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The identity theorem states that if two holomorphic functions on a connected open set agree at a sequence of distinct points with an accumulation point inside that set, then they agree everywhere on it. Equivalently, the zeros of a holomorphic function which is not identically zero cannot accumulate inside its domain.

Because $f,g$ never vanish, their logarithmic derivatives are holomorphic. The holomorphic functionvanishes at every $a_n$. Since the distinct $a_n$ converge to $a\in U$, the identity theorem gives $H\equiv 0$. Applying the quotient rule now givesA holomorphic function with zero derivative on a domain is constant. One can see this locally from its Taylor series, then propagate the constant using connectedness. Hence $f/g=c$ on $U$, with $c\ne 0$, andThe conclusion can fail if the accumulation occurs outside $U$. Take the punctured complex plane $U=\mathbb{C}\setminus \{0\}$ andBoth functions are nowhere vanishing and holomorphic on $U$, and $a_n\to 0\notin U$. The chain rule givesso these logarithmic derivatives agree at every $a_n$. However, $f$ is not constant: for example $f(1/\pi )=e^{-1}$ and $f(1/(2\pi ))=e$. Therefore no constant $c$ gives $f=cg$.

Use the univalent functions on $D$, namely the univalent functions $F:D\to \mathbb{C}$. Each has nonzero derivative, so its image $F(D)$ is a domain conformally equivalent to $D$, and is therefore simply connected. This image is proper: otherwise $F^{-1}:\mathbb{C}\to D$ would be a bounded nonconstant entire function, contrary to the Liouville theorem.

Conversely, every proper simply connected domain $G\subset \mathbb{C}$ is $F(D)$ for some such $F$, by taking the inverse of a Riemann mapping theorem bijection. Thus these functions parameterize all proper simply connected domains by their images. The complex plane itself is the exceptional simply connected domain and must be listed separately.

The parameterization has redundancy: two univalent functions $F_1,F_2$ have the same image if and only if $F_2=F_1\circ \varphi$ for an automorphism of the unit disk $\varphi$. All such automorphisms have the formIf an interior point $w_0\in G$ is prescribed, the normalization $F(0)=w_0$ and $F^{\prime }(0)>0$ real removes this redundancy. Equivalently one can describe these maps through their convergent Taylor series on $D$, imposing univalence on the resulting holomorphic functions.

The identity theorem states: if $V\subseteq \mathbb{C}$ is a nonempty connected open set and $p,q$ are holomorphic functions on $V$, and their equality set has a limit point in $V$, thenThe interior location of the limit point and connectedness are essential hypotheses.

For the proof put $h=p-q$. At any $z_0\in V$, its convergent Taylor series either has every coefficient zero, making $h$ zero on a neighbourhood, or has a first nonzero coefficient, of degree $m\ge 0$. In the latter case,where $k$ is holomorphic near $z_0$. By continuity, $k$ is nonzero on a smaller disc. Thus any zero at $z_0$ is isolated unless $h$ vanishes on a neighbourhood. An accumulation of distinct zeros at the assumed interior limit point therefore forces the neighbourhood alternative.

Let $E$ be the set of points of $V$ with a neighbourhood on which $h=0$. We have just proved $E\ne \varnothing$, and $E$ is an open set. It is also a closed set relative to $V$: a point of its closure outside $E$ would be approached by distinct points of $E$, hence by zeros, and the same Taylor series argument would place it in $E$. Since $V$ is connected, $E=V$. Thus $h=0$ on $V$, proving the theorem.

A holomorphic function has a zero of order $n$ at $a$ when it is holomorphic on a neighbourhood of $a$ and factors asEquivalently, $f(a)=f^{\prime }(a)=\cdots =f^{(n-1)}(a)=0$ and $f^{(n)}(a)\ne 0$. The order of a zero of a holomorphic function is the degree of the first nonzero term of its Taylor series.

We prove the Casorati-Weierstrass theorem by contradiction. Fix an allowed $\rho$. If the stated image were not a dense set in $\mathbb{C}$, there would be a disc $B(w,ϵ)$ disjoint from it. HenceThe holomorphic function $h(z)=1/(f(z)-w)$ is bounded by $1/ϵ$ on that punctured disc. By the permitted Riemann removable singularity theorem, it extends to a holomorphic function $H$ on the full disc.

If $H(a)\ne 0$, then $w+1/H$ is holomorphic near $a$ and agrees with $f$ away from $a$. This makes $a$ a removable singularity of $f$.

If $H(a)=0$, then $H$ is not identically zero, since $h$ is nowhere zero away from $a$. Its Taylor series therefore has a finite first nonzero coefficient: otherwise the identity theorem would make it identically zero. Write $H(z)=(z-a{)}^{m}K(z)$ with $m\ge 1$ and $K(a)\ne 0$. Thenhas a pole of order $m$ at $a$, since multiplying by $(z-a{)}^m$ gives the nonzero limit $1/K(a)$.

Both cases contradict an essential singularity. Thus

Adjoin $\dot{\mu }=0$ and put $w=z+\mu$. The extended centre manifold is a graph $w=H(x,y,\mu )$ tangent to $w=0$ at the origin. The equations becomeHere $ϵ>0$ is fixed. The centre-manifold invariance equation, obtained by the chain rule, isWrite $H=H_2+O({\rho }^3)$, where $\rho =|x|+|y|+|\mu |$ and $H_2$ is a homogeneous quadratic polynomial. To quadratic order the centre-manifold invariance equation isThe terms $\mu x,\mu y,{\mu }^2$ have zero coefficients: their equations are homogeneous, and the operator $1+y{\partial }_x-x{\partial }_y$ is invertible on these quadratic polynomials. For $H_2=a{x}^2+bxy+c{y}^2$, comparison of coefficients givesHence $a=3/5$, $b=-2/5$ and $c=2/5$, givingSubstitution into the $y$ equation givesThe constant in the big O notation may depend on fixed $ϵ$. Thus the equations printed in part (b) are the retained terms of the local Taylor series, rather than an exact identity for the full reduced vector field.

For the Hopf bifurcation classification in part (a), set $\mu =0$ and use $x=r\cos \theta$, $y=-r\sin \theta$. The radial derivative is $\dot{r}=(x\dot{x}+y\dot{y})/r$. Its cubic angular mean isHere $⟨\cdot ⟩$ denotes the mean over one revolution. Removing the oscillatory cubic terms by a near-identity transformation gives this same cubic radial coefficient in the Hopf normal form. The linear radial coefficient is $5ϵ\mu /2$.

Write ${\tilde{p}}_1$ and $\tilde{q}$ for the quantities computed from the perturbed second sample. The endpoint linear interpolation and its derivative areCompare them on their interpolation interval $0\le x\le h$, with fixed $a>0$. Put $x=th$, $0\le t\le 1$. The Taylor series of the sine gives, uniformly in $t$,Here the second equality compares with $f^{\prime }(0)=a$. More generally, the exact derivative is $f^{\prime }(th)=a\cos (ath)$ andThe input perturbation itself changes the value interpolant by exactly $aϵx/h$, and changes its derivative by exactly $aϵ/h$. ThusFor fixed nonzero $ϵ$, the value approximation has a bounded noise floor: its maximum error approaches $a|ϵ|$, and at $x=h$ that error is exactly $a|ϵ|$. It does not converge uniformly to the exact values as $h\to 0$. However, it does not amplify the absolute sample perturbation on $[0,h]$. The derivative error instead grows without bound like $a|ϵ|/h$.

With exact samples, $ϵ=0$, both approximations do converge on the shrinking interval: the value error is $O(h^3)$ for this particular sine, and the derivative error is $O(h^2)$. With step-dependent noise, value convergence requires $ϵ(h)\to 0$, whereas derivative convergence requires the stronger $ϵ(h)=o(h)$, where $o$ denotes little o notation.

This is noise amplification by differentiation of a linear interpolant. Value interpolation has an absolute perturbation bound independent of $h$ on the sampled interval. Numerical differentiation has a perturbation factor $1/h$, expressing the ill-posedness of numerical differentiation from values with fixed absolute uncertainty. The amplification follows from the mathematical data map even in exact arithmetic; it is distinct from additional roundoff error in a particular implementation. For fixed $x$ outside $[0,h]$, the operation becomes extrapolation, and the value perturbation $aϵx/h$ can also diverge.

The electric field within a conductor in electrostatic equilibrium vanishes. The normal electric field boundary condition, with outward normal vector $+\hat{\mathbf{z}}$, therefore gives $\sigma ={ϵ}_0{E}_z(0^{+})$. Directly below the monopole, $a=0$:For the electric dipole,The exact finite-separation answer iswhose Taylor series agrees with the boxed leading term. Although the real point charges have zero total electric charge, their unequal distances leave negative induced surface charge density immediately beneath them.

Introduce the dimensionless quantities $\tau =\mu t/r_0$ and $w=\mu v/(g{r}_0)$. ThenThe Taylor series at $t=0$ givesThus the initial slope is $g$, as for free fall. At late times,The oblique asymptote is $v=gt/4+g{r}_0/(4\mu )$, approached from below. Direct differentiation givesThe speed increases monotonically with decreasing slope, from $g$ towards $g/4$. It does not reach a terminal velocity in this drag-free model. Without mass accumulation, and still without drag, a drop released from rest has $v=gt$: its acceleration remains $g$, whereas the growing drop continually shares its momentum with newly acquired material.

First justify that the moments being used exist. For every $r>0$,The bound $|y{|}^k\le k!r^{-k}{e}^{r|y|}$ gives $\mathbb{E}|Y{|}^k<\infty$ for all $k$.

If $Y$ has a symmetric distribution, then $Y$ and $-Y$ have the same probability distribution, so for odd $k$,which forces $\mathbb{E}{Y}^k=0$.

Conversely, suppose every odd moment vanishes. For fixed $t\in \mathbb{R}$, the absolute values of the partial Taylor series for $e^{tY}$ are bounded by $e^{|t||Y|}$, whose expected value is finite by the preceding inequality. The dominated convergence theorem therefore permits termwise expectation:The uniqueness theorem for moment-generating functions implies equality of the probability distributions of $Y$ and $-Y$. Hence

It suffices that the moment-generating function be finite on some open interval containing zero:Equivalently, some two-sided exponential moment exists: $\mathbb{E}{e}^{r|Y|}<\infty$ for some $r>0$. For the forward implication, choose $0<r<\delta$ and use $e^{r|Y|}\le e^{rY}+e^{-rY}$; the converse follows from $e^{tY}\le e^{r|Y|}$ for $|t|\le r$.

Under this weaker hypothesis, all moments exist. For $|t|<r$, dominated convergence justifies the Taylor series argument in (i). Vanishing odd moments gives $M_Y(t)=M_{-Y}(t)$ near zero, and the local uniqueness theorem for moment-generating functions gives the symmetric distribution. The reverse implication follows directly from equality of the two probability distributions.

More generally, it suffices that all moments exist and the probability distribution have moment determinacy: $Y$ and $-Y$ then have identical moments and hence identical probability distributions. Existence of all moments by itself does not provide that uniqueness.

At $N=M$, the two spin occupations per site are $x$ and $1-x$. With $\tau =|t|>0$, the trial energy per site isIts derivative is $e^{\prime }(x)=-4\tau \cos (\pi x)+V[4x-2-(2/\pi )\sin (2\pi x)]$. Since $x=0$ is a boundary point, its local minimum test is one-sided: $e^{\prime }(0)=-4\tau -2V$. At $V=-2\tau$, where this derivative vanishes, its Taylor series begins $e(x)-e(0)=(2{\pi }^2\tau /3)x^3+O(x^4)>0$ for small positive $x$. HenceThe same applies to $x=1$ by spin interchange.

The symmetry $e(x)=e(1-x)$ makes $x=1/2$ a stationary point for every $V$. Here $e^{\prime \prime }(1/2)=4\pi \tau +8V$, so the second-derivative test settles all cases except equality. Writing $y=x-1/2$, at $V=-\pi \tau /2$ the Taylor series is $e(1/2+y)-e(1/2)=({\pi }^3\tau /6)y^4+O(y^6)$. ThusThe minimizing branches interpolate continuously. To find all of them, put $u=\pi (1/2-x)$ on $0<x<1/2$. An interior stationary point obeysThe right side decreases strictly from $-\pi /2$ to $-2$, sinceThe positivity follows from $u-\sin u\cos u={\int }_0^{u}2{\sin }^{2}vdv$. There is therefore exactly one such stationary point on each side of $1/2$ when $-2<V/\tau <-\pi /2$. The signs of $e^{\prime }$ show that both are global minima, whereas $1/2$ is then a local maximum. The complete answer isThe pair merges through a supercritical pitchfork bifurcation at $V/\tau =-\pi /2$. Comparing only $e(0)$ and $e(1/2)$ would miss the intervening ground states with partial spin polarization.

For $N\ge 2$, each adjacent pair of walkways has relative rapidity ${\varphi }_N=\mathrm{artanh}(1/N)$. There are $N$ such steps between $P_0$ and $P_N$, so the equal-rapidity chain of collinear Lorentz boosts has total rapidity $N{\varphi }_N$. Thus the actual velocity isIt is strictly below the speed of light for every finite $N\ge 2$. Adding the adjacent velocities as ordinary numbers ignores the relativistic velocity-addition formula.

For large $N$, the Taylor series of the inverse hyperbolic tangent givesConsequentlyMore precisely, $v_N/c=\tanh 1+{\mathrm{sech}}^2(1)/(3N^2)+O(N^{-4})$, so the limit is approached from above. The case $N=1$ would prescribe an adjacent relative velocity equal to the speed of light; it does not define an admissible walkway inertial frame.

The apparent singularity at zero cancels: the Taylor series of the sine givesThus zero is a removable singularity, with extension $f(0)=1/2$. At $z=-2$, the other terms are holomorphic, whereas $(z+2{)}^{-1}$ has a simple pole. ThereforeThere are no other finite singularities. If “all” includes the point at infinity on the Riemann sphere, it is an essential singularity: $f(1/w)=w/(1+2w)+w^2\sin (1/w)-w$ has infinitely many negative powers in its Laurent series at $w=0$.

The geometric series for $(2+z{)}^{-1}$ converges for $|z|<2$, while the cancelled sine terms have an everywhere convergent Taylor series. Hence the requested Laurent series has no negative powers:For example,The extension through zero also makes this a Taylor series. The pole at $-2$ fixes its radius of convergence at two.

Singularities. The complex exponential factors as $F(z)=e^{z/2}{e}^{1/(2z)}$. The second factor has an essential singularity at zero, and multiplication by the nonvanishing holomorphic function $e^{z/2}$ does not remove it. There are no other finite singularities. On the Riemann sphere, infinity is also an essential singularity, since $F(1/\zeta )=F(\zeta )$.

Symmetry and integral formula. The identity $F(1/z)=F(z)$ and uniqueness of the Laurent series on $0<|z|<\infty$ give $F_{-k}=F_k$ for every integer $k$. The Laurent coefficient formula, on the positively oriented unit circle, givesThe imaginary part vanishes by the substitution $\theta \mapsto 2\pi -\theta$. Hence

Polynomial contour integral. For an integer $n\ge 0$ and integer $p$, the binomial theorem gives $(1+z^2{)}^n={\sum }_{j=0}^n(\genfrac{}{}{0ex}{}{n}{j})z^{2j}$. The residue theorem selects $2j-n-p-1=-1$, that is $j=(n+p)/2$. Therefore the correct complete statement isThe printed nonnegative-integer condition alone misses $j\le n$. For example $n=0,p=2$ satisfies that condition, but the integral is zero and the displayed factorial $(-1)!$ is undefined.

Explicit coefficients. Both Taylor series inconverge absolutely and uniformly on every compact annulus away from zero, so their product can be regrouped. For $k\ge 0$, put $r=s+k$ and $s=m$. For arbitrary integer $k$, use the already proved symmetry:Here $I_j$ is a modified Bessel function. The printed series is directly meaningful for $k\ge 0$; for negative $k$ the absolute values, or an appropriately shifted lower summation limit, are essential.

Round-off error arises from representing numbers and carrying out arithmetic at finite precision: storing $1/3$ or subtracting nearly equal numbers can lose information even when the mathematical algorithm is exact. Truncation error arises from replacing a continuous or infinite mathematical operation by a finite approximation, such as discarding higher Taylor series terms in a finite difference. Refining a grid usually reduces truncation error but may increase accumulated round-off error; the two have different origins.

Choose the principal logarithm, so $L(z)=\mathrm{Log}(1-z)$ has branch cut $[1,\infty )$. Near zero its Taylor series givesThus zero is a simple pole of residue one, and one is a logarithmic branch point. The function is holomorphic on $\mathbb{C}\setminus (\{0\}\cup [1,\infty ))$ with this branch. Infinity is also a branch point in the extended-plane description; the function's decay there along the cut plane does not make it a single-valued holomorphic function through infinity.

Ideal and self-avoiding walks. An ideal chain has independent isotropic coarse-grained steps and permits self-intersections in its mathematical model. For $N$ steps of length $b$, the cross terms average to zero, so its root-mean-square end-to-end distance of a polymer chain obeysA self-avoiding walk excludes repeated occupancy or intersection at the segment scale. This represents positive excluded volume in a dilute good solvent. The nonlocal constraint correlates otherwise distant portions of the chain and swells it:The Flory approximation replaces this exponent by $3/5$, by balancing stretching free energy $R^2/(N{b}^2)$ with an interaction contribution $v{N}^2/R^3$. The result is a more extended and less internally dense coil than the ideal walk; self-avoidance changes the statistics, not merely the local step length.

Two ideal-chain regimes. At the theta temperature for a dilute polymer–solvent pair, leading effective segment attractions and repulsions cancel. Large-scale chain statistics are approximately ideal, even though microscopic atoms still have finite size. In a homogeneous polymer melt, surrounding chains screen long-range excluded volume interactions of a chosen chain, again making its equilibrium large-scale size approximately ideal. Local bond-angle correlations must be included in an effective statistical length such as the Kuhn length. Polymer entanglements can slow the dynamics without invalidating this leading equilibrium size law.

Semiflexible DNA. The nucleic acid double helix gives double-stranded DNA a bending stiffness and substantial local orientational persistence. Independent step orientations are therefore a poor description on short scales, while self-avoidance alone does not specify the energy cost of bending. The worm-like chain describes the molecule as a continuously bendable, locally inextensible curve $\mathbf{r}(s)$ parameterized by arc length $s$, with unit tangent $\mathbf{t}(s)=d\mathbf{r}/ds$. Its bending energy iswhere $A$ is bending stiffness. In the free three-dimensional model, the persistence length is ${\ell }_p=A/(k_{B}T)$ andThus ${\ell }_p$ is the contour separation over which tangent alignment falls by the factor $e^{-1}$; it quantifies stiffness relative to thermal bending. The model bridges locally rod-like and globally coil-like conformations. It omits long-range self-avoidance in its simplest form, so sufficiently long dilute DNA can require excluded-volume corrections.

Flexible and rigid limits. Integrating the tangent correlation over the two contour positions gives the end-to-end distance of a worm-like chain:For a very flexible chain on the scale of its total length, $L/{\ell }_p\gg 1$, the exponential is negligible andThis is the ideal chain result with Kuhn length $b_K=2{\ell }_p$ and effective step count $N_K=L/(2{\ell }_p)$. Many persistent sections have decorrelated orientations. The flexible limit means a large ratio $L/{\ell }_p$ within the continuum model's range, not an extrapolation of the persistence length below all molecular length scales.

For a nearly rigid chain, $L/{\ell }_p\ll 1$, the constant and linear terms in the exponential cancel those outside it. The Taylor series givessoThis is the expected rod limit: all tangents are nearly aligned, and the end-to-end distance approaches the arc length.

Helix–coil curves. A helix-coil transition model allows ordered helical runs and disordered coil regions. Use $s$ and $\sigma$ with exactly the free-energy conventions defined in the paper. A negative $\Delta F_{\mathrm{hc}}$ gives $s>1$ and favors the helical state; a positive junction cost gives $0<\sigma <1$ and penalizes starting new domains. The supplied helical fraction is the long-chain form associated with the Zimm-Bragg model.

Its main features follow directly without assuming a sharp transition. Put $Q=(s-1{)}^2+4s\sigma$. For every $\sigma >0$,and differentiating givesThus all curves rise monotonically and cross at $(1,1/2)$. Increasing the boundary free-energy cost reduces $\sigma$, suppresses frequent coil–helix boundaries and makes longer runs favorable. The transition region becomes narrower and steeper: balancing $(s-1{)}^2$ against $4s\sigma$ gives a width of order $\sqrt{\sigma }$ near $s=1$.

For $\sigma =1$, the fraction simplifies to $\boxed{f_h=s/(1+s)}$, the noncooperative two-state result. For $0<\sigma \ll 1$, the steep curve describes cooperative helix formation. In the singular limit $\sigma \downarrow 0$, it approaches zero for $s<1$ and one for $s>1$; its midpoint remains $1/2$ when approached through positive $\sigma$. Setting $\sigma =0$ directly makes the formula indeterminate at $s=1$. For any fixed positive $\sigma$, the given long-chain curve is smooth: it is a cooperative crossover, not a true discontinuity. The plot varies $s$ at fixed $\sigma$; in an actual temperature scan both parameters may change, and a finite chain can further round the crossover.

Choose a coordinate $z$ measured from the sphere centre towards increasing distance from the block, so an atom in the sphere is at height $L+z$. Summing the atom–block potential energy with the sphere's number density givesFor $R/L\ll 1$, a uniformly valid Taylor series over the sphere isThe odd moments vanish by reflection symmetry. The sphere volume is $V_s=4\pi R^3/3$, and its second moment isConsequentlyThe first finite-radius correction is therefore $-4{\pi }^{2}B{n}^2{R}^5/(15L^5)$. There is no relative $R/L$ correction because the first moment about the centre is zero. The leading term is just the number of sphere atoms times the potential at its centre; the correction increases the attraction because the nearer atoms contribute more strongly than the farther atoms lose.

This is the far-field sphere-plane dispersion energy, with Hamaker constant $A_H={\pi }^{2}B{n}^2$. It must be distinguished from the near-contact sphere-plane dispersion force, whose small parameter is the surface gap divided by the radius. Here $L$ is explicitly the centre-to-plane distance.