The determinant $\det (M-tI)$ is a polynomial in $t$ with leading term $(-t{)}^n$, so it is not the zero polynomial and has only finitely many roots. Hence one may choose $\alpha >0$ smaller than every positive root, if any. Then

Apply the root test to the terms $a_n{z}^{n^2}$:This tends to zero when $|z|<1$ and to infinity when $|z|>1$. On $|z|=1$, the series converges absolutely because $\sum |a_n|$ converges: the point $1$ lies strictly inside the original radius two. Thus the lacunary power series has radiusand in fact converges on its entire boundary circle.

The series can be written asThe original power series converges when $|z{|}^k<2$ and diverges when $|z{|}^k>2$. Hence its radius as a power series in $z$ is

Let $L={\mathrm{limsup}}_{n\to \infty }|a_n{|}^{1/n}$. The Cauchy-Hadamard theorem, proved by applying the root test, gives a radius of convergencewith the usual conventions. For $|z|<R$, choose $q<1$ eventually bounding $|a_n{z}^n{|}^{1/n}$, which gives absolute convergence by comparison with a geometric series. For $|z|>R$, infinitely many terms have $n$th root greater than one, so the terms fail to tend to zero and the series diverges.

Here $R=2$ and the assumed limit gives $|a_n{|}^{1/n}\to 1/2$. ThereforeThus

Here the full F-test has $p=0.9834$ and $R^2\approx 0$. There is no evidence against the joint null hypothesis that both slopes vanish, and hence no evidence in this model that either covariate explains the response.

The full F-test has $p<2.2\times 10^{-16}$ and tests the null hypothesis that both slope coefficients vanish. It decisively rejects that hypothesis, while $R^2=0.8126$ says the fitted model explains about $81\%$ of the sample variation. Together with the insignificant individual t-tests, this is characteristic of multicollinearity: the two covariates explain the response jointly, but the output does not identify either one as having a separately estimable effect.

The individual t-tests only test each regression coefficient after adjusting for the other covariate. Here neither covariate is individually significant at conventional levels, but from the abbreviated coefficient table alone one cannot conclude that neither explains the response: strong multicollinearity can make both standard errors large even when the covariates are jointly useful.

Let $\tau$ be the time at which a right-running characteristic curve leaves the piston. The boundary velocity and the characteristic it emits areWiththese relations becomeBecause $d\varphi /d\theta =1-ϵ\cos \theta >0$ and $\varphi (\theta +2\pi )=\varphi (\theta )+2\pi$, $U$ is single-valued and $2\pi$-periodic behind the wavefront. Ahead of the first characteristic, where $\varphi <0$, the fluid remains undisturbed and $U=0$. The implicit relation is Kepler equation; unless $ϵ=0$, its inverse is not a pure sinusoid.

Series reversion givesExpanding the cosine then givesand hence

Forthe adiabatic sound speed isSince $\int (c/\rho )d\rho =-B/\rho =-c$, choose invariants that vanish in the equilibrium state $(u,c)=(0,c_0)$:ConsequentlyIn a right-moving simple wave with $R_{-}=0$, the $R_{+}$ characteristic speed is therefore $u+c=c_0$, independent of amplitude. The transport equation becomes $({\partial }_t+c_0{\partial }_x)R_{+}=0$, so the whole velocity profile translates rigidly at the equilibrium sound speed $c_0$ without nonlinear distortion.

A real transmitted normal wavenumber exists only ifso the critical angle for total internal reflection isFor ${\theta }_I>{\theta }_I^{\mathrm{cr}}$, putFor incident amplitude $E_0\hat{\mathbf{y}}$, the transverse-electric Fresnel reflection coefficient and transmission coefficient areA boundary-matched solution iswith each magnetic component obtained from ${\mathbf{B}}_X={\mathbf{k}}_X\times {\mathbf{E}}_X/\omega$, now using the complex transmitted wavevector $(i\kappa ,0,k_z)$. Since $|r|=1$, the wave undergoes total internal reflection; the transmitted field is an evanescent wave that propagates parallel to the interface and decays exponentially into $x>0$.

The condition ${\mathbf{E}}_I\cdot \hat{\mathbf{x}}=0$, together with transversality ${\mathbf{k}}_I\cdot {\mathbf{E}}_I=0$, selects polarization along $\hat{\mathbf{y}}$ away from the normal-incidence degeneracy. Phase matching at a planar wave interface givesThe two dispersion relations give $|{\mathbf{k}}_T{|}^2=({\epsilon }_{+}/{\epsilon }_{-})k_I^2$. Choosing propagation into $x>0$ therefore yieldsEquivalently, this is Snell law for electromagnetic waves with refractive index proportional to $\sqrt{\epsilon }$ because the two media have equal magnetic permeability.

For each $X\in \{I,R,T\}$, substitution of a complex plane electromagnetic wave in a linear medium into the source-free Maxwell equations in a linear medium requireswhere ${\epsilon }_I={\epsilon }_R={\epsilon }_{-}$ and ${\epsilon }_T={\epsilon }_{+}$. The accompanying magnetic field isThe incident and transmitted waves have $k_{Ix}>0$ and $k_{Tx}>0$, while the reflected wave has $k_{Rx}<0$. Matching the fields at the stationary interface additionally imposes the phase matching at a planar wave interface conditions ${\omega }_I={\omega }_R={\omega }_T$ and equality of all tangential wavevector components.

For a time-independent field, Faraday's law gives $\nabla \times \mathbf{E}=0$. Integrating around a thin rectangular loop crossing $x=0$ and shrinking its height gives continuity of the tangential component:A thin Gaussian pillbox and $\nabla \cdot \mathbf{D}=0$ give continuity of the normal electric displacement field:These are the electrostatic cases of the dielectric-interface boundary conditions.

For $f(w)=w^M=r^M{e}^{iM\phi }$, the angular momentum operator $L_z=-i\hslash {\partial }_{\phi }$ givesThe probability density is proportional towhose maximum occurs atIn a large disk of area $A=\pi R^2$, the states that fit inside satisfy $M\lesssim qB{R}^2/(2\hslash )$. Thus the degeneracy of each Landau level is asymptoticallywhere ${\Phi }_B=BA$ is the magnetic flux. There is one state per magnetic flux quantum $h/q$, up to boundary corrections.

Let $\mathbf{\Pi }=\mathbf{p}-q\mathbf{A}$ be the kinetic momentum. The quantum Hamiltonian isAssume $qB>0$ and zero longitudinal momentum, as in the displayed states. Sincewe have $[a,a^{†}]=1$ andFor $w=x+iy$,which annihilates everywith $f$ holomorphic. Hence all these states haveThe Landau levels are $E_n=\hslash {\omega }_B(n+1/2)$ for $n=0,1,\dots$; the arbitrary holomorphic function exhibits the degeneracy of the lowest level. For $qB<0$, the normalizable lowest-level states are anti-holomorphic instead.

These potentials give the uniform magnetic field $\mathbf{B}=B\hat{\mathbf{z}}$ and zero electric field. With the signed frequency ${\omega }_c=qB/m$, the Lorentz force law becomesThusand explicitlyThe transverse trajectory is a circle about the fixed guiding centre $(x+\dot{y}/{\omega }_c,y-\dot{x}/{\omega }_c)$. The positive cyclotron frequency is

For $0<\mu <1$, all orbits converge to one fixed point, and for $\mu =1$ every orbit is eventually fixed or period two, so no iterate has a horseshoe. For the symmetric tent map with slope magnitude $\mu >1$, the topological entropy is $\log \mu >0$. The interval-map positive-entropy horseshoe theorem says that a continuous interval map has positive topological entropy exactly when some iterate has a horseshoe. HenceFor $\mu \ge \sqrt{2}$ this already follows from part (iii); for $1<\mu <\sqrt{2}$, a higher iterate supplies the horseshoe.

The map $F$ itself has a horseshoe only for $\mu =2$. Indeed, its Lipschitz constant is $\mu$. If two intervals $J_0,J_1$ form a horseshoe and their convex hull has length $L$, thenwhich forces $\mu \ge 2$. At $\mu =2$, take $J=(-1,1)$, $K_0=(-1,0)$ and $K_1=(0,1)$; both $K_i$ map onto $J$.

The Lipschitz constant of $F^2$ is ${\mu }^2$, so the same length argument makes $\mu \ge \sqrt{2}$ necessary for an $F^2$ horseshoe. It is sufficient. PutFor $\mu \ge \sqrt{2}$, one has $a<1/\mu \le b\le 1$. Withthe two relevant monotone branches satisfyThus $K_0,K_1\subset J$ are disjoint and $F^2(K_0)=F^2(K_1)=J$. Therefore

For $0<\mu <1$, one iterate sends the whole interval into $[1-\mu ,1]$, whereThis is a contraction mapping with unique fixed pointEvery orbit converges to $x_{\ast }$, alternating around it because the multiplier is $-\mu$.

For $\mu =1$, the first iterate lies in $[0,1]$ and $F(x)=1-x$ there. Hence $F^2$ is the identity on $[0,1]$: $1/2$ is fixed, while every other point in that interval belongs to the period-two orbit $\{x,1-x\}$.

The graph of $F$ is the symmetric tent with verticesIf $0<\mu <1$, then $F\ge 0$ andso $F^2$ is an upward-pointing V with its minimum at $(0,1-\mu )$ and equal endpoint values $1-\mu +{\mu }^2$.

If $1<\mu <2$, the zeros of $F$ are $\pm 1/\mu$ andThus its five vertices are