Sequences / Real Analyses question

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The discussion focuses on the analysis of a sequence defined by the recurrence relation \(x_{n+1} = x_n^2 + k\), where \(k\) is a parameter in the quadratic equation \(x^2 - x + k = 0\) with roots \(a\) and \(b\) defined for \(0 < k < \frac{1}{4}\). The roots are calculated as \(a = \frac{1 - \sqrt{1 - 4k}}{2}\) and \(b = \frac{1 + \sqrt{1 - 4k}}{2}\). The sequence \(x_n\) is shown to be bounded by \(a < x_{n+1} < x_n < b\), and the limit of the sequence is determined through induction and properties of quadratic inequalities.

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Sequences / Real Analysis question

Homework Statement


a,b are the roots of the quadratic equation x2 - x + k = 0, where 0 < k < 1/4.
(Suppose a is the smaller root). Let h belong to (a,b). The sequence xn is defined by:
x_1 = h, x_{n+1} = x^2_n + k.

Prove that a < xn+1 < xn < b, and then determine the limit of xn.

Homework Equations


The Attempt at a Solution


I have no idea how to start, if you could help me.
Thanks.
 
Last edited:
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perhaps starts by determining a and b in terms of k?
 
Okay, so I got a = \frac{1 - \sqrt{1 - 4k}}{2}, b = \frac{1 + \sqrt{1 - 4k}}{2}.

And I was able to prove X_{n+1} &lt; X_n by induction. But, I'm stuck on the outer inequalities.

EDIT: X_{n+1} &lt; X_n means that X1 = h is the largest value of Xn for all n. And h belongs to (a,b), so X1 < b, and consequently Xn < b.

I still need to prove that a is a lower bound..
 
Last edited:
I think the basic idea is:
X_{n+1} &lt; X_n \Leftrightarrow X^2_n - X_n + k &lt; 0

Therefore, Xn must be between the roots for this equation to be negative.
But is there a more mathematical way to state it?
 
Good job! How about saying x^2-x+k=(x-a)(x-b) which is negative if and only if a<x<b.
 
Oh right! Thanks a lot :)
 

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