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Let $h=\Delta x$, $k=\Delta t$ and $\mu =k/h$. For the exact solution $u(x,t)=F(x+t)$ of the advection equation, the three left-hand samples are translations by $(\mu -1)h$, $\mu h$ and $(\mu +1)h$ relative to $(x_m,t_n)$; the right-hand samples are translations by zero and $h$.

Denote their coefficients by $a=\mu (1+\mu )/6$, $b=(2-\mu )(1+\mu )/3$, $c=(2-\mu )(1-\mu )/6$, $d=(2-\mu )/3$ and $e=(1+\mu )/3$. Taylor expansion compares momentswhere $0^0=1$ in the zeroth moment. Direct calculation givesThus the unscaled exact-solution residual isFor fixed nonzero $\mu$, dividing by the time step yieldsThe scheme is therefore third order under fixed-Courant refinement, with an unscaled one-step defect of order four. Stability then gives third-order accumulated error for smooth data with periodic boundary conditions. Calling the unscaled defect fourth order is a different convention, not fourth-order approximation of the differential equation.

At $\mu =-1,1,2$, the amplification factor in (b) becomes exactly $e^{i\mu \theta }$: the numerical step translates the grid data by that integer number of cells and is exact for this constant-speed advection equation. At $\mu =0$ it is the identity, a zero-time-step limit. These are exact special shifts, rather than merely a cancellation of one term of the Taylor expansion. This is the implicit advection scheme with exact integer shifts.

For a sufficiently smooth function, Taylor expansion of the four neighboring values givesThe reaction term $\lambda u$ is sampled exactly at the grid point. On an exact solution, therefore,The method has second-order local accuracy for the partial differential equation. The original printed equations multiply this expression by $h^2$, so their unscaled residual is $O(h^4)$; that scaling does not make the Laplacian approximation fourth order. Resonance and inverse-matrix bounds are separate issues from this local truncation calculation.

Let $S_{1}u$ be the sum of the four axial neighbors and $S_{2}u$ the sum of the four diagonal neighbors. The two PDF stencils defineThe reaction stencil has four axial weights $1/12$ and center weight $2/3$, so its weights sum to one. Taylor expansion givesFor example, the diagonal neighbors supply the $u_{xxyy}$ term needed to turn the fourth derivatives into the full squared Laplacian. Adding the reaction contribution producesOn a smooth solution of the constant-coefficient equation, both displayed leading terms vanish. Thus the compact fourth-order Helmholtz stencil hasThe cancellation uses the differential equation and constant $\lambda$; the nine-point finite-difference stencil alone still has an $O(h^2)$ operator error on an arbitrary smooth function. It also does not establish a uniform global fourth-order error without an appropriate inverse bound and compatible treatment of boundaries and resonances.

A fixed-step linear multistep method for $y^{\prime }=f(t,y)$ uses several previous solution and derivative values:The coefficients are fixed independently of the step size. It is explicit when ${\beta }_s=0$ and implicit otherwise. Its characteristic polynomials of a linear multistep method are $\rho (w)={\sum }_j{\alpha }_j{w}^j$ and $\sigma (w)={\sum }_j{\beta }_j{w}^j$. The contraction mapping theorem gives a uniquely solvable implicit step under a Lipschitz continuity bound $L$ whenever $k|{\beta }_s|L<|{\alpha }_s|$, by contraction of the new-value equation on the appropriate solution region. A suitable starting procedure must supply the first $s$ values; the method alone does not determine them.

Order is defined by inserting a smooth exact solution into the recurrence. If the residual is $O(k^{p+1})$ and its first nonzero coefficient occurs at that degree, the method has order $p$; dividing the residual by $k$ gives an $O(k^p)$ local truncation error. Taylor expansion gives the exact order conditions for a linear multistep methodwith the right side interpreted as zero at $q=0$, and failure at $q=p+1$ for exact order. Equivalently,The basic consistency of a numerical method conditions are $\rho (1)=0$, ${\rho }^{\prime }(1)=\sigma (1)$; for the ordinary nondegenerate consistent formulation this common value is nonzero. Exactness on constants and linear functions does not control propagation of numerical perturbations.

That propagation is the role of zero-stability. On $y^{\prime }=0$ the homogeneous recurrence is $\rho (E)y_n=0$. Its modes are $n^r{\xi }^n$, with powers of $n$ determined by root multiplicities. The theorem on the root condition for a multistep method states that zero-stability is equivalent to every root of $\rho$ having modulus at most one, with every unit-modulus root simple. Interior repeated roots are permitted, since geometric decay dominates their polynomial factors. An exterior root amplifies perturbations geometrically; a repeated unit root amplifies them polynomially and violates a step-independent perturbation bound.

The Dahlquist equivalence theorem states that a fixed-step consistent linear multistep method is convergent exactly when it is zero-stable. This concerns ordinary differential equations with a uniformly Lipschitz continuous right-hand side on the solution region over a fixed finite interval, convergent starting values, and the well-defined nearby branch of each small-step implicit solve. For order $p$ convergence of a numerical method, the solution must have the required smoothness and the starting errors must be $O(k^p)$.

The mechanism can be sketched explicitly. The root condition for a multistep method bounds the recurrence's discrete Green function. For an order-$p$ exact-solution defect $d_n=O(k^{p+1})$, the Lipschitz continuity error equation then gives an estimate of the formThe current-step term can be absorbed for small $k$. The discrete Gronwall inequality says that a bound $E_n\le A+Bk{\sum }_{j<n}{E}_j$ implies $E_n\le A(1+Bk{)}^n\le A{e}^{B{t}_n}$. There are $O(k^{-1})$ defects on a fixed interval, so their sum is $O(k^p)$ and this yields the claimed global order. Conversely, testing $y^{\prime }=0$ with vanishing starting perturbations in exterior-root or repeated-unit-root modes proves that convergence of a numerical method fails without the root condition for a multistep method.

For example, the two-step Adams-Bashforth method hasIt is explicit, order two and zero-stable, hence convergent with order-two starting values. By contrast, a recurrence with $\rho =(w-1{)}^2$ has solutions $y_n=y_0+n(y_1-y_0)$ on the zero differential equation. Starting differences $\sqrt{k}$ tend to zero yet become unbounded at a fixed positive time, since $n$ is of order $1/k$. A small local defect or formal high order cannot rescue this mode. Canceling a common factor may remove it only by changing the allowed starting relations, as the parameter endpoint in Question 2 illustrates.

Absolute stability addresses a different limit: apply the method to the Dahlquist test equation $y^{\prime }=\lambda y$ and put $z=k\lambda$. The amplification roots satisfyAll roots must satisfy the disk condition, with unit roots simple; one cannot inspect only the root that tends to one as $z\to 0$. The degree and solvability of the recurrence must also be preserved. The resulting set of $z$ is the domain of absolute stability. A-stability means that it includes the whole closed left half-plane, so the numerical solutions of every Dahlquist test equation with $\mathrm{Re}\lambda \le 0$ are stable for every positive step size. It is a linear test-equation property, not a proof of arbitrary nonlinear contractivity or high accuracy at large steps.

Forward Euler method has order one and factor $1+z$, so its stability region is the disk $|1+z|\le 1$. Backward Euler method has factor $(1-z{)}^{-1}$ and is A-stable. The trapezoidal rule is order two with factor $(1+z/2)/(1-z/2)$; its modulus is at most one throughout the left half-plane. These one-step methods are included as the simplest multistep examples. For the two-step Adams-Bashforth method, the negative-real stability interval is $-1\le z\le 0$: its polynomial is $w^2-(1+3z/2)w+z/2$, and the root reaches $-1$ at $z=-1$. It therefore imposes a step restriction when applied to strongly decaying modes.

The genuinely two-step BDF2 method isIt has order two, roots $1,1/3$ at zero step and is A-stable. For its normalized polynomials $\rho =3w^2/2-2w+1/2$, $\sigma =w^2$, the unit-circle boundary locus hasThe leading coefficient has no zero in the left half-plane, and the roots at a small negative $z$ are both inside the disk. Root continuity then proves open-half-plane stability; on the imaginary boundary the only possible unit root is the simple root one at $z=0$. This gives a direct proof of the stated A-stability rather than relying on a boundary plot.

Two sharp order barriers explain the tradeoff. The first Dahlquist barrier states that a consistent zero-stable real linear $s$-step method using only first derivatives has order at most $s+1$ for odd $s$ and $s+2$ for even $s$; an explicit method has order at most $s$. The Second Dahlquist barrier states that an irreducible A-stable real linear multistep method has order at most two. These statements concern fixed constant coefficients and the ordinary first-derivative multistep class, not multiderivative methods or arbitrary variable-step formulations. Thus higher order alone is not a route to unconditional stiff stability. Question 2 provides a concrete illustration: its third-order member is zero-stable but not A-stable.

No nontrivial consistent explicit multistep method is A-stable. Its amplification polynomial of a multistep method has a fixed highest coefficient, while at least one lower coefficient grows without bound as real $z\to -\infty$. Were all roots uniformly in the unit disk, their elementary symmetric polynomials would keep all the monic coefficients bounded, a contradiction. This is the coefficient argument behind explicit multistep methods cannot be A-stable.

In practice, selecting a method therefore involves accuracy, startup, the presence of a stiff differential equation, implicit-solve cost and step-size control. Zero-stability handles propagation as the step tends to zero; consistency of a numerical method supplies a vanishing defect; A-stability handles the numerical solutions of all decaying Dahlquist test equations at finite steps. None substitutes for the others, and variable steps or PDE discretizations require additional uniform stability arguments appropriate to their setting.

Let $d=U^{\prime }(0)>0$, $h=U^{\prime \prime \prime }(0)$, $p={\varphi }_0(0)\ne 0$, and $q={\varphi }_0^{\prime }(0)$. At a regular neutral Rayleigh equation for inviscid shear flow mode, multiply the equation by $U-c_0$ and evaluate it at the critical layer in a shear flow. The term multiplying ${\varphi }_0^{\prime \prime }-k_0^2{\varphi }_0$ vanishes, leaving $U^{\prime \prime }(0)p=0$. Therefore $U^{\prime \prime }(0)=0$. If this condition failed, the coefficient $U^{\prime \prime }/(U-c_0)$ would have a simple pole and a nonzero $p$ would force a logarithmically singular slope instead of a regular eigenfunction.

There is a sign error in the printed first-order equation. Direct expansion of the original Rayleigh equation for inviscid shear flow givesThe coefficient on the left must have a minus sign. Indeed, $1/(U-c)=1/(U-c_0)+\epsilon c_1/(U-c_0{)}^2+\cdots$, and moving both first-order parameter corrections to the right gives exactly this equation. The plus sign printed there is incompatible with the original equation and its neutral eigenfunction. The later matching and integral formulas follow from the corrected equation. The PDF also has $y^{3}h/6$ in the Taylor expansion of $U$, and $U^{\prime \prime \prime }(0)$ in the requested logarithmic coefficient; these are mistranscribed in the converted TeX.

Near zero, $U-c_0=dy+h{y}^3/6+\cdots$ and $U^{\prime \prime }=hy+O(y^2)$. Thus $U^{\prime \prime }/(U-c_0)\to h/d$, whereas $U^{\prime \prime }/(U-c_0{)}^2=h/(d^{2}y)+O(1)$. The singular part of the first-order equation isIntegrating twice on either side givesThe constant $a$ is common because the inner solution is continuous. The coefficient necessarily also contains $p$: rescaling the neutral eigenfunction rescales its correction. One may choose the normalization $p=1$ if a formula only in terms of the flow derivatives and $c_1$ is wanted.

The outer expansion of $1/(U-c)$ fails when $|dy|$ is comparable with $\epsilon |c_1|$. Therefore the distinguished limit isIn this inner variable, the original equation becomesAt orders one and $\epsilon$, its right side is zero. Matching the outer expansion therefore selects ${\Phi }_0=p$ and ${\Phi }_1=q\eta +a$. At order ${\epsilon }^2\log \epsilon$, ${\theta }_2^{\prime \prime }=0$. The outer term $\epsilon \beta y\log |y|$ becomes ${\epsilon }^2\beta \eta (\log \epsilon +\log |\eta |)$, so the required switchback term is ${\theta }_2=\beta \eta$. At order ${\epsilon }^2$,Integrating, or using the supplied hint, givesThe constants $B,C$ are homogeneous matching freedoms; they do not affect the slope jump. The quadratic term agrees with ${\varphi }_0^{\prime \prime }(0)=p(k_0^2+h/d)$, another check on the sign of $L_0$.

For $\mathrm{Im}{c}_1>0$, $d\eta -c_1$ runs below the real axis. Choose the continuous branch of the complex logarithm there. At positive infinity its argument tends to zero; at negative infinity it tends to $-\pi$. Thus the linear parts of the two large-$|\eta |$ expansions of ${\Phi }_2$ areThese expressions omit the already matched common quadratic term and terms only logarithmic or constant in $\eta$. Matching the linear coefficients to $b_{\pm }$ proves the neutral-mode critical-layer matching relationThe sign comes from passing below the pole; taking $\mathrm{Im}{c}_1<0$ would reverse it.

Finally use the Wronskian $W={\varphi }_1{\varphi }_0^{\prime }-{\varphi }_0{\varphi }_1^{\prime }$. Since $L_0{\varphi }_0=0$ and $L_0{\varphi }_1=H{\varphi }_0$, subtraction gives $W^{\prime }=-H{\varphi }_0^2$. The wall boundary conditions give $W(a)=W(b)=0$, soThis is the requested excised-interval identity. As $\delta \downarrow 0$, the equal logarithmic slope terms cancel andThe remaining real singular integral is interpreted as a Cauchy principal value, consistent with excision on both sides of the critical layer in a shear flow. DefineThen the neutral Rayleigh-mode dispersion correction iswhere the last form assumes a nonzero denominator. For real $k_1$, $\mathrm{Im}{c}_1=2k_0{k}_{1}I\pi M/(J^2+{\pi }^2{M}^2)$, so the assumed growing branch must have the corresponding sign. If $h=0$, the displayed logarithmic coefficient and jump vanish; further degeneracy can require higher-order analysis. No nonzero growth rate follows from that degenerate first-order calculation alone.

Let $p_j^n$ be the probability of occupying site $ja$ after $n$ steps, with $p_j^0=1_{\{j=0\}}$ and $\alpha +\beta =1$. The master equation isWrite $p_j^n\approx ap(ja,n\tau )$ so that the limiting $p$ is a probability density. A Taylor expansion of this master equation givesThe weakly biased diffusion limit of a weakly biased random walk keeps $a^2/\tau$ and $(\alpha -\beta )a/\tau$ finite as $a,\tau \to 0$, so $\alpha -\beta =O(a)$. The limiting advection-diffusion equation isIts initial condition is the Dirac delta function at the origin. At finite spacing, the exact variance gained per step is $a^2[1-(\alpha -\beta {)}^2]=4a^2\alpha \beta$. Thus a finite-step central limit theorem gives the variance-matched coefficient $2a^2\alpha \beta /\tau$; its difference from the displayed $D$ vanishes in the stated weak-bias diffusion limit of a weakly biased random walk. Keeping a fixed nonzero bias while taking diffusive space-time scaling would instead make $V$ diverge.

At a critical point $p$ of a twice continuously differentiable function, the linear term in the Taylor expansion vanishes:The real symmetric Hessian matrix has an orthonormal basis of eigenvectors. If all its eigenvalues are positive, its quadratic term is bounded below by a positive multiple of $\Vert h{\Vert }^2$, giving a strict local minimum. If all are negative, it gives a strict local maximum. If positive and negative eigenvalues both occur, directions along their eigenvectors give values on both sides of $f(p)$, so the point is a saddle point of a scalar function. The non-singularity assumption excludes zero eigenvalues; with them the quadratic test may be inconclusive and higher-order terms are needed.

Under the usual regular, identifiable, fixed-dimensional setting, the maximum-likelihood estimator is an interior point with vanishing log-likelihood gradient, and the observed information is positive definite. DefineThe second-order Taylor expansion of the Bayesian deviance isWith a prior density approximately constant over the likelihood's concentration region, the posterior density is therefore proportional toRecognizing the multivariate normal density givesTypically $J$ grows at rate $n$ and the posterior concentrates in a region of radius $n^{-1/2}$, making higher-order terms negligible under the regularity assumptions. The local flatness assumption concerns this region, not necessarily the entire parameter space. Boundary maxima, unidentified directions, or persistent multiple modes can invalidate this approximation.

The least-squares estimator minimizes $(Y-Xb{)}^T(Y-Xb)$. Differentiation gives $X^{T}X\hat{\beta }=X^{T}Y$, and the full column rank of $X$ makes $X^{T}X$ positive definite. HenceThe fitted values are $\hat{Y}=X\hat{\beta }=HY$, where the hat matrix is $H=X(X^{T}X{)}^{-1}{X}^T$. The regression residuals are $\hat{\epsilon }=Y-\hat{Y}=(I-H)Y$. Since $H$ is the orthogonal projection matrix onto the column space of $X$, $(I-H)X=0$ andThis is a singular multivariate normal distribution supported on the $(n-p)$-dimensional orthogonal complement of the column space: its components are generally correlated. The equality $\hat{Y}=X\hat{\beta }$ puts the fitted vector in that column space, while $X^T\hat{\epsilon }=0$ expresses fitted-residual orthogonality. No residual component lies in an explanatory direction that could improve the least-squares fit.

For the calibration model, write $R_i$ for a reading and $a_i$ for the known amount. The first fit assumes $R_i=\alpha +\beta a_i+{\epsilon }_i$ with independent $N(0,{\sigma }^2)$ errors. Its residual-versus-fitted plot has systematic changes in group means and an increasing within-group spread. These suggest both a nonlinear mean relationship and heteroscedasticity, rather than the random scatter expected from a satisfactory normal linear model.

The Box–Cox transformation plot is a profile log-likelihood for the power parameter $\lambda$. Its maximum is near $0.94$; the horizontal line gives an approximate 95% likelihood-ratio confidence interval, which excludes the untransformed value $1$. Since affine rescaling of the response does not change the fitted mean space, fitting the raw power $R_i^{0.94}$ is equivalent to using $(R_i^{0.94}-1)/0.94$ with an intercept. The second model isIts residual spread is more even, but observation 17 has a conspicuously negative residual near $-9.95$. This motivates checking for an outlier or recording error. The third fit omits that observation; omission should be justified by that investigation, since choosing exclusions from residuals can affect inference. Some group pattern still remains, so the transformation alone does not certify a perfect model.

At an amount of $3$ nanograms, the third fit predicts the transformed meanSimply applying the inverse power gives ${\hat{m}}^{1/0.94}=118.5187$. The requested expected reading needs bias correction after an inverse transformation, since $\mathbb{E}[g(Z)]\ne g(\mathbb{E}[Z])$ for nonlinear $g$. With $q=1/0.94$ and $s=2.832$, a second-order Taylor expansion givesHere the correction uses the residual variance, rather than the variance of the estimated fitted mean. The power-transformed Gaussian model is an approximation for positive readings; the fitted mean is over 31 residual standard deviations above zero, so the negative tail has negligible effect on this local approximation. The ordinary inverse-transformed prediction is therefore essentially $118.52$ at the displayed precision.

Put $h=\Delta x$. The spatial operator is the sum of a centred $x$ difference and a dissipative second-order forward advection stencil in $y$. For a smooth function, their Taylor expansions giveThus the leading spatial local truncation error is $h^2(u_{xxx}/6-u_{yyy}/3)$, generally nonzero. The semidiscretization is second order in space. Time has not yet been discretized, so there is no separate time order or time-step condition at this stage.

The tableau as printed is not a collocation tableau. Its second stage row is $(3/4,3/4)$, whose sum is $3/2$, while the corresponding printed node is $1$. This violates the collocation tableau row-sum identity. The original PDF confirms this coefficient; it is not only a TeX transcription defect.

The required integrated basis at nodes $1/3,1$ isso direct integration givesThus replacing the last stage coefficient by $1/4$ gives the intended two-stage Radau IIA collocation method. Its quadrature moments areIt integrates quadratics exactly but not cubics, so the collocation order theorem gives order three. Independently, $b^{T}Ac=1/6$, confirming the remaining third-order Butcher order condition.

For completeness the literal printed Runge-Kutta method has only order one. For the autonomous equation $y^{\prime }=y$, the Taylor expansion of its stability function beginswhereas $e^z=1+z+z^2/2+O(z^3)$. Since $b^{T}1=1$, it is consistent, but this discrepancy already rules out order two. Therefore neither the requested equivalence nor the third-order conclusion is true without the stated coefficient repair.

Take $e^{-i\omega t}$ time dependence and measure angles from the upward normal. Write $p_i=k\sin {\theta }_i$, $q_i=k\cos {\theta }_i>0$ and choose incident amplitude $A_i$. Separate the flat Dirichlet reflection from the rough correction:Taylor expansion of the boundary condition about $z=0$ givesSince ${\partial }_z{\psi }^{[0]}(x,0)=-2i{q}_i{A}_i{e}^{i{p}_{i}x}$, the induced trace isUse $\hat{h}(Q)=\int h(x)e^{-iQx}dx$ and $\beta (\xi )=\sqrt{k^2-{\xi }^2}$ with the outgoing branch: $\beta >0$ for $|\xi |<k$ and $\mathrm{Im}\beta >0$ outside. The outgoing angular spectrum solves the Helmholtz equation with that trace, giving the first-order rough-surface scattered fieldIf “scattered field” includes the specular reflection, add $-A_i{e}^{i{p}_{i}x+i{q}_{i}z}$ to this rough correction. Fourier transforms of an infinite random surface are generalized objects; finite-window transforms give an ordinary integral and are also needed for its far-field intensity. The expansion assumes a fixed regular surface shape scaled to small height. Highly oscillatory profiles can require additional control of slopes and spectral moments; small height alone is not a uniform error estimate for every profile.

Take the boundary-fitted grid $h=1/(M+1)$ and assume the standard consistency of a numerical method hypothesis $u\in C^4(\overline{\Omega })$ with bounded fourth partial derivatives. Taylor expansion along each coordinate, including a bounded fourth-order remainder, gives the five-point Laplacian truncation errorThe numerical equation and exact grid values therefore imply $A_h\mathbf{e}=\mathbf{\tau }$, where $A_h=-{\Delta }_h$ with zero boundary values. Its discrete sine transform eigenvectors are $\sin (\pi rih)\sin (\pi sjh)$, $1\le r,s\le M$, and its eigenvalues areThe one-dimensional sine vectors form an orthogonal basis, hence so do their tensor products. The least eigenvalue occurs at $r=s=1$. Since $\sin (\pi h/2)\ge h$ for $0<h\le 1$, ${\lambda }_{11}\ge 8$, so the Euclidean operator norm of $A_h^{-1}$ is at most $1/8$. There are $M^2$ residual components, and $Mh<1$, givingThe factor $M$ matters: pointwise second-order consistency of a numerical method becomes first order in the unweighted Euclidean grid norm.

The PDF does not explicitly impose regularity on the exact solution. Without such a hypothesis its unrestricted error claim is false. Here is a counterexample, with a classical solution smooth in the open square and continuous with zero boundary values:The sampled numerical solution separates as $v_i\sin (\pi jh)$. Its error $e_i=v_i-g(ih)$ solves $Ae=\tau$, where $A=-{\Delta }_{h,x}+{\lambda }_{y}I$ and ${\lambda }_y=4h^{-2}{\sin }^2(\pi h/2)$. The one-dimensional residual is ${\tau }_i={\Delta }_{h,x}g(ih)-g^{\prime \prime }(ih)+({\pi }^2-{\lambda }_y)g(ih)$. At the first node,The leading coefficient is strictly negative. Moreover $g^{\prime \prime \prime \prime }<0$ on $(0,1)$, so the triangular-average formula for the second difference and concavity of $g^{\prime \prime }$ give ${\Delta }_{h,x}g\le g^{\prime \prime }$. Thus the positive part of every residual is at most $C{h}^2$. The matrix $A$ has a nonnegative inverse, and its first diagonal inverse entry is at least $1/(2h^{-2}+{\lambda }_y)$, as follows from the nonnegative Neumann expansion of its tridiagonal matrix. The discrete comparison principle bounds the inverse image of the residual's positive part by $C{h}^2$, while the negative first residual gives $e_1\le -c\sqrt{h}+C{h}^2$. Hence $|e_1|\ge c^{\prime }\sqrt{h}$ for small $h$. Since ${\sum }_{j=1}^M{\sin }^2(\pi jh)=1/(2h)$, the full Euclidean error is bounded below by a positive constant. It cannot be $O(h)$. Thus the boxed estimate holds for the intended smooth solution class; a regularity condition is genuinely necessary.

Put $\tau =\Delta t$ and $h=\Delta x$, with $h^2=\tau /\mu$ for fixed $\mu >0$. Substitute a smooth exact solution of the heat equation into the Adams-Bashforth method. Define its one-step defect $R$ as the left side minus the right side of the unscaled update. The centered finite difference satisfiesThe Taylor expansion in time givesSince $u_t=u_{xx}$ and $u_{tt}=u_{xxt}$, the first two terms cancel, leavingThis is the requested local error of the unscaled update. If “local truncation error” denotes the defect divided by $\Delta t$, it is instead $O(\Delta t)$ under fixed $\mu$: the centered spatial error is then $O(h^2)=O(\Delta t)$. The scheme is second order in time at fixed spatial discretization, which is a separate order statement.

Let $S=X+Y$ and $D=X-Y$. Since they are independent, applying the product rule for moment-generating functions to $S+D=2X$ givesAll factors are finite and strictly positive for every real $t$ by the stated hypothesis. Applying the same identity to $-t$ and dividing yields $\psi (2t)=\psi (t{)}^2$, orThe centered unit-variance Taylor expansion from part (ii) gives $M(h)=1+h^2/2+o(h^2)$ and the same expansion for $M(-h)$. Consequently $\psi (h)=1+o(h^2)$.

For the first step of dyadic rigidity of a moment-generating function, take logarithms, which are allowed because $\psi >0$. For each fixed $t$ and every integer $m$,Since $\log \psi (h)=o(h^2)$, the right side tends to zero: it is $2^{m}o(t^2/4^m)$. Thus $\psi (t)=1$ for every $t$, and the moment-generating function is even.

The original identity now becomes $M(2t)=M(t{)}^4$. Iterating this time with the correct fourth-power scaling givesNear zero, $\log M(h)=h^2/2+o(h^2)$, so the right side tends to $t^2/2$. ThereforeThis is the moment-generating function of a standard normal variable. By the uniqueness theorem for moment-generating functions, each of $X$ and $Y$ has the standard normal distribution. The proof establishes the Gaussian characterization by independent sum and difference directly from the two functional identities and the first two moments.

Finiteness of the moment-generating function on an open neighborhood of zero gives its differentiability there. Using the permitted interchange of differentiation and expectation,Hence $M_X(0)=1$, $M_X^{\prime }(0)=\mu$ and $M_X^{\prime \prime }(0)=s^2$. The second-order Taylor expansion with its Peano remainder givesHere $s^2$ is the raw second moment, not the variance: $\mathrm{Var}(X)=s^2-{\mu }^2$. This distinction matters before specializing to a centered variable.

Let $g$ be a bounded function with $P_{f}g=0$. A bounded density tilt realizes this direction:These are nonnegative probability density functions, since $\int f_t=1+t{P}_{f}g=1$. The uniform Taylor expansion of the square root givesThe squared L2 norm of this remainder is $O(t^4{P}_f{g}^4)=o(t^2)$, since $g$ is bounded. Thus the statistical path is differentiable in quadratic mean with score function $g$.

Conversely every score function is centered by the mean-zero score identity under quadratic-mean differentiability. Choosing all these bounded density tilts therefore gives the statistical tangent setThis is a valid choice of statistical tangent set; it does not assert that every possible score function in the unrestricted density model is bounded.