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Put $v=x-a$ and decompose it into axial and perpendicular parts, $v=(v\cdot n)n+v_{\perp }$. For the nondegenerate semi-angle $0<\alpha <\pi /2$, the equation is equivalent toThus each plane perpendicular to $n$ at axial distance $s$ from $a$ meets the surface in a circle of radius $|s|\tan \alpha$. Both signs of $s$ are allowed, and $s=0$ gives the vertex. This proves that it is a double circular cone with the specified vertex and axis; $\alpha$ is the angle each generator makes with either direction of the axis.

For the intersection calculation, define the symmetric matrix $Q={\cos }^2\alpha I-n{n}^T$. The two equations are $(x-a_i{)}^{T}Q(x-a_i)=0$. Their difference givesWith $b=a_1-a_2$ and $m=(a_1+a_2)/2$, symmetry of $Q$ makes the last two terms $2m\cdot Qb$. Hence the intersection plane of parallel congruent double cones isThe unnormalized normal vector is $v=Qb=b{\cos }^2\alpha -n(n\cdot b)$. Expanding its squared norm givesEquivalently, if $b=b_{\perp }+(n\cdot b)n$, this is ${\cos }^4\alpha |b_{\perp }{|}^2+{\sin }^4\alpha (n\cdot b{)}^2>0$. Thus normalization is legitimate and yieldsThe assumption of a genuine cone matters: at the degenerate limits $\alpha =0$ or $\pi /2$, the surface collapses respectively to an axis or a plane and this normal need not exist.

Move the origin to the centroid and write the four vertex vectors as $r_1,r_2,r_3,r_4$. Their sum is zero. The midpoints of the first opposite-edge pair have vectors $(r_1+r_2)/2$ and $(r_3+r_4)/2=-(r_1+r_2)/2$, so their distances from the centroid are equal. The same argument applies to the other two opposite-edge pairs.

Choose the three midpoint distances as $u=|r_1+r_2|/2$, $v=|r_1+r_3|/2$, $w=|r_1+r_4|/2$. Expanding the dot products,Identifying these four norms with the vertex distances proves the tetrahedron centroid-to-midpoint sum of squares:The proof uses no orthogonality or regularity assumption on the tetrahedron.

Take $E$ to be continuously differentiable everywhere on ${\mathbb{R}}^3$, as in the question. Its zero curl says ${\partial }_i{E}_j={\partial }_j{E}_i$. Construct a potential of a conservative vector field explicitly byAll segments lie in ${\mathbb{R}}^3$. Differentiating under the integral and using the symmetry of the derivatives,Hence $E=-\nabla \varphi$. This radial integral potential for a curl-free field also works on a star-shaped domain after translating its centre. On an arbitrary domain, zero curl only guarantees local potentials; the everywhere-on-${\mathbb{R}}^3$ interpretation is important for this global assertion.

For the given field, put $q=e^{-x^{2}z}$. The three pairs of derivatives entering its curl areThus the field is irrotational. To find its potential using $E_y=-{\partial }_y\varphi$, integrate ${\partial }_y\varphi =2yq$ to get $\varphi =y^{2}q+\psi (x,z)$. Comparison with $E_x=-{\partial }_x\varphi$ gives ${\psi }_x=0$, and comparison with $E_z=-{\partial }_z\varphi$ gives ${\psi }_z=0$. ThereforeDifferentiating this expression reproduces all three components with the required minus sign; the only freedom on the connected domain is the additive constant.

Use suffix notation, summing repeated indices. The contraction of two Levi-Civita symbols givesAdding the directional derivative $(F\cdot \nabla )G_i=F_j{\partial }_j{G}_i$ cancels the second term. Interchanging $F,G$ gives another pair whose sum is $G_j{\partial }_i{F}_j$. The total right-hand side is thereforethe $i$th component of the gradient of the dot product. Since this holds for each component, the dot-product gradient identity is proved:It holds for any continuously differentiable vector fields; no irrotationality assumption is used.

The second component is resonant. Differentiating $t{e}^{{\lambda }_{2}t}$ supplies the term missing from the exponential ansatz:Thus a particular integral isThe linearly growing oscillatory envelope is forced resonance in a differential equation; it does not contradict bounded oscillation of the complementary function. The worked initial-value solution in the question's root Solution uses this same resonant mechanism.

Try a particular integral along each forcing eigenvector. Since $A{\mathbf{a}}_j={\lambda }_j{\mathbf{a}}_j$,With both denominators nonzero, add the two responses:Only the forcing component's own eigenvalue enters its denominator; a frequency matching a different eigenmode does not create resonance in this already diagonal forcing component.

After multiplication by ${\sec }^{2}x$, recognize both sides as derivatives:The nontrivial first integral is thereforeIt necessarily depends on the derivative, correcting the printed notation $h(x,y)$. A second integration giveson the specified interval. Since $y(0)=0$, $C_2=0$. A finite derivative at zero also implies $y{y}^{\prime }\to 0$ there, forcing $C_1=0$. A positive particular solution is consequentlyThe negative of this function is the other sign choice. Near zero, $-2\log (\cos x)=x^2+x^4/6+O(x^6)$, so the positive branch has $y=x+x^3/12+O(x^5)$ and $y^{\prime }(0)=1$; the negative branch has derivative $-1$.

For the positive branch, $y^{\prime }=\tan x/y>0$. Its second derivative has numerator $-2\log \cos x-{\sin }^{2}x$, multiplied by a positive factor. This numerator is zero at zero and has derivative $2{\sin }^{3}x/\cos x>0$, so the curve is convex for positive $x$. It starts tangent to $y=x$ and rises without bound as $x\uparrow \pi /2$, with $y\sim \sqrt{-2\log (\pi /2-x)}$. The third panel of the sketch above shows this branch.

An integrating factor for a differential one-form makes the multiplied form exact. If $dF=\mu gdx+\mu fdy$, then $F_x=\mu g$ and $F_y=\mu f$. Equality of the mixed partial derivatives yieldsThis is a local compatibility condition under the usual smoothness assumptions.

For the trigonometric equation, where its coefficients are defined and division is legitimate, the slope is $y^{\prime }=\tan x\tan y$. In the first quadrant it is positive, small near either axis and large near the upper-right corner. The direction field below uses normalized line segments, so steep directions remain visible.

Multiplication by $\sin x\cos y$ givesThus the multiplying function is an integrating factor, and the implicit general solution isIn the larger square, $\cos x\ge 0$. The positive-$C$ curves have positive $y$ and are symmetric about the $y$-axis; the negative-$C$ curves are their reflections. For $0<|C|<1$, a graph representation iswith endpoints tending to $y=\mathrm{sgn}(C)\pi /2$. The $C=0$ interior curve is $y=0$. The domain boundaries include singular coefficients or zeros of the integrating factor, so these endpoint limits and smooth continuations through $x=0$ describe the regularized relation; they must not be mistaken for additional solutions of the literal original equation at undefined points.

For independent random variables, their sum has the product of their probability generating functions:Comparing polynomial coefficients identifies a binomial distribution with parameters $n+m,p$:and the probability is zero at all other values. The common success probability is essential to this binomial form; independent binomial counts with different probabilities do not in general add to a binomial count.

For a nonnegative integer-valued random variable $X$, its probability generating function isinitially defined at least for $|z|\le 1$. A general real-valued random variable need not have a probability generating function in this sense. For a binomial distribution, the binomial theorem givesDifferentiating the generating function produces factorial moments:Since $X^2=X(X-1)+X$, the mean and variance areThese formulas include the degenerate endpoint probabilities and $n=0$.

For $|y|>1$, the separated integral isInverting givesThe upper branch has $x+C>0$, where $y>1$ decreases towards $1$ as $x\to \infty$. The lower branch has $x+C<0$, where $y<-1$ decreases to $-\infty$ as $x\uparrow -C$. These are separate maximal intervals: the pole is not part of a solution. For the domain $x\ge 0$, a lower-branch initial value below $-1$ necessarily reaches that pole in finite positive $x$. The direction field sketch above displays both exterior branches and a central hyperbolic tangent branch.

Separate variables and integrate on $|y|<1$:Hence the general solution in this region isEvery such solution increases from values near $-1$ towards $1$, remains strictly between the two equilibria, and is defined for all real $x$.

Independence requires each joint probability to equal the product of the relevant marginals. In particular it implies $d=(c+d)(b+d)$, equivalent to $ad=bc$.

Conversely let $p=c+d$ and $q=b+d$, and assume $ad=bc$. The covariance calculation gives $d=pq$. ThenAll four joint probabilities therefore factor, proving independence. HenceThis is Bernoulli independence from zero covariance, including degenerate marginals. It is special to two-point variables and is not a general equivalence for arbitrary random variables.

The probabilities satisfy $a,b,c,d\ge 0$ and $a+b+c+d=1$. Since the variables are indicators,Their covariance isThus the necessary and sufficient condition for uncorrelated random variables isThis uses zero covariance as the definition. If one variable is constant, a normalized correlation coefficient is undefined, but zero covariance and this criterion remain meaningful.

For the stated angle, $q=\cos (2\pi /3)=-1/2$, so $\Vert u+v\Vert =\sqrt{2+2q}=1$. The minimizing unit vector is therefore $w=-(u+v)$, andIndeed all three pairwise dot products equal $-1/2$, so the vectors lie in one plane, separated by $120$ degrees. This is also the global minimum pairwise dot-product sum of three unit vectors: for any three unit vectors,so $S\ge -3/2$, with equality exactly when $u+v+w=0$.

The triangle inequality for the two unit vectors gives $0<\Vert u+v\Vert \le 2$. Taking reciprocals and using the negative sign in $\lambda =-1/\Vert u+v\Vert$ yieldsIn fact equality would require $u=v$, by the equality condition in the Cauchy-Schwarz inequality. Since the two vectors here are linearly independent, the inequality is strict for every admissible pair.

Put $q=u\cdot v$ and $d=u+v$. The linear independence assumption implies $d\ne 0$. Since $w$ is a unit vector, the Cauchy-Schwarz inequality givesEquality holds exactly when $w$ points opposite to $d$. Its required unit length then fixes the scale:Here $\Vert u+v{\Vert }^2=2+2q$ because both vectors have unit norm. The minimizing vector is unique, and the minimum pairwise dot-product sum of three unit vectors for these fixed $u,v$ is

Place the cube's vertices at $(\pm 1,\pm 1,\pm 1)$. Every symmetry fixes its centre, and permutes the three perpendicular pairs of face normals, so it is a signed permutation matrix. Conversely every such matrix preserves the cube. The symmetry group of a cube acts transitively on its eight vertices: independent coordinate sign changes take $(1,1,1)$ to any vertex.

A symmetry fixing $(1,1,1)$ must permute its three incident edges. This gives an injective map from its stabilizer subgroup into $S_3$, since those three edge directions determine the linear transformation. All six permutations are realized by coordinate permutation matrices, which fix $(1,1,1)$. Thus the vertex stabilizer has order six. The orbit-stabilizer theorem now yieldsThe use of all symmetries includes reflections as well as the twenty-four elements of the rotational symmetry group of a cube.

For a group action of $G$ on a set $X$ and $x\in X$, the orbit-stabilizer theorem statesIf $G$ is finite, this becomes $\boxed{|G|=|Gx||G_x|}$. The underlying bijection sends the left coset $g{G}_x$ to $gx$. Equality of two image points is equivalent to $g_2^{-1}{g}_1\in G_x$, so this map is both well-defined and injective, and it is surjective by the definition of the group orbit.