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Finding the roots of a 2 variable polynomial
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$begingroup$
Consider the polynomial $x^{n+1}+y^{n+1}+yx^n$ where $ninmathbb{N}$ is odd. I would like to find the real roots of this polynomial. I believe the only root is $(x,y)=(0,0)$. So far, I've been able to show that if we have a root different than $(0,0)$ then either $x>0,y<0$ or $x<0,y>0$. One would only have to check for one of the cases, since the polynomial is homogeneous. However, I'm completely stuck. Any ideas? Thank you in advance.
polynomials homogeneous-equation
asked Mar 9 at 17:03
Ray BernRay Bern
15813
$endgroup$
add a comment |
$begingroup$
Consider the polynomial $x^{n+1}+y^{n+1}+yx^n$ where $ninmathbb{N}$ is odd. I would like to find the real roots of this polynomial. I believe the only root is $(x,y)=(0,0)$. So far, I've been able to show that if we have a root different than $(0,0)$ then either $x>0,y<0$ or $x<0,y>0$. One would only have to check for one of the cases, since the polynomial is homogeneous. However, I'm completely stuck. Any ideas? Thank you in advance.
polynomials homogeneous-equation
asked Mar 9 at 17:03
Ray BernRay Bern
15813
$endgroup$
add a comment |
$begingroup$
Consider the polynomial $x^{n+1}+y^{n+1}+yx^n$ where $ninmathbb{N}$ is odd. I would like to find the real roots of this polynomial. I believe the only root is $(x,y)=(0,0)$. So far, I've been able to show that if we have a root different than $(0,0)$ then either $x>0,y<0$ or $x<0,y>0$. One would only have to check for one of the cases, since the polynomial is homogeneous. However, I'm completely stuck. Any ideas? Thank you in advance.
polynomials homogeneous-equation
asked Mar 9 at 17:03
Ray BernRay Bern
15813
$endgroup$
Consider the polynomial $x^{n+1}+y^{n+1}+yx^n$ where $ninmathbb{N}$ is odd. I would like to find the real roots of this polynomial. I believe the only root is $(x,y)=(0,0)$. So far, I've been able to show that if we have a root different than $(0,0)$ then either $x>0,y<0$ or $x<0,y>0$. One would only have to check for one of the cases, since the polynomial is homogeneous. However, I'm completely stuck. Any ideas? Thank you in advance.
polynomials homogeneous-equation
polynomials homogeneous-equation
asked Mar 9 at 17:03
Ray BernRay Bern
15813
asked Mar 9 at 17:03
Ray BernRay Bern
15813
asked Mar 9 at 17:03
Ray BernRay Bern
15813
asked Mar 9 at 17:03
Ray BernRay Bern
15813
asked Mar 9 at 17:03
Ray BernRay Bern
15813
15813
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using the well-known inequality $ab le frac{a^p}{p} + frac{b^q}{q}$ for $a, , b ge 0$, where $p^{-1} + q^{-1} = 1$, one can estimate
$$
yx^n ge - |y||x|^n ge - frac{|y|^{n+1}}{n+1} - frac{n|x|^{n+1}}{n+1}
$$
and therefore, since $n+1$ is even,
$$
y^{n+1} +y^{n+1} + yx^n ge frac{n}{n+1}y^{n+1} + frac{1}{n+1}x^{n+1} , .
$$
This is positive except if $x = y = 0$.
answered Mar 9 at 17:10
Hans EnglerHans Engler
10.5k11836
$endgroup$
$begingroup$
Wow, did not expect to take this path, thank you!
$endgroup$
– Ray Bern
Mar 9 at 17:19
add a comment |
$begingroup$
As the polynomial is homogeneous, you solve
$$t^{n+1}+t+1=0$$ and all $(x,tx)$ are solutions.
But it doesn't take long to see that the polynomial has no real roots.
answered Mar 9 at 17:19
Yves DaoustYves Daoust
130k676227
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Using the well-known inequality $ab le frac{a^p}{p} + frac{b^q}{q}$ for $a, , b ge 0$, where $p^{-1} + q^{-1} = 1$, one can estimate
$$
yx^n ge - |y||x|^n ge - frac{|y|^{n+1}}{n+1} - frac{n|x|^{n+1}}{n+1}
$$
and therefore, since $n+1$ is even,
$$
y^{n+1} +y^{n+1} + yx^n ge frac{n}{n+1}y^{n+1} + frac{1}{n+1}x^{n+1} , .
$$
This is positive except if $x = y = 0$.
answered Mar 9 at 17:10
Hans EnglerHans Engler
10.5k11836
$endgroup$
$begingroup$
Wow, did not expect to take this path, thank you!
$endgroup$
– Ray Bern
Mar 9 at 17:19
add a comment |
$begingroup$
Using the well-known inequality $ab le frac{a^p}{p} + frac{b^q}{q}$ for $a, , b ge 0$, where $p^{-1} + q^{-1} = 1$, one can estimate
$$
yx^n ge - |y||x|^n ge - frac{|y|^{n+1}}{n+1} - frac{n|x|^{n+1}}{n+1}
$$
and therefore, since $n+1$ is even,
$$
y^{n+1} +y^{n+1} + yx^n ge frac{n}{n+1}y^{n+1} + frac{1}{n+1}x^{n+1} , .
$$
This is positive except if $x = y = 0$.
answered Mar 9 at 17:10
Hans EnglerHans Engler
10.5k11836
$endgroup$
$begingroup$
Wow, did not expect to take this path, thank you!
$endgroup$
– Ray Bern
Mar 9 at 17:19
add a comment |
$begingroup$
Using the well-known inequality $ab le frac{a^p}{p} + frac{b^q}{q}$ for $a, , b ge 0$, where $p^{-1} + q^{-1} = 1$, one can estimate
$$
yx^n ge - |y||x|^n ge - frac{|y|^{n+1}}{n+1} - frac{n|x|^{n+1}}{n+1}
$$
and therefore, since $n+1$ is even,
$$
y^{n+1} +y^{n+1} + yx^n ge frac{n}{n+1}y^{n+1} + frac{1}{n+1}x^{n+1} , .
$$
This is positive except if $x = y = 0$.
answered Mar 9 at 17:10
Hans EnglerHans Engler
10.5k11836
$endgroup$
Using the well-known inequality $ab le frac{a^p}{p} + frac{b^q}{q}$ for $a, , b ge 0$, where $p^{-1} + q^{-1} = 1$, one can estimate
$$
yx^n ge - |y||x|^n ge - frac{|y|^{n+1}}{n+1} - frac{n|x|^{n+1}}{n+1}
$$
and therefore, since $n+1$ is even,
$$
y^{n+1} +y^{n+1} + yx^n ge frac{n}{n+1}y^{n+1} + frac{1}{n+1}x^{n+1} , .
$$
This is positive except if $x = y = 0$.
answered Mar 9 at 17:10
Hans EnglerHans Engler
10.5k11836
edited Mar 9 at 17:40
answered Mar 9 at 17:10
Hans EnglerHans Engler
10.5k11836
answered Mar 9 at 17:10
Hans EnglerHans Engler
10.5k11836
answered Mar 9 at 17:10
Hans EnglerHans Engler
10.5k11836
10.5k11836
$begingroup$
Wow, did not expect to take this path, thank you!
$endgroup$
– Ray Bern
Mar 9 at 17:19
add a comment |
$begingroup$
Wow, did not expect to take this path, thank you!
$endgroup$
– Ray Bern
Mar 9 at 17:19
$begingroup$
Wow, did not expect to take this path, thank you!
$endgroup$
– Ray Bern
Mar 9 at 17:19
$begingroup$
Wow, did not expect to take this path, thank you!
$endgroup$
– Ray Bern
Mar 9 at 17:19
add a comment |
$begingroup$
As the polynomial is homogeneous, you solve
$$t^{n+1}+t+1=0$$ and all $(x,tx)$ are solutions.
But it doesn't take long to see that the polynomial has no real roots.
answered Mar 9 at 17:19
Yves DaoustYves Daoust
130k676227
$endgroup$
add a comment |
$begingroup$
As the polynomial is homogeneous, you solve
$$t^{n+1}+t+1=0$$ and all $(x,tx)$ are solutions.
But it doesn't take long to see that the polynomial has no real roots.
answered Mar 9 at 17:19
Yves DaoustYves Daoust
130k676227
$endgroup$
add a comment |
$begingroup$
As the polynomial is homogeneous, you solve
$$t^{n+1}+t+1=0$$ and all $(x,tx)$ are solutions.
But it doesn't take long to see that the polynomial has no real roots.
answered Mar 9 at 17:19
Yves DaoustYves Daoust
130k676227
$endgroup$
As the polynomial is homogeneous, you solve
$$t^{n+1}+t+1=0$$ and all $(x,tx)$ are solutions.
But it doesn't take long to see that the polynomial has no real roots.
answered Mar 9 at 17:19
Yves DaoustYves Daoust
130k676227
answered Mar 9 at 17:19
Yves DaoustYves Daoust
130k676227
answered Mar 9 at 17:19
Yves DaoustYves Daoust
130k676227
answered Mar 9 at 17:19
Yves DaoustYves Daoust
130k676227
130k676227
add a comment |
add a comment |
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