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Topic: Low precision exponentiation
Replies: 5   Last Post: Feb 19, 2013 6:51 PM

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 Dana DeLouis Posts: 24 Registered: 11/18/12
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 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

Date Subject Author
2/18/13 Sseziwa Mukasa
2/18/13 Andrzej Kozlowski
2/18/13 Bob Hanlon
2/18/13 Murray Eisenberg
2/19/13 Dana DeLouis