Is there a function that satisfies $f(log frac{p}{1-p}) = p(1-p)$?












2












$begingroup$



Let $pin (0,1)$ I am wondering there exists a function $f:mathbb{R}to[0,{1over 2}]$ that satisfies
$$
fleft(log frac{p}{1-p}right) = p(1-p)
$$




I've tried messing around with exponentials but nothing has worked so far.






functions logarithms exponential-function functional-equations







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Let $pin (0,1)$ I am wondering there exists a function $f:mathbb{R}to[0,{1over 2}]$ that satisfies
    $$
    fleft(log frac{p}{1-p}right) = p(1-p)
    $$




    I've tried messing around with exponentials but nothing has worked so far.






    functions logarithms exponential-function functional-equations







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      Let $pin (0,1)$ I am wondering there exists a function $f:mathbb{R}to[0,{1over 2}]$ that satisfies
      $$
      fleft(log frac{p}{1-p}right) = p(1-p)
      $$




      I've tried messing around with exponentials but nothing has worked so far.






      functions logarithms exponential-function functional-equations







      share|cite|improve this question











      $endgroup$





      Let $pin (0,1)$ I am wondering there exists a function $f:mathbb{R}to[0,{1over 2}]$ that satisfies
      $$
      fleft(log frac{p}{1-p}right) = p(1-p)
      $$




      I've tried messing around with exponentials but nothing has worked so far.






      functions logarithms exponential-function functional-equations




      functions logarithms exponential-function functional-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 16:36









      greedoid

      39.2k114797




      39.2k114797










      asked Dec 1 '18 at 16:04









      5d41402abc45d41402abc4

      182




      182






















          1 Answer
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          $begingroup$

          Write $x= log frac{p}{1-p}$ then $p = {e^xover e^x+1}$ so $$f(x) = {e^xover e^x+1}left(1- {e^xover e^x+1}right) = {e^xover (e^x+1)^2}$$



          and clearly $f(x)in [0,{1over 2}]$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why are you unconventionally assuming that $log$ is base $10$ instead of base $e$?
            $endgroup$
            – Git Gud
            Dec 1 '18 at 16:17






          • 3




            $begingroup$
            Because in my country it is convenctonaly. Logarithm with base $e$ is denoted by $ln$ from latin logarithmus naturalis
            $endgroup$
            – greedoid
            Dec 1 '18 at 16:20








          • 1




            $begingroup$
            Ok, I have no problem with that.
            $endgroup$
            – greedoid
            Dec 1 '18 at 16:24






          • 2




            $begingroup$
            @GitGud: I think it is stretching it quite a bit to call using base-10 logarithms "noncompliance with a standard definition". It is one of the common meanings of the $log$ symbol, and the difference is more related to which field you're working in than what country you are in.
            $endgroup$
            – Henning Makholm
            Dec 1 '18 at 16:25






          • 1




            $begingroup$
            The ISO's stands on this doesn't comply with the mathematical culture. The standard, among mathematicians, is that $log$ is base $e$. This shouldn't be new for anyone with a (relatively) large exposure to mathematics. @callculus
            $endgroup$
            – Git Gud
            Dec 1 '18 at 16:30













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          1 Answer
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          active

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          1 Answer
          1






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          active

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          3












          $begingroup$

          Write $x= log frac{p}{1-p}$ then $p = {e^xover e^x+1}$ so $$f(x) = {e^xover e^x+1}left(1- {e^xover e^x+1}right) = {e^xover (e^x+1)^2}$$



          and clearly $f(x)in [0,{1over 2}]$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why are you unconventionally assuming that $log$ is base $10$ instead of base $e$?
            $endgroup$
            – Git Gud
            Dec 1 '18 at 16:17






          • 3




            $begingroup$
            Because in my country it is convenctonaly. Logarithm with base $e$ is denoted by $ln$ from latin logarithmus naturalis
            $endgroup$
            – greedoid
            Dec 1 '18 at 16:20








          • 1




            $begingroup$
            Ok, I have no problem with that.
            $endgroup$
            – greedoid
            Dec 1 '18 at 16:24






          • 2




            $begingroup$
            @GitGud: I think it is stretching it quite a bit to call using base-10 logarithms "noncompliance with a standard definition". It is one of the common meanings of the $log$ symbol, and the difference is more related to which field you're working in than what country you are in.
            $endgroup$
            – Henning Makholm
            Dec 1 '18 at 16:25






          • 1




            $begingroup$
            The ISO's stands on this doesn't comply with the mathematical culture. The standard, among mathematicians, is that $log$ is base $e$. This shouldn't be new for anyone with a (relatively) large exposure to mathematics. @callculus
            $endgroup$
            – Git Gud
            Dec 1 '18 at 16:30


















          3












          $begingroup$

          Write $x= log frac{p}{1-p}$ then $p = {e^xover e^x+1}$ so $$f(x) = {e^xover e^x+1}left(1- {e^xover e^x+1}right) = {e^xover (e^x+1)^2}$$



          and clearly $f(x)in [0,{1over 2}]$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why are you unconventionally assuming that $log$ is base $10$ instead of base $e$?
            $endgroup$
            – Git Gud
            Dec 1 '18 at 16:17






          • 3




            $begingroup$
            Because in my country it is convenctonaly. Logarithm with base $e$ is denoted by $ln$ from latin logarithmus naturalis
            $endgroup$
            – greedoid
            Dec 1 '18 at 16:20








          • 1




            $begingroup$
            Ok, I have no problem with that.
            $endgroup$
            – greedoid
            Dec 1 '18 at 16:24






          • 2




            $begingroup$
            @GitGud: I think it is stretching it quite a bit to call using base-10 logarithms "noncompliance with a standard definition". It is one of the common meanings of the $log$ symbol, and the difference is more related to which field you're working in than what country you are in.
            $endgroup$
            – Henning Makholm
            Dec 1 '18 at 16:25






          • 1




            $begingroup$
            The ISO's stands on this doesn't comply with the mathematical culture. The standard, among mathematicians, is that $log$ is base $e$. This shouldn't be new for anyone with a (relatively) large exposure to mathematics. @callculus
            $endgroup$
            – Git Gud
            Dec 1 '18 at 16:30
















          3












          3








          3





          $begingroup$

          Write $x= log frac{p}{1-p}$ then $p = {e^xover e^x+1}$ so $$f(x) = {e^xover e^x+1}left(1- {e^xover e^x+1}right) = {e^xover (e^x+1)^2}$$



          and clearly $f(x)in [0,{1over 2}]$.






          share|cite|improve this answer











          $endgroup$



          Write $x= log frac{p}{1-p}$ then $p = {e^xover e^x+1}$ so $$f(x) = {e^xover e^x+1}left(1- {e^xover e^x+1}right) = {e^xover (e^x+1)^2}$$



          and clearly $f(x)in [0,{1over 2}]$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 '18 at 16:35

























          answered Dec 1 '18 at 16:09









          greedoidgreedoid

          39.2k114797




          39.2k114797












          • $begingroup$
            Why are you unconventionally assuming that $log$ is base $10$ instead of base $e$?
            $endgroup$
            – Git Gud
            Dec 1 '18 at 16:17






          • 3




            $begingroup$
            Because in my country it is convenctonaly. Logarithm with base $e$ is denoted by $ln$ from latin logarithmus naturalis
            $endgroup$
            – greedoid
            Dec 1 '18 at 16:20








          • 1




            $begingroup$
            Ok, I have no problem with that.
            $endgroup$
            – greedoid
            Dec 1 '18 at 16:24






          • 2




            $begingroup$
            @GitGud: I think it is stretching it quite a bit to call using base-10 logarithms "noncompliance with a standard definition". It is one of the common meanings of the $log$ symbol, and the difference is more related to which field you're working in than what country you are in.
            $endgroup$
            – Henning Makholm
            Dec 1 '18 at 16:25






          • 1




            $begingroup$
            The ISO's stands on this doesn't comply with the mathematical culture. The standard, among mathematicians, is that $log$ is base $e$. This shouldn't be new for anyone with a (relatively) large exposure to mathematics. @callculus
            $endgroup$
            – Git Gud
            Dec 1 '18 at 16:30




















          • $begingroup$
            Why are you unconventionally assuming that $log$ is base $10$ instead of base $e$?
            $endgroup$
            – Git Gud
            Dec 1 '18 at 16:17






          • 3




            $begingroup$
            Because in my country it is convenctonaly. Logarithm with base $e$ is denoted by $ln$ from latin logarithmus naturalis
            $endgroup$
            – greedoid
            Dec 1 '18 at 16:20








          • 1




            $begingroup$
            Ok, I have no problem with that.
            $endgroup$
            – greedoid
            Dec 1 '18 at 16:24






          • 2




            $begingroup$
            @GitGud: I think it is stretching it quite a bit to call using base-10 logarithms "noncompliance with a standard definition". It is one of the common meanings of the $log$ symbol, and the difference is more related to which field you're working in than what country you are in.
            $endgroup$
            – Henning Makholm
            Dec 1 '18 at 16:25






          • 1




            $begingroup$
            The ISO's stands on this doesn't comply with the mathematical culture. The standard, among mathematicians, is that $log$ is base $e$. This shouldn't be new for anyone with a (relatively) large exposure to mathematics. @callculus
            $endgroup$
            – Git Gud
            Dec 1 '18 at 16:30


















          $begingroup$
          Why are you unconventionally assuming that $log$ is base $10$ instead of base $e$?
          $endgroup$
          – Git Gud
          Dec 1 '18 at 16:17




          $begingroup$
          Why are you unconventionally assuming that $log$ is base $10$ instead of base $e$?
          $endgroup$
          – Git Gud
          Dec 1 '18 at 16:17




          3




          3




          $begingroup$
          Because in my country it is convenctonaly. Logarithm with base $e$ is denoted by $ln$ from latin logarithmus naturalis
          $endgroup$
          – greedoid
          Dec 1 '18 at 16:20






          $begingroup$
          Because in my country it is convenctonaly. Logarithm with base $e$ is denoted by $ln$ from latin logarithmus naturalis
          $endgroup$
          – greedoid
          Dec 1 '18 at 16:20






          1




          1




          $begingroup$
          Ok, I have no problem with that.
          $endgroup$
          – greedoid
          Dec 1 '18 at 16:24




          $begingroup$
          Ok, I have no problem with that.
          $endgroup$
          – greedoid
          Dec 1 '18 at 16:24




          2




          2




          $begingroup$
          @GitGud: I think it is stretching it quite a bit to call using base-10 logarithms "noncompliance with a standard definition". It is one of the common meanings of the $log$ symbol, and the difference is more related to which field you're working in than what country you are in.
          $endgroup$
          – Henning Makholm
          Dec 1 '18 at 16:25




          $begingroup$
          @GitGud: I think it is stretching it quite a bit to call using base-10 logarithms "noncompliance with a standard definition". It is one of the common meanings of the $log$ symbol, and the difference is more related to which field you're working in than what country you are in.
          $endgroup$
          – Henning Makholm
          Dec 1 '18 at 16:25




          1




          1




          $begingroup$
          The ISO's stands on this doesn't comply with the mathematical culture. The standard, among mathematicians, is that $log$ is base $e$. This shouldn't be new for anyone with a (relatively) large exposure to mathematics. @callculus
          $endgroup$
          – Git Gud
          Dec 1 '18 at 16:30






          $begingroup$
          The ISO's stands on this doesn't comply with the mathematical culture. The standard, among mathematicians, is that $log$ is base $e$. This shouldn't be new for anyone with a (relatively) large exposure to mathematics. @callculus
          $endgroup$
          – Git Gud
          Dec 1 '18 at 16:30




















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