Power Series Expansion for Exponential of x by Sine of x

Theorem
for all $x \in \R$.

Proof
Let $\map f x = e^x \sin x$.

By definition of Maclaurin series:


 * $(1): \quad \map f x \sim \ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} \map {f^{\paren n} } 0$

where $\map {f^{\paren n} } 0$ denotes the $n$th derivative of $f$ $x$ evaluated at $x = 0$.

It remains to be shown that:
 * $\map {f^{\paren n} } 0 = 2^{n / 2} \map \sin {\dfrac {n \pi} 4}$

The proof proceeds by induction.

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

Basis for the Induction
$\map P 0$ is the case:

Thus $\map P 0$ is seen to hold.

This is the basis for the induction.

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

So this is the induction hypothesis:
 * $\map {f^{\paren k} } x = 2^{k / 2} e^x \, \map \sin {x + \dfrac {k \pi} 4}$

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

Induction Step
This is the induction step:

So $\map P k \implies \map P {k + 1}$ and by the Principle of Mathematical Induction:


 * $\forall n \in \Z_{\ge 0}: \map {f^{\paren n} } x = 2^{n / 2} e^x \, \map \sin {x + \dfrac {n \pi} 4}$

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