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 $\mid x-x_{0}\mid<\rho$,divergentforallxwhere$\midx-x_{0}\mid>\rho$
"Who keeps stealing the whiteboard erases? (jokingly) It's a useless object, anyways"
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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 $-1<x<3$,divergenton$\midx-1\mid>2$
"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)
$a_{k+1}=\frac{2}{k+1}a_{k-1}$ 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
>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."
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).
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## 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:
$\sum_{n=2}^\infty a_{n}n(n-1)t^{n-2}+\left( \sum_{n=0}^\infty(-1)^n \frac{t^{2n+1}}{(2n+1)!} \right)\left( \sum_{n=0}^\infty a_{n}t^n \right)$ remember, sin is odd so its infinite series has odd powers.