Contract the loops in the Pochhammer contour onto the interval $(0,1)$. When $\mathrm{Re}z,\mathrm{Re}b>0$, the small-circle contributions tend to zero. The two clockwise circuits change the branches by $e^{-2\pi ib}$ and $e^{-2\pi iz}$, and the return circuits undo them. The four interval contributions sum toUsing the Beta function and $1-e^{-2\pi iu}=2i{e}^{-\pi iu}\sin (\pi u)$ gives

$J$ is entire in $z$. Along the fixed Pochhammer contour, the continued logarithms of $t$ and $1-t$ are bounded, and the contour stays away from both branch points. The integrand $\exp ((z-1)\log t+(b-1)\log (1-t))$ is entire in $z$ with uniform bounds on each compact parameter set. Differentiation under the contour integral is therefore valid to every order.

Let $T_i^{+}=\inf \{n\ge 1:X_n=i\}$ be the first positive return time. A recurrent state is one for whichThe subscript means that the Markov chain starts at $i$. The positive-time condition distinguishes a genuine return from the automatic visit at time zero.

For a time-homogeneous Markov chain with transition matrix $P$, two states are communicating states when each can be reached from the other with positive probability in finitely many steps:The convention $P^0=I$ includes communication of a state with itself.

For each fixed $x$, $x/n\to 0$, so the pointwise limit is $1$. Moreover $1-\cos u=2{\sin }^2(u/2)\le u^2/2$ givesThus the limit is $1$, with uniform convergence on $(0,1)$.

For each fixed $x>0$, continuity of the logarithm gives the pointwise convergence $f_n(x)\to \log x$. But at $x_n=1/n$, which belongs to $(0,1)$ for $n>1$,Consequently the uniform error cannot tend to zero. In fact $f_n(x)-\log x=\log (1+1/(nx))$ is unbounded as $x\downarrow 0$ for every fixed $n$. The pointwise limit is $\log x$, and convergence is not uniform.

Write the amended primal as $\min c^{\mathsf{T}}x$ with $Ax\le b$, $x\ge 0$. A convenient linear programming duality convention uses $y\le 0$ and gives $\max b^{\mathsf{T}}y$ subject to $A^{\mathsf{T}}y\le c$. Explicitly the dual isThe optimal dual vector is the negative of the three payoff-row slack coefficients:Indeed substitution gives $A^{\mathsf{T}}{y}^{\ast }=(2,-3,-5{)}^{\mathsf{T}}=c$, so all dual inequalities hold. To certify optimality directly by weak duality, for any feasible primal and dual vectors,where the second inequality uses $Ax\le b$ and $y\le 0$. Our feasible primal and dual vectors have equal values $-349/40$, so neither can be improved. This is a linear programming optimality certificate.

The amended payoff is $P_{\mathrm{new}}=P+3x_3$. Using the previous simplex dictionary,Thus $x_3$ enters. The positive pivot-column coefficients give ratios $3/(1/2)=6$ from $x_2$ and $(29/2)/10=29/20$ from $z_3$; the coefficient $-3/2$ in the $x_1$ row gives no bound. The simplex ratio test makes $z_3$ leave. Dividing that row by $10$ and eliminating $x_3$ givesThe positive objective-row coefficients and nonnegative slack variables prove optimality at $z_1=z_2=z_3=0$. ThereforeAll three constraints are equalities at this point. Strict positivity of the three objective-row coefficients also forces every optimal point to have all three slacks zero; the displayed dictionary then makes the optimizer unique.

Use payoff $P=-(2x_1-3x_2-2x_3)$, so maximization of $P$ is equivalent to the required minimization. In the given simplex tableau, $x_1$ can increase the payoff. The simplex ratio test has candidate ratios $3/6=1/2$ in the $z_2$ row and $15/1=15$ in the $z_3$ row; the negative coefficient in the $x_2$ row imposes no upper bound. Thus $x_1$ enters the simplex basis and $z_2$ leaves.

Divide the pivot row by $6$, then eliminate $x_1$ from the other rows. The resulting simplex dictionary isAll slack variables and original variables are nonnegative. The last row therefore proves $P\le 8$ for every feasible point; setting the nonbasic variables $x_3,z_1,z_2$ to zero attains it and leaves all basic variables nonnegative. HenceThe slack vector is $(0,0,29/2)$.

Under the null hypothesis, the standardized intercept $Z=\sqrt{n}(\hat{\alpha }-{\alpha }_0)/\sigma$ is standard normal, and $V=\mathrm{RSS}/{\sigma }^2$ is independent with chi-squared distribution of $\nu =n-2$ degrees of freedom. Consequently the Student t test for a simple-regression intercept usesChoose a significance level $\gamma \in (0,1)$ before observing the test result. For the two-sided alternative, reject $H_0$ when $|T|>t_{n-2,1-\gamma /2}$, where the subscript denotes the indicated quantile of the Student t-distribution. The null rejection probability is exactly $\gamma$, independent of the nuisance slope and variance. Equivalently report the p-value $2[1-F_{t_{n-2}}(|T_{\mathrm{obs}}|)]$. The test assumes the normal independent-error model and a positive residual variance estimate. In the intercept-only case with $n>1$, use $s^2=\mathrm{RSS}/(n-1)$ and $t_{n-1}$ instead.

Set $\mathbf{e}=1/\sqrt{n}$ and $\mathbf{f}=\mathbf{x}/\sqrt{S_{xx}}$. These are orthonormal because ${\sum }_i{x}_i=0$. Complete them to an orthonormal basis $\mathbf{e},\mathbf{f},{\mathbf{q}}_3,\dots ,{\mathbf{q}}_n$ of ${\mathbb{R}}^n$. An orthogonal transformation of the standard normal error vector $\mathbf{\epsilon }/\sigma$ has independent normal random variables as coordinates $Z_1,\dots ,Z_n$: its joint probability density remains $(2\pi {)}^{-n/2}\exp (-\Vert z{\Vert }^2/2)$ because the transformation preserves squared length and has Jacobian determinant of absolute value one.

Projection onto these basis directions givesTherefore the joint distribution of least-squares and variance estimators isThe last law is the chi-squared distribution of a sum of $n-2$ squared independent normal random variables with mean zero and variance one, and independence follows because the three statistics use disjoint independent coordinate sets. When all $x_i=0$, omit the slope direction: $\hat{\alpha }$ and $n{\hat{\sigma }}^2/{\sigma }^2\sim {\chi }_{n-1}^2$ are independent, and there is no identifiable slope estimator.

Let $\mathrm{RSS}={\sum }_i(Y_i-\hat{\alpha }-\hat{\beta }{x}_i{)}^2$. Differentiating the log-likelihood in $v={\sigma }^2$ givesFor $\mathrm{RSS}>0$ this changes from positive to negative at $v=\mathrm{RSS}/n$, soTo find its bias, substitute the model and use the orthogonal fit directions $1$ and $\mathbf{x}$, obtainingThe three expectations are $n{\sigma }^2$, ${\sigma }^2$, and ${\sigma }^2$, respectively: $\mathrm{Var}(\bar{\epsilon })={\sigma }^2/n$ and $\mathrm{Var}({\sum }_i{x}_i{\epsilon }_i)={\sigma }^2{S}_{xx}$. Hence $\mathbb{E}\mathrm{RSS}=(n-2){\sigma }^2$, giving the unbiased estimatorWith $n>2$, a zero residual sum of squares is a probability-zero event in the model. For that exceptional dataset the likelihood is unbounded as ${\sigma }^2\downarrow 0$, rather than having a maximum at a positive variance. In the intercept-only case $S_{xx}=0$, the corresponding unbiased multiplier is $n/(n-1)$ for $n>1$.

Write $S_{xx}={\sum }_i{x}_i^2$ and $\bar{Y}=n^{-1}{\sum }_i{Y}_i$. For an identifiable two-parameter linear regression assume $S_{xx}>0$; for the variance estimate and test below also assume $n>2$. The least-squares normal equations areUsing ${\sum }_i{x}_i=0$ givesThe residual sum of squares has positive-definite quadratic part $n{\alpha }^2+S_{xx}{\beta }^2$, so these equations give its unique minimum. The normal distribution of the independent errors gives the log-likelihoodFor every fixed positive ${\sigma }^2$, maximizing this over $\alpha ,\beta$ is exactly minimizing the same residual sum of squares. Thus the ordinary least squares estimators are also the maximum-likelihood estimators.

The printed hypotheses do not explicitly require $S_{xx}>0$ or $n>2$. If $S_{xx}=0$, all $x_i$ vanish and $\beta$ is unidentifiable; this is an intercept-only model with $n-1$ residual degrees of freedom. For $n=1$ its residual is identically zero, so the likelihood has no finite maximizer with positive variance. If $S_{xx}>0$ but $n=2$, the fit is saturated, leaving no residual variance estimate or Student test with positive degrees of freedom. The following nondegenerate formulas use $S_{xx}>0$ and $n>2$.

In the pion rest frame, conservation of three-momentum makes the Electron and massless neutrino momenta equal and opposite, of common magnitude $p$. Their energies obeySquaring $E_e=m_{\pi }{c}^2-pc$ and cancelling $p^2{c}^2$ givesThe Electron four-momentum satisfies $p=E_{e}v/c^2$. Thus, for the physically allowed decay $m_{\pi }>m_e$,

Let the photon momenta be ${\mathbf{p}}_1,{\mathbf{p}}_2$, with $|{\mathbf{p}}_i|=E_i/c$. Four-momentum conservation gives total energy $E_1+E_2$ and total momentum ${\mathbf{p}}_1+{\mathbf{p}}_2$ for the pion. Its mass shell relation therefore givesSince $0\le \theta \le \pi$, the sine is nonnegative, and taking the positive square root yields

Write $F=\mathcal{L}f$. Using the Laplace transform of a derivative and the initial data givesThe transformed equation is $(s^4+2s^3+s^2)F=1$, so $F=1/[s^2(s+1{)}^2]$. The preceding inverse Laplace transform therefore givesFor a direct check, $f^{\prime }=1-(t+1)e^{-t}$, $f^{\prime \prime }=t{e}^{-t}$, $f^{(3)}=(1-t)e^{-t}$, and $f^{(4)}=(t-2)e^{-t}$. These satisfy both the ordinary differential equation and all four initial conditions.

Part (i) gives $\mathcal{L}(t)=1/s^2$ and $\mathcal{L}(t{e}^{-t})=1/(s+1{)}^2$. The convolution theorem for Laplace transforms therefore givesRepeated integration by parts yieldsSubstitution gives $f(t)=t-2+(t+2)e^{-t}$, agreeing with the contour integration result.

By the Bromwich inversion formula, for $t>0$ and $\gamma >0$,Close the upward vertical segment by a left semicircle of radius $R$ centred at $\gamma$. On that arc $\mathrm{Re}s\le \gamma$, so $|e^{st}|\le e^{\gamma t}$, while $|s^2(s+1{)}^2|$ is bounded below by a constant times $R^4$ for large $R$. Its length is $\pi R$, hence the arc integral tends to zero as $O(R^{-3})$. The contour is positively oriented and encloses the double poles at $0$ and $-1$.

Their residues areandThe residue theorem therefore yieldsThe value at zero follows by continuity and is $f(0)=0$.

For an integer $n\ge 0$ and $\mathrm{Re}s>0$, the definition of the Laplace transform gives $F_0(s)={\int }_0^{\infty }{e}^{-st}dt=1/s$. For $n\ge 1$, integration by parts givesThe endpoint term vanishes because exponential decay dominates every fixed power of $t$. Induction therefore gives $F_n(s)=n!/s^{n+1}$. Multiplication by $e^{at}$ replaces $e^{-st}$ by $e^{-(s-a)t}$ in the defining integral, soThe second formula holds for $\mathrm{Re}(s-a)>0$ and is the Laplace transform shift theorem.

For each integer $n\ge 1$, the open balls $\{B(x,1/n):x\in M\}$ cover the compact metric space $M$. Compactness supplies a finite subcover with centres forming a finite set $S_n$. Define $S={\bigcup }_{n\ge 1}{S}_n$, a countable union of finite sets and therefore countable.

For any $x\in M$ and $\epsilon >0$, choose $n$ with $1/n<\epsilon$. One of the covering balls contains $x$, so there is $s\in S_n\subseteq S$ with $d(x,s)<1/n<\epsilon$. Every open neighbourhood of $x$ therefore meets $S$, giving $x\in \mathrm{cl}(S)$. Thus $\mathrm{cl}(S)=M$, and $S$ is a countable dense subset. If $M$ is empty, simply take $S=\varnothing$.