Understanding Engineering Mathematics

Answer

− 3 p − 2 p −p 0 p 2 p 3 p t

Even;

π

2

;−

4

π

;0ifnis even, and−

4

π( 2 r+ 1 )^2

ifn= 2 r+1 is odd

17.10 The Fourier coefficients

There are, in general, an infinite number of the coefficientsan,bnof the Fourier series.
So we cannot expect to determine them ‘all at once’. We had a similar situation with
power series in Section 14.11 (435

➤
). There we found the coefficients one at a time by
successive differentiation to remove all terms except the one that we wanted to determine
at each stage. Such an approach won’t work in this case because no matter how many
times you differentiate a sinusoidal function, it won’t go away! However, the integral
formulae or ‘orthogonality relations’ given in Section 17.8 can serve a similar purpose in
enabling us to eliminate all coefficients except the one we want, one at a time. We do this
by multiplying through by cosntor sinnt, integrating over a single period and using the
orthogonality relations to remove all but the desired coefficient. An example will illustrate
this more clearly.
Suppose we want to finda 3 in the series

f(t)=

a 0

2

+

∑∞

n= 1

ancosnt+

∑∞

n= 1

bnsinnt

Multiply through by cos 3t, the sinusoid that goes witha 3 , and integrate over (−π,π)
(any interval of length 2πwill do). By the orthogonality relations of Section 17.8 every
resulting integral on the right-hand side of the series will vanishexceptthat corresponding
to cos 3titself, and we will obtain

∫π

−π

f(t)cos( 3 t)dt=a 3

∫π

−π

cos^2 ( 3 t)dt

=πa 3 from Section 17.8

So

a 3 =

1

π

∫−π

π

f(t)cos( 3 t)dt