Solving binomial coefficient inequality
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I have the following inequality,
$$binombinomnkabinombinomnk-lb<binombinomnka+btag 1$$
I was wondering how one solves this exact / approximately for l?
combinatorics inequality binomial-coefficients
asked Sep 4 at 13:00
William
131111
 |Â
show 1 more comment
up vote
0
down vote
favorite
I have the following inequality,
$$binombinomnkabinombinomnk-lb<binombinomnka+btag 1$$
I was wondering how one solves this exact / approximately for l?
combinatorics inequality binomial-coefficients
asked Sep 4 at 13:00
William
131111
1
Just checking - we assume that all variables are known except $l$, and we want to find the values of $l$ for which the inequality is true, right?
– Aleksejs Fomins
Sep 4 at 13:10
@AleksejsFomins yes that is true.
– William
Sep 4 at 13:11
does my answer work for you?
– Aleksejs Fomins
Sep 4 at 14:46
@AleksejsFomins unfortunately not, as I will need an analytical form.
– William
Sep 4 at 15:22
2
what makes you think that it is at all possible? Searching for zeros of high-dimensional polynomials is a notoriously hard problem which is unsolvable analytically for dimensions 5 and higher, and you can't solve inequalities if you can't solve for zeros. I can refer you to a few publications if you wish. Unless there is some significant symmetry in the above expression that I missed, it is very unlikely that it is solvable analytically
– Aleksejs Fomins
Sep 4 at 15:31
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the following inequality,
$$binombinomnkabinombinomnk-lb<binombinomnka+btag 1$$
I was wondering how one solves this exact / approximately for l?
combinatorics inequality binomial-coefficients
asked Sep 4 at 13:00
William
131111
I have the following inequality,
$$binombinomnkabinombinomnk-lb<binombinomnka+btag 1$$
I was wondering how one solves this exact / approximately for l?
combinatorics inequality binomial-coefficients
combinatorics inequality binomial-coefficients
asked Sep 4 at 13:00
William
131111
asked Sep 4 at 13:00
William
131111
edited Sep 4 at 20:20
asked Sep 4 at 13:00
William
131111
asked Sep 4 at 13:00
William
131111
asked Sep 4 at 13:00
William
131111
131111
1
Just checking - we assume that all variables are known except $l$, and we want to find the values of $l$ for which the inequality is true, right?
– Aleksejs Fomins
Sep 4 at 13:10
@AleksejsFomins yes that is true.
– William
Sep 4 at 13:11
does my answer work for you?
– Aleksejs Fomins
Sep 4 at 14:46
@AleksejsFomins unfortunately not, as I will need an analytical form.
– William
Sep 4 at 15:22
2
what makes you think that it is at all possible? Searching for zeros of high-dimensional polynomials is a notoriously hard problem which is unsolvable analytically for dimensions 5 and higher, and you can't solve inequalities if you can't solve for zeros. I can refer you to a few publications if you wish. Unless there is some significant symmetry in the above expression that I missed, it is very unlikely that it is solvable analytically
– Aleksejs Fomins
Sep 4 at 15:31
 |Â
show 1 more comment
1
Just checking - we assume that all variables are known except $l$, and we want to find the values of $l$ for which the inequality is true, right?
– Aleksejs Fomins
Sep 4 at 13:10
@AleksejsFomins yes that is true.
– William
Sep 4 at 13:11
does my answer work for you?
– Aleksejs Fomins
Sep 4 at 14:46
@AleksejsFomins unfortunately not, as I will need an analytical form.
– William
Sep 4 at 15:22
2
what makes you think that it is at all possible? Searching for zeros of high-dimensional polynomials is a notoriously hard problem which is unsolvable analytically for dimensions 5 and higher, and you can't solve inequalities if you can't solve for zeros. I can refer you to a few publications if you wish. Unless there is some significant symmetry in the above expression that I missed, it is very unlikely that it is solvable analytically
– Aleksejs Fomins
Sep 4 at 15:31
1
1
Just checking - we assume that all variables are known except $l$, and we want to find the values of $l$ for which the inequality is true, right?
– Aleksejs Fomins
Sep 4 at 13:10
Just checking - we assume that all variables are known except $l$, and we want to find the values of $l$ for which the inequality is true, right?
– Aleksejs Fomins
Sep 4 at 13:10
@AleksejsFomins yes that is true.
– William
Sep 4 at 13:11
@AleksejsFomins yes that is true.
– William
Sep 4 at 13:11
does my answer work for you?
– Aleksejs Fomins
Sep 4 at 14:46
does my answer work for you?
– Aleksejs Fomins
Sep 4 at 14:46
@AleksejsFomins unfortunately not, as I will need an analytical form.
– William
Sep 4 at 15:22
@AleksejsFomins unfortunately not, as I will need an analytical form.
– William
Sep 4 at 15:22
2
2
what makes you think that it is at all possible? Searching for zeros of high-dimensional polynomials is a notoriously hard problem which is unsolvable analytically for dimensions 5 and higher, and you can't solve inequalities if you can't solve for zeros. I can refer you to a few publications if you wish. Unless there is some significant symmetry in the above expression that I missed, it is very unlikely that it is solvable analytically
– Aleksejs Fomins
Sep 4 at 15:31
what makes you think that it is at all possible? Searching for zeros of high-dimensional polynomials is a notoriously hard problem which is unsolvable analytically for dimensions 5 and higher, and you can't solve inequalities if you can't solve for zeros. I can refer you to a few publications if you wish. Unless there is some significant symmetry in the above expression that I missed, it is very unlikely that it is solvable analytically
– Aleksejs Fomins
Sep 4 at 15:31
 |Â
show 1 more comment
1 Answer
1
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oldest
votes
up vote
1
down vote
If you substitute all combination expressions inside and open all the brackets, you will end up with a fraction $fracP(l)Q(l) < 0$, where $P(l)$ and $Q(l)$ are polynomials. You can then attempt to factorize each polynomial into linear multipliers and apply the standard high-school polynomial inequality solution method to determine for which intervals of the real line is the LHS of the equation positive and for which negative, and then determine which integer values of $l$ fit into the positive intervals.
However, if you do that, you will get into a giant mess, because factoring a polynomial containing products of products of factorials will give you thousands of zeros on your real line. Since $l$ is integer, why not just write a computer program checking if the inequality holds for each value of $l$ from $0$ to $k$?
answered Sep 4 at 13:34
Aleksejs Fomins
388111
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If you substitute all combination expressions inside and open all the brackets, you will end up with a fraction $fracP(l)Q(l) < 0$, where $P(l)$ and $Q(l)$ are polynomials. You can then attempt to factorize each polynomial into linear multipliers and apply the standard high-school polynomial inequality solution method to determine for which intervals of the real line is the LHS of the equation positive and for which negative, and then determine which integer values of $l$ fit into the positive intervals.
However, if you do that, you will get into a giant mess, because factoring a polynomial containing products of products of factorials will give you thousands of zeros on your real line. Since $l$ is integer, why not just write a computer program checking if the inequality holds for each value of $l$ from $0$ to $k$?
answered Sep 4 at 13:34
Aleksejs Fomins
388111
add a comment |Â
up vote
1
down vote
If you substitute all combination expressions inside and open all the brackets, you will end up with a fraction $fracP(l)Q(l) < 0$, where $P(l)$ and $Q(l)$ are polynomials. You can then attempt to factorize each polynomial into linear multipliers and apply the standard high-school polynomial inequality solution method to determine for which intervals of the real line is the LHS of the equation positive and for which negative, and then determine which integer values of $l$ fit into the positive intervals.
However, if you do that, you will get into a giant mess, because factoring a polynomial containing products of products of factorials will give you thousands of zeros on your real line. Since $l$ is integer, why not just write a computer program checking if the inequality holds for each value of $l$ from $0$ to $k$?
answered Sep 4 at 13:34
Aleksejs Fomins
388111
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you substitute all combination expressions inside and open all the brackets, you will end up with a fraction $fracP(l)Q(l) < 0$, where $P(l)$ and $Q(l)$ are polynomials. You can then attempt to factorize each polynomial into linear multipliers and apply the standard high-school polynomial inequality solution method to determine for which intervals of the real line is the LHS of the equation positive and for which negative, and then determine which integer values of $l$ fit into the positive intervals.
However, if you do that, you will get into a giant mess, because factoring a polynomial containing products of products of factorials will give you thousands of zeros on your real line. Since $l$ is integer, why not just write a computer program checking if the inequality holds for each value of $l$ from $0$ to $k$?
answered Sep 4 at 13:34
Aleksejs Fomins
388111
If you substitute all combination expressions inside and open all the brackets, you will end up with a fraction $fracP(l)Q(l) < 0$, where $P(l)$ and $Q(l)$ are polynomials. You can then attempt to factorize each polynomial into linear multipliers and apply the standard high-school polynomial inequality solution method to determine for which intervals of the real line is the LHS of the equation positive and for which negative, and then determine which integer values of $l$ fit into the positive intervals.
However, if you do that, you will get into a giant mess, because factoring a polynomial containing products of products of factorials will give you thousands of zeros on your real line. Since $l$ is integer, why not just write a computer program checking if the inequality holds for each value of $l$ from $0$ to $k$?
answered Sep 4 at 13:34
Aleksejs Fomins
388111
answered Sep 4 at 13:34
Aleksejs Fomins
388111
answered Sep 4 at 13:34
Aleksejs Fomins
388111
answered Sep 4 at 13:34
Aleksejs Fomins
388111
388111
add a comment |Â
add a comment |Â
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1
Just checking - we assume that all variables are known except $l$, and we want to find the values of $l$ for which the inequality is true, right?
– Aleksejs Fomins
Sep 4 at 13:10
@AleksejsFomins yes that is true.
– William
Sep 4 at 13:11
does my answer work for you?
– Aleksejs Fomins
Sep 4 at 14:46
@AleksejsFomins unfortunately not, as I will need an analytical form.
– William
Sep 4 at 15:22
2
what makes you think that it is at all possible? Searching for zeros of high-dimensional polynomials is a notoriously hard problem which is unsolvable analytically for dimensions 5 and higher, and you can't solve inequalities if you can't solve for zeros. I can refer you to a few publications if you wish. Unless there is some significant symmetry in the above expression that I missed, it is very unlikely that it is solvable analytically
– Aleksejs Fomins
Sep 4 at 15:31