Use the concavity of $log (1-x)+x$ to show that $log (1-x)leq -x$












1












$begingroup$


I've been reading this post and its accepted answer here. The OP (accepted answer) made a comment that $log(1-x)leq -x$ but I've been having issues proving it.



FULL PROOF (EDIT)



With credits to Kavi Rama Murthy, I provide a full proof.



Let $f: ]0,1[longrightarrow Bbb{R},;x mapsto log (1-x)+x$. Then,



begin{align}f''(x) = -frac{1}{(x-1)^2}leq 0,;forall ;xin ;]0,1[.end{align}
Hence, $f$ is monotone non-increasing and $f(x)leq f(0),forall ;xin ;]0,1[,$ that is
begin{align}log(1-x)leq - x end{align}






real-analysis analysis logarithms







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  • 3




    $begingroup$
    $log(1-x)$ is not defined for $xgeq 1$
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 8:53










  • $begingroup$
    To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
    $endgroup$
    – Omojola Micheal
    Jan 3 at 9:13


















1












$begingroup$


I've been reading this post and its accepted answer here. The OP (accepted answer) made a comment that $log(1-x)leq -x$ but I've been having issues proving it.



FULL PROOF (EDIT)



With credits to Kavi Rama Murthy, I provide a full proof.



Let $f: ]0,1[longrightarrow Bbb{R},;x mapsto log (1-x)+x$. Then,



begin{align}f''(x) = -frac{1}{(x-1)^2}leq 0,;forall ;xin ;]0,1[.end{align}
Hence, $f$ is monotone non-increasing and $f(x)leq f(0),forall ;xin ;]0,1[,$ that is
begin{align}log(1-x)leq - x end{align}






real-analysis analysis logarithms







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $log(1-x)$ is not defined for $xgeq 1$
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 8:53










  • $begingroup$
    To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
    $endgroup$
    – Omojola Micheal
    Jan 3 at 9:13
















1












1








1


1



$begingroup$


I've been reading this post and its accepted answer here. The OP (accepted answer) made a comment that $log(1-x)leq -x$ but I've been having issues proving it.



FULL PROOF (EDIT)



With credits to Kavi Rama Murthy, I provide a full proof.



Let $f: ]0,1[longrightarrow Bbb{R},;x mapsto log (1-x)+x$. Then,



begin{align}f''(x) = -frac{1}{(x-1)^2}leq 0,;forall ;xin ;]0,1[.end{align}
Hence, $f$ is monotone non-increasing and $f(x)leq f(0),forall ;xin ;]0,1[,$ that is
begin{align}log(1-x)leq - x end{align}






real-analysis analysis logarithms







share|cite|improve this question











$endgroup$




I've been reading this post and its accepted answer here. The OP (accepted answer) made a comment that $log(1-x)leq -x$ but I've been having issues proving it.



FULL PROOF (EDIT)



With credits to Kavi Rama Murthy, I provide a full proof.



Let $f: ]0,1[longrightarrow Bbb{R},;x mapsto log (1-x)+x$. Then,



begin{align}f''(x) = -frac{1}{(x-1)^2}leq 0,;forall ;xin ;]0,1[.end{align}
Hence, $f$ is monotone non-increasing and $f(x)leq f(0),forall ;xin ;]0,1[,$ that is
begin{align}log(1-x)leq - x end{align}






real-analysis analysis logarithms




real-analysis analysis logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Jan 3 at 9:25







Omojola Micheal

















asked Jan 3 at 8:51









Omojola MichealOmojola Micheal

1,969324




1,969324








  • 3




    $begingroup$
    $log(1-x)$ is not defined for $xgeq 1$
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 8:53










  • $begingroup$
    To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
    $endgroup$
    – Omojola Micheal
    Jan 3 at 9:13
















  • 3




    $begingroup$
    $log(1-x)$ is not defined for $xgeq 1$
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 8:53










  • $begingroup$
    To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
    $endgroup$
    – Omojola Micheal
    Jan 3 at 9:13










3




3




$begingroup$
$log(1-x)$ is not defined for $xgeq 1$
$endgroup$
– Kavi Rama Murthy
Jan 3 at 8:53




$begingroup$
$log(1-x)$ is not defined for $xgeq 1$
$endgroup$
– Kavi Rama Murthy
Jan 3 at 8:53












$begingroup$
To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
$endgroup$
– Omojola Micheal
Jan 3 at 9:13






$begingroup$
To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
$endgroup$
– Omojola Micheal
Jan 3 at 9:13












2 Answers
2






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oldest

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$begingroup$

The function is concave on $(0,1)$ because $f''(x)=-frac 1 {(x-1)^{2}} leq 0$. It is not defined for $x geq 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+1) But sorry for the stress!
    $endgroup$
    – Omojola Micheal
    Jan 3 at 9:07





















1












$begingroup$

Another way, is to consider the Maclaurin series expansion



begin{align} log(1-x)=-sum^{infty}_{n=1}dfrac{x^n}{n}=-x-dfrac{x^2}{2}-dfrac{x^3}{3}cdots leq -x,;text{for fixed};xin;]0,1[end{align}






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    The function is concave on $(0,1)$ because $f''(x)=-frac 1 {(x-1)^{2}} leq 0$. It is not defined for $x geq 1$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      (+1) But sorry for the stress!
      $endgroup$
      – Omojola Micheal
      Jan 3 at 9:07


















    3












    $begingroup$

    The function is concave on $(0,1)$ because $f''(x)=-frac 1 {(x-1)^{2}} leq 0$. It is not defined for $x geq 1$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      (+1) But sorry for the stress!
      $endgroup$
      – Omojola Micheal
      Jan 3 at 9:07
















    3












    3








    3





    $begingroup$

    The function is concave on $(0,1)$ because $f''(x)=-frac 1 {(x-1)^{2}} leq 0$. It is not defined for $x geq 1$.






    share|cite|improve this answer











    $endgroup$



    The function is concave on $(0,1)$ because $f''(x)=-frac 1 {(x-1)^{2}} leq 0$. It is not defined for $x geq 1$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 3 at 9:08

























    answered Jan 3 at 8:56









    Kavi Rama MurthyKavi Rama Murthy

    65.8k42867




    65.8k42867












    • $begingroup$
      (+1) But sorry for the stress!
      $endgroup$
      – Omojola Micheal
      Jan 3 at 9:07




















    • $begingroup$
      (+1) But sorry for the stress!
      $endgroup$
      – Omojola Micheal
      Jan 3 at 9:07


















    $begingroup$
    (+1) But sorry for the stress!
    $endgroup$
    – Omojola Micheal
    Jan 3 at 9:07






    $begingroup$
    (+1) But sorry for the stress!
    $endgroup$
    – Omojola Micheal
    Jan 3 at 9:07













    1












    $begingroup$

    Another way, is to consider the Maclaurin series expansion



    begin{align} log(1-x)=-sum^{infty}_{n=1}dfrac{x^n}{n}=-x-dfrac{x^2}{2}-dfrac{x^3}{3}cdots leq -x,;text{for fixed};xin;]0,1[end{align}






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Another way, is to consider the Maclaurin series expansion



      begin{align} log(1-x)=-sum^{infty}_{n=1}dfrac{x^n}{n}=-x-dfrac{x^2}{2}-dfrac{x^3}{3}cdots leq -x,;text{for fixed};xin;]0,1[end{align}






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Another way, is to consider the Maclaurin series expansion



        begin{align} log(1-x)=-sum^{infty}_{n=1}dfrac{x^n}{n}=-x-dfrac{x^2}{2}-dfrac{x^3}{3}cdots leq -x,;text{for fixed};xin;]0,1[end{align}






        share|cite|improve this answer









        $endgroup$



        Another way, is to consider the Maclaurin series expansion



        begin{align} log(1-x)=-sum^{infty}_{n=1}dfrac{x^n}{n}=-x-dfrac{x^2}{2}-dfrac{x^3}{3}cdots leq -x,;text{for fixed};xin;]0,1[end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 10:29









        Omojola MichealOmojola Micheal

        1,969324




        1,969324






























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