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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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An exponential best-error rate transferred to Chebyshev interpolation

Example

Let f(x)=exp⁡(x) on [−1,1]. For every n≥1, En(f)≤e(n+1)!,∥f−Inσf∥∞≤(1+Clog⁡(n+1))e(n+1)!,σ∈{R,L}, where C=4+(π+4)/log⁡2. In particular, both Chebyshev interpolants converge uniformly for this function with the displayed rate.

Facts & Assumptions

Given: n≥1, the function f(x)=exp⁡(x), and either Chebyshev node family.

[F1]

Best uniform polynomial error and Chebyshev interpolation operators: En(f)=inf⁡p∈Pn∥f−p∥∞, and Inσ is its specified interpolation operator.

[F2]

Chebyshev interpolation error is controlled by best polynomial error: ∥f−Inσf∥∞≤(1+Λnσ)En(f).

[F3]

Chebyshev-root and Lobatto interpolation norms have logarithmic upper bounds: Λnσ≤Clog⁡(n+1) for n≥1.

[F4]

The real exponential function and the number e by a power series: exp⁡(x)=∑k=0∞xk/k! and e=exp⁡(1)=∑k=0∞1/k!.

[F5]

The exponential series converges absolutely for every real argument: the exponential series converges absolutely for every real argument.

[F9]

The canonical natural ι(n)=n⋅1F of a field and Canonical naturals are positive and strictly increasing: natural indices and factorials in real inequalities are read through their canonical real images, and every positive natural has positive real image.

[F10]

The exponential function is strictly increasing: exp⁡ is continuous on R, so f∈C[−1,1] and the quantities in [F1]–[F3] apply.

Verification

1.1givenF1F4F5F6F10algebra

By [F10], f=exp⁡ is continuous on [−1,1]. Put pn(x)=∑k=0nxk/k!, which belongs to Pn. For ∣x∣≤1, [F4]–[F6] give ∣exp⁡(x)−pn(x)∣≤∑m=0∞1/(n+1+m)!≤1(n+1)!∑m=0∞1/m!=e/(n+1)!. The middle inequality follows from (n+2)⋯(n+1+m)≥1⋯m=m!, including the empty product for m=0. Taking the infimum over Pn proves En(exp⁡)≤e/(n+1)!.

2.1givenF2F3step 1.1algebra

Apply the near-best estimate [F2] and then the norm bound [F3] to step 1.1. This gives ∥exp⁡−Inσexp⁡∥∞≤(1+Λnσ)En(exp⁡)≤(1+Clog⁡(n+1))e/(n+1)! for each σ∈{R,L}.

3.1givenF7F8F9step 2.1algebra∎

By [F7], log⁡(n+1)=∫1n+1dt/t≤n and log⁡2=∫12dt/t>0, so C>0. For n≥1, (n+1)!=2⋅3⋯(n+1)≥2n because this product has n factors, each at least 2, so the bound in step 2.1 is at most e(1+Cn)/2n. For n≥2, the ratio of successive terms of n2−n is (n+1)/(2n)≤3/4, hence induction gives n2−n≤2(3/4)n−2. The geometric-sequence limit [F8] now shows both 2−n and n2−n tend to zero, so the interpolation errors tend to zero.

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