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Re: Re: Differences between recursions and limit


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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