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A Mehrstellen method gains accuracy by exploiting the differential equation in the local truncation error of a compact finite difference method stencil. The extra nearby values alone do not guarantee higher order: one must also modify the source term. Consider $\Delta u=f$ on a square with prescribed Dirichlet boundary data, using the same spacing $h$ in both directions.

For comparison, the five-point Laplacian has expansionIts leading error is not a multiple of ${\Delta }^{2}u$, because the mixed fourth derivative is missing. To obtain an isotropic leading error, consider a symmetric nine-point finite-difference stencil with weight $a$ on each axial neighbor, $b$ on each diagonal neighbor, and $c$ on the center, all divided by $h^2$. Vanishing on constants and consistency of a numerical method require $4a+4b+c=0$ and $a+2b=1$. Taylor expansion then gives the fourth-derivative contributionFor this to equal $h^2{\Delta }^{2}u/12$ we must choose $b=1/6$, hence $a=2/3$, $c=-10/3$. The resulting nine-point finite-difference stencil operator isIts expansion, retaining the next term for later use, isThus $D_9$ by itself is second order on a general function. On a Poisson equation solution, however, ${\Delta }^{2}u=\Delta f$. Approximating this known source derivative by $D_{5}f$ gives the fourth-order Mehrstellen method equationEquivalently, its two stencils areThe normalized local truncation error is $O(h^4)$, since $D_{5}f-\Delta f=O(h^2)$. If the equation is multiplied by $6h^2$, the row residual is $O(h^6)$; this change of normalization does not make the solution sixth order. The fourth-order correction of the nine-point Poisson stencil retains only the closest axial and diagonal unknowns. A fourth-order one-dimensional second derivative instead uses $(-U_{i+2}+16U_{i+1}-30U_i+16U_{i-1}-U_{i-2})/(12h^2)$, widening the unknown stencil and introducing negative outer weights. Mehrstellen method accuracy is obtained by correcting known source values rather than widening that unknown stencil.

The compact scheme has useful stability and solvability properties. With homogeneous Dirichlet boundary data, $-D_9$ is symmetric, has positive diagonal and nonpositive off-diagonal entries, and is a positive definite symmetric operator. Indeed its quadratic form is a sum of positive edge weights times squared differences, with boundary values set to zero. Vanishing of this sum would force a constant on the connected grid and then zero by connection to the boundary. Consequently the Dirichlet boundary condition linear system has a unique solution. The same weights give a discrete maximum principle: if $D_{9}v\ge 0$ at every interior node and $v\le 0$ on the boundary, a positive maximum inside would make every weighted difference to its neighbors nonpositive. Equality forces the maximum to propagate to the boundary, which is impossible. Thus $v\le 0$ everywhere.

This also turns consistency of a numerical method into a quantitative convergence result, rather than merely suggesting an order. Let $B_{h}f=f+h^2{D}_{5}f/12$, let $\tau =D_{9}u-B_{h}f$ be the normalized residual of the exact solution, and set $e=U-u$. Exact Dirichlet boundary data imply $e=0$ on the boundary and $D_{9}e=-\tau$. On the unit square the quadratic barrierhas $D_{9}q=-1$ exactly, is nonnegative on the boundary, and has maximum $1/8$. If $\Vert \tau {\Vert }_{\infty }\le \epsilon$, then $D_9(e-\epsilon q)\ge 0$ and $D_9(-e-\epsilon q)\ge 0$, while both comparison functions are nonpositive on the boundary. The discrete maximum principle provesThis is maximum-norm convergence of the corrected nine-point Poisson scheme. It requires sufficient smoothness of the solution up to the boundary for the uniform truncation bound. On other domains the source and boundary closures must retain the required accuracy; boundary singularities can invalidate a smooth-solution order estimate.

The same expansion reveals how to add still more accuracy. Differentiating $\Delta u=f$ givesThus the sixth derivatives appearing in the $h^4$ term can also be expressed solely through the source. With exact source derivatives,is the sixth-order source correction of the nine-point Poisson stencil, with normalized residual $O(h^6)$. If source derivatives are replaced by differences, $\Delta f$ must be approximated through fourth order and the fourth derivatives through second order to retain that accuracy. The unknowns still occupy the same nine-point finite-difference stencil; only evaluation of known source data becomes more elaborate. The same inverse bound yields sixth-order convergence when these approximations and the boundary data have matching accuracy.

A particularly instructive case is the Laplace equation, $f=0$. Both displayed error corrections automatically vanish: ${\Delta }^{2}u=0$ and $u_{xxxxyy}+u_{xxyyyy}=\Delta u_{xxyy}=0$, which also makes the pure sixth derivatives sum to zero. Expanding the stencil two orders further gives, for a sufficiently smooth harmonic function,For verification, the three eighth-derivative contributions are $(u_{xxxxxxxx}+u_{yyyyyyyy})/20160$, $(u_{xxxxxxyy}+u_{xxyyyyyy})/2160$ and $u_{xxxxyyyy}/864$. Because $u$ is a harmonic function, their sum is $u_{xxxxxxxx}(1/10080-1/1080+1/864)=u_{xxxxxxxx}/3024$. The example $u=\mathrm{Re}(x+iy{)}^8$ has $D_{9}u(0,0)=40h^6/3$, so this term is genuinely present. The unmodified harmonic nine-point scheme therefore has sixth-order nodal convergence under the same smoothness and boundary assumptions. This harmonic superconvergence of the nine-point stencil is special to harmonic functions; it should not be confused with the fourth-order source-corrected scheme for general Poisson equation data.

The appeal of Mehrstellenverfahren is this combination of a compact sparse matrix, favorable discrete maximum principle properties, and high accuracy obtained by using the governing equation. The derivation also shows precisely which accuracy belongs to the operator on arbitrary functions, which belongs to its action on solutions, and which survives in the computed boundary-value solution.

Write $n=km$, with positive integers $k,m$, and put $Q=P^m$. Polynomial substitution respects congruence, so $Q(z)\equiv z(\mathrm{mod}Q(z)-z)$ implies inductively $Q^j(z)\equiv z$ for every $j\ge 1$. Since $Q^k=P^n$,Equivalently one can factor each $Q^{j+1}(z)-Q^j(z)$ by $Q(z)-z$ and telescope. If $Q(z)-z$ is identically zero, then $Q$ is the identity and $P^n-z$ is also zero, so the assertion still holds in the sense that the latter is a polynomial multiple of the former.

For the holomorphic cycle, the chain rule giveswith indices cyclically understood. These are the same product at every point. If its modulus is less than one at one point, it is less than one at every point. Taylor expansion of each return map gives a contracting neighborhood of its fixed point, exactly as in question 1. Hence every point of the holomorphic cycle is attracting for $P^n$ whenever one is; a zero factor makes every return periodic-orbit multiplier zero.

For the quadratic family, factor the second-iterate fixed-point equation:The first factor gives fixed points. A root of the second is also fixed only if subtracting the two factors gives $2z+1=0$, so $z=-1/2$ and $c=-3/4$. If $c\ne -3/4$, the second factor has discriminant $-4c-3\ne 0$ and has two distinct roots, neither fixed. They are exchanged by $P$, since each is fixed by $P^2$ but not by $P$, and form an exact two-cycle. Therefore the unique parameter with no exact two-cycle isAt this parameter,Thus the four finite solutions counted with multiplicity are $3/2$ once and $-1/2$ three times: there are only two distinct finite solutions, both already fixed by $P$. The periodic-orbit multiplier at $3/2$ is three (nine for $P^2$), so it is repelling. At $-1/2$ the periodic-orbit multiplier for $P$ is minus one; for $P^2$ it is one. Indeed, with $u=z+1/2$, the local first-return expression for $P$ is $-u+u^2$, and its square is $u-2u^3+u^4$. This is a parabolic return with a triple fixed-point root. The two-cycle has merged into this fixed point; the triple root is not three distinct periodic points. In the Mandelbrot set picture this is the period-two bulb's attachment to the main cardioid.

Choose centred charts so the coordinate expression $F$ satisfies $F(0)=0$. On a component where the map is nonconstant, the identity theorem makes $F$ nonconstant. Its Taylor expansion factors as $F(z)=z^{n}h(z)$, with $n\ge 1$ and $h(0)\ne 0$. In a sufficiently small disc, choose a holomorphic logarithm of $h$, and hence a holomorphic $n$th root $q$ with $q^n=h$. The function $zq(z)$ has nonzero derivative at zero. The holomorphic inverse function theorem, which says that a holomorphic function with nonzero derivative has a holomorphic local inverse, makes it a valid new source coordinate. In that coordinate,If the map is bijective, local injectivity forces $n=1$: for $n>1$, a small nonzero value has $n$ distinct preimages in this local disc. The inverse is therefore holomorphic in a neighbourhood of every point, and the local inverses agree, proving the map is biholomorphic. The nonconstant hypothesis must apply to the component containing $p$ if disconnected Riemann surfaces are allowed; otherwise a constant component is an exception to the proposed local form.

The argument principle states that if $h$ is meromorphic on a neighbourhood of a positively oriented simple closed contour and its interior, with neither zeros nor poles on the contour, thenwhere $N$ and $P$ count its interior zeros and poles with multiplicity. This integer is also the winding number of $h(\gamma )$ about zero.

Fix $z_0\in U$. Injectivity prevents $f$ from being constant on a neighbourhood of $z_0$, so its Taylor expansion has the formChoose a small closed disk $D\subset U$, centred at $z_0$, such that $h$ and $mh+(z-z_0)h^{\prime }$ are nonzero throughout $D$. Then $f-f(z_0)$ has its sole zero at $z_0$ in $D$, of multiplicity $m$, while $f^{\prime }$ has no zero in $D\setminus \{z_0\}$. On its boundary let $\delta =\min |f-f(z_0)|>0$.

Choose $0<|w|<\delta$. For every $0\le t\le 1$, the holomorphic function $f-f(z_0)-tw$ is nonzero on $\partial D$. By the argument principle, its number of zeros is an integer given by a contour integral varying continuously with $t$, hence remains $m$. At $t=1$ none of these zeros is $z_0$, so their derivatives are nonzero and they are simple zeros. Thus there are $m$ distinct points taking the value $f(z_0)+w$. Injectivity forces $m=1$, proving $f^{\prime }(z_0)\ne 0$ everywhere in $U$.

Let $L(f)=f^{\prime }(0)-\mu (f)$. Direct evaluation on $1,x,x^2$ gives $L=0$, so the Peano kernel theorem givesThis representation also follows by applying $L$ to the Taylor expansion of $f$ with integral remainder. The derivative at zero of the truncated quadratic is zero for $t\ge 0$, and its sampled values giveThe Peano kernel is nonnegative, andHence $|L(f)|\le \frac{1}{3}\Vert f^{(3)}{\Vert }_{\infty }$. For $f(x)=x^3/6$, the third derivative is $1$, while $f^{\prime }(0)=0$ and $\mu (f)=-1/3$. Thus equality holds, proving the least constant is $c=1/3$, attained by $f(x)=x^3/6$.

At a target $t$, local polynomial regression fits a polynomial of degree at most $p$ by weighted least squares:Define the response vector $Y=(Y_1,\dots ,Y_n{)}^T$, the design matrix $X$ by $X_{i,j+1}=(x_i-t{)}^j$ for $j=0,\dots ,p$, and the diagonal matrix $W$ by $W_{ii}=K((x_i-t)/h)$. The normal equations for linear least squares giveThis formula assumes the local polynomial Gram matrix $X^{T}WX$ is invertible. It is invertible whenever at least $p+1$ distinct sites have positive weights: a nonzero degree-$p$ polynomial cannot vanish at all of them. With fixed $p$, a nonzero continuous regression kernel, and $nh\to \infty$, this condition holds eventually at the target points considered here. Using weights $h^{-1}K((x_i-t)/h)$ instead gives exactly the same fit.

Use the usual unit-integral normalization ${\mu }_0(K)=1$. A regression kernel need not intrinsically have unit integral, but multiplying all weights by a positive constant leaves local polynomial regression unchanged. With unnormalized raw moments the interior coefficient below is ${\mu }_2(K)/{\mu }_0(K)$; the version written using ${\mu }_2(K)$ alone requires this normalization. A zero regression kernel cannot define an estimator.

Write the local moment sums asThe local constant estimator is ${\hat{m}}_h(t;0)={\sum }_{i}K((x_i-t)/h)Y_i/{\sum }_{i}K((x_i-t)/h)$. Since the errors have zero mean, its bias of an estimator is the corresponding weighted average of $m(x_i)-m(t)$. By the second-order Taylor expansion, uniformly for contributing sites with $|x_i-t|\le h$,where ${\eta }_n\to 0$. This uniform remainder follows from uniform continuity of $m^{\prime \prime }$ on $[0,1]$. Nonnegativity of the regression kernel therefore giveswhenever $A_0$ tends to a positive limit and $A_2=O(h^2)$.

For fixed interior $x$, the moment approximations give $A_0(x)=1+o(1)$, $A_1(x)=O(n^{-1})$, and $A_2(x)=h^2{\mu }_2(K)+o(h^2)$. The hypothesis $n{h}^2\to \infty$ makes $n^{-1}=o(h^2)$, so the odd-moment contribution is negligible at second order. ThusThe symmetry of the regression kernel is what removes the first-order interior bias of an estimator.

At $z_n=\alpha h$, the rescaled observation window is truncated to $[-\alpha ,1]$. Put ${\mu }_{r,\alpha }={\int }_{-\alpha }^1{u}^{r}K(u)du$ and ${\beta }_r={\mu }_{r,\alpha }/{\mu }_{0,\alpha }$. Symmetry, nonnegativity, and unit mass give ${\mu }_{0,\alpha }\ge 1/2$. The supplied moment approximations already imply the leading boundary bias of local constant regression:To retain the second-order term rigorously, we need slightly more precise control of the first moment. The rescaled grid $u_i=i/(nh)-\alpha$ has spacing $1/(nh)$. Each truncated function $u^{r}K(u)1_{[-\alpha ,1]}$ has bounded variation, because $K$ is continuously differentiable on its support. The Riemann sum error is bounded by this variation times the grid spacing: summing the oscillation bounds on the grid cells provesEndpoint cells, including the omitted site at zero, contribute the same error order. ConsequentlyBoth discretization errors are $o(h^2)$ under $n{h}^2\to \infty$. The preceding Taylor expansion therefore proves the second-order boundary expansion for local constant regression:Equivalently, expanding the derivatives about zero givesIn general the truncated first moment is nonzero, so the boundary bias of an estimator has order $h$, although it can be smaller when $m^{\prime }(0)=0$ or the omitted part of the regression kernel carries no mass.

For local linear regression, the polynomial reproduction property of local polynomial regression removes the constant and linear terms exactly. Let $\Delta =A_0{A}_2-A_1^2$. Its effective kernel weights areso ${\sum }_i{l}_i=1$, ${\sum }_i{l}_i(x_i-t)=0$, andAt either an interior or the specified boundary target, $\Delta$ is asymptotic to a positive constant times $h^2$. At the boundary that constant is ${\mu }_{0,\alpha }{\mu }_{2,\alpha }-{\mu }_{1,\alpha }^2>0$, the mass squared times the variance of the continuous normalized truncated regression kernel. Moreover ${\sum }_i|l_i|=O(1)$, as follows by bounding $|A_2-A_1(x_i-t)|=O(h^2)$ in the window. Thus the Taylor remainder still contributes $o(h^2)$, even though some effective kernel weights may be negative. In particular,At an interior point this coefficient reduces to ${\mu }_2(K)$ by symmetry and normalization. Hence local linear regression has bias $O(h^2)$ at both interior and boundary points; its linear reproduction removes the generic order-$h$ boundary term.

Write the unit-amplitude incident plane wave as ${\psi }_i=e^{i{p}_{i}x-i{q}_{i}z}$, with $q_i=\sqrt{k^2-p_i^2}>0$. The flat Dirichlet reflected field is ${\psi }_s^{[0]}=-e^{i{p}_{i}x+i{q}_{i}z}$, so the flat total field vanishes at $z=0$ and has derivative ${\partial }_z{\psi }^{[0]}(x,0)=-2i{q}_i{e}^{i{p}_{i}x}$.

For a controlled height expansion take $f=\epsilon h$ with fixed regular bounded shape $h$. Taylor expansion at the moving boundary yieldsThus the first-order scattered trace is ${\psi }_s^{[1]}(x,0)=2i{q}_{i}f(x)e^{i{p}_{i}x}$, and its transform follows by the Translation property of the Fourier transform:The first formula is the roughness correction; the second includes the zeroth-order reflection. The small-height assumption controls the Taylor expansion for a fixed profile; arbitrarily fine surface scales can require additional derivative control, not merely a small height at fixed $k$.

If the mean roughness height is a constant $\bar{f}$, as for stationary roughness, then $⟨\hat{f}(\nu -p_i)⟩=2\pi \bar{f}\delta (\nu -p_i)$. Consequently the specular first-order mean reflection isIts tangential wavenumber is unchanged and its normal wavenumber is reversed relative to incidence, which is precisely specular reflection. For zero mean height the first-order mean correction is zero, but the full mean reflected field is not zero: it still contains the flat reflection.

Small height alone does not imply the statistical conclusion. For example, deterministic mean height $\bar{f}(x)=a\cos Kx$ with $|ka|\ll 1$ produces first-order mean sidebands at $p_i\pm K$ and is not purely specular. Constant mean is sufficient at first order; stationarity supplies it and, with suitable translation-invariant statistics, constrains higher-order coherent reflection as well.

Take transverse length units in which the wave phase modulation has wavenumber one. Set $E_0(z)=e^{i\cos z}$. A Taylor expansion of the free parabolic propagator givesFor a general real wave phase $\varphi$,Multiplying the envelope expansion by its complex conjugate and collecting powers of $x$ gives the phase-to-intensity conversion near a deterministic phase screen,For $\varphi =\cos z$, ${\varphi }^{\prime }=-\sin z$, ${\varphi }^{\prime \prime }=-\cos z$ and ${\varphi }^{\prime \prime \prime }=\sin z$, henceThe first-order brightening occurs where the wave phase curvature is negative. Both correction terms average to zero over a transverse period, in agreement with power conservation. If a dimensional modulation $\cos (qz)$ is used instead, the two terms are $(q^{2}x/k)\cos (qz)$ and $(q^4{x}^2/k^2)\cos (2qz)$.

The Gumbel distribution for maxima has distribution function $G(x)=\exp (-e^{-x})$ on the whole real line. For $n\ge 2$ setBoth the centering and scale are elementary, and the scale is positive. For any fixed real $x$, the normalized threshold ${\beta }_n+{\alpha }_{n}x$ is positive for all sufficiently large $n$. Writing $L=\log n$, a Taylor expansion givesConsequently,The logarithm calculation in part (i) now givesThus $F$ belongs to the maximum domain of attraction of the standard Gumbel distribution. The scale also has the Gumbel auxiliary function interpretation: the reciprocal hazard function is $2\sqrt{t}$, whose derivative $1/\sqrt{t}$ tends to zero at the infinite right endpoint. This agrees with the Von Mises conditions for extreme values, and at $t={\beta }_n$ gives exactly ${\alpha }_n=2\log n$.

Iterating the recurrence yieldsThus every finite collection of the $X_j$ is a linear transformation of independent standard normal random variables, and has a multivariate Gaussian distribution. The expected values are zero. Independence of the $Y_k$ givesFor $i<j$, the recurrence writes $X_j={\rho }^{j-i}{X}_i$ plus a sum of future innovations independent of $X_i$. Hence $\mathrm{Cov}(X_i,X_j)={\rho }^{j-i}$. The required vector law isThe covariance matrix is positive definite: the lower-triangular transformation from $(Y_1,\dots ,Y_n)$ has nonzero diagonal entries $1,\sqrt{1-{\rho }^2},\dots ,\sqrt{1-{\rho }^2}$. This also identifies a stationary autoregressive process of order one, with marginal density $f=\varphi$.

With the standard normal density as the kernel for density estimation, the two characteristic functions are $\psi (t)=e^{-t^2/2}$ and ${\psi }_K(ht)=e^{-h^2{t}^2/2}$. The Gaussian random variable $X_1-X_{j+1}$ has mean zero and variance $2(1-{\rho }^j)$, soThis expression is real. Substitution in the given integrated covariance expression, followed by the Gaussian integral ${\int }_{\mathbb{R}}{e}^{-a{t}^2}dt=\sqrt{\pi /a}$ for $a>0$, gives

To bound the total Gaussian autoregressive kernel variance correction, apply the mean value theorem to $u\mapsto u^{-1/2}$. For $j\ge 1$, the whole interval between $1+h^2-{\rho }^j$ and $1+h^2$ is bounded below by $1-\rho >0$. ThereforeSumming the geometric series gives the uniform boundIt follows from the supplied variance identity thatThe constant depends on the fixed $\rho$, but not on $h$ or $n$. In particular this applies to the requested shrinking-bandwidth regime; the covariance bound itself does not need $nh\to \infty$.

All observations have the same marginal density $f$, so dependence changes the variance but not the bias of a kernel density estimator: $\mathbb{E}{\hat{f}}_h=K_h\ast f$. Write $R(q)={\int }_{\mathbb{R}}{q}^2$ and ${\mu }_2(K)=\int u^{2}K(u)du$. For a smooth density with $f^{\prime \prime }\in L^2$, the integrated variance of a kernel density estimator in the independent case is $R(K)/(nh)+O(n^{-1})$. A second-order Taylor expansion of the convolution gives $K_h\ast f-f=(h^2{\mu }_2(K)/2)f^{\prime \prime }+o(h^2)$ in the L2 norm. Thus the bias-variance decomposition of mean squared error givesThe first two terms are the asymptotic mean integrated squared error. The additional dependence term is smaller than $n^{-4/5}$, so the leading optimal bandwidth and optimal rate are the same as for independent data:Here both $K$ and $f$ are the standard normal density. Since $f^{\prime \prime }(x)=(x^2-1)\varphi (x)$, direct Gaussian integrals giveConsequently the explicit leading bandwidth isAt this bandwidth the leading integrated variance is four times the leading integrated squared bias, which is also a check on the minimization constant.

Let the summands be independent and identically distributed random variables with mean $\mu$, variance ${\sigma }^2>0$, and standardized third cumulant ${\gamma }_1=\mathbb{E}[(Y_1-\mu {)}^3]/{\sigma }^3$. DefineAn Edgeworth expansion refines the central limit theorem using the higher cumulants of one summand. Some smoothness and moment assumptions are essential: finite variance alone does not supply a density expansion. A convenient sufficient set for the expansions here is a smooth density whose derivatives decay faster than every inverse power, together with $\mathbb{E}{e}^{\delta |Y_1|}<\infty$ for some $\delta >0$. These assumptions give all required moments, Fourier smoothing and the nonlattice characteristic-function condition. Less restrictive sufficient conditions are possible, but this set covers the logistic model below.

Write $\varphi$ and $\Phi$ for the standard normal density and standard normal distribution function. The first-order density Edgeworth expansion isThe remainder is uniform on every fixed compact interval under the stated assumptions. To see the origin of the correction, for $Z=(Y_1-\mu )/\sigma$ the cumulant expansion of its characteristic function givesExponentiating and using Fourier inversion yields the displayed correction, since the inverse transform of $(it{)}^3{e}^{-t^2/2}$ is $-{\varphi }^{(3)}(x)=(x^3-3x)\varphi (x)$. The Probabilists' Hermite polynomial convention is being used: $H_2(x)=x^2-1$ and $H_3(x)=x^3-3x$.

The corresponding distribution function expansion isuniformly on fixed compact intervals. The sign follows from $\frac{d}{dx}[H_2(x)\varphi (x)]=-H_3(x)\varphi (x)$. These are expansions of both the density and distribution function; a uniform density error alone would not justify integrating an error bound over the entire real line without the accompanying tail control.

For a fixed $\alpha \in (0,1)$, write $z=z_{\alpha }={\Phi }^{-1}(\alpha )$. At leading order the proposed quantile expansion gives $\Phi (p_0(z))=\Phi (z)$, and strict monotonicity gives $p_0(z)=z$. Substitute $y_{\alpha }=z+p_1(z)n^{-1/2}+O(n^{-1})$ into the distribution function expansion. A Taylor expansion at $z$ givesSince $\varphi (z)>0$, the coefficient of $n^{-1/2}$ must vanish. Thus the first Cornish-Fisher expansion coefficients are