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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!
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