Power Series Expansion for Exponential of x by Sine of x

Theorem
for all $x \in \R$.

Proof
Let $f \left({x}\right) = e^x \sin x$.

By definition of Maclaurin series:


 * $(1): \quad f \left({x}\right) \sim \displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} f^{\left({n}\right)} \left({0}\right)$

where $f^{\left({n}\right)} \left({0}\right)$ denotes the $n$th derivative of $f$ $x$ evaluated at $x = 0$.

It remains to be shown that:
 * $f^{\left({n}\right)} \left({0}\right) = 2^{n / 2} \sin \left({\dfrac {n \pi} 4}\right)$

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $f^{\left({n}\right)} \left({x}\right) = 2^{n / 2} e^x \sin \left({x + \dfrac {n \pi} 4}\right)$

Basis for the Induction
$P \left({0}\right)$ is the case:

Thus $P \left({0}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $f^{\left({k}\right)} \left({x}\right) = 2^{k / 2} e^x \sin \left({x + \dfrac {k \pi} 4}\right)$

from which it is to be shown that:
 * $f^{\left({k + 1}\right)} \left({x}\right) = 2^{\left({k + 1}\right) / 2} e^x \sin \left({x + \dfrac {\left({k + 1}\right) \pi} 4}\right)$

Induction Step
This is the induction step:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and by the Principle of Mathematical Induction:


 * $\forall n \in \Z_{\ge 0}: f^{\left({n}\right)} \left({x}\right) = 2^{n / 2} e^x \sin \left({x + \dfrac {n \pi} 4}\right)$

The result follows by setting $x = 0$ and substituting for $f^{\left({n}\right)} \left({0}\right)$ in $(1)$.