Let f be a real-valued function such that f(x+y) = f(x) + f(y) for all x, y belongs to R. Define a function?

F by F(x) = c + f(x), x belongs to R(real number's set), where c is a real constant. Show that for every positive integer n,


F^n (x) = (c + f (c) + f^2 (c) + ... + f^(n-1) (c)) + f^n (x);


where, for a real-valued function G^n (x) G, is defined by


G^0 (x) = 0, G^1 (x) = G (x), G^(k+1) (x) = G(G^k (x)).

Let f be a real-valued function such that f(x+y) = f(x) + f(y) for all x, y belongs to R. Define a function?
Proof by induction.





[ Also, G^0(x) = x, normally. ]





n=1: F^1(x) = c + f(x) = c + f^1(x).





It true for n, then:





F^(n+1)(x) = F(F^n(x))





We know that F^n(x) = c + f(c) + ... + f^(n-1)(c) + f^n(x)





So F^(n+1)(x) = F(c+f(c)+...+f^(n-1)(c) + f^n(x))





= c + f( c+f(c)+...+f^(n-1)(c)+f^n(x) )





But f(x+y)=f(x) + f(y) extends to f(x1+x2+...xn) = f(x1)+....f(xn)





So f( c+f(c)+...f^(n-1)(c)+f^n(x)) = f(c)+f^2(c) + ... + f^n(c) + f^(n+1)(x)





So:





F^(n+1)(x) = c + f(c) + f^2(c) + .... + f^n(c) + f^(n+1)(x)





So it is true for all positive integers.