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

**References**:**Re: Differences between recursions and limit***From:*albert <awnl@arcor.de>