# 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:

 $\ds f^{\paren {k + 1} }$ $=$ $\ds \map {\dfrac \d {\d z} } {\sum_{n \mathop = k}^\infty a_n \paren {z - \xi}^{n - k} n^{\underline k} }$ Induction Hypothesis $\ds$ $=$ $\ds \map {\dfrac \d {\d z} } {\sum_{n \mathop = 0}^\infty a_{n + k} \paren {z - \xi}^n \paren {n + k}^{\underline k} }$ subtracting $k$ from $n$ $\ds$ $=$ $\ds \sum_{n \mathop = 1}^\infty n a_{n + k} \paren {z - \xi}^{n - 1} \paren {n + k}^{\underline k}$ Derivative of Complex Power Series $\ds$ $=$ $\ds \sum_{n \mathop = 1}^\infty a_{n + k} \paren {z - \xi}^{n - 1} \paren {n + k}^{\underline {k + 1} }$ Definition of Falling Factorial $\ds$ $=$ $\ds \sum_{n \mathop = k + 1}^\infty a_n \paren {z - \xi}^{n - \paren {k + 1} } n^{\underline {k + 1} }$ adding $k$ to $n$

$\Box$

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

 $\ds \map {f^{\paren k} } \xi$ $=$ $\ds \sum_{n \mathop = k}^\infty a_n \paren {\xi - \xi}^{n - k} n^{\underline k}$ $\ds$ $=$ $\ds a_k 0^0 k^{\underline k}$ $\ds$ $=$ $\ds a_k k!$

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 $\xi$.

$\blacksquare$