The printed product omits the node $x_0$: its lower index must be zero for the stated formula to hold. For example, with $n=1$, nodes $0,1$, and data $f_0=0,f_1=1,g_0=g_1=0$, the printed $L_{1,1}=1$ gives $h=1$, which fails the first value condition. Use the corrected Lagrange basis polynomialsThey satisfy $L_k(x_j)={\delta }_{kj}$. Define $H_k=L_k^2[1-2(x-x_k)L_k^{\prime }(x_k)]$ and $K_k=L_k^2(x-x_k)$. At a node with $j\ne k$, the double zero of $L_k^2$ gives $H_k(x_j)=H_k^{\prime }(x_j)=K_k(x_j)=K_k^{\prime }(x_j)=0$. At $x_k$, direct differentiation givesBoth polynomials have degree at most $2n+1$. Consequently $h={\sum }_k(f_k{H}_k+g_k{K}_k)$ has the specified values and derivatives, and is the required Hermite interpolating polynomial. The source product is retained literally in the private transcription; this solution identifies its necessary correction.

The state-zero stationary distribution equation is ${\pi }_0=q{\pi }_0+q{\pi }_1$, so ${\pi }_1=(p/q){\pi }_0$. The state-one equation is ${\pi }_1=p{\pi }_0+q{\pi }_2$, giving ${\pi }_2=(p/q){\pi }_1$. More generally, the stationary probability flow across the cut $\{0,\dots ,i\}$ is balanced, so $p{\pi }_i=q{\pi }_{i+1}$. This also follows inductively from ${\pi }_i=p{\pi }_{i-1}+q{\pi }_{i+1}$.

Hence ${\pi }_i=(p/q{)}^i{\pi }_0$. Since $p<q$, the geometric series is summable, and normalization givesThese probabilities satisfy detailed balance and therefore all the stationarity equations. The Markov chain is an irreducible Markov chain, so this normalized stationary distribution is unique.

A Poisson process of rate $\lambda$ is a counting process starting at zero, with independent stationary increments, such that an increment over an interval of length $t$ has the Poisson distribution of parameter $\lambda t$. It has right-continuous paths with unit jumps almost surely.

Independence of the two processes makes increments of $A+B$ independent on disjoint intervals. The sum of independent Poisson random variables with parameters $\alpha t$ and $\beta t$ is Poisson with parameter $(\alpha +\beta )t$, as follows by convolution:Thus $A+B$ is a Poisson process of rate $\alpha +\beta$. Its first red and green waiting times are independent exponential random variables of rates $\alpha ,\beta$, so

The functional is unbounded above. Let $\ell$ be the positive affine interpolant of the endpoint values and choose a nonnegative, nonzero, continuously differentiable $h$ vanishing at both endpoints. The admissible functions $y_M=\ell +Mh$ satisfyThis excludes a global maximum, but does not prove that a stationary curve is a minimum. The concluding assertion is false without additional conditions.

An explicit counterexample uses $x_1=-2$, $x_2=2$, $y_1=y_2=\cosh 2$ and the stationary function $y=\cosh x$. Its second variation, for a perturbation $h$ vanishing at the endpoints, isChoose $h=1-x^2/4$. Since ${\mathrm{sech}}^{2}x\le 4e^{-2|x|}$, the positive contribution is at most ${\int }_{\mathbb{R}}{x}^2{e}^{-2|x|}dx=1/2$. On $[-1,1]$, $h^2\ge 9/16$, so the magnitude of the negative contribution is at least $(9/8)\tanh 1>9/16>1/2$. Thus ${\delta }^{2}J[h]<0$. Small perturbations in this direction preserve positivity and the endpoint data while decreasing surface area. Consequently this catenoid is not a local minimum. The valid conclusion from part (ii) is a necessary stationary form; minimization requires a separate stability and existence analysis.

For $F(y,y^{\prime })=y\sqrt{1+y^{\prime 2}}$, the Euler-Lagrange equation isSince $F$ has no explicit $x$ dependence, the Beltrami identity givesSolving this first integral, or substituting into the second-order equation, yieldsThe constants satisfy $y_1=C\cosh ((x_1-K)/C)$ and $y_2=C\cosh ((x_2-K)/C)$. There need not be a solution for arbitrary endpoint data, and when there is one these equations can have more than one pair of constants. These are the possible stationary catenoids, so the axisymmetric second variation of a catenoid still matters when deciding whether they minimize surface area.

Parametrize the surface of revolution byThe magnitude of the cross product ${\mathbf{r}}_x\times {\mathbf{r}}_{\theta }$ is $y\sqrt{1+y^{\prime 2}}$. Integrating the surface area element over $\theta$ givesPositivity of $y$ removes an otherwise necessary absolute value in the area element.

Treat the two squares as a single edge-pairing construction, including the identifications between different squares. Number the corners of the left square counterclockwise from its lower left corner by $A,B,C,D$, and similarly use $E,F,G,H$ on the right. The single arrows identify $D$ with $H$ and $C$ with $E$. The double arrows identify $G$ with $F$ and $H$ with $G$. The triple arrows identify $A$ with $E$ and $B$ with $F$. The quadruple arrows identify $A$ with $C$ and $D$ with $B$.

Thus there are precisely two vertex classes, $\{A,C,E\}$ and $\{B,D,F,G,H\}$. There are four edge classes and two two-dimensional cells, so the resulting closed connected surface has Euler characteristicIt is a nonorientable surface: on the left square the quadruple-arrow sides have the same orientation relative to a traversal of its boundary, giving a twisted band after gluing. Equivalently, no choice of orientations for the two faces makes the induced directions opposite on every paired edge. The vertex links are circles, so these edge identifications produce a closed surface, rather than a singular cell space. The classification of compact surfaces now givesA torus also has Euler characteristic zero, but is excluded by the failed orientation test.

For the canonical surjection $p:X\to X/\sim$, define a subset $W$ of $X/\sim$ to be open exactly when $p^{-1}(W)$ is open in $X$. This is the quotient topology. The empty set and the whole quotient have open inverse images. Moreover inverse images commute with arbitrary unions and finite intersections, so the corresponding axioms of a topology follow directly from those on $X$.

The map $p$ need not be an open map. For example, on $X=\mathbb{R}$ identify every point of $[0,1]$ to one point and leave all other classes singletons. The set $U=(-1/2,1/2)$ is open, butis not open. The definition of the quotient topology therefore shows that $p(U)$ is not open. The obstruction is that saturating an open set by equivalence classes need not preserve openness.

False. Let $X$ be a two-point set with the indiscrete topology, let $Y$ be a one-point space, and let $f$ be the unique map. The space $X$ is finite and hence compact; its only nonempty open subset is all of $X$, so it is connected. It is not Hausdorff, whereas $Y$ is Hausdorff and $f$ is a continuous surjection.

False. Take $X=[0,1]$, $Y=S^1$ and the surjection $f(t)=e^{2\pi it}$. The interval is compact and connected, and the circle is Hausdorff. But $f^{-1}(1)=\{0,1\}$ is a disconnected two-point subspace. Thus a continuous surjection need not have connected fibers.

True. The Hausdorff assumption makes $\{y\}$ closed in $Y$. Its inverse image under the continuous map $f$ is closed in the compact space $X$, and a closed subset of a compact space is compact. This is the compactness of the fiber with its subspace topology.

True. A separation of $Y$ into two disjoint nonempty open subsets would pull back under the continuous surjection $f$ to a separation of $X$. The inverse images are nonempty by surjectivity, contradicting connectedness of $X$. Thus a continuous image of a connected space is connected.

True. If open sets cover $Y$, their inverse images under the continuous surjection $f$ cover $X$. Compactness of $X$ gives a finite subcover; surjectivity then makes the corresponding finite family cover $Y$. This proves the compactness of a continuous image of a compact space.

For fixed $x>0$, $f_n(x)\to e^{-x}\sin x$; for $x\le 0$, the limit is zero. On $0<x<n$, the supplied inequality gives $(1-x/n{)}^n\le e^{-x}$, soThis is the given integrable dominating function. By the dominated convergence theorem,

The dominated convergence theorem states: if measurable $f_n$ converge almost everywhere to $f$ and $|f_n|\le g$ almost everywhere for an integrable $g\ge 0$, then $f$ is integrable and $\int f_n\to \int f$. In fact $\int |f_n-f|\to 0$.

To derive it from the monotone convergence theorem, first prove Fatou's lemma. For nonnegative measurable $h_n$, the functions $k_N={\inf }_{n\ge N}{h}_n$ increase to $\mathrm{liminf}{h}_n$ and satisfy $\int k_N\le {\inf }_{n\ge N}\int h_n$. Monotone convergence yields $\int \mathrm{liminf}{h}_n\le \mathrm{liminf}\int h_n$.

Now $|f|\le g$ almost everywhere, so $f$ is integrable. Apply Fatou's lemma to the nonnegative functions $2g-|f_n-f|$, which converge almost everywhere to $2g$:Hence $\int |f_n-f|\to 0$, and $|\int f_n-\int f|\le \int |f_n-f|$ proves the assertion. Common exceptional null sets may be removed at the beginning since the sequence is countable.

Put $u=x-y$ and $v=x+y$. The square becomes $(-1,1{)}^2$, with $x=(u+v)/2$, $y=(v-u)/2$ and absolute Jacobian determinant $1/2$. On its support the denominator $v+2$ exceeds one and $|u|<1$, so the nonnegative integrand is bounded by one on a set of finite area; it is therefore Lebesgue integrable. The change of variables formula and Fubini's theorem give

The change of variables formula for a Lebesgue integrable function isThe absolute Jacobian determinant accounts for area regardless of orientation. The formula also holds for nonnegative measurable functions with possibly infinite integral.

A coordinate transformation is a continuously differentiable bijection from the open set $U$ onto an open subset $T(U)\subset {\mathbb{R}}^2$, with continuously differentiable inverse. Equivalently, it is a $C^1$ diffeomorphism; its Jacobian determinant never vanishes. The inverse function theorem supplies the local inverse, while global injectivity is an additional requirement.

In characteristic two, the binomial theorem gives $X^8-1=(X-1{)}^8$. The only root is already in ${\mathbb{F}}_2$, so the splitting field is the original field, despite the repeated root:

The polynomial $X^3+X+1$ has no root in ${\mathbb{F}}_2$, so it is an irreducible polynomial. The quotient ring of ${\mathbb{F}}_2[X]$ by the ideal it generates is therefore a field with eight elements. The seven nonzero elements form a group of order seven, so each satisfies $x^7=1$ by Lagrange's theorem. Thus this field splits $X^7-1$. A root different from one has order seven; it cannot lie in ${\mathbb{F}}_2$ and generates this degree-three field extension, since three is prime and the tower law applies. Therefore