Is there an injective, monotonically increasing, strictly concave function from the reals, to the reals?












2












$begingroup$


I can't come up with a single one.



The range should be the whole of the reals. The best I have is $log(x)$ but that's only on the positive real line. And there's $f(x) = x$, but this is not strictly concave. And $-e^{-x}$ only maps to half of the real line.



Any ideas?






real-analysis functions recreational-mathematics real-numbers







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    $f(x) = -e^{-x}$?
    $endgroup$
    – Daniel Schepler
    4 hours ago






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    4 hours ago










  • $begingroup$
    Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
    $endgroup$
    – cammil
    3 hours ago








  • 1




    $begingroup$
    @cammil a surjection (i.e. a function whose range is equal to its codomain).
    $endgroup$
    – Jake
    3 hours ago










  • $begingroup$
    If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
    $endgroup$
    – Ross Millikan
    2 hours ago
















2












$begingroup$


I can't come up with a single one.



The range should be the whole of the reals. The best I have is $log(x)$ but that's only on the positive real line. And there's $f(x) = x$, but this is not strictly concave. And $-e^{-x}$ only maps to half of the real line.



Any ideas?






real-analysis functions recreational-mathematics real-numbers







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    $f(x) = -e^{-x}$?
    $endgroup$
    – Daniel Schepler
    4 hours ago






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    4 hours ago










  • $begingroup$
    Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
    $endgroup$
    – cammil
    3 hours ago








  • 1




    $begingroup$
    @cammil a surjection (i.e. a function whose range is equal to its codomain).
    $endgroup$
    – Jake
    3 hours ago










  • $begingroup$
    If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
    $endgroup$
    – Ross Millikan
    2 hours ago














2












2








2





$begingroup$


I can't come up with a single one.



The range should be the whole of the reals. The best I have is $log(x)$ but that's only on the positive real line. And there's $f(x) = x$, but this is not strictly concave. And $-e^{-x}$ only maps to half of the real line.



Any ideas?






real-analysis functions recreational-mathematics real-numbers







share|cite|improve this question











$endgroup$




I can't come up with a single one.



The range should be the whole of the reals. The best I have is $log(x)$ but that's only on the positive real line. And there's $f(x) = x$, but this is not strictly concave. And $-e^{-x}$ only maps to half of the real line.



Any ideas?






real-analysis functions recreational-mathematics real-numbers




real-analysis functions recreational-mathematics real-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago







cammil

















asked 4 hours ago









cammilcammil

1264




1264








  • 5




    $begingroup$
    $f(x) = -e^{-x}$?
    $endgroup$
    – Daniel Schepler
    4 hours ago






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    4 hours ago










  • $begingroup$
    Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
    $endgroup$
    – cammil
    3 hours ago








  • 1




    $begingroup$
    @cammil a surjection (i.e. a function whose range is equal to its codomain).
    $endgroup$
    – Jake
    3 hours ago










  • $begingroup$
    If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
    $endgroup$
    – Ross Millikan
    2 hours ago














  • 5




    $begingroup$
    $f(x) = -e^{-x}$?
    $endgroup$
    – Daniel Schepler
    4 hours ago






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    4 hours ago










  • $begingroup$
    Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
    $endgroup$
    – cammil
    3 hours ago








  • 1




    $begingroup$
    @cammil a surjection (i.e. a function whose range is equal to its codomain).
    $endgroup$
    – Jake
    3 hours ago










  • $begingroup$
    If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
    $endgroup$
    – Ross Millikan
    2 hours ago








5




5




$begingroup$
$f(x) = -e^{-x}$?
$endgroup$
– Daniel Schepler
4 hours ago




$begingroup$
$f(x) = -e^{-x}$?
$endgroup$
– Daniel Schepler
4 hours ago




1




1




$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
4 hours ago




$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
4 hours ago












$begingroup$
Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
$endgroup$
– cammil
3 hours ago






$begingroup$
Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
$endgroup$
– cammil
3 hours ago






1




1




$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
3 hours ago




$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
3 hours ago












$begingroup$
If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
$endgroup$
– Ross Millikan
2 hours ago




$begingroup$
If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
$endgroup$
– Ross Millikan
2 hours ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

How about



$f(x)=left{begin{array}{cc} ln(x+1)& &xge 0\1-e^{-x}& &x<0end{array}right.$






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    $$
    f(x) = x-e^{-x}
    $$
    is such a function. Since $f''(x) = -e^{-x}$ is always negative, it is strictly concave, and it's not hard to show it hits every real.



    Even better,
    $$
    f(x) = 2x -sqrt{1+3x^2}
    $$
    has $f''(x) = -3(1+3x^2)^{-3/2} < 0$ everywhere and the explicit inverse $f^{-1}(x) = 2x+sqrt{1+3x^2}$, clearly defined for all $x$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
      $endgroup$
      – Calum Gilhooley
      2 hours ago





















    0












    $begingroup$

    $F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      How about



      $f(x)=left{begin{array}{cc} ln(x+1)& &xge 0\1-e^{-x}& &x<0end{array}right.$






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        How about



        $f(x)=left{begin{array}{cc} ln(x+1)& &xge 0\1-e^{-x}& &x<0end{array}right.$






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          How about



          $f(x)=left{begin{array}{cc} ln(x+1)& &xge 0\1-e^{-x}& &x<0end{array}right.$






          share|cite|improve this answer









          $endgroup$



          How about



          $f(x)=left{begin{array}{cc} ln(x+1)& &xge 0\1-e^{-x}& &x<0end{array}right.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          paw88789paw88789

          29.4k12349




          29.4k12349























              4












              $begingroup$

              $$
              f(x) = x-e^{-x}
              $$
              is such a function. Since $f''(x) = -e^{-x}$ is always negative, it is strictly concave, and it's not hard to show it hits every real.



              Even better,
              $$
              f(x) = 2x -sqrt{1+3x^2}
              $$
              has $f''(x) = -3(1+3x^2)^{-3/2} < 0$ everywhere and the explicit inverse $f^{-1}(x) = 2x+sqrt{1+3x^2}$, clearly defined for all $x$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
                $endgroup$
                – Calum Gilhooley
                2 hours ago


















              4












              $begingroup$

              $$
              f(x) = x-e^{-x}
              $$
              is such a function. Since $f''(x) = -e^{-x}$ is always negative, it is strictly concave, and it's not hard to show it hits every real.



              Even better,
              $$
              f(x) = 2x -sqrt{1+3x^2}
              $$
              has $f''(x) = -3(1+3x^2)^{-3/2} < 0$ everywhere and the explicit inverse $f^{-1}(x) = 2x+sqrt{1+3x^2}$, clearly defined for all $x$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
                $endgroup$
                – Calum Gilhooley
                2 hours ago
















              4












              4








              4





              $begingroup$

              $$
              f(x) = x-e^{-x}
              $$
              is such a function. Since $f''(x) = -e^{-x}$ is always negative, it is strictly concave, and it's not hard to show it hits every real.



              Even better,
              $$
              f(x) = 2x -sqrt{1+3x^2}
              $$
              has $f''(x) = -3(1+3x^2)^{-3/2} < 0$ everywhere and the explicit inverse $f^{-1}(x) = 2x+sqrt{1+3x^2}$, clearly defined for all $x$.






              share|cite|improve this answer











              $endgroup$



              $$
              f(x) = x-e^{-x}
              $$
              is such a function. Since $f''(x) = -e^{-x}$ is always negative, it is strictly concave, and it's not hard to show it hits every real.



              Even better,
              $$
              f(x) = 2x -sqrt{1+3x^2}
              $$
              has $f''(x) = -3(1+3x^2)^{-3/2} < 0$ everywhere and the explicit inverse $f^{-1}(x) = 2x+sqrt{1+3x^2}$, clearly defined for all $x$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 3 hours ago

























              answered 3 hours ago









              eyeballfrogeyeballfrog

              6,664630




              6,664630












              • $begingroup$
                +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
                $endgroup$
                – Calum Gilhooley
                2 hours ago




















              • $begingroup$
                +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
                $endgroup$
                – Calum Gilhooley
                2 hours ago


















              $begingroup$
              +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
              $endgroup$
              – Calum Gilhooley
              2 hours ago






              $begingroup$
              +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
              $endgroup$
              – Calum Gilhooley
              2 hours ago













              0












              $begingroup$

              $F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.






                  share|cite|improve this answer









                  $endgroup$



                  $F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  zhw.zhw.

                  74.5k43175




                  74.5k43175






























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