Re: 9^(9^(9^9))

• To: mathgroup at smc.vnet.net
• Subject: [mg22900] Re: [mg22884] 9^(9^(9^9))
• From: Bojan Bistrovic <bojanb at python.physics.odu.edu>
• Date: Tue, 4 Apr 2000 22:41:04 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```> I am using Mathematica 4.0 and need help getting a few of the front digits of
> this large number.Your method would also be greatly appreciated.
>

These might NOT be good ideas, but here they are anyway:
1.) Use the identity x=Exp[Log[x]] to rewrite 9^(9^(9^9)) as
Exp[Log[9] Exp[Log[9] Exp[ 9 Log[9]]]] and then expand the Exp's in a
power series; the error for each Exp will be of the order x^(n+1)/(n+1)!
You'll have to do a little experimenting to see when will a few of the front
machine-precision numbers since you're onlt interested in a few main ones.
2.) write it as (1+8)^( (1+8)^( (1+8)^9 ) ) and then use the binomial formula
for expansion of (a+b)^n discard smaller terms and hope you're not introducing
too large error by doing it.
Comining these two approaches and experimenting a little will probably produce
even better results. It could be a good idea to lookup the derivation of
Stirling's approzimation and Stirling's series since they deal with large
numbers and might (or might not) give you a new idea; here's a link for start:
http://mathworld.wolfram.com/StirlingsSeries.html

Bye, Bojan

--
---------------------------------------------------------------------
Bojan Bistrovic,                                      bojanb at jlab.org
Old Dominion University, Norfolk VA & Jefferson Lab, Newport News, VA
---------------------------------------------------------------------

```

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