Derivative of Complex Power Series/Proof 1

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Theorem

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

Define:

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

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

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

Define:

$\displaystyle M = \paren {R - \epsilon - \cmod {z - \xi} }^{-2} \sum_{n \mathop = 2}^\infty \cmod {a_n} \paren {R - \epsilon}^n$

Suppose that $\cmod h \le R - \epsilon - \cmod {z - \xi}$.

Then, by the Binomial Theorem and the Triangle Inequality:

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

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

$\blacksquare$


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