.

Note that

= ,

= ,

= = .

Here is an empirical solution estimate for to . To estimate the value of , simulate m sets of n random numbers and for each set, let equal the number of n -tuplets such that

+ . . . + .

An estimate of is .

> randomize():
for n from 2 to 6 do
m := 10000: #This is the number of repetitions
Y[n] := 0: #Y will equal the number of "successes"
for k from 1 to m do
s := 0:
for j from 1 to n do
s := s + rng()^2: #sum of squares of n RNs
od:
if s < 1 then Y[n] := Y[n] + 1 fi:
od:
od:

> for n from 2 to 6 do
V[n] := 2^n*(Y[n]/m):
od:

> V[2] := evalf(V[2]);
V[3] := evalf(V[3]);
V[4] := evalf(V[4]);
V[5] := evalf(V[5]);
V[6] := evalf(V[6]);

The following procedure finds the "volume of a ball" of radius r in n -space. It finds the value of , a ball of radius r in n -space by integration using "volumes" of cross sections which are balls of radius in ( n - 1)-space:

This procedure was written by John Krueger when he was a student at Hope College .

> Volume := proc(r, n:: integer)
local x;
options remember;
if n = 0 then
RETURN(1):
else
RETURN(int(Volume(sqrt(r^2 - x^2), n - 1), x = -r .. r)):
fi:
end;

Volume(1,0);

> Volume(1,1);

> Volume(1,2);

> Volume(1,3);

> Volume(1,4);

> Volume(1,5);

> Volume(1,6);

> Volume(1,7);

> Volume(1,8);

> Volume(1,9);

> Volume(1,10);

Note that when n is even, say n = 2 k , then

.

Thus

.