Wavetable and Nyquist: which size do I need? I was reading this tutorial by the master earlevel, while trying to build some sort of wavetable.
He says First, let’s back up and figure out how long our tables need to be. Recalling that we need to sample a signal at greater than twice the highest frequency component, that means that for 368 harmonics, we need at least 368 x 2 + 1 samples, or a table length of at least 737 samples., refering to a 40hz (lowest wavetable frequency) and the following harmonics (up to 14720 hz).
Not really sure about those 737 samples keeping the sampling theorem by Nyquist.
Should be 40 * 368 * 2 + 1 samples needed? That's the real &quot;highest frequency component&quot; of the signal.
Why 368 * 2 + 1?
Not really sure why I just need a 737 samples to express a signal that reach ~15khz.
The wavetable is ignorant or agnostic about the playback rate and the sample rate in Hz.  What the wavetable knows about is how many points or samples define the waveform, $N$.  All it knows is the amplitude and phase of every harmonic (which are two numbers for each harmonic), and with a little bit of nuance regarding the DC component and, if the number of wavetable samples is even, the Nyquist component.
There is no phase property of the DC component, just a bipolar amplitude, which is fully described by one real number.
Because of aliasing, the Nyquist component must be a cosine component with no sine component to it at all.
So if $N$ is odd, you can fully define the DC component and both the amplitude and phase of $\frac{N-1}{2}$ harmonics, all below Nyquist.
If $N$ is even, you can fully define the DC component, the amplitude and phase of $\frac{N}{2}-1$ harmonics below Nyquist, and only the bipolar amplitude of the harmonic at Nyquist.
In all cases, $N$ real numbets fully define the wavetable and the DFT maps these $N$ numbers to the harmonic amplitude and phase data which are also fully defined by $N$ real numbers.