Mixing
Prove that the functional is non-negative for $x_igeq 0$
$begingroup$
Consider the following functional $Phi:mathbb R^ntomathbb R $:
$$
Phi(x)=sum_{i=1}^{n-1}(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1).
$$
The computer experiments show that it is non-negative for all $x_igeq 0$. I need to prove this. Note that we have both $Phi(x)=0$ and $nabla Phi(x)=0$ for all $x$ with equal coordinates. The proof should be simple, but I can't manage to find it. Any ideas?
inequality
asked Jan 10 at 22:16
Nasa MomdeleNasa Momdele
254
$endgroup$
add a comment |
$begingroup$
Consider the following functional $Phi:mathbb R^ntomathbb R $:
$$
Phi(x)=sum_{i=1}^{n-1}(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1).
$$
The computer experiments show that it is non-negative for all $x_igeq 0$. I need to prove this. Note that we have both $Phi(x)=0$ and $nabla Phi(x)=0$ for all $x$ with equal coordinates. The proof should be simple, but I can't manage to find it. Any ideas?
inequality
asked Jan 10 at 22:16
Nasa MomdeleNasa Momdele
254
$endgroup$
1
$begingroup$
$Phi$ is not nonnegative on $mathbb R^n$. E.g. when $x_1=x_2=cdots=x_{n-1}=1$, $Phi(x)=6(n-1)(x_n-1)^2(x_n+1)$, which is negative when $x_n+1<0$.
$endgroup$
– user1551
Jan 11 at 20:00
$begingroup$
@user1551 thank you very much, you are right. I corrected the question.
$endgroup$
– Nasa Momdele
Jan 11 at 20:19
add a comment |
$begingroup$
Consider the following functional $Phi:mathbb R^ntomathbb R $:
$$
Phi(x)=sum_{i=1}^{n-1}(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1).
$$
The computer experiments show that it is non-negative for all $x_igeq 0$. I need to prove this. Note that we have both $Phi(x)=0$ and $nabla Phi(x)=0$ for all $x$ with equal coordinates. The proof should be simple, but I can't manage to find it. Any ideas?
inequality
asked Jan 10 at 22:16
Nasa MomdeleNasa Momdele
254
$endgroup$
Consider the following functional $Phi:mathbb R^ntomathbb R $:
$$
Phi(x)=sum_{i=1}^{n-1}(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1).
$$
The computer experiments show that it is non-negative for all $x_igeq 0$. I need to prove this. Note that we have both $Phi(x)=0$ and $nabla Phi(x)=0$ for all $x$ with equal coordinates. The proof should be simple, but I can't manage to find it. Any ideas?
inequality
inequality
asked Jan 10 at 22:16
Nasa MomdeleNasa Momdele
254
asked Jan 10 at 22:16
Nasa MomdeleNasa Momdele
254
edited Jan 11 at 20:19
Nasa Momdele
asked Jan 10 at 22:16
Nasa MomdeleNasa Momdele
254
asked Jan 10 at 22:16
Nasa MomdeleNasa Momdele
254
asked Jan 10 at 22:16
Nasa MomdeleNasa Momdele
254
254
1
$begingroup$
$Phi$ is not nonnegative on $mathbb R^n$. E.g. when $x_1=x_2=cdots=x_{n-1}=1$, $Phi(x)=6(n-1)(x_n-1)^2(x_n+1)$, which is negative when $x_n+1<0$.
$endgroup$
– user1551
Jan 11 at 20:00
$begingroup$
@user1551 thank you very much, you are right. I corrected the question.
$endgroup$
– Nasa Momdele
Jan 11 at 20:19
add a comment |
1
$begingroup$
$Phi$ is not nonnegative on $mathbb R^n$. E.g. when $x_1=x_2=cdots=x_{n-1}=1$, $Phi(x)=6(n-1)(x_n-1)^2(x_n+1)$, which is negative when $x_n+1<0$.
$endgroup$
– user1551
Jan 11 at 20:00
$begingroup$
@user1551 thank you very much, you are right. I corrected the question.
$endgroup$
– Nasa Momdele
Jan 11 at 20:19
1
1
$begingroup$
$Phi$ is not nonnegative on $mathbb R^n$. E.g. when $x_1=x_2=cdots=x_{n-1}=1$, $Phi(x)=6(n-1)(x_n-1)^2(x_n+1)$, which is negative when $x_n+1<0$.
$endgroup$
– user1551
Jan 11 at 20:00
$begingroup$
$Phi$ is not nonnegative on $mathbb R^n$. E.g. when $x_1=x_2=cdots=x_{n-1}=1$, $Phi(x)=6(n-1)(x_n-1)^2(x_n+1)$, which is negative when $x_n+1<0$.
$endgroup$
– user1551
Jan 11 at 20:00
$begingroup$
@user1551 thank you very much, you are right. I corrected the question.
$endgroup$
– Nasa Momdele
Jan 11 at 20:19
$begingroup$
@user1551 thank you very much, you are right. I corrected the question.
$endgroup$
– Nasa Momdele
Jan 11 at 20:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$Phi$ is not always nonnegative. Let $f(x_i)=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)$. When $x_1=7,,x_2=x_3=cdots=x_{n-1}=1$ and $x_n=0$, we have, for $2le ile n-1$,
begin{aligned}
f(x_i)
&=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)\
&=2(1)^2left(2(1+1+0)+0-7right)\
&=-6.
end{aligned}
Therefore $Phi(x)=sum_{i=1}^{n-1}f(x_i)=f(x_1)-6(n-2)$ is negative when $n$ is sufficiently large.
answered Jan 11 at 20:34
user1551user1551
72.6k566127
$endgroup$
$begingroup$
Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
$endgroup$
– Nasa Momdele
Jan 15 at 22:16
add a comment |
$begingroup$
Not an answer, merely recording a partial result, found trying to find the least $n$ with a $Phi_n(x) < 0$.
begin{align*}
Phi_{78}(frac{13018}{3}, 1626, dots, 1626,0) = frac{-868,706,947,328}{81}
end{align*}
In fact, $Phi_{78}left( frac{25979 b}{9736}, b, dots, b, 0 right)$ has constant negative fourth derivative and has negative first, second, and third derivatives as soon as $b > 1200.487dots$, so $Phi_{78}$ decreases like $Theta(-b^4)$ starting there. (More on $Theta$.)
answered Jan 11 at 23:04
Eric TowersEric Towers
32.6k22370
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$Phi$ is not always nonnegative. Let $f(x_i)=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)$. When $x_1=7,,x_2=x_3=cdots=x_{n-1}=1$ and $x_n=0$, we have, for $2le ile n-1$,
begin{aligned}
f(x_i)
&=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)\
&=2(1)^2left(2(1+1+0)+0-7right)\
&=-6.
end{aligned}
Therefore $Phi(x)=sum_{i=1}^{n-1}f(x_i)=f(x_1)-6(n-2)$ is negative when $n$ is sufficiently large.
answered Jan 11 at 20:34
user1551user1551
72.6k566127
$endgroup$
$begingroup$
Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
$endgroup$
– Nasa Momdele
Jan 15 at 22:16
add a comment |
$begingroup$
$Phi$ is not always nonnegative. Let $f(x_i)=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)$. When $x_1=7,,x_2=x_3=cdots=x_{n-1}=1$ and $x_n=0$, we have, for $2le ile n-1$,
begin{aligned}
f(x_i)
&=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)\
&=2(1)^2left(2(1+1+0)+0-7right)\
&=-6.
end{aligned}
Therefore $Phi(x)=sum_{i=1}^{n-1}f(x_i)=f(x_1)-6(n-2)$ is negative when $n$ is sufficiently large.
answered Jan 11 at 20:34
user1551user1551
72.6k566127
$endgroup$
$begingroup$
Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
$endgroup$
– Nasa Momdele
Jan 15 at 22:16
add a comment |
$begingroup$
$Phi$ is not always nonnegative. Let $f(x_i)=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)$. When $x_1=7,,x_2=x_3=cdots=x_{n-1}=1$ and $x_n=0$, we have, for $2le ile n-1$,
begin{aligned}
f(x_i)
&=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)\
&=2(1)^2left(2(1+1+0)+0-7right)\
&=-6.
end{aligned}
Therefore $Phi(x)=sum_{i=1}^{n-1}f(x_i)=f(x_1)-6(n-2)$ is negative when $n$ is sufficiently large.
answered Jan 11 at 20:34
user1551user1551
72.6k566127
$endgroup$
$Phi$ is not always nonnegative. Let $f(x_i)=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)$. When $x_1=7,,x_2=x_3=cdots=x_{n-1}=1$ and $x_n=0$, we have, for $2le ile n-1$,
begin{aligned}
f(x_i)
&=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)\
&=2(1)^2left(2(1+1+0)+0-7right)\
&=-6.
end{aligned}
Therefore $Phi(x)=sum_{i=1}^{n-1}f(x_i)=f(x_1)-6(n-2)$ is negative when $n$ is sufficiently large.
answered Jan 11 at 20:34
user1551user1551
72.6k566127
answered Jan 11 at 20:34
user1551user1551
72.6k566127
answered Jan 11 at 20:34
user1551user1551
72.6k566127
answered Jan 11 at 20:34
user1551user1551
72.6k566127
72.6k566127
$begingroup$
Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
$endgroup$
– Nasa Momdele
Jan 15 at 22:16
add a comment |
$begingroup$
Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
$endgroup$
– Nasa Momdele
Jan 15 at 22:16
$begingroup$
Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
$endgroup$
– Nasa Momdele
Jan 15 at 22:16
$begingroup$
Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
$endgroup$
– Nasa Momdele
Jan 15 at 22:16
add a comment |
$begingroup$
Not an answer, merely recording a partial result, found trying to find the least $n$ with a $Phi_n(x) < 0$.
begin{align*}
Phi_{78}(frac{13018}{3}, 1626, dots, 1626,0) = frac{-868,706,947,328}{81}
end{align*}
In fact, $Phi_{78}left( frac{25979 b}{9736}, b, dots, b, 0 right)$ has constant negative fourth derivative and has negative first, second, and third derivatives as soon as $b > 1200.487dots$, so $Phi_{78}$ decreases like $Theta(-b^4)$ starting there. (More on $Theta$.)
answered Jan 11 at 23:04
Eric TowersEric Towers
32.6k22370
$endgroup$
add a comment |
$begingroup$
Not an answer, merely recording a partial result, found trying to find the least $n$ with a $Phi_n(x) < 0$.
begin{align*}
Phi_{78}(frac{13018}{3}, 1626, dots, 1626,0) = frac{-868,706,947,328}{81}
end{align*}
In fact, $Phi_{78}left( frac{25979 b}{9736}, b, dots, b, 0 right)$ has constant negative fourth derivative and has negative first, second, and third derivatives as soon as $b > 1200.487dots$, so $Phi_{78}$ decreases like $Theta(-b^4)$ starting there. (More on $Theta$.)
answered Jan 11 at 23:04
Eric TowersEric Towers
32.6k22370
$endgroup$
add a comment |
$begingroup$
Not an answer, merely recording a partial result, found trying to find the least $n$ with a $Phi_n(x) < 0$.
begin{align*}
Phi_{78}(frac{13018}{3}, 1626, dots, 1626,0) = frac{-868,706,947,328}{81}
end{align*}
In fact, $Phi_{78}left( frac{25979 b}{9736}, b, dots, b, 0 right)$ has constant negative fourth derivative and has negative first, second, and third derivatives as soon as $b > 1200.487dots$, so $Phi_{78}$ decreases like $Theta(-b^4)$ starting there. (More on $Theta$.)
answered Jan 11 at 23:04
Eric TowersEric Towers
32.6k22370
$endgroup$
Not an answer, merely recording a partial result, found trying to find the least $n$ with a $Phi_n(x) < 0$.
begin{align*}
Phi_{78}(frac{13018}{3}, 1626, dots, 1626,0) = frac{-868,706,947,328}{81}
end{align*}
In fact, $Phi_{78}left( frac{25979 b}{9736}, b, dots, b, 0 right)$ has constant negative fourth derivative and has negative first, second, and third derivatives as soon as $b > 1200.487dots$, so $Phi_{78}$ decreases like $Theta(-b^4)$ starting there. (More on $Theta$.)
answered Jan 11 at 23:04
Eric TowersEric Towers
32.6k22370
answered Jan 11 at 23:04
Eric TowersEric Towers
32.6k22370
answered Jan 11 at 23:04
Eric TowersEric Towers
32.6k22370
answered Jan 11 at 23:04
Eric TowersEric Towers
32.6k22370
32.6k22370
add a comment |
add a comment |
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$begingroup$
$Phi$ is not nonnegative on $mathbb R^n$. E.g. when $x_1=x_2=cdots=x_{n-1}=1$, $Phi(x)=6(n-1)(x_n-1)^2(x_n+1)$, which is negative when $x_n+1<0$.
$endgroup$
– user1551
Jan 11 at 20:00
$begingroup$
@user1551 thank you very much, you are right. I corrected the question.
$endgroup$
– Nasa Momdele
Jan 11 at 20:19