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The square-root endpoint does not have Romberg's even-power order

Example

For f(x)=x on [0,1], let TN be the composite trapezoidal value with mesh h=1/N, and set SN:=T2N+(T2N−TN)/3=(4T2N−TN)/3, the first-column Richardson value formed from the N- and 2N-panel levels. For N=2k with k∈N, this is Rk+1,1 of Finite Romberg table from nested composite trapezoid levels whenever its level cap L satisfies L≥k+1. For every integer N≥1, 82−11144h3/2≤∫01f(x) dx−SN. The coefficient is positive, so this error is not O(h2) as N→∞.

Facts & Assumptions

Given: N≥1, h=1/N, and f(x)=x on [0,1].

[A1]

N is a positive integer and h=1/N.

[F1]

Composite trapezoidal rule: TN is the sum of panel trapezoids on the uniform mesh; its individual panel values add to TN.

[F2]

Second-derivative error bound for the composite trapezoidal rule: on a panel of width h where f∈C2, the absolute difference between its integral and its trapezoid is at most h3sup⁡∣f′′∣/12.

[F3]

Finite Romberg table from nested composite trapezoid levels: its first-column recurrence is T2N+(T2N−TN)/3 when applied to the adjacent N- and 2N-panel levels; for dyadic N this is the corresponding table entry.

[F8]

Romberg cancels a proved finite even-power expansion: the cancellation conclusion requires the finite even-power expansion hypothesis stated there; it does not assert second-order accuracy for an endpoint-singular input without that expansion.

Proof

1.1givenA1F1F4F5F6algebra

Continuity from [F4] permits the substitution [F5] on [0,1] and on the first panel [0,h]. Thus ∫01x dx=2/3 and the first-panel integral is 2hh/3, while its trapezoid is hh/2; its excess is therefore hh/6=h3/2/6. For every panel, the chord joining endpoint values lies below x: for u,v≥0 and 0≤λ≤1, the square of (1−λ)u+λv is at most (1−λ)u+λv, since the difference is λ(1−λ)(u−v)2. Thus each panel integral is at least its trapezoid and eN:=2/3−TN≥h3/2/6.

2.1givenA1F2F6F7step 1.1algebra

For panels indexed by i=1,…,N−1, [F2] and [F7] bound each nonnegative panel excess by h3/(48ihih)=h3/2/(48i3/2), using the rational-power laws in Laws of rational exponents. The finite sum satisfies ∑i=1N−1i−3/2≤3: for N=1 it is empty; for N=2 it is 1; and for N≥3, if i−1≤x≤i then x3/2≤i3/2 by Monotonicity of r↦ar and of a↦ar. If x<i, inverse order Inverses of positives are positive, and reciprocation reverses order gives x−3/2>i−3/2, and if x=i equality holds. Integral monotonicity If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a) gives i−3/2≤∫i−1ix−3/2 dx for 2≤i≤N−1. Summing and using Truncated integrals of rational powers yields ∑i=1N−1i−3/2≤1+∫1N−1x−3/2 dx≤3. Adding the first-panel excess h3/2/6 gives eN≤(1/6+3/48)h3/2=11h3/2/48. Hence h3/2/6≤eN≤11h3/2/48.

3.1givenF3F8step 2.1algebra∎

The function is not C1 at 0, because its right difference quotient is x/x=1/x and is unbounded as x↓0; in particular it does not meet the smoothness assumptions behind the finite even-power expansion used for [F8]. Nevertheless the one-step value SN is defined. Writing e2N=2/3−T2N and using the preceding bounds at mesh h/2 and h gives ∫01f−SN=4e2N−eN3≥4(h/2)3/2/6−11h3/2/483=82−11144h3/2. The constant is positive since (82)2=128>121=112. Dividing by h2 makes the lower bound grow as a positive constant times h−1/2, so the error cannot be O(h2). This illustrates why [F8] cannot be applied without its expansion hypothesis.

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