A detached exercise in my course asks us to be able to simplify the following summation, where j is an imaginary number. The current material involves lossless and lossy compression, however I am unable to figure out how to make the jump from the starting summation:
$X(k) = \sum_{n=0}^{N-1} e^{-j2\pi k n / N}$
The professors provided solution involves separating the n exponent:
$X(k) = \sum_{n=0}^{N-1} (e^{-j2\pi k / N})^n$
And then proceeded to substitute for r, where:
$r = \begin{cases} r = N &\text{if} \mod(N, k) = 0 \\ r=0 &\text{otherwise} \end{cases}$
Giving us:
$X(k) = \sum_{n=0}^{N-1} r^n$
To the final result of:
$X(k) =\delta(mod(k, N))$
I am confused specifically on how they came to the conclusion of subbing $e^{-j2\pi k / N}$ for r, and how they defined r.
There is somewhere a mistake in the writeup
For instance
$r = \begin{cases} r = N &\text{if} \mod(N, k) = 0 \\ r=0 &\text{otherwise} \end{cases}$
does not really make any sense. You cant define $r$ in terms of $r$. I have no idea what is intended here, but I am very sure there is the error.
I think the intended derivation utilizes the geometric sum. This is
$\sum_{n=0}^{N-1} x^n = \begin{cases} N & x = 1 \\ \frac{x^N-1}{x-1} & x \neq 1 \end{cases}$.
Now if you plug in $x=e^{−j2πk/N}$ one will find that
$x^N = (e^{−j2πk/N})^N = e^{−j2πk} = 1$
(so numerator $x^N-1=0$) and for the $x=1$ condition we get
$e^{−j2πk/N} = 1 \Leftrightarrow \text{mod}(k,N) = 0$ (i.e. $k$ being an integer multiple of $N$)
Putting everything together one arrives at
$\sum_{n=0}^{N-1} e^{−j2πnk/N} = \begin{cases} N & \text{mod}(k,N) = 0 \\ 0 & \text{otherwise} \end{cases} = N \delta( \text{mod}(k,N) )$
which is amost your end result. A factor $N$ is missing though and in the above "definition" of $r$ you have swapped the $N$ and $k$ argument to mod, which indicates there are multiple additional inaccuracies in the writeup (in addition to the messed up definition of $r$).