For hitting zero before four, the success probabilities areConditioning each transition on eventual success givesThus the conditional expected hitting times satisfySolving this linear system gives

Part (i) gives $\mathbb{P}(A)=2/3$. Since $\{X_1=0\}\subseteq A$ and has probability $1/2$,Conditioned on $A$, a walk at one moves to zero with probability $3/4$ and to two with probability $1/4$. A walk at two must next move to one, since a move to three would violate $A$. Iffirst-step analysis givesSolving,

Let $h(x)$ be the probability that the simple symmetric random walk starting at $x\in \{0,\dots ,n\}$ hits zero before $n$. The first-step recurrence isThe recurrence says that $h$ is affine, and the boundary values determine it:

This can be false. The functionis a differentiable bijection of $\mathbb{R}$, buthas an unbounded difference quotient at zero and is not differentiable there. The inverse function theorem requires the additional local condition $f^{\prime }(x)\ne 0$.

True. Part (a) shows that $f$ is strictly monotone. Suppose it is increasing and let $y_0=f(x_0)$. Given $ϵ>0$,Choose $\delta >0$ smaller than both distances from $y_0$ to the two outer values. If $|y-y_0|<\delta$, monotonicity forcesThus $f^{-1}$ is continuous at $y_0$. The decreasing case is identical with inequalities reversed.

This can be false because a strictly increasing $g$ need not be surjective. TakeThen $g$ is strictly increasing and $f\circ g=0$ is continuous, while $f$ is discontinuous at zero.

True. If $x_n\to x$, continuity of $g$ gives $g(x_n)\to g(x)$, and continuity of $f$ then givesEquivalently, this is the composition of continuous functions theorem.

A plane reflection has determinant $-1$, while a rotation matrix has determinant $+1$. The determinant of their product, in either order, is therefore $-1$. Since every rotation has determinant $+1$, such a product can never be another rotation.

Choose the $z$-axis along the line of intersection of the two planes. Their unit normals lie in the $xy$-plane; after choosing the $x$-axis, takewhere $\theta$ is the oriented angle between the planes. The two reflection matrices restrict toon the $xy$-plane and both fix the $z$-axis. Their product isThus the composition of two plane reflections is a rotation about the intersection line through angle $2\theta$.

Writing $n$ as a column vector, the reflection isso its reflection matrix is

Each entry of $\mathrm{adj}(M-tI)$ is, up to sign, the determinant of an $(n-1)\times (n-1)$ minor whose entries are affine polynomials in $t$. The Leibniz formula for determinants therefore makes each adjugate entry a polynomial in $t$ of degree at most $n-1$.

For arbitrary $A,B$, choose positive sequences $s_j,t_j\to 0$ for which $A-s_{j}I$ and $B-t_{j}I$ are nonsingular. The nonsingular identity from part (b) applies to their product:Every entry is a polynomial, hence continuous, in the matrix entries. Letting $j\to \infty$ provesfor all square matrices.

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.