What is the correct Fourier Series for f(x) = sinx on the interval 0 < x < ∏?

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SUMMARY

The correct Fourier Series for the function f(x) = sin(x) on the interval 0 < x < π is f(x) = 1/π + sin(x)/2 + (1/π) Σ [(-1)^n + 1] / [π(1 - n^2)] cos(nx), with n starting from 2 to infinity. The initial calculation incorrectly omitted the sin(x)/2 term, which arises from the evaluation of the Fourier coefficients. The user clarified that the sine term must be included due to the specific behavior of the function at n=1, which requires reevaluation of the series.

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Homework Statement


I must calculate the Fourier Series of

f(x) = 0, when -∏< x < 0 and f(x) = sinx, 0 < x < ∏


Homework Equations





The Attempt at a Solution


Using the formulae, I calculated a0 = 2/pi, an = [ (-1)^n + 1 ] / [ ∏(1 - n^2) ], and bn = 0, so my Fourier series goes like this:

f(x) = 1/∏ + (1/∏) Ʃ [ (-1)^n + 1 ] / [ ∏(1 - n^2) ] cos nx, but since n = 1 is not an option, I start Ʃ from n = 2 to infinity.

However the book says that the answer is

f(x) = 1/∏ + sin x / 2 + (1/∏) Ʃ [ (-1)^n + 1 ] / [ ∏(1 - n^2) ] cos nx, where n also goes from n = 2 to infinity.

Where did that sin x / 2 came from?? I don't know there a sin term since bn = 0!
 
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The fact that you threw out the n=1 cosine term because it wasn't defined at n=1 suggests you made the same mistake for the sine term.

If you do the calculation for a general value of n, and find that there is a value of n for which your calculation wasn't defined (at ANY step) then you have to go back and do it again with that specific value of n to check what it is supposed to be.
 
Ah I get it! Now I got the correct answer if i do the integral. Thank you very much
 

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