082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

23.5 Fourier series 737
but cos(−nπ) = cosnπ and hence a n
= 0, for all positive integers n. Using Equation
(23.5) we find
∫ π
b n
= 1 (1+t)sinntdt
π −π
= 1 ([
−(1+t) cosnt
π n
= 1 π
] π
−π
∫ π
)
cosnt
+ dt
−π n
(
−(1+π) cosnπ +(1−π) cos(−nπ) [ ] sinnt π )
+
n n n 2 −π
= 1 πn (−2πcosnπ)
since sin ±nπ = 0.Hence,
b n
=− 2 n cosnπ=−2 n (−1)n
Wefindb 1
=2,b 2
=−1,b 3
= 2 ,.... Thus the Fourier series representation is given
3
fromEquation (23.2) as
f(t)=1+2sint−sin2t+ 2 3 sin3t+···
which wecan writeconcisely as
f(t)=1−
∞∑
n=1
2
n (−1)n sinnt
Example23.15 Find the Fourier series representation of the function with period 2π defined by
f(t)=t 2 ,0<t2π.
Solution Asusualwesketch f (t),asshowninFigure23.16.HereT = 2πandweshallintegrate,
forconvenience, over the interval [0,2π]. UsingEquation (23.3)wefind
a 0
= 1 π
∫ 2π
0
t 2 dt = 1 π
[ t
3
3
] 2π
0
= 8π2
3
f(t)
4p 2
2p
t
Figure23.16
Graph forExample 23.15.