Can the graph of $tan^{-1}{left(frac{sin x}{x}right)}$ be expressed as $Ce^{-kx}cos(omega x + phi)$?












2














After graphing $sin x$, I thought of trying something interesting. I wanted to plot the angle $theta$ that a point $(x,sin x )$ makes with the origin on the $y$-axis, against $x$ on the $x$-axis.



$$tantheta = frac{sin x}{x}Rightarrow theta=tan^{-1}{left(frac{sin x}{x}right)}$$



Graphing $y = 20timestheta$ (multiply by 20 for graphical purposes):
Diagram 1



Part of it reminded me of the graph for the damped oscillator (specifically, the $x>0$ part).
That made me wonder if it was possible to find constants $C,k,omega,$ and $phi$ such that $$theta = Ce^{-kx}cos(omega x + phi)$$



However, after toying with Grapher for a while, $y = theta$ didn't seem to decrease exponentially.



That led me to this question: is there any analytical way to find real constants $C,k,omega,$ and $phi$ such that $theta = Ce^{-kx}cos(omega x + phi)$?



Furthermore, are there any complex constants $C,k,omega, text{and } phi$?






real-analysis complex-analysis trigonometry exponential-function graphing-functions







share|cite|improve this question
























  • No for real constants: the derivatives are quite different.
    – Bernard
    Apr 5 '17 at 8:54










  • Certainly not, there is a serious mismatch for negative $x$.
    – Yves Daoust
    Apr 5 '17 at 14:47


















2














After graphing $sin x$, I thought of trying something interesting. I wanted to plot the angle $theta$ that a point $(x,sin x )$ makes with the origin on the $y$-axis, against $x$ on the $x$-axis.



$$tantheta = frac{sin x}{x}Rightarrow theta=tan^{-1}{left(frac{sin x}{x}right)}$$



Graphing $y = 20timestheta$ (multiply by 20 for graphical purposes):
Diagram 1



Part of it reminded me of the graph for the damped oscillator (specifically, the $x>0$ part).
That made me wonder if it was possible to find constants $C,k,omega,$ and $phi$ such that $$theta = Ce^{-kx}cos(omega x + phi)$$



However, after toying with Grapher for a while, $y = theta$ didn't seem to decrease exponentially.



That led me to this question: is there any analytical way to find real constants $C,k,omega,$ and $phi$ such that $theta = Ce^{-kx}cos(omega x + phi)$?



Furthermore, are there any complex constants $C,k,omega, text{and } phi$?






real-analysis complex-analysis trigonometry exponential-function graphing-functions







share|cite|improve this question
























  • No for real constants: the derivatives are quite different.
    – Bernard
    Apr 5 '17 at 8:54










  • Certainly not, there is a serious mismatch for negative $x$.
    – Yves Daoust
    Apr 5 '17 at 14:47
















2












2








2


1





After graphing $sin x$, I thought of trying something interesting. I wanted to plot the angle $theta$ that a point $(x,sin x )$ makes with the origin on the $y$-axis, against $x$ on the $x$-axis.



$$tantheta = frac{sin x}{x}Rightarrow theta=tan^{-1}{left(frac{sin x}{x}right)}$$



Graphing $y = 20timestheta$ (multiply by 20 for graphical purposes):
Diagram 1



Part of it reminded me of the graph for the damped oscillator (specifically, the $x>0$ part).
That made me wonder if it was possible to find constants $C,k,omega,$ and $phi$ such that $$theta = Ce^{-kx}cos(omega x + phi)$$



However, after toying with Grapher for a while, $y = theta$ didn't seem to decrease exponentially.



That led me to this question: is there any analytical way to find real constants $C,k,omega,$ and $phi$ such that $theta = Ce^{-kx}cos(omega x + phi)$?



Furthermore, are there any complex constants $C,k,omega, text{and } phi$?






real-analysis complex-analysis trigonometry exponential-function graphing-functions







share|cite|improve this question















After graphing $sin x$, I thought of trying something interesting. I wanted to plot the angle $theta$ that a point $(x,sin x )$ makes with the origin on the $y$-axis, against $x$ on the $x$-axis.



$$tantheta = frac{sin x}{x}Rightarrow theta=tan^{-1}{left(frac{sin x}{x}right)}$$



Graphing $y = 20timestheta$ (multiply by 20 for graphical purposes):
Diagram 1



Part of it reminded me of the graph for the damped oscillator (specifically, the $x>0$ part).
That made me wonder if it was possible to find constants $C,k,omega,$ and $phi$ such that $$theta = Ce^{-kx}cos(omega x + phi)$$



However, after toying with Grapher for a while, $y = theta$ didn't seem to decrease exponentially.



That led me to this question: is there any analytical way to find real constants $C,k,omega,$ and $phi$ such that $theta = Ce^{-kx}cos(omega x + phi)$?



Furthermore, are there any complex constants $C,k,omega, text{and } phi$?






real-analysis complex-analysis trigonometry exponential-function graphing-functions




real-analysis complex-analysis trigonometry exponential-function graphing-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 at 4:36









user1101010

7551630




7551630










asked Apr 5 '17 at 8:46









Ujkan Sulejmani

1586




1586












  • No for real constants: the derivatives are quite different.
    – Bernard
    Apr 5 '17 at 8:54










  • Certainly not, there is a serious mismatch for negative $x$.
    – Yves Daoust
    Apr 5 '17 at 14:47




















  • No for real constants: the derivatives are quite different.
    – Bernard
    Apr 5 '17 at 8:54










  • Certainly not, there is a serious mismatch for negative $x$.
    – Yves Daoust
    Apr 5 '17 at 14:47


















No for real constants: the derivatives are quite different.
– Bernard
Apr 5 '17 at 8:54




No for real constants: the derivatives are quite different.
– Bernard
Apr 5 '17 at 8:54












Certainly not, there is a serious mismatch for negative $x$.
– Yves Daoust
Apr 5 '17 at 14:47






Certainly not, there is a serious mismatch for negative $x$.
– Yves Daoust
Apr 5 '17 at 14:47












2 Answers
2






active

oldest

votes


















1














The damping that you see is



$$frac{arctandfrac{sin x}x}{sin x}.$$



It is very close to the hyperbola $dfrac1x$, as you remain in the linear part of the arc tangent.






share|cite|improve this answer





















  • Is there an explanation as to why the damping/envelope is represented by that expression?
    – Ujkan Sulejmani
    Apr 5 '17 at 16:49










  • @UjkanSulejmani: look again, this is obvious.
    – Yves Daoust
    Apr 5 '17 at 17:40



















3














The envelope of $frac{sin x}{x}$ is $pm frac{1}{x}$, and for large $x$, $arctan frac{1}{x} approx frac{1}{x}$. So the decay of the function is inverse to $x$, not exponential, and no constant $k$ exists (real or complex; an imaginary part to $k$ will just introduce a sinusoidal oscillation in the envelope, in addition to the exponential decay or growth).






share|cite|improve this answer























    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%2f2218917%2fcan-the-graph-of-tan-1-left-frac-sin-xx-right-be-expressed-as-ce%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









    1














    The damping that you see is



    $$frac{arctandfrac{sin x}x}{sin x}.$$



    It is very close to the hyperbola $dfrac1x$, as you remain in the linear part of the arc tangent.






    share|cite|improve this answer





















    • Is there an explanation as to why the damping/envelope is represented by that expression?
      – Ujkan Sulejmani
      Apr 5 '17 at 16:49










    • @UjkanSulejmani: look again, this is obvious.
      – Yves Daoust
      Apr 5 '17 at 17:40
















    1














    The damping that you see is



    $$frac{arctandfrac{sin x}x}{sin x}.$$



    It is very close to the hyperbola $dfrac1x$, as you remain in the linear part of the arc tangent.






    share|cite|improve this answer





















    • Is there an explanation as to why the damping/envelope is represented by that expression?
      – Ujkan Sulejmani
      Apr 5 '17 at 16:49










    • @UjkanSulejmani: look again, this is obvious.
      – Yves Daoust
      Apr 5 '17 at 17:40














    1












    1








    1






    The damping that you see is



    $$frac{arctandfrac{sin x}x}{sin x}.$$



    It is very close to the hyperbola $dfrac1x$, as you remain in the linear part of the arc tangent.






    share|cite|improve this answer












    The damping that you see is



    $$frac{arctandfrac{sin x}x}{sin x}.$$



    It is very close to the hyperbola $dfrac1x$, as you remain in the linear part of the arc tangent.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Apr 5 '17 at 14:51









    Yves Daoust

    124k671221




    124k671221












    • Is there an explanation as to why the damping/envelope is represented by that expression?
      – Ujkan Sulejmani
      Apr 5 '17 at 16:49










    • @UjkanSulejmani: look again, this is obvious.
      – Yves Daoust
      Apr 5 '17 at 17:40


















    • Is there an explanation as to why the damping/envelope is represented by that expression?
      – Ujkan Sulejmani
      Apr 5 '17 at 16:49










    • @UjkanSulejmani: look again, this is obvious.
      – Yves Daoust
      Apr 5 '17 at 17:40
















    Is there an explanation as to why the damping/envelope is represented by that expression?
    – Ujkan Sulejmani
    Apr 5 '17 at 16:49




    Is there an explanation as to why the damping/envelope is represented by that expression?
    – Ujkan Sulejmani
    Apr 5 '17 at 16:49












    @UjkanSulejmani: look again, this is obvious.
    – Yves Daoust
    Apr 5 '17 at 17:40




    @UjkanSulejmani: look again, this is obvious.
    – Yves Daoust
    Apr 5 '17 at 17:40











    3














    The envelope of $frac{sin x}{x}$ is $pm frac{1}{x}$, and for large $x$, $arctan frac{1}{x} approx frac{1}{x}$. So the decay of the function is inverse to $x$, not exponential, and no constant $k$ exists (real or complex; an imaginary part to $k$ will just introduce a sinusoidal oscillation in the envelope, in addition to the exponential decay or growth).






    share|cite|improve this answer




























      3














      The envelope of $frac{sin x}{x}$ is $pm frac{1}{x}$, and for large $x$, $arctan frac{1}{x} approx frac{1}{x}$. So the decay of the function is inverse to $x$, not exponential, and no constant $k$ exists (real or complex; an imaginary part to $k$ will just introduce a sinusoidal oscillation in the envelope, in addition to the exponential decay or growth).






      share|cite|improve this answer


























        3












        3








        3






        The envelope of $frac{sin x}{x}$ is $pm frac{1}{x}$, and for large $x$, $arctan frac{1}{x} approx frac{1}{x}$. So the decay of the function is inverse to $x$, not exponential, and no constant $k$ exists (real or complex; an imaginary part to $k$ will just introduce a sinusoidal oscillation in the envelope, in addition to the exponential decay or growth).






        share|cite|improve this answer














        The envelope of $frac{sin x}{x}$ is $pm frac{1}{x}$, and for large $x$, $arctan frac{1}{x} approx frac{1}{x}$. So the decay of the function is inverse to $x$, not exponential, and no constant $k$ exists (real or complex; an imaginary part to $k$ will just introduce a sinusoidal oscillation in the envelope, in addition to the exponential decay or growth).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 5 '17 at 14:37

























        answered Apr 5 '17 at 14:07









        Connor Harris

        4,303723




        4,303723






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f2218917%2fcan-the-graph-of-tan-1-left-frac-sin-xx-right-be-expressed-as-ce%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

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more