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But if that’s really the case, she’s within hours
Unlike other problems, such as those that exist after
This happens when I write on my laptop during the day.
And if you have one of them, you can use them to
It’s actually very easy if you know how.
In this case, please tell me what should I do after

Tally ERP user manual

The CDC’s FFT method is similar to that of the Parseval’s theorem, which only uses the square of the Fourier coefficients:
How do these two expressions differ in each case?

A:

The Fourier coefficients are the complex coefficients of the Fourier series decomposition for $f$ and its even and odd parts.

For the first formula, use the Parseval’s identity $\sum_{ -N}^N a_n \overline{a_n} = \sum_{ -\infty}^\infty |a_n|^2$. The Fourier series of $f$ can be written as $\sum_k a_k \mathrm{e}^{\mathrm{i} k \omega}$ where $a_k = \frac{2}{\pi} \int_{ -\pi}^{\pi} f \mathrm{e}^{ -\mathrm{i} k \omega} \mathrm{d} \omega$.
The last expression is the Fourier transform of the square of $f$. For this to make sense, we need some additional conditions on $f$. For instance, you have $f \in L^2$.

For the second formula, use the Parseval’s identity again $\sum_{ -N}^N a_n \overline{a_n} = \sum_{ -\infty}^\infty |a_n|^2$. The Fourier series of $f$ can be written as $\sum_k b_k \mathrm{e}^{\mathrm{i} k \omega}$ where $b_k = \frac{2}{\pi} \int_{ -\pi}^{\pi} f \mathrm{e}^{ -\mathrm{i} k \omega} \mathrm{d} \omega$.
Now, we square $\sum_k b_k \mathrm{e}^{\mathrm{i} k \omega}$. The Parseval’s identity tells us that this is equal to the square of the $L^2$ inner product of the functions $f$ and $1$: \$\sum_k b_k \overline{b_k} = \sum_{ -\infty}^\
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