# Thread: limits in double sum

1. ## limits in double sum

I have this notation:

$\sum_{a=0}^3\sum_{b=0}^aax^{a+b}$

As I want $x$ in the notation to be raised to the power of only one variable, can I set $m=a+b$ ?

What have I to change in the limits of double sum?

I think the first one must be $\sum_{m=0}^6$ -I am not sure- but the other???

And would the notation become inside the sum $(m-b)x^m$?

2. ## Re: limits in double sum

Originally Posted by grafan
I have this notation:
$\sum_{a=0}^3\sum_{b=0}^aax^{a+b}$
$\sum_{a=0}^3\sum_{b=0}^aax^{a+b}=\sum\limits_{b = 0}^1 {1x^{1 + b} } + \sum\limits_{b = 0}^2 {2x^{2 + b} } + \sum\limits_{b = 0}^3 {3x^{3 + b} }$

3. ## Re: limits in double sum

Originally Posted by grafan
I have this notation:

$\sum_{a=0}^3\sum_{b=0}^aax^{a+b}$

As I want $x$ in the notation to be raised to the power of only one variable, can I set $m=a+b$ ?

What have I to change in the limits of double sum?

I think the first one must be $\sum_{m=0}^6$ -I am not sure- but the other???

And would the notation become inside the sum $(m-b)x^m$?
The possible values of (a,b), with the corresponding values of a+b, are:

$\begin{array}{c|cccccccccc} (a,b)&(0,0)&(1,0)&(1,1)&(2,0)&(2,1)&(2,2)&(3,0)&(3 ,1)&(3,2)&(3,3)\\\hline a+b&0&1&2&2&3&4&3&4&5&6\end{array}.$
You can see from this that a+b takes each of the values 0, 1, 5, 6 once only, and each of the values 2, 3, 4 twice. That should be enough to make you realise that taking a+b as one of the summation variables is probably not a good idea.

The best thing to do here is probably to write out the whole double sum explicitly. After all, there are only ten terms altogether, one of which is zero. The other nine are as given by Plato.

4. ## Re: limits in double sum

My problem is that I have to solve a number of notations, and my final goal is to find the factors of $x$, not the result of the notation. So I always want the power only with one variable.

A lot of my notations have $x$ raised to the power of more variables (3 or 4) or I have more summations. I posted one of the simplest form to understand the logic.

So, any idea about the changes - and the logic - in the limits of double sum if I swap $m$ with $a+b$?

5. ## Re: limits in double sum

Originally Posted by grafan
My problem is that I have to solve a number of notations, and my final goal is to find the factors of $x$, not the result of the notation. So I always want the power only with one variable.
A lot of my notations have $x$ raised to the power of more variables (3 or 4) or I have more summations. I posted one of the simplest form to understand the logic.
So, any idea about the changes - and the logic - in the limits of double sum if I swap $m$ with $a+b$?
Frankly, I do not understand how you want us to help.
In particular, what does "in the limits of double sum if I swap $m$ with $a+b$" mean?

It seems to be about the use of the notation.
\begin{align*} \sum\limits_{b = 0}^1 {1x^{1 + b} } &= x+x^2 \\ \sum\limits_{b = 0}^2 {2x^{2 + b} }&=2x^2+2x^3+2x^4 \\ \sum\limits_{b = 0}^3 {3x^{3 + b} }&= 3x^3+3x^4+3x^5+3x^6 \end{align*}

So we get $\sum_{a=0}^3\sum_{b=0}^aax^{a+b}=x+3x^2+5x^3+5x^4+ 3x^5+3x^6$

6. ## Re: limits in double sum

Originally Posted by Plato
Frankly, I do not understand how you want us to help.
In particular, what does "in the limits of double sum if I swap $m$ with $a+b$" mean?

It seems to be about the use of the notation.
\begin{align*} \sum\limits_{b = 0}^1 {1x^{1 + b} } &= x+x^2 \\ \sum\limits_{b = 0}^2 {2x^{2 + b} }&=2x^2+2x^3+2x^4 \\ \sum\limits_{b = 0}^3 {3x^{3 + b} }&= 3x^3+3x^4+3x^5+3x^6 \end{align*}
So we get $\sum_{a=0}^3\sum_{b=0}^aax^{a+b}=x+3x^2+5x^3+5x^4+ 3x^5+3x^6$
Sorry if I become annoying.
I am looking for a general formula to find the factors of $x$. This is a simply notation.
Imagine that I have $x^{a+b+c+d}$ inside and the summations are five!
I am trying to understand the logic because I am confused about the changes to the limits of the summations.

If anyone could show me a general type for this notation, I would be grateful.
Ηowever, thank you very much for the time you spent.