Power Series Expansion for Exponential of x by Sine of x

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Theorem

\(\ds e^x \sin x\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac {2^{n / 2} \, \map \sin {n \pi / 4} x^n} {n!}\)
\(\ds \) \(=\) \(\ds x + x^2 + \frac {x^3} 3 - \frac {x^5} {30} - \frac {x^6} {90} + \cdots\)


for all $x \in \R$.


Proof

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

By definition of Maclaurin series:

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

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

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


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:


\(\ds \map {f^{\paren {k + 1} } } x\) \(=\) \(\ds \frac \d {\d x} \map {f^{\paren k} } x\)
\(\ds \) \(=\) \(\ds \frac \d {\d x} 2^{k / 2} e^x \, \map \sin {x + \dfrac {k \pi} 4}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds 2^{k / 2} \paren {e^x \frac \d {\d x} \map \sin {x + \dfrac {k \pi} 4} + \map \sin {x + \dfrac {k \pi} 4} \frac \d {\d x} e^x}\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds 2^{k / 2} \paren {e^x \, \map \cos {x + \dfrac {k \pi} 4} + \map \sin {x + \dfrac {k \pi} 4} \frac \d {\d x} e^x}\) Derivative of Sine Function, Derivatives of Function of $a x + b$
\(\ds \) \(=\) \(\ds 2^{k / 2} e^x \paren {\map \cos {x + \dfrac {k \pi} 4} + \map \sin {x + \dfrac {k \pi} 4} }\) Derivative of Exponential Function and simplification
\(\ds \) \(=\) \(\ds 2^{k / 2} e^x \sqrt 2 \, \map \sin {x + \dfrac {k \pi} 4 + \dfrac \pi 4}\) Sine of x plus Cosine of x: Sine Form
\(\ds \) \(=\) \(\ds 2^{\paren {k + 1} / 2} e^x \, \map \sin {x + \dfrac {\paren {k + 1} \pi} 4}\) 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

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