Tuesday, April 5, 2016

find sum of n terms of series $sum cos(ntheta)$


Use the result $1 + z + z^2...+z^n=\frac{z^{n+1}-1}{z-1}$ to sum the series to n terms



$1+\cos\theta+\cos2\theta+...$


also show that partial sums of series $\sum \cos (n\theta)$ is bounded when $0<\theta<\pi/2$


My attempt


so z can be written as $e^{i\theta}$ which means:


$1+ \cos \theta + \cos 2\theta ....+\cos n\theta + i(\sin \theta+\sin 2\theta+....+\sin n\theta)=\frac{z^{n+1}-1}{z-1}$


after this.. i dont know


Answer



Remember that $$ e^{it}=\cos t+i\sin t\;\;\;\;\forall t\in\Bbb C $$ and that $$ \sum_{j=0}^{n}z^j=\frac{1-z^{n+1}}{1-z}\;\;\forall z\in\Bbb C,\;\;|z|<1. $$ Thus $$ \sum_{j=0}^{n}\cos(j\theta)= \sum_{j=0}^{n}\Re{(e^{ij\theta})}= \Re\left(\sum_{j=0}^{n}(e^{ij\theta})\right)= \Re\left(\frac{1-e^{i\theta(n+1)}}{1-e^{i\theta}}\right) $$ The last term I wrote can be handled easily in order to be written explicitly and get the results you wanted.


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