A polynomial $p(x) in Bbb R_{2n-1}[x]$, $p(0) = 0$, $p(x) geq 0 forall x geq 0$, can be written as $p(x) =...
Let $p(x) in Bbb R_{2n-1}[x]$ be a polynomial such that $p(0)= 0$
and $p(x) geq 0 forall x geq 0$. Then there exists $q_1, q_2 in
Bbb R_{n-1}[x]$ such that $p(x) = xq_1(x)^2 + q_2(x)^2$.
Here, $Bbb R_{k}[x]$ is the set of polynomials with real coefficients of degree less than or equal to $k$.
Note that we always have $p(x) = xp_o(x^2) + p_e(x^2)$ for some $p_o, p_e in
Bbb R_{n-1}[x]$ by separating the terms of even and odd degree of $p$.
It is related to Prove that a positive polynomial function can be written as the squares of two polynomial functions but does not seem to follow from it.
This is stated without proof and reference at page 77 from The Classical Moment Problem, Oliver and Boyd Ltd, 1965, by Akhiezer.
polynomials roots
asked Dec 2 '18 at 14:32
Desura
901514
add a comment |
Let $p(x) in Bbb R_{2n-1}[x]$ be a polynomial such that $p(0)= 0$
and $p(x) geq 0 forall x geq 0$. Then there exists $q_1, q_2 in
Bbb R_{n-1}[x]$ such that $p(x) = xq_1(x)^2 + q_2(x)^2$.
Here, $Bbb R_{k}[x]$ is the set of polynomials with real coefficients of degree less than or equal to $k$.
Note that we always have $p(x) = xp_o(x^2) + p_e(x^2)$ for some $p_o, p_e in
Bbb R_{n-1}[x]$ by separating the terms of even and odd degree of $p$.
It is related to Prove that a positive polynomial function can be written as the squares of two polynomial functions but does not seem to follow from it.
This is stated without proof and reference at page 77 from The Classical Moment Problem, Oliver and Boyd Ltd, 1965, by Akhiezer.
polynomials roots
asked Dec 2 '18 at 14:32
Desura
901514
math.stackexchange.com/questions/2966594/…
– greedoid
Dec 4 '18 at 22:09
@greedoid Thank you for directing me to this.
– Desura
Dec 4 '18 at 22:10
add a comment |
Let $p(x) in Bbb R_{2n-1}[x]$ be a polynomial such that $p(0)= 0$
and $p(x) geq 0 forall x geq 0$. Then there exists $q_1, q_2 in
Bbb R_{n-1}[x]$ such that $p(x) = xq_1(x)^2 + q_2(x)^2$.
Here, $Bbb R_{k}[x]$ is the set of polynomials with real coefficients of degree less than or equal to $k$.
Note that we always have $p(x) = xp_o(x^2) + p_e(x^2)$ for some $p_o, p_e in
Bbb R_{n-1}[x]$ by separating the terms of even and odd degree of $p$.
It is related to Prove that a positive polynomial function can be written as the squares of two polynomial functions but does not seem to follow from it.
This is stated without proof and reference at page 77 from The Classical Moment Problem, Oliver and Boyd Ltd, 1965, by Akhiezer.
polynomials roots
asked Dec 2 '18 at 14:32
Desura
901514
Let $p(x) in Bbb R_{2n-1}[x]$ be a polynomial such that $p(0)= 0$
and $p(x) geq 0 forall x geq 0$. Then there exists $q_1, q_2 in
Bbb R_{n-1}[x]$ such that $p(x) = xq_1(x)^2 + q_2(x)^2$.
Here, $Bbb R_{k}[x]$ is the set of polynomials with real coefficients of degree less than or equal to $k$.
Note that we always have $p(x) = xp_o(x^2) + p_e(x^2)$ for some $p_o, p_e in
Bbb R_{n-1}[x]$ by separating the terms of even and odd degree of $p$.
It is related to Prove that a positive polynomial function can be written as the squares of two polynomial functions but does not seem to follow from it.
This is stated without proof and reference at page 77 from The Classical Moment Problem, Oliver and Boyd Ltd, 1965, by Akhiezer.
polynomials roots
polynomials roots
asked Dec 2 '18 at 14:32
Desura
901514
asked Dec 2 '18 at 14:32
Desura
901514
edited Dec 4 '18 at 22:05
asked Dec 2 '18 at 14:32
Desura
901514
asked Dec 2 '18 at 14:32
Desura
901514
asked Dec 2 '18 at 14:32
Desura
901514
901514
math.stackexchange.com/questions/2966594/…
– greedoid
Dec 4 '18 at 22:09
@greedoid Thank you for directing me to this.
– Desura
Dec 4 '18 at 22:10
add a comment |
math.stackexchange.com/questions/2966594/…
– greedoid
Dec 4 '18 at 22:09
@greedoid Thank you for directing me to this.
– Desura
Dec 4 '18 at 22:10
math.stackexchange.com/questions/2966594/…
– greedoid
Dec 4 '18 at 22:09
math.stackexchange.com/questions/2966594/…
– greedoid
Dec 4 '18 at 22:09
@greedoid Thank you for directing me to this.
– Desura
Dec 4 '18 at 22:10
@greedoid Thank you for directing me to this.
– Desura
Dec 4 '18 at 22:10
add a comment |
1 Answer
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The key to the solution is the identity
$$ (a^2x+b^2)(c^2x+d^2) = (ad-bc)^2x + (acx+bd)^2. tag1$$
You can put polynomials for $a,b,c,d$.
Let $mathcal{P} = Big{ xq_1(x)^2+q_2(x)^2 Big}$; we need to prove $p(x)inmathcal{P}$.
Notice that all complete squares and all nonnegative constants are elements of $mathcal{P}$.
The identity $(1)$ shows that $mathcal{P}$ is closed for multiplication.
Consider the prime factorization of $p(x)$:
$$ p(x) = A cdot prod(x+a_i)^{alpha_i} prod (x^2+c_ix+d_i). $$
From $xtoinfty$ we find that the leading coefficient $A$ is nonnegative, so $Ainmathcal{P}$.
All prime factors $x+a_i$ with $age0$ are in $mathcal{P}$.
In every irreducible factor $x^2+cx+d=(x-sqrt{d})^2+(2sqrt{d}-c)x$ we have $2sqrt{d}-c>0$, so they are in $mathcal{P}$.
Finally, the positive roots of $p(x)$ must have even multiplicities, so
every factor $(x+a_i)^{alpha_i}$ with $a_i<0$ is a complete square.
Hence $p(x)$, being the product of some elements of $mathcal{P}$, is an element of $mathcal{P}$.
answered Dec 5 '18 at 12:50
user141614
12.2k925
Nice and simple $(+1).$
– Yuri Negometyanov
Dec 10 '18 at 9:48
add a comment |
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1 Answer
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1 Answer
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The key to the solution is the identity
$$ (a^2x+b^2)(c^2x+d^2) = (ad-bc)^2x + (acx+bd)^2. tag1$$
You can put polynomials for $a,b,c,d$.
Let $mathcal{P} = Big{ xq_1(x)^2+q_2(x)^2 Big}$; we need to prove $p(x)inmathcal{P}$.
Notice that all complete squares and all nonnegative constants are elements of $mathcal{P}$.
The identity $(1)$ shows that $mathcal{P}$ is closed for multiplication.
Consider the prime factorization of $p(x)$:
$$ p(x) = A cdot prod(x+a_i)^{alpha_i} prod (x^2+c_ix+d_i). $$
From $xtoinfty$ we find that the leading coefficient $A$ is nonnegative, so $Ainmathcal{P}$.
All prime factors $x+a_i$ with $age0$ are in $mathcal{P}$.
In every irreducible factor $x^2+cx+d=(x-sqrt{d})^2+(2sqrt{d}-c)x$ we have $2sqrt{d}-c>0$, so they are in $mathcal{P}$.
Finally, the positive roots of $p(x)$ must have even multiplicities, so
every factor $(x+a_i)^{alpha_i}$ with $a_i<0$ is a complete square.
Hence $p(x)$, being the product of some elements of $mathcal{P}$, is an element of $mathcal{P}$.
answered Dec 5 '18 at 12:50
user141614
12.2k925
Nice and simple $(+1).$
– Yuri Negometyanov
Dec 10 '18 at 9:48
add a comment |
The key to the solution is the identity
$$ (a^2x+b^2)(c^2x+d^2) = (ad-bc)^2x + (acx+bd)^2. tag1$$
You can put polynomials for $a,b,c,d$.
Let $mathcal{P} = Big{ xq_1(x)^2+q_2(x)^2 Big}$; we need to prove $p(x)inmathcal{P}$.
Notice that all complete squares and all nonnegative constants are elements of $mathcal{P}$.
The identity $(1)$ shows that $mathcal{P}$ is closed for multiplication.
Consider the prime factorization of $p(x)$:
$$ p(x) = A cdot prod(x+a_i)^{alpha_i} prod (x^2+c_ix+d_i). $$
From $xtoinfty$ we find that the leading coefficient $A$ is nonnegative, so $Ainmathcal{P}$.
All prime factors $x+a_i$ with $age0$ are in $mathcal{P}$.
In every irreducible factor $x^2+cx+d=(x-sqrt{d})^2+(2sqrt{d}-c)x$ we have $2sqrt{d}-c>0$, so they are in $mathcal{P}$.
Finally, the positive roots of $p(x)$ must have even multiplicities, so
every factor $(x+a_i)^{alpha_i}$ with $a_i<0$ is a complete square.
Hence $p(x)$, being the product of some elements of $mathcal{P}$, is an element of $mathcal{P}$.
answered Dec 5 '18 at 12:50
user141614
12.2k925
Nice and simple $(+1).$
– Yuri Negometyanov
Dec 10 '18 at 9:48
add a comment |
The key to the solution is the identity
$$ (a^2x+b^2)(c^2x+d^2) = (ad-bc)^2x + (acx+bd)^2. tag1$$
You can put polynomials for $a,b,c,d$.
Let $mathcal{P} = Big{ xq_1(x)^2+q_2(x)^2 Big}$; we need to prove $p(x)inmathcal{P}$.
Notice that all complete squares and all nonnegative constants are elements of $mathcal{P}$.
The identity $(1)$ shows that $mathcal{P}$ is closed for multiplication.
Consider the prime factorization of $p(x)$:
$$ p(x) = A cdot prod(x+a_i)^{alpha_i} prod (x^2+c_ix+d_i). $$
From $xtoinfty$ we find that the leading coefficient $A$ is nonnegative, so $Ainmathcal{P}$.
All prime factors $x+a_i$ with $age0$ are in $mathcal{P}$.
In every irreducible factor $x^2+cx+d=(x-sqrt{d})^2+(2sqrt{d}-c)x$ we have $2sqrt{d}-c>0$, so they are in $mathcal{P}$.
Finally, the positive roots of $p(x)$ must have even multiplicities, so
every factor $(x+a_i)^{alpha_i}$ with $a_i<0$ is a complete square.
Hence $p(x)$, being the product of some elements of $mathcal{P}$, is an element of $mathcal{P}$.
answered Dec 5 '18 at 12:50
user141614
12.2k925
The key to the solution is the identity
$$ (a^2x+b^2)(c^2x+d^2) = (ad-bc)^2x + (acx+bd)^2. tag1$$
You can put polynomials for $a,b,c,d$.
Let $mathcal{P} = Big{ xq_1(x)^2+q_2(x)^2 Big}$; we need to prove $p(x)inmathcal{P}$.
Notice that all complete squares and all nonnegative constants are elements of $mathcal{P}$.
The identity $(1)$ shows that $mathcal{P}$ is closed for multiplication.
Consider the prime factorization of $p(x)$:
$$ p(x) = A cdot prod(x+a_i)^{alpha_i} prod (x^2+c_ix+d_i). $$
From $xtoinfty$ we find that the leading coefficient $A$ is nonnegative, so $Ainmathcal{P}$.
All prime factors $x+a_i$ with $age0$ are in $mathcal{P}$.
In every irreducible factor $x^2+cx+d=(x-sqrt{d})^2+(2sqrt{d}-c)x$ we have $2sqrt{d}-c>0$, so they are in $mathcal{P}$.
Finally, the positive roots of $p(x)$ must have even multiplicities, so
every factor $(x+a_i)^{alpha_i}$ with $a_i<0$ is a complete square.
Hence $p(x)$, being the product of some elements of $mathcal{P}$, is an element of $mathcal{P}$.
answered Dec 5 '18 at 12:50
user141614
12.2k925
edited Dec 10 '18 at 9:30
answered Dec 5 '18 at 12:50
user141614
12.2k925
answered Dec 5 '18 at 12:50
user141614
12.2k925
answered Dec 5 '18 at 12:50
user141614
12.2k925
12.2k925
Nice and simple $(+1).$
– Yuri Negometyanov
Dec 10 '18 at 9:48
add a comment |
Nice and simple $(+1).$
– Yuri Negometyanov
Dec 10 '18 at 9:48
Nice and simple $(+1).$
– Yuri Negometyanov
Dec 10 '18 at 9:48
Nice and simple $(+1).$
– Yuri Negometyanov
Dec 10 '18 at 9:48
add a comment |
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math.stackexchange.com/questions/2966594/…
– greedoid
Dec 4 '18 at 22:09
@greedoid Thank you for directing me to this.
– Desura
Dec 4 '18 at 22:10