Re: Low precision exponentiation

*To*: mathgroup at smc.vnet.net*Subject*: [mg129866] Re: Low precision exponentiation*From*: Dana DeLouis <dana01 at icloud.com>*Date*: Tue, 19 Feb 2013 18:52:14 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net

> 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 UTC-5, 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