The introduction of chaos, fractal, and dynamical system could be found in many classical textbooks, such as Scheinerman [1] . A dynamical system has two parts, a state and a function. The second part of a dynamical system is a rule which tell us how the system changes over time. According to the time, we have the discrete and continuous system. The discrete dynamical system, in which we are interested, always does not have an analytical solution. Therefore, the behaviors of fixed points are very important. They play a vital role in the chaos, bifurcation, Julia sets problem in the dynamical system (see [2] [3] ). Those problems have been studied for last thirty years. Using the dynamics of functions near the real fixed points, the dynamics of functions in complex

plane were induced by the following researchers: The dynamics of families of entire functions, were studied by Prasad [4] , Kapoor and Prasad [2] , Sajid and Kapoor [5] , respectively. The dynamics of is found in Devaney [6] . Recently, Sajid [7] [8] gave the results about the fixed points of one parameter family of function. His work is motivated by the relationship of the function with the well-known generating functions on base b by choosing and in the generalized Bernoulli generating function The proofs in [7] and [8] are too

complicated. In this paper, we not only give a simple proof of the work of Sajid [7] , but also generalize his work.

We will determine the fixed points of where

i.e., we will solve the equation Moreover, we also discuss the multiplicity and the behavior of the fixed points for two parameters b and n. For simplicity of notation, we denote and by and, respectively.

For, the following results in Theorem 1 were in Sajid [7] , but we have a simpler proof.

Theorem 1. Let where is given in (2-1). Then

(1) The function has a unique fixed point for any

(2) The unique fixed point is negative if and positive if Moreover, we have is decreasing if increasing if as is increasing and

(3) There exists such that the fixed point of the function is (i) attracting, i.e., for, (ii) rationally indifferent, i.e., at, and (iii) repelling, i.e., for

Proof. Suppose that and. Statements (1) and (2) can be proved directly as follows. The expression if and only if because Let be the fixed

point of the function for Then the fixed point

is unique. Moreover, (2-3) easily implies statement (2).

Next, we proved statement (3). It is easy that

and the function is contionus on Therefore, substituting (2-3) into (2-4), we have

Hence,

Therefore, statement (3) are true by (2-6), (2-7), and (2-8).

The results about depend on the parameters n and b. If n is odd, then the behavior of the

fixed points is similar to the case If n is even, the behavior of the fixed points depends on the parameter b. We have the simple facts about the functions f and g. It is easy that

Hence, if the integer n is even, then If the integer n is odd, then

Suppose that the fixed point of exists. If n is odd and, then the fixed point is negative. Otherwise, it is positive. Moreover,

and is continuous in Hence,

Lemma 2. Let, where, and Then (1) is concave downward in if is odd and (2) is concave upward in otherwise.

Proof. Suppose that and.

Therefore, (2-9) implies that and have the opposite sign if and only if is odd and. Moreover, since Thus, it suffices to prove that in where

(2-4) implies that We have

and

where

Let Equation (2-13) is transformed to

Moreover, let (2-14) is transformed to

In fact, the graph of has 2 critical points, including and is the global minimum.

By the algorithm of bisection, , and It is easy to check that and Therefore, is the minimum of H on. This completes the proof of Lemma 2.

To study the behavior of the fixed points in Theorem 5, we need Lemma 3 and Lemma 4 as follows.

Lemma 3. Suppose that

Then (1) and (2) The function h is decreasing, and concave upward in.

Proof. The statement (1) is easy. (2-16) implies Specially, we have. Therefore, the function h is decreasing in. Moreover,

Let Then We have for, and for. Therefore, , for, and the function h is concave upward in

. The proof of Lemma 3 is completed.

Lemma 4. Suppose that

and such that

and

Then there is a unique, such that

Moreover, if then intersects at exactly one point. If then does not intersect If then intersects at exactly two points.

Proof. Let Then (2-18) implies The Intermediate Value Theorem of the continuous function implies such that i.e., Equation (2-20) holds

Suppose to the contrary that is not unique.

There exist such that

Then by (2-17).

For the Mean Value Theorem and (2-17) imply such that

Therefore, the Equation (2-21) implies that. It is impossible.

Suppose that and intersects at. Then we have (2-20) and the uniqueness of imply

Suppose to the contrary that but intersects at the smallest on. Let Then and. This implies

and there exists the minimum of on.

Let the minimum occurs at. Then and. This contradicts to the unique- ness of in (2-20).

Finally, suppose that, we prove that intersects at exactly two points. Let Then and. Intermediate Value Theorem and (2-19) imply that there exists at which intersects.

Suppose to the contrary that does not intersect at exactly two points. By the previous proofs, the line will intersect at more than two points. Let intersect at the three consecutive points, then for Without loss of generality, we may

suppose that. Then and. It is impossible that by (2-17). The proof of Lemma 4 is completed.

Theorem 5. Let be even and Then

(1) There exists a unique such that has a unique fixed point at and has two fixed points, say, and for, and no fixed points for

(2) Let n be fixed. If, then is decreasing, and is increasing as decreases.

(3) Let b be fixed. Then is decreasing to 0 as n increases.

(4) Suppose exist for fixed. If, then is increasing, and is decreasing as n increases. If, then there exists a unique, such and for any even n.

Moreover, is decreasing if it is less than, and is increasing if it is less than, is increasing if it is greater than, and is decreasing if it is greater than, as n increases.

(5) The fixed points are attracting, rationally indifferent, and repelling, respectively.

Proof. Let be even and

(1) We want to solve the equation where and and is given as in (2-1). Note that the equation is equivalent to

Because of and we only consider

By for Lemma 4 implies that the number of

intersections of and are exactly two for, unique at, and none for. i.e., has two roots say, and for, one root, say at, and no root for. Therefore,

(2) The statement (2) is easy by Lemma 2 and Part (1).

(3) In fact, can be determined by solving the equation (2-4) implies to solve

(2-1) and (2-4) imply to solve

Let (2-24) is equivalent to

Lemma 3 and (2-25) imply that is decreasing to 0 as n is increasing for any

(4) Suppose that We have Then is increasing as even number increases by (2-9). Suppose that Then Let be the solution of. Therefore,. Since is increasing, and concave upward, Statement (4) holds.

(5) Let be the fixed point of Then

Let Then

Lemma 3, (2-21) and imply i.e., the state- ment (5) holds.

Theorem 6. Let be even, Then

(1) The function has a unique fixed point which is increasing as increases.

(2) Let and be fixed. If then is increasing as n increases. If then there exists a unique number, such for any even n, and is increasing if, and decreasing if as n increases.

(3) There exists such that the fixed point of the function is (i) attracting, i.e., for, (ii) rationally indifferent, i.e., at, and (iii) repelling, i.e., for

Proof. The proof of Theorem 6 is similar to that of Theorem 5. We just mention some key points. The function f is positive, decreasing, and concave upward. Let be the fixed point of Then

Let Then for and

Lemma 3 and (2-27) imply that there exists a unique such that, and if and if

Theorem 7. Let n be odd. Then

(1) has a unique fixed point for any The fixed point is negative if and positive if Moreover, we have

(2) Let the parameter n be fixed. Then is increasing if and decreasing if as increases.

(3) Let the parameter be fixed. If then is increasing as n increases. If then there exists

a unique number such for any odd n, and is increasing if, and decreasing if as n increases. If, then is increasing as n increases. If, then there exists a unique such that for any odd n. Furthermore, is decreasing if it is less than, and increasing if it is greater than as n increases.

(4) There exists such that the fixed point of the function is (i) attracting, i.e., for, (ii) rationally indifferent, i.e., at, and (iii) repelling, i.e., for

Proof. The proof of Theorem 7 is similar to that of Theorem 5. We just also mention some key points. The function f is decreasing if and f is positive, concave upward if, and negative, concave

downward if Let be the fixed point of Then by (2-26). Let Then for by statement (1). Therefore, for. Lemma 3 and (2-28) imply that there exists a unique such that, and

if and if

The Sarkovskii’s theorem said that let the function be continuous and it has points of prime period 3, then the function f has points of period k for all positive integer k. We also know that the dynamical behavior

of is very complicated, see Scheinerman [1] . In our problem, has points of prime period 3 if, , and. We anticipate the problem we studied will have complicated dynamical behavior as that of. On the other hand, the bifurcation of will be interesting.