The equation merely stating that a pendulum maxes out at a quarter-period leads to the identities JacobiAmplitude[y EllipticK[y^2], 1/y^2] == JacobiAmplitude[InverseJacobiSN[y, m], m] == ArcSin[y], at least for y and m in some neighborhood of 0. Taking sin, JacobiSN[y EllipticK[y^2], 1/y^2] == y, So for every K special value we get one for sn. Along the way, -(\[Pi]/2) + 2 ArcTan[(E^ArcTanh[y]) FullSimplified to Gudermannian[ArcTanh[y]] instead of ArcSin[y] ! --rwg 2011/4/11 Bill Gosper <billgosper@gmail.com>
2011/4/11 Bill Gosper <billgosper@gmail.com>
2011/4/11 Bill Gosper <billgosper@gmail.com>
(One of the few entries where Apple OS-X's built-in Oxford American Dictionary beats the PC MW Unabridged, which neglects the interjection. Though it does have gardyloo!, which OAD lacks.)
Once Neil seemed safely distracted by Minskyspace and sliding piece puzzle enumeration, I asked him how long it took for a pole at angle alpha from vertical to fall flat onto a nonskid surface, but he recognized it immediately as just another pendulum and we didn't work out the details. Once I was safely distracted by Neil's investigations, I forgot the factor of 2 on the pendulum equation of motion, and imagined the answer would be a pretty difference of complete EllipticKs. In truth, (measuring g as >0)
T==Sqrt[2/3]*Sqrt[L/g]*(2*EllipticK[Cos[α/2]^2] - EllipticF[Pi/4, Sec[α /2]^2]*Sec[α/2])
2 L α 2 Pi α 2 α T == Sqrt[-] Sqrt[-] (2 EllipticK[Cos[-] ] - EllipticF[--, Sec[-] ] Sec[-]) 3 g 2 4 2 2
(Because a stick pendulum acts as if all its mass were 2/3 of the way to the free end.)
Oops, no that doesn't vanish as α -> π/2 -- I used the semiperiod instead of the quarterperiod!
T==Sqrt[2/3]*Sqrt[L/g]*(EllipticK[Cos[α/2]^2] - EllipticF[Pi/4, Sec[α /2]^2]*Sec[α/2])
2 L α 2 Pi α 2 α T == Sqrt[-] Sqrt[-] (EllipticK[Cos[-] ] - EllipticF[--, Sec[-] ] Sec[-])
3 g 2 4 2 2
This, of course, raises
and answers
the question of whether EllipticF[Pi/4,...] has nice special values.
EllipticF[Pi/4, 2] == EllipticK[1/2]/Sqrt[2] == (4*Sqrt[2]*Pi^(3/2))/Gamma[-(1/4)]^2
1 EllipticK[-] 3/2 Pi 2 4 Sqrt[2] Pi EllipticF[--, 2] == ------------ == --------------- 4 Sqrt[2] 1 2 Gamma[-(-)] 4
More generally, EllipticF.nb in functions.wolfram.com gives
In[380]:= EllipticF[x, Csc[x]^2] == EllipticK[Sin[x]^2] Sin[x]
E.g., In[385]:=
EllipticF[ArcSin[3 - 2 Sqrt[2]], (1 + Sqrt[2])^4] == (2 (2 - Sqrt[2]) π ^(3/2))/Gamma[-(1/4)]^2
Out[385]=
3/2 4 2 (2 - Sqrt[2]) Pi EllipticF[ArcSin[3 - 2 Sqrt[2]], (1 + Sqrt[2]) ] == --------------------- 1 2 Gamma[-(-)] 4
Are there any not of this form?
--rwg