LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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An explicit rational envelope for the scaled Householder QR error

Statement

Let $q \geq n \geq 1$ , $u > 0$ , and let $κ_{q}$ be the local scaled-Householder coefficient of The scaled computed Householder step satisfies the local backward error premise at active row dimension $q$ . Set $β_{q, n} = (1 + κ_{q})^{n} - 1, C_{q, n} = 200 q n .$ If $C_{q, n} u < 1$ , then $β_{q, n} \leq η_{q, n} : = \frac{C_{q, n} u}{1 - C_{q, n} u} .$ Thus the QR backward-error factor can be bounded by the stated rational dimension-dependent expression in this precision range.

Facts & Assumptions

Given: Integers $q \geq n \geq 1$ , $u > 0$ , and $200 q n u < 1$ .

[F1]

For active row dimension $r$ , the local constants are $e_{r} = (1 + u)^{r / 2 + 3} - 1, d_{r} = e_{r} + u (2 + e_{r}), a_{r} = \frac{1 + u}{(1 - u)^{r}} - 1, g_{r} = \frac{r u}{1 - r u},$ $c_{r} = 2 ((1 + a_{r}) (1 + u)^{2} (1 + g_{r}) - 1), κ_{r} = \max {e_{r} + 8 d_{r}, (1 + u) c_{r} + 3 u} .$ The scaled computed Householder step satisfies the local backward error premise

Proof

Given: The hypotheses in the Statement.

1.1F1givenalgebra

Put $x = q u$ . The precision hypothesis gives $u \leq 1 / (200 q)$ and $0 < x < 1 / 200$ . Since $q / 2 + 3 \leq 7 q / 2$ , set $s = (q / 2 + 3) u \leq 7 x / 2 < 7 / 400$ . The power series gives $1 + u \leq e^{u}$ and, for $0 \leq z < 1$ , $e^{z} - 1 = \sum_{j \geq 1} z^{j} / j! \leq z / (1 - z)$ . Therefore $e_{q} \leq e^{s} - 1 \leq \frac{s}{1 - s} \leq \frac{1400}{393} x < \frac{18}{5} x .$ Also $u e_{q} \leq (9 / 500) u$ , so $d_{q} = e_{q} + 2 u + u e_{q} \leq \frac{18}{5} x + \frac{1009}{500} u \leq \frac{2809}{500} x < \frac{57}{10} x .$ Hence the first branch in [F1] satisfies $e_{q} + 8 d_{q} < (18 / 5 + 8 \cdot 57 / 10) x < 50 x$ .

1.2F1givenalgebra

Bernoulli's inequality gives $(1 - u)^{q} \geq 1 - q u = 1 - x$ ; for integer $q$ , this follows by induction from $(1 - u) (1 - q u) = 1 - (q + 1) u + q u^{2} \geq 1 - (q + 1) u$ . Since $u \leq 1 / 200$ and $x \leq 1 / 200$ , $(1 + u)^{3} \leq 1 + 4 u$ by expansion, and $(1 - x)^{- 2} \leq 1 + 3 x$ because $(1 + 3 x) (1 - x)^{2} - 1 = x (1 - 5 x + 3 x^{2}) \geq 0$ . Thus $(1 + a_{q}) (1 + u)^{2} (1 + g_{q}) \leq \frac{(1 + u)^{3}}{(1 - x)^{2}} \leq (1 + 4 u) (1 + 3 x) .$ As $u \leq x$ and $u \leq 1 / 200$ , the last expression minus $1$ is at most $4 x + 3 x + 12 u x \leq (7 + 3 / 50) x < 8 x$ . Consequently $c_{q} < 16 x$ and $(1 + u) c_{q} + 3 u < (16 + 16 / 200 + 3) x < 20 x$ . Both branches in [F1] are therefore bounded by $50 x$ , so $κ_{q} \leq 50 q u$ .

2.1F1step 1.1step 1.2givenalgebra∎

Put $t = C_{q, n} u < 1$ and $y = n κ_{q}$ . Step 1.2 gives $0 \leq y \leq 50 q n u = t / 4 < 1 / 4$ . Since $(1 + κ_{q})^{n} \leq e^{y}$ and $e^{y} - 1 \leq y / (1 - y)$ by the same power-series estimate as in step 1.1, $β_{q, n} \leq \frac{y}{1 - y} \leq \frac{t / 4}{1 - t / 4} = \frac{t}{4 - t} \leq \frac{t}{1 - t} = η_{q, n} .$ This proves the stated rational envelope.

Depends on

The scaled computed Householder step satisfies the local backward error premise

Used by

Product and Householder QR errors bound the represented joint covariance Lemma

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Kevin S. Tracy, A Square-Root Kalman Filter Using Only QR Decompositions (standard reference, not scraped)
M. A. Salzmann, Some Aspects of Kalman Filtering, Technical Report 140 (standard reference, not scraped)