[math-fun] EllipticE valuation [Was: EllipticK valuation [Was: Elliptic K π approximation]]
On Wed, Nov 7, 2012 at 3:05 PM, Bill Gosper <billgosper@gmail.com> wrote:
The ("complete") first kind elliptic integral K(m) and parameter m are both expressible in 𝜗 constants, and those three+ 𝜗 constants are all expressible in terms of Dedekind η. So
Out[840]= EllipticK[(16 η[q]^8 η[q^4]^16)/η[q^2]^24] == (π η[q^2]^10)/(2 η[q]^4 η[q^4]^4)
Crude but effective: d/dq and eliminate EllipticK to get EllipticE[(16 \[Eta][q^4]^8)/(\[Eta][q]^8 + 16 \[Eta][q^4]^8)] == -( 1/(\[Eta][q]^8 + 16 \[Eta][q^4]^8))(-((\[Pi] \[Eta][q]^4 \[Eta][q^2]^10)/( 2 \[Eta][ q^4]^4)) - (\[Pi] \[Eta][ q^2]^9 (6 Sqrt[3] q^(3/2) logderiveta[q] \[Eta][q] \[Eta][q^2] \[Eta][q^4] - 30 Sqrt[3] q^(5/2) logderiveta[q^2] \[Eta][q] \[Eta][q^2] \[Eta][q^4] + 24 Sqrt[3] q^(9/2) logderiveta[q^4] \[Eta][q] \[Eta][q^2] \[Eta][ q^4]) (\[Eta][q]^8 + 16 \[Eta][q^4]^8))/(2 q^( 1/3) \[Eta][q]^4 \[Eta][ q^4]^4 (6 Sqrt[3] q^(7/6) logderiveta[q] \[Eta][q] \[Eta][q^4] - 24 Sqrt[3] q^(25/6) logderiveta[q^4] \[Eta][q] \[Eta][q^4])) The remaining problem is those logderivetas, but lo, ((1/(4*x) - ((logderiveta[(1/(E^x))]))/(E^x))/((DedekindEta[((x*I)/(2*Pi))])^4)) is algebraic for (x/π)^2 rational. And for such x, we have many DedekindEta[((x*I)/(2*Pi)) in our collection. In fact, people.math.carleton.ca/~*williams*/papers/*pdf*/*299*.*pdf claims to give a complete solution. Early on, it appears to omit half the odd denominators, but I assume this is remedied later in the derivation. Anyway, as a fairly strenuous example: EllipticE[1/(1 + (-1 + GoldenRatio^(3/2))^8/(16 GoldenRatio^4))] == ( 2 5^(3/4) GoldenRatio^4 π^(3/2))/( Sqrt[1 + Sqrt[5]] (1525 + 682 Sqrt[5])^(1/4) Gamma[1/20] Gamma[9/ 20]) + ((-7 - 3 Sqrt[5] + Sqrt[190 + 85 Sqrt[5]]) Gamma[1/ 20] Gamma[9/20])/( 8 (1525 + 682 Sqrt[5])^(1/4) Sqrt[5 (1 + Sqrt[5]) π]) I'm having no luck finding a table of special values of EllipticE. This may be due to a simpler formula than above to produce them from EllipticK. It looks like they're something K + something/K. At the other extreme, has anyone seen *EllipticK[(-1)^(1/3)] == ((-1)^(1/12) 3^(1/4) Gamma[1/3]^3)/( 4 2^(1/3) π) ? --rwg
On 2012-11-10 19:54, Bill Gosper wrote:
On Wed, Nov 7, 2012 at 3:05 PM, Bill Gosper <billgosper@gmail.com> wrote:
The ("complete") first kind elliptic integral K(m) and parameter m are both expressible in 𝜗 constants, and those three+ 𝜗 constants are all expressible in terms of Dedekind η. So Out[840]= EllipticK[(16 η[q]^8 η[q^4]^16)/η[q^2]^24] == (π η[q^2]^10)/(2 η[q]^4 η[q^4]^4)
Crude but effective: d/dq and eliminate EllipticK to get EllipticE[(16 [Eta][q^4]^8)/([Eta][q]^8 + 16 [Eta][q^4]^8)] == -( 1/([Eta][q]^8 +
16 [Eta][q^4]^8))(-(([Pi] [Eta][q]^4 [Eta][q^2]^10)/(
2 [Eta][
q^4]^4)) - ([Pi] [Eta][
q^2]^9 (6 Sqrt[3] q^(3/2) logderiveta[q] [Eta][q] [Eta][q^2] [Eta][q^4] - 30 Sqrt[3] q^(5/2) logderiveta[q^2] [Eta][q] [Eta][q^2] [Eta][q^4] + 24 Sqrt[3] q^(9/2) logderiveta[q^4] [Eta][q] [Eta][q^2] [Eta][ q^4]) ([Eta][q]^8 + 16 [Eta][q^4]^8))/(2 q^( 1/3) [Eta][q]^4 [Eta][ q^4]^4 (6 Sqrt[3] q^(7/6)
logderiveta[q] [Eta][q] [Eta][q^4] -
24 Sqrt[3] q^(25/6) logderiveta[q^4] [Eta][q] [Eta][q^4]))
The remaining problem is those logderivetas, but lo, ((1/(4*x) -
((logderiveta[(1/(E^x))]))/(E^x))/((DedekindEta[((x*I)/(2*Pi))])^4))
is algebraic for (x/π)^2 rational.
Unfortunately, (x/π)^2=
(i/2-1/√12)^2 (as in K((-1)^(1/3))) isn't rational, so
EllipticE[(-1)^(1/3)] ==
E^(-(I π/12)) ((2 2^(1/3) π^2)/(3^(3/4) Gamma[1/3]^3) + Gamma[1/3]^3/(4 2^(1/3) 3^(1/4) π))
required Legendre's relation plus numerical empiricism.
This, with the K((-1)^(1/3)) result gives closed forms for all
2F1[a+1/2,b+1/2,c;(-1)^(1/3)], a,b,c integers.
--rwg And for such
x, we have many DedekindEta[((x*I)/(2*Pi)) in our collection. In fact, people.math.carleton.ca/~*williams*/papers/*pdf*/*299*.*pdf claims to give a complete solution. Early on, it appears to omit half the odd denominators, but I assume this is remedied later in the derivation. Anyway, as a fairly strenuous example: EllipticE[1/(1 + (-1 + GoldenRatio^(3/2))^8/(16 GoldenRatio^4))] == ( 2 5^(3/4) GoldenRatio^4 π^(3/2))/( Sqrt[1 + Sqrt[5]] (1525 + 682 Sqrt[5])^(1/4) Gamma[1/20] Gamma[9/ 20]) + ((-7 - 3 Sqrt[5] + Sqrt[190 + 85 Sqrt[5]]) Gamma[1/ 20] Gamma[9/20])/( 8 (1525 + 682 Sqrt[5])^(1/4) Sqrt[5 (1 + Sqrt[5]) π]) I'm having no luck finding a table of special values of EllipticE. This may be due to a simpler formula than above to produce them from EllipticK. It looks like they're something K + something/K. At the other extreme, has anyone seen *EllipticK[(-1)^(1/3)] == ((-1)^(1/12) 3^(1/4) Gamma[1/3]^3)/( 4 2^(1/3) π) ? --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun [1] Links: ------ [1] http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Hi, Just a quick thought I had the other day, as I understand it Fermat's last theorem (now proved) basically says that for: a^p + b^p = c^p Then where all variables are integers there are no solutions for a, b and c where p>2. My thought was has anyone considered: a^p + b^p + c^p = d^p or indeed: a1^p + a2^p + a3^p + ..... an^p = b^p And is it possible that for the case of: a^p + b^p + c^p = d^p Then there is a solution for a,b,c,d for p=3 but not for p>3 and generally for: a1^p + a2^p + a3^p + ..... an^p = b^p there's a solution for a1..an and b if p=n but not for p>n ?
David Makin:
My thought was has anyone considered: a^p + b^p + c^p = d^p or indeed: a1^p + a2^p + a3^p + ..... an^p = b^p
I was thinking more in terms of general proof for all cases ;) On 12 Nov 2012, at 12:47, Hans Havermann wrote:
David Makin:
My thought was has anyone considered: a^p + b^p + c^p = d^p or indeed: a1^p + a2^p + a3^p + ..... an^p = b^p
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Counterexample found by Lander & Parkin in 1966: 27^5 + 84^5 + 110^5 + 133^5 = 144^5 http://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture http://mathworld.wolfram.com/EulersSumofPowersConjecture.html http://euler.free.fr/index.htm Christian. -----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de David Makin Envoyé : lundi 12 novembre 2012 13:30 À : math-fun Objet : [math-fun] Fermat's last theorem Hi, Just a quick thought I had the other day, as I understand it Fermat's last theorem (now proved) basically says that for: a^p + b^p = c^p Then where all variables are integers there are no solutions for a, b and c where p>2. My thought was has anyone considered: a^p + b^p + c^p = d^p or indeed: a1^p + a2^p + a3^p + ..... an^p = b^p And is it possible that for the case of: a^p + b^p + c^p = d^p Then there is a solution for a,b,c,d for p=3 but not for p>3 and generally for: a1^p + a2^p + a3^p + ..... an^p = b^p there's a solution for a1..an and b if p=n but not for p>n ? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Fourth power counterexample: 95800^4 + 217519^4 + 414560^4 = 422481^4 . Not sure if this is Noam Elkies's counterexample or a minimal fourth power counterexample found later. On 11/12/2012 8:14 AM, Christian Boyer wrote:
Counterexample found by Lander & Parkin in 1966: 27^5 + 84^5 + 110^5 + 133^5 = 144^5
http://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture http://mathworld.wolfram.com/EulersSumofPowersConjecture.html http://euler.free.fr/index.htm
Christian.
Perhaps this link would be of interest: https://sites.google.com/site/tpiezas/Home On Mon, Nov 12, 2012 at 6:29 AM, David Makin <makinmagic@tiscali.co.uk> wrote:
Hi,
Just a quick thought I had the other day, as I understand it Fermat's last theorem (now proved) basically says that for:
a^p + b^p = c^p
Then where all variables are integers there are no solutions for a, b and c where p>2.
My thought was has anyone considered:
a^p + b^p + c^p = d^p
or indeed:
a1^p + a2^p + a3^p + ..... an^p = b^p
And is it possible that for the case of:
a^p + b^p + c^p = d^p
Then there is a solution for a,b,c,d for p=3 but not for p>3 and generally for:
a1^p + a2^p + a3^p + ..... an^p = b^p
there's a solution for a1..an and b if p=n but not for p>n ? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (7)
-
Bill Gosper -
Christian Boyer -
David Makin -
David Wilson -
Hans Havermann -
James Buddenhagen -
rwg