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An affine combination is a weighted point $P={\sum }_{i=0}^k{\lambda }_i{P}_i$ with ${\sum }_{i=0}^k{\lambda }_i=1$. The coefficients can be negative. The sum is meaningful in an affine space without selecting a preferred origin: relative to any point $O$, it means $P=O+{\sum }_i{\lambda }_i(P_i-O)$. Affine maps preserve these weighted points. ^[ref]

For an original example in the Euclidean plane, take $A=(0,0)$, $B=(2,0)$ and $C=(0,2)$. Then $P=\frac{1}{4}A+\frac{1}{2}B+\frac{1}{4}C=(1,\frac{1}{2})$ lies inside their triangle. Its coefficients are nonnegative, making it a convex combination. In contrast, $R=-\frac{1}{2}A+\frac{3}{2}B=(3,0)$ is beyond $B$: it is an affine combination but not a convex combination of these three points. Negative weights allow extrapolation, rather than only interpolation inside the convex hull.

For the affine map $T(x,y)=(2x+y+1,y-2)$, the same weights produce $T(P)=(\frac{7}{2},-\frac{3}{2})$ from $T(A),T(B),T(C)$, and produce $T(R)=(7,-2)$ from $T(A),T(B)$. Thus one can compute a weighted point before transforming the vertices or afterwards, obtaining exactly the same point.

This is useful whenever an object is built from weighted control points: its construction commutes with an affine map, so transforming the control points transforms the constructed object as well. The sum of the weights is the essential condition; no assumption about lengths or angles is needed.

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]

A control point is an editable geometric handle used to specify a curve or surface through a basis or refinement rule. The handle is part of the representation; it need not be a point on the resulting curve. In a Bézier curve, the control points are the geometric coefficients of the Bernstein basis. Joining them in order gives a control polygon, which makes the handles easy to manipulate without drawing the curve as a chain of line segments.^[ref]

For example, use the cubic Bézier curve with control pointsIts Bernstein basis representation givesThe greatest height is $1.5$, so neither interior control point, both at height $2$, lies on this curve. The endpoint control points do lie on it: $C(0)=P_0$ and $C(1)=P_3$. This separates interpolation, which requires passage through prescribed data, from an approximating representation, which uses handles to shape a curve without requiring passage through every handle. If measured points must be interpolated, choose control points to satisfy those constraints; simply using the measured points as the handles of a Bézier curve does not enforce interpolation.

Now move only $P_1$ by the vector $h=(0,1)$, giving ${\tilde{P}}_1=(1,3)$. With the Bernstein basis and parameter $t$ unchanged, the new curve isAt $t=1/2$ this gives $\tilde{C}(1/2)=(2,15/8)$, a displacement $(0,3/8)$ despite moving the handle by a full unit. Both endpoints stay fixed because the coefficient $3t(1-t{)}^2$ vanishes there. Thus dragging a control point shapes a whole range of the curve; it does not generally drag an arbitrary selected point on the curve by the same vector. The control-point influence on a curve tells us exactly how much each parameter value moves.

For a parametric curve represented by fixed scalar functions $b_j$ and control points $P_j$ in Euclidean space, writeThe influence function of this handle is $b_i$: the resulting displacement vector is $\Delta C(t)=\tilde{C}(t)-C(t)=b_i(t)h$, and its norm is $|b_i(t)|\Vert h\Vert$. These are exact finite changes, not only a first-order estimate. Where $b_i$ is negative the displacement points opposite to $h$. Where $b_i(t)=0$, the curve is unchanged at that parameter.

For the cubic Bézier curve above, the Bernstein basis influence function is $b_{1,3}(t)=3t(1-t{)}^2$. Its derivative isIt increases up to $t=1/3$ and decreases thereafter on $[0,1]$. Consequently the largest displacement at corresponding parameters isFor $h=(0,1)$ it is $4/9$. This is a comparison at the same parameter. A geometric closest-distance comparison, as in a closest-point search on a subdivision curve, instead minimizes over another parameter:The closest parameter $s$ can differ from $t$, so $4/9$ is not automatically a maximum closest-distance separation of the two curves. Neither parameter is assumed to be arc length.

The support of a function $b_i$ is the closure of the parameter values where $b_i\ne 0$. It bounds where a nonzero handle move can act, but may include boundary points where the actual displacement is zero. Here the support of a function $b_{1,3}$ is all of $[0,1]$ relative to the parameter domain, while the endpoints are unaffected. In contrast, a B-spline of order $k$ has support within $[t_i,t_{i+k}]$ in its spline knot sequence, so its handle affects only that parameter range; the support and Gram bandwidth of B-splines describes the same locality in another application. Changing the spline knots or the basis along with the handle requires a different comparison.

When the basis functions form a partition of unity, ${\sum }_j{b}_j(t)=1$, each evaluation is an affine combination of the control points. Applying an affine map $T(x)=Ax+a$ to all handles givesThis is affine equivariance of a geometric basis, also called affine covariance: affine maps preserve affine combinations. In particular, translating every handle by $h$ translates the whole curve by $h$, whereas moving just one handle produces $b_i(t)h$. The Bernstein basis has this partition of unity property.^[ref] Nonnegative weights further make evaluations convex combinations, giving the convex hull bound illustrated by the control polygon; nonnegativity is not needed for affine covariance.

For a proof outline of the influence formula, subtract the two basis representations: every unchanged coefficient cancels, leaving $b_i(t)(P_i+h)-b_i(t)P_i=b_i(t)h$. The same subtraction with several moved handles gives $\Delta C(t)={\sum }_i{b}_i(t)h_i$. No claim about interpolation, closest-point search on a subdivision curve or endpoint motion follows until the actual basis values are examined.

An exposure-averaged light curve models what a finite exposure measures. For a constant response during an exposure of duration $\Delta t_i>0$, centred at $t_i$, the expected mean brightness isThe integral has units of brightness times time; division restores the units of $F$. Observed points fluctuate around this mean. A response that changes during the exposure must instead be included inside the integral, as in the first figure's known drift. Exposure duration and the interval between samples are distinct: there may be dead time or missing exposures.

A short dip can be blurred even with perfectly noiseless measurements. For a toy rectangular dip of fractional depth $\delta$ and duration $T<\Delta t$, an exposure containing the whole dip has fractional depth $\delta T/\Delta t$: averaging hides part of its depth. A $1\%$ dip lasting ten minutes therefore appears only $0.333\%$ deep in a thirty-minute exposure. No interpolation recovers its shape from that single number without additional assumptions. Forward modelling should average the predicted light curve over each exposure before comparing it with the measurements. Kipping's analysis of exoplanet transit photometry demonstrates how ignoring this averaging biases inferred system parameters. ^[ref].

Estimating a coordinate transformation that aligns corresponding content in images. The transformation family, similarity objective, interpolation and regularization are separate modeling choices. Intensity agreement alone does not guarantee physically correct correspondence.

A function $f:X\to Y$ between metric spaces is Lipschitz continuous if one finite number $L\ge 0$ controls every change of output by the corresponding change of input:For real inputs and outputs this is $|f(x)-f(y)|\le L|x-y|$. The same bound must work for every pair in the specified domain of a function. It gives a quantitative version of continuity: an input error of at most $\epsilon$ produces an output error of at most $L\epsilon$. The least possible bound is the Lipschitz constant. Lebl presents the real and metric-space definitions and their connection with uniform continuity. ^[ref]

Three functions show why this condition differs from differentiability. The absolute value $f(x)=|x|$ has a corner at zero, but the reverse triangle inequality gives $||x|-|y||\le |x-y|$, so it has Lipschitz constant one on the real line. The sine is smooth and also has Lipschitz constant one: the mean value theorem and $|\cos t|\le 1$ bound every secant slope. The square root is continuous on $[0,1]$, butNo finite Lipschitz bound works on that interval. At $h=1,1/4,1/16,1/64$, the ratios are $1,2,4,8$. Smaller input changes near zero therefore require larger amplification factors.

For a real function, Lipschitz continuity means that the absolute slopes of all secant lines are bounded. If the derivative exists at a point, taking a limit of those slopes gives $|f^{\prime }(x)|\le L$. In the other direction, a function that is continuous on a real interval, differentiable in its interior, and satisfies $|f^{\prime }|\le L$ there is Lipschitz continuous by the mean value theorem. The interval hypothesis matters: the theorem uses the segment connecting the two inputs.

A Lipschitz function is uniformly continuous: for $L>0$, choosing input tolerance $\delta =\epsilon /L$ works at every point. If $L=0$, the function is constant. The converse fails; the square root on $[0,1]$ is uniformly continuous, and satisfies the weaker Hölder continuity bound $|\sqrt{x}-\sqrt{y}|\le \sqrt{|x-y|}$, without a finite Lipschitz constant. These distinctions are developed in Lebl's worked examples. ^[ref]

The usefulness is controlled sensitivity. Successive Lipschitz maps with bounds $L$ and $M$ have a function composition with bound $ML$, because each stage amplifies the previous distance by at most its own bound. For an ordinary differential equation, Lipschitz continuity in the unknown state is the condition that makes nearby trial solutions controllably close in the Picard-Lindelöf theorem. For incomplete data, the McShane extension theorem fills in unobserved values while preserving the same bound. This one inequality connects error estimates, uniqueness, and interpolation.

For the centered Cardinal cubic B-spline $\beta$ with knot values $1/6,4/6,1/6$, write $s(x)={\sum }_{j=-1}^{m+1}{c}_j\beta (x/h-j)$ on $[0,mh]$. interpolation gives $c_{i-1}+4c_i+c_{i+1}=6u(ih)$ for $i=0,\dots ,m$. Since ${\beta }^{\prime \prime }(0)=-2$ and ${\beta }^{\prime \prime }(\pm 1)=1$, the natural endpoint conditions $s^{\prime \prime }(0)=s^{\prime \prime }(mh)=0$ give the displayed extra two rows. These supply the two equations missing from interpolation alone.

For linear interpolation at zero and $h>0$, an error $\delta$ in the second sampled value, with the first exact, changes the linear interpolant by $\Delta p(x)=\delta x/h$. On $[0,h]$, its uniform norm is exactly $|\delta |$, whereas the derivative change is the constant $\delta /h$. With independent absolute bounds $|{\delta }_0|,|{\delta }_1|\le \delta$ on both samples, the value change remains at most $\delta$ and the derivative change is at most $2\delta /h$, attained for opposite endpoint errors.

Thus bounded absolute sample noise can remain bounded under value interpolation while becoming unbounded under numerical differentiation as $h\to 0$. Fixed nonzero sample noise also prevents exact convergence of the value approximation; stability against perturbations and convergence to exact data are different assertions. For a second-sample error $aϵ(h)$ with fixed $a>0$, value convergence requires $ϵ(h)\to 0$, while derivative convergence requires $ϵ(h)=o(h)$ in little o notation, in addition to vanishing truncation error.

Use the usual approximation-space convention $P_n=\{p:\mathrm{deg}p\le n\}$. The source's phrase “degree $n$” must include lower degrees for its later space and interpolation statements; the consequence of a literal exact-degree reading is addressed in part (d).

Order the alternating points as $x_0<\cdots <x_{n+1}$. If $q$ had strictly smaller error than $E_n$, the valueswould have the alternating signs of $(u-p)(x_j)$: the second term has magnitude strictly less than $E_n$. The intermediate value theorem would then give at least $n+1$ distinct zeros of $q-p$, one in each gap. A nonzero polynomial of degree at most $n$ cannot have that many zeros. Thus no such $q$ exists, provingThe remainder of the equioscillation theorem says that a best uniform approximating polynomial exists, is unique, and is characterized by this $n+2$-point alternating extreme-error condition. For positive error, alternation is necessary as well as sufficient. If $u$ itself lies in $P_n$, the unique best polynomial is $u$ with zero error.

Let $\omega (t)={\prod }_{r=0}^n(t-x_r)$. For $x$ not one of the nodes, put $c=(u(x)-p_n(x))/\omega (x)$ and defineThis function has zeros at the $n+1$ interpolation nodes and at the additional point $x$. Repeated Rolle's theorem gives a point $\eta$ between the outermost of these points such that $F^{(n+1)}(\eta )=0$. Since $p_n^{(n+1)}=0$ and ${\omega }^{(n+1)}=(n+1)!$,Substitution yieldsAt an interpolation node both sides are zero, so any $\eta \in [-1,1]$ gives the identity there. The assumed $C^{n+1}$ regularity permits all applications of Rolle's theorem.

Choose the nodes as the $n+1$ roots of the Chebyshev polynomial $T_{n+1}$. Its leading coefficient is $2^n$, so the monic nodal product is $\omega =2^{-n}{T}_{n+1}$. Since $|T_{n+1}|\le 1$ on $[-1,1]$, part (b) bounds this particular interpolant byThe best error cannot exceed the error of a particular member of $P_n$. ThereforeThe choice of Chebyshev roots controls the maximum of the nodal product, not the size of any individual interpolation coefficient.

Use the Taylor expansion $\sqrt{1-x}=1-x/2-x^2/8+O(x^3)$. Expanding the rational form givesMatching through degree two yields $a_0=1$, $a_1-b_1=-1/2$, and $b_1^2-a_1{b}_1=-1/8$. The last equation becomes $b_1/2=-1/8$, soThis is the $[1/1]$ Padé approximant about zero. Its pole is at four, outside the interpolation interval; its order of matching is local and does not assert uniform minimax optimality.

Let $\beta (q)$ be the centered nondimensional cubic B-spline, even and supported on $[-2,2]$. The outer cubic must be $b(2-|q|{)}^3$ to make the function and its first two derivatives vanish at the support edge. Write its inner part as $A+B{q}^2+C|q{|}^3$. Matching the value and first two derivatives at $|q|=1$ gives $A=4b$, $B=-6b$, $C=3b$. Its nonzero knot values are then $4b,b,b$, so requiring their sum to be one fixes $b=1/6$. ThusIn particular $\beta (0)=2/3$, $\beta (\pm 1)=1/6$.

With $h=1/m$, represent the spline on $[0,1]$ as $s(x)={\sum }_{j=-1}^{m+1}{c}_j\beta (x/h-j)$. There are $m+3$ unknown coefficients. interpolation at $x_i=ih$, $0\le i\le m$, givesThe natural endpoint conditions, using ${\beta }^{\prime \prime }(0)=-2$ and ${\beta }^{\prime \prime }(\pm 1)=1$, giveThese $m+3$ equations are the required matrix system: the $m+1$ interpolation rows have coefficients $(1,4,1)$ at consecutive positions and right sides $6u(x_i)$; the two endpoint rows have $(1,-2,1)$ and zero right sides. No solution of that system is required. The scaling uses $\beta (x/h-j)$, without an extra $1/h$, because normalization is by knot values rather than by integral.

The weak equation implies $u^{\prime \prime }=x^{2}u-f\in L^2(a,b)$ in distributions, hence $u\in H^2(a,b)$. Its nodal interpolant satisfiesThese follow interval by interval: the interpolation error vanishes at both endpoints, and its derivative has mean zero, so two Poincaré inequalities apply. If $M$ and $\gamma$ are the continuity and coercivity constants, Céa's lemma givesAll constants are independent of the mesh size.

Put $e_h=u-u_h$ and $w=z-I_{h}z$. Galerkin orthogonality gives $a(e_h,I_{h}z)=0$, so the dual identity becomesOn $I_i=(x_{i-1},x_i)$, $u_h^{\prime }$ is constant and $w$ is zero at both endpoints. Consequently ${\int }_{I_i}{u}_h^{\prime }{w}^{\prime }=u_h^{\prime }[w{]}_{x_{i-1}}^{x_i}=0$. Defining the element residual bygivesNo derivative-jump terms survive, because this interpolation error vanishes at every mesh node. This is the mechanism of the explicit one-dimensional dual residual bound in L2.

A $C^2$ piecewise cubic supported on $[0,ph]$ has the truncated-power representation with indices $0,\dots ,p$. Vanishing for $x\ge ph$ imposesafter expanding its cubic exterior polynomial; powers of $h$ cancel. For $p\le 3$, the Vandermonde matrix columns are independent, forcing every coefficient to be zero. For $p=4$, coefficients proportional to $(1,-4,6,-4,1)$ satisfy these conditions because a fourth finite difference annihilates cubics.

Normalize to get the Cardinal cubic B-splineAt its five support knots its values are $0,1/6,2/3,1/6,0$, so their sum is one. Thus four intervals are necessary and sufficient.

For $h=1/2$, the five translates active in the interior of $[0,1]=[0,2h]$ are indexed $-3,-2,-1,0,1$. Write $s(x)={\sum }_{j=-3}^1{c}_j{N}_j^3(x)$. Nodal interpolation givesAt the left endpoint the first derivatives of the three active functions are $-1/(2h),0,1/(2h)$, and their second derivatives are $1/h^2,-2/h^2,1/h^2$. The two remaining equations areTogether these five equations form the requested system; insert $h=1/2$ if numerical coefficients are desired. The first, fourth and fifth determine the first three coefficients, and the other two then determine $c_0,c_1$, so the system is nonsingular.

The source only assumes $u\in C[0,1]$, which need not have either endpoint derivative. The Hermite conditions require those derivatives to exist, or to be supplied as separate interpolation data. The nodal conditions alone remain meaningful for every continuous $u$.

Put $e=u-u_h$ and $V=\{v\in H^1(0,1):v(1)=0\}$. The primal solution and its finite element approximation lie in the same affine space with value one at the right endpoint, so $e\in V$. UseDefine the dual solution byEquivalently $z\in V$ satisfies $a(v,z)=(e,v)$ for every $v\in V$. Positivity of $p,q$ gives a unique solution, and one-dimensional regularity gives $z\in H^2$.

Then $\Vert e{\Vert }_{L^2}^2=a(e,z)$. The nodal interpolant $I_{h}z$ belongs to the homogeneous finite element test space, so Galerkin orthogonality gives $a(e,I_{h}z)=0$. Set $w=z-I_{h}z$. Using the primal weak equation,Integrate the derivative term separately on each element. Since $w$ vanishes at every mesh node, all endpoint flux and flux-jump terms vanish. ThereforeThe absence of jump terms here follows specifically from one-dimensional nodal interpolation; it is not an assumption that the finite element flux itself is continuous.

Let $p_{\ast }=\min p>0$, $q_{\ast }=\min q>0$, $q^{\ast }=\max q$, and $P_1=\Vert p^{\prime }{\Vert }_{\infty }$. Testing the dual equation with $z$ givesConsequently $\Vert z{\Vert }_2\le \Vert e{\Vert }_2/q_{\ast }$ and $\Vert z^{\prime }{\Vert }_2\le \Vert e{\Vert }_2/\sqrt{p_{\ast }{q}_{\ast }}$. From $p{z}^{\prime \prime }=qz-e-p^{\prime }{z}^{\prime }$,On an element of length $h_i$, the interpolation error vanishes at both endpoints and its derivative has zero mean. Two Poincaré inequalities giveApply Cauchy-Schwarz to the residual identity, with factors $h_i^2$ and $h_i^{-2}$:If the error is nonzero, cancel its norm after the regularity estimate; if it is zero the conclusion is immediate. One explicit reliable constant isThis proves the requested dual residual bound entirely in terms of the coefficient functions.

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.