Mixing
Closed form expression for summation of P(n, k)
$begingroup$
I was wondering if there is a closed form expression for the following:
begin{equation}
sumlimits_{k=1}^d P(n,k) cdot n^{n-k}
end{equation}
This expression came up while analyzing an algorithm. The analysis is rather lengthy, but if asked, I can provide that. Thanks for your help!
combinatorics permutations
asked Jan 18 at 22:34
Novice GeekNovice Geek
3811
$endgroup$
|
show 3 more comments
$begingroup$
I was wondering if there is a closed form expression for the following:
begin{equation}
sumlimits_{k=1}^d P(n,k) cdot n^{n-k}
end{equation}
This expression came up while analyzing an algorithm. The analysis is rather lengthy, but if asked, I can provide that. Thanks for your help!
combinatorics permutations
asked Jan 18 at 22:34
Novice GeekNovice Geek
3811
$endgroup$
1
$begingroup$
Note that you can write $$ sum_{k=1}^d frac{n!}{k!} n^{n-k} = n! n^n sum_{k=1}^d frac{(1/n)^{k}}{k!}. $$ The latter sum has no closed form I believe, but it approaches $e^{1/n}-1$ relatively quickly as $d$ grows larger.
$endgroup$
– MisterRiemann
Jan 18 at 22:38
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@MisterRiemann See wolframalpha.com/input/…! for a closed form of the sum in your comment.
$endgroup$
– Noble Mushtak
Jan 18 at 22:40
$begingroup$
@NobleMushtak Note the factorial in the denominator.
$endgroup$
– MisterRiemann
Jan 18 at 22:40
2
$begingroup$
@MisterRiemann The comment formatted improperly, which left out the exclamation point in the link. Try this link. Basically, it says $$sum_{i=1}^n frac{(1/n)^k}{k!}=frac{e^{1/n} Gamma(d+1,1/n)}{Gamma(d+1)}-1$$
$endgroup$
– Noble Mushtak
Jan 18 at 22:41
2
$begingroup$
@MisterRiemann You have $P(n,k)=n!/k!$ in your first comment, I think it should be $n!/(n-k)!$.
$endgroup$
– Mike Earnest
Jan 18 at 23:49
|
show 3 more comments
$begingroup$
I was wondering if there is a closed form expression for the following:
begin{equation}
sumlimits_{k=1}^d P(n,k) cdot n^{n-k}
end{equation}
This expression came up while analyzing an algorithm. The analysis is rather lengthy, but if asked, I can provide that. Thanks for your help!
combinatorics permutations
asked Jan 18 at 22:34
Novice GeekNovice Geek
3811
$endgroup$
I was wondering if there is a closed form expression for the following:
begin{equation}
sumlimits_{k=1}^d P(n,k) cdot n^{n-k}
end{equation}
This expression came up while analyzing an algorithm. The analysis is rather lengthy, but if asked, I can provide that. Thanks for your help!
combinatorics permutations
combinatorics permutations
asked Jan 18 at 22:34
Novice GeekNovice Geek
3811
asked Jan 18 at 22:34
Novice GeekNovice Geek
3811
asked Jan 18 at 22:34
Novice GeekNovice Geek
3811
asked Jan 18 at 22:34
Novice GeekNovice Geek
3811
asked Jan 18 at 22:34
Novice GeekNovice Geek
3811
3811
1
$begingroup$
Note that you can write $$ sum_{k=1}^d frac{n!}{k!} n^{n-k} = n! n^n sum_{k=1}^d frac{(1/n)^{k}}{k!}. $$ The latter sum has no closed form I believe, but it approaches $e^{1/n}-1$ relatively quickly as $d$ grows larger.
$endgroup$
– MisterRiemann
Jan 18 at 22:38
$begingroup$
@MisterRiemann See wolframalpha.com/input/…! for a closed form of the sum in your comment.
$endgroup$
– Noble Mushtak
Jan 18 at 22:40
$begingroup$
@NobleMushtak Note the factorial in the denominator.
$endgroup$
– MisterRiemann
Jan 18 at 22:40
2
$begingroup$
@MisterRiemann The comment formatted improperly, which left out the exclamation point in the link. Try this link. Basically, it says $$sum_{i=1}^n frac{(1/n)^k}{k!}=frac{e^{1/n} Gamma(d+1,1/n)}{Gamma(d+1)}-1$$
$endgroup$
– Noble Mushtak
Jan 18 at 22:41
2
$begingroup$
@MisterRiemann You have $P(n,k)=n!/k!$ in your first comment, I think it should be $n!/(n-k)!$.
$endgroup$
– Mike Earnest
Jan 18 at 23:49
|
show 3 more comments
1
$begingroup$
Note that you can write $$ sum_{k=1}^d frac{n!}{k!} n^{n-k} = n! n^n sum_{k=1}^d frac{(1/n)^{k}}{k!}. $$ The latter sum has no closed form I believe, but it approaches $e^{1/n}-1$ relatively quickly as $d$ grows larger.
$endgroup$
– MisterRiemann
Jan 18 at 22:38
$begingroup$
@MisterRiemann See wolframalpha.com/input/…! for a closed form of the sum in your comment.
$endgroup$
– Noble Mushtak
Jan 18 at 22:40
$begingroup$
@NobleMushtak Note the factorial in the denominator.
$endgroup$
– MisterRiemann
Jan 18 at 22:40
2
$begingroup$
@MisterRiemann The comment formatted improperly, which left out the exclamation point in the link. Try this link. Basically, it says $$sum_{i=1}^n frac{(1/n)^k}{k!}=frac{e^{1/n} Gamma(d+1,1/n)}{Gamma(d+1)}-1$$
$endgroup$
– Noble Mushtak
Jan 18 at 22:41
2
$begingroup$
@MisterRiemann You have $P(n,k)=n!/k!$ in your first comment, I think it should be $n!/(n-k)!$.
$endgroup$
– Mike Earnest
Jan 18 at 23:49
1
1
$begingroup$
Note that you can write $$ sum_{k=1}^d frac{n!}{k!} n^{n-k} = n! n^n sum_{k=1}^d frac{(1/n)^{k}}{k!}. $$ The latter sum has no closed form I believe, but it approaches $e^{1/n}-1$ relatively quickly as $d$ grows larger.
$endgroup$
– MisterRiemann
Jan 18 at 22:38
$begingroup$
Note that you can write $$ sum_{k=1}^d frac{n!}{k!} n^{n-k} = n! n^n sum_{k=1}^d frac{(1/n)^{k}}{k!}. $$ The latter sum has no closed form I believe, but it approaches $e^{1/n}-1$ relatively quickly as $d$ grows larger.
$endgroup$
– MisterRiemann
Jan 18 at 22:38
$begingroup$
@MisterRiemann See wolframalpha.com/input/…! for a closed form of the sum in your comment.
$endgroup$
– Noble Mushtak
Jan 18 at 22:40
$begingroup$
@MisterRiemann See wolframalpha.com/input/…! for a closed form of the sum in your comment.
$endgroup$
– Noble Mushtak
Jan 18 at 22:40
$begingroup$
@NobleMushtak Note the factorial in the denominator.
$endgroup$
– MisterRiemann
Jan 18 at 22:40
$begingroup$
@NobleMushtak Note the factorial in the denominator.
$endgroup$
– MisterRiemann
Jan 18 at 22:40
2
2
$begingroup$
@MisterRiemann The comment formatted improperly, which left out the exclamation point in the link. Try this link. Basically, it says $$sum_{i=1}^n frac{(1/n)^k}{k!}=frac{e^{1/n} Gamma(d+1,1/n)}{Gamma(d+1)}-1$$
$endgroup$
– Noble Mushtak
Jan 18 at 22:41
$begingroup$
@MisterRiemann The comment formatted improperly, which left out the exclamation point in the link. Try this link. Basically, it says $$sum_{i=1}^n frac{(1/n)^k}{k!}=frac{e^{1/n} Gamma(d+1,1/n)}{Gamma(d+1)}-1$$
$endgroup$
– Noble Mushtak
Jan 18 at 22:41
2
2
$begingroup$
@MisterRiemann You have $P(n,k)=n!/k!$ in your first comment, I think it should be $n!/(n-k)!$.
$endgroup$
– Mike Earnest
Jan 18 at 23:49
$begingroup$
@MisterRiemann You have $P(n,k)=n!/k!$ in your first comment, I think it should be $n!/(n-k)!$.
$endgroup$
– Mike Earnest
Jan 18 at 23:49
|
show 3 more comments
1 Answer
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$begingroup$
$$sum_{k=1}^d frac{n! }{(n-k)!}n^{n-k}=e^n Gamma (n+1,n)-e^n n!frac{ Gamma (n-d,n)}{Gamma(n-d)}-n^n$$ where appears the complete and incomplete gamma functions.
answered Jan 19 at 6:19
Claude LeiboviciClaude Leibovici
123k1157134
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$begingroup$
$$sum_{k=1}^d frac{n! }{(n-k)!}n^{n-k}=e^n Gamma (n+1,n)-e^n n!frac{ Gamma (n-d,n)}{Gamma(n-d)}-n^n$$ where appears the complete and incomplete gamma functions.
answered Jan 19 at 6:19
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
$begingroup$
$$sum_{k=1}^d frac{n! }{(n-k)!}n^{n-k}=e^n Gamma (n+1,n)-e^n n!frac{ Gamma (n-d,n)}{Gamma(n-d)}-n^n$$ where appears the complete and incomplete gamma functions.
answered Jan 19 at 6:19
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
$begingroup$
$$sum_{k=1}^d frac{n! }{(n-k)!}n^{n-k}=e^n Gamma (n+1,n)-e^n n!frac{ Gamma (n-d,n)}{Gamma(n-d)}-n^n$$ where appears the complete and incomplete gamma functions.
answered Jan 19 at 6:19
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
$$sum_{k=1}^d frac{n! }{(n-k)!}n^{n-k}=e^n Gamma (n+1,n)-e^n n!frac{ Gamma (n-d,n)}{Gamma(n-d)}-n^n$$ where appears the complete and incomplete gamma functions.
answered Jan 19 at 6:19
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 6:19
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 6:19
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 6:19
Claude LeiboviciClaude Leibovici
123k1157134
123k1157134
add a comment |
add a comment |
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1
$begingroup$
Note that you can write $$ sum_{k=1}^d frac{n!}{k!} n^{n-k} = n! n^n sum_{k=1}^d frac{(1/n)^{k}}{k!}. $$ The latter sum has no closed form I believe, but it approaches $e^{1/n}-1$ relatively quickly as $d$ grows larger.
$endgroup$
– MisterRiemann
Jan 18 at 22:38
$begingroup$
@MisterRiemann See wolframalpha.com/input/…! for a closed form of the sum in your comment.
$endgroup$
– Noble Mushtak
Jan 18 at 22:40
$begingroup$
@NobleMushtak Note the factorial in the denominator.
$endgroup$
– MisterRiemann
Jan 18 at 22:40
2
$begingroup$
@MisterRiemann The comment formatted improperly, which left out the exclamation point in the link. Try this link. Basically, it says $$sum_{i=1}^n frac{(1/n)^k}{k!}=frac{e^{1/n} Gamma(d+1,1/n)}{Gamma(d+1)}-1$$
$endgroup$
– Noble Mushtak
Jan 18 at 22:41
2
$begingroup$
@MisterRiemann You have $P(n,k)=n!/k!$ in your first comment, I think it should be $n!/(n-k)!$.
$endgroup$
– Mike Earnest
Jan 18 at 23:49