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Stokes theorem equates the circulation around the induced boundary orientation of an oriented surface to the normal flux of curl:The usual version requires a piecewise smooth oriented surface and a vector field continuously differentiable on a neighbourhood of it. We use the upward orientation wherever the surface is a graph; each outer circle is counterclockwise viewed from above, and an inner circle is clockwise.

Write $\rho =(x^2+y^2{)}^{1/2}$. The surface of revolution has $z=\sqrt{{\rho }^2+1-\lambda }$ with $0\le z\le 1$. For $0<\lambda <1$, it is a disk-shaped cap of a two-sheeted hyperboloid, running from $\rho =0,z=\sqrt{1-\lambda }$ to $\rho =\sqrt{\lambda },z=1$. For $\lambda =1$, it is a right circular cone, $z=\rho$, including its apex. For $\lambda >1$, it is an annular strip of a one-sheet hyperboloid, between the circles $\rho =\sqrt{\lambda -1},z=0$ and $\rho =\sqrt{\lambda },z=1$.

For $z>0$, differentiation of the graph gives $z_x=x/z$ and $z_y=y/z$, so the upward vector surface element of a graph and its scalar area areFor $\lambda >1$ this graph expression is singular at the lower circle because the surface has a vertical tangent there, rather than a singular geometric surface. A regular parametrization isIt proves that the graph surface integral is interpreted by its integrable endpoint limit. The same parametrization works away from the pole or apex in the other regimes.

Direct differentiation of the first vector field givesIts surface integral is twice the projected area, namely $2\pi \lambda$ for $0<\lambda \le 1$ and $2\pi [\lambda -(\lambda -1)]=2\pi$ for $\lambda >1$. On an oriented circle of radius $R$, $\mathbf{F}\cdot d\mathbf{x}=R^{2}d\varphi$. The outer circulation is $2\pi \lambda$; an inner circle, when present, contributes $-2\pi (\lambda -1)$. ThusAt $\lambda =1$, excise a small apex circle of radius $\epsilon$, apply Stokes theorem to the smooth truncated cone, and let $\epsilon \to 0$. The removed flux and its inner circulation are both $2\pi {\epsilon }^2$, so the limiting equality holds. This avoids silently applying a smooth-surface theorem at the cone's nonsmooth apex.

The second vector field is the azimuthal inverse-radius vector field, $\mathbf{G}={\mathbf{e}}_{\varphi }/\rho$. Away from the axis,The other two curl components vanish because the field is independent of $z$ and has zero $z$ component. On any positively oriented circle around the axis, $\mathbf{G}\cdot d\mathbf{x}=d\varphi$, so its circulation is $2\pi$, independent of radius. ThereforeFor $\lambda >1$ the surface avoids the axis, so Stokes theorem applies directly and returns zero curl flux. For $0<\lambda <1$ the disk cap meets the axis, and for $\lambda =1$ the apex lies on it: $\mathbf{G}$ is undefined there, so the neighbourhood hypothesis of Stokes theorem fails. Removing a small circle gives cancelling inner and outer circulations; the inner circulation remains $-2\pi$ as its radius tends to zero. Hence it cannot be discarded as it could for $\mathbf{F}$. A nonzero circulation here is compatible with zero curl away from the excluded axis.

Let $N=a+b$ and measure Lionel's fortune in unit stakes. This is gambler's ruin: the fortune moves by $+1$ with probability $p$ and $-1$ with probability $q$, stopping at $0$ or $N$. Let $m_i$ be the expected value of the absorption time starting at $i$. It is finite: from any interior state, a run of $N$ wins or losses has a fixed positive probability of absorption within $N$ steps, giving a geometric bound on survival over successive blocks. First-step analysis therefore givesFor $p\ne q$, the homogeneous solution of the recurrence is $A+B(q/p{)}^i$, and a particular solution is $-i/(p-q)$. Put $\rho =q/p$. Imposing the two endpoint boundary conditions gives the expected duration of biased gambler's ruinConsequently the expected number of games isThe denominator and numerator have matching signs, so this is positive. As a check its continuous limit at $p=q=1/2$ is $ab$, the expected duration of symmetric gambler's ruin.

Work on an interval where the change of variables is smooth and invertible, with $s=\dot{x}\ne 0$. If $v(t)=u(x(t))$, the chain rule gives $\dot{v}=s{u}_x$ and $\ddot{v}=s^2{u}_{xx}+\dot{s}{u}_x$. Eliminating $u_{xx}$ using the original second-order linear differential equation givesFor the real transformation as printed, take $s>0$. Set $v=s^{1/2}U$, which is the Liouville transformation of a second-order equation. Differentiating twice makes the coefficient of $\dot{U}$ cancel. Dividing by $s^{1/2}$ leavesThe additional term is exactly $s^{-1/2}[(\dot{s}/s)d(s^{1/2})/dt-d^2(s^{1/2})/d{t}^2]$, as required.

On an interval with $f>0$, choose $s=f^{-1/2}$ and therefore $t={\int }^x\sqrt{f(\zeta )}d\zeta$ up to a constant. Primes in the following expression mean derivatives with respect to $x$. We have $\dot{s}=-f^{\prime }/(2f^2)$ and $\ddot{s}=-f^{\prime \prime }/(2f^{5/2})+(f^{\prime }{)}^2/f^{7/2}$, giving the WKB correction termNeglecting $F$ gives $U=c_{+}{e}^t+c_{-}{e}^{-t}$. Since $u=s^{1/2}U$, the WKB approximation isChanging the lower integration limit only rescales the constants. Small $|F|$ is the stated local approximation criterion; rapid variation or a turning point can invalidate it, and small local residual alone need not imply uniform accuracy over an arbitrarily long interval.

The printed hypothesis only says $f\ne 0$. If real $f<0$ on the interval, the printed square roots require a consistent complex branch. Equivalently a real WKB approximation isThis supplies the oscillatory interpretation for the negative-$f$ case instead of silently assuming that the original nonzero hypothesis means positive. Smoothness through the derivatives used above and absence of zeros are required locally.

The chain rule gives ${\partial }_x={\partial }_{\xi }+\beta {\partial }_{\eta }$ and ${\partial }_y=\alpha {\partial }_{\xi }+{\partial }_{\eta }$. The coefficients of $u_{\xi \xi }$ and $u_{\eta \eta }$ are $2+3{\alpha }^2-7\alpha$ and $2{\beta }^2+3-7\beta$. Their integer roots are $\alpha =2$ and $\beta =3$. Thus choose characteristic coordinates $\xi =x+2y$, $\eta =3x+y$, whose Jacobian determinant is $-5\ne 0$. The coefficient of $u_{\xi \eta }$ is $4\beta +6\alpha -7(1+\alpha \beta )=-25$, so the equation becomes $-25u_{\xi \eta }=0$, equivalently $u_{\xi \eta }=0$.

Integrating the transformed hyperbolic partial differential equation gives $u=F(\xi )+G(\eta )$. Along $x=-2y$ the coordinates are $(0,-5y)$, so the zero boundary condition gives $G(\eta )=-F(0)$ for every real $\eta$. Along $x=0$ they are $(2y,y)$, so $F(2y)-F(0)=4y^2$, hence $F(\xi )={\xi }^2+F(0)$. The constants cancel and the unique resulting solution isIt directly satisfies both boundary conditions and $2u_{xx}+3u_{yy}-7u_{xy}=4+24-28=0$.

Introduce $\omega =\dot{\theta }$. The pendulum with quadratic damping is the autonomous planar systemIts equilibrium points are $(k\pi ,0)$. Since the derivative of $\omega |\omega |$ is zero at zero, the linearization of a dynamical system at each equilibrium point isAt an odd multiple of $\pi$ its eigenvalues are $\pm 1$, so it is a saddle equilibrium; the stable and unstable tangents are $\omega =-(\theta -k\pi )$ and $\omega =\theta -k\pi$. At an even multiple the eigenvalues are $\pm i$. This linearization is a centre and does not decide the nonlinear stability. In particular it would be incorrect to infer a linear damped focus from this Jacobian.

For the full system the mechanical energy is a Lyapunov function near an even equilibrium point:A sufficiently small sublevel set is compact and contains only that equilibrium point. The only invariant motion with $\dot{H}=0$ has $\omega =0$ and $\sin \theta =0$, hence is the equilibrium point itself. The LaSalle invariance principle proves asymptotic stability. Near it the trajectories turn clockwise and spiral inward, as the near-linear angular motion persists while energy decreases. Thus even equilibria are nonhyperbolic asymptotically stable spiral points; odd equilibria remain saddles.

For explicit trajectories put $W={\omega }^2$ on a branch where $\omega \ne 0$. The chain rule gives $dW/d\theta =2\dot{\omega }$, so the two signs yield the first-order linear differential equations $W^{\prime }+2cW=-2\sin \theta$ for $\omega >0$, and $W^{\prime }-2cW=-2\sin \theta$ for $\omega <0$. Their integrating factors giveOnly portions with a nonnegative right-hand side are physical. At a turning point the constants of the two branches are matched using $\omega =0$; neither expression is one global trajectory through both signs. When $c=0$ they reduce to energy contours, with closed orbits around the even equilibrium points. For every $c>0$, the quadratic damping destroys those closed orbits and makes the centres attracting spirals. Its effect is higher order in amplitude, so decay is weaker near rest than with linear damping. The saddle equilibria remain unstable, though their separatrices change shape.

Put $a=P(A)$, $b=P(B)$ and $j=P(A\cap B)$. Because $a,b>0$, the conditional probability inequality $P(A\mid B)>P(A)$ is equivalent to $j>ab$. But that same inequality gives $j/a>b$, or $\boxed{P(B\mid A)>P(B)}$. Thus positive association of two events is symmetric.

For the complement of an event, $P(B^c)=1-b>0$, andTherefore $B^c$ strictly repels $A$; equivalently $P(B^c\mid A)=1-j/a<1-b=P(B^c)$. Both directions are stated because the wording switches the grammatical roles of attraction and repulsion. The substantive conclusion is the same: the conditional probabilities involving the complement are strictly smaller than their unconditional counterparts.

The equilibrium points are $y=0,1,-1$. To obtain the other solutions of this separable differential equation without losing their signs, set $z=y^{-2}$ on an interval where $y\ne 0$. Then $z^{\prime }=1-z$, so $z=1+K{e}^{-t}$. Therefore all real solutions areon intervals where $1+K{e}^{-t}>0$. This family includes $y\equiv \pm 1$ when $K=0$. For a prescribed finite nonzero $y(0)=y_0$, $s=\mathrm{sgn}{y}_0$ and $K=y_0^{-2}-1>-1$, so the solution exists for all future time. The cubic logistic differential equation hasThese are the only possible limits. The sign cannot change by uniqueness, and the denominator tends to one. A nonzero initial condition producing a constant solution is $y(0)=1$; $y(0)=-1$ also works. The central equilibrium point is unstable and the two outer equilibrium points attract solutions from their respective half-lines.

The integrating factor is $e^{-2x}$, so $(e^{-2x}y{)}^{\prime }=e^{(\lambda -2)x}$. Integrating gives the general solutionThe particular solution is only defined up to a homogeneous solution. Subtract $e^{2x}/(\lambda -2)$ from it, absorbing that multiple into the arbitrary constant. The same family can then be writtenHolding $\tilde{C}$ fixed, the quotient tends to $x{e}^{2x}$, the derivative of $e^{\lambda x}$ with respect to $\lambda$ at $2$. Thus the resonant family is $y=e^{2x}(\tilde{C}+x)$. Direct differentiation gives $y^{\prime }-2y=e^{2x}$, so it is the full general solution at resonance. This resonant exponential forcing in a first-order equation limit concerns a reparametrized family, not a fixed value of the original divergent constant $C$. The initial condition gives $\tilde{C}=2e^{-2}-1$, hence

Interpret random choice of coin as equal prior probabilities, and the tosses as independent conditional on the chosen coin. Let $H_2$ denote two heads. The conditional probabilities are $P(H_2\mid A)=1/16$ and $P(H_2\mid B)=9/16$. For posterior coin selection from repeated tosses, Bayes' theorem givesThe tosses need not be unconditionally independent, because the same hidden coin is used twice. If the coin-choice prior were instead $P(B)=\pi$, the answer would be $9\pi /(1+8\pi )$; equal choice gives the stated $9/10$.

For nondecreasing uniform samples in the lottery, all $n^r$ ordered samples have equal probability by the discrete uniform distribution and independence. A nondecreasing sample is uniquely specified by its multiplicities $k_1,\dots ,k_n\ge 0$, with $k_1+\cdots +k_n=r$. The stars and bars argument places $n-1$ separators among $r$ identical markers, giving $\left(\genfrac{}{}{0ex}{}{r+n-1}{n-1}\right)$ multiplicity vectors. There is exactly one nondecreasing ordered sample for each such vector, so the probability isThis counts samples with ties correctly; multiplying a single strictly increasing count by $r!$ would not handle the repeated values.

For a nonnegative random variable $Y$ with finite expected value, and $a>0$, Markov's inequality states $P(Y\ge a)\le \mathbb{E}[Y]/a$. Indeed $Y\ge a{1}_{\{Y\ge a\}}$, and taking expected values proves the bound. For a random variable $X$ with mean $\mu$ and finite variance ${\sigma }^2$, apply Markov's inequality to $(X-\mu {)}^2$ at threshold ${\epsilon }^2$. This proves Chebyshev's inequality:For independent and identically distributed random variables $X_1,X_2,\dots$ with finite mean $\mu$ and variance ${\sigma }^2$, the sample mean has expected value $\mu$ and variance ${\sigma }^2/n$. ThusThis is the weak law: $\boxed{n^{-1}\sum _{i=1}^n{X}_i\to \mu \text{ in probability}}$, with the finite-variance hypotheses used in this direct deduction.

The more general weak law of large numbers needs only $\mathbb{E}|X_1|<\infty$. To obtain that version, truncate $Y_i^{(K)}=X_i{1}_{\{|X_i|\le K\}}$. For fixed $K$ the bounded variables obey the result just proved. Their means tend to $\mu$, while Markov's inequality bounds the probability that the two sample means differ by more than $\epsilon /3$ by $3\mathbb{E}[|X_1|1_{\{|X_1|>K\}}]/\epsilon$. First take $n\to \infty$ and then $K\to \infty$; the integrable tail vanishes, proving the full integrable i.i.d. version as well.

Finally assume $\mathbb{E}[X]=0$ and $\mathrm{Var}(X)={\sigma }^2$. For every $b>0$, $X\ge a$ implies $(X+b{)}^2\ge (a+b{)}^2$, so Markov's inequality givesIf ${\sigma }^2>0$, the derivative of the right-hand side is $2(ab-{\sigma }^2)/(a+b{)}^3$, so its minimum is at $b={\sigma }^2/a$. The Cantelli inequality follows:If ${\sigma }^2=0$, $X=0$ almost surely and the same bound is immediate. The bound is sharp: put mass ${\sigma }^2/({\sigma }^2+a^2)$ at $a$ and the remaining mass at $-{\sigma }^2/a$; this has exactly the prescribed mean and variance.

The proposed probability density function is nonnegative. To check its integral, put $I={\int }_{\mathbb{R}}{e}^{-x^2/2}dx$. The Gaussian integral can be evaluated by squaring and using polar coordinates:Since $I>0$, $I=\sqrt{2\pi }$ and $\int \varphi =1$. Thus $X$ has the standard normal distribution. Completing the square in its moment-generating function gives, for every real $t$,Differentiation under the integral is justified near every finite $t$ by Gaussian decay. Expanding $e^{t^2/2}={\sum }_{m\ge 0}{t}^{2m}/(2^{m}m!)$ gives all moments:

To obtain the two-term bounds for Mills ratio, use ${\varphi }^{\prime }(t)=-t\varphi (t)$. Integration by parts gives, for $x>0$,The positive remainder proves $r(x)<1/x$. A second integration by parts givesSubstitution and division by the strictly positive $\varphi (x)$ showThe last integral is strictly positive for every finite $x>0$, proving $\boxed{1/x-1/x^3<r(x)<1/x}$. Both remainders are finite for every fixed positive $x$; the lower bound remains valid even when it is negative.