Does anyone know why this is true? The sum from n = -infinity to infinity of delta(t-n)/2? Not sure if LaTeX works on this so I'll write it out below.

Question

anonymous · Answer

$$f(t) = \sum_{n =-\infty}^{\infty} \Delta(t-n)/2$$

anonymous · Answer

Sorry here is the equation I need explained: 

$$f(t) = \sum_{n = -\infty}^{\infty} (t-n)/2 = 1$$

anonymous · Answer

Sorry again :/

$$f(t) =\sum_{n = -\infty}^{\infty} \Delta(t-n)/2 = 1$$

experimentx · Answer

what's that delta

anonymous · Answer

I think it just means "change in", like a differential. So my interpretation of this is the infinite sum of changes in the function (t-n)/2... I just don't understand how that all adds up to one though.

anonymous · Answer

eh http://www.wolframalpha.com/input/?i=sum_%7Bn%3D-infinity%7D%5Einfinity+DifferenceDelta%28%28t-n%29%2F2%2Ct%29