If a univariate polynomial is greater than another, is their difference a square?
Let $f(x)$ and $g(x)$ be polynomials, and suppose that for all $x$, $f(x) ge g(x)$. Does this imply that there exists a polynomial $h(x)$ such that $f(x) - g(x) = h(x)^2$? If so, is this easy to prove, and if not, is there a simple counterexample?
polynomials
asked Jan 4 at 19:13
user54038user54038
1108
add a comment |
Let $f(x)$ and $g(x)$ be polynomials, and suppose that for all $x$, $f(x) ge g(x)$. Does this imply that there exists a polynomial $h(x)$ such that $f(x) - g(x) = h(x)^2$? If so, is this easy to prove, and if not, is there a simple counterexample?
polynomials
asked Jan 4 at 19:13
user54038user54038
1108
There are two polynomials $h_1$ and $h_2$ sucht that $f(x) - g(x)= h_1(x)^2 + h_2(x)^2$ since $f(x) - g(x)$ is a non-negative polynomial.
– An_876_Joke
Jan 4 at 19:18
1
You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
– Martin R
Jan 4 at 19:21
add a comment |
Let $f(x)$ and $g(x)$ be polynomials, and suppose that for all $x$, $f(x) ge g(x)$. Does this imply that there exists a polynomial $h(x)$ such that $f(x) - g(x) = h(x)^2$? If so, is this easy to prove, and if not, is there a simple counterexample?
polynomials
asked Jan 4 at 19:13
user54038user54038
1108
Let $f(x)$ and $g(x)$ be polynomials, and suppose that for all $x$, $f(x) ge g(x)$. Does this imply that there exists a polynomial $h(x)$ such that $f(x) - g(x) = h(x)^2$? If so, is this easy to prove, and if not, is there a simple counterexample?
polynomials
polynomials
asked Jan 4 at 19:13
user54038user54038
1108
asked Jan 4 at 19:13
user54038user54038
1108
asked Jan 4 at 19:13
user54038user54038
1108
asked Jan 4 at 19:13
user54038user54038
1108
asked Jan 4 at 19:13
user54038user54038
1108
1108
There are two polynomials $h_1$ and $h_2$ sucht that $f(x) - g(x)= h_1(x)^2 + h_2(x)^2$ since $f(x) - g(x)$ is a non-negative polynomial.
– An_876_Joke
Jan 4 at 19:18
1
You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
– Martin R
Jan 4 at 19:21
add a comment |
There are two polynomials $h_1$ and $h_2$ sucht that $f(x) - g(x)= h_1(x)^2 + h_2(x)^2$ since $f(x) - g(x)$ is a non-negative polynomial.
– An_876_Joke
Jan 4 at 19:18
1
You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
– Martin R
Jan 4 at 19:21
There are two polynomials $h_1$ and $h_2$ sucht that $f(x) - g(x)= h_1(x)^2 + h_2(x)^2$ since $f(x) - g(x)$ is a non-negative polynomial.
– An_876_Joke
Jan 4 at 19:18
There are two polynomials $h_1$ and $h_2$ sucht that $f(x) - g(x)= h_1(x)^2 + h_2(x)^2$ since $f(x) - g(x)$ is a non-negative polynomial.
– An_876_Joke
Jan 4 at 19:18
1
1
You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
– Martin R
Jan 4 at 19:21
You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
– Martin R
Jan 4 at 19:21
add a comment |
3 Answers
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Take $f(x)=x^2+2$ and $g(x)=2x$. Then, for each $xinmathbb R$,$$f(x)-g(x)=x^2-2x+2=(x-1)^2+1>0.$$However, there is no polynomial $h(x)$ such that $h^2(x)=x^2-2x+2$ because the degree of $h(x)$ could only be $1$ and therefore it would have to have a real root. But $x^2-2x+2$ has none.
answered Jan 4 at 19:19
José Carlos SantosJosé Carlos Santos
152k22123225
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There should be a very simple counterexample; $f(x)=x^4+x^2+1$ and $g(x)=x^4$. Then $f(x)-g(x)=x^2+1$, which is not a square.
answered Jan 4 at 19:19
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
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I don't think so, for example if you take $$f(x)=x^2>g(x)=x-1$$
so you can see: $$f(x)-g(x)=x^2-x+1$$ and you cant find $$h(x)^2 $$which equal to that,
if that's what you mean..
answered Jan 4 at 19:21
Yanir ElmYanir Elm
693
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3 Answers
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3 Answers
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Take $f(x)=x^2+2$ and $g(x)=2x$. Then, for each $xinmathbb R$,$$f(x)-g(x)=x^2-2x+2=(x-1)^2+1>0.$$However, there is no polynomial $h(x)$ such that $h^2(x)=x^2-2x+2$ because the degree of $h(x)$ could only be $1$ and therefore it would have to have a real root. But $x^2-2x+2$ has none.
answered Jan 4 at 19:19
José Carlos SantosJosé Carlos Santos
152k22123225
add a comment |
Take $f(x)=x^2+2$ and $g(x)=2x$. Then, for each $xinmathbb R$,$$f(x)-g(x)=x^2-2x+2=(x-1)^2+1>0.$$However, there is no polynomial $h(x)$ such that $h^2(x)=x^2-2x+2$ because the degree of $h(x)$ could only be $1$ and therefore it would have to have a real root. But $x^2-2x+2$ has none.
answered Jan 4 at 19:19
José Carlos SantosJosé Carlos Santos
152k22123225
add a comment |
Take $f(x)=x^2+2$ and $g(x)=2x$. Then, for each $xinmathbb R$,$$f(x)-g(x)=x^2-2x+2=(x-1)^2+1>0.$$However, there is no polynomial $h(x)$ such that $h^2(x)=x^2-2x+2$ because the degree of $h(x)$ could only be $1$ and therefore it would have to have a real root. But $x^2-2x+2$ has none.
answered Jan 4 at 19:19
José Carlos SantosJosé Carlos Santos
152k22123225
Take $f(x)=x^2+2$ and $g(x)=2x$. Then, for each $xinmathbb R$,$$f(x)-g(x)=x^2-2x+2=(x-1)^2+1>0.$$However, there is no polynomial $h(x)$ such that $h^2(x)=x^2-2x+2$ because the degree of $h(x)$ could only be $1$ and therefore it would have to have a real root. But $x^2-2x+2$ has none.
answered Jan 4 at 19:19
José Carlos SantosJosé Carlos Santos
152k22123225
edited Jan 4 at 19:41
answered Jan 4 at 19:19
José Carlos SantosJosé Carlos Santos
152k22123225
answered Jan 4 at 19:19
José Carlos SantosJosé Carlos Santos
152k22123225
answered Jan 4 at 19:19
José Carlos SantosJosé Carlos Santos
152k22123225
152k22123225
add a comment |
add a comment |
There should be a very simple counterexample; $f(x)=x^4+x^2+1$ and $g(x)=x^4$. Then $f(x)-g(x)=x^2+1$, which is not a square.
answered Jan 4 at 19:19
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
661
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There should be a very simple counterexample; $f(x)=x^4+x^2+1$ and $g(x)=x^4$. Then $f(x)-g(x)=x^2+1$, which is not a square.
answered Jan 4 at 19:19
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
661
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ItsJustSomeOrdinalsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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There should be a very simple counterexample; $f(x)=x^4+x^2+1$ and $g(x)=x^4$. Then $f(x)-g(x)=x^2+1$, which is not a square.
answered Jan 4 at 19:19
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
661
New contributor
ItsJustSomeOrdinalsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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There should be a very simple counterexample; $f(x)=x^4+x^2+1$ and $g(x)=x^4$. Then $f(x)-g(x)=x^2+1$, which is not a square.
answered Jan 4 at 19:19
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
661
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ItsJustSomeOrdinalsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered Jan 4 at 19:19
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
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answered Jan 4 at 19:19
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
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answered Jan 4 at 19:19
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
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ItsJustSomeOrdinalsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I don't think so, for example if you take $$f(x)=x^2>g(x)=x-1$$
so you can see: $$f(x)-g(x)=x^2-x+1$$ and you cant find $$h(x)^2 $$which equal to that,
if that's what you mean..
answered Jan 4 at 19:21
Yanir ElmYanir Elm
693
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Yanir Elm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I don't think so, for example if you take $$f(x)=x^2>g(x)=x-1$$
so you can see: $$f(x)-g(x)=x^2-x+1$$ and you cant find $$h(x)^2 $$which equal to that,
if that's what you mean..
answered Jan 4 at 19:21
Yanir ElmYanir Elm
693
New contributor
Yanir Elm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I don't think so, for example if you take $$f(x)=x^2>g(x)=x-1$$
so you can see: $$f(x)-g(x)=x^2-x+1$$ and you cant find $$h(x)^2 $$which equal to that,
if that's what you mean..
answered Jan 4 at 19:21
Yanir ElmYanir Elm
693
New contributor
Yanir Elm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I don't think so, for example if you take $$f(x)=x^2>g(x)=x-1$$
so you can see: $$f(x)-g(x)=x^2-x+1$$ and you cant find $$h(x)^2 $$which equal to that,
if that's what you mean..
answered Jan 4 at 19:21
Yanir ElmYanir Elm
693
New contributor
Yanir Elm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 4 at 19:21
Yanir ElmYanir Elm
693
New contributor
Yanir Elm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 4 at 19:21
Yanir ElmYanir Elm
693
answered Jan 4 at 19:21
Yanir ElmYanir Elm
693
693
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New contributor
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There are two polynomials $h_1$ and $h_2$ sucht that $f(x) - g(x)= h_1(x)^2 + h_2(x)^2$ since $f(x) - g(x)$ is a non-negative polynomial.
– An_876_Joke
Jan 4 at 19:18
1
You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
– Martin R
Jan 4 at 19:21