$-loglvert P(x)rvert$ is convex on an interval between two consecutive roots
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Let $P(x)=a_nx^n+dots+a_0$ ($n>1,a_nneq0$) be a polynomial with real coefficients that has only real roots. Then the function $f(x)=-loglvert P(x)rvert$ is convex on any open interval between two consecutive roots of $P(x)$.
I have tried to use $f''geq 0$ and induction on $n$ without much success. Thanks for any help!
real-analysis
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Let $P(x)=a_nx^n+dots+a_0$ ($n>1,a_nneq0$) be a polynomial with real coefficients that has only real roots. Then the function $f(x)=-loglvert P(x)rvert$ is convex on any open interval between two consecutive roots of $P(x)$.
I have tried to use $f''geq 0$ and induction on $n$ without much success. Thanks for any help!
real-analysis
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $P(x)=a_nx^n+dots+a_0$ ($n>1,a_nneq0$) be a polynomial with real coefficients that has only real roots. Then the function $f(x)=-loglvert P(x)rvert$ is convex on any open interval between two consecutive roots of $P(x)$.
I have tried to use $f''geq 0$ and induction on $n$ without much success. Thanks for any help!
real-analysis
Let $P(x)=a_nx^n+dots+a_0$ ($n>1,a_nneq0$) be a polynomial with real coefficients that has only real roots. Then the function $f(x)=-loglvert P(x)rvert$ is convex on any open interval between two consecutive roots of $P(x)$.
I have tried to use $f''geq 0$ and induction on $n$ without much success. Thanks for any help!
real-analysis
real-analysis
edited Nov 18 at 19:57
Bernard
117k637109
117k637109
asked Nov 18 at 19:25
user64066
1,626715
1,626715
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By factoring out the roots, we may write $P(x)=prod_i (x-r_i)$, where we have used the assumption that all roots are real. Then $f(x)=-sum_ilog(|x-r_i|)$. Since the sum of convex functions is convex, it only remains to check that $-log(|x-r_i|)$ is convex on any interval not containing one of the roots, which can be done by verifying that the second derivative is positive.
$P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
– Paul Frost
Nov 18 at 22:47
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
By factoring out the roots, we may write $P(x)=prod_i (x-r_i)$, where we have used the assumption that all roots are real. Then $f(x)=-sum_ilog(|x-r_i|)$. Since the sum of convex functions is convex, it only remains to check that $-log(|x-r_i|)$ is convex on any interval not containing one of the roots, which can be done by verifying that the second derivative is positive.
$P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
– Paul Frost
Nov 18 at 22:47
add a comment |
up vote
1
down vote
accepted
By factoring out the roots, we may write $P(x)=prod_i (x-r_i)$, where we have used the assumption that all roots are real. Then $f(x)=-sum_ilog(|x-r_i|)$. Since the sum of convex functions is convex, it only remains to check that $-log(|x-r_i|)$ is convex on any interval not containing one of the roots, which can be done by verifying that the second derivative is positive.
$P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
– Paul Frost
Nov 18 at 22:47
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
By factoring out the roots, we may write $P(x)=prod_i (x-r_i)$, where we have used the assumption that all roots are real. Then $f(x)=-sum_ilog(|x-r_i|)$. Since the sum of convex functions is convex, it only remains to check that $-log(|x-r_i|)$ is convex on any interval not containing one of the roots, which can be done by verifying that the second derivative is positive.
By factoring out the roots, we may write $P(x)=prod_i (x-r_i)$, where we have used the assumption that all roots are real. Then $f(x)=-sum_ilog(|x-r_i|)$. Since the sum of convex functions is convex, it only remains to check that $-log(|x-r_i|)$ is convex on any interval not containing one of the roots, which can be done by verifying that the second derivative is positive.
answered Nov 18 at 22:09
Mike Hawk
1,402110
1,402110
$P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
– Paul Frost
Nov 18 at 22:47
add a comment |
$P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
– Paul Frost
Nov 18 at 22:47
$P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
– Paul Frost
Nov 18 at 22:47
$P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
– Paul Frost
Nov 18 at 22:47
add a comment |
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