Approximating $sumlimits_{rsubset S}|r|!prodlimits_{xin r}x$












1












$begingroup$


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 $nlog(n)$.




I'm actually trying to do this:



$$frac{sumlimits_{rsubset S}|r|!prodlimits_{xin r}x}{sumlimits_{rsubset S, |r|ge 1}(|r|-1)!prodlimits_{xin r}x}$$



Which in my problems context: $sumlimits_{rsubset S}|r|!prodlimits_{xin r}x=1-sumlimits_{rsubset S, |r|ge 1}(|r|-1)!prodlimits_{xin r}x$
But I don't think there could be any shortcut.






combinatorics discrete-mathematics combinations approximation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are all the $x_i$s (positive) integers?
    $endgroup$
    – Cardioid_Ass_22
    Jan 12 at 17:45










  • $begingroup$
    @Cardioid_Ass_22 yes
    $endgroup$
    – Anais
    Jan 12 at 22:12










  • $begingroup$
    Perhaps this lower bound may be of some use to you $p(S)geq 1+sum_{xin S}x+sum_{rsubset S,|r|>1}2^{|r|}prod_{xin r} xgeq 1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}sum_{nin 2^r}prod_{xin n}x=1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}sum_{nsubset r}prod_{xin n}x=1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}prod_{xin r}(1+x)geq 1+sum_{xin S}x+sum_{rsubset S}prod_{xin r}(1+x)-1-sum_{xin S}(1+x)=sum_{rsubset S}prod_{xin r}(1+x)-|S|=sum_{tsubset S+1}prod_{yin t}y-|S|=prod_{yin S+1}(1+y)-|S|=prod_{xin S}(2+x)-|S|$
    $endgroup$
    – Cardioid_Ass_22
    Jan 12 at 22:23








  • 1




    $begingroup$
    In your last comment, when you have for the term where $r=emptyset$, you have $(-1)!$, could you explain what that is ? Also from since it is combinatorics, I would say that it is just that you should add a condition that $|r|geq 1$ right ?
    $endgroup$
    – P. Quinton
    Jan 13 at 12:37






  • 1




    $begingroup$
    @Anais I'm pretty sure I screwed something, if I can fix it I will undelete it. You didn't get the right result right ?
    $endgroup$
    – P. Quinton
    Jan 13 at 14:53


















1












$begingroup$


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 $nlog(n)$.




I'm actually trying to do this:



$$frac{sumlimits_{rsubset S}|r|!prodlimits_{xin r}x}{sumlimits_{rsubset S, |r|ge 1}(|r|-1)!prodlimits_{xin r}x}$$



Which in my problems context: $sumlimits_{rsubset S}|r|!prodlimits_{xin r}x=1-sumlimits_{rsubset S, |r|ge 1}(|r|-1)!prodlimits_{xin r}x$
But I don't think there could be any shortcut.






combinatorics discrete-mathematics combinations approximation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are all the $x_i$s (positive) integers?
    $endgroup$
    – Cardioid_Ass_22
    Jan 12 at 17:45










  • $begingroup$
    @Cardioid_Ass_22 yes
    $endgroup$
    – Anais
    Jan 12 at 22:12










  • $begingroup$
    Perhaps this lower bound may be of some use to you $p(S)geq 1+sum_{xin S}x+sum_{rsubset S,|r|>1}2^{|r|}prod_{xin r} xgeq 1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}sum_{nin 2^r}prod_{xin n}x=1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}sum_{nsubset r}prod_{xin n}x=1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}prod_{xin r}(1+x)geq 1+sum_{xin S}x+sum_{rsubset S}prod_{xin r}(1+x)-1-sum_{xin S}(1+x)=sum_{rsubset S}prod_{xin r}(1+x)-|S|=sum_{tsubset S+1}prod_{yin t}y-|S|=prod_{yin S+1}(1+y)-|S|=prod_{xin S}(2+x)-|S|$
    $endgroup$
    – Cardioid_Ass_22
    Jan 12 at 22:23








  • 1




    $begingroup$
    In your last comment, when you have for the term where $r=emptyset$, you have $(-1)!$, could you explain what that is ? Also from since it is combinatorics, I would say that it is just that you should add a condition that $|r|geq 1$ right ?
    $endgroup$
    – P. Quinton
    Jan 13 at 12:37






  • 1




    $begingroup$
    @Anais I'm pretty sure I screwed something, if I can fix it I will undelete it. You didn't get the right result right ?
    $endgroup$
    – P. Quinton
    Jan 13 at 14:53
















1












1








1


3



$begingroup$


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 $nlog(n)$.




I'm actually trying to do this:



$$frac{sumlimits_{rsubset S}|r|!prodlimits_{xin r}x}{sumlimits_{rsubset S, |r|ge 1}(|r|-1)!prodlimits_{xin r}x}$$



Which in my problems context: $sumlimits_{rsubset S}|r|!prodlimits_{xin r}x=1-sumlimits_{rsubset S, |r|ge 1}(|r|-1)!prodlimits_{xin r}x$
But I don't think there could be any shortcut.






combinatorics discrete-mathematics combinations approximation







share|cite|improve this question











$endgroup$




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 $nlog(n)$.




I'm actually trying to do this:



$$frac{sumlimits_{rsubset S}|r|!prodlimits_{xin r}x}{sumlimits_{rsubset S, |r|ge 1}(|r|-1)!prodlimits_{xin r}x}$$



Which in my problems context: $sumlimits_{rsubset S}|r|!prodlimits_{xin r}x=1-sumlimits_{rsubset S, |r|ge 1}(|r|-1)!prodlimits_{xin r}x$
But I don't think there could be any shortcut.






combinatorics discrete-mathematics combinations approximation




combinatorics discrete-mathematics combinations approximation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 20:59







Anais

















asked Jan 5 at 8:55









AnaisAnais

336




336








  • 1




    $begingroup$
    Are all the $x_i$s (positive) integers?
    $endgroup$
    – Cardioid_Ass_22
    Jan 12 at 17:45










  • $begingroup$
    @Cardioid_Ass_22 yes
    $endgroup$
    – Anais
    Jan 12 at 22:12










  • $begingroup$
    Perhaps this lower bound may be of some use to you $p(S)geq 1+sum_{xin S}x+sum_{rsubset S,|r|>1}2^{|r|}prod_{xin r} xgeq 1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}sum_{nin 2^r}prod_{xin n}x=1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}sum_{nsubset r}prod_{xin n}x=1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}prod_{xin r}(1+x)geq 1+sum_{xin S}x+sum_{rsubset S}prod_{xin r}(1+x)-1-sum_{xin S}(1+x)=sum_{rsubset S}prod_{xin r}(1+x)-|S|=sum_{tsubset S+1}prod_{yin t}y-|S|=prod_{yin S+1}(1+y)-|S|=prod_{xin S}(2+x)-|S|$
    $endgroup$
    – Cardioid_Ass_22
    Jan 12 at 22:23








  • 1




    $begingroup$
    In your last comment, when you have for the term where $r=emptyset$, you have $(-1)!$, could you explain what that is ? Also from since it is combinatorics, I would say that it is just that you should add a condition that $|r|geq 1$ right ?
    $endgroup$
    – P. Quinton
    Jan 13 at 12:37






  • 1




    $begingroup$
    @Anais I'm pretty sure I screwed something, if I can fix it I will undelete it. You didn't get the right result right ?
    $endgroup$
    – P. Quinton
    Jan 13 at 14:53
















  • 1




    $begingroup$
    Are all the $x_i$s (positive) integers?
    $endgroup$
    – Cardioid_Ass_22
    Jan 12 at 17:45










  • $begingroup$
    @Cardioid_Ass_22 yes
    $endgroup$
    – Anais
    Jan 12 at 22:12










  • $begingroup$
    Perhaps this lower bound may be of some use to you $p(S)geq 1+sum_{xin S}x+sum_{rsubset S,|r|>1}2^{|r|}prod_{xin r} xgeq 1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}sum_{nin 2^r}prod_{xin n}x=1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}sum_{nsubset r}prod_{xin n}x=1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}prod_{xin r}(1+x)geq 1+sum_{xin S}x+sum_{rsubset S}prod_{xin r}(1+x)-1-sum_{xin S}(1+x)=sum_{rsubset S}prod_{xin r}(1+x)-|S|=sum_{tsubset S+1}prod_{yin t}y-|S|=prod_{yin S+1}(1+y)-|S|=prod_{xin S}(2+x)-|S|$
    $endgroup$
    – Cardioid_Ass_22
    Jan 12 at 22:23








  • 1




    $begingroup$
    In your last comment, when you have for the term where $r=emptyset$, you have $(-1)!$, could you explain what that is ? Also from since it is combinatorics, I would say that it is just that you should add a condition that $|r|geq 1$ right ?
    $endgroup$
    – P. Quinton
    Jan 13 at 12:37






  • 1




    $begingroup$
    @Anais I'm pretty sure I screwed something, if I can fix it I will undelete it. You didn't get the right result right ?
    $endgroup$
    – P. Quinton
    Jan 13 at 14:53










1




1




$begingroup$
Are all the $x_i$s (positive) integers?
$endgroup$
– Cardioid_Ass_22
Jan 12 at 17:45




$begingroup$
Are all the $x_i$s (positive) integers?
$endgroup$
– Cardioid_Ass_22
Jan 12 at 17:45












$begingroup$
@Cardioid_Ass_22 yes
$endgroup$
– Anais
Jan 12 at 22:12




$begingroup$
@Cardioid_Ass_22 yes
$endgroup$
– Anais
Jan 12 at 22:12












$begingroup$
Perhaps this lower bound may be of some use to you $p(S)geq 1+sum_{xin S}x+sum_{rsubset S,|r|>1}2^{|r|}prod_{xin r} xgeq 1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}sum_{nin 2^r}prod_{xin n}x=1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}sum_{nsubset r}prod_{xin n}x=1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}prod_{xin r}(1+x)geq 1+sum_{xin S}x+sum_{rsubset S}prod_{xin r}(1+x)-1-sum_{xin S}(1+x)=sum_{rsubset S}prod_{xin r}(1+x)-|S|=sum_{tsubset S+1}prod_{yin t}y-|S|=prod_{yin S+1}(1+y)-|S|=prod_{xin S}(2+x)-|S|$
$endgroup$
– Cardioid_Ass_22
Jan 12 at 22:23






$begingroup$
Perhaps this lower bound may be of some use to you $p(S)geq 1+sum_{xin S}x+sum_{rsubset S,|r|>1}2^{|r|}prod_{xin r} xgeq 1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}sum_{nin 2^r}prod_{xin n}x=1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}sum_{nsubset r}prod_{xin n}x=1+sum_{xin S}{x}+sum_{rsubset S,|r|>1}prod_{xin r}(1+x)geq 1+sum_{xin S}x+sum_{rsubset S}prod_{xin r}(1+x)-1-sum_{xin S}(1+x)=sum_{rsubset S}prod_{xin r}(1+x)-|S|=sum_{tsubset S+1}prod_{yin t}y-|S|=prod_{yin S+1}(1+y)-|S|=prod_{xin S}(2+x)-|S|$
$endgroup$
– Cardioid_Ass_22
Jan 12 at 22:23






1




1




$begingroup$
In your last comment, when you have for the term where $r=emptyset$, you have $(-1)!$, could you explain what that is ? Also from since it is combinatorics, I would say that it is just that you should add a condition that $|r|geq 1$ right ?
$endgroup$
– P. Quinton
Jan 13 at 12:37




$begingroup$
In your last comment, when you have for the term where $r=emptyset$, you have $(-1)!$, could you explain what that is ? Also from since it is combinatorics, I would say that it is just that you should add a condition that $|r|geq 1$ right ?
$endgroup$
– P. Quinton
Jan 13 at 12:37




1




1




$begingroup$
@Anais I'm pretty sure I screwed something, if I can fix it I will undelete it. You didn't get the right result right ?
$endgroup$
– P. Quinton
Jan 13 at 14:53






$begingroup$
@Anais I'm pretty sure I screwed something, if I can fix it I will undelete it. You didn't get the right result right ?
$endgroup$
– P. Quinton
Jan 13 at 14:53












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