 Eden Solomon

2022-06-19

I wish to solve exactly this formula involving sums and products
$⟨{n}_{l}⟩={\left[1+\sum _{k\ne l}\left({e}^{bN\left(l-k\right)}\frac{\prod _{j\ne l}\left(1-{e}^{b\left(l-j\right)}\right)}{\prod _{j\ne k}\left(1-{e}^{b\left(k-j\right)}\right)}\right)\right]}^{-1}\left(N+\sum _{h\ne l}\frac{1}{1-{e}^{b\left(h-l\right)}}\right)-\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\sum _{h\ne l}\left\{{\left[1+\sum _{k\ne h}\left({e}^{bN\left(h-k\right)}\frac{\prod _{j\ne h}\left(1-{e}^{b\left(h-j\right)}\right)}{\prod _{j\ne k}\left(1-{e}^{b\left(k-j\right)}\right)}\right)\right]}^{-1}\left(\frac{1}{1-{e}^{b\left(l-h\right)}}\right)\right\}$
where b is a positive real number, N is a natural number, and all the sums and products are understood to run from 0 to $M-1$, where M is yet another natural number (all the l, h, j, k indices are therefore bound to this interval, with the appropriate restrictions written under the sums and the products). Eventually I would also like to take the limit as M goes to infinity, but I'm sure how well I can do that.
Do you think there is a hope to massage this formula to make it a little less ugly? For the moment, I've been trying to solve it numerically... Ryan Fitzgerald

Step 1
If I understand correctly your need, you are searching for a compact representation of the term $⟨{n}_{l}⟩$.
A possible way to deal with the product terms is to use log. Consider the Toeplitz matrix B with elements ${B}_{ij}=1-{e}^{b\left(i-j\right)}$ and compute the vector $\mathbf{v}=\mathrm{log}\left(\mathbf{B}+{\mathbf{I}}_{M}\right){\mathbf{1}}_{M}$.
Step 2
You will discover that ${p}_{k}=\prod _{j\ne k}\left(1-{e}^{b\left(k-j\right)}\right)$ can be computed as the k-th element of exp v.
Note here log and exp are computed elementwise and log returns complex values in case of negative values. So it holds ${p}_{k}={\mathbf{e}}_{k}^{T}\mathrm{log}\left(\mathbf{B}+{\mathbf{I}}_{M}\right){\mathbf{1}}_{M}$

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