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Let $f(X,Y)=X[r/(1+b{Y}^2)-\delta ]$. At the positive steady state, $1+b(N^{\ast }{)}^2=r/s$, where $s=\delta +1$. The two partial derivatives areThe first-order Taylor expansion with $N_t=N^{\ast }+n_t$ therefore givesEquivalently, the linearization on $(n_t,n_{t-1})$ has matrix $\left(\begin{array}{cc}1& -a\\ 1& 0\end{array}\right)$ and characteristic polynomial ${\mu }^2-\mu +a$.

For $x=h\nu /(k_{B}T)\ll 1$, the Taylor expansion $e^x-1=x+O(x^2)$ gives the Rayleigh-Jeans law for equilibrium radiation,This is the classical limit: each electromagnetic mode has mean thermal energy approximately $k_{B}T$. Extending it to arbitrarily large frequencies would give a divergent integrated energy density, the ultraviolet catastrophe.

For $x\gg 1$, $e^x-1\simeq e^x$, giving the Wien approximationPlanck's quantum hypothesis assigns equally spaced mode energies $E_n=nh\nu$. When the spacing is small relative to $k_{B}T$, many levels are occupied and the classical approximation works. When $h\nu \gg k_{B}T$, exciting even one quantum is exponentially unlikely, suppressing the high-frequency tail. The single expression $⟨E⟩=h\nu /(e^x-1)$ contains both limits. Comparing temperature directly to $h\nu$ would be dimensionally incorrect; the low-temperature condition means $k_{B}T\ll h\nu$.

Put $x=h\nu /(k_{B}T)=hc/(\lambda k_{B}T)$. In the long-wavelength limit $x\ll 1$, the Taylor expansion $e^x-1=x+O(x^2)$ in Planck's law gives the Rayleigh-Jeans law,This is the classical equipartition theorem result: each electromagnetic mode carries mean energy $k_{B}T$. Extending it to arbitrarily high frequency would make the total energy density diverge, the ultraviolet catastrophe.

In the short-wavelength limit $x\gg 1$, $e^x-1\simeq e^x$, giving the Wien approximation,Large photon energies are thermally suppressed, removing the classical high-frequency divergence.

If the requested “form” is expressed per unit wavelength, the change of variables includes its Jacobian determinant: $U_{\lambda }=U_{\nu }|d\nu /d\lambda |=U_{\nu }c/{\lambda }^2$. ThusSubstituting $\nu =c/\lambda$ without the Jacobian determinant would give an incorrect spectral density.

For fixed nonzero $l=p_{\varphi }$, the radial effective potential is $V_{\mathrm{eff}}(s)=l^2/(2m{s}^2{\sin }^2\theta )+mgs\cos \theta$. Its stationary point isThe equilibrium is therefore stable. A Taylor expansion with $x=s-s_0$ and $p_x=p_s$ givesQualification: the printed “any” $p_{\varphi }$ excludes a regular interior equilibrium when $p_{\varphi }=0$. In that case $V_{\mathrm{eff}}=mgs\cos \theta$ has no stationary point for $s>0$. Its minimum is the nonsmooth apex, where these coordinates and the smooth-cone dynamics fail.

Write $x=\Delta E/(k_{\mathrm{B}}T)$ and $r=N_{\uparrow }/N_{\downarrow }=e^{-x}$. Since $N=N_{\uparrow }+N_{\downarrow }$, eliminate the individual populations:The Taylor expansion $e^{-x}=1-x+O(x^2)$ gives the same result, and the odd expansion of the hyperbolic tangent yieldsEquivalently, $N_{\uparrow }\simeq (N/2)(1-x/2)$ and $N_{\downarrow }\simeq (N/2)(1+x/2)$. Their sum is still $N$. The negative sign follows directly from the stated higher-energy up-state population, and does not mean a negative magnitude of the equilibrium magnetization.

The canonical momentum is $p=\gamma m\dot{r}$, where $\gamma =(1-{\dot{r}}^2/c^2{)}^{-1/2}$. Eliminating velocity givesIts Hamilton equations are $\dot{r}=c^{2}p/H$ and $\dot{p}=0$. In the limit $|p|\ll mc$, the Taylor expansion is $H=m{c}^2+p^2/(2m)+O(p^4/(m^3{c}^2))$, and $\dot{r}\simeq p/m$. The additive rest energy does not change the classical free-particle equations.

For small displacement gradients write $F=I+H$. Taylor expansion around the stress-free reference state givesThe linear term vanishes because $DW(I)=0$. Drop the irrelevant constant and retain the quadratic term:The major symmetry $\boxed{C_{i\alpha j\beta }=C_{j\beta i\alpha }}$ follows from equality of mixed second derivatives. For a skew matrix $K$, stress covariance along $e^{tK}$ givesDifferentiation at zero yields $D\hat{T}(I)[K]=C[K]=0$. Thus $C_{i\alpha j\beta }{K}_{j\beta }=0$ for every skew matrix, which forces the input minor symmetry $C_{i\alpha j\beta }=C_{i\alpha \beta j}$. Combining it with the major symmetry gives the output minor symmetry as well:These identify reference and spatial indices at $F=I$. Decompose $H=e+K$ into symmetric and skew parts. Since $C[K]=0$ and the major symmetry also eliminates the other cross term,Thus infinitesimal rotations cost no energy in the quadratic approximation.

Availability and total entropy. The reservoir changes by $d{U}_R=-dU$ and $d{V}_R=-dV$. Its particle number and the system's particle number are fixed. The reservoir's first law of thermodynamics therefore givesAdd the system's entropy change. Because $T_R,p_R$ are constant,Thus define the availabilityAn additive constant in $A$ is immaterial; its value above the equilibrium reference is the exergy. The Second law of thermodynamics gives $\Delta S_{\mathrm{tot}}\ge 0$, hence $\Delta A\le 0$. A stable equilibrium maximizes total entropy over the allowed variations and therefore minimizes $A$. The environmental $T_R,p_R$ enter this definition even when the system has not yet reached their values. Because both thermal and mechanical exchanges are allowed, equilibrium here includes mechanical balance; equality of temperatures alone does not ensure the minimum.

Probability and curvature. For a fixed value of a macroscopic coordinate $x$, let $S(x)$ count the compatible system microstates. The product of the system and reservoir multiplicities isAll factors independent of $x$ enter a normalization constant, givingHere $A(x)$ is the constrained or effective availability for this coordinate, including the multiplicity of any eliminated variables. A nonuniform coordinate measure must likewise be retained or incorporated in that effective availability.

At a stable interior minimum, $A^{\prime }(x_0)=0$ and $A^{\prime \prime }(x_0)>0$. Its Taylor expansion givesKeeping the quadratic term yields a normal distribution,This is availability curvature and equilibrium fluctuations. The quoted fluctuation formula holds at quadratic order in the Gaussian fluctuation approximation; it is not an exact identity for an arbitrary nonlinear $A(x)$. Higher terms can shift the mean away from its most probable value $x_0$.

Bubble equilibrium. Keep the number of air molecules fixed and include the surface energy in the system's total internal energy. For constant surface tension $\Gamma$, this contribution is $\Gamma \Omega =\Gamma \alpha V^{2/3}$, with no surface entropy under the stated approximation. Its differential isHenceIndependent thermal and volume variations give $T=T_R$ andThe air pressure is evaluated at $V_0$; since the bubble is closed it generally depends on volume, so this equation is implicit until an air equation of state is supplied. For a sphere, $\alpha =(36\pi {)}^{1/3}$, $\Omega =4\pi R^2$, and the pressure jump becomes $2\Gamma /R$, the Young–Laplace equation for a single gas–liquid interface. There is no second surface as in a thin soap film.

Isothermal volume fluctuations. With the air maintained at $T_R$, eliminate its entropy in favor of its Helmholtz free energy:The identity $(\partial F_{\mathrm{air}}/\partial V{)}_{T,N}=-p_{\mathrm{air}}$ givesUsing the curvature result,These are the volume fluctuations of a closed gas bubble. Stability requires the denominator to be positive. The negative surface contribution reflects the concavity of $V^{2/3}$; gas compression supplies the positive restoring curvature. For ideal air, ${\kappa }_T^{-1}=p_{\mathrm{air}}$, and mechanical equilibrium reduces the denominator toThe units are consistent: $D$ is pressure, so $k_B{T}_R{V}_0/D$ has units ${\mathrm{m}}^6$, the units of a volume variance. Holding $p_{\mathrm{air}}$ externally fixed while varying $V$ would remove the gas-compression term and change this stability problem; that is not the fixed-particle-number situation here.

Numerical size of the fluctuations. The volume conversion is important:Use the specified $\alpha =4.84$, $\Gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $p_R=10^5\mathrm{Pa}$. ThenWith the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$,Thus the fractional root mean square fluctuation and fractional variance are respectivelyThe absolute root mean square fluctuation is $2.01\times 10^{-19}{\mathrm{m}}^3$. The very small fractional value supports the local Gaussian fluctuation approximation. This estimate neglects exchange of dissolved air, variations of surface tension, shape fluctuations and temperature fluctuations, consistently with the prescribed fixed-number, isothermal bubble model.

The fixed common temperature, rigid external container and fixed particle numbers on either side make the total Helmholtz free energy the appropriate thermodynamic potential. Let the barrier have cross-sectional area $A$, and let increasing $x$ expand the left compartment. Then $d{V}_1=Adx$ and $d{V}_2=-Adx$. At fixed temperature and particle number, $d{F}_j=-p_{j}d{V}_j$, soThus mechanical equilibrium, $p_1=p_2$, makes $F$ stationary. The physical force on the barrier towards increasing $x$ is $f=A(p_1-p_2)=-F^{\prime }(x)$. If the barrier is displaced slightly from $x_{\ast }$, strict mechanical stability requires a restoring force:ConsequentlyThe Taylor expansion $F(x_{\ast }+\delta x)=F(x_{\ast })+\frac{1}{2}{F}^{\prime \prime }(x_{\ast })\delta x^2+O(\delta x^3)$ proves that thermodynamic equilibrium is a strict local minimum of the Helmholtz free energy, rather than merely a stationary point.

For individually stable gases, the positive isothermal compressibilities ${\kappa }_{T,j}=-(1/V_j)(\partial V_j/\partial p_j{)}_{T,N_j}$ make the sign especially transparent:If these inequalities hold throughout the accessible interval, $F$ is a strictly convex function and an interior stationary point is the unique global minimum of the Helmholtz free energy. Stability at one point alone proves only the local statement. A vanishing restoring derivative describes marginal stability and requires examining higher-order terms.

Put $F(N)=f(N,1/2)$. A cobweb plot alternates a vertical move to the function graph $y=F(N)$ with a horizontal move to the diagonal $y=N$. Beginning at $N_0=3/2$, the first iterates areThe diagram shows the ensuing approach to $1/2$ with alternating deviations. More precisely, if $e_t=N_t-1/2$, Taylor expansion at the fixed point givesFor sufficiently small nonzero deviations the sign reverses each step and their magnitudes contract. The local approach is an alternating, geometrically damped oscillation, with asymptotic contraction factor $3/4$ per iterate.

In the first-order Taylor expansion correction it suffices to use the zeroth-order derivatives. They giveThe delayed product is consequentlyMultiplying by $r$ and substituting into the two transfer equations producesThe opposite signs of $f$ and $g$ are also required by conservation law of $S+I+R$: this correction changes the timing of the transfer between $S$ and $I$, while leaving the instantaneous death term unchanged.

Let $N_{\ast }>0$ be a fixed point and put $N_t=N_{\ast }+{\eta }_t$. Taylor expansion gives ${\eta }_{t+1}=g^{\prime }(N_{\ast }){\eta }_t+O({\eta }_t^2)$, so the linear perturbation is ${\eta }_t=g^{\prime }(N_{\ast }{)}^t{\eta }_0$. Thereforeis the condition for linear asymptotic stability. A multiplier in $(0,1)$ gives monotone decay; one in $(-1,0)$ gives alternating decay. If its magnitude exceeds one, perturbations grow. Magnitude one is nonhyperbolic and needs nonlinear analysis.

Strict concavity and $g(N_{\ast })/N_{\ast }=1$ already give $g^{\prime }(N_{\ast })<1$. Thus the remaining stability requirement is $g^{\prime }(N_{\ast })>-1$. If $g$ is additionally nonnegative on the entire half-line, strict concavity prevents its derivative becoming negative there, since it would then force $g$ eventually negative. Under that stronger global population-map assumption the positive fixed point is automatically stable. The printed local slope and concavity information, without that global nonnegativity requirement, does not by itself rule out a flip instability.

A uniform steady state satisfies $f(u^{\ast },v^{\ast })=g(u^{\ast },v^{\ast })=0$. Set $u=u^{\ast }+\tilde{u}$, $v=v^{\ast }+\tilde{v}$ and keep only first-order terms in the perturbations. Since the base state is constant, its Laplacian vanishes. Taylor expansion givesThe zero-flux boundary conditions linearize to homogeneous Neumann conditions, ${\partial }_n\tilde{u}={\partial }_n\tilde{v}=0$. The off-diagonal entries of the Jacobian describe the cross-effects of the two chemicals, while the diagonal entries describe local self-response.

Put $h=x-1$ and $k=y-1$. At $(1,1)$, the value is $5+4\alpha$, the gradient is $(5+4\alpha ,4+2\alpha )$, and the Hessian matrix is $\left(\begin{array}{cc}8& 2\alpha \\ 2\alpha & 2\end{array}\right)$. Thus the quadratic Taylor expansion isThe exact remainder is $h^3$, since the original polynomial has only the cubic term $x^3$ beyond degree two.

The requested partial derivatives areThe last printed derivative has a misplaced superscript on $f$; the second-derivative list and subsequent Taylor expansion use $f_{yy}$.

Suppose that newly infected members of a biological population pass through a noninfectious stage of fixed duration $\tau$ before becoming infectious. This delay is a latent period, whereas the incubation period ends at symptom onset. The two endpoints need not coincide.^[ref]

Let $b(t)=rI(t)S(t)$ be the rate of new infections in the specified compartmental model. Assume no death or other removal during the noninfectious stage. At an intervention time $T$, all infections acquired during the preceding $\tau$ time units are still pending entry into the infectious class, soFor a smooth recent incidence history, a Taylor expansion about $T$ givesThus the leading pipeline correction is proportional to the recent infection rate, rather than simply to the number currently infectious. A rapidly varying incidence history requires the integral instead of this short-delay expansion.

An ideal intervention stopping all new transmission at $T$ does not reverse these already acquired infections. Under the assumptions above, exactly $E(T)$ additional entries into the infectious class occur after $T$, over the next $\tau$ time units. They must be added when the reported cumulative count at $T$ records infectious-class entries. If that count already records infections at acquisition, the pending individuals are already included and must not be counted twice.

The definition of the susceptible individual compartment also matters: a genuinely uninfected susceptible class loses an individual at infection, while a model grouping not-yet-infectious individuals with susceptible ones postpones that recorded class transfer. The same biological delay therefore appears at different places in the bookkeeping. The assumption of perfect transmission removal is a model idealization rather than a clinical intervention rule.

Let $\zeta =g(z)$ be a conformal map, and let a point vortex of circulation $\Gamma$ lie at $z=c$. Suppose the mapped complex potential of a two-dimensional potential flow iswhere $R$ is a holomorphic function near $g(c)$ and represents the other point vortices, boundaries, and any imposed flow. The physical complex velocity is $w^{\prime }=g^{\prime }{W}^{\prime }$. The point vortex follows its regular velocity, found by removing $\Gamma /[2\pi i(z-c)]$ before taking the limit. A Taylor expansion proveswhich gives the title formula. The complex conjugate occurs because the complex velocity convention is $w^{\prime }=u-iv$, whereas $\dot{c}=u+iv$.

For a mapped upper half-plane with no imposed flow, the method of images gives $R(\zeta )=-\Gamma \log (\zeta -\overline{g(c)})/(2\pi i)$, henceThe first term is essential: excluding a mapped singularity and then transforming its regular velocity does not in general equal subtracting the physical singularity after transformation.

For a Poisson random variable $S_L$ of mean $L\lambda >0$ and $0<\beta <1$, the scaled cumulant-generating function is $\lambda L^{1-\beta }(e^{\theta L^{(\beta -1)/2}}-1-\theta L^{(\beta -1)/2})$. Its Taylor expansion tends to $\lambda {\theta }^2/2$ for every real $\theta$. The Gärtner–Ellis theorem applies because the limit is finite and differentiable everywhere. Its Legendre-Fenchel transform is the displayed quadratic good rate function.

With ${\mu }_i=e^{\beta x_i}$, define $U={\sum }_i{x}_i(Y_i-{\mu }_i)$ and $I={\sum }_i{x}_i^2{\mu }_i$. The estimating equation and a Taylor expansion give $\hat{\beta }-{\beta }_0\approx I({\beta }_0{)}^{-1}U({\beta }_0)$. If responses are independent with correct means and $\mathrm{Var}(Y_i)=\varphi {\mu }_i$, then $EU=0$ and $\mathrm{Var}U=\varphi I$. Consequently the first-order estimator variance is $\varphi /I$, even if the working Poisson distribution is wrong. Regular asymptotics require growing information and control of leverage and the Taylor remainder.

Quadratic energy loss from centred Gaussian gate timing error is the leading decrease in collected energy when a symmetric gate is displaced slightly from the centre of a Gaussian pulse. At the optimum the first-order change vanishes, but the second-order change is negative. The Gaussian pulse time-gate energy fraction gives the exact displaced-gate integral; here the local loss and its effect on jitter are derived explicitly.

Let $p(t)=e^{-t^2/(2{\sigma }^2)}/(\sqrt{2\pi }\sigma )$ be the centred energy-normalized pulse profile, let $a=T_g/2>0$, and setThe Leibniz integral rule gives $F^{\prime }(d)=p(d+a)-p(d-a)$. Since $p$ is even and $p^{\prime }(t)=-tp(t)/{\sigma }^2$,The Taylor expansion therefore yieldsThis is an absolute fraction of the full pulse energy. The relative loss compared with the centred gate is instead $[F(0)-F(d)]/F(0)$.

For a gate as wide as the power FWHM $\tau =2\sqrt{2\ln 2}\sigma$, $F(0)=0.760968$ and $C{\tau }^2=1.302337$. An offset $d=0.05\tau$ gives leading loss $0.00325584$, or about $0.428$ percent of the centred gate's collected energy. Neither the full pulse energy nor the gate width has changed.

A zero mean offset does not eliminate this loss. For a sufficiently narrow random offset $D$ with mean ${\mu }_D$ and variance ${\sigma }_D^2$, the leading expectation isIf $D$ has a centred normal distribution, the quadratic model also gives $\mathrm{SD}(F(D))\simeq \sqrt{2}C{\sigma }_D^2$. At ${\sigma }_D=0.05\tau$, the leading mean loss is $0.00325584$ and this standard deviation is $0.00460446$. These formulas neglect fourth-order effects; they are local approximations, not bounds for large timing excursions.

An apparatus check scans the gate delay about its optimum and compares the measured decrease with the model, while monitoring gate width and the pulse. A gate on the detector's electrical output need not act directly on optical power: the detector impulse response and readout convolution must be included. NIST's impulse-response measurements illustrate that distinction.^[ref]