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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.

A positive, counterclockwise rotation takes the standard basis vectors $(1,0{)}^T$ and $(0,1{)}^T$ to $(\cos \theta ,\sin \theta {)}^T$ and $(-\sin \theta ,\cos \theta {)}^T$. These images are the columns of its rotation matrix, soIt fixes the origin, preserves the dot product and has determinant one.

The absolute ratio of successive terms is $|z|n^2/(n+1{)}^2\to |z|$. The ratio test gives convergence for $|z|<1$ and divergence for $|z|>1$, so the radius of convergence is $\boxed{R=1}$. On the boundary, $|n^{-2}{z}^n|=n^{-2}$ andThe positive partial sums are bounded, and thus converge. By the comparison test for series, the original power series converges absolutely at every point of $|z|=1$.

If $y=gx{g}^{-1}$, cancellation of successive $g^{-1}g$ factors gives $y^k=g{x}^k{g}^{-1}$ for every positive integer $k$. Hence $y^k$ is the identity element exactly when $x^k$ is, proving that conjugate group elements have the same order of a group element. For permutations, the sign group homomorphism similarly givesThus conjugate permutations have the same sign of a permutation.

The converse is false: permutation order and sign do not determine conjugacy. In the symmetric group $S_6$, take $x=(123)$ and $y=(123)(456)$. Both have order three and positive sign. However, $x$ fixes three symbols whereas $y$ fixes none. If $x$ fixes a symbol $j$, then $gx{g}^{-1}$ fixes $g(j)$, and applying the inverse conjugation gives a bijection between the fixed-symbol sets. Their different fixed-point counts therefore prove that $x,y$ are not conjugate. Equivalently their cycle types $31^3$ and $3^2$ differ. Equal order and sign are necessary, but not sufficient, for conjugacy.

Compose permutations from right to left. Tracking the four symbols through the two permutation cycles gives $1\mapsto 2\mapsto 1$ and $3\mapsto 4\mapsto 3$. Thus the cycle decomposition of a permutation isThe disjoint transpositions commute, and squaring their product gives the identity element, while the product itself is nonidentity. Its order of a group element is therefore two. Each transposition has negative sign of a permutation, so the product has sign $(-1{)}^2=\boxed{+1}$: it is an even permutation.

Compare with the divergent series from part (i). The ratio of its new summand to $1/(n\log n)$ isFor completeness, $\log n={\int }_1^n{x}^{-1}dx\le {\int }_1^n{x}^{-1/2}dx=2(\sqrt{n}-1)$, so $(\log n)/n\to 0$. The limit comparison test now gives divergence. Although the exponent $1+1/n$ exceeds one for each summand, it approaches one and does not produce a convergent p-series with a fixed exponent greater than one.

Because $3>e$, each of $x$, $\log x$ and $\log \log x$ is positive and increasing for $x\ge 3$. Hence $h(x)=1/[x\log x(\log \log x{)}^2]$ is positive, continuous and decreasing. With $u=\log \log x$, $du=dx/(x\log x)$, soThe integral test for convergence implies that the series is convergent.