Re: Timing in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg74745] Re: Timing in Mathematica*From*: "dimitris" <dimmechan at yahoo.com>*Date*: Tue, 3 Apr 2007 00:29:59 -0400 (EDT)*References*: <euqnqt$8br$1@smc.vnet.net>

It is very easy. Let for example foo = {x^n, Cos[x], Log[x]/x, Log[Sin[x]^2]*Tan[x], BesselJ[n, x], (2*x)/((x + 1)*(x^3 + 3*x^2 + 2*x + 1))} {x^n, Cos[x], Log[x]/x, Log[Sin[x]^2]*Tan[x], BesselJ[n, x], (2*x)/((1 + x)*(1 + 2*x + 3*x^2 + x^3))} Then InputForm[(Timing[Integrate[#1, x]] & ) /@ foo] {{0.031*Second, x^(1 + n)/(1 + n)}, {0.015999999999999986*Second, Sin[x]}, {0.*Second, Log[x]^2/2}, {0.9530000000000001*Second, Log[Sec[x/2]^2]^2 + 2*Log[Sec[x/ 2]^2]*Log[(Cos[x]*Sec[x/2]^2)/2] + Log[Sec[x/2]^2]*Log[Sin[x]^2] - 2*Log[Sec[x/2]^2]*Log[-1 + Tan[x/ 2]^2] - Log[Sin[x]^2]*Log[-1 + Tan[x/2]^2] + Log[Tan[x/2]^2]*Log[-1 + Tan[x/2]^2] + 2*PolyLog[2, Sec[x/2]^2/2] + PolyLog[2, Cos[x]*Sec[x/2]^2] + PolyLog[2, -Tan[x/2]^2]}, {0.030999999999999917*Second, 2^(-1 - n)*x^(1 + n)*Gamma[(1 + n)/2]* HypergeometricPFQRegularized[{(1 + n)/2}, {1 + n, (3 + n)/2}, - (x^2/4)]}, {0.016000000000000014*Second, 2*(-Log[1 + x] + RootSum[1 + 2*#1 + 3*#1^2 + #1^3 & , (Log[x - #1] + 2*Log[x - #1]*#1 + Log[x - #1]*#1^2)/(2 + 6*#1 + 3*#1^2) & ])}} Regards Dimitris =CF/=C7 amitsoni.1984 at gmail.com =DD=E3=F1=E1=F8=E5: > Hi, > > I am using Timing[.......] to calculate the time taken by an > expression in Mathematica. But it is not showing values below 0.01 > seconds. For values below 0.01s it just shows 0. Second > > Is it possible to get the value of time lower than 0.01s by somehow > lowering the least count. I am using Mathematica 5.2 on Windows. I was > getting lower values of time on Mathematica(5.2) on a Linux computer. > > Thank you, > Amit