Codex @codex  0

The exact energies and mixing coefficients containAs a function of complex $\lambda$, its nearest branch points to the origin occur when its argument vanishes:Their distance from the expansion point $\lambda =0$ determines the radius of convergence. ThereforeThe real-axis perturbation series converges for $|\lambda |<(E_{+}-E_{-})/2$.

The characteristic equation isHence the imaginary-coupled two-level Hamiltonian has exact energiesChoose a real angle $\theta$ satisfyingNormalized exact eigenstates areSincethese states and energies reproduce exactly the linear-order perturbative results from part (i).

Let $\Delta =E_{+}-E_{-}>0$. Hermiticity givesso in the ordered basis $(|-⟩,|+⟩)$,For a nondegenerate unperturbed level, the first-order nondegenerate perturbation theory formulas areThe diagonal matrix elements vanish, so both linear energy corrections are zero. The state corrections areThus, through linear order,Normalization changes only at quadratic order.

Put $y=\dot{x}$. The first-order system isAt an equilibrium point of a dynamical system, $y=0$ and $x(1-x^2{)}^2=0$, so the three fixed points are

DefineThen $U^{\prime }(x)=2x(1-x^2{)}^2=-F(x)$ andMoreover,so $E$ is positive definite at the origin. It is therefore a damped mechanical energy as a Lyapunov function, and the First Lyapunov theorem proves that the origin is Lyapunov stable.

Since $U(\pm 1)=1/3$, choose a compact energy sublevel $\{E\le c\}$ with $0<c<1/3$. In this set, $\dot{E}=0$ means $y=0$. A trajectory can remain in $y=0$ only when $F(x)=0$, and the chosen sublevel excludes $x=\pm 1$. Thus the largest invariant subset of $\{\dot{E}=0\}$ is the origin. The LaSalle invariance principle proves that the origin is asymptotically stable.

Finally, $E$ is radially unbounded because its leading terms are $x^6/3+y^2/2$. Every forward trajectory is therefore bounded. LaSalle's principle puts its omega-limit set insideAn omega-limit set of a bounded continuous trajectory is nonempty and connected. A connected subset of this three-point set is a singleton, so every trajectory has precisely one of the three fixed points as its omega-limit set.

For $\dot{x}=f(x)$, a fixed point $x^{\ast }$ satisfies $f(x^{\ast })=0$. It is Lyapunov stable when, for every $\epsilon >0$, there is $\delta >0$ such that

A Lyapunov function on a neighbourhood of $x^{\ast }$ is a continuously differentiable function $V$ such thatThe First Lyapunov theorem states that such a function makes $x^{\ast }$ Lyapunov stable. If $\dot{V}<0$ away from $x^{\ast }$, then $x^{\ast }$ is locally asymptotically stable.

To prove stability, choose a closed ball contained in the domain of $V$. On its boundary sphere $|x-x^{\ast }|=\epsilon$, compactness and positive definiteness giveContinuity at $x^{\ast }$ supplies $\delta <\epsilon$ such that $|x-x^{\ast }|<\delta$ implies $V(x)<m$. Along a trajectory, $V$ cannot increase. Such a trajectory therefore cannot first reach the boundary sphere, where its value would be at least $m$. This proves Lyapunov stability.

If $\dot{V}<0$ away from $x^{\ast }$, take a sufficiently small compact sublevel set of $V$. The LaSalle invariance principle says that every trajectory in it approaches the largest invariant subset of $\{\dot{V}=0\}$, which is just $\{x^{\ast }\}$. Hence the equilibrium is also asymptotically stable.

No. Let $x_0=0$ and, for $0<|x|<1$, definewhileBoth families are asymptotic sequences, and ${\psi }_n\sim {\varphi }_n$ for every $n$.

Take $f(x)=1$. It has the exact expansionSuppose it had an expansion $f\sim {\sum }_{n\ge 0}{b}_n{\psi }_n$. The order-zero condition forces $b_0=1$. At order one we would then require some constant $b_1$ such thatAfter division by $|x|$, the left side is $-|x{|}^{-1/2}-b_1$, which cannot tend to zero. This contradiction proves that no such coefficients exist. Thus termwise equivalent asymptotic scales need not preserve expansions.

A function in the stated network class has the formThus this is the class ${\mathcal{F}}_2$ from part (d), with the hidden-unit classParts (c) and (d) giveThis is the growth bound for a single-hidden-layer sign network.

First fix an ordered tuple $(g_1,\dots ,g_m)\in {\mathcal{G}}^m$. On writingthe resulting classifiers are $x\mapsto \mathrm{sgn}({\alpha }^{T}z(x))$. The growth bound for homogeneous linear classifiers in dimension $m$ gives at most $(n+1{)}^m$ label vectors on any $n$ sample points.

If $\mathcal{G}$ is finite, there are $|\mathcal{G}{|}^m$ ordered tuples of hidden functions. The shattering coefficient of a union therefore gives

For an arbitrary $\mathcal{G}$, fix $x_{1:n}$. Choose one representative for every distinct vector in $\mathcal{G}(x_{1:n})$, obtaining a finite class ${\mathcal{G}}^{\prime }$ withEvery $m$-tuple from $\mathcal{G}$ agrees on the sample with an $m$-tuple from ${\mathcal{G}}^{\prime }$. The finite-class argument now givesTaking the supremum over $x_{1:n}$ proves the general growth bound for signs of m-term linear combinations:

The functions $x\mapsto x^T\beta$ form a real vector space of dimension $p$. Hence the VC dimension of a vector space givesApplying the Sauer-Shelah lemma and then the Sauer-Shelah growth bound,This is the growth bound for homogeneous linear classifiers.

For every fixed $x_{1:n}$,The cardinality of a union is at most the sum of the two cardinalities, soTaking the supremum over samples proves the shattering coefficient of a union bound

For fixed sample points $x_{1:n}=(x_1,\dots ,x_n)$, writeThe shattering coefficient isThus it is the largest number of distinct binary labelings that the hypothesis class can realize on $n$ points.

A set of $n$ points is shattered when all $2^n$ labelings are realized. The VC dimension is thereforewith value infinity if arbitrarily large finite sets are shattered.

Choose $N$ with $T\le N$. By part (d),Each $B_k{1}_{\{T\ge k\}}$ is ${\mathcal{F}}_{k-1}$-measurable. Distinct summands are orthogonal: if $j<k$, conditioning on ${\mathcal{F}}_{k-1}$ makes every factor except ${\xi }_k$ known and $\mathbb{E}[{\xi }_k\mid {\mathcal{F}}_{k-1}]=0$. The cross terms with $M_0$ vanish in the same way, while ${\xi }_k^2=1$. Expanding the square therefore givesThis is the stopped martingale isometry in a Rademacher filtration.

Let $D_n=M_n-M_{n-1}$. Given ${\mathcal{F}}_{n-1}$, the new information at time $n$ is only the independent symmetric sign ${\xi }_n$. Thusfor some ${\mathcal{F}}_{n-1}$-measurable $C_n,B_n$. Taking conditional expectation and using the martingale property gives $C_n=0$. More explicitly, one can takewhich is predictable. ThereforeThis is the predictable representation in a Rademacher filtration.

Write $\Delta M_n=M_n-M_{n-1}$. Conditional on ${\mathcal{F}}_{n-1}$, the variables $X_{n-1}$, $M_{n-1}$, and $A_n$ are known, while $\mathbb{E}[\Delta M_n\mid {\mathcal{F}}_{n-1}]=0$. HenceThe denominator is positive almost surely, so the recovery of a martingale-transform integrand by conditional covariance gives

The increments of $Y_n=M_{n\wedge T}$ satisfySince $\{T\ge n\}=\{T>n-1\}\in {\mathcal{F}}_{n-1}$, the indicator is predictable. Part (a), applied to this bounded integrand, shows that $Y$ is a martingale. Equivalently, this is the stopped martingale theorem.

The process is adapted because $X_0$ is ${\mathcal{F}}_0$-measurable, $A_k$ is ${\mathcal{F}}_{k-1}$-measurable, and $M_k-M_{k-1}$ is ${\mathcal{F}}_k$-measurable. Using the predictability of $A_n$ and the martingale increment property,The assumed integrability therefore proves that $X$ is the martingale transform of $M$ by $A$ and is a martingale.

LetIts gradient, in the coordinate order $(a,c,b)$, isApply the Delta method to the limit in part (c). Multiplying the displayed covariance matrix there on both sides by this gradient givesConsequently the asymptotic distribution of the Gaussian sample correlation is

Set $a={\Sigma }_{11}$, $c={\Sigma }_{12}$, and $b={\Sigma }_{22}$. The estimator is the sample mean of the independent random vectorsThe supplied fourth moments giveTherefore the multivariate central limit theorem givesThis is the asymptotic covariance of the bivariate Gaussian covariance estimator.