# Recursive sequence and a quadratic equation related inequality proof

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 $0<k<\frac{1}{4}$ and $h$ lies between the roots $a$ and $b$ of the equation $$x^2 -x +k = 0$$ Then show that $$a < x_{n+1}<x_n<b$$ 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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If $x_n \to x$ then $x = x^2 + k$, hence $x = a$ or $b$, and from your observations it can only converge to $a$. – TMM May 7 '12 at 20:44
@TMM I am not sure if i follow why it can only converge to $a$ ? – Comic Book Guy May 7 '12 at 20:49
If $x_n \to x$ for some $x$, then also $x_{n+1} \to x$. Taking the limit of $n \to \infty$ in $x_{n+1} = x_n^2 + k$ we get $x = x^2 + k$. This is exactly your equation $x^2 - x + k = 0$, so $x = a$ or $x = b$ are the only solutions. And since $x_{n+1} (< x_n < b)$ moves away from $b$ and towards $a$, it can only converge to $a$. – TMM May 7 '12 at 20:53
@TMM I am unsure if we can assume $x_{n+1} < x_{n}$ what if absolute values of $x_{n+1}$ was less than 1 then the inequality would reverse. – Comic Book Guy May 7 '12 at 21:02
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 $n$. – TMM May 7 '12 at 21:05

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 $\,f(x_0)<0 \Longleftrightarrow a<x_0<b\,$ .
Clearly $\,f(x_1)=f(h)<0\,$, and we also have $$x_2:=x_1^2+k<x_1\Longleftrightarrow f(x_1)=x_1^2-x_1+k<0\Longleftrightarrow a<x_1=h<b$$ Thus, we see that, in general, $$a<x_n<b\Longleftrightarrow f(x_n)=x_n^2-x_n+k<0\Longleftrightarrow x_{n+1}:=x_n^2+k<x_n$$
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 $\,i=n$: $$x_{n+1}=x_n^2+k<x_n\Longleftrightarrow f(x_n)=x_n^2-x_n+k\stackrel {ind. hypothesis}<0$$
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.