Page 306 - Matrix Analysis & Applied Linear Algebra

302 Chapter 5 Norms, Inner Products, and Orthogonality

and illustrated in Figure 5.4.2, satisfies these conditions. The value of f at t =0
is irrelevant—it’s not even necessary that f(0) be defined.

1
−π π

−1

Figure 5.4.2

To find the Fourier series expansion for f, compute the coefficients in (5.4.6) as

π 0 π
1 1 1
a n = f(t) cos nt dt = − cos nt dt + cos nt dt
π −π π −π π 0
=0,
1 π 1 0 1 π
b n = f(t) sin nt dt = − sin nt dt + sin nt dt
π −π π −π π 0

2 0 when n is even,
= (1 − cos nπ)=
nπ 4/nπ when n is odd,

so that

∞
4 4 4 4
F(t)= sin t + sin 3t + sin 5t + ··· = sin(2n − 1)t.
π 3π 5π (2n − 1)π
n=1
For each t ∈ (−π, π), except t =0, it must be the case that F(t)= f(t), and

+
f(0 )+ f(0 )
−
F(0) = =0.
2
Not only does F(t) agree with f(t)everywhere f is defined, but F also pro-
vides a periodic extension of f in the sense that the graph of F(t)is the entire
square wave depicted in Figure 5.4.2—the values at the points of discontinuity
(the jumps) are F(±nπ)=0.