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$$\delta(n) - \frac{1}{2} \mbox{sinc} \left(\frac{n}{2}\right) = (-1)^n \frac{1}{2} \mbox{sinc} \left( \frac{n}{2} \right)$$

I tried to split it into two sequences: one for n even and one for n odd. I also tried to make use of the fact that $sin(-\frac{n\pi}{2}) = (-1)^nsin(\frac{n\pi}{2}), \forall n \in \mathbb Z$

But so far I haven't managed to tie everything together. Here is a photo showing what I've tried so far. enter image description here

M.Bore
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  • Hi, please, could you read https://math.stackexchange.com/help/how-to-answer. Then what have you tried ? – EDX May 05 '20 at 11:10
  • @EDX thank you for that. I have added a bit more context of what I have tried to do so far, to no avail. – M.Bore May 05 '20 at 11:15
  • And about which $\delta$ do you refer ? Is it zero centered delta ? – EDX May 05 '20 at 11:24
  • @EDX yes that is the one I mean – M.Bore May 05 '20 at 11:28
  • So you equality is an equality of distribution – EDX May 05 '20 at 11:29
  • @EDX please check the photo I just uploaded, maybe you can provide some guidance from there – M.Bore May 05 '20 at 11:41
  • Ok, I need to things to help you. The first is the rigorous definition of the $\delta$ you're using (there are several way of using a $\delta$ function, sometimes mathematicaly, sometimes with a more "physicists" approach (less rigourous). Furthermore there exists several notations used. So I need precision on this .Secondly the last equality isn't really true since $\dfrac{\sin(\frac{n\pi}{2})}{\frac{n\pi}{2}} = sinc(\frac{n\pi}{2}) \neq sinc(\frac{n}{2} )$ – EDX May 05 '20 at 11:48
  • I am using the engineering delta,I don't know if that helps. $ \delta (t) = 1 when t = 0 and \delta (t) = 0 when t \neq 0 $ – M.Bore May 05 '20 at 11:59
  • So when $n \neq 0$ you can avoid delta. (Did you forget some $\pi$ ? – EDX May 05 '20 at 12:05
  • $ \delta(n) - \frac{1}{2} sinc (\frac{n}{2}) = (-1)^n \frac{1}{2} sinc( \frac{n}{2}) $ Given that $ \sin{\frac{-n\pi}{2}} = (-1)^n \sin{\frac{n\pi}{2}} $ \ $ \delta(n) - \frac{\frac{1}{2}\sin{\frac{\pi n}{2}} }{\pi \frac{n}{2}} $ = $ \frac{\frac{1}{2}\sin{\frac{-\pi n}{2}}}{\pi \frac{n}{2}} $ = $ \frac{1}{2}(-1)^n \frac{\sin{\frac{\pi n}{2}}}{\frac{\pi n}{2}} $ \ = $ \frac{1}{2}(-1)^n \sin{\frac{n}{2}} $ \QED – M.Bore May 05 '20 at 12:31
  • The result is true (assuming that $sinc$ is defined here by $\sin(\pi x)/(\pi x)$ and not $\sin(x)/x$ as usual. Its why I wasn't understanding your process. – EDX May 05 '20 at 16:26

1 Answers1

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$ \delta(n) - \frac{1}{2} sinc (\frac{n}{2}) = (-1)^n \frac{1}{2} sinc( \frac{n}{2}) $

Given that

$ \sin{\frac{-n\pi}{2}} = (-1)^n \sin{\frac{n\pi}{2}} $

$ \delta(n) - \frac{\frac{1}{2}\sin{\frac{\pi n}{2}} }{\pi \frac{n}{2}} $ =

$ \frac{\frac{1}{2}\sin{\frac{-\pi n}{2}}}{\pi \frac{n}{2}} $

$ \frac{1}{2}(-1)^n \frac{\sin{\frac{\pi n}{2}}}{\frac{\pi n}{2}} $

= $ \frac{1}{2}(-1)^n \sin{\frac{n}{2}} $

QED

M.Bore
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