Function Challenge (Posted on 2007-01-27)

	Submitted by atheron
Rating: 3.5000 (2 votes)
Solution:	(Hide)
If F(x) has a root r>1 then r^2 or 2r-1 is a root. Since r^2>r and 2r-1>r we would get endless series of roots which is impossible. Same applies to roots r<-1. So we now know that all of F(x)'s roots belong to closed set [-1,1]. Let's presume that r is F(x)'s greatest root. If r<1, we set x=sqrt((r+1)/2). We now have: F(x)*F(r)=F((r+1)/2)*F(2*sqrt((r+1)/2)-1). Now (r+1)/2>r and 2*sqrt((r+1)/2)-1>r and at least one of these two is a root of F(x). This however is impossible since r had to be the greatest root of F(x). So 1 is the only real root possible for F(x). So F(x) has to be of form F(x)=G(x)*(1-x)^k where k is an positive integer and G(x) is a polynomial with real coefficcients for which 1 is not root of G(x). Using the equation given in the problem we get: G(x)*G(2x^2-1) = G(x^2)*G(2x-1). As we have already proven this equation cannot have real roots since 1 couldn't be G(x)'s root. This means that: G(2x^2-1)/G(x^2) = G(2x-1)/G(x) Now let S(x)=G(2x-1)/G(x). From the previous eguation we get S(x)=S(x^2)=S(x^4)=... (|x| != 1) This means that rational function S(x) gets a value A in infinitely many points. This is also impossible unless S(x) is a constant. Because S(1)=1 this constant has to be 1. So G(x)=G(2x-1) for all x. Since G is a polynomial this is only possible if G(x)=c for all x(c being a real constant). So F(x) has to be of form F(x)=c*(x-1)^k (c is a real number and k is a positive integer)

	Subject	Author	Date
	Puzzle Thoughts	K Sengupta	2023-07-15 23:26:20
	Solution (without full proof)	Joel	2007-01-27 16:50:24
	A feasibility check	Federico Kereki	2007-01-27 11:14:00