Find the Fourier series for the given function. f(x) 9 200 f(x) ) = ²/²2 - ²2/2 2² 1 2m ² - 1 ² (2m - 1)лx L (m1)лx f(x) = 9 --> 100 9 -sin πm=1m - 1 L 9 (2m 1)лx f(x): - 20⁰ 9 · Σ πm=12m-1 -COS 2 L 9 200 9 ° f(x) = 2 + ²2₁ 22² sin((2m = 1) πm=12m - 1 L 9 2⁰0 9 (2m 1)лx f(x) (€ 2 πm=12m - 1 L = + - Σ - -sin -COS = f9, -L≤ x < 0, 0, 0 < x < L; f(x + 2L) = f(x)
Find the Fourier series for the given function. f(x) 9 200 f(x) ) = ²/²2 - ²2/2 2² 1 2m ² - 1 ² (2m - 1)лx L (m1)лx f(x) = 9 --> 100 9 -sin πm=1m - 1 L 9 (2m 1)лx f(x): - 20⁰ 9 · Σ πm=12m-1 -COS 2 L 9 200 9 ° f(x) = 2 + ²2₁ 22² sin((2m = 1) πm=12m - 1 L 9 2⁰0 9 (2m 1)лx f(x) (€ 2 πm=12m - 1 L = + - Σ - -sin -COS = f9, -L≤ x < 0, 0, 0 < x < L; f(x + 2L) = f(x)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Find the Fourier series for the given function.
f(x) =
9 2⁰⁰ 9
2 πm=12m
-
f(x) = 9 - - Σ
100 9
Im=1M - 1
f(x) =
20⁰ 9
Σ
πm=12m - 1
f(x): =
-
200 9
Σ
πm=12m
2
9
9
°ƒ(x) = ²/1 + ² £;
Σ
2
πm=12m -
9|N9|N
2
-sin
1
-sin
f(x) = {%
(2m 1)лx
L
(m1)лx
L
(2m — 1)лx)
L
(2m — 1)лx
-
L
(2m − 1)лx)
L
9, L≤ x < 0,
0 ≤ x < L;
¡cos((2m.
-sin
1
-7cos (12m-
-COS
1
f(x + 2L) = f(x)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa2e162b1-7cac-4840-92c7-248141e59cef%2Fd90fc098-58ce-4c57-a273-0264c52241c8%2F3d2327_processed.png&w=3840&q=75)
Transcribed Image Text:Find the Fourier series for the given function.
f(x) =
9 2⁰⁰ 9
2 πm=12m
-
f(x) = 9 - - Σ
100 9
Im=1M - 1
f(x) =
20⁰ 9
Σ
πm=12m - 1
f(x): =
-
200 9
Σ
πm=12m
2
9
9
°ƒ(x) = ²/1 + ² £;
Σ
2
πm=12m -
9|N9|N
2
-sin
1
-sin
f(x) = {%
(2m 1)лx
L
(m1)лx
L
(2m — 1)лx)
L
(2m — 1)лx
-
L
(2m − 1)лx)
L
9, L≤ x < 0,
0 ≤ x < L;
¡cos((2m.
-sin
1
-7cos (12m-
-COS
1
f(x + 2L) = f(x)
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