Prove that if $pin{mathbb{R}},;;p>1$ then $(1+x)^p > 1+x^p$ with $x>0$












0












$begingroup$


I have to prove that if $pin{mathbb{R}},;;p>1$, then $(1+x)^p > 1+x^p$ with $x>0$.



I have got it using derivatives but I am looking for a smarter proof.



Can anyone help me?



Thanks.






real-analysis analysis







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Could you share your proof using derivatives? It is hard to know what you mean by "smarter" without seeing it
    $endgroup$
    – pwerth
    Dec 24 '18 at 17:24






  • 1




    $begingroup$
    I mean that I looking for a prove without using derivatives.
    $endgroup$
    – mathlife
    Dec 24 '18 at 17:27






  • 1




    $begingroup$
    See the proof using geometric series for the generalized form of Bernulli's inequality.
    $endgroup$
    – zipirovich
    Dec 24 '18 at 18:01
















0












$begingroup$


I have to prove that if $pin{mathbb{R}},;;p>1$, then $(1+x)^p > 1+x^p$ with $x>0$.



I have got it using derivatives but I am looking for a smarter proof.



Can anyone help me?



Thanks.






real-analysis analysis







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Could you share your proof using derivatives? It is hard to know what you mean by "smarter" without seeing it
    $endgroup$
    – pwerth
    Dec 24 '18 at 17:24






  • 1




    $begingroup$
    I mean that I looking for a prove without using derivatives.
    $endgroup$
    – mathlife
    Dec 24 '18 at 17:27






  • 1




    $begingroup$
    See the proof using geometric series for the generalized form of Bernulli's inequality.
    $endgroup$
    – zipirovich
    Dec 24 '18 at 18:01














0












0








0





$begingroup$


I have to prove that if $pin{mathbb{R}},;;p>1$, then $(1+x)^p > 1+x^p$ with $x>0$.



I have got it using derivatives but I am looking for a smarter proof.



Can anyone help me?



Thanks.






real-analysis analysis







share|cite|improve this question











$endgroup$




I have to prove that if $pin{mathbb{R}},;;p>1$, then $(1+x)^p > 1+x^p$ with $x>0$.



I have got it using derivatives but I am looking for a smarter proof.



Can anyone help me?



Thanks.






real-analysis analysis




real-analysis analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 16:40









amWhy

1




1










asked Dec 24 '18 at 17:23









mathlifemathlife

629




629








  • 4




    $begingroup$
    Could you share your proof using derivatives? It is hard to know what you mean by "smarter" without seeing it
    $endgroup$
    – pwerth
    Dec 24 '18 at 17:24






  • 1




    $begingroup$
    I mean that I looking for a prove without using derivatives.
    $endgroup$
    – mathlife
    Dec 24 '18 at 17:27






  • 1




    $begingroup$
    See the proof using geometric series for the generalized form of Bernulli's inequality.
    $endgroup$
    – zipirovich
    Dec 24 '18 at 18:01














  • 4




    $begingroup$
    Could you share your proof using derivatives? It is hard to know what you mean by "smarter" without seeing it
    $endgroup$
    – pwerth
    Dec 24 '18 at 17:24






  • 1




    $begingroup$
    I mean that I looking for a prove without using derivatives.
    $endgroup$
    – mathlife
    Dec 24 '18 at 17:27






  • 1




    $begingroup$
    See the proof using geometric series for the generalized form of Bernulli's inequality.
    $endgroup$
    – zipirovich
    Dec 24 '18 at 18:01








4




4




$begingroup$
Could you share your proof using derivatives? It is hard to know what you mean by "smarter" without seeing it
$endgroup$
– pwerth
Dec 24 '18 at 17:24




$begingroup$
Could you share your proof using derivatives? It is hard to know what you mean by "smarter" without seeing it
$endgroup$
– pwerth
Dec 24 '18 at 17:24




1




1




$begingroup$
I mean that I looking for a prove without using derivatives.
$endgroup$
– mathlife
Dec 24 '18 at 17:27




$begingroup$
I mean that I looking for a prove without using derivatives.
$endgroup$
– mathlife
Dec 24 '18 at 17:27




1




1




$begingroup$
See the proof using geometric series for the generalized form of Bernulli's inequality.
$endgroup$
– zipirovich
Dec 24 '18 at 18:01




$begingroup$
See the proof using geometric series for the generalized form of Bernulli's inequality.
$endgroup$
– zipirovich
Dec 24 '18 at 18:01










2 Answers
2






active

oldest

votes


















2












$begingroup$

Without loss of generality, we can take $0<xleq1$. Then
$$
(1+x)^pgeq1+pxgeq1+px^p>1+x^p ,
$$
where we used a generalized Bernoulli's inequality in the first step,
the simple comparison $x^pleq x$ in the second step,
and $p>1$ in the last step.



Update: The linked Wiki page does not seem to have an explicit proof of the generalized Bernoulli inequality for real $p>1$, so let me include a proof here that uses properties of logarithm. We have
$$
frac1plog(1+x)
=frac{p-1}plog1+frac1plog(1+x)
leqlogbig(frac{p-1}p+frac{1+x}pbig)
=logbig(1+frac{x}pbig) ,
$$
where we have used the concavity of log in the second step. Now the monotonicity of log gives
$$
(1+x)^{frac1p}leq1+frac{x}p ,
$$
and the substitution $y=frac{x}p$ finishes the job.



General comment: Since this question involves real exponent $p$, and $x^p$ can be defined only for rational $p$ by using a finite process, the proof must involve an infinite process in some way. Thus I would argue that in this situation, using a derivative is not more "shameful" than any other proof. In the particular proof above, the infinite process is hidden under the generalized Bernoulli inequality.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Hint:



    Use Bernoulli's inequality:




    • If $0<xle1$, then $;(1+x)^p>1+px >1+(x^{p-1}),x$ since $p>1ge x^{p-1}$.

    • If $x>1$, set $;x=smash{dfrac1u}$ ($0<u<1$). It amounts to checking $;(1+u)^{p}>1+u^p$, which is the previous case.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      That only works for integer $p$, while the OP says that $pinmathbb{R}$.
      $endgroup$
      – zipirovich
      Dec 24 '18 at 17:54










    • $begingroup$
      Oops! I missed that detail!
      $endgroup$
      – Bernard
      Dec 24 '18 at 18:01










    • $begingroup$
      upvoted to balance the existing downvote
      $endgroup$
      – timur
      Dec 24 '18 at 18:02






    • 2




      $begingroup$
      Thanks, @Timur! Anyway, I modified the answer. Happy Xmas for the maniac downvoter!
      $endgroup$
      – Bernard
      Dec 24 '18 at 18:11













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    2 Answers
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    2 Answers
    2






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    active

    oldest

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    2












    $begingroup$

    Without loss of generality, we can take $0<xleq1$. Then
    $$
    (1+x)^pgeq1+pxgeq1+px^p>1+x^p ,
    $$
    where we used a generalized Bernoulli's inequality in the first step,
    the simple comparison $x^pleq x$ in the second step,
    and $p>1$ in the last step.



    Update: The linked Wiki page does not seem to have an explicit proof of the generalized Bernoulli inequality for real $p>1$, so let me include a proof here that uses properties of logarithm. We have
    $$
    frac1plog(1+x)
    =frac{p-1}plog1+frac1plog(1+x)
    leqlogbig(frac{p-1}p+frac{1+x}pbig)
    =logbig(1+frac{x}pbig) ,
    $$
    where we have used the concavity of log in the second step. Now the monotonicity of log gives
    $$
    (1+x)^{frac1p}leq1+frac{x}p ,
    $$
    and the substitution $y=frac{x}p$ finishes the job.



    General comment: Since this question involves real exponent $p$, and $x^p$ can be defined only for rational $p$ by using a finite process, the proof must involve an infinite process in some way. Thus I would argue that in this situation, using a derivative is not more "shameful" than any other proof. In the particular proof above, the infinite process is hidden under the generalized Bernoulli inequality.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Without loss of generality, we can take $0<xleq1$. Then
      $$
      (1+x)^pgeq1+pxgeq1+px^p>1+x^p ,
      $$
      where we used a generalized Bernoulli's inequality in the first step,
      the simple comparison $x^pleq x$ in the second step,
      and $p>1$ in the last step.



      Update: The linked Wiki page does not seem to have an explicit proof of the generalized Bernoulli inequality for real $p>1$, so let me include a proof here that uses properties of logarithm. We have
      $$
      frac1plog(1+x)
      =frac{p-1}plog1+frac1plog(1+x)
      leqlogbig(frac{p-1}p+frac{1+x}pbig)
      =logbig(1+frac{x}pbig) ,
      $$
      where we have used the concavity of log in the second step. Now the monotonicity of log gives
      $$
      (1+x)^{frac1p}leq1+frac{x}p ,
      $$
      and the substitution $y=frac{x}p$ finishes the job.



      General comment: Since this question involves real exponent $p$, and $x^p$ can be defined only for rational $p$ by using a finite process, the proof must involve an infinite process in some way. Thus I would argue that in this situation, using a derivative is not more "shameful" than any other proof. In the particular proof above, the infinite process is hidden under the generalized Bernoulli inequality.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Without loss of generality, we can take $0<xleq1$. Then
        $$
        (1+x)^pgeq1+pxgeq1+px^p>1+x^p ,
        $$
        where we used a generalized Bernoulli's inequality in the first step,
        the simple comparison $x^pleq x$ in the second step,
        and $p>1$ in the last step.



        Update: The linked Wiki page does not seem to have an explicit proof of the generalized Bernoulli inequality for real $p>1$, so let me include a proof here that uses properties of logarithm. We have
        $$
        frac1plog(1+x)
        =frac{p-1}plog1+frac1plog(1+x)
        leqlogbig(frac{p-1}p+frac{1+x}pbig)
        =logbig(1+frac{x}pbig) ,
        $$
        where we have used the concavity of log in the second step. Now the monotonicity of log gives
        $$
        (1+x)^{frac1p}leq1+frac{x}p ,
        $$
        and the substitution $y=frac{x}p$ finishes the job.



        General comment: Since this question involves real exponent $p$, and $x^p$ can be defined only for rational $p$ by using a finite process, the proof must involve an infinite process in some way. Thus I would argue that in this situation, using a derivative is not more "shameful" than any other proof. In the particular proof above, the infinite process is hidden under the generalized Bernoulli inequality.






        share|cite|improve this answer











        $endgroup$



        Without loss of generality, we can take $0<xleq1$. Then
        $$
        (1+x)^pgeq1+pxgeq1+px^p>1+x^p ,
        $$
        where we used a generalized Bernoulli's inequality in the first step,
        the simple comparison $x^pleq x$ in the second step,
        and $p>1$ in the last step.



        Update: The linked Wiki page does not seem to have an explicit proof of the generalized Bernoulli inequality for real $p>1$, so let me include a proof here that uses properties of logarithm. We have
        $$
        frac1plog(1+x)
        =frac{p-1}plog1+frac1plog(1+x)
        leqlogbig(frac{p-1}p+frac{1+x}pbig)
        =logbig(1+frac{x}pbig) ,
        $$
        where we have used the concavity of log in the second step. Now the monotonicity of log gives
        $$
        (1+x)^{frac1p}leq1+frac{x}p ,
        $$
        and the substitution $y=frac{x}p$ finishes the job.



        General comment: Since this question involves real exponent $p$, and $x^p$ can be defined only for rational $p$ by using a finite process, the proof must involve an infinite process in some way. Thus I would argue that in this situation, using a derivative is not more "shameful" than any other proof. In the particular proof above, the infinite process is hidden under the generalized Bernoulli inequality.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 24 '18 at 19:26

























        answered Dec 24 '18 at 18:10









        timurtimur

        12.1k2144




        12.1k2144























            1












            $begingroup$

            Hint:



            Use Bernoulli's inequality:




            • If $0<xle1$, then $;(1+x)^p>1+px >1+(x^{p-1}),x$ since $p>1ge x^{p-1}$.

            • If $x>1$, set $;x=smash{dfrac1u}$ ($0<u<1$). It amounts to checking $;(1+u)^{p}>1+u^p$, which is the previous case.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              That only works for integer $p$, while the OP says that $pinmathbb{R}$.
              $endgroup$
              – zipirovich
              Dec 24 '18 at 17:54










            • $begingroup$
              Oops! I missed that detail!
              $endgroup$
              – Bernard
              Dec 24 '18 at 18:01










            • $begingroup$
              upvoted to balance the existing downvote
              $endgroup$
              – timur
              Dec 24 '18 at 18:02






            • 2




              $begingroup$
              Thanks, @Timur! Anyway, I modified the answer. Happy Xmas for the maniac downvoter!
              $endgroup$
              – Bernard
              Dec 24 '18 at 18:11


















            1












            $begingroup$

            Hint:



            Use Bernoulli's inequality:




            • If $0<xle1$, then $;(1+x)^p>1+px >1+(x^{p-1}),x$ since $p>1ge x^{p-1}$.

            • If $x>1$, set $;x=smash{dfrac1u}$ ($0<u<1$). It amounts to checking $;(1+u)^{p}>1+u^p$, which is the previous case.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              That only works for integer $p$, while the OP says that $pinmathbb{R}$.
              $endgroup$
              – zipirovich
              Dec 24 '18 at 17:54










            • $begingroup$
              Oops! I missed that detail!
              $endgroup$
              – Bernard
              Dec 24 '18 at 18:01










            • $begingroup$
              upvoted to balance the existing downvote
              $endgroup$
              – timur
              Dec 24 '18 at 18:02






            • 2




              $begingroup$
              Thanks, @Timur! Anyway, I modified the answer. Happy Xmas for the maniac downvoter!
              $endgroup$
              – Bernard
              Dec 24 '18 at 18:11
















            1












            1








            1





            $begingroup$

            Hint:



            Use Bernoulli's inequality:




            • If $0<xle1$, then $;(1+x)^p>1+px >1+(x^{p-1}),x$ since $p>1ge x^{p-1}$.

            • If $x>1$, set $;x=smash{dfrac1u}$ ($0<u<1$). It amounts to checking $;(1+u)^{p}>1+u^p$, which is the previous case.






            share|cite|improve this answer











            $endgroup$



            Hint:



            Use Bernoulli's inequality:




            • If $0<xle1$, then $;(1+x)^p>1+px >1+(x^{p-1}),x$ since $p>1ge x^{p-1}$.

            • If $x>1$, set $;x=smash{dfrac1u}$ ($0<u<1$). It amounts to checking $;(1+u)^{p}>1+u^p$, which is the previous case.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 24 '18 at 18:09

























            answered Dec 24 '18 at 17:41









            BernardBernard

            121k740116




            121k740116












            • $begingroup$
              That only works for integer $p$, while the OP says that $pinmathbb{R}$.
              $endgroup$
              – zipirovich
              Dec 24 '18 at 17:54










            • $begingroup$
              Oops! I missed that detail!
              $endgroup$
              – Bernard
              Dec 24 '18 at 18:01










            • $begingroup$
              upvoted to balance the existing downvote
              $endgroup$
              – timur
              Dec 24 '18 at 18:02






            • 2




              $begingroup$
              Thanks, @Timur! Anyway, I modified the answer. Happy Xmas for the maniac downvoter!
              $endgroup$
              – Bernard
              Dec 24 '18 at 18:11




















            • $begingroup$
              That only works for integer $p$, while the OP says that $pinmathbb{R}$.
              $endgroup$
              – zipirovich
              Dec 24 '18 at 17:54










            • $begingroup$
              Oops! I missed that detail!
              $endgroup$
              – Bernard
              Dec 24 '18 at 18:01










            • $begingroup$
              upvoted to balance the existing downvote
              $endgroup$
              – timur
              Dec 24 '18 at 18:02






            • 2




              $begingroup$
              Thanks, @Timur! Anyway, I modified the answer. Happy Xmas for the maniac downvoter!
              $endgroup$
              – Bernard
              Dec 24 '18 at 18:11


















            $begingroup$
            That only works for integer $p$, while the OP says that $pinmathbb{R}$.
            $endgroup$
            – zipirovich
            Dec 24 '18 at 17:54




            $begingroup$
            That only works for integer $p$, while the OP says that $pinmathbb{R}$.
            $endgroup$
            – zipirovich
            Dec 24 '18 at 17:54












            $begingroup$
            Oops! I missed that detail!
            $endgroup$
            – Bernard
            Dec 24 '18 at 18:01




            $begingroup$
            Oops! I missed that detail!
            $endgroup$
            – Bernard
            Dec 24 '18 at 18:01












            $begingroup$
            upvoted to balance the existing downvote
            $endgroup$
            – timur
            Dec 24 '18 at 18:02




            $begingroup$
            upvoted to balance the existing downvote
            $endgroup$
            – timur
            Dec 24 '18 at 18:02




            2




            2




            $begingroup$
            Thanks, @Timur! Anyway, I modified the answer. Happy Xmas for the maniac downvoter!
            $endgroup$
            – Bernard
            Dec 24 '18 at 18:11






            $begingroup$
            Thanks, @Timur! Anyway, I modified the answer. Happy Xmas for the maniac downvoter!
            $endgroup$
            – Bernard
            Dec 24 '18 at 18:11




















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