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.

22.9k32656




22.9k32656










asked Dec 29 '18 at 20:21









Radu GabrielRadu Gabriel

1115




1115








  • 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

                27k22158




                27k22158












                • $begingroup$
                  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

                    123k741117




                    123k741117























                        -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

                            5,9731929




                            5,9731929






























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                                Wiesbaden

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                                Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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                                Marschland

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                                Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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                                Dieringhausen

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                                Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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