Parseval's theorem claims:
$$\sum_{n=-\infty}^\infty \left| \hat{f}(n) \right|^2 = \|f\|^2$$
Isn't the absolute value redundant, because of the square?
2026-04-12 00:00:14.1775952014
Question about Parseval's theorem
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No, because $\hat{f}(n) \in \mathbb{C}$ . Therefore $\left(\hat{f}(n)\right)^2 \neq \left|\hat{f}(n)\right|^2 $