Sequences / Real Analyses question

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In summary, we have a sequence defined by x_1 = h, x_{n+1} = x^2_n + k, where a and b are the roots of the quadratic equation x2 - x + k = 0, with 0 < k < 1/4 and a is the smaller root. We can prove that a < xn+1 < xn < b and that x_{n+1} < x_n by induction, and also that a is a lower bound for Xn. This is because x^2-x+k=(x-a)(x-b) which is negative if and only if a<x<b. The limit of xn can be determined to be between a and b.
  • #1
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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:
[tex]x_1 = h, x_{n+1} = x^2_n + k. [/tex]

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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  • #2
perhaps starts by determining a and b in terms of k?
 
  • #3
Okay, so I got [tex]a = \frac{1 - \sqrt{1 - 4k}}{2}, b = \frac{1 + \sqrt{1 - 4k}}{2}[/tex].

And I was able to prove [tex]X_{n+1} < X_n [/tex] by induction. But, I'm stuck on the outer inequalities.

EDIT: [tex]X_{n+1} < X_n [/tex] 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..
 
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  • #4
I think the basic idea is:
[tex]X_{n+1} < X_n \Leftrightarrow X^2_n - X_n + k < 0[/tex]

Therefore, Xn must be between the roots for this equation to be negative.
But is there a more mathematical way to state it?
 
  • #5
Good job! How about saying x^2-x+k=(x-a)(x-b) which is negative if and only if a<x<b.
 
  • #6
Oh right! Thanks a lot :)
 

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