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Use the canonical Poisson bracket convention $\{H,G\}={\sum }_k(H_{q_k}{G}_{p_k}-H_{p_k}{G}_{q_k})$. The Hamiltonian flow generated by $G$ has ${\dot{q}}_k=G_{p_k}$ and ${\dot{p}}_k=-G_{q_k}$. The chain rule therefore givesThus $H$ is invariant under every such trajectory exactly when $\boxed{\{H,G\}=0}$ identically. Antisymmetry then gives $dG/dt=\{G,H\}=0$ along the flow of $H$, so $G$ is a constant of motion.

For $G=\alpha {\sum }_k{p}_k$, the generated equations are ${\dot{q}}_k=\alpha$, ${\dot{p}}_k=0$, hence $q_k(t)=q_k(0)+\alpha t$, $p_k(t)=p_k(0)$. Invariance under this simultaneous translation implies conservation of $\boxed{\sum _k{p}_k}$ when $\alpha \ne 0$; the zero generator alone conveys no additional conservation law.

Away from zero and the zeros of $\cos (\pi /x)$, the sign function is locally constant and the absolute value function is differentiable at the value of the cosine. The product rule and chain rule giveThus the triangle inequality and the bounds on sine and cosine imply $|f^{\prime }(x)|\le 2|x|+\pi$. Part (c) rules out differentiability at every nonzero cosine zero, and part (b) gives $f^{\prime }(0)=0$, so this estimate covers every point where the derivative exists.

For a finite interval $I$, choose $L<\infty$ such that $|x|\le L$ throughout $I$. ThenThe apparent $x^{-2}$ factor in the derivative of the oscillation cancels against the multiplying $x^2$, which is what makes a uniform bound possible even near zero.

The first-order chain rule, with $x,y$ regarded as functions of $u,v$, isApply the product rule when differentiating the first identity in $v$, and apply the chain rule to $f_x,f_y$. Equality of mixed partial derivatives for a $C^2$ function givesThus $\boxed{H=x_{uv},K=y_{uv}}$. These terms record the second derivatives of the coordinate transformation, not of $f$.

For the given transformation,The mixed-derivative coefficient is zero, while $x_u{x}_v=x$, $y_u{y}_v=-y^2/x$, and $1/v^2=y/x$. The required partial differential equation is therefore exactly $g_{uv}=0$. Integrating first in $u$ and then in $v$ on a coordinate rectangle giveswhere $A,B$ are arbitrary twice continuously differentiable functions. The Jacobian determinant of the transformation is $-2u/v$, so this calculation uses patches with $u,v\ne 0$. On a positive-quadrant patch we can take $u=\sqrt{xy}$ and $v=\sqrt{x/y}$, obtainingChanging branches merely redefines the arbitrary functions.

For completeness, the original equation also makes sense in quadrants with $xy<0$, where that particular real square-root transformation is unavailable. Multiplying the equation by $x$ givesOn any quadrant with $xy\ne 0$, put $a=\log |x|$, $b=\log |y|$; then set $\xi =a+b$, $\eta =a-b$. The equation becomes $4f_{\xi \eta }=0$. Thus an equivalent general local answer on every such quadrant isHere $\Phi ,\Psi$ are arbitrary $C^2$ functions on the relevant intervals. If a solution is required across $y=0$, its quadrant expressions must extend with matching derivatives; on that line the equation itself imposes $x{f}_{xx}(x,0)+f_x(x,0)=0$, so its trace is $a_0\log |x|+b_0$ on each interval with $x\ne 0$. The displayed coordinate transformations do not assert invertibility at $y=0$.

Set $v=u^{\prime \prime }$. The resulting second-order linear differential equation is $v^{\prime \prime }+2v=4t^2$. A polynomial particular solution $v_p=a{t}^2+b$ requires $2a=4$ and $2a+2b=0$, hence $v_p=2t^2-2$. The homogeneous characteristic roots of a constant-coefficient differential equation are $\pm i\sqrt{2}$, soIntegrating twice and renaming constants gives the general real solutionThe four independent homogeneous terms account for the four initial data of the fourth-order linear ordinary differential equation.

For the substitution $u(t)=y(t^2)$, repeated use of the chain rule, with $x=t^2$, givesThereforeFor $x>0$, the map $t=\sqrt{x}$ is invertible, and the transformed equation is exactly the equation just solved. Hence the general real solution on $x>0$ isThere is no requirement that $u$ extend as an even function through $t=0$: only the branch $t>0$ is needed, so the sine and square-root terms must be retained.

Commutation of the two local flows means ${\varphi }_t{\psi }_s={\psi }_s{\varphi }_t$ for all sufficiently small independent parameters $s,t$, on their common local domains. Differentiate this identity in $s$ at zero. The chain rule givesIn a coordinate chart, differentiating again in $t$ at zero gives $D{X}_p{Y}_p=D{Y}_p{X}_p$. But the Lie bracket of vector fields has coordinate expressionso commuting local flows imply $[X,Y]=0$.

For the converse, suppose the Lie bracket of vector fields vanishes. Write $u(t)={\varphi }_t(p)$ in a coordinate chart, and put $J(t)=d{\varphi }_t{|}_p$. Differentiating the local flow equation with respect to the initial point gives the variational ordinary differential equationConsequently $W(t)=Y_{u(t)}-J(t)Y_p$ satisfieswhere the zero Lie bracket of vector fields was used in the second equality. Uniqueness for this linear ordinary differential equation gives $W=0$. This is a local argument and can be continued through successive coordinate charts, proving $d{\varphi }_t(Y)=Y\circ {\varphi }_t$ wherever the local flow is defined.

For fixed $t$, the curve $s\mapsto {\varphi }_t({\psi }_s(p))$ therefore has derivative $Y$ at its current point and starts at ${\varphi }_t(p)$. Uniqueness of the integral curve of a vector field $Y$ identifies it with $s\mapsto {\psi }_s({\varphi }_t(p))$. Hence $\boxed{{\varphi }_t{\psi }_s={\psi }_s{\varphi }_t⟺[X,Y]=0}$ locally. No complete vector field hypothesis is needed.

The expression $d\varphi (X)$ is in general a vector field along a map, rather than a vector field on $M$. Its intended meaning in the required formula is pointwise evaluation of the vector-bundle-valued differential form $\nabla s$:using the canonical identification $({\varphi }^{\ast }E{)}_p=E_{\varphi (p)}$. This interpretation works for every smooth map, including a constant map or a map with self-intersections.

Choose a frame of a vector bundle over $U$, written as a row $e=(e_1,\dots ,e_r)$. Define its connection one-form $\Omega$ byfor component columns $u$. The pulled-back frame $e^{\varphi }=({\varphi }^{\ast }{e}_1,\dots ,{\varphi }^{\ast }{e}_r)$ is a frame of a vector bundle over ${\varphi }^{-1}U$. For any component column $v$ of smooth functions on that open set, defineHere the matrix entries of ${\varphi }^{\ast }\Omega$ are ordinary pullbacks of a differential form. The displayed operator is linear and satisfies the Leibniz rule, hence is locally a connection on a vector bundle.

To prove these local operators glue, change the original frame of a vector bundle to $\tilde{e}=eh$. Expanding $\nabla (ehu)$ gives the change of frame of a vector-bundle connectionThe pullback of a differential form and the chain rule giveThis is precisely the required transformation law for the pulled-back frame ${\tilde{e}}^{\varphi }=e^{\varphi }(h\circ \varphi )$. Thus the definitions agree and give a global pullback connection.

For $s=eu$, substitute $v=u\circ \varphi$ in the defining formula and use $d(u\circ \varphi )(X)=du(d\varphi (X))$. This proves the required identity. Conversely that identity fixes ${\nabla }_X^{\prime }({\varphi }^{\ast }{e}_a)$; the Leibniz rule then fixes the derivative of every ${\sum }_a{v}^a{\varphi }^{\ast }{e}_a$, proving uniqueness. If the identity is initially imposed only on global sections of a vector bundle, multiply each local frame section by a smooth bump function equal to one near the point in question and supported in $U$, and extend by zero. These global sections agree with the frame locally, so the same uniqueness argument applies. The pullback connection exists and is unique.

Let $m=\dim M$, $n=\dim N$ and $k={\mathrm{codim}}_{N}S$. Fix $p\in f^{-1}(S)$ and choose the allowed slice chart for an embedded submanifold on a neighborhood $U$ of $f(p)$, so that $S\cap U$ is given by $x^1=\cdots =x^k=0$. On $f^{-1}(U)$ define the smooth mapThe derivative $L=d(x^1,\dots ,x^k)$ at a point of $S\cap U$ is surjective and has kernel $TS$, by the slice description. The printed sum-of-spaces condition givesIndeed any target tangent vector is a sum of an image vector and a tangent vector to $S$, and $L$ kills the latter. By the chain rule, $d{G}_r=L\circ d{f}_r$ is therefore surjective. This condition is precisely transversality of a map to a submanifold; the original sum need not be direct.

We have $G^{-1}(0)=f^{-1}(S)\cap f^{-1}(U)$. Part (b), applied to $G$, gives slice charts for an embedded submanifold of codimension $k$ around every point of this zero set. Since the construction is available around every $p\in f^{-1}(S)$, the charts give its global structure as an embedded submanifold with the subspace topology. Its dimension is $\boxed{m-k=m-n+\dim S}$. Moreover,This follows either from the slice coordinates or from $T_p{G}^{-1}(0)=\ker d{G}_p$ and $\ker L=T_{f(p)}S$. No injectivity of $df$, closedness of $S$, or direct-sum hypothesis is required.

We use $\Delta =\mathrm{div}\nabla$, whose spectrum on a compact manifold is nonpositive; write $P=-\Delta$ for the positive Laplace-Beltrami operator. This convention will also make the heat operator in Question 3 equal to ${\partial }_t-\Delta$.

Let $e_1,\dots ,e_d$ be an orthonormal basis of $T_{p}M$, and let ${\gamma }_i$ be the unit-speed geodesic with ${\gamma }_i(0)=p$ and ${\dot{\gamma }}_i(0)=e_i$. Since a geodesic has vanishing acceleration for the Levi-Civita connection, differentiation twice along it gives the Riemannian Hessian:Taking its metric trace yields the geodesic trace formula for the Laplace-Beltrami operatorThe positive-sign convention $P$ puts a minus sign in front of this sum.

For the unit sphere, take $p\in S^d$ and tangent unit vectors $e_i$ orthogonal to $p$. Its great circle geodesics are ${\gamma }_i(t)=p\cos t+e_i\sin t$. For a smooth ambient function $F$ near the sphere, the ordinary chain rule givesThe ambient orthonormal basis is $e_1,\dots ,e_d,p$. Thus, writing $F_r(p)={\partial }_{r}F(rp){|}_{r=1}$ and similarly for $F_{rr}$,This proves the ambient restriction formula for the spherical Laplacian and shows why ambient derivatives normal to the sphere must be subtracted. The equivalent formula in spherical coordinates is

Let ${\mathcal{P}}_{\ell }$ be the vector space of homogeneous polynomials of degree $\ell$ in $d+1$ variables, and ${\mathcal{H}}_{\ell }=\ker \tilde{\Delta }\cap {\mathcal{P}}_{\ell }$ its harmonic polynomials. If $H\in {\mathcal{H}}_{\ell }$, homogeneous polynomial scaling gives $H(r\omega )=r^{\ell }H(\omega )$. Substitution into the spherical coordinates formula givesThe restrictions are the degree-$\ell$ spherical harmonics. A homogeneous polynomial vanishing on the sphere vanishes on every nonzero ray and hence identically, so restriction is injective on ${\mathcal{H}}_{\ell }$.

We prove the harmonic decomposition of homogeneous polynomials that both counts these eigenfunctions and establishes completeness. For an ambient harmonic homogeneous polynomial $H_m$ of degree $m$ and $j\ge 1$, the spherical coordinates formula givesFor $d\ge 1$ its coefficient is nonzero. Inductively assume homogeneous polynomials of degree $\ell -2$ have been decomposed into sums of $r^{2j}{H}_m$. Apply this decomposition to $\tilde{\Delta }Q$ for $Q\in {\mathcal{P}}_{\ell }$, and lift each summand by multiplying by $r^2$ and dividing by the displayed nonzero coefficient. The resulting $R\in r^2{\mathcal{P}}_{\ell -2}$ satisfies $\tilde{\Delta }R=\tilde{\Delta }Q$, so $Q-R\in {\mathcal{H}}_{\ell }$. The same coefficient formula shows that no nonzero element of $r^2{\mathcal{P}}_{\ell -2}$ can be harmonic. Starting at degrees zero and one therefore proves the direct decompositionSince $\dim {\mathcal{P}}_{\ell }=\left(\genfrac{}{}{0ex}{}{\ell +d}{d}\right)$, the multiplicity iswhere the second term is zero for $\ell <2$.

On $r=1$, iterating the decomposition writes every polynomial restriction as a finite sum of spherical harmonics. The Stone-Weierstrass theorem makes these polynomial restrictions dense in $C(S^d)$, since coordinate functions separate points and constants are included; they are consequently dense in $L^2(S^d)$. Integration by parts makes $P$ symmetric on smooth functions and gives orthogonality between distinct eigenvalues. Normalizing within each finite-dimensional eigenspace gives a complete orthonormal eigenbasis $Y_{\ell a}$. To check the full operator spectrum, let $c_{\ell a}=⟨f,Y_{\ell a}⟩$. For smooth $f$, integration by parts gives the coefficients of $Pf$ as ${\lambda }_{\ell }{c}_{\ell a}$. Therefore finite harmonic partial sums approximate both $f$ and $Pf$ in $L^2$, hence approximate in the graph norm. Conversely every coefficient sequence with ${\sum }_{\ell ,a}{\lambda }_{\ell }^2|c_{\ell a}{|}^2<\infty$ is the graph norm limit of its finite harmonic sums. Thus the closure of $P$ has exactly this operator domain and is the real diagonal self-adjoint operator with entries ${\lambda }_{\ell }=\ell (\ell +d-1)$. If $z$ is not one of these entries, the inverse has entries $({\lambda }_{\ell }-z{)}^{-1}$ and is bounded because ${\lambda }_{\ell }\to \infty$. This proves that there are no missing spectral values. The spectrum of the Laplacian on a sphere iswith the above multiplicities; for $\Delta$, negate every eigenvalue. For example $S^1$ has eigenvalues ${\ell }^2$, with multiplicities one at zero and two for $\ell \ge 1$, and $S^2$ has multiplicity $2\ell +1$. The zero-dimensional sphere consists of two isolated points and has only the zero eigenvalue, with multiplicity two.

Let $P(z)$, $F(z)$ and $G(z)$ be the probability generating functions of $N$, one severity, and $S$, respectively. Positivity of the severities gives $F(0)=0$ and $g_0=p_0$; the random-sum transform identity gives $G=P\circ F$. The count recurrence in the Panjer claim-count class impliesUse the chain rule and substitute $F(z)$ to obtainEquating the coefficient of $z^{r-1}$ givesThe factor multiplying $f_j{g}_{r-j}$ after rearrangement is $a(r-j)+(a+b)j=ar+bj$. Therefore the Panjer recursion isThese power-series calculations are valid inside the convergence discs of the probability generating functions, and hence justify the coefficient identities without assumptions about positive exponential moments.

For a Poisson distribution count, $p_n/p_{n-1}=\lambda /n$, so $a=0$ and $b=\lambda$. The aggregate recursion becomesTo obtain a moment recursion, write ${\mu }_j=\mathbb{E}{X}^j$, $m_j=\mathbb{E}{S}^j$, and $m_0=1$. For nonnegative measurable $h$, conditioning on $N=n$ and using exchangeability of the severities giveswhere $X$ on the right is independent of $S$. The second equality uses $n{p}_n=\lambda p_{n-1}$, the Poisson size-bias identity. Choosing $h(s)=s^{k-1}$, applying the binomial theorem, and using independence proves the raw moment recursion for a compound Poisson distribution:For any particular $k$, a finite severity $k$th moment suffices: $({\sum }_{i=1}^n{X}_i{)}^k\le n^{k-1}{\sum }_{i=1}^n{X}_i^k$, and the Poisson distribution has finite moments of every order. No positive moment-generating function domain is required.

The first three iterations giveSubtracting the appropriate powers of the expected value yieldsIn particular, the aggregate variance and third central moment involve the raw second and third severity moments, rather than their centered counterparts.

Let $A=V^{1/2}$ be the symmetric positive square root of the covariance matrix, and write $S_1=\mu +AZ$ with $Z$ having the multivariate normal distribution $N_d(0,I)$. Bounded gradient makes $g$ globally Lipschitz, so $g(S_1)$ has at most linear growth and is square integrable. This also justifies the permitted Gaussian integration by parts formula for each coordinate.

By the chain rule, ${\nabla }_{z}g(\mu +Az)=A^T\nabla g(\mu +Az)$. Applying Gaussian integration by parts givesThus the Gaussian quadratic hedge holds ${\pi }^{\ast }=\mathbb{E}[\nabla g(S_1)]$. Substitution into its initial capital formula yieldsOnly the hedge calculation is being used here; the Gaussian model need not possess a positive pricing weight for this least-squares identity.

For an autonomous ordinary differential equation, the chain rule gives $y^{\prime \prime }=f^{\prime }(y)f(y)$. The formula is therefore a multiderivative multistep method. Insert the exact solution and expand the two earlier values about $t=t_{n+2}$:The unscaled local truncation error isThe coefficients through $h^3$ vanish, but the fourth-order coefficient does not. Hence the method has order three, not four: the one-step residual is $O(h^{p+1})$ for an order-$p$ multistep formula. Its zero-stability polynomial is $(\zeta -1)(\zeta -1/7)$, with a simple unit root and the other root inside the unit disk. Thus the stated order is also the convergence order for smooth solutions, starting errors $O(h^3)$, and the nearby implicit solution branch, by convergence of a zero-stable multiderivative method.

Away from zero, the product and composition of the displayed smooth functions are continuous and differentiable. At zero,so $f$ remains continuous. Hence the continuity set is $\boxed{\mathbb{R}}$.

The difference quotient at zero is $f(h)/h=\sin (1/h)$. For $h_n=(\pi /2+2\pi n{)}^{-1}$ it equals one, and for $k_n=(3\pi /2+2\pi n{)}^{-1}$ it equals minus one. Both sequences tend to zero, so no derivative exists there. For $x\ne 0$, the product rule and chain rule giveThus the differentiability set is $\boxed{\mathbb{R}\setminus \{0\}}$.

The fundamental theorem of calculus connects the Riemann integral with differentiation. Its first part says that for continuous $h:[a,b]\to \mathbb{R}$, the accumulation function $H(x)={\int }_a^{x}h(t)dt$ is continuous on $[a,b]$ and differentiable inside it, withTo prove this, boundedness of $h$ gives $|H(y)-H(x)|\le \Vert h{\Vert }_{\infty }|y-x|$, proving continuity. For an interior point and a sufficiently small nonzero increment $u$,Its absolute value is at most ${\sup }_{|t-x|\le |u|}|h(t)-h(x)|$, which tends to zero by continuity of $h$ at $x$. This proves the derivative formula. The same estimate gives the appropriate one-sided endpoint derivatives.

The second part says that if $Q$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $Q^{\prime }=h$ with $h$ continuous, thenIndeed $(Q-H{)}^{\prime }=0$, so the mean value theorem makes $Q-H$ constant, and evaluating at the endpoints proves the formula. A useful stronger version only requires $Q^{\prime }$ to be Riemann integrable. For a partition $a=x_0<\cdots <x_N=b$, the mean value theorem on each subinterval provides ${\xi }_j\in (x_{j-1},x_j)$ such thatAs the mesh tends to zero, these tagged Riemann sums converge to ${\int }_a^b{Q}^{\prime }$, proving that version too. Its integrability hypothesis must not be omitted.

An integrable integrand need not give an everywhere differentiable accumulation function. For a counterexample, let $f(t)=0$ for $t<1/2$ and $f(t)=1$ for $t\ge 1/2$. This bounded step function is Riemann integrable, butIts left and right derivatives at $1/2$ are zero and one. Thus $F$ need not be differentiable at every interior point. The first part of the fundamental theorem of calculus does apply at every point where the integrand is continuous.

A derivative need not have a Riemann integral. Define on all of $\mathbb{R}$At zero $f(h)/h=h\sin (1/h^2)\to 0$, so $f^{\prime }(0)=0$; elsewhere the product rule and chain rule giveAt $x_n=(2\pi n{)}^{-1/2}$, this equals $-2/x_n\to -\infty$. Hence $f^{\prime }$ is unbounded on $[0,1]$, whereas a properly Riemann integrable function on a compact interval must be bounded. The requested Riemann integral therefore need not exist, even though $f$ is differentiable everywhere. This example is an unbounded derivative obstruction to Riemann integrability; it does not contradict the theorem's version that explicitly assumes integrability of the derivative.

With $t=1-x^2$ the chain rule gives $y^{\prime }=-2x{u}^{\prime }$ and $y^{\prime \prime }=-2u^{\prime }+4x^2{u}^{\prime \prime }$. Substitution, initially for $x>0$, givesThe transformed coefficients are $P(t)=1/t$ and $Q(t)=1/[4t(1-t)]$. At $t=0$, both $tP$ and $t^{2}Q$ are analytic; at $t=1$, both $(t-1)P$ and $(t-1{)}^{2}Q$ are analytic. Each point is singular because a normalized coefficient has a pole, but satisfies the regular singular point criterion for a second-order equation. Thus both points are regular singular points.

For a Frobenius method substitution $u=t^{\rho }{\sum }_{n\ge 0}{a}_n{t}^n$, the lowest power gives the indicial equationThe repeated indicial root is zero. Setting $\rho =0$ and equating the coefficient of $t^n$ givesConsequently the analytic power-series solution of a differential equation isTaking $a_0=2\pi$ reproduces the ellipse perimeter expansion.

For a second independent solution, normalize $u_1(0)=1$. Reduction of order and $P=1/t$ giveSince $u_1(s{)}^{-2}=1+O(s)$, this has the Logarithmic Frobenius solution formThe logarithm is $\log |t|$ on the real negative side, or a chosen complex logarithm on a slit complex neighborhood. Its nonzero logarithmic coefficient establishes independence. The physical perimeter is finite at $t=0$, so it selects the analytic branch rather than this logarithmic one.

In Cartesian coordinates, independent coordinate variables satisfywhere ${\delta }_{ij}$ is the Kronecker delta. Applying the product rule and chain rule to $u_i=r^{\alpha }{x}_i$ gives the Jacobian matrixThis tensor is symmetric in $i,j$. Hence its contraction with the antisymmetric Levi-Civita symbol vanishes:Similarly, $v_i={ϵ}_{ijk}{k}_j{u}_k$ and the constancy of $\mathbf{k}$ giveThe trace of the Jacobian matrix iswhich vanishes for $\alpha =-3$. The curl of a cross product identity, with $\mathbf{k}$ constant, now givesFor $\alpha =-3$,soAll identities for this singular case hold on $r>0$. In particular, zero divergence here is a pointwise assertion away from the origin, not an assertion that the flux through a sphere enclosing the singularity is zero.

The Poisson bracket in canonical coordinates isUsing Hamilton's equations ${\dot{q}}_i=H_{p_i}$ and ${\dot{p}}_i=-H_{q_i}$ in the chain rule yieldsThe partial time derivative holds phase-space coordinates fixed; the total derivative follows the actual trajectory.

For fixed $u\in (0,1)$, $x=1-u\in (0,1)$ andAs $v$ runs from zero to one, $y$ decreases continuously from one to zero. Thus the map is onto the open unit square and is one-to-one, with smooth inverseIts denominator is positive in the open square. This is a rational square diffeomorphism. Its Jacobian determinant isWriting $A=1-(1-x)y$, the inverse relation gives $1-uv=x/A$. ConsequentlyThe second transformation also maps the open unit square bijectively onto the open unit square: for fixed $w$, $u=(1-t)/(1-wt)$ decreases strictly from one to zero as $t$ goes from zero to one, while $v=1-w$ independently spans $(0,1)$. Its positive Jacobian determinant isTo express this in $x,y$, the composite transformation gives $y=w/[x+(1-x)w]$ and $x=t(1-w)/(1-wt)$. Put $B=1-(1-x^2)y$. Solving these relations givesThe chain rule for Jacobian determinants therefore givesThe requested integrand is exactly the reciprocal of this positive determinant. Apply the change of variables formula using the composite diffeomorphism:Possible singular behavior at the square's boundary does not invalidate the calculation: first integrate over the images of compact interior squares, where all transformations are smooth with nonzero Jacobian determinant, and then increase these domains to the full square. Positivity and monotone convergence theorem justify this limit.