Talk:Summation of i from 1 to n of Summation of j from 1 to i

I had to stop and think about this result for a few moments, so I just wanted to write up some intuition for this result as two different orders of summing the lower triangle of a matrix.

We interpret $a_{i,j}$ as elements of a matrix and consider the expression $$\sum^n_{i = 1} \sum^i_{j = 1} a_{i,j}$$

The outside summation is summing over the rows of the matrix, and the inner summation is summing over the first $i$ elements of row $i$.

Graphically (where steps are iterations of $i$):

# -> #  -> #   -> #
     ##    ##     ##
           ###    ###
                  ####

Then consider $$\sum^n_{j = 1} \sum^n_{i = j} a_{i,j}$$

The outside summation is summing over the columns of the matrix, and the inner summation is summing over the elements of column $j$ starting from row $i$.

Graphically (where steps are iterations of $j$):

# -> #  -> #   -> #
#    ##    ##     ##
#    ##    ###    ###
#    ##    ###    ####

Having considering the row-wise and column-wise iteration methods for summing, one can reinforce this graphical idea with a diagonal-wise approach:

Graphically (where steps are iterations of diagonal length $d$):

                  #
           #      ##
     #     ##     ###
# -> ## -> ### -> ####

Corresponding to: $$\sum^n_{d = 1} \sum^d_{k = 1} a_{n - d + k, k}$$

Apologies is this is unwanted prose --Ybab321 (talk) 08:49, 16 May 2018 (EDT)

We haven't really taken on board the idea of providing "intuitive understanding" pages on $\mathsf{Pr} \infty \mathsf{fWiki}$, but I can see how they may be useful. --prime mover (talk) 11:50, 16 May 2018 (EDT)