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For a small opposing force, the relevant parameter is $q=\beta F\delta \ll 1$. A Taylor series givessoIn particular the zero-force rate is finite, despite the numerator and denominator of the original expression both vanishing.

For $q\gg 1$, the exponential dominates the denominator:Opening a monomer-sized gap is exponentially rare at high opposing force. The algebraic $F^2$ prefactor belongs to this first-passage model and must not be discarded when quoting the leading asymptotic rate.

The displacement field is $u(x)=y(x)-x$, and its gradient is $Du=Dy-I$. Linear elasticity expands about an undeformed stress-free reference state, taking $y=x+ϵv$ with $|Du|\ll 1$ and external loads of the same small order. A rigid translation need not itself be small; it is the deformation gradient that controls the infinitesimal strain tensor.

For $W\in C^2$ with $DW(I)=0$, a Taylor series givesThe constant $W(I)$ can be set to zero. The tensor $\mathbb{A}$ is the elasticity tensor. Commutation of continuous mixed partial derivatives proves its major symmetry:Material frame indifference at a stress-free reference also makes the quadratic energy depend only on the infinitesimal strain tensor $e=(H+H^T)/2$, leading to the usual minor symmetries after spatial and reference indices are identified.

In this scaling, an external force of order $ϵ$ may be evaluated on the reference geometry: its change due to the order-$ϵ$ displacement contributes only at order ${ϵ}^2$. Thus the leading forces are fixed dead loads. Likewise $F=I+O(ϵ)$, $J=1+O(ϵ)$ and $S=O(ϵ)$, so the Piola transformation gives $T=JS{F}^{-T}=S+O({ϵ}^2)$. Hence Cauchy and first Piola stresses agree to linear order. This reasoning uses the usual small-load, stress-free linearization; an order-one prestress or leading follower-load effect requires a different expansion.

The small-argument Taylor series of the single-site response isLet $K=4\beta J$. For a small nonzero order parameter at zero field, the self-consistency equation becomesSince $K_c=3/2$ and $K=K_c/(1+t)$, where $t=T/T_c-1$, this givesConsequentlyThe positive branch is the expression printed in the question, while its negative partner is required by spin inversion symmetry. The square-root behavior gives the mean-field critical exponent $1/2$ for the spontaneous magnetization.

The one-step master equation for the biased random walk isExpand in a Taylor series and retain the first time derivative and the first two space derivatives. Since $w_{+}+w_{-}=1$ and $w_{-}-w_{+}=2ϵ$,Thus the usual continuum truncation gives the requested Fokker-Planck equation,The diffusion coefficient has units $\mathrm{lengt}{\mathrm{h}}^2/time$, while $C$ has units of velocity. The physical drift velocity is $v=-C=a(w_{+}-w_{-})/\tau$, because the standard advection term is $-v{\partial }_{x}P$. A rightward bias therefore means $ϵ<0$, $C<0$, and $v>0$.

The approximation must be stated carefully. The truncation is a controlled diffusion limit of a weakly biased random walk when $a\to 0$, $\tau \propto a^2$, and $ϵ\propto a$, keeping $D_0$ and $C$ finite. It is also the spatial diffusion approximation for a continuous-time jump process with exponentially distributed waiting times of mean $\tau$.

If the original fixed-step-time walk has a finite bias, its exact variance from (a) instead gives the diffusion coefficient of the Gaussian envelope about the drifting mean:One obtains the same correction by retaining $\frac{1}{2}{\tau }^2{\partial }_t^{2}P$ in the time expansion and using the leading advection equation to replace ${\partial }_t^{2}P$ by $v^2{\partial }_x^{2}P$. The result is $D_{\mathrm{fixed}}=D_0-v^2\tau /2$. In particular, a deterministic one-way walk has zero variance, as this expression requires. Thus $D_0$ is the usual intended first-time-derivative approximation, not an exact variance-growth coefficient at arbitrary bias. The later continuum estimates use that convention, with the fixed-time alternative identified where relevant.

At the origin the needed partial derivatives areThe Taylor series through total degree two is thereforewhich givesHere total degree is what matters: $x^2{y}^2$ is quartic, not quadratic. The exact omitted polynomial is $y^3+x^2{y}^2+x^2{y}^4$, which verifies the stated Big O notation remainder directly.

Near zero, $x^3-x^5=x^3(1-x^2)$ has the sign of $x$, so $f(x)<1$ just to the left and $f(x)>1$ just to the right. Thus the critical point is neither a local minimum nor a local maximum. Moreover the Taylor series gives $f^{\prime \prime }(x)=6x+O(x^3)$, which changes sign at zero. ThereforeThis is the point called a “saddle point” in the question; stationary inflection point specifies the one-variable meaning.

Use the Taylor series $e^u=1+u+u^2/2+\cdots$ with $u=x^3-x^5$. Since $u$ begins at cubic order and $u^2$ begins at sixth order, through cubic terms the Taylor series isThe linear and quadratic coefficients are zero. The stronger remainder $O(x^5)$, in big O notation, follows because the fourth-order term is also zero; the next term is $-x^5$.

One standard finite form of Taylor's theorem is the following. Suppose $f$ has continuous derivatives through order $m+1$ on an interval containing $a$ and $x$. Then, if $x\ne a$, there is some $\xi$ strictly between them such thatThe final term is the Lagrange remainder. At $x=a$ the identity holds directly with zero remainder. The finite expansion requires no assumption that an infinite Taylor series converges to $f$.

Use the reciprocal-factorial sum for increasing accuracy. Let $d$ be the requested number of decimal places and use absolute error tolerance proportional to $10^{-d}$. Expanding $\log (1+1/n)$ gives $e_a(n)=e-e/(2n)+O(n^{-2})$ and $e_c(n)=e+e/(24n^2)+O(n^{-3})$, so these are genuine first nonzero error orders. The first approximation needs $n$ of order $10^d$. Even its fast implementation therefore uses $\Theta (d)$ multiplications, while the linear version uses exponentially many in $d$. The third approximation needs $n$ of order $10^{d/2}$, but its fast powering still uses $\Theta (d)$ multiplications, with two parallel accumulators and a cancellation-prone subtraction.

For the Taylor series sum, Stirling's formula gives $\log (n!)=\Theta (n\log n)$, so a sufficient term count grows as $n=\Theta (d/\log d)$. Its divisions therefore grow more slowly than the multiplication counts of either fast-power alternative. More concretely, its positive tail satisfiesThe first omitted term also gives a lower bound $1/(n+1)!$. This is the reciprocal-factorial series remainder bound, supplying an explicit stopping certificate rather than relying only on Big O notation. It follows by bounding every successive tail denominator factor below by $n+2$.

The sum also avoids subtraction of nearly equal large values and has small positive terms, making it numerically preferable. For a particular small accuracy request, constants and actual term counts may be compared directly; an asymptotic error bound alone does not prove a universal finite-input operation ranking. With ordinary Double, sufficiently many tiny positive terms stop changing the rounded sum, so beyond machine precision a higher-precision version is necessary regardless of the formula.

Put $h=\Delta x$ and suppose $L=(n+1)h$, so there are $n$ interior grid points $x_j=jh$, $j=1,\dots ,n$. Write $u_j(t)\simeq u(x_j,t)$. The fixed ends impose homogeneous Dirichlet boundary conditions,For a sufficiently smooth displacement field, the two Taylor series at fixed time givewhere all displayed derivatives on the right are evaluated at $(x_j,t)$. Adding cancels the odd terms, leaving the second-order central differenceSubstitute this finite difference into the wave equation and retain continuous time. The resulting method of lines gives coupled second-order ordinary differential equations,In particular, the first and last equations are ${\ddot{u}}_1=(c^2/h^2)(-2u_1+u_2)$ and ${\ddot{u}}_n=(c^2/h^2)(u_{n-1}-2u_n)$ for $n\ge 2$. With the interior displacement vector only,Equivalently, $A_{jj}=-2c^2/h^2$, $A_{j,j+1}=A_{j+1,j}=c^2/h^2$, and all other entries vanish. This entrywise definition also covers small $n$, including $A=(-2c^2/h^2)$ when $n=1$. The matrix is a symmetric matrix and a tridiagonal matrix. The spatial approximation has error $O(h^2)$ in Big O notation; time has not yet been discretized.

For real constants $a,c$, apply Picard iteration to the initial value problem $y^{\prime }=ay/(1-x)$, $y(0)=c$, starting with $y_0(x)=c$. The recurrence isSince ${\int }_0^x{\log }^k(1-t)/(1-t)dt=-{\log }^{k+1}(1-x)/(k+1)$, mathematical induction gives the displayed partial sums. The exponential series converges at every finite argument, so $y_n(x)\to c(1-x{)}^{-a}$ for every real $x<1$, uniformly on each compact interval in that range. This is not a Taylor series in $x$, so its convergence is not restricted to $|x|<1$. The differential equation's coefficient is singular at $x=1$ even when a special solution has a removable singularity there.

Extend this expression by zero at zero. Gaussian decay dominates every power, so the resulting function and its derivative difference quotients vanish to every order at zero. Away from zero, differentiation replaces $p(t)$ by $2t^{3}p(t)-t^2{p}^{\prime }(t)$, another polynomial. Induction therefore gives a smooth function flat at zero. If $p$ is nonzero, its Taylor series is zero while the function is nonzero at every sufficiently small nonzero argument, so it is not real analytic there.

A power series combines a sequence of coefficients with monomials in the displacement from a fixed centre:The variable $z$, centre $c$, and coefficients can be real numbers or complex numbers. Its degree-$N$ partial sum $P_N(z)={\sum }_{n=0}^N{a}_n(z-c{)}^n$ is a polynomial, so a convergent power series can turn a complicated function into increasingly accurate finite calculations. Convergence, and the error after stopping, are essential parts of that claim.

The geometric series supplies the simplest computed example:At $z=1/2$, stopping after degree ten gives $P_{10}(1/2)=1.9990234375$, differing from the exact value $2$ by $1/1024=0.0009765625$. At $z=0.99$, the same degree gives only about $10.4662$ instead of $100$. The same formula has very different practical accuracy at different points; the geometric power-series remainder quantifies it.

The structure is useful beyond numerical evaluation. Within its radius of convergence, termwise differentiation of a power series turns a question about a derivative into a question about coefficients. A Taylor series chooses $a_n=f^{(n)}(c)/n!$ when it represents the function locally. A formal power series instead uses the coefficients algebraically, without asserting numerical convergence; the existing generating function examples use that distinction to encode counting problems.

The quantum harmonic oscillator describes a particle of mass $m>0$ in a quadratic potential energy with angular frequency $\omega >0$. In the position representation its Quantum Hamiltonian isThe model is useful far beyond a literal spring: a Taylor series about a stable equilibrium with positive potential curvature has this quadratic term as its leading nonconstant approximation. The pure quartic oscillator has zero curvature at its minimum, so that particular approximation does not apply. ^[ref].

In units with $\hslash =1$, creation and annihilation operators giveHere $N$ is the number operator. Restoring the reduced Planck constant, the energy eigenvalues are $E_n=\hslash \omega (n+1/2)$ for $n=0,1,2,\dots$: neighboring levels are separated by $\hslash \omega$, and the ground-state energy is the zero-point energy $\hslash \omega /2$.

With length scale $\ell =\sqrt{\hslash /(m\omega )}$ and dimensionless coordinate $u=x/\ell$, the ground state is the normalized Gaussian wave packetFor $\hslash =m=\omega =1$, this is precisely the $a=1$ trial wavefunction in the quantum variational principle. The balance of kinetic energy and potential energy therefore recovers the exact ground-state energy using only Gaussian integrals.

In any finite-dimensional Euclidean space, this extension is a smooth function and all its partial derivatives at the origin vanish. Away from zero, repeated differentiation gives finite sums of coordinate polynomials divided by powers of $|x|$, multiplied by $e^{-1/|x{|}^2}$. Every such term is bounded in absolute value by $C|x{|}^{-N}{e}^{-1/|x{|}^2}$ near zero. Since Gaussian decay dominates every power, this bound tends to zero faster than any prescribed positive power of $|x|$. Induction using difference quotients then proves smoothness and zero derivatives of every order at zero.

The origin is nevertheless a strict global minimum, because $F(x)>0$ for $x\ne 0$. All its Taylor series coefficients vanish, so the function is not real analytic at zero. Adding a constant preserves the stationary minimum but changes the zeroth-order value. An expression written only for $x\ne 0$ has no stationary point at the excluded origin until an extension is specified.

A function $f$ is real analytic at $y$ when its Taylor seriesconverges to $f(x)$ throughout some neighborhood of $y$.

The reciprocal chart makes infinity an ordinary coordinate origin. Its domain is the Riemann sphere minus the point $z=0$; $w(\infty )=0$ and $w(z)=1/z$ elsewhere. A neighbourhood $U_R$ of infinity becomes the open disc $|w|<1/R$. In sphere coordinates this is exactly the capFor $R=2$, the exterior $|z|>2$ together with infinity becomes $Z>3/5$ and $|w|<1/2$. The boundary itself is excluded from each open set.

The point $z=2+i$ lies in this neighbourhood because $|z|=\sqrt{5}>2$. Its reciprocal coordinate is $w=(2-i)/5$, with $|w|=1/\sqrt{5}<1/2$, and its sphere height is $Z=2/3>3/5$. The negative imaginary part of $w$ is the axis reversal required by the holomorphic stereographic atlas of the sphere.

For a complex-valued holomorphic function $f$ defined outside a large disc, having a removable singularity at infinity means that $g(w)=f(1/w)$ extends to a holomorphic function at $w=0$. This definition reduces questions at infinity to familiar questions at a finite point.^[ref] For example,The geometric series gives a Taylor series at the new origin, so $f(\infty )=1$ and $g^{\prime }(0)=-5$. The coefficient $-5$ is the derivative in the specified $w$ coordinate. In contrast, $f(z)=z^2$ becomes $g(w)=w^{-2}$, displaying a pole of order two at infinity when the target is the complex plane. As a holomorphic map to the Riemann sphere, it uses the target reciprocal coordinate $1/f(1/w)=w^2$, which is regular at zero.

For holomorphic $N,D$ with $N(z_0)D(z_0)\ne 0$, write the quotient as $h(z)/(z-z_0)$ where $h=N/D$. Its Taylor series immediately gives the displayed Laurent series. The residue is $h(z_0)$ and the constant regular term is $h^{\prime }(z_0)$; the latter requires differentiating both numerator and remaining denominator. Higher regular coefficients follow from the higher Taylor coefficients of $h$.

The residue of a function $f$ at an isolated singularity $a$ is its Laurent coefficient of degree $-1$. More explicitly, if $f$ is holomorphic for $0<|z-a|<\rho$, write its Laurent series asThis one Laurent coefficient determines the contour integral around a sufficiently small counterclockwise circle about $a$. A complicated local Laurent series can therefore leave a remarkably simple answer after one complete trip around the point.

For a simple pole, multiplying by $z-a$ removes the pole, leaving a holomorphic extension whose value at $a$ is the residue:There is a useful shortcut for a quotient $f=g/h$. If $g$ and $h$ are holomorphic near $a$, $h$ has a simple zero there, and $g(a)\ne 0$, then $f$ has a simple pole andThe first-order Taylor series of $h$ explains the denominator: dividing $h(z)$ by $z-a$ leaves its derivative at $a$ in the limit. This computes a residue without finding the full Laurent series, which is especially useful when the numerator or denominator is complicated. ^[ref]

For example, the rational function $f(z)=1/(z^2+1)$ has simple poles at $i$ and $-i$. Factoring the denominator makes both residues immediately visible:The signs matter: these residues cancel when both poles are enclosed, but either one contributes when enclosed alone. The residue theorem below turns this distinction into different contour integrals of the same function.

A residue is not the same thing as a pole, nor does it measure the order of a pole. The function $1/z^2$ has a double pole at zero but residue zero. Conversely,has an essential singularity at zero and residue one. Infinitely many negative powers cause no difficulty in reading off that particular Laurent coefficient. At a removable singularity the residue is zero. ^[ref]

Why does only this Laurent coefficient survive? Parametrize a counterclockwise circle by $z=a+r{e}^{it}$, $0\le t\le 2\pi$, with $0<r<\rho$. For every integer $n$,When $n\ne -1$, the complex exponential completes a whole number of turns and its contributions cancel; when $n=-1$, it is constant. Uniform convergence of the Laurent series on this circle justifies termwise integration, giving $\oint f(z)dz=2\pi i\mathrm{Res}(f,a)$. Thus the residue isolates exactly the local information that a closed contour integral detects. ^[ref]

This selection rule explains why a very large value near an isolated singularity need not produce a large contour integral. The double pole $1/z^2$ grows faster near zero than $1/z$, yet its contributions around a circle cancel completely. What matters is the arrangement of complex values along the path, including the differential $dz$, rather than their moduli alone. A residue records that surviving contribution, not the size of the function near the point.

Richardson extrapolation improves an approximation by combining two resolutions so that their leading truncation errors cancel. The useful surprise is that the size of that leading error need not be known: its scaling under refinement is enough. This lets an existing finite difference method or quadrature rule produce a more accurate answer from two runs, without first deriving a new method from scratch.

Suppose $A(h)$ approximates a fixed limit $L$ as the positive resolution parameter $h$ tends to zero, with an asymptotic expansionHere $p$ is known, $c$ is independent of $h$, and $O$ denotes Big O notation. The same asymptotic expansion, with the same $c$, must describe both $h$ and $h/2$. Nested grids make it natural to reuse samples, although nesting itself does not establish this error model. Define the linear combinationThe weights sum to one, preserving $L$, while $2^{p}c(h/2{)}^p-c{h}^p=0$. Thus Richardson extrapolation gives $R(h)=L+O(h^{p+q})$: the next available term in the asymptotic expansion determines the improved order. Justin Solomon's numerical analysis text develops this cancellation for finite differences and applies it to numerical integration.^[ref]

Geometrically, regard the approximation as a function of $t=h^p$. Its first two terms are a straight line $L+ct$. The two computed points lie at positive $t$; extending the line through them to $t=0$ gives $R(h)$. This is why the operation is extrapolation: it estimates a value outside the sampled parameter interval. The remaining curvature corresponds to the uncancelled terms of the asymptotic expansion.

Worked example: a centered derivative of sine. For the smooth function $f(x)=\sin x$, the exact derivative at $x=0$ is $L=\cos 0=1$. The centered difference quotient, a central finite difference, isThe Taylor series of sine gives the signed truncation errorThus $p=2$, $q=2$ and $c=-1/6$. Halving $h$ divides the leading truncation error by four. Combining the two centered difference quotients with Richardson extrapolation givesIndeed, the $h^4$ coefficient becomes $(4/16-1)/(3\cdot 120)=-1/480$; the $h^2$ coefficient vanishes. This is the even-power case of Richardson extrapolation discussed in Persson's authored lecture notes.^[ref]

For $h=0.4$ and $h/2=0.2$, the actual computations areThe absolute error drops from about $2.6454\times 10^{-2}$ for $A(0.4)$ to $5.3080\times 10^{-5}$ for $R(0.4)$, an improvement by roughly a factor of 498. Refining again gives $R(0.2)\approx 0.999996670632607$, whose absolute error is about $3.3294\times 10^{-6}$. The refined truncation errors decrease by nearly four for $A$ and nearly sixteen for $R$, reflecting their second and fourth orders.

The logarithmic scale on each axis turns power laws into straight lines. The dashed reference curves are $h^2/6$ and $h^4/480$, the magnitudes of the leading truncation errors. The plotted errors are computed with 70-digit decimal arithmetic, rather than clipped to a display floor; this isolates the asymptotic expansion from roundoff error.

Worked example: integrating an exponential. Approximate the definite integral of the exponential function on $[0,1]$:For numerical integration, the composite trapezoidal rule adds the areas of trapezoids between successive samples; this is the quadrature use of the endpoint average that also occurs in the ODE trapezoidal rule. On $n$ equal intervals, with $h=1/n$, defineSumming the geometric series gives $T_n=(e-1)\frac{h}{2}\frac{e^h+1}{e^h-1}$. Expanding the exponential function near zero yieldsThe leading truncation error again has $p=2$, $q=2$. On the uniform nested grids with $n$ and $2n$ intervals, Richardson extrapolation therefore givesThis linear combination is exactly composite Simpson's rule on the finer grid, whose spacing is $h/2$. On each coarse panel, an old endpoint receives weight $h/6$ and the new midpoint weight $2h/3$; shared old endpoints receive weight $h/3$. These are the usual Simpson's rule weights $1,4,2,\dots ,4,1$ multiplied by $h/6$. The equivalence is explained directly in LeMesurier's university notes.^[ref]

For two coarse panels and four fine panels,The absolute errors are about $0.0356493$, $0.00894008$ and $0.0000370135$, respectively. Reusing the fine-grid samples makes Richardson extrapolation especially attractive here: the resulting Simpson's rule needs no samples beyond those already used by $T_{2n}$.

When the improvement is justified. Richardson extrapolation requires the stated asymptotic expansion and the correct positive $p$. If the actual leading exponent is $r\ne p$, its coefficient after combination is $c(2^{p-r}-1)/(2^p-1)$, so the leading truncation error generally remains. Merely knowing $A(h)-L=O(h^p)$ does not supply a common coefficient $c$ or a higher-order remainder. A discontinuity of a function, a singular endpoint, or unrelated changes between grids can invalidate the assumed asymptotic expansion; refinement alone does not guarantee improvement for every numerical problem.

Roundoff error and errors in the sampled values survive the cancellation of truncation error. If the coarse and fine values have absolute perturbations bounded by $\epsilon$, the perturbation in $R(h)$ is bounded by $(2^p+1)\epsilon /(2^p-1)$; for $p=2$ this is $5\epsilon /3$. In floating-point arithmetic, a centered difference quotient can also amplify sample errors through division by $h$, so an indefinitely smaller $h$ need not be more accurate. Persson's notes describe this balance between truncation error and roundoff error.^[ref]

The same cancellation idea appears in step-doubling error estimation. For an order-$p$ one-step method for an ordinary differential equation, two half steps each contribute local truncation error of order $h^{p+1}$, so their accumulated leading local error has ratio $2^{-p}$ to one full step. This explains the denominator $2^p-1$ there; the appropriate exponent follows from that local error model, rather than automatically from a generic use of Richardson extrapolation.