Power Series is Taylor Series

Theorem
Let $\ds \map f z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ 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 $n \in \N$:


 * $a_n = \dfrac {\map {f^{\paren n} } \xi} {n!}$

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


 * $\ds \forall z \in \C, \size {z - \xi} < R: \quad \map f z = \sum_{n \mathop = 0}^\infty \dfrac {\paren {z - \xi}^n} {n!} \map {f^{\paren n} } \xi$

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


 * $\ds \map {f^{\paren k} } z = \sum_{n \mathop = k}^\infty a_n \paren {z - \xi}^{n - k} n^{\underline k}$

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

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


 * $\ds \map {f^{\paren k} } z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$

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 $\size {z - \xi} < R$.

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


 * $\ds \map {f^{\paren k} } z = \sum_{n \mathop = k}^\infty a_n \paren {z - \xi}^{n - k} n^{\underline k}$

and $f^{\paren k}$ has radius of convergence $R$.

Induction Step
Note that $f^{\paren {k + 1} }$ is the derivative of $f^{\paren k}$.

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

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

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

Hence, it follows that $a_n = \dfrac {\map {f^{\paren n} } \xi} {n!}$.

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