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For $c>0$, consider the quadratic equation ${x^2-x-c=0, x>0}$. Define the sequence $\{x_n\}$ recursively by fixing $x_1>0$ and then, if n is an index for which $x_n$ has been defined, defining ${x_{n+1}=\sqrt{(c+x_n)}}$. Prove that the sequence $\{x_n\}$ converges monotonically to the solution of the above equation.

So, I know that a sequence must be bounded and monotone in order for for it to converge monotonically. I also have the solution to the quadratic equation, $x'=\frac{1+\sqrt{1+4c}}{2}$. My professor says that there will be two cases, which I understand as we have to consider it increasing or decreasing. But, the other hint he gives which is to use algebra to establish that $x_1\leq x_2 = \sqrt{c+x_1}$ is confusing to me. He also says to state case 1 as $x_1 \leq x'$, and I am just very confused about everything.

Wouldn't I be able to just prove it is monotonically increasing as $x,c\geq 0$ and then somehow show it is bounded? (Not that I would know how to do that either)

Thanks for any help

CleoFroPollo
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1 Answers1

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Your polynomial $x^2-x-c$ has two roots, of which the one that you mentioned is the only positive one. I will call it $\gamma$ and will call $\delta$ the other one.

If $x_1\geqslant \gamma$, then the sequence is decreasing. In fact\begin{align}x_{n+1}\leqslant x_n&\iff c+x_n\leqslant {x_n}^2\\&\iff {x_n}^2-x_n-c\geqslant0.\end{align}From this, together with that fict that $x^2-x-c\leqslant0$ if and only if $x\in[\delta,\gamma]$, it is easy to establish by induction that $(x_n)_{n\in\mathbb N}$ is indeed decreasing and that furthermore $(\forall n\in\mathbb N):x_n\geqslant\gamma$. So, $(x_n)_{n\in\mathbb N}$ converges and its limit $l$ is greater than or equal to $\gamma$. But\begin{align}l&=\lim_{n\to\infty}x_{n+1}\\&=\lim_{n\to\infty}\sqrt{c+x_n}\\&=\sqrt{c+l},\end{align}and $l=\sqrt{c+l}$ is equivalent to $l\geqslant0$ and $l$ is a root of $x^2-x-c$; in other words, $l=\gamma$.

The case in which $x_1\leqslant\gamma$ is similar.

José Carlos Santos
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  • Thank you for your response. It makes a lot more sense now, but I do not know how to do a proof by induction on a recursively defined equation. Specially, one that also has to do with the quadratic equation give. Are you able to give me a hint on how to prove that it is in fact decreasing? I know it is as I have worked out the first few terms of the sequence. – CleoFroPollo Aug 28 '18 at 15:34
  • I proved that $x\geqslant\gamma\implies\sqrt{c+x}\geqslant\gamma$. Therefore,$$x_1\geqslant\gamma\implies x_2=\sqrt{c+x_1}\geqslant\gamma.$$Now, take $n\in\mathbb N$ and assume that $x_n\geqslant\gamma$. Then$$x_{n+1}=\sqrt{c+x_n}\geqslant\gamma.$$ – José Carlos Santos Aug 28 '18 at 15:39