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Cancellation of the leading Simpson error coefficient

Example

For f(x)=x6−5x4 on [0,1], the leading coefficient −(1/180)∫01f(4) in Conditional asymptotic validity of the Simpson /15 indicator vanishes, but the one-level indicator and the refined Simpson error are nonzero. In fact, the indicator is 21/5 times the true refined error in this exact example.

Facts & Assumptions

Given: The polynomial f(x)=x6−5x4 on [0,1], the exact integral, and composite Simpson grids with n=2 and n=4.

[F1]

Composite Simpson rule on an even uniform grid: on an even uniform grid, QS=(h/3)∑j<n/2(f(x2j)+4f(x2j+1)+f(x2j+2)).

[F2]

Embedded Simpson refinement indicator: η=∣Qn=4−Qn=2∣/15 for this parent and its two-half refinement.

[F3]

Conditional asymptotic validity of the Simpson /15 indicator: the leading coefficient is −(1/180)∫01f(4).

[F6]

Every continuous function on a closed nondegenerate rectangle in Rm is Riemann integrable: continuous polynomials on [0,1] are Riemann integrable.

[F8]

Higher derivatives and the classes Ck and C∞: C4 means the derivatives through order four are continuous on the closed interval.

The polynomial is C4 on [0,1].

Verification

1.1givenF3F4F5F6F7F8algebra

By [F4] and [F5], f(4)(x)=360x2−120 and this derivative is continuous. An antiderivative is 120x3−120x, so [F6]–[F7] give ∫01f(4)(x) dx=0. Hence f∈C4([0,1]) by [F8], and the leading coefficient in [F3] is zero.

1.2givenF1F4F5F6F7algebra

By [F4]–[F5], x7/7−x5 has derivative f and f is continuous, so [F6]–[F7] give the exact integral 1/7−1=−6/7. Also f(0)=0, f(1/2)=−19/64, and f(1)=−4; substitution in [F1] gives Qn=2=−83/96. The quarter-point values are f(1/4)=−79/4096 and f(3/4)=−5751/4096, so [F1] also gives Qn=4=112(4(−794096)+2(−1964)+4(−57514096)−4)=−52676144. Thus the refined error is −6/7−Qn=4=5/43008, and the coarse error is −6/7−Qn=2=5/672.

2.1givenF2step 1.1step 1.2algebra∎

From [F2], η=∣Qn=4−Qn=2∣/15=1/2048. Hence η/(5/43008)=21/5. The nonzero refined error and zero leading coefficient show why the asymptotic /15 relation supplies no relative finite-level guarantee in the cancelled case.

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