15.7 Exercises 441
The real part of this is (see Example 15.10)
(15.86)
If we replace(x,y,a,b)by(t, ω, 1 2 T, ω
0
)in (15.86), we obtain the pair of Fourier
transforms
The second term ofg(ω)is in practice small compared with the first, and the
function then reduces to a Lorentzian centred atω 1 = 1 ω
0
15.7 Exercises
Section 15.2
1.Given the power seriesf(x) 1 = 1 a
0
1 + 1 a
1
x 1 + 1 a
2
x
2
1 +1-, use equation (15.9) to find the
coefficientc
1
ofP
1
(x)in the expansion (15.3) off(x)in Legendre polynomials.
- (i) Find the first three terms of the expansion of the function
(Figure 15.16) in Legendre polynomials. (ii) Sketch
graphs of the one-term, two-term, and three-term
representations off(x).
Section 15.3
3.Show that the coefficient ofP
1
(x)in the expansion
isc
1
1 = 13 ij
1
(t)where.
4.Expand in terms of Legendre polynomials (i)cos 1 tx, (ii)sin 1 tx.
5.Find the first nonzero term in the expansion in powers of
12 Rof the potential at point P for the system of three
charges shown in Figure 15.17.
jt
t
t
t
t
1
1
()
sin
=−cos
ecPx
itx
l
ll
=
=
∑
0
∞
()
fx
x
xx
()=
−< <
<<
010
01
if
if
ft e t g
T
T
T
tT
() cos ( )
() (
=,=
+−
−/
ωω
ωω
0
0
22
1
22
11
π
ωωω+
0
22
)T
Regy
a
aby
a
aby
()
() ()
=
+−
++
1
22
2222
π
...................................................................................................................................................................
.
...
...
...
...
..
...
...
...
...
..
...
...
...
...
...
..
...
...
...
..
...
...
...
...
..
...
...
...
...
...
..
...
...
...
..
...
...
...
...
..
...
...
...
..
...
...
...
...
...
..
...
..
- •
o
+q
+q
− 2 q
p
R
r
r
..
..
...
..
...
...
....
....
θ
Figure 15.17
...........
..
..
..
...
..
.
.
..
...
..
..
.
....................................................................................................................................................................................................
.......
........
.......
.......
.......
.......
.......
........
......
.......
........
.......
.......
.......
.......
.......
........
.......
......
........
.......
.......
.......
.......
.......
........
.......
......
........
.......
.......
........
......
.......
........
.......
.......
......
....
- 1
− 1 +1
0
Figure 15.16