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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: <9j396v$iiv$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi, 
Mathematica can't calculate exactly 2^3^4^5 not only because 
there's no place in the whole Universe to save these some 
1.124021466*10^488 digits.

But She can't even evaluate it as it is much nore than
$MaxNumber which in my case is only of some  
3.2322801015848*10^8 digits.

Is it so?

%%%%%%%%%%%%%%%%%%%
Subject: [mg30258]      Re: Exponents and notation
Author:       Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
Organization: Universitaet Leipzig

Hi,

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 

2^3^4^5

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

In[]:=
x = 2.^3^4;
xx = x*x*x*x*x

Out[]=
  8.263199609878108*^121

but we can get an other result with

In[]:=x^5
Out[]=8.263199609878108*^121

*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.

Regards
  Jens

> 
> Hmm, this is the first I've seen the 'General::"ovfl": "Overflow
occurred in
> computation."' message.  If I try a different calculation, say
8^8^8, I
> don't get an overflow, but instead Mathematica just happily tries to
find
> the results (but I'm not holding my breath).
> 
> > Is the HP wrong ?
> >
> >
> 
> Well it would seem to be grouping the operators differently, but
there may
> be a good reason for this?
> 
> > L.DERUYCK
> > mailto://dr01202 at pophost.mediring.be
> >
> 
> Michael




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