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A subdivision curve is the limit of repeated refinement of a control polygon, when that limit exists. Each finite stage still consists of line segments; the limiting curve can have a continuous derivative. Linear stationary rules, which use the same fixed weighted averages at every stage, can be studied through their subdivision masks and often represent splines. An interpolation rule retains the old control points, as in binary linear interpolatory subdivision; an approximating rule such as Chaikin subdivision generally moves away from them. Increasing the number of control points alone does not prove convergence or smoothness.

// Target: analysis.bigb

For backward differences $\Delta P_i=P_i-P_{i-1}$, a scalar stationary arity-$a$ subdivision mask has symbol $M(z)={\sum }_i{M}_i{z}^i$. The identity $(1-z)M(z)=B(z)(1-z^a)$ gives the displayed intertwining whenever the quotient is a finite Laurent polynomial. Iterating supplies higher-difference masks. The coefficients $a^{r\ell }{\Delta }^r{P}_i^{\ell }$ are candidate $r$th-derivative controls. A convergent derivative scheme, proved by contraction of its first differences, establishes corresponding smoothness of the original limit under the standard consistent parameterization.

Interrogation of a subdivision limit needs patch/topology navigation, refinement, limit evaluation and derivatives where they exist, with certified bounds and geometric error estimates. On regular patches one may use the corresponding spline representation. At extraordinary subdivision vertices, local subdivision matrices determine limit positions and tangents under the scheme's smoothness conditions. A normal vector may exist even when second derivatives and Gaussian curvature do not; finite refined polygons are approximations rather than exact limit queries.

A periodic orbit of a differentiable interval map is superstable when its periodic-orbit multiplier vanishes. A cycle passing through a critical point has zero multiplier because the derivative product includes $f^{\prime }(c)=0$. For the nonconstant quadratic family $f_{\mu }(x)=1-\mu x^2$, a superstable $2^n$ cycle satisfies $f_{\mu }^{2^n}(0)=0$ and does not have a smaller critical period.

// Target: dynamical-systems.bigb

A tangential boundary derivative differentiates the boundary restriction of a function along a tangent vector. Prescribing it fixes the boundary value only up to a constant on each connected boundary component. This differs from a Neumann boundary condition, which specifies a normal derivative.

For a planar tangential charge-one nematic disclination, $\nabla \cdot Q=-({\lambda }^{\prime }/2+\lambda /r){\mathbf{e}}_r$. This follows by differentiating the tensor dyad, using $\nabla \cdot {\mathbf{e}}_{\theta }=0$ and $({\mathbf{e}}_{\theta }\cdot \nabla ){\mathbf{e}}_{\theta }=-{\mathbf{e}}_r/r$. With bulk density $a{\lambda }^2/2+b{\lambda }^4/4$ and divergence elastic density $K({\lambda }^{\prime }/2+\lambda /r{)}^2/2$, radial variation of $2\pi \int rfdr$ givesFor $a<0$, a regular melted core has $\lambda =O(r^2)$ and an ordered far field has ${\lambda }_0=\sqrt{-a/b}$. Linearizing the bulk derivative gives $\lambda /{\lambda }_0=1-K/(2|a|r^2)+O(r^{-4})$. The core scale is of order $\sqrt{K/|a|}$; the amplitude disturbance has an integrable algebraic tail, rather than exactly compact support.

The tangent line to a differentiable function at $a$ is its affine first-order approximation $f(a)+f^{\prime }(a)(x-a)$. For a differentiable concave function, it is an upper bound throughout a convex interval: the secant-slope inequality and passage to the derivative give $f(x)\le f(a)+f^{\prime }(a)(x-a)$. This upper bound constructs valid envelopes in adaptive rejection sampling.

A tangent vector at a point $p$ of a smooth manifold without boundary records an instantaneous velocity through $p$. Its base point is part of the data; measuring its length additionally requires a Riemannian metric. Intrinsically, a smooth curve $\gamma$ with $\gamma (0)=p$ defines its tangent vector by how every local smooth function changes along it:This is a derivation at a point. Two smooth curves through $p$ represent the same tangent vector if these derivatives agree for every such smooth function; equivalently, their coordinate velocity vectors agree in one manifold chart. The chain rule makes that agreement independent of the chosen manifold chart. ^[ref]

For an embedded submanifold of Euclidean space, the intrinsic tangent vector is represented by the ordinary derivative ${\gamma }^{\prime }(0)$, with its base point $p$ retained. A drawn arrow starting at $p$ represents that tangent vector; its endpoint $p+{\gamma }^{\prime }(0)$ need not lie on the embedded submanifold. Nor is a tangent vector required to be a unit tangent vector: a stationary smooth curve gives the zero vector, which belongs to the tangent space but has no direction to normalize.

Here are explicit calculations. In ${\mathbb{R}}^2$, $\gamma (t)=(t,t^2)$ and $\eta (t)=(t,2t^2)$ have different routes but the same tangent vector $(1,0)$ at $(0,0)$. For $f(x,y)=3x+y$, both give $v(f)=3$. For a circle of radius $R$, $c(t)=(R\cos (\omega t),R\sin (\omega t))$ has tangent vector $(0,R\omega )$ at $(R,0)$: changing $\omega$ changes the speed or reverses travel without changing the circle.

A reparametrization $\beta (s)=\gamma (h(s))$ with $h(0)=0$ gives, by the chain rule, ${\beta }^{\prime }(0)=h^{\prime }(0){\gamma }^{\prime }(0)$. A regular change of curve parameter has $h^{\prime }(0)\ne 0$. The singular choice $h(s)=s^3$ instead givesBoth describe the same parabola, whose tangent line at the origin is still the horizontal axis. The zero velocity diagnoses this singular change of curve parameter, whose inverse is not differentiable at zero; it does not show that the geometric tangent line is absent. The regular curve $\gamma$ supplies its nonzero direction.

The diagram is an analytical illustration of these exact parametric curves, with equal parameter increments marked on each route. It separates geometry from traversal: the tangent line depends on the route, while a velocity vector also depends on how the curve parameter advances.

For a smooth real function satisfying the displayed equation with positive integer $r$, differentiating makes $f^{(j+r)}=f^{(j)}$. On any compact interval, all derivatives are therefore bounded by the maximum of the bounds of $f,f^{\prime },\dots ,f^{(r-1)}$. The Taylor formula with integral remainder consequently has error at most $M{R}^{n+1}/(n+1)!$, which tends to zero uniformly for $|x|\le R$. Hence the function equals its globally convergent Taylor series. For $r=3$ and initial derivatives $(1,0,0)$, it is ${\sum }_{k\ge 0}{x}^{3k}/(3k)!$.

A Taylor polynomial replaces a function near a chosen point by a polynomial with matching derivatives. If $f$ has derivatives through order $n$ at $a$, defineThe factorial denominator makes $T_{n,a}^{(j)}(a)=f^{(j)}(a)$ for $0\le j\le n$. This determines a unique polynomial of degree of a polynomial at most $n$. Order zero matches the value, order one also matches the tangent line, and order two also incorporates the second derivative. Matching more local information is useful because polynomials are easy to evaluate, differentiate, and integrate. The distinction between the finite Taylor polynomial and the infinite Taylor series is developed in OpenStax, Taylor and Maclaurin series.

For a computed example, let $f(x)=\sqrt{1+x}$ and expand at zero. Its first three derivatives giveAt $x=0.1$, the linear approximation is $1.05$ and the quadratic approximation is $1.04875$, compared with $\sqrt{1.1}\approx 1.048808848$. The quadratic correction captures the downward bending of the square root and reduces the absolute difference from about $0.00119115$ to $0.0000588482$.

The Taylor remainder turns this convenient approximation into a controlled calculation. On $[0,0.1]$, $f^{\prime \prime \prime }(x)=3/[8(1+x{)}^{5/2}]$ is positive and at most $3/8$. The Taylor theorem with Lagrange remainder therefore givesConsequently $1.04875<\sqrt{1.1}\le 1.0488125$ follows without using the displayed decimal evaluation as evidence. The Taylor polynomial and the Taylor remainder play different roles: the former supplies an answer that is easy to compute, while the latter specifies what accuracy is justified. Farther from the expansion point, the same low-degree polynomial need not remain accurate.

For a function with derivatives of every order at $a$, its Taylor series is the power seriesIts partial sums are Taylor polynomials. At $a=0$ it is called a Maclaurin series. Forming the coefficients, proving that its partial sums form a convergent sequence, and proving that their limit equals $f(x)$ are separate tasks. A smooth function provides all the coefficients, but a real analytic function is additionally represented locally by the resulting power series.

For the exponential function, every derivative of $e^x$ at zero equals one, givingHere equality holds for every real $x$: the Taylor remainder is bounded by $e^{|x|}|x{|}^{N+1}/(N+1)!$, which tends to zero for each fixed $x$. At $x=1$, the degree-five Taylor polynomial is $163/60\approx 2.71666667$, and the absolute difference from $e$ is less than $e/720$. The factorial denominator eventually overwhelms every fixed power, allowing local derivative information to reconstruct this function throughout the real line.

The geometric series shows why the region of validity matters. For $f(x)=1/(1-x)$, the Taylor series at zero is ${\sum }_{k\ge 0}{x}^k$, with radius of convergence one. For $x\ne 1$, its finite Taylor polynomial satisfies the exact identityAt $x=1/2$ and $N=5$, the remainder is $1/32$. At $x=0.9$, the same degree-five polynomial has value $4.68559$ and remainder $5.31441$, although the Taylor series still converges to $10$ there. Convergence of the infinite expression does not guarantee accuracy from a particular small number of terms. For $|x|>1$, even the terms fail to tend to zero, so this power series cannot represent the function. The standard expansions and the conditions for convergent series are discussed in OpenStax, working with Taylor series.

A stronger caution comes from the flat functionIt is a smooth function with $h^{(n)}(0)=0$ for every $n$, so every Taylor polynomial at zero is zero and its Taylor series converges everywhere to zero. Nevertheless $h(1/2)=e^{-4}>0$. Off zero, repeated differentiation produces a polynomial in $1/x$ times $e^{-1/x^2}$; the exponential decay makes each required limit at zero vanish. This explains how all the local coefficients can miss a positive function arbitrarily close to the centre. Real analyticity supplies the additional condition that rules out this behavior.

Taylor's theorem expresses a sufficiently differentiable function as a finite Taylor polynomial plus a Taylor remainder. One convenient version assumes $f$ is $n+1$ times continuously differentiable on the segment between $a$ and $x$. Thenfor some intermediate point $\xi$, when $x\ne a$. Thus a bound $|f^{(n+1)}|\le M$ on that segment gives $|R_{n,a}(x)|\le M|x-a{|}^{n+1}/(n+1)!$. The intermediate point usually need not be computed: bounding the derivative throughout the segment is enough. This is why local approximation can yield reliable numerical bounds. A finite Taylor polynomial does not require an infinite Taylor series to represent the function.

Historically, the Royal Society holds Brook Taylor's letter of 26 July 1712 to John Machin. Its catalogue connects his work on Kepler's second law with the expansion later published in the 1715 Methodus incrementorum directa et inversa. This is a documented route from an orbital calculation to a general approximation method, rather than a claim that every earlier use of such expansions began with Taylor. See Royal Society archive, letter MS/82/2.

Nonnegative tensor-product surface basis factors yield nonnegative product weights. Their two partition of unity sums multiply to one. If the factors are $C^r$ and $C^s$, mixed derivatives through separate orders $r,s$ are continuous, so the resulting finite or locally finite surface representation is jointly $C^{\min (r,s)}$. Control coefficients can cancel derivative jumps and yield more smoothness; inherited continuity is a guaranteed lower bound, not a claim that every surface has exactly the basis's worst regularity.