Stuck on a recursively defined converging sequence problem.

Question

I am given a simple quadratic equation $$x^2-x-c=0, x>0, c>0$$ and then we define a sequence $\{x_n\}$ with $x_1>0$ fixed and then, if $n$ is an index for which $x_n$ has been defined, we define $$x_{n+1}=\sqrt{c+x_n}$$.

With that I am asked to prove that $\{x_n\}$ converges monotonically to the solution of the polynomial.

I've done quite a bit of scratch work. Obviously we can solve the quadratic and the positive solution is $\frac{1+\sqrt{5}}{2}$. I have an inkling that the equation is decreasing and so I tried working with $x_{n+1}-x_{n+2}$ to show that the difference is positive but I didn't come up with anything useful.

I did realize that if I simply write out the limit we see that $$\lim_{n\rightarrow\infty}[(\sqrt{c+x_{n+1}})^2)-\sqrt{c+x_{n+1}}-c]=\lim_{n\rightarrow\infty}[x_{n+1}-\sqrt{c+x_{n+1}}]$$

So if we want this final limit to go to zero then all I really need is $x_n$ to be monotonically decreasing since it is clearly bounded below by zero since $c$ and $x_1$ were taken to be positive.

Answer 1

Hint

$$x_{n+1}^2-x_n=c$$ $$x_{n+2}^2-x_{n+1}=c$$ so,

$$x_{n+2}^2-x_{n+1}^2-(x_{n+1}-x_n)=0$$

$$(x_{n+2}-x_{n+1})(x_{n+2}+x_{n+1})=(x_{n+1}-x_n)$$

suppose that $x_{N+1}<x_{N}$ for some $N$. What can you conclude?

After that, you have to study the relation between $x_1$ and $x_2$, which will depends on $c$.

Answer 2

$\begin{array}\\ x_{n+1}-x_n &=\sqrt{x_n+c}-x_n\\ &=(\sqrt{x_n+c}-x_n)\dfrac{\sqrt{x_n+c}+x_n}{\sqrt{x_n+c}+x_n}\\ &=\dfrac{x_n+c-x_n^2}{\sqrt{x_n+c}+x_n}\\ \end{array} $

If $f(x) = x^2-x-c$, $f'(x) = 2x-1$, so $f'(x) > 0$ for $x > \frac12$.

The roots of $f(x)$ are $x =\dfrac{1\pm\sqrt{1+4c}}{2} $, so the positive root $x_c$ satisfies $1 < x_c \lt 1+c$.

If $\frac12 < x_n < x_c$, then $x_n^2-x_n-c < 0$ so $x_{n+1} > x_n$.

If $ x_n > x_c$, then $x_n^2-x_n-c > 0$ so $x_{n+1} < x_n$.

Similarly,

$\begin{array}\\ x_{n+1}-x_c &=\sqrt{x_n+c}-x_c\\ &=(\sqrt{x_n+c}-x_c)\dfrac{\sqrt{x_n+c}+x_c}{\sqrt{x_n+c}+x_c}\\ &=\dfrac{x_n+c-x_c^2}{\sqrt{x_n+c}+x_c}\\ &=\dfrac{x_n+c-(x_c+c)}{\sqrt{x_n+c}+x_c} \qquad\text{since } x^c_2 = x_c+c\\ &=\dfrac{x_n-x_c}{\sqrt{x_n+c}+x_c}\\ \end{array} $

Therefore $x_{n+1}-x_c$ has the same sign and is smaller in absolute value than $x_n-x_c$.

Therefore $x_n \to x_c$ since $\sqrt{x_n+c}+x_c \gt 1+c$.