FOURIER SERIES
apply forr= 0 as well asr>0. The relations (12.5) and (12.6) may be derived
as follows.
Suppose the Fourier series expansion off(x) can be written as in (12.4),f(x)=a 0
2+∑∞r=1[
arcos(
2 πrx
L)
+brsin(
2 πrx
L)]
.Then, multiplying by cos(2πpx/L), integrating over one full period inxand
changing the order of the summation and integration, we get
∫x 0 +Lx 0f(x)cos(
2 πpx
L)
dx=a 0
2∫x 0 +Lx 0cos(
2 πpx
L)
dx+∑∞r=1ar∫x 0 +Lx 0cos(
2 πrx
L)
cos(
2 πpx
L)
dx+∑∞r=1br∫x 0 +Lx 0sin(
2 πrx
L)
cos(
2 πpx
L)
dx.(12.7)We can now find the Fourier coefficients by considering (12.7) asptakes different
values. Using the orthogonality conditions (12.1)–(12.3) of the previous section,
we find that whenp= 0 (12.7) becomes
∫x 0 +Lx 0f(x)dx=a 0
2L.Whenp= 0 the only non-vanishing term on the RHS of (12.7) occurs when
r=p,andso
∫x 0 +Lx 0f(x)cos(
2 πrx
L)
dx=ar
2L.The other Fourier coefficientsbrmay be found by repeating the above process
but multiplying by sin(2πpx/L) instead of cos(2πpx/L) (see exercise 12.2).
Express the square-wave function illustrated in figure 12.2 as a Fourier series.Physically this might represent the input to an electrical circuit that switches between a
high and a low state with time periodT. The square wave may be represented by
f(t)={
−1for−^12 T≤t< 0 ,
+1 for 0≤t<^12 T.In deriving the Fourier coefficients, we note firstly that the function is an odd function
and so the series will contain only sine terms (this simplification is discussed further in the