Derivative of Complex Power Series/Proof 1

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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:

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



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

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:

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

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.