Henri . Cohen on Sat, 06 Feb 2010 00:42:57 +0100 |
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Re: PARI stable release 2.3.5 (for real now) |
No, not quite. There are indeed some of problems with the method, but here you do not have to do anything: the [[1],-3/2] does not at all mean a decrease in Z^(-3/2) (which is in fact in Z^(-2)), but a decrease like exp(-(3/2)Z). So it must simply be omitted: intnum(Z=[0,-1/2],[1],1/sqrt(Z)*(1+Z)^(-3/2)) 2.000000000000000000000000000 Henri Cohen ============================ Bill Allombert <Bill.Allombert@math.u-bordeaux1.fr> a écrit :
On Wed, Feb 03, 2010 at 12:14:57AM -0800, Ilya Zakharevich wrote:*) intnum() is still borken (same as in 3.0): ? intnum(Z=[0,-1/2],[[1],-3/2],1/sqrt(Z)*(1+Z)^(-3/2)) %1 = 7.150393303262919781690330041 + 1018.215229301285883563073358 I (I presume this is integral from 0 to infinity...)You need to split the integral:? intnum(Z=[0,-1/2],1,1/sqrt(Z)*(1+Z)^(-3/2))+intnum(Z=1,[[1],-3/2],1/sqrt(Z)*(1+Z)^(-3/2)) %36 = 2.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000This belongs to a class of problem inherent to the method. Bill.