When a is irrational,
takes on an infinite number of values, making it difficult
to group the terms in V_{n} and arrive at a limit for the sum. However, notice that any
irrational number can be approximated as closely as desired by a sequence of rationals with
increasing denominators. That is, for every irrational number a, there exist sequences of
integers (p_{m}) and (q_{m}), with (q_{m}) strictly increasing, such that p_{m} and q_{m}are relatively prime for all m and
Figure 1 shows the function for several values of q. Notice that as q gets larger, the curves seem to be approaching the line f=l. Indeed, this limit can be shown by using Stirling's approximation for N! when N is large. Stirling's formula is
Remark. Note that this is not a proof. The limit of
E[X_{n,a}] will be l for
every irrational point a if and only if f(a,l) is continuous whenever a is irrational;
however, it is not known for certain that this is the case. One way to prove that
(69) gives the correct limit would be first to show that
E[X_{n,a}] (with a
given n) is continuous at an irrational a, so that
There is, however, some evidence to support the result in (69). Computer simulations indicated that E[X_{n,a}] tended toward l when a was irrational, although this may merely reflect the limit in (69) since computers cannot process true irrational numbers. Another indication that this limit might be correct comes from a comparison with some of the results in [4] and [5] involving the number of eigenvalues in a fixed interval. Here, too, the limit depended only on the denominators of the endpoints when they were rational, and in this case, letting the denominators approach infinity did produce the correct result for irrational endpoints under certain circumstances.