Calculation of Complex Trigonometric Summation












2












$begingroup$



Evaluation of $$sum^{n}_{k=1}frac{tan(x/2^k)}{2^{k-1}cdot cos(x/2^{k-1})}.$$




Try:Let $$S=sum^{n}_{k=1}frac{sin(x/2^k)}{2^{k-1}cos(x/2^{k-1})cdot cos(x/2^k)}$$



$$S=sum^{n}_{k=1}frac{sinbigg(frac{x}{2^{k-1}}-frac{x}{2^k}bigg)}{2^{k-1}cos(x/2^{k-1})cdot cos(x/2^k)}$$



So $$S =sum^{n}_{k=1}bigg[frac{1}{2^{k-1}}tan(x/2^{k-1})-frac{1}{2^{k-1}}tan(x/2^k)bigg]$$



But is is not in Telescopic sum.



I did not understand how to find that sum.



Could some help me how to sole it, Thanks






trigonometric-series







share|cite|improve this question









$endgroup$

















    2












    $begingroup$



    Evaluation of $$sum^{n}_{k=1}frac{tan(x/2^k)}{2^{k-1}cdot cos(x/2^{k-1})}.$$




    Try:Let $$S=sum^{n}_{k=1}frac{sin(x/2^k)}{2^{k-1}cos(x/2^{k-1})cdot cos(x/2^k)}$$



    $$S=sum^{n}_{k=1}frac{sinbigg(frac{x}{2^{k-1}}-frac{x}{2^k}bigg)}{2^{k-1}cos(x/2^{k-1})cdot cos(x/2^k)}$$



    So $$S =sum^{n}_{k=1}bigg[frac{1}{2^{k-1}}tan(x/2^{k-1})-frac{1}{2^{k-1}}tan(x/2^k)bigg]$$



    But is is not in Telescopic sum.



    I did not understand how to find that sum.



    Could some help me how to sole it, Thanks






    trigonometric-series







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      2



      $begingroup$



      Evaluation of $$sum^{n}_{k=1}frac{tan(x/2^k)}{2^{k-1}cdot cos(x/2^{k-1})}.$$




      Try:Let $$S=sum^{n}_{k=1}frac{sin(x/2^k)}{2^{k-1}cos(x/2^{k-1})cdot cos(x/2^k)}$$



      $$S=sum^{n}_{k=1}frac{sinbigg(frac{x}{2^{k-1}}-frac{x}{2^k}bigg)}{2^{k-1}cos(x/2^{k-1})cdot cos(x/2^k)}$$



      So $$S =sum^{n}_{k=1}bigg[frac{1}{2^{k-1}}tan(x/2^{k-1})-frac{1}{2^{k-1}}tan(x/2^k)bigg]$$



      But is is not in Telescopic sum.



      I did not understand how to find that sum.



      Could some help me how to sole it, Thanks






      trigonometric-series







      share|cite|improve this question









      $endgroup$





      Evaluation of $$sum^{n}_{k=1}frac{tan(x/2^k)}{2^{k-1}cdot cos(x/2^{k-1})}.$$




      Try:Let $$S=sum^{n}_{k=1}frac{sin(x/2^k)}{2^{k-1}cos(x/2^{k-1})cdot cos(x/2^k)}$$



      $$S=sum^{n}_{k=1}frac{sinbigg(frac{x}{2^{k-1}}-frac{x}{2^k}bigg)}{2^{k-1}cos(x/2^{k-1})cdot cos(x/2^k)}$$



      So $$S =sum^{n}_{k=1}bigg[frac{1}{2^{k-1}}tan(x/2^{k-1})-frac{1}{2^{k-1}}tan(x/2^k)bigg]$$



      But is is not in Telescopic sum.



      I did not understand how to find that sum.



      Could some help me how to sole it, Thanks






      trigonometric-series




      trigonometric-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 11 '18 at 15:27









      DXTDXT

      5,7582731




      5,7582731






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          HINT: The trick is to find:



          $$f(x)=sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}}$$



          Introduce function:



          $$F(x)=int_0^x f(x) dx=int_0^x sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}} dx$$



          $$F(x)=sum_{k=1}^{n}int_0^x frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}} dx$$



          $$F(x)=-sum_{k=1}^{n}ln cosfrac{x}{2^{k-1}}$$



          $$F(x)=-lnprod_{k=0}^{n-1}cosfrac{x}{2^{k}}tag{1}$$



          On the other side it's easy to prove that:



          $$sin x = {2^n}sin frac{x}{{{2^n}}}prodlimits_{k = 1}^n {cos frac{x}{{{2^k}}}}$$



          $$prodlimits_{k = 1}^n {cos frac{x}{{{2^k}}}}=frac{sin x}{{2^n}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^n {cos frac{x}{{{2^k}}}}=frac{sin x cos x}{{2^n}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^n {cos frac{x}{{{2^k}}}}=frac{sin 2x}{{2^{n+1}}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^{n-1} {cos frac{x}{{{2^k}}}}=frac{sin 2x}{{2^{n}}sin frac{x}{{{2^{n-1}}}}}tag{2}$$



          Now replace (2) into (1) and you get:



          $$F(x)=-lnfrac{sin 2x}{{2^{n}}sin frac{x}{{{2^{n-1}}}}}$$



          The first sum in your final expression is the first derivative:



          $$f(x)=F'(x)=frac{1}{2^{n-1}}cotfrac{x}{2^{n-1}}-2cot2x$$



          ...and you can tackle the second sum by using the result for the first one.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            We can use $$cot y-tan y=2cot2yifftan y=cot y-2cot2y$$ to find $$sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}}$$ But here we need something else, right?
            $endgroup$
            – lab bhattacharjee
            Dec 12 '18 at 7:10



















          0












          $begingroup$

          Thanks Friends Got it.



          $$sum^{n}_{k=1}frac{tan(x/2^k)}{2^{k-1} cos(x/2^{k-1})} = sum^{n}_{k=1}frac{sin^2(x/2^k)}{2^{k-1}sin (x/2^k)cos(x/2^k)cos(x/2^{k-1})}$$



          $$=2sum^{n}_{k=1}frac{1-cos(x/2^{k-1})}{2^{k-1}sin(x/2^{k-1})cos(x/2^{k-1})}$$
          $$= 2sum^{n}_{k=1}bigg[frac{1}{2^{k-2}sin(x/2^{k-2})}-frac{1}{2^{k-1}sin(x/2^{k-1})}bigg]$$



          $$ = 2sum^{n}_{k=1}bigg[frac{2}{sin x}-frac{2}{2^n sin (2x/2^n)}bigg].$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            math.stackexchange.com/questions/2406433/…
            $endgroup$
            – DXT
            Dec 16 '18 at 7:37











          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%2f3035402%2fcalculation-of-complex-trigonometric-summation%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












          $begingroup$

          HINT: The trick is to find:



          $$f(x)=sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}}$$



          Introduce function:



          $$F(x)=int_0^x f(x) dx=int_0^x sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}} dx$$



          $$F(x)=sum_{k=1}^{n}int_0^x frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}} dx$$



          $$F(x)=-sum_{k=1}^{n}ln cosfrac{x}{2^{k-1}}$$



          $$F(x)=-lnprod_{k=0}^{n-1}cosfrac{x}{2^{k}}tag{1}$$



          On the other side it's easy to prove that:



          $$sin x = {2^n}sin frac{x}{{{2^n}}}prodlimits_{k = 1}^n {cos frac{x}{{{2^k}}}}$$



          $$prodlimits_{k = 1}^n {cos frac{x}{{{2^k}}}}=frac{sin x}{{2^n}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^n {cos frac{x}{{{2^k}}}}=frac{sin x cos x}{{2^n}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^n {cos frac{x}{{{2^k}}}}=frac{sin 2x}{{2^{n+1}}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^{n-1} {cos frac{x}{{{2^k}}}}=frac{sin 2x}{{2^{n}}sin frac{x}{{{2^{n-1}}}}}tag{2}$$



          Now replace (2) into (1) and you get:



          $$F(x)=-lnfrac{sin 2x}{{2^{n}}sin frac{x}{{{2^{n-1}}}}}$$



          The first sum in your final expression is the first derivative:



          $$f(x)=F'(x)=frac{1}{2^{n-1}}cotfrac{x}{2^{n-1}}-2cot2x$$



          ...and you can tackle the second sum by using the result for the first one.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            We can use $$cot y-tan y=2cot2yifftan y=cot y-2cot2y$$ to find $$sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}}$$ But here we need something else, right?
            $endgroup$
            – lab bhattacharjee
            Dec 12 '18 at 7:10
















          3












          $begingroup$

          HINT: The trick is to find:



          $$f(x)=sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}}$$



          Introduce function:



          $$F(x)=int_0^x f(x) dx=int_0^x sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}} dx$$



          $$F(x)=sum_{k=1}^{n}int_0^x frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}} dx$$



          $$F(x)=-sum_{k=1}^{n}ln cosfrac{x}{2^{k-1}}$$



          $$F(x)=-lnprod_{k=0}^{n-1}cosfrac{x}{2^{k}}tag{1}$$



          On the other side it's easy to prove that:



          $$sin x = {2^n}sin frac{x}{{{2^n}}}prodlimits_{k = 1}^n {cos frac{x}{{{2^k}}}}$$



          $$prodlimits_{k = 1}^n {cos frac{x}{{{2^k}}}}=frac{sin x}{{2^n}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^n {cos frac{x}{{{2^k}}}}=frac{sin x cos x}{{2^n}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^n {cos frac{x}{{{2^k}}}}=frac{sin 2x}{{2^{n+1}}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^{n-1} {cos frac{x}{{{2^k}}}}=frac{sin 2x}{{2^{n}}sin frac{x}{{{2^{n-1}}}}}tag{2}$$



          Now replace (2) into (1) and you get:



          $$F(x)=-lnfrac{sin 2x}{{2^{n}}sin frac{x}{{{2^{n-1}}}}}$$



          The first sum in your final expression is the first derivative:



          $$f(x)=F'(x)=frac{1}{2^{n-1}}cotfrac{x}{2^{n-1}}-2cot2x$$



          ...and you can tackle the second sum by using the result for the first one.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            We can use $$cot y-tan y=2cot2yifftan y=cot y-2cot2y$$ to find $$sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}}$$ But here we need something else, right?
            $endgroup$
            – lab bhattacharjee
            Dec 12 '18 at 7:10














          3












          3








          3





          $begingroup$

          HINT: The trick is to find:



          $$f(x)=sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}}$$



          Introduce function:



          $$F(x)=int_0^x f(x) dx=int_0^x sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}} dx$$



          $$F(x)=sum_{k=1}^{n}int_0^x frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}} dx$$



          $$F(x)=-sum_{k=1}^{n}ln cosfrac{x}{2^{k-1}}$$



          $$F(x)=-lnprod_{k=0}^{n-1}cosfrac{x}{2^{k}}tag{1}$$



          On the other side it's easy to prove that:



          $$sin x = {2^n}sin frac{x}{{{2^n}}}prodlimits_{k = 1}^n {cos frac{x}{{{2^k}}}}$$



          $$prodlimits_{k = 1}^n {cos frac{x}{{{2^k}}}}=frac{sin x}{{2^n}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^n {cos frac{x}{{{2^k}}}}=frac{sin x cos x}{{2^n}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^n {cos frac{x}{{{2^k}}}}=frac{sin 2x}{{2^{n+1}}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^{n-1} {cos frac{x}{{{2^k}}}}=frac{sin 2x}{{2^{n}}sin frac{x}{{{2^{n-1}}}}}tag{2}$$



          Now replace (2) into (1) and you get:



          $$F(x)=-lnfrac{sin 2x}{{2^{n}}sin frac{x}{{{2^{n-1}}}}}$$



          The first sum in your final expression is the first derivative:



          $$f(x)=F'(x)=frac{1}{2^{n-1}}cotfrac{x}{2^{n-1}}-2cot2x$$



          ...and you can tackle the second sum by using the result for the first one.






          share|cite|improve this answer











          $endgroup$



          HINT: The trick is to find:



          $$f(x)=sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}}$$



          Introduce function:



          $$F(x)=int_0^x f(x) dx=int_0^x sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}} dx$$



          $$F(x)=sum_{k=1}^{n}int_0^x frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}} dx$$



          $$F(x)=-sum_{k=1}^{n}ln cosfrac{x}{2^{k-1}}$$



          $$F(x)=-lnprod_{k=0}^{n-1}cosfrac{x}{2^{k}}tag{1}$$



          On the other side it's easy to prove that:



          $$sin x = {2^n}sin frac{x}{{{2^n}}}prodlimits_{k = 1}^n {cos frac{x}{{{2^k}}}}$$



          $$prodlimits_{k = 1}^n {cos frac{x}{{{2^k}}}}=frac{sin x}{{2^n}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^n {cos frac{x}{{{2^k}}}}=frac{sin x cos x}{{2^n}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^n {cos frac{x}{{{2^k}}}}=frac{sin 2x}{{2^{n+1}}sin frac{x}{{{2^n}}}}$$



          $$prodlimits_{k = 0}^{n-1} {cos frac{x}{{{2^k}}}}=frac{sin 2x}{{2^{n}}sin frac{x}{{{2^{n-1}}}}}tag{2}$$



          Now replace (2) into (1) and you get:



          $$F(x)=-lnfrac{sin 2x}{{2^{n}}sin frac{x}{{{2^{n-1}}}}}$$



          The first sum in your final expression is the first derivative:



          $$f(x)=F'(x)=frac{1}{2^{n-1}}cotfrac{x}{2^{n-1}}-2cot2x$$



          ...and you can tackle the second sum by using the result for the first one.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 17:07

























          answered Dec 11 '18 at 16:50









          OldboyOldboy

          8,4521936




          8,4521936








          • 1




            $begingroup$
            We can use $$cot y-tan y=2cot2yifftan y=cot y-2cot2y$$ to find $$sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}}$$ But here we need something else, right?
            $endgroup$
            – lab bhattacharjee
            Dec 12 '18 at 7:10














          • 1




            $begingroup$
            We can use $$cot y-tan y=2cot2yifftan y=cot y-2cot2y$$ to find $$sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}}$$ But here we need something else, right?
            $endgroup$
            – lab bhattacharjee
            Dec 12 '18 at 7:10








          1




          1




          $begingroup$
          We can use $$cot y-tan y=2cot2yifftan y=cot y-2cot2y$$ to find $$sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}}$$ But here we need something else, right?
          $endgroup$
          – lab bhattacharjee
          Dec 12 '18 at 7:10




          $begingroup$
          We can use $$cot y-tan y=2cot2yifftan y=cot y-2cot2y$$ to find $$sum_{k=1}^{n}frac{1}{2^{k-1}}{tanfrac{x}{2^{k-1}}}$$ But here we need something else, right?
          $endgroup$
          – lab bhattacharjee
          Dec 12 '18 at 7:10











          0












          $begingroup$

          Thanks Friends Got it.



          $$sum^{n}_{k=1}frac{tan(x/2^k)}{2^{k-1} cos(x/2^{k-1})} = sum^{n}_{k=1}frac{sin^2(x/2^k)}{2^{k-1}sin (x/2^k)cos(x/2^k)cos(x/2^{k-1})}$$



          $$=2sum^{n}_{k=1}frac{1-cos(x/2^{k-1})}{2^{k-1}sin(x/2^{k-1})cos(x/2^{k-1})}$$
          $$= 2sum^{n}_{k=1}bigg[frac{1}{2^{k-2}sin(x/2^{k-2})}-frac{1}{2^{k-1}sin(x/2^{k-1})}bigg]$$



          $$ = 2sum^{n}_{k=1}bigg[frac{2}{sin x}-frac{2}{2^n sin (2x/2^n)}bigg].$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            math.stackexchange.com/questions/2406433/…
            $endgroup$
            – DXT
            Dec 16 '18 at 7:37
















          0












          $begingroup$

          Thanks Friends Got it.



          $$sum^{n}_{k=1}frac{tan(x/2^k)}{2^{k-1} cos(x/2^{k-1})} = sum^{n}_{k=1}frac{sin^2(x/2^k)}{2^{k-1}sin (x/2^k)cos(x/2^k)cos(x/2^{k-1})}$$



          $$=2sum^{n}_{k=1}frac{1-cos(x/2^{k-1})}{2^{k-1}sin(x/2^{k-1})cos(x/2^{k-1})}$$
          $$= 2sum^{n}_{k=1}bigg[frac{1}{2^{k-2}sin(x/2^{k-2})}-frac{1}{2^{k-1}sin(x/2^{k-1})}bigg]$$



          $$ = 2sum^{n}_{k=1}bigg[frac{2}{sin x}-frac{2}{2^n sin (2x/2^n)}bigg].$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            math.stackexchange.com/questions/2406433/…
            $endgroup$
            – DXT
            Dec 16 '18 at 7:37














          0












          0








          0





          $begingroup$

          Thanks Friends Got it.



          $$sum^{n}_{k=1}frac{tan(x/2^k)}{2^{k-1} cos(x/2^{k-1})} = sum^{n}_{k=1}frac{sin^2(x/2^k)}{2^{k-1}sin (x/2^k)cos(x/2^k)cos(x/2^{k-1})}$$



          $$=2sum^{n}_{k=1}frac{1-cos(x/2^{k-1})}{2^{k-1}sin(x/2^{k-1})cos(x/2^{k-1})}$$
          $$= 2sum^{n}_{k=1}bigg[frac{1}{2^{k-2}sin(x/2^{k-2})}-frac{1}{2^{k-1}sin(x/2^{k-1})}bigg]$$



          $$ = 2sum^{n}_{k=1}bigg[frac{2}{sin x}-frac{2}{2^n sin (2x/2^n)}bigg].$$






          share|cite|improve this answer









          $endgroup$



          Thanks Friends Got it.



          $$sum^{n}_{k=1}frac{tan(x/2^k)}{2^{k-1} cos(x/2^{k-1})} = sum^{n}_{k=1}frac{sin^2(x/2^k)}{2^{k-1}sin (x/2^k)cos(x/2^k)cos(x/2^{k-1})}$$



          $$=2sum^{n}_{k=1}frac{1-cos(x/2^{k-1})}{2^{k-1}sin(x/2^{k-1})cos(x/2^{k-1})}$$
          $$= 2sum^{n}_{k=1}bigg[frac{1}{2^{k-2}sin(x/2^{k-2})}-frac{1}{2^{k-1}sin(x/2^{k-1})}bigg]$$



          $$ = 2sum^{n}_{k=1}bigg[frac{2}{sin x}-frac{2}{2^n sin (2x/2^n)}bigg].$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 13:00









          DXTDXT

          5,7582731




          5,7582731












          • $begingroup$
            math.stackexchange.com/questions/2406433/…
            $endgroup$
            – DXT
            Dec 16 '18 at 7:37


















          • $begingroup$
            math.stackexchange.com/questions/2406433/…
            $endgroup$
            – DXT
            Dec 16 '18 at 7:37
















          $begingroup$
          math.stackexchange.com/questions/2406433/…
          $endgroup$
          – DXT
          Dec 16 '18 at 7:37




          $begingroup$
          math.stackexchange.com/questions/2406433/…
          $endgroup$
          – DXT
          Dec 16 '18 at 7:37


















          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%2f3035402%2fcalculation-of-complex-trigonometric-summation%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

          Plaza Victoria

          Image
          Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
          Read more

          Brian Clough

          Image
          Brian Clough Datos personales Nombre completo Brian Howard Clough Apodo(s) Old Big 'Ead, Cloughie [ 1 ] ​ Nacimiento Middlesbrough, Inglaterra, Reino Unido 21 de marzo de 1935 Nacionalidad(es) Fallecimiento Derby, Reino Unido 20 de septiembre de 2004 (69 años) Altura 1,78 [ 1 ] ​ metros Carrera Deporte Fútbol Debut deportivo 1955 (Como jugador) 1965 (Como entrenador) (Middlesbrough F. C. (Como jugador) Hartlepool United F. C. (Como entrenador) ) Posición Delantero Goles en clubes 269 [ 2 ] ​ Retirada deportiva 1964 (Como jugador) 1993 (Como entrenador) (Sunderland A. F. C. (Como jugador) Nottingham Forest (Como entrenador) ) Carrera internacional Part. 2 [editar datos en Wikidata] Brian Howard Clough , OBE [ 3 ] ​ (Middlesbrough, Inglaterra, Reino Unido, 21 de marzo de 1935 – Derby, Inglaterra, Reino Unido, 20 de septiembre de 2004) fue un jugador y entrenador británico de fútbol. Es conocido por...
          Read more

          Cáceres

          Image
          Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
          Read more