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The unit-ball volume is maximal in dimension five

Statement

Among positive integer dimensions, the unit-ball volume is uniquely maximal at $n = 5$ .

Facts & Assumptions

Given: Unit-ball volumes $V_{n} : = V_{n} (1)$ for positive integers $n$ .

[F1]

For every $n \geq 1$ , $V_{n} = π^{n / 2} / Γ (n / 2 + 1)$ (The closed form for the volume of the unit $n$ -ball).

[F2]

For every natural $N$ , the Gregory--Leibniz formula writes $π / 4$ as its partial sum through $N$ plus a signed remainder of magnitude at most $1 / (2 N + 3)$ (The Gregory-Leibniz series: pi over four equals 1-1/3+1/5-1/7+...).

[F3]

For every $s > 0$ , $Γ (s + 1) = s Γ (s)$ , and $Γ (1) = 1$ (The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ ).

[F4]

$Γ (1 / 2) = \sqrt{π}$ ( $Γ (1 / 2) = \sqrt{π}$ from the Gaussian integral).

Proof

technique · direct

1.1F2algebra

In [F2], the partial sum through $N = 7$ is $33976 / 45045 > 3 / 4$ and its remainder is positive, so $π > 3$ . The partial sum through $N = 18$ is $133330680156299 / 166966608033225 < 4 / 5$ and its remainder is negative, so $π < 16 / 5$ .

2.1F1F3step 1.1algebra

Facts [F1] and [F3] give $V_{n + 2} / V_{n} = 2 π / (n + 2)$ . Using step 1.1, the odd chain increases through $V_{5}$ and then decreases, while the even chain increases through $V_{6}$ and then decreases.

2.2F1F3F4step 1.1algebra

From [F3], $Γ (7 / 2) = (5 / 2) (3 / 2) (1 / 2) Γ (1 / 2) = (15 / 8) \sqrt{π}$ by [F4], and $Γ (4) = 3 \cdot 2 \cdot 1 \cdot Γ (1) = 6$ . Hence [F1] gives $V_{5} = π^{5 / 2} / ((15 / 8) \sqrt{π}) = 8 π^{2} / 15$ and $V_{6} = π^{3} / 6$ . The upper bound $π < 16 / 5$ from step 1.1 gives $V_{5} > V_{6}$ .

3.1step 2.1step 2.2∎

Step 2.1 identifies the unique maximum within each parity chain, and step 2.2 compares the two candidates. Therefore $V_{5}$ is the unique global maximum.

Depends on

Used by

FALSE: unit-ball volume increases with dimension False statement

Dependency tree · two levels

29 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sheldon Axler, Measure, Integration & Real Analysis, §5C, Exercise 12(b) (standard reference, not scraped)