At a point where $df$ is surjective, choose charts and reorder the domain coordinates so that an $\ell \times \ell$ minor of the derivative is invertible. The map $F=(f_1,\dots ,f_{\ell },x_{\ell +1},\dots ,x_k)$ has invertible derivative. By the inverse function theorem it gives local coordinates, in which $f$ is the projection $(u_1,\dots ,u_k)\mapsto (u_1,\dots ,u_{\ell })$. This is the submersion theorem.

The genus-two surface has fundamental group $G=⟨a_1,b_1,a_2,b_2\mid [a_1,b_1][a_2,b_2]=1⟩$. Connected double covering spaces correspond to index-two subgroups. Each is the kernel of a nonzero homomorphism $G\to C_2$; conjugacy creates no identifications because these kernels are normal. The commutator relation is automatic in $C_2$, so there are $2^4-1$ choices:To obtain a surjection onto the free group $F_2=⟨x,y⟩$, send $a_1\mapsto x$, $a_2\mapsto y$ and $b_1,b_2\mapsto 1$. The defining relation again vanishes, and the images generate $F_2$.

The polygon presentation of the Klein bottle, or Seifert-van Kampen theorem applied to its cell structure, givesSend $a$ to $(12)$ and $b$ to $(123)$ in $S_3$. Conjugation by $(12)$ inverts $(123)$, so this defines a surjective group homomorphism. Since $S_3$ is nonabelian, the fundamental group is nonabelian.

Modulo five, $X^2+26=(X-2)(X+2)$. Choose ${\mathfrak{p}}_5=(5,2+\alpha )$; then $(5)={\mathfrak{p}}_5{\overline{\mathfrak{p}}}_5$, and $N(2+\alpha )=4+26=30$. The element $2+\alpha$ belongs to ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$ and ${\mathfrak{p}}_5$. Their product has norm $2\cdot 3\cdot 5=30$, soThe imaginary-quadratic Minkowski bound for ideal classes is $(2/\pi )\sqrt{104}<7$. Every ideal class has an integral representative of norm at most six, whose prime ideal factors lie over two, three or five. Thus these prime ideals generate the ideal class group. Conjugate classes are inverses, and the displayed relation eliminates the class over five. The remaining generators have orders two and three, so their product has order six and generates the whole group:

The ideal ${\mathfrak{p}}_2=(2,\alpha )$ has ideal norm two. Its square is contained in $(2)$ and has norm four, so $(2)={\mathfrak{p}}_2^2$. A generator would have field norm $u^2+26v^2=2$, impossible for integers $u,v$. Thus its ideal class has order two.

Modulo three, $X^2+26=(X-1)(X+1)$, giving $(3)={\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$ with ${\mathfrak{p}}_3=(3,1-\alpha )$. A generator of ${\mathfrak{p}}_3$ would have norm three, again impossible. If ${\mathfrak{p}}_3^2$ were principal, its generator would have norm nine. The only possibilities are $\pm 3$, whose ideal is ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, not ${\mathfrak{p}}_3^2$.

Finally $N(1-\alpha )=27$. This element lies in ${\mathfrak{p}}_3$ but not ${\overline{\mathfrak{p}}}_3$, because $\alpha \equiv -1$ in the latter quotient. Unique unique factorization of ideals in a number field and norms therefore give

Write $\alpha =\sqrt{-26}$. Since the squarefree radicand is not congruent to one modulo four, the ring of integers of a quadratic field is ${\mathcal{O}}_K=\mathbb{Z}[\alpha ]$. The discriminant of the basis $(1,\alpha )$ is $4(-26)$, giving

Because $b$ is not an integer, the coefficient $\sin (\pi b)$ is nonzero. The expression of $B$ in terms of the entire function $J$ can therefore have poles only at integral $z$, and at most simple ones. Positive integers are removable: $B(z,b)$ is already analytic there for $\mathrm{Re}b>0$, and analytic continuation in $b$ givesThus $J$ vanishes at those integers, cancelling the denominator zero. The Gamma function formula $B(z,b)=\Gamma (z)\Gamma (b)/\Gamma (z+b)$ identifies the remaining singularities:For nonintegral $b$, these residues are nonzero, so every listed singularity is a genuine simple pole.

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 $n>2$.