Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-29
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  • 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.
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The computed floating-point dot product obeys componentwise and normwise error bounds

Statement

Let n≥1 and let x=(xj),y=(yj)∈Rn, with the dot product x⋅y=∑j<nxjyj computed from left to right in an arithmetic model with unit roundoff u satisfying nu<1 (The standard relative floating-point model and unit roundoff), where every product xjyj, every rounded product pj:=fl⁡(xjyj), and every intermediate sum lies in the normal range. Let s1:=p0 and, for 1≤k<n, let sk+1:=fl⁡(sk+pk). Then:

  1. Componentwise form. There are reals δ0,…,δn−1 with ∣δj∣≤γn:=nu/(1−nu) such that sn=∑j<nxjyj(1+δj).
  2. Normwise form. ∣sn−x⋅y∣  ≤  γn∑j<n∣xjyj∣  ≤  γn ∥x∥1∥y∥1, where ∥⋅∥1 is the published 1-norm (The p-norms ∥x∥p for rational p≥1, and ∥x∥∞, Each ∥⋅∥p is a norm on Rn, and the induced metrics are exactly d1, d2 and d∞ of the published metric-spaces page) and ∣⋅∣ the absolute value (Absolute value in an ordered field).

Facts & Assumptions

Given: A natural number n≥1, vectors x,y∈Rn, a unit roundoff u with nu<1, the rounded products pj=fl⁡(xjyj) for j<n, and the left-to-right computed sum sn above.

[L1]

Each product and each sum obeys the relative model: fl⁡(t)=t(1+ε) with ∣ε∣≤u for every real t in the normal range (The standard relative floating-point model and unit roundoff).

[L2]

A product of factors (1+εi)i<m with m≤n and ∣εi∣≤u for every i<m equals 1+δ with ∣δ∣≤γm≤γn, where γn=nu/(1−nu) (If |delta_i| <= u and n u < 1, then prod (1+delta_i) = 1 + theta_n with |theta_n| <= gamma_n).

[L3]

∥x∥1=∑j<n∣xj∣, and ∣xjyj∣=∣xj∣ ∣yj∣ (The p-norms ∥x∥p for rational p≥1, and ∥x∥∞, Absolute value in an ordered field).

Proof

technique · direct
1.1givenL1

By [L1], for each j<n there is πj with pj=xjyj(1+πj) and ∣πj∣≤u; and for each k with 1≤k<n there is αk with sk+1=(sk+pk)(1+αk) and ∣αk∣≤u.

2.1step 1.1algebra

Unrolling the recursion of step 1.1 gives sn=x0y0(1+π0)∏k=1n−1(1+αk)+∑j=1n−1xjyj(1+πj)∏k=jn−1(1+αk), where every empty product is 1; each coefficient is the product of the error factor from the rounded multiplication pj and the error factors from the later additions through which that term passes.

3.1step 2.1L2algebra

The coefficient of x0y0 in step 2.1 is a product of exactly n factors of the form 1+ε with ∣ε∣≤u, and for each j with 1≤j<n the coefficient of xjyj is a product of n−j+1≤n such factors. By [L2], each coefficient therefore equals 1+δj with ∣δj∣≤γn, so step 2.1 becomes sn=∑j<nxjyj(1+δj), which is claim 1.

4.1step 3.1L3algebra

Subtracting x⋅y=∑jxjyj from claim 1 gives sn−x⋅y=∑jxjyjδj, so by the triangle inequality and [L3] ∣sn−x⋅y∣≤∑j∣xjyj∣ ∣δj∣≤γn∑j∣xjyj∣.

5.1step 4.1L3algebra

Since ∑j∣xjyj∣=∑j∣xj∣ ∣yj∣≤(∑j∣xj∣)(∑k∣yk∣)=∥x∥1∥y∥1 by [L3], step 4.1 gives the normwise form of claim 2.

6.1step 3.1step 4.1step 5.1∎

Claim 1 is step 3.1 and claim 2 is steps 4.1 and 5.1.

Depends on

Used by

Dependency tree · two levels

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Sources