(b) Find the Fourier series for the given function. The equation is in the first picture. My question is why the textbook wrote the answer like the second picture show. Why, for the cos part, why you have coefficient 2/(2n-1)^2? Would you please give me a clear explanation? I am bewildered. I was thinking we need to get an and bn
(b) Find the Fourier series for the given function. The equation is in the first picture. My question is why the textbook wrote the answer like the second picture show. Why, for the cos part, why you have coefficient 2/(2n-1)^2? Would you please give me a clear explanation? I am bewildered. I was thinking we need to get an and bn
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
This question is about:
In each of Problems 11 through 16:
(a) Sketch the graph of the given function for three periods.
(b) Find the Fourier series for the given function.
The equation is in the first picture.
My question is why the textbook wrote the answer like the second picture show. Why, for the cos part, why you have coefficient 2/(2n-1)^2? Would you please give me a clear explanation? I am bewildered. I was thinking we need to get an and bn then plug into the original equation, then will be fine.
Transcribed Image Text:Sæ+L, -L < r < 0,
0 < x < L;
15. f (x) =
f (x + 2L) = f (æ)
||
Answer
Solution
(a) The figure shows the case L = 1.
-2
(b) The Fourier coefficients are calculated using the Euler-Fourier formulas:
L
1
3L
1
(x + L) dx
1
+ "
dx
L dx
L
L
2
-L
-L
Transcribed Image Text:J-L
as = 12) dz =(z + L) dz + Ldz = "
Le+L) dz +
L dx
-L
For n > 0,
1
(+ L) cos
dx +
L
CoS
dx
L cos
dx
L(1 -
cos nn)
Likewise,
+ f(2) sin
1
dx
L
1
dx +
L
bn
(x + L) sin
L sin
dx
L
--
L cos nT
Note that cos nT =
(-1)". It follows that the Fourier series for the given function is
- ()
(2n — 1) тӕ
(-1)"7
sin
3L
2
f(x) =
4
Σ
(2n – 1)?
CoS
L
L
-
n=1
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