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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Exercise (3-2): find Fourier series on [0,2n]
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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
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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