Jacobi Elliptic Functions Special CaseCan this integral $int_0^2pi fracdtheta(a^2 cos^2 theta +b^2sin^2theta)^3/2$ be written in the form of a elliptic integralIntegral involving Complete Elliptic Integral of the First Kind K(k)How to compute elliptic integrals in MATLABDoes $int_0^2 pi sqrt1-(a+b sinphi)^2 dphi $ have a closed form in terms of elliptic integrals?Elliptic IntegralsAn elliptic integral?Curious integrals for Jacobi Theta Functions $int_0^1 vartheta_n(0,q)dq$Jacobian Elliptic FunctionsCan $int_0^inftyfracdxsqrt[leftroot-1uproot1n]x(x+a^n)(x+1)^n-1$ be expressed with an elliptic integral?How does one express the length of curves in terms of the elliptic integral

Why "Having chlorophyll without photosynthesis is actually very dangerous" and "like living with a bomb"?

How old can references or sources in a thesis be?

Smoothness of finite-dimensional functional calculus

Show that if two triangles built on parallel lines, with equal bases have the same perimeter only if they are congruent.

Why are electrically insulating heatsinks so rare? Is it just cost?

Problem of parity - Can we draw a closed path made up of 20 line segments...

What's the point of deactivating Num Lock on login screens?

Why Is Death Allowed In the Matrix?

Service Entrance Breakers Rain Shield

What does it mean to describe someone as a butt steak?

How to test if a transaction is standard without spending real money?

Email Account under attack (really) - anything I can do?

LaTeX closing $ signs makes cursor jump

How do we improve the relationship with a client software team that performs poorly and is becoming less collaborative?

How is it possible to have an ability score that is less than 3?

Can divisibility rules for digits be generalized to sum of digits

Have astronauts in space suits ever taken selfies? If so, how?

How does one intimidate enemies without having the capacity for violence?

Is it legal for company to use my work email to pretend I still work there?

Today is the Center

The Two and the One

Prove that NP is closed under karp reduction?

Theorems that impeded progress

"You are your self first supporter", a more proper way to say it



Jacobi Elliptic Functions Special Case


Can this integral $int_0^2pi fracdtheta(a^2 cos^2 theta +b^2sin^2theta)^3/2$ be written in the form of a elliptic integralIntegral involving Complete Elliptic Integral of the First Kind K(k)How to compute elliptic integrals in MATLABDoes $int_0^2 pi sqrt1-(a+b sinphi)^2 dphi $ have a closed form in terms of elliptic integrals?Elliptic IntegralsAn elliptic integral?Curious integrals for Jacobi Theta Functions $int_0^1 vartheta_n(0,q)dq$Jacobian Elliptic FunctionsCan $int_0^inftyfracdxsqrt[leftroot-1uproot1n]x(x+a^n)(x+1)^n-1$ be expressed with an elliptic integral?How does one express the length of curves in terms of the elliptic integral













3












$begingroup$


I have spent some time analysing the pendulum problem, and hence the Jacobi elliptic functions recently, and have come across what seems to me to be a slight inconsitency.
I define my $mathrmam(t|k)$ as the inverse to the integral
$$int_0^t (1-k^2 sin^2x)^-1/2 , mathrmdx = mathrmam^-1(t|k) $$
And I'm interested in the special case of $k^2 = 2$



I understand the Jacobi elliptic functions to be defined as parametrising the ellipse
$$x^2 + fracy^2b^2 = 1$$
and $k^2 = 1-b^-2$. Then parametrise the ellipse by $x = mathrmcn(t|k)$, $y = b ,mathrmsn(t|k)$. In the special case of $k = 0$ it's simple to show that the ellipse becomes a circle, and these functions reduce to standard trig functions, and for $k = 1$ that the ellipse becomes two parallel lines, and that the elliptic functions reduce to $tanh$ and $mathrmsech$.



Now let $b = i beta$, and the ellipse becomes a hyperbola. $k^2 = 1 + beta^2$, and the special case of $beta = 1$ is the rectangular hyperbola
$$x^2 - y^2 = 1$$



This corresponds to $k^2 = 2$, so my claim is that it should be possible to solve the integrals for the elliptic functions in this case to give
$$ mathrmsn(t|sqrt2) = sinh(t)$$
$$ mathrmcn(t|sqrt2) = cosh(t)$$
Which would then agree with the parametrisation of the original ellipse equation, with $b = i$



However the defining integral reduces to
$$int_0^t sqrtsec(2x) , mathrmdx = mathrmam^-1(t|sqrt2) $$
Which I don't believe to be integrable in terms of non-special functions. (I have spent quite a long time trying different substitutions and trig rearrangements on this integral and not had much luck).



Does anyone have any thoughts on this?










share|cite|improve this question











$endgroup$
















    3












    $begingroup$


    I have spent some time analysing the pendulum problem, and hence the Jacobi elliptic functions recently, and have come across what seems to me to be a slight inconsitency.
    I define my $mathrmam(t|k)$ as the inverse to the integral
    $$int_0^t (1-k^2 sin^2x)^-1/2 , mathrmdx = mathrmam^-1(t|k) $$
    And I'm interested in the special case of $k^2 = 2$



    I understand the Jacobi elliptic functions to be defined as parametrising the ellipse
    $$x^2 + fracy^2b^2 = 1$$
    and $k^2 = 1-b^-2$. Then parametrise the ellipse by $x = mathrmcn(t|k)$, $y = b ,mathrmsn(t|k)$. In the special case of $k = 0$ it's simple to show that the ellipse becomes a circle, and these functions reduce to standard trig functions, and for $k = 1$ that the ellipse becomes two parallel lines, and that the elliptic functions reduce to $tanh$ and $mathrmsech$.



    Now let $b = i beta$, and the ellipse becomes a hyperbola. $k^2 = 1 + beta^2$, and the special case of $beta = 1$ is the rectangular hyperbola
    $$x^2 - y^2 = 1$$



    This corresponds to $k^2 = 2$, so my claim is that it should be possible to solve the integrals for the elliptic functions in this case to give
    $$ mathrmsn(t|sqrt2) = sinh(t)$$
    $$ mathrmcn(t|sqrt2) = cosh(t)$$
    Which would then agree with the parametrisation of the original ellipse equation, with $b = i$



    However the defining integral reduces to
    $$int_0^t sqrtsec(2x) , mathrmdx = mathrmam^-1(t|sqrt2) $$
    Which I don't believe to be integrable in terms of non-special functions. (I have spent quite a long time trying different substitutions and trig rearrangements on this integral and not had much luck).



    Does anyone have any thoughts on this?










    share|cite|improve this question











    $endgroup$














      3












      3








      3





      $begingroup$


      I have spent some time analysing the pendulum problem, and hence the Jacobi elliptic functions recently, and have come across what seems to me to be a slight inconsitency.
      I define my $mathrmam(t|k)$ as the inverse to the integral
      $$int_0^t (1-k^2 sin^2x)^-1/2 , mathrmdx = mathrmam^-1(t|k) $$
      And I'm interested in the special case of $k^2 = 2$



      I understand the Jacobi elliptic functions to be defined as parametrising the ellipse
      $$x^2 + fracy^2b^2 = 1$$
      and $k^2 = 1-b^-2$. Then parametrise the ellipse by $x = mathrmcn(t|k)$, $y = b ,mathrmsn(t|k)$. In the special case of $k = 0$ it's simple to show that the ellipse becomes a circle, and these functions reduce to standard trig functions, and for $k = 1$ that the ellipse becomes two parallel lines, and that the elliptic functions reduce to $tanh$ and $mathrmsech$.



      Now let $b = i beta$, and the ellipse becomes a hyperbola. $k^2 = 1 + beta^2$, and the special case of $beta = 1$ is the rectangular hyperbola
      $$x^2 - y^2 = 1$$



      This corresponds to $k^2 = 2$, so my claim is that it should be possible to solve the integrals for the elliptic functions in this case to give
      $$ mathrmsn(t|sqrt2) = sinh(t)$$
      $$ mathrmcn(t|sqrt2) = cosh(t)$$
      Which would then agree with the parametrisation of the original ellipse equation, with $b = i$



      However the defining integral reduces to
      $$int_0^t sqrtsec(2x) , mathrmdx = mathrmam^-1(t|sqrt2) $$
      Which I don't believe to be integrable in terms of non-special functions. (I have spent quite a long time trying different substitutions and trig rearrangements on this integral and not had much luck).



      Does anyone have any thoughts on this?










      share|cite|improve this question











      $endgroup$




      I have spent some time analysing the pendulum problem, and hence the Jacobi elliptic functions recently, and have come across what seems to me to be a slight inconsitency.
      I define my $mathrmam(t|k)$ as the inverse to the integral
      $$int_0^t (1-k^2 sin^2x)^-1/2 , mathrmdx = mathrmam^-1(t|k) $$
      And I'm interested in the special case of $k^2 = 2$



      I understand the Jacobi elliptic functions to be defined as parametrising the ellipse
      $$x^2 + fracy^2b^2 = 1$$
      and $k^2 = 1-b^-2$. Then parametrise the ellipse by $x = mathrmcn(t|k)$, $y = b ,mathrmsn(t|k)$. In the special case of $k = 0$ it's simple to show that the ellipse becomes a circle, and these functions reduce to standard trig functions, and for $k = 1$ that the ellipse becomes two parallel lines, and that the elliptic functions reduce to $tanh$ and $mathrmsech$.



      Now let $b = i beta$, and the ellipse becomes a hyperbola. $k^2 = 1 + beta^2$, and the special case of $beta = 1$ is the rectangular hyperbola
      $$x^2 - y^2 = 1$$



      This corresponds to $k^2 = 2$, so my claim is that it should be possible to solve the integrals for the elliptic functions in this case to give
      $$ mathrmsn(t|sqrt2) = sinh(t)$$
      $$ mathrmcn(t|sqrt2) = cosh(t)$$
      Which would then agree with the parametrisation of the original ellipse equation, with $b = i$



      However the defining integral reduces to
      $$int_0^t sqrtsec(2x) , mathrmdx = mathrmam^-1(t|sqrt2) $$
      Which I don't believe to be integrable in terms of non-special functions. (I have spent quite a long time trying different substitutions and trig rearrangements on this integral and not had much luck).



      Does anyone have any thoughts on this?







      integration geometry functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 22 at 3:50









      clathratus

      5,0991439




      5,0991439










      asked Dec 17 '14 at 15:38









      JoeJoe

      1456




      1456




















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Hmm, something's wrong here... The function $f(z)=operatornamesn(z,sqrt2)$ is definitely not equal to $sinh z$. There's a transformation formula which gives $f(z)=operatornamesn(zsqrt2,1/sqrt2)/sqrt2$, and this is a Jacobi sn function with $0<k<1$, which (as you probably know) implies that it is periodic on the real axis and also doubly periodic in the complex plane (unlike $sinh z$).



          I haven't thought about what exactly goes wrong in your reasoning, but
          I think that maybe you are putting too much emphasis on the formula $x^2 + fracy^2b^2=1$. That formula alone isn't enough to define the elliptic functions, since for example $(x,y)=(cos t,b sin t)$ is another perfectly good parametrization of the ellipse, with good old trig functions.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Yes I think I can see that it can't be equal to $sinh$, I had been looking at that transformation formula, and also if the Jacobi functions were equal to the hyperbolic functions, they couldn't satisfy $sn^2 + cn^2 = 1$ either. But I still feel like the ellipse equation reducing to a rectangular hyperbola for $k^2 = 2$ is enough to signify that there should be a simplification for Jacobi functions though
            $endgroup$
            – Joe
            Dec 18 '14 at 21:59











          • $begingroup$
            Well, the Jacobi sn and cn functions parametrize the unit circle for any $k$, but this doesn't imply that they can be simplified to elementary functions...
            $endgroup$
            – Hans Lundmark
            Dec 19 '14 at 11:11










          • $begingroup$
            No most definitely not, but the fact that the same parametrization that they use for the ellipse reduces to the circle for $k = 0$ does imply that they can be simplified to elementary functions in this case, and to me this is analagous to the hyperbola becoming a rectangular hyperbola for $k^2 = 2$
            $endgroup$
            – Joe
            Dec 19 '14 at 21:38










          • $begingroup$
            Well, to be honest I don't understand your argument at all. The fact that sn and cn reduce to trig functions for $k=0$ does not follow merely from the fact that they happen to parametrize the circle in that case, you need to look at their definitions. After all, they parametrize the circle for every $k$, but for $k neq 0$ (or $1$) they don't simplify.
            $endgroup$
            – Hans Lundmark
            Dec 19 '14 at 22:17











          • $begingroup$
            Ah, I think in previous comment, 'suggests' would have been a better word than 'implies', (although I think it might be an implication). I am defining the Jacobi Elliptic functions in terms of properties on the ellipse. I define the elliptic angular measure, $am(phi|k)$, which is like an extension of the radian to `elliptical angular units'. Then you take $sin$ and $cos$ (and the $dn$ function) of it to give the elliptic functions. Then as $k->0$ we recover the circle, and $am(phi|0) = phi$.
            $endgroup$
            – Joe
            Dec 20 '14 at 1:05












          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%2f1072098%2fjacobi-elliptic-functions-special-case%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Hmm, something's wrong here... The function $f(z)=operatornamesn(z,sqrt2)$ is definitely not equal to $sinh z$. There's a transformation formula which gives $f(z)=operatornamesn(zsqrt2,1/sqrt2)/sqrt2$, and this is a Jacobi sn function with $0<k<1$, which (as you probably know) implies that it is periodic on the real axis and also doubly periodic in the complex plane (unlike $sinh z$).



          I haven't thought about what exactly goes wrong in your reasoning, but
          I think that maybe you are putting too much emphasis on the formula $x^2 + fracy^2b^2=1$. That formula alone isn't enough to define the elliptic functions, since for example $(x,y)=(cos t,b sin t)$ is another perfectly good parametrization of the ellipse, with good old trig functions.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Yes I think I can see that it can't be equal to $sinh$, I had been looking at that transformation formula, and also if the Jacobi functions were equal to the hyperbolic functions, they couldn't satisfy $sn^2 + cn^2 = 1$ either. But I still feel like the ellipse equation reducing to a rectangular hyperbola for $k^2 = 2$ is enough to signify that there should be a simplification for Jacobi functions though
            $endgroup$
            – Joe
            Dec 18 '14 at 21:59











          • $begingroup$
            Well, the Jacobi sn and cn functions parametrize the unit circle for any $k$, but this doesn't imply that they can be simplified to elementary functions...
            $endgroup$
            – Hans Lundmark
            Dec 19 '14 at 11:11










          • $begingroup$
            No most definitely not, but the fact that the same parametrization that they use for the ellipse reduces to the circle for $k = 0$ does imply that they can be simplified to elementary functions in this case, and to me this is analagous to the hyperbola becoming a rectangular hyperbola for $k^2 = 2$
            $endgroup$
            – Joe
            Dec 19 '14 at 21:38










          • $begingroup$
            Well, to be honest I don't understand your argument at all. The fact that sn and cn reduce to trig functions for $k=0$ does not follow merely from the fact that they happen to parametrize the circle in that case, you need to look at their definitions. After all, they parametrize the circle for every $k$, but for $k neq 0$ (or $1$) they don't simplify.
            $endgroup$
            – Hans Lundmark
            Dec 19 '14 at 22:17











          • $begingroup$
            Ah, I think in previous comment, 'suggests' would have been a better word than 'implies', (although I think it might be an implication). I am defining the Jacobi Elliptic functions in terms of properties on the ellipse. I define the elliptic angular measure, $am(phi|k)$, which is like an extension of the radian to `elliptical angular units'. Then you take $sin$ and $cos$ (and the $dn$ function) of it to give the elliptic functions. Then as $k->0$ we recover the circle, and $am(phi|0) = phi$.
            $endgroup$
            – Joe
            Dec 20 '14 at 1:05
















          2












          $begingroup$

          Hmm, something's wrong here... The function $f(z)=operatornamesn(z,sqrt2)$ is definitely not equal to $sinh z$. There's a transformation formula which gives $f(z)=operatornamesn(zsqrt2,1/sqrt2)/sqrt2$, and this is a Jacobi sn function with $0<k<1$, which (as you probably know) implies that it is periodic on the real axis and also doubly periodic in the complex plane (unlike $sinh z$).



          I haven't thought about what exactly goes wrong in your reasoning, but
          I think that maybe you are putting too much emphasis on the formula $x^2 + fracy^2b^2=1$. That formula alone isn't enough to define the elliptic functions, since for example $(x,y)=(cos t,b sin t)$ is another perfectly good parametrization of the ellipse, with good old trig functions.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Yes I think I can see that it can't be equal to $sinh$, I had been looking at that transformation formula, and also if the Jacobi functions were equal to the hyperbolic functions, they couldn't satisfy $sn^2 + cn^2 = 1$ either. But I still feel like the ellipse equation reducing to a rectangular hyperbola for $k^2 = 2$ is enough to signify that there should be a simplification for Jacobi functions though
            $endgroup$
            – Joe
            Dec 18 '14 at 21:59











          • $begingroup$
            Well, the Jacobi sn and cn functions parametrize the unit circle for any $k$, but this doesn't imply that they can be simplified to elementary functions...
            $endgroup$
            – Hans Lundmark
            Dec 19 '14 at 11:11










          • $begingroup$
            No most definitely not, but the fact that the same parametrization that they use for the ellipse reduces to the circle for $k = 0$ does imply that they can be simplified to elementary functions in this case, and to me this is analagous to the hyperbola becoming a rectangular hyperbola for $k^2 = 2$
            $endgroup$
            – Joe
            Dec 19 '14 at 21:38










          • $begingroup$
            Well, to be honest I don't understand your argument at all. The fact that sn and cn reduce to trig functions for $k=0$ does not follow merely from the fact that they happen to parametrize the circle in that case, you need to look at their definitions. After all, they parametrize the circle for every $k$, but for $k neq 0$ (or $1$) they don't simplify.
            $endgroup$
            – Hans Lundmark
            Dec 19 '14 at 22:17











          • $begingroup$
            Ah, I think in previous comment, 'suggests' would have been a better word than 'implies', (although I think it might be an implication). I am defining the Jacobi Elliptic functions in terms of properties on the ellipse. I define the elliptic angular measure, $am(phi|k)$, which is like an extension of the radian to `elliptical angular units'. Then you take $sin$ and $cos$ (and the $dn$ function) of it to give the elliptic functions. Then as $k->0$ we recover the circle, and $am(phi|0) = phi$.
            $endgroup$
            – Joe
            Dec 20 '14 at 1:05














          2












          2








          2





          $begingroup$

          Hmm, something's wrong here... The function $f(z)=operatornamesn(z,sqrt2)$ is definitely not equal to $sinh z$. There's a transformation formula which gives $f(z)=operatornamesn(zsqrt2,1/sqrt2)/sqrt2$, and this is a Jacobi sn function with $0<k<1$, which (as you probably know) implies that it is periodic on the real axis and also doubly periodic in the complex plane (unlike $sinh z$).



          I haven't thought about what exactly goes wrong in your reasoning, but
          I think that maybe you are putting too much emphasis on the formula $x^2 + fracy^2b^2=1$. That formula alone isn't enough to define the elliptic functions, since for example $(x,y)=(cos t,b sin t)$ is another perfectly good parametrization of the ellipse, with good old trig functions.






          share|cite|improve this answer









          $endgroup$



          Hmm, something's wrong here... The function $f(z)=operatornamesn(z,sqrt2)$ is definitely not equal to $sinh z$. There's a transformation formula which gives $f(z)=operatornamesn(zsqrt2,1/sqrt2)/sqrt2$, and this is a Jacobi sn function with $0<k<1$, which (as you probably know) implies that it is periodic on the real axis and also doubly periodic in the complex plane (unlike $sinh z$).



          I haven't thought about what exactly goes wrong in your reasoning, but
          I think that maybe you are putting too much emphasis on the formula $x^2 + fracy^2b^2=1$. That formula alone isn't enough to define the elliptic functions, since for example $(x,y)=(cos t,b sin t)$ is another perfectly good parametrization of the ellipse, with good old trig functions.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '14 at 15:55









          Hans LundmarkHans Lundmark

          36.1k564115




          36.1k564115











          • $begingroup$
            Yes I think I can see that it can't be equal to $sinh$, I had been looking at that transformation formula, and also if the Jacobi functions were equal to the hyperbolic functions, they couldn't satisfy $sn^2 + cn^2 = 1$ either. But I still feel like the ellipse equation reducing to a rectangular hyperbola for $k^2 = 2$ is enough to signify that there should be a simplification for Jacobi functions though
            $endgroup$
            – Joe
            Dec 18 '14 at 21:59











          • $begingroup$
            Well, the Jacobi sn and cn functions parametrize the unit circle for any $k$, but this doesn't imply that they can be simplified to elementary functions...
            $endgroup$
            – Hans Lundmark
            Dec 19 '14 at 11:11










          • $begingroup$
            No most definitely not, but the fact that the same parametrization that they use for the ellipse reduces to the circle for $k = 0$ does imply that they can be simplified to elementary functions in this case, and to me this is analagous to the hyperbola becoming a rectangular hyperbola for $k^2 = 2$
            $endgroup$
            – Joe
            Dec 19 '14 at 21:38










          • $begingroup$
            Well, to be honest I don't understand your argument at all. The fact that sn and cn reduce to trig functions for $k=0$ does not follow merely from the fact that they happen to parametrize the circle in that case, you need to look at their definitions. After all, they parametrize the circle for every $k$, but for $k neq 0$ (or $1$) they don't simplify.
            $endgroup$
            – Hans Lundmark
            Dec 19 '14 at 22:17











          • $begingroup$
            Ah, I think in previous comment, 'suggests' would have been a better word than 'implies', (although I think it might be an implication). I am defining the Jacobi Elliptic functions in terms of properties on the ellipse. I define the elliptic angular measure, $am(phi|k)$, which is like an extension of the radian to `elliptical angular units'. Then you take $sin$ and $cos$ (and the $dn$ function) of it to give the elliptic functions. Then as $k->0$ we recover the circle, and $am(phi|0) = phi$.
            $endgroup$
            – Joe
            Dec 20 '14 at 1:05

















          • $begingroup$
            Yes I think I can see that it can't be equal to $sinh$, I had been looking at that transformation formula, and also if the Jacobi functions were equal to the hyperbolic functions, they couldn't satisfy $sn^2 + cn^2 = 1$ either. But I still feel like the ellipse equation reducing to a rectangular hyperbola for $k^2 = 2$ is enough to signify that there should be a simplification for Jacobi functions though
            $endgroup$
            – Joe
            Dec 18 '14 at 21:59











          • $begingroup$
            Well, the Jacobi sn and cn functions parametrize the unit circle for any $k$, but this doesn't imply that they can be simplified to elementary functions...
            $endgroup$
            – Hans Lundmark
            Dec 19 '14 at 11:11










          • $begingroup$
            No most definitely not, but the fact that the same parametrization that they use for the ellipse reduces to the circle for $k = 0$ does imply that they can be simplified to elementary functions in this case, and to me this is analagous to the hyperbola becoming a rectangular hyperbola for $k^2 = 2$
            $endgroup$
            – Joe
            Dec 19 '14 at 21:38










          • $begingroup$
            Well, to be honest I don't understand your argument at all. The fact that sn and cn reduce to trig functions for $k=0$ does not follow merely from the fact that they happen to parametrize the circle in that case, you need to look at their definitions. After all, they parametrize the circle for every $k$, but for $k neq 0$ (or $1$) they don't simplify.
            $endgroup$
            – Hans Lundmark
            Dec 19 '14 at 22:17











          • $begingroup$
            Ah, I think in previous comment, 'suggests' would have been a better word than 'implies', (although I think it might be an implication). I am defining the Jacobi Elliptic functions in terms of properties on the ellipse. I define the elliptic angular measure, $am(phi|k)$, which is like an extension of the radian to `elliptical angular units'. Then you take $sin$ and $cos$ (and the $dn$ function) of it to give the elliptic functions. Then as $k->0$ we recover the circle, and $am(phi|0) = phi$.
            $endgroup$
            – Joe
            Dec 20 '14 at 1:05
















          $begingroup$
          Yes I think I can see that it can't be equal to $sinh$, I had been looking at that transformation formula, and also if the Jacobi functions were equal to the hyperbolic functions, they couldn't satisfy $sn^2 + cn^2 = 1$ either. But I still feel like the ellipse equation reducing to a rectangular hyperbola for $k^2 = 2$ is enough to signify that there should be a simplification for Jacobi functions though
          $endgroup$
          – Joe
          Dec 18 '14 at 21:59





          $begingroup$
          Yes I think I can see that it can't be equal to $sinh$, I had been looking at that transformation formula, and also if the Jacobi functions were equal to the hyperbolic functions, they couldn't satisfy $sn^2 + cn^2 = 1$ either. But I still feel like the ellipse equation reducing to a rectangular hyperbola for $k^2 = 2$ is enough to signify that there should be a simplification for Jacobi functions though
          $endgroup$
          – Joe
          Dec 18 '14 at 21:59













          $begingroup$
          Well, the Jacobi sn and cn functions parametrize the unit circle for any $k$, but this doesn't imply that they can be simplified to elementary functions...
          $endgroup$
          – Hans Lundmark
          Dec 19 '14 at 11:11




          $begingroup$
          Well, the Jacobi sn and cn functions parametrize the unit circle for any $k$, but this doesn't imply that they can be simplified to elementary functions...
          $endgroup$
          – Hans Lundmark
          Dec 19 '14 at 11:11












          $begingroup$
          No most definitely not, but the fact that the same parametrization that they use for the ellipse reduces to the circle for $k = 0$ does imply that they can be simplified to elementary functions in this case, and to me this is analagous to the hyperbola becoming a rectangular hyperbola for $k^2 = 2$
          $endgroup$
          – Joe
          Dec 19 '14 at 21:38




          $begingroup$
          No most definitely not, but the fact that the same parametrization that they use for the ellipse reduces to the circle for $k = 0$ does imply that they can be simplified to elementary functions in this case, and to me this is analagous to the hyperbola becoming a rectangular hyperbola for $k^2 = 2$
          $endgroup$
          – Joe
          Dec 19 '14 at 21:38












          $begingroup$
          Well, to be honest I don't understand your argument at all. The fact that sn and cn reduce to trig functions for $k=0$ does not follow merely from the fact that they happen to parametrize the circle in that case, you need to look at their definitions. After all, they parametrize the circle for every $k$, but for $k neq 0$ (or $1$) they don't simplify.
          $endgroup$
          – Hans Lundmark
          Dec 19 '14 at 22:17





          $begingroup$
          Well, to be honest I don't understand your argument at all. The fact that sn and cn reduce to trig functions for $k=0$ does not follow merely from the fact that they happen to parametrize the circle in that case, you need to look at their definitions. After all, they parametrize the circle for every $k$, but for $k neq 0$ (or $1$) they don't simplify.
          $endgroup$
          – Hans Lundmark
          Dec 19 '14 at 22:17













          $begingroup$
          Ah, I think in previous comment, 'suggests' would have been a better word than 'implies', (although I think it might be an implication). I am defining the Jacobi Elliptic functions in terms of properties on the ellipse. I define the elliptic angular measure, $am(phi|k)$, which is like an extension of the radian to `elliptical angular units'. Then you take $sin$ and $cos$ (and the $dn$ function) of it to give the elliptic functions. Then as $k->0$ we recover the circle, and $am(phi|0) = phi$.
          $endgroup$
          – Joe
          Dec 20 '14 at 1:05





          $begingroup$
          Ah, I think in previous comment, 'suggests' would have been a better word than 'implies', (although I think it might be an implication). I am defining the Jacobi Elliptic functions in terms of properties on the ellipse. I define the elliptic angular measure, $am(phi|k)$, which is like an extension of the radian to `elliptical angular units'. Then you take $sin$ and $cos$ (and the $dn$ function) of it to give the elliptic functions. Then as $k->0$ we recover the circle, and $am(phi|0) = phi$.
          $endgroup$
          – Joe
          Dec 20 '14 at 1:05


















          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.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1072098%2fjacobi-elliptic-functions-special-case%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

          How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

          random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

          Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye