Complete the square in the joint probability density function:The density therefore factors as the standard normal distribution density for $Y$ times a normal distribution density in $x$ with mean $\rho y$ and variance $1-{\rho }^2$. It follows that $\mathbb{E}[Y]=0$, $\mathbb{E}[Y^2]=1$, and $\mathbb{E}[X\mid Y]=\rho Y$. Symmetry gives $\mathbb{E}[X]=0$. By the law of total expectation,Consequently the covariance of this bivariate normal distribution is $\mathrm{Cov}(X,Y)=\rho$. This calculation derives the result from the given density.

The update in Newton's method in optimization is ${\mathbf{z}}_{k+1}={\mathbf{z}}_k-H_f({\mathbf{z}}_k{)}^{-1}\nabla f({\mathbf{z}}_k)$. Substituting the previous formulas givesThus the distance from the unique minimum obeys $r_{k+1}=r_k^3$, where $r_k=\Vert {\mathbf{z}}_k\Vert$. In particular,For $0<r_0<1$, the error ratio $r_{k+1}/r_k^3$ is exactly $1$, proving exact cubic convergence. If $r_0=0$, the initial point already attains the minimum. All updates are defined because the Hessian is invertible everywhere. This is Newton minimization of the square-root norm; its special update also shows why the given initial condition matters: $r_0=1$ produces an alternating two-cycle, and $r_0>1$ diverges.

For iterates approaching ${\mathbf{x}}_{\ast }$, write $e_k=\Vert {\mathbf{x}}_k-{\mathbf{x}}_{\ast }\Vert$. Exact order of convergence $p$ means $e_{k+1}/e_k^p\to C$ for a finite constant $C>0$, with nonzero errors. Thus quadratic convergence has $p=2$, and cubic convergence has $p=3$. The weaker eventual bound $e_{k+1}\le C{e}_k^p$ means at least order $p$. Finite termination is a separate case.^[ref]

The assertion about being always faster is false. For example, the fixed-point iterations with errors $e_{k+1}=e_k^2$ and $d_{k+1}=50d_k^3$ have exact quadratic convergence and cubic convergence, respectively. Starting both at $0.1$, their next errors are $0.01$ and $0.05$: the cubic convergence method is worse on that step, although both converge. Error constants, starting points, and work per step all matter. The higher order of convergence describes asymptotic behavior within a convergence region, not a uniform comparison of all computations.

Let $P=\mathbf{z}{\mathbf{z}}^T$, so $P^2=r^{2}P$ and $f^2=1+r^2$. Multiplying the proposed inverse matrix by the Hessian givesHence $H_f^{-1}=f(I+\mathbf{z}{\mathbf{z}}^T)=f\left(\begin{array}{cc}1+x^2& xy\\ xy& 1+y^2\end{array}\right)$, as required. Its existence also follows from the Hessian being a positive-definite matrix.

Put $\mathbf{z}=(x,y{)}^T$, $r^2={\mathbf{z}}^T\mathbf{z}$ and $f=\sqrt{1+r^2}$. The chain rule yieldsFor every nonzero vector $\mathbf{v}$, the Cauchy-Schwarz inequality givesThus the Hessian is a positive-definite matrix everywhere. The Hessian criterion for convexity implies that $f$ is strictly convex on ${\mathbb{R}}^2$. Its unique global minimum is at $\mathbf{z}=0$, where $f=1$.

For a differentiable scalar function of $n$ real variables, its derivative is the linear mapwhere the column gradient is $\nabla f=(\partial f/\partial x_1,\dots ,\partial f/\partial x_n{)}^T$. Its Hessian is the $n\times n$ matrix $H_f=({\partial }^{2}f/\partial x_i\partial x_j{)}_{i,j}$. For a $C^2$ function, equality of mixed partial derivatives makes this matrix symmetric.

Choose equation (1), evaluated by refined composite Simpson's rule. It gives $y$ directly as a one-dimensional definite integral with fixed endpoints and a smooth integrand. It therefore avoids root finding altogether. Before evaluating, use the trigonometric identityThis equivalent integrand is nonnegative throughout $[0,\pi /2]$, avoiding cancellation between the original two terms. Equation (2) instead subtracts two potentially large exponential functions, while equation (3) would ordinarily put a numerical integration inside a root finding loop.

For even $n$, let $h=(\pi /2)/n$, $\varphi (x)=2\sin x\cos ((\pi /4)\cos x)$, and computeThen double $n$. The integrand has bounded derivatives of every fixed order on the integration interval; in the fourth-order error regime, the estimated remaining truncation error in $Y_{2n}$ is $|Y_{2n}-Y_n|/15$. The factor $15$ follows by eliminating the leading term $C{h}^4$ between the two mesh widths.^[ref]

Stop when this estimate is comparable to ${\epsilon }_{\mathrm{mach}}\max (1,|Y_{2n}|)$ and further refinement shows stagnation at the roundoff error scale. Check that the differences previously decreased consistently with fourth-order convergence; differences alone are not a rigorous error certificate if cancellation or roundoff error dominates. Compensated summation helps keep the summation error near the precision limit. Refining indefinitely after this point increases work without reliably adding meaningful digits. This answers the requested choice and termination strategy without requiring an explicit value of $y$.