
Re: Low precision exponentiation
Posted:
Feb 19, 2013 6:51 PM


> I am trying to evaluate 2.5^125 to high precision. > 5.527147875260445183346e+49
Hi. Not sure, but you may find this interesting on a large number. Here, I'll use 50 Integer part + 125 fractional part + 5 extra zero's at end = 180)
n=Power[5/2,125];
NumberForm[N[n,180],ExponentFunction>(Null&)]
= 55271478752604445602472651921922557255142402332392.2008641517022090789875402395331710176480222226446499875026812553578470207686332597244588393792241731716785579919815063476562500000
If one like comma's:
NumberForm[N[n,180],DigitBlock>{3,\[Infinity]},ExponentFunction>(Null&)] =
55,271,478,752,604,445,602, . . . etc
= = = = = = = = = = HTH :>) Dana DeLouis Mac & Mathematica 9 = = = = = = = = = =
On Sunday, February 17, 2013 4:08:38 AM UTC5, Blaise F Egan wrote: > I am trying to evaluate 2.5^125 to high precision. > > > > R gives 5.527147875260445183346e+49 as the answer but Mathematica with N[2.5^125,30] gives 5.52715*10^49 and says that is to machine precision. > > > > I am inexperienced at Mathematica. Am I doing something silly? > > > > Blaise

