Sum notation - lower/upper limits

[JFo]

This is a simple problem which I'm having trouble finding an answer.
What would \sum_{n = 0}^{-1} 1 be?
Would this be undefined? 0? 2? or ?
The reason this came up in the first place is that I was trying to prove that the convolution sum is commutative, that is h*x = x*h.
I started with h*x
\sum_{n = - \infty}^{infty} h(n-m)x(m)
making the substitution k = n-m, i get
\sum_{k = \infty}^{- \infty} x(k-m)h(k)
The problem I have is witht the upper/lower limits of the sum. Does this mean the sum "decrements" through values of k?

 

[JFo]

It won't let me edit my original post, so I have to repost to make corrections.

This is a simple problem which I'm having trouble finding an answer.
What would
\\ \sum_{n = 0}^{-1} 1 \\
be?
Would this be undefined? 0? 2? or ?

The reason this came up in the first place is that I was trying to prove that the convolution sum is commutative, that is h*x = x*h.
I started with h*x
\\ \sum_{n = - \infty}^{\infty} h(n-m)x(m) \\
making the substitution k = n-m, i get
\\ \sum_{k = \infty}^{- \infty} x(k-m)h(k) \\
The problem I have is witht the upper/lower limits of the sum. Does this mean the sum "decrements" through values of k? How do I get this to be

\\ \sum_{k =- \infty}^{\infty} x(k-m)h(k) \\

 

[Hurkyl]

The usual way of assigning meaning to \sum_{n=0}^{-1} always assigns it the value of zero, whatever the summand. Also, \sum_{n=0}^{-2} f(n) = -f(-1).

However, this isn't quite what you want. You're not doing this sort of "directed summation"¹... you're merely summing over a set of integers, and the bounds are just a convenient way to write that set. Always write them with the right orientation unless you know what you're doing.

Now, if you really want to deal with the analogy to integrals, then your sum has an orientation which we usually don't bother writing. When you made the substitution k = n-m, then your bounds should flip, but the orientation also gets reversed, which contributes a negative sign. You can consume that sign to flip the bounds again. (it might help to work through the same substitution with an ordinary integral)


1: quotes because this is not a standard term.

 

[JFo]

Thanks Hurkyl for the informative reply. That is a very interesting about the conventions you mentioned involving sums. I wish to learn more about them, is there a site, or book I can read?

By the way, I noticed several errors in my post, but you seemed to understand what I was getting at anyway. Just for reference to other readers, I have added corrections below.

.
This is a simple problem which I'm having trouble finding an answer.
What would
\\ \sum_{n = 0}^{-1} 1 \\
be?
Would this be undefined? 0? 2? or ?
The reason this came up in the first place is that I was trying to prove that the convolution sum is commutative, that is h*x = x*h.
I started with h*x
\\ \sum_{n = - \infty}^{\infty} h(m-n)x(n) \\ \
making the substitution k = m-n, i get
\\ \sum_{k = \infty}^{- \infty} x(k-m)h(k) \\ \
The problem I have is witht the upper/lower limits of the sum. Does this mean the sum "decrements" through values of k? How do I get this to be
\\ \sum_{k =- \infty}^{\infty} x(m-k)h(k) \\

 

[JFo]

errr... it should be x(m-k) in the second to last summand as well. I wish the edit button was working.