Let 0<a<b<1 be reals numbers and
g(x)=\left\{\begin{array}{cc}x+1-a,&\mbox{ if } 0n+1 reals numbers
0<x0<x1<...<xn<1, for which
g[n](xi)=xi (0≤i≤n). Prove that there exists a positive integer
N, such that
g[N](x)=x for all
0<x<1.(
g[n](x)=n timesg(g(....(g(x))....)))Official solution [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=365714&p=2011659#p2011659]here