12.3 Differentiation of Power Series

If $\sum\{c_nz^n\}=\{c_0,c_0+c_1z,c_0+c_1z+c_2z^2,\cdots\}$ is a power series, then the series obtained by differentiating the terms of $\sum\{c_nz^n\}$ is

$$\sum\{c_n n z^{n-1}\}=\{0,c_1,c_1+2c_2z,c_1+2c_2z+3c_3z^2,\cdots\}.$$

This is not a power series, but its translate

$$\sum\{c_{n+1}(n+1)z^n\}=\{c_1,c_1+2c_2z,c_1+2c_2z+3c_3z^2,\cdots\}$$

is.

12.15   Definition (Formal derivative.) If $\sum\{c_nz^n\}$ is a power series, then the formal derivative of $\sum\{c_nz^n\}$ is

$$D(\sum\{c_nz^n\})= \sum\{c_{n+1}(n+1)z^n\}.$$

I will sometimes write $D(\sum\{c_nz^n\})=\sum\{c_nnz^{n-1}\}$ when I think this will cause no confusion.

12.16   Examples.

$$\begin{eqnarray*} D(\sum\{z^n\})&=&\sum\{ nz^{n-1}\}=\sum\{(n+1)z^n\} \\ &=&\{1,1+2z,1+2z+3z^2,\cdots\}. \end{eqnarray*}$$

$$\begin{eqnarray*} D(S)&=& D\left(\sum \left\{ {(-1)^nz^{2n+1} \over (2n+1)!}\rig... ...} \\ &=& \sum \left\{ {(-1)^n z^{2n}\over (2n)!}\right\} = C\\ \end{eqnarray*}$$

or

$$D\left(\sum \left\{ {(-1)^nz^{2n+1} \over (2n+1)!}\right\}\ri... ...\right\} =\sum \left\{ {(-1)^n z^{2n} \over (2n)!}\right\} = C.$$

Our fundamental theorem on power series is:

12.17   Theorem (Differentiation theorem.) Let $\sum\{a_nz^n\}$ be a power series. Then $D(\sum\{a_nz^n\})$ and $(\sum\{a_nz^n\})$ have the same radius of convergence. The function $f$ associated with $\sum\{a_nz^n\}$ is differentiable in the disc of convergence, and the function represented by $D(\sum\{a_nz^n\})$ agrees with $f'$ on the disc of convergence.

The proof is rather technical, and I will postpone it until section 12.8. I will derive some consequences of it before proving it (to convince you that it is worth proving).

12.18   Example. We know that the geometric series $\sum\{z^n\}$ has radius of convergence $1$ and $${f(z)=\sum_{n=0}^\infty z^n={1\over {1-z}}}$$ for $\vert z\vert<1$. The differentiation theorem says $D(\sum\{z^n\})=\sum\{nz^{n-1}\}$ also has radius of convergence $1$, and

$$f'(z)=\sum_{n=0}^\infty nz^{n-1}=\sum_{n=0}^\infty (n+1)z^n\mbox{ for } \vert z\vert<1;$$

i.e.,

$$\sum_{n=0}^\infty(n+1)z^n={1\over {(1-z)^2}}\mbox{ for } \vert z\vert<1.$$

We can apply the theorem again and get

$$\sum_{n=0}^\infty(n+1)(n)z^{n-1}={2\over {(1-z)^3}}\mbox{ for } \vert z\vert<1,$$

or

$$\sum_{n=0}^\infty{{(n+1)(n+2)}\over 2}z^n={1\over {(1-z)^3}}\mbox{ for } \vert z\vert<1.$$

Another differentiation gives us

$$\sum_{n=0}^\infty {{n(n+1)(n+2)z^{n-1}}\over 2}={3\over {(1-z)^4}}\mbox{ for } \vert z\vert<1,$$

or

$$\sum_{n=0}^\infty {{(n+1)(n+2)(n+3)}\over {3!}}z^n={1\over {(1-z)^4}}\mbox{ for } \vert z\vert<1.$$

The pattern is clear, and I omit the induction proof that for all $k \in\mbox{{\bf N}}$

$$\begin{eqnarray*} {1\over {(1-z)^{k+1}}}&=&\sum_{n=0}^\infty {{(n+1)(n+2)(n+3)\c... ...n=0}^\infty{{(n+k)!}\over {n!k!}}z^n\mbox{ for } \vert z\vert<1. \end{eqnarray*}$$

12.19   Exercise. By assuming the differentiation theorem, we've shown that the series $${\sum\left\{\left({{(n+k)!}\over {n!k!}}\right) z^n\right\}}$$ has radius of convergence $1$ for all $k \in\mbox{{\bf N}}$. Verify this directly.

12.20   Exercise. A Find formulas for $${\sum_{n=0}^\infty nz^n}$$ and $${\sum_{n=0}^\infty n^2z^n}$$ that are valid for $\vert z\vert<1$. (You may assume the differentiation theorem.)

12.21   Example. By the differentiation theorem, if

$$C(z)=\sum_{n=0}^\infty {{(-1)^nz^{2n}}\over {(2n)!}}\mbox{ and } S(z)=\sum_{j=0}^\infty{{(-1)^nz^{2n+1}}\over {(2n+1)!}},$$

then $C$ and $S$ are differentiable on $\mbox{{\bf C}}$ and $C'(z)=-S(z)$, and $S'(z)=C(z)$. (We saw in earlier examples that both series have radius of convergence $\infty$, and that the formal derivatives satisfy $DS=C$ and $DC=-S$.) Also, clearly $C(z)$ and $S(z)$ are real when $z$ is real. The discussion in example 10.3 then shows that for real $z$, $C$ and $S$ agree with the cosine and sine functions you discussed in your previous calculus course, and in particular that

$$\sin^2z+\cos^2z=1 \mbox{ for all }z\in\mbox{{\bf R}}.$$