We are interested in the average length of the vector *X* sum of *N* unit vectors
randomly chosen in *D* dimensions, ie, unformly distributed on the unit *N* sphere and independent.

In one dimension, *D=1*, the identities of Quiz 23, can be used to compute the
average length of the vector, and its average square length,

E[X] = const sqrt(N)

E[X^2] = N

Similar relations hold in any dimension:

The relation *E[X^2] = N* is easy to prove:
the *N* unit vectors are "decoupled" in the sum *X ^{2}*.

Prove (or disprove) the relations for higher even powers:

*E[X ^{4}] = N^{2} ( 1 + 2/D )*

**Exercise**

*a(k+2,D) = (D+k)/D a(k,D)*.

Given that *a(2,D)*=1 for every *D*, this recursion solves the problem once the values for
*k=1* and *k=2* are known.

What are the values *a(1,D)* ?

D = 2 3 4 5 6 k=-2 3/1 4/2 5/3 6/4 0 1 1 1 1 1 2 1 1 1 1 1 4 4/2 5/3 6/4 7/5 8/6 6 24/4 35/9 48/16 63/25 80/36

The values for *k=1* are (integral from 0 to infinity)

The first few values are ( P = sqrt(pi) )

D = 2 3 4 5 6 k= 1 P/2 sqrt(2/3)*2/P sqrt(2/4)*3P/8 sqrt(2/5)*8/3P sqrt(2/6)*15P/16

Marco Corvi - 2016