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Re: Mathematica:recursion with 2 arguments?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg64441] Re: Mathematica:recursion with 2 arguments?
*From*: "Scout" <Scout at nodomain.com>
*Date*: Fri, 17 Feb 2006 04:11:50 -0500 (EST)
*References*: <dspfp8$cfh$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
"ddjjjkkkk" <noble_c at hotmail.com>
> Hi,everybody:
> My English is a little poor.But I have a very flustered problem to consult
> you.Who can help me? Here is the prob:
> IF function f(m,n)(m,n are non-negative integers)satisfying 3 conditions:
> f(0,n)=n+1,f(m+1,0)=f(m,1) and f(m+1,n+1)=f[m,f(m+1,n)],then how can we
> use Wolfram's Mathematica to solve f(m,1),f(m,2),f(m,3) and so on.I can
> get f(1,n)=n+2,f(2,n)=n+3,f(3,n)=2^(n+3)-3 by hand and pen(but how by
> Mathematica?).Surely,f(4,n) is very complicated.
> Who can use Mathematica's language (recursion or iteration etc.) to solve
> them out? Friends,help me--a poor person,please!!!!!!
>
That is an example of a doubly-recursive function that is not reducible to
primitive recursion,
it's not so reducible.
The function grows extremely fast!
In fact f(4,1)=65533 while f(4,2) has 19729 digits!
More advanced mathematics is needed.
This type of function occurs as "Wilhelm Ackermann" 's function.
~Scout~
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