3

I have the following sum:

$$\sum_{i=0}^n\sum_{j=0}^{n-i}a_{i,j}r^is^jt^{n-i-j}u^{2n-2i-j},$$

which I'd like to get into the form

$$\sum_{k=0}^{2n}\lambda_ku^k.$$

In the case $n=3$ I have formed the following analysis of the indices table:

$$\begin{array}{c|c|c} i&j&6-2i-j\\ \hline 0&0&6\\ 0&1&5\\ 0&2&\textbf{4}\\ 0&3&\textbf{3}\\ \hline 1&0&\textbf{4}\\ 1&1&\textbf{3}\\ 1&2&\textit{2}\\ \hline 2&0&\textit{2}\\ 2&1&1\\ \hline 3&0&0 \end{array}$$

The bold and italic terms could clearly be grouped together, so obviously yes it can be written in the form described, but is there a nice formula for any $n$ ?

pshmath0
  • 10,565
  • 1
    Note that you can reverse the summation order of first and second sum, this gives you $u^{i+j} $ Then the rest should be simple regrouping. – SK19 Dec 03 '17 at 23:09
  • Thank you. I think I got it. Will post my update tomorrow. Zzz time. – pshmath0 Dec 03 '17 at 23:42

1 Answers1

1

Given S.Koch's comment above I was able to do the following, which I should've been able to do anyway as I used to use these sums quite a bit but a long time ago. We have:

$$\sum_{i=0}^n\sum_{j=0}^{n-i}a_{i,j}r^is^jt^{n-i-j}u^{2n-2i-j},$$ changing the indexing in the inner sum slightly gives $$\sum_{i=0}^n\sum_{j=i}^na_{i,j-i}r^is^{j-i}t^{n-j}u^{2n-(i+j)}.$$ Now, taking a look at the index set for $n=3$ we see the following: $$\begin{array}{l}i\\\hline0,1,2,3,4,5,6\\1,2,3,4,5,6\\2,3,4,5,6\\3,4,5,6\\4,5,6\\5,6\\6\end{array},$$ with $j$ running down the rows. We can traverse these instead over the diagonals such that $i+j=k$ with $k=0,1,2,3,4,5,6$, hence $$\sum_{k=0}^{2n}\left(\sum_{i+j=k}a_{i,j-i}r^is^{j-i}t^{n-j}\right)u^{2n-k}.$$ Expanding the index of the inner sum, $$\sum_{k=0}^{2n}\left(\sum_{i=0}^ka_{i,k-2i}r^is^{k-2i}t^{n-k+i}\right)u^{2n-k},$$ then substituting $k$ with $2n-k$ finally gives $$\sum_{k=0}^{2n}\left(\sum_{i=0}^{2n-k}a_{i,2n-k-2i}r^is^{2n-k-2i}t^{i+k-n}\right)u^k,$$ as required.

pshmath0
  • 10,565