# If $x_1=1$ and $x_{n+1}=\frac{1}{2} (x_n+\frac{a}{x_n})$ for all $n\in \mathbb{R}$ where $a>0$, Prove that $x_n$ convergent.

If $$x_1=1$$ and $$x_{n+1}=\frac{1}{2} (x_n+\frac{a}{x_n})$$ for all $$n\in \mathbb{R}$$ where $$a>0$$, Prove that $$x_n$$ convergent.

I want to understand this solution.

Notice that $$x_n>0$$ for all $$n\in \mathbb{N}$$ and $$x_n$$ is a solution of this equation $$t^2-2x_{n+1} t+1=0$$ "How can I find this equation?"

Then $$\Delta=4x^2_{n+1}-4a$$ is non-negative "Is that because if it is negative then the roots of this equation well be complex? and the sequence $$(x_n)$$ is a real sequence" Then $$x^2_{n+1} \geq a$$ for all $$n\in \mathbb{N}$$.

That's mean $$(x_n)$$ is bounded below.

Also we have $$x_{n+1}-x_n=\frac{1}{2} x_n+\frac{a}{2x_n}-x_n$$

$$=\frac{a}{2x_n}-\frac{x_n}{2}$$

$$=\frac{a-x^2_n}{2x_n}\leq 0$$ for all $$n\geq 2$$

Hence $$(x_n)$$ is decreasing, and therefore convergence.

So I only want to understand How can I find the equation (*), And why $$\Delta$$ is non-negative, Please.

Thanks.

It is given that $$x_{n+1}=\tfrac{1}{2} \left(x_n+\frac{a}{x_n}\right)$$. Take $$x_n=t$$. Then $$x_{n+1}=\tfrac{1}{2} \left(x_n+\frac{a}{x_n}\right)$$ can be written as $$x_{n+1}=\frac{1}{2} \left(t+\frac{a}{t}\right).$$Now if you rearrange the equation, you will get a quadratic in $$t$$,$$t^2-2x_{n+1} t+a=0$$ and $$t=x_n$$ is a solution.