Use the result 1+z+z2...+zn=zn+1−1z−1 to sum the series to n terms
1+cosθ+cos2θ+...
also show that partial sums of series ∑cos(nθ) is bounded when 0<θ<π/2
My attempt
so z can be written as eiθ which means:
1+cosθ+cos2θ....+cosnθ+i(sinθ+sin2θ+....+sinnθ)=zn+1−1z−1
after this.. i dont know
Answer
Remember that eit=cost+isint∀t∈C and that n∑j=0zj=1−zn+11−z∀z∈C,|z|<1. Thus n∑j=0cos(jθ)=n∑j=0ℜ(eijθ)=ℜ(n∑j=0(eijθ))=ℜ(1−eiθ(n+1)1−eiθ) The last term I wrote can be handled easily in order to be written explicitly and get the results you wanted.
No comments:
Post a Comment