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uk=10k/2,k∈{1,2,3}u_k = 10^{k/2}, \qquad k \in \{1, 2, 3\}

Why this formula appears here

Epoch AI, a research organisation that tracks compute at the frontier, publishes exactly this kind of reconstruction for Colossus and states its own confidence in bands rather than single numbers: an estimate graded “Confident” carries roughly ±3x uncertainty, “Likely” roughly ±10x, and “Speculative” roughly ±31x [ 4 ] . Those three multipliers are not arbitrary; they sit at half-order-of-magnitude steps, uk=10k/2,k∈{1,2,3}u_k = 10^{k/2}, \qquad k \in \{1, 2, 3\}. giving u1u_1 ≈\approx 3.16 , u2u_2 = 10 , and u3u_3 ≈\approx 31.6 — which is a formal way of saying that even a well-resourced independent estimate of a frontier training run can be off by an order of magnitude or more, and everyone doing the estimating knows it going in.…

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Published contexts (1)

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uk=10k/2,k∈{1,2,3},u_k = 10^{k/2}, \qquad k \in \{1, 2, 3\},

Equation 1 · AI Industry

The Hardest Unsolved Problems in Evaluating xAI's Claims

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

Epoch AI, a research organisation that tracks compute at the frontier, publishes exactly this kind of reconstruction for Colossus and states its own confidence in bands rather than single numbers: an estimate graded “Confident” carries roughly ±3x uncertainty, “Likely” roughly ±10x, and “Speculative” roughly ±31x [ 4 ] . Those three multipliers are not arbitrary; they sit at half-order-of-magnitude steps, uk=10k/2,k∈{1,2,3}u_k = 10^{k/2}, \qquad k \in \{1, 2, 3\}. giving u1u_1 ≈\approx 3.16 , u2u_2 = 10 , and u3u_3 ≈\approx 31.6 — which is a formal way of saying that even a well-resourced independent estimate of a frontier training run can be off by an order of magnitude or more, and everyone doing the estimating knows it going in.…

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