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Home » Total favorable roll sequences (before sum constraint): - inBeat

Total favorable roll sequences (before sum constraint): - inBeat

📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 But $ d(t) = t^3 $ has no minimum. Unless minimum depth refers to **local minimum**? But still, $ t^3 $ has no local minimum.

📌 Hence, the only resolution is that the model is not $ d(t) = t^3 $, so perhaps there's a typo? But wait — reconsider: the values $ d(1)=1, d(2)=8, d(3)=27, d(4)=64 $ **force** $ d(t) = t^3 $, due to interpolation.

📌 The condition that a cubic achieves a minimum must be compatible. But $ t^3 $ has no minimum. Therefore, the only possibility is that the model is still $ d(t) = t^3 $, and the phrase minimum depth refers to the **lowest recorded value in a physical window**, but mathematically, over $ \mathbb{R} $, $ t^3 $ has no minimum.

📌 But perhaps we misread: the problem says d depicts the depth... as a cubic polynomial and achieves its minimum depth exactly once. So the polynomial must have a unique critical point where $ d'(t) = 0 $ and $ d''(t) > 0 $.

📌 But $ d(t) = t^3 $ has $ d'(t) = 3t^2 \geq 0 $, zero only at $ t=0 $, but second derivative $ 6t $, so $ d''(0)=0 $ — not a strict minimum.

📌 So $ d(t) = t^3 $ fails the minimum condition.

📌 But $ d(t) $ agrees with $ t^3 $ at $ t=1,2,3,4 $, yet is a cubic. So define $ p(t) = d(t) - t^3 $. Then $ p(t) $ is a polynomial of degree at most 3 (since $ d(t) $ is cubic, $ t^3 $ is cubic), and $ p(1) = p(2) = p(3) = p(4) = 0 $.

📌 So $ p(t) $ has four roots and is degree ≤ 3 → $ p(t) \equiv 0 $. Hence $ d(t) = t^3 $ is the only possibility.

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