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<< P.3. Composition of Functions
P.4. Odd and Even Functions
Some functions remain unchanged when reflected about the y-axis. In other words, these functions are symmetric about the y-axis. Such functions are called even functions. Mathematically speaking, a function f(x) is even if f(-x) = f(x) for all x in the domain of f. The reason for this name is that a polynomial f(x) is even if and only if all its nonzero coefficients correspond to even powers of x, i.e. if and only if f(x) has the form f(x) = a2nx2n + a2n-2x2n-2 + ... + a4x4 + a2x2 + a0 for some nonnegative integer n. The absolute value function f(x) = |x| is another example of an even function.
An equally important class of functions are odd functions. These are functions f(x) satisfying f(-x) = -f(x) for all x in the domain of f. Odd functions are symmetric about the origin, i.e. if the point (x,y) lies on the graph of f, then the point (-x, -y) also lies on the graph of f. The reason for this name is that a polynomial f(x) is odd if and only if all its nonzero coefficients correspond to odd powers of x, i.e. if and only if f(x) has the form f(x) = a2n+1x2n+1 + a2n-1x2n-1 + ... + a5x5 + a3x3 + a1x for some nonnegative integer n.
It turns out that every function defined over all real numbers has a unique decomposition into even and odd parts. The decomposition is as follows. Given a function f(x) defined over the reals, let fe(x) = [f(x) + f(-x)] / 2 and fo(x) = [f(x) - f(-x)] / 2. Then it is completely straightforward to verify that fe is even, fo is odd, and that f = fe + fo. We leave this as an exercise.
Example 1: For each of the following functions, state whether they are odd, even, or neither. In cases in which the function is neither odd nor even, decomposite it into even and odd parts.
- (a) f(x) = x2
- (b) f(x) = 1/x
- (c) f(x) = x2 + x + 1
- (d) f(x) = sin x
- (e) f(x) = cos x
- (f) f(x) = cos (x - π/4)
Solution:
- (a) f(x) = x2 is even since f(-x) = (-x)2 = x2 = f(x).
- (b) f(x) = 1/x is odd since f(-x) = 1/(-x) = -1/x = -f(x).
- (c) f(x) = x2 + x + 1 is neither odd nor even since f(-x) = x2 - x + 1 is not equal to f(x) or -f(x). We have
fe(x) = [f(x) + f(-x)] / 2 = (x2 + x + 1 + x2 - x + 1) / 2 = x2 + 1;
fe(x) = [f(x) - f(-x)] / 2 = (x2 + x + 1 - x2 + x - 1) / 2 = x.
- (d) f(x) = sin x is odd since f(-x) = sin(-x) = -sin x = -f(x).
- (e) f(x) = cos x is even since f(-x) = cos(-x) = cos x = f(x).
- (f) f(x) = cos (x - π/4) is neither odd nor even since f(-x) = cos(-x - π/4) = cos(x + π/4) is not equal to ±cos (x - π/4) = ±f(x). We have
fe(x) = [f(x) + f(-x)] / 2 = [cos (x - π/4) + cos (-x - π/4)] / 2 = [cos (x - π/4) + cos (x + π/4)] / 2 = cos x cos(π/4)
= (cos x) / √2;
fo(x) = [f(x) - f(-x)] / 2 = [cos (x - π/4) - cos (-x - π/4)] / 2 = [cos (x - π/4) - cos (x + π/4)] / 2 = sin x sin(π/4)
= (sin x) / √2.
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