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Home » We want $ \mathbbE[\textoutput] $. - 500apps

We want $ \mathbbE[\textoutput] $. - 500apps

📅 March 13, 2026 👤 scraface

Mar 13, 2026

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📌 = 1^3,\quad 8 = 2^3,\quad 27 = 3^3,\quad 64 = 4^3.

📌 So $ d(n) = n^3 $ for $ n = 1,2,3,4 $. Since $ d(t) $ is a cubic polynomial and agrees with $ t^3 $ at four distinct points, by uniqueness of interpolation,

📌 d(t) = t^3.

📌 But $ d(t) = t^3 $ is strictly convex and has no minimum (it decreases on $ (-\infty, 0) $, increases on $ (0, \infty) $), so it does not achieve a minimum. However, the problem states that the **minimum depth is achieved exactly once**, which implies $ d(t) $ has a unique global minimum. But $ t^3 $ has no such minimum. Contradiction?

📌 Wait — unless our initial assumption that $ d(t) = t^3 $ is forced is correct, but it violates the minimum condition. So perhaps $ d(t) $ is not exactly $ t^3 $, but fits it at four points. But a cubic is uniquely determined by four points, so $ d(t) = t^3 $ is the only cubic polynomial satisfying the conditions.

📌 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.

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