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If $\sigma$ has ${\alpha }_r$ cycles of length $r$, contraction of matrix entries around its cycles gives $\mathrm{tr}(\sigma {\xi }^{\otimes n})={\prod }_r\mathrm{tr}({\xi }^r{)}^{{\alpha }_r}$. The formula holds for every endomorphism, not only diagonalizable ones. The Schur–Weyl duality decomposition equates it with a sum of products of Specht and Schur characters.

A column of length greater than $m$ antisymmetrizes more than $m$ vectors, giving zero. When there are at most $m$ rows, place the $i$th basis vector in every tensor position of row $i$. Row symmetrization multiplies by a nonzero factorial product, and column antisymmetrization is nonzero because the vectors in every column are distinct. This proves the precise nonvanishing criterion for a Schur module.

For a bounded operator on a real Hilbert space, $⟨Lv,v⟩=⟨Sv,v⟩$ with $S=(L+L^{\ast })/2$. Consequently the Euler-Lagrange equation of $⟨Lv,v⟩-2⟨f,v⟩$ is $Sv=f$. It is $Lv=f$ only when $L$ is self-adjoint or when the relevant solution also annihilates the skew part. For example $L=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ has positive quadratic form $\Vert v{\Vert }^2$, but that form contains no information about its skew part.

At the critical field,The quasiparticle gap closes and the low-energy branch is linear, unlike a massive excitation. This is the Ising-chain quantum critical point of the Transverse-field Ising model; the cusp in the nonnegative dispersion represents the two opposite propagation directions.

Expanding $r_k$ at large positive $g$ gives $r_k=g-\cos k+{\sin }^{2}k/(2g)+O(g^{-2})$, soThe branch is strongly gapped and relatively flat, with width approaching $4J$ against a mean energy of order $2Jg$. These are field-polarized spin-flip quasiparticles with a small relative exchange modulation. Averaging the expansion over momentum also gives $E_0/N=-J[g+1/(4g)+O(g^{-3})]$.

At zero field, $r_k=1$ and ${ϵ}_k=2J$ for every $k$. The Ising-chain quasiparticle dispersion is flat: the bulk quadratic excitation costs no momentum-dependent energy. In spin language these are domain-wall excitations. A periodic finite spin configuration has an even number of domain walls, so the single bulk quasiparticle energy $2J$ does not imply an allowed isolated one-wall spin state; the first such closed-chain spin excitation costs $4J$.

For any degree-at-most-$n$ trigonometric polynomial $t_n$, linearity and reproduction giveThe operator norm bound in the preceding part yieldsTake the infimum over all such trigonometric polynomials. By the definition of best uniform approximation,No choice of a minimizer is needed for this argument. It is an instance of the polynomial reproduction error bound: a bounded linear reproducing operator has error at most $1+\Vert P\Vert$ times the optimal error.

The indexing of the Fejér sums givesSubtracting removes precisely the initial $n$ Fourier partial sums. ThusFor $n=0$, omit the second term, so that no undefined ${\sigma }_0$ is needed. Apply the triangle inequality and the uniform-norm contraction of Fejér summation estimate to getHence the operator norm of the de la Vallée Poussin sum is at most $1+2n/m$.

Let $m$ be a positive integer and let the trigonometric polynomial $t_n$ have frequencies only in $[-n,n]$. For every $j\ge n$, its Fourier partial sum is the polynomial itself: $s_j(t_n)=t_n$. Every term in the defining average of the de la Vallée Poussin sum therefore equals $t_n$, givingThis is exact reproduction of the degree-at-most-$n$ trigonometric polynomials, irrespective of the positive averaging length $m$.

For no central accretion, $A=0$ andThis nonaccreting constant-torque disk has a finite inner stress and normally a nonzero inner surface density. An external inner torque supplies the angular momentum transported to an outer sink, although mass does not flow. If both this zero-mass-flux condition and a zero inner torque are imposed, $B=0$; there is no nontrivial positive-viscosity steady disk satisfying both.

For a zero-torque inner boundary condition at finite $r_{\mathrm{in}}$,With positive finite inner kinematic viscosity, $\Sigma$ tends to zero there. At fixed nonzero inward mass flux, the corresponding $v_r=-\dot{M}/(2\pi r\Sigma )$ becomes large, signaling the breakdown of the nearly circular thin disc approximation in the inner transition region.

For marginal utility pricing with proportional transaction costs, if $X-\epsilon \ge 0$ almost surely or $-X-\epsilon \ge 0$ almost surely, the source's second alternative holds. Otherwise $\mathbb{P}(X<\epsilon )>0$ and $\mathbb{P}(X>-\epsilon )>0$. The Fatou lemma argument now gives $G(\theta )\to -\infty$ at both ends, since a large positive holding loses on the first event and a large negative holding loses on the second. Let ${\theta }_{\ast }$ be a finite maximizer.

If ${\theta }_{\ast }>0$, set $Z_0=U^{\prime }({\theta }_{\ast }(X-\epsilon ))$. The derivative condition gives $\mathbb{E}[X{Z}_0]=\epsilon \mathbb{E}{Z}_0$. As before, split according to $|X-\epsilon |\ge 1$ to obtain integrability of $Z_0$ from the absolutely integrable derivative, with a compact-set bound on the complement. If ${\theta }_{\ast }<0$, use $Z_0=U^{\prime }({\theta }_{\ast }(X+\epsilon ))$ and obtain $\mathbb{E}[X{Z}_0]=-\epsilon \mathbb{E}{Z}_0$. In either case normalize $Z=Z_0/\mathbb{E}{Z}_0$; then $Z>0$, $\mathbb{E}Z=1$, and $XZ$ is absolutely integrable.

If ${\theta }_{\ast }=0$, a concave maximum has $G_{-}^{\prime }(0)\ge 0\ge G_{+}^{\prime }(0)$, so $\mathbb{E}X\in [-\epsilon ,\epsilon ]$. Choose $Z=1$. In every case of this first alternative, $\boxed{\mathbb{E}[XZ]\in [-\epsilon ,\epsilon ]}$. The normalization establishes a stronger, economically meaningful statement than the literal unnormalized condition: the expected gain under the resulting density lies inside the transaction-cost spread.

On each open half-line the sample objective is a utility of a linear gain: $U(\theta (X-\epsilon ))$ on the positive half-line, and $U(\theta (X+\epsilon ))$ on the negative half-line. On a compact subinterval wholly inside either half-line, the secant domination argument from the frictionless part ii applies verbatim, using finiteness of $G$ at four surrounding points. The dominated convergence theorem therefore gives the derivativesThey are continuous on their respective half-lines. Near zero, the same outer secants of the globally concave sample objective bound its one-sided slopes. Dominated convergence consequently givesHere $\mathbb{E}|X|<\infty$: the frictionless derivative at zero already proves it, since $U^{\prime }(0)>0$. Thus zero generally has a transaction-cost kink; when $\epsilon =0$ the two formulas agree.

Under a proportional transaction cost, the sample gain $h_x(\theta )=\theta x-\epsilon |\theta |$ is concave when $\epsilon \ge 0$. The utility function is increasing and concave, soTake finite expectations. Thus $\boxed{G\text{ is concave}}$. The monotonicity of $U$ is essential to composing it with the concave transaction-cost payoff.

Among patients still alive at the same time after surgery, chemotherapy has an estimated lung-cancer death rate 3.71 times that of radiotherapy, assuming proportional cause-specific hazards. The estimated 95% confidence interval for this hazard ratio is $(1.583,8.697)$. Thus the fit suggests a higher instantaneous lung-cancer death rate on chemotherapy, with appreciable uncertainty in its size.

The null hypothesis of no treatment effect on that hazard is $H_0:{\gamma }_1=0$ against ${\gamma }_1\ne 0$. Its Wald test uses $Z=1.3113/0.4346\simeq 3.018$, approximately standard normal under the null hypothesis, with two-sided $p=0.00255$. This is evidence against equal lung-cancer death hazards in the fitted model. It is not a statement that five-year lung-cancer death probability is multiplied by 3.71 or that survival duration is divided by 3.71. The cumulative incidence function depends on both cause-specific hazards, and the model needs proportional-hazards and independent observation-censoring assumptions. In a randomised comparison the assigned treatment contrast is protected against baseline confounding, but the instantaneous comparison concerns the groups who have survived to the time in question; it does not identify a treatment effect within a common latent survivor subgroup.

Every regular radial solution has outward normal derivative $u^{\prime }(a)=a/3$, which is nonzero since $a>0$. Therefore there is no solution with the stated homogeneous Neumann boundary condition. This also follows without imposing spherical symmetry: the compatibility condition from part (i) would requirewhich is impossible. Thus the full-ball Poisson equation has no classical solution for these boundary data, radial or otherwise. A singular $1/r$ term could formally cancel the derivative at $r=a$, but would introduce a source at the origin and violate the problem's equation there.

The Dirichlet boundary condition gives $B=-a^2/6$, so the unique radial solution isIts radial Laplacian is one, it is smooth at the origin, and its boundary value is zero. The uniqueness result in part (i) shows that this is also the only solution among all classical functions on the ball, not merely among spherically symmetric ones.

The symmetry between the last two coordinates gives the eigenvector $(0,1,-1{)}^T$ with eigenvalue $2$. On vectors $(s,t,t{)}^T$, the remaining eigenvalue equation iswhose characteristic polynomial is ${\lambda }^2-5\lambda +4=(\lambda -1)(\lambda -4)$. Corresponding eigenvectors are $(1,-1,-1{)}^T$ and $(2,1,1{)}^T$. Thus all eigenvalues and eigenspaces areNonzero scalar multiples give all eigenvectors in each eigenspace. The three displayed vectors are mutually orthogonal and have squared norms $3,2,6$, respectively.

For any fixed real $x\ne 0$, the ratios between consecutive nonzero terms in the absolute-value series areandThe ratio test proves absolute convergence of both power series at every such $x$; convergence at zero is immediate. Both radii of convergence are therefore $\boxed{\infty }$. Applying the test to the nonzero terms avoids the undefined coefficient ratios caused by alternate zero coefficients when these are viewed as full power series in $x$.

For a series $\sum u_n$ with $u_n>0$, suppose $L={\lim }_{n\to \infty }{u}_{n+1}/u_n$ exists, allowing $L=+\infty$. The ratio test says the series converges if $L<1$ and diverges if $L>1$. In the first case an eventual ratio bound by some $q<1$ compares the tail to a geometric series. In the second case an eventual ratio above some $q>1$ prevents the terms tending to zero.

If $L=1$, the test is inconclusive: both $\sum 1/n$ and $\sum 1/n^2$ have limiting ratio one, while the first diverges and the second converges. If the ratio has no limit, this particular limit formulation gives no conclusion.