Sum of Series of Product of Power and Sine

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Theorem

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

Then:

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


Proof

From Euler's Formula:

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

Hence:

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

$\blacksquare$


Also see


Sources

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