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












3












$begingroup$


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











$endgroup$








  • 6




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






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    5 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
    4 hours ago








  • 1




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






  • 1




    $begingroup$
    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
    3 hours ago
















3












$begingroup$


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











$endgroup$








  • 6




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






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    5 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
    4 hours ago








  • 1




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






  • 1




    $begingroup$
    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
    3 hours ago














3












3








3





$begingroup$


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 4 hours ago







cammil

















asked 5 hours ago









cammilcammil

1314




1314








  • 6




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






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    5 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
    4 hours ago








  • 1




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






  • 1




    $begingroup$
    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
    3 hours ago














  • 6




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






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    5 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
    4 hours ago








  • 1




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






  • 1




    $begingroup$
    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
    3 hours ago








6




6




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




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




1




1




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




$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
5 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
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
4 hours ago






1




1




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




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




1




1




$begingroup$
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
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
3 hours ago










3 Answers
3






active

oldest

votes


















6












$begingroup$

$$
f(x) = x-e^{-x}
$$
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
    3 hours ago








  • 1




    $begingroup$
    @CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^{-x}$), then fiddled with the parameters until both the function and its inverse came out looking nice.
    $endgroup$
    – eyeballfrog
    45 mins ago












  • $begingroup$
    @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
    $endgroup$
    – Apass.Jack
    15 mins ago





















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









$endgroup$





















    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









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      $$
      f(x) = x-e^{-x}
      $$
      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
        3 hours ago








      • 1




        $begingroup$
        @CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^{-x}$), then fiddled with the parameters until both the function and its inverse came out looking nice.
        $endgroup$
        – eyeballfrog
        45 mins ago












      • $begingroup$
        @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
        $endgroup$
        – Apass.Jack
        15 mins ago


















      6












      $begingroup$

      $$
      f(x) = x-e^{-x}
      $$
      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
        3 hours ago








      • 1




        $begingroup$
        @CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^{-x}$), then fiddled with the parameters until both the function and its inverse came out looking nice.
        $endgroup$
        – eyeballfrog
        45 mins ago












      • $begingroup$
        @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
        $endgroup$
        – Apass.Jack
        15 mins ago
















      6












      6








      6





      $begingroup$

      $$
      f(x) = x-e^{-x}
      $$
      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 4 hours ago

























      answered 4 hours ago









      eyeballfrogeyeballfrog

      6,709630




      6,709630












      • $begingroup$
        +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
        3 hours ago








      • 1




        $begingroup$
        @CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^{-x}$), then fiddled with the parameters until both the function and its inverse came out looking nice.
        $endgroup$
        – eyeballfrog
        45 mins ago












      • $begingroup$
        @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
        $endgroup$
        – Apass.Jack
        15 mins 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
        3 hours ago








      • 1




        $begingroup$
        @CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^{-x}$), then fiddled with the parameters until both the function and its inverse came out looking nice.
        $endgroup$
        – eyeballfrog
        45 mins ago












      • $begingroup$
        @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
        $endgroup$
        – Apass.Jack
        15 mins 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
      3 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
      3 hours ago






      1




      1




      $begingroup$
      @CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^{-x}$), then fiddled with the parameters until both the function and its inverse came out looking nice.
      $endgroup$
      – eyeballfrog
      45 mins ago






      $begingroup$
      @CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^{-x}$), then fiddled with the parameters until both the function and its inverse came out looking nice.
      $endgroup$
      – eyeballfrog
      45 mins ago














      $begingroup$
      @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
      $endgroup$
      – Apass.Jack
      15 mins ago






      $begingroup$
      @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
      $endgroup$
      – Apass.Jack
      15 mins ago













      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 4 hours ago









          paw88789paw88789

          29.4k12349




          29.4k12349























              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 4 hours ago









                  zhw.zhw.

                  74.5k43175




                  74.5k43175






























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