d(t) = t^3.

February 22, 2026

d(t) = t^3.

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Observe:
= 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,
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.

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