Infinite Summation Question (Euler's Identity)












4














So one of my past exam papers has the question:



Show:
$sum_{n=0}^infty frac{sin(ntheta)}{2^n} = frac{sin(theta)}{5-4cos(theta)}$



My working:



$e^{intheta} = cos(ntheta) + isin(ntheta)$



So $sin(ntheta)$ is the imaginary part of $e^{intheta}$



Thus we get: $mathbb I(sum_{n=0}^infty (frac{e^{itheta}}{2})^n)$



Using the summation formula for infinite geometric series:



$=frac{1}{1-frac{e^{itheta}}{2}}$



And after multiplying by the complex conjugate I get the imaginary part to be



$= frac{2sin(theta)}{5-4cos(theta)}$



Any help would be great, thanks!






complex-numbers summation







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    4














    So one of my past exam papers has the question:



    Show:
    $sum_{n=0}^infty frac{sin(ntheta)}{2^n} = frac{sin(theta)}{5-4cos(theta)}$



    My working:



    $e^{intheta} = cos(ntheta) + isin(ntheta)$



    So $sin(ntheta)$ is the imaginary part of $e^{intheta}$



    Thus we get: $mathbb I(sum_{n=0}^infty (frac{e^{itheta}}{2})^n)$



    Using the summation formula for infinite geometric series:



    $=frac{1}{1-frac{e^{itheta}}{2}}$



    And after multiplying by the complex conjugate I get the imaginary part to be



    $= frac{2sin(theta)}{5-4cos(theta)}$



    Any help would be great, thanks!






    complex-numbers summation







    share|cite|improve this question

























      4












      4








      4







      So one of my past exam papers has the question:



      Show:
      $sum_{n=0}^infty frac{sin(ntheta)}{2^n} = frac{sin(theta)}{5-4cos(theta)}$



      My working:



      $e^{intheta} = cos(ntheta) + isin(ntheta)$



      So $sin(ntheta)$ is the imaginary part of $e^{intheta}$



      Thus we get: $mathbb I(sum_{n=0}^infty (frac{e^{itheta}}{2})^n)$



      Using the summation formula for infinite geometric series:



      $=frac{1}{1-frac{e^{itheta}}{2}}$



      And after multiplying by the complex conjugate I get the imaginary part to be



      $= frac{2sin(theta)}{5-4cos(theta)}$



      Any help would be great, thanks!






      complex-numbers summation







      share|cite|improve this question













      So one of my past exam papers has the question:



      Show:
      $sum_{n=0}^infty frac{sin(ntheta)}{2^n} = frac{sin(theta)}{5-4cos(theta)}$



      My working:



      $e^{intheta} = cos(ntheta) + isin(ntheta)$



      So $sin(ntheta)$ is the imaginary part of $e^{intheta}$



      Thus we get: $mathbb I(sum_{n=0}^infty (frac{e^{itheta}}{2})^n)$



      Using the summation formula for infinite geometric series:



      $=frac{1}{1-frac{e^{itheta}}{2}}$



      And after multiplying by the complex conjugate I get the imaginary part to be



      $= frac{2sin(theta)}{5-4cos(theta)}$



      Any help would be great, thanks!






      complex-numbers summation




      complex-numbers summation






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      asked 2 hours ago









      PolynomialC

      436




      436






















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          Testing... if $theta=frac{pi}{2}$, the original series is $frac1{2^1}-frac1{2^3}+frac1{2^5}-frac1{2^7}+cdots =left(frac12-frac18right)left(1+frac1{2^4}+frac1{2^8}+cdotsright)$ which becomes $frac38cdotfrac{16}{15}=frac25$



          The official answer claims $frac{1}{5-4cdot 0}=frac15$. Yours claims $frac{2}{5-4cdot 0}=frac25$. Your calculation is right and the official answer is wrong.






          share|cite|improve this answer





















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            1 Answer
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            1 Answer
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            active

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            oldest

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            active

            oldest

            votes









            5














            Testing... if $theta=frac{pi}{2}$, the original series is $frac1{2^1}-frac1{2^3}+frac1{2^5}-frac1{2^7}+cdots =left(frac12-frac18right)left(1+frac1{2^4}+frac1{2^8}+cdotsright)$ which becomes $frac38cdotfrac{16}{15}=frac25$



            The official answer claims $frac{1}{5-4cdot 0}=frac15$. Yours claims $frac{2}{5-4cdot 0}=frac25$. Your calculation is right and the official answer is wrong.






            share|cite|improve this answer


























              5














              Testing... if $theta=frac{pi}{2}$, the original series is $frac1{2^1}-frac1{2^3}+frac1{2^5}-frac1{2^7}+cdots =left(frac12-frac18right)left(1+frac1{2^4}+frac1{2^8}+cdotsright)$ which becomes $frac38cdotfrac{16}{15}=frac25$



              The official answer claims $frac{1}{5-4cdot 0}=frac15$. Yours claims $frac{2}{5-4cdot 0}=frac25$. Your calculation is right and the official answer is wrong.






              share|cite|improve this answer
























                5












                5








                5






                Testing... if $theta=frac{pi}{2}$, the original series is $frac1{2^1}-frac1{2^3}+frac1{2^5}-frac1{2^7}+cdots =left(frac12-frac18right)left(1+frac1{2^4}+frac1{2^8}+cdotsright)$ which becomes $frac38cdotfrac{16}{15}=frac25$



                The official answer claims $frac{1}{5-4cdot 0}=frac15$. Yours claims $frac{2}{5-4cdot 0}=frac25$. Your calculation is right and the official answer is wrong.






                share|cite|improve this answer












                Testing... if $theta=frac{pi}{2}$, the original series is $frac1{2^1}-frac1{2^3}+frac1{2^5}-frac1{2^7}+cdots =left(frac12-frac18right)left(1+frac1{2^4}+frac1{2^8}+cdotsright)$ which becomes $frac38cdotfrac{16}{15}=frac25$



                The official answer claims $frac{1}{5-4cdot 0}=frac15$. Yours claims $frac{2}{5-4cdot 0}=frac25$. Your calculation is right and the official answer is wrong.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                jmerry

                1,69218




                1,69218






























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