2^3^4^5 in Mathematica?
- To: mathgroup at smc.vnet.net
- Subject: [mg30258] 2^3^4^5 in Mathematica?
- From: seidov at yahoo.com (Zakir F. Seidov)
- Date: Sat, 4 Aug 2001 01:14:25 -0400 (EDT)
- References: <firstname.lastname@example.org>
- Sender: owner-wri-mathgroup at wolfram.com
Mathematica can't calculate exactly 2^3^4^5 not only because
there's no place in the whole Universe to save these some
But She can't even evaluate it as it is much nore than
$MaxNumber which in my case is only of some
Is it so?
Subject: [mg30258] Re: Exponents and notation
Author: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
Organization: Universitaet Leipzig
there is a good reason. Every high precision arithmetic must store
the number of the digits somewhere (in a base you like).
Typical the digit number is not a high precision number itself.
The overflow comes from the fact that for
Mathematica needs more than the number of allowed digits.
May be, that a 64-Bit Mathematica will help (for Sun's).
BTW the HP is wrong because
x = 2.^3^4;
xx = x*x*x*x*x
but we can get an other result with
*and* it is a *bug* that none of the results above generate an
Overflow (as it should) but 2.^3^4^5 does.
Only the Overflow result is correct.
> Hmm, this is the first I've seen the 'General::"ovfl": "Overflow
> computation."' message. If I try a different calculation, say
> don't get an overflow, but instead Mathematica just happily tries to
> the results (but I'm not holding my breath).
> > Is the HP wrong ?
> Well it would seem to be grouping the operators differently, but
> be a good reason for this?
> > L.DERUYCK
> > mailto://dr01202 at pophost.mediring.be
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