Timeline for On $x^k+y^k=1$ and the Dixonian elliptic functions
Current License: CC BY-SA 3.0
18 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jan 16, 2017 at 16:18 | comment | added | Tito Piezas III | @Nemo: Now that has been clarified, so it turns out the Weierstrass form for $r=4$ just simply uses a Pythagorean triple, $$\mathrm{sin}_4^2\,z = \frac{2\lambda}{\lambda^2+1},\quad \mathrm{cos}_4^2\,z = \frac{\lambda^2-1}{\lambda^2+1}$$ with $\lambda=2\wp(z;1,0)$. | |
| Jan 16, 2017 at 16:09 | comment | added | Tito Piezas III | @Nemo: I finally figured out the issue. It's this choice between $m = k^2$. I was plugging in $k=1/\sqrt{2}$, when Mathematica demands $m = k^2 = 1/2$. I made some edits so it will be instantly clear for other readers. | |
| Jan 16, 2017 at 16:06 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Syntax
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| Jan 16, 2017 at 14:20 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Clarified for questions
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| Jan 15, 2017 at 15:33 | history | edited | Nemo | CC BY-SA 3.0 |
corrected typo
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| Jan 15, 2017 at 15:27 | comment | added | J. M. isn't a mathematician | @Nemo, sounds right; in any event, one could use $\wp^\prime(z;1,0)^2=4\wp(z;1,0)^3-\wp(z;1,0)$ to simplify things further. | |
| Jan 15, 2017 at 15:18 | comment | added | Nemo | @J.M. I think one can not write $\sin_{\frac43,4}z$ as a rational function $f(\wp(z),\wp'(z))$ for the following reason. Let $a$ be the root of $4\wp(z;1,0)^2+1=0$. Since $a$ is not a double root of this equation it follows from $\sin_{\frac43,4}z={\frac{2 \wp (z;1,0)}{\sqrt{\wp(z;1,0)(4\wp(z;1,0)^2+1)}}}$ that $\sin_{\frac43,4}z$ has a singularity $\frac{1}{\sqrt{z-a}}$ near $z=a$. But no rational function $f(\wp(z),\wp'(z))$ can have such a singularity. | |
| Jan 12, 2017 at 16:43 | comment | added | J. M. isn't a mathematician | I could derive it @Tito, but I don't foresee it looking just as nice. Let me get back to you on that... | |
| Jan 12, 2017 at 16:40 | comment | added | Tito Piezas III | @J.M.: Would you happen to know the closed-form of the partner for $r=4$ in terms of the Weierstrass function? | |
| Jan 12, 2017 at 16:29 | comment | added | J. M. isn't a mathematician | Whoops, typo; that should be "lemniscatic". | |
| Jan 12, 2017 at 16:25 | vote | accept | Tito Piezas III | ||
| Jan 12, 2017 at 16:09 | comment | added | J. M. isn't a mathematician | Also, note that $$\frac{1+\operatorname{cn}(u,k)}{1-\operatorname{cn}(u,k)}=\frac{\operatorname{cn}^2\left(\frac{u}{2},k\right)}{\operatorname{sn}^2\left(\frac{u}{2},k\right)\operatorname{dn}^2\left(\frac{u}{2},k\right)}$$ | |
| Jan 12, 2017 at 16:06 | comment | added | J. M. isn't a mathematician | I note that those Weierstrass functions for $r=4$ are of the "lemnisactic" type. That's why it's relatively easy to interconvert between them and the Jacobi ones. | |
| Jan 11, 2017 at 18:39 | history | edited | Nemo | CC BY-SA 3.0 |
Weierstrass form is added
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| Jan 11, 2017 at 7:25 | comment | added | Tito Piezas III | For completeness, since we already have $r=3$, kindly edit your answer and include the Weierstrass formulas for $r=4$ and $r=6$. Thanks. | |
| Jan 10, 2017 at 22:18 | history | edited | Nemo | CC BY-SA 3.0 |
added 493 characters in body
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| Jan 10, 2017 at 19:27 | history | answered | Nemo | CC BY-SA 3.0 |