Find a function which when applied to the inverse of an argument, only changes sign












4












$begingroup$


So basically, a function $f$ with $f(frac{1}{x}) = - f(x)$. Additionally, it should also be strictly increasing.



I know that the logarithm has this property, but I'm looking for a function with different boundary conditions. Namely: f(0) = -1 (and $f(x -> infty) = 1$).



I know one solution to this: $f(x) = frac{x-1}{x+1}$, but I am wondering:




  1. Is there a general method to find such functions?


  2. Is my given solution unique?







functions







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    So basically, a function $f$ with $f(frac{1}{x}) = - f(x)$. Additionally, it should also be strictly increasing.



    I know that the logarithm has this property, but I'm looking for a function with different boundary conditions. Namely: f(0) = -1 (and $f(x -> infty) = 1$).



    I know one solution to this: $f(x) = frac{x-1}{x+1}$, but I am wondering:




    1. Is there a general method to find such functions?


    2. Is my given solution unique?







    functions







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      So basically, a function $f$ with $f(frac{1}{x}) = - f(x)$. Additionally, it should also be strictly increasing.



      I know that the logarithm has this property, but I'm looking for a function with different boundary conditions. Namely: f(0) = -1 (and $f(x -> infty) = 1$).



      I know one solution to this: $f(x) = frac{x-1}{x+1}$, but I am wondering:




      1. Is there a general method to find such functions?


      2. Is my given solution unique?







      functions







      share|cite|improve this question











      $endgroup$




      So basically, a function $f$ with $f(frac{1}{x}) = - f(x)$. Additionally, it should also be strictly increasing.



      I know that the logarithm has this property, but I'm looking for a function with different boundary conditions. Namely: f(0) = -1 (and $f(x -> infty) = 1$).



      I know one solution to this: $f(x) = frac{x-1}{x+1}$, but I am wondering:




      1. Is there a general method to find such functions?


      2. Is my given solution unique?







      functions




      functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 9 at 14:26







      Thomas Kehrenberg

















      asked Jan 9 at 13:40









      Thomas KehrenbergThomas Kehrenberg

      234




      234






















          1 Answer
          1






          active

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          5












          $begingroup$

          For $x>0$, substitute $x = e^t$ or $t = log x$. Then, we have
          $$
          g(t):=f(e^t) = -f(e^{-t})=-g(-t),quad forall tinmathbb{R},
          $$          and $g(infty) = 1$, $g(-infty)=-1$ (assuming continuity of $f$ at $0$.) Hence $fbig|_{(0,infty)}$ corresponds to an odd function $g$ satisfying the above boundary condition at $pminfty$. Similarly, $fbig|_{(-infty,0)}$ corresponds to some odd $h$ in the same manner.



          This shows there are as many solutions as there are odd functions $g,h$ for which $limlimits_{xtoinfty} g(x)=limlimits_{xtoinfty}h(x)=1$. One example other than $f(x) = frac{x-1}{x+1}$ is
          $$
          f(x) = frac{2}{pi}arctan(log |x|),quad xneq 0
          $$           and $f(0)=-1$. Also notice that the OP's solution corresponds to $g(t) = frac{e^t-1}{e^t+1}=tanh frac{t}{2}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            It might be mentioned that $f(x) = frac{x-1}{x+1}$ corresponds to $g(t)=tanh(frac12 t)$.
            $endgroup$
            – Henning Makholm
            Jan 9 at 14:17










          • $begingroup$
            @HenningMakholm Thanks for pointing out :)
            $endgroup$
            – Song
            Jan 9 at 14:18










          • $begingroup$
            And $g(t)=tanh t$ leads to $f(x)=frac{x^2-1}{x^2+1}$ which satisfies $f'(0)=0$ (which might either be desirable or not for the OP's purposes).
            $endgroup$
            – Henning Makholm
            Jan 9 at 14:24











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          $begingroup$

          For $x>0$, substitute $x = e^t$ or $t = log x$. Then, we have
          $$
          g(t):=f(e^t) = -f(e^{-t})=-g(-t),quad forall tinmathbb{R},
          $$          and $g(infty) = 1$, $g(-infty)=-1$ (assuming continuity of $f$ at $0$.) Hence $fbig|_{(0,infty)}$ corresponds to an odd function $g$ satisfying the above boundary condition at $pminfty$. Similarly, $fbig|_{(-infty,0)}$ corresponds to some odd $h$ in the same manner.



          This shows there are as many solutions as there are odd functions $g,h$ for which $limlimits_{xtoinfty} g(x)=limlimits_{xtoinfty}h(x)=1$. One example other than $f(x) = frac{x-1}{x+1}$ is
          $$
          f(x) = frac{2}{pi}arctan(log |x|),quad xneq 0
          $$           and $f(0)=-1$. Also notice that the OP's solution corresponds to $g(t) = frac{e^t-1}{e^t+1}=tanh frac{t}{2}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            It might be mentioned that $f(x) = frac{x-1}{x+1}$ corresponds to $g(t)=tanh(frac12 t)$.
            $endgroup$
            – Henning Makholm
            Jan 9 at 14:17










          • $begingroup$
            @HenningMakholm Thanks for pointing out :)
            $endgroup$
            – Song
            Jan 9 at 14:18










          • $begingroup$
            And $g(t)=tanh t$ leads to $f(x)=frac{x^2-1}{x^2+1}$ which satisfies $f'(0)=0$ (which might either be desirable or not for the OP's purposes).
            $endgroup$
            – Henning Makholm
            Jan 9 at 14:24
















          5












          $begingroup$

          For $x>0$, substitute $x = e^t$ or $t = log x$. Then, we have
          $$
          g(t):=f(e^t) = -f(e^{-t})=-g(-t),quad forall tinmathbb{R},
          $$          and $g(infty) = 1$, $g(-infty)=-1$ (assuming continuity of $f$ at $0$.) Hence $fbig|_{(0,infty)}$ corresponds to an odd function $g$ satisfying the above boundary condition at $pminfty$. Similarly, $fbig|_{(-infty,0)}$ corresponds to some odd $h$ in the same manner.



          This shows there are as many solutions as there are odd functions $g,h$ for which $limlimits_{xtoinfty} g(x)=limlimits_{xtoinfty}h(x)=1$. One example other than $f(x) = frac{x-1}{x+1}$ is
          $$
          f(x) = frac{2}{pi}arctan(log |x|),quad xneq 0
          $$           and $f(0)=-1$. Also notice that the OP's solution corresponds to $g(t) = frac{e^t-1}{e^t+1}=tanh frac{t}{2}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            It might be mentioned that $f(x) = frac{x-1}{x+1}$ corresponds to $g(t)=tanh(frac12 t)$.
            $endgroup$
            – Henning Makholm
            Jan 9 at 14:17










          • $begingroup$
            @HenningMakholm Thanks for pointing out :)
            $endgroup$
            – Song
            Jan 9 at 14:18










          • $begingroup$
            And $g(t)=tanh t$ leads to $f(x)=frac{x^2-1}{x^2+1}$ which satisfies $f'(0)=0$ (which might either be desirable or not for the OP's purposes).
            $endgroup$
            – Henning Makholm
            Jan 9 at 14:24














          5












          5








          5





          $begingroup$

          For $x>0$, substitute $x = e^t$ or $t = log x$. Then, we have
          $$
          g(t):=f(e^t) = -f(e^{-t})=-g(-t),quad forall tinmathbb{R},
          $$          and $g(infty) = 1$, $g(-infty)=-1$ (assuming continuity of $f$ at $0$.) Hence $fbig|_{(0,infty)}$ corresponds to an odd function $g$ satisfying the above boundary condition at $pminfty$. Similarly, $fbig|_{(-infty,0)}$ corresponds to some odd $h$ in the same manner.



          This shows there are as many solutions as there are odd functions $g,h$ for which $limlimits_{xtoinfty} g(x)=limlimits_{xtoinfty}h(x)=1$. One example other than $f(x) = frac{x-1}{x+1}$ is
          $$
          f(x) = frac{2}{pi}arctan(log |x|),quad xneq 0
          $$           and $f(0)=-1$. Also notice that the OP's solution corresponds to $g(t) = frac{e^t-1}{e^t+1}=tanh frac{t}{2}$.






          share|cite|improve this answer











          $endgroup$



          For $x>0$, substitute $x = e^t$ or $t = log x$. Then, we have
          $$
          g(t):=f(e^t) = -f(e^{-t})=-g(-t),quad forall tinmathbb{R},
          $$          and $g(infty) = 1$, $g(-infty)=-1$ (assuming continuity of $f$ at $0$.) Hence $fbig|_{(0,infty)}$ corresponds to an odd function $g$ satisfying the above boundary condition at $pminfty$. Similarly, $fbig|_{(-infty,0)}$ corresponds to some odd $h$ in the same manner.



          This shows there are as many solutions as there are odd functions $g,h$ for which $limlimits_{xtoinfty} g(x)=limlimits_{xtoinfty}h(x)=1$. One example other than $f(x) = frac{x-1}{x+1}$ is
          $$
          f(x) = frac{2}{pi}arctan(log |x|),quad xneq 0
          $$           and $f(0)=-1$. Also notice that the OP's solution corresponds to $g(t) = frac{e^t-1}{e^t+1}=tanh frac{t}{2}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 9 at 17:29

























          answered Jan 9 at 14:07









          SongSong

          8,689625




          8,689625












          • $begingroup$
            It might be mentioned that $f(x) = frac{x-1}{x+1}$ corresponds to $g(t)=tanh(frac12 t)$.
            $endgroup$
            – Henning Makholm
            Jan 9 at 14:17










          • $begingroup$
            @HenningMakholm Thanks for pointing out :)
            $endgroup$
            – Song
            Jan 9 at 14:18










          • $begingroup$
            And $g(t)=tanh t$ leads to $f(x)=frac{x^2-1}{x^2+1}$ which satisfies $f'(0)=0$ (which might either be desirable or not for the OP's purposes).
            $endgroup$
            – Henning Makholm
            Jan 9 at 14:24


















          • $begingroup$
            It might be mentioned that $f(x) = frac{x-1}{x+1}$ corresponds to $g(t)=tanh(frac12 t)$.
            $endgroup$
            – Henning Makholm
            Jan 9 at 14:17










          • $begingroup$
            @HenningMakholm Thanks for pointing out :)
            $endgroup$
            – Song
            Jan 9 at 14:18










          • $begingroup$
            And $g(t)=tanh t$ leads to $f(x)=frac{x^2-1}{x^2+1}$ which satisfies $f'(0)=0$ (which might either be desirable or not for the OP's purposes).
            $endgroup$
            – Henning Makholm
            Jan 9 at 14:24
















          $begingroup$
          It might be mentioned that $f(x) = frac{x-1}{x+1}$ corresponds to $g(t)=tanh(frac12 t)$.
          $endgroup$
          – Henning Makholm
          Jan 9 at 14:17




          $begingroup$
          It might be mentioned that $f(x) = frac{x-1}{x+1}$ corresponds to $g(t)=tanh(frac12 t)$.
          $endgroup$
          – Henning Makholm
          Jan 9 at 14:17












          $begingroup$
          @HenningMakholm Thanks for pointing out :)
          $endgroup$
          – Song
          Jan 9 at 14:18




          $begingroup$
          @HenningMakholm Thanks for pointing out :)
          $endgroup$
          – Song
          Jan 9 at 14:18












          $begingroup$
          And $g(t)=tanh t$ leads to $f(x)=frac{x^2-1}{x^2+1}$ which satisfies $f'(0)=0$ (which might either be desirable or not for the OP's purposes).
          $endgroup$
          – Henning Makholm
          Jan 9 at 14:24




          $begingroup$
          And $g(t)=tanh t$ leads to $f(x)=frac{x^2-1}{x^2+1}$ which satisfies $f'(0)=0$ (which might either be desirable or not for the OP's purposes).
          $endgroup$
          – Henning Makholm
          Jan 9 at 14:24


















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