Nonnegative coefficients of a product of polynomials
Let $P(x_1,dots,x_n)inmathbb{R}[x_1,dots,x_n]$. Does there exist an algorithm to decide whether there is a nonzero polynomial $Q(x_1,dots,x_n)inmathbb{R}[x_1,dots,x_n]$ such that the product $PQ$ has nonnegative coefficients? (The case $n=1$ is well-known and not difficult.)
polynomials positivity
asked 6 hours ago
Richard Stanley
28.2k8113184
add a comment |
Let $P(x_1,dots,x_n)inmathbb{R}[x_1,dots,x_n]$. Does there exist an algorithm to decide whether there is a nonzero polynomial $Q(x_1,dots,x_n)inmathbb{R}[x_1,dots,x_n]$ such that the product $PQ$ has nonnegative coefficients? (The case $n=1$ is well-known and not difficult.)
polynomials positivity
asked 6 hours ago
Richard Stanley
28.2k8113184
2
mathoverflow.net/a/314433/2384 and mathoverflow.net/questions/301022/…
– Gjergji Zaimi
4 hours ago
add a comment |
Let $P(x_1,dots,x_n)inmathbb{R}[x_1,dots,x_n]$. Does there exist an algorithm to decide whether there is a nonzero polynomial $Q(x_1,dots,x_n)inmathbb{R}[x_1,dots,x_n]$ such that the product $PQ$ has nonnegative coefficients? (The case $n=1$ is well-known and not difficult.)
polynomials positivity
asked 6 hours ago
Richard Stanley
28.2k8113184
Let $P(x_1,dots,x_n)inmathbb{R}[x_1,dots,x_n]$. Does there exist an algorithm to decide whether there is a nonzero polynomial $Q(x_1,dots,x_n)inmathbb{R}[x_1,dots,x_n]$ such that the product $PQ$ has nonnegative coefficients? (The case $n=1$ is well-known and not difficult.)
polynomials positivity
polynomials positivity
asked 6 hours ago
Richard Stanley
28.2k8113184
asked 6 hours ago
Richard Stanley
28.2k8113184
asked 6 hours ago
Richard Stanley
28.2k8113184
asked 6 hours ago
Richard Stanley
28.2k8113184
asked 6 hours ago
Richard Stanley
28.2k8113184
28.2k8113184
2
mathoverflow.net/a/314433/2384 and mathoverflow.net/questions/301022/…
– Gjergji Zaimi
4 hours ago
add a comment |
2
mathoverflow.net/a/314433/2384 and mathoverflow.net/questions/301022/…
– Gjergji Zaimi
4 hours ago
2
2
mathoverflow.net/a/314433/2384 and mathoverflow.net/questions/301022/…
– Gjergji Zaimi
4 hours ago
mathoverflow.net/a/314433/2384 and mathoverflow.net/questions/301022/…
– Gjergji Zaimi
4 hours ago
add a comment |
1 Answer
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Assuming $P(1,1,dots,1) >0$, the condition (necessary and sufficient) on $P$ is that $P_F$ is strictly positive on the strictly positive orthant for all faces of dimension one or more of the Newton polyhedron of $P$ (where $P_F$ is the subpolynomial of $P$ obtained by discarding all the monomials whose exponent does not lie in $F$; we of course have to permit $F$ to be the improper face), and when that occurs, we can choose $Q$ to be some (unknown) power of $Q_1= sum x^w$ where $w$ runs over the lattice points in the Newton polytope (yielding a pseudo-algorithm; we don't know if it eventually results in no negative coefficients; it will only do this if a $Q$ exists, and even then, there does not appear to be a way of estimating the power of $Q_1$ required).
Is there an algorithm to decide whether a polynomial is strictly positive on the positive orthant? If so, then it can be applied to all the $P_F$.
If however, as I suspect, there is no algorithm (to decide strict positivity on an orthant), then there is a problem. In the one variable case, if $Q$ exists, it can be chosen of the form $(1+x)^n$, and so we obtain a pseudo-algorithm (we don't know if it will terminate) by repeatedly adding pairs of consecutive coefficients). With more than one variable, no such single poly powers of which will do (in fact, infinitely many are required).
We can get some inkling of what $Q$ should look like, by restricting to faces; assuming the Newton polyhedron of $Q$ is $k$ times that of $P$ (which we can assume for some $k$, except we don't know what the minimum $k$ will be), then $P_F Q_{kF}$ will also have no negative coefficients.
answered 2 hours ago
David Handelman
3,66211428
Thanks for this useful answer. For one variable there is no problem finding the exponent of $1+x$ by looking at the real linear and quadratic factors of $P(x)$ separately. The condition for $Q(x)$ to exist is that $P(x)$ has no positive real zeros.
– Richard Stanley
1 hour ago
add a comment |
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Assuming $P(1,1,dots,1) >0$, the condition (necessary and sufficient) on $P$ is that $P_F$ is strictly positive on the strictly positive orthant for all faces of dimension one or more of the Newton polyhedron of $P$ (where $P_F$ is the subpolynomial of $P$ obtained by discarding all the monomials whose exponent does not lie in $F$; we of course have to permit $F$ to be the improper face), and when that occurs, we can choose $Q$ to be some (unknown) power of $Q_1= sum x^w$ where $w$ runs over the lattice points in the Newton polytope (yielding a pseudo-algorithm; we don't know if it eventually results in no negative coefficients; it will only do this if a $Q$ exists, and even then, there does not appear to be a way of estimating the power of $Q_1$ required).
Is there an algorithm to decide whether a polynomial is strictly positive on the positive orthant? If so, then it can be applied to all the $P_F$.
If however, as I suspect, there is no algorithm (to decide strict positivity on an orthant), then there is a problem. In the one variable case, if $Q$ exists, it can be chosen of the form $(1+x)^n$, and so we obtain a pseudo-algorithm (we don't know if it will terminate) by repeatedly adding pairs of consecutive coefficients). With more than one variable, no such single poly powers of which will do (in fact, infinitely many are required).
We can get some inkling of what $Q$ should look like, by restricting to faces; assuming the Newton polyhedron of $Q$ is $k$ times that of $P$ (which we can assume for some $k$, except we don't know what the minimum $k$ will be), then $P_F Q_{kF}$ will also have no negative coefficients.
answered 2 hours ago
David Handelman
3,66211428
Thanks for this useful answer. For one variable there is no problem finding the exponent of $1+x$ by looking at the real linear and quadratic factors of $P(x)$ separately. The condition for $Q(x)$ to exist is that $P(x)$ has no positive real zeros.
– Richard Stanley
1 hour ago
add a comment |
Assuming $P(1,1,dots,1) >0$, the condition (necessary and sufficient) on $P$ is that $P_F$ is strictly positive on the strictly positive orthant for all faces of dimension one or more of the Newton polyhedron of $P$ (where $P_F$ is the subpolynomial of $P$ obtained by discarding all the monomials whose exponent does not lie in $F$; we of course have to permit $F$ to be the improper face), and when that occurs, we can choose $Q$ to be some (unknown) power of $Q_1= sum x^w$ where $w$ runs over the lattice points in the Newton polytope (yielding a pseudo-algorithm; we don't know if it eventually results in no negative coefficients; it will only do this if a $Q$ exists, and even then, there does not appear to be a way of estimating the power of $Q_1$ required).
Is there an algorithm to decide whether a polynomial is strictly positive on the positive orthant? If so, then it can be applied to all the $P_F$.
If however, as I suspect, there is no algorithm (to decide strict positivity on an orthant), then there is a problem. In the one variable case, if $Q$ exists, it can be chosen of the form $(1+x)^n$, and so we obtain a pseudo-algorithm (we don't know if it will terminate) by repeatedly adding pairs of consecutive coefficients). With more than one variable, no such single poly powers of which will do (in fact, infinitely many are required).
We can get some inkling of what $Q$ should look like, by restricting to faces; assuming the Newton polyhedron of $Q$ is $k$ times that of $P$ (which we can assume for some $k$, except we don't know what the minimum $k$ will be), then $P_F Q_{kF}$ will also have no negative coefficients.
answered 2 hours ago
David Handelman
3,66211428
Thanks for this useful answer. For one variable there is no problem finding the exponent of $1+x$ by looking at the real linear and quadratic factors of $P(x)$ separately. The condition for $Q(x)$ to exist is that $P(x)$ has no positive real zeros.
– Richard Stanley
1 hour ago
add a comment |
Assuming $P(1,1,dots,1) >0$, the condition (necessary and sufficient) on $P$ is that $P_F$ is strictly positive on the strictly positive orthant for all faces of dimension one or more of the Newton polyhedron of $P$ (where $P_F$ is the subpolynomial of $P$ obtained by discarding all the monomials whose exponent does not lie in $F$; we of course have to permit $F$ to be the improper face), and when that occurs, we can choose $Q$ to be some (unknown) power of $Q_1= sum x^w$ where $w$ runs over the lattice points in the Newton polytope (yielding a pseudo-algorithm; we don't know if it eventually results in no negative coefficients; it will only do this if a $Q$ exists, and even then, there does not appear to be a way of estimating the power of $Q_1$ required).
Is there an algorithm to decide whether a polynomial is strictly positive on the positive orthant? If so, then it can be applied to all the $P_F$.
If however, as I suspect, there is no algorithm (to decide strict positivity on an orthant), then there is a problem. In the one variable case, if $Q$ exists, it can be chosen of the form $(1+x)^n$, and so we obtain a pseudo-algorithm (we don't know if it will terminate) by repeatedly adding pairs of consecutive coefficients). With more than one variable, no such single poly powers of which will do (in fact, infinitely many are required).
We can get some inkling of what $Q$ should look like, by restricting to faces; assuming the Newton polyhedron of $Q$ is $k$ times that of $P$ (which we can assume for some $k$, except we don't know what the minimum $k$ will be), then $P_F Q_{kF}$ will also have no negative coefficients.
answered 2 hours ago
David Handelman
3,66211428
Assuming $P(1,1,dots,1) >0$, the condition (necessary and sufficient) on $P$ is that $P_F$ is strictly positive on the strictly positive orthant for all faces of dimension one or more of the Newton polyhedron of $P$ (where $P_F$ is the subpolynomial of $P$ obtained by discarding all the monomials whose exponent does not lie in $F$; we of course have to permit $F$ to be the improper face), and when that occurs, we can choose $Q$ to be some (unknown) power of $Q_1= sum x^w$ where $w$ runs over the lattice points in the Newton polytope (yielding a pseudo-algorithm; we don't know if it eventually results in no negative coefficients; it will only do this if a $Q$ exists, and even then, there does not appear to be a way of estimating the power of $Q_1$ required).
Is there an algorithm to decide whether a polynomial is strictly positive on the positive orthant? If so, then it can be applied to all the $P_F$.
If however, as I suspect, there is no algorithm (to decide strict positivity on an orthant), then there is a problem. In the one variable case, if $Q$ exists, it can be chosen of the form $(1+x)^n$, and so we obtain a pseudo-algorithm (we don't know if it will terminate) by repeatedly adding pairs of consecutive coefficients). With more than one variable, no such single poly powers of which will do (in fact, infinitely many are required).
We can get some inkling of what $Q$ should look like, by restricting to faces; assuming the Newton polyhedron of $Q$ is $k$ times that of $P$ (which we can assume for some $k$, except we don't know what the minimum $k$ will be), then $P_F Q_{kF}$ will also have no negative coefficients.
answered 2 hours ago
David Handelman
3,66211428
edited 1 hour ago
answered 2 hours ago
David Handelman
3,66211428
answered 2 hours ago
David Handelman
3,66211428
answered 2 hours ago
David Handelman
3,66211428
3,66211428
Thanks for this useful answer. For one variable there is no problem finding the exponent of $1+x$ by looking at the real linear and quadratic factors of $P(x)$ separately. The condition for $Q(x)$ to exist is that $P(x)$ has no positive real zeros.
– Richard Stanley
1 hour ago
add a comment |
Thanks for this useful answer. For one variable there is no problem finding the exponent of $1+x$ by looking at the real linear and quadratic factors of $P(x)$ separately. The condition for $Q(x)$ to exist is that $P(x)$ has no positive real zeros.
– Richard Stanley
1 hour ago
Thanks for this useful answer. For one variable there is no problem finding the exponent of $1+x$ by looking at the real linear and quadratic factors of $P(x)$ separately. The condition for $Q(x)$ to exist is that $P(x)$ has no positive real zeros.
– Richard Stanley
1 hour ago
Thanks for this useful answer. For one variable there is no problem finding the exponent of $1+x$ by looking at the real linear and quadratic factors of $P(x)$ separately. The condition for $Q(x)$ to exist is that $P(x)$ has no positive real zeros.
– Richard Stanley
1 hour ago
add a comment |
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2
mathoverflow.net/a/314433/2384 and mathoverflow.net/questions/301022/…
– Gjergji Zaimi
4 hours ago