I have the following algorithm for producing rational approximations of $\sqrt x$. We take an initial $x_0$ and apply the following steps:
$$x_n=\sqrt{\mu_n}+\sqrt{x+\mu_n}\tag1$$ $$x_{n+1}=\sqrt{x+\mu_n}\tag2$$ $$x_{n+1}=\sqrt{\mu_{n+1}}+\sqrt{x+\mu_{n+1}}\tag 3$$ etc. Since $\mu_{t+1}<\mu_t$, this converges on $\sqrt x$
This formula $(1)$ can be algebraically manipulated:
$$2\mu_n+x+2\sqrt{\mu_n}\sqrt{x+\mu_n}=x_n^2$$ $$\to\mu_n+\sqrt{\mu_n}\sqrt{x+\mu_n}=\frac{x_n^2-x}{2}$$ $$\to x_n\sqrt{\mu_n}=\frac{x_n^2-x}{2}$$ $$\mu_n=\bigg(\frac{x_n^2-x}{2x_n}\bigg)^2$$ $$\to x_{n+1}=\sqrt{x+\mu_n}=\frac{x_n^2+x}{2x_n}$$ Which is the well known Babylonian Method.
We know, as the algorithm I present proves, that this is an overestimate of $\sqrt x$. I tried switching up my algorithm to: $$x_n=\sqrt{x-\nu_n}-\sqrt{\nu_n}\tag 1$$ $$x_{n+1}=\sqrt{x-\nu_n}\tag2$$ $$x_{n+1}=\sqrt{x-\nu_{n+1}}-\sqrt{\nu_{n+1}}\tag3$$ etc in order to find an underestimate with roughly the same accuracy scaling, but couldn't find a similar generalisation. I squared $(1)$ as before and got to: $$x_n=x-2\sqrt{\nu_n}\sqrt{x-\nu_n}$$ Then applied $\sqrt{x-\nu_n}=x_n+\nu_n$ which led to: $$-2x_n\sqrt{\nu_n}-2\nu_n=x_n-x$$ but this would require solving a quadratic (taking $y=\sqrt{\nu_n}$ gives us $y^2+x_ny-\frac12(x-x_n)=0$), leaving irrationals all over the place.
Is it possible to manipulate the second system to $x_{n+1}=f(x_n)$ where $f:\Bbb Q \mapsto \Bbb Q$, as I have done with the first system, and if so, which steps have I overlooked towards achieving this?
Thanks for any assistance.