Exercise (3-2): find Fourier series on [0,2n] Question Answer (1) f(x) = x/2 sin nx 2 2) f(x)=-x sin nx
Exercise (3-2): find Fourier series on [0,2n] Question Answer (1) f(x) = x/2 sin nx 2 2) f(x)=-x sin nx
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 3E
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Question
![Exercise (3-2): find Fourier series on [0,2n]
Question
Answer
(1) f(x) = x/2
sin nx
2) f(x)=-x
2sin nr
(3) f(x)=sin x
sin x
(4) f(x)=cos x
COS X
(5) f(x) =xsin x
-1+E2
COs nx
n- n -1
5 2n(-1)"
H(n-1)(n +1)
(6) f(x)=xcos x
sin(nx)
0<x<て
5l-(-1)"
1
sin nx
(7) f(x) =.
2
|1 くx<2
-1
0<x<T
-1+(-1)"
sin nx
(8) f(x) =-
1
11
1
(9) f(x)=.
0<x<T
3
1-(-1)"
sin nx
2 <x< 2T
- T<x<0
パ-1+(-1)"
sin nx
(10) f(x)=
元/4
0<x<だ
0<x<7
(-1)" -1
sin nx
(11) f(x) =
|T Tくx<2元
4
T-n
|オーX
(12) f(x) = {
0<x<T
*5-1)" -1
sin nx
COS nX +
T<x< 27
4
オn
0<x<T
(13) f(x)={
-25 -1) - cos nx
27-x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F935fe33c-ee4b-4e3e-b637-8e482c70d047%2Fe361e803-9bc1-42f3-b0c3-1630b7bbd143%2Fla7c6jk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Exercise (3-2): find Fourier series on [0,2n]
Question
Answer
(1) f(x) = x/2
sin nx
2) f(x)=-x
2sin nr
(3) f(x)=sin x
sin x
(4) f(x)=cos x
COS X
(5) f(x) =xsin x
-1+E2
COs nx
n- n -1
5 2n(-1)"
H(n-1)(n +1)
(6) f(x)=xcos x
sin(nx)
0<x<て
5l-(-1)"
1
sin nx
(7) f(x) =.
2
|1 くx<2
-1
0<x<T
-1+(-1)"
sin nx
(8) f(x) =-
1
11
1
(9) f(x)=.
0<x<T
3
1-(-1)"
sin nx
2 <x< 2T
- T<x<0
パ-1+(-1)"
sin nx
(10) f(x)=
元/4
0<x<だ
0<x<7
(-1)" -1
sin nx
(11) f(x) =
|T Tくx<2元
4
T-n
|オーX
(12) f(x) = {
0<x<T
*5-1)" -1
sin nx
COS nX +
T<x< 27
4
オn
0<x<T
(13) f(x)={
-25 -1) - cos nx
27-x
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