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







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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:


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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









    $endgroup$



    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.



















      -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









      $endgroup$




      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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      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.







      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.






















          1 Answer
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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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            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









            $endgroup$


















              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









              $endgroup$
















                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









                $endgroup$



                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












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 18 at 7:23









                Eric WofseyEric Wofsey

                187k14215344




                187k14215344















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