The equations $3m=4l$ and $2n=5k$ are equivalent tofor nonnegative integers $r,s$. Hence the language iswhich is the nested two-pair count language generated byIt is therefore context-free.

This language is not context-free. Suppose it had pumping length $N$, and choose a prime $p>N$. Apply the pumping lemma to $w=0^p{1}^p$. Let $\alpha$ and $\beta$ be the total numbers of zeros and ones in the pumped pieces $v$ and $y$. If $\alpha \ne \beta$, pumping with $i=0$ produces unequal block lengths and leaves the language. If $\alpha =\beta$, then $|vy|>0$ forces $\alpha =\beta =d>0$. Pumping with $i=p+1$ produces equal block lengthswhich are composite. Every possible pumping decomposition fails, proving the equal prime block language is not context-free.

This language is context-free. It is the language of even-length palindromes over $\{a,b\}$, generated by the context-free grammarEvery derivation adds matching outer symbols, and removing the matching first and last symbols gives the converse induction.

HereWrite $A=(\gamma +1)/2$. Substituting $x=(c_0-a)t$ into the $C_{+}$ characteristic equation givesFor $0<a<A$ and $0<\alpha \ll 1$, the relevant emission time is exponentially small. The terms containing an explicit factor $s$ are therefore beyond all algebraic orders, leavingThus the singular small-power piston emission time isAlthough $s\to 0$, the limiting state is nontrivial:which is the centered-rarefaction state on the ray $x/t=c_0-a$ when $\alpha =0$.

At $t=t_c$, the piston velocity reachesThe simple-wave relation then gives $c=0$ and $\rho =0$. For later times the formal continuation would require negative sound speed, so the gas can no longer follow the piston. By vacuum formation behind a receding piston, a vacuum region opens between $X_p(t)$ and the gas edgeThe earlier rarefaction solution remains to the right of $x_v(t)$, while $X_p(t)<x<x_v(t)$ is vacuum.

The undisturbed gas has $u=0$, $c=c_0$, and hence $R_{-}=0$. The disturbance from the receding piston is a right-moving simple wave, so this invariant remains zero throughout it:

Let $s$ be the time at which the relevant $C_{+}$ characteristic leaves the piston. The boundary condition is $u={\dot{X}}_p(s)$ there. Both $u$ and $c$ are constant along this characteristic, whose speed isHence $s=s(x,t)$ is determined implicitly by the receding-piston rarefaction wave equationWithin $X_p(t)<x<c_{0}t$, the solution is thereforeFor $x\ge c_{0}t$, the gas remains undisturbed: $u=0$ and $\rho ={\rho }_0$. The speed bound keeps $c$ and $\rho$ positive, and monotonic increase of $|{\dot{X}}_p|$ makes this an expanding rarefaction wave rather than a characteristic-crossing compression.

Assume (iii). Fix $n\ge 0$ and $A\in {\mathcal{F}}_n$, and defineThis is a stopping time: $\{T\le n\}=A\in {\mathcal{F}}_n$. Applying (iii) at time $n+1$ first to $T$ and then to the deterministic stopping time $n+1$ givesSubtracting yieldsBy the defining characterization of conditional expectation, this says $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. Thus $M$ is a martingale, proving (i). Together the three implications establish the characterization of a martingale by stopped expectations.

Assume (ii). A martingale has constant expected value by the tower property of conditional expectation, so for every stopping time $\tau$ and every $n$,This is (iii).

Assume (i), and let $\tau$ be a stopping time. The increments of the stopped process satisfyBecause $\{\tau >n\}\in {\mathcal{F}}_n$, taking conditional expectation given ${\mathcal{F}}_n$ makes the right-hand side zero. Integrability follows froma finite sum of integrable random variables. Hence $M^{\tau }$ is a martingale, proving (ii).

Because $J_m\to \infty$ almost surely, no infinite portion of the jump chain is hidden inside a finite time interval. After either chain leaves a state $i$, the event that it ever returns is therefore the same event for both chains. Hence irreducibility and recurrence pass between a CTMC and its jump chain gives

The jump chain of $(X_t)$ is exactly $(Y_m)$: from $i$ it visits $j$ next with probability $p_{ij}$. Consequently the two chains have exactly the same directed paths of positive-probability transitions. Thus

Combining the preceding parts,This is the characteristic function of $N(0,2\alpha I_d)$. The uniqueness theorem for characteristic functions therefore givesWhen $\alpha =0$, this means the degenerate distribution concentrated at the origin. This proves the Maxwell characterization of the normal distribution.

First, $f(0)=1$. For every $r\ge 0$,It cannot vanish: if $f(r_0)=0$ for some $r_0>0$, thenwould imply $f(r_0/n)=0$ for every positive integer $n$, contradicting continuity at zero. Thus $f(r)>0$. Since every characteristic function has modulus at most one,

Now set $h(r)=-\log f(r)$. It is continuous, nonnegative, and additive:The classification of a continuous additive function on the nonnegative real numbers gives $h(r)=\alpha r$, where $\alpha =h(1)\ge 0$. Consequently

Independence of the coordinates gives factorization of their joint characteristic function of a random vector. Orthogonal invariance under coordinate permutations makes all coordinate marginals identical. Hence, for $r_1,r_2\ge 0$ and $d\ge 2$,Therefore

For an orthogonal matrix $A$, invariance in distribution givesThe orthogonal group is transitive on each Euclidean sphere, so ${\phi }_X(u)$ depends only on $\Vert u{\Vert }^2$. DefineThe continuity of a characteristic function makes $f$ continuous. Taking $A=-I$ also gives $X\stackrel{d}{=}-X$, and henceso $f$ is real-valued. Thusas for every orthogonally invariant random vector.

The irreducible affine hypersurface $y^2-z^3=0$ in the $(y,z)$-plane has dimension one, while a closed point has dimension zero. Therefore

The curve $C$ has dimension one. Its tangent has dimension one at every point except the origin and dimension two at the origin. By the smooth locus of a variety,and the origin is its unique singular point of an algebraic variety. Each reduced point component has dimension zero and zero-dimensional tangent space, so both $P$ and $Q$ are smooth on their respective components.

Let $(a,b,c)$ be a tangent vector in the $x,y,z$ directions. At a point $(0,y,z)\in C$, the two isolated components are absent in a Zariski neighbourhood, so the Zariski tangent space is that of $C$. Differentiating $x$ and $y^2-z^3$ givesFor $(y,z)\ne (0,0)$ this is a line. At the cusp,which is two-dimensional.

Each of $P$ and $Q$ is a reduced isolated component with local ring $k$, soThis last conclusion must use all functions in $I(V)$, as required by the definition of the Zariski tangent space. The Jacobian of only the three displayed defining polynomials has an extra spurious tangent direction at $Q$, because those polynomials do not generate the radical ideal there; for example, $x(x^2-1)\in I(V)$ removes that direction.

WriteThe equation $xy=0$ splits the analysis into two cases. If $x=0$, the remaining equation is $y^2=z^3$, giving the cuspidal cubicIt is irreducible because $y^2-z^3$ is irreducible in $k[y,z]$: as a quadratic in $y$, it could factor only if $z^3$ were a square in $k[z]$.

If $y=0$, the other equations becomeTheir solutions are $(0,0,0)$, which already lies on $C$, and the two pointsThus the irreducible components areEquivalently, the radical ideal of the cusp with two isolated-point components is

When $n$ is odd there is no alternating sign mode. The same cosine-sine construction gives invariant irreducible planes $W_k$ foragain pairwise nonisomorphic because their generator traces are distinct. The real Fourier decomposition isIts dimensions sum to $1+2(n-1)/2=n$. Thus the real regular representation of a finite cyclic group has one one-dimensional irreducible summand and $(n-1)/2$ nonisomorphic two-dimensional irreducible summands.