My question concerns this exercise in Tao's notes on the semi-circular law for random matrices. We are trying to solve equation (22) in the notes, which is a quadratic equation whose coefficients are given up to $o(1)$ accuracy:
$$\displaystyle {\bf E} s_n(z) = - \frac{1}{z + {\bf E} s_n(z)} + o(1). \ \ \ \ \ $$
Here the $s_n$ are the Stieltjes transforms of a sequence of measures, and we would like to obtain $$\displaystyle {\bf E}s_n(z)=\frac{-z + \sqrt{z^2-4}}{2} + o(1)\ \ \ \ \ $$
Tao mentions we can do this using a "robust version of the quadratic formula." What does he mean by this?
It's tempting for me to just say that the roots of a quadratic are continuous functions of the coefficients, and so the given statement is obvious. Is it really that easy?
Given any $z,c \in \mathbb{C}$ such that $f(z) = -\frac1{f(z)+z} + c$:
$f(z)^2 + z f(z) + 1 = c (f(z)+z)$.
$4f(z)^2 + 4(z-c) f(z) + 4(1-cz) = 0$.
$( 2f(z) + z-c )^2 = (z-c)^2 - 4 + 4cz = z^2 - 4 + 2cz + c^2$.
Note that (using principal branch for $\sqrt{}$): $\def\wi{\subseteq}$
$\sqrt{a+o(1)} \wi \sqrt{a} \sqrt{1+o(1)} \wi \sqrt{a}(1+o(1)) \wi \sqrt{a}+o(1)$ for any $a \in \mathbb{C}$.
As $n \to \infty$, and given $z \in \mathbb{C}$ such that $f(z) = -\frac1{f(z)+z} + o(1)$:
$( 2f(z) + z-o(1) )^2 \wi z^2 - 4 + o(1)$.
Thus $2f(z) + z-o(1) \wi \sqrt{z^2-4} + o(1)$.
Thus $f(z) \in \dfrac{-z+\sqrt{z^2-4}}{2} + o(1)$.
Note that if you want more precise error bounds, you will have to use the Taylor expansion for $\sqrt{}$, in which case you have to distinguish two cases:
If $z^2-4$ is bounded away from $0$, then the error term is of the same order.
If not, then the error term can be up to the square-root of the original.