=2}^\infty \underbrace{ (a_{k+2}(k+1)(k+2)-a_{k-1}k)x^k }_{ =0 }=0$
$a_{2}=0 \qquad a_{3}=\frac{a_{0}}{6}$
$a_{k+2}=\frac{k}{(k+1)(k+2)}a_{k-1}$ where $k=2,3,4,\dots$
Finally, a recursive relation!
We easily deduce from the recursive relation and $a_{2}=0$ that:
$a_{2}, a_{5}, a_{8}, \dots=0$
ie: $a_{3k-1}=0$ where $k=1,2,\dots$
$a_{4}=\frac{2}{3\cdot 4}a_{1}$
$a_{7}=\frac{5}{6\cdot 7} \frac{2}{3\cdot 4}a_{1}$
realize if we multiply here by 5 and 2:
$a_{7}=\frac{5^2}{5\cdot6\cdot 7} \frac{2^2}{2\cdot3\cdot 4}a_{1}$
$a_{7}=\frac{(2\\dot{c} 5)^2}{7!}a_{1}$
$a_{4}=\frac{2^2}{4!}a_{1}$
the pattern leads us to:
$a_{3k+1}=\frac{(2\cdot 5 \dots(3k-1))^2}{(3k+1)!}$ where k=1,2,3, ...
$a_{3k}=\frac{(1\cdot 4\cdot \dots(3k-2))^2}{(3k)!}a_{0}$ k=1,2,...
$z(x)=a_{0}\left( 1+\sum_{k=1}^\infty \frac{(1\cdot4\cdot\dots(3k-2))^2}{(3k)!}x^{3k} \right)$
$a_{1}\left( x+\sum_{k=1}^\infty \frac{(2\cdot 5\cdot \dots(3k-1))^2}{(3k+1)!} x^{3k+1}\right)$
there we go, $z$ is a linear combination of those two expressions
class done at 1:56 (a lil late but the journey is worth it)
#end of lec 23 #start of lec 24
<i>midterms have been marked and returned today.</i>

we consider:
$$y''+p(x)y'+q(x)y=0$$
this is in standard form, it's a second order linear equation
Definition:
if $p(x)$ and $q(x)$ are <u>analytic</u> functions in a vicinity of $x_{0}$ then $x_0$ is <u>ordinary</u>. Otherwise, $x_{0}$ is <u>singular</u>.
we expect that the solution y can be represented by a power series. This is true according to the following theorem:
Theorem: If $x_{0}$ is ordinary point then the differential equation above has two linearly independent solution of the form $\sum_{n=0} ^\infty a_{n}(x-x_{0})^n, \qquad\sum_{n=0}^\infty b_{n}(x-x_{0})^n$.
The radius of convergence for them is at least as large as the distance between $x_{0}$ and the closest singular point (which can be real or complex).
![[Drawing 2023-10-30 13.12.57.excalidraw.png]]

## Examples for calculating $\rho$
#ex
$$(x+1)y''-3xy'+2y=0 \quad x_{0}=1$$
put it in standard form:
$y''-\frac{3xy'}{x+1}+\frac{2y}{x+1}=0$
the only singular point for this equation is $x=-1$
so the minimum value of radius convergence is $\rho=2$ (distance between -1 and x_0)
we are guaranteed that the power series will converge <i>at least</i> in $(-1,3)$, possibly more. You can try solving for y as a power series.

#ex 
$$y''-\tan xy'+y=0 \quad x_{0}=0$$
notice the coefficient beside y is 1, 1 is analytic and differentiable everywhere, obviously!
Same goes for any polynomial, it's obvious that any polynomial is infinitely differentiable but it's important to know.
What about tan x?
$\tan x=\frac{\sin x}{\cos x}$ is not defined on $x=\frac{\pi}{2}\pm n\pi, \qquad n=0,1,2,\dots$
the closest singular points are $\frac{\pi}{2}$ and $\frac{-\pi}{2}$ so our radius of convergence is the minimum distance of x_0 to these two points:
$\rho\geq\mid x_{0}-\frac{\pi}{2}\mid=\frac{\pi}{2}$
convergence could be larger, but we are guaranteed convergence on $x=x_{0}-\rho$ to $x_{0}+\rho$

#ex 
$$(x^2+1)y''+xy'+y=0 \qquad x_{0}=1$$
put it in standard form:
$y''+\frac{x}{x^2+1}y'+\frac{y}{x^2+1}=0$
remember singular points can be complex the two singular points are:
$x^2=1=0 \qquad x=\pm i$
now we have to compute the two distances of these singular points to x=1
![[Drawing 2023-11-03 13.40.54.excalidraw.png]]
To calculate distance: $\alpha_{1}+\beta_{1}i, \qquad \alpha_{2}+\beta_{2}i$
$\sqrt{ (\alpha_{1}-\alpha_{2})^2+(\beta_{1}-\beta_{2})^2 }$
$\rho\geq \sqrt{ 1^2+1^2 }=\sqrt{ 2 }$
#end of lec 24
#start of lec 25
find the zeros of
$xy''-y'+y=0 \qquad x_{0}=2$
is this function analytic about x_0=2?
DONT FORGET! put it in standard form:
$y''-\frac{1}{x}y'+\frac{y}{x}=0$ <- now we can see that there are singular points at x=0
so we have a radius convergence of $\rho=2$
$y(x)=\sum_{n=0}^\infty a_{n}(x-2)^n \quad x\in(0,4)$
$x-2=t \qquad t\in(-2,2)$
$y(t)=\sum_{n=0}^\infty a_{n}t^n$
$y'(t)=\sum_{n=1}^\infty a_{n}nt^{n-1}$
$y''(t)=\sum_{n=2}^\infty a_{n}n(n-1)t^{n-2}$
$\sum_{n=2}^\infty a_{n}n(n-1)t^{n-1}+2\sum_{n=2}^\infty a_{n} n(n-1)t^{n-2}-\sum_{n=1}^\infty a_{n}nt^{n-1}+\sum_{n=0}^\infty a_{n}t^n=0$
n-1=k n-2=k n-1=k 
first 5 non-zeros:
$\sum_{k=1}^\infty a_{k+1}(k+1)kt^k+\sum_{k=0}^\infty 2(k+2)(k+1)a_{k+2}t^k-\sum_{k=0}^\infty a_{k+1}(k+1)t^k+\sum_{k=0}^\infty a_{k}t^k$
$\underbrace{ 4a_{2}-a_{1}+a_{0} }_{ =0 }+\sum_{k=1}^\infty \underbrace{ (a_{k+1}(k+1)k+4a_{k+2}-a_{k+1}(k+1)+a_{k}) }_{ =0 }t^k=0$
$a_{2}=\frac{a_{1}-a_{0}}{4}$
$12a_{3}+2a_{2}-2a_{2}+a_{1}=0$
$a_{3}=-\frac{a_{1}}{12}$
$a_{4}=-\frac{1}{24}(3a_{3}+a_{2})=\frac{1}{96}a_{1}-\frac{a_{1}}{96}+\frac{a_{0}}{96}=\frac{a_{0}}{96}$
$y(x)=a_{0}+a_{1}(x-2)+\frac{a_{1}-a_{0}}{4}(x-2)^2-\frac{a_{1}}{12}(x-2)^3+\frac{a_{0}}{96}(x-2)^4+\dots$
in this case we cant go much further, cant explicitly find the coefficients for each term. in the last lecture's example we were lucky.
So we are done.

#ex 
Find first four non-zero terms of the power series for $y(x)$ about $x_{0}=\pi$
of the IVP:
$$y''-\sin (x)y=0 \qquad y(\pi)=1 \qquad y'(\pi)=0$$
This is already in standard form.
clearly this is analytical over the entire real axis, sin(x) and 1 are both infinitely differentiable. no singular points real or complex.
$y(x)=\sum_{n=0}^\infty a_{n}(x-\pi)^n \qquad x-\pi=t$
$y(t)=\sum_{n=0}^\infty a_{n}t^n$ <- we are abusing notation, the y here isn't the same as above. But it's all good.
$y''-\sin(t+\pi)\sum_{n=0}^\infty a_{n}t^n=0$
$y''+\sin(t)\sum_{n=0}^\infty a_{n}t^n=0$
$\sum_{n=2}^\infty a_{n}n(n-1)t^{n-2}+\left( \sum_{n=0}^\infty(-1)^n \frac{t^{2n+1}}{(2n+1)!} \right)\left( \sum_{n=0}^\infty a_{n}t^n \right)$ remember, sin is odd so its infinite series has odd powers.
now from:

$y''+\sin(t)\sum_{n=0}^\infty a_{n}t^n=0$
$y(t)=\sum_{n=0}^\infty a_{n}t^n$
this implies $y(0)=1=a_{0} \quad y'(0)=0=a_{1}$

from the big summ-y equation: 
$(2a_{2}+6a_{3}t+12a_{4}t^2+20a_{5}t^3+\dots)+\left( t-\frac{t^3}{6}+\frac{t^5}{120}-\dots \right)(a_{0}+a_{1}t+a_{2}t^2+a_{3}t^3+\dots)=0$
the only constant factor is $a_{2}$
this implies: $2a_{2}=0 \implies a_{2}=0$
what about the factors of $t$?
$(6a_{3}+a_{0})t=0$
$a_{3}=-\frac{a_{0}}{6}=-\frac{1}{6}$

$t^2$:
$(12a_{4}+a_{1})t^2=0$
$a_{2}=-\frac{a_{1}}{12}$

$t^3$:
$\left( 20a_{5}+a_{2}-\frac{a_{0}}{6} \right)=0 \implies a_{5}=\frac{1}{120}$

$t^4$:
$\left( 30a_{6}+a_{3}-\frac{a_{1}}{6} \right)t^4=0 \implies a_{6}=\frac{1}{180}$
$y(x)=1-\frac{1}{6}(x-\pi)^3+\frac{1}{120}(x-\pi)^5+\frac{1}{180}(x-\pi)^6+\dots$
theres no general formula here for the constants? (or maybe no formula for y(x)?), but we can write the solution in the following form^.

#ex 
$$y'-xy=e^x \qquad x_{0}=0$$
$y(x)=\sum_{n=0}^\infty a_{n}x^n$
$y'(x)=\sum_{n=1}^\infty a_{n}nx^{n-1}$
$\sum_{k=0}^\infty a_{k+1}(k+1)x^k-\sum_{k=1}^\infty a_{k-1}x^k-\sum_{k=0}^\infty \frac{x^k}{k!}=0$

$a_{1}-1=0 \implies a_{1}=1$
$a_{k+1}=\frac{a_{k-1}+\frac{1}{k!}}{k+1}$
$k=1\implies a_{2}=\frac{a_{0}}{2}+\frac{1}{2}$
$k=2\implies a_{3}=\frac{1}{2}$
$k=3\implies a_{4}=\frac{ \left( \frac{a_{0}}{2}+\frac{1}{2} \right)+\frac{1}{6}}{4}$
We are lucky, in this course fubini's method is not needed. (what?)
and with that, we are finished this chapter on power series.
#end of lec 25