Calculating Hyperbolic Sin faster than using a standard power series












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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.






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    $begingroup$


    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.







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      4












      4








      4





      $begingroup$


      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.
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      edited 3 hours ago









      MPW

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      asked 4 hours ago









      Bill BollingerBill Bollinger

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          $begingroup$

          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|improve this answer









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              2 Answers
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              2 Answers
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              7












              $begingroup$

              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









              $endgroup$


















                7












                $begingroup$

                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









                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  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 3 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|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|improve this answer









                        $endgroup$
















                          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|improve this answer









                          $endgroup$



                          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|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 1 hour ago









                          Claude LeiboviciClaude Leibovici

                          123k1157135




                          123k1157135






















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