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
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