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(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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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. 

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: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
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
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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