Nonnegative coefficients of a product of polynomials












7














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







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  • 2




    mathoverflow.net/a/314433/2384 and mathoverflow.net/questions/301022/…
    – Gjergji Zaimi
    4 hours ago
















7














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







share|cite|improve this question


















  • 2




    mathoverflow.net/a/314433/2384 and mathoverflow.net/questions/301022/…
    – Gjergji Zaimi
    4 hours ago














7












7








7







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









Richard Stanley

28.2k8113184




28.2k8113184








  • 2




    mathoverflow.net/a/314433/2384 and mathoverflow.net/questions/301022/…
    – Gjergji Zaimi
    4 hours ago














  • 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










1 Answer
1






active

oldest

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2














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.






share|cite|improve this answer























  • 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













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1 Answer
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active

oldest

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active

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active

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2














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.






share|cite|improve this answer























  • 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


















2














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.






share|cite|improve this answer























  • 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
















2












2








2






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 1 hour 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




















  • 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




















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