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Even and Odd Functions
DeÖnition. Saying that f is an even function means that f ( x) = f (x) for all x in the domain of f: Saying that f is an odd function means that f ( x) = f (x) or f (x) = f ( x) for all x in the domain of f.
Note. The graph of an even function is symmetric about the y -axis.
-5 -4 -3 -2 -1 0 1 2 3 4 5
5
10
15
20
25
x
y
An Even Function
The graph of an odd function is symmetric about the origin. (x; y) is on the graph if and only if ( x; y)is on the graph.
-5 -4 -3 -2 -1 1 2 3 4 5
20
40
60
80
100
120
x
y
An Odd Function
Note. If f (x) = xn^ then f is an even function when n is an even integer and f is an odd function when f is an odd integer. The cosine function is even and the sine function is odd.
Theorem. Suppose that each of f and g is an even function and each of u and v is an odd function all with the same domain D.
- f + g is an even function.
- u + v is an odd function (unlike with integers).
- f g is an even function.
- u v is an even function (unlike with integers).
- f u is an odd function (unlike with integers).
Proof of (5).
(f u)( x) = f ( x)u( x) = f (x) ( u(x)) = f (x)u(x) = (f u)(x)
for all x in D.
Suggested Problem. Prove Parts (1) - (4).
Note. Most functions are neither even nor odd. For example, if
f (x) = x + x^2
then f ( 1) = 0 while f (1) = 2:
Of course, 0 6 = 2 and 0 6 = 2 : So f is neither even nor odd.
However we do have the following fact.
Theorem. If the domain of f is symmetric about 0 (meaning x is in the domain if and only if x is in the domain) then f is the sum of an even function and an odd function.
Proof. Let fe(x) =
[f (x) + f ( x)] and fo(x) =
[f (x) f ( x)]
Note. From Calculus, we have Z (^) a
b
f (x)dx =
Z (^) b
a
f (x)dx
and (^) Z (^) h(b)
h(a)
f (x)dx =
Z (^) b
a
f (h(x))h^0 (x)dx:
Theorem. If f is an even function, then Z (^) L
L
f (x)dx = 2
Z L
0
f (x)dx:
Proof. Let h(x) = x:Then Z (^) L
L
f (x)dx =
Z 0
L
f (x)dx +
Z L
0
f (x)dx =
Z (^) h(0)
h(L)
f (x)dx +
Z L
0
f (x)dx
Z 0
L
f (h(x))h^0 (x)dx +
Z L
0
f (x)dx
Z 0
L
f ( x)( 1)dx +
Z L
0
f (x)dx
Z 0
L
f (x)dx +
Z L
0
f (x)dx =
Z L
0
f (x)dx +
Z L
0
f (x)dx
Z L
0
f (x)dx
Theorem. If f is an odd function, then Z (^) L
L
f (x)dx = 0:
Proof. Let h(x) = x: Z (^) L
L
f (x)dx =
Z 0
L
f (x)dx +
Z L
0
f (x)dx =
Z (^) h(0)
h(L)
f (x)dx +
Z L
0
f (x)d
Z 0
L
f (h(x))h^0 (x)dx +
Z L
0
f (x)dx
Z 0
L
f ( x)( 1)dx +
Z L
0
f (x)dx
Z 0
L
( f (x))dx +
Z L
0
f (x)dx
Z 0
L
f (x)dx +
Z L
0
f (x)dx =
Z L
0
f (x)dx +
Z L
0
f (x)dx = 0
Theorem. If f is an even function that is integrable over [ L; L], the Fourier Series for f is fSng where
Sn(x) = A 0 +
X^ n
k=
Ak cos kx L
in which
A 0 =
L
Z L
0
f (x)dx
and
Ak =
L
Z L
0
f (x) cos kx L dx
for k = 1; 2 ; : : :.
Proof. According to the deÖnition of a Fourier Series,
Sn(x) = A 0 +
X^ n
k=
Ak cos kx L
where
A 0 =
2 L
Z L
L
f (x)dx;
Ak =
L
Z L
L
f (x) cos kx L dx for k = 1; 2 ; : : : ; and
Bk =
L
Z L
L
f (x) sin kx L dx for k = 1; 2 ; : : : :
Since the integrand is even
A 0 =
2 L
Z L
0
f (x)dx =
L
Z L
0
f (x)dx
and
Ak =
L
Z L
L
f (x) cos kx L dx =
L
Z L
0
f (x) cos kx L dx =
L
Z L
0
f (x) cos kx L dx:
Since the integrand is odd (the product of an even function and an odd function is an odd function),
Bk =
L
Z L
L
f (x) sin kx L dx = 0
Example. If f (x) = jxj for L x L;since f is even, the Fourier Series for f is given by fSng where
Sn(x) = A 0 +
X^ n
k=
Ak cos
kx L
Since the integrand is odd, A 0 = 0
and Ak = 0
for k = 1; 2 ; : : :.Since the integrand is even,
Bk =
L
Z L
L
f (x) sin kx L
dx
L
Z L
0
f (x) sin
kx L dx =
L
Z L
0
x sin
kx L dx
L
[[x (
L
k ) cos kx L ]xx==0L +
Z L
0
L
k cos kx L dx]
=
L
[
L^2
k ( 1)k^ +
L^2
k^2 ^2 [sin kx L ]xx==0L ] 2 L k ( 1)k+
So the Fourier Series is fSng where
Sn(x) =
2 L
X^ n
k=
( 1)k k
sin kx L
