dft - Why is X(0) the DC component

Why exactly is X(0) the DC component of a signal?

How is it equal to N times x(n)'s average value and why it is at X(0)?

Answer

Follows from the DFT definition. It's defined as

$$\begin{equation} X(k) = \sum_{n=0}^{N-1} x(n) e^{-j2\pi \frac{kn}{N}} \end{equation}$$

So $X(0)$ is

$$\begin{equation} X(0) = \sum_{n=0}^{N-1} x(n) e^{-j2\pi \frac{0 \cdot n}{N}} \end{equation}$$

Having $k=0$ gives $e^0=1$ all the time so that

$$\begin{equation} X(0) = \sum_{n=0}^{N-1} x(n) 1 \end{equation}$$

Comparing this to the average

$$\begin{equation} \overline{x} = \frac{1}{N} \sum_{n=0}^{N-1} x(n) \end{equation}$$

shows that $X(0) = N \overline{x}$