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I would like to change the order so that I know the coefficients of each $x^i$.
How do I change the order of the summation in the following sum?

$$ \sum_{k=0}^m\sum_{j=0}^n\alpha_k\beta_j\sum_{s=0}^{\min(j,k)}c_s(j,k)x^{k+j-2s} $$

The constants $c_s(j,k)$ depends on j,k and s.

Thanks for your time.

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can you tell us anything more? are the constants positive? do we know any sizes of $\alpha_k$, $\beta_j$, or $c_s(j,k)$? – robjohn Dec 1 '11 at 16:10
The constants $c_s(j,k)$ are positive. The constant $\alpha_k$ and $\beta_j$ can be, but generally are not. They can all be any size. Does that matter when we want to swap finite sums? – alext87 Dec 1 '11 at 16:16
What order would you like to obtain? – Brian M. Scott Dec 1 '11 at 17:59
I would like to know the general formula for the coordinates in the basis 1,$x$,$x^2$,...,$x^{m+n}$. – alext87 Dec 1 '11 at 18:21
up vote 2 down vote accepted

It’s convenient to define $\alpha_i=0$ and $\beta_i=0$ when $i$ is out of bounds.

Let $a_i$ be the coefficient of $x^i$. The only solutions to $k+j-2s=0$ with $s\le\min(j,k)$ are $k=j=s$, so $$a_0=\sum_{i\ge 0}\alpha_i\beta_ic_i(i,i)\;.$$

If $k+j-2s=1$, either $k=j+1$ and $s=j$, or $j=k+1$ and $s=k$, so

$$a_1=\sum_{i\ge 0}\bigg(\alpha_i\beta_{i+1}c_i(i+1,i)+\alpha_{i+1}\beta_ic_i(i,i+1)\bigg)\;.$$

More generally, if $k+j-2s=d$, the only possibilities are

$$\begin{align*}k&=s,j=s+d\;;\\ k&=s+1,j=s+d-1\;;\\ k&=s+2,j=s+d-2\;;\\ &\;\vdots\\ k&=s+d,j=s\;, \end{align*}$$


$$a_d=\sum_{s\ge 0}\sum_{i=0}^d\alpha_{s+i}\beta_{s+d-i}c_s(s+d-i,s+i)$$

As a quick check, this yields

$$a_{m+n}=\sum_{s\ge 0}\sum_{i=0}^{m+n}\alpha_{s+i}\beta_{s+m+n-i}c_s(s+m+n-i,s+i)\;,$$

and it’s not hard to see that the only non-zero term is $\alpha_m\beta_nc_0(n,m)$, when $s=0,i=m$, as it should be: the indices of $\alpha$ and $\beta$ sum to $s+m+n$, so if $s>0$, at least one of $\alpha_{s+i}$ and $\beta_{s+m+n-i}$ is zero, and the same is true if $s=0$ and $i\ne m$.

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Thanks! That's excellent! – alext87 Dec 2 '11 at 8:50

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