# What is the form of the spectral derivative in the all-positive-frequency notation in DFT?

The Discrete Fourier Transform (DFT) of a function $$u:[0,2\pi] \to \mathbb R$$ sampled over $$N$$ equidistant points $$\theta_j = 2\pi j/N,\, j = 0, \dots, N-1,$$ is defined by

$$\tilde U_k = \frac1N \sum_{j=0}^{N-1} u_j e^{-ik\theta_j}\,, \qquad \text{where } k \in [k]_N \in \frac{\mathbb Z}{N\mathbb Z}\,.$$

The indices $$k$$ could go from $$-N/2+1$$ to $$N/2$$ in one convention and from $$0$$ to $$N-1$$ in another. Since $$\tilde U_{k + N} = \tilde U_k\,, \forall k \in \mathbb Z$$, the negative frequencies can be pushed all the way over to the other side of the positive frequencies to get the all-positive convention. This means that the negative frequencies $$k \in \{ -1, -2, \dots, -N/2 + 1\}$$ in one convention map to the frequencies $$k + N \in \{N-1, N-2, \dots, N/2 + 1\}$$ in the other convention.

We know that the derivative of the function $$u$$ can be calculated as follows:

$$u'(\theta) = \sum_{k=-N/2 + 1}^{N/2} ik\, \tilde U_k e^{ik\theta}\,.$$

This would mean that in the all-positive convention,

$$u'(\theta) = \sum_{k=0}^{N/2} ik\, \tilde U_k e^{ik\theta} + \sum_{k = N/2 + 1}^{N-1} i(k-N)\, \tilde U_k e^{ik\theta} \,.$$

Is that correct? If yes, why is it that the spectral derivative requires the use of negative frequencies? What goes wrong in the claim that

$$u'(\theta) = \sum_{k=0}^{N-1} ik\, \tilde U_k e^{ik\theta}\,?$$