Applied Mathematics
Vol.5 No.15(2014), Article ID:48885,26 pages
DOI:10.4236/am.2014.515227
An Asymptotic Distribution Function of the Three-Dimensional Shifted van der Corput Sequence
Jana Fialová1, Ladislav Mišk2, Oto Strauch1
1Mathematical Institute, Slovak Academy of Sciences, Bratislava, Slovakia
2Department of Mathematics, University of Ostrava, Ostrava, Czech Republic
Email: jana.fialova@mat.savba.sk, ladislav.misik@osu.cz, oto.strauch@mat.savba.sk
Copyright © 2014 by authors and Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY).
http://creativecommons.org/licenses/by/4.0/
Received 25 May 2014; revised 1 July 2014; accepted 15 July 2014
Abstract
In this paper, we apply the Weyl's limit relation to calculate the limit
where
is the van der Corput sequence in base
,
is the asymptotic distribution function of
, and
,
and
, respectively.
Keywords:Sequences, Arithmetic Means, Riemann-Stieltjes Integraion
1. Introduction
In this paper we apply the Weyl’s limit relation [1] (p. 1-61)
(1.1)
to the sequence, where
is the van der Corput sequence in base
and
is the asymptotic distribution function (abbreviated a.d.f.) of
and
. The van der Corput sequence in base
is defined as follows: Let
be the
-adic expression of a positive integer
. Then
(1.2)
It is well-known that this sequence is uniformly distributed (abbreviated u.d.), see [1] (2.11, p. 2-102), [2] (Theorem 3.5, p. 127), [3] (p. 41).
For
a motivation for the study of the distribution function (abbreviated d.f.)
of
,
is a result of Pillichshammer and Steinerberger in [4] which states that
(1.3)
while in J. Fialová and O. Strauch [5] the relation (1.3) was proved applying (1.1) as
Moreover, in the Unsolved Problems [6] (1.12),
the following problem is stated: Find the d.f.
of the sequence
,
, in
. Ch. Aistleitner and M. Hofer
[7] gave the following theoretical solution:
Theorem 1 Let
denote the von Neuman-Kakutani transformation described in
Figure 1. Define the
- dimensional curve
, where
. Then the searched
a.d.f. is
where
is the Lebesgue measure of a set
.
The paper consists of the following parts: After definitions (Part 2) we derive
the a.d.f. of
Figure 1.
Line segments containing
The graph of the von Neumann-Kakutani transformation
.
Haoshangban (Part 3), the a.d.f. of
(Part 4), intervals containing
in diagonals (Part 5) and an explicit form of a.d.f.
(Part 6). As an application (Part 7) we compute
the limit
(1.4)
for,
and
, respectively, see (0.39), (0.41)
and (0.46).
2. Definitions and Notations
Let,
be a sequence in the unit interval
. Denote
the step distribution function (step d.f.) of the finite sequence
in
, while
.
A function
is a distribution function (d.f.) if
(i)
is nondecreasing;
(ii)
and
.
A d.f.
is a d.f. of the sequence
,
if an increasing sequence of positive integers
exists such that
a.e. on
.
A d.f.
is an asymptotic d.f. (a.d.f.) of the sequence
,
if
a.e. on
.
The sequence
is uniformly distributed (abbreviating u.d.) if its a.d.f. is
.
Similar definitions take place for
and
-dimensional sequence
,
, in
, cf.
[1] (1.11, pp. 1-60).
In the sequel the
-dimensional interval
we denote by
, where
are projections on
axes, respectively.
3. a.d.f. of
Let
be an integer.
Lemma 1 Every point,
, lies on the diagonals of intervals
(1.5)
(1.6)
Proof. Express an integer
in the base
where
and
. We consider the following two cases:
,
.
Let
Then
,
and by (0.2)
. In this case
Thus such
lies on the line-segment
(1.7)
Let
Then
and
, where
. Then
. Thus
,
, and we have
and
and
. Thus such
lies on the segment
(1.8)
Thus, for, terms of the sequence
lie on the diagonal of the interval
(1.9)
and for, after reduction
, terms of the sequence
lie on the diagonals of the intervals
(1.10)
These intervals are maximal with respect to inclusion.
Adding the maps (1.7) and (1.8) we found the so-called von Neumann-Kakutani transformation, see Figure 1.
Because
is u.d., the sequence
has a.d.f.
of the form1
(1.11)
where
is the projection of a two dimensional set to the
-axis.
The sum (1.11) implies
(1.12)
From (1.12) it follows
(1.13)
and for, the mean equality misses.
4. a.d.f. of
Let
be an integer.
Lemma 2 All terms of the sequence,
, lie in the diagonals of the following intervals
(1.14)
(1.15)
(1.16)
Proof. Express an integer
in the base
(1.17)
where
and
. We consider three following cases:
,
,
.
Let
Then
,
and
. In this case
and thus such
lies on the line-segment
(1.18)
Let
. Then
and
, then
. Thus
,
, and we have
.
Furthermore
and
.
Thus in this case
lies on the line-segment
(1.19)
Let.
Then
and
, then
. Thus
,
, and we have
.
Furthermore
and
.
This gives
(1.20)
Summary, if the
satisfies
, then
is contained in the diagonal of
(1.14)
for
in the diagonal of
(1.15)
and for
in the diagonal of
(1.16).
Proof. Express an integer
in the base
(1.17)
where
and
. We consider three following cases:
,
,
.
Let
Then
,
and
. In this case
and thus such
lies on the linesegment
(1.18)
Let
. Then
and
, then
. Thus
,
, and we have
.
Furthermore
and
.
Thus in this case
lies on the line-segment
(1.19)
Let
. then
and
, then
. Thus
,
, and we have
.
Furthermore
and
.
This gives
(1.20)
Summary, if the
satisfies
, then
is contained in the diagonal of
(1.14) for
in the diagonal of
(1.15)
and for
in the diagonal of
(1.16).
Composition of the maps (0.18), (0.19) and (0.20) of
forms the second iteration
of the von Neumann-Kakutani transformation
. The diagonals of (1.14),
(1.16) and (1.15) yield the following graph of
in Figure 2.
Here the interval
on
-axis is decomposed in
,
, and the interval
is decomposed in
,
. On
-axis the interval
is decomposed in
,
and the interval
is decomposed in
,
. Note that for
, the interval
has a zero length and is missing.
Exchange for a moment the axis
by
. Similarly as in (1.11), we have that the a.d.f.
of the sequence
is
(1.21)
Decompose
as the following figure shows:
Then by Figure 3 we have
Figure 2. Straight lines
containing
,
.
Figure 3. Decomposition of the Vunit square to parts with
fixed expression of.
(1.22)
Let.
In this case we find a.d.f
from (1.22) omitting
In Part 7. Applications we need to find
from
in (1.22):
For
(1.23)
For
(1.24)
For
(1.25)
Note that for
the term
is omitted.
5. a.d.f. of
Let
be an integer.
Lemma 3 Every point
is contained in diagonals of the intervals
(1.26)
(1.27)
(1.28)
where
if
. These intervals are maximal with respect to inclusion.
Proof. Every maximal
-dimensional interval
containing points
will be written as
, where
are projections of
to the
, axes, respectively. Moreover if
then
and
. From u.d. of
follows that the lengths
. Combining intervals (1.5),
(1.14), (1.15), (1.16), (1.6) of equal lengths by following
Figure 3.
We find (1.26), (1.27), and (1.28).
Now, let
be the union of diagonals of (1.27), (1.28) and (1.26). Again, as in (1.11), the
a.d.f.
has2 the form
(1.29)
and it can be rewritten as
(1.30)
To calculate minimums in (1.30) we can use the following
Figure 4 (here):
As an example of application of (1.30) and Figure 4,
we compute
for
without using the knowledge of
,3
(1.31)
Proof.
1. Let.
Then,
,
, consequently
.
2. Let. Then
,
,
, consequently
.
3. Let. Then
,
, consequently
.
4. Let.
Specify,
. Then
,
for
. Thus (1.30) implies
For
we have
(1.32)
6. Explicit Form of
Let
be an integer.
Motivated by the Figure 4 we decompose the unit
interval
on
,
and
axes in the Figure 5
Figure 4. Projections of intervals
on axes
.
Figure 5. Divisions of the unit intervals.
intervals (here):
In this decomposition, for, we have
possibilities. We shall order choices of
from the left to the right. Detailed proofs are included only in non-trivial cases.
1. Let
. Then
.
Proof. We have,
,
. Then, by (1.30),
.
Similarly, in the following cases 2-9.
2. Let. Then
.
3. Let. Then
.
4. Let. Then
.
5. Let. Then
.
6. Let. Then
.
7. Let. Then
.
8. Let. Then
.
9. Let. Then
.
Proof. We use.
10. Let. Then
.
Proof. We use,
,
. Similarly11. Let
. Then
.
12. Let. Then
.
Proof. We use,
,
.
13. Let. Then
.
14. Let. Then
.
15. Let. Then
.
16. Let.
Specify,
,
. Then
Proof. First observe that, and
. Thus
and
Further, for, we have
Thus, using (1.30), we find
.
17. Let.
Specify,
. Then
Proof. We have,
,
, then
18. Let.
Specify,
. Then
Proof. We have
19. Let. Then
.
20. Let.
Specify,
,
. Then we have
Proof. We have,
and
21. Let.
Specify,
. Then
Proof.,
,
,
22. Let. Then
.
23. Let.
Specify,
. Then
Proof.,
,
,
24. Let. Then
.
Proof. We have. Specify
. Then
The first term is
and the second is
.
25. Let.
Specify,
. Then
Proof.
26. Let.
Specify,
and
. Then
Proof. We have,
,
,
. Moreover
which gives
.
The final equation holds if
and
. It can be seen that it holds also for
and
. For
and
we need to compute this sum separately.
27. Let. Then
.
Proof. First observe
,
,
. Thus
.
New specify
and
. Then we have
and,
,
. Then for the sums in (1.30)
we have
which gives.
The above computation of
holds for
.
Let.
We have
and
(1.33)
Thus
for
directly follows from
for
if we use only such items in
for which
,
,
. These are
, i.e.,
.
The non-zero values of
can also be seen in the following table.
In all other cases.
7. Applications
The knowledge of the a.d.f.
of the sequence
allows us to compute the following limit by the Weyl limit relation (1.1) in dimension
.
(1.34)
where
is an arbitrary continuous function defined in
. For computing (1.34)
we use the following two methods.
7.1. Method I
In the first method in the Riemann-Stieltjes integral (1.34) we apply integration by parts.
Lemma 4 Assume that
is a continuous in
and
is a.d.f. Then
(1.35)
Here
(1.36)
Note that
if the partial derivatives exist.
Exercise 1 Put. We have
,
The differential
is non-zero if and only if
and in this case
.
Proof: For every interval
and every continuous
the differential
is defined as
(1.37)
Putting,
,
we have
Then by (1.35)
(1.38)
For
and by (1.31) we have
For
and by (1.32) we have
Therefore for, by (1.34) and by (1.38) we
have
(1.39)
Note that the same result follows from (1.44).
Exercise 2 Put. Since
,
if
,
if
,
if
, and
if
and
otherwiseapplying (1.34) and (1.35) we have
(1.40)
Here we have
in (1.13) for
. For
we use
in (1.23) if
, (1.24) if
and (1.25) if
and for
we use (1.31) if
and (1.32) if
. Thus we have:
a)
for
;
b)
for
;
c)
for
;
d)
for
.
e)
for
;
f)
for
.
Putting a)-f) into (1.40) yields
(1.41)
7.2. Method II
In the second method we compute the differential
directly. It is nonzero only for
. For such
the a.d.f.
has the form
(1.42)
Thus
and moreover
only on the following straight lines
Considering three possible cases, calculate
(1.43)
Summary
(1.44)
Exercise 3 Put. By Method I, we
have
, similarly
, etc., and by
(1.35) we have
(1.45)
Since a computation of (1.45) is complicated we use Method II, by (1.44) we have
Inserting these formulas into (1.44) we have
(1.46)
for.
8. Conclusion
The problems solved in this paper is significantly more complicated in higher dimensions. For example, in dimension
, to compute the d.f.
of the sequence
, it is necessary to
investigate
cases analogous to
cases for the explicit form
in the part 0.6. Also Figure 3 would have to be
converted to the dimension
. Finally, we would need
the third iteration of von Neumann-Kakutani transformation.
Acknowledgements
The paper is sponsored by the project P201/12/2351 of GA Czech Republic.
References
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NOTES
1 is a copula.