Inequality In Real Number [closed]

-1

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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

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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.

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-1

$begingroup$

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

share|cite|improve this question

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 | 

-1

-1

-1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

asked Jan 18 at 7:18

AndreAndre

23

share|cite|improve this question

asked Jan 18 at 7:18

AndreAndre

23

share|cite|improve this question

share|cite|improve this question

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

active

oldest

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3

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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.

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answered Jan 18 at 7:23

Eric WofseyEric Wofsey

187k14215344

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

3

$begingroup$

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.

share|cite|improve this answer

answered Jan 18 at 7:23

Eric WofseyEric Wofsey

187k14215344

$endgroup$

add a comment | 

3

$begingroup$

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.

share|cite|improve this answer

answered Jan 18 at 7:23

Eric WofseyEric Wofsey

187k14215344

$endgroup$

add a comment | 

3

3

3

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Jan 18 at 7:23

Eric WofseyEric Wofsey

187k14215344

share|cite|improve this answer

share|cite|improve this answer

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 | 

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