Saturday, 16 April 2016

Geometric series and complex numbers




I'm new to this site, english is not my mother tongue, and I'm just learning LaTeX. I'm basically a noob, so please be indulgent if I break any rule or habits.



I'm stuck at proving the following equation. I suppose I should use the formula for geometric series (nk=0qk=1qn+11q), and also use somewhere that eîθ=cos(θ)+isin(θ)



So here is the equation I have to prove :
12+cos(θ)+cos(2θ)+...+cos(nθ)=sin(n+12θ)2sin(θ2)




Thanks for your help


Answer




12+nr=1cos(rθ)={12+nr=1eirθ}={12+nr=0eirθ}={12+1e(n+1)θ1eiθ}={12e(n+1)θ+eiθ2(1eiθ)}={eiθ/22e(n+1/2)θ+eiθ/22(eiθ/2eiθ/2)}={2cos(θ2)2cos(n+1/2)θ2isin(n+1/2)θ2isin(θ2)}




Multiply through by 1=ii, take the real part and there you have it. ^_^


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