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A successful rounded de Boor evaluation has a degree-dependent forward bound

Statement

Use the operationwise model in Operationwise real and complex floating point arithmetic with rounded square roots and fused multiply-add with unit roundoff 0<u<1. Suppose the floating-point mode of de Boor coefficient evaluation on a specified knot span completes successfully on stored coefficients ck−p,…,ck and every exact intermediate is zero or normal. Put η=(1+u)21−u−1,e=η+u(2+η),W=1+e,E=e+u(2+u)W. If S(x)=∑iciBi,p(x) is the exact B-spline value on the supplied span and S^(x) is the returned value, then ∣S^(x)−S(x)∣≤((1+2E)p−1)max⁡k−p≤i≤k∣ci∣. For p=0 the error is zero. The conclusion is conditional on success and does not certify that the intermediate range conditions will hold.

Facts & Assumptions

Given: A successful operationwise floating-point evaluation of the de Boor recurrence on a nonempty span, with 0<u<1.

[F1]

Operationwise real and complex floating point arithmetic with rounded square roots and fused multiply-add states that every successful real addition, subtraction, multiplication, and division has relative error at most u for a nonzero normal exact result, while exact zeros are returned exactly; other nonzero range cases return failure.

[F2]

de Boor coefficient evaluation on a specified knot span first computes the rounded numerator subtraction. If it is zero, the algorithm copies the left coefficient; otherwise it computes the denominator subtraction and division, complement subtraction, two multiplications, and addition in that order.

[F3]

de Boor evaluation equals the B-spline coefficient expansion identifies the exact recurrence output with S(x).

[A1]

Every operation in the floating-point run succeeds, as required in the Statement.

Proof

1.1A1F1F2algebra

At one blend let α∈[0,1] be the exact ratio. If its numerator is zero, the algorithm's exact-zero branch returns the left coefficient copy. Otherwise the two subtractions and division in [F2] give α^=α(1+δ1)(1+δ3)/(1+δ2), with ∣δi∣≤u by [F1]. Hence ∣α^−α∣≤η. The complement subtraction gives ∣β^−(1−α)∣≤η+u(2+η)=e, because ∣1−α^∣≤2+η. Also ∣α^∣,∣β^∣≤1+e=W.

1.2A1F1F2algebra

Write the two rounded products and final rounded sum as y^=fl⁡(fl⁡(β^a)+fl⁡(α^b)). Subtract the exact convex blend (1−α)a+αb. The coefficient differences contribute at most e(∣a∣+∣b∣); the two product roundings contribute at most uW(∣a∣+∣b∣); and the final addition contributes at most u(1+u)W(∣a∣+∣b∣). Thus ∣y^−((1−α)a+αb)∣≤E(∣a∣+∣b∣). If M is the maximum magnitude in the preceding computed column, this is at most 2EM. Since the exact blend is a convex combination, its magnitude is at most M, so the next computed column has maximum at most (1+2E)M.

2.1step 1.2algebrainduction

Let Mr be the maximum magnitude of the computed column after r stages and let εr be the maximum difference from the exact column. Step 1.2 gives Mr≤(1+2E)rM0. Since the exact weights are nonnegative and sum to one, the previous-column discrepancy contributes at most εr−1, while the local rounding contributes at most 2EMr−1. Therefore εr≤εr−1+2E(1+2E)r−1M0. Starting with ε0=0 and summing this geometric series gives εp≤((1+2E)p−1)M0.

3.1F3step 2.1algebra∎

By [F3], the final exact column is S(x); the final computed column is S^(x), and M0=max⁡k−p≤i≤k∣ci∣. Substitution in step 2.1 proves the bound.

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