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ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-29
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A short floating-point dot product obeys the gamma_n error bound

Example

Take x=(1,2), y=(3,4) and the binary64 unit roundoff u=2−53, for which γ2=2u/(1−2u)≈2.2⋅10−16. The left-to-right computed dot product s2 satisfies

s2=3(1+δ0)+8(1+δ1)with ∣δ0∣,∣δ1∣≤γ2,

so the computed value is the exact dot product of slightly perturbed summands, and

∣s2−11∣  ≤  γ2 (3+8)=11γ2  ≤  γ2 ∥x∥1∥y∥1=21γ2≈4.7⋅10−15.

Facts & Assumptions

Given: The vectors x=(1,2), y=(3,4), the exact dot product x⋅y=11, the unit roundoff u=2−53 with 2u<1, and the left-to-right computed sum s2 of the two products x0y0=3 and x1y1=8.

[L1]

The computed dot product obeys the componentwise form sn=∑j<nxjyj(1+δj) with ∣δj∣≤γn=nu/(1−nu), and the normwise bound ∣sn−x⋅y∣≤γn∑j∣xjyj∣≤γn∥x∥1∥y∥1 (The computed floating-point dot product obeys componentwise and normwise error bounds).

Verification

technique · direct
1.1L1algebra

With n=2 and 2u<1, [L1] applies and yields s2=x0y0(1+δ0)+x1y1(1+δ1) with ∣δ0∣,∣δ1∣≤γ2, which for x=(1,2) and y=(3,4) reads s2=3(1+δ0)+8(1+δ1).

1.2algebra

The exact dot product is x⋅y=1⋅3+2⋅4=11, and the absolute summand total is ∣x0y0∣+∣x1y1∣=3+8=11, while ∥x∥1∥y∥1=(1+2)(3+4)=21.

2.1L1step 1.2algebra

By the normwise bound of [L1] applied to the numbers of step 1.2, ∣s2−11∣≤γ2⋅11≤γ2⋅21; substituting u=2−53 gives γ2=2−52/(1−2−52), so ∣s2−11∣≤11⋅2−52/(1−2−52), a bound of order 10−15.

3.1step 1.1step 2.1∎

Step 1.1 verifies the componentwise perturbation form and step 2.1 the normwise bound for this short vector pair: the computed dot product agrees with the exact dot product of slightly perturbed summands and lies within 11γ2 of 11.

Depends on

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Sources