1) --- Suppose that you have an “operator”, denoted by ‘, which takes any function f:R-->R and converts it to another function f’: R-->R. 1 Suppose also that ′ is linear: i.e. (f + g)′ = f ′ + g′.Let f: R-->R and g:R-->R be two functions and DEFINE h(x):=f(x)*g(x) for all ∈ R. Suppose that the operator ’ satisfies the following formula:

h’(x)= f’(x)g(x) +f(x)g’(x)

We use the notation f^(2):=(f)’, f^(3):=(f^(2))’, and generally f^(n+1):=(f^(n))’ for natural number n.

1i)--- h^(2)(x) = f^(2)(x)g(x) + 2f '(x)g′(x) + f(x)g^(2)(x)

and

h^(3)(x) = f^(3)(x)g(x)+3f^(2)(x)g′(x)+3f′(x)g^(2)(x)+f(x)(g)^(3)(x)

1ii) --- Prove that the following is true for all n ∈ N :

h^(n)= ∑ (k=0 to n) n!/k!(n-k)! f^(n-k) * g^(k)(x)

( Hint: You can assume Pascal’s Identity: for 1 ≤ k ≤ m.)