Topic: Review of summations

## Question

Simplify this summation

$\sum_{n=-\infty}^\infty n \delta [n]$

The delta function is zero everywhere except when n=0 and since we are multiplying the delta by n the answer would thus be 0.

Instructor's comment: Yes, that's the idea. Now can you justify your answer "in math" instead of "in words"? -pm

The answer is zero since impulse function is 0 everywhere except n = 0.

Instructor's comments: Ok, I guess I am going to have to be a bit more specific. I would like to see a way to answer this question as a sequence of small changes to this expression until you get to zero. Something like
\begin{align} \sum_{n=-\infty}^\infty n \delta [n] &= \text{ blah } \\ &= \text{ blih} \\ &= 0 \end{align}
Can you try that? -pm

\begin{align} \sum_{n=-\infty}^\infty n \delta [n] &= ...(-2)\delta[-2] + (-1)\delta[-1] + 0\delta + \delta + 2\delta ... \\ &= ...(-2)\cdot 0 + (-1)\cdot 0 + 0\cdot 1 + 0 + 2\cdot 0... \\ &= ...0 + 0 + 0 + 0 + 0... \\ &= 0 \end{align}
Instructor's comments: It's awesome to see Purdue alums participate in this collective problem solving! As expected from a graduate of our ECE program, this solution does not contain any mistake. Now a challenge: can anybody do it without using any "dot dot dot"? -pm

Unless you want a deeper proof of the discrete sifting property:

\begin{align} \sum_{n=-\infty}^\infty f[n] \delta [k-n] &= f[k] \Rightarrow \sum_{n=-\infty}^\infty f[n] \delta [k-n] \Bigg|_{f[n]=n, \ k=0} = \sum_{n=-\infty}^\infty n \delta [n] = 0 \end{align}
Instructor's comments: You correctly observed that the sifting property applies in this case. Good job! Your answer is correct, but perhaps it would flow better (more logically) if you changed the other of the first two inequalities:
\begin{align} \sum_{n=-\infty}^\infty f[n] \delta [k-n] &= f[k] \text{ (sifting property)}\\ \Rightarrow\sum_{n=-\infty}^\infty n \delta [n] &= \sum_{n=-\infty}^\infty f[n] \delta [k-n] \Bigg|_{f[n] =n, \ k=0} &= 0 \end{align}

Instructor's comment: Can anybody write a solution based on this little "trick" I have used several times in class already? -pm

We know that $x[n]\delta [n-k] = x[k]\delta [n-k]$

Then: using $x[n]=n$ and $k=0$ we get: $\sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0$

Instructor's comments: I think this solution is nice and sweet. It does use the "trick" I showed in class. To get full credit, it would be sufficient to write:
$\sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0$
-pm

$\sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0$ because everywhere else besides when n=0, $\delta[n] = 0$

$\delta[n] = 1$ only when n = 0.

Since $\delta[n] = 0$ when n is not equal to 0,
$\sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0$

When $n\neq0$, $\delta[n] = 0$

Also when n = 0, $\delta[n] = 1$ so...
$\sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0$

$\delta[n] = 0$ everywhere except when n = 0. Therefore, at n = 0, the value is 0.
$\sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0$

when n=0, $n\delta[n]=0$ when $n\neq0$, $\delta[n]=0, n\delta[n]=0$ therefore $\sum_{n=-\infty}^\infty n \delta [n]$ for all n

$\sum_{n=-\infty}^\infty n \delta [n] = n \cdot 1 \Bigg|_{n =0} = 0$

Since the range is from $-\infty$ to $\infty$,the range is totally symmetrical.When we get an $n\delta[n]$,we will get an $-n\delta[-n]$ with the same amplitude but reverse direction, to which,the sum will be zero. And when n=0,the value of $n\delta[n]$ is zero. Thus the summation in the question is zero.

TA's comments: Are you arguing that $n\delta [n]$ is an odd function of n so that the summation over the range of $(-\infty,\ \infty)$ equals zero? -pm

$\sum_{n=-\infty}^\infty n \delta [n] = 0 \delta  = 0$

Evaluate:

$\sum_{n=-\infty}^\infty n \delta [n]$

Given:

$\delta [n] = 0 \; \forall n \neq 0$

Therefore:

$\sum_{n=-\infty}^\infty n \delta [n] = 0\delta  = 0$

## Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy Francisco Blanco-Silva