# Is there a way we can compute my sum involving subsets more efficiently?

Suppose we have a countably infinite $$A$$ and $$F_1,F_2,\cdot\cdot\cdot$$ are an infinite sequence of finite sets (denoted $$\left\{F_n\right\}_{n=1}^{\infty}$$) such that $$\bigcup\limits_{n=1}^{\infty}F_n=A$$, $$F_1\subset F_2\subset \cdot\cdot\cdot$$ and $$F_t$$ is a structure of $$A$$ where $$t$$ is between one and infinity.

## Core Question

I want to come up with a code that computes the following for specific $$F_t$$ and find the $$F_t$$ that maximizes the following.

$$d=\sum\limits_{X\in \mathcal{P}(\Delta F_t/||F_t||)\setminus\emptyset}\left(\prod\limits_{x\in X}x \right)$$

• Where $$\Delta{F_t}$$ is a * multiset * that arranges the elements in $$F_t$$ from least to greatest and takes the difference between consecutive pairs.

For example, if $$F_3=\left\{1,2,4,3\right\}$$ then arrange the elements from least to greatest

$$\left\{1,2,3,4\right\}$$

Then get $$\Delta F_3=\left\{2-1,3-2,4-3\right\}=\left\{1,1,1\right\}$$

• $$||F_t||=\sum\limits_{x\in \Delta F_t}x$$ normalizes the elements in $$\Delta F_t$$ so the sum of all the elements is $$1$$ (similar to a probability distribution.)

• $$\Delta F_t/||F_t||$$ divides every element in $$\Delta F_t$$ by $$||F_t||$$.

• $$\mathcal{P}(\Delta F_t/||F_t||)$$ takes all subsets of $$\Delta F_t/||F_t||$$ in a way that repeated elements are distinctive.

So for our example where $$F_3=\left\{1,2,3\right\}$$, $$||F_3||=1+1+1=3$$, $$\Delta F_3/||F_3||=\left\{1/3,1/3,1/3\right\}$$,

$$\mathcal{P}\left(\Delta F_3/||F_3||\right)\setminus \emptyset=\left\{\left\{1/3\right\},\left\{1/3\right\},\left\{1/3\right\},\left\{1/3,1/3\right\},\left\{1/3,1/3\right\},\left\{1/3,1/3\right\},\left\{1/3,1/3,1/3\right\}\right\}$$

hence

$$\sum\limits_{X\in \mathcal{P}(\Delta F_3/||F_3||)\setminus\emptyset}\left(\prod\limits_{x\in X}x\right)=\prod\limits_{x\in \left\{1/3\right\}} x+\prod\limits_{x\in\left\{1/3\right\}}x+\prod\limits_{x\in\left\{1/3\right\}}x+\prod\limits_{x\in\left\{1/3,1/3\right\}}x+\prod\limits_{x\in\left\{1/3,1/3\right\}}x+\prod\limits_{x\in\left\{1/3,1/3\right\}}+\prod\limits_{x\in\left\{1/3,1/3,1/3\right\}}x= 1/3+1/3+1/3+1/9+1/9+1/9+1/27=37/27$$

## Example

If $$A=\left\{\frac{1}{n}:n\in\mathbb{N}\right\}$$ and $$F_t=\left\{\frac{1}{n}:n\in\mathbb{N},n\le t\right\}$$, I did the following to compute $$d$$:

F[t_] := F[t] =
Sort[DeleteDuplicates[
Select[Flatten[Table[1/m, {m, 1, t}]], Between[#, {0, 1}] &]]]
F1[t_] := F1[t] = Total[Differences[F[t]]]
F2[t_] := F2[t] = Sort[Differences[F[t]]/F1[t]]
d1[t_] := d1[t] = Subsets[F2[t], {1, Length[F2[t]]}];
d[t_] := d[t] =
N[Total[Table[Times @@ d1[t][[s]], {s, 1, Length[d1[t]]}]]]


Once t goes beyond $$20$$ it takes more time to compute and even if it converges it converges extremely slow.

Is there a way we can compute my example of $$d$$ more efficiently? Is there a way we can generalize this for any $$F_t$$?

Edit: As a user pointed out, generalizing for any $$F_t$$ is impossible. However, I wish to compute $$d$$, where $$A=\left\{\frac{1}{n}:n\in\mathbb{N}\right\}$$, for the following $$F_t$$:

1. $$F_t=\left\{\frac{1}{n}:n\in\mathbb{N},n\le t\right\}$$

2. $$F_t=\left\{{1}/{\left[2^t/m\right]}:m\in\mathbb{N}, m\le 2^t \right\}$$ (where $$[\cdot]$$ is the nearest integer function)

I hope 1. and 2. will give different results for $$d$$.

I also wish to calculate $$d$$, where $$A=\mathbb{Q}\cap[0,1]$$, for the following:

1. $$F_t=\left\{\frac{m}{n!}: m,n\in\mathbb{N}, m\le n!\le t\right\}$$

2. $$F_t=\left\{\frac{m}{n}:m,n\in\mathbb{N},m\le n\le t\right\}$$

I also hope, in this case, that 1. and 2. give different results with $$d$$.

• Is the empty set not a member of the power set in your notation? Sep 22 at 14:58
• No but I should have used the set difference notation. Sep 22 at 15:44

Just a cheap observation because I did not follow the whole notation, but maybe you can use the fact that for any set $$S$$ (with distinct elements) $$\sum_{X \in \mathcal{P}(S)} (\prod_{x\in X} x) = \prod_{x\in S} (1+x)$$ holds.
• Since $\Delta F_t/||F_t||$ is a multi-set, this equation may not work. Is there a way we can manipulate this? Sep 22 at 14:53