forked from Sasserisop/MATH201
73 lines
4.2 KiB
Markdown
73 lines
4.2 KiB
Markdown
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#start of lec 22
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Finished chapter 7 of the course textbook, Let's begin chapter 8!
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# Power series
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A power series is defined by:
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$$\sum_{n=0}^\infty a_{n}(X-X_{0})^n=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots$$
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It is convergent if:
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$$\sum_{n=0} ^ \infty a_{n}(x-x_{0})^n<\infty \text{ at a given x}$$
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Otherwise, it is divergent.
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If $\sum_{n=0}^\infty \mid a_{n}(x-x_{0})^n\mid$ is convergent
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$\implies\sum_{n=0}^\infty a_{n}(x-x_{0})^n$ is absolutely convergent
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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).
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Theorem: With each $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ we can associate $0\leq \rho\leq \infty$ such that
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$\sum_{n=0} ^\infty a_{n}(x-x_{0})^n$ is absolutely convergent
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for all x such that $\mid x-x_{0}\mid<\rho$, divergent for all x where $\mid x-x_{0}\mid>\rho$
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"Who keeps stealing the whiteboard erases? (jokingly) It's a useless object, anyways"
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![[Drawing 2023-10-30 13.12.57.excalidraw.png]]
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how can we find $\rho$?
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Ratio test: If $\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=L$
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then $\rho=\frac{1}{L}$
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## Examples:
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#ex
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is this convergent? Divergent? and where so?
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$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}(x-1)^n$
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determine the convergent set.
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Use ratio test:
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$\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=\lim_{ n \to \infty } \frac{2^{-(n+1)}}{n+2} \frac{n+1}{2^{-n}}=\frac{1}{2}\implies \rho=2$
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so it's convergent on $-1<x<3$, divergent on $\mid x-1\mid>2$
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But what about on the points $-1$ and $3$?
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plug in $x_{0}=-1$
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$\sum_{n=0}^{\infty} \frac{2^{-n}}{n+1}(-2)^n=\sum_{n=0}^\infty \frac{(-1)^n}{n+1}<\infty$ <- That is the alternating harmonic series, it is convergent.
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plug in $x_{0}=3$:
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$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}2^n=\sum_{n=0}^\infty \frac{1}{n+1}>\infty$ <- harmonic series, this diverges.
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so the power series is convergent on $[-1,3)$ divergent otherwise.
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$$\text{ converges only on: } [-1,3)$$
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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 $\rho>0$
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Then:
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1.) $\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=0}^{\infty}b_{n}(x-x_{0})^n=\sum_{n=0}^\infty(a_{n}+b_{n})(x-x_{0})^n$
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That has a radius of convergence of at least $\rho$.
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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)
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$=(a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots)(b_{0}+b_{1}(x-x_{0})+b_{2}(x-x_{0})^2+\dots)$
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$=a_{0}b_{0}+(a_{0}b_{1}+a_{1}b_{0})(x-x_{0})+(a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{1})(x-x_{0})^2+\dots$ (Cauchy multiplication)
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more Definitions of power series:
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If $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ is convergent with $\rho>0$
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$\mid x-x_{0}\mid<\rho$
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we can differentiate this infinite sum and get:
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$\implies y'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
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$y''(x)=\sum_{n=2}^\infty a_{n}n(n-1)(x-x_{0})^{n-2}$
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Theorem: If $y(x)$ is infinitely many times differentiable on some interval: $\mid x-x_{0}\mid<\rho$
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then: $\sum_{n=0}^\infty \frac{y^{(n)}(x_{0})}{n!}(x-x_{0})^n$ (Taylor series)
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"believe me, taylor series is the most important theorem in engineering."
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"I mean engineering is all about approximations, do you know how your calculator computes ...? Taylor series!"
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"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!)
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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 $(x_{0}-\rho,x_{0}+\rho)$
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btw analytic functions are very important in complex calculus MATH301. (i don't have that next term)
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$f(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n<\infty$
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$f'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
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$f(x)+f'(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
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let $n-1=k$
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$=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{k=0}^\infty a_{n}(k+1)(x-x_{0})^{k}$
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$=\sum_{n=0}^\infty(a_{n}+a_{n}(n+1))(x-x_{0})^n$
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Last theorem fo' da day:
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If $\sum_{n=0}^\infty a_{n}(x-x_{0})^n=0$ for all x$\in(x_{0}-\rho,x_{0}+\rho)$ where $\rho>0$
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$\implies a_{n}=0$, $n=0,1,2,\dots$
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#end of lec 22
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