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

Source for Cremation Specifically Not Jewish

Called into a meeting and told we are being made redundant (laid off) and "not to share outside". Can I tell my partner?

Six real numbers so that product of any five is the sixth one

Calculating Hyperbolic Sin faster than using a standard power series

How to make a *empty* field behaves like a *null* field when it comes to standard values?

Dystopian novel where telepathic humans live under a dome

Are small insurances worth it

Misplaced tyre lever - alternatives?

Why do members of Congress in committee hearings ask witnesses the same question multiple times?

What could trigger powerful quakes on icy world?

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

Wrap all numerics in JSON with quotes

School performs periodic password audits. Is my password compromised?

Are paired adjectives bad style?

The need of reserving one's ability in job interviews

Should we avoid writing fiction about historical events without extensive research?

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

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

Citing contemporaneous (interlaced?) preprints

Get length of the longest sequence of numbers with the same sign

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

How do you say "powers of ten"?

Inverse of the covariance matrix of a multivariate normal distribution

I can't die. Who am I?



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 3 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|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
              });


              }
              });






              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|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 22 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

                              Щит и меч (фильм) Содержание Названия серий | Сюжет |...

                              Image
                              Фильмы по алфавитуФильмы 1968 годаФильмы СССР 1968 годаЧёрно-белые фильмы СССРФильмы студии «Мосфильм»Лидеры советского кинопрокатаФильмы-драмы СССРФильмы о Второй мировой войнеФильмы о шпионахЭкранизации произведений Вадима КожевниковаФильмы Владимира БасоваФильмы о фашизме 1968 годуодноимённому романуВадима КожевниковаВладимиром БасовымСтанислав ЛюбшинГерманиюАбверегауптштурмфюрераССБерлинслужбу СДSchwarzbraun ist die HaselnussXVIII векаЭрикиХермса НиляГитлерюгендаАбвераАх, мой милый АвгустинXVII векаHorst WesselDeutschland über alles У этого термина существуют и другие значения, см. Щит и меч. Щит и меч Жанр драма военный фильм Режиссёр Владимир Басов Автор сценария Владимир Басов Вадим Кожевников В главных ролях Станислав Любшин Валентина Титова Олег Янковский Георгий Мартынюк Владимир Басов Оператор Тимофей Лебешев Композитор Вениамин Баснер Кинокомпания Киностудия «Мосфильм». Творческое объединение писателей и киноработников ...
                              Read more

                              Венесуэла на летних Олимпийских играх 2000 Содержание Состав...

                              Image
                              БангладешБахрейнБутанБрунейВьетнамГонконгИндияИндонезияИорданияИракИранЙеменКазахстанКамбоджаКатарКитайКиргизияКНДРКувейтЛаосЛиванМалайзияМальдивыМонголияМьянмаНепалОАЭОман ПакистанГосударство ПалестинаСаудовская АравияСингапурСирияТаджикистанТаиландТайваньТуркменияУзбекистанФилиппиныШри-ЛанкаЮжная КореяЯпонияИндивидуальные олимпийские спортсменыАлжирАнголаБенинБотсванаБуркина-ФасоБурундиГабонГамбияГанаГвинеяГвинея-БисауДемократическая Республика КонгоДжибутиЕгипетЗамбияЗимбабвеКабо-ВердеКамерунКенияКоморские островаКот-д’ИвуарЛесотоЛиберияЛивияМаврикийМавританияМадагаскарМалавиМалиМароккоМозамбикНамибияНигерНигерияРеспублика КонгоРуандаСан-Томе и ПринсипиСвазилендСейшелыСенегалСомалиСуданСьерра-ЛеонеТанзанияТогоТунисУгандаЦентральноафриканская РеспубликаЧадЭкваториальная ГвинеяЭритреяЭфиопияЮАРАвстралияАмериканское СамоаВануатуГуамМикронезияНауруНовая ЗеландияОстрова КукаПалауПапуа — Новая ГвинеяСамоаСоломоновы ОстроваТонгаФиджи Венесуэла на Олимпийских играх2000 год в ВенесуэлеСтран...
                              Read more

                              Meter-Bus Содержание Параметры шины | Стандартизация |...

                              Image
                              RS-232IrDAHARTмодемтоковая петляИРПСMIDIIO-LinkRS-422RS-423RS-485LINM-BusKansas City StandardКомпакт-кассетагрампластинкаPPPSLIPIrDAHARTISO 7816MIDIIO-LinkModbusLINDMX-512P-NETProfibusIEC 608708250 UART16550 UART Сетевые протоколыПоследовательные шины физического уровняполевой шиныасинхронного интерфейсаOpen Metering SystemAESздесь У этого термина существуют и другие значения, см. M-bus (значения). M-Bus (Meter-Bus) — стандарт физического уровня для полевой шины на основе асинхронного интерфейса. Также под этим названием понимают коммуникационный протокол, используемый для связи устройств по этой шине. M-bus преимущественно применяется для приборов учета электрической энергии (электросчётчики), тепловой энергии (теплосчётчики), расходомеров воды и газа. Данные передаются на компьютерную станцию (сервер) напрямую или через концентраторы шины M-Bus, а также усилители-повторители сигнала. Содержание 1 Параметры шины 2 Стандартизация 2.1 Open Metering System ...
                              Read more