If a univariate polynomial is greater than another, is their difference a square?
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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
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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
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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.
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– An_876_Joke
Jan 4 at 19:18
1
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You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
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– Martin R
Jan 4 at 19:21
add a comment |
$begingroup$
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
$endgroup$
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
1508
1508
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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.
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– An_876_Joke
Jan 4 at 19:18
1
$begingroup$
You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
$endgroup$
– Martin R
Jan 4 at 19:21
add a comment |
$begingroup$
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.
$endgroup$
– An_876_Joke
Jan 4 at 19:18
1
$begingroup$
You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
$endgroup$
– Martin R
Jan 4 at 19:21
$begingroup$
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.
$endgroup$
– An_876_Joke
Jan 4 at 19:18
$begingroup$
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.
$endgroup$
– An_876_Joke
Jan 4 at 19:18
1
1
$begingroup$
You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
$endgroup$
– Martin R
Jan 4 at 19:21
$begingroup$
You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
$endgroup$
– 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.
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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.
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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..
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3 Answers
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3 Answers
3
active
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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.
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add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Jan 4 at 19:41
answered Jan 4 at 19:19
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
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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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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
661
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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..
$endgroup$
add a comment |
$begingroup$
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..
$endgroup$
add a comment |
$begingroup$
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..
$endgroup$
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
693
add a comment |
add a comment |
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$begingroup$
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.
$endgroup$
– An_876_Joke
Jan 4 at 19:18
1
$begingroup$
You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
$endgroup$
– Martin R
Jan 4 at 19:21