Does there exist a function which satisfies follow conditions?












-1












$begingroup$


Does there exist a function which satisfies following conditions?



1) $F(x)=frac{G(x)}{H(x)}$ is a sigmoid function (or S-shaped function);



2) $G(x)$ is strictly concave;



3) $H(x)$ is not a constant.






linear-algebra







share|cite|improve this question











$endgroup$

















    -1












    $begingroup$


    Does there exist a function which satisfies following conditions?



    1) $F(x)=frac{G(x)}{H(x)}$ is a sigmoid function (or S-shaped function);



    2) $G(x)$ is strictly concave;



    3) $H(x)$ is not a constant.






    linear-algebra







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1


      0



      $begingroup$


      Does there exist a function which satisfies following conditions?



      1) $F(x)=frac{G(x)}{H(x)}$ is a sigmoid function (or S-shaped function);



      2) $G(x)$ is strictly concave;



      3) $H(x)$ is not a constant.






      linear-algebra







      share|cite|improve this question











      $endgroup$




      Does there exist a function which satisfies following conditions?



      1) $F(x)=frac{G(x)}{H(x)}$ is a sigmoid function (or S-shaped function);



      2) $G(x)$ is strictly concave;



      3) $H(x)$ is not a constant.






      linear-algebra




      linear-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 30 '18 at 13:46







      Lee White

















      asked Nov 30 '18 at 12:35









      Lee WhiteLee White

      86




      86






















          1 Answer
          1






          active

          oldest

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          0












          $begingroup$

          G(x) = log(x+1)



          H(x) = 1+log(x+1)



          F(x) = log(x+1)/(1+log(x+1))



          That works grand



          edit: you edited your question. Your new answer is



          H = log(x+1)/(tanh(x+4)+1)



          G = log(x+1)



          F = tanh(x+4)+1






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
            $endgroup$
            – Lee White
            Nov 30 '18 at 13:23










          • $begingroup$
            Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
            $endgroup$
            – D'Arcy
            Nov 30 '18 at 13:31






          • 1




            $begingroup$
            Cool, man! I get it! I will try to work as you said, it's a nice method.
            $endgroup$
            – Lee White
            Nov 30 '18 at 13:43













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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          G(x) = log(x+1)



          H(x) = 1+log(x+1)



          F(x) = log(x+1)/(1+log(x+1))



          That works grand



          edit: you edited your question. Your new answer is



          H = log(x+1)/(tanh(x+4)+1)



          G = log(x+1)



          F = tanh(x+4)+1






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
            $endgroup$
            – Lee White
            Nov 30 '18 at 13:23










          • $begingroup$
            Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
            $endgroup$
            – D'Arcy
            Nov 30 '18 at 13:31






          • 1




            $begingroup$
            Cool, man! I get it! I will try to work as you said, it's a nice method.
            $endgroup$
            – Lee White
            Nov 30 '18 at 13:43


















          0












          $begingroup$

          G(x) = log(x+1)



          H(x) = 1+log(x+1)



          F(x) = log(x+1)/(1+log(x+1))



          That works grand



          edit: you edited your question. Your new answer is



          H = log(x+1)/(tanh(x+4)+1)



          G = log(x+1)



          F = tanh(x+4)+1






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
            $endgroup$
            – Lee White
            Nov 30 '18 at 13:23










          • $begingroup$
            Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
            $endgroup$
            – D'Arcy
            Nov 30 '18 at 13:31






          • 1




            $begingroup$
            Cool, man! I get it! I will try to work as you said, it's a nice method.
            $endgroup$
            – Lee White
            Nov 30 '18 at 13:43
















          0












          0








          0





          $begingroup$

          G(x) = log(x+1)



          H(x) = 1+log(x+1)



          F(x) = log(x+1)/(1+log(x+1))



          That works grand



          edit: you edited your question. Your new answer is



          H = log(x+1)/(tanh(x+4)+1)



          G = log(x+1)



          F = tanh(x+4)+1






          share|cite|improve this answer











          $endgroup$



          G(x) = log(x+1)



          H(x) = 1+log(x+1)



          F(x) = log(x+1)/(1+log(x+1))



          That works grand



          edit: you edited your question. Your new answer is



          H = log(x+1)/(tanh(x+4)+1)



          G = log(x+1)



          F = tanh(x+4)+1







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 30 '18 at 13:30

























          answered Nov 30 '18 at 12:52









          D'ArcyD'Arcy

          263




          263












          • $begingroup$
            Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
            $endgroup$
            – Lee White
            Nov 30 '18 at 13:23










          • $begingroup$
            Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
            $endgroup$
            – D'Arcy
            Nov 30 '18 at 13:31






          • 1




            $begingroup$
            Cool, man! I get it! I will try to work as you said, it's a nice method.
            $endgroup$
            – Lee White
            Nov 30 '18 at 13:43




















          • $begingroup$
            Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
            $endgroup$
            – Lee White
            Nov 30 '18 at 13:23










          • $begingroup$
            Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
            $endgroup$
            – D'Arcy
            Nov 30 '18 at 13:31






          • 1




            $begingroup$
            Cool, man! I get it! I will try to work as you said, it's a nice method.
            $endgroup$
            – Lee White
            Nov 30 '18 at 13:43


















          $begingroup$
          Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
          $endgroup$
          – Lee White
          Nov 30 '18 at 13:23




          $begingroup$
          Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
          $endgroup$
          – Lee White
          Nov 30 '18 at 13:23












          $begingroup$
          Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
          $endgroup$
          – D'Arcy
          Nov 30 '18 at 13:31




          $begingroup$
          Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
          $endgroup$
          – D'Arcy
          Nov 30 '18 at 13:31




          1




          1




          $begingroup$
          Cool, man! I get it! I will try to work as you said, it's a nice method.
          $endgroup$
          – Lee White
          Nov 30 '18 at 13:43






          $begingroup$
          Cool, man! I get it! I will try to work as you said, it's a nice method.
          $endgroup$
          – Lee White
          Nov 30 '18 at 13:43




















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