How can you determine if the following recursive sequence converges:
$$x_{n+1}=\frac{1}{2}(a+x_n^2)$$
where $0\le a \le 1$ and $x_1=0$
I know that the limit x (if it exists) satisfies the following equation:
$$x=\frac{1}{2}(a+x^2)$$
since $\lim_{x\rightarrow \infty} x_n = \lim_{x\rightarrow \infty} x_{n+1}$. Therefore $x=1\pm \sqrt{1-a}$
Thank you in advance 🙂
Best Answer
By induction we can show that $0\leq x_n<1$.
Now, for $a=1$ we obtain $$x_{n+1}-x_n=\frac{1}{2}(x_n-1)^2\geq0$$ and the rest is smooth.
But for $a<1$ we obtain $0<r=\frac{2-\sqrt{1-a}}{2}<1$ and $$|x_{n+1}-1+\sqrt{1-a}|=\left|\frac{1}{2}x_n^2+\frac{1}{2}a-1+\sqrt{1-a}\right|=$$ $$=\frac{1}{2}\left|x_n^2-\left(1-\sqrt{1-a}\right)^2\right|=\frac{1}{2}\left|x_n-1+\sqrt{1-a}\right|\cdot\left|x_n+1-\sqrt{1-a}\right|\leq$$ $$\leq r\left|x_n-1+\sqrt{1-a}\right|\leq...\leq r^n\left|x_1-1+\sqrt{1-a}\right|\rightarrow0.$$