## Homework Statement

I must calculate the Fourier Series of

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

## 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!!

Office_Shredder
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