[math-fun] bizarre response from http://isc.carma.newcastle.edu.au/standardCalc
I unwittingly gave it .39269908169872415480783042290994 == N[π/8, 32] and got three answers: Best guess: sol of (1-tan(x))/log(x) sol of (1-tan(x))/log(x) 3926990816987241 sol of cos(x)/(W(x)+exp(x)) sum(1/(16*n^2-16*n+3),n=1..inf) <http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html> I just realized that the first two are satisfied by 10π/8. And would remain so without those exotic denominators. Looks like they need to hire Rob Munafo. --rwg
I imagine it has been this way for 20 years! Did you try the advanced version? J in cyberspace
On 13 Dec 2014, at 1:28 pm, Bill Gosper <billgosper@gmail.com> wrote:
I unwittingly gave it .39269908169872415480783042290994 == N[π/8, 32] and got three answers: Best guess: sol of (1-tan(x))/log(x)
sol of (1-tan(x))/log(x) 3926990816987241 sol of cos(x)/(W(x)+exp(x)) sum(1/(16*n^2-16*n+3),n=1..inf)
I just realized that the first two are satisfied by 10π/8. And would remain so without those exotic denominators. Looks like they need to hire Rob Munafo. --rwg
On Fri, Dec 12, 2014 at 6:38 PM, Jon Borwein <jon.borwein@gmail.com> wrote:
I imagine it has been this way for 20 years! Did you try the advanced version?
J in cyberspace
Oh yes. Tons more batbleep, but now the 3rd hit is 9/8 1/(√π)²/9 . --rwg
On 13 Dec 2014, at 1:28 pm, Bill Gosper <billgosper@gmail.com> wrote:
I unwittingly gave it .39269908169872415480783042290994 == N[π/8, 32] and got three answers: Best guess: sol of (1-tan(x))/log(x) sol of (1-tan(x))/log(x) 3926990816987241 sol of cos(x)/(W(x)+exp(x)) sum(1/(16*n^2-16*n+3),n=1..inf) <http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html> I just realized that the first two are satisfied by 10π/8. And would remain so without those exotic denominators. Looks like they need to hire Rob Munafo. --rwg
On Fri, Dec 12, 2014 at 7:19 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Dec 12, 2014 at 6:38 PM, Jon Borwein <jon.borwein@gmail.com> wrote:
I imagine it has been this way for 20 years! Did you try the advanced version?
J in cyberspace
Oh yes. Tons more batbleep, but now the 3rd hit is 9/8 1/(√π)²/9 .
Oops, I mean 9/8 1/9(√π)² .
--rwg
On 13 Dec 2014, at 1:28 pm, Bill Gosper <billgosper@gmail.com> wrote:
I unwittingly gave it .39269908169872415480783042290994 == N[π/8, 32] and got three answers: Best guess: sol of (1-tan(x))/log(x) sol of (1-tan(x))/log(x) 3926990816987241 sol of cos(x)/(W(x)+exp(x)) sum(1/(16*n^2-16*n+3),n=1..inf) <http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html> I just realized that the first two are satisfied by 10π/8. And would remain so without those exotic denominators. Looks like they need to hire Rob Munafo. --rwg
Hello, the version of the ISC in australia is a copy of my inverter which has tables that were in Vancouver in 1998. I have this output from my local inverter with 14.083 billion entries including table expression ====================================== s001 sum(1/10^n*A091473[n],n=1..infinity) s001 sum(1/10^(n-1)*A091473[n],n=1..infinity) g010 Re[Beta[3/2,3/2]] m002 Pi/8 s001 sum(1/10^n*A019675[n],n=1..infinity) v024 Sum[1/(3+16*n^2-16*n),{n,1,Infinity}] w006 Weierstrass(1/2) w007 Real solution of -2+2*cos(x)^2+1/4/cos(x)^2 w012 Real solution of -1+2*cos(x)^2-2*sin(x)*cos(x) m004 (5*Pi)/4 s001 sum(1/10^(n-1)*A019675[n],n=1..infinity) by using the information of solutions of tables w007 and w012 I get : arctan(2*(1/2*(2+2^(1/2))^(3/2)-3/2*(2+2^(1/2))^(1/2))/(2+2^(1/2))^(1/2)) for Pi/8. the sequence A019675 is Pi/8 and the entry of A091473 is an approximation of Pi/8 related to that strange integral which is Pi/8 up to 42 digits, I use 32 digits validation on each case. The pismart algorithm returns approximately 1400 answers for that number alone that includes approximations like this one : -Pi^105*2^100*bernoulli(104)/104! which gives Pi/8 to 31 digits. Best regards, Simon Plouffe
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Bill Gosper -
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Simon Plouffe