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Use the usual assumption that the two die scores are independent random variables. A fair score has expected value $7/2$ and variance $35/12$. Linearity of expectation and the variance of a sum therefore giveCount the ordered pairs of scores to obtain the probability mass functionsOn these attainable probability supports, the maxima occur at $x=7$ and $y=0$, each with probability $1/6$; the minima occur at $x=2,12$ and $y=-5,5$, each with probability $1/36$. Outside the respective probability support, the mass is zero.

They are not independent: $Y=0$ implies equal die scores and hence $X$ is an even number, so $P(X=7,Y=0)=0$, whereas $P(X=7)P(Y=0)=1/36>0$. In fact $\mathrm{Cov}(X,Y)=\mathrm{Var}(S_1)-\mathrm{Var}(S_2)=0$: this is an example of uncorrelated random variables that are dependent.

Apply differentiation under the integral sign on the moving drop:The moving-boundary term vanishes because $h(R,t)=0$. At the origin, symmetry and regularity give $h_r(0,t)=0$. At the front, the stated $h\propto (R-r{)}^{1/3}$ gives $h^3{h}_r=O((R-r{)}^{1/3})\to 0$. Thus $\boxed{dM/dt=0}$, expressing volume conservation for this axisymmetric viscous gravity current.

With $\eta =r/t^{\alpha }$ and $R={\eta }_0{t}^{\alpha }$, substitute the self-similar solution into the conserved integral:For a nonzero finite drop volume, the power of $t$ must vanish, so $\boxed{\alpha =1/8}$. Balancing powers in the evolution equation gives the same result: its left side scales as $t^{-5/4}$ and its right side as $t^{-1-2\alpha }$. The front-edge exponent alone would not determine this spreading exponent without the conserved volume.

The chain rule gives ${\partial }_x={\partial }_u+{\partial }_v$ and ${\partial }_t={\partial }_u-{\partial }_v$. Hence ${\partial }_x^2-{\partial }_t^2=4{\partial }_u{\partial }_v$, and the forced wave equation becomes $y_{uv}=1$. Integrating twice gives $y=uv+F(u)+G(v)$. At $t=0$ the initial conditions areThus $F-G$ is constant, which cancels from the final expression. Combining the remaining sum yieldsDifferentiation checks the forcing and both initial conditions.

If $\rho =0$, the bivariate normal density in part (a) factorizes asIntegrating over any product of measurable sets proves independence of random variables. Conversely, independent variables with finite second moments have $\mathbb{E}[X_1{X}_2]=\mathbb{E}[X_1]\mathbb{E}[X_2]$, so their covariance is zero and $\rho =\mathrm{Cov}(X_1,X_2)/({\sigma }_1{\sigma }_2)=0$. This proves independence of uncorrelated jointly normal variables; zero correlation alone would not imply independence without the joint normal assumption.

With ${\sigma }_1,{\sigma }_2>0$ and $|\rho |<1$, the bivariate normal distribution has probability density functionThe strict correlation bound makes its covariance matrix positive definite, so this is an ordinary two-dimensional density rather than a distribution supported on a line.

The Galton-Watson extinction fixed point equation isIts smallest root in $[0,1]$ is $q=(3\sqrt{3}-5)/4$, not the other allowed fixed point $1$. Consequently the large-depth escape probability isFor finite $N$, the exact escape probability is $1-q_N$ and is slightly larger. Indeed ${\varphi }^{\prime }(q)=2-\sqrt{3}<1$, and the mean value theorem on $[q_N,q]$ gives $0\le q-q_N\le q(2-\sqrt{3}{)}^N$. This also quantifies how quickly the finite-depth answer approaches the displayed limit.

View the accessible rooms as the open cluster of bond percolation on a rooted ternary tree. Each accessible room has three forward corridors, independently open with probability $2/3$, so its number of accessible children has the binomial distribution $\mathrm{Bin}(3,2/3)$. The counts at successive levels form a binomial branching process with offspring probability generating functionLet $q_N$ be the probability that the root has no open path to level $N$. Then $q_0=0$, and independence of the three descendant subtrees givesThe sequence increases to the smallest fixed point $q=\varphi (q)$ in $[0,1]$, the branching extinction probability. Thus $q_N$ is close to that solution for large $N$.

Connectivity is the relevant escape event: if no window is reachable, escape is impossible; if one is reachable, ordinary random wandering reaches a window eventually with probability one. Before reaching level $N$ the guest moves within a finite connected graph, so an accessible exit cannot be avoided forever. This interpretation does not require the guest to move monotonically outward.

For fixed real $x$, define $u_0=x$ and $u_n=\sin (0.99u_{n-1})$. This makes $u_n$ exactly the summand containing $n$ nested sines. The elementary inequality $|\sin t|\le |t|$ gives inductivelyThe fact that an iterated contraction gives a summable orbit applies here. Comparison with a geometric series therefore givesThus the nested-sine series converges absolutely for every real $x$; there are no divergent or merely conditionally convergent cases.

Let $r=|\sin x|\in [0,1]$. For $r<1$, the absolute summand $r^n/\sqrt{n}$ is at most $r^n$, so a geometric series comparison proves absolute convergence. For $r=1$, the series is $\sum (-1{)}^n/\sqrt{n}$: its positive term sizes decrease to zero, so the alternating series test proves convergence. Its absolute series is a divergent p-series with exponent $1/2$. HenceIt converges for every real $x$ and never diverges.

Put $s=\sin x$, so $-1\le s\le 1$. The printed summand is $(3s^n+s^{2n})/n$. If $|s|<1$, its absolute value is at most $3|s{|}^n+|s{|}^{2n}$, and the two geometric series converge. Thus there is absolute convergence.

For $s=1$ the series becomes $\sum 4/n$, a divergent multiple of the harmonic series. For $s=-1$ it becomesThe first series converges by the alternating series test, while the second diverges to $+\infty$, so their partial sums tend to $+\infty$. In particular, alternation of some terms does not imply convergence here. The complete classification isThere are no conditionally convergent cases.

A member of both groups satisfies $A^{T}A=I$ and $A^{T}JA=J$. Multiplication of the second identity by $A$ on the left, using $A{A}^T=I$, gives $JA=AJ$. A real two-by-two matrix commuting with $J=\mathrm{diag}(1,-1)$ must be diagonal, since the two off-diagonal entries change sign in this equation. Its diagonal entries are $\pm 1$ by the orthogonal matrix identity, and the restriction $a>0$ forces the first to be $1$. Therefore the intersection is the two-element group

For the Minkowski metric $J=\mathrm{diag}(1,-1)$, preservation means $(Ax{)}^{T}J(Ay)=x^{T}Jy$ for all real vectors $x,y$. This is equivalent to $A^{T}JA=J$, or, entrywise,The restriction $a>0$ puts the first column on the right branch of the unit hyperbola, so uniquely $a=\cosh t$, $c=\sinh t$ for a real parameter $t$. The third equation makes the second column a scalar multiple of $(\sinh t,\cosh t{)}^T$: this also covers $t=0$, when $b=0$. The second equation makes that multiple $\epsilon =\pm 1$. Thus all the required matrices areThe hyperbolic function addition formulas give $H(t)H(s)=H(t+s)$, while $D_{\epsilon }H(s)=H(\epsilon s)D_{\epsilon }$. ConsequentlyThe product remains of the required form and its upper-left entry is positive. The identity is $H(0)D_1$, and the inverse is $H(-\epsilon t)D_{\epsilon }$. Together with associativity of matrix multiplication, these identities establish the group property. This is the time-orientation-preserving Lorentz group in one spatial dimension.

Preserving the Euclidean scalar product means $(Ax{)}^T(Ay)=x^{T}y$ for every pair of real vectors $x,y$. Since these bilinear expressions agree on all pairs of basis vectors, this is equivalent to $A^{T}A=I$: the columns of $A$ form an orthonormal basis.

The first column is a unit vector, so it is $(\cos \theta ,\sin \theta {)}^T$ for some real $\theta$. A perpendicular unit vector is one of the two choices $\epsilon (-\sin \theta ,\cos \theta {)}^T$, where $\epsilon \in \{1,-1\}$. Hence all the required orthogonal matrices areHere $R(\theta )$ is a planar rotation, and $\theta$ is determined modulo $2\pi$. To check the group laws explicitly, $D_{\epsilon }R(\varphi )=R(\epsilon \varphi )D_{\epsilon }$, soThis proves closure. The identity is $R(0)D_1=I$, the inverse of $R(\theta )D_{\epsilon }$ is $R(-\epsilon \theta )D_{\epsilon }$, and associativity is inherited from matrix multiplication. Thus the set is the orthogonal group $O(2)$.

There is no such universal constant. For a real parameter $M$, consider the real matrixIts characteristic polynomial is $(t-1{)}^2$, so both eigenvalues are real and ${\lambda }_{\max }(B_M)=1$. ButFor any proposed finite $C$, choosing $M>2(C-1)$ violates the proposed bound. This is an explicit illustration that nonnormal eigenvalues do not bound a quadratic quotient.

Expand $x={\sum }_j{c}_j{v}_j$ in the orthonormal basis from part (a). ThenFor $x\ne 0$, the weights $|c_j{|}^2/{\sum }_k|c_k{|}^2$ are nonnegative and sum to one. The Rayleigh quotient therefore lies between the smallest and largest eigenvalues. Conversely, normalized extreme eigenvectors $v_{\min },v_{\max }$ give, for $0\le t\le 1$,Thus the exact Hermitian Rayleigh quotient range isWhen $n=1$, the range is the singleton containing the sole eigenvalue.

Use the conventionfor the characteristic polynomial. Choosing $\det (A-tI)$ instead multiplies it by $(-1{)}^n$ and leaves the requested zero identity unchanged. For each eigenvector $v_j$, induction gives $A^r{v}_j={\lambda }_j^r{v}_j$, including $r=0$. Thusbecause ${\lambda }_{j}I-A$ has the nonzero kernel vector $v_j$ and therefore has zero determinant. Part (a) supplies a basis consisting of these $n$ eigenvectors. A linear map vanishing on every basis vector vanishes everywhere. ConsequentlyThis proves the Cayley-Hamilton theorem identity directly in the distinct-eigenvalue case assumed here.

Take eigenvectors $v_i,v_j$ for distinct eigenvalues ${\lambda }_i,{\lambda }_j$. Since the eigenvalues are real and $A$ is a Hermitian matrix,Hence $({\lambda }_j-{\lambda }_i)v_i^{†}{v}_j=0$, so eigenvectors for distinct eigenvalues are orthogonal vectors. They are therefore linearly independent; after normalization the given $n$ eigenvectors form an orthonormal basis of ${\mathbb{C}}^n$.

Separate off the integer part of the phase. Set $d_n=\sqrt{n^2+n}-n$. Rationalization givesSince squaring removes the sign in $\cos (\pi n+u)=(-1{)}^n\cos u$, the given sequence satisfies $b_n={\cos }^2(\pi d_n)$. The continuity of the cosine now gives

Rationalizing avoids cancellation between the two large terms of the sequence:By continuity of the square root and the reciprocal at the limiting values,

The shifted matrix $N=D-\lambda I$ is still a normal matrix, sinceand $N{N}^{†}$ has the same expansion except that its first term is $D{D}^{†}$, which equals $D^{†}D$. For the given eigenvector, $Ne=0$. Part (b), applied to $N$, therefore gives $N^{†}e=0$. Since $N^{†}=D^{†}-\overline{\lambda }I$,The corresponding eigenvalue of the Hermitian conjugate is the complex conjugate of the original eigenvalue.