[math-fun] Amazing hit by ries
Integrate[((2 + 1/Sqrt[3 + t])*ArcSec[t])/(Sqrt[1 + t]*(2 + t)), {t, 1, 2}] == Pi^2/30 + (1/6)*Pi*ArcSin[(1/16)*(7 - 3*Sqrt[5])] The integral {t,1,6} = 2 π^2/15 = Coxeter's still-mysterious orthoscheme integral. --rwg
I didn’t know about Coxeter’s orthoscheme integral. Why 15? Someone should write an article on the combinatorics and numerology of fractions that keep company with pi and its powers. Jim Propp On Wednesday, July 4, 2018, Bill Gosper <billgosper@gmail.com> wrote:
Integrate[((2 + 1/Sqrt[3 + t])*ArcSec[t])/(Sqrt[1 + t]*(2 + t)), {t, 1, 2}] == Pi^2/30 + (1/6)*Pi*ArcSin[(1/16)*(7 - 3*Sqrt[5])]
The integral {t,1,6} = 2 π^2/15 = Coxeter's still-mysterious orthoscheme integral. --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
https://wikivividly.com/wiki/Schl%C3%A4fli_orthoscheme ; paper #9 in https://books.google.ie/books?isbn=0471010030 ; WFL On 7/4/18, James Propp <jamespropp@gmail.com> wrote:
I didn’t know about Coxeter’s orthoscheme integral. Why 15?
Someone should write an article on the combinatorics and numerology of fractions that keep company with pi and its powers.
Jim Propp
On Wednesday, July 4, 2018, Bill Gosper <billgosper@gmail.com> wrote:
Integrate[((2 + 1/Sqrt[3 + t])*ArcSec[t])/(Sqrt[1 + t]*(2 + t)), {t, 1, 2}] == Pi^2/30 + (1/6)*Pi*ArcSin[(1/16)*(7 - 3*Sqrt[5])]
The integral {t,1,6} = 2 π^2/15 = Coxeter's still-mysterious orthoscheme integral. --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On 04/07/2018 14:56, Fred Lunnon wrote:
So far as I can see, the difference between https://wikivividly.com/wiki/X and https://en.wikipedia.org/wiki/X is that the former is festooned with scammy advertisements and images from vaguely related Wikipedia pages. May I suggest that Wikipedia links should be to the One True Wikipedia? -- g
They usually are --- this one must have sneaked through under cover of too many late nights! WFL On 7/18/18, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On 04/07/2018 14:56, Fred Lunnon wrote:
So far as I can see, the difference between
https://wikivividly.com/wiki/X
and
https://en.wikipedia.org/wiki/X
is that the former is festooned with scammy advertisements and images from vaguely related Wikipedia pages. May I suggest that Wikipedia links should be to the One True Wikipedia?
-- g
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I suspect there are more of these, but they seem just out of reach. E.g., Integrate[ArcSec@t/Sqrt[2 + t]/Sqrt[(t + 1)^3], {t, 1, 2}] = (1/Sqrt[2] - 4/(3 Sqrt[3])) π + Sqrt[2] ArcSin[17/81], which Mathematica can do indefinitely(!), but -l8 runs me out of memory(?): ries/ries -l8 .1020759204661688653138 [. . .] ln(tanpi(6 x)^2)-1 = e-e^(1/ln(2 pi)) for x = T - 4.14884e-15 {175} sqrt(tanpi(sqrt(9"/2 x))) = 1-ln(ln(sqrt(e+phi))) for x = T + 1.38825e-15 {180} ln(tanpi(sqrt(x)+1/9^2)) = 1/((log_(5 phi)(6))+1) for x = T + 1.69075e-16 {179} Killed: 9 -l7 reaches the first of these, but can't quite express the answer. --rwg On Wed, Jul 4, 2018 at 5:34 AM Bill Gosper <billgosper@gmail.com> wrote:
Integrate[((2 + 1/Sqrt[3 + t])*ArcSec[t])/(Sqrt[1 + t]*(2 + t)), {t, 1, 2}] == Pi^2/30 + (1/6)*Pi*ArcSin[(1/16)*(7 - 3*Sqrt[5])]
The integral {t,1,6} = 2 π^2/15 = Coxeter's still-mysterious orthoscheme integral. --rwg
On Wed, Jul 4, 2018 at 5:34 AM Bill Gosper <billgosper@gmail.com> wrote:
Integrate[((2 + 1/Sqrt[3 + t])*ArcSec[t])/(Sqrt[1 + t]*(2 + t)), {t, 1, 2}] == Pi^2/30 + (1/6)*Pi*ArcSin[(1/16)*(7 - 3*Sqrt[5])]
The integral {t,1,6} = 2 π^2/15 = Coxeter's still-mysterious orthoscheme integral. --rwg
More generally, Integrate[((a + b/Sqrt[3 + t])*ArcSec[t])/(Sqrt[1 + t]*(2 + t)), {t, 1, 2}] == (a*Pi^2)/72 + (b*Pi^2)/180 + (1/6)*b*Pi*ArcSin[(1/16)*(7 - 3*Sqrt[5])] But AargH! Now I can't reconstruct the derivation from the Mathematica and ries transcripts! Mathematica overwrote a _crucial_ command line, and ries can't handle the b≠0 case, except when b magically = -5a/2. The breakthrough came when ries quickly (level 6) identified 0.411233516712190204444136 = 1/24 (π + Cos[π Log[GoldenRatio]]^10)^2 but I can't figure out what the heck I integrated to get .4112... ! --rwg On Wed, Jul 4, 2018 at 5:34 AM Bill Gosper <billgosper@gmail.com> wrote:
Integrate[((2 + 1/Sqrt[3 + t])*ArcSec[t])/(Sqrt[1 + t]*(2 + t)), {t, 1, 2}] == Pi^2/30 + (1/6)*Pi*ArcSin[(1/16)*(7 - 3*Sqrt[5])]
The integral {t,1,6} = 2 π^2/15 = Coxeter's still-mysterious orthoscheme integral. --rwg
participants (4)
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Bill Gosper -
Fred Lunnon -
Gareth McCaughan -
James Propp