# variable substitution in multiple summation

I have a question about variable substitution in summation and I don't know the answer. Didn't find the answer by searching, thought of asking it here.

Assume a polynomial matrix $P(\alpha)$ is written as:

$P(\alpha) = \sum_{k=0}^{g} P_k (\sum_{j=0}^{k} c(j,k) \alpha^j)$

where $c(j,k)$ and $P_k$ are coefficients depending on both indices $j,k$ while $\alpha$ is the monomial. Now in order to simplify the equation and write it explicitly with $\alpha$ being the monomial, I need to write it in the form of:

$P(\alpha) = \sum_{j} \alpha^j (\sum_{k} c(j,k) P_k)$.

So, now my question is:

how can I do this? and what are the lower and upper value of the new $k,j$?

Thanks

Cheers, Keivan

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Yes, you can do this just like you've written it. Both $j$ and $k$ run from $0$ to $g$, with the restriction $j\le k$, which now becomes $k\ge j$, so

$$P(\alpha)=\sum_{j=0}^g\alpha^j\left(\sum_{k=j}^gc(j,k)P_k\right)\;.$$

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Thanks for your reply. I already figured what the answer is. I just don't know how to come up with it! – Keivan Feb 5 '13 at 15:22

I figured it out. What Joriki has written is completely correct and here is how you can derive it. If you increase the indices values you can write

$k=0 \rightarrow$ $0 \quad$

$k=1 \rightarrow$ $0 \quad 1$

$k=2 \rightarrow$ $0 \quad 1 \quad 2$

$k=3 \rightarrow$ $0 \quad 1 \quad 2 \quad 3$

$\vdots \quad \quad \quad\quad\quad\quad\quad\quad\quad \ddots$

$k=g \rightarrow$ $0 \quad 1 \quad 2 \quad 3 \quad \dots \quad g$

where the columns are $j$.

If you look at the triangle from the bottom,

$j=0 \rightarrow k=0:g$

$j=1 \rightarrow k=1:g$

$j=2 \rightarrow k=2:g$

which is:

$$\sum_{j=0}^{g} \sum_{k=j}^{g}$$

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