__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)

Hence

f(x) = g(x) + h(x)

where g(x) is even and h(x) is odd and are defined in terms of f(x) above.

__Question 2:__

Express f(x) = 2x^{4} - 5 x^{3} + 2x^{2} + 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 x^{4} + 2 x^{2} - 4) + (- 5 x^{3} + x)

where 2 x ^{4} + 2 x ^{2} - 4 is a n even function and -5 x^{3} + 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 / (x^{2} - 1)

h(x) = (1 / 2)(1 / (x - 1) - 1 / (- x - 1)) = x / (x^{2} - 1)

We now check our answer. We add g(x) and h(x) and see if the addition gives f(x)

g(x) + h(x) = 1 / (x^{2} - 1) + x / (x^{2} - 1)

= ( 1 + x ) / (x^{2} - 1)

= (1 + x) / [(x - 1)(x + 1)]

= 1 / x - 1 = f(x)

More references on calculus

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