# Double sum of same variable simplification help

1. Mar 6, 2015

### perplexabot

Hello all,
I have been meditating on this for a while, but can't seem to understand how this simplification came to be. Any help will be greatly appreciated.

$\mathop{\sum_{k=0}^m\sum_{l=0}^n}_{m{\geq}n} x(k,l)$
We also know that: l (lower case L) = n-m+k

and here is what my book ends with:
$\sum\limits_{k=m-n}^m x(k,n-m+k)$

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Here is my attempt, using the fact that l = n-m+k and substituting, I get:
$\mathop{\sum_{k=0}^m\sum_{n-m+k=0}^n}_{m{\geq}n} x(k,n-m+k)$

$= \mathop{\sum_{k=0}^m\sum_{k=m-n}^n}_{m{\geq}n} x(k,n-m+k)$

I then have no idea where to go from here : ( I tried substituting numbers for m and n and looked at what was happening, but that didn't help me much. Can someone please provide some insight please!!! I have been at this for a bit.

2. Mar 6, 2015

### Stephen Tashi

If that is true then for a given value of $k$ there is only a single value of $l$. So for a given value of $k$, the expression $\sum_{l=0}^{n} x(k,l)$ doesn't make sense since it implies that for a given $k$ you can vary $l$ over several different values. If we adopt the convention that $\sum _{l=0}^{n} x(k,l)$ shall mean the single term $x(k,n-m+k)$ then you could get the desired result.

When you substitute, you can't use $k$ as the name of the variable whose values are summed by the second summation $\sum_{l=0}^n$ because the name $k$ is already in use as the variable for the first summation $\sum_{k=0}^m$. The first argument of $x(k,l)$ refers to the $k$ that is summed by $\sum_{k=0}^m$.

3. Mar 6, 2015

### perplexabot

Hey, thank you for your reply. I still can't figure this out.

I didn't think of this, but now that you mentioned it, wouldn't $l$ have infinite possible values? If you set $k$ to $a$ then you can have any value for $n$ and $m$ which results in $l$ having many values. What is wrong with this?

I'm not sure what you mean here : ( I'm sorry, but I really am trying... I have to admit, the equation I supplied is modified from the original shown in the book. I will attach a copy of what I am struggling with maybe it will be better. Here it is (I understand everything until where it says "Thus for m ≥ n"):

Hmmm, ok that kind of makes sense. So what then shall we do in such a situation?

Thank you for your help : )

4. Mar 7, 2015

### Stephen Tashi

There's nothing to indicate that $m$ and $n$ are variables. They appear to denote constants.

To make an analogy with calculus, if you are finding $\int ( \int f(x,y) dy) dx$ and you use integration by subsitution to find $\int f(x,y) dy$ then you can do a substitution like $y = 3u^2$, but you can't do the substitution $y = 3x^2$ because the variable $x$ is already in use to denote a quantity that is independent of $y$.

In your example, the variable $k$ is already in use in the first summation. You can't introduce it as meaning a different variable in the second summation since the $k$ in $x(k,l)$ refers to the $k$ used in the first summation.

The correct analogy to what is given in the text is the statement "$x(k,l) = 0$ except when $l = n - m + k$ " (which is different than asserting " $l = n - m + k$ ").

Then $\sum_{l=0}^n x(k,l)$ has only one nonzero term, which is $x(k,n-m+k)$.

Perhaps there is also information in the text that $x(k,n-m+k)$ is zero for $k < n-m$, but I'd have to think about it further to see why.

5. Mar 7, 2015

### perplexabot

Thank you Mr.Tashi. I am still trying to figure out how they got those last two lines. I will report back if I find anything.

Last edited: Mar 7, 2015
6. Mar 7, 2015

### perplexabot

I think I may have understood how they achieved the second to last line. So we know $l = n - m + k$ (else the equation will be zero). And we also know from the second summation that $0 ≤ l ≤ n$. So I did the following to kind of see what was going on:

for $l = 0, l = 0 = n-m+k => k = m-n$ this will make the first summation to be $\sum\limits_{k=m-n}^m x(k,n-m+k)$
for $l = n, l = n = n-m+k => k = m$ this will make the first summation to be $\sum\limits_{k=m}^m x(k,n-m+k)$
for $l = n-1, l = n-1 = n-m+k => k = m-1$ this will make the first summation to be $\sum\limits_{k=m-1}^m x(k,n-m+k)$

So after viewing these summations, I see it is for $l =0$ in which the first summation takes into account all possible values (or has the greatest summation intervals).

What do you think of this?

EDIT1: I think the last line is achieved from the one before by using the fact that $\sum\limits_{k=0}^m x(k) = \sum\limits_{k=0+n}^{m+n} x(k-n)$
EDIT2: My first edit (EDIT1) may be wrong. I am still checking.

Last edited: Mar 7, 2015