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The Mean When a is Irrational

When a is irrational, $\{a\}$ takes on an infinite number of values, making it difficult to group the terms in Vn 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 (pm) and (qm), with (qm) strictly increasing, such that pm and qmare relatively prime for all m and

\begin{displaymath}\lim_{m\rightarrow\infty} {p_m \over q_m}=a.
\end{displaymath}

Thus, we could take the formula for $f\left({p \over q},l\right)$ given in Theorem 1 and let $q\rightarrow\infty$. The hope is that this might provide an idea of what happens for irrational a.

Figure 1 shows the function $f\left({p \over q},l\right)$ 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


 \begin{displaymath}N! \approx N^N e^{-N} \sqrt{2\pi N}.
\end{displaymath} (63)

Applying (63) to (62), with $N=\lfloor ql \rfloor$,
 
$\displaystyle \lim_{n\rightarrow\infty} E\left[X_{n,{p \over q}}\right]$ = $\displaystyle \frac{1}{q} \ln \left(\frac{(ql)^{\lfloor ql \rfloor}}
{\lfloor ql \rfloor !}\right)$ (64)
  $\textstyle \approx$ $\displaystyle {1 \over q}\ln \left( {(ql)^{\lfloor ql \rfloor}
\over {\lfloor q...
...loor ql \rfloor} e^{-\lfloor ql \rfloor}
\sqrt{2\pi \lfloor ql \rfloor}}\right)$ (65)
  = $\displaystyle {1 \over q}\ln \left( {(ql) \over \lfloor ql \rfloor} \right)^
{\...
... \rfloor \over q} - {1 \over q}\ln
\left(\sqrt{2\pi \lfloor ql \rfloor}\right).$ (66)

Now, taking the limit for large q,
 
$\displaystyle \lim_{m\rightarrow\infty} \lim_{n\rightarrow\infty} E\left[X_{n,{p_m \over q_m}}\right]$ = $\displaystyle \lim_{q\rightarrow\infty} \left[ {\lfloor ql \rfloor \over q}
\ln...
...r \over q} - {1 \over q}\ln
\left(\sqrt{2\pi \lfloor ql \rfloor}\right) \right]$ (67)
  = $\displaystyle l \cdot 0 + l - 0$ (68)
  = l. (69)

This result seems to suggest that the limit of the mean of Xn,a for irrational a is simply l.

Remark. Note that this is not a proof. The limit of E[Xn,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[Xn,a] (with a given n) is continuous at an irrational a, so that

\begin{displaymath}E[X_{n,a}] = \lim_{m\rightarrow\infty}E\left[X_{n,{p_m \over q_m}}\right].
\end{displaymath}

Since $f(a,l)=\lim_{n\rightarrow\infty}E[X_{n,a}]$, it would then need to be shown that

\begin{displaymath}\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}E\left[X_{n...
...}\lim_{n\rightarrow\infty}E\left[X_{n,{p_m \over q_m}}\right].
\end{displaymath}

It is not difficult to determine where E[Xn,a] is continuous when n is fixed, but so far, there is no justification for switching the order of the limits.

There is, however, some evidence to support the result in (69). Computer simulations indicated that E[Xn,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.


next up previous
Next: Conclusion Up: Calculation of the Mean Previous: The Mean When a

2000-09-25