Multiplying shifts of Kronecker Delta

Question

Can someone please explain to me these two equations?

$$\delta_{n}\delta_{n-2k}=\delta_{n}$$ and why $$\delta_{n-1}\delta_{n-2k}=0$$ and why $$\delta_{n}\delta_{n-2k-1}=0$$

Any help would be appreciated! Thanks guys

Answer 1

This answer presumes $$\delta_n=\left\{\begin{matrix}1 & \text{if $n=0$,}\\0 & \text{otherwise.}\end{matrix}\right. \tag 1$$ Using $(1)$, it is easy to verify that $\delta_{n}\delta_{m}=1 \implies n=m=0$. Consequently, the last equations are correct, but the first equation does not hold except for $k=0$.

However, $$\sum_{k=-\infty}^{\infty} \delta_{n}\delta_{n-2k} = \delta_n(\cdots+\delta_{n-2}+\delta_n+\delta_{n+2} + \cdots)=\delta_n. \tag 2$$ Also, $$\sum_{k=-\infty}^{\infty} \delta_{n-1}\delta_{n-2k} = \delta_{n-1}(\cdots+\delta_{n-2}+\delta_n+\delta_{n+2} + \cdots)=0, \tag 3$$ and $$\sum_{k=-\infty}^{\infty} \delta_{n}\delta_{n-2k-1} = \delta_{n}(\cdots+\delta_{n-1}+\delta_{n+1}+\delta_{n+3} + \cdots)=0. \tag 4$$