[math-fun] (no subject)
If you split Ramanujan's 1/pi = 4F3[1/64] (search for Ramanujan in www.tweedledum.com/rwg/idents.htm) into two 3F2s, Mma 6.0 recognizes one of them, producing the identity expand(Hyper_F[P,Q]([1,1,3]/2,[1,1],1/64)=16/(21*%Pi) + (64*Elliptic_Kc(1/2*(1-3*Sqrt(7)/8))^2)/(21*%Pi^2)) 3 1 1 1 hyper_f ([-, -, -], [1, 1], --) = 3, 2 2 2 2 64 2 1 3 sqrt(7) 64 elliptic_kc (- - ---------) 2 16 16 ------------------------------ + ------ 2 21 %pi 21 %pi ~ 1.01336493939236. If we can find one more, we might get the whole contiguous family, since 3F2[z] obeys four term recurrences. Embarrassments: Mma 6.0 can't do FullSimplify[ FunctionExpand[ HypergeometricPFQ[{1, a, b}, {c, 3 - c + b + a},1]]] nor even the special case HypergeometricPFQ[{1,1,1},{5,5}/2,1] = 9 pi - 27, even though the general sum telescopes. My enhanced Macsyma also fails unless you first convert to Sigma notation (makesum). --rwg PS, overall, I'm favorably impressed with Mma 6.0, although I did lose a session when something in the help system cleared all my definitions.
(Sorry about the missing 3F2[1/64] previous subject line.) I just (re?)found another misprint in my ancient (1974) AI Memo 304 series acceleration paper, dspace.mit.edu/handle/1721.1/6088 : p 73, 2^j-1 should be 2^j+1 in the quasidouble sum for Euler's constant: inf ==== j + 1 \ 6 k + 2 + 3 > ---------------------------------- inf / k + 1 j ==== ==== 4 binomial(2 k + 2 + 1, 2 k) \ k = 0 %gamma = > ---------------------------------------- / j j ==== 2 (2 + 1) j = 1 (Quasi because you only need ~ n log n terms for n digits.) I'll post a corrected version on gosper.org when I figure out how to mark it up with Acrobat. Oddity: inf ==== \ n 1 n > (psi (2 + -) - psi (2 )) / 0 2 0 ==== n = 0 %gamma = ------------------------------- 2 inf ==== \ n + 1 n = > (psi (x ) - psi (x ) - log(x)), / 0 0 ==== n = 0 x>1, where psi[0](k) can be replaced by H[k-1]. It must be independent of x, since grouping terms pairwise is the same as squaring x. --rwg PS: Substituting x=1, we have the surprising result that %gamma=0.-)
(from both my Macsyma and Mma 6.0): 3F2[a,b,c;d,e] = R(e) + r(e) Psi[1](e), for positive integer a,b,c, and d, where R and r are rational functions, and r(e) = 0 when max(a,b,c)>=d. E.g., hyper_f ([1, 1, 1], [3, e + 1], 1) = 2 e (e Psi (e) - 1). 3, 2 1 --rwg
(from both my Macsyma and Mma 6.0): 3F2[a,b,c;d,e] = R(e) + r(e) Psi[1](e), for positive integer a,b,c, and d, where R and r are rational functions, and r(e) = 0 when max(a,b,c)>=d. E.g.,
hyper_f ([1, 1, 1], [3, e + 1], 1) = 2 e (e Psi (e) - 1). 3, 2 1 --rwg
Well, this is odd. As I went to implement this case it spontaneously started working: (c228) (hyper_f[1,1]([2,2,2],[3,x]),%% = hypersimp(%%)) (d228) hyper_f ([2, 2, 2], [3, x]) 3, 2 2 (x - 2) (x - 1) ((x - 3) Psi (x - 2) - 1) 1 = - ------------------------------------------- . x - 3 Note that the 3F2 series definition wants Re(x)>3 for convergence. (It's not obvious why it diverges for x=3+%i: The terms decrease like 1/n and the signs of the real and imaginary parts fluctuate. But as the term ratio approaches 1, its imaginary part fades away and with it the fluctuation frequency. The ever longer bursts of "parallel" terms slew the partial sums at decreasing frequency and nonvanishing amplitude.) But convergence is a red herring, since the numerical method (3 by 3 matrix product) gives consistent values for all x: (c232) expand(subst(1/2,x,d228)) 2 1 3 %pi 109 (d232) hyper_f ([2, 2, 2], [3, -]) = - ------ - --- 3, 2 2 4 15 (c233) dfloat(%) (d233) - 14.6688699674837d0 = - 14.6688699674837d0 (Note that the series terms are all positive.) Another EulerGamma series with degrees of freedom: For nonnegative integers p and m, (d197) Psi (n) = 0 k ==== \ j > (- 1) log(n - j - 1) binomial(p + k, p + j) m - 1 / ==== ==== \ j = 0 > -------------------------------------------------- / binomial(p + k + 1, p + 1) ==== k = 0 -------------------------------------------------------- p + 1 m + k ==== \ j > (- 1) binomial(m + k, j) log(n - m + j) inf / ==== ==== m \ j = 0 (- 1) > ---------------------------------------------- / binomial(m + k + 1, m + 1) ==== k = 0 + ----------------------------------------------------------- m + 1 m ==== \ j > (- 1) binomial(m + k, k + j) log(n - j) p - 1 / ==== ==== \ j = 1 - > ----------------------------------------------. / (k + 1) binomial(m + k + 1, k + 1) ==== k = 0 Some combinations of p and m converge faster than others. Note that replacing log by some other slowly growing (e.g., log^2) or decreasing (e.g., 1/x) function will compute some other f(n), independent of p and m. Random afterblurt: Does the sequence A081464 = 1,2,4,29,95,153,532,613,840,2033,..., = the n for which fractionpart((3/2)^n) form a decreasing sequence, characterize 3/2? If not, what set? 1,2,3,4,..., characterizes an infinite set of algebraics, (and 0<x<1, but I'd think most sequences would characterize the empty set. --rwg
The (mostly impressive) Mma 6.0 doc gives, under applications of EllipticF, "Parametrization of a mylar balloon (two flat sheets of plastic sewn together at their circumference and then inflated):", followed by three definitions and a call to ParametricPlot3D that makes an "M&M". Inelastic mylar would crinkle at the equator, so the true solution would be more like two shallow paper cupcake molds. More convincing, the doc of LambertW (which they insist on calling ProductLog, probably because a bully named Lambert beat up Wolfram in grade school or something), claims that the equipotentials of the fringing field of a plate capacitor are given by \[Phi][{x_, y_}] := With[{z = x + I y}, Im[z - 1 - ProductLog[Ceiling[(y - Pi)/(2 Pi)], Exp[z - 1]]]] ContourPlot[\[Phi][{x, y}], {x, -2, 2}, {y, -4, 4}, Epilog -> {Red, Thickness[0.02], Line[{{-2, Pi}, {0, Pi}}], Line[{{-2, -Pi}, {0, -Pi}}]}, ContourShading -> False, Contours -> 20] This would make a nice closure. AT MIT, they always told us to neglect fringing because it was too difficult. --rwg
This subject line got me erroneously excited into believing that someone had proved RH with a polylogarithm identity. It's not far-fetched. Zeilberger thinks that some generalization of RH will drop out of an identity some day. (Oral communication, probably in an elevator.)
(I promised Cheny Xu, proprietor of Jade Palace Restaurant, that I'd solve a problem with the help of some fish he fed me last night.) It's not obvious how to extract a tractable halfangle formula from the many double angle formulae in W&W. After you eliminate the three undesired species, the resultant polynomial is too hairy. Actually, it would probably simplify if Macsyma or Mma only knew how, but I can't even get Mma 6.0 to EllipticTheta[1, Pi, q] -> 0 ! (Hey guys, need a programmer?) Fortunately, I (re?)conjectured a family of theta determinant identities, http://www.tweedledum.com/rwg/thet.PNG (too tall for IE 7), arxiv.org/pdf/math/0703470, the first few (including the one used here) proved by Rich (and Hilarie?). These are nice when actually typeset: x theta (-, q) = 1 2 3 3 3 - theta (0, q) theta (x, q) + theta (0, q) theta (x, q) - theta (0, q) theta (x, q) 4 4 3 3 2 2 1/4 (-----------------------------------------------------------------------------------) , 2 x theta (-, q) = 2 2 3 3 3 - theta (0, q) theta (x, q) + theta (0, q) theta (x, q) + theta (0, q) theta (x, q) 4 4 3 3 2 2 1/4 (-----------------------------------------------------------------------------------) , 2 x theta (-, q) = 3 2 3 3 3 theta (0, q) theta (x, q) + theta (0, q) theta (x, q) + theta (0, q) theta (x, q) 4 4 3 3 2 2 1/4 (---------------------------------------------------------------------------------) , 2 x theta (-, q) = 4 2 3 3 3 theta (0, q) theta (x, q) + theta (0, q) theta (x, q) - theta (0, q) theta (x, q) 4 4 3 3 2 2 1/4 (---------------------------------------------------------------------------------) . 2 E.g., pi pi theta (--, q) = theta (--, q) = 1 4 2 4 2 2 theta (0, q) theta (0, q) (theta (0, q) - theta (0, q)) 3 4 3 4 1/4 (-------------------------------------------------------) , 2 pi pi theta (--, q) = theta (--, q) = 3 4 4 4 2 2 theta (0, q) theta (0, q) (theta (0, q) + theta (0, q)) 3 4 4 3 1/4 (-------------------------------------------------------) . 2 Finally, this must be in Tannery & Molk, but I don't recall seeing a Lambert type series come out as a pure theta instead of a log derivative theta: inf 2 ==== j 2 j - 1 theta (0, q) - 1 \ (- 1) q 4 > --------------- = ----------------. / 2 j - 1 4 ==== q + 1 j = 1 --rwg
The formula used by Mma for the fringing field of a capacitor was first given in Valluri, Jeffrey, Corless: Canadian J. Physics, vol 78, pp823-831 (2000) rwg@sdf.lonestar.org wrote:
The (mostly impressive) Mma 6.0 doc gives, under applications of EllipticF, "Parametrization of a mylar balloon (two flat sheets of plastic sewn together at their circumference and then inflated):", followed by three definitions and a call to ParametricPlot3D that makes an "M&M". Inelastic mylar would crinkle at the equator, so the true solution would be more like two shallow paper cupcake molds.
More convincing, the doc of LambertW (which they insist on calling ProductLog, probably because a bully named Lambert beat up Wolfram in grade school or something), claims that the equipotentials of the fringing field of a plate capacitor are given by
\[Phi][{x_, y_}] := With[{z = x + I y}, Im[z - 1 - ProductLog[Ceiling[(y - Pi)/(2 Pi)], Exp[z - 1]]]]
ContourPlot[\[Phi][{x, y}], {x, -2, 2}, {y, -4, 4}, Epilog -> {Red, Thickness[0.02], Line[{{-2, Pi}, {0, Pi}}], Line[{{-2, -Pi}, {0, -Pi}}]}, ContourShading -> False, Contours -> 20]
This would make a nice closure. AT MIT, they always told us to neglect fringing because it was too difficult. --rwg
participants (2)
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David Jeffrey -
rwg＠sdf.lonestar.org