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
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.