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I have to compute the coefficient $b_3$ of the odd Fourier Series associated with the function $y=2-x$ in the interval $(0,1)$, period $2$. By using the formula $$ b_k = \frac{1}{T}\int_{-T}^{T} f(t)\sin\left(\frac{k\pi t}{T}\right)dt $$ I get $$ b_3 = \frac{1}{2}\int_{-2}^{2} (2-x)\sin\left(\frac{2\pi 3}{2}\right)dt = -\frac{4}{3\pi} $$ but the text of the exercise tells me it should be 2/$\pi$. Is my answer wrong?

Ludovico L
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    The answer in the text can't be right, because the Fourier coefficient is a number, not a function of $x$. However, I'm a little confused by your formulas. I'm a bit rusty, so I won't say that the first formula is incorrect, but if you are taking the function $y=2-x$ on the interval $(0,1)$ and extending it to a function of period $2$ somehow, you shouldn't be using $f(x)=2-x$, but rather a piecewise defined function. – Aaron Jan 29 '16 at 11:08
  • Sorry, I meant to write $2/\pi$ – Ludovico L Jan 29 '16 at 11:13
  • @Aaron also, apparently the coefficient for $b_2$ is $2/\pi$, so that might actually be the problem – Ludovico L Jan 29 '16 at 11:17
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    Ahh. Well, regardless,if I'm understanding the situation correctly, I think $T=1$, not $2$ (becuase you are going from $-T$ to $T$, so your period is $2T$. Additionally, I think $f(t)=-2-x$ on $(-1,0)$ and $2-x$ on $(0,1)$. Otherwise, I am misinterpreting the problem and formulas. – Aaron Jan 29 '16 at 11:21

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Thanks to Aaron up there for pointing me to the right direction.

Being the period $2$, the formula becomes $$ \frac{1}{1}\int_{-1}^1 (2-t)\sin(3\pi t)dt = 2\int_0^1(2-t)\sin(3\pi t)dt = \frac{2}{\pi} $$

Ludovico L
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