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Equispaced interpolation of Runge's function diverges

Example

Let f(x)=1/(1+25x2) on [−1,1]. For each integer n≥2, take the n+1 equispaced nodes xj=−1+2jn,0≤j≤n, let ℓj,n(x)=∏k≠j(x−xk)/(xj−xk), and put pn(x)=∑j=0nf(xj)ℓj,n(x). Then, at the evaluation point yn=1−1/n, ∣f(yn)−pn(yn)∣⟶∞. Thus the uniform interpolation errors diverge, although the best polynomial errors En(f) tend to zero by Best polynomial errors vanish for continuous functions.

Facts & Assumptions

Given: An integer n≥2, the equispaced nodes above, a=1/5, i∈C with i2=−1, and f(x)=1/(1+25x2).

[F1]

Best uniform polynomial error and Chebyshev interpolation operators: Pn, C[−1,1], and ∥⋅∥∞ have the stated meanings.

[F2]

Best polynomial errors vanish for continuous functions: En(g)→0 for every g∈C[−1,1].

[F3]

A nonzero real polynomial of degree n has no more than n distinct real roots: a nonzero real polynomial of degree at most n has at most n distinct real roots.

[F7]

The binomial theorem in R: (x+y)n=∑k<n+1ι ⁣(nk) xky n−k: (x+y)n=∑k=0nι ⁣((nk))xkyn−k in R, where ι:N→R is the canonical embedding of [F9].

[F9]

The canonical natural ι(n)=n⋅1F of a field and Canonical naturals are positive and strictly increasing: throughout all real expressions, including identities and inequalities, natural numbers, factorials, and binomial coefficients abbreviate their canonical real images under ι. This embedding preserves addition, multiplication, and order, sends 0 to 0, and sends positive naturals to positive reals.

[F11]

Positive rational bases have rational powers as in Rational powers ar of a positive base. By Monotonicity of r↦ar and of a↦ar, powers decrease with the rational exponent when 0<a<1, increase when a>1, and increase with the positive base for a positive rational exponent. Laws of rational exponents gives (ar)s=ars. Integer indices and natural counts embed into Q by The integers embed in the rationals, and rational comparisons use Order on the rationals.

[F12]

For every real x, x2=0 if x=0 by Integer powers am and Multiplication by zero: 0⋅a=0, while x2>0 if x≠0 by Squares of nonzero elements are positive. The positive integer 25 and the real number 1 are positive by [F9]; the ordered-field axioms Ordered field therefore give 1+25x2>0.

[F14]

If d is continuous and nowhere zero, its reciprocal is continuous on its domain (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 4). By [F12], d(x)>0 for every real x, so f=1/d belongs to C[−1,1].

Proof

1.1givenF1F3F10constructalgebra

For 0≤j<n, xj+1−xj=2/n>0, so the nodes are distinct. Each product ℓj,n(x) has n linear factors and degree at most n; at xj all its factors equal 1, while at any other node xk the factor indexed by k is zero. Thus pn∈Pn and pn(xj)=f(xj). Define ωn(x)=∏j=0n(x−xj) and for nodal data define the complex-linear Lagrange map Jng=∑jg(xj)ℓj,n. For non-real z, put gz(x)=1/(x−z) and qn,z(x)=(1−ωn(x)/ωn(z))/(x−z). The denominator is nonzero because the roots of ωn are real and C is a field [F10]. Expanding ωn(x)−ωn(z) termwise and using xk−zk=(x−z)∑r=0k−1xk−1−rzr shows qn,z is a complex polynomial of degree at most n; at each node xj it equals 1/(xj−z). The difference qn,z−Jngz is a complex polynomial of degree at most n with n+1 distinct real roots. By unique complex coordinates [F10], its real and imaginary coefficient polynomials are real polynomials; both vanish at all the nodes, so [F3] forces both to be zero. Since gz−qn,z=ωn(x)/((x−z)ωn(z)), this proves gz(x)−Jngz(x)=ωn(x)/((x−z)ωn(z)) for real x.

1.2givenF4F5F6F7F8F9algebra

At yn=1−1/n, direct multiplication gives ∣ωn(yn)∣=∏k=1n(2k−1)/nn+1=(2nn)n!/(2nnn+1). By [F8], (2nn) is the largest of the 2n+1 binomial coefficients, and [F7] says their sum is 4n, so (2nn)≥4n/(2n+1). Also e<3: for k≥2, k!=2⋅3⋯k≥2k−1 because each of these k−1 factors is at least 2, with strict inequality for k≥3; therefore [F4]–[F6] give e=2+∑k=2∞1/k!<3. By [F8], (nk)k!=n(n−1)⋯(n−k+1)≤nk, hence (nk)/nk≤1/k!; applying [F7] gives (1+1/n)n≤∑k=0n1/k!<e<3. Induction then yields n!≥(n/3)n: it holds at n=1, and the ratio of successive values of n!/(n/3)n is 3/(1+1/n)n>1. Therefore ∣ωn(yn)∣≥(2/3)n/[n(2n+1)].

1.3givenF1F10F11algebra

The positive node radii form an arithmetic progression in (0,1] of spacing δ=2/n; for even n there is also the central node 0. Pairing r,−r in ωn(ia) gives the factor ∣(ia−r)(ia+r)∣=a2+r2 by [F10], and the even-grid central factor is ∣ia∣=a≤1. Each bin has length 1/4, and the radius progression begins within δ of 0 and ends at 1. Thus for each bin its endpoint gaps are at most δ and its internal gaps equal δ: if it contains mb≥1 radii, its length lies between (mb−1)δ and (mb+1)δ; if it contains none, its length is at most one progression gap, so the same count bounds hold. Therefore n/8−1≤mb≤n/8+1. Partition the radii into (0,1/4],(1/4,1/2],(1/2,3/4],(3/4,1]. The paired factors in these bins are bounded above by B1=41/400, B2=29/100, B3=241/400, and B4=26/25. By [F11], for b=1,2,3 the bound Bb<1 and mb≥n/8−1 imply Bbmb≤Bbn/8−1; for b=4, B4>1 and m4≤n/8+1 imply B4m4≤B4n/8+1. Multiplying these bounds and the central factor gives ∣ωn(ia)∣≤A(B1B2B3B4)n/8, where A=B1−1B2−1B3−1B4. Direct rational multiplication gives B1B2B3B4<1/50<(5/8)8; base monotonicity and the iterated-power law in [F11] then give (B1B2B3B4)n/8<(5/8)n.

2.1givenF1F10step 1.1algebra

With a=1/5, direct subtraction gives f(x)=a2i(1/(x−ia)−1/(x+ia)), and the finite Lagrange sum is complex-linear by field distributivity [F10], so it preserves the identity. Since xn−j=−xj, reindexing the factors gives ωn(−x)=(−1)n+1ωn(x); thus ωn is odd for even n and even for odd n. Every factor ia−xj is nonzero, so ωn(ia)≠0; real coefficients and parity give ωn(ia)=iW for a nonzero real W when n is even, and ωn(ia)=ωn(−ia)=W for a nonzero real W when n is odd. For even n, substituting the pole-error identities from step 1.1 and using 1/(x−ia)+1/(x+ia)=2x/(x2+a2) gives f(x)−pn(x)=axωn(x)/(i(x2+a2)ωn(ia)). For odd n, using 1/(x−ia)−1/(x+ia)=2ia/(x2+a2) gives f(x)−pn(x)=a2ωn(x)/((x2+a2)ωn(ia)). Taking moduli and using [F10] gives, for real x, ∣f(x)−pn(x)∣={a∣x∣x2+a2∣ωn(x)∣∣ωn(ia)∣,n even,a2x2+a2∣ωn(x)∣∣ωn(ia)∣,n odd.

3.1givenF2F9F10F12F13F14step 2.1step 1.2step 1.3algebra∎

For n≥2, yn≥1/2 and yn2+a2≤26/25, so the coefficient in step 2.1 is at least 5/52 for even n and 1/26 for odd n, hence at least 1/26. By [F10], the complex-modulus identity in step 2.1 is the stated real absolute error; steps 1.2 and 1.3 then give ∣f(yn)−pn(yn)∣>126A(16/15)n/[n(2n+1)]. The right-hand side tends to infinity because the ratio of successive terms tends to 16/15>1 and is eventually bounded below by a fixed ratio greater than 1. Therefore ∥f−pn∥∞→∞. By [F12]–[F14], f∈C[−1,1], so [F2] applies and En(f)→0.

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