Page 223 - Numerical Analysis Using MATLAB and Excel

Chapter 6 Fourier, Taylor, and Maclaurin Series

Next, we multiply both sides of the above expression by dt , and we integrate over the period to
0
2π . Then,

2π 1 2π 2π 2π
∫ ft()sin 2t t = --a 0 ∫ sin 2t t + a 1 ∫ cos tsin 2t t + a 2 ∫ cos 2tsin 2t t
-
d
d
d
d
2
0 0 0 0
2π 2π
d
+ a 3 ∫ cos 3tsin 2t t + a 4 ∫ cos 4tsin 2t t + …
d
0 0 (6.11)
2π 2π 2π
2
)
d
d
+ b 1 ∫ sin tsin 2t t + b 2 ∫ ( sin 2t d + b 3 ∫ sin 3tsin 2t t
t
0 0 0
2π
d
+ b 4 ∫ sin 4tsin 2t t + …
0
We observe that every term on the right side of (6.11) except the term
2π
)
2
b 2 ∫ ( sin 2t d t
0
is zero as we found in (6.6) and (6.7). Therefore, (6.11) reduces to
2π 2π
∫ ft()sin 2t t = b 2 ∫ ( sin 2t d = b π
2
)
d
t
2
0 0
or
1 2π
b = --- ∫ 0 ft()sin 2t t
d
2
π
and thus we can evaluate this integral for any given function ft() . The remaining coefficients can
be evaluated similarly.
The coefficients a 0 , a n , and b n are found from the following relations.
1 1 2π
-
d
--a = ------ ∫ ft() t (6.12)
2 0 2π 0
1 2π
a = --- ∫ 0 ft()cos nt t (6.13)
d
n
π
1 2π
b = --- ∫ 0 ft()sin nt t (6.14)
d
n
π

6−6 Numerical Analysis Using MATLAB® and Excel®, Third Edition
Copyright © Orchard Publications