Calculating Hyperbolic Sin faster than using a standard power seriesIs there a way to get trig functions...

Skis versus snow shoes - when to choose which for travelling the backcountry?

A bug in Excel? Conditional formatting for marking duplicates also highlights unique value

Where is the line between being obedient and getting bullied by a boss?

I can't die. Who am I?

What are all the squawk codes?

Roots of 6th chords on the guitar for different inversions/voicings

Is there a frame of reference in which I was born before I was conceived?

What happened to QGIS 2.x LTR?

Levi-Civita symbol: 3D matrix

Where is the fallacy here?

Is there any relevance to Thor getting his hair cut other than comedic value?

Reason why dimensional travelling would be restricted

Toast materialize

What is a term for a function that when called repeatedly, has the same effect as calling once?

If a set is open, does that imply that it has no boundary points?

Can I become debt free or should I file for bankruptcy? How do I manage my debt and finances?

How to evaluate the limit where something is raised to a power of x?

How can I be pwned if I'm not registered on the compromised site?

Rationale to prefer local variables over instance variables?

What is better: yes / no radio, or simple checkbox?

Why can't we make a perpetual motion machine by using a magnet to pull up a piece of metal, then letting it fall back down?

Giving a talk in my old university, how prominently should I tell students my salary?

Citing contemporaneous (interlaced?) preprints

How to substitute values from a list into a function?



Calculating Hyperbolic Sin faster than using a standard power series


Is there a way to get trig functions without a calculator?Broken Calculator: only certain unary functions work.How do we define $sin(theta)$ or $cos(theta)$Help with Trigonometric FunctionsHow would Taylor Series work?Understanding power series and their representation of functionsSeries Convergence of $sumlimits_{n=1}^{infty} frac{cos(npi/3)}{n!}$ by Ratio TestOrthogonal Function Dirac Delta SeriesHow do calculators calculate the value of trigonometric functions?How to calculate a “rational” sine/cosine to specific digits of precision?Calculating trig functions faster than with standard power series













3












$begingroup$


Using $$ sinh x = x + tfrac{x^3}{3!}+ tfrac{x^5}{5!} + tfrac{x^7}{7!}+ cdots$$ as the Standard Power Series. This series takes a very long time to run. Can it be written without using the exponentials divided by a huge factorial. The example functions in Is there a way to get trig functions without a calculator? using the "Tailored Taylor" series representation for sin and cosine are very fast and give the same answers. I want to use it within my calculator program.



Thank you very much.






sequences-and-series trigonometry







share|cite|improve this question









New contributor




Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$

















    3












    $begingroup$


    Using $$ sinh x = x + tfrac{x^3}{3!}+ tfrac{x^5}{5!} + tfrac{x^7}{7!}+ cdots$$ as the Standard Power Series. This series takes a very long time to run. Can it be written without using the exponentials divided by a huge factorial. The example functions in Is there a way to get trig functions without a calculator? using the "Tailored Taylor" series representation for sin and cosine are very fast and give the same answers. I want to use it within my calculator program.



    Thank you very much.






    sequences-and-series trigonometry







    share|cite|improve this question









    New contributor




    Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      3












      3








      3





      $begingroup$


      Using $$ sinh x = x + tfrac{x^3}{3!}+ tfrac{x^5}{5!} + tfrac{x^7}{7!}+ cdots$$ as the Standard Power Series. This series takes a very long time to run. Can it be written without using the exponentials divided by a huge factorial. The example functions in Is there a way to get trig functions without a calculator? using the "Tailored Taylor" series representation for sin and cosine are very fast and give the same answers. I want to use it within my calculator program.



      Thank you very much.






      sequences-and-series trigonometry







      share|cite|improve this question









      New contributor




      Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Using $$ sinh x = x + tfrac{x^3}{3!}+ tfrac{x^5}{5!} + tfrac{x^7}{7!}+ cdots$$ as the Standard Power Series. This series takes a very long time to run. Can it be written without using the exponentials divided by a huge factorial. The example functions in Is there a way to get trig functions without a calculator? using the "Tailored Taylor" series representation for sin and cosine are very fast and give the same answers. I want to use it within my calculator program.



      Thank you very much.






      sequences-and-series trigonometry




      sequences-and-series trigonometry






      share|cite|improve this question









      New contributor




      Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      MPW

      30.5k12157




      30.5k12157






      New contributor




      Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 3 hours ago









      Bill BollingerBill Bollinger

      312




      312




      New contributor




      Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          Note that $$sinh x=frac{e^x-e^{-x}}2$$
          So all you need is a fast way to calculate the exponential $e^x$. You can use the regular Taylor series, but that's slow. So you can use the definition $$e^x=lim_{ntoinfty}left(1+frac xnright)^n$$
          For calculation purposes, use $n$ as a power of $2$, $n=2^k$. You calculate first $y=1+frac x{2^k}$, then you repeat the $y=ycdot y$ operation $k$ times. I've got the idea about calculating the fast exponential from this article.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Let me consider the problem from a computing point of view assumin that you do not know how to compute $e^x$.



            The infinite series is
            $$sinh(x)=sum_{n=0}^infty frac{x^{2n+1}}{(2n+1)!}$$ If you compute each term independently of the other, for sure, it is expensive since you have to compute each power of $x$ as well as each factorial.



            But suppose that you write instead
            $$sinh(x)=sum_{n=0}^infty T_n qquad text{where} qquad T_n=frac{x^{2n+1}}{(2n+1)!}qquad text{and} qquad T_0=x$$ then
            $$T_{n+1}= frac {t,, T_n}{(2n+2)(2n+3)}qquad text{where} qquad t=x^2$$ This would be much less expensive in terms of basic operations.



            You could use the same trick for most functions expressed as infinite series.






            share|cite









            $endgroup$













              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
              });


              }
              });






              Bill Bollinger is a new contributor. Be nice, and check out our Code of Conduct.










              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3137004%2fcalculating-hyperbolic-sin-faster-than-using-a-standard-power-series%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









              6












              $begingroup$

              Note that $$sinh x=frac{e^x-e^{-x}}2$$
              So all you need is a fast way to calculate the exponential $e^x$. You can use the regular Taylor series, but that's slow. So you can use the definition $$e^x=lim_{ntoinfty}left(1+frac xnright)^n$$
              For calculation purposes, use $n$ as a power of $2$, $n=2^k$. You calculate first $y=1+frac x{2^k}$, then you repeat the $y=ycdot y$ operation $k$ times. I've got the idea about calculating the fast exponential from this article.






              share|cite|improve this answer









              $endgroup$


















                6












                $begingroup$

                Note that $$sinh x=frac{e^x-e^{-x}}2$$
                So all you need is a fast way to calculate the exponential $e^x$. You can use the regular Taylor series, but that's slow. So you can use the definition $$e^x=lim_{ntoinfty}left(1+frac xnright)^n$$
                For calculation purposes, use $n$ as a power of $2$, $n=2^k$. You calculate first $y=1+frac x{2^k}$, then you repeat the $y=ycdot y$ operation $k$ times. I've got the idea about calculating the fast exponential from this article.






                share|cite|improve this answer









                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  Note that $$sinh x=frac{e^x-e^{-x}}2$$
                  So all you need is a fast way to calculate the exponential $e^x$. You can use the regular Taylor series, but that's slow. So you can use the definition $$e^x=lim_{ntoinfty}left(1+frac xnright)^n$$
                  For calculation purposes, use $n$ as a power of $2$, $n=2^k$. You calculate first $y=1+frac x{2^k}$, then you repeat the $y=ycdot y$ operation $k$ times. I've got the idea about calculating the fast exponential from this article.






                  share|cite|improve this answer









                  $endgroup$



                  Note that $$sinh x=frac{e^x-e^{-x}}2$$
                  So all you need is a fast way to calculate the exponential $e^x$. You can use the regular Taylor series, but that's slow. So you can use the definition $$e^x=lim_{ntoinfty}left(1+frac xnright)^n$$
                  For calculation purposes, use $n$ as a power of $2$, $n=2^k$. You calculate first $y=1+frac x{2^k}$, then you repeat the $y=ycdot y$ operation $k$ times. I've got the idea about calculating the fast exponential from this article.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  AndreiAndrei

                  12.7k21128




                  12.7k21128























                      0












                      $begingroup$

                      Let me consider the problem from a computing point of view assumin that you do not know how to compute $e^x$.



                      The infinite series is
                      $$sinh(x)=sum_{n=0}^infty frac{x^{2n+1}}{(2n+1)!}$$ If you compute each term independently of the other, for sure, it is expensive since you have to compute each power of $x$ as well as each factorial.



                      But suppose that you write instead
                      $$sinh(x)=sum_{n=0}^infty T_n qquad text{where} qquad T_n=frac{x^{2n+1}}{(2n+1)!}qquad text{and} qquad T_0=x$$ then
                      $$T_{n+1}= frac {t,, T_n}{(2n+2)(2n+3)}qquad text{where} qquad t=x^2$$ This would be much less expensive in terms of basic operations.



                      You could use the same trick for most functions expressed as infinite series.






                      share|cite









                      $endgroup$


















                        0












                        $begingroup$

                        Let me consider the problem from a computing point of view assumin that you do not know how to compute $e^x$.



                        The infinite series is
                        $$sinh(x)=sum_{n=0}^infty frac{x^{2n+1}}{(2n+1)!}$$ If you compute each term independently of the other, for sure, it is expensive since you have to compute each power of $x$ as well as each factorial.



                        But suppose that you write instead
                        $$sinh(x)=sum_{n=0}^infty T_n qquad text{where} qquad T_n=frac{x^{2n+1}}{(2n+1)!}qquad text{and} qquad T_0=x$$ then
                        $$T_{n+1}= frac {t,, T_n}{(2n+2)(2n+3)}qquad text{where} qquad t=x^2$$ This would be much less expensive in terms of basic operations.



                        You could use the same trick for most functions expressed as infinite series.






                        share|cite









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Let me consider the problem from a computing point of view assumin that you do not know how to compute $e^x$.



                          The infinite series is
                          $$sinh(x)=sum_{n=0}^infty frac{x^{2n+1}}{(2n+1)!}$$ If you compute each term independently of the other, for sure, it is expensive since you have to compute each power of $x$ as well as each factorial.



                          But suppose that you write instead
                          $$sinh(x)=sum_{n=0}^infty T_n qquad text{where} qquad T_n=frac{x^{2n+1}}{(2n+1)!}qquad text{and} qquad T_0=x$$ then
                          $$T_{n+1}= frac {t,, T_n}{(2n+2)(2n+3)}qquad text{where} qquad t=x^2$$ This would be much less expensive in terms of basic operations.



                          You could use the same trick for most functions expressed as infinite series.






                          share|cite









                          $endgroup$



                          Let me consider the problem from a computing point of view assumin that you do not know how to compute $e^x$.



                          The infinite series is
                          $$sinh(x)=sum_{n=0}^infty frac{x^{2n+1}}{(2n+1)!}$$ If you compute each term independently of the other, for sure, it is expensive since you have to compute each power of $x$ as well as each factorial.



                          But suppose that you write instead
                          $$sinh(x)=sum_{n=0}^infty T_n qquad text{where} qquad T_n=frac{x^{2n+1}}{(2n+1)!}qquad text{and} qquad T_0=x$$ then
                          $$T_{n+1}= frac {t,, T_n}{(2n+2)(2n+3)}qquad text{where} qquad t=x^2$$ This would be much less expensive in terms of basic operations.



                          You could use the same trick for most functions expressed as infinite series.







                          share|cite












                          share|cite



                          share|cite










                          answered 7 mins ago









                          Claude LeiboviciClaude Leibovici

                          123k1157135




                          123k1157135






















                              Bill Bollinger is a new contributor. Be nice, and check out our Code of Conduct.










                              draft saved

                              draft discarded


















                              Bill Bollinger is a new contributor. Be nice, and check out our Code of Conduct.













                              Bill Bollinger is a new contributor. Be nice, and check out our Code of Conduct.












                              Bill Bollinger is a new contributor. Be nice, and check out our Code of Conduct.
















                              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%2f3137004%2fcalculating-hyperbolic-sin-faster-than-using-a-standard-power-series%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

                              Why do type traits not work with types in namespace scope?What are POD types in C++?Why can templates only be...

                              Image
                              What does it mean to express a gate in Dirac notation? A Strange Latex Symbol How can Republicans who favour free markets, consistently express anger when they don't like the outcome of that choice? What happened to Captain America in Endgame? How to solve constants out of the internal energy equation? How can I place the product on a social media post better? Will tsunami waves travel forever if there was no land? Unexpected email from Yorkshire Bank What is the relationship between spectral sequences and obstruction theory? How can I practically buy stocks? Will a top journal at least re
                              Read more

                              Will tsunami waves travel forever if there was no land?Why do tsunami waves begin with the water flowing away...

                              Image
                              How much cash can I safely carry into the USA and avoid civil forfeiture? Is there really no use for MD5 anymore? Is contacting this expert in the field something acceptable or would it be discourteous? How to pronounce 'C++' in Spanish How to creep the reader out with what seems like a normal person? Critique of timeline aesthetic Sci fi novel series with instant travel between planets through gates. A river runs through the gates How did Captain America manage to do this? Are Boeing 737-800’s grounded? Error message with tabularx How to make a pipeline wait for end-of-file or stop af
                              Read more

                              Simple Scan not detecting my scanner (Brother DCP-7055W)Brother MFC-L2700DW printer can print, can't...

                              Image
                              Can someone publish a story that happened to you? Will tsunami waves travel forever if there was no land? What is the strongest case that can be made in favour of the UK regaining some control over fishing policy after Brexit? Controversial area of mathematics Packing rectangles: Does rotation ever help? Is there a way to get a compiler for the original B programming language? Using a Lyapunov function to classify stability and sketching a phase portrait Do I have an "anti-research" personality? Reducing vertical space in stackrel What does KSP mean? Pulling the rope with one hand is
                              Read more