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The Liouville approximation theorem states that if a real irrational algebraic number $\alpha$ has degree $d$, there is $c>0$ such that $|\alpha -p/q|\ge c/q^d$ for every integer $p$ and positive integer $q$.

Choose its irreducible polynomial $F\in \mathbb{Z}[X]$ of degree $d$. Since $\alpha$ is irrational, $F(p/q)\ne 0$; after multiplication by $q^d$ it is a nonzero integer. Hence $|F(p/q)|\ge q^{-d}$. On $[\alpha -1,\alpha +1]$ the polynomial derivative has a finite bound $M>0$. For $|p/q-\alpha |<1$, the mean value theorem gives $|F(p/q)|\le M|p/q-\alpha |$, proving the required bound with $c=1/M$. For more distant rationals, the bound holds with $c\le 1$. Taking $c=\min \{1,1/M\}$ covers every rational and proves the theorem.

Divide the supplied scheme by $\omega$ and takeThen $A=M-N$, and $M$ is invertible because it is lower triangular with positive diagonal $D/\omega$. Symmetry gives $U=L^T$, soThe Householder-John theorem therefore gives convergence.

For a direct spectral proof, let $B=M^{-1}N=I-M^{-1}A$. Expansion verifies $A-B^{\ast }AB=(I-B{)}^{\ast }W(I-B)$. If $Bv=\lambda v$ with $v\ne 0$, thenThe value $\lambda =1$ is impossible since $(I-B)v=M^{-1}Av$ and $A$ is invertible. Both quadratic forms are positive, hence $|\lambda |<1$. The error obeys $e^{(k+1)}=B{e}^{(k)}$ and tends to zero, proving convergence to the unique solution $A^{-1}b$ for every initial iterate. This is the successive-over-relaxation iteration, with no assertion of convergence at the excluded endpoints.

One form of the Householder-John theorem is as follows. Suppose $A=A^{\ast }$ and $A=M-N$, with $M$ invertible and $W=M+M^{\ast }-A$ positive definite. Then the stationary iteration $M{x}^{(k+1)}=N{x}^{(k)}+b$ converges for every initial vector exactly when $A$ is positive definite; its iteration matrix is $B=M^{-1}N$ and convergence means spectral radius below one.

For a known positive definite system, design a splitting with $M$ easy to solve, such as diagonal or triangular, and arrange $M+M^{\ast }-A>0$. The theorem then guarantees convergence. For example $M=sI$ has $W=2sI-A>0$ when $2s$ exceeds the largest eigenvalue of $A$. In the converse direction, the discrete energy identity $A-B^{\ast }AB=(I-B{)}^{\ast }W(I-B)$ explains the criterion: stable $B$ gives $A$ as a convergent sum of positive matrices, while positive $A,W$ force every iteration eigenvalue inside the unit disk.

Let $x$ increase away from the tip along the wall and $y$ point normally into the fluid. Put $u=u_x$, $v=u_y$ for velocity components and $U(x)=-A/x^2$ for the outer tangential velocity. The local Cartesian momentum and pressure equations areFor a two-dimensional flat layer the usual continuity equation is ${\partial }_{x}u+{\partial }_{y}v=0$. For this axisymmetric cone, however, the circumference is proportional to $x$, so its leading continuity equation isThe circumference term is of the same order as ${\partial }_{x}u$ and cannot be omitted when deriving the cone's similarity equation. Conditions at the stationary wall are $u=v=0$; at the layer edge $u\to U$.

Balance $U^2/x\sim \nu |U|/{\delta }^2$. With $|U|=A/x^2$ this givesThe choice of multiplicative constant fixes the normalization of the similarity variable. To enforce axisymmetric continuity set $xu={\Psi }_y$, $xv=-{\Psi }_x$, withThenDifferentiation gives $u{\partial }_{x}u+v{\partial }_{y}u=-(A^2/x^5)[2F^{\prime 2}+F{F}^{\prime \prime }/2]$, whereas $U{U}^{\prime }=-2A^2/x^5$ and $\nu u_{yy}=-A^2{F}^{\prime \prime \prime }/x^5$. Consequently the conical sink boundary layer satisfiesThe first condition is no penetration, the second no slip, and the last matches the inward core flow. If one instead makes a strictly planar approximation and omits the circumference term, the streamfunction is $-\sqrt{A\nu }{x}^{-1/2}F$ and the equation becomes $F^{\prime \prime \prime }+F{F}^{\prime \prime }/2+2(1-F^{\prime 2})=0$. The thickness exponent is the same, but that planar equation is not the axisymmetric cone equation.

Across a cap near $R=x$, the inward flux in a layer truncated at $y=Y$ is, to leading order,To match to the core, use the finite displacement-flux deficit rather than integrate the uniform outer flow to infinite $Y$. If $I={\int }_0^{\infty }(1-F^{\prime })d\eta ={\lim }_{\eta \to \infty }(\eta -F)$, thenConservation of the fixed total extraction rate requires an extra core flux of this order. Dividing by cap area, proportional to $R^2$, provesThe correction is a relative $O(\sqrt{\nu R/A})$ contribution. Its angular structure requires outer matching; the requested radial power follows from the displacement flux alone. Its cap-average direction increases the inward core speed to compensate for the wall deficit.

Using local speed $A/R^2$ and length $R$, the Reynolds number isInertia dominates and a thin viscous boundary layer is possible when this is large, namely $R\ll A/\nu$. The crossover scale is $R\sim Q/[2\pi (1-\cos \alpha )\nu ]$, or $Q/\nu$ up to the fixed-angle factor. The inverse dependence on $R$ explains why this inertial sink regime occurs near the tip rather than far away. The fixed cone angle must also allow the wall layer to be small relative to the local cone radius.

The outward flux through a spherical cap of radius $R$ inside the cone isThus the inward extraction flux is $Q$ at leading order. The thin no-slip wall layer is omitted from this approximation; its smaller displacement-flux effect supplies the later correction.

For $g\in \mathcal{S}({\mathbb{R}}^3)$ define convolution byThis is smooth and has at most polynomial growth, hence defines a tempered distribution. Its equivalent pairing is $⟨T\ast g,\varphi ⟩=⟨T,\tilde{g}\ast \varphi ⟩$ with $\tilde{g}(y)=g(-y)$. Fourier duality and the usual Schwartz-function convolution calculation giveThe product on the right is distributional multiplication by the smooth Schwartz function $\hat{g}$. With the convention in part (i), no extra factor occurs. One can verify it first for functions by changing $x=y+z$ inside the two integrals, then use the displayed pairing to perform the same calculation for arbitrary $T$.

The unrestricted categorical MLE is ${\tilde{\pi }}_j=N_j/n$, so the multinomial combinatorial factor cancels from the likelihood ratio andAssume the true cell probabilities are positive and the null has identifiable dimension $d$ with full-rank information. With probability tending to one, all $N_j$ are positive. Put ${\delta }_j=n{\hat{\pi }}_j-N_j$. Regular estimation gives ${\delta }_j=O_p(\sqrt{n})$ and $N_j$ of order $n$. Expanding $-2N_j\log (1+{\delta }_j/N_j)$ yieldsThe linear terms cancel exactly because ${\sum }_j{\delta }_j=0$. ThereforeThe full probability simplex has dimension $K-1$, explaining the degrees of freedom. Existence and uniqueness of an MLE alone do not supply all Wilks regularity assumptions; positivity, smoothness and the local rank condition are also needed. The displayed approximation is not a finite-sample prescription for division by a zero observed count.

The likelihood-ratio statistic for nested parameter spaces isUnder the null, Wilks theorem gives $W_n\Rightarrow {\chi }_{d_1-d_0}^2$ when the full model and null submodel are smooth and identifiable, the true parameter is an interior regular point, Fisher information is nonsingular in the identifiable directions, and the usual differentiability and likelihood-asymptotic conditions hold. A level-$\alpha$ test rejects for $W_n$ exceeding the upper $\alpha$ quantile of that chi-square law. The approximation is for large samples; boundary parameters, singularities or nonidentifiability can give a different limiting distribution.

At a state with positive departure rate ${\lambda }_i=R-r_i$, the next-jump probabilities areIf at least two rates are positive, every state has positive departure rate and the normalizing constant $Z={\sum }_i{r}_i(R-r_i)=R^2-{\sum }_i{r}_i^2$ is positive. The jump-chain equilibrium law isIndeed ${\hat{\pi }}_i{\hat{p}}_{ij}=r_i{r}_j/Z$ is symmetric, proving both stationarity and reversibility. Jump observations weight the continuous-time equilibrium by the departure rate, so $\hat{\pi }$ is generally different from $\pi$.

If exactly one rate, $r_k$, is positive, state $k$ is absorbing and the denominator $R-r_k$ vanishes. There is no next jump from that state: the genuine embedded jump sequence terminates there. Under the usual extended-chain convention set ${\hat{p}}_{kk}=1$; every other state jumps to $k$ with probability one. Its equilibrium law is ${\delta }_k$ and it is trivially reversible there. The normalized $r_i(R-r_i)$ formula cannot be used in that absorbing case.

The stationary equation $\pi Q=0$ reads $r^T-R\pi =0$, because $\pi 1=1$. ThereforeIt is the unique stationary law and vanishes on the open classes. For distinct $i,j$, ${\pi }_i{q}_{ij}=r_i{r}_j/R={\pi }_j{q}_{ji}$. Thus the chain is reversible in its equilibrium law, including the reducible cases; on the closed class this is the usual reversible irreducible chain. This structure is the destination-dependent reset chain.

Let $S_{+}=\{j:r_j>0\}$, nonempty since $R>0$. Every pair of distinct states in $S_{+}$ communicates directly, and no transition leaves $S_{+}$ for a zero-rate state. Hence $S_{+}$ is the unique closed communicating class. Each zero-rate state forms an open singleton: it has total departure rate $R>0$, goes into $S_{+}$, and can never be revisited. This includes the case that $S_{+}$ itself is a single absorbing state.

For $R>0$, the chain has one communicating class exactly when every $r_j>0$. Sufficiency follows from the positive direct transition rate between each distinct pair of states. For necessity, if $r_j=0$ and there is another state, no other state can jump to $j$, either directly or as the final step of a path, so $j$ cannot communicate with them. The one-state case also satisfies the condition because $R=r_1>0$. A single class containing the entire state space is automatically closed.

When $R=0$ all rates vanish. Every state is absorbing and the communicating classes are the $s$ singletons, all closed. There are no open communicating classes. Every probability distribution is stationary in this degenerate case.

For $i\ne j$ the generator entry is $q_{ij}=r_j$. Its diagonal is the negative total departure rate, $q_{ii}=-(R-r_i)$. ThusEvery row sums to zero, as required for a continuous-time Markov-chain generator.

Let $U$ be one uniform random variable on $(0,1)$, and defineFor every $U>0$ the sequence is eventually zero, so $X_n\to 0$ almost surely. But $\mathrm{Pr}(|X_n|>1/2)=1/n$, whose sum diverges. This sequence is not completely convergent; the exceedance events are nested and dependent.

Almost sure convergence implies convergence in probability, so part (b) identifies the limit as a constant $c$ almost surely. For each $ϵ>0$, the events $A_n=\{|X_n-c|>ϵ\}$ are independent. Almost sure convergence gives $\mathrm{Pr}(A_n\text{ infinitely often})=0$. If their probability sum diverged, the second Borel-Cantelli lemma would instead give probability one. Hence the sum is finite for every $ϵ$, proving almost sure convergence implies complete convergence for independent sequences.

Under complete convergence of random variables, for each integer $k\ge 1$ the first Borel-Cantelli lemma gives $\mathrm{Pr}(|X_n-X|>1/k\text{ infinitely often})=0$. Intersecting these probability-one conclusions over the countably many $k$ shows that every tolerance is eventually satisfied. Therefore complete convergence implies almost sure convergence, without independence.

Convergence in probability supplies a subsequence $X_{n_k}$ converging almost surely to $X$: choose it so $\mathrm{Pr}(|X_{n_k}-X|>2^{-k})<2^{-k}$, then apply the first Borel-Cantelli lemma. For each fixed $N$, the tail of this subsequence lies in $\sigma (X_n:n\ge N)$, and its almost sure limit is $X$. Thus $X$ has a version measurable with respect to the completed tail sigma-field of the independent sequence.

By Kolmogorov's zero–one law, every event in that tail sigma-field has probability zero or one. Consequently $\mathrm{Pr}(X\le r)\in \{0,1\}$ for every rational $r$. Since $X$ is a finite real random variable, its distribution function has limits zero and one at the two infinities; monotonicity then identifies a single threshold $c$ with $\mathrm{Pr}(X=c)=1$. Hence $X$ is almost surely constant. Completion only adds null events and does not alter the zero-one conclusion.

The first Borel-Cantelli lemma says that ${\sum }_n\mathrm{Pr}(A_n)<\infty$ implies $\mathrm{Pr}(A_n\text{ infinitely often})=0$, without an independence assumption. Indeed, for every $N$,The decreasing intersection of these tail unions is the infinitely-often event; continuity of probability from above proves the assertion. The second lemma says that for independent $A_n$, divergence of the probability sum implies $\mathrm{Pr}(A_n\text{ infinitely often})=1$.