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Use $\hat{f}(t)=\int f(x)e^{-itx}dx$ and its extension to square-integrable functions. For $T_{k}f(x)=f(x+k)$, changing variables gives $\widehat{T_{k}f}(t)=e^{ikt}\hat{f}(t)$. The Plancherel theorem therefore givesFor an arbitrary square-integrable $f$, the translation identity follows by approximation by integrable square-integrable functions: translations are isometries in $L^2$ and the Fourier transform is continuous in the Plancherel norm. The Cauchy-Schwarz inequality makes the product on the left integrable, and $|\hat{f}{|}^2$ is integrable on the right.

Put $m=l-k$. Every unequal integer pair gives a nonzero integer $m$, and every nonzero integer arises, for example with $k=0,l=m$. The identity thus proves both directions of the requested equivalence. It is the orthogonality of integer translates criterion. A different Fourier normalization changes only the nonzero constant factor, not the vanishing conditions.

Direct differentiation of the velocity field gives $(\mathbf{u}\cdot \nabla )\mathbf{u}=(-{\Omega }^{2}x,-{\Omega }^{2}y,0)$. Since $u^2={\Omega }^2(x^2+y^2)$, this equals $\nabla (-u^2/2)$ as required.

In cylindrical coordinates, steady rigid rotation has $u_r=u_z=0$, $u_{\theta }=\Omega r$. The radial and vertical Euler equations for constant-density fluid are $p_r=\rho {\Omega }^{2}r$ and $p_z=-\rho g$, hence $p=\rho ({\Omega }^2{r}^2/2-gz)+C$. Constant atmospheric pressure on the free surface gives $z_s(r)=H+{\Omega }^2{r}^2/(2g)$. Conservation of fluid volume determines $H$:ThereforeThe lowest point is the axis, where $z_s(0)=h-{\Omega }^2{a}^2/(4g)$. The free surface first meets the bottom whenThis assumes no overflow and uses the full disk of wetted bottom until first contact. Once a dry central region forms, the volume constraint and geometry must be modified.

For a linear map $T:U\to V$, its rank of a linear map is $\dim \mathrm{im}T$, and its nullity is $\dim \ker T$. The rank-nullity theorem states $\dim U=\mathrm{rank}T+\mathrm{nullity}T$. One obtains it by extending a basis of the kernel to a basis of $U$; the images of the added vectors form a basis of the image.

Restrict $\beta$ to $A=\mathrm{im}\alpha$. Its image is $\mathrm{im}(\beta \alpha )$ and its kernel is $A\cap \ker \beta$. ThusThe intersection dimension is at most $\dim \ker \beta =\dim V-\mathrm{rank}\beta$, proving the lower bound. It is nonnegative, proving the upper bound by $\mathrm{rank}\alpha$, while $\mathrm{im}(\beta \alpha )\subseteq \mathrm{im}\beta$ proves the other upper bound. Consequently the rank inequality for a composition is

Consider a finite directed flow network with source $s$, sink $t$, and nonnegative finite flow network edge capacities $c_e$. A feasible flow assigns $0\le f_e\le c_e$ to each arc and satisfies flow conservation at every vertex except $s,t$. Its value $|f|$ is net outflow from $s$, equivalently net inflow to $t$. A maximal flow here means a flow of greatest value, also called a maximum flow. A source–sink cut of a flow network is a partition $(S,S^c)$ with $s\in S$, $t\notin S$; its cut capacity is $c(S,S^c)={\sum }_{u\in S,v\notin S}{c}_{uv}$, summing directed arcs from $S$ to its complement.

The max-flow min-cut theorem statesFor any feasible flow and cut, summing flow conservation over vertices in $S$ cancels internal arcs and gives $|f|={\sum }_{S\to S^c}{f}_e-{\sum }_{S^c\to S}{f}_e\le c(S,S^c)$. This proves the upper bound by every cut.

For the reverse bound, a maximum exists: the feasible set is a nonempty closed subset of the bounded product ${\prod }_e[0,c_e]$, hence compact, and $|f|$ is continuous. Choose a maximizing flow $f$. Its residual network has a forward arc for unused capacity $c_e-f_e$ and a reverse arc for cancellable flow $f_e$, retaining original-arc identities if necessary. If a residual path joined $s$ to $t$, the minimum of its positive residual capacities would be positive. Increasing forward flows and cancelling reverse flows by this amount would preserve all constraints while increasing $|f|$, a contradiction.

Let $S$ consist of vertices reachable from $s$ through positive residual arcs. Then $t\notin S$. An original arc from $S$ to $S^c$ must be saturated, since otherwise its forward residual arc would make its endpoint reachable. An original arc from $S^c$ to $S$ must carry zero flow, since a positive reverse residual arc would again extend reachability. ThusThe upper bound and this equality prove the theorem and exhibit a residual reachability certificate for maximum flow. The argument works for real capacities and does not assume that an arbitrary augmenting-path algorithm terminates in finitely many steps.

Let $X_j$ be the number of failures in mix $j$. The total observed failures are $32+2(12)+3(2)+4(1)=66$ among $600$ blocks. Equal marginal failure probabilities give $E(X_j)=6p$, regardless of dependence within a mix. Thus the unbiased estimator based on grouped Bernoulli failure counts isFor a conventional binomial confidence interval, additionally assume independent block outcomes. Then the total $S$ has binomial distribution $\mathrm{Bin}(600,p)$ and the estimated standard error is $\sqrt{\hat{p}(1-\hat{p})/600}=0.01277$. The normal approximation gives the approximate 95 percent intervalOne can instead obtain an exact, generally conservative binomial interval by solving $P_{p_L}(S\ge 66)=0.025$ and $P_{p_U}(S\le 66)=0.025$ for the two endpoints. These equations invert binomial tail tests.

Equal marginal probabilities alone do not imply independent blocks. If the mixes are independent but blocks within a mix can be dependent, use the sample variance of the 100 counts. Here $\sum X_j^2=114$ and $\overline{X}=0.66$, so $s_X^2=(114-100(0.66{)}^2)/99=0.7115$. The estimated variance of $\hat{p}=\overline{X}/6$ is $s_X^2/(100\times 36)$; its standard error is $0.01406$, giving the large-sample interval approximately $(0.0824,0.1376)$. If even between-mix independence is absent, no stated model justifies either numerical coverage claim. The estimate remains unbiased, but the uncertainty calculation needs its sampling assumptions.

Take $a,b,c>0$ as the semiaxis lengths. For a centered axis-aligned box with positive half-edge lengths $x,y,z$, the volume is $V=8xyz$. Maximize $xyz$ under $g=x^2/a^2+y^2/b^2+z^2/c^2=1$. The Lagrange multiplier equations areMultiplying by $x,y,z$, respectively, shows that $x^2/a^2=y^2/b^2=z^2/c^2$. The constraint makes each value $1/3$. All boundary boxes with a zero edge have zero volume, while the positive stationary box exists. Compactness therefore gives the maximum box volume in an ellipsoidAllowing a displaced or differently oriented rectangular box cannot improve this result. Let $A=\mathrm{diag}(a,b,c)$, let $c_0$ be its center, and let $u_1,u_2,u_3$ be its perpendicular half-edge vectors. Averaging $|A^{-1}(c_0+\sum {\epsilon }_i{u}_i){|}^2\le 1$ over the eight signs gives $|A^{-1}{c}_0{|}^2+\sum |A^{-1}{u}_i{|}^2\le 1$. The determinant bound $|\det (A^{-1}{u}_1,A^{-1}{u}_2,A^{-1}{u}_3)|\le \prod |A^{-1}{u}_i|$, followed by the arithmetic-geometric mean inequality, makes its volume at most $8abc/(3\sqrt{3})$. The box found above attains the bound.

Use the standard constant-density incompressible flow model, with conservative body force. Write $\mathbf{u}=(u,v,0)$ independent of $z$, and $\omega =v_x-u_y$. Differentiate the $y$-momentum Euler equations with respect to $x$ and subtract the $y$-derivative of the $x$-momentum equation. Mixed pressure derivatives and the curl of a conservative body force cancel, givingThe last term vanishes by incompressibility, so the material derivative of vorticity is zero. Thus $D\omega /Dt=0$. Incompressibility is important: inviscidness alone does not suffice; in a compressible barotropic planar flow the equation is $D\omega /Dt=-\omega \nabla \cdot \mathbf{u}$, conserving $\omega /\rho$ instead.

Take vortex strength $\kappa$ to denote circulation, positive counterclockwise. A point vortex at ${\mathbf{x}}_0$ generatesIt has zero vorticity away from its core, with circulation $\kappa$ around the core. Each vortex moves with the other's regular induced velocity, not its own singular field. With $\mathbf{d}={\mathbf{x}}_1-{\mathbf{x}}_2$,Multiplication by the two strengths proves $\dot{\mathbf{q}}={\kappa }_1{\dot{\mathbf{x}}}_1+{\kappa }_2{\dot{\mathbf{x}}}_2=0$. Also $\dot{\mathbf{d}}=({\kappa }_1+{\kappa }_2)J\mathbf{d}/(2\pi |\mathbf{d}{|}^2)$, so $d|\mathbf{d}{|}^2/dt=2\mathbf{d}\cdot \dot{\mathbf{d}}=0$.

For equal strengths $\kappa$ and the specified initial positions, $\mathbf{q}=0$ and $|\mathbf{d}|=2a$. The motion of two point vortices is rotation with angular velocity ${\Omega }_v=\kappa /(4\pi a^2)$:Negative circulation reverses the rotation. If vortex strength is defined instead as the coefficient of $1/r$ velocity, absorb the factor $2\pi$ consistently into its definition.

The dual space $V^{\ast }$ consists of all real linear functionals on $V$. If $v_1,\dots ,v_d$ is a basis, its dual basis ${\epsilon }^1,\dots ,{\epsilon }^d$ is characterized by ${\epsilon }^i(v_j)={\delta }_{ij}$; explicitly ${\epsilon }^i$ extracts the $i$th coordinate. The natural map $J:V\to V^{\ast \ast }$ is $J(v)(\varphi )=\varphi (v)$. It is linear and injective, because the coordinate functionals separate nonzero vectors, and both spaces have dimension $d$. Hence it is an isomorphism. Its definition uses no choice of basis, which is the meaning of “naturally” here.

Evaluation $e_x(p)=p(x)$ is linear, so belongs to $V_n^{\ast }$. For distinct nodes $x_i$, define the Lagrange interpolation polynomialsThey lie in $V_n$ and obey $e_{x_j}({\ell }_i)={\delta }_{ij}$. Applying a putative relation $\sum a_i{e}_{x_i}=0$ to ${\ell }_j$ gives $a_j=0$, so the $n+1$ evaluations are independent and form a basis of $V_n^{\ast }$. The polynomials ${\ell }_i$ give their dual basis in $V_n$, using the canonical identification with $V_n^{\ast \ast }$.

Every $p\in V_n$ has the interpolation expansion $p(t)={\sum }_{i}p(x_i){\ell }_i(t)$: the difference has $n+1$ distinct roots but degree at most $n$. Integrate to get the numerical integration weights ${\lambda }_i={\int }_{-1}^1{\ell }_i(t)dt$. For the specified symmetric five nodes, symmetry gives weights $(a,b,c,b,a)$. Exactness on $1,t^2,t^4$ givesOdd moments vanish automatically. Solving yields Boole's rule

First take $s_i>0$, the hypothesis required for the printed ratios and their optimization formula to be well-defined. Since $\sum p_i=1$,The maximizing and minimizing problems therefore have exactly the same solutions.

Let $\tau >0$ and minimize the separable optimization LagrangianEach summand is convex, and its derivative is $-p_i{s}_i/(s_i+x_i{)}^2+\tau$. Its minimizer is $x_i(\tau )=(\sqrt{p_i{s}_i/\tau }-s_i{)}^{+}$. The sum of these coordinates is continuous in $\tau$, decreases from infinity to zero, and is strictly decreasing wherever positive. Thus there is a unique $\tau$ for which the sum is $b$. For every feasible $y$, coordinatewise minimization gives $L(y,\tau )\ge L(x(\tau ),\tau )$; both constraint terms vanish. The Lagrangian sufficiency theorem proves global optimality, not just a necessary stationary condition.

Put $r_i=p_i/s_i$ in decreasing order. A positive bet occurs exactly when $r_i>\tau$. Hence the positive bets form an initial block. To interpret the maximal index in the question, one may include any subsequent indices tied at $r_i=\tau$: their bets are zero but their marginal-return equality still holds. Take $k=\max \{i:r_i\ge \tau \}$. ThenFor this block, $\sqrt{\tau }=({\sum }_{i\le k}\sqrt{p_i{s}_i})/(b+{\sum }_{i\le k}{s}_i)$, with the threshold inequalities determining $k$. Zero-probability horses receive zero bets.

Let $H$ be the set attaining $\rho =\max r_i$, and write $S_H={\sum }_{i\in H}{s}_i$, $P_H={\sum }_{i\in H}{p}_i=\rho S_H$. For sufficiently small $b$, the solution uses only $H$, with $x_i=b{s}_i/S_H$ there and $\tau =\rho [S_H/(S_H+b){]}^2$. If the next ratio is $r_{\ast }<\rho$, this remains valid when $b\le S_H(\sqrt{\rho /r_{\ast }}-1)$; if all outside ratios are zero there is no finite threshold. Put $S={\sum }_i{s}_i$. The fixed pool is $S+b$, so the expected gross payout and net gain areTied best horses receive bets proportional to their existing stakes; betting all on one tied horse is not generally optimal for finite $b$. This is pari-mutuel expected-return allocation.

The literal allowance $s_i=0$ needs qualification. At $s_i=x_i=0$ the printed objective contains $0/0$ and the ordered ratios are undefined. Even defining an unbacked horse's payout to be zero does not always yield an attained optimum. For example, take $b=1$, $p_1=p_2=1/2$, $s_1=0$, $s_2=1$. For $x_1>0$, the normalized expected payout is $1/2+(1/2)(1-x_1)/(2-x_1)$, tending to $3/4$ as $x_1\downarrow 0$ but never attaining it; at $x_1=0$ that zero-payout convention gives $1/4$. Thus the proved allocation formula requires positive existing stakes, or an explicitly modified market model. The zero-stake case cannot be justified by silently dividing by zero.