Is this expression true?
$begingroup$
Let $a_1=b_1/h,...,a_n=b_n/hinmathbb{R}$ with $hinmathbb{R}$ small. It's true that, given a $alphainmathbb{R}$:
begin{eqnarray}
(a_1+...+a_n)^alpha=sum_{i=1}^n (a_i)^alpha+mathcal{O}left(frac{1}{h^2}right)
end{eqnarray}
?? Is there any (good) reference where there is proof of some theorem / lemma that I can use to justify this?
reference-request real-analysis
asked Jan 4 at 23:20
Leonardo S. VieiraLeonardo S. Vieira
11
$endgroup$
migrated from mathoverflow.net Jan 11 at 14:41
This question came from our site for professional mathematicians.
add a comment |
$begingroup$
Let $a_1=b_1/h,...,a_n=b_n/hinmathbb{R}$ with $hinmathbb{R}$ small. It's true that, given a $alphainmathbb{R}$:
begin{eqnarray}
(a_1+...+a_n)^alpha=sum_{i=1}^n (a_i)^alpha+mathcal{O}left(frac{1}{h^2}right)
end{eqnarray}
?? Is there any (good) reference where there is proof of some theorem / lemma that I can use to justify this?
reference-request real-analysis
asked Jan 4 at 23:20
Leonardo S. VieiraLeonardo S. Vieira
11
$endgroup$
migrated from mathoverflow.net Jan 11 at 14:41
This question came from our site for professional mathematicians.
$begingroup$
This is false if $b_1=cdots =b_n=1$ and $alpha >2$.
$endgroup$
– abx
Jan 5 at 5:21
$begingroup$
You are asking in the wrong forum, I think.
$endgroup$
– GEdgar
Jan 5 at 14:03
$begingroup$
There is a problem when $b_1 = h$, $b_2 = -h$, and $alpha = -1$. I think you're missing some constraints if you want this to be true.
$endgroup$
– Eric Towers
Jan 11 at 15:00
add a comment |
$begingroup$
Let $a_1=b_1/h,...,a_n=b_n/hinmathbb{R}$ with $hinmathbb{R}$ small. It's true that, given a $alphainmathbb{R}$:
begin{eqnarray}
(a_1+...+a_n)^alpha=sum_{i=1}^n (a_i)^alpha+mathcal{O}left(frac{1}{h^2}right)
end{eqnarray}
?? Is there any (good) reference where there is proof of some theorem / lemma that I can use to justify this?
reference-request real-analysis
asked Jan 4 at 23:20
Leonardo S. VieiraLeonardo S. Vieira
11
$endgroup$
Let $a_1=b_1/h,...,a_n=b_n/hinmathbb{R}$ with $hinmathbb{R}$ small. It's true that, given a $alphainmathbb{R}$:
begin{eqnarray}
(a_1+...+a_n)^alpha=sum_{i=1}^n (a_i)^alpha+mathcal{O}left(frac{1}{h^2}right)
end{eqnarray}
?? Is there any (good) reference where there is proof of some theorem / lemma that I can use to justify this?
reference-request real-analysis
reference-request real-analysis
asked Jan 4 at 23:20
Leonardo S. VieiraLeonardo S. Vieira
11
asked Jan 4 at 23:20
Leonardo S. VieiraLeonardo S. Vieira
11
asked Jan 4 at 23:20
Leonardo S. VieiraLeonardo S. Vieira
11
asked Jan 4 at 23:20
Leonardo S. VieiraLeonardo S. Vieira
11
asked Jan 4 at 23:20
Leonardo S. VieiraLeonardo S. Vieira
11
11
migrated from mathoverflow.net Jan 11 at 14:41
This question came from our site for professional mathematicians.
migrated from mathoverflow.net Jan 11 at 14:41
This question came from our site for professional mathematicians.
$begingroup$
This is false if $b_1=cdots =b_n=1$ and $alpha >2$.
$endgroup$
– abx
Jan 5 at 5:21
$begingroup$
You are asking in the wrong forum, I think.
$endgroup$
– GEdgar
Jan 5 at 14:03
$begingroup$
There is a problem when $b_1 = h$, $b_2 = -h$, and $alpha = -1$. I think you're missing some constraints if you want this to be true.
$endgroup$
– Eric Towers
Jan 11 at 15:00
add a comment |
$begingroup$
This is false if $b_1=cdots =b_n=1$ and $alpha >2$.
$endgroup$
– abx
Jan 5 at 5:21
$begingroup$
You are asking in the wrong forum, I think.
$endgroup$
– GEdgar
Jan 5 at 14:03
$begingroup$
There is a problem when $b_1 = h$, $b_2 = -h$, and $alpha = -1$. I think you're missing some constraints if you want this to be true.
$endgroup$
– Eric Towers
Jan 11 at 15:00
$begingroup$
This is false if $b_1=cdots =b_n=1$ and $alpha >2$.
$endgroup$
– abx
Jan 5 at 5:21
$begingroup$
This is false if $b_1=cdots =b_n=1$ and $alpha >2$.
$endgroup$
– abx
Jan 5 at 5:21
$begingroup$
You are asking in the wrong forum, I think.
$endgroup$
– GEdgar
Jan 5 at 14:03
$begingroup$
You are asking in the wrong forum, I think.
$endgroup$
– GEdgar
Jan 5 at 14:03
$begingroup$
There is a problem when $b_1 = h$, $b_2 = -h$, and $alpha = -1$. I think you're missing some constraints if you want this to be true.
$endgroup$
– Eric Towers
Jan 11 at 15:00
$begingroup$
There is a problem when $b_1 = h$, $b_2 = -h$, and $alpha = -1$. I think you're missing some constraints if you want this to be true.
$endgroup$
– Eric Towers
Jan 11 at 15:00
add a comment |
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$begingroup$
This is false if $b_1=cdots =b_n=1$ and $alpha >2$.
$endgroup$
– abx
Jan 5 at 5:21
$begingroup$
You are asking in the wrong forum, I think.
$endgroup$
– GEdgar
Jan 5 at 14:03
$begingroup$
There is a problem when $b_1 = h$, $b_2 = -h$, and $alpha = -1$. I think you're missing some constraints if you want this to be true.
$endgroup$
– Eric Towers
Jan 11 at 15:00