Re: Bug in Mathematica 7.0?

*To*: mathgroup at smc.vnet.net*Subject*: [mg99151] Re: Bug in Mathematica 7.0?*From*: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>*Date*: Tue, 28 Apr 2009 04:48:34 -0400 (EDT)*References*: <gt3tla$non$1@smc.vnet.net>

Nobody is infallible. So, show your proof that A=B. Having said that, it seems that Mathematica has some problems with BarnesG. Given that BarnesG[a + 1] == Gamma[a] BarnesG[a] the following table should contain only zeros but it doesn't: Table[BarnesG[a + 1] - Gamma[a] BarnesG[a], {a, 1, 15, .1}] It's probably due to the enormous numbers involved BarnesG[15] == 792786697595796795607377086400871488552960000000000000. So, I can imagin the difference between large numbers like that isn't going to be very precise. Given the low M values you're trying that can't be the case Cheers -- Sjoerd On Apr 27, 11:24 am, Not An Expert <not_an_exp... at directbox.com> wrote: > Hello Mathematica-Experts, > > I'm using Mathematica 7.0 on a Windows XP machine and might have come acr= oss > a bug in Mathematica 7.0. > > I define the following function as a product of Gamma functions: > > A[M_, r_] := Product[Gamma[M + j]*Gamma[1/2 + j]*Gamma[r + 1/2 + j]/ > Gamma[r + M + j], {j, 0, M - 1}] > > To allow non-integer values M we re-write the above expression in terms > of the Barnes G-function and so we obtain (we've checked the algebra > many times!): > > B[M_, r_] := BarnesG[2 + M]/BarnesG[2]*BarnesG[1/2 + M]/BarnesG[1/2] > *BarnesG[r + 1/2 + M]/ > BarnesG[r + 1/2]*BarnesG[r + M]/BarnesG[r + 2*M] > > As a test I've chosen M = 2 and r = 1 > and indeed I obtain > A[2, 1] = Pi^2/32 > and > B[2, 1] = Pi^2/32 > and so agreement in both functions. > > But choosing M = 3 and r = 1 gives > A[3, 1] = 9*Pi^3/2048 > B[3, 1] = 3^Pi^3/16384 > > So two different values for A[3, 1] and B[3, 1]. > Also A[3, 0] and B[3, 0] would give different answers - and there are > many other examples. > > Do you have any thoughts on what went wrong? > > Cheerio!