Annoying Polynomial Inequality
Clash Royale CLAN TAG#URR8PPP
up vote
14
down vote
favorite
Suppose we have a polynomial satisfying $p+p''' geq p'+p''$ for all $x$. Then prove that $p(x)geq 0$ for all $x$.
I've been stuck on this problem for weeks. The best I can do is supposing there exists $x$ so that $p(x) < 0$ (for a contradiction), that we can find a point where $p'=0$, $p''>0$, and $p'''>0$, but I can't use this (nor think of anything new) to show it violates the inequality!
Is there a line of thinking I am missing, or am I only missing something little? Is there a better way to go about proving this?
inequality polynomials
edited Aug 29 at 5:18
Robson
47320
asked Jul 31 '13 at 16:35
FireGarden
2,71921525
add a comment |Â
up vote
14
down vote
favorite
Suppose we have a polynomial satisfying $p+p''' geq p'+p''$ for all $x$. Then prove that $p(x)geq 0$ for all $x$.
I've been stuck on this problem for weeks. The best I can do is supposing there exists $x$ so that $p(x) < 0$ (for a contradiction), that we can find a point where $p'=0$, $p''>0$, and $p'''>0$, but I can't use this (nor think of anything new) to show it violates the inequality!
Is there a line of thinking I am missing, or am I only missing something little? Is there a better way to go about proving this?
inequality polynomials
edited Aug 29 at 5:18
Robson
47320
asked Jul 31 '13 at 16:35
FireGarden
2,71921525
1
maybe try rearranging into $(p''' - p'') geq (p' - p)$ to get some symmetry in the picture?
– Scaramouche
Jul 31 '13 at 16:43
Another thing to note is that $q:=p+p'''-p'-p''$ has the same degree and leading coefficient as $p$; this shows you that $p$ must be of even degree and have positive leading coefficient.
– Nick Peterson
Jul 31 '13 at 17:02
1
Why does this show $p$ is of even degree and has positive leading coefficient?
– Spine Feast
Aug 2 '13 at 21:05
add a comment |Â
up vote
14
down vote
favorite
up vote
14
down vote
favorite
Suppose we have a polynomial satisfying $p+p''' geq p'+p''$ for all $x$. Then prove that $p(x)geq 0$ for all $x$.
I've been stuck on this problem for weeks. The best I can do is supposing there exists $x$ so that $p(x) < 0$ (for a contradiction), that we can find a point where $p'=0$, $p''>0$, and $p'''>0$, but I can't use this (nor think of anything new) to show it violates the inequality!
Is there a line of thinking I am missing, or am I only missing something little? Is there a better way to go about proving this?
inequality polynomials
edited Aug 29 at 5:18
Robson
47320
asked Jul 31 '13 at 16:35
FireGarden
2,71921525
Suppose we have a polynomial satisfying $p+p''' geq p'+p''$ for all $x$. Then prove that $p(x)geq 0$ for all $x$.
I've been stuck on this problem for weeks. The best I can do is supposing there exists $x$ so that $p(x) < 0$ (for a contradiction), that we can find a point where $p'=0$, $p''>0$, and $p'''>0$, but I can't use this (nor think of anything new) to show it violates the inequality!
Is there a line of thinking I am missing, or am I only missing something little? Is there a better way to go about proving this?
inequality polynomials
edited Aug 29 at 5:18
Robson
47320
asked Jul 31 '13 at 16:35
FireGarden
2,71921525
edited Aug 29 at 5:18
Robson
47320
edited Aug 29 at 5:18
Robson
47320
edited Aug 29 at 5:18
Robson
47320
47320
asked Jul 31 '13 at 16:35
FireGarden
2,71921525
asked Jul 31 '13 at 16:35
FireGarden
2,71921525
asked Jul 31 '13 at 16:35
FireGarden
2,71921525
2,71921525
1
maybe try rearranging into $(p''' - p'') geq (p' - p)$ to get some symmetry in the picture?
– Scaramouche
Jul 31 '13 at 16:43
Another thing to note is that $q:=p+p'''-p'-p''$ has the same degree and leading coefficient as $p$; this shows you that $p$ must be of even degree and have positive leading coefficient.
– Nick Peterson
Jul 31 '13 at 17:02
1
Why does this show $p$ is of even degree and has positive leading coefficient?
– Spine Feast
Aug 2 '13 at 21:05
add a comment |Â
1
maybe try rearranging into $(p''' - p'') geq (p' - p)$ to get some symmetry in the picture?
– Scaramouche
Jul 31 '13 at 16:43
Another thing to note is that $q:=p+p'''-p'-p''$ has the same degree and leading coefficient as $p$; this shows you that $p$ must be of even degree and have positive leading coefficient.
– Nick Peterson
Jul 31 '13 at 17:02
1
Why does this show $p$ is of even degree and has positive leading coefficient?
– Spine Feast
Aug 2 '13 at 21:05
1
1
maybe try rearranging into $(p''' - p'') geq (p' - p)$ to get some symmetry in the picture?
– Scaramouche
Jul 31 '13 at 16:43
maybe try rearranging into $(p''' - p'') geq (p' - p)$ to get some symmetry in the picture?
– Scaramouche
Jul 31 '13 at 16:43
Another thing to note is that $q:=p+p'''-p'-p''$ has the same degree and leading coefficient as $p$; this shows you that $p$ must be of even degree and have positive leading coefficient.
– Nick Peterson
Jul 31 '13 at 17:02
Another thing to note is that $q:=p+p'''-p'-p''$ has the same degree and leading coefficient as $p$; this shows you that $p$ must be of even degree and have positive leading coefficient.
– Nick Peterson
Jul 31 '13 at 17:02
1
1
Why does this show $p$ is of even degree and has positive leading coefficient?
– Spine Feast
Aug 2 '13 at 21:05
Why does this show $p$ is of even degree and has positive leading coefficient?
– Spine Feast
Aug 2 '13 at 21:05
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
12
down vote
accepted
The statement can be slightly stronger:
Claim: If a real polynomial $p$ satisfies that $p+p'''ge p'+p''$, then
either $p>0$ or $pequiv 0$.
Proof 1:
Define
$$f(x)=e^xleft(p(x)-2p'(x)+p''(x)right).$$
Then $limlimits_xto-inftyf(x)=0$ and
$$f'(x)=e^xleft(p(x)-p'(x)-p''(x)+p'''(x)right)ge 0,$$
i.e. $f$ is increasing. It follows that $fge 0$, and hence
$$p-2p'+p''ge 0.$$
Define
$$g(x)=e^-xleft(p(x)-p'(x)right).$$
Then $limlimits_xto+inftyg(x)=0$ and
$$g'(x)=-e^-xleft(p(x)-2p'(x)+p''(x)right)le 0,$$
i.e. $g$ is decreasing. It follows that $gge 0$, and hence
$$p-p'ge 0.$$
Define
$$h(x)=e^-xp(x).$$
Then $limlimits_xto+inftyh(x)=0$ and
$$h'(x)=-g(x)le 0,$$
i.e. $h$ is decreasing. Therefore, either $h>0$ or $hequiv 0$. The conclusion follows.
Proof 2:
Denote $q=p-p'$. Then
$$p+p'''ge p'+p''iff q-q''ge 0.$$
It implies that
$deg(q-q'')=deg q=deg p$ must be even. Therefore, $q$ has a global minimum value point, say $x_q$. Then
$$q(x)ge q(x_q)ge q''(x_q)ge 0,$$
i.e.
$$pge p'.$$
Similarly, $p$ also has a global minimum value point, say $x_p$. Then
$$p(x_p)ge p'(x_p)= 0.$$
If $p(x_p)>0$, we are done. To complete the proof, suppose that $p(x_p)=0$ and let us show that $pequiv 0$ by reduction to absurdity. If $p(x_p)=0$ and $p$ is not constant, then there exists an integer $kge 1$, such that $p(x)=(x-x_p)^k r(x)$ for some polynomial $r$ with $r(x_p)ne 0$. Since $x_p$ is a minimum point of $p$, $k$ must be even. Then
$$p(x)-p'(x)=-kr(x_p)(x-x_p)^k-1+O((x-x_p)^k)$$
cannot be always non-negative, a contradiction.
answered Jul 31 '13 at 17:37
23rd
13.8k2754
@GitGud: Since $f$ is increasing and $f(-infty)=0$, $fge 0$.
– 23rd
Jul 31 '13 at 19:42
Of course, thanks.
– Git Gud
Jul 31 '13 at 19:44
@GitGud: You are welcome.
– 23rd
Jul 31 '13 at 19:47
@23rd: From $h$ decreasing and $lim _x to infty h(x)=0$ ,how do you conclude either $h equiv 0$ or $h>0$ ? I seem to only conclude $h ge 0$ ....
– Souvik Dey
Jan 3 '15 at 12:41
@SouvikDey: Please note that: (1) $h(x)=0$ if and only if $p(x)=0$; (2) $p$ is a polynomial.
– 23rd
Jan 6 '15 at 16:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
The statement can be slightly stronger:
Claim: If a real polynomial $p$ satisfies that $p+p'''ge p'+p''$, then
either $p>0$ or $pequiv 0$.
Proof 1:
Define
$$f(x)=e^xleft(p(x)-2p'(x)+p''(x)right).$$
Then $limlimits_xto-inftyf(x)=0$ and
$$f'(x)=e^xleft(p(x)-p'(x)-p''(x)+p'''(x)right)ge 0,$$
i.e. $f$ is increasing. It follows that $fge 0$, and hence
$$p-2p'+p''ge 0.$$
Define
$$g(x)=e^-xleft(p(x)-p'(x)right).$$
Then $limlimits_xto+inftyg(x)=0$ and
$$g'(x)=-e^-xleft(p(x)-2p'(x)+p''(x)right)le 0,$$
i.e. $g$ is decreasing. It follows that $gge 0$, and hence
$$p-p'ge 0.$$
Define
$$h(x)=e^-xp(x).$$
Then $limlimits_xto+inftyh(x)=0$ and
$$h'(x)=-g(x)le 0,$$
i.e. $h$ is decreasing. Therefore, either $h>0$ or $hequiv 0$. The conclusion follows.
Proof 2:
Denote $q=p-p'$. Then
$$p+p'''ge p'+p''iff q-q''ge 0.$$
It implies that
$deg(q-q'')=deg q=deg p$ must be even. Therefore, $q$ has a global minimum value point, say $x_q$. Then
$$q(x)ge q(x_q)ge q''(x_q)ge 0,$$
i.e.
$$pge p'.$$
Similarly, $p$ also has a global minimum value point, say $x_p$. Then
$$p(x_p)ge p'(x_p)= 0.$$
If $p(x_p)>0$, we are done. To complete the proof, suppose that $p(x_p)=0$ and let us show that $pequiv 0$ by reduction to absurdity. If $p(x_p)=0$ and $p$ is not constant, then there exists an integer $kge 1$, such that $p(x)=(x-x_p)^k r(x)$ for some polynomial $r$ with $r(x_p)ne 0$. Since $x_p$ is a minimum point of $p$, $k$ must be even. Then
$$p(x)-p'(x)=-kr(x_p)(x-x_p)^k-1+O((x-x_p)^k)$$
cannot be always non-negative, a contradiction.
answered Jul 31 '13 at 17:37
23rd
13.8k2754
@GitGud: Since $f$ is increasing and $f(-infty)=0$, $fge 0$.
– 23rd
Jul 31 '13 at 19:42
Of course, thanks.
– Git Gud
Jul 31 '13 at 19:44
@GitGud: You are welcome.
– 23rd
Jul 31 '13 at 19:47
@23rd: From $h$ decreasing and $lim _x to infty h(x)=0$ ,how do you conclude either $h equiv 0$ or $h>0$ ? I seem to only conclude $h ge 0$ ....
– Souvik Dey
Jan 3 '15 at 12:41
@SouvikDey: Please note that: (1) $h(x)=0$ if and only if $p(x)=0$; (2) $p$ is a polynomial.
– 23rd
Jan 6 '15 at 16:34
add a comment |Â
up vote
12
down vote
accepted
The statement can be slightly stronger:
Claim: If a real polynomial $p$ satisfies that $p+p'''ge p'+p''$, then
either $p>0$ or $pequiv 0$.
Proof 1:
Define
$$f(x)=e^xleft(p(x)-2p'(x)+p''(x)right).$$
Then $limlimits_xto-inftyf(x)=0$ and
$$f'(x)=e^xleft(p(x)-p'(x)-p''(x)+p'''(x)right)ge 0,$$
i.e. $f$ is increasing. It follows that $fge 0$, and hence
$$p-2p'+p''ge 0.$$
Define
$$g(x)=e^-xleft(p(x)-p'(x)right).$$
Then $limlimits_xto+inftyg(x)=0$ and
$$g'(x)=-e^-xleft(p(x)-2p'(x)+p''(x)right)le 0,$$
i.e. $g$ is decreasing. It follows that $gge 0$, and hence
$$p-p'ge 0.$$
Define
$$h(x)=e^-xp(x).$$
Then $limlimits_xto+inftyh(x)=0$ and
$$h'(x)=-g(x)le 0,$$
i.e. $h$ is decreasing. Therefore, either $h>0$ or $hequiv 0$. The conclusion follows.
Proof 2:
Denote $q=p-p'$. Then
$$p+p'''ge p'+p''iff q-q''ge 0.$$
It implies that
$deg(q-q'')=deg q=deg p$ must be even. Therefore, $q$ has a global minimum value point, say $x_q$. Then
$$q(x)ge q(x_q)ge q''(x_q)ge 0,$$
i.e.
$$pge p'.$$
Similarly, $p$ also has a global minimum value point, say $x_p$. Then
$$p(x_p)ge p'(x_p)= 0.$$
If $p(x_p)>0$, we are done. To complete the proof, suppose that $p(x_p)=0$ and let us show that $pequiv 0$ by reduction to absurdity. If $p(x_p)=0$ and $p$ is not constant, then there exists an integer $kge 1$, such that $p(x)=(x-x_p)^k r(x)$ for some polynomial $r$ with $r(x_p)ne 0$. Since $x_p$ is a minimum point of $p$, $k$ must be even. Then
$$p(x)-p'(x)=-kr(x_p)(x-x_p)^k-1+O((x-x_p)^k)$$
cannot be always non-negative, a contradiction.
answered Jul 31 '13 at 17:37
23rd
13.8k2754
@GitGud: Since $f$ is increasing and $f(-infty)=0$, $fge 0$.
– 23rd
Jul 31 '13 at 19:42
Of course, thanks.
– Git Gud
Jul 31 '13 at 19:44
@GitGud: You are welcome.
– 23rd
Jul 31 '13 at 19:47
@23rd: From $h$ decreasing and $lim _x to infty h(x)=0$ ,how do you conclude either $h equiv 0$ or $h>0$ ? I seem to only conclude $h ge 0$ ....
– Souvik Dey
Jan 3 '15 at 12:41
@SouvikDey: Please note that: (1) $h(x)=0$ if and only if $p(x)=0$; (2) $p$ is a polynomial.
– 23rd
Jan 6 '15 at 16:34
add a comment |Â
up vote
12
down vote
accepted
up vote
12
down vote
accepted
The statement can be slightly stronger:
Claim: If a real polynomial $p$ satisfies that $p+p'''ge p'+p''$, then
either $p>0$ or $pequiv 0$.
Proof 1:
Define
$$f(x)=e^xleft(p(x)-2p'(x)+p''(x)right).$$
Then $limlimits_xto-inftyf(x)=0$ and
$$f'(x)=e^xleft(p(x)-p'(x)-p''(x)+p'''(x)right)ge 0,$$
i.e. $f$ is increasing. It follows that $fge 0$, and hence
$$p-2p'+p''ge 0.$$
Define
$$g(x)=e^-xleft(p(x)-p'(x)right).$$
Then $limlimits_xto+inftyg(x)=0$ and
$$g'(x)=-e^-xleft(p(x)-2p'(x)+p''(x)right)le 0,$$
i.e. $g$ is decreasing. It follows that $gge 0$, and hence
$$p-p'ge 0.$$
Define
$$h(x)=e^-xp(x).$$
Then $limlimits_xto+inftyh(x)=0$ and
$$h'(x)=-g(x)le 0,$$
i.e. $h$ is decreasing. Therefore, either $h>0$ or $hequiv 0$. The conclusion follows.
Proof 2:
Denote $q=p-p'$. Then
$$p+p'''ge p'+p''iff q-q''ge 0.$$
It implies that
$deg(q-q'')=deg q=deg p$ must be even. Therefore, $q$ has a global minimum value point, say $x_q$. Then
$$q(x)ge q(x_q)ge q''(x_q)ge 0,$$
i.e.
$$pge p'.$$
Similarly, $p$ also has a global minimum value point, say $x_p$. Then
$$p(x_p)ge p'(x_p)= 0.$$
If $p(x_p)>0$, we are done. To complete the proof, suppose that $p(x_p)=0$ and let us show that $pequiv 0$ by reduction to absurdity. If $p(x_p)=0$ and $p$ is not constant, then there exists an integer $kge 1$, such that $p(x)=(x-x_p)^k r(x)$ for some polynomial $r$ with $r(x_p)ne 0$. Since $x_p$ is a minimum point of $p$, $k$ must be even. Then
$$p(x)-p'(x)=-kr(x_p)(x-x_p)^k-1+O((x-x_p)^k)$$
cannot be always non-negative, a contradiction.
answered Jul 31 '13 at 17:37
23rd
13.8k2754
The statement can be slightly stronger:
Claim: If a real polynomial $p$ satisfies that $p+p'''ge p'+p''$, then
either $p>0$ or $pequiv 0$.
Proof 1:
Define
$$f(x)=e^xleft(p(x)-2p'(x)+p''(x)right).$$
Then $limlimits_xto-inftyf(x)=0$ and
$$f'(x)=e^xleft(p(x)-p'(x)-p''(x)+p'''(x)right)ge 0,$$
i.e. $f$ is increasing. It follows that $fge 0$, and hence
$$p-2p'+p''ge 0.$$
Define
$$g(x)=e^-xleft(p(x)-p'(x)right).$$
Then $limlimits_xto+inftyg(x)=0$ and
$$g'(x)=-e^-xleft(p(x)-2p'(x)+p''(x)right)le 0,$$
i.e. $g$ is decreasing. It follows that $gge 0$, and hence
$$p-p'ge 0.$$
Define
$$h(x)=e^-xp(x).$$
Then $limlimits_xto+inftyh(x)=0$ and
$$h'(x)=-g(x)le 0,$$
i.e. $h$ is decreasing. Therefore, either $h>0$ or $hequiv 0$. The conclusion follows.
Proof 2:
Denote $q=p-p'$. Then
$$p+p'''ge p'+p''iff q-q''ge 0.$$
It implies that
$deg(q-q'')=deg q=deg p$ must be even. Therefore, $q$ has a global minimum value point, say $x_q$. Then
$$q(x)ge q(x_q)ge q''(x_q)ge 0,$$
i.e.
$$pge p'.$$
Similarly, $p$ also has a global minimum value point, say $x_p$. Then
$$p(x_p)ge p'(x_p)= 0.$$
If $p(x_p)>0$, we are done. To complete the proof, suppose that $p(x_p)=0$ and let us show that $pequiv 0$ by reduction to absurdity. If $p(x_p)=0$ and $p$ is not constant, then there exists an integer $kge 1$, such that $p(x)=(x-x_p)^k r(x)$ for some polynomial $r$ with $r(x_p)ne 0$. Since $x_p$ is a minimum point of $p$, $k$ must be even. Then
$$p(x)-p'(x)=-kr(x_p)(x-x_p)^k-1+O((x-x_p)^k)$$
cannot be always non-negative, a contradiction.
answered Jul 31 '13 at 17:37
23rd
13.8k2754
edited Aug 2 '13 at 18:36
answered Jul 31 '13 at 17:37
23rd
13.8k2754
answered Jul 31 '13 at 17:37
23rd
13.8k2754
answered Jul 31 '13 at 17:37
23rd
13.8k2754
13.8k2754
@GitGud: Since $f$ is increasing and $f(-infty)=0$, $fge 0$.
– 23rd
Jul 31 '13 at 19:42
Of course, thanks.
– Git Gud
Jul 31 '13 at 19:44
@GitGud: You are welcome.
– 23rd
Jul 31 '13 at 19:47
@23rd: From $h$ decreasing and $lim _x to infty h(x)=0$ ,how do you conclude either $h equiv 0$ or $h>0$ ? I seem to only conclude $h ge 0$ ....
– Souvik Dey
Jan 3 '15 at 12:41
@SouvikDey: Please note that: (1) $h(x)=0$ if and only if $p(x)=0$; (2) $p$ is a polynomial.
– 23rd
Jan 6 '15 at 16:34
add a comment |Â
@GitGud: Since $f$ is increasing and $f(-infty)=0$, $fge 0$.
– 23rd
Jul 31 '13 at 19:42
Of course, thanks.
– Git Gud
Jul 31 '13 at 19:44
@GitGud: You are welcome.
– 23rd
Jul 31 '13 at 19:47
@23rd: From $h$ decreasing and $lim _x to infty h(x)=0$ ,how do you conclude either $h equiv 0$ or $h>0$ ? I seem to only conclude $h ge 0$ ....
– Souvik Dey
Jan 3 '15 at 12:41
@SouvikDey: Please note that: (1) $h(x)=0$ if and only if $p(x)=0$; (2) $p$ is a polynomial.
– 23rd
Jan 6 '15 at 16:34
@GitGud: Since $f$ is increasing and $f(-infty)=0$, $fge 0$.
– 23rd
Jul 31 '13 at 19:42
@GitGud: Since $f$ is increasing and $f(-infty)=0$, $fge 0$.
– 23rd
Jul 31 '13 at 19:42
Of course, thanks.
– Git Gud
Jul 31 '13 at 19:44
Of course, thanks.
– Git Gud
Jul 31 '13 at 19:44
@GitGud: You are welcome.
– 23rd
Jul 31 '13 at 19:47
@GitGud: You are welcome.
– 23rd
Jul 31 '13 at 19:47
@23rd: From $h$ decreasing and $lim _x to infty h(x)=0$ ,how do you conclude either $h equiv 0$ or $h>0$ ? I seem to only conclude $h ge 0$ ....
– Souvik Dey
Jan 3 '15 at 12:41
@23rd: From $h$ decreasing and $lim _x to infty h(x)=0$ ,how do you conclude either $h equiv 0$ or $h>0$ ? I seem to only conclude $h ge 0$ ....
– Souvik Dey
Jan 3 '15 at 12:41
@SouvikDey: Please note that: (1) $h(x)=0$ if and only if $p(x)=0$; (2) $p$ is a polynomial.
– 23rd
Jan 6 '15 at 16:34
@SouvikDey: Please note that: (1) $h(x)=0$ if and only if $p(x)=0$; (2) $p$ is a polynomial.
– 23rd
Jan 6 '15 at 16:34
add a comment |Â
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1
maybe try rearranging into $(p''' - p'') geq (p' - p)$ to get some symmetry in the picture?
– Scaramouche
Jul 31 '13 at 16:43
Another thing to note is that $q:=p+p'''-p'-p''$ has the same degree and leading coefficient as $p$; this shows you that $p$ must be of even degree and have positive leading coefficient.
– Nick Peterson
Jul 31 '13 at 17:02
1
Why does this show $p$ is of even degree and has positive leading coefficient?
– Spine Feast
Aug 2 '13 at 21:05