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" ! 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)

## 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 $\rho$.
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: $=(a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots)(b_{0}+b_{1}(x-x_{0})+b_{2}(x-x_{0})^2+\dots)$ $=a_{0}b_{0}+(a_{0}b_{1}+a_{1}b_{0})(x-x_{0})+(a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{0})(x-x_{0})^2+\dots$
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: $\implies y'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$ $y''(x)=\sum_{n=2}^\infty a_{n}n(n-1)(x-x_{0})^{n-2}$
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!)
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.
### 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 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). !

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 ! 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