Foundations of Data Science

integral gives
∫ ∞
0
e −r2 r d−1 dr =
∫ ∞
0
e −t t d−1
2
(
1
2 t− 1 2 dt
)
= 1 2
∫ ∞
0
e −t t d 2 − 1 dt = 1 2 Γ ( d
2
and hence, I(d) = A(d) 1 2 Γ ( d
2)
where the Gamma function Γ (x) is a generalization of the
factorial function for noninteger values of x. Γ (x) = (x − 1) Γ (x − 1), Γ (1) = Γ (2) = 1,
and Γ ( 1
2)
=
√ π. For integer x, Γ (x) = (x − 1)!.
Combining I (d) = π d 2 with I (d) = A (d) 1 2 Γ ( d
2)
yields
)
establishing the following lemma.
A (d) = π d 2
1
Γ ( )
d
2 2
Lemma 2.6 The surface area A(d) and the volume V (d) of a unit-radius ball in d dimensions
are given by
A (d) = 2π d 2
Γ ( ) and V (d) = 2 d
d
2
π d 2
Γ ( d
2
To check the formula for the volume of a unit ball, note that V (2) = π and V (3) =
2 π 3 2
3 Γ( 3 2) = 4 π, which are the correct volumes for the unit balls in two and three dimensions.
To check the formula for the surface area of a unit ball, note that A(2) = 2π
3
and
A(3) = 2π 3 2
1 √
2 π
= 4π, which are the correct surface areas for the unit ball in two and three
dimensions. Note that π d 2 is an exponential in d and Γ ( )
d
2 2 grows as the factorial of
d
. 2
This implies that lim V (d) = 0, as claimed.
d→∞
).
2.4.2 Most of the Volume is Near the Equator
An interesting fact about the unit ball in high dimensions is that most of its volume is
concentrated near its equator no matter what direction one uses to define the North Pole
and hence the “equator”. Arbitrarily letting x 1 denote “north”, most of the volume of the
unit ball has |x 1 | = O(1/ √ d). Using this fact, we will show that two random points in the
unit ball are with high probability nearly orthogonal, and also give an alternative proof
from the one in Section 2.4.1 that the volume of the unit ball goes to zero as d → ∞.
Theorem 2.7 For c ≥ 1 and d ≥ 3, at least a 1 − 2 /2 fraction of the volume of the
c e−c2
d-dimensional unit ball has |x 1 | ≤ √ c
d−1
.
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