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


$=\frac{a-x^2_n}{2x_n}\leq 0$ 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.



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.

Your reasoning about the discriminant is correct.


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