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First, the discrete Fourier transform(DFT) of a sequences $x=\{x_n\}_{n=1}^{N-1}$ is defined by $$ \mathcal{F}_k(x)= \sum_{n=0}^{N-1} x_ne^{-\frac{2\pi ikn}{N} } $$ for $k=0,1,..., N-1$. I have studied about Fourier transform, not DFT, which is defined by $$ \mathcal{F}(f)(\xi) = \hat{f}(\xi) = \int_{\mathbb{R}^n} f(x)e^{-2\pi i x\cdot\xi}\,dx, $$ (we suppose this Fourier transform is well-defined regardless of where $f$ is in any classes.)

My question is here. Roughly speaking, is DFT discrete version of Fourier transform? If it is, why is dividing by $N$ in $e^{-\frac{2\pi ikn}{N} }$ considered? I think that the $x_n, k, n$ and $\sum$ are correspond to $f, \xi,x$ and $\int$, respectively. But, what $N$ do? I need any help.

Idkwhat
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  • have a look at the questions which appear on the "related" tab. They may have an answer to your question. https://math.stackexchange.com/questions/189667/using-the-discrete-fourier-transform-to-approximate-the-regular-fourier-transfor?rq=1 https://math.stackexchange.com/questions/1585060/sampling-fourier-transform-and-discrete-fourier-transform?rq=1 – kodlu Oct 21 '22 at 16:01
  • Yeah, I understand you. Event if I have a strong background in maths and understand well FF, it still took me a month to understand the difference between DFT and FF last year. Now I forget all, so I cant help sorry. However, your question will help me recall what I forget about DFT – NN2 Oct 21 '22 at 16:02
  • kodlu's comment was very helpful. Simply, splitting domain of $f$ into $N$ pieces (each mapped point is said $x_k$ for $k=0,1,...,N-1$) occurs $1/N$. – Idkwhat Oct 21 '22 at 17:48

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