Mixing
Prove the identity with Stirling numbers.
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
For arbitrary integer $m, t, s<n$ prove the identity
$$
sum_i=s+1^n(i!)^2 left n atop iright left m atop iright binomsibinomti=0,
$$
here $displaystyle left n atop iright$ is the Stirling numbers of the second kind.
I come across by accident to the identity as secondary result.
I have checked it for many values, it is true. What about a proof in the general case?
combinatorics summation binomial-coefficients stirling-numbers
edited Jul 16 at 16:46
Tiwa Aina
2,576319
asked Jul 16 at 16:40
Leox
5,0921323
add a comment |Â
up vote
1
down vote
favorite
For arbitrary integer $m, t, s<n$ prove the identity
$$
sum_i=s+1^n(i!)^2 left n atop iright left m atop iright binomsibinomti=0,
$$
here $displaystyle left n atop iright$ is the Stirling numbers of the second kind.
I come across by accident to the identity as secondary result.
I have checked it for many values, it is true. What about a proof in the general case?
combinatorics summation binomial-coefficients stirling-numbers
edited Jul 16 at 16:46
Tiwa Aina
2,576319
asked Jul 16 at 16:40
Leox
5,0921323
1
Note that $binomsi=0$ if $i>s$.
– Markus Scheuer
Jul 16 at 16:45
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For arbitrary integer $m, t, s<n$ prove the identity
$$
sum_i=s+1^n(i!)^2 left n atop iright left m atop iright binomsibinomti=0,
$$
here $displaystyle left n atop iright$ is the Stirling numbers of the second kind.
I come across by accident to the identity as secondary result.
I have checked it for many values, it is true. What about a proof in the general case?
combinatorics summation binomial-coefficients stirling-numbers
edited Jul 16 at 16:46
Tiwa Aina
2,576319
asked Jul 16 at 16:40
Leox
5,0921323
For arbitrary integer $m, t, s<n$ prove the identity
$$
sum_i=s+1^n(i!)^2 left n atop iright left m atop iright binomsibinomti=0,
$$
here $displaystyle left n atop iright$ is the Stirling numbers of the second kind.
I come across by accident to the identity as secondary result.
I have checked it for many values, it is true. What about a proof in the general case?
combinatorics summation binomial-coefficients stirling-numbers
edited Jul 16 at 16:46
Tiwa Aina
2,576319
asked Jul 16 at 16:40
Leox
5,0921323
edited Jul 16 at 16:46
Tiwa Aina
2,576319
edited Jul 16 at 16:46
Tiwa Aina
2,576319
edited Jul 16 at 16:46
Tiwa Aina
2,576319
2,576319
asked Jul 16 at 16:40
Leox
5,0921323
asked Jul 16 at 16:40
Leox
5,0921323
asked Jul 16 at 16:40
Leox
5,0921323
5,0921323
1
Note that $binomsi=0$ if $i>s$.
– Markus Scheuer
Jul 16 at 16:45
add a comment |Â
1
Note that $binomsi=0$ if $i>s$.
– Markus Scheuer
Jul 16 at 16:45
1
1
Note that $binomsi=0$ if $i>s$.
– Markus Scheuer
Jul 16 at 16:45
Note that $binomsi=0$ if $i>s$.
– Markus Scheuer
Jul 16 at 16:45
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Since $s + 1 leq i leq n$, we see that $i>s , , , forall s < n$. Hence $binomsi=0$. But each term of your sum is the product of $binomsi$ and other values, which means each term will be zero. So the entire sum will be zero.
answered Jul 16 at 16:51
Tiwa Aina
2,576319
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since $s + 1 leq i leq n$, we see that $i>s , , , forall s < n$. Hence $binomsi=0$. But each term of your sum is the product of $binomsi$ and other values, which means each term will be zero. So the entire sum will be zero.
answered Jul 16 at 16:51
Tiwa Aina
2,576319
add a comment |Â
up vote
2
down vote
accepted
Since $s + 1 leq i leq n$, we see that $i>s , , , forall s < n$. Hence $binomsi=0$. But each term of your sum is the product of $binomsi$ and other values, which means each term will be zero. So the entire sum will be zero.
answered Jul 16 at 16:51
Tiwa Aina
2,576319
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since $s + 1 leq i leq n$, we see that $i>s , , , forall s < n$. Hence $binomsi=0$. But each term of your sum is the product of $binomsi$ and other values, which means each term will be zero. So the entire sum will be zero.
answered Jul 16 at 16:51
Tiwa Aina
2,576319
Since $s + 1 leq i leq n$, we see that $i>s , , , forall s < n$. Hence $binomsi=0$. But each term of your sum is the product of $binomsi$ and other values, which means each term will be zero. So the entire sum will be zero.
answered Jul 16 at 16:51
Tiwa Aina
2,576319
edited Jul 16 at 17:02
answered Jul 16 at 16:51
Tiwa Aina
2,576319
answered Jul 16 at 16:51
Tiwa Aina
2,576319
answered Jul 16 at 16:51
Tiwa Aina
2,576319
2,576319
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853582%2fprove-the-identity-with-stirling-numbers%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Note that $binomsi=0$ if $i>s$.
– Markus Scheuer
Jul 16 at 16:45