Number of Forts in Iterated Logistic Mapping

Using the theory of complete discrimination system and the computer algebra system MAPLE V.17, we compute the number of forts for the logistic mapping fλ(x) = λx(1 − x) on [0, 1] parameterized by λ ∈ (0, 4]. We prove that if 0 < λ ≤ 2 then the number of forts does not increase under iteration and that if λ > 2 then the number of forts is not bounded under iteration. Furthermore, we focus on the case of λ > 2 and give for each k = 1, . . . , 7 some critical values of λ for the change of numbers of forts.


Introduction
Iteration is the act of repeating a process with the aim of approaching a desired goal, target, or result.In mathematical sense, for a fixed integer  ≥ 1, the th iterate   of a mapping  :  → , where  is a nonempty set, is defined recursively by where ∘ presents the composition of functions and id denote the identity mapping; that is, id() ≡  for all  ∈ .Being indispensable in the computer era, iteration brings many interesting but difficult problems to mathematics.Only from one-dimensional case, one can simply notice that an iterate of a linear function of any order remains linear but the degree of a polynomial may increase drastically, which shows that the nonlinear complexity is amplified by iteration.
Actually, in the one-dimensional case, the complexity of nonlinear functions is related to nonmonotonicity.For a continuous nonmonotonic self-mapping  :  → , where  is an interval, a point  0 ∈  is called a monotone point of  if  is strictly monotone in a neighborhood of  0 ; otherwise,  0 is called a nonmonotone point or simply a fort of .Obviously, a linear function does not have a fort generically.In 1980s, Zhang and Yang (see [1]) investigated the number of forts for a class of nonmonotonic functions called strictly piecewise monotone functions and simply PM functions, which are selfmapping on a compact interval and have at most finitely many forts each.Let () denote the set of all forts of  and let () denote the cardinality of ().It is shown in [1,2] that (2) that is, the number (  ) of forts is nondecreasing as  is increasing.One can similarly prove that (2) also holds for functions defined on the whole R. It is easy to find nonlinear functions whose number of forts, regarded as the damagers of monotonicity, increases rapidly under iteration.Consider the quadratic function for example.Computing derivatives of   ,  = 1, . . ., 5, and counting the number of real zeros with odd multiplicity for the derivatives (  )  (as done in [3]), we get () = 1, ( 2 ) = 3, ( 3 ) = 7, ( 4 ) = 15, and ( 5 ) = 31.
From the increasing tendency, without continuing the tedious computation, we have the following question: Does (  ) have a bound or approach infinity as  tends to ∞? How can we compute the number of forts for nonmonotonic functions?Polynomials, a special class of nonmonotonic functions, possess the advantage that each fort of a polynomial of degree ≥ 1 is either a peak or a valley although the notion is not true in general.In this paper, we focus on the family of logistic mappings: where  ∈ (0, 4] is a parameter, which is one of the simplest polynomial mappings, and a typical example used to show chaos and some complicated dynamics, for those problems.First of all, we introduce the theory of complete discrimination system (see [4]) and then use it to give a method for the computation of () with  polynomial in Section 2. In Section 3, we employ the method in the computer algebra system MAPLE V.17 for the family of logistic mappings.We prove in Theorem 4 that (   ) = (  ) = 1 for all integer  ≥ 2 if 0 <  ≤ 2 and that (   ) approaches ∞ as  → ∞ if 2 <  ≤ 4. Furthermore, for various choices of  ∈ (2,4], we compute the number (   ) for each fixed  = 2, 3, . . ., 7 in Theorem 5.

Preliminaries
In general, for polynomial where  ≥ 2 and   ̸ = 0. (  ),  = 1, 2, . .., is decided by real zeros of the derivatives (  )  .Lemma 1 (see [3,Lemma 2.1]). 0 is a fort of a real polynomial  if and only if  0 is a real zero of the derivative   of odd multiplicity.Moreover, () is odd (resp., even) if the degree  of  is even (resp., odd).
Actually, the above lemma shows how a real zero of the derivative   can be a fort of .Note that (  )  = (  ∘  −1 )(  ∘  −2 ) ⋅ ⋅ ⋅ (  ∘ )  .Then the set of zeros of (  )  is a union of the set of zeros of   ∘  −1 and the set of zeros of (  ∘  −2 ) ⋅ ⋅ ⋅ (  ∘ )  .Therefore, in order to know if   have more forts than  −1 , we need to judge if   ∘  −1 have real zeros different from (  ∘  −2 ) ⋅ ⋅ ⋅ (  ∘ )  's with odd multiplicities; the following lemma gives the answer.
Lemma 2 (see [3,Lemma 2.5]).Let  and  be real polynomials and () ̸ ≡ 0. Then the composition  ∘  and the derivative   do not have a common real zero with odd multiplicity.
Taking  =   and  =  −1 , by Lemmas 1 and 2, we see that (  )  has more real zeros of odd multiplicities than ( −1 )  if and only if   ∘  −1 has real zeros with odd multiplicities.Hence, in the process of computing (  ), we only need to find out the number of real zeros for   ∘  −1 with odd multiplicities.For this reason, we first introduce some notations of the theory of complete discrimination system (see [4,5]) which will lead us to solve this problem.
The following lemma tells us how to find the number of distinct zeros by means of the revised sign list.

Lemma 3 (see [4, Theorem 1]).
Let  be a real polynomial and suppose that the number of the sign changes in the revised sign list of  is ].Then the number of pairs of distinct conjugate complex zeros of  equals ].Furthermore, if the number of nonvanishing members in the revised sign list is , then  has  − 2] distinct real zeros.Now, we are ready to apply the above lemmas to detail the process in computing (  ),  = 1, 2, . ... First of all, we compute   and   ∘  −1 and the discriminant sequence for   ∘  −1 .Secondly, under algebraic relations among coefficients   's in the discriminant sequence, discuss the sign of each component of the discriminant sequence and list the sign lists.Then, compute the revised sign lists for   ∘  −1 through the sign lists.According to the revised sign lists, we find out the number of real zeros of   ∘  −1 with odd multiplicities and finally obtain (  ).
The above idea can be implemented in the computer algebra system MAPLE V.17, and we will use this method for the logistic mappings up to iteration index  = 7 in next section.

Number of Forts
In this section, we first draw a conclusion for the logistic mappings which describe that the numbers of forts can be preserved or approach ∞ as  varies under iteration and then compute (   ) for   () up to iteration index  = 7 with different choice of .Proof.In order to obtain the condition for (   ) = (  ) = 1, from the method mentioned in the end of Section 2, we need to compute ( 2  ).Simple computation shows that Then computing the discriminant sequence of ( 7), we have From ( 8), if 0 <  < 2, the revised sign list is (1, −1), implying that    ∘   has one pair of complex zeros, which leads to ( 2  ) = 1; if  = 2, the revised sign list is (1, 0), implying that    ∘   has a double real zero, which leads to ( 2  ) = 1; therefore, (  ) = (   ) = 1 if and only if 0 <  ≤ 2. Without loss of generality, we can turn the general form So the vertex of the parabola is ) .
Theorem 4 shows that the number (   ) approaches ∞ as  → ∞ for each fixed  ∈ (2,4].It is also interesting to see for each fixed  how the number (   ) varies as the parameter  changes in (2,4].The following theorem shows the change of numbers (   ) as  varies for each  = 2, . . ., 7 (but larger  can be considered if the computational capacity of our computer is better).It gives a sequence of parameter values at which new forts arise.Furthermore, in order to obtain ( 3  ), we compute

Theorem 5. 𝑁(𝑓
As shown in Section 2, we give the discriminant sequence for ( 15): Then, the revised sign list for ( 16) is (ii) (1, 1, 1, 0), if  =  3,2 , which implies that    ∘  2  has 3 distinct real zeros, 2 of which are simple zeros and the remaining one is a double zero; has one pair of complex zeros and 2 distinct simple real zeros.
A related work can be found from [8], but the question is not answered yet.