Finding $cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+…+cos(theta+nalpha)$ with complex variable analysis ...

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Finding $cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+…+cos(theta+nalpha)$ with complex variable analysis



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Finding the integral $int_0^pidfrac{dtheta}{(2+costheta)^2}$ by complex analysisCalculating the following integral using complex analysis: $int_{0}^{pi}e^{acos(theta)}cos(asin(theta)), dtheta$complex analysis - differentiabiliityHow to use complex analysis to find the integral $int^pi_{−pi} frac 1 {1+sin^2(theta)} dtheta$?Trigonometric Expression for $1 + cos alpha + cos 2alpha + cdots + cos n alpha$ using complex numbersComplex Analysis: why does $cos(3theta)$ = $cos^3theta - 3costheta sin^2theta$.$int^{pi/2}_{0}frac{theta cos(theta)}{1+sin^{2}(theta)}$ through complex analysisUse the Maclaurin series to prove that $e^{itheta} = cos(theta) + isin(theta)$Show (via Complex Numbers): $frac{cosalphacosbeta}{cos^2theta}+frac{sinalphasinbeta}{sin^2theta}+1=0$ under given conditionsProving complex series $1 + costheta + cos2theta +… + cos ntheta $












2












$begingroup$


We have a series as



$cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+...+cos(theta+nalpha)=U$



How can we make use of complex variable analysis to arrive at the term below which is equivalent to the above series?



$U=frac{sin(frac{n+1}{2}alpha)}{sin(frac{1}{2}alpha)}cos(theta+frac{1}{2}nalpha)$






sequences-and-series complex-analysis complex-numbers







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    We have a series as



    $cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+...+cos(theta+nalpha)=U$



    How can we make use of complex variable analysis to arrive at the term below which is equivalent to the above series?



    $U=frac{sin(frac{n+1}{2}alpha)}{sin(frac{1}{2}alpha)}cos(theta+frac{1}{2}nalpha)$






    sequences-and-series complex-analysis complex-numbers







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      We have a series as



      $cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+...+cos(theta+nalpha)=U$



      How can we make use of complex variable analysis to arrive at the term below which is equivalent to the above series?



      $U=frac{sin(frac{n+1}{2}alpha)}{sin(frac{1}{2}alpha)}cos(theta+frac{1}{2}nalpha)$






      sequences-and-series complex-analysis complex-numbers







      share|cite|improve this question











      $endgroup$




      We have a series as



      $cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+...+cos(theta+nalpha)=U$



      How can we make use of complex variable analysis to arrive at the term below which is equivalent to the above series?



      $U=frac{sin(frac{n+1}{2}alpha)}{sin(frac{1}{2}alpha)}cos(theta+frac{1}{2}nalpha)$






      sequences-and-series complex-analysis complex-numbers




      sequences-and-series complex-analysis complex-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 5 mins ago









      YuiTo Cheng

      2,58641037




      2,58641037










      asked 1 hour ago









      UnbelievableUnbelievable

      1163




      1163






















          1 Answer
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          $begingroup$

          Use the fact that this is almost a geometric series. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$






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            $begingroup$

            Use the fact that this is almost a geometric series. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$






            share|cite|improve this answer









            $endgroup$


















              5












              $begingroup$

              Use the fact that this is almost a geometric series. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$






              share|cite|improve this answer









              $endgroup$
















                5












                5








                5





                $begingroup$

                Use the fact that this is almost a geometric series. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$






                share|cite|improve this answer









                $endgroup$



                Use the fact that this is almost a geometric series. $$begin{align}U&=mathfrak R[e^{itheta} +e^{itheta+ialpha}+cdots+e^{itheta+inalpha}]\&=mathfrak Rleft[e^{itheta}sum_{j=0}^n e^{ijalpha}right]\&=mathfrak Rleft[e^{itheta}frac{1-e^{i(n+1)alpha}}{1-e^{ialpha}}right]\&=mathfrak Rleft[e^{itheta}frac{e^{-i(n+1)alpha/2}-e^{i(n+1)alpha/2}}{e^{-ialpha/2}-e^{ialpha/2}}e^{inalpha/2}right]\&=mathfrak Rleft[e^{i(nalpha/2+theta)}frac{sin[(n+1)alpha/2]}{sin[alpha/2]}right]\&=cos(theta+tfrac{nalpha}{2})frac{sinleft(frac12(n+1)alpharight)}{sinleft(frac12alpharight)}end{align}$$







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                share|cite|improve this answer










                answered 49 mins ago









                dialogdialog

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