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

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

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

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

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

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

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 $${f(z)=\sum_{n=0}^\infty c_nz^n}$$ for all $z$ in the disc of convergence for $f$ and let $${Df(z)=\sum_{n=1}^\infty nc_nz^{n-1}}$$ 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

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

Let

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

Then

$$f(z)-f(a)=(z-a)D_af(z),$$

and since

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

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

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

$$D_af(z)-D_af(a) = \sum_{n=1}^\infty c_n\sum_{j=0}^{n-1}z^{n-1-j}a^j -\sum_{n=1}^\infty c_n\sum_{j=0}^{n-1}a^{n-1-j}a^j$$

$$= \sum_{n=2}^\infty c_n\sum_{j=0}^{n-1}a^j(z^{n-1-j}-a^{n-1-j})$$

$$= \sum_{n=2}^\infty c_n\sum_{j=0}^{n-2}a^j(z-a)\sum_{k=0}^{n-2-j}z^{n-2-j-k}a^k$$

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

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

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

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

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

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

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

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

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

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

$$\vert D_af(w_n)-D_af(a)\vert\leq\vert w_n-a\vert\cdot M$$

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 $a$. $\mid\!\mid\!\mid$