Oranges in Many Dimensions

Peeling oranges in high dimensionality $k ≫ 3$ is hard. This post discusses a fundamental phenomenon in asymptotic geometry and explains at which rate you should peel an orange.

Suppose we have a special branch of hyper-orange existing in $d \geq 1$ and we would like to peel such that $95 %$ of pulp is preserved. We will denote the volume of an $L^{p}$ norm in $d$ dimensions, that is $C = {x \in ℝ^{d} ∣ {‖ x ‖}_{p} \leq r}$ , as $V_{C} = V_{p, d}$ . Because we first deal with balls we have $p = 2$ and set by default $V_{2, d} = V_{d}$ .

We first look at ratios of volume of balls radii scaled by $φ$ such that $V_{d} (φr) ∕ V_{d} (r) = 0.95$ . For $d \in {1, 2, 3}$ we know from middle school that $V_{1} (r) = 2 r$ , $V_{2} (r) = π r^{2}$ and $V_{3} (r) = \frac{4}{3} π r^{3}$ , and hence $φ_{1, 2, 3} = \sqrt[1, 2, 3]{0.95}$ . For $d > 3$ we need some material from real analysis to work out the volume.

Volume of n-ball Exponential Function

One proof works with two observations: first scaling area identity $A_{n - 1} (r) = r^{n - 1} A_{n - 1} (1)$ and evaluates the integral for rotational invariant function $f (x_{1}, \dots, x_{d}) = \exp (- x_{1}^{2} \dots - x_{d}^{2})$ .

On the one side, we have

\begin{array}{l} \int_{ℝ^{d}} f (x_{1}, \dots, x_{d}) d x_{1} \dots d x_{d} = \int_{ℝ} \exp (- x_{1}^{2}) d x_{1} \dots \int_{ℝ} \exp (- x_{d}^{2}) d x_{d} = π^{d ∕ 2} \end{array}

On the other side, isolating one variable and applying the area identity

\begin{array}{l} \int_{ℝ^{d}} f (x_{1}, \dots, x_{d}) d x_{1} \dots d x_{d} & = \int_{0}^{\infty} \int_{S^{n - 1} (r)} \exp (- r^{2}) d A d r \\ = \int_{0}^{\infty} A_{n - 1} (1) r^{d - 1} \exp (- r^{2}) d r \\ = A_{n - 1} (1) \int_{0}^{\infty} \exp (- r^{2}) r^{d - 1} d r = \frac{1}{2} A_{n - 1} (r) Γ (\frac{d}{2}) . \end{array}

Solving for the $d - 1$ dimensional area we get $A_{d - 1} (1) = \frac{2 π^{d ∕ 2}}{Γ (d ∕ 2)}$ and substituting into the volume formula, we get the desired $V_{d} (R) = \frac{π^{d ∕ 2} R^{d}}{\frac{d}{2} Γ (d ∕ 2)}$ .

Consequence for Peeling Oranges

Now consider the volume of two balls $V_{d} (1)$ , $V_{d} (1 - 𝜀)$ and compare their volume

\begin{array}{l} \frac{V_{d} (1 - 𝜀)}{V_{d} (1)} = {(1 - 𝜀)}^{d} . \end{array}

Then, for every constant $𝜀$ the ratio tends to zero, and we are left with no pulp at all! Realizing that for $𝜀 \sim d^{- 1}$ , the limit yields exponential function

\begin{array}{l} \frac{V_{d} (1 - 𝜀)}{V_{d} (1)} = {(1 - \frac{M}{d})}^{d} \to \exp (- M) for d \to \infty, \end{array}

we should peel $(- \log 0.95) ∕ d$ of the skin to get a fair share of pulp for high dimensions.

Consequence for Approximating Norms

Let’s pack the orange into a box. For simplicity we assume that the box has side length $1$ and recognize that an $d$ -cube is described with the maximum norm. The volume of such a box is $V_{\infty, d} = 1^{d} = 1$ , and comparing that to the volume of our orange

\begin{array}{l} \frac{V_{2, d}}{V_{\infty, d}} = \frac{π^{d ∕ 2}}{\frac{d}{2} Γ (d ∕ 2) 1} \end{array}

Express this in log-space and use log-gamma approximation $\log Γ (z) \approx (z - \frac{1}{2}) \log z - z + \frac{1}{2} \log (2 π)$

\begin{array}{l} \log \frac{V_{2, d}}{V_{\infty, d}} & = \frac{d}{2} \log π - \log \frac{d}{2} - \log Γ (\frac{d}{2}) \\ \sim \frac{d}{2} \log \frac{2 πe}{d} - \frac{1}{2} \log 2 πd \\ \sim \frac{d}{2} \log \frac{2 πe}{d} \end{array}

which turns negative once $d > 2 πe$ and grows linear in the dimensionality. Hence, the fraction diminishes exponentially fast. That means for large $d$ we can pack exponential many oranges into a single box!

29 March 2025