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I am trying to evaluate the following sum

$$S(m) = \sum_{n=0}^{-\infty}\theta(m+n)\theta(-n)$$ Where $m$ and $n$ are both integers and $\theta$ is a Heaviside step function. I am trying to understand whether $S$ has any finite value. Can anyone shed some light on this? Thanks.

felix
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  • Is the $-\infty$ in the top index of your sum supposed to indicate that $n$ values are $0,-1,-2,\ldots$? – BSplitter Jun 28 '21 at 22:32
  • @BSplitter, yes – felix Jun 28 '21 at 22:33
  • Then isn't the step function irrelevant here, since both $m+n$ and $-n$ will always be integer values? Or is the step function non-standard? Like is it the floor function $\theta(x) = \lfloor x \rfloor$? – BSplitter Jun 28 '21 at 22:35
  • @BSplitter, yes indeed. I guess, the sum is something finite. – felix Jun 28 '21 at 22:38
  • If $\theta$ is the Heaviside theta function, $S(0)=\theta(0)\theta(0)$, $S(m)=2\theta(0)+(m-1)$ for $m>0$ and $S=0$ for $m<0$ – Sal Jun 28 '21 at 23:03
  • @Sal, can you please elaborate your answer? – felix Jun 28 '21 at 23:05
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    First change $n \to -n$, the sum is then from $0$ to $+\infty$. Writing out a few terms: $S(m)=\theta(m)\theta(0)+\theta(m-1)\theta(1)+\theta(m-2)\theta(2)+\dots+\theta(m-m)\theta(m)+0 + \dots$. The second theta function returns unity for $n>0$. All terms with $n>m$ vanish. There are $(m-1)$ occurrences of $\theta(\text{positive number})$, and two occurrences (at $n=0$ and $n=m$) of $\theta(0)$. – Sal Jun 28 '21 at 23:14
  • @Sal, Just to know, can we set $\theta(0)=1$? – felix Jun 29 '21 at 15:54
  • $\theta(0)$ is a matter of convention. Choosing $\theta(0)=1$ is certainly a valid definition (as is $1/2$ or even $0$). Of course, once chosen you should be consistent with the definition for the whole calculation – Sal Jun 29 '21 at 16:35
  • @Sal, thank you the clarification. – felix Jun 29 '21 at 16:50

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Since $\theta$ is the Heaviside step function, we know that

$$\theta(x)=\begin{cases} 1, &x>0 \\ \frac{1}{2}, &x=0\\ 0, &x<0 \end{cases}$$

Then we have that

$$\begin{align*} S(m) &= \sum_{n=0}^{-\infty} \theta(m+n)\theta(-n) \\ &= \sum_{n=0}^\infty \theta(m-n)\theta(n)\\ &= \theta(m-0)\underbrace{\theta(0)}_{=1/2} + \theta(m-1)\theta(1) + \cdots + \theta(m-(m-1))\theta(m-1) + \underbrace{\theta(m-m)}_{=1/2}\theta(m) + \underbrace{\theta(m-(m+1))}_{=0}\theta(m+1) + \cdots\\ &= \frac{1}{2} + \underbrace{1 + 1 + \cdots + 1}_{m-1 \text{ times}} + \frac{1}{2} + 0 \\ &= \frac{1}{2} + m-1+\frac{1}{2}\\ &=m \end{align*}$$

BSplitter
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