1.3 Solutions 41which is consistent with Dirichlet’s theorem. Similar behavior is exhibited at
x=π,± 2 π...Figure 1.8 shows first four partial sums with equations
y=π/ 2
y=π/ 2 +2sinx
y=π/ 2 +2(sinx+(1/3) sin 3x)
y=π/ 2 +2(sinx+(1/3) sin 3x+(1/5) sin 5x)
1.19 By Problem 1.18,y=π
2+ 2
(
sinx+(
1
3
)
sin 3x+(
1
5
)
sin 5x+(
1
7
)
sin 7x+···)
Putx=π/2 in the above seriesy=π=π
2+ 2
(
1 −
1
3
+
1
5
−
1
7
+···
)
Henceπ 4 = 1 −^13 +^15 −^17 +···1.20 The Fourier transform off(x)is
T(u)=1
√
2 π∫a−aeiuxf(x)dx=
1
√
2 π∫a−a1 .eiuxdx=1
√
2 πeiux
iu∣
∣
∣
∣
a−a=
1
√
2 π(
eiua−e−iua
iu)
=
√
2
πsinua
u,u
= 0Foru=0,T(u)=√
2
πu.
The graphs of f(x) andT(u)foru=3 are shown in Fig. 1.9a, b, respec-
tively
Note that the above transform finds an application in the FraunHofer
diffraction.
̃f(ω)=Asinα/αThis is the basic equation which describes the Fraunhofer’s diffraction pat-
tern due to a single slit.Fig. 1.9Slit function and its Fourier transform