MATH201/content/Power series (lec 22-25).md

#start of lec 22
Finished chapter 7 of the course textbook, Let's begin chapter 8!
# Power series
A power series is defined by:
$$\sum_{n=0}^\infty a_{n}(X-X_{0})^n=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots$$
It is convergent if:
$$\sum_{n=0} ^ \infty a_{n}(x-x_{0})^n<\infty \text{ at a given x}$$
Otherwise, it is divergent.
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 absolutely convergent doesn't mean it is conditionally convergent. think of the harmonic series. It is absolutely convergent but also divergent (conditionally divergent).

Theorem: With each $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ we can associate $0\leq \rho\leq \infty$ such that
$\sum_{n=0} ^\infty a_{n}(x-x_{0})^n$ is absolutely convergent 
for all x such that $\mid x-x_{0}\mid<\rho$, 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 $\rho$?
Ratio test: If $\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=L$
then $\rho=\frac{1}{L}$

## Examples:
#ex 
is this convergent? Divergent? and where so?
$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}(x-1)^n$
determine the convergent set.
Use ratio test:
$\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=\lim_{ n \to \infty } \frac{2^{-(n+1)}}{n+2} \frac{n+1}{2^{-n}}=\frac{1}{2}\implies \rho=2$
so it's convergent on $-1<x<3$, divergent on $\mid x-1\mid>2$
But what about on the points $-1$ and $3$?
plug in $x_{0}=-1$
$\sum_{n=0}^{\infty} \frac{2^{-n}}{n+1}(-2)^n=\sum_{n=0}^\infty \frac{(-1)^n}{n+1}<\infty$ <- That is the alternating harmonic series, it is convergent.
plug in $x_{0}=3$:
$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}2^n=\sum_{n=0}^\infty \frac{1}{n+1}>\infty$ <- harmonic series, this diverges.
so the power series is convergent on $[-1,3)$ divergent otherwise.
$$\text{ converges only on: } [-1,3)$$

Assume that $\sum_{n=0}^\infty  a_{n}(x-x_{0})^n$ and $\sum_{n=0}^\infty  b_{n}(x-x_{0})^n$ are converget 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 $\rho$.
2.) $\left( \sum_{n=0}^\infty a_{n}(x-x_{0})^n \right)\left( \sum_{n=0}^\infty b_{n}(x-x_{0})^n \right) \qquad c_n=\sum_{k=0}^n a_{k}b_{n-k}$(Cauchy)
$=(a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots)(b_{0}+b_{1}(x-x_{0})+b_{2}(x-x_{0})^2+\dots)$
$=a_{0}b_{0}+(a_{0}b_{1}+a_{1}b_{0})(x-x_{0})+(a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{1})(x-x_{0})^2+\dots$ (Cauchy multiplication)

more Definitions of power series:
If $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ is convergent with $\rho>0$
$\mid x-x_{0}\mid<\rho$
we can differentiate this infinite sum and get:
$\implies y'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
$y''(x)=\sum_{n=2}^\infty a_{n}n(n-1)(x-x_{0})^{n-2}$

Theorem: If $y(x)$ is infinitely many times differentiable on some interval: $\mid x-x_{0}\mid<\rho$
then: $\sum_{n=0}^\infty \frac{y^{(n)}(x_{0})}{n!}(x-x_{0})^n$ (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!)

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 MATH301. (i don't have that next term)

$f(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n<\infty$
$f'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
$f(x)+f'(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
let $n-1=k$
$=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{k=0}^\infty a_{n}(k+1)(x-x_{0})^{k}$
$=\sum_{n=0}^\infty(a_{n}+a_{n}(n+1))(x-x_{0})^n$

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 $\rho>0$
$\implies a_{n}=0$, $n=0,1,2,\dots$
#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!
No magic formulas we need to memorize when solving equations using power series (Yay!)
#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)
so we should expect we can represent it 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!
$k=n-1,\ k=n+1$
$\sum_{k=0}^\infty a_{k+1}(k+1)x^{k}-\sum_{k=1}^\infty 2a_{k-1}x^k=0$


$a_{1}+\sum_{k=1}^\infty (\underbrace{ a_{k+1}(k+1)-2a_{k-1} }_{ =0 })x^k=0$
The whole series equals zerro,
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 this equation:
$a_{1}, a_{3}, a_{5}, \dots=0$
$a_{2k+1}=0, k=0,1,2,\dots$
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
$a_{2}=\frac{2}{2}a_{0}$ ($k=1$)
$a_{4}=\frac{2}{3+1}a_{2}=\frac{a_{0}}{2}$ ($k=2$)
$a_{6}=\frac{2}{5+1} \frac{a_{0}}{2}=\frac{a_{0}}{6}$ ($k=3$)
$a_{8}=\frac{1}{4} \frac{a_{0}}{6}=\frac{a_{0}}{24}$ ($k=4$)
you might start noticing a factorial-y pattern:
$a_{2k}=\frac{1}{k!}a_{0}$ where $k=0,1,2,\dots$

$y(x)=a_{0}\sum_{k=0} ^\infty \frac{1}{k!}x^{2k}=a_{0}\sum_{k=0} ^\infty \frac{1}{k!}(x^2)^k$
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" -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:
$z(x)=\sum_{n=0}^\infty a_{n}nx^{n-1}$
$z''(x)=\sum_{n=2}^\infty a_{n}n(n-1)x^{n-2}$
$\underset{ n-2=k }{ \sum_{n=2}^\infty a_{n}n(n-1)x^{n-2} }-\underset{ n+1=k }{ \sum_{n=1}^\infty a_{n} nx^{n+1} }-\underset{ n+1=k }{ \sum_{n=0}^\infty a_{n}x^{n+1} }=0$

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."

$\sum_{k=0}^\infty a_{k+2}(k+2)(k+1)x^k-\sum_{k=2}^\infty a_{k-1}x^k-\sum_{k=1}^\infty a_{k-1}x^k=0$
$2a_{2}+6a_{3}+\sum_{k=2}^\infty a_{k+2}(k+2)(k+1)x^k-\sum_{k=2}^\infty a_{k-1}(k-1)x^k-a_{0}x-\sum_{k=2}^\infty a_{k-1}x^k=0$
$6a_{2}+(6a_{3}-a_{0})x+\sum_{k=2}^\infty (a_{k+2}(k+1)(k+2)-a_{k-1}k)x^k=0$
$a_{2}=0 \qquad a_{3}=\frac{a_{0}}{6}$
$a_{k+2}=\frac{k}{(k+1)(k+2)}a_{k-1}$ where $k=2,3,4,\dots$
Finally, a recursive relation!
it should be clear that each step of 3 starting from $a_{2}$ should all equal 0.
$a_{2}, a_{5}, a_{8}, \dots=0$
$a_{3k-1}=0$ 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,...
$z(x)=a_{0}\left( 1+\sum_{k=1}^\infty \frac{(1*4*\dots(3k-2))^2}{(3k)!}x^{3k} \right)$
$a_{1}\left( x+\sum_{k=1}^\infty \frac{(2\cdot 5\cdot \dots(3k-1))^2}{(3k+1)!} x^{3k+1}\right)$
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
*midterms have been marked and returned today.*

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 **analytic** functions in a vicinity of $x_{0}$ then $x_0$ is **ordinary**. Otherwise, $x_{0}$ is **singular**.
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 *at least* 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