I'm trying to understand the change of variable for this double summation:
$$\sum_{n=0}^{\infty} \sum_{m=0}^{n} \xi(m,n)=\sum_{\ell=0}^{\infty} \sum_{m=0}^{[\frac{\ell}{2}]} \xi(m,n=\ell-m)$$
where $\ell=m+n$ and $[\frac{\ell}{2}]$ means summing up to the largest integer that is less or equal than $\frac{\ell}{2}.$ So based on this question Double summation, index change clarification. I was trying to use Iverson bracket notation but I'm not sure of my deduction.
What I have so far is
$$\begin{align}\sum_{n=0}^{\infty} \sum_{m=0}^{n} \xi(m,n) &= \sum_{n,m} [0\leq n<\infty]\cdot [0\leq m \leq n] \xi(m,n) \\ &= \sum_{n,m} [0\leq m \leq n < \infty] \xi(m,n) \end{align}$$
From here I could write
$$\sum_{n,m} [0\leq m <\infty]\cdot [m\leq n < \infty] \xi(m,n) = \sum_{m=0}^{\infty} \sum_{n=m}^{\infty} \xi(m,n)$$
So to get my result I'm not sure if I can also write
$$\begin{align}[0\leq m \leq n < \infty] &\rightarrow [0\leq 2m \leq n + m < \infty] \\ &\rightarrow [0\leq 2m \leq \ell < \infty]\cdot [\ell=m+n] \\ &\rightarrow [0\leq m \leq \frac{\ell}{2}]\cdot[0 \leq \ell < \infty]\cdot [\ell=m+n]\end{align}$$
What also confuses me is if in the last line I should write the first bracket as either $2m\leq\ell$ or $2m < \infty$ to separate the conditions.
Best Answer
The work with Iverson brackets looks (besides notational aspects) fine. Note that in the derivation we not only have to show implications $\rightarrow$ but equivalence relations of the index regions we want to sum up.
Here is a derivation based upon index transformations without using Iverson brackets.
Comment:
In (1) we write the index region somewhat more conveniently.
In (2) we do an equivalence transformation of the inequality chain by adding $m$ as preparation for the next step.
In (3) we introduce $l$ and substitute $l=m+n$.
In (4) we rewrite the index region, so that summing over $l$ becomes the outer series.