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7/30/2019 Solucionario parte 4 Matemticas Avanzadas para Ingeniera - 2da Edicin - Glyn James
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4
Fourier series
Exercises 4.2.9
1(a)
a0 =1
0dt +
0
tdt
=1
(t)0 +
t22
0
=
1
2 +
2
2
=
2
an =1
0 cos ntdt +
0
t cos ntdt
=1
nsin nt
0
+
t
nsin nt +
1
n2cos nt
0
= 1n2
(cos n 1) = 2
n2, n odd
0, n even
bn =1
0 sin ntdt +
0
t sin ntdt
=1
n
cos nt0
+ t
ncos nt +
1
n2sin nt
0
=1
n(1 2cos n) =
3
n, n odd
1
n
, n even
Thus the Fourier expansion of f(t) is
f(t) = 4
+n odd
2n2
cos nt +
n odd
3
nsin nt
n even
1
nsin nt
i.e. f(t) = 4 2
n=1
cos(2n 1)t(2 1)2 + 3
n=1
sin(2n 1)t(2n 1)
n=1
sin2nt
2n
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192 Glyn James: Advanced Modern Engineering Mathematics, Third edition
1(b)
a0 =1
0
(t + )dt =1
t2
2+ t
0
=
2
an =1
0
(t + )cos ntdt =1
(t + )
sin nt
n+
cos nt
n2
0
=1
n2(1 cos n) =
0, n even
2
n2, n odd
bn = 1
0
(t + )sin ntdt = 1
(t + ) cos ntn
+ sin ntn2
0
= 1n
Thus the Fourier expansion of f(t) is
f(t) =
4+
n odd
2
n2cos nt
n=1
1
nsin nt
i.e. f(t) =
4+
2
n=1
cos(2n 1)t(2n
1)2
n=1
sin nt
n
1(c) From its graph we see that f(t) is an odd function so it has Fourier
expansion
f(t) =n=1
bn sin nt
with
bn = 2
0
f(t)sin nt = 2
0
1 t
sin ntdt
=2
1
n
1 t
cos nt 1
n2sin nt
0
=2
n
Thus the Fourier expansion of f(t) is
f(t) =2
n=1
sin nt
n
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 193
1(d) From its graph f(t) is seen to be an even function so its Fourier
expansion is
f(t) =a02
+n=1
an cos nt
with
a0 =2
0
f(t)dt =2
/20
2cos tdt =2
[2sin t]
/20 =
4
an =2
0
f(t)cos ntdt =2
/20
2cos t cos ntdt
=2
/20
[cos(n + 1)t + cos(n 1)t]dt
=2
sin(n + 1)t
(n + 1)+
sin(n 1)t(n 1)
/20
=2
1
(n + 1)sin(n + 1)
2+
1
(n 1) sin(n 1)
2
=
0, n odd
4
1
(n2 1) , n = 4, 8, 12, . . .4
1
(n2 1), n = 2, 6, 10, . . .
Thus the Fourier expansion of f(t) is
f(t) =2
+
4
n=1
(1)n+1 cos2nt4n2 1
1(e)
a0 =
1
cos
t
2 dt =
1
2sin
t
2 =
4
an =1
cost
2cos ntdt =
1
2
cos(n +
1
2)t + cos(n 1
2)t
dt
=2
2
2
(2n + 1)sin(n +
1
2) +
2
(2n 1) sin(n1
2)
=
4
(4n2 1) , n = 1, 3, 5, . . .
4(4n2 1) , n = 2, 4, 6, . . .
bn = 0
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194 Glyn James: Advanced Modern Engineering Mathematics, Third edition
Thus the Fourier expansion of f(t) is
f(t) =2
+
4
n=1
(1)n+1 cos nt(4n2 1)
1(f) Since f(t) is an even function it has Fourier expansion
f(t) =a02
+n=1
an cos nt
with
a0 =2
0
| t | dt = 2
0
tdt =
an =2
0
t cos ntdt =2
t
nsin nt +
1
n2cos nt
0
=2
n2(cos n 1) =
0, n even
4n2
, n odd
Thus the Fourier expansion of f(t) is
f(t) = 2 4
n odd
1n2
cos nt
i.e. f(t) =
2 4
n=1
cos(2n 1)t(2n 1)2
1(g)
a0 =1
0
(2t
)dt =1
t2 t
0
= 0
an =1
0
(2t )cos ntdt = 1
(2t )
nsin nt +
2
n2cos nt
0
=2
n2(cos n 1) =
4
n2, n odd
0, n even
bn =1
0
(2t )sin ntdt = 1
(2t )
ncos nt +
2
n2sin nt
0
=
1
n
(cos n + 1) = 0, n odd
2
n , n even
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 195
Thus the Fourier expansion of f(t) is
f(t) =n odd
4n2
cos nt +n even
2n
sin nt
i.e. f(t) = 4
n=1
cos(2n 1)t(2n 1)2
n=1
sin2nt
n
1(h)
a0 =1
0
(t + et)dt +0
(t + et)dt
=1
t22
+ et0
+ t2
2+ et
0
=1
2 + (e e) = + 2
sinh
an =
1
0(t + e
t
)cos ntdt +0 (t + e
t
)cos ntdt
=1
t
nsin nt +
1
n2cos nt
0
+1
(n2 + 1)
net sin nt + et cos nt
0
+
t
nsin nt +
1
n2cos nt
0
+1
(n2 + 1)
net sin nt + et cos nt
0
=2
n2(1 + cos n) + 2cos n
(n2 + 1)
e e2
=
2
(cos 1)
n2+
cos n
(n2 + 1)sinh , cos n = (1)
n
bn =1
0
(t + et)sin ntdt +0
(t + et)sin ntdt
=1
tn
cos nt 1n2
sin nt0
+ t
ncos nt +
1
n2sin nt
0
+n2
2 + 1
et cos nt
n+
et sin nt
n2
= n
(n2 + 1) cos n(e
e
) = 2n
(n2 + 1) cos n sinh , cos n = (1)n
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196 Glyn James: Advanced Modern Engineering Mathematics, Third edition
Thus the Fourier expansion of f(t) is
f(t) =
2+
1
sinh
+
2
n=1
(1)n 1
n2+
(1)n sinh n2 + 1
cos nt
2
n=1
n(1)nn2 + 1
sinh sin nt
2 Since the periodic function f(t) is an even function its Fourier expansion is
f(t) =a02
+n=1
an cos nt
with
a0 =2
0
( t)2dt = 2
1
3( t)3
0
=2
32
an =2
0
( t)2 cos ntdt = 2
( t)2
nsin nt 2( t)
n2cos nt 2
n3sin nt
0
=4
n2
Thus the Fourier expansion of f(t) is
f(t) =2
3+ 4
n=1
1
n2cos nt
Taking t = gives
0 =2
3+ 4
n=1
1
n2(1)n
so that
1
122 =
n=1(1)n+1
n2
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 197
3 Since q(t) is an even function its Fourier expansion is
q(t) =a02
+
n=1
an cos nt
with
a0 =2
0
Qt
dt = Q
an =2
0
Qt
cos ntdt =
2Q
2
t
nsin nt +
1
n2cos nt
0
= 2Q2n2
(cos n 1) = 0, n even 4Q2n2
, n odd
Thus the Fourier expansion of q(t) is
q(t) = Q
1
2 4
2
n=1
cos(2n 1)t(2n 1)2
4
a0 =
1
0 5sin tdt =
1
[5cos t]0 =
10
an =5
0
sin t cos ntdt =5
2
0
[sin(n + 1)t sin(n 1)t]dt
=5
2
cos(n + 1)t
(n + 1)+
cos(n 1)t(n 1)
0
, n = 1
=5
2
cos nn + 1
cos n(n 1)
1n + 1
+1
n 1
=
5
(n2
1)(cos n + 1) =
0, n odd, n = 1
10
(n2 1), n even
Note that in this case we need to evaluate a1 separately as
a1 =1
0
5sin t cos tdt =5
2
0
sin2tdt = 0
bn =5
0
sin t sin ntdt = 52
0
[cos(n + 1)t cos(n 1)t]dt
= 52
sin(n + 1)t
(n + 1) sin(n 1)t
(n 1)0
, n = 1
= 0 , n = 1c
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198 Glyn James: Advanced Modern Engineering Mathematics, Third edition
Evaluating b1 separately
b1 =5
0
sin t sin tdt =5
2
0
(1 cos2t)dt
=5
2
t 1
2sin2t
0
=5
2
Thus the Fourier expansion of f(t) is
f(t) =5
+
5
2sin t 10
n=1
cos2nt
4n2 1
5
a0 =1
0
2dt +
0
(t )2dt=
1
2t0 +
13
(t )30
=
4
32
an =1
0
2 cos ntdt +
0
(t )2 cos ntdt
=1
2n
sin nt0
+ (t )2
nsin nt +
2(t )n2
cos nt 2n3
sin nt0
=2
n2
bn =1
0
2 sin ntdt +
0
(t )2 sin ntdt
=1
2n
cos nt0
+ (t )2
ncos nt + 2
(t )n2
sin nt +2
n3cos nt
0
=1
2n
+2
n(1)n
=
n(1)n 2
n3[1 (1)n]
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 199
Thus the Fourier expansion of f(t) is
f(t) =2
32 +
n=1
2
n2cos nt +
(1)nn
sin nt
4
n=1
sin(2n 1)t(2n 1)3
5(a) Taking t = 0 gives
2 + 2
2=
2
32 +
n=1
2
n2
and hence the required resultn=1
1
n2=
1
62
5(b) Taking t = gives
2 + 0
2
=2
3
2 +
n=1
2
n2
(
1)n
and hence the required result
n=1
(1)n+1n2
=1
122
6(a)
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200 Glyn James: Advanced Modern Engineering Mathematics, Third edition
6(b)
The Fourier expansion of the even function (a) is given by
f(t) =a02
+n=1
an cos nt
with
a0 =2
/20
tdt +
/2
( t)dt
= 2
12 t2/20
+12 ( t)2
/2= 2
an =2
/20
t cos ntdt +
/2
( t)cos ntdt
=2
tn
sin nt +1
n2cos nt
/20
+ t
nsin nt 1
n2cos nt
/2
=2
2
n2cos
n
2 1
n2(1 + (1)n)
=
0, n odd
8n2
, n = 2, 6, 10, . . .
0, n = 4, 8, 12, . . .
Thus the Fourier expansion of f(t) is
f(t) =
4 2
n=1
cos(4n 2)t(2n 1)2
Taking t = 0 where f(t) = 0 gives the required result.
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 201
7
a0 =1
0
(2 t
)dt +
2
t/dt
=1
2t t
2
2
0
+ t2
2
2
= 3
an =1
0
(2 t
)cos ntdt +
2
t
cos ntdt
=1
2n
sin nt tn
sin nt 1n2
cos nt0
+ t
nsin nt +
1
n2cos nt
2
=2
2n2 [1 (1)n
]
=
0, n even
4
2n2, n odd
bn =1
0
(2 t
)sin ntdt +
2
t
sin ntdt
=1
2n
cos nt +t
ncos nt 1
n2sin nt
0
+ t
ncos nt +
1
n2sin nt
2
= 0
Thus the Fourier expansion of f(t) is
f(t) =3
2+
4
2
n=1
cos(2n 1)t(2n 1)2
Replacing t by t 12 gives
f(t
1
2
) =3
2
+4
2
n=1
cos(2n 1)(t )(2n 1)2
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202 Glyn James: Advanced Modern Engineering Mathematics, Third edition
Since
cos(2n 1)(t 12
) = cos(2n 1)t cos(2n 1) 2
+ sin(2n 1)t sin(2n 1) 2
= (1)n+1 sin(2n 1)t
f(t 12
) 32
=4
2
n=1
(1)n+1 sin(2n 1)t(2n 1)2
The corresponding odd function is readily recognised from the graph of f(t).
Exercises 4.2.11
8 Since f(t) is an odd function the Fourier expansion is
f(t) =n=1
bn sinnt
with
bn =2
0
t sinnt
dt =2
t
n
cosnt
+ n2
sinnt
0
= 2n
cos n
Thus the Fourier expansion of f(t) is
f(t) =2
n=1
(1)n+1n
sinnt
9 Since f(t) is an odd function (readily seen from a sketch of its graph) its
Fourier expansion is
f(t) =
n=1
bn sinnt
with
bn =2
0
K
( t)sin nt
tdt
=2
K
ncos
nt
+
Kt
ncos
nt
K
(n)2sin
nt
0
=
2K
n
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 203
Thus the Fourier expansion of f(t) is
f(t) =2K
n=1
1
nsin
nt
10
a0 =1
5
50
3dt = 3
an =1
5 5
0
3cosnt
5dt =
1
515
nsin
nt
5 5
0
= 0
bn =1
5
50
3sinnt
5dt =
1
5
15
ncos
nt
5
50
=3
n[1 (1)n] =
6
n, n odd
0, n even
Thus the Fourier expansion of f(t) is
f(t) =3
2+
6
n=11
(2n
1)
sin(2n 1)
5t
11
a0 =2
2
/0
A sin tdt =
A
cos t
/0
=2A
an =A
/0
sin t cos ntdt =A
2
/0
[sin(n + 1)t sin(n 1)t]dt
=A
2
cos(n + 1)t
(n + 1)+
cos(n 1)t(n 1)
/
0
, n= 1
=A
2
2(1)n+1
n2 1 2
n2 1
=A
(n2 1) [(1)n+1 1]
=
0, n odd , n = 1 2A
(n2 1) , n even
Evaluating a1 separately
a1 =
A
/0 sin t cos tdt =
A
2/0 sin2tdt = 0
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204 Glyn James: Advanced Modern Engineering Mathematics, Third edition
bn = A
/
0
sin t sin ntdt = A2
/
0
[cos(n + 1)t cos(n 1)t]dt
= A2
sin(n + 1)t
(n + 1) sin(n 1)t
(n 1)/0
, n = 1
= 0, n = 1
b1 =A
/0
sin2 tdt =A
2
/0
(1 cos2t)dt = A2
Thus the Fourier expansion of f(t) is
f(t) =A
1 +
2sin t 2
n=1
cos2nt
4n2 1
12 Since f(t) is an even function its Fourier expansion is
f(t) =a02
+
n=1
an cosnt
T
with
a0 =2
T
T0
t2dt =2
T
1
3t3T0
=2
3T2
an =2
TT
0
t2 cosnt
Tdt =
2
TT t2
nsin
nt
T+
2tT2
(n)2cos
nt
T 2T3
(n)3sin
nt
TT
0
=4T2
(n)2(1)n
Thus the Fourier series expansion of f(t) is
f(t) =T2
3+
4T2
2
n=1
(1)nn2
cosnt
T
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 205
13
a0 = 2T
T0
ET
tdt = 2ET2
12
t2T0
= E
an =2
T
T0
E
Tt cos
2nt
Tdt
=2E
T2
tT
2nsin
2nt
T+ T
2n
2cos
2nt
T
T0
= 0
bn =2E
T2
T0
t sin2nt
Tdt
= 2ET2 tT2n cos 2ntT + T2n
2
sin 2ntTT0
= EnThus the Fourier expansion of e(t) is
e(t) =E
2 E
n=1
1
nsin
2nt
T
Exercises 4.3.3
14 Half range Fourier sine series expansion is given by
f(t) =n=1
bn sin nt
with
bn =2
0
1sin ntdt =2
1
ncos nt
0
=
2
n
[(
1)n
1]
=
0, n even
4
n, n odd
Thus the half range Fourier sine series expansion of f(t) is
f(t) =4
n=1
sin(2n 1)t(2n 1)
Plotting the graphs should cause no problems.
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15 Half range Fourier cosine series expansion is given by
f(t) =a02
+
n=1
an cos nt
with
a0 =2
1
10
(2t 1)dt = 0
an = 210
(2t 1)cos ntdt
= 2
(2t 1)
nsin nt +
2
(n)2cos nt
10
=4
(n)2[(1)n 1]
=
0, n even
8(n)2
, n odd
Thus the half range Fourier cosine series expansion of f(t) is
f(t) = 82
n=1
1
(2n 1)2 cos(2n 1)t
Again plotting the graph should cause no problems.
16(a)
a0 = 2
10
(1 t2)dt = 2t 13
t310
=4
3
an = 2
10
(1 t2)cos2ntdt
= 2
(1 t2)
2nsin2nt 2t
(2n)2cos2nt +
2
(2n)3sin2nt
10
= 1
(n)2
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 207
bn = 210 (1 t
2
)sin2ntdt
= 2
(1 t
2)
2ncos2nt 2t
(2n)2sin2nt 2
(2n)3cos2nt
10
=1
n
Thus the full-range Fourier series expansion for f(t) is
f(t) = f1(t) =2
3 1
2
n=1
1
n2cos2nt +
1
n=1
1
nsin2nt
16(b) Half range sine series expansion is
f2(t) =n=1
bn sin nt
with
bn = 210 (1 t
2
)sin ntdt
= 2
(1 t
2)
ncos nt 2t
(n)2sin nt 2
(n)3cos nt
10
= 2
2
(n)3(1)n + 1
n+
2
(n)3
=
2
n, n even
2
1
n+
4
(n)3 , n odd
Thus half range sine series expansion is
f2(t) =1
n=1
1
nsin2nt +
2
n=1
1
(2n 1) +4
2(2n 1)3
sin(2n 1)t
16(c) Half range cosine series expansion is
f3(t) =a0
2
+
n=1
an cos nt
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208 Glyn James: Advanced Modern Engineering Mathematics, Third edition
with
a0 = 2
1
0
(1 t2)dt = 43
an = 2
10
(1 t2)cos ntdt
= 2
(1 t2)
nsin nt 2t
(n)2cos nt +
2
(n)3sin nt
10
=4(1)n
(n)2
Thus half range cosine series expansion is
f3(t) =2
3+
4
2
n=1
(1)n+1n2
cos nt
Graphs of the functions f1(t), f2(t), f3(t) for 4 < t < 4 are as follows
17 Fourier cosine series expansion is
f1(t) =a0
2
+
n=1
an cos nt
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 209
witha0 =
2
0
(t t2)dt = 13
2
an =2
0
(t t2)cos ntdt
=2
(t t2)
nsin nt +
( 2t)n2
cos nt +2
n3sin nt
0
= 2n2
[1 + (1)n]
= 0, n odd 4
n2, n even
Thus the Fourier cosine series expansion is
f1(t) =1
62
n=1
1
n2cos2nt
Fourier sine series expansion is
f2(t) =
n=1
bn sin nt
with
bn =2
0
(t t2)sin ntdt
=2
(t t
2)
ncos nt +
( 2t)n2
sin nt 2n3
cos nt
0
= 4n3 [1 (1)n]
=
0, n even
8
n3, n odd
Thus the Fourier sine series expansion is
f2(t) =8
n=1
1
(2n 1)3 sin(2n 1)t
cPearson Education Limited 2004
7/30/2019 Solucionario parte 4 Matemticas Avanzadas para Ingeniera - 2da Edicin - Glyn James
20/76
210 Glyn James: Advanced Modern Engineering Mathematics, Third edition
Graphs of the functions f1(t) and f2(t) for 2 < t < 2 are:
18
f(x) =2a
x , 0 < x