# Express a Function as the Sum of an Even and an Odd Functions

 Question 1: Show that any function f may be expressed as the sum of an even and an odd functions. Solution to Question 1: Let us write f(x) as follows f(x) = (1 / 2) f(x) + (1 / 2) f(x) + (1 / 2) f(-x) - (1 / 2) f(-x) = (1 / 2) (f(x) + f(-x)) + (1 / 2)(f(x) - f(-x)) Let . g(x) = (1 / 2) (f(x) + f(-x)) Check that g(x) is even g(-x) = (1 / 2) (f(-x) + f(x)) = g(x) Let . h(x) = (1 / 2) (f(x) - f(-x)) Check that h(x) is odd h(-x) = (1 / 2) (f(-x) - f(x)) = - (1 / 2) (f(x) - f(-x)) = - h(x) Question 2: Express f(x) = 2x4 - 5 x3 + 2x2 + x - 4 as the sum of an even and an odd functions. Solution to Question 2: f(x) is a polynomial and it is therefore straightforward to separate even and odd parts of the polynomial as follows f(x) = (2 x4 + 2 x2 - 4) + (- 5 x3 + x) where 2 x 4 + 2 x 2 - 4 is a n even function and -5 x3 + x is an odd function. Question 3: Express f(x) = 1 / (x - 1) as the sum of an even and an odd functions and check your answer. Solution to Question 3: In exercise 1, we showed that any function f may be expressed as the sum of an even and an odd functions as follows f(x) = (1 / 2) (f(x) + f(-x)) + (1 / 2)(f(x) - f(-x)) where g(x) = (1 / 2) (f(x) + f(-x)) is an even function and h(x) = (1 / 2)(f(x) - f(-x)) is an odd function. Hence if f(x) = 1 /(x - 1), then g(x) = (1 / 2)(1 / (x - 1) + 1 / (- x - 1)) = 1 / (x2 - 1) h(x) = (1 / 2)(1 / (x - 1) - 1 / (- x - 1)) = x / (x2 - 1) We now check our answer. We add g(x) and h(x) and see if the addistion gives f(x) g(x) + h(x) = 1 / (x2 - 1) + x / (x2 - 1) = ( 1 + x ) / (x2 - 1) = (1 + x) / [(x - 1)(x + 1)] = 1 / x - 1 = f(x) More on calculus questions with answers, tutorials and problems .