From: Christian.Radoux@skynet.be Newsgroups: sci.math.num-analysis Subject: Re: anybody see this formula? Date: Sat, 06 Jun 1998 19:52:43 GMT In article , Francis Sergeraert wrote: > > george.barwood@dial.pipex.com (George Barwood) writes: > > > Here's another: > > > > If we have a rectangle 30' x 130' then the diagonal > > is 133' 4.999688". Amazing? I don't think so. > > Maybe in this thread, the Ramanujan example > > e^{\pi \sqrt{163}} = > > 262537412640768744 - 0.000000000000749927... > > should be quoted. It was guessed by Ramanujan without any computer > because of some related properties around modular functions, I think ; > this claim is only a vague souvenir of a discussion with a true > specialist in arithmetic. Maybe someone could correct and/or precise. > In this case, usual floating computer arithmetic does not allow you to > disprove this number is integer, but with Maple or Mathematica, it's > now easy. > > Here is an explanation. Let us call sigma(3,n) the sum of the cubes of all positive integers (including 1 and n itself) dividing n. Let us call q = exp(2*i*pi*tau), with Im(tau) > 0 (so |q| < 1) The Jacobi J function is defined by j(tau) = (1 + 240 * sum, from n = 1 to infinite, of sigma(3,n) * q^n)^3 divided by q * product, for n = 1 to infinite, of (1 - q^n)^24 N.B. The Taylor expansion of the denominator is the generating function of the Ramanujan's tau function. It is well known, in the theory of modular functions, that j(tau) is invariant under the group of unimodular transformations (tau has for image (alpha * tau + beta)/(gamma * tau + delta), with alpha, beta, gamma, delta belonging to Z and alpha * delta - beta * gamma = 1). Let us now consider the quadratic forms a * x^2 + b * x * y + c * y^2 of given negative discriminant d. We use now the classical notation h(d) for the class number. A theorem of Heegner (1952), Baker and Stark (1966) says that for d negative, h(d) = 1 if, ond only if d is one of the numbers -1, -2, -3, -4, -7, -8, -11, -19, -43, -67, -163 (here it is !) On the other hand, Weber has proved that j((-b+sqrt(d))/(2*a)) is an algebraic integer of degree h(d). In particular, j((-b+sqrt(d))/(2*a)) is a rational integer if, and only if h(d) = 1. For example : j(i) = 1728, j((1+sqrt(3))/2) = 0, j(2*i) = 287496, j(i*sqrt(2)) = 8000,... Now, we take the special values a = 1, b = 1, c = 41 (so, d = -163). We get at once : j((-1+i*sqrt(163))/2) is a rational integer (-640220^3). But, as a meromorphic function of q, J(q) defined by J(q) = j(tau) has the Laurent expansion series : J(q) = 1/q + 744 + 196884 * q + 21493760 */ q^2 + 864299970 * q^3 + 20245856256 * q^4 + 33202640600 * q^5 + ... when |q| < 1. (N.B. the coefficients have beautiful arithmetic properties, but that's another question...) And now, we have : the INTEGER j((-1+i*sqrt(163))/2) = exp(pi*sqrt(163) + 744 - 196884 * exp(-pi*sqrt(163)) + ... But exp(-pi*sqrt(163)) = 7.5 * 10^(-13) and the following terms are "still more negligeable". I hope this dreadful ASII is readable, but you asked for it (without jokes, now : beautiful question !) With best regards ! e-mail : Christian.Radoux@skynet.be URL : http://users.skynet.be/radoux P.S. 1 From this URL (follow the link "gratuits", you can download (for free of course) many DOS programs dealing in multiprecision with several celebrated Ramanujan's functions. P.S. 2 If you like them, please use the page code number 437 to avoid misprints with the greek letters (page code 850 is not optimal, changing letters such pi, tau,...). -----== Posted via Deja News, The Leader in Internet Discussion ==----- http://www.dejanews.com/ Now offering spam-free web-based newsreading