Derivative of Complex Power Series

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Let $\xi \in \C$ be a complex number.

Let $\sequence {a_n}$ be a sequence in $\C$.

Let $\displaystyle f \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a power series in a complex variable $z \in \C$ about $\xi$.

Let $R$ be the radius of convergence of the series defining $f \paren z$.

Let $\cmod {z - \xi} < R$.

Then $f$ is complex-differentiable and its derivative is:

$\displaystyle f' \paren z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$

Proof 1


$\displaystyle g \left({z}\right) = \sum_{n \mathop =1}^\infty na_n \left({z-\xi}\right)^{n-1}$

From Radius of Convergence of Derivative of Complex Power Series, it follows that $g$ has radius of convergence $R$.

Fix an $\epsilon > 0$ satisfying $\epsilon < R - \left\vert{ z-\xi }\right\vert$.


$\displaystyle M=\left({ R - \epsilon - \left\vert{ z-\xi }\right\vert }\right)^{-2} \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \left({R - \epsilon}\right)^n$

Suppose that $\left\vert{ h}\right\vert \le R - \epsilon - \left\vert{ z-\xi}\right\vert$.

Then, by the Binomial Theorem and the Triangle Inequality:

\(\displaystyle \left\vert{ \dfrac{f \left({z+h}\right) - f \left({z}\right) }{h} - g \left({z}\right) }\right\vert\) \(=\) \(\displaystyle \left\vert{ \dfrac 1 h \left({\sum_{n \mathop =0}^\infty a_n \left({z + h - \xi}\right)^n - \sum_{n \mathop =0}^\infty a_n \left({z - \xi}\right)^n }\right) - \sum_{n \mathop =1}^\infty n a_n \left({z - \xi}\right)^{n-1} }\right\vert\)
\(\displaystyle \) \(=\) \(\displaystyle \left\vert{ \sum_{n \mathop =0}^\infty a_n \sum_{k \mathop =0}^n \binom n k \left({z - \xi}\right)^k h^{n - k - 1} - \sum_{n \mathop =0}^\infty \dfrac{a_n}{h} \left({z - \xi}\right)^n - \sum_{n \mathop =1}^\infty n a_n \left({z - \xi}\right)^{n-1} }\right\vert\) Binomial Theorem
\(\displaystyle \) \(=\) \(\displaystyle \left\vert{ \sum_{n \mathop =1}^\infty a_n \left({ \sum_{k \mathop =0}^n \binom n k \left({z - \xi}\right)^k h^{n - k - 1} - \dfrac 1 h \left({z - \xi}\right)^n - n \left({z - \xi}\right)^{n-1} }\right) }\right\vert\) Difference of Absolutely Convergent Series
\(\displaystyle \) \(=\) \(\displaystyle \left\vert{\sum_{n \mathop =2}^\infty a_n \sum_{k \mathop =0}^{n-2} \binom{n}{k} \left({z-\xi}\right)^k h^{n-k-1} }\right\vert\) by algebraic manipulations
\(\displaystyle \) \(\le\) \(\displaystyle \left\vert{ h }\right\vert \sum_{n \mathop =2}^\infty \left\vert{ a_n }\right\vert \sum_{k \mathop =0}^{n-2} \binom{n}{k} \left\vert{ z-\xi}\right\vert^k \left\vert{ h}\right\vert^{n-k-2}\) Triangle Inequality
\(\displaystyle \) \(\le\) \(\displaystyle \left\vert{ h}\right\vert \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \sum_{k \mathop =0}^{n-2} \binom{n}{k} \left\vert{ z-\xi}\right\vert^k \left({R - \epsilon - \left\vert{ z-\xi}\right\vert }\right)^{n-k-2}\) by assumption
\(\displaystyle \) \(\le\) \(\displaystyle \left\vert{ h}\right\vert \left({R - \epsilon - \left\vert{ z -\xi}\right\vert }\right)^{-2} \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \sum_{k \mathop =0}^n \binom{n}{k} \left\vert{ z-\xi}\right\vert^k \left({R - \epsilon - \left\vert{ z-\xi}\right\vert }\right)^{n-k}\)
\(\displaystyle \) \(=\) \(\displaystyle \left\vert{ h}\right\vert \left({R - \epsilon - \left\vert{ z-\xi}\right\vert }\right)^{-2} \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \left({R -\epsilon}\right)^n\) by the binomial theorem
\(\displaystyle \) \(=\) \(\displaystyle M \left\vert{ h}\right\vert\)

Letting $h\to 0$ we see that $f' \left({z}\right)$ exists and $f' \left({z}\right) = g \left({z}\right)$, as desired.


Proof 2


$\displaystyle \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} = 1$



$\displaystyle g \left({z}\right) = \sum_{n \mathop =1}^\infty n a_n \left({z-\xi}\right)^{n-1}$

Fix an $\epsilon > 0$ satisfying $\epsilon < R-\left\vert{ z-\xi}\right\vert$.


$\displaystyle M = \sum_{n \mathop =2}^\infty \dfrac{n \left({n-1}\right) }{2} \left\vert{ a_n}\right\vert \left({R - \epsilon}\right)^{n-2}$

We use the root test to prove convergence of this series:

\(\displaystyle \limsup_{n \to \infty} \left[{ \dfrac {n\left({n-1}\right) }{2} \left\vert{ a_n }\right\vert \left({R - \epsilon}\right)^{n-2} }\right]^{1/n}\) \(=\) \(\displaystyle \lim_{n \to \infty} \left[{ \dfrac {n \left({n-1}\right) }{2} }\right]^{1/n} \lim_{n \to \infty} \left[{ \left({R - \epsilon}\right)^{n-2} }\right]^{1/n} \limsup_{n \to \infty} \left\vert{ a_n }\right\vert^{1/n}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac{R - \epsilon}{R}\)

The last equality follows from the lemma and:

$\displaystyle \limsup_{n \to \infty} \left \vert{ a_n }\right \vert^{1/n} = \dfrac{1}{R}$

Suppose that $\left\vert{ h}\right\vert \le R - \epsilon - \left\vert{ z-\xi}\right\vert$.

It follows by the triangle inequality that $\left\vert{ z - \xi + h }\right\vert \le \left\vert{ z-\xi}\right\vert + \left\vert{ h}\right\vert \le R-\epsilon$.

By the triangle inequality, Difference of Two Powers, and Closed Form for Triangular Numbers, the following holds:

\(\displaystyle \left\vert{\dfrac{f \left({z+h}\right) - f \left({z}\right) }{h} -g \left({z}\right) }\right\vert\) \(=\) \(\displaystyle \left\vert{\sum_{n \mathop =2}^\infty a_n \left[{\dfrac{\left({z - \xi + h}\right)^n - \left({z-\xi}\right)^n }{h} - n \left({z-\xi}\right)^{n-1} }\right] }\right\vert\)
\(\displaystyle \) \(\le\) \(\displaystyle \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \left\vert{\dfrac{\left({z - \xi + h}\right)^n - \left({z-\xi}\right)^n }{h} - n \left({z-\xi}\right)^{n-1} }\right\vert\) by the triangle inequality
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \left\vert{ \left[{\sum_{u \mathop =0}^{n-1} \left({z - \xi + h}\right)^u \left({z-\xi}\right)^{n-u-1} }\right] - n \left({z-\xi}\right)^{n-1} }\right\vert\) by Difference of Two Powers
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \left\vert{ \sum_{u \mathop =0}^{n-1} \left[{ \left({z - \xi + h}\right)^u \left({z-\xi}\right)^{n-u-1} - \left({z-\xi}\right)^{n-1} }\right] }\right\vert\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \left\vert{\sum_{u \mathop =0}^{n-1} \left({z-\xi}\right)^{n-u-1} \left[{\left({z - \xi + h}\right)^u - \left({z-\xi}\right)^u }\right] }\right\vert\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \left\vert{ h\sum_{u \mathop =0}^{n-1} \left({z-\xi}\right)^{n-u-1} \sum_{v \mathop =0}^{u-1} \left({z - \xi + h}\right)^v \left({z-\xi}\right)^{u-v-1} }\right\vert\) by Difference of Two Powers
\(\displaystyle \) \(\le\) \(\displaystyle \left\vert{ h}\right\vert \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \sum_{u \mathop =0}^{n-1} \left\vert{ z-\xi}\right\vert^{n-u-1} \sum_{v \mathop =0}^{u-1}\left\vert{ z - \xi + h}\right\vert^v \left\vert{ z-\xi}\right\vert^{u-v-1}\) by the triangle inequality
\(\displaystyle \) \(\le\) \(\displaystyle \left\vert{ h}\right\vert \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \sum_{u \mathop =0}^{n-1} \left({R-\epsilon}\right)^{n-u-1} \sum_{v \mathop =0}^{u-1} \left({R-\epsilon}\right)^v \left({R-\epsilon}\right)^{u-v-1}\) by assumption and $\left\vert{ z - \xi + h}\right\vert \le R - \epsilon$, which was proven earlier
\(\displaystyle \) \(=\) \(\displaystyle \left\vert{ h}\right\vert \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \left({R-\epsilon}\right)^{n-2} \sum_{u \mathop =0}^{n-1} \sum_{v \mathop =0}^{u-1}1\)
\(\displaystyle \) \(=\) \(\displaystyle \left\vert{ h}\right\vert \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \left({R-\epsilon}\right)^{n-2} \sum_{u \mathop =0}^{n-1}u\)
\(\displaystyle \) \(=\) \(\displaystyle \left\vert{ h}\right\vert \sum_{n \mathop =2}^\infty \dfrac{n \left({n-1}\right) }{2} \left\vert{ a_n}\right\vert \left({R-\epsilon}\right)^{n-2}\) by Closed Form for Triangular Numbers
\(\displaystyle \) \(=\) \(\displaystyle M\left\vert{ h}\right\vert\)

Letting $h\to 0$ we see that $f'\left({z}\right)$ exists and $f' \left({z}\right)=g \left({z}\right)$, as desired.



The proof for real power series is identical.