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Put $A_j=\left(\begin{array}{cc}{a}_j& b_j\\ 1& 0\end{array}\right)$. The identityfollows by induction: multiplying the formula for $n-1$ by $A_n$ gives the two stated recurrences in the first column and $b_n(p_{n-1},q_{n-1}{)}^T$ in the second. The associated Möbius transformations $z\mapsto a_j+b_j/z$, applied from the right to $a_n$, show that the first-column ratio is the displayed positive generalized continued fraction.

Now set every $b_j=1$. Taking determinants in the matrix identity givesConsequently adjacent convergents differ byThe recurrence and $a_j\ge 1$ imply $q_{n+1}\ge q_n+q_{n-1}$, so $q_n\to \infty$. The determinant signs show that the even convergents increase, the odd convergents decrease, and every even one is below every odd one. Their adjacent separation tends to zero, so both subsequences have a common limit $x$. Since $x$ lies between each adjacent pair, the convergence of a positive simple continued fraction gives exactly

If all $a_j=1$ as well, the recurrence and initial values giveThe characteristic roots of $F_{n+2}=F_{n+1}+F_n$ are
$\varphi =(1+\sqrt{5})/2$ and $\psi =(1-\sqrt{5})/2=-{\varphi }^{-1}$, henceTherefore $F_{n+2}/F_{n+1}\to \varphi$. Moreover,Using $F_n{\varphi }^{-n}\to 1/\sqrt{5}$ yieldsandThis is the all-one continued fraction and Fibonacci ratios case.

Let ${\hat{\theta }}_n$ be the MLE and let $I(\theta )$ be the one-observation Fisher information matrix. One standard form of the Wald statistic isAny consistent information estimator gives the same asymptotics. Under $H_0$,
$W_n({\theta }_0)\Rightarrow {\chi }_p^2$. If
$c_{p,\alpha }$ is the $(1-\alpha )$ quantile of ${\chi }_p^2$, reject whenThe asymptotic type-I error is $\alpha$.

By part (a), each $A{\phi }_{\pm }+{\partial }_t{\phi }_{\pm }$ lies in the same two-dimensional eigenspace. Reality and conjugacy again force the coefficient matrix to have the formBecause $u$ and hence $A$ are periodic, ${\partial }_t+A$ commutes with translation by $2\pi$. Applying it to
$\Psi (x+2\pi )=\hat{T}\Psi (x)$ givesThe basis $\Psi$ is independent, so the periodic KdV transfer matrix obeysTaking traces,Hence $\boxed{\mathrm{Re}a}$ is independent of time.

Translation by $2\pi$ preserves the scattering equation, so it acts linearly on its two-dimensional solution space. Because $u$ is real and the initial conditions are conjugate,
${\phi }_{-}=\overline{{\phi }_{+}}$, forcing the transfer matrix to have the formThe Wronskian of two solutions of
$-{\varphi }^{\prime \prime }+u\varphi =k^2\varphi$ is independent of $x$. Translation therefore preserves it, soConsequently

A density operator satisfiesAn $N\times N$ Hermitian matrix has $N^2$ real parameters, and the trace condition removes one, leavingreal free parameters.

If $A_H(t)=U(t{)}^{†}{A}_{S}U(t)$, equality of expectation values requiresTherefore

The kinetic matrix is diagonal:For the linear potential,andThus, in units ${\hslash }^2/(m{a}^2)$,The lower eigenvalue gives the two-mode variational bound for a linearly tilted square well

If the Schmidt rank is one, the decomposition is $|\psi ⟩=|\alpha ⟩|\beta ⟩$ after absorbing its sole coefficient, so the state is a product. Conversely, a product state has coefficient matrix $X=a{b}^T$, which has rank one. Therefore a bipartite pure state is entangled exactly when its Schmidt rank is at least two.

The reduction map sends $H_k$ onto $H_{k-1}$ with kernel of order $p$, so $|H_k|=p^{k-1}$. More explicitly, the lifting-the-exponent calculationshows that $1+p$ has order $p^{k-1}$ modulo $p^k$. Hence $H_k=⟨1+p⟩$ is cyclic of that order.

For the matrix claim, suppose $A\ne I$ and choose $m$ maximal such that $A=I+p^{m}B$, so some entry of $B$ is nonzero modulo $p$. Replacing $A$ by a suitable power, we may assume its order is a prime $q$. If $q\ne p$, binomial expansion modulo $p^{m+1}$ givesIf $q=p$, expansion modulo $p^{m+2}$ givesbecause $p$ is odd. Both contradict $A^q=I$, so $A=I$.

For $p=2$, $r=1$ fails because $-I=I+2(-I)$ has order two. If $r=2$, the same argument works; in the order-two case, for maximal $m\ge 2$,since the parenthesis is nonzero modulo two. Thus the smallest value is $r=2$.

Consider $F(A)=A^{T}A$ with values in the symmetric matrices. At the identity,which is onto the $n(n+1)/2$-dimensional space of symmetric matrices. The same rank holds at every orthogonal matrix. Thus $O(n)=F^{-1}(I)$ is a manifold, and its determinant-one component, the special orthogonal group, has dimensionThe tangent equation at the identity is $X^T+X=0$, so

For $-j\le m<j$, the matrix element of $[O,J_{+}]$ between $|j,m+1⟩$ and $|j,m⟩$ givesThe ladder coefficient is nonzero, so adjacent diagonal entries agree. Induction makes the diagonal entry independent of $m$.

Taking a matrix element of $[O,J_3]=0$ givesThus the matrix element vanishes unless $m=m^{\prime }$.

Since $O$ commutes with every angular momentum operator, it commutes with $J^2$. Taking a matrix element of $[O,J^2]=0$ givesThe allowed nonnegative values of $j$ make $j(j+1)$ injective, so the matrix element vanishes unless $j=j^{\prime }$.

The coupled basis consists of the tripletand the singletThese equations list all nonzero Clebsch--Gordan coefficients, each coefficient being $1$, $1/\sqrt{2}$, or $-1/\sqrt{2}$ as displayed, andBecause each irreducible total-spin representation occurs once, an operator commuting with total angular momentum isIn the product basis, the nonzero matrix elements areand

For real $E$ and real $V$, choose a real fundamental pair of solutions. The corresponding Floquet matrix has real entries, so $\mathrm{tr}F(E)$ is real. The trace is basis independent, so this holds for every fundamental pair.

For two solutions of the Schrodinger equation, their Wronskian is constant because the equation has no first-derivative term. The determinant of the Floquet matrix is the ratio of the Wronskian at $x+a$ to that at $x$, hence

In a linear model containing an intercept, the normal equations make the residual vector $e$ orthogonal to every design column. In particular $e\perp 1$, soup to floating-point rounding.

The second design matrix has $49$ random Gaussian columns plus the intercept, hence is a random $50\times 50$ matrix. It has full rank with probability one because the determinant vanishes only on a measure-zero algebraic set. Its column space is then all of ${\mathbb{R}}^{50}$, so the fitted vector equals $Y$, the residual sum of squares is zero, and $R^2=1$.

The matrix $U_{\theta }$ is real orthogonal. Part (a) givesExpanding the left side in the computational basis yields