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Re: Re: Differences between recursions and limit
*To*: mathgroup at smc.vnet.net
*Subject*: [mg60585] Re: [mg60553] Re: Differences between recursions and limit
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Wed, 21 Sep 2005 03:20:18 -0400 (EDT)
*References*: <dgm01k$njm$1@smc.vnet.net> <200509200919.FAA17855@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
On 20 Sep 2005, at 18:19, albert wrote:
> for your other question -- I don't remember the correct conditions
> for the
> following to work, but in your case the result (0.4142...) you are
> after
> can be received with (I get a negative result and 1 as other
> solutions of
> the NSolve, but these can be excluded with your starting value):
>
> NSolve[x == .25*(1 + x + (x)^2 + (x)^3), x]
>
No need to remember any conditions since this easy to prove directly.
Consider the relation
x[0] = 0;
x[k] == 0.25*(1 + x[k - 1] + (x[k - 1])^2 + (x[k - 1])^3);
Note that we can prove by induction that for all k
1. x[k] <= x[k+1]
2. x[k]<= 1 (x[k] is one fourth of the sum of 4 numbers each <=1)
These two conditions imply that Limit[x[k],k->Infinity] exists. Now
you can just take the limit of the recursive relation to get the
equation
x==0.25 (1+x+x^2+x^3) where x is the limit.
Andrzej Kozlowski
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