Power Series is Taylor Series

Theorem
Let $\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^\infty a_n \left({z - \xi}\right)^n$ be a complex power series about $\xi \in \C$.

Let $R$ be the radius of convergence of $f$.

Then, $f$ is of differentiability class $C^\infty$.

For all $n \in \N$:


 * $a_n = \dfrac{f^{\left({n}\right) } \left({\xi}\right) }{ n! }$

Hence, $f$ is equal to its Taylor series expansion about $\xi$:


 * $\displaystyle \forall z \in \C, \left\vert{z - \xi}\right\vert < R: \quad f \left({z}\right) = \sum_{n \mathop = 0}^\infty \dfrac{\left({z - \xi}\right)^n}{n!} f^{\left({n}\right) } \left({\xi}\right)$

Proof
First, we prove by induction over $k \in \N_{\ge 1}$ that:


 * $\displaystyle f^{\left({k}\right) } \left({z}\right) = \sum_{n \mathop = k}^\infty a_n \left({z - \xi}\right)^{n - k} n^{\underline k}$

where $n^{\underline k}$ denotes the falling factorial.

Basis for the Induction
For $k = 1$, it follows from Derivative of Complex Power Series that


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

As $n = n^{\underline 1}$, this proves the hypothesis.

From Radius of Convergence of Derivative of Complex Power Series, it follows that the equation holds for all $z \in \C$ with $\left\vert{z - \xi}\right\vert < R$.

Induction Hypothesis
For fixed $k \in \N_{\ge 1}$, the hypothesis is that:


 * $\displaystyle f^{\left({k}\right) } \left({z}\right) = \sum_{n \mathop = k}^\infty a_n \left({z - \xi}\right)^{n-k} n^{\underline k}$

and $f^{\left({k}\right) }$ has radius of convergence $R$.

Induction Step
Note that $f^{\left({k+1}\right) }$ is the derivative of $f^{\left({k}\right) }$.

From Radius of Convergence of Derivative of Complex Power Series, it follows that $f^{\left({k+1}\right) }$ has radius of convergence $R$.

For $z \in \C$ with $\left\vert{z - \xi}\right\vert < R$, we have:

With $k \in \N$, we have:

Hence, it follows that $a_n = \dfrac{f^{\left({n}\right)} \left({\xi}\right)} {n!}$.

By definition of Taylor series, it follows that $f$ is equal to its Taylor series expansion about $\xi$.