- #1

TranscendArcu

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## Homework Statement

## The Attempt at a Solution

So after I have properly extended the series, I want to find the A_n coefficients. In the case of A_0 I get 1/2 -- good, great. That's what the book gets.

Now for the general A_n:

[itex]A_n = \frac{1}{2} \int^2_{-2} (1)cos(\frac{n\pi x}{2}) dx [/itex]

because, in the piecewise function, the portions from -2 to -1 and from 1 to 2 are zero, we can just write this as

[itex]A_n = \frac{1}{2} \int^1_{-1} (1)cos(\frac{n\pi x}{2}) dx [/itex]

[itex]= \frac{1}{2}(\frac{2}{n\pi} sin(\frac{n\pi x}{2}) |^1_{-1} = \frac{2}{n\pi} sin(\frac{n\pi}{2}) n ≥ 1[/itex]

and because integers n's will just cause the sin term to alternate between -1 and 1:

[itex] = \frac{(2)(-1)^{n+1}}{n\pi} n≥1[/itex], and this is my general term for the coefficients. But this isn't what the book gets, so what have I done wrong?