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I'm trying to find the Fourier coefficients of $ \sin(4 \pi t) $

I thought I knew how to do it, working backwards with Euler's formula, but when I check my answer I'm off by a negative.

I said that $a_1 = a^* _{-1} = -\frac{1}{2} i$

because $ -\frac{1}{2} i e ^{i (4 \pi) t}$

gives $-\frac{1}{2} \cos(4 \pi t) - \frac{1}{2} (i)^2\sin(4 \pi t) + \frac{1}{2}\cos(4 \pi t) -\frac{1}{2} (i)^2 \sin(4 \pi t)$

which simplifies to $- (i)^2 \sin(4 \pi t)$

which is just $\sin(4 \pi t)$

The answer in the book however says that it should be just $a_1 = a^* _{-1} = \frac{1}{2} i$ with no negative. I'm confused as to why this negative gets dropped, is there something I'm missing from the formula?

Thanks!

Bobson Dugnutt
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Indigo
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    It turns out, after checking with the professor, that this is an error in the textbook and the correct answer does include the negative. – Indigo Mar 18 '16 at 20:38

1 Answers1

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It depends on which interval you are. If we are on $[0,1]$ then the ONB is $(e^{2k\pi it})_{k\in\mathbb Z}$. Since $$ \sin(4\pi t) = \frac{e^{4\pi it}-e^{-4\pi it}}{2i} = \frac 1 {2i}e^{2\cdot 2\pi it} + \frac{-1}{2i}e^{2\cdot(-2)k\pi it}, $$ we have that the Fourier coefficients are $c_k = 0$ for all $k\in\mathbb Z$, except $c_2 = 1/(2i) = -i/2$ and $c_{-2} = i/2$.

Friedrich Philipp
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