Mixing
Inequality In Real Number [closed]
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Can we find a constant $C$ independent on $a$ such that
$$(a+tau)^{mu}-a^{mu}leq C tau^{mu},$$
where $ainmathbb{R}^+,tauin(0,1)$ and $muin(1,2)$? I have found $C=1$ for $muin(0,1]$, but I couldn't find $C$ when $muin(1,2)$.
real-analysis inequality
asked Jan 18 at 7:18
AndreAndre
23
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closed as off-topic by Did, Martin R, Cesareo, A. Pongrácz, RRL Jan 20 at 18:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Martin R, Cesareo, A. Pongrácz, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Can we find a constant $C$ independent on $a$ such that
$$(a+tau)^{mu}-a^{mu}leq C tau^{mu},$$
where $ainmathbb{R}^+,tauin(0,1)$ and $muin(1,2)$? I have found $C=1$ for $muin(0,1]$, but I couldn't find $C$ when $muin(1,2)$.
real-analysis inequality
asked Jan 18 at 7:18
AndreAndre
23
$endgroup$
closed as off-topic by Did, Martin R, Cesareo, A. Pongrácz, RRL Jan 20 at 18:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Martin R, Cesareo, A. Pongrácz, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Can we find a constant $C$ independent on $a$ such that
$$(a+tau)^{mu}-a^{mu}leq C tau^{mu},$$
where $ainmathbb{R}^+,tauin(0,1)$ and $muin(1,2)$? I have found $C=1$ for $muin(0,1]$, but I couldn't find $C$ when $muin(1,2)$.
real-analysis inequality
asked Jan 18 at 7:18
AndreAndre
23
$endgroup$
Can we find a constant $C$ independent on $a$ such that
$$(a+tau)^{mu}-a^{mu}leq C tau^{mu},$$
where $ainmathbb{R}^+,tauin(0,1)$ and $muin(1,2)$? I have found $C=1$ for $muin(0,1]$, but I couldn't find $C$ when $muin(1,2)$.
real-analysis inequality
real-analysis inequality
asked Jan 18 at 7:18
AndreAndre
23
asked Jan 18 at 7:18
AndreAndre
23
asked Jan 18 at 7:18
AndreAndre
23
asked Jan 18 at 7:18
AndreAndre
23
asked Jan 18 at 7:18
AndreAndre
23
23
closed as off-topic by Did, Martin R, Cesareo, A. Pongrácz, RRL Jan 20 at 18:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Martin R, Cesareo, A. Pongrácz, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Martin R, Cesareo, A. Pongrácz, RRL Jan 20 at 18:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Martin R, Cesareo, A. Pongrácz, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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No such constant exists for any $mu>1$. To see this, let $f(x)=x^mu$ and observe that $f'(x)=mu x^{mu-1}$ goes to $infty$ as $xtoinfty$. It follows that for fixed $tau>0$, $f(a+tau)-f(a)$ goes to $infty$ as $atoinfty$ and so cannot be bounded above.
answered Jan 18 at 7:23
Eric WofseyEric Wofsey
187k14215344
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No such constant exists for any $mu>1$. To see this, let $f(x)=x^mu$ and observe that $f'(x)=mu x^{mu-1}$ goes to $infty$ as $xtoinfty$. It follows that for fixed $tau>0$, $f(a+tau)-f(a)$ goes to $infty$ as $atoinfty$ and so cannot be bounded above.
answered Jan 18 at 7:23
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
No such constant exists for any $mu>1$. To see this, let $f(x)=x^mu$ and observe that $f'(x)=mu x^{mu-1}$ goes to $infty$ as $xtoinfty$. It follows that for fixed $tau>0$, $f(a+tau)-f(a)$ goes to $infty$ as $atoinfty$ and so cannot be bounded above.
answered Jan 18 at 7:23
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
No such constant exists for any $mu>1$. To see this, let $f(x)=x^mu$ and observe that $f'(x)=mu x^{mu-1}$ goes to $infty$ as $xtoinfty$. It follows that for fixed $tau>0$, $f(a+tau)-f(a)$ goes to $infty$ as $atoinfty$ and so cannot be bounded above.
answered Jan 18 at 7:23
Eric WofseyEric Wofsey
187k14215344
$endgroup$
No such constant exists for any $mu>1$. To see this, let $f(x)=x^mu$ and observe that $f'(x)=mu x^{mu-1}$ goes to $infty$ as $xtoinfty$. It follows that for fixed $tau>0$, $f(a+tau)-f(a)$ goes to $infty$ as $atoinfty$ and so cannot be bounded above.
answered Jan 18 at 7:23
Eric WofseyEric Wofsey
187k14215344
answered Jan 18 at 7:23
Eric WofseyEric Wofsey
187k14215344
answered Jan 18 at 7:23
Eric WofseyEric Wofsey
187k14215344
answered Jan 18 at 7:23
Eric WofseyEric Wofsey
187k14215344
187k14215344
add a comment |
add a comment |
