Prove that $x_{n+1}=1/2(x_n+\frac{c}{x_n})$ converges to $\sqrt{c}$

Question

Suppose that $x_1=1$ and $x_{n+1}=1/2(x_n+\frac{c}{x_n})$, show the sequence converges to $\sqrt{c}$

Attempt:

Let $r$ be the limit of the sequence. I think I need break into two cases: $0\leq x_n<r$ and $x_n>r$ to reach $0<x_n<x_{n+1}<r$ and $x_n>x_{n+1}>r$ so that I can apply the monotone convergent theorem.

I think I need to show $x_{n+1}=1/2(x_n+\frac{c}{x_n})\leq x_n$ but I don't quite sure how to show this is true for $0\leq x_n<r$. To show $x_{n+1}<r$, I don't see any information I can apply. Can anyone give me a hit or suggestion to do this question? Thanks

Hint: $x_{n+1}$ is the arithmetic mean of $x_n$ and $\frac{c}{x_n}$. What's their geometric mean? — Steven Stadnicki, Oct 30 '15 at 03:45
@StevenStadnicki sorry, I don't what is geometric mean in English — Simple, Oct 30 '15 at 03:50
Simple: https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means — Steven Stadnicki, Oct 30 '15 at 03:50
@StevenStadnicki $\sqrt{c}$, but the monotone convergent theorem is the most recent theorem I learn to show a sequence converges to real number — Simple, Oct 30 '15 at 03:56
@Simple My point is that you can use the inequality in the link to derive a relevant result between $x_{n+1}$ and $\sqrt{c}$. — Steven Stadnicki, Oct 30 '15 at 04:24

score 0 · Answer 1 · answered Oct 30 '15 at 03:50

Consider $f(x)=1/2(x_{n}+c\frac{1}{x_n})$, the only posibily for x>0 that $x=f(x)$ is $x=\sqrt c$, and as $x_n \geq 0$ (show it), evaluate the sign of $x_{n+1}-x_n$ and show that it is growing depending a certain value of c and $x_n \leq \sqrt c$ so it converge and the limit could only be $\sqrt c$.

Prove that $x_{n+1}=1/2(x_n+\frac{c}{x_n})$ converges to $\sqrt{c}$

1 Answers1