It is immediate from the cos(n acos(x)) definition that ChebyshevT[n, ChebyshevT[m, x]] == ChebyshevT[n*m, x] which extends to complex m,n. E.g., T[1/n](x) is the inverse of T[n](x). By "cleverly" rewriting sqrt(1-x^2) as sin(one*acos(x)), I got ChebyshevU[n, ChebyshevT[m, x]] == ChebyshevU[-1 + m (1 + n), x]/ChebyshevU[-1 + m, x] when all I needed was d/dx. --rwg Reminder: ChebyshevU[n, I/2]/I^n == Fibonacci[n+1] World's shortest fibula? Er, fib formula?