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12.1 Definition and Examples

12.1   Definition (Power Series.) Let $\{a_n\}$ be a sequence of complex numbers. A series of the form $\sum\{a_nz^n\}$ is called a power series.

We think of a power series as a sequence of polynomials

\begin{displaymath}\{a_0,a_0+a_1 z ,a_0+a_1z+a_2z^2,a_0+a_1z+a_2z^2+a_3z^3,\cdots\}.\end{displaymath}

In general, this sequence will converge for certain complex numbers, and diverge for other numbers. A power series $\sum\{a_nz^n\}$ determines a function whose domain is the set of all $z\in\mbox{{\bf C}}$ such that $\sum\{a_nz^n\}$ converges.

12.2   Examples. The geometric series $\sum\{z^n\}$ is a power series that converges to $\displaystyle {{1\over
{1-z}}}$ for $\vert z\vert<1$ and diverges for $\vert z\vert\geq 1$.

The series $\displaystyle {C=\sum\left\{ {{(-1)^nz^{2n}}\over {(2n)!}}\right\}}$ and $\displaystyle {S=\sum\left\{ {{(-1)^nz^{2n+1}}\over {(2n+1)!}}\right\}}$ are power series that converge for all $z\in\mbox{{\bf C}}$. $C$ corresponds to the sequence

\begin{displaymath}\displaystyle {\{a_n\}=\left\{1,0,-{1\over 2},0,{1\over 24},\cdots\right\}}\end{displaymath}

and $S$ corresponds to

\begin{displaymath}\displaystyle {\{a_n\}=\left\{0,1,0,-{1\over 6},0,{1\over
{120}},\cdots\right\}}.\end{displaymath}

The limits are $\cos z$ and $\sin z$, respectively (by definition 11.43.)


Every power series $\sum\{a_nz^n\}$ converges at $z=0$. (The limit is $a_0$.)


The series $\sum\{n!z^n\}$ converges only when $z=0$ (see exercise 12.5).

12.3   Notation ($a^{b^c}$) The expression $a^{b^c}$ is ambiguous. Since

\begin{displaymath}2^{(2^3)} = 2^8 = 256,\end{displaymath}

and

\begin{displaymath}\big({2^2}\big)^3 = 4^3 = 64,\end{displaymath}

we see that in general $a^{(b^c)} \neq {\big({a^b}\big)}^c$. We make the convention that

\begin{displaymath}a^{b^c} \mbox{ means } a^{(b^c)}.\end{displaymath}

The expression ${(a^b)}^c$ is usually simplified and written without parentheses by use of exercise 3.64:

\begin{displaymath}{(a^b)}^c = a^{(bc)} = a^{bc}.\end{displaymath}

12.4   Example. I would like to consider the series $\displaystyle {\sum\left\{ {{z^{n^2}}\over
{n^2}}\right\}_{n\geq 1}}$ to be a power series. This series corresponds to $\sum\{c_nz^n\}$ where

\begin{eqnarray*}
\{c_n\}&= &\{0,1,0,0,{1\over 4},0,0,0,0,{1\over 9},\cdots\} \\...
...nz^n\} &= &\{0,z,z,z,z+{{z^4}\over 4},z+{{z^4}\over 4},\cdots\},
\end{eqnarray*}



which is not identical with

\begin{displaymath}\sum\left\{ {{z^{n^2}}\over {n^2}}\right\}_{n\geq 1}=\left\{z,z+{{z^4}\over
4},z+{{z^4}\over 4}+{{z^9}\over 9},\cdots\right\},\end{displaymath}

but you should be able to see that one series converges if and only if the other does, and that they have the same limits. In the future I will sometimes blur the distinctions between two series like this.

For $z\neq 0$, let $\displaystyle {a_n={{z^{n^2}}\over {n^2}}}$. Then

\begin{displaymath}{{\vert a_{n+1}\vert}\over {\vert a_n\vert}}=\left\vert {{z^{...
...}\right\vert=\vert z\vert^{2n+1}\left( {n\over {n+1}}\right)^2.\end{displaymath}

If $\vert z\vert<1$, then $\displaystyle {\vert z\vert^{2n+1}\left({n\over {n+1}}\right)^2\leq\vert z\vert^{2n+1}}$ and $\displaystyle { \lim\left\{ {{\vert a_{n+1}\vert}\over {\vert a_n\vert}} \right\}_{n\geq 1}=0<1}$, so by the ratio test, $\displaystyle {\sum\left\{ {{z^{n^2}}\over
{n^2}}\right\}_{n\geq 1}}$ converges absolutely for $\vert z\vert<1$.

If $\vert z\vert>1$ and $n\geq 1$, then

\begin{displaymath}\displaystyle { \left\{\left\vert {{a_{n+1}}\over
{a_n}}\righ...
...over
{(n+1)}}\right)^2\geq\vert z\vert^{2n+1}\cdot {1\over 4}},\end{displaymath}

so $\displaystyle { {{\vert a_{n+1}\vert}\over {\vert a_n\vert}} >1}$ for large $n$, and the series diverges. If $\vert z\vert=1$, then $\displaystyle {\vert a_n\vert={1\over {n^2}}}$, so $\sum\{\vert a_n\vert\}$ converges by the comparison test, and $\displaystyle {\sum\left\{ {{z^{n^2}}\over
{n^2}}\right\}_{n\geq 1}}$ converges absolutely. This shows that the function

\begin{displaymath}f(z)=\sum_{n=1}^\infty {{z^{n^2}}\over {n^2}}\end{displaymath}

is defined for all $z\in\bar D(0,1)$, and determines a function from $\bar D(0,1)$ into $\mbox{{\bf C}}$.

The figure below 12.1 shows the images under $f$ of circles of radius $\displaystyle {
{j\over {10}}}$ for $1\leq j\leq 10$ and of rays that divide the disc into twelve equal parts. The images of the interior circles are nice differentiable curves. The image of the boundary circle seems to have interesting properties that I do not know how to demonstrate.

12.5   Exercise.
a) Show that $\sum\{n!z^n\}$ converges only for $z=0$.
b) Show that $\displaystyle {\sum\left\{ {{z^{2^n}}\over {2^n}}\right\}}$ converges if and only if $\vert z\vert\leq 1$. $\mid\!\mid\!\mid$

Image of $\displaystyle { \sum\Big\{{z^{n^2}\over n^2}\Big\}_{n\geq 1} }$



Let $\displaystyle {g(z)=\sum_{n=0}^\infty {{z^{2^n}}\over {2^n}}}$ for $\vert z\vert\leq 1$.

The figure below 12.1 shows the images under $g$ of circles of radius $\displaystyle {
{j\over {10}}}$ for $1\leq j\leq 10$, and of rays that divide the disc into 12 equal parts.

12.6   Exercise. A Let $\displaystyle {g(z)=\sum_{n=0}^\infty {{z^{2^n}}\over {2^n}}}$ for $\vert z\vert\leq 1$. It appears from figure below 12.1 that $g(-1)=0$, and $g(i)$ is pure imaginary. Show that this is the case.

\psfig{file=pow2.ps,width=5in}



Image of $\displaystyle { \sum\Big\{{z^{2^n}\over 2^n}\Big\} }$

12.7   Entertainment. It appears from the image of $\displaystyle {g(z)=\sum_{n=0}^\infty {{z^{2^n}}\over {2^n}}}$ that if $\displaystyle {w={1\over 2}+i{{\sqrt 3}\over 2}}$ (a cube root of $-1$), then $g(w)$ is pure imaginary, and has length a little larger than the length of $g(i)$. Show that this is the case. (From the fact that $w^3=-1$, notice that

\begin{displaymath}\{w^1,w^2,w^4,w^8,w^{16},w^{32},w^{64},\cdots\}=\{w,w^2,-w,w^2,-w,w^2,-w,\cdots\}.)\end{displaymath}


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Next: 12.2 Radius of Convergence Up: 12. Power Series Previous: 12. Power Series   Index