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Barycentric interpolation identities

Statement

Let $n \in N$ , $n \geq 0$ , let $x_{0} < \dots < x_{n}$ be distinct real nodes, let $y_{0}, \dots, y_{n} \in R$ , and define $ω_{n}, w_{j}, I_{n} y$ as in Cardinal-polynomial interpolation. For every real $x$ that is not a node, $ω_{n} (x) \sum_{j = 0}^{n} \frac{w_{j}}{x - x_{j}} = 1, ω_{n} (x) \sum_{j = 0}^{n} \frac{y_{j} w_{j}}{x - x_{j}} = I_{n} y (x) .$ In particular, the denominator sum is nonzero and $I_{n} y (x) = \frac{\sum_{j = 0}^{n} y_{j} w_{j} / (x - x_{j})}{\sum_{j = 0}^{n} w_{j} / (x - x_{j})} .$ At a node $x_{i}$ , the interpolant is defined by the separate value $I_{n} y (x_{i}) = y_{i}$ ; the quotient is not used there.

Facts & Assumptions

Given: Distinct real nodes $x_{0} < \dots < x_{n}$ , real data $y_{0}, \dots, y_{n}$ , and a real query $x$ not equal to any node.

[F1]

Cardinal-polynomial interpolation: $ω_{n} (X) = \prod_{k = 0}^{n} (X - x_{k})$ , $w_{j} = (\prod_{k \neq j} (x_{j} - x_{k}))^{- 1}$ , $ℓ_{j} (X) = \prod_{k \neq j} (X - x_{k}) / (x_{j} - x_{k})$ , and $I_{n} y (X) = \sum_{j = 0}^{n} y_{j} ℓ_{j} (X)$ .

[F2]

Existence and uniqueness of the cardinal-polynomial interpolant: for any data on these nodes there is exactly one polynomial of degree at most $n$ interpolating them, and it is $I_{n} y$ .

[F3]

Cardinal interpolation functions are polynomials of the stated degrees: $ℓ_{j} (x_{i}) = 1$ when $i = j$ and $0$ otherwise, and $ω_{n} (x) \neq 0$ off the nodes.

[F4]

The reals form a field and Laws of finite sums and finite products: real division by a nonzero number is defined; finite products split into consecutive factors and satisfy $\prod_{k < n} (a_{k} b_{k}) = (\prod_{k < n} a_{k}) (\prod_{k < n} b_{k})$ , including the empty-product case.

Proof

1.1givenF1F4algebra

Split the product for $ω_{n} (x)$ before and after index $j$ using [F4]: it is $A (x - x_{j}) B$ , where $A = \prod_{k < j} (x - x_{k})$ and $B = \prod_{j < k \leq n} (x - x_{k})$ ; the omitted numerator product is $A B$ . In the increasing omitted-index order of [F1], the product law [F4] gives $ℓ_{j} (x) = (\prod_{k \neq j} (x - x_{k})) (\prod_{k \neq j} (x_{j} - x_{k})^{- 1})$ . The second product is $w_{j}$ : multiplying it by $\prod_{k \neq j} (x_{j} - x_{k})$ gives the product of $1$ 's, including the empty case, so it is that product's inverse. Thus $ℓ_{j} (x) = A B w_{j} = ω_{n} (x) w_{j} / (x - x_{j})$ , using $x \neq x_{j}$ to cancel the nonzero factor.

1.2givenF1F2F3algebra

By [F1], $S (X) : = \sum_{j = 0}^{n} ℓ_{j} (X) = I_{n} (1, \dots, 1) (X)$ , so it has degree at most $n$ by [F2]. At each node $x_{i}$ , [F3] gives $S (x_{i}) = 1$ . The constant polynomial $1$ is another degree-at-most- $n$ interpolant of these data; uniqueness in [F2] gives $S (X) = 1$ .

2.1givenF3F4step 1.1step 1.2algebra

Summing the identities of step 1.1 over $j$ gives $ω_{n} (x) \sum_{j = 0}^{n} w_{j} / (x - x_{j}) = \sum_{j = 0}^{n} ℓ_{j} (x) = 1$ by step 1.2. Since $ω_{n} (x) \neq 0$ by [F3], the denominator sum equals $1 / ω_{n} (x)$ and is nonzero by the field laws [F4].

3.1givenF1F2F4step 1.1step 2.1algebra∎

Multiplying step 1.1 by $y_{j}$ and summing gives $ω_{n} (x) \sum_{j = 0}^{n} y_{j} w_{j} / (x - x_{j}) = \sum_{j = 0}^{n} y_{j} ℓ_{j} (x) = I_{n} y (x)$ by [F1]. Dividing this identity by the nonzero identity in step 2.1 proves the quotient formula. At $x_{i}$ , [F2] gives $I_{n} y (x_{i}) = y_{i}$ , which is the separate node value stated above.

Depends on

Used by

Finite barycentric interpolation evaluator Algorithm

Dependency tree · two levels

33 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Mark Embree, Lecture Notes on Numerical Analysis, lecture 1 (standard reference, not scraped)
Jean-Paul Berrut and Lloyd N. Trefethen, Barycentric Lagrange Interpolation (standard reference, not scraped)