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In units with $\hslash =1$, the quantum harmonic oscillator Hamiltonian and number operator areThe nontrivial commutators among the creation and annihilation operators and $N$ areTheir reversed commutators have the opposite signs, while the commutator of any operator with itself is zero.

The operator $N$ is self-adjoint becauseIts eigenvalues are therefore real. Moreover, for every state $|\psi ⟩$,so every eigenvalue is nonnegative.

If $N|\nu ⟩=\nu |\nu ⟩$, the commutator with $A$ givesandRepeated application of $A$ lowers the eigenvalue by one. It must terminate before producing a negative eigenvalue. If $m$ is the last occupied rung, then $A|\nu -m⟩=0$, and the norm identity forces $\nu -m=0$. HenceThis is the integer spectrum of the number operator.

At an equilibrium point of a dynamical system, the second equation givesThe choice $x=0$ is impossible because then $\dot{x}=r>0$. Hence $y=s/x$, and substitution into the first equation givesThus the unique fixed point is

Choosewhich satisfies $0<\alpha <r$. LetWe check the direction of the vector field on its four edges.

On $x=\alpha$,On $y=0$,On $y=s/\alpha$, where $x\ge \alpha$,Finally, throughout the system,On the sloping edge $x+y=r+s/\alpha$, the bound $y\le s/\alpha$ implies $x\ge r$, and hence $\dot{x}+\dot{y}\le 0$. Every boundary edge therefore points inward. Violations of the corresponding four inequalities are driven toward the boundary in the same order, so positive trajectories enter this compact trapping region and then remain there. This is the Brusselator trapping region.

The Jacobian matrix at the unique fixed point isIts determinant and trace areWhen $s-1>r^2$, the trace is positive, so linear stability of a planar equilibrium shows that the fixed point is a repeller. It lies in the interior of $R$.

Choose a trajectory in $R$ other than the equilibrium. Compactness of $R$ gives a nonempty compact omega-limit set, and the repelling equilibrium cannot belong to that set. Since there are no other equilibria, the Poincaré-Bendixson theorem gives a periodic orbit. HenceThis is the Brusselator periodic-orbit criterion.

Dulac theorem states the following. Let $D$ be a simply connected region for the planar system $\dot{z}=f(z)$, and suppose there is a continuously differentiable function $B:D\to \mathbb{R}$ for whichhas one strict sign throughout $D$. Then there is no periodic orbit contained in $D$.

To prove it, suppose a periodic orbit $\Gamma$ existed and let $A$ be its interior. Simple connectedness ensures $A\subset D$. By the planar divergence theorem,The vector field $f$ is tangent to its trajectory $\Gamma$, so its scalar product with the outward normal $n$ vanishes. The right-hand side is therefore zero. The left-hand side cannot be zero because its integrand has one strict sign, a contradiction.

The Poincaré-Bendixson theorem states that if a forward trajectory of a smooth planar system remains in a compact set and its nonempty omega-limit set contains no equilibrium point, then that omega-limit set is a periodic orbit.

Using the harmless normalization $1/(2n)$, the squared-loss empirical risk for linear prediction iswith explicit gradientStarting from any ${\beta }_0\in C$, projected gradient descent with step sizes ${\eta }_k>0$ is

For a fully explicit update, write ${\tilde{\beta }}_{k+1}=(u_k,t_k)$ and ${\rho }_k=\Vert u_k{\Vert }_2$. The projection from part (e) isEvery iterate therefore lies in the prescribed hypothesis class, and any convergent run under the standard convex-optimization step-size conditions targets its empirical-risk minimizer.

The set $C$ is the second-order cone. PutWhen $\rho \ge |t|$ and $\rho >0$, one has $0\le \alpha \le 1$, and the proposed pointlies on the boundary of $C$ because $\Vert \alpha u{\Vert }_2=\alpha \rho$.

Set $\delta =(\rho -t)/2\ge 0$. Sincefor any $(v,s)\in C$ we obtainwhere the first inequality is the Cauchy-Schwarz inequality and the second uses $\Vert v{\Vert }_2\le s$. Part (d) therefore giveswhenever $\Vert u{\Vert }_2\ge |t|$ and $u\ne 0$. At $(u,t)=(0,0)$ the projection is plainly $(0,0)$, consistent with the limiting formula.

If $\rho \le -t$, then for every $(v,s)\in C$,Applying part (d) with $\pi =0$ givesTogether with the trivial case $\rho \le t$, this is the full projection onto the second-order cone.

Assume $\pi \in C$ andfor every $z\in C$. Expanding the squared Euclidean distance givesHence $\pi$ minimizes the distance from $x$ over $C$, so $\pi ={\pi }_C(x)$. This proves the sufficient direction of the variational characterization of convex projection.

For $x\in {\mathbb{R}}^d$, the euclidean projection onto a convex set $C$ isThe squared distance $z\mapsto \Vert x-z{\Vert }_2^2$ is a strictly convex function. Its restriction to the convex set $C$ can therefore have at most one minimizer by part (b). Closedness, together with finite dimensionality, supplies existence, although existence was not required to be proved.

A function $f:C\to \mathbb{R}$ is strictly convex when, for distinct $x,y\in C$ and $0<t<1$,

Suppose distinct points $x$ and $y$ were both minimizers, with minimum value $m$. Since $C$ is a convex set, their midpoint belongs to $C$, while strict convexity givesa contradiction. Thus any minimizer is unique, which is the uniqueness of a minimizer of a strictly convex function.

A set $C\subseteq {\mathbb{R}}^d$ is convex when every line segment between two of its points remains in the set: for every $x,y\in C$ and $0\le t\le 1$,

Condition on the information available before time $n$. The current wealth is then fixed, while ${\xi }_n\sim N(b,\Sigma )$ is independent of that information. PutFor any portfolio $\theta$, use the inner product induced by the positive-definite matrix $\Sigma$ to writeThe linear image of a multivariate normal vector shows that the two portfolio returns are jointly normal, andThus independence of uncorrelated jointly normal variables makes the residual return ${\eta }^T{\xi }_n$ independent of the aligned return; it also has expected value zero.

The continuation value $V(n,\cdot )$ is concave by part (c). Conditional Jensen inequality therefore shows that adding the independent centered residual cannot improve the objective:where $m=(1+r)X_{n-1}$. Since the optimal portfolio is unique, its residual must be zero.

It remains to determine the sign. If $\lambda <0$, the portfolios $\lambda {\Sigma }^{-1}b$ and $-\lambda {\Sigma }^{-1}b$ have the same return variance, namely ${\lambda }^{2}B$, while their means are $\lambda B$ and $-\lambda B$. The latter return has the distribution of the former plus the positive constant $-2\lambda B$. Since $V(n,\cdot )$ is increasing, replacing $\lambda$ by $-\lambda$ cannot reduce the objective, contradicting uniqueness. Hence $\lambda \ge 0$.

Applying this conditional argument at every time gives nonnegative, past-measurable random variables ${\lambda }_n$ such thatThis is the Gaussian one-fund theorem.

Let the value function at time $n$ be the greatest conditional expected terminal utility attainable from wealth $x$. Independence of the innovations gives the Bellman equationfor $0\le n<N$. A policy obtained from the maximizing choices is admissible because each time-$n$ holding is a predictable process, hence is known before ${\xi }_n$ is observed.

We prove the claimed properties by backward induction. They hold at $n=N$ because $V(N,\cdot )=U$ is increasing and concave. Suppose they hold at time $n+1$. For fixed $\theta$, the mapis increasing because $1+r>0$, and taking a supremum preserves this inequality.

For concavity, let ${\theta }_x$ and ${\theta }_y$ be optimal at wealths $x$ and $y$. For $0\le t\le 1$, use the admissible portfolioThe next wealth is the same convex combination of the next wealths generated from $(x,{\theta }_x)$ and $(y,{\theta }_y)$. ThereforeThis completes the induction and proves monotonicity and concavity of a portfolio value function.

For $0\le t\le 1$, concavity of $U$ impliesso $G$ is concave.

Since $\mathbb{E}Z=0$, Jensen inequality givesIf $0\le s<t$, concavity and $s=(1-s/t)0+(s/t)t$ yieldHence $G$ is decreasing on $[0,\infty )$; this is scaled centered risk under concave utility.

If $y_1\le y_2$, then the utility function is increasing, so for every $X\in \mathcal{X}$,Taking suprema gives $F(y_1)\le F(y_2)$.

For $0\le t\le 1$, let $X_1$ and $X_2$ attain the suprema at $y_1$ and $y_2$. Because $\mathcal{X}$ is a vector space,The concavity of $U$ now givesThus $F$ is increasing and concave, as stated by optimized affine shift of concave utility.

Since the positive series converges, $a_n\to 0$. We construct the subsequence recursively. Choose $n(1)$ arbitrarily. Once $n(1)<\cdots <n(k-1)$ have been chosen, write the reduced partial sums asChoose $n(k)>n(k-1)$ so far out thatThis is possible because $a_n\to 0$.

LetFor every fixed $j$, each later choice includes $q_j^{-j}$ in its minimum, so

First, $\alpha$ is irrational. If $\alpha =u/v$ were reduced and rational, the strictly increasing rational sequence $s_j\to \alpha$ would have unbounded denominators $q_j$; only finitely many reduced fractions in a bounded interval have bounded denominator. Choose $j>2$ with $q_j>v$. Part (b) would givecontrary to $|\alpha -s_j|<2^{-j}{q}_j^{-j}<q_j^{-2}$.

If $\alpha$ were algebraic of degree $d$, part (a) would give a constant $c>0$ withFor $j>d$ our construction instead giveswhich contradicts the lower bound once $2^{-j}<c$. Hence $\alpha$ is transcendental. This proves the transcendental subseries of a positive rational convergent series construction.

Because both fractions are reduced and $q>v$, they are distinct. Their difference isThe numerator is a nonzero integer, so its absolute value is at least one. Thereforewhere the strict inequality uses $v<q$. This is the separation of distinct reduced fractions.

The Liouville approximation theorem states that if $\alpha$ is an irrational algebraic number of degree $d\ge 2$, then there is a constant $c(\alpha )>0$ such that every reduced rational $p/q$ satisfies

Let $P\in \mathbb{Z}[X]$ be the minimal polynomial of $\alpha$, of degree $d$. Since $P$ is irreducible of degree greater than one, $P(p/q)\ne 0$. Moreover,and henceOn the compact interval $I=[\alpha -1,\alpha +1]$, putIf $p/q\in I$, the mean value theorem gives a point $\xi$ between $\alpha$ and $p/q$ such thatThus $|\alpha -p/q|\ge M^{-1}{q}^{-d}$. If $p/q\notin I$, then $|\alpha -p/q|>1\ge q^{-d}$. Taking anyproves the stated strict inequality.

LetThen the null hypothesis is $R{\theta }_0=r$. Define the Wald statistic for linear restrictions

Under $H_0$, part (a) and the continuous mapping theorem givewhereThe matrix $V_0$ is positive definite because $I({\theta }_0)$ is positive definite and $R$ has full row rank. Consistency givesTherefore Slutsky theorem yieldsand applying the continuous squared-norm map proves rigorously that

Let $q_{k,1-\alpha }$ be the $(1-\alpha )$ quantile of ${\chi }_k^2$. Rejecting $H_0$ whenhas rejection probability tending to $\alpha$ under every fixed parameter satisfying $H_0$, and is therefore an asymptotically valid level-$\alpha$ test.

The multivariate Wald statistic for a candidate parameter $\theta \in {\mathbb{R}}^p$ isReplacing $I({\hat{\theta }}_n)$ by any consistent estimator of $I({\theta }_0)$ gives the same limit.

Under $P_{{\theta }_0}$, part (a) and consistency of the information matrix imply, by Slutsky theorem, thatConsequentlyusing the characterization of the chi-squared distribution as the squared norm of a standard normal vector.

If $q_{p,1-\alpha }$ is the $(1-\alpha )$ quantile of ${\chi }_p^2$, an asymptotic $(1-\alpha )$ confidence region is