# Power Series Expansion for Exponential of x by Sine of x

## Contents

## Theorem

\(\displaystyle e^x \sin x\) | \(=\) | \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {2^{n / 2} \, \map \sin {n \pi / 4} x^n} {n!}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x + x^2 + \frac {x^3} 3 - \frac {x^5} {30} - \frac {x^6} {90} + \cdots\) |

for all $x \in \R$.

## Proof

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

By definition of Maclaurin series:

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

where $\map {f^{\paren n} } 0$ denotes the $n$th derivative of $f$ with respect to $x$ 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:

\(\displaystyle \map f x\) | \(=\) | \(\displaystyle e^x \sin x\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2^{0 / 2} e^x \, \map \sin {x + \dfrac {0 \pi} 4}\) |

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:

\(\displaystyle \map {f^{\paren {k + 1} } } x\) | \(=\) | \(\displaystyle \frac \d {\d x} \map {f^{\paren k} } x\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac \d {\d x} 2^{k / 2} e^x \, \map \sin {x + \dfrac {k \pi} 4}\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2^{k / 2} \paren {e^x \frac \d {\d x} \map \sin {x + \dfrac {k \pi} 4} + \map \sin {x + \dfrac {k \pi} 4} \frac \d {\d x} e^x}\) | Product Rule | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2^{k / 2} \paren {e^x \, \map \cos {x + \dfrac {k \pi} 4} + \map \sin {x + \dfrac {k \pi} 4} \frac \d {\d x} e^x}\) | Derivative of Sine Function, Derivatives of Function of $a x + b$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2^{k / 2} e^x \paren {\map \cos {x + \dfrac {k \pi} 4} + \map \sin {x + \dfrac {k \pi} 4} }\) | Derivative of Exponential Function and simplification | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2^{k / 2} e^x \sqrt 2 \, \map \sin {x + \dfrac {k \pi} 4 + \dfrac \pi 4}\) | Sine of x plus Cosine of x: Sine Form | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2^{\paren {k + 1} / 2} e^x \, \map \sin {x + \dfrac {\paren {k + 1} \pi} 4}\) | simplification |

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 $(1)$.

$\blacksquare$

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 20$: Miscellaneous Series: $20.46$