Pattern Recognition and Machine Learning

62 1. INTRODUCTION

1.17 (

) www The gamma function is defined by

Γ(x)≡

∫∞

0

ux−^1 e−udu. (1.141)

Using integration by parts, prove the relationΓ(x+1) =xΓ(x). Show also that

Γ(1) = 1and hence thatΓ(x+1)=x!whenxis an integer.

1.18 (

) www We can use the result (1.126) to derive an expression for the surface

areaSD, and the volumeVD, of a sphere of unit radius inDdimensions. To do this,

consider the following result, which is obtained by transforming from Cartesian to

polar coordinates

∏D

i=1

∫∞

−∞

e−x

(^2) i
dxi=SD
∫∞
0
e−r
2
rD−^1 dr. (1.142)
Using the definition (1.141) of the Gamma function, together with (1.126), evaluate
both sides of this equation, and hence show that

SD=

2 πD/^2

Γ(D/2)

. (1.143)

Next, by integrating with respect to radius from 0 to 1 , show that the volume of the

unit sphere inDdimensions is given by

VD=

SD

D

. (1.144)

Finally, use the resultsΓ(1) = 1andΓ(3/2) =

√

π/ 2 to show that (1.143) and

(1.144) reduce to the usual expressions forD=2andD=3.

1.19 (

) Consider a sphere of radiusainD-dimensions together with the concentric

hypercube of side 2 a, so that the sphere touches the hypercube at the centres of each

of its sides. By using the results of Exercise 1.18, show that the ratio of the volume

of the sphere to the volume of the cube is given by

volume of sphere

volume of cube

=

πD/^2

D 2 D−^1 Γ(D/2)

. (1.145)

Now make use of Stirling’s formula in the form

Γ(x+1)(2π)^1 /^2 e−xxx+1/^2 (1.146)

which is valid forx 

1 , to show that, asD→∞, the ratio (1.145) goes to zero.

Show also that the ratio of the distance from the centre of the hypercube to one of

the corners, divided by the perpendicular distance to one of the sides, is

√

D, which

therefore goes to∞asD→∞. From these results we see that, in a space of high

dimensionality, most of the volume of a cube is concentrated in the large number of

corners, which themselves become very long ‘spikes’!