← All parts of this equation

Equation 9 · Part 5 · Embeddings and the Geometry of Similarity

Symbol d^1/k - 1/2

lim⁡d→∞E ⁣[Dmax⁡k−Dmin⁡kd1/k−1/2]=C⋅1(k+1)1/k12k+1,\lim_{d \to \infty} \mathbb{E}\!\left[ \frac{D_{\max}^{k} - D_{\min}^{k}}{d^{1/k - 1/2}} \right] = C \cdot \frac{1}{(k+1)^{1/k}} \sqrt{\frac{1}{2k+1}},
d1/k−1/2d^{1/k - 1/2}

What this part means

d1d^1/k - 1/2 occurs below the fraction bar. The quantity above the bar is divided by this expression; zero is excluded as a denominator.

Its job in the formula

d1d^1/k - 1/2 occurs below the fraction bar. The quantity above the bar is divided by this expression; zero is excluded as a denominator.

The passage around this formula

Aggarwal, Hinneburg and Keim then showed the rate depends on the norm. For uniform data under an LkL_k metric, the expected gap between the farthest and nearest distances scales as lim⁡d→∞E ⁣[Dmax⁡k−Dmin⁡kd1/k−1/2]=C⋅1(k+1)1/k12k+1\lim_{d \to \infty} \mathbb{E}\!\left[ \frac{D_{\max}^{k} - D_{\min}^{k}}{d^{1/k - 1/2}} \right] = C \cdot \frac{1}{(k+1)^{1/k}} \sqrt{\frac{1}{2k+1}}. with C constant, while typical distances themselves grow like d1/kd^{1/k} . Relative contrast therefore decays with dimension, and the constant shrinks as k rises — which is why they conclude that L1L_1 is preferable to L2L_2 in high dimensions, L2L_2 to L3L_3 , and that fractional norms preserve contrast better still [ 12 ] .

Read this part in the article →

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.

Open the illustrated exponents: repeated multiplication and powers guide →

See this notation across published equations →

Sources cited in the surrounding passage

These citations provide research context; check each source for the exact claim it supports.