Currently, I'm faced with this problem:
Prove that this property holds when $N_s$ is an even integer, for any integers $n$, $k'$, $k$: $\sum_{i=0}^{N_s - 1} e^{-j2\pi n(k'-k)/N_s} = N_s\delta(k'-k)$
I initially saw this problem as one that could be solved by proof by induction. For example, the base case:
Let $N_s = 2$
$$\sum_{i=0}^{2 - 1} e^{-j2\pi n(k'-k)/N_s} = 2\delta(k'-k)$$ $$e^{-j2\pi (0)(k'-k)/N_s} + e^{-j2\pi (1)(k'-k)/N_s}= 2\delta(k'-k)$$ $$1 + e^{-j2\pi (0)(k'-k)/N_s} = 2\delta(k'-k)$$ Case 1: Let $k'-k = 0$: $$1+e^{0} = 2\delta(0)$$ $$2 = 2$$ Case 2: Let $\delta(k'- k) = 1$ $$1+e^{-j\pi} = 2\delta(1)$$ $$1 + cos(-\pi) + jsin(-\pi)) = 2\delta(1)$$ $$1 + (-1 + 0) = 2\delta(1)$$ $$0 = 0$$
However, when it comes to the induction step, I haven't gotten the proof to actually work out. In particular, what I did was Let $N_s = 2m$ for some $m$ in the Integers, and try and prove $N_s = 2(m+1)$. But based on the base case you might be able to spot a pattern that proved problematic in my own proof. For example in the first case, there will always be $1 + (2m-1)(1) = (2m+2)\delta(0)$. Which is certainly not equivalent, and makes me think that induction isn't the best course of action for a proof in this case, and I am way overthinking it. I'm thinking that some usage of Euler's formula in a direct proof might be a better route, but I believe I should be close on the logic with two different cases for k' and k.
Would someone be able to help me out with this proof? Thanks in advanced!