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












1














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




    $log(1-x)$ is not defined for $xgeq 1$
    – Kavi Rama Murthy
    2 hours ago










  • To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
    – Mike
    1 hour ago


















1














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




















  • 3




    $log(1-x)$ is not defined for $xgeq 1$
    – Kavi Rama Murthy
    2 hours ago










  • To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
    – Mike
    1 hour ago
















1












1








1


1





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















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






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share|cite|improve this question













share|cite|improve this question




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edited 1 hour ago

























asked 2 hours ago









Mike

1,366221




1,366221








  • 3




    $log(1-x)$ is not defined for $xgeq 1$
    – Kavi Rama Murthy
    2 hours ago










  • To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
    – Mike
    1 hour ago
















  • 3




    $log(1-x)$ is not defined for $xgeq 1$
    – Kavi Rama Murthy
    2 hours ago










  • To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
    – Mike
    1 hour ago










3




3




$log(1-x)$ is not defined for $xgeq 1$
– Kavi Rama Murthy
2 hours ago




$log(1-x)$ is not defined for $xgeq 1$
– Kavi Rama Murthy
2 hours ago












To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
– Mike
1 hour ago






To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
– Mike
1 hour ago












2 Answers
2






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oldest

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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























  • (+1) But sorry for the stress!
    – Mike
    1 hour ago





















0














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














    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























    • (+1) But sorry for the stress!
      – Mike
      1 hour ago


















    3














    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























    • (+1) But sorry for the stress!
      – Mike
      1 hour ago
















    3












    3








    3






    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














    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 1 hour ago

























    answered 2 hours ago









    Kavi Rama Murthy

    50.6k31854




    50.6k31854












    • (+1) But sorry for the stress!
      – Mike
      1 hour ago




















    • (+1) But sorry for the stress!
      – Mike
      1 hour ago


















    (+1) But sorry for the stress!
    – Mike
    1 hour ago






    (+1) But sorry for the stress!
    – Mike
    1 hour ago













    0














    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


























      0














      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
























        0












        0








        0






        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












        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 37 mins ago









        Mike

        1,366221




        1,366221






























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