Consider the power series below. Given the center and find radius of convergence for each.
Solution
1) Comparing to form \({\displaystyle \sum \limits _{n=1}^{\infty }} a_{n}\left ( z-z_{0}\right ) ^{n}\) then center is \(-i\sqrt{2}\). Now, \begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{a_{n+1}}{a_{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( n+1\right ) }{n}\right \vert \\ & =1 \end{align*}
Hence \(R=1\)
2) Comparing to form \({\displaystyle \sum \limits _{n=1}^{\infty }} a_{n}\left ( z-z_{0}\right ) ^{n}\) then center is \(i\pi \). Now,
\begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{a_{n+1}}{a_{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( \frac{a}{b}\right ) ^{n+1}}{\left ( \frac{a}{b}\right ) ^{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \left ( \frac{a}{b}\right ) \right \vert \\ & =\frac{a}{b} \end{align*}
Hence \(R=\frac{b}{a}\)
3) Comparing \({\displaystyle \sum \limits _{n=0}^{\infty }} \frac{\left ( 3n\right ) !}{2^{n}\left ( n!\right ) ^{3}}z^{n}\) to \({\displaystyle \sum \limits _{n=1}^{\infty }} a_{n}\left ( z-z_{0}\right ) ^{n}\) then center is \(0\). Now
\begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{a_{n+1}}{a_{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\frac{\left ( 3\left ( n+1\right ) \right ) !}{2^{n+1}\left ( \left ( n+1\right ) !\right ) ^{3}}}{\frac{\left ( 3n\right ) !}{2^{n}\left ( n!\right ) ^{3}}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 3\left ( n+1\right ) \right ) !2^{n}\left ( n!\right ) ^{3}}{\left ( 3n\right ) !2^{n+1}\left ( \left ( n+1\right ) !\right ) ^{3}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 3n+3\right ) !\left ( n!\right ) ^{3}}{\left ( 3n\right ) !\left ( \left ( n+1\right ) !\right ) ^{3}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 3n+1\right ) \left ( 3n+2\right ) \left ( 3n+1\right ) \left ( 3n\right ) !\left ( n!\right ) ^{3}}{\left ( 3n\right ) !\left ( \left ( n+1\right ) !\right ) ^{3}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 3n+1\right ) \left ( 3n+2\right ) \left ( 3n+1\right ) \left ( n!\right ) ^{3}}{\left ( \left ( n+1\right ) !\right ) ^{3}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 3n+1\right ) \left ( 3n+2\right ) \left ( 3n+1\right ) \left ( n!\right ) ^{3}}{\left ( \left ( n+1\right ) n!\right ) ^{3}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 3n+1\right ) \left ( 3n+2\right ) \left ( 3n+1\right ) \left ( n!\right ) ^{3}}{\left ( n+1\right ) ^{3}\left ( n!\right ) ^{3}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 3n+1\right ) \left ( 3n+2\right ) \left ( 3n+1\right ) }{\left ( n+1\right ) ^{3}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 3n+1\right ) \left ( 3n+2\right ) \left ( 3n+1\right ) }{\left ( n+1\right ) ^{3}}\right \vert \end{align*}
Hence the above becomes
\begin{align*} L & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{27n^{3}+36n^{2}+15n+2}{n^{3}+3n^{2}+3n+1}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{27+36\frac{1}{n}+15\frac{1}{n^{2}}+\frac{2}{n^{3}}}{1+3\frac{1}{n}+3\frac{1}{n^{2}}+\frac{1}{n^{3}}}\right \vert \\ & =\frac{27}{2} \end{align*}
Hence \(R=\frac{2}{27}.\)
4) Comparing \({\displaystyle \sum \limits _{n=0}^{\infty }} \frac{1}{\left ( 1+i\right ) ^{n}}\left ( z-\left ( -2+i\right ) \right ) ^{n}\) to \({\displaystyle \sum \limits _{n=1}^{\infty }} a_{n}\left ( z-z_{0}\right ) ^{n}\) shows that center is \(z_{0}=-2+i\). Now
\begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{a_{n+1}}{a_{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\frac{1}{\left ( 1+i\right ) ^{n+1}}}{\frac{1}{\left ( 1+i\right ) ^{n}}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 1+i\right ) ^{n}}{\left ( 1+i\right ) ^{n+1}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{1}{\left ( 1+i\right ) }\right \vert \\ & =\left \vert \frac{1}{\left ( 1+i\right ) }\right \vert \\ & =\frac{1}{\sqrt{2}} \end{align*}
Hence \(R=\sqrt{2}\)
Find radius of convergence using both 1) \(R=\frac{1}{L}\) where \(L=\lim _{n\rightarrow \infty }\left \vert \frac{a_{n+1}}{a_{n}}\right \vert \) and 2) the termwise differentiation/integration properties of power series. Do this for
Solution
1) First method. The center is \(i\). And
\begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{a_{n+1}}{a_{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\frac{6^{n+1}}{n+1}}{\frac{6^{n}}{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 6^{n+1}\right ) n}{6^{n}\left ( n+1\right ) }\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{6n}{\left ( n+1\right ) }\right \vert \\ & =6\lim _{n\rightarrow \infty }\left \vert \frac{n}{\left ( n+1\right ) }\right \vert \\ & =6 \end{align*}
Hence \(R=\frac{1}{6}\).
Second method: Taking termwise differentiation gives \begin{align*} f^{\prime }\left ( z\right ) & =\sum _{n=1}^{\infty }\frac{6^{n}}{n}n\left ( z-i\right ) ^{n-1}\\ & =6\sum _{n=1}^{\infty }6^{n-1}\left ( z-i\right ) ^{n-1} \end{align*}
Changing the indexing gives
\begin{align*} f^{\prime }\left ( z\right ) & =6\sum _{n=0}^{\infty }6^{n}\left ( z-i\right ) ^{n}\\ & =6\sum _{n=0}^{\infty }\left ( 6\left ( z-i\right ) \right ) ^{n} \end{align*}
Comparing to Binomial series \({\displaystyle \sum \limits _{n=0}^{\infty }} r^{n}\), the above is \(6\frac{1}{1-r}\) where \(r=6\left ( z-i\right ) \). Hence this converges for \(\left \vert r\right \vert <1\) or \(\left \vert 6\left ( z-i\right ) \right \vert <1\) or \(\left \vert \left ( z-i\right ) \right \vert <\frac{1}{6}\) and diverges for \(\left \vert z-i\right \vert >\frac{1}{6}\). Since termwise differentiated series has same radius of convergence, then \(R=\frac{1}{6}\) as using first method.
2) First method. Comparing \(\sum _{n=0}^{\infty }\frac{3^{n}\left ( n+1\right ) n}{5^{n}}\left ( z^{2}\right ) ^{n}\) to \(\sum _{n=0}^{\infty }a_{n}\left ( z-z_{0}\right ) ^{n}\) then center is zero. And
TODO
Show that \(\frac{1}{\left ( 1-z\right ) ^{2}}=\sum _{n=0}^{\infty }\left ( n+1\right ) z^{n}\,\), using (a) the Cauchy product. (b) By differentiating a suitable series.
Solution (a)
\begin{align*} \frac{1}{\left ( 1-z\right ) ^{2}} & =\frac{1}{\left ( 1-z\right ) }\frac{1}{\left ( 1-z\right ) }\\ & =\left ( 1+z+z^{2}+z^{3}+\cdots \right ) \left ( 1+z+z^{2}+z^{3}+\cdots \right ) \\ & =\left ( 1+z+z^{2}+z^{3}+\cdots \right ) +z\left ( 1+z+z^{2}+z^{3}+\cdots \right ) +z^{2}\left ( 1+z+z^{2}+z^{3}+\cdots \right ) +\cdots \\ & =\left ( 1+z+z^{2}+z^{3}+\cdots \right ) +\left ( z+z^{2}+z^{3}+\cdots \right ) +\left ( z^{2}++z^{3}+z^{4}+\cdots \right ) +\cdots \\ & =1+2z+3z^{2}+4z^{4}+\cdots \qquad \left \vert z\right \vert <1 \end{align*}
But \(\sum _{n=0}^{\infty }\left ( n+1\right ) z^{n}=1+2z+3z^{2}+4z^{4}+\cdots \). Hence the same.
Solution (b) Observing that
\[ \left ( n+1\right ) z^{n}=\frac{d}{dz}z^{n+1}\]
Then \begin{align*} \sum _{n=0}^{\infty }\left ( n+1\right ) z^{n}\, & =\sum _{n=0}^{\infty }\frac{d}{dz}z^{n+1}\\ & =\frac{d}{dz}\sum _{n=0}^{\infty }z^{n+1}\\ & =\frac{d}{dz}\sum _{n=1}^{\infty }z^{n}\\ & =\frac{d}{dz}\left ( z+z^{2}+z^{3}+\cdots \right ) \\ & =\frac{d}{dz}\left ( z\left ( 1+z+z^{2}+\cdots \right ) \right ) \\ & =\frac{d}{dz}\left ( \frac{z}{1-z}\right ) \end{align*}
But \(\frac{d}{dz}\frac{A\left ( z\right ) }{B\left ( z\right ) }=\frac{A^{\prime }B-AB^{\prime }}{B^{2}}\), hence the above becomes, where \(A=z,B=1-z\)
\begin{align*} \sum _{n=0}^{\infty }\left ( n+1\right ) z^{n}\, & =\frac{\left ( 1-z\right ) -z\left ( -1\right ) }{\left ( 1-z\right ) ^{2}}\\ & =\frac{1-z+z}{\left ( 1-z\right ) ^{2}}\\ & =\frac{1}{\left ( 1-z\right ) ^{2}} \end{align*}
If \(f\left ( z\right ) \) is an even function, where \(f\left ( z\right ) =\sum _{n=0}^{\infty }a_{n}z^{n}\), show that \(a_{n}=0\) when \(n\) is odd. And if \(f\left ( z\right ) \) is odd function, show that \(a_{n}=0\) when \(n\) is even.
Solution
If \(f\left ( z\right ) \) is even, then \(f\left ( -z\right ) =f\left ( z\right ) \). Therefore
\begin{align*} \sum _{n=0}^{\infty }a_{n}\left ( -z\right ) ^{n} & =\sum _{n=0}^{\infty }a_{n}z^{n}\\ a_{0}-a_{1}z+a_{2}z^{2}-a_{3}z^{3}+a_{4}z^{4}-\cdots & =a_{0}+a_{1}z+a_{2}z^{2}+a_{3}z^{3}+a_{4}z^{4}+\cdots \end{align*}
Since power series is unique, then we must have \(a_{1}=-a_{1}\) which means \(a_{1}=0\), the same for \(a_{3}=-a_{3}\), which gives \(a_{3}=0\) and so on for all odd \(a_{n}\).
If \(f\left ( z\right ) \) is odd, then \(f\left ( -z\right ) =-f\left ( z\right ) \). Therefore
\begin{align*} \sum _{n=0}^{\infty }a_{n}\left ( -z\right ) ^{n} & =-\sum _{n=0}^{\infty }a_{n}z^{n}\\ a_{0}-a_{1}z+a_{2}z^{2}-a_{3}z^{3}+a_{4}z^{4}-\cdots & =-\left ( a_{0}+a_{1}z+a_{2}z^{2}+a_{3}z^{3}+a_{4}z^{4}+\cdots \right ) \\ & =-a_{0}-a_{1}z-a_{2}z^{2}-a_{3}z^{3}-a_{4}z^{4}+\cdots \end{align*}
Since power series is unique, then we must have \(a_{0}=-a_{0}\) which means \(a_{0}=0\), the same for \(a_{2}=-a_{2}\), which gives \(a_{2}=0\) and so on for all even \(a_{n}\).
Develop the functions below in Maclaurin’s series and determine the radius of convergence \(R\) for each. (a) \(\cos \left ( 2z^{2}\right ) \), (b) \(\frac{z+2}{1-z^{2}}\)
Solution (a)
\[ \cos \left ( x\right ) =1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots \]
Replacing \(x=2z^{2}\) gives
\begin{align*} \cos \left ( 2z^{2}\right ) & =1-\frac{\left ( 2z^{2}\right ) ^{2}}{2!}+\frac{\left ( 2z^{2}\right ) ^{4}}{4!}-\frac{\left ( 2z^{2}\right ) ^{6}}{6!}+\cdots \\ & =1-\frac{2^{2}z^{4}}{2!}+\frac{2^{4}z^{8}}{4!}-\frac{2^{6}z^{12}}{6!}+\cdots \\ & =1-\frac{4z^{4}}{2!}+\frac{4^{2}z^{8}}{4!}-\frac{4^{3}z^{12}}{6!}+\cdots \\ & =\sum _{n=0}^{\infty }\left ( -1\right ) ^{n}\frac{4^{n}z^{2n}}{\left ( 2n\right ) !}\\ & =\sum _{n=0}^{\infty }\left ( -1\right ) ^{n}\frac{\left ( 4z^{2}\right ) ^{n}}{\left ( 2n\right ) !} \end{align*}
Hence \begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{a_{n+1}}{a_{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\frac{\left ( 4z^{2}\right ) ^{n+1}}{\left ( 2\left ( n+1\right ) \right ) !}}{\frac{\left ( 4z^{2}\right ) ^{n}}{\left ( 2n\right ) !}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 4z^{2}\right ) ^{n+1}\left ( 2n\right ) !}{\left ( 4z^{2}\right ) ^{n}\left ( 2\left ( n+1\right ) \right ) !}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{4z^{2}\left ( 2n\right ) !}{\left ( 2n+2\right ) !}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{4z^{2}\left ( 2n\right ) !}{\left ( 2n+2\right ) \left ( 2n+1\right ) \left ( 2n\right ) !}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{4z^{2}}{\left ( 2n+2\right ) \left ( 2n+1\right ) }\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{4z^{2}}{4n^{2}+6n+2}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\frac{4}{n^{2}}z^{2}}{4+6\frac{1}{n}+\frac{2}{n}}\right \vert \\ & =\left \vert z^{2}\right \vert \lim _{n\rightarrow \infty }\left \vert \frac{0}{4}\right \vert \\ & =0 \end{align*}
Hence \(R=\frac{1}{L}=\infty \)
(b) \(\frac{z+2}{1-z^{2}}\). Apply partial fractions. Obtain two binomial series and combine.
Develop (a) \(f\left ( z\right ) =\frac{1}{z}\) in Taylor series around \(z_{0}=i\). (b) \(g\left ( z\right ) =e^{z}\) around \(z_{0}=a\). What is radius of convergence?
Solution (a)
\[ f\left ( z\right ) =f\left ( i\right ) +\left ( z-i\right ) f^{\prime }\left ( i\right ) +\frac{\left ( z-i\right ) ^{2}f^{\prime \prime }\left ( i\right ) }{2!}+\frac{\left ( z-i\right ) ^{3}f^{\prime \prime \prime }\left ( i\right ) }{3!}+\cdots \]
But \(f^{\prime }\left ( z\right ) =-\frac{1}{z^{2}},f^{\prime \prime }\left ( z\right ) =\frac{2}{z^{3}},f^{\prime \prime \prime }\left ( z\right ) =-\frac{\left ( 2\right ) \left ( 3\right ) }{z^{4}},\cdots \), hence the above becomes
\begin{align*} f\left ( z\right ) & =\frac{1}{i}-\left ( z-i\right ) \frac{1}{i^{2}}+\frac{\left ( z-i\right ) ^{2}}{2!}\frac{2}{i^{3}}+\frac{\left ( z-i\right ) ^{3}}{3!}\left ( -\frac{2\left ( 3\right ) }{i^{4}}\right ) +\cdots \\ & =-i+\left ( z-i\right ) +2i\frac{\left ( z-i\right ) ^{2}}{2!}-2\left ( 3\right ) \frac{\left ( z-i\right ) ^{3}}{3!}+\cdots \\ & =-i+\left ( z-i\right ) +i\left ( z-i\right ) ^{2}-\left ( z-i\right ) ^{3}+\cdots \\ & =\sum _{n=0}^{\infty }\frac{\left ( -1\right ) ^{n}}{i^{n+1}}\left ( z-i\right ) ^{n} \end{align*}
Hence this convergence for \(\left \vert z-i\right \vert <1\).
Solution (b)
\[ g\left ( z\right ) =g\left ( a\right ) +\left ( z-a\right ) g^{\prime }\left ( a\right ) +\frac{\left ( z-a\right ) ^{2}g^{\prime \prime }\left ( a\right ) }{2!}+\frac{\left ( z-a\right ) ^{3}g^{\prime \prime \prime }\left ( a\right ) }{3!}+\cdots \]
But \(g^{\prime }\left ( z\right ) =e^{z},g^{\prime \prime }\left ( z\right ) =e^{z},g^{\prime \prime \prime }\left ( z\right ) =e^{z},\cdots \), hence the above becomes
\begin{align*} g\left ( z\right ) & =e^{a}+\left ( z-a\right ) e^{a}+\frac{\left ( z-a\right ) ^{2}e^{a}}{2!}+\frac{\left ( z-a\right ) ^{3}e^{a}}{3!}+\cdots \\ & =e^{a}\left ( 1+\left ( z-a\right ) +\frac{\left ( z-a\right ) ^{2}}{2!}+\frac{\left ( z-a\right ) ^{3}}{3!}+\cdots \right ) \\ & =e^{a}\sum _{n=0}^{\infty }\frac{\left ( z-a\right ) ^{n}}{n!} \end{align*}
Where \(L=\lim _{n\rightarrow \infty }\left \vert \frac{a_{n+1}}{a_{n}}\right \vert =\lim _{n\rightarrow \infty }\left \vert \frac{\frac{1}{\left ( n+1\right ) !}}{\frac{1}{n!}}\right \vert =\lim _{n\rightarrow \infty }\left \vert \frac{n!}{\left ( 1+n\right ) !}\right \vert =\lim _{n\rightarrow \infty }\frac{n!}{n!\left ( 1+n\right ) }=\lim _{n\rightarrow \infty }\frac{1}{1+n}=0\). Hence \(R=\frac{1}{L}=\infty \). Converges everywhere.
Show that \(\sum _{n=0}^{\infty }\frac{\left ( n!\right ) ^{2}}{\left ( 2n\right ) !}z^{n}\) converges uniformly in \(\left \vert z\right \vert \leq 3\)
Solution:
To find if it converges uniformly for \(\left \vert z\right \vert \leq 3\), we need to find \(R\), the radius of converges using normal method, then it \(R>3\), then it will converge uniformly for \(\left \vert z\right \vert \leq 3\).\begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{a_{n+1}}{a_{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\frac{\left ( \left ( n+1\right ) !\right ) ^{2}}{\left ( 2\left ( n+1\right ) \right ) !}}{\frac{\left ( n!\right ) ^{2}}{\left ( 2n\right ) !}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( \left ( n+1\right ) !\right ) ^{2}\left ( 2n\right ) !}{\left ( n!\right ) ^{2}\left ( 2\left ( n+1\right ) \right ) !}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( \left ( n+1\right ) n!\right ) ^{2}\left ( 2n\right ) !}{\left ( n!\right ) ^{2}\left ( 2\left ( n+1\right ) \right ) !}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( n+1\right ) ^{2}\left ( 2n\right ) !}{\left ( 2n+2\right ) !}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( n+1\right ) ^{2}\left ( 2n\right ) !}{\left ( 2n+2\right ) \left ( 2n+1\right ) \left ( 2n\right ) !}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( n+1\right ) ^{2}}{\left ( 2n+2\right ) \left ( 2n+1\right ) }\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{n^{2}+2n+1}{4n^{2}+6n+2}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{1+\frac{2}{n}+\frac{1}{n^{2}}}{4+\frac{6}{n^{2}}+\frac{2}{n^{2}}}\right \vert \\ & =\frac{1}{4} \end{align*}
Hence Radius of convergence \(R=4\). Since \(3<4\), then it converges uniformly for \(R<3\).
Where does \(\sum _{n=1}^{\infty }\left ( \frac{n+2}{5n-3}\right ) ^{n}z^{n}\) converges uniformly?
Solution We first find \(R\). Since the series of the form\(\ \sum _{n=1}^{\infty }A^{n}z^{n}\) then it is easier to use \begin{align*} L & =\lim _{n\rightarrow \infty }\sqrt [\frac{1}{n}]{\left \vert A^{n}\right \vert }\\ & =\lim _{n\rightarrow \infty }\sqrt [\frac{1}{n}]{\left \vert A^{n}\right \vert }\\ & =\lim _{n\rightarrow \infty }\sqrt [\frac{1}{n}]{\left \vert \frac{n+2}{5n-3}\right \vert ^{n}}\\ & =\lim _{n\rightarrow \infty }\sqrt [\frac{1}{n}]{\left \vert \frac{1+\frac{2}{n}}{5-\frac{3}{n}}\right \vert ^{n}}\\ & =\frac{1}{5} \end{align*}
Hence \(R=5.\) Therefore it converges uniformly for \(\left \vert z\right \vert \leq r<5\)