On Iteration of Bijective Functions with Discontinuities

Abstract We present three different types of bijective functions f : I → I on a compact interval I with finitely many discontinuities where certain iterates of these functions will be continuous. All these examples are strongly related to permutations, in particular to derangements in the first case, and permutations with a certain number of successions (or small ascents) in the second case. All functions of type III form a direct product of a symmetric group with a wreath product. It will be shown that any iterative root F : J → J of the identity of order k on a compact interval J with finitely many discontinuities is conjugate to a function f of type III, i.e., F = φ−1 ∘ f ∘ φ where φ is a continuous, bijective, and increasing mapping between J and [0, n] for some integer n.


Introduction
During the ISFE54 Zygfryd Kominek raised discussion about the behavior of iterates of real functions with discontinuities. "Is it possible that the k-th iterate of such a function is continuous?" During the problems and remarks sessions there were some remarks concerning this topic by Roman Ger, Peter Stadler and myself (cf. [6, Problem 2.5 and Problem 2.9]). Finally it turned out that only surjective functions are interesting.
In order to obtain nice results it will also be assumed that these functions are injective. In the present paper three different types of bijective functions defined on a compact interval with finitely many removable and/or jump discontinuities will be presented, where certain iterates of these functions will be continuous. As a matter of fact, functions of type III are generalizations of functions of type I or II. We will see that these examples of bijective functions are strongly related to permutations of finite sets. Therefore, we consider these functions also as discrete structures, and in addition to analyzing their properties we will also try to enumerate them. This way we obtain an overview on how many different types of these functions can be constructed.

Functions of type I
Let n ≥ 2 be an integer, I = [0, n + 1] be the closed real interval, f : I → I be a bijective function with n removable discontinuities in the points belonging to n := {1, . . . , n}. From the context it should always be clear whether n denotes a positive integer or a set of positive integers. Then f is a function of type I, iff f (x) = x for x ∈ I \ n. Since f is bijective, for all j ∈ n there exists some i ∈ n such that i = j and f (i) = j. Thus f restricted to n is a permutation π = π f ∈ S n , the symmetric group of n. It is free of fixed points, thus it is a derangement. We call it the derangement obtained from f . Conversely, to each derangement there corresponds exactly one function of type I.
Some relations between f and π are collected in Lemma 2.1. Let f be a function of type I and π the derangement obtained from f . Then 1. f k is continuous, iff f k = id. 2. f k (i) = π k (i), i ∈ n, k ∈ N. 3. f k is continuous, iff π k = id, iff the order ord(π) of π is a divisor of k.
There are various formulae known concerning the enumeration of derangements. Let d n be the number of derangements in S n , then it is also the number of functions of type I having n discontinuities. E.g., following [2, page 182 and 180] there is a recursive formula d 0 = 1, d 1 = 0, d n = (n − 1)(d n−1 + d n−2 ), n ≥ 2, and a formula based on the inclusion-exclusion principle These numbers d n can be found as A000166 in the On-Line Encyclopedia of Integer Sequences (OEIS).
Some numerical values are presented in Table 1. Approximately 37% of all permutations are derangements. Actually, it is easy to prove that If π ∈ S n decomposes into a i disjoint cycles of length i, for i ∈ n, we call a = (a 1 , . . . , a n ) the cycle type of π. The order of π depends only on the cycle type of π since it is the least common multiple of all cycle lengths occurring in the decomposition of π. We can express ord(π) as the lcm{i ∈ n | a i = 0} where a = (a 1 , . . . , a n ) is the cycle type of π. In general, a sequence of n non-negative integer (a 1 , . . . , a n ) is a cycle type of a permutation π in S n , iff i∈n ia i = n.
Such sequences are sometimes called cycle types of n. From these considerations and the example above it is clear that the following lemma holds true.
Lemma 2.2. Consider a positive integer k. Let f be a function of type I and π the derangement obtained from f . The discontinuities of f corresponding to any cycle of length i of π disappear in the k-th iterate f k of f , iff i | k. Therefore, the number of discontinuities of f k is We call the least positive integer k such that f k is continuous the order of f written as ord(f ). Thus ord(f ) = ord(π f ) where π f is associated with f .
What is the maximum order of a function of type I with n discontinuities? The maximum possible order of permutations in S n is given by the Landau function g(n) := max{ord(π) | π ∈ S n }. It satisfies g(n) ≤ g(n + 1) for all n. Furthermore, letg(n) := max{ord(π) | π ∈ S n is a derangement} be the maximum order of a derangement of n. It satisfies g(n − 2) ≤g(n) ≤ g(n) for all n ≥ 4. Whenever g(n − 1) < g(n), then necessarilyg(n) = g(n). Obviously, g(n) is the maximum order of a function of type I with n discontinuities.
For example we list some values of g(n) andg(n) in Table 2. The numbers g(n) andg(n) can be fund in the OEIS as A000793 and A123131 respectively.
In order to get an overview over all functions of type I with n discontinuities it is enough to study functions where the associated derangements belong to different conjugacy classes in S n . The different conjugacy classes in S n correspond to the different cycle types of n. Consider two functions f i , i = 1, 2, of type I where the associated derangements π i , i = 1, 2, are conjugate in S n . Then the π i have the same cycle types and according to Lemma 2.2 the number of discontinuities of f k 1 and f k 2 , k ≥ 1, coincide. Therefore functions of type I, the associated derangements are conjugate in S n , show similar behavior. From Lemma 2.2 we deduce that the number of discontinuities of f k can be described in terms of the cycle type of π f . Thus the behavior of f depends only on the conjugacy class of π f .  Cycle types of derangements in S n correspond to partitions of the integer n having no parts of size 1. A partition of n is a sequence α = (α 1 , . . . , α h ) of integers α 1 ≥ . . . ≥ α h ≥ 1 with α 1 + · · · + α h = n.
Given a positive integer n the set of orders of functions of type I having n discontinuities is finite. It is a subset of {2, . . . ,g(n)}. E.g., for n = 1 it is the empty set, for n = 8 it is {8, 6, 15, 4, 2}. There are no functions of type I with n removable discontinuities such that ord(f ) >g(n). E.g., there are no functions with 2 removable discontinuities such that f 3 is continuous.
Considering just conjugacy classes reduces the combinatorial complexity. The numbers p n of all partitions of n, andp n , the partition numbers without 1, can be found in the OEIS as A002865 and A000041, see Table 3 for some values.
There are no functions of type I with exactly one removable discontinuity. The iterates f k have at most as many discontinuities as f .

Functions of type II
Now we consider bijective functions f : [0, n] → [0, n], n ≥ 2, such that for each i ∈ n there exists one j ∈ n such that and f (n) = n. Therefore, f is continuous in each interval I i := [i − 1, i) (in i − 1 continuous from the right), i ∈ n. Discontinuities can appear only in the positions 1, . . . , n.
Since f is bijective, it defines a permutation π ∈ S n given by For example f = Lemma 3.1. If f is a function of type II and π ∈ S n is obtained from f , then: An element i ∈ n − 1 is called a succession (or a small ascent) of π, iff π(i + 1) = π(i) + 1. The f above has exactly one succession namely 2. The number of discontinuities of f among {1, . . . , n − 1} is the number of i-s which are not successions of π. A permutation π without successions satisfying π(n) < n defines a function with n discontinuities. These are the functions of type II having the maximum number of discontinuities. E.g., π = (1, n)(2, n − 1) . . . or σ = (1, n, 2, n − 1, . . .) lead to n discontinuities of f , n ≥ 2. Hence, we try to enumerate permutations without successions. Let a n be the number of permutations in S n having no successions and b n the number of permutations in S n having exactly one succession, then it is easy to prove that and a n = (n − 1)a n−1 + b n−1 , n ≥ 2, b n = (n − 1)a n−1 , n ≥ 2, thus a n = (n − 1)a n−1 + (n − 2)a n−2 Consequently, b n = d n , n ≥ 1, the number of derangements of n objects.
For a n see A000255 in the OEIS. Let c n be the number of permutations π in S n having no successions and satisfying π(n) = n. Then c n = a n−1 − c n−1 , n ≥ 2.
Therefore a n−1 = c n + c n−1 , n ≥ 2, and since The number of permutations π in S n having no successions and satisfying π(n) < n is therefore a n − c n = a n − b n−1 = b n = (n − 1)a n−1 , n ≥ 2.
In what follows we construct functions of type II with certain properties.
Consider as above a cycle π = (1, 2, . . . , k) be the function of type II determined by this π, then the iterates f j 1,k have where the two discontinuities of f j 1,k occur in k − (j mod k) and k. By j mod k we indicate the unique element i ∈ {0, . . . , k − 1} satisfying i ≡ j mod k.
The iterates f j s,k of the functions f s,k : [0, sk] → [0, sk] corresponding to the product of s cycles of length k where the 2s discontinuities of f j s,k occur in rk − (j mod k) and rk for 1 ≤ r ≤ s.
Similarly we consider the iterates g j s,k of the functions g s,k : [0, sk + 1] → [0, sk + 1] corresponding to the product of s cycles and one fixed point where the 2s + 1 discontinuities of g j s,k occur in 1 and rk + 1 − (j mod k) and rk + 1 for 1 ≤ r ≤ s. Theorem 3.3. For any n ≥ 2 and k ≥ 2 the iterates f j n/2,k (for even n) or g j (n−1)/2,k (for odd n) of the functions f n/2,k , or g (n−1)/2,k have Since f and g are bijective and f (n) = n, the concatenation f • g is bijective, and f • g is of type II. If, furthermore, f is continuous in n and g(0) = 0, then f • g is continuous in n since g is continuous from the right side in 0. The function f • g is not continuous in n, iff f is not continuous in n or g(0) = 0.
Theorem 3.4. Consider f and g of type II having r respectively s discontinuities. Then the number of discontinuities of f • g is r + s + 1 if f is continuous in n and g(0) = 0, r + s else.
2. Even though f 1,k (0) = 0 the function f s,k has 2s (and g s,k has 2s + 1) discontinuities since f s−1,k and g s−1,k are not continuous at the end of their domains. 3. The functions g s,k satisfy g s,k (0) = 0, thus the j-th iterate of the concatenation of g s 1 , discontinuities. Concatenation of g s,k does not introduce new discontinuities. 4. Concatenation of the functions f s,k is more complicated, since f s,k (0) = 2 = 0, and f j s,k (0) = 0 whenever j is a multiple of k.
E.g., the numbers of discontinuities of (f 1,2 • f 1,3 ) j and (f 1,3 • f 1,2 ) j are j 1 2 3 4 5 6 number of discontinuities of (f 1,2 • f 1,3 ) j 4 3 2 3 4 0 number of discontinuities of (f 1,3 • f 1,2 ) j 4 2 3 2 4 0 In the next examples we restrict ourselves to functions which are continuous in 0, the left end of their domains. We already know the functions g s,k with this property whose iterates have either 2s + 1 or no discontinuities. Hence we are looking for functions whose iterates either have an even number > 0 or 0 discontinuities.
What about functions with 2 or 4 discontinuities among 1, . . . , n − 1? It is easy to see that there is no function f : [0, n] → [0, n] of type II such that f (0) = 0 having exactly two discontinuities. These functions have at least three discontinuities. An example for n = 5 is given by Concerning permutations with exactly four discontinuities we obtain: The permutation π = (1)(2, 4)(3) has order 2 and yields 4 discontinuities.

d d d
A new phenomenon occurs with these functions. There exist iterates of f having more discontinuities than f itself. The number of discontinuities of the iterates f j , f corresponding to π 3 , are: Probably these results can be generalized for arbitrary k.
As a generalization of functions of type I and type II we introduce

Functions of type III
A bijective function f : [0, n] → [0, n] is a function of type III, iff f permutes the integers {0, 1, . . . , n}, and for each i ∈ n there exists exactly one j ∈ n such that either

This means that f permutes the open intervals
In the first case f is strictly increasing on I i , in the second case strictly decreasing on I i .
f is continuous in 0, iff either (λ(1)) = 1 and π(0) = λ(1) − 1 or (λ(1)) = −1 and π(0) = λ(1). In a similar way the continuity of f in n can be described. There are two possibilities that the k-th iterate of a function f of type III is continuous, either f k = id or f k = n − id.
Now we show that the pairs ( , λ) are elements of a wreath product. This is a particular form of a semidirect product (cf.   ↔ (π , ( , λ )). Then their composition yields Thus the set of all functions of type III is the direct product where the factor on the right side is a wreath product Consequently, the number of functions of type III on [0, n] is n!(n + 1)!2 n , see Table 5. Functions of type I or type II are particular cases of these functions.
A function of the first form is the concatenation of id n 1 and n 2 − id n 2 both of which are continuous, and the discontinuity disappears with the second iteration.
Also the discontinuity of functions of the second form disappears with the second iteration.
The behaviour of functions of the third and forth form can be studied. There exist functions of type III with exactly two discontinuities in the interior of the interval. They can be constructed from the four different forms of functions with exactly one discontinuity. E.g., the functions have exactly two discontinuities in the interior of the interval and are of order 2. They correspond to the third and fourth form. These examples again can be generalized by partitioning the interval [0, n] into 3 parts [0, n 1 ], [n 1 , n 2 ], [n 2 , n] with 0 < n 1 < n 2 < n ∈ N.
Finally we study some relations between functions of type III and iterative roots of the identity. Let k be the order of a function f of type III. Then f k = id, and f is an iterative root of the identity. By applying a continuous, bijective, and increasing function ϕ we obtain is bijective, has r discontinuities, and satisfies F k = id J , thus F is an iterative root of the identity of order k.
Conversely, consider an iterative root F : J → J of the identity of order k on a compact interval J with finitely many discontinuities. We will prove that it is always possible to find some n ∈ N, a continuous, bijective, and increasing function ϕ : J → [0, n] and a function f : [0, n] → [0, n] of type III so that It is obvious that if J is a compact interval and F : J → J is a bijective mapping with finitely many discontinuities, then they must be removable or jump discontinuities.
In general the integer n is not uniquely determined, so we are looking for the smallest n possible. Assume that F k = id and F has r discontinuities ξ 1 , . . . , ξ r ∈ J = [a, b]. Consider the union of orbits then U is finite and we determine n by n = |U | − 1.
This particular n will be called n(F ). The n+1 elements of U will be labeled by For all i ∈ n it is obvious that F is continuous on J i , and there exists some j ∈ n so that F (J i ) = J j , thus F permutes the intervals J i .
The function ϕ will be constructed in two steps: First we determine some then ϕ is continuous in J i , and lim x→x + Therefore, ϕ is continuous on J. Moreover, ϕ is strictly increasing and bijective. IfF denotes the function ϕ•F •ϕ −1 : [0, n] → [0, n], then •F is bijective, •F is an iterative root of the identity of order k, •F has discontinuities in ϕ(ξ i ), i ∈ r, •F (i) ∈ {0, . . . , n}, i ∈ {0, . . . , n},F permutes these elements, In a second step we try to find some ψ : If f is strictly increasing, then there exists some ψ j : I j → I j bijective and increasing, so that If f is strictly decreasing, then there exists some ψ j : I j → I j bijective and increasing, so that then ψ j is a bijective and increasing mapping I j → I j , and Let ψ j (x) = x for x ∈ I j , then ψ j is bijective and increasing on [0, n]. We had just seen thatF is a permutation of the intervals I i , i ∈ n. Consider a cycle Composition of two increasing or two decreasing functions yields an increasing function, composition of one increasing and one decreasing function yields a decreasing function. Therefore, ifF is decreasing on an even number of intervals in this cycle, thenF is increasing on all I i j , otherwiseF is decreasing on all I i j .
Since ψ j restricted to I i is a bijective mapping I i → I i , i ∈ n, the restriction Continuous iterative roots of the identity on an interval I are continuous solutions of the Babbage equation. According to [5,Theorem 11.7.1] they are either the identity on I or they are strictly decreasing involutions. The graph of a strictly decreasing involution of an interval is symmetric with respect to the line {(x, x) | x ∈ R} (cf. [5,Theorem 11.7.2]).
In the first case we assume thatF contains a cycle of intervals of length with an even number of decreasing functions in this cycle, thenF is continuous and strictly increasing on each of these intervals which means that F | I i j = id| I i j for all j ∈ .
In the case = 1 the functionF is already the identity on I i 1 . Assume that ≥ 2. Then the functionF | I i 1 maps I i 1 → I i 2 . According to Lemma 4.4 there exists a bijective and increasing mapping ψ i 2 on [0, n] so that ψ i 2 •F | I i 1 is affine, i.e. it is either x → i 2 +x−i 1 or x → i 2 −1+i 1 −x. Then also ψ i 2 •F •ψ −1 i 2 is affine on I i 1 since ψ −1 i 2 does not influence the function restricted to I i 1 . If > 2, then the function ψ i 2 •F • ψ −1 i 2 | I i 2 maps I i 2 → I i 3 . According to Lemma 4.4 there exists a bijective and increasing mapping ψ i 3 on [0, n] so that Continuing in the same way, the function There exists a bijective and increasing mapping The term between [ and ] is a composition of affine functions, thus it is affine, In the second case assume thatF contains a cycle of intervals of length with an odd number of decreasing functions in this cycle, thenF restricted to I i j is a decreasing involution on I i j , j ∈ , but it need not be affine on these intervals. Then there exists a bijective and increasing functionψ so that (ψ •F •ψ −1 )| I i j = (ψ •F •ψ −1 ) | I i j is also affine, i.e. of the form x → 2i j − 1 − x, x ∈ I i j for each j ∈ . Without loss of generality we assume thatF | I i j is affine for each j ∈ .
Similar to the first case, there exists ψ = ψ i • . . . • ψ i 2 so that ψ •F • ψ| I i j is affine on I i j for j ∈ − 1. By construction ψ(x) = x for x ∈ I i 1 . Therefore . The term between [ and ] is a composition of affine functions, thus it is affine, whence also ψ •F | I i • ψ −1 | I i is affine. Consequently, ψ •F | I i j • ψ −1 is affine on I i j for each j ∈ . This finishes the proof of Theorem 4.5. Let J be a compact interval, F : J → J an iterative root of the identity of order k with finitely many discontinuities. Then there exists some positive integer n and a continuous, bijective, and increasing function ϕ : J → [0, n] so that f = ϕ • F • ϕ −1 is a function of type III with ord(f ) = k.
Two bijective functions F 1 : J 1 → J 1 and F 2 : J 2 → J 2 defined on compact intervals J 1 and J 2 are considered to be equivalent F 1 ∼ F 2 iff there exists a bijective increasing function ϕ : J 1 → J 2 so that It is easy to prove that for all bijective functions F i : Theorem 4.6. Consider f 1 , f 2 functions of type III corresponding to elements of S n+1 × ({±1} S n ), with n = n(f 1 ) = n(f 2 ). Then Theorem 4.7. Consider an iterative root F : J → J of the identity of order k on a compact interval J with finitely many discontinuities. Let n = n(F ). Then there exists exactly one function f ∈ S n+1 ×({±1} S n ) of type III so that F ∼ f.
How many functions f of type III exist with n = n(f )? Their number is the number of non-equivalent functions f of type III with n = n(f ). So far the author does not know an explicit formula in order to enumerate them. For small values of n it is possible to check all functions of type III. Table 6 contains numerical data (computed with SYMMETRICA [7]) comparing the numbers of all functions of type III for small n, with the numbers of functions with n(f ) < n and n(f ) = n.
Consider e.g. functions of type III which are of the form