Re: Low precision exponentiation
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- Subject: [mg129866] Re: Low precision exponentiation
- From: Dana DeLouis <dana01 at icloud.com>
- Date: Tue, 19 Feb 2013 18:52:14 -0500 (EST)
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> 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.
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> 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.
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> I am inexperienced at Mathematica. Am I doing something silly?
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> Blaise