Sum of two infinite complex geometric sums












4












$begingroup$



let $$C=frac{costheta}{2}-frac{cos2theta}{4}+frac{cos3theta}{8}+...$$
$$S=frac{sintheta}{2}-frac{sin2theta}{4}+frac{sin3theta}{8}+...$$
I want to find the sum of the series $C+iS$ and thus find expressions for $C$ and $S$.




so the sum of $C+iS$ is
$$C+iS=(frac{costheta}{2}-frac{cos2theta}{4}+frac{cos3theta}{8}+...)+i(frac{sintheta}{2}-frac{sin2theta}{4}+frac{sin3theta}{8}+...)$$
$$=frac{1}{2}(costheta+isintheta)-frac{1}{4}(cos2theta+isin2theta)+frac{1}{8}(cos3theta+isin3theta) + ...$$
$$=frac{1}{2}(costheta+isintheta)-frac{1}{4}(costheta+isintheta)^2+frac{1}{8}(costheta+isintheta)^3 + ...$$



So I know i have to now put his into a geometric series where i can find the first term and common ratio as i will want to sum to infinity, but I'm not sure where to go from here?



So I've continued and got the series to here:
$$frac{1}{2}[e^{itheta}-frac{1}{2}(e^{itheta})^2+frac{1}{4}(e^{itheta})^3...]$$ But I'm still not sure where to continue.






sequences-and-series complex-numbers







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$endgroup$












  • $begingroup$
    Use $costheta+isintheta=e^{itheta}$ and factor out $frac{1}{2}$.
    $endgroup$
    – Yadati Kiran
    Nov 15 '18 at 11:21


















4












$begingroup$



let $$C=frac{costheta}{2}-frac{cos2theta}{4}+frac{cos3theta}{8}+...$$
$$S=frac{sintheta}{2}-frac{sin2theta}{4}+frac{sin3theta}{8}+...$$
I want to find the sum of the series $C+iS$ and thus find expressions for $C$ and $S$.




so the sum of $C+iS$ is
$$C+iS=(frac{costheta}{2}-frac{cos2theta}{4}+frac{cos3theta}{8}+...)+i(frac{sintheta}{2}-frac{sin2theta}{4}+frac{sin3theta}{8}+...)$$
$$=frac{1}{2}(costheta+isintheta)-frac{1}{4}(cos2theta+isin2theta)+frac{1}{8}(cos3theta+isin3theta) + ...$$
$$=frac{1}{2}(costheta+isintheta)-frac{1}{4}(costheta+isintheta)^2+frac{1}{8}(costheta+isintheta)^3 + ...$$



So I know i have to now put his into a geometric series where i can find the first term and common ratio as i will want to sum to infinity, but I'm not sure where to go from here?



So I've continued and got the series to here:
$$frac{1}{2}[e^{itheta}-frac{1}{2}(e^{itheta})^2+frac{1}{4}(e^{itheta})^3...]$$ But I'm still not sure where to continue.






sequences-and-series complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Use $costheta+isintheta=e^{itheta}$ and factor out $frac{1}{2}$.
    $endgroup$
    – Yadati Kiran
    Nov 15 '18 at 11:21
















4












4








4





$begingroup$



let $$C=frac{costheta}{2}-frac{cos2theta}{4}+frac{cos3theta}{8}+...$$
$$S=frac{sintheta}{2}-frac{sin2theta}{4}+frac{sin3theta}{8}+...$$
I want to find the sum of the series $C+iS$ and thus find expressions for $C$ and $S$.




so the sum of $C+iS$ is
$$C+iS=(frac{costheta}{2}-frac{cos2theta}{4}+frac{cos3theta}{8}+...)+i(frac{sintheta}{2}-frac{sin2theta}{4}+frac{sin3theta}{8}+...)$$
$$=frac{1}{2}(costheta+isintheta)-frac{1}{4}(cos2theta+isin2theta)+frac{1}{8}(cos3theta+isin3theta) + ...$$
$$=frac{1}{2}(costheta+isintheta)-frac{1}{4}(costheta+isintheta)^2+frac{1}{8}(costheta+isintheta)^3 + ...$$



So I know i have to now put his into a geometric series where i can find the first term and common ratio as i will want to sum to infinity, but I'm not sure where to go from here?



So I've continued and got the series to here:
$$frac{1}{2}[e^{itheta}-frac{1}{2}(e^{itheta})^2+frac{1}{4}(e^{itheta})^3...]$$ But I'm still not sure where to continue.






sequences-and-series complex-numbers







share|cite|improve this question











$endgroup$





let $$C=frac{costheta}{2}-frac{cos2theta}{4}+frac{cos3theta}{8}+...$$
$$S=frac{sintheta}{2}-frac{sin2theta}{4}+frac{sin3theta}{8}+...$$
I want to find the sum of the series $C+iS$ and thus find expressions for $C$ and $S$.




so the sum of $C+iS$ is
$$C+iS=(frac{costheta}{2}-frac{cos2theta}{4}+frac{cos3theta}{8}+...)+i(frac{sintheta}{2}-frac{sin2theta}{4}+frac{sin3theta}{8}+...)$$
$$=frac{1}{2}(costheta+isintheta)-frac{1}{4}(cos2theta+isin2theta)+frac{1}{8}(cos3theta+isin3theta) + ...$$
$$=frac{1}{2}(costheta+isintheta)-frac{1}{4}(costheta+isintheta)^2+frac{1}{8}(costheta+isintheta)^3 + ...$$



So I know i have to now put his into a geometric series where i can find the first term and common ratio as i will want to sum to infinity, but I'm not sure where to go from here?



So I've continued and got the series to here:
$$frac{1}{2}[e^{itheta}-frac{1}{2}(e^{itheta})^2+frac{1}{4}(e^{itheta})^3...]$$ But I'm still not sure where to continue.






sequences-and-series complex-numbers




sequences-and-series complex-numbers






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 15 '18 at 15:39







H.Linkhorn

















asked Nov 15 '18 at 11:14









H.LinkhornH.Linkhorn

473113




473113












  • $begingroup$
    Use $costheta+isintheta=e^{itheta}$ and factor out $frac{1}{2}$.
    $endgroup$
    – Yadati Kiran
    Nov 15 '18 at 11:21




















  • $begingroup$
    Use $costheta+isintheta=e^{itheta}$ and factor out $frac{1}{2}$.
    $endgroup$
    – Yadati Kiran
    Nov 15 '18 at 11:21


















$begingroup$
Use $costheta+isintheta=e^{itheta}$ and factor out $frac{1}{2}$.
$endgroup$
– Yadati Kiran
Nov 15 '18 at 11:21






$begingroup$
Use $costheta+isintheta=e^{itheta}$ and factor out $frac{1}{2}$.
$endgroup$
– Yadati Kiran
Nov 15 '18 at 11:21












2 Answers
2






active

oldest

votes


















2












$begingroup$

HINT



We have that



$$C+iS=sum_{k=1}^inftyfrac{(-1)^{k+1}}{2^k}e^{left(ikthetaright)}
=-sum_{k=1}^inftyleft(-frac{e^{left(ithetaright)}}{2}right)^k$$



then refer to geometric series which holds also for $r$ complex $|r|<1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $frac{exp(itheta)+1}{exp(itheta)+2exp(-itheta)+3} $which isn't the correct answer. I was told i should be $frac{2exp(itheta)+1}{5+4costheta}$
    $endgroup$
    – H.Linkhorn
    Nov 17 '18 at 18:31












  • $begingroup$
    @H.Linkhorn We have that $$-sum_{k=1}^inftyleft(-frac{e^{left(ithetaright)}}{2}right)^k=-frac{1}{1+frac{e^{itheta}}{2}}+1=frac{e^{itheta}}{2+e^{itheta}}$$ Then multiply by $2+e^{-itheta}$ denominator and numerator to obtain teh result.
    $endgroup$
    – gimusi
    Nov 17 '18 at 21:36





















3












$begingroup$

Hint:



Using Intuition behind euler's formula and the special case $e^{ipi}=-1$



$$dfrac{(cos t+isin t)^n(-1)^{n-1}}{2^n}=-dfrac{e^{int}(e^{ipi})^n}{2^n}=-left(dfrac{e^{i(t+pi)}}2right)^n$$



Now for the common ratio$(r)$ of Geometric Series,



$|r|=left|dfrac{e^{i(t+pi)}}2right|=dfrac12<1$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    HINT



    We have that



    $$C+iS=sum_{k=1}^inftyfrac{(-1)^{k+1}}{2^k}e^{left(ikthetaright)}
    =-sum_{k=1}^inftyleft(-frac{e^{left(ithetaright)}}{2}right)^k$$



    then refer to geometric series which holds also for $r$ complex $|r|<1$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $frac{exp(itheta)+1}{exp(itheta)+2exp(-itheta)+3} $which isn't the correct answer. I was told i should be $frac{2exp(itheta)+1}{5+4costheta}$
      $endgroup$
      – H.Linkhorn
      Nov 17 '18 at 18:31












    • $begingroup$
      @H.Linkhorn We have that $$-sum_{k=1}^inftyleft(-frac{e^{left(ithetaright)}}{2}right)^k=-frac{1}{1+frac{e^{itheta}}{2}}+1=frac{e^{itheta}}{2+e^{itheta}}$$ Then multiply by $2+e^{-itheta}$ denominator and numerator to obtain teh result.
      $endgroup$
      – gimusi
      Nov 17 '18 at 21:36


















    2












    $begingroup$

    HINT



    We have that



    $$C+iS=sum_{k=1}^inftyfrac{(-1)^{k+1}}{2^k}e^{left(ikthetaright)}
    =-sum_{k=1}^inftyleft(-frac{e^{left(ithetaright)}}{2}right)^k$$



    then refer to geometric series which holds also for $r$ complex $|r|<1$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $frac{exp(itheta)+1}{exp(itheta)+2exp(-itheta)+3} $which isn't the correct answer. I was told i should be $frac{2exp(itheta)+1}{5+4costheta}$
      $endgroup$
      – H.Linkhorn
      Nov 17 '18 at 18:31












    • $begingroup$
      @H.Linkhorn We have that $$-sum_{k=1}^inftyleft(-frac{e^{left(ithetaright)}}{2}right)^k=-frac{1}{1+frac{e^{itheta}}{2}}+1=frac{e^{itheta}}{2+e^{itheta}}$$ Then multiply by $2+e^{-itheta}$ denominator and numerator to obtain teh result.
      $endgroup$
      – gimusi
      Nov 17 '18 at 21:36
















    2












    2








    2





    $begingroup$

    HINT



    We have that



    $$C+iS=sum_{k=1}^inftyfrac{(-1)^{k+1}}{2^k}e^{left(ikthetaright)}
    =-sum_{k=1}^inftyleft(-frac{e^{left(ithetaright)}}{2}right)^k$$



    then refer to geometric series which holds also for $r$ complex $|r|<1$.






    share|cite|improve this answer











    $endgroup$



    HINT



    We have that



    $$C+iS=sum_{k=1}^inftyfrac{(-1)^{k+1}}{2^k}e^{left(ikthetaright)}
    =-sum_{k=1}^inftyleft(-frac{e^{left(ithetaright)}}{2}right)^k$$



    then refer to geometric series which holds also for $r$ complex $|r|<1$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 15 '18 at 11:28

























    answered Nov 15 '18 at 11:21









    gimusigimusi

    93k84494




    93k84494












    • $begingroup$
      I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $frac{exp(itheta)+1}{exp(itheta)+2exp(-itheta)+3} $which isn't the correct answer. I was told i should be $frac{2exp(itheta)+1}{5+4costheta}$
      $endgroup$
      – H.Linkhorn
      Nov 17 '18 at 18:31












    • $begingroup$
      @H.Linkhorn We have that $$-sum_{k=1}^inftyleft(-frac{e^{left(ithetaright)}}{2}right)^k=-frac{1}{1+frac{e^{itheta}}{2}}+1=frac{e^{itheta}}{2+e^{itheta}}$$ Then multiply by $2+e^{-itheta}$ denominator and numerator to obtain teh result.
      $endgroup$
      – gimusi
      Nov 17 '18 at 21:36




















    • $begingroup$
      I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $frac{exp(itheta)+1}{exp(itheta)+2exp(-itheta)+3} $which isn't the correct answer. I was told i should be $frac{2exp(itheta)+1}{5+4costheta}$
      $endgroup$
      – H.Linkhorn
      Nov 17 '18 at 18:31












    • $begingroup$
      @H.Linkhorn We have that $$-sum_{k=1}^inftyleft(-frac{e^{left(ithetaright)}}{2}right)^k=-frac{1}{1+frac{e^{itheta}}{2}}+1=frac{e^{itheta}}{2+e^{itheta}}$$ Then multiply by $2+e^{-itheta}$ denominator and numerator to obtain teh result.
      $endgroup$
      – gimusi
      Nov 17 '18 at 21:36


















    $begingroup$
    I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $frac{exp(itheta)+1}{exp(itheta)+2exp(-itheta)+3} $which isn't the correct answer. I was told i should be $frac{2exp(itheta)+1}{5+4costheta}$
    $endgroup$
    – H.Linkhorn
    Nov 17 '18 at 18:31






    $begingroup$
    I tried to sum to infinity using your sum with $a=0.5e^itheta$ and $r=-0.5e^itheta$ but i obtained $frac{exp(itheta)+1}{exp(itheta)+2exp(-itheta)+3} $which isn't the correct answer. I was told i should be $frac{2exp(itheta)+1}{5+4costheta}$
    $endgroup$
    – H.Linkhorn
    Nov 17 '18 at 18:31














    $begingroup$
    @H.Linkhorn We have that $$-sum_{k=1}^inftyleft(-frac{e^{left(ithetaright)}}{2}right)^k=-frac{1}{1+frac{e^{itheta}}{2}}+1=frac{e^{itheta}}{2+e^{itheta}}$$ Then multiply by $2+e^{-itheta}$ denominator and numerator to obtain teh result.
    $endgroup$
    – gimusi
    Nov 17 '18 at 21:36






    $begingroup$
    @H.Linkhorn We have that $$-sum_{k=1}^inftyleft(-frac{e^{left(ithetaright)}}{2}right)^k=-frac{1}{1+frac{e^{itheta}}{2}}+1=frac{e^{itheta}}{2+e^{itheta}}$$ Then multiply by $2+e^{-itheta}$ denominator and numerator to obtain teh result.
    $endgroup$
    – gimusi
    Nov 17 '18 at 21:36













    3












    $begingroup$

    Hint:



    Using Intuition behind euler's formula and the special case $e^{ipi}=-1$



    $$dfrac{(cos t+isin t)^n(-1)^{n-1}}{2^n}=-dfrac{e^{int}(e^{ipi})^n}{2^n}=-left(dfrac{e^{i(t+pi)}}2right)^n$$



    Now for the common ratio$(r)$ of Geometric Series,



    $|r|=left|dfrac{e^{i(t+pi)}}2right|=dfrac12<1$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Hint:



      Using Intuition behind euler's formula and the special case $e^{ipi}=-1$



      $$dfrac{(cos t+isin t)^n(-1)^{n-1}}{2^n}=-dfrac{e^{int}(e^{ipi})^n}{2^n}=-left(dfrac{e^{i(t+pi)}}2right)^n$$



      Now for the common ratio$(r)$ of Geometric Series,



      $|r|=left|dfrac{e^{i(t+pi)}}2right|=dfrac12<1$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Hint:



        Using Intuition behind euler's formula and the special case $e^{ipi}=-1$



        $$dfrac{(cos t+isin t)^n(-1)^{n-1}}{2^n}=-dfrac{e^{int}(e^{ipi})^n}{2^n}=-left(dfrac{e^{i(t+pi)}}2right)^n$$



        Now for the common ratio$(r)$ of Geometric Series,



        $|r|=left|dfrac{e^{i(t+pi)}}2right|=dfrac12<1$






        share|cite|improve this answer









        $endgroup$



        Hint:



        Using Intuition behind euler's formula and the special case $e^{ipi}=-1$



        $$dfrac{(cos t+isin t)^n(-1)^{n-1}}{2^n}=-dfrac{e^{int}(e^{ipi})^n}{2^n}=-left(dfrac{e^{i(t+pi)}}2right)^n$$



        Now for the common ratio$(r)$ of Geometric Series,



        $|r|=left|dfrac{e^{i(t+pi)}}2right|=dfrac12<1$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 15 '18 at 11:21









        lab bhattacharjeelab bhattacharjee

        226k15158275




        226k15158275






























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