Proving $sin(2x) = 2sin(x)cos(x)$ using Cauchy product












1












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I've stumbled upon this question. I can't understand why all even terms of the Cauchy product are $0$, since we add only positive numbers:
begin{equation*}
c_n = begin{cases} sumlimits_{k=0}^{m} frac{(-1)^{k}(-1)^{m-k}}{(2k+1)!(2m-2k)!} & n = 2m + 1 \ hspace{32 pt} 0 & n = 2m end{cases}
end{equation*}






trigonometry cauchy-product







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  • 3




    $begingroup$
    All even terms of the Cauchy product are 0 because $sin(2x)$ is an odd function.
    $endgroup$
    – Robert Z
    Dec 17 '18 at 14:55










  • $begingroup$
    @RobertZ but how does this result from the Cauchy product?
    $endgroup$
    – Dreikäsehoch
    Dec 17 '18 at 14:58
















1












$begingroup$


I've stumbled upon this question. I can't understand why all even terms of the Cauchy product are $0$, since we add only positive numbers:
begin{equation*}
c_n = begin{cases} sumlimits_{k=0}^{m} frac{(-1)^{k}(-1)^{m-k}}{(2k+1)!(2m-2k)!} & n = 2m + 1 \ hspace{32 pt} 0 & n = 2m end{cases}
end{equation*}






trigonometry cauchy-product







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    All even terms of the Cauchy product are 0 because $sin(2x)$ is an odd function.
    $endgroup$
    – Robert Z
    Dec 17 '18 at 14:55










  • $begingroup$
    @RobertZ but how does this result from the Cauchy product?
    $endgroup$
    – Dreikäsehoch
    Dec 17 '18 at 14:58














1












1








1


1



$begingroup$


I've stumbled upon this question. I can't understand why all even terms of the Cauchy product are $0$, since we add only positive numbers:
begin{equation*}
c_n = begin{cases} sumlimits_{k=0}^{m} frac{(-1)^{k}(-1)^{m-k}}{(2k+1)!(2m-2k)!} & n = 2m + 1 \ hspace{32 pt} 0 & n = 2m end{cases}
end{equation*}






trigonometry cauchy-product







share|cite|improve this question









$endgroup$




I've stumbled upon this question. I can't understand why all even terms of the Cauchy product are $0$, since we add only positive numbers:
begin{equation*}
c_n = begin{cases} sumlimits_{k=0}^{m} frac{(-1)^{k}(-1)^{m-k}}{(2k+1)!(2m-2k)!} & n = 2m + 1 \ hspace{32 pt} 0 & n = 2m end{cases}
end{equation*}






trigonometry cauchy-product




trigonometry cauchy-product






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share|cite|improve this question











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share|cite|improve this question










asked Dec 17 '18 at 14:53









DreikäsehochDreikäsehoch

3048




3048








  • 3




    $begingroup$
    All even terms of the Cauchy product are 0 because $sin(2x)$ is an odd function.
    $endgroup$
    – Robert Z
    Dec 17 '18 at 14:55










  • $begingroup$
    @RobertZ but how does this result from the Cauchy product?
    $endgroup$
    – Dreikäsehoch
    Dec 17 '18 at 14:58














  • 3




    $begingroup$
    All even terms of the Cauchy product are 0 because $sin(2x)$ is an odd function.
    $endgroup$
    – Robert Z
    Dec 17 '18 at 14:55










  • $begingroup$
    @RobertZ but how does this result from the Cauchy product?
    $endgroup$
    – Dreikäsehoch
    Dec 17 '18 at 14:58








3




3




$begingroup$
All even terms of the Cauchy product are 0 because $sin(2x)$ is an odd function.
$endgroup$
– Robert Z
Dec 17 '18 at 14:55




$begingroup$
All even terms of the Cauchy product are 0 because $sin(2x)$ is an odd function.
$endgroup$
– Robert Z
Dec 17 '18 at 14:55












$begingroup$
@RobertZ but how does this result from the Cauchy product?
$endgroup$
– Dreikäsehoch
Dec 17 '18 at 14:58




$begingroup$
@RobertZ but how does this result from the Cauchy product?
$endgroup$
– Dreikäsehoch
Dec 17 '18 at 14:58










2 Answers
2






active

oldest

votes


















3












$begingroup$

Note that
$$c_n=sum_{k=0}^n a_kb_{n-k}$$
where $a_k$ is the coefficient of $x^k$ of the series of $sin(x)$ and $b_{n-k}$ is the coefficient of $x^{n-k}$ of the series of $cos(x)$. Now if $n$ is even then



i) if $k$ is odd then $n-k$ is odd and $b_{n-k}=0$ which implies that $a_kb_{n-k}=0$;



ii) if $k$ is even then $a_{k}=0$ which implies that $a_kb_{n-k}=0$.



Hence $c_n=sum_{k=0}^n 0=0$.



By using the same argument we can show that the product of an "even" series with and an "odd" series is an "odd" series.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Since$$sin x=0+x+0times x^2-frac1{3!}x^3+cdots$$and$$cos x=1+0times x-frac1{2!}x^2+0times x^3+cdots$$when you compute a term of even order of the Cauchy product of these series, you get a sum with zeros everywhere; for instance, the coeffiecient of $x^2$ is$$0timesleft(-frac1{2!}right)+1times0+0times1=0.$$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Note that
      $$c_n=sum_{k=0}^n a_kb_{n-k}$$
      where $a_k$ is the coefficient of $x^k$ of the series of $sin(x)$ and $b_{n-k}$ is the coefficient of $x^{n-k}$ of the series of $cos(x)$. Now if $n$ is even then



      i) if $k$ is odd then $n-k$ is odd and $b_{n-k}=0$ which implies that $a_kb_{n-k}=0$;



      ii) if $k$ is even then $a_{k}=0$ which implies that $a_kb_{n-k}=0$.



      Hence $c_n=sum_{k=0}^n 0=0$.



      By using the same argument we can show that the product of an "even" series with and an "odd" series is an "odd" series.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Note that
        $$c_n=sum_{k=0}^n a_kb_{n-k}$$
        where $a_k$ is the coefficient of $x^k$ of the series of $sin(x)$ and $b_{n-k}$ is the coefficient of $x^{n-k}$ of the series of $cos(x)$. Now if $n$ is even then



        i) if $k$ is odd then $n-k$ is odd and $b_{n-k}=0$ which implies that $a_kb_{n-k}=0$;



        ii) if $k$ is even then $a_{k}=0$ which implies that $a_kb_{n-k}=0$.



        Hence $c_n=sum_{k=0}^n 0=0$.



        By using the same argument we can show that the product of an "even" series with and an "odd" series is an "odd" series.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Note that
          $$c_n=sum_{k=0}^n a_kb_{n-k}$$
          where $a_k$ is the coefficient of $x^k$ of the series of $sin(x)$ and $b_{n-k}$ is the coefficient of $x^{n-k}$ of the series of $cos(x)$. Now if $n$ is even then



          i) if $k$ is odd then $n-k$ is odd and $b_{n-k}=0$ which implies that $a_kb_{n-k}=0$;



          ii) if $k$ is even then $a_{k}=0$ which implies that $a_kb_{n-k}=0$.



          Hence $c_n=sum_{k=0}^n 0=0$.



          By using the same argument we can show that the product of an "even" series with and an "odd" series is an "odd" series.






          share|cite|improve this answer











          $endgroup$



          Note that
          $$c_n=sum_{k=0}^n a_kb_{n-k}$$
          where $a_k$ is the coefficient of $x^k$ of the series of $sin(x)$ and $b_{n-k}$ is the coefficient of $x^{n-k}$ of the series of $cos(x)$. Now if $n$ is even then



          i) if $k$ is odd then $n-k$ is odd and $b_{n-k}=0$ which implies that $a_kb_{n-k}=0$;



          ii) if $k$ is even then $a_{k}=0$ which implies that $a_kb_{n-k}=0$.



          Hence $c_n=sum_{k=0}^n 0=0$.



          By using the same argument we can show that the product of an "even" series with and an "odd" series is an "odd" series.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 15:05

























          answered Dec 17 '18 at 15:00









          Robert ZRobert Z

          98.3k1067139




          98.3k1067139























              2












              $begingroup$

              Since$$sin x=0+x+0times x^2-frac1{3!}x^3+cdots$$and$$cos x=1+0times x-frac1{2!}x^2+0times x^3+cdots$$when you compute a term of even order of the Cauchy product of these series, you get a sum with zeros everywhere; for instance, the coeffiecient of $x^2$ is$$0timesleft(-frac1{2!}right)+1times0+0times1=0.$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Since$$sin x=0+x+0times x^2-frac1{3!}x^3+cdots$$and$$cos x=1+0times x-frac1{2!}x^2+0times x^3+cdots$$when you compute a term of even order of the Cauchy product of these series, you get a sum with zeros everywhere; for instance, the coeffiecient of $x^2$ is$$0timesleft(-frac1{2!}right)+1times0+0times1=0.$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Since$$sin x=0+x+0times x^2-frac1{3!}x^3+cdots$$and$$cos x=1+0times x-frac1{2!}x^2+0times x^3+cdots$$when you compute a term of even order of the Cauchy product of these series, you get a sum with zeros everywhere; for instance, the coeffiecient of $x^2$ is$$0timesleft(-frac1{2!}right)+1times0+0times1=0.$$






                  share|cite|improve this answer









                  $endgroup$



                  Since$$sin x=0+x+0times x^2-frac1{3!}x^3+cdots$$and$$cos x=1+0times x-frac1{2!}x^2+0times x^3+cdots$$when you compute a term of even order of the Cauchy product of these series, you get a sum with zeros everywhere; for instance, the coeffiecient of $x^2$ is$$0timesleft(-frac1{2!}right)+1times0+0times1=0.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 17 '18 at 15:03









                  José Carlos SantosJosé Carlos Santos

                  162k22130233




                  162k22130233






























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