I understand that when we compute a DFT, there are $N$ data points and the indices $n$ and $m$ go from zero to $N-1$ where $y_m$ are the data points in the time domain, and $Y_n$ are the amplitudes in the frequency domain.
My first question is more straightforward. Why do the indices count from zero rather than from one? If this is true, it seems to me that $Y_0$ is the amplitude of a zero Hz signal, which I cannot make sense of.
My second question probably reflects an even greater lack of understanding, but: If we are able to obtain the amplitudes for $N-1$ frequency components, and the Nyquist Frequency is half the sampling rate, do we not in practice take $0\le m\le N/2$ for $N$ even, and $0\le m\le N\pm 1$ for $N$ odd? Furthermore is this the reason that $Y_{N/2-n}=\overline{Y_{N/2+n}}$?
Also if it looks like I am missing any key concepts please let me know.
Thanks
Edit: (3 years later) Since encountering the material in a few undergraduate Physics courses, the concept is a lot clearer to me now and I'm not even remotely as confused about it.
Best Answer
To your first question: Yes, $Y_0$ is the amplitude of the component with zero frequency, which is a constant: $Y_0 \mathrm{e}^{2\pi\mathrm{i}m0/N}=Y_0 \mathrm{e}^0=Y_0$. You can see that you need a component like this because all the non-constant components have an average of $0$, so $Y_0$ is the only component that can give the sum a non-zero average, and in fact it simply measures the average of your signal.
To your second question: I'm not sure I understand it correctly. First, we don't obtain amplitudes for $N-1$ frequency components, but for $N$ frequency components -- this might be related to your first question. Second, why do you write $N/2$ for $N$ even but $N\pm1$ for $N$ odd? Did you mean $(N\pm1)/2$? The reason that $Y_{N/2-n}=\overline{Y_{N/2+n}}$ is that $\mathrm{e}^{2\pi\mathrm{i}m(N/2-n)/N}=\mathrm{e}^{2\pi\mathrm{i}m(N/2-n)/N}\mathrm{e}^{-2\pi\mathrm{i}m}=\mathrm{e}^{2\pi\mathrm{i}m(-N/2-n)/N}=\overline{\mathrm{e}^{2\pi\mathrm{i}m(N/2+n)/N}}$.