From: israel@math.ubc.ca (Robert Israel) Subject: Re: Semigroups of polynomials Date: 2 Jan 2001 00:32:42 GMT Newsgroups: sci.math Summary: Sets of polynomials forming semigroups under composition In article <92msu5$qjf$1@nnrp1.deja.com>, wrote: >I am sorry, I forgot to mention that I am looking for a set of >polynomials P_n n = 1,2,3... with degree(P_n) = n . >In article <92msbp$q8c$1@nnrp1.deja.com>, > noam_katz@my-deja.com wrote: >> The set of polynomials x^n n=1,2,3... forms a semigroup with respect to >> composition i.e (x^n)^m = x^(n*m) and so does the set of >> Chebyshev polynomials P_n = cos(n*arccos(x)) ,again >> P_(m*n) = P_m(P_n) = P_n(P_m). >> >> Is there another set of polynomials that forms a semigroup like that ? There is a two-parameter symmetry group acting on the solutions: if P_n is a semigroup, then so is Q_n(x) = (P_n(a x + b)-b)/a for any complex a<>0 and b. The only possible semigroups are equivalent by these symmetries to the monomials and Chebyshev polynomials. Consider P_2 and P_3. By using the symmetry, we can assume P_2(x) = a x + x^2; write P_3(x)=c_0+c_1 x + c_2 x^2 + c_3 x^3. Now consider the equation P_2(P_3(x))=P_3(P_2(x)). Matching the coefficients of each power of x, we get 7 equations in the 5 unknowns a, c_0, c_1, c_2, c_3. According to Maple's "gsolve" command, there are only 4 solutions with c_3 <> 0: 1) a=0, c_0=c_1=c_2=0, c_3=1 2) a=-2, c_0=3, c_1=0, c_2=-3, c_3=1 3) a=2, c_0=0, c_1=3, c_2=3, c_3=1 4) a=4, c_0=0, c_1=9, c_2=6, c_3=1 The monomials correspond to solutions 1) and (via a symmetry) 3), while the Chebyshev polynomials correspond (via symmetries) to 2) and 4). Theorem: If Q(x) is a polynomial that commutes with x^2 (i.e. Q(x^2) = Q(x)^2), then Q(x) is either 0 or a monomial x^d. Proof: Use induction on the degree d of Q(x). For d=0 it is clear. So suppose Q has degree d >= 1, and the statement is true for polynomials of degree 1, and consider the sequence z_0, z_1, z_2, ... with z_{j+1} = z_j^2. We have z_j -> 0 so Q(z_j) -> Q(0). But Q(z_{j+1}) = Q(z_j^2) = Q(z_j)^2. Since |Q(z_0)| <> 1, the only possible finite limit of this sequence is 0, so Q(0)=0. Now we can write Q(x)=x^k P(x) where k >= 1 and P(x) is a polynomial of degree d-k < d. We have Q(x^2) = x^(2k) P(x^2) = Q(x)^2 = x^(2k) P(x)^2 so P also commutes with x^2. By the induction hypothesis, P is a monomial, and therefore so is Q. Theorem: The only nonconstant polynomials that commute with the Chebyshev polynomial T_2 are Chebyshev polynomials. Proof: Suppose P commutes with T_2. T_2 has the property that T_2([-1,1]) = [-1,1], and [-1,1] is the only interval with this property. For any complex z not in [-1,1], the iterates of T_2 on z go to infinity in absolute value. From this it is easy to see that P([-1,1]) = [-1,1]. In particular P has real coefficients. Now T_2(P(1)) = P(T_2(1))=P(1), so P(1) is a fixed point of T_2. There are two of these: -1/2 and 1. But if P(1) = -1/2 we would have P(1+epsilon) in [-1,1] for some epsilon > 0, and this is impossible since the iterates of T_2 on P(1+epsilon) go to infinity in absolute value. So P(1) = 1. Now let F(x) = arccos(P(cos(x))). This is continuous on R with F(0) = arccos(P(1)) = 0. Moreover, cos(F(2 x)) = P(cos(2 x)) = P(T_2(cos x)) = T_2(P(cos(x))) = T_2(cos(F(x))) = cos(2 F(x)). We conclude that F(2 x) = 2 F(x) on some interval [0, epsilon]. Then on this interval, F(x) = x g(log_2 x) where g is a positive periodic function with period 1. So P(cos(x)) = cos(g(log_2 x) x). But, expanding in powers of x, 1 - P'(1) x^2/2 + ... = 1 - g(log_2 x)^2 x^2/2 + ... implying that g^2 (and thus g) is constant. Thus P(cos x) = cos(g x) on the interval (and by analytic continuation for all x). This must be periodic with period dividing 2 pi, so g must be an integer, and then P(x) = T_g(x). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2