# How do I prove that delta - sinc function is the same as an (-1)^n times the sinc

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

The picture shows what I've tried

• I think your picture is wrong. $\delta(n)$ is only non-zero for $n=0$ – Hilmar May 5 '20 at 13:05

Hint : Use the fact that, $$\sin(-\frac{n\pi}{2}) = (-1)^n \sin(\frac{n\pi}{2}), \forall n \in \mathbb Z$$

• I am still failing to prove it using your suggestion. Please check my picture. – M.Bore May 5 '20 at 11:39
• In your picture , in the end you are getting right result. What is the problem? I could not understand. Can you please give a little more detail. – DSP Rookie May 5 '20 at 11:41
• I am sceptical because the graphs don't quite show it – M.Bore May 5 '20 at 11:45
• $\delta(n) - \frac{1}{2} sinc (\frac{n}{2}) = (-1)^n \frac{1}{2} sinc( \frac{n}{2})$ \\Given that $\sin{\frac{-n\pi}{2}} = (-1)^n \sin{\frac{n\pi}{2}}$ \\ $\delta(n) - \frac{\frac{1}{2}\sin{\frac{\pi n}{2}} }{\pi \frac{n}{2}}$ = $\frac{\frac{1}{2}\sin{\frac{-\pi n}{2}}}{\pi \frac{n}{2}}$ = $\frac{1}{2}(-1)^n \frac{\sin{\frac{\pi n}{2}}}{\frac{\pi n}{2}}$ \\ = $\frac{1}{2}(-1)^n \sin{\frac{n}{2}}$ can you confirm if that is true – M.Bore May 5 '20 at 12:33
• yes this is absolutely right. – DSP Rookie May 5 '20 at 12:39

You wrote "show" rather than "prove". Here's how you "show" your equality using MATLAB:

n = -7: 1 : 7;

delta = zeros(1, length(n));

delta (8) = 1

Left = delta-sinc(n/2)/2;

Right = (-1).^n.*sinc(n/2)/2;

figure(1), clf, hold on

plot(n, Left, '-bo', 'markersize', 6)

plot(n, Right, ':rd', 'markersize', 9); grid on, zoom on

title('Blue squares = Left side, Red dots = Right side'); xlabel('n');

• true! I corrected my question. Thank you for showing it, I'm still struggling to prove it though. – M.Bore May 5 '20 at 11:00

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

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

Hint: split it into two sequences: one for $$n$$ even and one for $$n$$ odd