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I am trying to show that if a sequence of number $x_{n}$ is defined by $x_1 = h$, $x_{n+1}=x_n^2 + k$, where $0 and $h$ lies between the roots $a$ and $b$ of the equation $x^2 -x +k = 0$ Then show that $a < x_{n+1} and i am also interested in evaluating the limit of $x_n$.

Analysis towards a solution

I suspect that geometrically this sequence may have tendencies to converge or intersect this quadratic equation's parabola although i am unsure how to exploit this hunch. What else do I know $x^2 -x +k = 0 = (x-a)(x-b)$ hence $a + b =1$ and $0 < k = ab < \frac{1}{4}$

Although i am unsure how to proceed from here any help would be much appreciated.

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    Well, x_{n+1} - x_n = x_n^2 - x_n + k < 0 for $x_n \in (a,b)$ so $x_n$ is strictly decreasing, and from $a + b = 1$ and 0 < ab < \frac{1}{4} we know $a$ and $b$ must be positive. So also $x_n$ is positive for all $n$.2012-05-07

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Let us try to rescue one more damsel-question from the unanswered questions dragon's den.

First, the roots of the quadratic equation $\,x^2-x+k=0\,$ are $\frac{1\pm\sqrt{1-4k}}{2}\Longrightarrow a:=\frac{1-\sqrt{1-4k}}{2} < h< \frac{1+\sqrt{1-4k}}{2}=:b$ Observing the geometric interpretation of the above, we have the upwards parabola $\,f(x)=x^2-x+k\,$ with two intersection points with the $\,x-$axis, both with positive abscissa, and such that $f(h)=h^2-h+k<0\,$ , since $\,f(x_0)<0 \Longleftrightarrow a .

Clearly $\,f(x_1)=f(h)<0\,$, and we also have $x_2:=x_1^2+k Thus, we see that, in general, $a

So it is enough to prove now inductively on the index of $\,\{x_n\}\,$ that $\,x_{i+1}>x_i\,\,,\,\forall i\in\mathbb N\,$ ; assuming for $\,i< n\,$ we prove it for $\,i=n$: $x_{n+1}=x_n^2+k

Thus, $\,\{x_n\}\,$ is a monotonically decreasing sequence bounded below by $\,a\,$ , so its limit exists, call it $\,\alpha\,$. Using arithmetic of limits and the recursion $\,x_{n+1}=x_n^2+k\,$ we get $\alpha\xleftarrow [\infty\leftarrow n]{}{\color{red} {x_{n+1}=x_n^2+k}}\xrightarrow [n\to\infty]{} \alpha+k\,\,\Longrightarrow \alpha^2-\alpha+k=0\Longrightarrow \alpha=a$ since the sequence decreases.