3.12 Radius of Convergence

Show that the following series of functions converge uniformly on \(\R\)
1. \(\sum^\infty_{n=0}\frac{\sin ^2(x)}{2^n+1}\)
  
  Since \(f_{n}(x)=\frac{\sin^{2}(x)}{2^{n}+1}\), then \(|f_{n}|=|\frac{\sin^{2}(x)}{2^{n}+1}|\leq|\frac{1}{2^{n}+1}|<\frac{1}{2^{n}}=M_{n}\).
  
  We know \(\sum M_n\) is convergent, then \(\sum^\infty_{n=0}\frac{\sin ^2(x)}{2^n+1}\) converges uniformly by M-test 2. \(\sum\frac{e^{-nx^2}}{n^{2}+x^{2}}\)
  
  Since \(f_{n}(x)=\frac{e^{-nx^2}}{n^{2}+x^{2}}\), then \(|f_{n}|=\left|\frac{e^{-nx^2}}{n^{2}+x^{2}}\right|\leq\left|\frac{1}{n^{2}+x^{2}} \right|<\frac{1}{n^{2}}=M_{n}\).
  
  We know \(\sum M_n\) is convergent, then \(\sum^\infty_{n=0}\frac{\sin ^2(x)}{2^n+1}\) converges uniformly
Show that \(\sum\frac{\cos(2^n x)}{2^n}\) is continuous in \(\R\) and \(\sum\frac{x^{n}}{n^{2}}\) is continuous on \([-1,1]\)

First, we know \(\frac{\cos(2^nx)}{2^n}\) is continuous

Next, we need to show that \(\sum\frac{\cos(2^{n}x)}{2^{n}}\) converges uniformly on \(\R\)

Easily, use M-test, then it is convergent uniformly

First, we know \(\frac{x^{n}}{n^{2}}\) is continuous

Next, we need to show that \(\sum\frac{x^{n}}{n^{2}}\) converges uniformly on \(\R\)

Easily, use M-test, then it is convergent uniformly
At most, on how many points can a power series converge conditionally

Two points, which is \(x=-R,x=R\) since in the interval it is absolutely convergent, and outside is divergent
For each of the following series, find the interval of convergence and the radius of convergence
1. \(\sum (-1)^n\frac{10^n}{n!}(x-10)^n\)
  
  Use DA criterium: \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim_{n\to\infty}\frac{1}{n+1} \left(x-10\right)=0\), then \(x\in \R\) 2. \(\sum(-1)^{n}\frac{1}{n\times10^{n}}(x-2)^{n}\)
  
  Use DA criterium: \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\left|-\left(x-2\right)\right |\lim_{n\to\infty}\frac{n}{10\left(n+1\right)}=\left|x-2\left|<10\Rightarrow-8<x< 12\right.\right.\)
  
  When \(x=-8\), we have \(\sum(-1)^{n}\frac{1}{n\times10^{n}}(-10)^{n}=\sum\frac{1}{n}\) is divergent
  
  When \(x=12\), then we have \(\sum(-1)^{n}\frac{1}{n\times10^{n}}(10)^{n}=\sum\frac{\left(-1\right)^{n}}{n}\) is convergent
  
  Thus it is convergent on \((-8,12]\)
Suppose that a power series has radius of convergence \(R\). Then the series \(\sum na_{n}(x)^{n-1}\) also has radius of convergence \(R\)

Take \(x\neq 0\) with \(|x|<R\). Choose \(\rho\) such that \(|x|<\rho<R\) and \(r=\frac{|x|}{\rho}\), then \(0<r<1\)

To estimate the terms of the differentiated series by the terms of the original series, we rewrite their absolute value as follow

\(|na_{n}x^{n-1}|=\frac{n}{\rho}\left(\frac{|x|}{\rho}\right)^{n-1}|a_{n}\rho^{n}| =\frac{nr^{n-1}}{\rho}\left|a_{n}\rho^{n}\right|\)

Then we consider the series \(\sum nr^{n-1}\) converges(D'A test), then the sequence \(nr^{n-1}\) bounded, say by \(M\)

Then \(|na_{n}x^{n-1}|=\frac{nr^{n-1}}{\rho}\left|a_{n}\rho^{n}\right|\leq\frac{M}{\rho} \left|a_{n}\rho^{n}\right|\) and we know \(\sum a_n\rho^n\) is absolutely convergent since \(\rho<R\)

By comparison test, we have that \(\sum na_{n}(x)^{n-1}\) is absolutely convergent

We still need to check that series is divergent outside the \(R\) by definition of \(R\)

Now assume \(|x|>R\), then \(\sum |a_nx^n|\) diverges, then \(|na_{n}x^{n-1}|\geq\frac{ 1 }{|x|}|a_nx^n|\)

Since \(\sum |na_nx^{n-1}|\) diverges then by comparison test we know that \(\sum|na_nx^{n-1}|\) diverges \(\forall x\in \R\) such that \(|x|>R\)

In general, if power series \(f(x)=\sum_{n=0}^{\infty}a_{n}\left(x-x_{0}\right)^{n}\) with positive radius of convergent \(R\) has derivatives of all orders in it's interval of convergence, we have that \(f^{(k)}(x)=\sum_{n=k}^{\infty}n\left(n-1\right)\left(n-2\right)\ldots\left(n-k+1 \right)a_{n}\left(x-x_{0}\right)^{n-k}\) and the radius of convergence for each of these series is \(R\)

Even more, if \(f(x)=\sum a_{n}\left(x-x_{0}\right)^{n}\) where \(|x-x_{0}|<R\), then \(a_{n}=\frac{f^{(n)}(x_{0})}{n!}\) since we can replace \(x=x_0\)

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