Eden Solomon

2022-06-19

I wish to solve exactly this formula involving sums and products

$\u27e8{n}_{l}\u27e9={[1+\sum _{k\ne l}\left({e}^{bN(l-k)}\frac{\prod _{j\ne l}(1-{e}^{b(l-j)})}{\prod _{j\ne k}(1-{e}^{b(k-j)})}\right)]}^{-1}(N+\sum _{h\ne l}\frac{1}{1-{e}^{b(h-l)}})-\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\sum _{h\ne l}\left\{{[1+\sum _{k\ne h}\left({e}^{bN(h-k)}\frac{\prod _{j\ne h}(1-{e}^{b(h-j)})}{\prod _{j\ne k}(1-{e}^{b(k-j)})}\right)]}^{-1}\left(\frac{1}{1-{e}^{b(l-h)}}\right)\right\}$

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...

$\u27e8{n}_{l}\u27e9={[1+\sum _{k\ne l}\left({e}^{bN(l-k)}\frac{\prod _{j\ne l}(1-{e}^{b(l-j)})}{\prod _{j\ne k}(1-{e}^{b(k-j)})}\right)]}^{-1}(N+\sum _{h\ne l}\frac{1}{1-{e}^{b(h-l)}})-\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\sum _{h\ne l}\left\{{[1+\sum _{k\ne h}\left({e}^{bN(h-k)}\frac{\prod _{j\ne h}(1-{e}^{b(h-j)})}{\prod _{j\ne k}(1-{e}^{b(k-j)})}\right)]}^{-1}\left(\frac{1}{1-{e}^{b(l-h)}}\right)\right\}$

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

Beginner2022-06-20Added 17 answers

Step 1

If I understand correctly your need, you are searching for a compact representation of the term $\u27e8{n}_{l}\u27e9$.

A possible way to deal with the product terms is to use log. Consider the Toeplitz matrix B with elements ${B}_{ij}=1-{e}^{b(i-j)}$ and compute the vector $\mathbf{v}=\mathrm{log}(\mathbf{B}+{\mathbf{I}}_{M}){\mathbf{1}}_{M}$.

Step 2

You will discover that ${p}_{k}=\prod _{j\ne k}(1-{e}^{b(k-j)})$ 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 ${p}_{k}={\mathbf{e}}_{k}^{T}\mathrm{log}(\mathbf{B}+{\mathbf{I}}_{M}){\mathbf{1}}_{M}$

If I understand correctly your need, you are searching for a compact representation of the term $\u27e8{n}_{l}\u27e9$.

A possible way to deal with the product terms is to use log. Consider the Toeplitz matrix B with elements ${B}_{ij}=1-{e}^{b(i-j)}$ and compute the vector $\mathbf{v}=\mathrm{log}(\mathbf{B}+{\mathbf{I}}_{M}){\mathbf{1}}_{M}$.

Step 2

You will discover that ${p}_{k}=\prod _{j\ne k}(1-{e}^{b(k-j)})$ 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 ${p}_{k}={\mathbf{e}}_{k}^{T}\mathrm{log}(\mathbf{B}+{\mathbf{I}}_{M}){\mathbf{1}}_{M}$