Demonstrating an equivalent formula for $sin(x)/x$ using power series












1












$begingroup$


Would appreciate some ideas for the following:



"Prove that $frac{sin{x}}{x}=prod_{n=1}^{infty}cos{frac{x}{2^n}}$ using power series."



I'm aware this identity can be shown using trig identities and a telescoping product. Also, you can get another proof using the infinite product expressions $sin{x} = xprod_{k=1}^{infty} (1-frac{x^2}{k^2 pi^2})$ and $cos{x}=prod_{k=1, k text{odd}}^infty (1-frac{4x^2}{k^2 pi^2})$.



However, since the question explicitly mentions power series, I was wondering if there is a proof that directly uses power series? I've tried calculating some derivatives and coefficients but they seem to get pretty nasty.






power-series infinite-product







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  • $begingroup$
    use Taylor series for sin at zero?
    $endgroup$
    – Dole
    Dec 23 '18 at 6:26












  • $begingroup$
    yeah, but the RHS is giving me trouble
    $endgroup$
    – Merk Zockerborg
    Dec 23 '18 at 6:28










  • $begingroup$
    Is there a nice power series for $log(cos x)$?
    $endgroup$
    – mathworker21
    Dec 23 '18 at 6:44
















1












$begingroup$


Would appreciate some ideas for the following:



"Prove that $frac{sin{x}}{x}=prod_{n=1}^{infty}cos{frac{x}{2^n}}$ using power series."



I'm aware this identity can be shown using trig identities and a telescoping product. Also, you can get another proof using the infinite product expressions $sin{x} = xprod_{k=1}^{infty} (1-frac{x^2}{k^2 pi^2})$ and $cos{x}=prod_{k=1, k text{odd}}^infty (1-frac{4x^2}{k^2 pi^2})$.



However, since the question explicitly mentions power series, I was wondering if there is a proof that directly uses power series? I've tried calculating some derivatives and coefficients but they seem to get pretty nasty.






power-series infinite-product







share|cite|improve this question











$endgroup$












  • $begingroup$
    use Taylor series for sin at zero?
    $endgroup$
    – Dole
    Dec 23 '18 at 6:26












  • $begingroup$
    yeah, but the RHS is giving me trouble
    $endgroup$
    – Merk Zockerborg
    Dec 23 '18 at 6:28










  • $begingroup$
    Is there a nice power series for $log(cos x)$?
    $endgroup$
    – mathworker21
    Dec 23 '18 at 6:44














1












1








1


1



$begingroup$


Would appreciate some ideas for the following:



"Prove that $frac{sin{x}}{x}=prod_{n=1}^{infty}cos{frac{x}{2^n}}$ using power series."



I'm aware this identity can be shown using trig identities and a telescoping product. Also, you can get another proof using the infinite product expressions $sin{x} = xprod_{k=1}^{infty} (1-frac{x^2}{k^2 pi^2})$ and $cos{x}=prod_{k=1, k text{odd}}^infty (1-frac{4x^2}{k^2 pi^2})$.



However, since the question explicitly mentions power series, I was wondering if there is a proof that directly uses power series? I've tried calculating some derivatives and coefficients but they seem to get pretty nasty.






power-series infinite-product







share|cite|improve this question











$endgroup$




Would appreciate some ideas for the following:



"Prove that $frac{sin{x}}{x}=prod_{n=1}^{infty}cos{frac{x}{2^n}}$ using power series."



I'm aware this identity can be shown using trig identities and a telescoping product. Also, you can get another proof using the infinite product expressions $sin{x} = xprod_{k=1}^{infty} (1-frac{x^2}{k^2 pi^2})$ and $cos{x}=prod_{k=1, k text{odd}}^infty (1-frac{4x^2}{k^2 pi^2})$.



However, since the question explicitly mentions power series, I was wondering if there is a proof that directly uses power series? I've tried calculating some derivatives and coefficients but they seem to get pretty nasty.






power-series infinite-product




power-series infinite-product






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 6:30









Eevee Trainer

6,1531936




6,1531936










asked Dec 23 '18 at 6:17









Merk ZockerborgMerk Zockerborg

18310




18310












  • $begingroup$
    use Taylor series for sin at zero?
    $endgroup$
    – Dole
    Dec 23 '18 at 6:26












  • $begingroup$
    yeah, but the RHS is giving me trouble
    $endgroup$
    – Merk Zockerborg
    Dec 23 '18 at 6:28










  • $begingroup$
    Is there a nice power series for $log(cos x)$?
    $endgroup$
    – mathworker21
    Dec 23 '18 at 6:44


















  • $begingroup$
    use Taylor series for sin at zero?
    $endgroup$
    – Dole
    Dec 23 '18 at 6:26












  • $begingroup$
    yeah, but the RHS is giving me trouble
    $endgroup$
    – Merk Zockerborg
    Dec 23 '18 at 6:28










  • $begingroup$
    Is there a nice power series for $log(cos x)$?
    $endgroup$
    – mathworker21
    Dec 23 '18 at 6:44
















$begingroup$
use Taylor series for sin at zero?
$endgroup$
– Dole
Dec 23 '18 at 6:26






$begingroup$
use Taylor series for sin at zero?
$endgroup$
– Dole
Dec 23 '18 at 6:26














$begingroup$
yeah, but the RHS is giving me trouble
$endgroup$
– Merk Zockerborg
Dec 23 '18 at 6:28




$begingroup$
yeah, but the RHS is giving me trouble
$endgroup$
– Merk Zockerborg
Dec 23 '18 at 6:28












$begingroup$
Is there a nice power series for $log(cos x)$?
$endgroup$
– mathworker21
Dec 23 '18 at 6:44




$begingroup$
Is there a nice power series for $log(cos x)$?
$endgroup$
– mathworker21
Dec 23 '18 at 6:44










2 Answers
2






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1












$begingroup$

Using the fact that $sin(2x)=2sin(x)cos(x)$ we have $cos(x)=frac{sin(2x)}{2sin(x)}$.



Define $p_n(x):=prodlimits_{j=1}^n cos(frac{x}{2^j})$ and note that $cos(frac{x}{2^j})=dfrac{sin(frac{x}{2^{j-1}})}{2sin(frac{x}{2^j})}$ for each $j$.



Now, $p_n(x)=frac{1}{2^n}frac{sin(x)}{sin(frac{x}{2^n})}=frac{sin(x)}{x}dfrac{frac{x}{2^n}}{sin(frac{x}{2^n})}$ and $p_n(x)to_n frac{sin(x)}{x}$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    HINT



    $cos z= e^{iz}(1+e^{-2iz})/2$ so $ln cos z=cdots$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Using the fact that $sin(2x)=2sin(x)cos(x)$ we have $cos(x)=frac{sin(2x)}{2sin(x)}$.



      Define $p_n(x):=prodlimits_{j=1}^n cos(frac{x}{2^j})$ and note that $cos(frac{x}{2^j})=dfrac{sin(frac{x}{2^{j-1}})}{2sin(frac{x}{2^j})}$ for each $j$.



      Now, $p_n(x)=frac{1}{2^n}frac{sin(x)}{sin(frac{x}{2^n})}=frac{sin(x)}{x}dfrac{frac{x}{2^n}}{sin(frac{x}{2^n})}$ and $p_n(x)to_n frac{sin(x)}{x}$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Using the fact that $sin(2x)=2sin(x)cos(x)$ we have $cos(x)=frac{sin(2x)}{2sin(x)}$.



        Define $p_n(x):=prodlimits_{j=1}^n cos(frac{x}{2^j})$ and note that $cos(frac{x}{2^j})=dfrac{sin(frac{x}{2^{j-1}})}{2sin(frac{x}{2^j})}$ for each $j$.



        Now, $p_n(x)=frac{1}{2^n}frac{sin(x)}{sin(frac{x}{2^n})}=frac{sin(x)}{x}dfrac{frac{x}{2^n}}{sin(frac{x}{2^n})}$ and $p_n(x)to_n frac{sin(x)}{x}$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Using the fact that $sin(2x)=2sin(x)cos(x)$ we have $cos(x)=frac{sin(2x)}{2sin(x)}$.



          Define $p_n(x):=prodlimits_{j=1}^n cos(frac{x}{2^j})$ and note that $cos(frac{x}{2^j})=dfrac{sin(frac{x}{2^{j-1}})}{2sin(frac{x}{2^j})}$ for each $j$.



          Now, $p_n(x)=frac{1}{2^n}frac{sin(x)}{sin(frac{x}{2^n})}=frac{sin(x)}{x}dfrac{frac{x}{2^n}}{sin(frac{x}{2^n})}$ and $p_n(x)to_n frac{sin(x)}{x}$






          share|cite|improve this answer









          $endgroup$



          Using the fact that $sin(2x)=2sin(x)cos(x)$ we have $cos(x)=frac{sin(2x)}{2sin(x)}$.



          Define $p_n(x):=prodlimits_{j=1}^n cos(frac{x}{2^j})$ and note that $cos(frac{x}{2^j})=dfrac{sin(frac{x}{2^{j-1}})}{2sin(frac{x}{2^j})}$ for each $j$.



          Now, $p_n(x)=frac{1}{2^n}frac{sin(x)}{sin(frac{x}{2^n})}=frac{sin(x)}{x}dfrac{frac{x}{2^n}}{sin(frac{x}{2^n})}$ and $p_n(x)to_n frac{sin(x)}{x}$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 23 '18 at 7:00









          Martín Vacas VignoloMartín Vacas Vignolo

          3,816623




          3,816623























              0












              $begingroup$

              HINT



              $cos z= e^{iz}(1+e^{-2iz})/2$ so $ln cos z=cdots$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                HINT



                $cos z= e^{iz}(1+e^{-2iz})/2$ so $ln cos z=cdots$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  HINT



                  $cos z= e^{iz}(1+e^{-2iz})/2$ so $ln cos z=cdots$






                  share|cite|improve this answer









                  $endgroup$



                  HINT



                  $cos z= e^{iz}(1+e^{-2iz})/2$ so $ln cos z=cdots$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 23 '18 at 10:08









                  G CabG Cab

                  19.6k31239




                  19.6k31239






























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