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The balanced Carothers equation follows immediately from linear step-growth molecule count and functional-group conversion in step growth:The same Carothers equation extends to unequal group counts once the limiting-group convention is specified.^[ref]

For an original balanced feed of $2000$ monomer molecules, $1900$, $1960$, and $1990$ intermolecular joins give $p=0.95$, $0.98$, and $0.995$. The remaining counts are $100$, $40$, and $10$, hence number-average degree of polymerization values $20$, $50$, and $200$. The last thirty joins quadruple the average even though they consume only another $1.5\%$ of the initial groups of either type. These are exact synthetic counts, not experimental measurements.

The derivative $d{X}_n/dp=(1-p{)}^{-2}$ explains the sensitivity: near $p=0.99$, an absolute conversion uncertainty of $0.001$ changes the predicted degree of polymerization substantially. The exact balanced values at $p=0.989$ and $0.991$ are approximately $90.9$ and $111.1$, instead of $100$. An instrument reporting rounded “complete conversion” cannot distinguish those chain lengths. In a finite no-ring system, even one chain retains two ends; the maximum $p$ is $1-1/N_0$ and $X_n=N_0$. The formal divergence at $p=1$ is a large-system limit, not an infinite chain in a finite flask.

Catastrophic cancellation occurs when subtraction of nearly equal approximations removes their matching leading digits and exposes earlier roundoff error. The subtraction can itself be exact on the stored operands: the information may already have been lost while obtaining them. Goldberg distinguishes this from cancellation of accurately known quantities, which can be harmless.^[ref]

For $x>0$, consider the single real-valued functionThe rationalization of a square-root difference proves equality in exact arithmetic. The two evaluations use the same input $x$, but the direct version first forms a value near one and then subtracts one. The rearranged version keeps the small input in its numerator and divides by a denominator near two.

For binary64 x = 1e-16, the rounded sum 1.0 + x is one, so sqrt(1.0 + x) - 1.0 is zero. The rationalized evaluation returns approximately $5\times 10^{-17}$. A 100-digit Decimal calculation using the exact stored input gives relative errors of one and about $2.50\times 10^{-17}$ respectively. At x = 1e-12, the direct result is about $5.0004445029\times 10^{-13}$, with relative error $8.89\times 10^{-5}$; the rearranged result is about $4.99999999999875\times 10^{-13}$, with relative error $1.21\times 10^{-17}$.

The plot uses Decimal.from_float(x) to preserve the binary input exactly, computes $x/(\sqrt{1+x}+1)$ at 100 decimal digits, and verifies the reference at 160 digits. Exact conversion of a stored binary value is documented by Python.^[ref] Each error is $|\hat{f}-f|/|f|$, with the computed result converted exactly to Decimal before comparison. Hence the curves measure evaluation error for one common input, rather than a mixture of input conversion error and evaluation error. The plotted interval $10^{-16}\le x\le 10^{-1}$ is far above the subnormal number range.

The original problem is well conditioned with respect to relative changes in $x$. Using the derivative $f^{\prime }(x)=1/(2\sqrt{1+x})$ and rationalization of a square-root difference, its relative condition number isIt tends to one as $x$ tends to zero. The direct evaluation instead introduces the intermediate difference $s-1$ with $s=\sqrt{1+x}$; an error of order unit roundoff in $s$ is large relative to $f(x)\approx x/2$. This is a failure of the chosen algorithm's accuracy despite a small condition number for the same real-valued function. The rationalized evaluation avoids that intermediate catastrophic cancellation, retaining relative accuracy near the unit roundoff scale throughout the plotted interval. Recognizing this structure can improve a calculation far more than merely printing extra digits.

An axisymmetric smooth connected soap film spanning equal radius-$R$ rings at heights $\pm H$ has stationary profile $r=k^{-1}\cosh (kz)$, $k>0$. The boundary equation is $R/H=\cosh u/u$, $u=kH$. Its unique minimum occurs at $u\tanh u=1$, because the derivative numerator is $u\sinh u-\cosh u$. If $A$ is this root, the minimum is $\cosh A/A=\sinh A$. There are no catenoids below that ratio, one at equality, and two above it. Existence of stationary connected films does not assert global minimality against pinching or disconnected competitors.

Suppose $f$ is a holomorphic function on an open set containing the closed disc $|z-a|\le R$, with $R>0$. For any nonnegative integer $n$ and $|w-a|<R$, the counterclockwise boundary contour satisfiesThe case $n=0$ is the Cauchy integral formula. On a smaller closed disc, the denominator stays uniformly away from zero, allowing repeated differentiation under the contour integral. This proves that a holomorphic function has complex derivatives of every order. ^[ref]

At the centre, if $M={\max }_{|z-a|=R}|f(z)|$, taking absolute values gives the Cauchy estimate $|f^{(n)}(a)|\le n!M/R^n$. Boundary size therefore limits interior variation. For $f(z)=z^2$ on the unit disc, $M=1$, and the $n=2$ estimate is exact: $|f^{\prime \prime }(0)|=2$. This is one quantitative reason that holomorphic functions are so rigid.

For $a<b$ and real-valued functions $f,g$ that are continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, there exists $\xi \in (a,b)$ withApply Rolle's theorem to the linear combination $h(x)=[f(b)-f(a)]g(x)-[g(b)-g(a)]f(x)$, which has equal endpoint values. The displayed product identity permits zero derivatives. For the ratio form, additionally assume $g^{\prime }(x)\ne 0$ for every $x\in (a,b)$. Then Rolle's theorem ensures $g(b)\ne g(a)$, so both denominators below are nonzero:Neither derivative needs to be continuous, and $f^{\prime }$ may vanish. Choosing $g(x)=x$ recovers the mean value theorem; the ratio form also gives the zero-over-zero endpoint case of L'Hôpital's rule. ^[ref]

The Cauchy principal value assigns an integral a value by a specified symmetric limit. Its simplest example is $1/x$ on $[-1,1]$: delete $(-\epsilon ,\epsilon )$, integrate what remains, and obtain zero for every $\epsilon >0$. The positive signed area under a graph cancels the negative one, although each grows without bound as the gap closes. This is a useful cancellation rule, but it does not make the ordinary improper integral exist.

Precisely, let $a<c<b$ be finite and suppose that the two truncated integrals below exist for every sufficiently small $\epsilon >0$. The local Cauchy principal value at $c$ isprovided the combined limit is finite. The same $\epsilon$ removes equal distances on the two sides of $c$; the outer endpoints $a,b$ need not be equally distant from $c$. By contrast, the ordinary improper integral requires both one-sided limits to be finite separately before adding them. Whenever those separate limits exist, the Cauchy principal value agrees with their sum.^[ref]

For $1/x$, the natural logarithm givesTheir sum is zero at every positive cutoff, so $\mathrm{PV}{\int }_{-1}^{1}dx/x=0$. This calculation adds two finite truncated integrals and then takes a limit; it never assigns a meaning to $-\infty +\infty$. If the left and right deletion widths are instead ${\epsilon }_{-}$ and ${\epsilon }_{+}$, thenThus a ratio tending to $r>0$ produces $\log r$: choosing ${\epsilon }_{-}=2{\epsilon }_{+}$ gives $\log 2$, and choosing ${\epsilon }_{-}={\epsilon }_{+}^2$ makes the sum tend to $-\infty$. Both widths can tend to zero without selecting a unique answer. The local Cauchy principal value selects the ratio $1$.

Unequal outer endpoints leave a finite contribution even with symmetric local deletion. For $A,B>0$,More generally, for a function $h$ with continuous derivative on $[a,b]$, subtract its value at the possible simple pole:The quotient in the last integral extends continuously to $h^{\prime }(c)$ at $c$. This formula explains both the cancellation and a practical way to compute it: integrate a continuous remainder instead of subtracting two large, nearly equal divergent contributions.

A finite Cauchy principal value is weaker than absolute convergence. Indeed,Taking the absolute value destroys the cancellation. Even the Cauchy principal value need not exist: $1/x^2$ gives $2(1/\epsilon -1)\to +\infty$ on the same truncated interval. Absolute integrability makes the cutoff prescription unnecessary; an ordinary improper integral can also converge without absolute convergence, as the Dirichlet integral illustrates. The three notions should therefore be checked separately.

There is an independent Cauchy principal value at infinity. For a function integrable on every finite interval, definewhen the limit is finite. The ordinary improper integral instead requires convergence of the two tails separately.^[ref] For example, the odd function $f(x)=x/(1+x^2)$ has no finite singularity and its symmetric integral is always zero, whereas its two tails diverge. Unequal truncations giveLetting $B/A\to r>0$ as $A,B\to \infty$ yields $\log r$. Symmetry at infinity is therefore a separate choice from symmetry near a finite pole. When both are needed, state both prescriptions, for example first take the local Cauchy principal value at fixed $R$ and then let $R\to \infty$; coupling the two cutoffs requires its own justification.

The local cancellation also defines the principal-value reciprocal distribution. Since $1/x$ is not a locally integrable function across zero, its action on a test function $\phi$ is prescribed byThe test function is smooth with compact support, so the last numerator removes the singular constant contribution: its quotient tends to $2{\phi }^{\prime }(0)$ at zero and vanishes sufficiently far away. This defines a distribution, not a finite pointwise value of $1/x$ at zero. It extends to Schwartz functions as a tempered distribution, and is also the distributional derivative of the logarithmic modulus.^[ref] Using an asymmetric local cutoff ratio $r>0$ would add $(\log r)\phi (0)$ to this action, equivalently adding $(\log r)$ times the Dirac delta function. Thus the cutoff choice matters even after passing to distributions.

Why keep this prescription? A Fourier transform naturally encounters such a kernel. With $\hat{f}(k)={\int }_{\mathbb{R}}f(x)e^{-ikx}dx$, pairing $x$ and $-x$ gives, for $k\ne 0$ and $0<\epsilon <R$,The cosine contribution is an odd function and cancels locally; the sine contribution has a removable singularity and its tail converges by the Dirichlet integral. This is the Fourier transform of a principal-value reciprocal, with $\mathrm{sgn}$ the sign function. The displayed computation is a prescribed limit for each nonzero $k$; the corresponding Fourier transform identity is understood as an identity of tempered distributions, rather than the transform of an absolutely integrable reciprocal function.

The Hilbert transform uses exactly this principal-value reciprocal distribution as a convolution kernel. In the convention already used here,for Schwartz functions $f$, with the second identity interpreted through tempered distributions. Hence the singular integral has a simple description as a Fourier multiplier.^[ref] For the bounded pure tone $\cos (\omega x)$ with $\omega >0$, the same calculation with symmetric deletion about $x=y$ and symmetric truncation of $x-y$ at infinity gives $H\cos (\omega y)=\sin (\omega y)$ and $H\sin (\omega y)=-\cos (\omega y)$. These tones lie outside the Schwartz space; their prescribed Cauchy principal values explain the quarter-cycle shift in the figure. Both figures are calculated mathematical illustrations. Reversing the Hilbert denominator reverses these signs; NIST uses that opposite convention.^[ref]

Intersecting a parametric curve with a plane reduces to isolating zeros of a scalar function. A certified range for $F$ rejects intervals that do not contain zero. On an interval whose derivative range excludes zero, a sign-changing endpoint bracket gives exactly one root, which bisection method or a safeguarded Newton root-finding iteration refines. Other intervals are subdivided conservatively; tangencies and coplanar pieces need multiplicity or identity tests supported by the representation. An unresolved small interval is a possible contact, not a proof of an intersection. With finitely many transverse intersections and convergent enclosures, isolation terminates to any prescribed tolerance.

This centered quadratic cardinal B-spline weights each original control ordinate at its uniformly spaced abscissa. Its translates sum to one. It and its first derivative match at half-integer knots, while the second derivative can jump. The limit graph is ${\sum }_i{y}_i\beta ((x-x_i)/h)$; one ordinate influences exactly three original intervals. Refinement of a quadratic adds $a{h}^2/4$ in the limit, yielding this basis by extending each three-point neighborhood quadratically.

Chaikin subdivision generates a rounded subdivision curve by repeatedly cutting the corners off a control polygon. Each edge contributes two new control points, one a quarter of the way along the edge and the other three quarters of the way along it. Connecting these new control points in edge order replaces the old corners by shorter line segments. This makes Chaikin subdivision a useful way to draw a smooth outline from a small, editable set of control points, without evaluating B-spline basis functions at each drawing step. Joy presents both the corner-cutting rule and its closed version.^[ref]

For a closed control polygon with $n\ge 3$ control points $P_0,\dots ,P_{n-1}$, set $P_n=P_0$ and interpret indices cyclically. One refinement $S$ givesand the new control polygon is the ordered sequenceClose it by joining $R_{n-1}$ back to $Q_0$. Use the old control points for every average in the same refinement; updating them in place would define a different algorithm. One refinement produces $2n$ stored control points, and $k$ refinements produce $2^{k}n$, counting entries even if some coordinates coincide. Open control polygons need an explicit endpoint convention: applying only the interior edge rule discards the original endpoints, whereas retaining endpoints changes the boundary rule. The closed analysis below uses cyclic indices throughout.

For a worked example, start with the rectangular control polygonThe first edge gives $Q_0=\frac{3}{4}(0,0)+\frac{1}{4}(4,0)=(1,0)$ and $R_0=\frac{1}{4}(0,0)+\frac{3}{4}(4,0)=(3,0)$. Applying Chaikin subdivision to all four edges, including the closing edge, gives exactly eight control points in this order:Thus the rectangle becomes an octagonal control polygon. Two refinements have $16$ control points, and five have $128$. The illustration shows finite control polygons at stages $0,1,2,5$; the last is already visually rounded but still consists of line segments.

The convex hull property follows directly from the weights. Each new control point is a convex combination of the endpoints of one old edge, because the weights are nonnegative and sum to one. Consequently, if $H_k$ is the convex hull of the stage-$k$ control points, then $H_{k+1}\subseteq H_k$. Every later line segment, and every limit point, stays in the original convex hull. This gives a useful geometric enclosure. It also explains why Chaikin subdivision is approximating rather than an interpolation rule: the rectangle's four original corners are outside the first octagon's convex hull, so the limiting curve cannot pass through them. A refinement control polygon generally lies off the limiting curve as well.

For the cyclic uniform rule, the closed limit is the periodic quadratic uniform B-spline curve determined by the original control points. Uniform refers to equal spacing in the spline knot sequence, not equal physical lengths of edges of the control polygon. An infinite uniform control sequence has the corresponding local description; arbitrary open endpoint conventions need separate treatment. Joy derives the quarter-point refinement by splitting quadratic uniform B-spline spans.^[ref]

Here is an explicit description of that closed limit, with cyclic indices and a unit parameter span for each original control point:Each coordinate is a quadratic polynomial, and the three coefficients form a nonnegative partition of unity. The span runs from $(P_{i-1}+P_i)/2$ to $(P_i+P_{i+1})/2$. Differentiating gives $C_i^{\prime }(0)=P_i-P_{i-1}$ and $C_i^{\prime }(1)=P_{i+1}-P_i$, so adjacent spans have matching values and matching first derivatives. Their second derivatives generally differ. Thus the parameterized curve is $C^1$, generally not $C^2$; degenerate control points can still make its tangent vector vanish. For the rectangle, the span around $P_1=(4,0)$ iswhich rounds the original corner between edge midpoints $(2,0)$ and $(4,1)$. For equally spaced abscissae, the centered Chaikin basis function gives the scalar graph version of this quadratic B-spline representation and has support of width three original grid intervals.

Another useful property is affine equivariance of a geometric basis: refining and applying an affine map commute. For $T(x)=Ax+b$,and the same calculation holds for $R_i$. The sum-to-one weights account for the translation. Therefore $T(S(P))=S(T(P))$ at every refinement, and also for the limit. A designer can apply a shear mapping, translation, or other affine map to the original control polygon and obtain exactly the transformed outline.

For example, compose the shear mapping $(x,y)\mapsto (x+y/2,y)$ with the translation $(1,-1)$ to obtain $T(x,y)=(x+y/2+1,y-1)$. The rectangle's control points become $(1,-1),(5,-1),(6,1),(2,1)$. On its right edge, cutting first gives $Q_1=(4,1/2)$ and then $T(Q_1)=(21/4,-1/2)$. Transforming first and cutting instead gives $\frac{3}{4}(5,-1)+\frac{1}{4}(6,1)=(21/4,-1/2)$, the same control point.

Historically, George M. Chaikin published “An algorithm for high-speed curve generation” in Computer Graphics and Image Processing 3(4), December 1974, pages 346–349. Its emphasis was practical curve generation through a recursive algorithm using simple arithmetic, with applications including machine control and three-dimensional curves. The lasting appeal of Chaikin subdivision is how an elementary local average creates an entire smooth B-spline curve.^[ref]

The chain rule calculates the derivative of a function composition by combining local rates. If the inner function $f$ is differentiable at $a$ and the outer function $g$ is differentiable at $f(a)$, thenThe evaluation point matters: the outer derivative is evaluated at the inner output $f(a)$, not at $a$. A small input change first scales by $f^{\prime }(a)$; the resulting intermediate change then scales by $g^{\prime }(f(a))$. In terms of linear approximations, composing those two local descriptions multiplies their scale factors. ^[ref].

For example, $h(x)=\sin (x^2)$ applies a quadratic function and then the sine. With $f(x)=x^2$ and $g(u)=\sin u$, the chain rule givesHere all angles are in radians. Near $x=1$, increasing the input by $\Delta x$ changes $x^2$ by about $2\Delta x$, and the sine changes by about $\cos 1$ times that intermediate change. For $\Delta x=0.01$, the linear approximation predicts an output increase of $0.0108060$; the exact increase is $\sin (1.0201)-\sin 1\approx 0.0106894$.

The same rule handles several nested stages. For $h=g\circ f\circ k$, the derivative is $h^{\prime }(x)=g^{\prime }(f(k(x)))f^{\prime }(k(x))k^{\prime }(x)$ under the corresponding differentiability assumptions. Writing each intermediate output prevents evaluating a derivative at the wrong stage. It also makes the chain rule useful for propagating sensitivities through a mathematical model, where one measured quantity influences another through several transformations.

For a quick proof idea, replace $f$ near $a$ by its linear approximation, then replace $g$ near $f(a)$ by its linear approximation. The composite leading term is their product; differentiability ensures the remaining error is smaller than the original input change. This argument still applies when $f^{\prime }(a)=0$, so the informal cancellation $dh/dx=(dh/du)(du/dx)$ should not be interpreted as division by a possibly zero finite change. The hypotheses are sufficient, but are not necessary for every composite to be differentiable, as the nondifferentiable inner function with a differentiable nondegenerate composite example shows.

The change of variables formula expresses an integral in coordinates that make its integrand or its region simpler. Three ingredients change together: the region, the function being integrated, and the element of area or volume. The last change is measured by the absolute Jacobian determinant; forgetting it amounts to counting transformed cells as though their sizes had stayed the same.

In one dimension, substitution is the integrated chain rule. If $g$ is continuously differentiable on $[a,b]$ and $f$ is continuous on its image, thenThese are directed definite integrals: if $g$ decreases, the transformed endpoints reverse. This identity does not require $g$ to be injective, because an antiderivative composed with $g$ already accounts for any backtracking.^[ref]

For a concrete computation, set $u=1+x^2$, so $du=2xdx$. The endpoints $x=0,1$ become $u=1,2$:The numerator is precisely the derivative of the expression in the denominator. Substitution reveals the familiar natural logarithm hidden inside this rational function. Keeping the original endpoints after changing the integrand would describe a different integral.

For the elementary two-dimensional theorem, let $T:U\to V$ be a $C^1$ diffeomorphism between open sets in the plane, let $S\subset U$ be a bounded region with piecewise smooth boundary, and let $f$ be continuous on $T(S)$, with the displayed integrals finite. ThenHere $DT$ is the Jacobian matrix. The diffeomorphism assumption means a bijection with a continuously differentiable inverse, and ensures a nonzero Jacobian determinant. This excludes overlapping sheets; the more general area formula handles multiplicity.^[ref]

Take $S=[0,1{]}^2$ and the linear mapIts inverse is $(u,v)=((x-y)/2,y)$. The square becomes a parallelogram with vertices $(0,0),(2,0),(3,1),(1,1)$: base two, height one, area two. Consequently ${\iint }_{T(S)}1dxdy={\iint }_{S}2dudv=2$. A nonconstant integrand changes too: ${\iint }_{T(S)}xdxdy={\int }_0^1{\int }_0^1(2u+v)2dudv=3$.

The direction of the map matters when selecting the factor. Here $T$ goes from the new coordinates to the original coordinates, so the factor is two. Its inverse has Jacobian determinant $1/2$, appropriate for converting area in the opposite direction. Writing down the map before computing its Jacobian matrix prevents a reciprocal-factor error.

Now reflect the image vertically: $\tilde{T}(u,v)=(2u+v,-v)$ has Jacobian determinant $-2$, but still produces area two. The sign records reversal of the orientation of a vector space; ordinary area uses its absolute value. For oriented differential forms, the pullback of a differential form instead retains the signed determinant.^[ref]

The geometric beauty is that a nonlinear differentiable map has a local linear approximation: tiny rectangles behave approximately like these parallelograms. Summing their scaled contributions gives the change of variables formula. In polar coordinates this factor is $r$, explaining $dxdy=rdrd\theta$; equal angular steps cover more area farther from the origin. Choosing suitable coordinates can simplify both the boundary and the integrand.^[ref]