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!
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 $c_{-2} = i/2$.