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












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












$begingroup$


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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044043%2fproving-sin2x-2-sinx-cosx-using-cauchy-product%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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

          97.4k1066137




          97.4k1066137























              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

                  161k22127232




                  161k22127232






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044043%2fproving-sin2x-2-sinx-cosx-using-cauchy-product%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Quarter-circle Tiles

                      Image
                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
                      Read more

                      build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

                      Image
                      0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
                      Read more

                      Mont Emei

                      Image
                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
                      Read more