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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

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

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