gabarito da lista de s erie de fourier...gabarito da lista de s erie de fourier 1 : ( a ) oa cogr de...

GABARITO DA LISTA DE S�ERIE DE FOURIER

1:

(a) O gr�a�co de f �e esbo�cado na Figura 1:

-1

-0.5

0

0.5

1

1.5

2

-2 -1 0 1 2 3 4 5 6

y

x

Figura 1

(b)

� C�alculo de a0:

Temos que:

a0 =1

2

Z 2

�2

f (x)dx =) a0 =1

2

Z 2

0

dx =) a0 =1

2[x]

2

0=) a0 = 1: ((1))

� C�alculo de an:

Temos que:

an =1

2

Z 2

�2

f (x) cosn�x

2dx =) an =

1

2

Z 2

0

cosn�x

2dx =) an =

1

n�

hsin

n�x

2

i20

=) an = 0: ((2))

� C�alculo de bn:

Temos que:

bn =1

2

Z 2

�2

f (x) sinn�x

2dx =) bn =

1

2

Z 2

0

sinn�x

2dx =) bn = � 1

n�

hcos

n�x

2

i20

:

Logo:

bn =1

n�[1� (�1)n] :

((3))

Portanto, a s�erie de Fourier da fun�c~ao f �e da forma:

1

2+

1

�

1Pn=1

[1� (�1)n]n

sinn�x

2:

((4))

1

(c)

Teorema de Fourier: Suponha que f e f 0 s~ao cont��nuas por partes no intervalo [�L;L] :Suponha tamb�em que f est�a de�nida fora do intervalo [�L;L] ; de modo a ser peri�odica

com per��odo T = 2L: Ent~ao, f tem uma s�erie de Fourier

a02

+

1Xn=1

an cosn�x

L+

1Xn=1

bn sinn�x

L: ((5))

Al�em disso, a s�erie de Fourier converge para f (x) em todos os pontos onde f �e

cont��nua e converge paraf (x+) + f (x�)

2em todos os pontos onde f �e descont��nua.

No nosso caso f e f 0 satisfazem as hip�oteses do teorema no intervalo [�2; 2] e T = 4:

A fun�c~ao f �e descont��nua em x = �2; x = 0 e x = 2: Ent~ao, pelo Terema de Fourier

temos que se:

(i) x = �2 =) a s�erie de Fourier converge paraf (�2+) + f (�2�)

2=

0 + 1

2=

1

2;

(ii) x = 0 =) a s�erie de Fourier converge paraf (0+) + f (0�)

2=

1 + 0

2=

1

2;

(iii) x = 2 =) a s�erie de Fourier converge paraf (2+) + f (2�)

2=

0 + 1

2=

1

2:

Por outro lado, nos pontos onde f �e cont��nua pelo Terema de Fourier temos que a s�erie

converge para f (x):

O gr�a�co da soma da s�erie �e esbo�cado na Figura 2:

-1

-0.5

0

0.5

1

1.5

2

-2 -1 0 1 2 3 4 5 6

y

x

Figura 2

2:

(a) O gr�a�co da f �e esbo�cado na Figura 3:

2

-1

-0.5

0

0.5

1

1.5

2

-1 -0.5 0 0.5 1 1.5 2 2.5 3

y

x

Figura 3

(b) A fun�c~ao f �e par. Portanto:

bn = 0: ((6))


Temos que:

a0 = 2

Z 1

0

f (x)dx =) a0 = 2

Z 1

0

(1� x)dx =) a0 = 2

�[x]

1

0� 1

2

�x2�10

�=) a0 = 1: ((7))


Temos que:

an = 2

Z 1

0

f (x) cosn�xdx =) an = 2

Z 1

0

(1� x) cosn�xdx: ((8))

� C�alculo daR 10(1� x) cosn�xdx:

Sejam:

��u = 1� x =) du = �dx;

dv = cosn�xdx =) v =1

n�sinn�x:

Logo:

Z 1

0

(1� x) cosn�xdx =1

n�[(1� x) sinn�x]

1

0+

1

n�

Z 1

0

sinn�xdx;

ou equivalentemente,

Z 1

0

(1 + x) cosn�xdx = � 1

n2�2[cosn�x]

1

0=

1

n2�2[1� (�1)n] ((9))

Substituindo (9) em (8) resulta que:

3

an =2

n2�2[1� (�1)n] :

((10))


1

2+

2

�2

1Pn=1

[1� (�1)n]n2

cosn�x:((11))

(c)



a02

+

1Xn=1

an cosn�x

L+

1Xn=1

bn sinn�x

L: ((12))




No nosso caso f �e cont��nua e f 0 �e cont��nua por partes no intervalo [�1; 1] e T = 2:

Como f em �e cont��nua [�1; 1] segue pelo Terema de Fourier que a s�erie converge para

f (x); ou seja:

f (x) =3

2+

2

�2

1Pn=1

[1� (�1)n]n2

cosn�x:((13))

O gr�a�co da soma da s�erie �e esbo�cado na Figura 4.

-1

-0.5

0

0.5

1

1.5

2

-1 -0.5 0 0.5 1 1.5 2 2.5 3

y

x

Figura 4

3:

(a) O gr�a�co da f �e esbo�cado na Figura 5:

4

-1

-0.5

0

0.5

1

1.5

2

-2 0 2 4 6 8

y

x

Figura 5

(b)


Temos que:

a0 =1

�

Z �

��

f (x)dx =) a0 =1

�

Z �=2

0

dx =) a0 =1

�[x]

�=20

=) a0 =1

2:

((14))


Temos que:

an =1

�

Z �

��

f (x) cosnxdx =) an =1

�

Z �=2

0

cosnxdx =) an =1

n�[sinnx]

�=20

=) an =1

n�sin

n�

2:

Logo:

a2n�1 =(�1)n+1(2n� 1)�

:((15))


Temos que:

bn =1

�

Z �

��

f (x) sinnxdx =) bn =1

�

Z �=2

0

sinnxdx =) bn = � 1

n�[cosnx]

�=20

:

Logo:

bn =[1� cosn�=2]

n� ((16))


1

4+

1

�

1Pn=1

(�1)n+1(2n� 1)

cosnx+1

�

1Pn=1

[1� cosn�=2]

nsinnx:

((17))

5

(c)



a02

+

1Xn=1

an cosn�x

L+

1Xn=1

bn sinn�x

L: ((18))




No nosso caso f e f 0 satisfazem as hip�oteses do teorema no intervalo [��; �] e T = 2�:

A fun�c~ao f �e descont��nua em x = 0 e x =�

2: Ent~ao, pelo Terema de Fourier temos que

se:

(i) x = 0 =) a s�erie de Fourier converge paraf (0+) + f (0�)

2=

1 + 0

2=

1

2;

(ii) x =�

2=) a s�erie de Fourier converge para

f (�=2+) + f (�=2�)2

=0 + 1

2=

1

2:




-1

-0.5

0

0.5

1

1.5

2

-2 0 2 4 6 8

y

x

Figura 6

4:

(a) O gr�a�co de f �e esbo�cado na Figura 7.

6

-3

-2

-1

0

1

2

3

-2 0 2 4 6 8

y

x

Figura 7

(b) A fun�c~ao f �e ��mpar a menos da origem. Portanto:

a0 = an = 0: ((19))


Temos que:

bn =2

�

Z �

0


2

Z �

0

sinnxdx =) bn = � 1

2n[cosnx]

�0;


bn =[1� (�1)n]

2n: ((20))


1

2

1Pn=1

[1� (�1)n]n

sinnx:((21))

(c)



a02

+

1Xn=1

an cosn�x

L+

1Xn=1

bn sinn�x

L: ((22))





A fun�c~ao f �e descont��nua em x = 0; x = �� e x = �: Ent~ao, pelo Terema de Fourier

7

temos que se:

(i) x = �� =) a s�erie de Fourier converge paraf (��+) + f (��)

2=��=4 + �=4

2= 0;


2=

�=4� �=4

2= 0;

(iii) x = � =) a s�erie de Fourier converge paraf (�+) + f (��)

2=��=4 + �=4

2= 0:




-3

-2

-1

0

1

2

3

-2 0 2 4 6 8

y

x

Figura 8

5:

(a) O gr�a�co de f �e esbo�cado na Figura 9:

-4-3-2-1 0 1 2 3 4

-2 0 2 4 6 8

y

x

Figura 9

(b)


Temos que:

a0 =1

�

Z �

��

f (x)dx =) a0 =1

�

��

Z 0

��

dx+

Z �

0

xdx

�;


8

a0 =1

�

�� [x]

0

�� +1

2

�x2��0

�=) a0 = ��

2:

((23))


an =1

�

Z �

��


�

��

Z 0

��

cosnxdx+

Z �

0

x cosnxdx

�;


an =1

�

��

n[sinnx]

0

�� +

Z �

0

x cosnxdx

�=) an =

1

�

Z �

0

x cosnxdx: ((24))

� C�alculo daR �0x cosnxdx:

Sejam:

��u = x =) du = dx;

dv = cosnxdx =) v =1

nsinnx:

Logo:

Z �

0

x cosnxdx =1

n[x sinnx]

�0+

1

n

Z �

0

sinnxdx;


Z �

0

x cosnxdx = � 1

n2[cosnx]

�0=

1

n2[1� (�1)n] ((25))


an =1

n2�[1� (�1)n] :

((26))


Temos que:

bn =1

�

Z �

��


�

��

Z 0

��

sinnxdx+

Z �

0

x sinnxdx

�;


bn =1

�

��

n[cosnx]

0

�� +

Z �

0

x sinnxdx

�=) bn =

1

�

�� [1� (�1)n]

n+

Z �

0

x sinnxdx:

�((27))

9

� C�alculo daR �0x sinnxdx:

Sejam:

��u = x =) du = dx;

dv = sinnxdx =) v = � 1

ncosnx:

Logo:

Z �

0

x sinnxdx = � 1

n[x cosnx]

�0+

1

n

Z �

0

cosnxdx;


Z �

0

x sinnxdx =�(�1)n+1

n+

1

n2[sinnx]

�0=

�(�1)n+1n

((28))


bn =[1� (�1)n] + (�1)n+1

n:

((29))


��

4+

1

�

1Pn=1

[1� (�1)n]n2

cosnx+1Pn=1

�[1� (�1)n] + (�1)n+1

n

�sinnx:

((30))

(c)



a02

+

1Xn=1

an cosn�x

L+

1Xn=1

bn sinn�x

L: ((31))





A fun�c~ao f �e descont��nua em x = ��; x = 0 e x = �: Ent~ao, pelo Terema de Fourier

temos que se:


2=�� + �

2= 0;

10


2=

0� �

2= ��

2;

(iii) x = � =) a s�erie de Fourier converge paraf (�+) + f (��)

2=�� + �

2= 0:




-4-3-2-1 0 1 2 3 4

-2 0 2 4 6 8

y

x

Figura 10

6:


0 2 4 6 8

10 12 14

-2 0 2 4 6 8

y

x

Figura 11

(b)


Temos que:

a0 =1

�

Z �

��

f (x)dx =) a0 =1

�

Z �

0

x2dx;


a0 =1

3�

�x3��0=) a0 =

�2

3:

((32))

11


an =1

�

Z �

��


�

Z �

0

x2 cosnxdx: ((33))

� C�alculo daR �0x2 cosnxdx:

Sejam:

��u = x2 =) du = 2xdx;


nsinnx:

Logo:

Z �

0

x2 cosnxdx =1

n

�x2 sinnx

��0� 2

n

Z �

0

x sinnxdx =)Z �

0

x2 cosnxdx = � 2

n

Z �

0

x sinnxdx ((34))

� C�alculo daR �0x sinnxdx:

Sejam:

��u = x =) du = dx;


ncosnx:

Portanto:

Z �

0

x sinnxdx = � 1

n[x cosnx]

�0+

1

n

Z �

0

cosnxdx;


Z �

0

x sinnxdx =�(�1)n+1

n+

1

n2[sinnx]

�0=

�(�1)n+1n

: ((35))


Z �

0

x2 cosnxdx =2�(�1)n+2

n2: ((36))


an =2(�1)n+2

n2:

((37))


Temos que:

12

bn =1

�

Z �

��


�

Z �

0

x2 sinnxdx: ((38))

� C�alculo daR �0x2 sinnxdx:

Sejam:

��u = x2 =) du = 2xdx;


ncosnx:

Logo:

Z �

0

x2 sinnxdx = � 1

n

�x2 cosnx

��0+

2

n

Z �

0

x cosnxdx =�2(�1)n+1

n+

2

n

Z �

0

x cosnxdx: ((39))

� C�alculo daR �0x cosnxdx:

Sejam: ��u = x =) du = dx;


nsinnx:

Logo:

Z �

0

x cosnxdx =1

n[x sinnx]

�0� 1

n

Z �

0

sinnxdx;


Z �

0

x cosnxdx =1

n2[cosnx]

�0=

[(�1)n � 1]

n2((40))


Z �

0

x2 sinnxdx =�2(�1)n+1

n+

2 [(�1)n � 1]

n3: ((41))

Portanto:

bn =�(�1)n+1

n+

2 [(�1)n � 1]

�n3:

((42))


�2

6+ 2

1Pn=1

(�1)n+2n2

cosnx+1Pn=1

��(�1)n+1

n+

2 [(�1)n � 1]

�n3

�sinnx:

((43))

13

(c)



a02

+

1Xn=1

an cosn�x

L+

1Xn=1

bn sinn�x

L: ((44))





A fun�c~ao f �e descont��nua em x = �� e x = �: Ent~ao, pelo Terema de Fourier

temos que se:


2=

0 + �2

2=

�2

2;

(ii) x = � =) a s�erie de Fourier converge paraf (�+) + f (��)

2=

0 + �2

2=

�2

2:




0 2 4 6 8

10 12 14

-2 0 2 4 6 8

y

x

Figura 12

7:


14

-1.5

-1

-0.5

0

0.5

1

1.5

-1 -0.5 0 0.5 1 1.5 2 2.5 3

y

x

Figura 13


bn = 0: ((45))


Temos que:

a0 =3

�

Z �=3

��=3

f (x)dx =) a0 =6

�

Z �=3

0

cos 2xdx;


a0 =3

�[sin 2x]

�=30

=) a0 =3p3

2�:

((46))


an =3

�

Z �=3

��=3

f (x) cosn�x

�=3dx =) an =

6

�

Z �=3

0

cos 2x cos 3nxdx: ((47))

� C�alculo daR �=30

cos 2x cos 3nxdx:

Sejam:

��u = cos 2x =) du = �2 sin 2xdx;

dv = cos 3nxdx =) v =1

3nsin 3nx:

Logo:

Z �=3

0

cos 2x cos 3nxdx =1

3n[cos 2x sin 3nx]

�=30

+2

3n

Z �=3

0

sin 2x sin 3nxdx;


15

Z �=3

0

cos 2x cos 3nxdx =2

3n

Z �=3

0

sin 2x sin 3nxdxdx: ((48))


sin 2x sin 3nxdxdx:

Sejam:

��u = sin 2x =) du = 2 cos 2xdx;

dv = sin 3nxdx =) v = � 1

3ncos 3nx:

Logo:

Z �=3

0

sin 2x sin 3nxdxdx = � 1

3n[sin 2x cos 3nx]

�=30

+2

3n

Z �=3

0

cos 2x cos 3nxdx;


Z �=3

0

sin 2x sin 3nxdxdx =(�1)n+1p3

6n+

2

3n

Z �=3

0



Z �=3

0

cos 2x cos 3nxdx =2

3n

((�1)n+1p3

6n+

2

3n

Z �=3

0

cos 2x cos 3nxdx

);


Z �=3

0

cos 2x cos 3nxdx� 2

3n

Z �=3

0

cos 2x cos 3nxdx =(�1)n+1p3

9n2;


Z �=3

0

cos 2x cos 3nxdx =(�1)n+1p3

3n(3n� 2): ((50))


an =2(�1)n+1p3

n�(3n� 2):

((51))


3p3

4�+

2p3

�

1Pn=1

(�1)n+1n(3n� 2)

cos 3nx:((52))

16

(c)



a02

+

1Xn=1

an cosn�x

L+

1Xn=1

bn sinn�x

L: ((53))




No nosso caso f e f 0 satisfazem as hip�oteses do teorema no intervalo [��=3; �=3] eT = 2�=3: A fun�c~ao f �e cont��nua em todos os pontos. Ent~ao, pelo Terema de Fourier

temos:

cos 2x =3p3

4�+

2p3

�

1Pn=1

(�1)n+1n(3n� 2)

cos 3nx:((54))


-1.5

-1

-0.5

0

0.5

1

1.5

-1 -0.5 0 0.5 1 1.5 2 2.5 3

y

x

Figura 14

8.


-0.5

0

0.5

1

1.5

-2 0 2 4 6 8

y

x

Figura 15

17


bn = 0: ((55))


Temos que:

a0 =1

�

Z �

��

f (x)dx =) a0 =2

�

(Z �=2

0

cosxdx�Z �

�=2

cosxdx

);


a0 =2

�[sinx]

�=20

� 2

�[sinx]

��=2 =)

a0 =4

�:

((56))

� C�alculo de an; para n 6= 1:

Temos que:

an =1

�

Z �

��


�

(Z �=2

0

cosx cosnxdx�Z �

�=2

cosx cosnxdx

): ((57))


cosx cosnxdx; para n 6= 1:

Sejam:

��u = cosx =) du = � sinxdx;


nsinnx:

Logo:

Z �=2

0

cosx cosnxdx =1

n[cosx sinnx]

�=20

+1

n

Z �=2

0

sinx sinnxdx;


Z �=2

0

cosx cosnxdx =1

n

Z �=2

0

sinx sinnxdx: ((58))


sinx sinnxdx:

Sejam:

��u = sinx =) du = cosxdx;


ncosnx:

18

Logo:

Z �=2

0

sinx sinnxdx = � 1

n[sinx cosnx]

�=20

+1

n

Z �=2

0

cosx cosnxdx;


Z �=2

0


ncos

n�

2+

1

n

Z �=2

0

cosx cosnxdx: ((59))


Z �=2

0

cosx cosnxdx = � 1

n2cos

n�

2+

1

n2

Z �=2

0

cosx cosnxdx;


Z �=2

0

cosx cosnxdx� 1

n2

Z �=2

0


n2cos

n�

2;


Z �=2

0

cosx cosnxdx =n2

(n2 � 1)

�� 1

n2cos

n�

2

�;


R �=20


(n2 � 1)cos

n�

2:

((60))

� C�alculo daR ��=2

cosx cosnxdx; para n 6= 1:

Do que foi exposto obtemos:

Z �

�=2

cosx cosnxdx =1

n[cosx sinnx]

��=2 +

1

n

Z �

�=2

sinx sinnxdx;


Z �

�=2

cosx cosnxdx =1

n

Z �

�=2

sinx sinnxdx: ((61))


sinx sinnxdx:

Do que foi exposto resulta que:

Z �

�=2


n[sinx cosnx]

��=2 +

1

n

Z �

�=2

cosx cosnxdx;

19


Z �

�=2

sinx sinnxdx =1

ncos

n�

2+

1

n

Z �

�=2

cosx cosnxdx: ((62))


Z �

�=2

cosx cosnxdx =1

n2cos

n�

2+

1

n2

Z �

�=2

cosx cosnxdx;

ou equivalentemente, Z �

�=2

cosx cosnxdx� 1

n2

Z �

�=2

cosx cosnxdx =1

n2cos

n�

2;


R ��=2

cosx cosnxdx =1

(n2 � 1)cos

n�

2:

((63))

Substituindo (60) e (63) em (57) resulta que:

an =2

�

�� 1

(n2 � 1)cos

n�

2� 1

(n2 � 1)cos

n�

2

�;


an = � 4

�(n2 � 1)cos

n�

2; n = 2; 3; :::

((64))


Temos que:

a1 =

Z �=2

0

cos2 xdx�Z �

�=2

cos2 xdx: ((65))


cos2 xdx:

Temos que:

Z �=2

0

cos2 xdx =1

2

(Z �=2

0

dx+

Z �=2

0

cos 2xdx

);

ou equivalentemente, Z �=2

0

cos2 xdx =1

2

�[x]

�=20

+1

2[sin 2x]

�=20

�;

20


Z �=2

0

cos2 xdx =�

4: ((66))


cos2 xdx:

Temos que:

Z �

�=2

cos2 xdx =1

2

�[x]

��=2 +

1

2[sin 2x]

��=2

�;

ou equivalentemente, Z �

�=2

cos2 xdx =�

4: ((67))


a1 =�

4� �

4= 0:

((68))


2

�+

4

�

1Pn=1

(�1)n+14n2 � 1

cos 2nx:((69))

(c)



a02

+

1Xn=1

an cosn�x

L+

1Xn=1

bn sinn�x

L: ((70))





A fun�c~ao f �e cont��nua em todos os pontos. Ent~ao, pelo Teorema de Fourier temos:

jcosxj = 2

�+

4

�

1Pn=1

(�1)n+14n2 � 1

cos 2nx:((71))


21

-0.5

0

0.5

1

1.5

-2 0 2 4 6 8

y

x

Figura 16

9.


0

5

10

15

20

25

-2 0 2 4 6 8

y

x

Figura 17

(b)


Temos que:

a0 =1

3

Z 3

�3

f (x)dx =) a0 =1

3

Z 3

�3

exdx;


a0 =1

3[ex]

3

�3=) a0 =

e3 � e�3

3:

((72))


an =1

3

Z 3

�3

f (x) cosn�x

3dx;


an =1

3

Z 3

�3

ex cosn�x

3dx ((73))

22

� C�alculo daR 3�3

ex cosn�x

3dx:

Sejam:

��u = ex =) du = exdx;

dv = cosn�x

3dx =) v =

3

n�sin

n�x

3:

Logo:

Z 3

�3

ex cosn�x

3dx =

3

n�

hex sin

n�x

3

i3�3

� 3

n�

Z 3

�3

ex sinn�x

3dx;


Z 3

�3

ex cosn�x

3dx = � 3

n�

Z 3

�3

ex sinn�x

3dx: ((74))


ex sinn�x

3dx:

Sejam:


dv = sinn�x

3dx =) v = � 3

n�cos

n�x

3:

Logo:

Z 3

�3

ex sinn�x

3dx = � 3

n�

hex cos

n�x

3

i3�3

+3

n�

Z 3

�3

ex cosn�x

3dx;


Z 3

�3

ex sinn�x

3dx =

3(�1)nn�

�e�3 � e3

�+

3

n�

Z 3

�3

ex cosn�x

3dx: ((75))


Z 3

�3

ex cosn�x

3dx = � 3

n�

�3(�1)nn�

�e�3 � e3

�+

3

n�

Z 3

�3

ex cosn�x

3dx

�;


Z 3

�3

ex cosn�x

3dx+

9

n2�2

Z 3

�3

ex cosn�x

3dx =

9(�1)n+1n2�2

�e�3 � e3

�;


23

Z 3

�3

ex cosn�x

3dx =

9(�1)n+1(n2�2 + 9)

�e�3 � e3

�: ((76))


an =3(�1)n+1(n2�2 + 9)

�e�3 � e3

�:

((77))


bn =1

3

Z 3

�3

f (x) sinn�x

3dx;


bn =1

3

Z 3

�3

ex sinn�x

3dx ((78))


ex sinn�x

3dx:

Sejam:


dv = sinn�x

3dx =) v = � 3

n�cos

n�x

3:

Logo:

Z 3

�3

ex sinn�x

3dx = � 3

n�

hex cos

n�x

3

i3�3

+3

n�

Z 3

�3

ex cosn�x

3dx;


Z 3

�3

ex sinn�x

3dx =

3(�1)nn�

�e�3 � e3

�+

3

n�

Z 3

�3

ex cosn�x

3dx ((79))


ex cosn�x

3dx:

Sejam:


dv = cosn�x

3dx =) v =

3

n�sin

n�x

3:

Logo:

24

Z 3

�3

ex cosn�x

3dx =

3

n�

hex sin

n�x

3

i3�3

� 3

n�

Z 3

�3

ex sinn�x

3dx;


Z 3

�3

ex cosn�x

3dx = � 3

n�

Z 3

�3

ex sinn�x

3dx: ((80))


Z 3

�3

ex sinn�x

3dx =

3(�1)nn�

�e�3 � e3

�� 9

n2�2

Z 3

�3

ex sinn�x

3dx;


Z 3

�3

ex sinn�x

3dx+

9

n2�2

Z 3

�3

ex sinn�x

3dx =

3(�1)nn�

�e�3 � e3

�;


Z 3

�3

ex sinn�x

3dx =

3n�(�1)n(n2�2 + 9)

�e�3 � e3

�: ((81))


bn =n�(�1)n+1(n2�2 + 9)

�e�3 � e3

�:

((82))


e3 � e�3�1

6� 3

1Pn=1

(�1)n+1(n2�2 + 9)

cosn�

3x� �

1Pn=1

n(�1)n+1(n2�2 + 9)

sinn�

3x

�:

((83))

(c)



a02

+

1Xn=1

an cosn�x

L+

1Xn=1

bn sinn�x

L: ((84))




25

No nosso caso f e f 0 satisfazem as hip�oteses do teorema no intervalo [�3; 3] e T = 6:

A fun�c~ao f �e descont��nua em x = �3 e x = 3: Ent~ao, pelo Terema de Fourier temos que

se:

(i) x = �3 =) a s�erie de Fourier converge paraf (�3+) + f (�3�)

2=

e�3 + e3

2;


2=

e�3 + e3

2:




0

5

10

15

20

25

-2 0 2 4 6 8

y

x

Figura 18

10.


-10

-5

0

5

10

-2 0 2 4 6 8

y

x

Figura 19

(b) A fun�c~ao f �e ��mpar. Portanto:

a0 = an = 0: ((85))


Temos que:

26

bn =1

�

Z �

��

f(x) sinnxdx;


bn = � 7

�

Z �

��

sin 15x sinnxdx: ((86))

Temos o seguinte resultado (ver Boyce-Diprima - se�c~ao 10:2):

Z L

�L

sinm�x

Lsin

n�x

Ldx =

�� 0; se m 6= n;L; se m = n:

((87))

No nosso caso m = n = 15 e L = �: Disto e de (87) resulta que:

b15 = �7 e b1 = b2 = ::: = b14 = b16 = ::: = 0 ((88))

Portanto, s�o o termo b15 �e diferente de zero. A s�erie �e igual a pr�opria fun�c~ao.

11.


-4

-2

0

2

4

-1 0 1 2 3 4

y

x

Figura 20


bn = 0: ((89))


Temos que:

a0 =2

�

Z �=2

��=2

f(x)dx;


27

a0 =8

�

Z �=2

0

cos 2xdx; ((90))

ou equivalentemente ,

a0 =4

�[sin 2x]

�=20

=) a0 = 0: ((91))


Temos que:

an =2

�

Z �=2

��=2

f(x) cos 2nxdx;


an =4

�

Z �=2

��=2


Temos o seguinte resultado (ver Boyce-Diprima - se�c~ao 10:2):

Z L

�L

cosm�x

Lcos

n�x

Ldx =

�� 0; se m 6= n;L; se m = n:

((93))

No nosso caso m = n = 1 e L = �=2: Disto e de (93) resulta que:

a1 = 4 e a2 = a3 = a4 = ::: = an = ::: = 0 ((94))

Portanto, s�o o termo a1 �e diferente de zero. A s�erie �e igual a pr�opria fun�c~ao.

12.


-0.5

0

0.5

1

1.5

-1 0 1 2 3 4

y

x

Figura 21

28

(b)

Os coe�cientes de Fourier de uma fun�c~ao de�nida em um intervalo [a; b] s~ao dados por:

a0 =2

b� a

Z b

a

f(x)dx; ((95))

an =2

b� a

Z b

a

f(x) cos2n�x

b� adx; ((96))

bn =2

b� a

Z b

a

f(x) sin2n�x

b� adx ((97))

No nosso caso o per��odo T = b� a = � =) L =�

2: A seguir calcularemos cada coe�ciente de

Fourier utilizando as f�ormulas (95); (96) e (97):


De (95) temos que:

a0 =2

�

Z �

0

sin2 xdx;


a0 =1

�

�Z �

0

dx�Z �

0

cos 2xdx

�;


a0 =1

�

�[x]

�0� 1

2[sin 2x]

�0

�=) a0 = 1: ((98))

� C�alculo de an; para n 6= 1:

De (96) temos que:

an =2

�

Z �

0

sin2 x cos 2nxdx;


an =1

�

�Z �

0

cos 2nxdx�Z �

0

cos 2x cos 2nxdx

�;


29

an =1

�

�1

2n[sin 2nx]

�0�Z �

0

cos 2x cos 2nxdx

�;


an = � 1

�

Z �

0


� C�alculo daR �0cos 2x cos 2nxdx; para n 6= 1:

Sejam:

��u = cos 2x =) du = �2 sin 2xdx;


2nsin 2nx:

Logo:

Z �

0

cos 2x cos 2nxdx =1

2n[cos 2x sin 2nx]

�0+

1

n

Z �

0

sin 2x sin 2nxdx;


Z �

0

cos 2x cos 2nxdx =1

n

Z �

0

sin 2x sin 2nxdx: ((100))

� C�alculo daR �0sin 2x sin 2nxdx:

Sejam:



2ncos 2nx:

Logo:

Z �

0

sin 2x sin 2nxdx = � 1

2n[sin 2x cos 2nx]

�0+

1

n

Z �

0

cos 2x cos 2nxdx;


Z �

0

sin 2x sin 2nxdx =1

n

Z �

0



Z �

0

cos 2x cos 2nxdx =1

n2

Z �

0

cos 2x cos 2nxdx;

30


Z �

0

cos 2x cos 2nxdx� 1

n2

Z �

0

cos 2x cos 2nxdx = 0;


Z �

0

cos 2x cos 2nxdx = 0: ((102))


an = 0; para n 6= 1: ((103))


Temos que:

a1 =2

�

Z �

0

sin2 x cos 2xdx;


a1 =1

�

�Z �

0

cos 2xdx�Z �

0

cos2 2xdx

�;


a1 =1

�

�1

2[sin 2x]

�0� 1

2

Z �

0

[1 + cos 4x] dx

�;


a1 = � 1

2�

�[x]

�0+

1

4[sin 4x]

�0

�;


a1 = �1

2:

((104))

� C�alculo de bn; para n 6= 1:

De (97) temos que:

bn =2

�

Z �

0

sin2 x sin 2nxdx;


31

bn =1

�

�Z �

0

sin 2nxdx�Z �

0

cos 2x sin 2nxdx

�;


bn =1

�

�� 1

2n[cos 2nx]

�0�Z �

0

cos 2x sin 2nxdx:

�; ((105))


bn = � 1

�

Z �

0

cos 2x sin 2nxdx:; ((106))

� C�alculo daR �0cos 2x sin 2nxdx; para n 6= 1:

Sejam:

��u = cos 2x =) du = �2 sin 2xdx;


2ncos 2nx:

Logo:

Z �

0

cos 2x sin 2nxdx = � 1

2n[cos 2x cos 2nx]

�0+

1

n

Z �

0

sin 2x cos 2nxdx;


Z �

0

cos 2x sin 2nxdx =1

n

Z �

0

sin 2x cos 2nxdx: ((107))

� C�alculo daR �0sin 2x cos 2nxdx:

Sejam:



2nsin 2nx sin 2nx:

Logo:

Z �

0

sin 2x cos 2nxdx =1

2n[sin 2x sin 2nx]

�0� 1

n

Z �

0

cos 2x sin 2nxdx;


Z �

0

sin 2x cos 2nxdx = � 1

n

Z �

0

cos 2x sin 2nxdx: ((108))

32


Z �

0

cos 2x sin 2nxdx = � 1

n2

Z �

0

cos 2x sin 2nxdx; ((109))


Z �

0

cos 2x sin 2nxdx� 1

n2

Z �

0

cos 2x sin 2nxdx = 0;


Z �

0

cos 2x sin 2nxdx = 0: ((110))


bn = 0; para n 6= 1: ((111))

� C�alculo de b1:

Temos que:

b1 =2

�

Z �

0

sin2 x sin 2xdx;


b1 =1

�

�Z �

0

sin 2xdx�Z �

0

cos 2x sin 2xdx

�;


b1 =1

�

��1

2[cos 2x]

�0+

1

2

�sin2 2x

��0

�;


b1 = 0:

Portanto, o gr�a�co da s�erie �e igual a pr�opria fun�c~ao, ou seja:

sin2 x =1

2� 1

2cos 2x:

((112))

13.

33

(a)

Temos pelo problema 4 que a s�erie de Fourier �e da forma:

1

2

1Xn=1

�1� (�1)n

n

�sinnx: ((113))

� Considere x =�

2: Ent~ao pelo teorema de Fourier e de (113) obtemos que:

�

4=

1

2

1Xn=1

�1� (�1)n

n

�sin

n�

2;


�

4=

1

2

1Xn=1

2(�1)n(2n� 1)

;


�

4=

1Xn=1

(�1)n(2n� 1)

= 1� 1

3+

1

5� 1

7+

1

9� 1

11:::+

(�1)n+1(2n� 1)

: ((114))

(b)



�

4=

1

2

1Xn=1

�1� (�1)n

n

�sin

n�

6;


�

4=

1

2

1Xn=1

2

(2n� 1)sin

(2n� 1)�

6


�

4= sin

�

6+

1

3sin

3�

6+

1

5sin

5�

6+

1

7sin

7�

6+

1

9sin

9�

6+

1

11sin

11�

6:::;


�

4=

1

2+

1

3+

1

2:5� 1

2:7� 1

9� 1

2:11+ ::: ((115))

Adicionando (115) �a (114) resulta que:

34

2�

4=

3

2+

3

2:5� 3

2:11+

3

2:13+

3

2:17� :::;


� = 3

�1 +

1

5� 1

11+

1

13+

1

17� :::;

�


�

3=

�1 +

1

5� 1

11+

1

13+

1

17� :::

�((116))

(c)



�

4=

1

2

1Xn=1

�1� (�1)n

n

�sin

n�

3;


�

4=

1

2

1Xn=1

2

(2n� 1)sin

(2n� 1)�

3;


�

4= sin

�

3+

1

3sin

3�

3+

1

5sin

5�

3+

1

7sin

7�

3+

1

9sin

9�

3+

1

11sin

11�

3:::;


�

4=

p3

2�p3

2:5+

p3

2:7�p3

2:11+

p3

2:13�p3

2:17:::; ((117))


�

2p3= 1� 1

5+

1

7� 1

11+

1

13� 1

17:::;


�p3

6= 1� 1

5+

1

7� 1

11+

1

13� 1

17::: ((118))

14.

35


��

4+

1

�

1Xn=1

[1� (�1)n]n2

cosnx+

1Xn=1

�[1� (�1)n] + (�1)n+1

n

�sinnx: ((119))

� Considere x = 0: Ent~ao pelo teorema de Fourier e de (119) obtemos que:

0 = ��

4+

1

�

1Xn=1

[1� (�1)n]n2

;


�

4=

1

�

1Xn=1

2

(2n� 1)2;


�2

8=

1Xn=1

1

(2n� 1)2: ((120))

15.


�2

6+ 2

1Xn=1

(�1)n+2n2

cosnx+

1Xn=1

��(�1)n+1

n+

2 [(�1)n � 1]

n3�

�sinnx: ((121))

� Considere x = �: Ent~ao pelo teorema de Fourier e de (121) obtemos que:

�2

2=

�2

6+ 2

1Xn=1

(�1)n+2(�1)nn2

;


�2

2� �2

6= 2

1Xn=1

(�1)2n+2n2

;


�2

6=

1Xn=1

1

n2: ((122))

36

16.

A fun�c~ao g(x) = x2 �e par e a fun�c~ao h(x) = sinx �e ��mpar. Logo, por um resultado conhecido a fun�c~aof(x) = g(x)h(x) = x2 sinx �e ��mpar.

17. A fun�c~ao g(x) = sin2 x �e par e a fun�c~ao h(x) = sinx �e ��mpar. Logo, por um resultado

conhecido a fun�c~ao f(x) = g(x)h(x) = (sinx)3 �e ��mpar.

18. Seja f(x) = x+ x2 + x3 =) f(�x) = �x+ (�x)2 + (�x)3 =) f(�x) = �x+ x2 � x3: Por

outro lado temos que �f(�x) = x� x2 + x3: Como, f(x) 6= f(�x) e f(x) 6= �f(�x); ent~aoa fun�c~ao f n~ao �e par e nem ��mpar.

19. Seja f(x) = sinx2 =) f(�x) = sin(�x)2 = sinx2: Como, f(x) = f(�x) a fun�c~ao f �e par.

20. Seja f(x) = ln1 + x

1� x=) f(x) = ln(1 + x)� ln(1� x) =) f(�x) = ln(1� x)� ln(1 + x) =)

=) f(x) = �f(�x): Logo, a fun�c~ao f �e ��mpar.

21. Seja f(x) = ex =) f(�x) = e�x: Como, f(x) 6= f(�x) e f(x) 6= �f(�x); ent~ao a fun�c~ao f

n~ao �e par e nem ��mpar.

22.

Denotemos por g(x) a extens~ao par e peri�odica de�nida por:

g(x) =

�� f(x); 0 < x 6 2;f(�x); � 2 < x < 0:

; g(x+ 4) = g(x)

Logo :

g(x) =

��2; 0 < x 6 1;0; 1 < x 6 2;2; � 1 6 x < 0;0; � 2 < x < �1:

; g(x+ 4) = g(x) ((123))

O gr�a�co de g(x) �e esbo�cado na Figura 22.

37

-1-0.5

0 0.5

1 1.5

2 2.5

3

-6 -4 -2 0 2 4 6

y

x

Figura 22

Denotemos por h(x) a extens~ao ��mpar e peri�odica de�nida por:

h(x) =

��f(x); 0 < x < 2;

0; x = 0 e x = 2;� f(�x); � 2 < x < 0:

; h(x+ 4) = h(x)

Logo :

h(x) =

��

2; 0 < x 6 1;0; 1 < x < 2;0; x = 0 e x = 2;

�2; � 1 < x < 0;0; � 2 < x < �1:

; h(x+ 4) = h(x) ((124))

O gr�a�co de h(x) �e esbo�cado na Figura 23.

-3

-2

-1

0

1

2

3

-6 -4 -2 0 2 4 6

y

x

Figura 23

23.


g(x) =

�� f(x); 0 6 x 6 1;f(�x); � 1 < x < 0:

; g(x+ 2) = g(x)

Logo :

38

g(x) =

��x+ 1=4; 0 < x < 1=2;x� 3=4; 1=2 6 x 6 1;x+ 1=4; � 1=2 < x < 0

�x� 3=4; � 1 < x 6 �1=2:; g(x+ 2) = g(x): ((125))


-0.4

-0.2

0

0.2

0.4

-3 -2 -1 0 1 2 3

y

x

Figura 24

24.


h(x) =

��f(x); 0 < x < �;

0; x = 0 e x = �;� f(�x); � � < x < 0:

; h(x+ 2�) = h(x)

Logo :

h(x) =

��ex; 0 < x < �;0; x = 0 e x = �;

�e�x; � � < x < 0:;h(x+ 2�) = g(x): ((128))


-20

-10

0

10

20

-8 -6 -4 -2 0 2 4 6 8

y

x

Figura 25

25.

39


g(x) =

�� f(x); 0 6 x 6 1;f(�x); � 1 < x < 0:

; g(x+ 2) = g(x)

Logo:

g(x) =

�� x2 � x+ 1=6; 0 6 x 6 1;x2 + x+ 1=6; � 1 < x < 0:

; g(x+ 2) = g(x) ((129))


-0.2

-0.1

0

0.1

0.2

0.3

-3 -2 -1 0 1 2 3

y

x

Figura 26

26.


g(x) =

�� f(x); 0 < x < �;f(�x); � � < x < 0:

; g(x+ 2�) = g(x)

Logo:

g(x) =

��x; 0 < x 6 �=2;� � x; �=2 < x < �;�x; � �=2 6 x < 0;� + x; � � < x < ��=2:

; g(x+ 2�) = g(x) ((131))


0

0.5

1

1.5

2

-8 -6 -4 -2 0 2 4 6 8

y

x

Figura 27

40


h(x) =

��f(x); 0 < x < �;

0; x = 0 e x = �;� f(�x); � � < x < 0:

; h(x+ 2�) = h(x)

Logo:

h(x) =

��

x; 0 < x 6 �=2;� � x; �=2 < x < �;0; x = 0 e x = �;x; � �=2 6 x < 0;�� x; � � < x < ��=2:

; h(x+ 2�) = h(x) ((132))


-2-1.5

-1-0.5

0 0.5

1 1.5

2

-8 -6 -4 -2 0 2 4 6 8

y

x

Figura 28

27.


Temos que:

a0 =

Z 2

0

f(x)dx =) a0 = 2

Z 1

0

dx =) a0 = 2 [x]1

0=) a0 = 2: ((133))


Temos que:

an =

Z 2

0

f(x) cosn�x

2dx =) an = 2

Z 1

0

cosn�x

2dx;


an =4

n�

hsin

n�x

2

i10

=) an =4

n�sin

n�

2 ((134))

41

De (133) e (134) resulta que a s�erie de cossenos da fun�c~ao f �e da forma:

1 +4

�

1Pn=1

sinn�

2n

:((135))


-1-0.5

0 0.5

1 1.5

2 2.5

3

-6 -4 -2 0 2 4 6

y

x

Figura 29

28.


Temos que:

bn = 2

Z 1

0

f(x) sinn�xdx =) bn = 2

(Z 1=2

0

�1

4� x

�sinn�xdx+

Z 1

1=2

�x� 3

4

�sinn�xdx

): ((136))

� C�alculo daR 1=20

�1

4� x

�sinn�xdx:

Sejam:

��u =

1

4� x =) du = �dx;

dv = sinn�xdx =) v = � 1

n�cosn�x:

Logo:

Z 1=2

0

�1

4� x

�sinn�xdx = � 1

n�

��1

4� x

�cosn�x

�1=20

� 1

n�

Z 1=2

0

cosn�xdx;


42

Z 1=2

0

�1

4� x

�sinn�xdx =

1

4ncos

n�

2+

1

4n�� 1

n2�2[sinn�x]

1=20

;


Z 1=2

0

�1

4� x

�sinn�xdx =

1

4n�cos

n�

2+

1

4n�� 1

n2�2sin

n�

2: ((137))


�x� 3

4

�sinn�xdx:

Sejam:

��u = x� 3

4=) du = dx;


n�cosn�x:

Logo:

Z 1

1=2

�x� 3

4

�sinn�xdx = � 1

n�

��x� 3

4

�cosn�x

�11=2

+1

n�

Z 1

1=2

cosn�xdx;


Z 1

1=2

�x� 3

4

�sinn�xdx =

1

4n�(�1)n+1 � 1

4n�cos

n�

2+

1

n2�2[sinn�x]

1

1=2 ;


Z 1

1=2

�x� 3

4

�sinn�xdx =

1

4n�(�1)n+1 � 1

4n�cos

n�

2� 1

n2�2sin

n�

2: ((138))


bn = 2

�1

4n�cos

n�

2+

1

4n�� 1

n2�2sin

n�

2+

1

4n�(�1)n+1 � 1

4n�cos

n�

2� 1

n2�2sin

n�

2

�;


bn =

�1 + (�1)n+1�

2n�� 4

n2�2sin

n�

2:

((139))

De (139) resulta que a s�erie de senos da fun�c~ao f �e da forma:

43

1Pn=1

(�1 + (�1)n+1�

2n�� 4

n2�2sin

n�

2

)sinn�x:

((140))


-0.4

-0.2

0

0.2

0.4

-3 -2 -1 0 1 2 3

y

x

Figura 30

29.


Temos que:

a0 = 2

Z 1

0

f(x)dx =) a0 = 2

(Z 1=2

0

�1

4� x

�dx+

Z 1

1=2

�x� 3

4

�dx;

)


a0 = 2

8<:�1

2

"�1

4� x

�2#1=20

+1

2

"�x� 3

4

�2#11=2

9=; =) a0 = 0: ((141))


Temos que:

an = 2

Z 1

0

f(x) cosn�xdx =) an = 2

(Z 1=2

0

�1

4� x

�cosn�xdx+

Z 1

1=2

�x� 3

4

�cosn�xdx

):

((142))


�1

4� x

�cosn�xdx:

Sejam:

��u =

1

4� x =) du = �dx;


n�sinn�x:

44

Logo:

Z 1=2

0

�1

4� x

�cosn�xdx =

1

n�

��1

4� x

�sinn�x

�1=20

+1

n�

Z 1=2

0

sinn�xdx;


Z 1=2

0

�1

4� x

�cosn�xdx = � 1

4n�sin

n�

2� 1

n2�2[cosn�x]

1=20

;


Z 1=2

0

�1

4� x

�cosn�xdx = � 1

4n�sin

n�

2� 1

n2�2cos

n�

2+

1

n2�2: ((143))


�x� 3

4

�cosn�xdx:

Sejam:

��u = x� 3

4=) du = dx;


n�sinn�x:

Logo:

Z 1

1=2

�x� 3

4

�cosn�xdx =

1

n�

��x� 3

4

�sinn�x

�11=2

� 1

n�

Z 1

1=2

sinn�xdx;


Z 1

1=2

�x� 3

4

�cosn�xdx =

1

4n�sin

n�

2+

1

n2�2[cosn�x]

1

1=2 ;


Z 1

1=2

�x� 3

4

�cosn�xdx =

1

4n�sin

n�

2+

(�1)nn2�2

� 1

n2�2cos

n�

2: ((144))


an = 2

�� 1

4n�sin

n�

2� 1

n2�2cos

n�

2+

1

n2�2+

1

4n�sin

n�

2+

(�1)nn2�2

� 1

n2�2cos

n�

2

�;


45

an =

�� 4

n2�2cos

n�

2+

2 [1 + (�1)n]n2�2

�:

((145))


1Pn=1

�� 4

n2�2cos

n�

2+

2 [1 + (�1)n]n2�2

��cosn�x:

((146))


-0.4

-0.2

0

0.2

0.4

-3 -2 -1 0 1 2 3

y

x

Figura 31

30.


Temos que:

bn =4

�

Z �=2

0

cosx sin 2nxdx: ((147))


cosx sin 2nxdx:

Sejam:

��u = cosx =) du = � sinxdx;


2ncos 2nx:

Logo:

Z �=2

0

cosx sin 2nxdx = � 1

2n[cosx cos 2nx]

�=20

� 1

2n

Z �=2

0

sinx cos 2nxdx;


Z �=2

0

cosx sin 2nxdx =1

2n� 1

2n

Z �=2

0

sinx cos 2nxdx: ((148))

46


sinx cos 2nxdx:

Sejam:

��u = sinx =) du = cosxdx;


2nsin 2nx:

Logo:

Z �=2

0

sinx cos 2nxdx =1

2n[sinx sin 2nx]

�=20

� 1

2n

Z �=2

0

cosx sin 2nxdx;


Z �=2

0

sinx cos 2nxdx = � 1

2n

Z �=2

0

cosx sin 2nxdx: ((149))


Z �=2

0

cosx sin 2nxdx =1

2n+

1

4n2

Z �=2

0

cosx sin 2nxdx;


Z �=2

0

cosx sin 2nxdx� 1

4n2

Z �=2

0

cosx sin 2nxdx =1

2n;


Z �=2

0

cosx sin 2nxdx =2n

4n2 � 1: ((150))


bn =8n

�(4n2 � 1):

((151))


8

�

1Pn=1

n

(4n2 � 1)sin 2nx:

((152))


47

-1.5

-1

-0.5

0

0.5

1

1.5

-4 -3 -2 -1 0 1 2 3 4

y

x

Figura 32

31.


Temos que:

a0 = 2

Z �

0

f(x)dx =) a0 = 2

(Z �=2

0

xdx+

Z �

�=2

(� � x)dx;

)


a0 =�x2��=20

� �(� � x)2��=2


a0 =�2

2:

((153))


Temos que:

an = 2

Z �

0

f(x) cosnxdx =) an = 2

(Z �=2

0

x cosnxdx+

Z 1

1=2

(� � x) cosnxdx

): ((154))


x cosnxdx:

Sejam:

��u = x =) du = dx;


nsinnx:

Logo:

Z �=2

0

x cosnxdx =1

n[x sinnx]

�=20

� 1

n

Z �=2

0

sinnxdx;

48


Z �=2

0

x cosnxdx =�

2nsin

n�

2+

1

n2[cosnx]

�=20

;


Z �=2

0

x cosnxdx =�

2nsin

n�

2+

1

n2

hcos

n�

2� 1

i: ((155))


(� � x) cosnxdx:

Sejam:

��u = � � x =) du = �dx;


nsinnx:

Logo:

Z �

�=2

(� � x) cosnxdx =1

n[(� � x) sinnx]

��=2 +

1

n

Z �

�=2

sinnxdx;


Z �

�=2

(� � x) cosnxdx = � �

2nsin

n�

2� 1

n2[cosnx]

��=2 ;


Z �

�=2

(� � x) cosnxdx = � �

2nsin

n�

2� 1

n2

h(�1)n � cos

n�

2

i: ((156))


an = 2

Z �

0

f(x) cosnxdx =) an =4

n2cos

n�

2� 2 [1� (�1)n]

n2:

((157))


1Pn=1

�4

n2cos

n�

2� 2 [1� (�1)n]

n2

�cosnx:

((158))


49

-0.5

0

0.5

1

1.5

2

-8 -6 -4 -2 0 2 4 6 8

y

x

Figura 33

32.


Temos que:

bn = 2

Z 1

0

f(x) sinn�xdx =) bn = 2

Z 1

0

x sinn�xdx: ((159))

� C�alculo daR 10x sinn�xdx:

Sejam:

��u = x =) du = dx;


n�cosn�x:

Logo:

Z 1

0

x sinn�xdx = � 1

n�[x cosn�x]

1

0+

1

n�

Z 1

0

cosn�xdx;


Z 1

0

x sinn�xdx =(�1)n+1

n�+

1

n2�2[sinn�x]

1

0;


Z 1

0

x sinn�xdx =(�1)n+1

n�: ((160))


bn =2(�1)n+1

n�:

((161))

50


2

�

1Pn=1

(�1)n+1n

sinn�x:((162))


-2-1.5

-1-0.5

0 0.5

1 1.5

2

-3 -2 -1 0 1 2 3

y

x

Figura 34

33.


Temos que:

a0 = 2

Z 1

0

f(x)dx =) a0 = 2

Z 1

0

xdx =) a0 =�x2�10=) a0 = 1: ((163))


Temos que:

an = 2

Z 1

0

f(x) cosn�xdx =) an = 2

Z 1

0

x cosn�xdx: ((164))

� C�alculo daR 10x cosn�xdx:

Sejam:

��u = x =) du = dx;


n�sinn�x:

Logo:

Z 1

0

x cosn�xdx =1

n�[x sinn�x]

1

0� 1

n�

Z 1

0

sinn�xdx;


51

Z 1

0

x cosn�xdx = � 1

n2�2[cosn�x]

1

0;


Z 1

0

x cosn�xdx =[1� (�1)n]

n2�2: ((165))


an =2 [1� (�1)n]

n2�2:

((166))


1

2+

2

�2

1Pn=1

[1� (�1)n]n2

cosn�x:((167))


-0.5

0

0.5

1

1.5

-3 -2 -1 0 1 2 3

y

x

Figura 35

34.


Temos que:

bn =2

�

Z �

0

f(x) sinnxdx =) bn =2

�

Z �

0

(2x+ 1) sinnxdx: ((168))

� C�alculo daR �0(2x+ 1) sinnxdx:

Sejam:

��u = (2x+ 1) =) du = 2dx;


ncosnx:

52

Logo:

Z �

0

(2x+ 1) sinnxdx = � 1

n[(2x+ 1) cosnx]

�0+

2

n

Z 1

0

cosnxdx;


Z �

0

(2x+ 1) sinnxdx =1� (2� + 1)(�1)n

n+

2

n2[sinnx]

�0;


Z �

0

(2x+ 1) sinnxdx =1� (2� + 1)(�1)n

n((169))


bn =2 [1� (2� + 1)(�1)n]

n�:

((170))


2

�

1Pn=1

[1� (2� + 1)(�1)n]n

sinnx:((171))


-10

-5

0

5

10

-8 -6 -4 -2 0 2 4 6 8

y

x

Figura 36

35.


Temos que:

a0 =2

�

Z �

0

f(x)dx =) a0 =2

�

Z �

0

(2x+ 1)dx;

53


a0 =1

2�

�(2x+ 1)2

��0=) a0 = 2� + 2: ((172))


Temos que:

an =2

�

Z �

0

f(x) cosnxdx =) an =2

�

Z �

0

(2x+ 1) cosnxdx: ((173))

� C�alculo daR �0(2x+ 1) cosnxdx:

Sejam:

��u = (2x+ 1) =) du = 2dx;


nsinnx:

Logo:

Z �

0

(2x+ 1) cosnxdx =1

n[(2x+ 1) sinnx]

�0� 2

n

Z 1

0

sinnxdx;


Z �

0

(2x+ 1) cosnxdx =2

n2[cosnx]

�0;


Z �

0

(2x+ 1) cosnxdx =2 [(�1)n � 1]

n2: ((174))


an =4 [(�1)n � 1]

�n2:

((175))


� + 1 +4

�

1Pn=1

[(�1)n � 1]

n2cosnx:

((176))


54

0

2

4

6

8

10

-8 -6 -4 -2 0 2 4 6 8

y

x

Figura 37

36.:

A express~ao para a S�erie de Fourier em senos de uma fun�c~ao peri�odica com per��odo 2� �e dada porP1

n=1 bn sinnx. Como sinx �e uma fun�c~ao cont��nua com derivada cont��nua temos, pelo Teorema deFourier, que a sua S�erie de Fourier coincide com a fun�c~ao em todos os pontos. Comparando sinx coma sua representa�c~ao como S�erie de Fourier em senos obtemos que c1 = 1 e cn = 0 para todo n > 1.Portanto, a S�erie de Fourier em senos de sinx �e a pr�opria fun�c~ao.


-1.5

-1

-0.5

0

0.5

1

1.5

-8 -6 -4 -2 0 2 4 6 8

y

x

Figura 38

Observe que se a fun�c~ao fosse, por exemplo, sinx

2a argumenta�c~ao acima n~ao se aplica e dever��amos

calcular os coe�cientes de Fourier realizando integra�c~oes similares �as dos exerc��os anteriores.

55

gabarito da lista de s erie de fourier...gabarito da lista de s erie de fourier 1 : ( a ) oa cogr de...

Documents