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
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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
$begingroup$
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?
integration geometry functions
$endgroup$
add a comment |
$begingroup$
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?
integration geometry functions
$endgroup$
add a comment |
$begingroup$
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?
integration geometry functions
$endgroup$
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
integration geometry functions
edited Mar 22 at 3:50
clathratus
5,0991439
5,0991439
asked Dec 17 '14 at 15:38
JoeJoe
1456
1456
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
|
show 5 more comments
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$begingroup$
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.
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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.
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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.
$endgroup$
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.
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
|
show 5 more comments
$begingroup$
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 $k^2 = 2$
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– Joe
Dec 19 '14 at 21:38
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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.
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– Hans Lundmark
Dec 19 '14 at 22:17
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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.
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– Hans Lundmark
Dec 19 '14 at 22:17
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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 $am(phi|0) = phi$.
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– Joe
Dec 20 '14 at 1:05
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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 $am(phi|0) = phi$.
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– Joe
Dec 20 '14 at 1:05
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