Mixing
Compare $|h|^k$ and $sum_i_1, ldots, i_k h_i_1cdots h_i_k$ when $h ll 0$
Clash Royale CLAN TAG#URR8PPP
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I have a question because when I search something on mathstackexchange, I've read something that I'm not sure it's true, and I would like to know if I'm right or I'm wrong.
Actually, we consider $h in mathbbR^n$, and the author of the answer of the topic I talk about claims that, for $h = (h_1, ldots, h_n)$ such that $|h| ll 0$, we have :
$$|h|^6 leq left( sum_i_1, i_2, i_3 in 0,ldots, n h_i_1 h_i_2 h_i_3right)^2$$
and
$$|h|^6 geq n^6 left(sum_i_1, i_2, i_3 h_i_1h_i_2h_i_3right)^2$$
For the second inequality, it's okay, but I'm not okay with the first one. Actually, If I consider the euclidean norm on $mathbbR^n$, then we have
$$|h|^6 leq left(sum_i_1, i_2, i_3 h_i_1h_i_2h_i_3right)^2$$
and as we have : $sum_i_1, i_2, i_3 h_i_1h_i_2h_i_3 = (h_1+h_2+cdots+h_n)^3$, we have
$$iff sqrt(h_1^2+h_2^2+cdots+h_n^2)^6 leq (h_1+h_2+cdots+h_n)^6 $$
$$iff (h_1^2+h_2^2+cdots+h_n^2)^3 leq (h_1+h_2+cdots+h_n)^6 $$
But if I consider, for $varepsilon > 0$, $(varepsilon, -varepsilon, 0, ldots, 0)$, then I have, if the equality is true : $8varepsilon^6 leq 0$, which is absurd. Then, the equality is wrong. Have I right ?
In general, I have the impression that we have $$sum_i_1, ldots, i_k h_i_1cdots h_i_k = O(|h|^k),$$ but not necessarily $|h|^k = Oleft(sum_i_1, ldots, i_k h_i_1cdots h_i_kright)$.
Have I right ?
Thank you for your help !
real-analysis inequality norm approximation
edited Jul 26 at 0:21
Michael Hardy
204k23186461
asked Jul 25 at 23:18
ChocoSavour
1017
add a comment |Â
up vote
0
down vote
favorite
I have a question because when I search something on mathstackexchange, I've read something that I'm not sure it's true, and I would like to know if I'm right or I'm wrong.
Actually, we consider $h in mathbbR^n$, and the author of the answer of the topic I talk about claims that, for $h = (h_1, ldots, h_n)$ such that $|h| ll 0$, we have :
$$|h|^6 leq left( sum_i_1, i_2, i_3 in 0,ldots, n h_i_1 h_i_2 h_i_3right)^2$$
and
$$|h|^6 geq n^6 left(sum_i_1, i_2, i_3 h_i_1h_i_2h_i_3right)^2$$
For the second inequality, it's okay, but I'm not okay with the first one. Actually, If I consider the euclidean norm on $mathbbR^n$, then we have
$$|h|^6 leq left(sum_i_1, i_2, i_3 h_i_1h_i_2h_i_3right)^2$$
and as we have : $sum_i_1, i_2, i_3 h_i_1h_i_2h_i_3 = (h_1+h_2+cdots+h_n)^3$, we have
$$iff sqrt(h_1^2+h_2^2+cdots+h_n^2)^6 leq (h_1+h_2+cdots+h_n)^6 $$
$$iff (h_1^2+h_2^2+cdots+h_n^2)^3 leq (h_1+h_2+cdots+h_n)^6 $$
But if I consider, for $varepsilon > 0$, $(varepsilon, -varepsilon, 0, ldots, 0)$, then I have, if the equality is true : $8varepsilon^6 leq 0$, which is absurd. Then, the equality is wrong. Have I right ?
In general, I have the impression that we have $$sum_i_1, ldots, i_k h_i_1cdots h_i_k = O(|h|^k),$$ but not necessarily $|h|^k = Oleft(sum_i_1, ldots, i_k h_i_1cdots h_i_kright)$.
Have I right ?
Thank you for your help !
real-analysis inequality norm approximation
edited Jul 26 at 0:21
Michael Hardy
204k23186461
asked Jul 25 at 23:18
ChocoSavour
1017
That norm might not be Euclidean. BTW, what does $|h|ll 0$ mean?
– xbh
Jul 26 at 1:29
Yes but as all the norm are equivalent in finite dimension, I can suppose it's the euclidean norm? We can still consider the inequality and multiplied one of the term by a constant, and it doesn't change anything, so it's not a problem, no? For the notation, it means for $h$ close to zero (as close as we want).
– ChocoSavour
Jul 26 at 9:31
Your thinking is fine. The first inequality is false.
– Kavi Rama Murthy
Jul 27 at 6:41
Okay, thank you! :)
– ChocoSavour
Jul 27 at 8:40
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a question because when I search something on mathstackexchange, I've read something that I'm not sure it's true, and I would like to know if I'm right or I'm wrong.
Actually, we consider $h in mathbbR^n$, and the author of the answer of the topic I talk about claims that, for $h = (h_1, ldots, h_n)$ such that $|h| ll 0$, we have :
$$|h|^6 leq left( sum_i_1, i_2, i_3 in 0,ldots, n h_i_1 h_i_2 h_i_3right)^2$$
and
$$|h|^6 geq n^6 left(sum_i_1, i_2, i_3 h_i_1h_i_2h_i_3right)^2$$
For the second inequality, it's okay, but I'm not okay with the first one. Actually, If I consider the euclidean norm on $mathbbR^n$, then we have
$$|h|^6 leq left(sum_i_1, i_2, i_3 h_i_1h_i_2h_i_3right)^2$$
and as we have : $sum_i_1, i_2, i_3 h_i_1h_i_2h_i_3 = (h_1+h_2+cdots+h_n)^3$, we have
$$iff sqrt(h_1^2+h_2^2+cdots+h_n^2)^6 leq (h_1+h_2+cdots+h_n)^6 $$
$$iff (h_1^2+h_2^2+cdots+h_n^2)^3 leq (h_1+h_2+cdots+h_n)^6 $$
But if I consider, for $varepsilon > 0$, $(varepsilon, -varepsilon, 0, ldots, 0)$, then I have, if the equality is true : $8varepsilon^6 leq 0$, which is absurd. Then, the equality is wrong. Have I right ?
In general, I have the impression that we have $$sum_i_1, ldots, i_k h_i_1cdots h_i_k = O(|h|^k),$$ but not necessarily $|h|^k = Oleft(sum_i_1, ldots, i_k h_i_1cdots h_i_kright)$.
Have I right ?
Thank you for your help !
real-analysis inequality norm approximation
edited Jul 26 at 0:21
Michael Hardy
204k23186461
asked Jul 25 at 23:18
ChocoSavour
1017
I have a question because when I search something on mathstackexchange, I've read something that I'm not sure it's true, and I would like to know if I'm right or I'm wrong.
Actually, we consider $h in mathbbR^n$, and the author of the answer of the topic I talk about claims that, for $h = (h_1, ldots, h_n)$ such that $|h| ll 0$, we have :
$$|h|^6 leq left( sum_i_1, i_2, i_3 in 0,ldots, n h_i_1 h_i_2 h_i_3right)^2$$
and
$$|h|^6 geq n^6 left(sum_i_1, i_2, i_3 h_i_1h_i_2h_i_3right)^2$$
For the second inequality, it's okay, but I'm not okay with the first one. Actually, If I consider the euclidean norm on $mathbbR^n$, then we have
$$|h|^6 leq left(sum_i_1, i_2, i_3 h_i_1h_i_2h_i_3right)^2$$
and as we have : $sum_i_1, i_2, i_3 h_i_1h_i_2h_i_3 = (h_1+h_2+cdots+h_n)^3$, we have
$$iff sqrt(h_1^2+h_2^2+cdots+h_n^2)^6 leq (h_1+h_2+cdots+h_n)^6 $$
$$iff (h_1^2+h_2^2+cdots+h_n^2)^3 leq (h_1+h_2+cdots+h_n)^6 $$
But if I consider, for $varepsilon > 0$, $(varepsilon, -varepsilon, 0, ldots, 0)$, then I have, if the equality is true : $8varepsilon^6 leq 0$, which is absurd. Then, the equality is wrong. Have I right ?
In general, I have the impression that we have $$sum_i_1, ldots, i_k h_i_1cdots h_i_k = O(|h|^k),$$ but not necessarily $|h|^k = Oleft(sum_i_1, ldots, i_k h_i_1cdots h_i_kright)$.
Have I right ?
Thank you for your help !
real-analysis inequality norm approximation
edited Jul 26 at 0:21
Michael Hardy
204k23186461
asked Jul 25 at 23:18
ChocoSavour
1017
edited Jul 26 at 0:21
Michael Hardy
204k23186461
edited Jul 26 at 0:21
Michael Hardy
204k23186461
edited Jul 26 at 0:21
Michael Hardy
204k23186461
204k23186461
asked Jul 25 at 23:18
ChocoSavour
1017
asked Jul 25 at 23:18
ChocoSavour
1017
asked Jul 25 at 23:18
ChocoSavour
1017
1017
That norm might not be Euclidean. BTW, what does $|h|ll 0$ mean?
– xbh
Jul 26 at 1:29
Yes but as all the norm are equivalent in finite dimension, I can suppose it's the euclidean norm? We can still consider the inequality and multiplied one of the term by a constant, and it doesn't change anything, so it's not a problem, no? For the notation, it means for $h$ close to zero (as close as we want).
– ChocoSavour
Jul 26 at 9:31
Your thinking is fine. The first inequality is false.
– Kavi Rama Murthy
Jul 27 at 6:41
Okay, thank you! :)
– ChocoSavour
Jul 27 at 8:40
add a comment |Â
That norm might not be Euclidean. BTW, what does $|h|ll 0$ mean?
– xbh
Jul 26 at 1:29
Yes but as all the norm are equivalent in finite dimension, I can suppose it's the euclidean norm? We can still consider the inequality and multiplied one of the term by a constant, and it doesn't change anything, so it's not a problem, no? For the notation, it means for $h$ close to zero (as close as we want).
– ChocoSavour
Jul 26 at 9:31
Your thinking is fine. The first inequality is false.
– Kavi Rama Murthy
Jul 27 at 6:41
Okay, thank you! :)
– ChocoSavour
Jul 27 at 8:40
That norm might not be Euclidean. BTW, what does $|h|ll 0$ mean?
– xbh
Jul 26 at 1:29
That norm might not be Euclidean. BTW, what does $|h|ll 0$ mean?
– xbh
Jul 26 at 1:29
Yes but as all the norm are equivalent in finite dimension, I can suppose it's the euclidean norm? We can still consider the inequality and multiplied one of the term by a constant, and it doesn't change anything, so it's not a problem, no? For the notation, it means for $h$ close to zero (as close as we want).
– ChocoSavour
Jul 26 at 9:31
Yes but as all the norm are equivalent in finite dimension, I can suppose it's the euclidean norm? We can still consider the inequality and multiplied one of the term by a constant, and it doesn't change anything, so it's not a problem, no? For the notation, it means for $h$ close to zero (as close as we want).
– ChocoSavour
Jul 26 at 9:31
Your thinking is fine. The first inequality is false.
– Kavi Rama Murthy
Jul 27 at 6:41
Your thinking is fine. The first inequality is false.
– Kavi Rama Murthy
Jul 27 at 6:41
Okay, thank you! :)
– ChocoSavour
Jul 27 at 8:40
Okay, thank you! :)
– ChocoSavour
Jul 27 at 8:40
add a comment |Â
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That norm might not be Euclidean. BTW, what does $|h|ll 0$ mean?
– xbh
Jul 26 at 1:29
Yes but as all the norm are equivalent in finite dimension, I can suppose it's the euclidean norm? We can still consider the inequality and multiplied one of the term by a constant, and it doesn't change anything, so it's not a problem, no? For the notation, it means for $h$ close to zero (as close as we want).
– ChocoSavour
Jul 26 at 9:31
Your thinking is fine. The first inequality is false.
– Kavi Rama Murthy
Jul 27 at 6:41
Okay, thank you! :)
– ChocoSavour
Jul 27 at 8:40