# Changing order of a double sum

1. Feb 8, 2009

### geor

[SOLVED] Changing order of a double sum

Hello everybody,

I am a bit confused here, any help would be greatly appreciated..
I have this double sum:

$$\sum_{i=0}^{n-1}a_i \sum_{j=0}^{i} {i \choose j} b^{i-j}x^j$$

How can I take x out of the inner sum?

Thank you very much in advance...

Last edited: Feb 9, 2009
2. Feb 8, 2009

### adriank

Would it be helpful to use the binomial theorem?

$$\sum_{j=0}^{i} {i \choose j} b^{i-j}x^j = (b + x)^i$$

3. Feb 8, 2009

### geor

Thanks for taking the time to answer!

Well, no, I started from there, I want to write this as a polynomial of x in the usual way, that is, in the form:

a_n*x^n+....+a_1*x+a_0

I want to have only x there...

4. Feb 8, 2009

### geor

It is possible, is it not?!

5. Feb 9, 2009

### adriank

Whoops, I thought I had hit the submit button hours ago, but apparently I didn't.

For changing order of sums, the Iverson bracket
http://xrl.us/befjqx
is a useful tool.

$$\sum_{i=0}^{n-1}a_i \sum_{j=0}^{i} \binom{i}{j} b^{i-j} x^j$$
$$= \sum_{i,j} [0 \le j \le i][0 \le i \le n-1] a_i \binom{i}{j} b^{i-j} x^j$$
$$= \sum_{i,j} [0 \le j \le i \le n-1] a_i \binom{i}{j} b^{i-j} x^j$$
$$= \sum_{j,i} [0 \le j \le n-1][j \le i \le n-1] a_i \binom{i}{j} b^{i-j} x^j$$
$$= \sum_{j=0}^{n-1} x^j \sum_{i=j}^{n-1} a_i \binom{i}{j} b^{i-j}.$$

6. Feb 9, 2009

### geor

Thanks so much for the help!!

7. Feb 9, 2009

### geor

What a nice tool! I was struggling for so much time trying to change that variables!

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