Re: Re: limit
- To: mathgroup at smc.vnet.net
- Subject: [mg78764] Re: Re: limit
- From: chuck009 <dmilioto at comcast.com>
- Date: Mon, 9 Jul 2007 01:32:11 -0400 (EDT)
> But the latter derivative is clearly wrong (probably
> due to some
> simplification rule, applied without our knowledge,
> that isn't correct at
> s=1):
>
Hello Dr. Bob,
I believe part of the problem here lies with the inability of Mathematica to calculate the derivative of HypergeometricPFQ when p>2 and q>1. For example, consider:
f[s_]:=HypergeometricPFQ[{1/4+s/4,1/4+s/4},{1/2},1]
which converges for s<0. I can rewrite this in terms of the Gamma function as:
f2[s_] := Sum[(Gamma[1/4 + s/4 + k]/
Gamma[1/4 + s/4])^2*(Gamma[1/2]/
(Gamma[1/2 + k]*k!)), {k, 0, Infinity}]
Just to check it:
f[-0.4]
f2[-0.4]
Out[128]=
1.25516
Out[129]=
1.25516
I can now take the derivative of both f and f2 and check the value at s=-0.4:
In[150]:=
fd[s_]=D[f[s],s];
fd2[s_]=D[f2[s],s];
fd[-0.4]
fd2[-0.4]
Out[152]=
1.454715739187473
Out[153]=
1.45472
Everything is ok up to this point. If I add one more term to {a} and another to {b} (this time converging for s<1):
In[158]:=
f[s_] := HypergeometricPFQ[{1/4 + s/4, 1/4 + s/4,1/4+s/4},
{1/2,1}, 1];
f2[s_] := Sum[(Gamma[1/4 + s/4 + k]/
Gamma[1/4 + s/4])^3*(Gamma[1/2]/
(Gamma[1/2 + k]*k!)(Gamma[1]/Gamma[1+k])), {k, 0, Infinity}]
f[-0.4]
f2[-0.4]
Out[160]=
1.01107
Out[161]=
1.01107
Still can calculate HypergeometricPFQ but if I try to take the derivative of either expression, Mathematica seems unable to complete the calculation, my machine starts humming indicating (for me anyway) a chronic load on the CPU and I end up interrupting the calculation with Alt ,
Note the cube and another term in the sum. Perhaps we're asking too much to calculate the derivative of such an expression.