I want to show that an iteration method converges locally quadratic to $\sqrt{a}$.

I have done the following:

We have that $|x_{n+1}-\sqrt a|=\frac1{2x_n}|x_n-\sqrt a|^2$.

To get local convergence we want that the starting point is near the root. Then this should hold for all approximations. So it must hold that $\left |x_n-\sqrt{a}\right |<\epsilon$, with $\epsilon>0$. Is this true for all $\epsilon$ ? Can we just pick one?

Then we get: $$\left |x_n-\sqrt{a}\right |<\epsilon \Rightarrow \frac{1}{2(\sqrt{a}+\epsilon)}<\frac{1}{2x_n}<\frac{1}{2(\sqrt{a}-\epsilon)}$$ So we get $$|x_{n+1}…

for $X \sim \mathcal{N}(0,I_d)$,
\begin{align*}
\E[\norm{\Sigma^{\frac{1}{2}}X}^2] &= \int_{\R^d} \norm{\Sigma^{\frac{1}{2}}x}^2\cdot \frac{1}{(2 \pi)^\frac{d}{2}} e^{-\frac{1}{2}x^Tx}\,\d x \\ &= \int_{\R^d} x^T \cdot \Sigma \cdot x\cdot \frac{1}{(2 \pi)^\frac{d}{2}} e^{-\frac{1}{2}x^Tx}\,\d x \\ &= \sum_{i,j} \sigma_{i,j} \int_{\R^d} x_ix_j \frac{1}{(2 \pi)^\frac{d}{2}} e^{-\frac{1}{2}x^Tx}\,\d x = \sigma_{1,1} + ... +\sigma_{d,d}, \end{align*}
this should be it if I don't overlook something ovious