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












2












$begingroup$


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











$endgroup$












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










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


















2












$begingroup$


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











$endgroup$












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










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
















2












2








2


1



$begingroup$


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











$endgroup$




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 '18 at 4:36









user1101010

8121730




8121730










asked Apr 5 '17 at 8:46









Ujkan SulejmaniUjkan Sulejmani

1606




1606












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










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




















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










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


















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




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












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






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












2 Answers
2






active

oldest

votes


















1












$begingroup$

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









$endgroup$













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










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



















3












$begingroup$

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











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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









    $endgroup$













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










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
















    1












    $begingroup$

    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









    $endgroup$













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










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














    1












    1








    1





    $begingroup$

    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









    $endgroup$



    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 DaoustYves Daoust

    126k672225




    126k672225












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










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


















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










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
















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




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












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




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











    3












    $begingroup$

    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











    $endgroup$


















      3












      $begingroup$

      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











      $endgroup$
















        3












        3








        3





        $begingroup$

        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











        $endgroup$



        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 HarrisConnor Harris

        4,365724




        4,365724






























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