Consider the sequence $x_{n+1}=x_n-\frac{x_n^2-a}{2}$, prove that $\forall a \in (0,1) $ the sequence converges to $\sqrt{a}$.
Considering a function $f(x)=x-\frac{x^2-a}{2}$ we have $$f'(x) = 1 - x,$$ which means the function is increasing $\forall x < 1$.
Next if we look at what happnes if $x<1$, we have $$f(x) \le f(1) = 1 - \frac{1-a}{2} = \frac{1}{2} +\frac{a}{2} <1.$$
But this is where I got stuck. I can't figure out such $Q$ that $f:Q \rightarrow Q$ and moreover I can't show that $f(x)$ is a contractor since Lagrange's theorem yields nothing useful:
$$|f(x) - f(y)| = |1-\xi |* |x-y|.$$
But I don't see a way of bounding $|1-\xi|$ above especially not knowing the domain $Q$.
Without those 2 conditions I can't use the fixed point theorem which allows me to find the limit by simply solving the following equation:
$$x=x-\frac{x^2-a}{2}$$ which sure enough yields $x^*=\sqrt{a}$.
How would one approach this ?
To get a contraction mapping, what is required is $$|f(x)-f(y)|=|x-y-\frac{x^2-y^2}{2}|=\left|1-\frac{x+y}{2}\right||x-y|\le c|x-y|$$ with $0\le c<1$. So one condition is $$1-c\le\frac{x+y}{2}\le 1+c\tag{1}$$ Let $Q=[\alpha,\beta]$, then the contraction condition would be ensured if $1-c\le\alpha$ and $\beta\le 1+c$.
The other condition is $f:Q\to Q$, that is $$\alpha\le x-\frac{x^2-a}{2}\le \beta\tag{2}$$ Given that $a\in(0,1)$ and $x-\frac{x^2}{2}\le\frac{1}{2}$, one can pick $\alpha=a/2$, $\beta=1$, to ensure both $(1)$ and $(2)$.