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

```

• Prev by Date: Re: Stephen Wolfram's recent blog
• Next by Date: Re: Ingolf Dahl's "SetFaceAndFont" palette broken in
• Previous by thread: Re: Low precision exponentiation
• Next by thread: Does MLGetUTF8String() return a null-terminated string?