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

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SUMMARY

This discussion focuses on using complex analysis to compute the sum of powers of sine functions, specifically the series $$\sin^3x + \sin^32x + \sin^33x + \ldots + \sin^3nx$$. The identities utilized include $\sin^{3} \alpha = \frac{3\ \sin \alpha - \sin 3\ \alpha}{4}$ and the exponential form of sine, $\sin \alpha = \frac{e^{i\ \alpha} - e^{- i\ \alpha}}{2\ i}$. The resulting formula for the sum involves complex exponentials and simplifies to a combination of geometric series, demonstrating the power of complex analysis in evaluating trigonometric sums.

PREREQUISITES
  • Understanding of complex analysis principles
  • Familiarity with trigonometric identities
  • Knowledge of geometric series summation
  • Proficiency in manipulating exponential functions
NEXT STEPS
  • Study the derivation of trigonometric identities, particularly $\sin^{3} \alpha$ and $\sin \alpha$ in exponential form
  • Learn about geometric series and their applications in complex analysis
  • Explore advanced topics in complex analysis, such as contour integration
  • Investigate other applications of complex analysis in summing series
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Mathematicians, physics students, and anyone interested in advanced calculus or complex analysis, particularly those looking to deepen their understanding of trigonometric series and their evaluations.

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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