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sum using complex analysis

suvadip

Member
Feb 21, 2013
69
How to find the sum using complex analysis
\(\displaystyle sin^3x+sin^32x+sin^33x+sin^34x+....+sin^3nx\)
 

chisigma

Well-known member
Feb 13, 2012
1,704
How to find the sum using complex analysis
\(\displaystyle sin^3x+sin^32x+sin^33x+sin^34x+....+sin^3nx\)
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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