Jtdylktuy

There is a set $S={x_1, x_2, ..., x_N}.$

I'm trying to approximate this:
$$p(S)=sum_{rsubset S}|r|!prod_{xin r}x$$

I know that:

$$sum_{rsubset S}prod_{xin r}x=prod_{xin S}(1+x)$$

I was wondering if there is a way to approximate $p(S)$ with something.

An idea:
Change $x$ in $prodlimits_{xin S}(1+x)$ to $a(x)x$ so that:

$$prod_{xin S}(1+a(x)x)simsum_{rsubset S}|r|!prod_{xin r}x$$

Stirling's approximation

There is: $$n! sim (2pi n)^frac{1}{2}(frac{n}{e})^n$$
which for my problem $n^n$ is troubling and I can't fiure out a way for $prod_{xin S}(1+a(x)x)$ to make $n^n$. It could also go up to a power of e:
$$n! sim e^{log(2pi)/2-n+nlog(n)}$$
But again, can't figure out to handle $nlog(n)$.

We can find an easy upper bound for $p(S)$ in the function $prod_{xin S}(1+x^2)$, as $$prod_{xin S}(1+x^2)=sum_{rsubset S}prod_{xin r}x^2=sum_{rsubset S}bigg(prod_{xin r}xbigg)^2=sum_{rsubset S}bigg(prod_{xin r}xbigg)prod_{xin r}xgeqsum_{rsubset S}|r|!prod_{xin r}x=p(S)$$

(the inequality holding as all $x$'s are positive integers)

For a lower bound, you could use the product $prod_{xin S}(1+x)$, as you have mentioned.

Correcting what I wrote in a comment, you could also use $prod_{xin S}(2+x)-R(S)$ where $R(S)$ is a remainder term.

$$p(S)=sum_{rsubset S}|r|!prod_{xin r}x=\sum_{rsubset S,|r|=0}0!prod_{xin r}x+sum_{rsubset S,|r|=1}1!prod_{xin r}x+sum_{rsubset S,|r|=2}2!prod_{xin r}x+sum_{rsubset S,|r|=3}3!prod_{xin r}x+sum_{rsubset S,|r|>3}|r|!prod_{xin r}x=\ 1+sum_{xin S}x+2sum_{x,yin S\xneq y}xy+6sum_{x,y,zin S\xneq yneq z}xyz+sum_{rsubset S,|r|>3}|r|!prod_{xin r}xgeq\ 1+sum_{xin S}x+2sum_{x,yin S\xneq y}xy+6sum_{x,y,zin S\xneq yneq z}xyz+sum_{rsubset S,|r|>3}2^{|r|}prod_{xin r}xgeq\ 1+sum_{xin S}x+2sum_{x,yin S\xneq y}xy+6sum_{x,y,zin S\xneq yneq z}xyz+sum_{rsubset S\|r|>3}sum_{nin 2^{|r|}}prod_{xin n}x=\1+sum_{xin S}x+2sum_{x,yin S\xneq y}xy+6sum_{x,y,zin S\xneq yneq z}xyz+sum_{rsubset S\|r|>3}sum_{nsubset r}prod_{xin n}x=\ 1+sum_{xin S}x+2sum_{x,yin S\xneq y}xy+6sum_{x,y,zin S\xneq yneq z}xyz+sum_{rsubset S\|r|>3}prod_{xin r}(1+x)=\bigg(1+sum_{xin S}x+2sum_{x,yin S\xneq y}xy+6sum_{x,y,zin S\xneq yneq z}xyzbigg)+sum_{rsubset S}prod_{xin r}(1+x)-bigg(1+sum_{xin S}(1+x)+sum_{x,yin S\ xneq y}(1+x)(1+y)+sum_{x,y,zin S\xneq yneq z}(1+x)(1+y)(1+z)bigg)=\sum_{rsubset S}prod_{xin r}(1+x)-|S|-binom{|S|}{2}-binom{|S|}{3}-frac{|S|(|S|-1)}{2}sum_{xin S}x-(|S|-3)sum_{x,yin S\xneq y}xy+\5sum_{x,y,zin S\xneq yneq z}xyz=\sum_{rsubset S}prod_{xin r}(1+x)-R(S)=sum_{tsubset S+1}prod_{yin t}y-R(S)=prod_{yin S+1}(1+y)-R(S)=\prod_{xin S}(2+x)-R(S)$$

(I have used expressions like $S+1$ above to denote the set consisting of elements of $S$ each incremented by $1$)

The upper bound represents setting $a(x)=x$ and the lower bound, setting $a(x)=1+frac{1}{x}$, both, of course, diverge from $p(S)$ vastly for large sizes of $S$.

I can see no way of finding a function $a(x)$ that would satisfy the constraints in your question unless some additional restrictions on $S$ were added.

(please edit or comment for any corrections)