Re: [math-fun] New(?) elliptic K transformation
On Mon, Nov 5, 2012 at 11:44 PM, Bill Gosper <billgosper@gmail.com> wrote: Purely empirically, for 0<r<1/2, EllipticK[4 E^(I ArcTan[1/2 Sqrt[-4 + 1/(1 - r) + 1/r]]) Sqrt[(1- r) r]] == E^(I ArcTan[Sqrt[r/(1 - r)]]) EllipticK[r] E.g., for r=1/4 EllipticK[Sqrt[3] E^(I*π/6)] == E^(I*π/6) EllipticK[1/4] For r=1/2, it gets the conjugate of the right answer, which is EllipticK[2] == -(-1)^(3/4) EllipticK[1/2] which can be found by listing all eight sign variations of the three square roots. Then we can try complex r: EllipticK[1/2 + I/2] == (-3 + 2 Sqrt[2])^(1/4) EllipticK[4 - 2 Sqrt[2]] which puts the lhs in polar form. [Added 2012-11-16 07:03 Namely, EllipticK[1/2 + I/2] == Sqrt[-1 + Sqrt[2]] E^(I π/4) EllipticK[4 - 2 Sqrt[2]] And for r = I, EllipticK[(8 + 8 I) - 4 Sqrt[-1 + 7 I]] == (2^(1/4) E^(-(I*π/8)) - I E^(I*π/4)) EllipticK[I] ] None of the eight sign guesses works for Re(r)>1/2. If no one finds a proof of this stuff, imagine some future Mathematica just trying everything and choosing the numerically plausible one. --rwg No one seemed to recognize it, so, grotesque though it is, NeilB and I decided it's time to write this up. So Neil bashed on it with Mathematica, and it got a simpler. Then I bashed on it so hard that it got unnervingly ungrotesque: (*) EllipticK[1/4 (2 + (-2 + t)/Sqrt[1 - t])] == (1 - t)^(1/4) EllipticK[t] How could anything this simple be new? It looks like a routine 2F1 quadratic transformation. But which? As before, it makes the strange little results EllipticK[1/4] == E^(-((I \[Pi])/6)) EllipticK[Sqrt[3] E^((I \[Pi])/6)] and EllipticK[I] == (E^((I \[Pi])/16) EllipticK[1/2 - 1/4 Sqrt[7/2 - I/2]])/2^(1/8) Are these familiar? And somehow, (*) has miraculously lost its Re[t]<1/2 restriction:. E.g. for t=2, EllipticK[1/2] == (-1)^(1/4) EllipticK[2], as one might hope. For t=(-1)^(1/6), EllipticK[1/4 (2 + (-2 + (-1)^(1/6))/Sqrt[1 - (-1)^(1/6)])] == (2 - Sqrt[3])^(1/8) E^(-((5 I \[Pi])/48)) EllipticK[E^((I \[Pi])/6)] Angling for a K'/K identity, solve 1/4 (2 + (-2 + t)/Sqrt[1 - t]) == 1 - t : t -> 1/32 (31 + 3 I Sqrt[7]) So, Log[EllipticNomeQ[1/32 (31 - 3 I Sqrt[7])]] == -(1/2) Sqrt[1/2 (-3 I + Sqrt[7])] E^((I \[Pi])/4) \[Pi] along with its formal conjugate. Stop me if you've heard this one. --rwg (Is there a (dlmf?) page of special values of nome? Or just garden varieties:-)
On Sun, May 24, 2015 at 2:48 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Mon, Nov 5, 2012 at 11:44 PM, Bill Gosper <billgosper@gmail.com> wrote:
Purely empirically, for 0<r<1/2, EllipticK[4 E^(I ArcTan[1/2 Sqrt[-4 + 1/(1 - r) + 1/r]]) Sqrt[(1- r) r]] == E^(I ArcTan[Sqrt[r/(1 - r)]]) EllipticK[r]
E.g., for r=1/4 EllipticK[Sqrt[3] E^(I*π/6)] == E^(I*π/6) EllipticK[1/4]
For r=1/2, it gets the conjugate of the right answer, which is EllipticK[2] == -(-1)^(3/4) EllipticK[1/2] which can be found by listing all eight sign variations of the three square roots. Then we can try complex r: EllipticK[1/2 + I/2] == (-3 + 2 Sqrt[2])^(1/4) EllipticK[4 - 2 Sqrt[2]] which puts the lhs in polar form.
[Added 2012-11-16 07:03 Namely, EllipticK[1/2 + I/2] == Sqrt[-1 + Sqrt[2]] E^(I π/4) EllipticK[4 - 2 Sqrt[2]]
And for r = I, EllipticK[(8 + 8 I) - 4 Sqrt[-1 + 7 I]] == (2^(1/4) E^(-(I*π/8)) - I E^(I*π/4)) EllipticK[I] ]
None of the eight sign guesses works for Re(r)>1/2.
If no one finds a proof of this stuff, imagine some future Mathematica just trying everything and choosing the numerically plausible one. --rwg
No one seemed to recognize it, so, grotesque though it is, NeilB and I decided it's time to write this up. So Neil bashed on it with Mathematica, and it got a simpler. Then I bashed on it so hard that it got unnervingly ungrotesque:
(*) EllipticK[1/4 (2 + (-2 + t)/Sqrt[1 - t])] == (1 - t)^(1/4) EllipticK[t]
How could anything this simple be new? It looks like a routine 2F1 quadratic transformation. But which? As before, it makes the strange little results EllipticK[1/4] == E^(-((I \[Pi])/6)) EllipticK[Sqrt[3] E^((I \[Pi])/6)] and EllipticK[I] == (E^((I \[Pi])/16) EllipticK[1/2 - 1/4 Sqrt[7/2 - I/2]])/2^(1/8)
Are these familiar? And somehow, (*) has miraculously lost its Re[t]<1/2 restriction:. E.g. for t=2,
EllipticK[1/2] == (-1)^(1/4) EllipticK[2], as one might hope.
For t=(-1)^(1/6),
EllipticK[1/4 (2 + (-2 + (-1)^(1/6))/Sqrt[1 - (-1)^(1/6)])] == (2 - Sqrt[3])^(1/8) E^(-((5 I \[Pi])/48)) EllipticK[E^((I \[Pi])/6)]
Angling for a K'/K identity, solve 1/4 (2 + (-2 + t)/Sqrt[1 - t]) == 1 - t :
t -> 1/32 (31 + 3 I Sqrt[7])
So, Log[EllipticNomeQ[1/32 (31 - 3 I Sqrt[7])]] == -(1/2) Sqrt[1/2 (-3 I + Sqrt[7])] E^((I \[Pi])/4) \[Pi] along with its formal conjugate.
Stop me if you've heard this one. --rwg
(Is there a (dlmf?) page of special values of nome? Or just garden varieties:-)
I have a computer. Why would I want a table? -(Pi/Sqrt[2]) EllipticNomeQ[-2 + 2 Sqrt[2]] == E 2 + Sqrt[3] -(Pi/Sqrt[3]) EllipticNomeQ[-----------] == E 4 -Pi/2 EllipticNomeQ[-16 + 12 Sqrt[2]] == {E } 1 -(Pi/Sqrt[5]) EllipticNomeQ[- + Sqrt[-2 + Sqrt[5]]] == {E } 2 -(Pi/Sqrt[6]) EllipticNomeQ[-34 - 24 Sqrt[2] + 20 Sqrt[3] + 14 Sqrt[6]] == {E } 8 + 3 Sqrt[7] -(Pi/Sqrt[7]) EllipticNomeQ[-------------] == {E } 16 -Pi/(2 Sqrt[2]) EllipticNomeQ[-112 - 80 Sqrt[2] + 4 Sqrt[2 (799 + 565 Sqrt[2])]] == {E } 1/4 3/4 8 + 32 Sqrt[2] 3 - 16 Sqrt[2] 3 -Pi/3 EllipticNomeQ[-------------------------------------] == {E } 16 -(Pi/Sqrt[10]) EllipticNomeQ[-322 + 228 Sqrt[2] - 144 Sqrt[5] + 102 Sqrt[10]] == {E } [Nome^-1[e^-(pi/rt11) under destruction] -Pi/(2 Sqrt[3]) EllipticNomeQ[-832 + 588 Sqrt[2] - 480 Sqrt[3] + 340 Sqrt[6]] == {E } 8 + 48 Sqrt[-18 + 5 Sqrt[13]] -(Pi/Sqrt[13]) EllipticNomeQ[-----------------------------] == {E } 16 EllipticNomeQ[-994 - 704 Sqrt[2] + 376 Sqrt[7] + 266 Sqrt[14] + -(Pi/Sqrt[14]) 4 Sqrt[247473 + 174990 Sqrt[2] - 93536 Sqrt[7] - 66140 Sqrt[14]]] == {E } 32768 + 14336 Sqrt[3] + 2048 Sqrt[15] -(Pi/Sqrt[15]) EllipticNomeQ[-------------------------------------] == {E } 65536 1/4 3/4 -Pi/4 EllipticNomeQ[-4480 + 3768 2 - 3168 Sqrt[2] + 2664 2 ] == {E } In the 50s-60s computers were divided into "Business" and "Scientific". It was believed that the world needed only a handful of scientific computers because their results could be tabulated in books such as Abramowitz & Stegun. Plus one more to predict global weather. As late as the 1972 SYMSAC in Yorktown Heights, W. Kahan ridiculed computer algebra as "The pen that writes under water", a mere parasite on the economies of scale created by the demand for business machines. He was comparing symbolic integration to the Fisher Space Pen, developed for the Mercury astronauts. He could only imagine computer algebra as a symbolic calculator, assisting pen and paper. Much like the MIT math department of the time. --rwg Warren wondered about other special K's. In a math-fun Date 2012-11-15 02:27I mentioned EllipticK[(1 + Sqrt[2])^4] == ((1/4 - I/2)π^(3/2))/ ((2 + Sqrt[2]) Gamma[3/4]^2) and EllipticK[8 2^(1/4) (-1 + Sqrt[2])^3* E^(I ArcTan[(3 (3 + Sqrt[2]))/(4 2^(1/4))])] == (I (1 + Sqrt[2]) E^(-2 I ArcCot[2^(1/4)]) π^(3/2))/ (4 Sqrt[2] Gamma[3/4]^2) and lamented the published scarcity of such results. Maybe NeilB & I should write something. Here's hoping Warren can fix his non-algebraic reformulation of the K relation.
For the nome table below EllipticNomeQ[ 1/2 + Sqrt[11]/12 - 1/6 (-21 Sqrt[3] + 11 Sqrt[11])^(1/3) - 1/6 (21 Sqrt[3] + 11 Sqrt[11])^(1/3)] == E^(-Sqrt[11] \[Pi]) On 2015-05-24 23:00, Bill Gosper wrote:
On Sun, May 24, 2015 at 2:48 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Mon, Nov 5, 2012 at 11:44 PM, Bill Gosper <billgosper@gmail.com> wrote: Purely empirically, for 0<r<1/2, EllipticK[4 E^(I ArcTan[1/2 Sqrt[-4 + 1/(1 - r) + 1/r]]) Sqrt[(1- r) r]] == E^(I ArcTan[Sqrt[r/(1 - r)]]) EllipticK[r]
E.g., for r=1/4 EllipticK[Sqrt[3] E^(I*π/6)] == E^(I*π/6) EllipticK[1/4]
For r=1/2, it gets the conjugate of the right answer, which is EllipticK[2] == -(-1)^(3/4) EllipticK[1/2] which can be found by listing all eight sign variations of the three square roots. Then we can try complex r: EllipticK[1/2 + I/2] == (-3 + 2 Sqrt[2])^(1/4) EllipticK[4 - 2 Sqrt[2]] which puts the lhs in polar form. [Added 2012-11-16 07:03
Namely, EllipticK[1/2 + I/2] == Sqrt[-1 + Sqrt[2]] E^(I π/4) EllipticK[4 - 2 Sqrt[2]]
And for r = I, EllipticK[(8 + 8 I) - 4 Sqrt[-1 + 7 I]] == (2^(1/4) E^(-(I*π/8)) - I E^(I*π/4)) EllipticK[I] ]
None of the eight sign guesses works for Re(r)>1/2.
If no one finds a proof of this stuff, imagine some future Mathematica just trying everything and choosing the numerically plausible one. --rwg
No one seemed to recognize it, so, grotesque though it is, NeilB and I decided it's time to write this up. So Neil bashed on it with Mathematica, and it got a simpler. Then I bashed on it so hard that it got unnervingly ungrotesque:
(*) EllipticK[1/4 (2 + (-2 + t)/Sqrt[1 - t])] == (1 - t)^(1/4) EllipticK[t]
How could anything this simple be new? It looks like a routine 2F1 quadratic transformation. But which? As before, it makes the strange little results EllipticK[1/4] == E^(-((I \[Pi])/6)) EllipticK[Sqrt[3] E^((I \[Pi])/6)] and EllipticK[I] == (E^((I \[Pi])/16) EllipticK[1/2 - 1/4 Sqrt[7/2 - I/2]])/2^(1/8)
Are these familiar? And somehow, (*) has miraculously lost its Re[t]<1/2 restriction:. E.g. for t=2,
EllipticK[1/2] == (-1)^(1/4) EllipticK[2], as one might hope.
For t=(-1)^(1/6),
EllipticK[1/4 (2 + (-2 + (-1)^(1/6))/Sqrt[1 - (-1)^(1/6)])] == (2 - Sqrt[3])^(1/8) E^(-((5 I \[Pi])/48)) EllipticK[E^((I \[Pi])/6)]
Angling for a K'/K identity, solve 1/4 (2 + (-2 + t)/Sqrt[1 - t]) == 1 - t :
t -> 1/32 (31 + 3 I Sqrt[7])
So, Log[EllipticNomeQ[1/32 (31 - 3 I Sqrt[7])]] == -(1/2) Sqrt[1/2 (-3 I + Sqrt[7])] E^((I \[Pi])/4) \[Pi] along with its formal conjugate.
Stop me if you've heard this one. --rwg
(Is there a (dlmf?) page of special values of nome? Or just garden varieties:-)
I have a computer. Why would I want a table? -(Pi/Sqrt[2]) EllipticNomeQ[-2 + 2 Sqrt[2]] == E 2 + Sqrt[3] -(Pi/Sqrt[3]) EllipticNomeQ[-----------] == E 4 -Pi/2 EllipticNomeQ[-16 + 12 Sqrt[2]] == {E } 1 -(Pi/Sqrt[5]) EllipticNomeQ[- + Sqrt[-2 + Sqrt[5]]] == {E } 2 -(Pi/Sqrt[6]) EllipticNomeQ[-34 - 24 Sqrt[2] + 20 Sqrt[3] + 14 Sqrt[6]] == {E } 8 + 3 Sqrt[7] -(Pi/Sqrt[7]) EllipticNomeQ[-------------] == {E } 16 -Pi/(2 Sqrt[2]) EllipticNomeQ[-112 - 80 Sqrt[2] + 4 Sqrt[2 (799 + 565 Sqrt[2])]] == {E } 1/4 3/4 8 + 32 Sqrt[2] 3 - 16 Sqrt[2] 3 -Pi/3 EllipticNomeQ[-------------------------------------] == {E } 16 -(Pi/Sqrt[10]) EllipticNomeQ[-322 + 228 Sqrt[2] - 144 Sqrt[5] + 102 Sqrt[10]] == {E } [Nome^-1[e^-(pi/rt11) under destruction] -Pi/(2 Sqrt[3]) EllipticNomeQ[-832 + 588 Sqrt[2] - 480 Sqrt[3] + 340 Sqrt[6]] == {E } 8 + 48 Sqrt[-18 + 5 Sqrt[13]] -(Pi/Sqrt[13]) EllipticNomeQ[-----------------------------] == {E } 16 EllipticNomeQ[-994 - 704 Sqrt[2] + 376 Sqrt[7] + 266 Sqrt[14] + -(Pi/Sqrt[14]) 4 Sqrt[247473 + 174990 Sqrt[2] - 93536 Sqrt[7] - 66140 Sqrt[14]]] == {E } 32768 + 14336 Sqrt[3] + 2048 Sqrt[15] -(Pi/Sqrt[15]) EllipticNomeQ[-------------------------------------] == {E } 65536 1/4 3/4 -Pi/4 EllipticNomeQ[-4480 + 3768 2 - 3168 Sqrt[2] + 2664 2 ] == {E } In the 50s-60s computers were divided into "Business" and "Scientific". It was believed that the world needed only a handful of scientific computers because their results could be tabulated in books such as Abramowitz & Stegun. Plus one more to predict global weather. As late as the 1972 SYMSAC in Yorktown Heights, W. Kahan ridiculed computer algebra as "The pen that writes under water", a mere parasite on the economies of scale created by the demand for business machines. He was comparing symbolic integration to the Fisher Space Pen, developed for the Mercury astronauts. He could only imagine computer algebra as a symbolic calculator, assisting pen and paper. Much like the MIT math department of the time. --rwg Warren wondered about other special K's. In a math-fun Date 2012-11-15 02:27I mentioned EllipticK[(1 + Sqrt[2])^4] == ((1/4 - I/2)π^(3/2))/ ((2 + Sqrt[2]) Gamma[3/4]^2) and EllipticK[8 2^(1/4) (-1 + Sqrt[2])^3* E^(I ArcTan[(3 (3 + Sqrt[2]))/(4 2^(1/4))])] == (I (1 + Sqrt[2]) E^(-2 I ArcCot[2^(1/4)]) π^(3/2))/ (4 Sqrt[2] Gamma[3/4]^2) and lamented the published scarcity of such results. Maybe NeilB & I should write something. Here's hoping Warren can fix his non-algebraic reformulation of the K relation. -- You received this message because you are subscribed to the Google Groups "mathfuneavesdroppers" group. To unsubscribe from this group and stop receiving emails from it, send an email to mathfuneavesdroppers+unsubscribe@googlegroups.com. For more options, visit https://groups.google.com/d/optout [1]. Links: ------ [1] https://groups.google.com/d/optout
The EllipticE analog of EllipticK[1/4 (2 - (1 + t)/Sqrt[t])] == t^(1/4) EllipticK[1 - t] is evidently 2 EllipticE[1/4 (2 - (1 +t)/Sqrt[t])] == EllipticE[1 - t]/t^(1/4) + t^(1/4) EllipticK[1 - t] For a test,see gosper.org/newExfrmation.png EllipticE and EllipticK are contiguous 2F1s. You always want both. --rwg (Does anyone have a clue how to prove these?) On Sun, May 24, 2015 at 2:48 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Mon, Nov 5, 2012 at 11:44 PM, Bill Gosper <billgosper@gmail.com> wrote:
Purely empirically, for 0<r<1/2, EllipticK[4 E^(I ArcTan[1/2 Sqrt[-4 + 1/(1 - r) + 1/r]]) Sqrt[(1- r) r]] == E^(I ArcTan[Sqrt[r/(1 - r)]]) EllipticK[r]
E.g., for r=1/4 EllipticK[Sqrt[3] E^(I*π/6)] == E^(I*π/6) EllipticK[1/4]
For r=1/2, it gets the conjugate of the right answer, which is EllipticK[2] == -(-1)^(3/4) EllipticK[1/2] which can be found by listing all eight sign variations of the three square roots. Then we can try complex r: EllipticK[1/2 + I/2] == (-3 + 2 Sqrt[2])^(1/4) EllipticK[4 - 2 Sqrt[2]] which puts the lhs in polar form.
[Added 2012-11-16 07:03 Namely, EllipticK[1/2 + I/2] == Sqrt[-1 + Sqrt[2]] E^(I π/4) EllipticK[4 - 2 Sqrt[2]]
And for r = I, EllipticK[(8 + 8 I) - 4 Sqrt[-1 + 7 I]] == (2^(1/4) E^(-(I*π/8)) - I E^(I*π/4)) EllipticK[I] ]
None of the eight sign guesses works for Re(r)>1/2.
If no one finds a proof of this stuff, imagine some future Mathematica just trying everything and choosing the numerically plausible one. --rwg
No one seemed to recognize it, so, grotesque though it is, NeilB and I decided it's time to write this up. So Neil bashed on it with Mathematica, and it got a simpler. Then I bashed on it so hard that it got unnervingly ungrotesque:
(*) EllipticK[1/4 (2 + (-2 + t)/Sqrt[1 - t])] == (1 - t)^(1/4) EllipticK[t]
How could anything this simple be new? It looks like a routine 2F1 quadratic transformation. But which? As before, it makes the strange little results EllipticK[1/4] == E^(-((I \[Pi])/6)) EllipticK[Sqrt[3] E^((I \[Pi])/6)] and EllipticK[I] == (E^((I \[Pi])/16) EllipticK[1/2 - 1/4 Sqrt[7/2 - I/2]])/2^(1/8)
Are these familiar? And somehow, (*) has miraculously lost its Re[t]<1/2 restriction:. E.g. for t=2,
EllipticK[1/2] == (-1)^(1/4) EllipticK[2], as one might hope.
For t=(-1)^(1/6),
EllipticK[1/4 (2 + (-2 + (-1)^(1/6))/Sqrt[1 - (-1)^(1/6)])] == (2 - Sqrt[3])^(1/8) E^(-((5 I \[Pi])/48)) EllipticK[E^((I \[Pi])/6)]
Angling for a K'/K identity, solve 1/4 (2 + (-2 + t)/Sqrt[1 - t]) == 1 - t :
t -> 1/32 (31 + 3 I Sqrt[7])
So, Log[EllipticNomeQ[1/32 (31 - 3 I Sqrt[7])]] == -(1/2) Sqrt[1/2 (-3 I + Sqrt[7])] E^((I \[Pi])/4) \[Pi] along with its formal conjugate.
Stop me if you've heard this one. --rwg
(Is there a (dlmf?) page of special values of nome? Or just garden varieties:-)
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rwg