Sum of Infinite Series of Product of Power and Sine

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $r \in \R$ such that $\size r < 1$.


Then:

\(\ds \sum_{k \mathop = 1}^\infty r^k \sin k x\) \(=\) \(\ds r \sin x + r^2 \sin 2 x + r^3 \sin 3 x + \cdots\)
\(\ds \) \(=\) \(\ds \dfrac {r \sin x} {1 - 2 r \cos x + r^2}\)


Proof

From Euler's Formula:

$e^{i \theta} = \cos \theta + i \sin \theta$

Hence:

\(\ds \sum_{k \mathop = 1}^\infty r^k \sin k x\) \(=\) \(\ds \map \Im {\sum_{k \mathop = 1}^\infty r^k e^{i k x} }\)
\(\ds \) \(=\) \(\ds \map \Im {\sum_{k \mathop = 0}^\infty \paren {r e^{i x} }^k}\) as $\map \Im {e^{i \times 0 \times x} } = \map \Im 1 = 0$
\(\ds \) \(=\) \(\ds \map \Im {\frac 1 {1 - r e^{i x} } }\) Sum of Infinite Geometric Sequence: valid because $\size r < 1$
\(\ds \) \(=\) \(\ds \map \Im {\frac {1 - r e^{-i x} } {\paren {1 - r e^{-i x} } \paren {1 - r e^{i x} } } }\)
\(\ds \) \(=\) \(\ds \map \Im {\frac {1 - r e^{-i x} } {1 - r \paren {e^{i x} + e^{- i x} } + r^2} }\)
\(\ds \) \(=\) \(\ds \map \Im {\frac {1 - r \paren {\cos x - i \sin x} } {1 - 2 r \cos x + r^2} }\) Euler's Formula: Corollary
\(\ds \) \(=\) \(\ds \dfrac {r \sin x} {1 - 2 r \cos x + r^2}\) after simplification

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Sum_of_Infinite_Series_of_Product_of_Power_and_Sine&oldid=653410"