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The integral is linear in its first argument and conjugate-linear in its second, and complex conjugation of the integrand shows $⟨g,f⟩=\overline{⟨f,g⟩}$. Thus it is a Hermitian form.

If $f(z)={\sum }_{j=0}^N{a}_j{z}^j$, Fourier orthogonality givesThis is nonnegative and vanishes only when every coefficient vanishes, that is, only for the zero polynomial. Hence the Hermitian form is positive definite.

If $\alpha =\beta$, $\psi =-\alpha (x-y{)}^2/2$. Its contours are the parallel lines $x-y=d$. On each such line,For $d\ne 0$ this is an unbounded straight streamline. Every point on $x=y$ has zero velocity, so that diagonal is a line of equilibria, rather than one moving streamline.

Rotate coordinates to $s=(x+y)/\sqrt{2}$ and $t=(x-y)/\sqrt{2}$. The stream function becomesFor $\alpha <\beta$, both coefficients are negative, so every nonzero contour is a bounded ellipse surrounding the stationary origin. For $\alpha >\beta$, the form is indefinite: nonzero contours are unbounded hyperbolas, and the zero contour consists of straight separatrices. Consequently all nonstationary streamlines are bounded exactly when $\alpha <\beta$. The origin is itself a stationary bounded trajectory in every case; the assertion about boundedness concerns the nonstationary streamlines. The equality case is handled separately below. These are the linear planar strain and rotation streamlines.

The velocity divergence is ${\partial }_x{u}_x+{\partial }_y{u}_y=\alpha -\alpha =0$, satisfying the constant-density incompressible flow condition. Choose the stream function convention $(u_x,u_y)=({\psi }_y,-{\psi }_x)$. Integration then givesIndeed, ${\psi }_y=\alpha x-\beta y$ and $-{\psi }_x=\beta x-\alpha y$. Streamlines are contours of this function. The vector field is also compatible with steady inviscid momentum balance without body forces: $(\mathbf{u}\cdot \nabla )\mathbf{u}=({\alpha }^2-{\beta }^2)(x,y)$ is balanced by pressure $p=p_0-\rho ({\alpha }^2-{\beta }^2)(x^2+y^2)/2$.

The vorticity is the curl of the velocity, perpendicular to the flow plane:The planar rotation rate is half this scalar vorticity, not the vorticity itself.

The uniform prior and likelihood give the posterior densitya Beta distribution with parameters $K+1,n-K+1$. Under squared-error loss for $\varphi$, the posterior expected loss is $\mathbb{E}[{\varphi }^2\mid K]-2a\mathbb{E}[\varphi \mid K]+a^2$, minimized uniquely at the posterior mean of $\varphi$, not at the product of the two posterior mean factors.

Using the supplied beta integral,Thus the Bayes estimate isThis remains positive even on the endpoint samples, unlike the extended MLE; it integrates over the posterior uncertainty in $\theta$.

Independence gives $\mathbb{E}[X_1(1-X_2)]=\theta (1-\theta )=\varphi$, proving unbiasedness. Conditional on $K=k$, all binary sample strings with $k$ ones are equally likely. ThereforeThe Rao-Blackwell theorem, or the tower property and total variance decomposition directly, yields the improved estimatorIt is unbiased by conditional expectation and is a function of the requested MLE. Its variance is strictly smaller, sinceFor $0<\theta <1$, the event $K=1$ has positive probability. On it, $X_1(1-X_2)$ has probability $1/n$ of being one, strictly between zero and one for $n\ge 2$. Thus the final expectation is strictly positive. This proves strict variance reduction, not merely a weak inequality.

Let $K={\sum }_i{X}_i$ and $\bar{X}=K/n$. The likelihood is ${\theta }^K(1-\theta {)}^{n-K}$, so for $0<K<n$ its unique maximizer is $\hat{\theta }=K/n$. By invariance of maximum likelihood estimation,For $K=0$ or $n$, the strict parameter interval $(0,1)$ has no attained likelihood maximum; the usual extended estimate takes the limiting endpoint and gives $\hat{\varphi }=0$. Thus the displayed estimator is the conventional extended MLE on those samples, an important boundary qualification. This is the first estimator in estimating Bernoulli variance.

At the cone's vertex height $h$, the first part gives $U(h)=U_0\sqrt{1-2gh/U_0^2}$ and $A(h)U(h)=A_0{U}_0$. Insert these into the assumed local impingement force from part (ii). Static support requires upward force equal to $Mg$, givingEquivalently,For a nonnegative support height the ratio in parentheses must not exceed one, and the prescribed $h\ll U_0^2/(2g)$ requires that ratio to be close to one. If the weight exceeds the lift available at the nozzle, this model has no supported height above the nozzle.

Choose a control volume containing the small impingement region, with an inlet section across the incident jet and an outlet annulus across the conical sheet. The remaining boundary follows free surfaces and the wetted cone. Subtract atmospheric pressure and neglect gravity. Bernoulli's equation gives the same speed $U$ in the outgoing sheet as in the incoming jet, and its total volume flux is $AU$.

The incoming axial momentum rate is $\rho A{U}^2$. The outgoing velocity makes angle $\alpha$ with the cone axis, so its axial momentum rate is $\rho A{U}^2\cos \alpha$; transverse components cancel by axisymmetry. The integral momentum equation says that the cone's axial force on the water is their difference, $-\rho A{U}^2(1-\cos \alpha )$. By equal and opposite reaction, the jet's force on the cone is along the incident jet, with magnitudeThe outlet control surface is across the sheet, not along it, so its entire outgoing volume flux is counted.

The free jet is at atmospheric pressure. Bernoulli's equation from the nozzle to height $z$ gives $U(z{)}^2/2+gz=U_0^2/2$. Conservation of volume flux gives $A(z)U(z)=A_0{U}_0$. HenceFor the prescribed well-below-apex regime, $U(z)=U_0[1-gz/U_0^2+O((gz/U_0^2{)}^2)]$ and $A(z)=A_0[1+gz/U_0^2+O((gz/U_0^2{)}^2)]$. The slowing jet therefore thickens. These expressions use the slender-jet, nearly uniform cross-sectional speed approximation underlying the question; the formal vanishing-speed apex is outside that approximation.

Split the exponential kernel at zero. For $a>0$,Thus its Fourier transform isBoth half-line integrals converge absolutely because $a>0$.

Assume (iii). Take $Y=X\times X$ and let $f,g$ be the two coordinate projections, continuous by the definition of the product topology. Their equality set is the diagonal, so it is closed. For distinct $x_1,x_2$, the point $(x_1,x_2)$ lies in its open complement. Choose a basic rectangle $U_1\times U_2$ containing it and contained in that complement. If $U_1\cap U_2$ were nonempty, a point $u$ in their intersection would give $(u,u)$ in the rectangle, impossible. Thus $U_1,U_2$ are disjoint open neighborhoods of $x_1,x_2$, proving that $X$ is Hausdorff. Hence (iii) implies (i), completing all three equivalences.

Assume the diagonal $\Delta \subset X\times X$ is closed. For continuous $f,g:Y\to X$, the map $h:Y\to X\times X$, $h(y)=(f(y),g(y))$, is continuous: the inverse image of a basic rectangle $U\times V$ is $f^{-1}(U)\cap g^{-1}(V)$. Thereforeis closed as the inverse image of a closed set under a continuous map. This proves (ii) implies (iii).

Assume $X$ is Hausdorff. For any $(x_1,x_2)$ outside the diagonal, the points differ, so choose disjoint open neighborhoods $U_1,U_2$. Their rectangle contains $(x_1,x_2)$ and misses the diagonal. Thus the complement of the diagonal is open in the product topology, proving (i) implies (ii). This is one direction of the closed diagonal theorem.

Let $G$ be a finite subgroup of $K^{\times }$. It is abelian since multiplication in the field commutes. The structure theorem therefore writes $G$ as a product of cyclic groups with invariant factors $d_1\mid \cdots \mid d_s$, omitting trivial factors. Every element satisfies $g^{d_s}=1$, so $G$ consists of roots of $t^{d_s}-1$. The root bound from part (ii) giveswhile the decomposition gives $|G|={\prod }_j{d}_j\ge d_s$. Equality forces every factor except the last to be trivial. Hence $G$ is cyclic. The trivial subgroup is cyclic as well. This is the finite multiplicative subgroup of a field is cyclic theorem, proved here using the preceding two parts.

The infinite case is false. In ${\mathbb{C}}^{\times }$, the group of all roots of unity is infinite, but every element has finite order. If it were cyclic, its generator would have finite order and the whole group would be finite, a contradiction.

A nonzero polynomial of degree $n$ over a field has at most $n$ distinct roots in that field. For a proof, the factor theorem gives $(t-a)\mid f(t)$ whenever $f(a)=0$. Distinct linear factors are coprime in $K[t]$, a unique factorization domain, so for distinct roots $a_1,\dots ,a_k$ their product divides $f$. Its degree is $k$, giving $k\le n$. One may equivalently remove one linear factor at a time and argue by induction.

If the field has at least $n$ elements, this bound is attained by a product of $n$ distinct linear factors. For a fixed finite field with fewer than $n$ elements, the largest possible number of distinct roots is $\min (n,|K|)$ instead; multiplicities do not create more distinct roots. The zero polynomial is excluded since it has no finite degree.

The Fundamental theorem of finitely generated abelian groups says that every finitely generated abelian group is isomorphic toThe rank of an abelian group $r$ and the invariant factors $d_j$ are uniquely determined, with the finite summand allowed to be absent. Equivalently, one can decompose the torsion subgroup into cyclic groups of prime-power orders. The infinite cyclic part and torsion part must both be included in the statement.