1 Even and odd functions

We have shown in the previous Section how to calculate, by integration, the coefficients a n ( n = 0 , 1 , 2 , 3 , ) and b n ( n = 1 , 2 , 3 , ) in a Fourier series. Clearly this is a somewhat tedious process and it is advantageous if we can obtain as much information as possible without recourse to integration. In the previous Section we showed that the square wave (one period of which shown in Figure 12) has a Fourier series containing a constant term and cosine terms only (i.e. all the Fourier coefficients b n are zero) while the function shown in Figure 13 has a more complicated Fourier series containing both cosine and sine terms as well as a constant.

Figure 12 :

{ Square wave}

Figure 13 :

{ Saw-tooth wave}

Task!

Contrast the symmetry or otherwise of the functions in Figures 12 and 13.

The square wave in Figure 12 has a graph which is symmetrical about the y -axis and is called an even function. The saw-tooth wave shown in Figure 13 has no particular symmetry.

In general a function is called even if its graph is unchanged under reflection in the y -axis. This is equivalent to

f ( t ) = f ( t ) for all t

Obvious examples of even functions are t 2 , t 4 , t , cos t , cos 2 t , sin 2 t , cos n t .

A function is said to be odd if its graph is symmetrical about the origin (i.e. it has rotational symmetry about the origin). This is equivalent to the condition

f ( t ) = f ( t )

Figure 14 shows an example of an odd function.

Figure 14

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Examples of odd functions are t , t 3 , sin t , sin n t . A periodic function which is odd is the saw-tooth wave in Figure 15.

Figure 15

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Some functions are neither even nor odd . The periodic saw-tooth wave of Figure 13 is an example; another is the exponential function e t .

Task!

State the period of each of the following periodic functions and say whether it is even or odd or neither.

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  1. is neither even nor odd (with period 2 π )
  2. is odd (with period π ).

A Fourier series contains a sum of terms while the integral formulae for the Fourier coefficients a n and b n contain products of the type f ( t ) cos n t and f ( t ) sin n t . We need therefore results for sums and products of functions.

Suppose, for example, g ( t ) is an odd function and h ( t ) is an even function.

Let F 1 ( t ) = g ( t ) h ( t ) ( product of odd and even functions ) so F 1 ( t ) = g ( t ) h ( t ) ( replacing t by t ) = ( g ( t ) ) h ( t ) ( since g is odd and h is even ) = g ( t ) h ( t ) = F 1 ( t )

So F 1 ( t ) is odd.

Now suppose F 2 ( t ) = g ( t ) + h ( t ) ( sum of odd and even functions ) F 2 ( t ) = g ( t ) + h ( t ) = g ( t ) + h ( t ) We  see that F 2 ( t ) F 2 ( t ) and F 2 ( t ) F 2 ( t )

So F 2 ( t ) is neither even nor odd.

Task!

Investigate the odd/even nature of sums and products of

  1. two odd functions g 1 ( t ) , g 2 ( t )
  2. two even functions h 1 ( t ) , h 2 ( t )

G 1 ( t ) = g 1 ( t ) g 2 ( t ) G 1 ( t ) = ( g 1 ( t ) ) ( g 2 ( t ) ) = g 1 ( t ) g 2 ( t ) = G 1 ( t )

so the product of two odd functions is even.

G 2 ( t ) = g 1 ( t ) + g 2 ( t ) G 2 ( t ) = g 1 ( t ) + g 2 ( t ) = g 1 ( t ) g 2 ( t ) = G 2 ( t )

so the sum of two odd functions is odd.

H 1 ( t ) = h 1 ( t ) h 2 ( t ) H 2 ( t ) = h 1 ( t ) + h 2 ( t )

A similar approach shows that

H 1 ( t ) = H 1 ( t ) H 2 ( t ) = H 2 ( t )

i.e. both the sum and product of two even functions are even.

These results are summarized in the following Key Point.

Key Point 5

Products of functions

(even) × (even) = (even) (even) × (odd) = (odd) (odd) × (odd) = (even)

Sums of functions

(even) + (even) = (even) (even) + (odd) = (neither) (odd) + (odd) = (odd)

Useful properties of even and of odd functions in connection with integrals can be readily deduced if we recall that a definite integral has the significance of giving us the value of an area:

Figure 16

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a b f ( t ) d t gives us the net value of the shaded area, that above the t -axis being positive, that below being negative.

Task!

For the case of a symmetrical interval ( a , a ) deduce what you can about

a a g ( t ) d t and a a h ( t ) d t

where g ( t ) is an odd function and  h ( t ) is an even function.

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We have

a a g ( t ) d t = 0 for an odd function

a a h ( t ) d t = 2 0 a h ( t ) d t for an even function

(Note that neither result holds for a function which is neither even nor odd.)