[math-fun] The hexadeciaml pi formula; Catalan &al.
Simon Plouffe was one of the discoverers of the nice pi formula inf 4 2 1 1 -k pi = sum (---- - ---- - ---- - ----) 16 k=0 8k+1 8k+4 8k+5 8k+6 which allows computing individual bits of pi without computing the preceding part. It seems to me that there should be another such formula, in which the mod 8 residues are rearranged a little. Perhaps with 8k+3 and 8k+7 replacing 1 & 5, or maybe just 1&3 instead of 1&5, or with 8k+2 in place of 8k+6. Has anyone seen variations like this? ----- The sums of 1/(2n+1)^k, for even k, give rational multiples of pi^k, and the alternating sums, for odd k, also give rational multiples of pi^k. So 1 - 1/27 + 1/125 - ... = pi^3 / 32. Perhaps there's some relationship between the leftovers: gamma, Catalan, zeta(odd), etc? Rich
Hello, there are many ways to reformulate the original series for Pi : inf 4 2 1 1 -k pi = sum (---- - ---- - ---- - ----) 16 k=0 8k+1 8k+4 8k+5 8k+6 the original finding on sept 19, 1995 was actualy different : Pi = 4*f(1/4) + 2 arctan (1/2) - log(5) where f(1/4) = sum((-1)^(n+1)/(4*n+1),n=1..infinity) OR f(1/4) = 2F2(1,1/4; 5/4; -1/4) another formulation found that same day was also : Pi = sum(c(n)/16^floor(n/8)/n,n=0..infinity) where c(n) = {4,0,0,-2, -1, -1, 0, 0} and periodic. The published formula for pi, called the BBP formula was re-arranged by Peter Borwein, 2 days after the initial discovery. <we> kept that one to be published, the rest of the story for the discovery was posted by me on june 23, 2003 for personal reasons on sci.math. Simon Plouffe
Simon Plouffe was one of the discoverers of the nice pi formula
inf 4 2 1 1 -k pi = sum (---- - ---- - ---- - ----) 16 k=0 8k+1 8k+4 8k+5 8k+6
which allows computing individual bits of pi without computing the preceding part.
It seems to me that there should be another such formula, in which the mod 8 residues are rearranged a little. Perhaps with 8k+3 and 8k+7 replacing 1 & 5, or maybe just 1&3 instead of 1&5, or with 8k+2 in place of 8k+6.
Has anyone seen variations like this?
You don't need lattice numerology for these. Here are the closed forms-- rub them together symbolically: inf ==== \ 1 sqrt(2)
------------- = (sqrt(2) log(sqrt(2) + 1) + 2 sqrt(2) atan(-------) / k 2 ==== (8 k + 1) 16 k = 0
- sqrt(2) log(sqrt(2) - 1) + log(5) + 2 atan(2))/8 inf ==== \ 1 log(3) + 2 acot(2)
------------- = ------------------ / k 4 ==== (8 k + 2) 16 k = 0
inf ==== \ 1 sqrt(2)
------------- = (sqrt(2) log(sqrt(2) + 1) - 2 sqrt(2) atan(-------) / k 2 ==== (8 k + 3) 16 k = 0
- sqrt(2) log(sqrt(2) - 1) - log(5) + 2 atan(2))/4 inf ==== \ 1
------------- = acoth(4) / k ==== (8 k + 4) 16 k = 0 inf ==== \ 1 sqrt(2) ------------- = (sqrt(2) log(sqrt(2) + 1) + 2 sqrt(2) atan(-------) / k 2 ==== (8 k + 5) 16 k = 0
- sqrt(2) log(sqrt(2) - 1) - log(5) - 2 atan(2))/2 inf ==== \ 1
------------- = log(3) - 2 acot(2) / k ==== (8 k + 6) 16 k = 0
inf ==== \ 1 sqrt(2)
------------- = sqrt(2) log(sqrt(2) + 1) - 2 sqrt(2) atan(-------) / k 2 ==== (8 k + 7) 16 k = 0
- sqrt(2) log(sqrt(2) - 1) + log(5) - 2 atan(2)
The sums of 1/(2n+1)^k, for even k, give rational multiples of pi^k, and the alternating sums, for odd k, also give rational multiples of pi^k. So 1 - 1/27 + 1/125 - ... = pi^3 / 32. Perhaps there's some relationship between the leftovers: gamma, Catalan, zeta(odd), etc?
Rich
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integrate((1/sqrt(t+3)+2)*asec(t)/(sqrt(t+1)*(t+2)),t,1,6)=2*%pi^2/15 6 1 / (----------- + 2) asec(t) 2 [ sqrt(t + 3) 2 pi I ------------------------- dt = ----- ] sqrt(t + 1) (t + 2) 15 / 1 (~ two screens into http://www.tweedledum.com/rwg/idents.htm) from a geometrical argument about orthoschemes, but, afaik, no one knows how to get it analytically. Integral_1^inf is empirically 3 pi^2/8. This suggests there may be similar such embarrassments (anyone?), and we're missing a piece of mathematics. Mma 7.0 *is* able to show Re(Integral_-1^1) = 4 pi^2/3. --rwg
rcs> Simon Plouffe was one of the discoverers of the nice pi formula
inf 4 2 1 1 -k pi = sum (---- - ---- - ---- - ----) 16 k=0 8k+1 8k+4 8k+5 8k+6
which allows computing individual bits of pi without computing the preceding part.
It seems to me that there should be another such formula, in which the mod 8 residues are rearranged a little. Perhaps with 8k+3 and 8k+7 replacing 1 & 5, or maybe just 1&3 instead of 1&5, or with 8k+2 in place of 8k+6.
Has anyone seen variations like this?
rwg> You don't need lattice numerology for these. Here are the closed forms--
rub them together symbolically:
inf ==== \ 1 sqrt(2)
------------- = (sqrt(2) log(sqrt(2) + 1) + 2 sqrt(2) atan(-------) / k 2 ==== (8 k + 1) 16 k = 0 [...]
I believe it was discussed here that Simon (et al?) later discovered that six of these series were bisections of inf ==== \ 1 log(5) + 2 atan(2) > ---------------- = ------------------, / k 4 ==== (4 k + 1) (- 4) k = 0 inf ==== \ 1 > ---------------- = acot(2), and / k ==== (4 k + 2) (- 4) k = 0 inf ==== \ 1 log(5) - 2 atan(2) > ---------------- = - ------------------, / k 2 ==== (4 k + 3) (- 4) k = 0 which are all you need. Also, apropos the 1st and 7th eqns in his inspired3.pdf, sum(n/(%e^(%pi*n)-1),n,1,inf) = gamma(1/4)^4/(64*%pi^3)-1/(4*%pi)+1/24 4 inf 1 ==== 4 (-)! \ n 1 4 1 > ---------- = - ---- + ------- + -- / n pi 4 pi 3 24 ==== %e - 1 pi n = 1 sum(n/(%e^(2*%pi*n)-1),n,1,inf) = 1/24-1/(8*%pi) inf ==== \ n 1 1 > ------------ = -- - ---- / 2 n pi 24 8 pi ==== %e - 1 n = 1 sum(n/(%e^(4*%pi*n)-1),n,1,inf) = -gamma(1/4)^4/(256*%pi^3)-1/(16*%pi)+1/24 4 inf 1 ==== (-)! \ n 1 4 1 > ------------ = - ----- - ----- + -- / 4 n pi 16 pi 3 24 ==== %e - 1 pi n = 1 and the middle of these is 1/8 of his 1st minus 1/24 of his 7th. His 1st can also be written 1/%pi = -16*sum(n/(%e^(2*%pi*n)+1),n,1,inf)+12*sum(n/(%e^(%pi*n)+1),n,1,inf)-4*sum(n/(%e^(%pi*n)-1),n,1,inf) inf inf inf ==== ==== ==== 1 \ n \ n \ n -- = - 16 > ----------- + 12 > --------- - 4 > --------- pi / 2 n pi / n pi / n pi ==== e + 1 ==== e + 1 ==== e - 1 n = 1 n = 1 n = 1 Also, in my old q-trig paper are a bunch of relations, for general q, between etas and logderivative(etas), e.g. sum((2*n-1)*q^(2*n-1)/(1-q^(2*n-1)),n,1,inf) = 2*eta(q^4)^8/(3*eta(q^2)^4)+(eta(q^2)^20/(eta(q)^8*eta(q^4)^8)-1)/24 20 2 eta (q ) inf ---------------- - 1 ==== 2 n - 1 8 4 8 8 4 \ (2 n - 1) q 2 eta (q ) eta (q) eta (q )
------------------ = ---------- + -------------------- / 2 n - 1 4 2 24 ==== 1 - q 3 eta (q ) n = 1
sum(n*q^n/(1-q^n),n,1,inf)-3*sum(n*q^(3*n)/(1-q^(3*n)),n,1,inf) = eta(q)^2*eta(q^3)^2*(3*eta(q^6)^8/eta(q^3)^4+eta(q^2)^8/eta(q)^4)^2/(12*eta(q^2)^4*eta(q^6)^4)-1/12 8 6 8 2 2 2 3 3 eta (q ) eta (q ) 2 inf inf eta (q) eta (q ) (---------- + --------) ==== n ==== 3 n 4 3 4 \ n q \ n q eta (q ) eta (q) 1
------ - 3 > -------- = ----------------------------------------- - -- / n / 3 n 4 2 4 6 12 ==== 1 - q ==== 1 - q 12 eta (q ) eta (q ) n = 1 n = 1
You probably don't recall, but the q=e^-pi case of the trisectands of this last series are improbably complicated. --rwg PS, I fixed the algebra drill color collisions in http://www.tweedledum.com/rwg/squares.htm . RISTOCETIN TRISECTION PREDYNASTIC CANDYSTRIPE SHATTERABLE BLAST-HEATER BARDOLATRIES LABRADORITES
A private conversation raised the following: a) continuous maps from 1D onto a 2D patch visit countably many points at least three times. What is the minimum dense revisitation count for continuous volume fillers, 1D onto 3D? It was seven for that "Treano" function I discussed a few months ago. b) What is the dimension of the boundary of the 2D quincunx fractal, the fixed point of surrounding one by four in a square, then dividing by 2+i? --rwg
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Simon Plouffe