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For a nonnegative random variable $Y$ and $a>0$, the Markov inequality isIndeed, $Y\ge a{1}_{\{Y\ge a\}}$ pointwise. Taking expectations and dividing by $a$ proves the claim.

Write $u=xv$ on an interval not containing zero. The equation becomes $x^2{v}^{\prime }=x$, so $v^{\prime }=1/x$. Hence

Since $1+\cos 2y=2{\cos }^{2}y$, separation givesThe limiting condition gives $C=1$. Taking ${\tan }^{-1}$ with range $(-\pi /2,\pi /2)$ and selecting the branch approaching $9\pi /4$ yieldsfor $x>0$.

Taylor's theorem about $1$ gives, for some $\xi \in [t,1]$,Since $G_X^{\prime \prime }$ is nondecreasing andthe stated inequality follows. At the fixed point $d<1$,Therefore the quadratic Galton-Watson extinction bound isThe denominator is at least $2(\mu -1)$ because $X$ is integer-valued and $\mathbb{E}[(X-1)(X-2)]\ge 0$, so $d^{\ast }\ge 0$.

Since $G_X(1)=1$ and $G_X^{\prime }(1)=\mu >1$, for $t<1$ sufficiently close to one,At zero, $G_X(0)-0\ge 0$. Continuity therefore gives a fixed point in $[0,1)$, and part (b) implies $d<1$.

Let $d_n$ be the probability of extinction by generation $n$. Conditioning on the initial individual's offspring givesThe events increase to eventual extinction, so $d_n\uparrow d$ and continuity gives $d=G_X(d)$. If $r\ge 0$ is any other fixed point, induction from $d_0=0\le r$ and monotonicity of $G_X$ gives $d_n\le r$ for every $n$. Hence $d\le r$, proving the Galton-Watson extinction fixed point characterization.

The probability generating function isTermwise differentiation givesThe coefficients are nonnegative, and $\mathbb{P}(X\ge 2)>0$, so both derivatives are positive and nondecreasing on $(0,1]$.

Take the spatial Fourier transform of the wave equation. The transformed initial-value problem isand hence

Part (c) and the convolution theorem turn the first term intoFor the second term, choose $p$ with $p^{\prime }=g$. Part (b) says $\tilde{g}=ik\tilde{p}$, soPart (c) now makes its inverse transformCombining the two terms gives the D'Alembert formula

Let $A=2V$ and $B=W-V$. Part (c) makes $A$ and $B$ independent identically distributed exponential variables of rate $\lambda$. The change of variables $(A,B)=(rs,(1-r)s)$ has Jacobian $s$, and integrating ${\lambda }^{2}s{e}^{-\lambda s}$ over $s>0$ gives density one for $0<r<1$. Therefore the uniform ratio of independent exponential variables gives

After the first of two equal-rate clocks rings, the residual lifetime of the other clock is, by memorylessness, a fresh exponential variable of rate $\lambda$, independent of the first ringing time and of which clock rang. Henceand these variables are independent.

The exponential memoryless property gives

The competing exponential clocks calculation givesso $V$ is exponential with rate ${\lambda }_1+{\lambda }_2$. Also

An Erlang$(n,1)$ variable is a sum of $n$ independent rate-one exponential variables, each with mean and variance one. The central limit theorem therefore givesConsequently, for every fixed $q\ge 0$,

Differentiating the moment-generating function at zero gives

This is the Erlang distribution with shape $n$ and rate one. Direct integration gives, for $t<1$,

Let $S_n={\sum }_{i\le n}{a}_i$ and $T_n={\sum }_{i\le n}|a_i|$. Conditional convergence gives $S_n\to S$ and $T_n\to \infty$. Since $P_n=T_n+S_n$ and $N_n=T_n-S_n$,

The alternating-series test says that $b_n\downarrow 0$ implies convergence of $\sum (-1{)}^n{b}_n$. The hypothesis implies $b_n/b_{n+1}>1$ eventually, so $b_n$ is eventually decreasing. For some $c>0$, eventually $b_n/b_{n+1}\ge 1+c/n$; the divergent product of these factors forces $b_n\to 0$. The test now applies.

The arithmetic--geometric mean inequality gives $x_n\ge 1$ from $n=2$ onward. For $x>1$,so the tail decreases to a limit $L\ge 1$. Passing to the recurrence gives $2L=L+L^{-1}$, hence $L=1$.

For the subadditive sequence, $0\le x_n/n\le x_1$, so it is bounded. The Fekete lemma can be proved directly here as follows. Let $\alpha ={\inf }_k{x}_k/k$. Fix $k$ and write $n=qk+r$, $0\le r<k$. ThenThus $\mathrm{limsup}{x}_n/n\le \alpha$, while the definition gives $\mathrm{liminf}{x}_n/n\ge \alpha$. Hence $x_n/n\to \alpha$.

Herewith eigenvalues $4,1$ and corresponding normalized vectors $(1,-1{)}^T/\sqrt{2}$ and $(1,1{)}^T/\sqrt{2}$. A compatible choice isDirect multiplication gives $V^{†}AU=D$.

Choose an orthonormal eigenbasis $e_i$ of $G$ and set $u_i=e_i$, $v_i=A{e}_i/\sqrt{{\lambda }_i}$. Part (a) shows that the $v_i$ form an orthonormal eigenbasis of $H$. If $U$ and $V$ have these vectors as columns, then they are unitary andThis is the singular value decomposition.

For $x\ne 0$,so the Hermitian matrix $G$ has positive real eigenvalues. If $G{e}_i={\lambda }_i{e}_i$, thenMoreover $|A{e}_i{|}^2={\lambda }_i|e_i{|}^2$, so $|f_i|/|e_i|=\sqrt{{\lambda }_i}$.

The Cayley-Hamilton theorem states that a square matrix satisfies its characteristic polynomial. If a $2\times 2$ matrix is diagonalisable, $A=S\mathrm{diag}({\lambda }_1,{\lambda }_2)S^{-1}$; applying $p(t)=(t-{\lambda }_1)(t-{\lambda }_2)$ to the diagonal matrix gives zero and hence $p(A)=0$.

If $B^k=0$, every eigenvalue $\lambda$ satisfies ${\lambda }^k=0$, so all eigenvalues are zero and the characteristic polynomial is $t^n$. Cayley--Hamilton then gives $B^n=0$.