Sasserisop
/
MATH201
1
0
Fork 1
Code Issues Pull Requests 1 Packages Projects Releases Wiki Activity

revised power series

This commit is contained in:
Sasserisop 2023-12-03 01:18:12 -07:00
parent e88b973271
commit ac079dbbfc
3 changed files with 110 additions and 70 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

176
content/Power series (lec 22-25).md
View File

@ -22,7 +22,7 @@ how can we find $\rho$?
Definition of ratio test: If $\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=L$ 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: $\rho=\frac{1}{L}$ then the radius of convergence $\rho$ is: $\rho=\frac{1}{L}$
## Examples: ## Convergence example:
#ex #powseries #ex #powseries
Is this infinite series convergent? divergent? and where so? Is this infinite series convergent? divergent? and where so?
$$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}(x-1)^n$$ $$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}(x-1)^n$$
@ -39,7 +39,7 @@ $\sum_{n=0}^\infty \frac{2^{-n}}{n+1}2^n=\sum_{n=0}^\infty \frac{1}{n+1}>\infty$
so the power series is convergent on so the power series is convergent on
$$x=[-1,3)$$and diverges otherwise. $$x=[-1,3)$$and diverges otherwise.
</br> </br>
## Theorems: ## 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 $\rho>0$ 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: Then:
1.) $\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=0}^{\infty}b_{n}(x-x_{0})^n=\sum_{n=0}^\infty(a_{n}+b_{n})(x-x_{0})^n$ 1.) $\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=0}^{\infty}b_{n}(x-x_{0})^n=\sum_{n=0}^\infty(a_{n}+b_{n})(x-x_{0})^n$
@ -59,9 +59,10 @@ $y''(x)=\sum_{n=2}^\infty a_{n}n(n-1)(x-x_{0})^{n-2}$
</br> </br>
Theorem: If $y(x)$ is infinitely many times differentiable on some interval $\mid x-x_{0}\mid<\rho$ 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) 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." >"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!" >"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!) >"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> </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 $(x_{0}-\rho,x_{0}+\rho)$ 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. btw analytic functions are very important in complex calculus.
@ -82,7 +83,7 @@ This is called a recursive relation and it will come in handy when solving diffe
#end of lec 22 #start of lec 23 #end of lec 22 #start of lec 23
Mid terms are almost done being marked! Mid terms are almost done being marked!
## Solving DE using series ## Solving DE using power series
Let's start using power series to start solving DE! Let's start using power series to start solving DE!
#ex #ex
$$y'-2xy=0 \qquad x_{0}=0$$ $$y'-2xy=0 \qquad x_{0}=0$$
@ -141,7 +142,7 @@ plug in:
$\underset{ n-2=k }{ \sum_{n=2}^\infty a_{n}n(n-1)x^{n-2} }-\underset{ n+1=k }{ \sum_{n=1}^\infty a_{n} nx^{n+1} }-\underset{ n+1=k }{ \sum_{n=0}^\infty a_{n}x^{n+1} }=0$ $\underset{ n-2=k }{ \sum_{n=2}^\infty a_{n}n(n-1)x^{n-2} }-\underset{ n+1=k }{ \sum_{n=1}^\infty a_{n} nx^{n+1} }-\underset{ n+1=k }{ \sum_{n=0}^\infty a_{n}x^{n+1} }=0$
shift the index to equalize the powers. 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." >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."
$\sum_{k=0}^\infty a_{k+2}(k+2)(k+1)x^k-\sum_{k=2}^\infty a_{k-1}(k-1)x^k-\sum_{k=1}^\infty a_{k-1}x^k=0$ $\sum_{k=0}^\infty a_{k+2}(k+2)(k+1)x^k-\sum_{k=2}^\infty a_{k-1}(k-1)x^k-\sum_{k=1}^\infty a_{k-1}x^k=0$
just as in the previous example, we take out the first terms so that each index starts at the same number. just as in the previous example, we take out the first terms so that each index starts at the same number.
@ -154,30 +155,39 @@ $a_{k+2}=\frac{k}{(k+1)(k+2)}a_{k-1}$ where $k=2,3,4,\dots$
Finally, a recursive relation! Finally, a recursive relation!
We easily deduce from the recursive relation and $a_{2}=0$ that: We easily deduce from the recursive relation and $a_{2}=0$ that:
$a_{2}, a_{5}, a_{8}, \dots=0$ $a_{2}, a_{5}, a_{8}, \dots=0$
ie: $a_{3k-1}=0$ where $k=1,2,\dots$ 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:
$a_{4}=\frac{2}{3\cdot 4}a_{1}$ $a_{4}=\frac{2}{3\cdot 4}a_{1}$
$a_{7}=\frac{5}{6\cdot 7} \frac{2}{3\cdot 4}a_{1}$ $a_{7}=\frac{5}{6\cdot 7} \frac{2}{3\cdot 4}a_{1}$
realize if we multiply here by 5 and 2: realize if we multiply here by 5 and 2:
$a_{7}=\frac{5^2}{5\cdot6\cdot 7} \frac{2^2}{2\cdot3\cdot 4}a_{1}$ $a_{7}=\frac{5^2}{5\cdot6\cdot 7} \frac{2^2}{2\cdot3\cdot 4}a_{1}$
$a_{7}=\frac{(2\\dot{c} 5)^2}{7!}a_{1}$ $a_{7}=\frac{(2\cdot 5)^2}{7!}a_{1}$
$a_{4}=\frac{2^2}{4!}a_{1}$ $a_{4}=\frac{2^2}{4!}a_{1}$
the pattern leads us to: the pattern leads us to:
$a_{3k+1}=\frac{(2\cdot 5 \dots(3k-1))^2}{(3k+1)!}$ where k=1,2,3, ... $a_{3k+1}=\frac{(2\cdot 5\cdot 8\cdot \dots(3k-1))^2}{(3k+1)!}a_{1}$ where $k=1,2,3, \dots$
$a_{3k}=\frac{(1\cdot 4\cdot \dots(3k-2))^2}{(3k)!}a_{0}$ k=1,2,... similarly can be done for $a_{3k}$
$z(x)=a_{0}\left( 1+\sum_{k=1}^\infty \frac{(1\cdot4\cdot\dots(3k-2))^2}{(3k)!}x^{3k} \right)$ $a_{3k}=\frac{(1\cdot 4\cdot \dots(3k-2))^2}{(3k)!}a_{0}$ $k=1,2,\dots$
$a_{1}\left( x+\sum_{k=1}^\infty \frac{(2\cdot 5\cdot \dots(3k-1))^2}{(3k+1)!} x^{3k+1}\right)$ Finally we plug these coefficients back into $z(x)=\sum_{n=0}^\infty a_{n}x^n$:
there we go, $z$ is a linear combination of those two expressions $z(x)=a_{0}(x)^0+a_{1}(x)^1+\sum_{k=1}^\infty \frac{(1\cdot4\cdot\dots(3k-2))^2}{(3k)!}a_{0}x^{3k}+\sum_{k=1}^\infty \frac{(2\cdot 5\cdot \dots(3k-1))^2}{(3k+1)!}a_{1} x^{3k+1}$
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) class done at 1:56 (a lil late but the journey is worth it)
#end of lec 23 #start of lec 24 #end of lec 23 #start of lec 24
<i>midterms have been marked and returned today.</i> <i>midterms have been marked and returned today.</i>
## Ordinary and singular points
we consider: we consider:
$$y''+p(x)y'+q(x)y=0$$ $$y''+p(x)y'+q(x)y=0$$
this is in standard form, it's a second order linear equation 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: 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>. 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: 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 $\sum_{n=0} ^\infty a_{n}(x-x_{0})^n, \qquad\sum_{n=0}^\infty b_{n}(x-x_{0})^n$. 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). 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]] ![[Drawing 2023-10-30 13.12.57.excalidraw.png]]
@ -187,17 +197,18 @@ $$(x+1)y''-3xy'+2y=0 \quad x_{0}=1$$
put it in standard form: put it in standard form:
$y''-\frac{3xy'}{x+1}+\frac{2y}{x+1}=0$ $y''-\frac{3xy'}{x+1}+\frac{2y}{x+1}=0$
the only singular point for this equation is $x=-1$ 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) 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. 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 #ex
$$y''-\tan xy'+y=0 \quad x_{0}=0$$ $$y''-\tan xy'+y=0 \quad x_{0}=0$$
notice the coefficient beside y is 1, 1 is analytic and differentiable everywhere, obviously! 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. Same goes for any polynomial, it's obvious that any polynomial is infinitely differentiable but it's important to know.
What about tan x? What about $\tan x$?
$\tan x=\frac{\sin x}{\cos x}$ is not defined on $x=\frac{\pi}{2}\pm n\pi, \qquad n=0,1,2,\dots$ $\tan x=\frac{\sin x}{\cos x}$ 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: 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:
$\rho\geq\mid x_{0}-\frac{\pi}{2}\mid=\frac{\pi}{2}$ $\rho\geq\mid x_{0}-\frac{\pi}{2}\mid$
$\rho\geq \frac{\pi}{2}$
convergence could be larger, but we are guaranteed convergence on $x=x_{0}-\rho$ to $x_{0}+\rho$ convergence could be larger, but we are guaranteed convergence on $x=x_{0}-\rho$ to $x_{0}+\rho$
#ex #ex
@ -211,79 +222,108 @@ now we have to compute the two distances of these singular points to x=1
To calculate distance: $\alpha_{1}+\beta_{1}i, \qquad \alpha_{2}+\beta_{2}i$ To calculate distance: $\alpha_{1}+\beta_{1}i, \qquad \alpha_{2}+\beta_{2}i$
$\sqrt{ (\alpha_{1}-\alpha_{2})^2+(\beta_{1}-\beta_{2})^2 }$ $\sqrt{ (\alpha_{1}-\alpha_{2})^2+(\beta_{1}-\beta_{2})^2 }$
$\rho\geq \sqrt{ 1^2+1^2 }=\sqrt{ 2 }$ $\rho\geq \sqrt{ 1^2+1^2 }=\sqrt{ 2 }$
#end of lec 24 #end of lec 24 #start of lec 25
#start of lec 25
find the zeros of #ex #powseries
$xy''-y'+y=0 \qquad x_{0}=2$ Find the first 5 non-zero terms of the general solution of:
is this function analytic about x_0=2? $$xy''-y'+y=0 \qquad x_{0}=2$$
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 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.
so we have a radius convergence of $\rho=2$ DONT FORGET! put it in standard form first:
$y(x)=\sum_{n=0}^\infty a_{n}(x-2)^n \quad x\in(0,4)$ $y''-\frac{1}{x}y'+\frac{y}{x}=0$ <- now we can see that there are singular points at $x=0$
$x-2=t \qquad t\in(-2,2)$ 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 $(t+2) \frac{d^2y}{dt^2}-\frac{dy}{dt}+y=0$
$y(t)=\sum_{n=0}^\infty a_{n}t^n$ $y(t)=\sum_{n=0}^\infty a_{n}t^n$
$y'(t)=\sum_{n=1}^\infty a_{n}nt^{n-1}$ $y'(t)=\sum_{n=1}^\infty a_{n}nt^{n-1}$
$y''(t)=\sum_{n=2}^\infty a_{n}n(n-1)t^{n-2}$ $y''(t)=\sum_{n=2}^\infty a_{n}n(n-1)t^{n-2}$
$\sum_{n=2}^\infty a_{n}n(n-1)t^{n-1}+2\sum_{n=2}^\infty a_{n} n(n-1)t^{n-2}-\sum_{n=1}^\infty a_{n}nt^{n-1}+\sum_{n=0}^\infty a_{n}t^n=0$ plug them into the DE:
n-1=k n-2=k n-1=k $\underset{ n-1=k }{ \sum_{n=2}^\infty a_{n}n(n-1)t^{n-1} }+\underset{ n-2=k }{ 2\sum_{n=2}^\infty a_{n}n(n-1)t^{n-2} }-\underset{ n-1=k }{ \sum_{n=1}^\infty a_{n}nt^{n-1} }+\underset{ n=k }{ \sum_{n=0}^\infty a_{n}t^n }=0$
first 5 non-zeros: shift the index to equalize powers
$\sum_{k=1}^\infty a_{k+1}(k+1)kt^k+\sum_{k=0}^\infty 2(k+2)(k+1)a_{k+2}t^k-\sum_{k=0}^\infty a_{k+1}(k+1)t^k+\sum_{k=0}^\infty a_{k}t^k$ $\sum_{k=1}^\infty a_{k+1}(k+1)kt^k+\sum_{k=0}^\infty 2(k+2)(k+1)a_{k+2}t^k-\sum_{k=0}^\infty a_{k+1}(k+1)t^k+\sum_{k=0}^\infty a_{k}t^k=0$
$\underbrace{ 4a_{2}-a_{1}+a_{0} }_{ =0 }+\sum_{k=1}^\infty \underbrace{ (a_{k+1}(k+1)k+4a_{k+2}-a_{k+1}(k+1)+a_{k}) }_{ =0 }t^k=0$ take out some terms to equalize the starting index
$a_{2}=\frac{a_{1}-a_{0}}{4}$ $\underbrace{ 4a_{2}-a_{1}+a_{0} }_{ =0 }+\sum_{k=1}^\infty \underbrace{ (a_{k+1}(k+1)k+2(k+2)(k+1)a_{k+2}-a_{k+1}(k+1)+a_{k}) }_{ =0 }t^k=0$
$12a_{3}+2a_{2}-2a_{2}+a_{1}=0$ left hand side tells us $a_{2}=\frac{a_{1}-a_{0}}{4}$
$a_{3}=-\frac{a_{1}}{12}$ and the rest of the coefficients can be written as a recursive relation:
$a_{4}=-\frac{1}{24}(3a_{3}+a_{2})=\frac{1}{96}a_{1}-\frac{a_{1}}{96}+\frac{a_{0}}{96}=\frac{a_{0}}{96}$ $a_{k+1}(k+1)k+2(k+2)(k+1)a_{k+2}-a_{k+1}(k+1)+a_{k}=0$
$y(x)=a_{0}+a_{1}(x-2)+\frac{a_{1}-a_{0}}{4}(x-2)^2-\frac{a_{1}}{12}(x-2)^3+\frac{a_{0}}{96}(x-2)^4+\dots$ This yields:
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. >for $k=1$:
So we are done.
#ex $\cancel{ 2a_{2} }+12a_{3}-\cancel{ 2a_{2} }+a_{1}=0$
$a_{3}=-\frac{a_{1}}{12}$
>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 $x_{0}=\pi$ Find first four non-zero terms of the power series for $y(x)$ about $x_{0}=\pi$
of the IVP: of the IVP:
$$y''-\sin (x)y=0 \qquad y(\pi)=1 \qquad y'(\pi)=0$$ $$y''-\sin (x)y=0 \qquad y(\pi)=1 \qquad y'(\pi)=0$$
This is already in standard form. 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. clearly this is analytical over the entire real axis, $\sin(x)$ and $1$ are both infinitely differentiable. No singular points, real or complex.
$y(x)=\sum_{n=0}^\infty a_{n}(x-\pi)^n \qquad x-\pi=t$ $y(x)=\sum_{n=0}^\infty a_{n}(x-\pi)^n \qquad x-\pi=t$
$y(t)=\sum_{n=0}^\infty a_{n}t^n$ <- we are abusing notation, the y here isn't the same as above. But it's all good. 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.
$y''-\sin(t+\pi)\sum_{n=0}^\infty a_{n}t^n=0$ $y''-\sin(t+\pi)\sum_{n=0}^\infty a_{n}t^n=0$
$y''+\sin(t)\sum_{n=0}^\infty a_{n}t^n=0$ $y''+\sin(t)\sum_{n=0}^\infty a_{n}t^n=0$
$\sum_{n=2}^\infty a_{n}n(n-1)t^{n-2}+\left( \sum_{n=0}^\infty(-1)^n \frac{t^{2n+1}}{(2n+1)!} \right)\left( \sum_{n=0}^\infty a_{n}t^n \right)$ remember, sin is odd so its infinite series has odd powers. $\sum_{n=2}^\infty a_{n}n(n-1)t^{n-2}+\left( \sum_{n=0}^\infty(-1)^n \frac{t^{2n+1}}{(2n+1)!} \right)\left( \sum_{n=0}^\infty a_{n}t^n \right)=0$ remember, $\sin$ is an odd function so its infinite series has only odd powers.
now from: using:
$y''+\sin(t)\sum_{n=0}^\infty a_{n}t^n=0$
$y(t)=\sum_{n=0}^\infty a_{n}t^n$ $y(t)=\sum_{n=0}^\infty a_{n}t^n$
this implies $y(0)=1=a_{0} \quad y'(0)=0=a_{1}$ $y'(t)=\sum_{n=1}^\infty a_{n}nt^{n-1}$
we can deduce that $y(0)=1=a_{0} \quad y'(0)=0=a_{1}$
from the big summ-y equation: from the big summ-y equation:
$(2a_{2}+6a_{3}t+12a_{4}t^2+20a_{5}t^3+\dots)+\left( t-\frac{t^3}{6}+\frac{t^5}{120}-\dots \right)(a_{0}+a_{1}t+a_{2}t^2+a_{3}t^3+\dots)=0$ $(2a_{2}+6a_{3}t+12a_{4}t^2+20a_{5}t^3+30a_{6}t^4+\dots)+\left( t-\frac{t^3}{6}+\frac{t^5}{120}-\dots \right)(a_{0}+a_{1}t+a_{2}t^2+a_{3}t^3+\dots)=0$
the only constant factor is $a_{2}$ the only constant factor is $a_{2}$
this implies: $2a_{2}=0 \implies a_{2}=0$ this implies: $2a_{2}=0 \implies a_{2}=0$
what about the factors of $t$? what about the factors of $t$?
$(6a_{3}+a_{0})t=0$ $(6a_{3}+a_{0})t=0$
$a_{3}=-\frac{a_{0}}{6}=-\frac{1}{6}$ $a_{3}=-\frac{a_{0}}{6}=-\frac{1}{6}$
</br>
$t^2$: factors of $t^2$:
$(12a_{4}+a_{1})t^2=0$ $(12a_{4}+a_{1})t^2=0$
$a_{2}=-\frac{a_{1}}{12}$ $a_{4}=-\frac{a_{1}}{12}=0$
</br>
$t^3$: $t^3$:
$\left( 20a_{5}+a_{2}-\frac{a_{0}}{6} \right)=0 \implies a_{5}=\frac{1}{120}$ $\left( 20a_{5}+\cancelto{ 0 }{ a_{2} }-\frac{a_{0}}{6} \right)=0 \implies a_{5}=\frac{1}{120}$
</br>
$t^4$: $t^4$:
$\left( 30a_{6}+a_{3}-\frac{a_{1}}{6} \right)t^4=0 \implies a_{6}=\frac{1}{180}$ $\left( 30a_{6}+a_{3}-\cancelto{ 0 }{ \frac{a_{1}}{6} } \right)t^4=0 \implies a_{6}=\frac{1}{180}$
$y(x)=1-\frac{1}{6}(x-\pi)^3+\frac{1}{120}(x-\pi)^5+\frac{1}{180}(x-\pi)^6+\dots$ </br>
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^. 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 #ex #powseries
Here's a non-homogenous example: (RHS$\ne 0$)
$$y'-xy=e^x \qquad x_{0}=0$$ $$y'-xy=e^x \qquad x_{0}=0$$
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.
$y(x)=\sum_{n=0}^\infty a_{n}x^n$ $y(x)=\sum_{n=0}^\infty a_{n}x^n$
$y'(x)=\sum_{n=1}^\infty a_{n}nx^{n-1}$ $y'(x)=\sum_{n=1}^\infty a_{n}nx^{n-1}$
shift the index and plug in to the equation.
$\sum_{k=0}^\infty a_{k+1}(k+1)x^k-\sum_{k=1}^\infty a_{k-1}x^k-\sum_{k=0}^\infty \frac{x^k}{k!}=0$ $\sum_{k=0}^\infty a_{k+1}(k+1)x^k-\sum_{k=1}^\infty a_{k-1}x^k-\sum_{k=0}^\infty \frac{x^k}{k!}=0$
$a_{1}-1+\sum_{k=1}^\infty a_{k+1}(k+1)x^k-\sum_{k=1}^\infty a_{k-1}x^k-\sum_{k=1}^\infty \frac{x^k}{k!}=0$
$\underbrace{ a_{1}-1 }_{ =0 }+\sum_{k=1}^\infty (\underbrace{ a_{k+1}(k+1)-a_{k-1}-\frac{1}{k!} }_{ =0 })x^k=0$
$a_{1}-1=0 \implies a_{1}=1$ $a_{1}-1=0 \implies a_{1}=1$
$a_{k+1}=\frac{a_{k-1}+\frac{1}{k!}}{k+1}$ $a_{k+1}=\frac{a_{k-1}+\frac{1}{k!}}{k+1}$
$k=1\implies a_{2}=\frac{a_{0}}{2}+\frac{1}{2}$ $k=1\implies a_{2}=\frac{a_{0}}{2}+\frac{1}{2}$
$k=2\implies a_{3}=\frac{1}{2}$ $k=2\implies a_{3}=\frac{a_{1}}{3}+\frac{\frac{1}{2}}{3}=\frac{1}{2}$
$k=3\implies a_{4}=\frac{ \left( \frac{a_{0}}{2}+\frac{1}{2} \right)+\frac{1}{6}}{4}$ $k=3\implies a_{4}=\frac{ \left( \frac{a_{0}}{2}+\frac{1}{2} \right)+\frac{1}{6}}{4}$
We are lucky, in this course fubini's method is not needed. (what?) We are lucky, in this course fubini's method is not needed. (what?)
and with that, we are finished this chapter on power series. and with that, we are finished this chapter on power series.

2
content/_index.md
View File

@ -22,7 +22,7 @@ I have written these notes for myself, I thought it would be cool to share them.
[Convolution (lec 19-20)](convolution-lec-19-20.html) [Convolution (lec 19-20)](convolution-lec-19-20.html)
[Dirak δ-function (lec 21)](dirak-δ-function-lec-21.html) [Dirak δ-function (lec 21)](dirak-δ-function-lec-21.html)
[Systems of linear equations (lec 21-22)](systems-of-linear-equations-lec-21-22.html) [Systems of linear equations (lec 21-22)](systems-of-linear-equations-lec-21-22.html)
[Power series (lec 22-25)](power-series-lec-22-25.html) (raw notes, not reviewed or revised yet.) [Power series (lec 22-25)](power-series-lec-22-25.html)
[Separation of variables & Eigen value problems (lec 26-28)](separation-of-variables-eigen-value-problems-lec-26-28.html) (raw notes, not reviewed or revised yet.) [Separation of variables & Eigen value problems (lec 26-28)](separation-of-variables-eigen-value-problems-lec-26-28.html) (raw notes, not reviewed or revised yet.)
[Fourier series (lec 28-29)](fourier-series-lec-28-29.html) (raw notes, not reviewed or revised yet.) [Fourier series (lec 28-29)](fourier-series-lec-28-29.html) (raw notes, not reviewed or revised yet.)
[Partial differential equations (lec 30-33)](partial-differential-equations-lec-30-33.html) (raw notes, not reviewed or revised yet.) [Partial differential equations (lec 30-33)](partial-differential-equations-lec-30-33.html) (raw notes, not reviewed or revised yet.)

BIN
content/drawings/powerseriessolution.png Normal file
View File

Binary file not shown.
Side by Side

After

Width:  |  Height:  |  Size: 230 KiB