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The chain rule along the specified straight line givesHere equality of mixed partial derivatives follows from smoothness. Apply Taylor theorem in the single variable $t$:Writing $h_1=x_1-a=\alpha t$ and $h_2=x_2-b=\beta t$ produces the linear and quadratic termswith all displayed partial derivatives evaluated at $(a,b)$. For a remainder valid uniformly over directions, apply the Taylor formula with integral remainder to $s\mapsto f((a,b)+sh)$:Continuity of the Hessian matrix makes the difference between the integral and $\frac{1}{2}{h}^{T}H(a,b)h$ be $o(|h{|}^2)$. The printed ellipsis is therefore justified as a finite Taylor expansion with a remainder. Smoothness alone does not assert convergence of an infinite Taylor series.

At a stationary point, the linear term vanishes. The symmetric Hessian matrix satisfiesIf $H_{11}>0$ and $\det H>0$, this is a positive-definite quadratic form. Its minimum over the unit circle is a positive number $c$, so $h^{T}Hh\ge c|h{|}^2$. The Taylor remainder is smaller than $c|h{|}^2/4$ for sufficiently small $h$, proving $f((a,b)+h)>f(a,b)$ for nonzero nearby $h$. Thus the stationary point is a strict local minimum.

For the given polynomial, the gradient equations reduce toThe stationary points are $(0,0),(1,-1),(-1,1)$, andAt both points with $x_1=\pm 1$, we have $H_{11}=10>0$ and $\det H=16>0$. ConsequentlyAt the origin the determinant of the Hessian matrix is $-8$, giving a saddle point. An independent check is the identity $f=(x_1^2-1{)}^2+(x_1+x_2{)}^2-1$, which proves that the two displayed local minima are also the only global minima.

An integrating factor for a differential one-form makes the multiplied form exact. If $dF=\mu gdx+\mu fdy$, then $F_x=\mu g$ and $F_y=\mu f$. Equality of the mixed partial derivatives yieldsThis is a local compatibility condition under the usual smoothness assumptions.

For the trigonometric equation, where its coefficients are defined and division is legitimate, the slope is $y^{\prime }=\tan x\tan y$. In the first quadrant it is positive, small near either axis and large near the upper-right corner. The direction field below uses normalized line segments, so steep directions remain visible.

Multiplication by $\sin x\cos y$ givesThus the multiplying function is an integrating factor, and the implicit general solution isIn the larger square, $\cos x\ge 0$. The positive-$C$ curves have positive $y$ and are symmetric about the $y$-axis; the negative-$C$ curves are their reflections. For $0<|C|<1$, a graph representation iswith endpoints tending to $y=\mathrm{sgn}(C)\pi /2$. The $C=0$ interior curve is $y=0$. The domain boundaries include singular coefficients or zeros of the integrating factor, so these endpoint limits and smooth continuations through $x=0$ describe the regularized relation; they must not be mistaken for additional solutions of the literal original equation at undefined points.

Differentiability at $(a,b)$ means that there is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ such thatNecessarily $L(h,k)=f_x(a,b)h+f_y(a,b)k$ when those partial derivatives exist. Under the stated assumptions, apply the one-variable mean value theorem separately to the two coordinate increments:for some intermediate $\theta ,\eta \in (0,1)$, omitting a term if its increment is zero. Existence of the partial derivatives in a neighborhood supplies the differentiability along each segment required for that theorem. Their continuity at $(a,b)$ makes the difference from $f_x(a,b)h+f_y(a,b)k$ bounded by $(|h|+|k|)\epsilon (h,k)$, where $\epsilon (h,k)\to 0$. Since $|h|+|k|\le \sqrt{2}\sqrt{h^2+k^2}$, this is the required little-o remainder and proves Fréchet differentiability.

For the particular function, $f(h,0)=f(0,k)=0$, so the defining difference quotients give $f_x(0,0)=f_y(0,0)=0$. However $f(t,t)=1/2$ for every $t\ne 0$. It is not continuous at the origin, and therefore not differentiable there. This illustrates why existence of partial derivatives alone is insufficient.

Choose an integral Weierstrass equation of an elliptic curve with unit elliptic-curve discriminant. For $P\in E({\mathbb{Q}}_p)$, choose homogeneous coordinates $[X:Y:Z]$ in ${\mathbb{Z}}_p$ with at least one unit by multiplying all three coordinates by a common power of $p$. DefineTwo primitive integral representatives differ by a unit, so this is well-defined. Every point of the reduced cubic is a nonsingular point of an algebraic curve, and the identity reduces to $[0:1:0]$. This definition includes points whose affine coordinates are nonintegral.

To prove the good-reduction map respects elliptic addition, use the smooth proper integral model. The elliptic curve group law is a morphism on this model, constructed by addition of degree-zero divisor classes; it is defined even where an affine chord formula has a vanishing denominator. Passing to the residue field commutes with this morphism. Henceso reduction is a homomorphism of abelian groups.

For surjectivity, the simple-root Hensel lemma says that if $g\in {\mathbb{Z}}_p[t]$, $g(t_0)\equiv 0(\mathrm{mod}p)$ and $g^{\prime }(t_0)\not\equiv 0(\mathrm{mod}p)$, then there is a unique $t\in {\mathbb{Z}}_p$ with $t\equiv t_0(\mathrm{mod}p)$ and $g(t)=0$. The point at infinity lifts to $O$. At an affine point $(\overline{x},\overline{y})$ of the reduced equation $F(x,y)=0$, being a nonsingular point of an algebraic curve means at least one partial derivative $F_x,F_y$ is nonzero. If $F_y\ne 0$, fix any integral lift $x_0$ of $\overline{x}$ and apply the Hensel lemma to $g(t)=F(x_0,t)$ to lift $\overline{y}$. If $F_x\ne 0$, fix $y_0$ and lift $\overline{x}$ instead. This proves the surjectivity of good reduction over a local field and gives

Use the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplying the block gamma matrices and using the Pauli matrix multiplication law givesThus these matrices satisfy the Clifford algebra relation $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }{I}_4$. Since partial derivatives commute, the antisymmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ drops out of its contraction with ${\partial }_{\mu }{\partial }_{\nu }$. Multiplying the Dirac equation on the left by $i{\gamma }^{\mu }{\partial }_{\mu }+m$ givesTherefore every component of the Dirac spinor obeys the Klein-Gordon equation with mass $m$.

For Lorentz covariance, take $x^{\prime }=Lx$ and ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$, with constant $L$ and $S$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. The given intertwining relation then impliesThis proves invariance of the Dirac equation under the joint coordinate and spinor transformation.

The infinitesimal identity follows directly from the spinor Lorentz generator commutator. To derive it, first move ${\gamma }^{\rho }$ through a two-matrix product using the Clifford algebra:Subtract the expression with $\mu$ and $\nu$ interchanged and multiply by $i/2$. ThusWriting $S=1-\frac{i}{4}{\sigma }^{\mu \nu }{\omega }_{\mu \nu }$ gives $S^{-1}=1+\frac{i}{4}{\sigma }^{\mu \nu }{\omega }_{\mu \nu }+O({\omega }^2)$, and henceThe last step uses ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$, including the metric signs in raising the first index.

For parity symmetry in quantum field theory, set $y=x_P=(t,-\mathbf{x})$. The time derivative of $\psi (y)$ is unchanged and each spatial derivative changes sign. Also ${\gamma }^0$ commutes with itself and anticommutes with every ${\gamma }^j$. ConsequentlyThis verifies the specified parity symmetry in quantum field theory action explicitly.

For charge conjugation, put $B=i{\gamma }^2$, so ${\psi }_C=B{\psi }^{\ast }$. In the Dirac representation of the gamma matrices, ${\gamma }^0,{\gamma }^1,{\gamma }^3$ are real and ${\gamma }^2$ is imaginary. The Clifford algebra therefore gives $B{\gamma }^{\mu \ast }{B}^{-1}=-{\gamma }^{\mu }$ for all four indices. Complex conjugation of the Dirac equation, followed by multiplication by $B$, yieldsThis is the antilinear charge-conjugation intertwiner property. No electromagnetic background is being held fixed in this free-field calculation.

Apply the specified parity symmetry in quantum field theory action before charge conjugation. Since ${\gamma }^0$ is real and $B{\gamma }^0=-{\gamma }^{0}B$,Thus the induced parity action on the charge-conjugate spinor has the opposite sign. In particular, if a spinor has intrinsic parity $\eta =\pm 1$, its charge conjugate has intrinsic parity $-\eta$. Equivalently, at rest the positive-frequency Dirac spinors occupy the upper two components and have ${\gamma }^0$ eigenvalue $+1$, whereas the negative-frequency Dirac spinors occupy the lower two components and have eigenvalue $-1$. The corresponding particle and antiparticle creation operators inherit these opposite signs under the specified parity convention. Hence electrons and positrons have opposite intrinsic parities; conventionally one assigns $+1$ to the electron and $-1$ to the positron. The common real overall parity sign may be reversed without changing this relative sign, as in opposite intrinsic parities of a Dirac particle and antiparticle.

Use $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ throughout. Vary the Maxwell action with respect to the electromagnetic four-potential. Antisymmetry of the electromagnetic field tensor gives $\delta \mathcal{L}=-F^{\mu \nu }{\partial }_{\mu }\delta A_{\nu }$. After integration by parts, the Euler-Lagrange field equation isA gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. It therefore leaves the Maxwell action and observable electromagnetic fields unchanged: different potentials represent the same physical field. This redundancy matters for canonical quantization: before gauge fixing, the momentum conjugate to $A_0$ vanishes and the four potential components cannot be treated as four independent physical oscillators.

Choose the Lorenz gauge $D={\partial }_{\mu }{A}^{\mu }=0$, attainable by solving $\square \chi =-D$ with suitable boundary data. The Euler-Lagrange field equation then reduces toResidual Lorenz gauge symmetry consists of transformations with $\square \chi =0$. The gauge condition and this residual redundancy ultimately leave two transverse polarizations of a free photon.

For the specified kinetic density ${\mathcal{L}}_C=-\frac{1}{2}{\partial }_{\mu }{A}_{\nu }{\partial }^{\mu }{A}^{\nu }$, one has $\partial {\mathcal{L}}_C/\partial ({\partial }_{\mu }{A}_{\nu })=-{\partial }^{\mu }{A}^{\nu }$. Its Euler-Lagrange field equation is therefore $-\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction, sinceThe divergence follows by expanding both sides and commuting the second derivatives. Thus the wave equation follows from a Feynman gauge action; the additional physical-state condition will implement the Lorenz gauge. One must not impose $D=0$ inside the original action before varying it.

Take the covariant components $A_{\nu }$ as canonical coordinates. The canonical momentum computed from ${\mathcal{L}}_C$ isThe equal-time canonical Poisson brackets are $\{A_{\mu }(t,\mathbf{x}),{\Pi }^{\nu }(t,\mathbf{y})\}={\delta }_{\mu }{}^{\nu }{\delta }^3(\mathbf{x}-\mathbf{y})$, with coordinate-coordinate and momentum-momentum brackets zero. Applying canonical quantization givesIn particular $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The negative temporal sign is the source of the indefinite state-space norm in covariant photon quantization.

The spatial Fourier transform of the wave equation gives an oscillator of frequency $\omega =|\mathbf{k}|$ for each component. The Hermitian adjoint condition on the electromagnetic four-potential then yields the mode expansion of a free fieldThese are massless on shell modes. Consistently with the preceding equal-time canonical commutation relation,The normalization is the same as for the scalar modes, with $-g_{\mu \nu }$ replacing the species identity matrix.

Gupta-Bleuler quantization starts with the Fock vacuum annihilated by every $a_{\mu }$ and imposes only the annihilation, or positive-frequency part of a quantum field, of the Lorenz gauge condition:Imposing the entire Hermitian field divergence as a strong operator identity would be incompatible with the four-component canonical commutation relation. The weaker condition suffices: between two physical states its positive-frequency part vanishes on the ket, and its negative-frequency part vanishes on the bra. Hence $⟨{\mathrm{phys}}^{\prime }|{\partial }_{\mu }{A}^{\mu }|\mathrm{phys}⟩=0$.

For a one-photon state $|\mathbf{k},ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(\mathbf{k})|0⟩$, commuting the constraint through the creator givesThus the allowed photon polarization vectors satisfy $\boxed{k\cdot ϵ=0}$. Their state inner product, with the common momentum-normalization factor suppressed, is $⟨{ϵ}^{\prime }|ϵ⟩=-{ϵ}^{\prime \ast }\cdot ϵ$. Momentum eigenstates are delta-normalized; the same statements hold with finite norms for normalizable wave packets.

Rotate the spatial axes so that $k=\omega (1,0,0,1)$. The constraint is ${ϵ}^0={ϵ}^3$, so the complete allowed family isThe first vector is the allowed longitudinal gauge polarization, proportional to the null four-momentum $k$. Its state has zero norm because $k^2=0$; it is also orthogonal to every constrained state because $k\cdot ϵ=0$. A purely spatial vector $(0,0,0,1)$ alone is not an allowed physical polarization: its time component must accompany it to satisfy the constraint. The other two vectors are transverse polarizations, perpendicular to $\mathbf{k}$, with positive unit norms. For the displayed family,The timelike negative-norm excitation has been excluded by the constraint, and the remaining form is positive semidefinite. The Gupta-Bleuler null-state quotient identifies $ϵ$ and $ϵ+\lambda k$, removing the null direction. It leaves exactly two transverse physical photon polarizations with a positive inner product; their bosonic Fock space is the physical photon state space.

Finally use Feynman slash notation, $/a={\gamma }^{\mu }{a}_{\mu }$, and impose four-momentum conservation $p+q=k+k^{\prime }$. The external Dirac spinors obeySet $r=p-k$ and $r^{\prime }=p-k^{\prime }$. Replacing $ϵ$ by $k$ in the first term of the electron-positron annihilation into two photons amplitude, the rightmost contracted vertex obeysSince the Clifford algebra implies $(/r+m)(/r-m)=(r^2-m^2)I_4$, that term becomesFor the second term, four-momentum conservation gives $k=q+r^{\prime }$. The contracted vertex is now adjacent to the outgoing positron spinor on the left, soConsequentlyThe two terms cancel, proving the two-photon fermion Ward identityExchanging the two photon labels proves the same result for ${ϵ}^{\prime }=k^{\prime }$. By linearity the amplitude is invariant under $ϵ\mapsto ϵ+\lambda k$ on either external photon leg. Therefore the null longitudinal gauge polarization has zero production amplitude: physical scattering produces only the two transverse photon polarizations. This cancellation also shows why both photon orderings in the tree-level Feynman diagrams are essential; either term alone fails the Ward identity.

The pullback of the flat Euclidean metric under $x^{\prime }{}^{\mu }=x^{\mu }+ϵ{\omega }^{\mu }(x)$ isFor a conformal transformation it must be a scalar multiple of ${\delta }_{\mu \nu }$. Taking the trace determines that multiple. With $\rho ={\partial }_{\alpha }{\omega }^{\alpha }$, the necessary and sufficient infinitesimal condition is the Conformal Killing equationDifferentiate its $(\alpha ,\beta )$ instance with respect to $x^{\gamma }$, its $(\alpha ,\gamma )$ instance with respect to $x^{\beta }$, and its $(\beta ,\gamma )$ instance with respect to $x^{\alpha }$. Adding the first two and subtracting the last, commuting partial derivatives, leaves $2{\partial }_{\beta }{\partial }_{\gamma }{\omega }_{\alpha }$. Hence flat conformal Killing integrability givesContracting $\beta$ with $\gamma$ gives $\Delta {\omega }_{\alpha }=(2-d){\partial }_{\alpha }\rho /d$. Its divergence is $\Delta \rho =(2-d)\Delta \rho /d$, so $2(d-1)\Delta \rho =0$: thus $\boxed{\Delta \rho =0\text{ for }d>1}$. Taking instead the $\alpha$ divergence of the displayed second-derivative identity givesConsequently $\boxed{{\partial }_{\beta }{\partial }_{\gamma }\rho =0\text{ for }d>2}$.

On a connected flat domain, write $\rho =d\lambda +2db\cdot x$, with constant $\lambda$ and $b$. Integrating the second-derivative identity, and imposing the Conformal Killing equation on its remaining affine part, gives the complete Conformal Killing vector fieldIndeed the quadratic part has precisely the required Hessian matrix, while the symmetric part of the remaining linear term must be $\lambda {\delta }_{\mu \nu }$. The constants generate translations, rotations, dilations and special conformal transformations, respectively. There are $d$, $d(d-1)/2$, $1$ and $d$ independent parameters, so the full conformal group has dimensionHere “full” means local transformations, or global transformations of the conformal compactification of Euclidean space; its connected group is locally $SO(d+1,1)$. Requiring every transformation to be nonsingular on all of uncompactified ${\mathbb{R}}^d$ would exclude the special conformal transformations with finite poles.

Set $r=\sqrt{x^2+y^2}$. The bound $|xy|\le r^2/2$ and $|x^2-y^2|\le r^2$ give $|f(x,y)|\le r^2/2$. Hence $f\to 0$ at the origin, proving continuity, and $|f(x,y)|/r\to 0$, proving Fréchet differentiability there with derivative equal to the zero linear map.

Along the coordinate axes, direct difference quotients giveThese formulas include the origin. Therefore the two mixed partial derivatives differ:The explicit operator order removes any ambiguity about the notation for mixed derivatives. Also $f_{xx}(0,0)=f_{yy}(0,0)=0$, since the first derivatives restricted to their own axes vanish identically.

All four second-order partial derivatives are discontinuous at the origin. Away from it,Along $x=y\ne 0$ these are one and minus one, rather than their zero values at the origin. Meanwhile ${\partial }_y{\partial }_{x}f(x,0)=1$ for $x\ne 0$, whereas its value at the origin is minus one; and ${\partial }_x{\partial }_{y}f(0,y)=-1$ for $y\ne 0$, whereas its origin value is one. This gives a concrete unequal mixed partial derivatives example despite differentiability of the original function.

The exterior derivative is an $\mathbb{R}$-linear operator $d:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ which agrees with the differential of a smooth function on degree zero, satisfies $d^2=0$, and obeys the graded Leibniz rule

For the uniqueness of the exterior derivative from its axioms, first note that these axioms force locality. If $\alpha$ vanishes near $p$, choose a smooth bump function $\chi$ equal to one near $p$ and supported where $\alpha =0$. Then $0=d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $d\alpha (p)=0$. Thus $d\alpha (p)$ depends only on the form near $p$.

In a coordinate chart write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$. Since $d(d{x}^i)=d^2{x}^i=0$, the axioms forceThis also proves existence. Define $d$ by that formula in each chart. The product rule for partial derivatives gives the graded Leibniz rule, and symmetry of mixed partial derivatives against antisymmetry of the wedge product of differential forms gives $d^2=0$. On overlaps it still agrees with the differential of every function, by the chain rule, so the preceding uniqueness argument in the second coordinate system makes the formulas agree. They therefore glue to a global exterior derivative.

For a differential one-form $\omega ={\sum }_i{a}_{i}d{x}^i$, the exterior derivative of a one-form evaluated on vector fields follows directly:In $X(\omega (Y))-Y(\omega (X))$, the additional terms are ${\sum }_i{a}_i[X(Y^i)-Y(X^i)]=\omega ([X,Y])$, where the bracket is the Lie bracket of vector fields. Hence

The de Rham cohomology iswith zero denominator in degree zero. The Poincaré lemma states that a closed differential form of positive degree on a star-shaped open subset of Euclidean space is an exact differential form; consequently every closed positive-degree form is locally exact on a smooth manifold.

For the suspension isomorphism for top de Rham cohomology of spheres, put $U=S^n\setminus \{\text{south pole}\}$ and $V=S^n\setminus \{\text{north pole}\}$. Both are diffeomorphic to ${\mathbb{R}}^n$, while $W=U\cap V$ deformation retracts onto the equatorial $S^{n-1}$. The Mayer--Vietoris sequence for de Rham cohomology comes from the short exact sequence of cochain complexesThe final map is surjective: a partition of unity ${\chi }_U+{\chi }_V=1$ lets a form $\eta$ on $W$ be lifted by extending ${\chi }_V\eta$ to $U$ and $-{\chi }_U\eta$ to $V$ by zero. The Poincaré lemma makes $H^{n-1}(U),H^{n-1}(V),H^n(U),H^n(V)$ vanish when $n>1$. Exactness givesIf $j:S^{n-1}\hookrightarrow W$ is the equatorial inclusion, homotopy invariance of de Rham cohomology makes $j^{\ast }$ an isomorphism. Thereforeis injective, indeed an isomorphism. Concretely, choose local primitives $d{\alpha }_U=\omega {|}_U$, $d{\alpha }_V=\omega {|}_V$, and send $[\omega ]$ to $[j^{\ast }({\alpha }_U-{\alpha }_V)]$. Their difference is closed; changing either primitive changes it by an exact form, so this agrees with the cohomological construction.

The function is Fréchet differentiable at $(a,b)$ when there is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ such thatIf it exists, $L(h,k)=f_x(a,b)h+f_y(a,b)k$, as is seen by taking increments along each coordinate axis.

To prove differentiability from continuous partial derivatives, take a small rectangle in the given neighbourhood and split the increment along its sides. The one-variable mean value theorem givesfor some $0<\theta ,\eta <1$ when the corresponding increment is nonzero. If an increment is zero, its term is simply omitted. The required one-variable continuity follows from existence of the corresponding partial derivative on the segment. Subtracting $L(h,k)$ gives a remainder whose absolute value, by continuity of both partial derivatives at $(a,b)$, is at mostfor sufficiently small increments. Since $ϵ$ is arbitrary, the normalized remainder tends to zero. Thus the function is differentiable at $(a,b)$, with the displayed derivative.

Take the missing strong-swimmer times $U_j=S_{j1}$, $j=4,\dots ,8$, as the latent variables. Their partners' times are $T_j-U_j$. The complete-data log-likelihood isFor the E-step of the expectation-maximization algorithm, use the conditional split of two exponential times. At the old rates, with $d_r={\lambda }_1^{(r)}-{\lambda }_2^{(r)}>0$,This is a truncated exponential distribution, whose conditional expectation isThe displayed mean follows by integrating $u{e}^{-d_{r}u}$ against the conditional density; the missing weak-swimmer mean is $T_j-m_j^{(r)}$. As $d_r\to 0$, the conditional density becomes uniform on $(0,T_j)$ and $m_j^{(r)}\to T_j/2$.

Write $U_r=A+{\sum }_j{m}_j^{(r)}$ and $V_r=B+{\sum }_j(T_j-m_j^{(r)})$. The E-step objective isSetting its partial derivatives to zero gives the unconstrained M-stepThese are eight divided by the completed expected totals for the corresponding swimmers; imputing only observed pair totals without the conditional split would not perform the E-step.

The stated ordering must also be enforced. The ordered-rate exponential M-step accepts these updates when $U_r<V_r$, so ${\lambda }_1^{(r+1)}>{\lambda }_2^{(r+1)}$. Otherwise its maximum on the closed constraint ${\lambda }_1\ge {\lambda }_2$ lies atUnder a strictly open constraint ${\lambda }_1>{\lambda }_2$, a boundary optimum is only a supremum; a finite interior maximizer is not assured for every possible sample. One should report that situation rather than swap the labeled swimmer groups. Starting from positive ordered rates, iterate the E-step and constrained M-step until the parameter changes and observed log-likelihood improvement are small. EM likelihood monotonicity applies to every exact constrained M-step. No numerical swimming times are supplied, so the data-dependent updates, rather than fabricated numerical estimates, are the answer.

Write $A_{ij}={\partial }_j{u}_i$ and use incompressibility, $A_{ii}=0$. The displayed vector can be expressed asOn taking its divergence, ${\partial }_i{A}_{ij}=0$. Commuting the remaining partial derivatives giveswhich cancels the derivative of the second term of $D_i$. The derivatives falling on $u_k$ leaveIn homogeneous turbulence, averaging commutes with differentiation and $⟨D_i⟩$ is position independent, provided these moments exist. Consequently $⟨\mathrm{tr}(A^3)⟩=0$.

Using the strain-rate tensor and vorticity decomposition supplied in the question now gives the Betchov relation:This uses homogeneity and incompressibility; isotropy is unnecessary.