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For a continuously differentiable vector field on all of ${\mathbb{R}}^3$, the necessary and sufficient condition for a conservative vector field is $\nabla \times \mathbf{F}=0$. Necessity follows from equality of mixed partial derivatives of a potential; sufficiency uses the ${\mathbb{R}}^3$ being a simply connected space. On a general domain the topology cannot be omitted.

Here the relevant mixed partial derivatives areThus the curl vanishes. Integrating the first component in $x$ gives $\Phi =x^2\cos y-2x{z}^3+A(y,z)$. Matching the second component gives $A_y=3+2y{e}^z$, hence $A=3y+y^2{e}^z+B(z)$. Matching the last component forces $B^{\prime }=0$. A potential of a conservative vector field with the convention $\mathbf{F}=\nabla \Phi$ is consequentlyIf a physical potential is defined instead through $\mathbf{F}=-\nabla V$, it is $V=-\Phi$.

A cyclic group has one generator of a group: $G=⟨a⟩=\{a^j:j\in \mathbb{Z}\}$. An abelian group has $xy=yx$ for every pair of elements. Powers of one element commute because $a^r{a}^s=a^{r+s}=a^{s+r}=a^s{a}^r$, so every cyclic group is Abelian. The Klein four-group $C_2\times C_2$ is Abelian but is not a cyclic group: every nonidentity element has order two, whereas a generator of a group for a four-element cyclic group would have order four.

Fix a generator of a group $x$ of $C_n$. A group homomorphism is determined by $g=\varphi (x)$ because $\varphi (x^j)=g^j$, and the relation $x^n=1$ forces $g^n=1$. Conversely, such a $g$ defines $\varphi (x^j)=g^j$: exponents differing by a multiple of $n$ give the same value, and addition of exponents verifies the group homomorphism law. Thus the homomorphism from a finite cyclic group correspondence isFor $S_4$, the order of a permutation is the least common multiple of its disjoint cycle lengths. The condition ${\sigma }^4=1$ allows identity, transpositions, two disjoint transpositions, and four-cycles. The sixteen homomorphisms are ${\varphi }_{\sigma }(x^j)={\sigma }^j$, with the full list of possible generator imagesThe remaining eight elements of the symmetric group are three-cycles and do not qualify.

Both factors being normal subgroups places cross-commutators in their intersection, forcing cross-commutation.

The product hypothesis supplies surjectivity of the multiplication map.

Coprimality makes the intersection trivial by the subgroup-order divisibility theorem.

For a Cartesian change of basis represented by an orthogonal matrix $Q$, vectors transform as $E_i^{\prime }=Q_{ia}{E}_a$ and $B_i^{\prime }=Q_{ia}{B}_a$. Their squared norms are invariant and $Q_{ia}{Q}_{jb}{\delta }_{ab}={\delta }_{ij}$. Substitution into the given expression therefore givesthe transformation law of a Cartesian second-rank tensor. The magnetic field's extra axial sign under an improper physical reflection, if included, appears twice and cancels in its quadratic contribution.

Write the Maxwell stress tensor as $T=\mathbf{EE}+\mathbf{BB}-\frac{1}{2}(E^2+B^2)I$. Its divergence isThe identity $(\mathbf{A}\cdot \nabla )\mathbf{A}-\frac{1}{2}\nabla A^2=-\mathbf{A}\times (\nabla \times \mathbf{A})$ follows by contracting two Levi-Civita symbols, or directly by differentiating components. Using Maxwell's equations consequently yieldsHence the local conservation of electromagnetic momentum isThe two subtracted terms are the Lorentz force density, while $\mathbf{E}\times \mathbf{B}$ is the electromagnetic momentum density in the units used here.

Take the normal vector outward from the cylinder-and-cone solid. The surface consists of the cylindrical side and conical roof, with no bottom disk; this convention fixes the otherwise unspecified orientation of a surface. The boundary is the radius-two circle at $z=-2$. Reversing every normal vector reverses the final sign.

For $\mathbf{F}=(y{z}^2,0,0)$, its curl is $(0,2yz,-z^2)$. For the cylindrical parametrized surface, use ${\mathbf{r}}_c=(2\cos \theta ,2\sin \theta ,z)$, with $-2\le z\le 2$, and use ${\mathbf{r}}_{c,\theta }\times {\mathbf{r}}_{c,z}=(2\cos \theta ,2\sin \theta ,0)$ as the outward area vector. Its flux isFor the cone, let $q=4-z$, $2\le z\le 4$ and ${\mathbf{r}}_k=(q\cos \theta ,q\sin \theta ,z)$. Its outward area vector is ${\mathbf{r}}_{k,\theta }\times {\mathbf{r}}_{k,z}=(q\cos \theta ,q\sin \theta ,q)$. ThusThe total surface integral is therefore $-16\pi$ with the outward convention, or $+16\pi$ for the opposite convention.

For Stokes theorem, the induced bottom-circle orientation is counterclockwise as viewed from above, opposite to the orientation of an outward bottom cap. Use $\mathbf{r}=(2\cos \theta ,2\sin \theta ,-2)$ with increasing $\theta$. Thenagreeing with the two parametrized fluxes. The seam at $z=2$ is internal and cancels between the two pieces; the cone tip does not supply another boundary curve.

The exponential distribution has survival probability $e^{-u}$ for $u\ge 0$. Conditional probability therefore giveswhich proves the memoryless property. Let $K=\lfloor Z\rfloor$ and $U=Z-K$. With $q=e^{-1}$,The geometric distribution here starts at zero. Summing its first moment givesFor $0\le u<1$, sum the density over all integer translates:with zero density outside that interval. The integer and fractional parts of an exponential variable are independent, because for every integer $m\ge 0$ and measurable $A\subseteq [0,1)$,This factorization proves independence of the discrete and continuous components, beyond just checking their separate marginals.

Let $\Delta ={\theta }_1-{\theta }_0>0$, with $\alpha ,{\beta }_{\max }>0$. Newton law of cooling gives the warming and cooling profilesIn the stipulated linear model, $\beta (t)={\beta }_{\max }(\theta (t)-{\theta }_0)/\Delta$. The bacterial count obeys $\dot{N}=-\beta (t)N$, so its logarithmic reduction is the cumulative thermal destruction under exponential warming and cooling:Writing $q=e^{-\alpha T}$, the warming integral divided by ${\beta }_{\max }$ is $T-(1-q)/\alpha$ and the cooling contribution is $(1-q{)}^2/\alpha$. Hence the required implicit equation isThe bracket is zero at zero and has derivative $(1-q)(1+2q)>0$ for $T>0$, tending to infinity with $T$. There is therefore one positive solution.

For the hardier species, $T$ is unchanged in this model if $\alpha$ and the achieved ${\beta }_{\max }$ remain the same. Raising the oven temperature changes $\Delta$, but its factor cancels against the changed slope of the linear destruction law. The entire normalized temperature history and hence the destruction integral remain the same; the comparison uses both the warming and equal-duration cooling stages.

Retain the derivative marks and coefficients from the printed system. In matrix form it is $M\dot{u}+Ku=b(t)$, withSince $\det M=11$, multiplication by its inverse givesThe linear transformation $u=x+y$, $v=x-y$ diagonalizes this system:Both start at zero. An integrating factor gives $u=v=e^{-t}-e^{-3t}$, soDirect substitution into the two original equations confirms the different forcing coefficients; the exact cancellation in $y$ depends on those coefficients.

Expand the function into separated terms: $f=g(x)+h(y)$, where $g=x^2-x^4$ and $h=y^8-y^4$. ThenPut $a=2^{-1/2}$ and $b=2^{-1/4}$. The nine critical points are all pairs from $x\in \{0,\pm a\}$ and $y\in \{0,\pm b\}$. The diagonal Hessian matrix has entries $2-12x^2$ and $56y^6-12y^2$.

At $(0,\pm b)$ both entries are positive, giving strict local minima of value $-1/4$. At $(\pm a,\pm b)$ they have opposite signs, giving four saddle points of value zero. At the three points on $y=0$ the Hessian is degenerate, so its sign alone does not classify them. Use the higher-order test for separated extrema: $h(y)=-y^4+O(y^8)$ is negative for small nonzero $y$. At $(0,0)$ the positive $x^2$ term and negative $y^4$ term give a saddle. At $(\pm a,0)$, $g-1/4=-(x^2-1/2{)}^2$, so every sufficiently small nonzero displacement decreases $f$ and these are strict local maxima.

For the level curves, the exact zero set consists of the two parabolas $x=\pm y^2$ and the oval $x^2+y^4=1$. They cross at the four nondegenerate saddle points. Small positive levels enclose each maximum; small negative levels enclose each minimum. The degenerate central saddle has quartic narrowing, locally $x^2-y^4=c$. Far along the $x$ axis the function is negative, while far along the $y$ axis it is positive. These signs and zero curves determine the connectivity shown in the sketch.

For the normalized second-order equation, an ordinary point is one where both coefficient functions $p,q$ are analytic. A singular point of a second-order linear ODE is a point that is not ordinary. It is a regular singular point if $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ extend analytically to $x_0$; a singular point failing that condition is irregular.

Here $p=0$, $q=1/x$. Thus zero is singular but regular, because $xp=0$ and $x^{2}q=x$ are analytic. Write the analytic solution as $y={\sum }_{n\ge 0}{a}_n{x}^n$. Its constant equation gives $a_0=0$, and for $n\ge 1$,Using $a_1=1$ givesThe factorial denominators show convergence for every finite $x$.

For the other branch, integrating the first correction equation twice and imposing its two conditions gives $y_1^{\prime }=-\log x$ andThen $y_2^{\prime \prime }=\log x-1$, so $y_2^{\prime }=x\log x-2x+2$ andEach correction tends to zero at the origin and its derivative vanishes at one. Although the second solution has a finite limiting value one, it has a logarithmically divergent endpoint derivative. Indeed the differential equation gives $y^{\prime \prime }\sim -1/x$, and integration givesA finite initial derivative cannot be prescribed for this second branch. The logarithm is consistent with the integer separation of the two exponents in the Frobenius method; this is not the repeated-exponent case.