The rule of subtraction isIn particular, if $B\subseteq A$ are finite sets, the set difference has cardinality $|A|-|B|$.

For pairwise disjoint finite sets, the rule of sum givesThe password sets of different lengths are disjoint, so the required number is $P_4+P_6+P_8$.

By the diagonal reduction in a two-variable Newton iteration, $y_n=x_n$ for $n\ge 1$. Put $e_n=x_n-1$. Direct factorization gives the exact Newton map in optimization error identityThe assumed convergence $x_n\to 1$ makes the final factor tend to $3/2$, so it is bounded for all sufficiently large $n$. Hence $|e_{n+1}|\le C|e_n{|}^2$. For the Euclidean norm error $E_n=\Vert (x_n,y_n)-(1,1)\Vert =\sqrt{2}|e_n|$, this becomes $E_{n+1}\le (C/\sqrt{2})E_n^2$, proving quadratic convergence. Unless the exact limit is reached in finitely many steps, $E_{n+1}/E_n^2\to 3/(2\sqrt{2})$.

For Newton's method in optimization, let $\Delta =H(f{)}^{-1}\nabla f$. Its second row equation is $-4{\Delta }_x+4{\Delta }_y=4y-4x$, so ${\Delta }_y={\Delta }_x+y-x$. Therefore $y_{n+1}=y_n-{\Delta }_y=x_n-{\Delta }_x=x_{n+1}$. The given first coordinate yieldsThe Newton step is defined where $\det H(f)=16(3x^2-1)\ne 0$; both coordinates after any defined step are equal.

For a generic integrand in 20 dimensions, Monte Carlo integration gives root mean square error $O(N^{-1/2})$, whereas tensor-product quadrature with composite Simpson's rule gives only $O(N^{-4/20})=O(N^{-1/5})$. One can sample uniformly in the Euclidean unit ball, for example by an independent uniform direction and radius $U^{1/20}$ with $U$ uniform on $[0,1]$.

For this particular integrand, the most efficient elementary choice uses radial integration over a ball first. The area of the unit 19-sphere is $2{\pi }^{10}/9!$, as follows from the volume of an n-ball. HenceThe reduced integrand is smooth. Apply composite Simpson's rule to obtain $O(N^{-4})$ deterministic error in big O notation, with one original radial integrand evaluation per node. If the selection is restricted to direct multidimensional methods, Monte Carlo integration with $O(N^{-1/2})$ root mean square error is the appropriate choice.

A convenient choice for this irregular region is Monte Carlo integration. The region is contained in $[-1,1{]}^2$. For independent uniform points $U_j$ in that square, letThe indicator random variable incorporates the curved boundary directly. This estimator is unbiased; the integrand is bounded on the square, so its variance is finite. The variance of a sum of independent terms gives root mean square error $O(N^{-1/2})$ in big O notation. This is a stochastic error rate. Applying a rectangular Simpson's rule to the indicator-weighted integrand would lose the smoothness required for its fourth-order error bound.

Use tensor-product quadrature with composite Simpson's rule in each coordinate. If there are $m$ subintervals in each direction, then $N=(m+1{)}^2$ and the quadrature rule error is $O(m^{-4})$: apply the one-dimensional error bound for composite Simpson's rule successively, using bounded fourth partial derivatives. Hence the error is $O(N^{-2})$ in big O notation. The rectangular domain makes the product grid convenient.

Use composite Simpson's rule on this smooth, bounded one-dimensional interval. Its fourth derivative is bounded on the interval, so the error bound for composite Simpson's rule gives $O(N^{-4})$ in big O notation as the number of function evaluations increases.

Use Simpson's rule: the cubic exactness of Simpson's rule includes every quadratic integrand, so three function evaluations already give the exact answer. Thus the error is zero. For a general smooth integrand the corresponding composite Simpson's rule has error $O(N^{-4})$ in big O notation, but that bound is unnecessarily weak here.

Writing $g=e^{x^{\mathsf{T}}x}$, the gradient and Hessian matrix areConsequently, for any nonzero $v$,In particular the Hessian matrix is a positive semidefinite matrix everywhere, so the Hessian criterion for convexity proves $g$ is convex. The strict inequality along every nonconstant line also proves strict convexity.

The Hessian criterion for convexity, in the usual twice-differentiable multivariate setting, isThus the Hessian matrix must be a positive semidefinite matrix everywhere. The criterion follows by applying the second-derivative criterion for convexity to each line restriction $t\mapsto g(x+tv)$, whose second derivative is $v^{\mathsf{T}}H(g)(x+tv)v$. Continuous second partials are a sufficient regularity hypothesis for this line-restriction argument.

Expand before taking the second derivative:The second-derivative criterion for convexity proves that $f$ is strictly convex, hence convex.

The second-derivative criterion for convexity isThe non-strict inequality is necessary and sufficient; the stronger condition $f^{\prime \prime }>0$ is sufficient for strict convexity.

For an axis halfway from $+z$ toward $+x$, let $|n+⟩=\cos (\pi /8)|\uparrow ⟩+\sin (\pi /8)|\downarrow ⟩$. The Born rule givesThe negative-outcome probabilities are complementary. Both outcomes can occur for both inputs, so one result cannot distinguish them perfectly. Repeated measurements on independently prepared copies distinguish their different statistics with arbitrarily high confidence. Reversing the diagonal direction exchanges the two numbers. This separates single-copy certainty from statistical distinguishability.

A horizontal Stern-Gerlach spin measurement measures $S_x$. The inputs are its orthogonal eigenstates, soThe output beam identifies the input with certainty in a single ideal measurement.

A vertical Stern-Gerlach spin measurement is a $z$-spin measurement. The Born rule givesThus this measurement cannot distinguish the two preparations, even by repeating the same measurement on many identically prepared copies: it discards their relative-phase difference in this basis.

Exactly $|a|>1$ works. Composition with the hyperbolic tangent is smooth for every $a$. If $|a|>1$, set $R=\mathrm{arctanh}(1/|a|)$. For $|x|\ge R$, $|a\tanh x|\ge 1$, so the result vanishes; it is therefore a test function with support of a function $[-R,R]$.

If $|a|\le 1$, every finite $x$ satisfies $|a\tanh x|<1$. The given positivity makes $\varphi (|a|\tanh x)>0$ for every $x$, so its support is all of $\mathbb{R}$. In particular $|a|=1$ is excluded even though the function tends to zero at infinity: decay is not compact support.

Exactly $a\ne 0$ works. For nonzero $a$, composition with $x\mapsto ax$ preserves smoothness, and the support of a function is $[-1/|a|,1/|a|]$, which is compact. For $a=0$ the result is the strictly positive constant $\varphi (0)$ on all of $\mathbb{R}$, so it lacks compact support and is not a test function.

Every $a\in \mathbb{R}$ works. Translation preserves a smooth function, and the support of a function becomes the compact interval $[a-1,a+1]$. Hence $\varphi (\cdot -a)\in C_c^{\infty }(\mathbb{R})$, the space of test functions.

Assume condition (2). Fix $a\in X$ and $\epsilon >0$. The superlevel setis a closed set and does not contain $a$. Its complement is therefore an open neighbourhood of $a$, so some open ball $B(a,r)$ avoids $F$. Consequently $f(x)<f(a)+\epsilon$ throughout that ball, givingThe same bound holds for every smaller radius. Taking $r\downarrow 0$ and then $\epsilon \downarrow 0$ yieldsTogether with the previous implication this proves the equivalence. A supremum may equal $f(a)+\epsilon$ even when every individual value is strictly smaller; the non-strict estimate here handles that possibility.