Given that the domain of definition of odd function f (x) is r, and for any real number x, f (x + 2) = - f (x), and f (1) = 4, then f [f (2015)]=

Given that the domain of definition of odd function f (x) is r, and for any real number x, f (x + 2) = - f (x), and f (1) = 4, then f [f (2015)]=

The odd function shows that - f (x) = f (- x) so f (- x + 2) = - f (- x) = f (x) = - f (x + 2) then f (x + 2) = - f (2-x) = f (X-2) so let x + 2 = t then f (T) = f (T-4) so f [f (2015)] = f [f (2015-4n)] = f [f (- 1)] = f (- 4) = - f (4) = - f (0) be symmetric about the origin, so f