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Equation 3 · Part 4 · Small and On-Device AI in 2035: Scenarios and Falsifiers

Symbol p^*

D(t)=1 ⁣[P(t)≥p∗]⋅1 ⁣[L(t)≥ℓ∗]D(t) = \mathbb{1}\!\left[P(t) \ge p^{*}\right] \cdot \mathbb{1}\!\left[L(t) \ge \ell^{*}\right]
p∗p^{*}

What this part means

p∗p^* is one factor in the product that computes the quantity on the left.

Its job in the formula

p∗p^* is one factor in the product that computes the quantity on the left.

The passage around this formula

…out. Becoming the default substrate needs both conditions at once, not an average of them: D(t)=1 ⁣[P(t)≥p∗]⋅1 ⁣[L(t)≥ℓ∗]D(t) = \mathbb{1}\!\left[P(t) \ge p^{*}\right] \cdot \mathbb{1}\!\left[L(t) \ge \ell^{*}\right]. D(t) stays at zero however high either term climbs alone. Axis A determines whether P(t) can plausibly clear p∗p^{*} within this article’s horizon; Axis B determines whether L(t) can. Neither can be inferred from the other, which is why they are kept as two axes rather than folded into one.

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Learn the underlying idea

An exponent tells how a base is used in multiplication. In x³, x is the base and 3 is the exponent: x³ = x × x × x.

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Sources cited in the article section

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