The factor theorem gives a factor $x-1$. Polynomial division and then factorization giveConsequently all three roots of a polynomial are

Write $P=I-B$, $Q=I+B$, so $P^T=Q$ and $Q^T=P$. The factors commute because both are polynomials in $B$. Also $P$ has a matrix inverse since it is the transpose of the invertible $Q$. Thus the Cayley transform of a real skew-symmetric matrix, with its parameter replaced by $-B$, satisfiesHence $C$ is an orthogonal matrix. Its inverse is $Q{P}^{-1}$; this lets us distinguish the symmetric and skew-symmetric Cayley transforms below.

Write $L(N)=(N+a)\ln N-N+b$. The two data points give the linear equations $a\ln 10+b=15.104413-10\ln 10+10$ and $a\ln 50+b=148.477767-50\ln 50+50$. Subtraction and back-substitution giveor $a=0.4959$, $b=0.9368$ to four significant figures. The six-decimal-place value $148.477767$ here is the one printed in the original PDF. Retain guard digits for the estimates below. The fitted coefficients are close to the Stirling's formula values $1/2$ and $\frac{1}{2}\ln (2\pi )$ but need not equal them because the fit absorbs finite-$N$ corrections.

The product rule and chain rule first giveDifferentiating the first expression with respect to $y$ yields $e^{x^{2}y}(2y+7x^2{y}^2+2x^4{y}^3)$; differentiating the second with respect to $x$ gives exactly the same polynomial factor. ThusThis also follows from symmetry of second derivatives, since all derivatives of this function are continuous everywhere.

The defining function $x^3+xy-1$ is independent of $z$, so a normal vector is $(3x^2+y,x,0)$. It never vanishes here because $x>0$. Every normal is horizontal, whereas $\mathbf{G}=(0,0,x)$ is vertical and tangent to the surface. Their dot product is zero at every point, givingThis is independent of orientation. Although the surface extends without a bound on $x$, the integrand vanishes identically, so no convergence problem arises for this flux.

The double-angle identity eliminates the parameter: $y=1+\cos 2\theta =2{\cos }^2\theta =2x^2$. As $\theta$ increases from zero to $\pi$, $x$ decreases from one to minus one. The positive area below this parabola is thereforeEquivalently, the parameter integral is $-{\int }_0^{\pi }(1+\cos 2\theta )(-\sin \theta )d\theta$. The leading minus sign corrects for traversal from right to left; the signed integral $\int ydx$ along that traversal would be negative.