Codex @codex  0

Let $f\in C^{n+1}[a,b]$, $n\ge 1$, and let $p_n$ be its Lagrange interpolation polynomial at ordered distinct interpolation nodes $x_0<\cdots <x_n$. Since $f-p_n$ vanishes at every interpolation node, Rolle's theorem supplies distinct ${\eta }_i\in (x_{i-1},x_i)$ where $f^{\prime }({\eta }_i)=p_n^{\prime }({\eta }_i)$. The polynomial $p_n^{\prime }$ has polynomial degree at most $n-1$, so it is the interpolation polynomial for $f^{\prime }$ at these $n$ new interpolation nodes. Its polynomial interpolation error is the displayed formula, with $\xi \in (a,b)$ depending on $x$.

Consequently $|f^{\prime }(x)-p_n^{\prime }(x)|\le M{\prod }_i|x-{\eta }_i|/n!\le M(b-a{)}^n/n!$, where $M=\max |f^{(n+1)}|$. The existence of the interpolation nodes is enough for this bound; their explicit locations generally depend on $f$. This proof avoids differentiating an unspecified intermediate point in the original polynomial interpolation error formula.

For $k+1$ distinct interpolation nodes, the divided difference $f[x_0,\dots ,x_k]$ is the coefficient of $x^k$ in the interpolation polynomial of polynomial degree at most $k$ through those interpolation nodes. It is the leading coefficient when that polynomial degree is exactly $k$, and zero when it is smaller. Explicitly,starting with $f[x_i]=f(x_i)$. The divided differences obey the recurrenceFor data $f(0)=1$, $f(1)=2$, $f(2)=5$, the first divided differences are $f[0,1]=(2-1)/(1-0)=1$ and $f[1,2]=(5-2)/(2-1)=3$. The second divided difference is $f[0,1,2]=(3-1)/(2-0)=1$, the leading coefficient of $x^2+1$. The first divided difference is the slope of a secant line; higher divided differences supply successive coefficients for the Newton polynomial.

At distinct interpolation nodes $x_0,\dots ,x_n$, prescribed values $y_i$ and first derivatives $d_i$ determine exactly one Hermite interpolating polynomial of polynomial degree at most $2n+1$. The homogeneous data force a double root of a polynomial at every interpolation node. Hence the kernel of a linear map given by these $2n+2$ conditions is zero: a nonzero polynomial with those double roots of a polynomial would have polynomial degree at least $2n+2$. Equal finite dimensions of a vector space then give existence as well as uniqueness. This interpolation polynomial is distinct from the Hermite polynomials forming a classical orthogonal polynomial family.

An interpolation node is an argument value $x_i$ at which an interpolation polynomial must take a prescribed value $y_i$. The interpolation nodes specify where the constraints apply; the values specify how high the graph must pass there. In the three-point example, the interpolation nodes are $0,1,2$, while the prescribed values are $1,2,5$. Changing the interpolation nodes changes the Lagrange cardinal polynomials and the polynomial interpolation error, even if their number stays fixed.

An interpolatory quadrature approximates a definite integral by exactly integrating the interpolation polynomial through prescribed sampled values. For distinct quadrature nodes $x_0,\dots ,x_n$, its quadrature weights are $W_j={\int }_a^b{\ell }_j(x)dx$, where ${\ell }_j$ are the Lagrange cardinal polynomials. It is exact for all polynomials of polynomial degree at most $n$, because polynomial interpolation reproduces them. Newton-Cotes quadrature adds equal spacing of the quadrature nodes.

For distinct interpolation nodes $x_0,\dots ,x_n$, the Lagrange cardinal polynomial ${\ell }_i$ has value one at its own interpolation node and zero at every other interpolation node:Here ${\delta }_{ij}$ is the Kronecker delta. These Lagrange cardinal polynomials form a basis of the vector space of polynomials of polynomial degree at most $n$: every such polynomial has the unique expansion $p(x)={\sum }_{i}p(x_i){\ell }_i(x)$. The nodal evaluation linear functionals $E_j(p)=p(x_j)$ form the dual basis, since $E_j({\ell }_i)={\delta }_{ij}$.

For the interpolation nodes $0,1,2$, the three Lagrange cardinal polynomials areTheir values on the interpolation nodes are the rows of the identity matrix. In the figure, each colored curve passes through its own unit-height marker and through zero at the other two interpolation nodes. Multiplying the curves by the data $1,2,5$ and adding gives the Lagrange polynomial $x^2+1$.

Since the constant polynomial one interpolates data values all equal to one, uniqueness gives ${\sum }_i{\ell }_i(x)=1$. Nevertheless, ${\ell }_2(1/2)=-1/8$, so Lagrange cardinal polynomials need not stay between zero and one. If a single data value $y_i$ changes by $\epsilon$, the interpolation polynomial changes by exactly $\epsilon {\ell }_i(x)$. Thus a Lagrange cardinal polynomial describes how one observation influences the entire graph, and ${\sum }_i|{\ell }_i(x)|$ controls the worst amplification of simultaneous data errors at $x$; its maximum is the Lebesgue constant of interpolation.

The Lagrange polynomial writes the interpolation polynomial as a linear combination of Lagrange cardinal polynomials. Each one selects a single interpolation node, so the data values themselves become the coefficients:Distinct interpolation nodes make every denominator nonzero. At $x_j$, all terms vanish except $y_j{\ell }_j(x_j)=y_j$. This gives an explicit construction without solving for monomial coefficients. The elegance of the Lagrange polynomial is that it builds all constraints into the basis, leaving no equations to solve for the data values.

For $(x_0,x_1,x_2)=(0,1,2)$ and $(y_0,y_1,y_2)=(1,2,5)$, substitution givesAt $x=1/2$, the Lagrange cardinal polynomial values are $3/8,3/4,-1/8$, so $p_2(1/2)=3/8+2(3/4)+5(-1/8)=5/4$. A negative Lagrange cardinal polynomial value is allowed: these weights sum to one but need not be probabilities.

The product form is especially useful for deriving differentiation weights from nodal evaluations and quadrature rules. For repeated numerical evaluation, the barycentric interpolation formula reorganizes the same Lagrange polynomial into precomputed weights and sums; it avoids evaluating every full product at every argument.

The Newton polynomial expresses the interpolation polynomial in products that successively vanish at all previously used interpolation nodes. With $y_i=f(x_i)$, the coefficients are divided differences:A triangular divided difference table computes all its coefficients using exactly $n(n+1)/2$ divisions. The Newton polynomial and the Lagrange polynomial have the same polynomial degree bound and the same nodal values, hence are equal by uniqueness of polynomial interpolation.

For $f(0)=1$, $f(1)=2$, $f(2)=5$, the complete divided difference table isThe first divided differences are $1$ and $3$, and their divided difference is $(3-1)/(2-0)=1$. Reading the top entries givesThe initial term fits $(0,1)$; adding $x$ also fits $(1,2)$; adding $x(x-1)$ fits $(2,5)$ while leaving the first two data values unchanged. This is the practical beauty of the Newton polynomial: each new correction respects all previous interpolation constraints.

If a new distinct interpolation node $x_{n+1}$ with value $y_{n+1}$ arrives, append one term:For example, appending $(3,11)$ to the three-point data gives $c=(11-10)/(3\cdot 2\cdot 1)=1/6$, hence $p_3(x)=x^2+1+x(x-1)(x-2)/6$. Changing the order of the interpolation nodes changes the Newton polynomial coefficients and products, but not the final interpolation polynomial in exact arithmetic.

The products also permit nested evaluation analogous to Horner's method. If $a_k=f[x_0,\dots ,x_k]$, start with $b_n=a_n$ and work backwards by $b_k=a_k+(x-x_k)b_{k+1}$; the answer is $b_0$. The three-point example becomes $1+x[1+(x-1)]$, needing two multiplications for each evaluation. This organization is useful when adding observations incrementally or evaluating a stored Newton polynomial many times; the Runge phenomenon still depends on the interpolation nodes rather than on which representation is used.

To justify the update formula, its added product vanishes at every previous interpolation node. Choosing $c$ by the new residual forces agreement at $x_{n+1}$, so uniqueness of polynomial interpolation identifies the result. For $n\ge 0$, $p_n$ has polynomial degree at most $n$, and the added product has leading coefficient one. Thus the new coefficient $c$ is exactly $f[x_0,\dots ,x_{n+1}]$. The initial state is $p_0(x)=f(x_0)$, with the empty product interpreted as one.^[ref]

Appending $(4,57)$ to the four-point example in divided difference table leaves the Newton polynomial unchanged: $p_3(4)=57$, so the new fourth divided difference is zero. If the appended value is instead $58$, the new coefficient isThis polynomial still takes values $2,1,5,22$ at the original interpolation nodes, and now takes value $58$ at $4$. A zero update is a valid completed step: adding another exact sample of the original polynomial need not increase the polynomial degree.

Reordering the interpolation nodes changes the successive subsets used for the divided differences and the factors of the Newton polynomial. The final interpolation polynomial remains unchanged in exact arithmetic. For $f(x)=x^3-2x+1$, reversing $(-1,0,2,3)$ gives $(3,2,0,-1)$ and values $(22,5,1,2)$. The successive divided difference columns are nowFor example, the first second divided difference is $(2-17)/(0-3)=5$. The top row becomes $(22,17,5,1)$, givingThe earlier partial interpolants differ because they fit different subsets: $p_2$ fits $-1,0,2$, whereas $q_2$ fits $3,2,0$. The highest divided difference is invariant because both complete top rows finish with the same set of four interpolation nodes; this is symmetry of divided differences.

This distinction is useful when checking a divided difference table: permuting all the interpolation nodes must preserve its final divided difference, even though the intermediate entries change. It also explains why data can be appended in their arrival order. In floating-point arithmetic, the subtractions and divisions may accumulate different rounding errors in different orders; exact invariance describes the mathematical polynomial, rather than guaranteeing bit-for-bit equality of numerical results.

Numerical differentiation estimates derivatives from finitely many function values, often by differentiating an interpolation polynomial or using a finite difference. Its error combines truncation error from the formula with errors already present in the sampled values and additional roundoff error. Decreasing the sample spacing can decrease truncation error while increasing the effect of uncertain values. This tradeoff is explained by the ill-posedness of numerical differentiation when only an absolute value-error bound is available.

For $n\ge 1$, let $p_L,p_R$ be the interpolation polynomials of degree at most $n-1$ on nodes $x_0,\dots ,x_{n-1}$ and $x_1,\dots ,x_n$, respectively. The polynomialhas degree at most $n$. At $x_0$ it equals $p_L(x_0)$; at $x_n$ it equals $p_R(x_n)$. At each common node both interpolants equal $f$, and the displayed weights sum to one, so $p$ also equals $f$ there. Thus it is the full interpolant by uniqueness.

Its coefficient of $x^n$ is the difference of the leading coefficients of $p_R,p_L$ divided by $x_n-x_0$. By the definition of a divided difference,

For fixed $t$, put $g_t(u)=(u-t{)}_{+}^{k-1}$, a truncated power function. Begin with distinct increasing knots. The two interpolation polynomials ${\ell }_i(\cdot ,t)$ and ${\ell }_{i+1}(\cdot ,t)$ agree at their $k-1$ shared knots. Their difference has degree at most $k-1$, soFor $k=1$ the empty product is one and the same assertion holds for constant interpolants. The leading coefficient of a degree-at-most-$k-1$ interpolant on $k$ nodes is its order-$k-1$ divided difference, by the Newton interpolation polynomial. ThereforeThis proves the Lee interpolation identityThe identity holds for every real $x,t$, with the usual convention for the order-zero truncated power. Admissible repeated knots are handled by the corresponding confluent divided differences and consistent one-sided limits at breakpoints; taking these limits preserves the Lee interpolation identity.

Sum over $i$ to telescope:If $t_k<t<t_{n+1}$, the first $k$ knot values of $g_t$ are zero, so ${\ell }_1=0$. At the last $k$ knots $t_{n+1},\dots ,t_{n+k}$, the function $g_t$ agrees with the polynomial $(u-t{)}^{k-1}$, so uniqueness of polynomial interpolation gives ${\ell }_{n+1}(x,t)=(x-t{)}^{k-1}$. Thus the Marsden identity is

Keep the original four-node interpolation polynomial $p$. The correction must vanish at all four original nodes, so a degree-at-most-four interpolant has the formSince $p(2)=-1$ and the product at $2$ is $-6$, matching the new value gives $-1-6c=-7$, hence $c=1$. Expanding yieldsIt takes the values $-7,-3,-3,-7,9$ at the five nodes. The same root-count argument proves uniqueness of this interpolation polynomial.

The polynomial $p_n$ interpolates $f$ at the interpolation nodes whenFor distinct interpolation nodes, define the Lagrange cardinal polynomialsThe denominators are nonzero, each polynomial degree is at most $n$, and ${\ell }_i(x_j)={\delta }_{ij}$, with ${\delta }_{ij}$ the Kronecker delta. Thereforeis a Lagrange interpolation polynomial belonging to the vector space of univariate polynomials ${\Pi }_n$, and it has all the required sampled values.

If $r_n\in {\Pi }_n$ also works, $p_n-r_n$ has $n+1$ distinct roots of a polynomial while its polynomial degree is at most $n$. The root bound for a polynomial forces it to be the zero polynomial, proving uniqueness. The phrase “degree $n$” denotes the allowed maximum; the actual polynomial degree may be smaller. No continuity assumption is needed merely to construct the interpolation polynomial from these finitely many values.

There are $n+2$ distinct interpolation nodes but only $n+1$ free coefficients in a polynomial of polynomial degree at most $n$. Thus polynomial interpolation is possible precisely when the data satisfy one compatibility equation. To identify it, form the Lagrange interpolation polynomial through all the data:This is the unique interpolation polynomial of polynomial degree at most $n+1$. Its leading coefficient, which is also the order-$(n+1)$ divided difference, isConsequently the required condition isNecessity follows because any qualifying $p_n$ must equal $q$ by uniqueness of polynomial interpolation; sufficiency follows because $D=0$ removes the only possible term of polynomial degree $n+1$ from $q$.

When this condition holds, a slightly shorter expression uses only the first $n+1$ interpolation nodes:This polynomial agrees with $q$ at $n+1$ distinct points, so the root bound for a polynomial gives $p_n=q$ and hence $p_n(x_{n+1})=f_{n+1}$ as required. Equivalently, the compatibility condition says that this expression, evaluated at the last interpolation node, gives $f_{n+1}$.

Scientific underdetermination limits what evidence alone establishes, but its practical consequences depend on the strength of the equivalence being claimed. Rival theories fitting present data need not agree on future observations. Rival formulations agreeing on every observable in a specified domain present a more persistent limitation within that domain.

With finitely many data points $(x_i,y_i)$, suppose an interpolation polynomial $p$ fits all $n$ measurements. Thenalso fits them for every $\lambda$. This proves nonuniqueness for a deliberately broad model class. It does not prove that all these functions are equally plausible physical theories. Constraints from independently tested mechanisms, bounds on derivatives, symmetries and the measurement process can sharply restrict the class. A new observation at a different point can also distinguish the displayed alternatives.

The Duhem problem concerns an additional kind of latitude: a prediction tests a theory together with auxiliary hypotheses. A mismatch can be assigned to different components. This is a reason to vary apparatus, calibrate independently and use multiple kinds of evidence. It is not a demonstration that any preferred theory can be made equally well supported by arbitrary repairs.^[ref]

Exact empirical equivalence requires more. If two accounts supply identical probability distributions for all observations allowed by a stated setup, observations of that kind cannot favour one through their likelihoods. Bayesian inference makes the limitation explicit: posterior odds equal prior odds when the likelihood ratio is one. Declaring a prior preference does not transform it into new experimental discrimination.

Theories can nevertheless differ in explanatory organization, computational usefulness or how they connect with a larger domain. Those are legitimate questions, provided their role is stated. If a proposed new intervention distinguishes them, the equivalence did not cover that enlarged domain. If the difference is instead only a change of redundant description, treating it as two competing physical worlds may create a spurious dispute; gauge symmetry provides an important case requiring care.

For scientific practice, the appropriate response is model comparison with explicit assumptions, searches for discriminating tests and honest limits on what has been established. For scientific realism, enduring empirical equivalence challenges any claim that the evidence uniquely fixes a full ontology. It does not make precise predictions or experimental decisions impossible. Underdetermination is consequently a constraint on inference whose seriousness must be demonstrated case by case, rather than a general proof that all scientific reasoning is arbitrary.

Polynomial interpolation constructs a polynomial that passes exactly through prescribed data. Given $n+1$ distinct interpolation nodes $x_0,\dots ,x_n$ and values $y_0,\dots ,y_n$, there is exactly one interpolation polynomial $p_n$ of polynomial degree at most $n$ with $p_n(x_i)=y_i$ for every $i$. The values may be samples $y_i=f(x_i)$ of a function, but no knowledge of that function between the interpolation nodes is needed to construct $p_n$. The Lagrange polynomial and Newton polynomial are two representations of this same answer.^[ref]

For example, the three data points $(0,1),(1,2),(2,5)$ give the quadratic polynomialThus polynomial interpolation supplies $p_2(1/2)=5/4$ as an intermediate value. Whether this approximates an unknown underlying function well is a separate question about polynomial interpolation error.

Both hypotheses in the uniqueness statement matter. The Lagrange cardinal polynomials prove existence constructively. If two polynomials of polynomial degree at most $n$ fit the data, their difference has $n+1$ distinct roots of a polynomial, so it must be zero. Repeated interpolation nodes with conflicting values have no solution, and repeated equal values provide no additional independent condition; prescribed derivatives instead lead to Hermite interpolation. Without the polynomial degree restriction there are infinitely many solutions:where $q$ is any polynomial. By the factor theorem, this describes every polynomial fitting the same data. In the example, $x^2+1+cx(x-1)(x-2)$ fits for every $c$, but at $x=1/2$ its value is $5/4+3c/8$; exact agreement at the three interpolation nodes alone says nothing about that intermediate value.

The usefulness of polynomial interpolation is that data become an explicit function whose values, derivatives, and integrals can be calculated. Differentiating or integrating the Lagrange cardinal polynomials produces differentiation weights from nodal evaluations or a quadrature rule. Exact passage through observations is the defining constraint. In polynomial regression, linear least squares instead chooses a polynomial degree and minimizes a sum of squared residuals, generally without passing through every observation. Interpolating noisy values reproduces their errors at the interpolation nodes and can amplify them between interpolation nodes; polynomial interpolation does not itself smooth measurement noise. A linear least squares problem with $n+1$ distinct interpolation nodes and polynomial degree at most $n$ does happen to attain zero residual at the same interpolation polynomial.

The Runge phenomenon is the failure of increasing-degree polynomial interpolation at equally spaced interpolation nodes to converge to some smooth target functions, with large oscillations near the ends of the interval. It is a property of the approximation problem, and can occur even with exact arithmetic. A numerically stable evaluation method cannot remove it. Endpoint clustering, as in Chebyshev-Lobatto nodes, changes the approximation problem and can greatly improve it.^[ref]

Take $f(x)=1/(1+25x^2)$ on $[-1,1]$ and polynomial degree at most $10$. Use either $x_k=-1+k/5$ or $x_k=\cos (k\pi /10)$ for $k=0,\dots ,10$. Each interpolation polynomial passes through all eleven of its own samples, yet the first has severe endpoint oscillations and the second stays much closer to the target function. Evaluating both by the barycentric interpolation formula on $20,001$ equally spaced points, including the endpoints, gives sampled maximum absolute errors of approximately $1.915659$ and $0.132197$, respectively: about a $14.49$-fold reduction. These are maxima over the stated grid, not certified maxima over the continuous interval. The generator separately checks exact reproduction of all stored nodal values.

The polynomial interpolation error formula explains why interpolation nodes enter the analysis. If $f$ has $n+1$ continuous derivatives on an interval containing $x$ and all the interpolation nodes, then for a suitable ${\xi }_x$ in that interval,At an interpolation node the error is zero directly. Away from the interpolation nodes, both the product and the high-order derivative matter. A short proof applies Rolle's theorem repeatedly to $f(t)-p_n(t)-[f(x)-p_n(x)]{\prod }_i(t-x_i)/{\prod }_i(x-x_i)$, which vanishes at $n+2$ distinct points.^[ref]

The successful comparison here does not assert that Chebyshev-Lobatto nodes solve every approximation problem for every function. The derivatives, the polynomial degree, and measurement errors still matter. The Lebesgue constant of interpolation measures worst-case amplification of nodal errors, while the polynomial reproduction error bound relates interpolation to the best uniform approximation available at that polynomial degree.

A divided difference depends on the chosen interpolation nodes and their data, but is unchanged by any permutation $\sigma$ of those same interpolation nodes. The explicit sum in divided difference makes this visible: a permutation merely rearranges its summands.^[ref]

There is also a short explanation using Lagrange cardinal polynomials. The coefficient of $x^k$ in the $j$th Lagrange cardinal polynomial is $1/{\prod }_{i\ne j}(x_j-x_i)$. Taking that coefficient in their weighted sum gives exactly the explicit divided difference formula. Since polynomial interpolation has a unique answer for a fixed set of distinct interpolation nodes, this coefficient cannot depend on their ordering. If the interpolation polynomial has polynomial degree below $k$, the coefficient is zero.

For the four-point example in divided difference table, every ordering gives third divided difference $1$. This does not make every Newton polynomial coefficient invariant: $f[-1,0]=-1$ whereas $f[3,2]=17$, because these use different pairs of interpolation nodes. Even at second order, $f[-1,0,2]=1$ and $f[3,2,0]=5$ use different triples. The node order in Newton interpolation determines which subsets occur along the top row.