Bounds of $frac{ln(x+1)}{x} forall x>0$












2












$begingroup$


$f:(0,infty)$,
$f(x)=frac{ln(x+1)}{x}$

Prove that for $forall x>0$ that $f(x) in(0,1)$. I calculated the derivative of $f(x)$: $f'(x)=frac{frac{x}{x+1}-ln(1+x)}{x^2}$ which I think simplifies to $frac{x^3}{x+1}-x^2(ln(1+x))$. I have no idea what to do next, I can't find the roots of this equation and I don't see any connection as of why it should be bounded by 0 and 1.



I hope I formatted this well, I don't usually post here but I am really curious how could I solve this kind of exercise.






calculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $0<frac{ln(x+1)}{x}<1$ $iff$ $0<ln(x+1)<x$.
    $endgroup$
    – A.Γ.
    Dec 29 '18 at 20:24












  • $begingroup$
    Note that if $x>0$ then $e^x>x+1$ by a simple Taylor expansion. This proves what you want
    $endgroup$
    – Shalop
    Dec 29 '18 at 20:43










  • $begingroup$
    @A.Γ. Wow I never thought to work from the statement that I have to prove. It almost seems too easy to be true. I can't thank you enough for this answer.
    $endgroup$
    – Radu Gabriel
    Dec 29 '18 at 21:13
















2












$begingroup$


$f:(0,infty)$,
$f(x)=frac{ln(x+1)}{x}$

Prove that for $forall x>0$ that $f(x) in(0,1)$. I calculated the derivative of $f(x)$: $f'(x)=frac{frac{x}{x+1}-ln(1+x)}{x^2}$ which I think simplifies to $frac{x^3}{x+1}-x^2(ln(1+x))$. I have no idea what to do next, I can't find the roots of this equation and I don't see any connection as of why it should be bounded by 0 and 1.



I hope I formatted this well, I don't usually post here but I am really curious how could I solve this kind of exercise.






calculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $0<frac{ln(x+1)}{x}<1$ $iff$ $0<ln(x+1)<x$.
    $endgroup$
    – A.Γ.
    Dec 29 '18 at 20:24












  • $begingroup$
    Note that if $x>0$ then $e^x>x+1$ by a simple Taylor expansion. This proves what you want
    $endgroup$
    – Shalop
    Dec 29 '18 at 20:43










  • $begingroup$
    @A.Γ. Wow I never thought to work from the statement that I have to prove. It almost seems too easy to be true. I can't thank you enough for this answer.
    $endgroup$
    – Radu Gabriel
    Dec 29 '18 at 21:13














2












2








2





$begingroup$


$f:(0,infty)$,
$f(x)=frac{ln(x+1)}{x}$

Prove that for $forall x>0$ that $f(x) in(0,1)$. I calculated the derivative of $f(x)$: $f'(x)=frac{frac{x}{x+1}-ln(1+x)}{x^2}$ which I think simplifies to $frac{x^3}{x+1}-x^2(ln(1+x))$. I have no idea what to do next, I can't find the roots of this equation and I don't see any connection as of why it should be bounded by 0 and 1.



I hope I formatted this well, I don't usually post here but I am really curious how could I solve this kind of exercise.






calculus







share|cite|improve this question











$endgroup$




$f:(0,infty)$,
$f(x)=frac{ln(x+1)}{x}$

Prove that for $forall x>0$ that $f(x) in(0,1)$. I calculated the derivative of $f(x)$: $f'(x)=frac{frac{x}{x+1}-ln(1+x)}{x^2}$ which I think simplifies to $frac{x^3}{x+1}-x^2(ln(1+x))$. I have no idea what to do next, I can't find the roots of this equation and I don't see any connection as of why it should be bounded by 0 and 1.



I hope I formatted this well, I don't usually post here but I am really curious how could I solve this kind of exercise.






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 20:28









A.Γ.

22.6k32656




22.6k32656










asked Dec 29 '18 at 20:21









Radu GabrielRadu Gabriel

463




463








  • 2




    $begingroup$
    $0<frac{ln(x+1)}{x}<1$ $iff$ $0<ln(x+1)<x$.
    $endgroup$
    – A.Γ.
    Dec 29 '18 at 20:24












  • $begingroup$
    Note that if $x>0$ then $e^x>x+1$ by a simple Taylor expansion. This proves what you want
    $endgroup$
    – Shalop
    Dec 29 '18 at 20:43










  • $begingroup$
    @A.Γ. Wow I never thought to work from the statement that I have to prove. It almost seems too easy to be true. I can't thank you enough for this answer.
    $endgroup$
    – Radu Gabriel
    Dec 29 '18 at 21:13














  • 2




    $begingroup$
    $0<frac{ln(x+1)}{x}<1$ $iff$ $0<ln(x+1)<x$.
    $endgroup$
    – A.Γ.
    Dec 29 '18 at 20:24












  • $begingroup$
    Note that if $x>0$ then $e^x>x+1$ by a simple Taylor expansion. This proves what you want
    $endgroup$
    – Shalop
    Dec 29 '18 at 20:43










  • $begingroup$
    @A.Γ. Wow I never thought to work from the statement that I have to prove. It almost seems too easy to be true. I can't thank you enough for this answer.
    $endgroup$
    – Radu Gabriel
    Dec 29 '18 at 21:13








2




2




$begingroup$
$0<frac{ln(x+1)}{x}<1$ $iff$ $0<ln(x+1)<x$.
$endgroup$
– A.Γ.
Dec 29 '18 at 20:24






$begingroup$
$0<frac{ln(x+1)}{x}<1$ $iff$ $0<ln(x+1)<x$.
$endgroup$
– A.Γ.
Dec 29 '18 at 20:24














$begingroup$
Note that if $x>0$ then $e^x>x+1$ by a simple Taylor expansion. This proves what you want
$endgroup$
– Shalop
Dec 29 '18 at 20:43




$begingroup$
Note that if $x>0$ then $e^x>x+1$ by a simple Taylor expansion. This proves what you want
$endgroup$
– Shalop
Dec 29 '18 at 20:43












$begingroup$
@A.Γ. Wow I never thought to work from the statement that I have to prove. It almost seems too easy to be true. I can't thank you enough for this answer.
$endgroup$
– Radu Gabriel
Dec 29 '18 at 21:13




$begingroup$
@A.Γ. Wow I never thought to work from the statement that I have to prove. It almost seems too easy to be true. I can't thank you enough for this answer.
$endgroup$
– Radu Gabriel
Dec 29 '18 at 21:13










4 Answers
4






active

oldest

votes


















0












$begingroup$

First of all if is easy to see that $dfrac{ln{(x+1)}}{x}>0$.



It's well known that $ln{x}leq x-1$ where the equality is valid only for $x=1$. (You can prove it if you define the function $f(x)=ln{x}-x+1$ and find the global maximum using simple calculus). Then by this we get that $ln{(x+1)}leq x$ so for $x>0$ we get $dfrac{ln{(x+1)}}{x}<1$.



Alexandros






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    For $x in (0,infty)$, you have
    $$ln (x+1) = int_0^x frac{dt}{1+t}$$



    Hence



    $$0 le f(x) = frac{1}{x}int_0^x frac{dt}{1+t} < frac{1}{x}int_0^x dt =1$$



    As all considered maps are continuous.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Why do you argue for $xin (0,1)$? The domain of the function is $(0, infty)$. Also, what you show seems to be insufficient to prove the desired inequalities.
      $endgroup$
      – A. Pongrácz
      Dec 29 '18 at 20:29












    • $begingroup$
      No, it is to be proven on the set of all positive reals (and it is true there). Also, as I said, your argument is not sufficient to show the upper estimation (even on the unit interval).
      $endgroup$
      – A. Pongrácz
      Dec 29 '18 at 20:34










    • $begingroup$
      @A.Pongrácz You’re right. Modified the answer.
      $endgroup$
      – mathcounterexamples.net
      Dec 29 '18 at 20:34



















    0












    $begingroup$

    $ln(x+1)$ is a concave function, so the curve is under each of it tangents. Now it happens that the line with equation $y=x$ is its tangent at origin. So for any $x 0$ of its domain, $ln(x+1)le x$. Furthermore, considering the function $x-ln(x+1)$, it is easy to see this function is increasing on [0,+infty), so
    $$ln(x+1)<xenspaceforall x>0iff frac{ln(x+1)}x<1enspaceforall x>0.$$
    The inequality $dfrac{ln(x+1)}x>0$ is obvious since $1+x>1$.






    share|cite|improve this answer









    $endgroup$





















      -1












      $begingroup$

      It is much simpler to show that the derivative of $ln(x+1)$ is always in $(0,1)$ (and it is continuous).
      Then $ln(x+1)$ is strictly between $0$ and $x$, think about it.






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        First of all if is easy to see that $dfrac{ln{(x+1)}}{x}>0$.



        It's well known that $ln{x}leq x-1$ where the equality is valid only for $x=1$. (You can prove it if you define the function $f(x)=ln{x}-x+1$ and find the global maximum using simple calculus). Then by this we get that $ln{(x+1)}leq x$ so for $x>0$ we get $dfrac{ln{(x+1)}}{x}<1$.



        Alexandros






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          First of all if is easy to see that $dfrac{ln{(x+1)}}{x}>0$.



          It's well known that $ln{x}leq x-1$ where the equality is valid only for $x=1$. (You can prove it if you define the function $f(x)=ln{x}-x+1$ and find the global maximum using simple calculus). Then by this we get that $ln{(x+1)}leq x$ so for $x>0$ we get $dfrac{ln{(x+1)}}{x}<1$.



          Alexandros






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            First of all if is easy to see that $dfrac{ln{(x+1)}}{x}>0$.



            It's well known that $ln{x}leq x-1$ where the equality is valid only for $x=1$. (You can prove it if you define the function $f(x)=ln{x}-x+1$ and find the global maximum using simple calculus). Then by this we get that $ln{(x+1)}leq x$ so for $x>0$ we get $dfrac{ln{(x+1)}}{x}<1$.



            Alexandros






            share|cite|improve this answer









            $endgroup$



            First of all if is easy to see that $dfrac{ln{(x+1)}}{x}>0$.



            It's well known that $ln{x}leq x-1$ where the equality is valid only for $x=1$. (You can prove it if you define the function $f(x)=ln{x}-x+1$ and find the global maximum using simple calculus). Then by this we get that $ln{(x+1)}leq x$ so for $x>0$ we get $dfrac{ln{(x+1)}}{x}<1$.



            Alexandros







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 29 '18 at 20:34









            AlexSygelakisAlexSygelakis

            11




            11























                0












                $begingroup$

                For $x in (0,infty)$, you have
                $$ln (x+1) = int_0^x frac{dt}{1+t}$$



                Hence



                $$0 le f(x) = frac{1}{x}int_0^x frac{dt}{1+t} < frac{1}{x}int_0^x dt =1$$



                As all considered maps are continuous.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Why do you argue for $xin (0,1)$? The domain of the function is $(0, infty)$. Also, what you show seems to be insufficient to prove the desired inequalities.
                  $endgroup$
                  – A. Pongrácz
                  Dec 29 '18 at 20:29












                • $begingroup$
                  No, it is to be proven on the set of all positive reals (and it is true there). Also, as I said, your argument is not sufficient to show the upper estimation (even on the unit interval).
                  $endgroup$
                  – A. Pongrácz
                  Dec 29 '18 at 20:34










                • $begingroup$
                  @A.Pongrácz You’re right. Modified the answer.
                  $endgroup$
                  – mathcounterexamples.net
                  Dec 29 '18 at 20:34
















                0












                $begingroup$

                For $x in (0,infty)$, you have
                $$ln (x+1) = int_0^x frac{dt}{1+t}$$



                Hence



                $$0 le f(x) = frac{1}{x}int_0^x frac{dt}{1+t} < frac{1}{x}int_0^x dt =1$$



                As all considered maps are continuous.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Why do you argue for $xin (0,1)$? The domain of the function is $(0, infty)$. Also, what you show seems to be insufficient to prove the desired inequalities.
                  $endgroup$
                  – A. Pongrácz
                  Dec 29 '18 at 20:29












                • $begingroup$
                  No, it is to be proven on the set of all positive reals (and it is true there). Also, as I said, your argument is not sufficient to show the upper estimation (even on the unit interval).
                  $endgroup$
                  – A. Pongrácz
                  Dec 29 '18 at 20:34










                • $begingroup$
                  @A.Pongrácz You’re right. Modified the answer.
                  $endgroup$
                  – mathcounterexamples.net
                  Dec 29 '18 at 20:34














                0












                0








                0





                $begingroup$

                For $x in (0,infty)$, you have
                $$ln (x+1) = int_0^x frac{dt}{1+t}$$



                Hence



                $$0 le f(x) = frac{1}{x}int_0^x frac{dt}{1+t} < frac{1}{x}int_0^x dt =1$$



                As all considered maps are continuous.






                share|cite|improve this answer











                $endgroup$



                For $x in (0,infty)$, you have
                $$ln (x+1) = int_0^x frac{dt}{1+t}$$



                Hence



                $$0 le f(x) = frac{1}{x}int_0^x frac{dt}{1+t} < frac{1}{x}int_0^x dt =1$$



                As all considered maps are continuous.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 29 '18 at 20:40

























                answered Dec 29 '18 at 20:27









                mathcounterexamples.netmathcounterexamples.net

                25.4k21953




                25.4k21953












                • $begingroup$
                  Why do you argue for $xin (0,1)$? The domain of the function is $(0, infty)$. Also, what you show seems to be insufficient to prove the desired inequalities.
                  $endgroup$
                  – A. Pongrácz
                  Dec 29 '18 at 20:29












                • $begingroup$
                  No, it is to be proven on the set of all positive reals (and it is true there). Also, as I said, your argument is not sufficient to show the upper estimation (even on the unit interval).
                  $endgroup$
                  – A. Pongrácz
                  Dec 29 '18 at 20:34










                • $begingroup$
                  @A.Pongrácz You’re right. Modified the answer.
                  $endgroup$
                  – mathcounterexamples.net
                  Dec 29 '18 at 20:34


















                • $begingroup$
                  Why do you argue for $xin (0,1)$? The domain of the function is $(0, infty)$. Also, what you show seems to be insufficient to prove the desired inequalities.
                  $endgroup$
                  – A. Pongrácz
                  Dec 29 '18 at 20:29












                • $begingroup$
                  No, it is to be proven on the set of all positive reals (and it is true there). Also, as I said, your argument is not sufficient to show the upper estimation (even on the unit interval).
                  $endgroup$
                  – A. Pongrácz
                  Dec 29 '18 at 20:34










                • $begingroup$
                  @A.Pongrácz You’re right. Modified the answer.
                  $endgroup$
                  – mathcounterexamples.net
                  Dec 29 '18 at 20:34
















                $begingroup$
                Why do you argue for $xin (0,1)$? The domain of the function is $(0, infty)$. Also, what you show seems to be insufficient to prove the desired inequalities.
                $endgroup$
                – A. Pongrácz
                Dec 29 '18 at 20:29






                $begingroup$
                Why do you argue for $xin (0,1)$? The domain of the function is $(0, infty)$. Also, what you show seems to be insufficient to prove the desired inequalities.
                $endgroup$
                – A. Pongrácz
                Dec 29 '18 at 20:29














                $begingroup$
                No, it is to be proven on the set of all positive reals (and it is true there). Also, as I said, your argument is not sufficient to show the upper estimation (even on the unit interval).
                $endgroup$
                – A. Pongrácz
                Dec 29 '18 at 20:34




                $begingroup$
                No, it is to be proven on the set of all positive reals (and it is true there). Also, as I said, your argument is not sufficient to show the upper estimation (even on the unit interval).
                $endgroup$
                – A. Pongrácz
                Dec 29 '18 at 20:34












                $begingroup$
                @A.Pongrácz You’re right. Modified the answer.
                $endgroup$
                – mathcounterexamples.net
                Dec 29 '18 at 20:34




                $begingroup$
                @A.Pongrácz You’re right. Modified the answer.
                $endgroup$
                – mathcounterexamples.net
                Dec 29 '18 at 20:34











                0












                $begingroup$

                $ln(x+1)$ is a concave function, so the curve is under each of it tangents. Now it happens that the line with equation $y=x$ is its tangent at origin. So for any $x 0$ of its domain, $ln(x+1)le x$. Furthermore, considering the function $x-ln(x+1)$, it is easy to see this function is increasing on [0,+infty), so
                $$ln(x+1)<xenspaceforall x>0iff frac{ln(x+1)}x<1enspaceforall x>0.$$
                The inequality $dfrac{ln(x+1)}x>0$ is obvious since $1+x>1$.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  $ln(x+1)$ is a concave function, so the curve is under each of it tangents. Now it happens that the line with equation $y=x$ is its tangent at origin. So for any $x 0$ of its domain, $ln(x+1)le x$. Furthermore, considering the function $x-ln(x+1)$, it is easy to see this function is increasing on [0,+infty), so
                  $$ln(x+1)<xenspaceforall x>0iff frac{ln(x+1)}x<1enspaceforall x>0.$$
                  The inequality $dfrac{ln(x+1)}x>0$ is obvious since $1+x>1$.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    $ln(x+1)$ is a concave function, so the curve is under each of it tangents. Now it happens that the line with equation $y=x$ is its tangent at origin. So for any $x 0$ of its domain, $ln(x+1)le x$. Furthermore, considering the function $x-ln(x+1)$, it is easy to see this function is increasing on [0,+infty), so
                    $$ln(x+1)<xenspaceforall x>0iff frac{ln(x+1)}x<1enspaceforall x>0.$$
                    The inequality $dfrac{ln(x+1)}x>0$ is obvious since $1+x>1$.






                    share|cite|improve this answer









                    $endgroup$



                    $ln(x+1)$ is a concave function, so the curve is under each of it tangents. Now it happens that the line with equation $y=x$ is its tangent at origin. So for any $x 0$ of its domain, $ln(x+1)le x$. Furthermore, considering the function $x-ln(x+1)$, it is easy to see this function is increasing on [0,+infty), so
                    $$ln(x+1)<xenspaceforall x>0iff frac{ln(x+1)}x<1enspaceforall x>0.$$
                    The inequality $dfrac{ln(x+1)}x>0$ is obvious since $1+x>1$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 29 '18 at 20:43









                    BernardBernard

                    119k639112




                    119k639112























                        -1












                        $begingroup$

                        It is much simpler to show that the derivative of $ln(x+1)$ is always in $(0,1)$ (and it is continuous).
                        Then $ln(x+1)$ is strictly between $0$ and $x$, think about it.






                        share|cite|improve this answer









                        $endgroup$


















                          -1












                          $begingroup$

                          It is much simpler to show that the derivative of $ln(x+1)$ is always in $(0,1)$ (and it is continuous).
                          Then $ln(x+1)$ is strictly between $0$ and $x$, think about it.






                          share|cite|improve this answer









                          $endgroup$
















                            -1












                            -1








                            -1





                            $begingroup$

                            It is much simpler to show that the derivative of $ln(x+1)$ is always in $(0,1)$ (and it is continuous).
                            Then $ln(x+1)$ is strictly between $0$ and $x$, think about it.






                            share|cite|improve this answer









                            $endgroup$



                            It is much simpler to show that the derivative of $ln(x+1)$ is always in $(0,1)$ (and it is continuous).
                            Then $ln(x+1)$ is strictly between $0$ and $x$, think about it.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 29 '18 at 20:27









                            A. PongráczA. Pongrácz

                            5,9381929




                            5,9381929






























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                                Human spaceflight

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                                Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
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                                Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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                                0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
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                                張江高科駅

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                                張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2
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