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looking for a calculator...
PostPosted: Sun Apr 03, 2005 3:22 am
by Slater
... that can do arbitrarily large numbers... like... a few million didgets long. anyone know where I can get one?
PostPosted: Sun Apr 03, 2005 4:38 am
by Fsiphskilm
?!
PostPosted: Sun Apr 03, 2005 8:51 am
by Technomancer
You'll never find a calculator that has that kind of precision, and you won't need one for any practical application. On the other hand, if you simply mean large numbers with truncated precision you should be able to find one easily. Most scientific calculators will allow numbers up to ~10.0^99 although their actual precision (# of digits stored) may vary.
What sort of applications do you need it for? Some calculators come with all kinds of bells and whistles (plotting, matrices, symbolics, etc) which can add to the cost. Personally, I found the Casio fx-6300G to be sufficient for most needs (it's inexpensive and has graphing/programmable capability) and used my desktop computer for more complicated tasks (IOW I used some mathematics software or wrote the program in C).
PostPosted: Sun Apr 03, 2005 1:51 pm
by Slater
I know. There are "only" about 10^100 atoms in the universe according to scientists nowadays.
Why do I want it? There was a small challenge on a forum that wondered if someone could answer the following equation (or something much like it):
N = A!^B!^C!^D! ... ^L!^14
where A = 1000!^1000!
B = A!^A!
C = B!^B!
etc.
That'd take a while on paper. I managed to solve for A to about 32 didgets of significant figures, but beyond that... hehe.
PostPosted: Sun Apr 03, 2005 3:04 pm
by Fsiphskilm
The ans
PostPosted: Sun Apr 03, 2005 3:47 pm
by Technomancer
The answer is definately not zero! The trick to the question is not to try and compute it directly. Instead you need to take the logarithm and use the properties of that operation. To get you started
N=A!^B!....L!^14
log N= 14*L!*log(A!^B!...^K!)
=14*L!*K!*....B!*log(A!)
From here on in it should be relatively straightforward.
PostPosted: Mon Apr 04, 2005 12:17 pm
by Godly Paladin
Technomancer, sometimes you truly are frightening.
How old are you?
PostPosted: Tue Apr 05, 2005 4:18 am
by Fsiphskilm
He is thousan
PostPosted: Tue Apr 05, 2005 1:18 pm
by Icarus
If I may say so, Tech, that is still a flipping large number.
PostPosted: Tue Apr 05, 2005 2:05 pm
by Technomancer
There's no real way around that unfortunately, given the question. The problem seems designed to test one's patience rather than mathematical insight. There are unfortunately no nifty transformations or identities with which one could easily reduce the problem, so a lot of hand calculation is still required.
PostPosted: Tue Apr 05, 2005 3:34 pm
by Mr. SmartyPants
technomancer is more wise than oldphil o_O
PostPosted: Wed Apr 06, 2005 5:48 am
by Technomancer
In truth you know, even with the logs there's no really compact way of writing the answer. You're pretty much restriced to the "power tower" that was used in the original question. The reason is that the actual number is well in excess of a googleplex, which would take more space to write than exists in the known universe.
Edit: with regards to wisdom, I really wouldn't know that. I do know about mathematics though.
PostPosted: Wed Apr 06, 2005 12:15 pm
by Godly Paladin
Hey, I could get fired up about a Technomancer vs. OldPhil bout. We should set one up, Mr. SP.
PostPosted: Wed Apr 06, 2005 12:25 pm
by Mr. SmartyPants
indeed so!
PostPosted: Wed Apr 06, 2005 12:30 pm
by Godly Paladin
Let's see... What style should it be in? An RP type fight, like the bouts for Zelda 27 and Sam*ron, or a test of intelligence? You've been around longer, SP...what do you think?
And hey, Technomancer, are you up to it?