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12.8 Proof of the Differentiation Theorem

12.73   Lemma. The power series $\sum\{nz^n\}$ has radius of convergence equal to $1$.

12.74   Exercise. Prove lemma 12.73. (We proved this lemma earlier using the differentiation theorem. Since we need this result to prove the differentiation theorem, we now want a proof that does not use the differentiation theorem.)

12.75   Lemma. Let $\sum\{a_nz^n\}$ be a power series. Then the two series $\sum\{a_nz^n\}$ and $\sum\{na_nz^{n-1}\}$ have the same radius of convergence.

Proof: I'll show that for all $w,v\in\mbox{{\bf C}}\backslash\{0\}$.

a) If $\sum\{\vert na_nw^{n-1}\vert\}$ converges, then $\{\sum\vert a_nw^n\vert\}$ converges.
b) If $\sum\{\vert a_nw^n\vert\}$ converges and $\vert v\vert<\vert w\vert$, then $\sum\{\vert na_nv^{n-1}\vert\}$ converges.
a) follows from the comparison test, since

\begin{displaymath}\vert a_nw^n\vert\leq \vert na_nw^{n-1}\vert\cdot \vert w\vert\mbox{ for all }n\in\mbox{${\mbox{{\bf Z}}}^{+}$}.\end{displaymath}

To prove b), suppose $\sum\{\vert a_nw^n\vert\}$ converges and $\vert v\vert<\vert w\vert$. By lemma 12.73, $\displaystyle {\sum\left\{n\left\vert{v\over w}\right\vert^n\right\}}$ converges, and hence $\displaystyle {\left\{n\left\vert{v\over w}\right\vert^n\right\}}$ is bounded. Choose $M\in\mbox{${\mbox{{\bf R}}}^{+}$}$ such that

\begin{displaymath}n\left\vert{v\over w}\right\vert^n\leq M\mbox{ for all }n\in\mbox{{\bf N}}.\end{displaymath}

Then $n\vert v\vert^n<M\vert w^n\vert$, and

\begin{displaymath}\left\vert a_nn\vert v\vert^{n-1}\right\vert\leq \vert a_nw^n...
...cdot {M\over {\vert v\vert}}\mbox{ for all }n\in\mbox{{\bf N}}.\end{displaymath}

By the comparison test, $\sum\{\vert a_n nv^{n-1}\vert\}$ converges. $\mid\!\mid\!\mid$

12.76   Corollary. $\sum\{a_nz^n\}$ and $\sum\{a_n n(n-1)z^{n-2}\}$ have the same radius of convergence.

Proof: Use the lemma twice. $\mid\!\mid\!\mid$

12.77   Theorem. Let $\sum\{c_nz^n\}$ be a power series with positive radius of convergence. Let $\displaystyle {f(z)=\sum_{n=0}^\infty c_nz^n}$ for all $z$ in the disc of convergence for $f$ and let $\displaystyle {Df(z)=\sum_{n=1}^\infty nc_nz^{n-1}}$ be the function corresponding to the formal derivative of $\sum\{c_nz^n\}$. Then $f$ is differentiable on its disc of convergence, and $f'(a)=Df(a)$ for all $a$ in the disc of convergence.

Proof: Let $a$ be a point in the disc of convergence, and let $z$ be a different point in the disc. Then

f(z)-f(a)&=&\sum_{n=0}^\infty c_nz^n-\sum_{n=0}^\infty c_na^n ...
...j \\
&=&(z-a)\sum_{n=1}^\infty c_n\sum_{j=0}^{n-1}z^{n-1-j}a^j.


\begin{displaymath}D_a f(z)=\sum_{n=1}^\infty c_n\sum_{j=0}^{n-1}z^{n-1-j}a^j.\end{displaymath}



and since

\begin{displaymath}D_af(a)=\sum_{n=1}^\infty c_n\sum_{j=0}^{n-1}a^{n-1}
=\sum_{j=1}^\infty c_n na^{n-1}=Df(a),\end{displaymath}

the theorem will follow if we can show that $D_af$ is continuous at $a$.

In the calculation below, I quietly use the following facts:

a) When $n=1$, $\displaystyle {\sum_{j=0}^{n-1}z^{n-1-j}a^j-\sum_{j=0}^{n-1}a^{n-1-j}a^j=0}$.
b) When $j=n-1$, $\displaystyle {z^{n-1-j}-a^{n-1-j}=0}$.

$\displaystyle D_af(z)-D_af(a)$ $\textstyle =$ $\displaystyle \sum_{n=1}^\infty c_n\sum_{j=0}^{n-1}z^{n-1-j}a^j
-\sum_{n=1}^\infty c_n\sum_{j=0}^{n-1}a^{n-1-j}a^j$  
  $\textstyle =$ $\displaystyle \sum_{n=2}^\infty c_n\sum_{j=0}^{n-1}a^j(z^{n-1-j}-a^{n-1-j})$  
  $\textstyle =$ $\displaystyle \sum_{n=2}^\infty c_n\sum_{j=0}^{n-2}a^j(z-a)\sum_{k=0}^{n-2-j}z^{n-2-j-k}a^k$  
  $\textstyle =$ $\displaystyle (z-a)\sum_{n=2}^\infty
c_n\sum_{j=0}^{n-2}\;\sum_{k=0}^{n-2-j}z^{n-2-j-k}a^{j+k}.$ (12.78)

Let the radius of convergence of our power series be $R$, and let $\displaystyle {\varepsilon={{R-\vert a\vert}\over 2}}$. Then

\vert z-a\vert<\varepsilon &\mbox{$\Longrightarrow$}& \vert z...
...vert a\vert+{{R-\vert a\vert}\over 2}={{R+\vert a\vert}\over 2}.

Let $\displaystyle {S={{R+\vert a\vert}\over 2}<R}$. Then $\vert a\vert<S$, and

\vert z-a\vert<\varepsilon & \mbox{$\Longrightarrow$}& \vert z...
... n\\
&=&S^{n-2}\cdot{{n(n-1)}\over 2}
\leq S^{n-2}\cdot n(n-1).

(Here I've used the fact that $n-1-j\leq n$ for $0\leq j\leq n-2$.) Thus

\begin{displaymath}\vert z-a\vert<\varepsilon\mbox{$\hspace{1ex}\Longrightarrow\...
...t\vert \leq\sum_{n=2}^\infty\vert c_n\vert S^{n-2}\cdot n(n-1).\end{displaymath}

We noticed in the corollary to lemma 12.75 that the series $\displaystyle {\sum\{n(n-1)c_nz^{n-2}\}}$ has radius of convergence $R$, and hence $\displaystyle {\sum\{\vert c_n\vert S^{n-2}n(n-1)\}_{n\geq 2}}$ converges to a limit $M$, and by (12.78),

\begin{displaymath}\vert D_af(z)-D_af(a)\vert\leq\vert z-a\vert\cdot M \mbox{ whenever }
\vert z-a\vert<\varepsilon.\end{displaymath}

If $\{w_n\}$ is a sequence in $\mbox{{\rm dom}}(D_af)$ such that $\{w_n\} \to a$, then

\begin{displaymath}\vert D_af(w_n)-D_af(a)\vert\leq\vert w_n-a\vert\cdot M\end{displaymath}

for all large $n$, and by the null-times bounded theorem and comparison theorem for null sequences, $\{D_af(w_n)\}\to D_af(a)$. Hence, $D_af$ is continuous at $a$. $\mid\!\mid\!\mid$

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Next: 12.9 Some XVIII-th Century Up: 12. Power Series Previous: 12.7 Special Values of   Index