Jacobi sn: Does $w=operatornamesn(z,m)$ send the line $operatornameIm(z)=K'/2$ to the circle $left |w^4 right |=1/m?$ The Next CEO of Stack Overflowholomorphic that send the unit circle to the triangle $0,1,i$rational function with special properties on unit disk.Proof of $left| operatornamecnleft( x(1+ i) mid m right)right|=1$ for $m=frac12$ and $x in mathbbR$The Jacobi nome $q$Does this conformal map from a rectangle exist?Does the elliptic function $operatornamecnleft(frac23Kleft(frac12right)big|frac12right)$ have a closed form?Algebraic equation concerning the Jacobi sn functionWhat is the Jacobi amplitude?A holomorphic function which has $|f(z)|>1$ for every $|z|=1$ and $|f(0)|<1$ has a fixed point inside the unit circleDoes the Möbius map $T(z) = fracz - i iz + 3$ maps circle to circle and line to line?
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Jacobi sn: Does $w=operatornamesn(z,m)$ send the line $operatornameIm(z)=K'/2$ to the circle $left |w^4 right |=1/m?$
The Next CEO of Stack Overflowholomorphic that send the unit circle to the triangle $0,1,i$rational function with special properties on unit disk.Proof of $left| operatornamecnleft( x(1+ i) mid m right)right|=1$ for $m=frac12$ and $x in mathbbR$The Jacobi nome $q$Does this conformal map from a rectangle exist?Does the elliptic function $operatornamecnleft(frac23Kleft(frac12right)big|frac12right)$ have a closed form?Algebraic equation concerning the Jacobi sn functionWhat is the Jacobi amplitude?A holomorphic function which has $|f(z)|>1$ for every $|z|=1$ and $|f(0)|<1$ has a fixed point inside the unit circleDoes the Möbius map $T(z) = fracz - i iz + 3$ maps circle to circle and line to line?
$begingroup$
I've been using Mathematica to study the image of a rectilinear grid in $mathbb C$ under the elliptic mapping $operatornamesn(z,m)$, as shown below. Everything I'm going to say here is suggested by numeric and visual experiments, and I'm looking for proofs (or references) because my elliptic function fu is very feeble. The questions here concern the highlighted circle.
Using the terminology of the Wikipedia article on Jacobi elliptic functions, let $min (0,1],K=operatornameK(m), K'=operatornameK(1-m),$ and $k=sqrt m.$
If $L$ is the line $operatornameIm(z)=tfrac 1 2K'$ and $C$ is the circle $left |z^4right |=1/m$ then the main claim here is that $C$ is the circle in the diagram, and that
$$operatorname sn_m: L rightarrow C.$$
so that for $xin mathbb R$
$$left |operatornamesn^4(x+tfrac 1 2K'i,m)right |=1/m.$$
Can anybody point me to a proof or exposition of the above?
complex-analysis elliptic-functions
asked Mar 19 at 21:44
brainjambrainjam
417314
$endgroup$
add a comment |
$begingroup$
I've been using Mathematica to study the image of a rectilinear grid in $mathbb C$ under the elliptic mapping $operatornamesn(z,m)$, as shown below. Everything I'm going to say here is suggested by numeric and visual experiments, and I'm looking for proofs (or references) because my elliptic function fu is very feeble. The questions here concern the highlighted circle.
Using the terminology of the Wikipedia article on Jacobi elliptic functions, let $min (0,1],K=operatornameK(m), K'=operatornameK(1-m),$ and $k=sqrt m.$
If $L$ is the line $operatornameIm(z)=tfrac 1 2K'$ and $C$ is the circle $left |z^4right |=1/m$ then the main claim here is that $C$ is the circle in the diagram, and that
$$operatorname sn_m: L rightarrow C.$$
so that for $xin mathbb R$
$$left |operatornamesn^4(x+tfrac 1 2K'i,m)right |=1/m.$$
Can anybody point me to a proof or exposition of the above?
complex-analysis elliptic-functions
asked Mar 19 at 21:44
brainjambrainjam
417314
$endgroup$
1
$begingroup$
With the lattice $L = 2KBbbZ+2iK' BbbZ$ and $f(x) = sum_l in L log(1-fracxl+iK'/2)$ $+log(1-fracxl+iK'/2+K)-log(1-fracxl-iK'/2)-log(1-fracxl-iK'/2+K)$ from the period and zero/pole location of $sn(x)$ then $log sn(x+iK'/2) - f(x)$ will be analytic $L$ periodic thus constant. As $L = overlineL$ then $f$ is purely imaginary for $x in BbbR$ thus $Re(sn(x+iK'/2))$ is constant for $x in BbbR$
$endgroup$
– reuns
Mar 19 at 23:53
$begingroup$
@reuns, are you saying that sn(𝑥+𝑖𝐾′/2,m) is not a circle?
$endgroup$
– brainjam
Mar 20 at 3:56
1
$begingroup$
I meant $Re(log sn(x+iK'/2))$ is constant
$endgroup$
– reuns
Mar 20 at 4:37
$begingroup$
It also appears that the diagram is symmetric under inversion in circle $C$, so for $x,yin mathbb R$ $$left | operatornamesn^2(x+(tfrac 1 2K'-y)i,m) operatornamesn^2(x+(tfrac 1 2K'+y)i,m) right | =1/m$$ or slightly more compactly: $$left | operatornamesn(x+(tfrac 1 2K'-y)i,m) operatornamesn(x+(tfrac 1 2K'+y)i,m) right | =1/k.$$ Also, $$ dfrac operatornamesn(x+(tfrac 1 2K'-y)i,m) operatornamesn(x+(tfrac 1 2K'+y)i,m) in mathbbR.$$
$endgroup$
– brainjam
Mar 22 at 16:29
add a comment |
$begingroup$
I've been using Mathematica to study the image of a rectilinear grid in $mathbb C$ under the elliptic mapping $operatornamesn(z,m)$, as shown below. Everything I'm going to say here is suggested by numeric and visual experiments, and I'm looking for proofs (or references) because my elliptic function fu is very feeble. The questions here concern the highlighted circle.
Using the terminology of the Wikipedia article on Jacobi elliptic functions, let $min (0,1],K=operatornameK(m), K'=operatornameK(1-m),$ and $k=sqrt m.$
If $L$ is the line $operatornameIm(z)=tfrac 1 2K'$ and $C$ is the circle $left |z^4right |=1/m$ then the main claim here is that $C$ is the circle in the diagram, and that
$$operatorname sn_m: L rightarrow C.$$
so that for $xin mathbb R$
$$left |operatornamesn^4(x+tfrac 1 2K'i,m)right |=1/m.$$
Can anybody point me to a proof or exposition of the above?
complex-analysis elliptic-functions
asked Mar 19 at 21:44
brainjambrainjam
417314
$endgroup$
I've been using Mathematica to study the image of a rectilinear grid in $mathbb C$ under the elliptic mapping $operatornamesn(z,m)$, as shown below. Everything I'm going to say here is suggested by numeric and visual experiments, and I'm looking for proofs (or references) because my elliptic function fu is very feeble. The questions here concern the highlighted circle.
Using the terminology of the Wikipedia article on Jacobi elliptic functions, let $min (0,1],K=operatornameK(m), K'=operatornameK(1-m),$ and $k=sqrt m.$
If $L$ is the line $operatornameIm(z)=tfrac 1 2K'$ and $C$ is the circle $left |z^4right |=1/m$ then the main claim here is that $C$ is the circle in the diagram, and that
$$operatorname sn_m: L rightarrow C.$$
so that for $xin mathbb R$
$$left |operatornamesn^4(x+tfrac 1 2K'i,m)right |=1/m.$$
Can anybody point me to a proof or exposition of the above?
complex-analysis elliptic-functions
complex-analysis elliptic-functions
asked Mar 19 at 21:44
brainjambrainjam
417314
asked Mar 19 at 21:44
brainjambrainjam
417314
edited Mar 22 at 16:32
brainjam
asked Mar 19 at 21:44
brainjambrainjam
417314
asked Mar 19 at 21:44
brainjambrainjam
417314
asked Mar 19 at 21:44
brainjambrainjam
417314
417314
1
$begingroup$
With the lattice $L = 2KBbbZ+2iK' BbbZ$ and $f(x) = sum_l in L log(1-fracxl+iK'/2)$ $+log(1-fracxl+iK'/2+K)-log(1-fracxl-iK'/2)-log(1-fracxl-iK'/2+K)$ from the period and zero/pole location of $sn(x)$ then $log sn(x+iK'/2) - f(x)$ will be analytic $L$ periodic thus constant. As $L = overlineL$ then $f$ is purely imaginary for $x in BbbR$ thus $Re(sn(x+iK'/2))$ is constant for $x in BbbR$
$endgroup$
– reuns
Mar 19 at 23:53
$begingroup$
@reuns, are you saying that sn(𝑥+𝑖𝐾′/2,m) is not a circle?
$endgroup$
– brainjam
Mar 20 at 3:56
1
$begingroup$
I meant $Re(log sn(x+iK'/2))$ is constant
$endgroup$
– reuns
Mar 20 at 4:37
$begingroup$
It also appears that the diagram is symmetric under inversion in circle $C$, so for $x,yin mathbb R$ $$left | operatornamesn^2(x+(tfrac 1 2K'-y)i,m) operatornamesn^2(x+(tfrac 1 2K'+y)i,m) right | =1/m$$ or slightly more compactly: $$left | operatornamesn(x+(tfrac 1 2K'-y)i,m) operatornamesn(x+(tfrac 1 2K'+y)i,m) right | =1/k.$$ Also, $$ dfrac operatornamesn(x+(tfrac 1 2K'-y)i,m) operatornamesn(x+(tfrac 1 2K'+y)i,m) in mathbbR.$$
$endgroup$
– brainjam
Mar 22 at 16:29
add a comment |
1
$begingroup$
With the lattice $L = 2KBbbZ+2iK' BbbZ$ and $f(x) = sum_l in L log(1-fracxl+iK'/2)$ $+log(1-fracxl+iK'/2+K)-log(1-fracxl-iK'/2)-log(1-fracxl-iK'/2+K)$ from the period and zero/pole location of $sn(x)$ then $log sn(x+iK'/2) - f(x)$ will be analytic $L$ periodic thus constant. As $L = overlineL$ then $f$ is purely imaginary for $x in BbbR$ thus $Re(sn(x+iK'/2))$ is constant for $x in BbbR$
$endgroup$
– reuns
Mar 19 at 23:53
$begingroup$
@reuns, are you saying that sn(𝑥+𝑖𝐾′/2,m) is not a circle?
$endgroup$
– brainjam
Mar 20 at 3:56
1
$begingroup$
I meant $Re(log sn(x+iK'/2))$ is constant
$endgroup$
– reuns
Mar 20 at 4:37
$begingroup$
It also appears that the diagram is symmetric under inversion in circle $C$, so for $x,yin mathbb R$ $$left | operatornamesn^2(x+(tfrac 1 2K'-y)i,m) operatornamesn^2(x+(tfrac 1 2K'+y)i,m) right | =1/m$$ or slightly more compactly: $$left | operatornamesn(x+(tfrac 1 2K'-y)i,m) operatornamesn(x+(tfrac 1 2K'+y)i,m) right | =1/k.$$ Also, $$ dfrac operatornamesn(x+(tfrac 1 2K'-y)i,m) operatornamesn(x+(tfrac 1 2K'+y)i,m) in mathbbR.$$
$endgroup$
– brainjam
Mar 22 at 16:29
1
1
$begingroup$
With the lattice $L = 2KBbbZ+2iK' BbbZ$ and $f(x) = sum_l in L log(1-fracxl+iK'/2)$ $+log(1-fracxl+iK'/2+K)-log(1-fracxl-iK'/2)-log(1-fracxl-iK'/2+K)$ from the period and zero/pole location of $sn(x)$ then $log sn(x+iK'/2) - f(x)$ will be analytic $L$ periodic thus constant. As $L = overlineL$ then $f$ is purely imaginary for $x in BbbR$ thus $Re(sn(x+iK'/2))$ is constant for $x in BbbR$
$endgroup$
– reuns
Mar 19 at 23:53
$begingroup$
With the lattice $L = 2KBbbZ+2iK' BbbZ$ and $f(x) = sum_l in L log(1-fracxl+iK'/2)$ $+log(1-fracxl+iK'/2+K)-log(1-fracxl-iK'/2)-log(1-fracxl-iK'/2+K)$ from the period and zero/pole location of $sn(x)$ then $log sn(x+iK'/2) - f(x)$ will be analytic $L$ periodic thus constant. As $L = overlineL$ then $f$ is purely imaginary for $x in BbbR$ thus $Re(sn(x+iK'/2))$ is constant for $x in BbbR$
$endgroup$
– reuns
Mar 19 at 23:53
$begingroup$
@reuns, are you saying that sn(𝑥+𝑖𝐾′/2,m) is not a circle?
$endgroup$
– brainjam
Mar 20 at 3:56
$begingroup$
@reuns, are you saying that sn(𝑥+𝑖𝐾′/2,m) is not a circle?
$endgroup$
– brainjam
Mar 20 at 3:56
1
1
$begingroup$
I meant $Re(log sn(x+iK'/2))$ is constant
$endgroup$
– reuns
Mar 20 at 4:37
$begingroup$
I meant $Re(log sn(x+iK'/2))$ is constant
$endgroup$
– reuns
Mar 20 at 4:37
$begingroup$
It also appears that the diagram is symmetric under inversion in circle $C$, so for $x,yin mathbb R$ $$left | operatornamesn^2(x+(tfrac 1 2K'-y)i,m) operatornamesn^2(x+(tfrac 1 2K'+y)i,m) right | =1/m$$ or slightly more compactly: $$left | operatornamesn(x+(tfrac 1 2K'-y)i,m) operatornamesn(x+(tfrac 1 2K'+y)i,m) right | =1/k.$$ Also, $$ dfrac operatornamesn(x+(tfrac 1 2K'-y)i,m) operatornamesn(x+(tfrac 1 2K'+y)i,m) in mathbbR.$$
$endgroup$
– brainjam
Mar 22 at 16:29
$begingroup$
It also appears that the diagram is symmetric under inversion in circle $C$, so for $x,yin mathbb R$ $$left | operatornamesn^2(x+(tfrac 1 2K'-y)i,m) operatornamesn^2(x+(tfrac 1 2K'+y)i,m) right | =1/m$$ or slightly more compactly: $$left | operatornamesn(x+(tfrac 1 2K'-y)i,m) operatornamesn(x+(tfrac 1 2K'+y)i,m) right | =1/k.$$ Also, $$ dfrac operatornamesn(x+(tfrac 1 2K'-y)i,m) operatornamesn(x+(tfrac 1 2K'+y)i,m) in mathbbR.$$
$endgroup$
– brainjam
Mar 22 at 16:29
add a comment |
0
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$begingroup$
With the lattice $L = 2KBbbZ+2iK' BbbZ$ and $f(x) = sum_l in L log(1-fracxl+iK'/2)$ $+log(1-fracxl+iK'/2+K)-log(1-fracxl-iK'/2)-log(1-fracxl-iK'/2+K)$ from the period and zero/pole location of $sn(x)$ then $log sn(x+iK'/2) - f(x)$ will be analytic $L$ periodic thus constant. As $L = overlineL$ then $f$ is purely imaginary for $x in BbbR$ thus $Re(sn(x+iK'/2))$ is constant for $x in BbbR$
$endgroup$
– reuns
Mar 19 at 23:53
$begingroup$
@reuns, are you saying that sn(𝑥+𝑖𝐾′/2,m) is not a circle?
$endgroup$
– brainjam
Mar 20 at 3:56
1
$begingroup$
I meant $Re(log sn(x+iK'/2))$ is constant
$endgroup$
– reuns
Mar 20 at 4:37
$begingroup$
It also appears that the diagram is symmetric under inversion in circle $C$, so for $x,yin mathbb R$ $$left | operatornamesn^2(x+(tfrac 1 2K'-y)i,m) operatornamesn^2(x+(tfrac 1 2K'+y)i,m) right | =1/m$$ or slightly more compactly: $$left | operatornamesn(x+(tfrac 1 2K'-y)i,m) operatornamesn(x+(tfrac 1 2K'+y)i,m) right | =1/k.$$ Also, $$ dfrac operatornamesn(x+(tfrac 1 2K'-y)i,m) operatornamesn(x+(tfrac 1 2K'+y)i,m) in mathbbR.$$
$endgroup$
– brainjam
Mar 22 at 16:29