For any, let and be defined as

where denote the integer part. And we define the sequence as follows:

where denotes the nth iterate of.

It is well known that from the algorithm (1), all can be developped uniquely into an infinite or finite series

In the literature [2], (3) is called the Alternating Balkema-Oppenheim expansion of x and denoted by for short. From the algorithm, one can see that T maps irrational element into irrational element, and the series is infinite. While for rational numbers, in fact, we have is rational if and only if its sequence of digits is terminate or periodic, see [1-3].

For any and, define

From the algorithm of (1), it is clear that

For any, let be its Alternating Sylvester expansion, then we have

for any. On the other hand, any of integer sequence satisfying

for all is a Sylvester admissible sequence, that is, there exists a unique such that for all, see [9].

The behaviors of the sequence are of interest and the metric and ergodic properties of the sequence and have been investigated by a number of authors, see [1-3].

For any, define

and we call the error-sum function of Alternating Sylvester series. By (4), since for all, then and is well defined. In this paper, we shall discuss some basic nature of, also the Hausdorff dimension of the graph of is determined.

In what follows, we shall often make use of the symbolic space.

For any, let

Define

For any, write

We use to denote the following subset of (0,1],

From theorem 4.14 of [8], we have when is even, and when is odd. Finally, define

Lemma 1. For any and1) (10)

2) (11)

3) (12)

Proof. 1) Since and , so when, we can get

accordingly

we write, so.

Now implies

for

Thus

let, we have and, thus

2) From 1) we know that

from the definition of we also know that, so

thus

3) Since as,

Thus

Let

Proposition 2. For any, if , then is left continuous but not right continuous. If, then is right continuous but not left continuous.

Proof. For any and, write, , where, are given by (6) and (7).

Case I, , then

and. For any, since when

This situation is included in Case II, so we can take and

i.e.

By (2),

which implies

and

Let, we get and, thus

and this implies is left continuous at.

Let

then

Let, we have

and this implies is not right continuous at. For

following the same line as above, we have

Case II

Let

Following the same line as above, we have

and is right continuous.

Corollary 3. For any and, write,. Then for any, if then

where.

From the corollary, for any

where is the Lebesgue measure of.

Theorem 4. is continuous on.

Proof: For any and, let be its Alternating Sylvester expansion. For any, write . By (Corollary 3), for any, we have

Write, where

Theorem 5. If, then there exists, such that

Proof. Set, then has the same continuity as. Write

trivially, , then the set is well defined.

If, then by the left continuity of, we have

As a result, there exists a such that for any.

If, since is not left continuous, then such that for any, , that is.

Following the same line as above, we can prove.

Now we shall prove that. We can choose such that, if, then

if, then

In both case. Following the same line as above, we can prove, and .

Therefore, there exists, such that

Theorem 6. and

Proof.

Let, then thus

thus,

Through the MATLAB program we can get the definite integration

Write

Theorem 7..

Proof. For any, is a covering of. From (Cor 3), can be covered by squares with side of length. For any,

Thus,

Since

then

so.

This work is supported by the Hunan Education Department Fund (11C671).