Power Series Expansion for Sine Function

From ProofWiki
Jump to navigation Jump to search

Theorem

The sine function has the power series expansion:

\(\ds \sin x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}\)
\(\ds \) \(=\) \(\ds x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7} {7!} + \cdots\)

valid for all $x \in \R$.


Proof

From Derivative of Sine Function:

$\dfrac \d {\d x} \sin x = \cos x$

From Derivative of Cosine Function:

$\dfrac \d {\d x} \cos x = -\sin x$


Hence:

\(\ds \dfrac {\d^2} {\d x^2} \sin x\) \(=\) \(\ds -\sin x\)
\(\ds \dfrac {\d^3} {\d x^3} \sin x\) \(=\) \(\ds -\cos x\)
\(\ds \dfrac {\d^4} {\d x^4} \sin x\) \(=\) \(\ds \sin x\)


and so for all $m \in \N$:

\(\ds m = 4 k: \ \ \) \(\ds \dfrac {\d^m} {\d x^m} \sin x\) \(=\) \(\ds \sin x\)
\(\ds m = 4 k + 1: \ \ \) \(\ds \dfrac {\d^m} {\d x^m} \sin x\) \(=\) \(\ds \cos x\)
\(\ds m = 4 k + 2: \ \ \) \(\ds \dfrac {\d^m} {\d x^m} \sin x\) \(=\) \(\ds -\sin x\)
\(\ds m = 4 k + 3: \ \ \) \(\ds \dfrac {\d^m} {\d x^m} \sin x\) \(=\) \(\ds -\cos x\)

where $k \in \Z$.


This leads to the Maclaurin series expansion:

\(\ds \sin x\) \(=\) \(\ds \sum_{r \mathop = 0}^\infty \paren {\frac {x^{4 k} } {\paren {4 k}!} \map \sin 0 + \frac {x^{4 k + 1} } {\paren {4 k + 1}!} \map \cos 0 - \frac {x^{4 k + 2} } {\paren {4 k + 2}!} \map \sin 0 - \frac {x^{4 k + 3} } {\paren {4 k + 3}!} \map \cos 0}\)
\(\ds \) \(=\) \(\ds \sum_{r \mathop = 0}^\infty \paren {\frac {x^{4 k + 1} } {\paren {4 k + 1}!} - \frac {x^{4 k + 3} } {\paren {4 k + 3}!} }\) Sine of Zero is Zero, Cosine of Zero is One
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}\) setting $n = 2 k$

From Series of Power over Factorial Converges, it follows that this series is convergent for all $x$.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Power_Series_Expansion_for_Sine_Function&oldid=616820"