Is the fundamental basis function of the DFT not an intregral number of cycles?

Question

When the DFT is defined and described I typically see the basis functions described as an integral number of sine and cosine waves.

But the basis functions are $\sin(2\pi k (n/N))$ or $\cos(2\pi k(n/N))$ where $k$ is the harmonic, $N$ is the number of samples, $n$ is the time domain sample number.

So the right hand boundary of this argument is $2\pi k(N-1)/N$, which seems to be always a bit short of a full cycle.

Could someone please confirm or explain my error?

The picture below shows a sine basis function, apparently ending short of a full cycle.

Answer 1

First, a small correction:
The basis functions of all Fourier transforms are of the the type $e^{2j\pi k\frac nN}$, which can also be interpreted as $\cos(2\pi k \frac nN) + i\cdot\sin(2\pi k \frac nN)$, but never only real-values sines or cosines; that'd lead to large problems on the algebraic structures around. But that really doesn't matter here:

You're absolutely right, the "completing" sample of the oscillation is always missing – but that's absolutely logical, if you think about it – when you periodically continue the signal you're transforming, you'd get twice the same value (the "start" value) after each other, and that mustn't happen.

Your picture is an excellent illustration of this:

The oscillation is observed for exactly two periods, and each period is 16 samples long; that means that for the $n$th sample, the $(n+16)$th sample must have the same value. For example, the $n=16$th has the same value as the $(16+16)=32$nd sample – which, by principle of 32-periodicity is the same as the $0$th sample.