x^2 times convex nonnegative polynomial is convex?
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If $p(x)$ is a convex nonnegative polynomial in $x$, is $x^2 p(x)$ convex?
Its second derivative is
$$
x^2 p''(x) + 4x p'(x) + 2 p(x).
$$
We know $x^2 p''(x)$ is nonnegative by convexity of $p$, and $2p(x)$ is nonnegative by assumption. I don't know how to conclude anything about $x p'(x)$ though.
If the general case is not convex, what about the special case $p(x) = (a + bx)^{2k}, k in mathbb{N}, a, b in mathbb{R}$?
polynomials convex-analysis
asked Nov 19 at 4:47
japreiss
1657
add a comment |
up vote
0
down vote
favorite
If $p(x)$ is a convex nonnegative polynomial in $x$, is $x^2 p(x)$ convex?
Its second derivative is
$$
x^2 p''(x) + 4x p'(x) + 2 p(x).
$$
We know $x^2 p''(x)$ is nonnegative by convexity of $p$, and $2p(x)$ is nonnegative by assumption. I don't know how to conclude anything about $x p'(x)$ though.
If the general case is not convex, what about the special case $p(x) = (a + bx)^{2k}, k in mathbb{N}, a, b in mathbb{R}$?
polynomials convex-analysis
asked Nov 19 at 4:47
japreiss
1657
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $p(x)$ is a convex nonnegative polynomial in $x$, is $x^2 p(x)$ convex?
Its second derivative is
$$
x^2 p''(x) + 4x p'(x) + 2 p(x).
$$
We know $x^2 p''(x)$ is nonnegative by convexity of $p$, and $2p(x)$ is nonnegative by assumption. I don't know how to conclude anything about $x p'(x)$ though.
If the general case is not convex, what about the special case $p(x) = (a + bx)^{2k}, k in mathbb{N}, a, b in mathbb{R}$?
polynomials convex-analysis
asked Nov 19 at 4:47
japreiss
1657
If $p(x)$ is a convex nonnegative polynomial in $x$, is $x^2 p(x)$ convex?
Its second derivative is
$$
x^2 p''(x) + 4x p'(x) + 2 p(x).
$$
We know $x^2 p''(x)$ is nonnegative by convexity of $p$, and $2p(x)$ is nonnegative by assumption. I don't know how to conclude anything about $x p'(x)$ though.
If the general case is not convex, what about the special case $p(x) = (a + bx)^{2k}, k in mathbb{N}, a, b in mathbb{R}$?
polynomials convex-analysis
polynomials convex-analysis
asked Nov 19 at 4:47
japreiss
1657
asked Nov 19 at 4:47
japreiss
1657
asked Nov 19 at 4:47
japreiss
1657
asked Nov 19 at 4:47
japreiss
1657
asked Nov 19 at 4:47
japreiss
1657
1657
add a comment |
add a comment |
1 Answer
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3
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No: for instance, $f(x)=x^2(x-1)^2$ is not convex. A quick way to see this is that $f(0)=f(1)=0$ but $f$ is strictly positive in between.
More generally, if $p(r)=0$ for some $rneq 0$ (and $p$ is not identically $0$), then $f(x)=x^2p(x)$ will not be convex since $f(0)=f(r)=0$ but $f$ is positive somewhere between $0$ and $r$.
answered Nov 19 at 4:54
Eric Wofsey
177k12202328
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
No: for instance, $f(x)=x^2(x-1)^2$ is not convex. A quick way to see this is that $f(0)=f(1)=0$ but $f$ is strictly positive in between.
More generally, if $p(r)=0$ for some $rneq 0$ (and $p$ is not identically $0$), then $f(x)=x^2p(x)$ will not be convex since $f(0)=f(r)=0$ but $f$ is positive somewhere between $0$ and $r$.
answered Nov 19 at 4:54
Eric Wofsey
177k12202328
add a comment |
up vote
3
down vote
No: for instance, $f(x)=x^2(x-1)^2$ is not convex. A quick way to see this is that $f(0)=f(1)=0$ but $f$ is strictly positive in between.
More generally, if $p(r)=0$ for some $rneq 0$ (and $p$ is not identically $0$), then $f(x)=x^2p(x)$ will not be convex since $f(0)=f(r)=0$ but $f$ is positive somewhere between $0$ and $r$.
answered Nov 19 at 4:54
Eric Wofsey
177k12202328
add a comment |
up vote
3
down vote
up vote
3
down vote
No: for instance, $f(x)=x^2(x-1)^2$ is not convex. A quick way to see this is that $f(0)=f(1)=0$ but $f$ is strictly positive in between.
More generally, if $p(r)=0$ for some $rneq 0$ (and $p$ is not identically $0$), then $f(x)=x^2p(x)$ will not be convex since $f(0)=f(r)=0$ but $f$ is positive somewhere between $0$ and $r$.
answered Nov 19 at 4:54
Eric Wofsey
177k12202328
No: for instance, $f(x)=x^2(x-1)^2$ is not convex. A quick way to see this is that $f(0)=f(1)=0$ but $f$ is strictly positive in between.
More generally, if $p(r)=0$ for some $rneq 0$ (and $p$ is not identically $0$), then $f(x)=x^2p(x)$ will not be convex since $f(0)=f(r)=0$ but $f$ is positive somewhere between $0$ and $r$.
answered Nov 19 at 4:54
Eric Wofsey
177k12202328
answered Nov 19 at 4:54
Eric Wofsey
177k12202328
answered Nov 19 at 4:54
Eric Wofsey
177k12202328
answered Nov 19 at 4:54
Eric Wofsey
177k12202328
177k12202328
add a comment |
add a comment |
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