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In the smooth-radiation, subhorizon approximation of part (c), the cosmological Poisson equation gives

From $y^{\prime }=y\mathcal{H}$ and part (b),The expanding branch has $y^{\prime }>0$. Applying the chain rule yieldsSubstitute these expressions and divide by $C(1+y)>0$:This is the Mészáros equation for $y>0$; it keeps radiation in the expansion rate and matter in the slowly varying self-gravity source.

Assume a time-independent spherical relative potential $\psi (r)$, with acceleration $\nabla \psi$. The relative energy and specific angular momentum areAlong an orbit, $\dot{E}=\nabla \psi \cdot \mathbf{v}-\mathbf{v}\cdot \nabla \psi =0$ and $\dot{\mathbf{\ell }}=\mathbf{r}\times \nabla \psi =0$, since a spherical gradient is radial. Thus both $E$ and $L$ are integrals of motion. Choose the escape level $\psi =0$ and take the bound galactic distribution function to vanish for $E\le 0$.

For an isotropic $f(E)$, integrate over a spherical coordinate system in velocity space:Differentiating gives ${\rho }^{\prime }(\psi )=2\pi \sqrt{2}{\int }_0^{\psi }f(E)/\sqrt{\psi -E}dE$. Convolve once more with the reciprocal-square-root kernel. Reversing the order of integration, under the usual integrability assumptions, usesTherefore ${\int }_0^{E_0}{\rho }^{\prime }(\psi )/\sqrt{E_0-\psi }d\psi =2\sqrt{2}{\pi }^2{\int }_0^{E_0}f(E)dE$. Differentiating proves Eddington inversion:Here ${\pi }^2$ is outside the square root, as in the original PDF; the TeX transcription wrongly places it inside. The assumptions include a locally integrable bound distribution and sufficient density regularity, with no extra unbound or boundary population. An inverted expression must additionally be nonnegative to be physical. Since the velocity measure and $f$ are invariant under all velocity rotations, $⟨v_r^2⟩=⟨v_{\theta }^2⟩=⟨v_{\varphi }^2⟩=⟨v^2⟩/3$ whenever these moments exist; mixed moments vanish.

For the half-anisotropic distribution function, write the polar velocity angle from the radial direction as $\eta$. Then $L=rv\sin \eta$, andThe apparent $1/L$ singularity is integrable because it cancels the factor $v\sin \eta$ in the velocity measure. Integrating $\eta \in [0,\pi ]$ and $\phi \in [0,2\pi ]$ givesThe density relation requires $r\rho \to 0$ at the escape boundary and $f_E\ge 0$ for a physical distribution. Consider a monotonically decreasing $\psi$. The largest allowed radius at fixed binding energy is the radial-orbit limit $r_E$ with $\psi (r_E)=E$; a nonzero-$L$ orbit normally has a smaller apocenter because of its centrifugal term. Changing from $\psi$ to $r$ givesThe numerator in the original PDF instead uses $\rho +d\rho /d\psi$. That is a genuine printed error: the chain rule for $d(r\rho )/d\psi$ requires $\rho +rd\rho /dr$. It is not legitimate to prove the printed expression as written. A concrete counterexample is the unit-parameter Hernquist model at $r=1$: ${\psi }^{\prime }=-1/4$, ${\rho }^{\prime }=-(5/2)\rho$ and $d\rho /d\psi =10\rho$. The printed quotient is $-44\rho$, whereas the correct $g$ is $6\rho >0$. The former would give a negative distribution.

The angular weight after cancellation is uniform in $\eta$, so the averages of ${\cos }^2\eta$ and ${\sin }^2\eta$ are both $1/2$. The azimuthal angle splits the tangential term equally. Hence $⟨v_r^2⟩=⟨v_{\theta }^2+v_{\varphi }^2⟩$ and each individual tangential moment is half the radial one. The velocity-anisotropy parameter is consequentlyThis is a radially biased constant-anisotropy distribution function, not an isotropic one or purely radial motion. It does not by itself prove dynamical stability; it preferentially weights small angular momentum while remaining integrable.

For the Hernquist model, use the Poisson equation for Newtonian gravity with the relative-potential sign, ${\nabla }^2\psi =-4\pi G\rho$. Since ${\psi }^{\prime }=-GM/(r+a{)}^2$,Its enclosed mass is $M{r}^2/(r+a{)}^2$, which tends to $M$. Expressing the augmented density in the relative potential gives $r\rho =a{\psi }^3/(2\pi G^3{M}^2)$. The preceding derivative therefore giveswith zero bound-model population outside the allowed energy domain. It is nonnegative and reproduces the density by direct integration. The Hernquist model has a central density cusp, so the formulas for the local $1/L$ distribution are interpreted at $r>0$. As a further check, the radial second moment obtained from the same integral is $\psi /4$ and each tangential second moment is $\psi /8$, consistent with $\beta =1/2$.

The multiplication of a distribution by a smooth function is $⟨\psi T,\phi ⟩=⟨T,\psi \phi ⟩$. If $\psi$ is a Schwartz function, the Leibniz rule shows that $\phi \mapsto \psi \phi$ is a continuous map $\mathcal{S}\to \mathcal{S}$, so $\psi T$ is a tempered distribution. When both $\psi$ and $T$ are radial, $\psi \circ R^t=\psi$, andThus Multiplication by a radial Schwartz function preserves radial tempered distributions.

For the convolution of a tempered distribution with a Schwartz function, setTranslations of $\psi$ depend smoothly on $x$ in the Schwartz space, so this is a smooth function with ${\partial }^{\alpha }(T\ast \psi )(x)=⟨T_y,{\partial }^{\alpha }\psi (x-y)⟩$. The bound for $T$ by finitely many seminorms, together with $1+|y|\le (1+|x|)(1+|x-y|)$, provesThus the function also defines a tempered distribution. For a radial function $\psi$, put $g_x(y)=\psi (x-y)$. Then $\psi (Rx-y)=\psi (x-R^{t}y)=g_x(R^{t}y)$, soHence Convolution with a radial Schwartz function preserves radial tempered distributions.

Smoothing alone need not give a Schwartz function: for the constant tempered distribution $T=1$ and a Schwartz function with integral one, $T\ast \psi =1$. The radial Schwartz approximation of tempered distributions therefore combines smoothing with a large-radius cutoff. Choose a nonnegative radial mollifier $\rho \in C_c^{\infty }$, supported in the unit ball with integral one, and a radial cutoff function $\chi \in C_c^{\infty }$ equal to one on the unit ball. PutEach $f_j$ is a smooth function of compact support, hence a Schwartz function, and the two invariance calculations above make it radial.

It remains to prove convergence, including the simultaneous changes of both scales. With $\check{\rho }(x)=\rho (-x)$, the distributional convolution pairing isFor every fixed $m$, the Leibniz rule, rapid decay outside the radius-$j$ ball, and the chain rule for $\chi (x/j)$ giveConvolution by ${\check{\rho }}_{\epsilon }$ is uniformly bounded in $q_m$ for $0<\epsilon \le 1$, because its shifts have size at most one. The mean value theorem applied to $\phi (x-y)-\phi (x)$ similarly givesSplitting the error into the convolved cutoff error and the approximate identity error provesIf $|⟨T,\theta ⟩|\le C{q}_m(\theta )$, this yieldsConsequentlyweakly, and even in the strong dual topology, since $q_{m+1}$ is uniformly bounded on every subset of the Schwartz space that is a bounded set in a topological vector space. No assertion that $T\ast {\rho }_{1/j}$ itself is rapidly decreasing is needed.

Interpret the malformed printed domain inequality as the first quadrant $x>0,y>0$, as indicated by the two axis boundary conditions. There is also a genuine missing hypothesis: decay of the boundary data does not require the harmonic function itself to be bounded. For zero data, both $q=0$ and $q=xy$ satisfy the Laplace equation and both edge values, but the Wirtinger derivative of the latter isThus the data do not determine an arbitrary solution's derivative without a growth restriction. We derive the integral representation for the bounded solution, which is also the standard decaying solution when the continuous edge data decay. Any other admissible unbounded solution can differ by a zero-boundary harmonic function.

The conformal map $w=z^2$ sends the quadrant bijectively to the complex upper half-plane: the positive real axis maps to $s>0$, and the positive imaginary axis maps to $s<0$. Define its real-axis boundary function byThe compatibility at the corner makes $f$ continuous there. The Dirichlet Poisson integral in a quadrant follows from the upper-half-plane Poisson kernel:It is bounded by the supremum of the bounded edge data, and the kernel's approximate identity property supplies both Dirichlet boundary conditions. More explicitly, for $z^2=a+ic$, where $a=x^2-y^2$ and $c=2xy>0$, this integral isIt displays directly which part of the real-line boundary corresponds to each original axis.

To obtain its derivative, first take real data and use the Schwarz integral formula. A holomorphic function whose real part is the Poisson solution in the $w$-plane isThe regularizing subtraction is independent of $w$ and affects only the imaginary normalization; it permits bounded data without an artificial integrability assumption. Then $q=\mathrm{Re}F(z^2)$. By the chain rule and the Wirtinger derivative,The factor is $z$, not $2z$: taking the real part contributes the factor $1/2$ in the complex derivative. Differentiation under the integral sign is justified uniformly on compact interior subsets because the denominator stays away from the real boundary and decays quadratically at infinity. Splitting the real integral at zero and substituting $s=r^2$ or $s=-r^2$ gives the holomorphic derivative of a quadrant Dirichlet solution:This converges for bounded data, and in particular for the stated sufficiently decaying data. No interior pole is encountered since $\mathrm{Im}(z^2)=2xy>0$. For complex-valued data the same formula holds by applying the real-data construction separately to their real and imaginary parts and using linearity.

An equivalent form, often convenient when the data's derivatives decay, comes from integration by parts. Sincethe two endpoint terms are $-g_2(0)/z^2$ and $+g_1(0)/z^2$. They cancel by the specified compatibility, leavingThis is the harmonic quadrant solution from tangential boundary derivatives applied to the actual derivatives $g_1^{\prime },g_2^{\prime }$; the supplied $g_1,g_2$ in this question are boundary values, not derivatives. Omitting those primes would solve a different boundary problem.

For completeness, bounded Dirichlet uniqueness in a quadrant selects the solution just constructed. The difference of two bounded solutions is zero on both axes. Odd reflection across each axis gives a bounded harmonic extension to the punctured plane; the removable singularity for a bounded harmonic function fills the corner. The harmonic Liouville theorem makes the extension constant, and oddness makes it zero. Decaying data give a decaying Poisson extension: split the real-axis integral into a fixed compact interval and a tail with arbitrarily small data; the compact contribution tends to zero as $|z^2|\to \infty$, while the tail is bounded by that small supremum.

As literally printed without boundedness or a growth condition, the most one can assert isThe zero-boundary harmonic growth in a quadrant includes $v=cxy$ and more generally $v=\mathrm{Im}(z^{2m})$. This is an actual nonuniqueness of the printed problem, not a defect in the integral for the normalized bounded solution.

Insert an exact smooth solution of a differential equation and expand about $t=t_{n+2}$. The unscaled local truncation error, with every term moved to the left, isHere the constant, first-derivative, and second-derivative terms all cancel, using the chain rule identity $y^{\prime \prime }=f^{\prime }(y)f(y)$. Hence the formal defect convention $d_h=O(h^{p+1})$ givesAt $\alpha =1/7$, the coefficient of $h^4{y}^{(4)}$ is $1/21$, so there is no further cancellation. At any other $\alpha$, the cubic coefficient is nonzero.

There is a degeneracy at $\alpha =1$: both the first-derivative coefficient and ${\rho }^{\prime }(1)$ vanish. The raw defect still has the displayed formal order two, but it cannot be normalized by $(1-\alpha )h$ into an ordinary first-order consistency of a numerical method test. Thus a convention requiring this nondegenerate normalization would leave the ODE order undefined at $\alpha =1$. In either convention, it is not a convergent approximation to arbitrary initial-value data; the next part explains why.

Assume the holonomic constraints have linearly independent gradients, so their common level set is locally a smooth constraint manifold parametrized by $q_1,\dots ,q_n$. The multiplier form of the Euler-Lagrange equations isAt fixed time ${\sum }_A{\partial }_A{f}_{\alpha }{\partial }_i{x}_A=0$. The chain rule for the reduced Lagrangian givesConversely, if these reduced equations hold, the Cartesian Euler-Lagrange residual annihilates every tangent direction. Linear independence of the constraint gradients implies that it is a linear combination of those gradients, giving the required Lagrange multipliers. This proves the local equivalence, including time-dependent constraints; singular constraints would need separate analysis.

For the bead, $\dot{y}=f^{\prime }(x)\dot{x}$, soThe autonomous conservation of energy gives $E=\frac{1}{2}(1+f^{\prime 2}){\dot{x}}^2$. If $E>0$, the motion has a fixed sign $s=\mathrm{sgn}\dot{x}$ andThe integral is the signed arc length divided by the constant speed, and is locally invertible. The special solution $E=0$ is a stationary bead, for which writing time as a function of its unchanging position is not meaningful.

The chain rule gives $dF(x,y(x))/dx=F_x+F_y{y}^{\prime }$. Equality with $P+Q{y}^{\prime }$ for every $y(x)$ therefore requires $F_x=P$ and $F_y=Q$, which implies $P_y=Q_x$. Conversely, on the plane this compatibility condition makes $Pdx+Qdy$ an exact differential, so a potential $F$ exists.

Here $P=4x^3+3y$, $Q=2y+3x$, and $P_y=Q_x=3$. Integrating gives $F=x^4+3xy+y^2$, hence

Along a curve $s\mapsto (x(s),y(s))$, the chain rule givesTogether with the second-order partial differential equation, these form a linear system for $(u_{xx},u_{xy},u_{yy})$ whose coefficient matrix isA characteristic curve is precisely one along which this system is singular, equivalently where the principal symbol vanishes on a normal covector. Taking the determinant yields

Let $u(t)$ be the signed air speed along the straw towards the bubble. Conservation of mass in a pipe for the incompressible plug flow givesThe unsteady Bernoulli equation along the straw, whose open end is at atmospheric pressure, givesby the Young–Laplace equation. Substitution yields

Set $w=a^2\dot{a}$. Since $a\dot{w}=-C$, the chain rule gives $wdw/da=-Ca$. The initial condition $\dot{a}(0)=0$ impliesWriting $a=a_0\sin \theta$ and integrating gives the implicit solutionAt disappearance $a=0$ and $\theta =0$, so

Taking logarithms of the Cox partial likelihood givesDifferentiate each term, using the chain rule for the logarithm:This score function is the sum over failures of the observed covariate minus its current risk-weighted mean. Define that mean as ${\overline{z}}_{\beta }(x_i)$; observations with $v_i=0$ have no direct event term, although they occur in the relevant risk sets.

For one observation, let ${\ell }_i={\sum }_{c=0}^1{t}_{ic}\log p_{ic}$ and define ${\delta }_{ic}=t_{ic}-p_{ic}$. Differentiating the softmax function and its log-likelihood gives $\partial {\ell }_i/\partial o_{ic}={\delta }_{ic}$. The sigmoid-softmax network gradients areThese equations are backpropagation: use the chain rule through the output affine map, then ${\sigma }^{\prime }(u)=\sigma (u)(1-\sigma (u))$ through the hidden units, then the input affine map.

For a learning rate $\eta >0$, select one observation and update every parameter by $\theta \leftarrow \theta +\eta \nabla {\ell }_i(\theta )$, computing all entries of the gradient from the old parameters before applying the update. A uniformly selected observation gives an unbiased gradient of the average log-likelihood; multiplying by 3000 would instead estimate the gradient of the sum. Equivalently perform stochastic gradient descent on the negative log-likelihood. Batch size one makes one update per observation, and five epochs make five passes through the data, typically with a fresh random ordering on each pass.

Evaluate all derivatives at zero long perturbation, where $\nu ={\delta }_c/{\sigma }_{G,s}$, and write $\bar{n}(\nu )$ for the unperturbed abundance. The peak-background split changes the halo peak height toThe chain rule in the specified bias expansion therefore givesNo explicit abundance law is specified, so these derivatives are the general answer. For example, if $\bar{n}$ has the Press-Schechter halo mass function dependence $\nu e^{-{\nu }^2/2}$, then $b_{10}=({\nu }^2-1)/{\delta }_c$ and $b_{01}=2f_{\mathrm{NL}}({\nu }^2-1)$.

For the linear long mode, the cosmological Poisson equation gives ${\phi }_{G,L}(\mathbf{k})={\delta }_{G,L}(\mathbf{k})/\alpha (k)$. Consequently the deterministic galaxy bias and power spectrum areHere $P_m$ is the linear cosmological density power spectrum; stochastic tracer noise has been omitted. This scale-dependent halo bias from local non-Gaussianity is proportional to $k^{-2}$ in the specified $\alpha (k)=2k^2/(3H^2{\Omega }_m)$ convention.

For $b_{10}\ne 0$, plot the ratio $P_g/(b_{10}^2{P}_m)$ to separate the effect from the shape of $P_m$. Positive $f_{\mathrm{NL}}$ increases the ratio as $k$ decreases. Negative $f_{\mathrm{NL}}$ first suppresses it, gives a zero when $\alpha (k)=2|f_{\mathrm{NL}}|{\delta }_c$, and then makes it rise again after the effective galaxy bias changes sign. The auto-power spectrum never becomes negative. A truncation $P_g\simeq b_{10}^2[1+4f_{\mathrm{NL}}{\delta }_c/\alpha (k)]P_m$ is valid only when $|2f_{\mathrm{NL}}{\delta }_c/\alpha (k)|\ll 1$ and must not be extrapolated through the zero. Within the deterministic expression, both signs give $P_g\propto k^{-4}{P}_m$ at sufficiently small $k$. For a nearly scale-invariant primordial spectrum, $P_m\propto k^{n_s}$ in the large-scale transfer limit, so this formal asymptote is $k^{n_s-4}$.

There is a convention issue in calling the specified abundance response an observed galaxy overdensity. The threshold response is naturally a linear Lagrangian halo bias. If $b_{10}$ instead denotes the physical linear Eulerian halo bias, the Lagrangian-to-Eulerian linear bias relation adds one to the density coefficient: $b_{01}=2f_{\mathrm{NL}}{\delta }_c(b_{10}-1)$, and $P_g=[b_{10}+b_{01}/\alpha {]}^2{P}_m$. The boxed formulas follow the abundance expansion explicitly supplied in the paper. The Newtonian treatment also cannot be extrapolated beyond its physical large-scale domain without relativistic projection effects.

For an autonomous Lagrangian, the Euler-Lagrange equation usesApply the chain rule to the Hamiltonian along an arbitrary smooth path, without assuming the unforced Euler-Lagrange equation:Consequently the energy balance for an autonomous Lagrangian isWith explicit time dependence, an additional $-{\partial }_{t}L$ enters $dH/dt$. For ideal Gaussian white noise, the identity is understood through smooth-noise regularization or the Stratonovich chain rule. The original PDF correctly differentiates $L$ with respect to $\dot{x}$ in $H$; the supplied TeX's derivative with respect to $x$, its endpoint $t^2$, and its ordinary derivative of $A$ are transcription errors.

Along a trajectory, the chain rule givesThus $\boxed{H=x^2+y^2-y^4}$ is a first integral, and every trajectory lies on one of its level sets.

A map $f:A\to {\mathbb{R}}^m$ is differentiable at $p\in A$ when there is a linear map $Df(p):{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe linear map $Df(p)$ is the Fréchet derivative of $f$ at $p$.

The multivariable chain rule states that if $f$ is differentiable at $p$ and $g$ is differentiable at $f(p)$, thenIndeed, write $f(p+h)=f(p)+Ah+r(h)$ and $g(f(p)+k)=g(f(p))+Bk+s(k)$, where $r(h)=o(|h|)$ and $s(k)=o(|k|)$. Since $k=Ah+r(h)=O(|h|)$,and the last two terms are $o(|h|)$. This proves the formula.

For matrix inversion near the identity matrix,The last term is $O(\Vert h{\Vert }^2)$ because inversion is bounded near $I$. Therefore the inversion map is differentiable at $I$ and

For every invertible matrix $A$,The maps $l_C$ and $r_C$ are linear. Differentiate this identity at $A=I$ and use the chain rule and part (a):Thus $Df(C)(Ch)C=-h$. Given an arbitrary increment $k$, set $h=C^{-1}k$ to obtain the derivative of matrix inversionThis also proves differentiability at every $C\in V$.

Let $M(t_1,\dots ,t_n)=\mathrm{diag}(t_1,\dots ,t_n)-A$. Expanding the determinant along row $i$ shows thatwhere $C_{ii}$ is the corresponding cofactor. Since a diagonal entry of the adjugate matrix is that same cofactor, the multivariable chain rule along the diagonal $t_1=\cdots =t_n=t$ yieldsTherefore

In a holomorphic coordinate chart, $T^{1,0}X$ has frame $\partial /\partial z^1,\dots ,\partial /\partial z^n$. Under a holomorphic coordinate change $w=w(z)$, the chain rule givesThe Jacobian is an invertible matrix of holomorphic functions. These frames therefore make $T^{1,0}X$ the holomorphic tangent bundle.

Write ${\psi }^P(x)={\eta }_P{\gamma }^0\psi (x_P)$. The chain rule gives ${\partial }_0\psi (x_P)=({\partial }_0\psi )(x_P)$ and ${\partial }_i\psi (x_P)=-({\partial }_i\psi )(x_P)$. Using the gamma matrix relations $({\gamma }^0{)}^2=1$ and ${\gamma }^i{\gamma }^0=-{\gamma }^0{\gamma }^i$,Thus the Dirac equation is invariant under parity symmetry in quantum field theory.