# Multiplying shifts of Kronecker Delta

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

• I think there is some context missing. Also note that Kronecker delta is a function of two variables: see for example en.wikipedia.org/wiki/Kronecker_delta – Andrei Nov 5 '17 at 18:52
• With $\delta_n$ I assume you mean $\delta_{n0}$ ? (i.e. $\delta_0=1$ and $\delta_n = 0$ if $n\not = 0$) – Winther Nov 5 '17 at 19:21

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