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)


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Updated: 2 April 2013

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