Power Series is Taylor Series

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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:

\(\displaystyle f^{\left({k+1}\right) }\) \(=\) \(\displaystyle \dfrac{\mathrm d}{\mathrm dz} \left({\sum_{n \mathop = k}^\infty a_n \left({z - \xi}\right)^{n-k} n^{\underline k} }\right)\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \dfrac{\mathrm d}{\mathrm dz} \left({\sum_{n \mathop = 0}^\infty a_{n+k} \left({z - \xi}\right)^{n} \left({n+k}\right)^{\underline k} }\right)\) subtracting $k$ from $n$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty n a_{n+k} \left({z - \xi}\right)^{n-1} \left({n+k}\right)^{\underline k}\) Derivative of Complex Power Series
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty a_{n+k} \left({z - \xi}\right)^{n-1} \left({n+k}\right)^{\underline{k+1} }\) Definition of Falling Factorial
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = k+1}^\infty a_{n} \left({z - \xi}\right)^{n - \left({k+1}\right) } n^{\underline{k+1} }\) adding $k$ to $n$

$\Box$


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

\(\displaystyle f^{\left({k}\right)} \left({\xi}\right)\) \(=\) \(\displaystyle \sum_{n \mathop = k}^\infty a_n \left({\xi - \xi}\right)^{n - k} n^{\underline k}\)
\(\displaystyle \) \(=\) \(\displaystyle a_k 0^0 k^{\underline k}\)
\(\displaystyle \) \(=\) \(\displaystyle a_k k!\)

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 $\xi$.

$\blacksquare$


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