Homework 11
Physics 2140
Methods in Theoretical Physics
Due: Wednesday April 20, 2005
Reading Assignment: Boas Ch. 7.5-7.9, Thornton and Marion, Ch. 3.8.
1.
(a) Boas 7.5.2.
Z an = 1 f (x) cos nxdx = 1
,
8
>
even n 6= 0;
< 0
= > n1 n = 1; 5; 9; : : :
:
, n1 n = 3; 7; 11; : : :
Z =2
0
1 sin nxj=2 = 1 sin n
cos nxdx = n
0
n 2
(2)
a0 = 1 f (x)dx = 1
dx = 21 ! a20 = 41 :
,
0
Z =2
1 cos nxj=2
sin nxdx = , n
bn = 1
0
8 0
1
>
< n odd n;
= > n2 n = 2; 6; 10; : : :
:
0 n = 4; 8; 12; : : :
Z (1)
Z =2
(3)
(4)
(5)
(6)
Thus
1
1
cos
3
x
cos
5
x
1
2
sin
2
x
sin
3
x
sin
5
x
f (x) = 4 + cos x , 3 + 5 : : : + sin x + 2 + 3 + 5 : : : :
(7)
(b) Boas 7.5.7.
an
Z 1
= x cos nxdx = n1 2 (cos nx + nx sin nx) j0 = n1 2 (cos n , 1)
(8)
=
(9)
(
0
even n =
6 0;
, n2 odd n:
Z a = 1 xdx = 2 ! a2 = 4 :
0
bn = 1
Thus
0
2
(10)
0
0
Z 0
x sin nxdx = n1 2 (sin nx , nx cos nx) j0 = , cosnn = (,1)n
n+1
cos
5
x
sin
2
x
2
cos
3
x
sin
3
x
f (x) = 4 , cos x + 32 + 52 : : : + sin x , 2 + 3 + : : : :
1
(11)
(12)
(13)
2.
3.
(a) Boas 7.6.2. At x = and at x = =2 f (x) is a continuous function so, at these points, the
Fourier series converges to the value of f (x), namely 0. At x = 2, x = 0, and x = =2, there is
a jump in f (x) from 0 to 1 so, at these points, the Fourier series converges to the mid-point of the
jump, namely 1/2.
(b) Boas 7.6.7. f (x) is continuous at x = 2, =2, and 0. Thus, at these point, the Fourier
series converges to the vlaue of f (x). At x = 2, ,=2, and 0, f (x) = 0 and at x = =2, f (x) = =2.
Boas 7.7.2.
Z Z =2
1
1
c0 = 2 f (x)dx = 2
dx = 14
,
0
Z Z =2
1
1 e,inx j=2
cn = 2 f (x)e,inxdx = 21
e,inxdx = 2,in
0
,
0
1 e,in=2 , 1 :
= 2,in
(14)
(15)
(16)
For n = 1; 2; 3; 4 exp(,in=2) = ,i; ,1; i; 1 and these four values repeat for increasing n. For
negative n, the coecients are equal to the complex conjugate of the positive n values, c,n = cjnj .
This insures that the function f (x) is real. Thus,
1 f,i , 1; ,1 , 1; i , 1; 1 , 1; and repeatg
(17)
cn = 2,in
1 f1 , i; ,2i; ,1 , i; 0; and repeatg
= 2n
(18)
(19)
c,n = 21jnj f1 + i; 2i; ,1 + i; 0; and repeatg
Thus,
1
1
2
i
2
i
ix
,
ix
i
2x
,
i
2x
f (x) = 4 + 2 (1 , i)e + (1 + i)e + , 2 e + 2 e
(20)
i) ei3x + (,1 + i) e,i3x + (1 , i) ei5x + (1 + i) e,i5x : : :
+ , (1 +
(21)
3
3
5
5
This answer is formally correct, but you can write the series in sine-cosine form by combining terms
as follows:
eix + e,ix = 2 cos x
,i(eix , e,ix) = 2 sin x:
(22)
(23)
Doing so we nd:
f (x) = 41 + 1 cos x , cos33x + cos55x : : : + 1 sin x + sin 2x + sin33x + sin55x : : : :
4.
(24)
Boas 7.8.7.
an
2 `
Z `
nx
1
`
nx
nx
nx
1
= ` x cos ` dx = ` n cos ` + ` sin ` 0
0
2
(25)
=
a0 =
bn =
=
c0 =
cn
(
` (cos n , 1) =
0 even n 6= 0;
2
2
, n22` 2 odd n:
n
1 Z ` xdx = ` ! a0 = ` :
` 0
2 2 4
Z `
1 x sin nx dx = 1 ` 2 sin nx , nx cos nx `
` 0
`
` n
`
`
` 0
` (sin n , n cos n) = , cos n = (,1)n+1 ` :
2
n Z2
n
n
1 ` xdx = ` :
2`
4
0
(26)
(27)
(28)
(29)
(30)
!
`
Z `
2
= 21` xe,inx=` dx = 21` ,n`2 2 e,inx=` ,inx
,
1
`
0
0
(
i`
h
i
even n 6= 0;
= , 2n`2 2 e,in (,in , 1) + 1 = , ` 2n+ i` odd n:
2n
n2 2
Thus for the sine-cosine Fourier series
1 (,1)n+1 nx X 1
X
2
`
nx
`
`
cos ` + f (x) = 4 , 2
2
n sin ` :
n=1
odd n n
and for the complex exponential Fourier series
1 1
1 (,1)n
X
X
inx=` + i`
e
einx=` :
f (x) = 4` , `2
2
n
2
n
n=,1
n=,1
(31)
(32)
(33)
(34)
(35)
n6=0
n odd
To see that these two series in equations (34) and (35) are equivalent, note that
1 1 einx=` + e,inx=`
1 1
1 cos nx = ,2` X
` X
inx=`
=
,
2 n=1 n2
`
2 n=1 n2
2
2 n=,1 n2 e
odd n
odd n
odd n
,2`
and
1
X
1 (,1)n+1 nx 1 (,1)n einx=` , e,inx=`
`X
`X
sin
=
,
n=1 n
`
n=1 n
2i
1 (,1)n
X
= 2i`
einx=`:
n
n=,1
(36)
(37)
(38)
n6=0
5.
. Find the Fourier Series expansions of the following functions on the specied intervals. State whether
they are even or odd.
a. f (t) = 1 if 0 t < and 0 if t < 2. P
sin(2n,1)t
Answer: f (t) = 12 + 2 sin t + 13 sin 3t + 15 sin 5t + ::: = 21 + 2 1
n=1 2n,1 .
f (t) is an odd function. See Figure 1 for a graph of it. Half the period of the function is T = .
Thus, !n = n=T = n= = n. Since it's an odd function, its Fourier Series will contain only sine
3
terms. Thus, we only need tond the bn coecients and, since the average value at the jumps is not
zero, we also need to nd a0 .
Z Z 1
1
f (t)(sin ! t)dt =
sin(nt)dt = ,1 cos(nt)j = ,1 [cos(n) , 1] ; (39)
b =
n
dt = = 1:
(40)
,
Note that the value of cos(n) depends on n. cos(n) = 1 if n is even and cos(n) = ,1 if n is odd.
n
Z,
a0 = 1
n
n
0
0
Thus,
2 if n is odd;
bn = n
bn = 0 if n is even:
(41)
(42)
Thus, we want to skip the even terms in the sum. To do so we have replace n with 2n , 1. When
n = 1; 2n , 1 = 1; n = 2; 2n , 1 = 3; n = 3; 2n , 1 = 5; :::. In this case:
!n = 2n , 1:
(43)
bn = (2n2, 1)
Therefore,
f (t) = a20 +
1
X
n=1
bn sin(!nt) = 12 + 2
1
X
n=1
1
2n , 1 sin(2n , 1)t:
(44)
b. g(t) = t2 if for ,1=2 < t < 1=2.
P
n+1 cos(2n)t .
Answer: 121 , 12 cos 2t , 212 cos 4t + 312 cos 6t + , + ::: = 121 , 12 1
n=1 (,1)
n2
g(t) is an even function. See Figure 2 for a graph of it. Half the period of the function is T = 1=2.
Thus, !n = n=T = 2n. Since it's an even function, its Fourier Series will contain only cosine
terms. Thus, we only need to nd the an coecients and, since the average value at the jumps is not
o, we also need to nd a0 .
an = 2
=
=
=
=
Z
1=2
,1=2
"
g(t)(cos !nt)dt = 2
Z
1 =2
,1=2
t2 (cos 2nt)dt
!
(45)
#
2 2
2 sin(2nt) j1=2
2 4n22 2 t cos(2nt) + 4n8n3 t ,
,1=2
3
"
!
#
2 2
1
n
,
1
2 4n2 2 cos(n) + 4n3 3 sin(n)
"
!
#
2 2
,
1
,
n
,
1
, 2 4n22 cos(,n) + 4n3 3 sin(,n)
1
1
2 4n2 2 cos(n) + 4n2 2 cos(,n)
1 cos(n)
2
n 2
1 (,1)n :
= n2 2
Z 1=2
t2 dt = 16 :
a0 = 2
(46)
(47)
(48)
(49)
(50)
(51)
(52)
,1=2
4
R
In equation (46) we have used the integral t2 sin(at)dt = 2t cos(at)=a2 + (a2 t , 2) sin(at)=a3 . In
equations (49) and (50) we use have used the facts that sin(n) = 0 and cos(,n) = cos(n).
Note, again, that the value of cos(n) depends on n. cos(n) = 1 if n is even and cos(n) = ,1 if n
is odd. That is, cos(n) = (,1)n , which we used in equation (51). Therefore,
f (t) = a20 +
6.
1
X
n=1
1 + 1
an cos(!nt) = 12
2
1
X
n=1
(,1)n cos 2nt:
n2
. Find the solution to the forced, damped harmonic oscillator equation:
x + 2 x_ + ! x = F (t) :
0
m
(53)
(54)
a. Let F (t) be f (t) from #5a.
!n = (2n , 1) and a0 and bn are given by equations (40) and (43).
1
X
2
sin((2n , 1)t + n )
1
xp(t) = 2m!2 + m
1=2 ;
2
0
n=1 (2n , 1) (!0 , (2n , 1)2 )2 + 4 2 (2n , 1)2
,
2
(2
n
,
1)
,
1
n = tan !2 , (2n , 1)2 )
0
(55)
(56)
(57)
b. Let F (t) be g(t) from #5b.
!n = 2n and a0 and an are given by equations (52) and (51).
1
1 + 1 X
cos(2nt + n )
n
xp (t) = 12m!
(
,
1)
;
2
2
2 (! 2 , 4n2 2 )2 + 16n2 2 2 1=2
m n=1
n
0
0
,
4
n
,
1
n = tan !2 , 4n22 )
0
5
(58)
(59)
(60)