MHB How Can Complex Analysis Be Used to Sum Powers of Sine Functions?

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Complex analysis can be applied to sum powers of sine functions, specifically for the series sin³x + sin³2x + sin³3x + ... + sin³nx. By utilizing the identity sin³α = (3 sinα - sin 3α)/4 and expressing sine in terms of exponential functions, the sum can be reformulated. The resulting expression involves the sums of exponential terms, allowing for simplification through geometric series. This approach provides a systematic method to compute the desired sum. The discussion emphasizes the effectiveness of complex analysis in tackling such trigonometric series.
Suvadip
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How to find the sum using complex analysis
$$sin^3x+sin^32x+sin^33x+sin^34x+...+sin^3nx$$
 
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suvadip said:
How to find the sum using complex analysis
$$sin^3x+sin^32x+sin^33x+sin^34x+...+sin^3nx$$

Using the identities $\displaystyle \sin^{3} \alpha = \frac{3\ \sin \alpha - \sin 3\ \alpha}{4}$ and $\displaystyle \sin \alpha = \frac{e^{i\ \alpha} - e^{- i\ \alpha}}{2\ i}$ You have...

$\displaystyle \sum_{k=1}^{n} \sin^{3} k\ x = \frac{3}{8\ i} (\sum_{k=1}^{n} e^{k\ i\ x} - \sum_{k=1}^{n} e^{- k\ i\ x}) - \frac{1}{8\ i}\ ( \sum_{k=1}^{n} e^{3\ k\ i\ x} - \sum_{k=1}^{n} e^{- 3\ k\ i\ x}) =$$\displaystyle = \frac{3}{8\ i}\ (e^{i\ x}\ \frac{1 - e^{n\ i\ x}}{1 - e^{i\ x}} - e^{- i\ x}\ \frac{1 - e^{- n\ i\ x}}{1- e^{- i\ x}}) - \frac{1}{8\ i}\ (e^{3\ i\ x}\ \frac{1 - e^{3\ n\ i\ x}}{1 - e^{3\ i\ x}} - e^{- 3\ i\ x}\ \frac{1 - e^{- 3\ n\ i\ x}}{1- e^{- 3\ i\ x}})$

Are You able to proceed?...

Kind regards

$\chi$ $\sigma$
 
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We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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