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

#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"
![draw](drawings/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: $\rho=\frac{1}{L}$

## Convergence example:
#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 regarding power series:
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 power 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, "What's so funny?" :D "I'm not a dictator" something about you are not forced to sit through and watch the lecture if you don't like to, "I don't 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: $a_{3k+2}=0$ where $k=0,1,2,\dots$
Well, now we need the rest of the terms $a_{3k+1}$ and $a_{3k}$
compute $a_{3k+1}$ terms:
$a_{4}=\frac{2}{3\cdot 4}a_{1}$ 
$a_{7}=\frac{5}{6\cdot 7} \frac{2}{3\cdot 4}a_{1}$
realize if we multiply here by 5 and 2:
$a_{7}=\frac{5^2}{5\cdot6\cdot 7} \frac{2^2}{2\cdot3\cdot 4}a_{1}$
$a_{7}=\frac{(2\cdot 5)^2}{7!}a_{1}$
$a_{4}=\frac{2^2}{4!}a_{1}$
the pattern leads us to:
$a_{3k+1}=\frac{(2\cdot 5\cdot 8\cdot \dots(3k-1))^2}{(3k+1)!}a_{1}$ where $k=1,2,3, \dots$
similarly can be done for $a_{3k}$
$a_{3k}=\frac{(1\cdot 4\cdot \dots(3k-2))^2}{(3k)!}a_{0}$ $k=1,2,\dots$
Finally we plug these coefficients back into $z(x)=\sum_{n=0}^\infty a_{n}x^n$:
$z(x)=a_{0}(x)^0+a_{1}(x)^1+\sum_{k=1}^\infty \frac{(1\cdot4\cdot\dots(3k-2))^2}{(3k)!}a_{0}x^{3k}+\sum_{k=1}^\infty \frac{(2\cdot 5\cdot \dots(3k-1))^2}{(3k+1)!}a_{1} x^{3k+1}$
rewriting gives:
$$z(x)=a_{0}\underbrace{ \left( 1+\sum_{k=1}^\infty \frac{(1\cdot4\cdot\dots(3k-2))^2}{(3k)!}x^{3k} \right) }_{ y_{1}(x)}+a_{1}\underbrace{ \left( x+\sum_{k=1}^\infty \frac{(2\cdot 5\cdot \dots(3k-1))^2}{(3k+1)!} x^{3k+1}\right) }_{ y_{2}(x) }$$
There we go, $z$ is a linear combination of $y_{1}$ and $y_{2}$
Let's say we were given the initial conditions $z(0)=1 \qquad z'(0)=1$
then $a_{0}=1$ because $z(0)=a_{0}(1+0)+a_{1}(0+0)$
and $a_{1}=1$ because $z'(0)=a_{0}(0+0)+a_{1}(1+0)$

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>
## Ordinary and singular points
we consider:
$$y''+p(x)y'+q(x)y=0$$
this is in standard form, it's a second order linear equation (btw linear means the coefficients of y and its derivatives are functions of x or constants only. Eg: $yy'=1$ and $y^2=1$ are both non-linear.)
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 an 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).

## 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 radius of convergence is $\rho=2$ (ie: the distance between $x_{0}$ and -1 which is the nearest singular point to $x_0$)
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 
$$y''-\tan xy'+y=0 \quad x_{0}=0$$
This linear equation is already in standard form so we don't need to adjust it. 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 at least the distance of $x_0$ to these two points:
$\rho\geq\mid x_{0}-\frac{\pi}{2}\mid$
$\rho\geq \frac{\pi}{2}$
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
![draw](drawings/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

#ex #powseries 
Find the first 5 non-zero terms of the general solution of:
$$xy''-y'+y=0 \qquad x_{0}=2$$

Is this function analytic about $x_0=2$? We can't expect to answer this problem using power series if the function is non analytic at $x_{0}=2$ so let's check.
DONT FORGET! put it in standard form first:
$y''-\frac{1}{x}y'+\frac{y}{x}=0$ <- now we can see that there are singular points at $x=0$
so $x_{0}$ is analytic with a radius convergence of $\rho=2$
We can begin solving:
$y(x)=\sum_{n=0}^\infty a_{n}(x-2)^n \qquad x\in(0,4)$
substitute for convenience:
$x-2=t \qquad t\in(-2,2)$ where $\frac{dy}{dt}=\frac{dy}{dx}$ and in general $\frac{d^ny}{dx}=\frac{d^ny}{dt}$
so our new DE is $(t+2) \frac{d^2y}{dt^2}-\frac{dy}{dt}+y=0$
$y(t)=\sum_{n=0}^\infty a_{n}t^n$
$y'(t)=\sum_{n=1}^\infty a_{n}nt^{n-1}$
$y''(t)=\sum_{n=2}^\infty a_{n}n(n-1)t^{n-2}$
plug them into the DE:
$\underset{ n-1=k }{ \sum_{n=2}^\infty a_{n}n(n-1)t^{n-1} }+\underset{ n-2=k }{ 2\sum_{n=2}^\infty a_{n}n(n-1)t^{n-2} }-\underset{ n-1=k }{ \sum_{n=1}^\infty a_{n}nt^{n-1} }+\underset{ n=k }{ \sum_{n=0}^\infty a_{n}t^n }=0$
shift the index to equalize powers
$\sum_{k=1}^\infty a_{k+1}(k+1)kt^k+\sum_{k=0}^\infty 2(k+2)(k+1)a_{k+2}t^k-\sum_{k=0}^\infty a_{k+1}(k+1)t^k+\sum_{k=0}^\infty a_{k}t^k=0$
take out some terms to equalize the starting index
$\underbrace{ 4a_{2}-a_{1}+a_{0} }_{ =0 }+\sum_{k=1}^\infty \underbrace{ (a_{k+1}(k+1)k+2(k+2)(k+1)a_{k+2}-a_{k+1}(k+1)+a_{k}) }_{ =0 }t^k=0$
left hand side tells us $a_{2}=\frac{a_{1}-a_{0}}{4}$
and the rest of the coefficients can be written as a recursive relation: 
$a_{k+1}(k+1)k+2(k+2)(k+1)a_{k+2}-a_{k+1}(k+1)+a_{k}=0$
This yields:
>for $k=1$:

$\cancel{ 2a_{2} }+12a_{3}-\cancel{ 2a_{2} }+a_{1}=0$
$a_{3}=-\frac{a_{1}}{12}$
>for $k=2$:

$a_{3}(3)2+2(4)(3)a_{4}-a_{3}(3)+a_{2}=0$
$a_{4}=\frac{1}{2\cdot 3\cdot 4}(-3a_{3}-a_{2})$

$a_{4}=-\frac{1}{24}(3a_{3}+a_{2})=-\frac{1}{24}\left( 3 \frac{-a_{1}}{12}+\frac{a_{1}-a_{0}}{4} \right)=\frac{a_{0}}{96}$

Okay we have $a_{2}, a_{3}, a_{4}$ as some multiple of $a_{0}$ and $a_{1}$ which are the two arbitrary constants to the solution of the DE. We have the first 5 non zero terms.
recall our solution is in the form $y(t)=\sum_{n=0}^\infty a_{n}t^n$
Sub back in $x-2=t$
$$y(x)=a_{0}+a_{1}(x-2)+\frac{a_{1}-a_{0}}{4}(x-2)^2-\frac{a_{1}}{12}(x-2)^3+\frac{a_{0}}{96}(x-2)^4+\dots$$
in this case we can't go much further, we can't explicitly find the coefficients for each term. In fact the solution is related to the Bessel function. In the last lecture's example we were lucky we could write a formula for the coefficients.
Here's a plot to give you an idea of what's happening, the blue line is one solution to the DE:
![powerseriessolution.png](drawings/powerseriessolution.png)
We are done.

#ex #powseries #IVP
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.
$y(x)=\sum_{n=0}^\infty a_{n}(x-\pi)^n \qquad x-\pi=t$
our new initial conditions are $y(0)=1 \quad y'(0)=0$
$y(t)=\sum_{n=0}^\infty a_{n}t^n$ <- we are abusing notation, the $y$ here isn't the same as above. But it's all good.
$y''-\sin(t+\pi)\sum_{n=0}^\infty a_{n}t^n=0$
$y''+\sin(t)\sum_{n=0}^\infty a_{n}t^n=0$
$\sum_{n=2}^\infty a_{n}n(n-1)t^{n-2}+\left( \sum_{n=0}^\infty(-1)^n \frac{t^{2n+1}}{(2n+1)!} \right)\left( \sum_{n=0}^\infty a_{n}t^n \right)=0$ remember, $\sin$ is an odd function so its infinite series has only odd powers.
using:
$y(t)=\sum_{n=0}^\infty a_{n}t^n$
$y'(t)=\sum_{n=1}^\infty a_{n}nt^{n-1}$
we can deduce that $y(0)=1=a_{0} \quad y'(0)=0=a_{1}$
from the big summ-y equation: 
$(2a_{2}+6a_{3}t+12a_{4}t^2+20a_{5}t^3+30a_{6}t^4+\dots)+\left( t-\frac{t^3}{6}+\frac{t^5}{120}-\dots \right)(a_{0}+a_{1}t+a_{2}t^2+a_{3}t^3+\dots)=0$
the only constant factor is $a_{2}$
this implies: $2a_{2}=0 \implies a_{2}=0$
what about the factors of $t$?
$(6a_{3}+a_{0})t=0$
$a_{3}=-\frac{a_{0}}{6}=-\frac{1}{6}$
</br>
factors of $t^2$:
$(12a_{4}+a_{1})t^2=0$
$a_{4}=-\frac{a_{1}}{12}=0$
</br>
$t^3$:
$\left( 20a_{5}+\cancelto{ 0 }{ a_{2} }-\frac{a_{0}}{6} \right)=0 \implies a_{5}=\frac{1}{120}$
</br>
$t^4$:
$\left( 30a_{6}+a_{3}-\cancelto{ 0 }{ \frac{a_{1}}{6} } \right)t^4=0 \implies a_{6}=\frac{1}{180}$
</br>
substitute back $x-\pi=t$
$$y(x)=1-\frac{1}{6}(x-\pi)^3+\frac{1}{120}(x-\pi)^5+\frac{1}{180}(x-\pi)^6+\dots$$
there's no general formula here for the constants (or maybe he said no formula for $y(x)$?), but we can write the solution in the following form^.

#ex #powseries 
Here's a non-homogenous example: (RHS$\ne 0$) 
$$y'-xy=e^x \qquad x_{0}=0$$
Find the first 5 coefficients of the power series representation.
The equation is already in standard form.
Clearly there are no singular points here. We expect the series representation of $y(x)$ to converge everywhere.
$y(x)=\sum_{n=0}^\infty a_{n}x^n$
$y'(x)=\sum_{n=1}^\infty a_{n}nx^{n-1}$
shift the index and plug in to the equation.
$\sum_{k=0}^\infty a_{k+1}(k+1)x^k-\sum_{k=1}^\infty a_{k-1}x^k-\sum_{k=0}^\infty \frac{x^k}{k!}=0$
$a_{1}-1+\sum_{k=1}^\infty a_{k+1}(k+1)x^k-\sum_{k=1}^\infty a_{k-1}x^k-\sum_{k=1}^\infty \frac{x^k}{k!}=0$
$\underbrace{ a_{1}-1 }_{ =0 }+\sum_{k=1}^\infty (\underbrace{ a_{k+1}(k+1)-a_{k-1}-\frac{1}{k!} }_{ =0 })x^k=0$
$a_{1}-1=0 \implies a_{1}=1$
$a_{k+1}=\frac{a_{k-1}+\frac{1}{k!}}{k+1}$
$k=1\implies a_{2}=\frac{a_{0}}{2}+\frac{1}{2}$
$k=2\implies a_{3}=\frac{a_{1}}{3}+\frac{\frac{1}{2}}{3}=\frac{1}{2}$
$k=3\implies a_{4}=\frac{ \left( \frac{a_{0}}{2}+\frac{1}{2} \right)+\frac{1}{6}}{4}$
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