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\section{Power Series}

\subsection{General Series}

We first recall some facts about series from Calculus.

A \emph{series}, $s = \sum_{n=0}^\infty u_n$, is defined as a limit, $s =
\lim_{n\to\infty} s_n$, where $s_n$ is the \emph{ $n$-th partial sum},
$s_n = u_0 + \cdots + u_n$. The series \emph{converges} if the limit $s$
exists and $|s|<\infty$; otherwise it \emph{diverges}. We usually write
simply $\sum u_n$ to denote a series with the starting limit implied by
context and the upper limit~$\infty$.

A series $\sum u_n$ converges \emph{absolutely} if $\sum |u_n|$ converges.
Recall that absolute convergence implies ordinary convergence. The terms
of an absolutely convergent series may be rearranged in any manner without
altering the sum. In contrast to this, the terms of a \emph{conditionally
convergent} series---convergent, but not absolutely convergent---may be
rearranged to sum to any given value.

\emph{Convergence Tests}

\begin{description}

\item[Simple Divergence Test] If the terms $u_n$ do not approach 0,
then the series $\sum u_n$ diverges. (Equivalently, if
$\sum u_n$ converges then $u_n\to 0$.)

\item[Comparison Test] Suppose $0 \le u_n \le w_n$, $\forall\, n\ge N$. Then
$\sum w_n$ converges $\Rightarrow$ $\sum u_n$ converges;
$\sum u_n$ diverges $\Rightarrow$ $\sum w_n$ diverges.

\item[Root/Ratio Tests] Assume $0 \le u_n$ and let $L = \lim_{n\to\infty}
u_{n+1}/u_n$ or $L = \lim_{n\to\infty}\sqrt[n]{u_n}$. If $L < 1$ then the
series $\sum u_n$ converges and if $L > 1$, the series diverges.

\item[Geometric Series] $\sum_{n=k}^\infty u^n$ converges to $u^k/(1-u)$
if $|u|<1$ and diverges if $|u| \ge 1$.

\end{description}

The convergence of geometric series follows from the algebraic identity
\[ 1 + u + u^2 + \cdots + u^n = \frac{1}{1-u} - \frac{u^{n+1}}{1-u} \]
that gives an explicit formula for the $n$-th partial sum. The Root and
Ratio Tests are derived from the geometric series and the Comparison Test.

\subsection{Convergence of Power Series}

\begin{definition} A \emph{power series} is a series of the form
\[ \sum_{n=0}^\infty a_n(x-x_0)^n = a_0 + a_1(x-x_0) + a_2(x-x_0)^2 + \dots
\]
\end{definition}

To keep the notation simple we shall assume without loss of generality
that $x_0 = 0$.

\begin{proposition}\label{03:UNIFORM}
If $\sum_{n=0}^\infty a_nx^n$ converges for one number, say $b$, then it
converges absolutely and uniformly on $[-r,r]$ for any $0 \le r < |b|$.
\end{proposition}

\proof Since $\sum a_nb^n$ converges, the terms must approach~0, and hence
$\exists\, M$ such that $|a_nb^n| \le M$, $\forall\, n$.  Then
\[|a_nx^n| = |a_nb^n||x^n/b^n| \le M|x/b|^n \]
If $|x|\le r < |b|$, then $|x/b|\le r/|b| <1$ and $\sum |a_nx^n|$ is
dominated by a convergent geometric series, $\sum M(r/|b|)^n$. Note that
the convergence depends only on $r/|b|<1$ and not on any particular value
of $x$. Therefore, $\sum a_nx^n$ converges absolutely and uniformly on
$[-r,r]$. \qed

This proposition shows if a power series converges, it must converge on a
symmetric \emph{interval} $(-R,R)$. The the largest such $R$ is given
a name.

\begin{definition}
$R$ is the \emph{radius of convergence} of $\sum a_nx^n$ if
the series converges for $|x| <R$ and diverges for $|x|>R$.
\end{definition}

In this definition we allow $R=0$ (the series converges only for $x=0$)
or $R = \infty$ (the series converges for all $x$). The series may or
may not converge at the endpoints $\pm R$. The value of $R$ can often
be conveniently found using the Root or Ratio Tests.

\emph{Examples:} Let $\alpha>0$.

1) Consider $\sum \alpha^n x^{2n}$.

We apply the Root Test to $\sum u_n$ where $u_n = (\alpha x^2)^n$. The
limit $L =\lim_{n\to\infty}\sqrt[n]{u_n} = \alpha x^2$ and so, by the
Root Test, the series converges if $|x| < 1/\sqrt{\alpha}$ and diverges
if $|x| > 1/\sqrt{\alpha}$. Therefore, by definition, the radius of
convergence is $R = 1/\sqrt{\alpha}$.

2) Consider $\sum {\alpha\choose n} x^n$ where $\alpha > 0$ is not an
integer.

We apply the Ratio Test to $\sum u_n$ where $u_n = |{\alpha\choose n}||x|^n$.
Since
\begin{eqnarray*}
\frac{u_{n+1}}{u_n} &=&
\frac{|\alpha(\alpha-1)\cdots(\alpha-n)||x|^{n+1}}{(n+1)!}\cdot
\frac{n!}{|\alpha(\alpha-1)\cdots(\alpha-n+1)||x|^n}\\
&=&\frac{|\alpha - n|}{n+1}|x| \longrightarrow |x|
\end{eqnarray*}
the Ratio Test implies the series converges if $|x|<1$ and diverges if
$|x|>1$. Therefore the radius of convergence is $R = 1$.

3) Consider $\sum a_n x^n$ where
\[ a_n = \left\{\begin{array}{ll}
1 & n\ \mbox{even}\\
\frac{1}{n^n} & n\ \mbox{odd}
\end{array}\right. \]
If we let $u_n = a_n |x|^n$, we find that neither $u_{n+1}/u_n$ nor
$\sqrt[n]{u_n}$ has a limit so the Root and Ratio Tests are not effective.

The next theorem shows a reliable way to obtain the radius of convergence
$R$ for any power series $\sum a_n x^n$.

\begin{theorem}\label{03:RADIUS}
$1/R = \limsup |a_n|^{1/n}$
\end{theorem}

\proof
Step 1): Show $\limsup |a_n|^{1/n} \le 1/R$.

The case $R = 0$ is trivial ($\limsup |a_n|^{1/n} \le \infty$), so
we assume $R>0$. Choose $0 < r < R$. Since $\sum a_nr^n$ converges
by the definition of $R$, $\exists\, M$ such that $|a_nr^n| \le M$,
$\forall\, n$. This implies \[\limsup|a_n|^{1/n} \le
\frac{1}{r}\limsup M^{1/n} = \frac{1}{r} \] Since this
inequality is true $\forall\, r < R$, $\limsup|a_n|^{1/n}
\le 1/R$.

Step 2): Show $\limsup |a_n|^{1/n} \ge 1/R$.

Define $R_0$ by $1/R_0 = \limsup |a_n|^{1/n}$ (we allow $R_0 = 0$
or $\infty$). The case $R_0 = 0$ is trivial ($\infty \ge 1/R$),
so we assume $R_0 > 0$. Fix $r$, $0 < r < R_0$, and choose $R_1$,
$r < R_1 < R_0$.  Since $1/R_0 < 1/R_1$, the definition of $\limsup$
implies $\exists\, N$ such that $|a_n|^{1/n} < 1/R_1$, $\forall\, n\ge
N$. This implies \[ |a_nr^n| \le \Big|\frac{r}{R_1}\Big|^n\quad \forall\, n
\ge N \] Since $|r/R_1| <1$, $\sum|a_nr^n|$ is dominated by the convergent
geometric series $\sum |r/R_1|^n$ and is itself convergent. Therefore,
by the definition of radius of convergence, $r \le R$. Consequently,
$1/R \le 1/r$, $\forall\, r<R_0$, and so $1/R \le 1/R_0$. \qed

\emph{Example:} Consider $\sum a_n x^x$ where $a_n$ is defined as in
example 3) above. Since $\limsup |a_n|^{1/n} = 1$, the radius of
convergence is $R=1$.

\begin{theorem}
Suppose $f(x) = \sum_{n=0}^\infty a_nx^n$ has radius of convergence~$R$.
Then $f(x)$ has derivatives of all orders. The derivatives are given by
differentiating the series term by term,
\[ f^{(k)}(x) = \sum_{n=k}^\infty n(n-1)\cdots(n-k+1) a_n x^{n-k} \]
and have radius of convergence~$R$. In particular,
\[ a_n = \frac{f^{(n)}(0)}{n!} \]
\end{theorem}

\proof By Proposition~\ref{03:UNIFORM}, the polynomials $f_k(x)
= \sum_{n=0}^k a_nx^n$ converge \emph{uniformly} to $f(x) =
\sum_{n=0}^\infty a_nx^n$ for $|x| \le r < R$ where $R$ is the radius
of convergence. Since $\limsup |na_n|^{1/n} = \limsup |a_n|^{1/n}$, the
radius of convergence of $g(x) = \sum_{n=1}^\infty n a_n x^{n-1}$ is
still $R$ by Theorem~\ref{03:RADIUS}.  Thus the polynomials $f_k'(x)
= \sum_{n=1}^k n a_n x^{n-1}$ also converge uniformly to $g(x)$ for
$|x| \le r < R$. By the Theorem~\ref{02:DERIVATIVES} (convergence of
derivatives), $f'(x) = g(x)$. We can repeat this argument to obtain
derivatives of all orders. Finally, since $r<R$ was arbitrary, the above
statements hold for $|x| < R$.  \qed

\subsection{Expansions of Power Series}

\begin{theorem}\label{03:EXPANSION}
Suppose $f(x) = \sum_{n=0}^\infty a_n(x-x_0)^n$ has radius of convergence
$R$. Then $f$ has a power series expansion about any point $x_1$ in the
interval $|x - x_0| < R$, and the radius of convergence of the new series
is $\ge R - |x_1 - x_0|$.
\end{theorem}

\[ \includegraphics{Pictures/03.fig1.eps} \]

\proof
For simplicity of notation, we shall assume $x_0 = 0$. We use the
Binomial Theorem to expand $x^n$ in terms of $(x-x_1)^k$,
\[ x^n = (x - x_1 + x_1)^n = \sum_{k=0}^n{n\choose k}x_1^{n-k}(x-x_1)^k \]
Then we plug this into the Taylor polynomial:
\begin{eqnarray*}
P_N(x) &=& \sum_{n=0}^N a_n x^n\\
&=& \sum_{n=0}^N a_n \sum_{k=0}^n{n\choose k}x_1^{n-k}(x-x_1)^k \\
&=& \sum_{k=0}^N b_k (x-x_1)^k\quad \mbox{[rearranging terms]}
\end{eqnarray*}
where $b_k = \sum_{n=k}^N {n \choose k}a_nx_1^{n-k} = \sum_{n=0}^{N-k}
{n+k \choose k}a_{n+k}x_1^n$. The following diagram helps to see how the
indexes have been rearranged. The indexing of the original sum follows the
vertical lines from $k=0$ to $k=n$ with $n$ running from 0 to~$N$. The
rearrangement indexes the same shaded region horizontally from $n=k$
to $n=N$ with $k$ running from 0 to~$N$.

\[ \includegraphics{Pictures/03.fig2.eps} \]

This suggests that $f(x) = \sum_{k=0}^\infty b_k(x-x_1)^k$ where
\[ b_k =\sum_{n=0}^\infty {n+k \choose k}a_{n+k}x_1^n \]
It is interesting to note that this formula for $b_k$ could be obtained
by computing derivatives:
\[ f^{(k)}(x_1) = \sum_{n=k}^\infty n(n-1)\dots(n-k+1)a_n x_1^{n-k} \]
\[ \frac{f^{(k)}(x_1)}{k!} = \sum_{n=k}^\infty {n \choose k}a_n x_1^{n-k}
= \sum_{n=0}^\infty {n+k \choose k}a_{n+k} x_1^n = b_k \]

Note that the above series for $b_k$ converges for $|x_1| < R$ since the
factor ${n+k \choose k} = (n+k)(n+k-1)\cdots(n+1)/k!$ is a polynomial in
$n$ and therefore ${n+k \choose k}^{1/n} \to 1$ as $n\to \infty$ (apply
Theorem~\ref{03:RADIUS}).

To prove that the series $\sum_{k=0}^\infty b_k(x-x_1)^k$ converges
to $f(x)$ we first show that the following series, indexed by the
\emph{pairs} $(n,k)$, $k\le n$, converges absolutely:
\[ s(x) = \sum_{n=0}^\infty\sum_{k=0}^n {n\choose k}a_nx_1^{n-k}(x-x_1)^k \]
Let $r = |x-x_1| + |x_1|$. Then
\[ \sum_{k=0}^n {n\choose k}|x-x_1|^{n-k}|x_1|^k = (|x-x_1| + |x_1|)^n =
r^n \]
Therefore,
\[ \sum_{n=0}^\infty\sum_{k=0}^n {n\choose k}|a_n||x_1|^{n-k}|x-x_1|^k
= \sum_{n=0}^\infty |a_n|r^n \]
converges for $r < R$ (equivalently for $|x-x_1| < R-|x_1|$).
Since the series $s(x)$ converges absolutely, we may rearrange the order of
summation. Summing $k$ first and then $n$ gives
\[ s(x) = \sum_{n=0}^\infty a_n \sum_{k=0}^n{n\choose k}x_1^{n-k}(x-x_1)^k
= f(x) \]
Summing $n$ first and then $k$ gives
\[ s(x) = \sum_{k=0}^\infty\Big(\sum_{n=k}^n{n\choose k}x_1^{n-k}\Big)(x-x_1)^k
= \sum_{k=0}^\infty b_k(x-x_1)^k \]
Therefore, $f(x) = \sum_{k=0}^\infty b_k(x-x_1)^k$ for $|x-x_1| < R-|x_1|$.
\qed

\begin{definition}
A function is \emph{analytic} if it has a power series expansion about
every point in its domain.
\end{definition}


\subsubsection{Facts from Calculus}

\begin{itemize}

\item Most important functions in mathematics are analytic, e.g.,
  \begin{itemize}
  \item polynomials
  \item trigonometric functions
  \item exponential and logarithm functions
  \end{itemize}

\item If $f$ and $g$ are analytic, so are $f \pm g$, $f\cdot g$,
$f/g$ (wherever $g(x) \ne 0$) and $f\circ g$ are analytic.

\item \emph{Taylor's Theorem}: If $f$ is any function with
derivatives up to order $n+1$ at $x_0$, then
\[ f(x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k\quad +\quad R_n(x) \]
where
\[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1} \]
for some $c$ between $x_0$ and $x$, $|c - x_0| < |x - x_0|$.
In particular, $f$ is analytic at $x_0$ if and only if $|R_n(x)|\to 0$
as $n\to\infty$ for $x$ in some interval around $x_0$.

\end{itemize}


\subsubsection{A Fact from Complex Analysis.} If $f$ is analytic, then
the radius of convergence for the power series expansion around $x_0$
is the distance of $x_0$ to the ``nearest singularity'' in the complex
plane~$\C$.  A series with complex numbers converges for all complex
numbers within a circle of radius~$R$ of the center, and that is the
origin of the term \emph{radius} of convergence.

\emph{Example:} $\frac{1}{1+x^2} = \sum_{n=0}^\infty (-1)^nx^{2n}$. The
radius of convergence is $R = 1$ even though $\frac{1}{1+x^2}$ is defined
for all $x\in\R$. But in $\C$, $\frac{1}{1+x^2}$ has singularities at $x =
\pm i$ and the distance from 0 to $\pm i$ is $1 = R$.

\[ \resizebox{2in}{!}{\includegraphics{Pictures/03.fig3.eps}} \]

\subsection{Analytic Continuation}

A remarkable fact about an analytic function $f$ is that its values on
an interval $(a,b)$ are determined by its values on any neighborhood
contained in $(a,b)$, no matter how small. We are \emph{not} assuming
that $f$ has one power series expansion whose interval of convergence
covers the entire interval $(a,b)$, so we cannot simply quote
Theorem~\ref{03:EXPANSION} to prove this statement.

What we do assume is that $f$ has a power series expansion at each point
of $(a,b)$ with varying, and perhaps very small, radii of convergence.
The process of linking these series together from one interval of
convergence to the next is called \emph{analytic continuation}.
We shall now sketch how to do this on $(a,b)$.

Let $(c,d)$ be an interval contained in the given neighborhood where
we know the values of $f(x)$. Let $p \in (a,b)\setminus(c,d)$. We must
find the value of $f(p)$ using the values of $f(x)$ for $x\in(c,d)$.

\[ \includegraphics{Pictures/03.fig4.eps} \]

Let us assume $p>d$ (a similar argument would handle the case $p<c$). Let
$R_1$ be the radius of convergence of the power series expansion of
$f$ about $x_1 = d$. Choose $x_0\in(x_1 - R_1, x_1)$ close to $x_1=d$.
By Theorem~\ref{03:EXPANSION}, the radius of convergence of the power
series expansion of $f$ about $x_0$ is $R_0 \ge R_1 - (x_1 - x_0)$. In
particular, $x_1 \in(x_0-R_0,x_0+R_0)$. Therefore, the known values of
$f^{(n)}(x_0)$ determine the values of $f^{(n)}(x_1)$, and these in turn
determine the values of $f$ (through the power series expansion at $x_1$)
in $(c,d)\cup(x_1-R_1,x_1+R_1)$.

\[ \includegraphics{Pictures/03.fig5.eps} \]

Thus, we can always extend the interval of the known values of $f$ past the
endpoint of an interval of previously known values (as long as we remain in
$(a,b)$). Repeating this process, we obtain a sequence of centers,
$x_i = x_{i-1}+R_{i-1}$ with radii of convergence $R_i$. We claim that
$p\in(x_i-R_i,x_i+R_i)$ for some $i$. Otherwise, the centers must
converge $x_i\to q$ and the radii approach 0, $R_i\to 0$.  But $f$
has a power series expansion at $q$ with radius of convergence~$Q$
and the intervals $(x_i-R_i,x_i+R_i)$ must eventually fall inside
$(q-Q, q+Q)$. For example, there is an $x_i$ with $|x_i-q|<Q/2$ and
$R_i<Q/2$. This contradicts Theorem~\ref{03:EXPANSION} which asserts
that the radius of $R_i$ should be at least $Q - |x_i-q| > Q/2$. \qed

 \end{document}
