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 digits stop changing. Good thing about this method is that you cna use 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 ---------------------------------------------------------------------