Mixing
Real solutions of a monic polynomial $f(x) in mathbb R[x]$ only containing even degree terms
$begingroup$
Suppose $f(x) = x^{2k} + a_{2k-2} x^{2k-2} + dots + a_2 x^2 + a_0 in mathbb R[x]$ with $k in mathbb N$. Let us assume $k ge 2$. What can be said about the number of distinct real roots for $f(x)$?
By Fundamental Theorem of Algebra, it could be at most $2k$. But on the other hand, when I was playing with $x^4 + ax^2 + b$, it seems that if the roots are all real, it must be factored as $(x^2 - c^2)^2$ for some $c$ which would only give us two distinct real roots.
abstract-algebra polynomials roots
asked Jan 11 at 22:58
user1101010user1101010
7391730
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add a comment |
$begingroup$
Suppose $f(x) = x^{2k} + a_{2k-2} x^{2k-2} + dots + a_2 x^2 + a_0 in mathbb R[x]$ with $k in mathbb N$. Let us assume $k ge 2$. What can be said about the number of distinct real roots for $f(x)$?
By Fundamental Theorem of Algebra, it could be at most $2k$. But on the other hand, when I was playing with $x^4 + ax^2 + b$, it seems that if the roots are all real, it must be factored as $(x^2 - c^2)^2$ for some $c$ which would only give us two distinct real roots.
abstract-algebra polynomials roots
asked Jan 11 at 22:58
user1101010user1101010
7391730
$endgroup$
add a comment |
$begingroup$
Suppose $f(x) = x^{2k} + a_{2k-2} x^{2k-2} + dots + a_2 x^2 + a_0 in mathbb R[x]$ with $k in mathbb N$. Let us assume $k ge 2$. What can be said about the number of distinct real roots for $f(x)$?
By Fundamental Theorem of Algebra, it could be at most $2k$. But on the other hand, when I was playing with $x^4 + ax^2 + b$, it seems that if the roots are all real, it must be factored as $(x^2 - c^2)^2$ for some $c$ which would only give us two distinct real roots.
abstract-algebra polynomials roots
asked Jan 11 at 22:58
user1101010user1101010
7391730
$endgroup$
Suppose $f(x) = x^{2k} + a_{2k-2} x^{2k-2} + dots + a_2 x^2 + a_0 in mathbb R[x]$ with $k in mathbb N$. Let us assume $k ge 2$. What can be said about the number of distinct real roots for $f(x)$?
By Fundamental Theorem of Algebra, it could be at most $2k$. But on the other hand, when I was playing with $x^4 + ax^2 + b$, it seems that if the roots are all real, it must be factored as $(x^2 - c^2)^2$ for some $c$ which would only give us two distinct real roots.
abstract-algebra polynomials roots
abstract-algebra polynomials roots
asked Jan 11 at 22:58
user1101010user1101010
7391730
asked Jan 11 at 22:58
user1101010user1101010
7391730
asked Jan 11 at 22:58
user1101010user1101010
7391730
asked Jan 11 at 22:58
user1101010user1101010
7391730
asked Jan 11 at 22:58
user1101010user1101010
7391730
7391730
add a comment |
add a comment |
3 Answers
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$begingroup$
Well, writing $g(x)=x^k+a_{2k-2}x^{k-1}+dots+a_2x+a_0$, we have $f(x)=g(x^2)$. So, the roots of $f$ are just the square roots of the roots of $g$. This means that $f$ has two real roots for every positive root of $g$, and one more real root if $0$ is a root of $g$.
In particular, $f$ could have any number of distinct real roots from $0$ to $2k$. Indeed, to get any even number $2m$ of real roots ($0leq mleq k$), just let $g$ be a polynomial with exactly $m$ positive roots and for which $0$ is not a root. To get an odd number $2m+1$ of real roots ($0leq m<k$), just let $g$ have $m$ positive roots and also have $0$ as a root.
answered Jan 11 at 23:08
Eric WofseyEric Wofsey
185k14214341
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add a comment |
$begingroup$
What can be said easily is that the real roots come in pairs, as obviously with $x_0$ also $-x_0$ will be a root, and if $0$ is a root, it has an even degree $ge 2$.
Also, $x^4-5x^2+4=(x^2-1)(x^2-4)=0$ has roots $pm1, pm2$.
Basically your equation has real roots $x_{2i-1}=-x_{2i}$ that correspond to the non-negative roots $y_i$ of $y^k+a_{2k-2}y^{k-1}+ldots+a_2y+a_0=0$ via $x^2_{2i-1}=x^2_{2i}=y_i$.
answered Jan 11 at 23:10
IngixIngix
3,869146
$endgroup$
add a comment |
$begingroup$
If $a_0ne 0$ then the number of real roots is even.
If $a_0=0$, then the number of real roots is odd.
Observe that $f(x)$ can be written as $g(x^2)$ for a polynomial $g$ (with the same coefficients $a_{2k}$).
Every positive root $s$ of $g$ determines two real solutions of $f$: $pmsqrt s$.
answered Jan 11 at 23:07
BerciBerci
60.7k23673
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, writing $g(x)=x^k+a_{2k-2}x^{k-1}+dots+a_2x+a_0$, we have $f(x)=g(x^2)$. So, the roots of $f$ are just the square roots of the roots of $g$. This means that $f$ has two real roots for every positive root of $g$, and one more real root if $0$ is a root of $g$.
In particular, $f$ could have any number of distinct real roots from $0$ to $2k$. Indeed, to get any even number $2m$ of real roots ($0leq mleq k$), just let $g$ be a polynomial with exactly $m$ positive roots and for which $0$ is not a root. To get an odd number $2m+1$ of real roots ($0leq m<k$), just let $g$ have $m$ positive roots and also have $0$ as a root.
answered Jan 11 at 23:08
Eric WofseyEric Wofsey
185k14214341
$endgroup$
add a comment |
$begingroup$
Well, writing $g(x)=x^k+a_{2k-2}x^{k-1}+dots+a_2x+a_0$, we have $f(x)=g(x^2)$. So, the roots of $f$ are just the square roots of the roots of $g$. This means that $f$ has two real roots for every positive root of $g$, and one more real root if $0$ is a root of $g$.
In particular, $f$ could have any number of distinct real roots from $0$ to $2k$. Indeed, to get any even number $2m$ of real roots ($0leq mleq k$), just let $g$ be a polynomial with exactly $m$ positive roots and for which $0$ is not a root. To get an odd number $2m+1$ of real roots ($0leq m<k$), just let $g$ have $m$ positive roots and also have $0$ as a root.
answered Jan 11 at 23:08
Eric WofseyEric Wofsey
185k14214341
$endgroup$
add a comment |
$begingroup$
Well, writing $g(x)=x^k+a_{2k-2}x^{k-1}+dots+a_2x+a_0$, we have $f(x)=g(x^2)$. So, the roots of $f$ are just the square roots of the roots of $g$. This means that $f$ has two real roots for every positive root of $g$, and one more real root if $0$ is a root of $g$.
In particular, $f$ could have any number of distinct real roots from $0$ to $2k$. Indeed, to get any even number $2m$ of real roots ($0leq mleq k$), just let $g$ be a polynomial with exactly $m$ positive roots and for which $0$ is not a root. To get an odd number $2m+1$ of real roots ($0leq m<k$), just let $g$ have $m$ positive roots and also have $0$ as a root.
answered Jan 11 at 23:08
Eric WofseyEric Wofsey
185k14214341
$endgroup$
Well, writing $g(x)=x^k+a_{2k-2}x^{k-1}+dots+a_2x+a_0$, we have $f(x)=g(x^2)$. So, the roots of $f$ are just the square roots of the roots of $g$. This means that $f$ has two real roots for every positive root of $g$, and one more real root if $0$ is a root of $g$.
In particular, $f$ could have any number of distinct real roots from $0$ to $2k$. Indeed, to get any even number $2m$ of real roots ($0leq mleq k$), just let $g$ be a polynomial with exactly $m$ positive roots and for which $0$ is not a root. To get an odd number $2m+1$ of real roots ($0leq m<k$), just let $g$ have $m$ positive roots and also have $0$ as a root.
answered Jan 11 at 23:08
Eric WofseyEric Wofsey
185k14214341
answered Jan 11 at 23:08
Eric WofseyEric Wofsey
185k14214341
answered Jan 11 at 23:08
Eric WofseyEric Wofsey
185k14214341
answered Jan 11 at 23:08
Eric WofseyEric Wofsey
185k14214341
185k14214341
add a comment |
add a comment |
$begingroup$
What can be said easily is that the real roots come in pairs, as obviously with $x_0$ also $-x_0$ will be a root, and if $0$ is a root, it has an even degree $ge 2$.
Also, $x^4-5x^2+4=(x^2-1)(x^2-4)=0$ has roots $pm1, pm2$.
Basically your equation has real roots $x_{2i-1}=-x_{2i}$ that correspond to the non-negative roots $y_i$ of $y^k+a_{2k-2}y^{k-1}+ldots+a_2y+a_0=0$ via $x^2_{2i-1}=x^2_{2i}=y_i$.
answered Jan 11 at 23:10
IngixIngix
3,869146
$endgroup$
add a comment |
$begingroup$
What can be said easily is that the real roots come in pairs, as obviously with $x_0$ also $-x_0$ will be a root, and if $0$ is a root, it has an even degree $ge 2$.
Also, $x^4-5x^2+4=(x^2-1)(x^2-4)=0$ has roots $pm1, pm2$.
Basically your equation has real roots $x_{2i-1}=-x_{2i}$ that correspond to the non-negative roots $y_i$ of $y^k+a_{2k-2}y^{k-1}+ldots+a_2y+a_0=0$ via $x^2_{2i-1}=x^2_{2i}=y_i$.
answered Jan 11 at 23:10
IngixIngix
3,869146
$endgroup$
add a comment |
$begingroup$
What can be said easily is that the real roots come in pairs, as obviously with $x_0$ also $-x_0$ will be a root, and if $0$ is a root, it has an even degree $ge 2$.
Also, $x^4-5x^2+4=(x^2-1)(x^2-4)=0$ has roots $pm1, pm2$.
Basically your equation has real roots $x_{2i-1}=-x_{2i}$ that correspond to the non-negative roots $y_i$ of $y^k+a_{2k-2}y^{k-1}+ldots+a_2y+a_0=0$ via $x^2_{2i-1}=x^2_{2i}=y_i$.
answered Jan 11 at 23:10
IngixIngix
3,869146
$endgroup$
What can be said easily is that the real roots come in pairs, as obviously with $x_0$ also $-x_0$ will be a root, and if $0$ is a root, it has an even degree $ge 2$.
Also, $x^4-5x^2+4=(x^2-1)(x^2-4)=0$ has roots $pm1, pm2$.
Basically your equation has real roots $x_{2i-1}=-x_{2i}$ that correspond to the non-negative roots $y_i$ of $y^k+a_{2k-2}y^{k-1}+ldots+a_2y+a_0=0$ via $x^2_{2i-1}=x^2_{2i}=y_i$.
answered Jan 11 at 23:10
IngixIngix
3,869146
answered Jan 11 at 23:10
IngixIngix
3,869146
answered Jan 11 at 23:10
IngixIngix
3,869146
answered Jan 11 at 23:10
IngixIngix
3,869146
3,869146
add a comment |
add a comment |
$begingroup$
If $a_0ne 0$ then the number of real roots is even.
If $a_0=0$, then the number of real roots is odd.
Observe that $f(x)$ can be written as $g(x^2)$ for a polynomial $g$ (with the same coefficients $a_{2k}$).
Every positive root $s$ of $g$ determines two real solutions of $f$: $pmsqrt s$.
answered Jan 11 at 23:07
BerciBerci
60.7k23673
$endgroup$
add a comment |
$begingroup$
If $a_0ne 0$ then the number of real roots is even.
If $a_0=0$, then the number of real roots is odd.
Observe that $f(x)$ can be written as $g(x^2)$ for a polynomial $g$ (with the same coefficients $a_{2k}$).
Every positive root $s$ of $g$ determines two real solutions of $f$: $pmsqrt s$.
answered Jan 11 at 23:07
BerciBerci
60.7k23673
$endgroup$
add a comment |
$begingroup$
If $a_0ne 0$ then the number of real roots is even.
If $a_0=0$, then the number of real roots is odd.
Observe that $f(x)$ can be written as $g(x^2)$ for a polynomial $g$ (with the same coefficients $a_{2k}$).
Every positive root $s$ of $g$ determines two real solutions of $f$: $pmsqrt s$.
answered Jan 11 at 23:07
BerciBerci
60.7k23673
$endgroup$
If $a_0ne 0$ then the number of real roots is even.
If $a_0=0$, then the number of real roots is odd.
Observe that $f(x)$ can be written as $g(x^2)$ for a polynomial $g$ (with the same coefficients $a_{2k}$).
Every positive root $s$ of $g$ determines two real solutions of $f$: $pmsqrt s$.
answered Jan 11 at 23:07
BerciBerci
60.7k23673
answered Jan 11 at 23:07
BerciBerci
60.7k23673
answered Jan 11 at 23:07
BerciBerci
60.7k23673
answered Jan 11 at 23:07
BerciBerci
60.7k23673
60.7k23673
add a comment |
add a comment |
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