Let $V$ be a bounded region with piecewise smooth closed boundary $\partial V$, and let $\mathbf{F}$ be a continuously differentiable vector field on an open neighbourhood of its closure. The divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal vector. Thus the volume integral of the divergence equals the outward boundary flux integral. For a region containing a hole, the outward unit normal vector on its inner boundary points into that hole.

Let $P=3x^4-5x^3+2x^2-3x+2$ and $Q=3x^3+x^2+x-2$. Applying polynomial long division gives the Euclidean algorithmThus their polynomial greatest common divisor is $3x-2$, up to a nonzero constant factor. Their unique common root of a polynomial is therefore

The Euclidean algorithm computes the greatest common divisor of two integers by repeated division with remainder. Start with $r_0=|A|$, $r_1=|B|$, interchanging them if necessary, and repeatedly writeuntil a zero remainder occurs. The positive remainders strictly decrease, so the process terminates. At each step the common divisors of $r_{j-1},r_j$ are exactly the common divisors of $r_j,r_{j+1}$, because $r_{j+1}=r_{j-1}-q_j{r}_j$. HenceIf one input is zero, the answer is the absolute value of the other; the usual convention sets $\gcd (0,0)=0$. Back-substitution also expresses the greatest common divisor as an integer linear combination of the inputs, as in Bézout's identity.

The partial derivatives of the parametrization areTheir cross product isTaking its euclidean norm gives the surface area elementThe two poles and duplicated seam have zero surface area, so integrating over the stated parameter domain counts the surface area once. With the change of variables $u=\cos \theta$ and the supplied definite integral,Here $-\ln (\sqrt{2}-1)=\ln (\sqrt{2}+1)$ because $(\sqrt{2}-1)(\sqrt{2}+1)=1$.

The ellipsoid is the level set $F(x,y,z)=2x^2+y^2+z^2=1$. Its gradient is $(4x,2y,2z)$, which never vanishes on the ellipsoid. Half this gradient gives the outward normal vectorThis normal vector remains valid at $\theta =0,\pi$, where the given parametrization is singular.

Expanding and taking partial derivatives gives the gradient and Hessian matrixAt $p_0=(-1,0{)}^T$, these are $\nabla f(p_0)=(-6,2{)}^T$ andThus the Newton step solvesThe second equation gives $s_y=1-2s_x$; substituting into the first gives $22s_x=10$. HenceAt $p_{\ast }=(-1/2,-1/2{)}^T$, however,The eigenvalues of this Hessian matrix are $-1$ in direction $(1,1)$ and $7$ in direction $(1,-1)$. It is invertible but indefinite. Thus the Newton step is zero, andThis illustrates that stationary points are fixed points of Newton's method. Here the fixed point is a saddle point of a scalar function, not a local minimum. Indeed, direct substitution along the two eigenvector directions givesFor sufficiently small nonzero $t$, one direction decreases $f$ and the other increases it. The global minimum is instead zero: the sum of squares vanishes at $(0,0)$ and $(-1,-1)$. These are the only real zeros, since $x=-y^2$ and $y=-x^2$ imply $y(y^3+1)=0$. Thus a zero gradient is an insufficient stopping certificate for minimization problem without classifying the stationary point.

For an unconstrained differentiable function of two real variables, an interior local minimum must solve the simultaneous stationary point equationsApply Newton's method in optimization to this gradient equation. The Jacobian matrix of the gradient is the Hessian matrixAt $p_n=(x_n,y_n{)}^T$, solve the linear system $H(p_n)s_n=-\nabla f(p_n)$, then set $p_{n+1}=p_n+s_n$. When the Hessian matrix is invertible, this isThis follows by setting the first order Taylor expansion $\nabla f(p_n+s_n)\simeq \nabla f(p_n)+H(p_n)s_n$ equal to zero. Solving the linear system avoids explicitly computing a matrix inverse. A stationary point need not be a local minimum: for example, a positive-definite matrix $H(p)$ suffices to classify a stationary point as a strict local minimum, whereas an indefinite Hessian matrix identifies a saddle point of a scalar function.

An infinitesimal parameter rectangle becomes a parallelogram with edges ${\mathbf{r}}_{s}ds$ and ${\mathbf{r}}_{t}dt$. Its area is the Euclidean norm of their cross product, giving the surface area elementThus, for a regular parametrization covering the surface once except possibly on sets of area zero,For several patches, add their surface integrals while avoiding overlap; repeated coverage would otherwise count area with multiplicity.

At a regular point of the parametrized surface, the partial derivatives ${\mathbf{r}}_s$ and ${\mathbf{r}}_t$ are linearly independent tangent vectors. Their cross product gives a nonzero normal vector:Dividing by its Euclidean norm gives a unit normal; changing the sign chooses the opposite orientation. If the cross product vanishes, this parametrization is singular there and cannot determine a normal vector by this formula.

No, not from the stated assumptions alone. The centered slope is a weighted sum,and the weighted errors need not be identically distributed. The ordinary central limit theorem therefore does not apply directly.

For a concrete counterexample, for every $n\ge 3$ chooseThese satisfy all three centring and orthogonality assumptions, and both explanatory columns are nonzero. Let the errors be independent with Rademacher distribution, taking values $\pm 1$ with equal probability. Thenwhich takes values $-1,0,1$ with probabilities $1/4,1/2,1/4$ for every $n$. Increasing the sample size adds no information about this slope, and the standardized estimator does not approach a normal distribution.

A useful additional assumption for asymptotic normality of a weighted regression slope is that no single design value dominates:Under this condition and the given finite error variance, the Lindeberg-Feller central limit theorem yieldsjustifying the approximation $N({\beta }_1,{\sigma }^2/S_{xx})$ for the slope as well.

The algebraic identity ${\hat{\beta }}_0-{\beta }_0=n^{-1}{\sum }_i{ϵ}_i$ does not require a normal distribution for the errors. Applying the central limit theorem to their sample mean givesThus, for large $n$,Its mean ${\beta }_0$ and variance ${\sigma }^2/n$ remain exact, even though its normal distribution is now only an approximation. With the error probability distribution unspecified, the formula is retained as an estimator; it is not being asserted to remain a maximum likelihood estimation formula.

The central limit theorem states that if $W_1,W_2,\dots$ are independent and identically distributed with finite mean $\mu$ and variance $0<{\tau }^2<\infty$, thenHere $\stackrel{d}{\to }$ denotes convergence in distribution.

Substitute the linear regression model into the estimators and use the centring and orthogonality conditions:A linear combination of independent normal random variables is normal. Their means are ${\beta }_0$ and ${\beta }_1$, while variance additivity for independent random variables givesConsequently

Treat the explanatory variables as fixed and put $S_{xx}={\sum }_i{x}_i^2$, $S_{zz}={\sum }_i{z}_i^2$. To identify both slopes, assume $S_{xx}>0$ and $S_{zz}>0$. The given centring and orthogonality conditions make the three columns $1,x,z$ pairwise orthogonal.

The log-likelihood function in the normal linear model, up to a constant independent of the coefficients, isThus maximum likelihood estimation minimizes the residual sum of squares. Setting its partial derivatives to zero givesUsing $\sum x_i=\sum z_i=\sum x_i{z}_i=0$ decouples these equations:The Hessian matrix of the residual sum of squares is $2\mathrm{diag}(n,S_{xx},S_{zz})$, which is positive definite. These are therefore the unique minimizers. If either squared sum is zero, the associated explanatory variable vanishes identically and its coefficient cannot be identified.

The pivotal quantity $U=M/\theta$ lies in $[0,1]$ and satisfies $\mathbb{P}(U\le t)=t^n$. ThusRearranging these inequalities yields the exact confidence intervalFor every $\theta >0$, its coverage probability is exactly $1-\alpha$; the result uses the finite-sample probability distribution, with no normal approximation.

Each observation from the uniform distribution has mean $\theta /2$. By linearity of expectation, $\mathbb{E}{\sum }_i{X}_i=n\theta /2$, so another unbiased estimator is

Differentiating the cumulative distribution function in part (ii) gives the probability density function $n{t}^{n-1}$ on $(0,1)$. HenceThus an unbiased estimator based on the sample maximum is

For $0\le t\le 1$, independence and the uniform distribution giveTherefore the cumulative distribution function isIt contains no $\theta$, so $M/\theta$ is a pivotal quantity.

Let $M={\max }_i{X}_i$. For an observed sample in the nonnegative orthant, the likelihood function under the uniform distribution isIt is zero for $\theta <M$, and decreases with $\theta$ on $[M,\infty )$. Thus the maximum likelihood estimation problem is solved at the smallest admissible value:Under the model $M>0$ with probability one. The all-zero sample is a null event on which the likelihood function has no finite maximizer in the parameter space $\theta >0$.

By Chebyshev's inequality and part (i),This actually proves the stronger statement $\boxed{S_n/n\stackrel{\mathbb{P}}{\to }1}$, namely convergence in probability, despite the statistical dependence between adjacent products.