Let f : S → S where S = 0 , ∞ be a twice differentiable function such that f x + 1 = x f x . If g : S → R be defined as g x = log e f x , then the value of g '' 5 - g '' 1 is equal to :

Step-by-step solution

l n f x + 1 = l n x f x l n f x + 1 = l n x + lnf x ⇒ g x + 1 = l n x + g x ⇒ g x + 1 - g ( x ) = l n x ⇒     g '' x + 1 - g '' x = - 1 x 2 Put x = 1 , 2 , 3 , 4 g '' 2 - g '' 1 = - 1 1 2   . . . 1 g '' 3 - g '' 2 = - 1 2 2   . . . 2 g '' 4 - g '' 3 = - 1 3 2   . . . 3 g '' 5 - g '' 4 = - 1 4 2   . . . 4 Add all the equations we get g '' 5 - g '' 1 = - 1 1 2 - 1 2 2 - 1 3 2