Is there a bijective, monotonically increasing, strictly concave function from the reals, to the reals?












5












$begingroup$


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
    Mar 22 at 17:58






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    Mar 22 at 17:59






  • 1




    $begingroup$
    @cammil a surjection (i.e. a function whose range is equal to its codomain).
    $endgroup$
    – Jake
    Mar 22 at 18:19






  • 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
    Mar 22 at 20:03






  • 2




    $begingroup$
    A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
    $endgroup$
    – Apass.Jack
    Mar 22 at 22:58


















5












$begingroup$


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
    Mar 22 at 17:58






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    Mar 22 at 17:59






  • 1




    $begingroup$
    @cammil a surjection (i.e. a function whose range is equal to its codomain).
    $endgroup$
    – Jake
    Mar 22 at 18:19






  • 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
    Mar 22 at 20:03






  • 2




    $begingroup$
    A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
    $endgroup$
    – Apass.Jack
    Mar 22 at 22:58
















5












5








5


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 Mar 23 at 8:30









Jack M

18.9k33882




18.9k33882










asked Mar 22 at 17:52









cammilcammil

1436




1436








  • 6




    $begingroup$
    $f(x) = -e^{-x}$?
    $endgroup$
    – Daniel Schepler
    Mar 22 at 17:58






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    Mar 22 at 17:59






  • 1




    $begingroup$
    @cammil a surjection (i.e. a function whose range is equal to its codomain).
    $endgroup$
    – Jake
    Mar 22 at 18:19






  • 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
    Mar 22 at 20:03






  • 2




    $begingroup$
    A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
    $endgroup$
    – Apass.Jack
    Mar 22 at 22:58
















  • 6




    $begingroup$
    $f(x) = -e^{-x}$?
    $endgroup$
    – Daniel Schepler
    Mar 22 at 17:58






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    Mar 22 at 17:59






  • 1




    $begingroup$
    @cammil a surjection (i.e. a function whose range is equal to its codomain).
    $endgroup$
    – Jake
    Mar 22 at 18:19






  • 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
    Mar 22 at 20:03






  • 2




    $begingroup$
    A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
    $endgroup$
    – Apass.Jack
    Mar 22 at 22:58










6




6




$begingroup$
$f(x) = -e^{-x}$?
$endgroup$
– Daniel Schepler
Mar 22 at 17:58




$begingroup$
$f(x) = -e^{-x}$?
$endgroup$
– Daniel Schepler
Mar 22 at 17:58




1




1




$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
Mar 22 at 17:59




$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
Mar 22 at 17:59




1




1




$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
Mar 22 at 18:19




$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
Mar 22 at 18:19




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
Mar 22 at 20:03




$begingroup$
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
Mar 22 at 20:03




2




2




$begingroup$
A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
$endgroup$
– Apass.Jack
Mar 22 at 22:58






$begingroup$
A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
$endgroup$
– Apass.Jack
Mar 22 at 22:58












3 Answers
3






active

oldest

votes


















16












$begingroup$

$$
f(x) = x-e^{-x}
$$
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 $x$.



EDIT: Since it was requested in the comments, here is a plot of this function and its inverse:
Plot



Note that even though the growth rate for positive $x$ is slow, the function is asymptotically linear (with slope $2-sqrt{3}approx 0.268$) and thus unbounded.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    +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
    Mar 22 at 19:44








  • 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
    Mar 22 at 22:22








  • 1




    $begingroup$
    @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
    $endgroup$
    – Apass.Jack
    Mar 22 at 22:51












  • $begingroup$
    I could’ve never come up with this.
    $endgroup$
    – Randall
    Mar 22 at 23:55










  • $begingroup$
    As in Ross Millikan's comment, $f$ is the lower branch of a hyperbola. Neatly, $f^{-1}$ is the upper branch. The lower branch passes through the points $(1,0)$, $(0,-1)$, $(4,1)$, $(-1,-4)$, $(15,4)$, $(-4,-15)$. (So the upper branch passes through the reflections of these six points in the line $x=y$.) This makes the hyperbola easy to plot (in GeoGebra, for instance). Its equation is $x^2-4xy+y^2=1$. The equation of its asymptotes is $x^2-4xy+y^2=0$.
    $endgroup$
    – Calum Gilhooley
    Mar 23 at 19:52





















5












$begingroup$

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









$endgroup$





















    3












    $begingroup$

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






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I would have preferred to incorporate the $pi x$ term into the integral, by adding a constant to the integrand. And using $pi$ is maybe not the most obvious choice of a number larger than$~fracpi2$; the essential point is having the primitive of an everywhere decreasing function of $t$ that moreover is positively bounded away from $0$ (i.e., remains ${}>c$ for some constant $c>0$).
      $endgroup$
      – Marc van Leeuwen
      Mar 25 at 17:57












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    16












    $begingroup$

    $$
    f(x) = x-e^{-x}
    $$
    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 $x$.



    EDIT: Since it was requested in the comments, here is a plot of this function and its inverse:
    Plot



    Note that even though the growth rate for positive $x$ is slow, the function is asymptotically linear (with slope $2-sqrt{3}approx 0.268$) and thus unbounded.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      +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
      Mar 22 at 19:44








    • 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
      Mar 22 at 22:22








    • 1




      $begingroup$
      @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
      $endgroup$
      – Apass.Jack
      Mar 22 at 22:51












    • $begingroup$
      I could’ve never come up with this.
      $endgroup$
      – Randall
      Mar 22 at 23:55










    • $begingroup$
      As in Ross Millikan's comment, $f$ is the lower branch of a hyperbola. Neatly, $f^{-1}$ is the upper branch. The lower branch passes through the points $(1,0)$, $(0,-1)$, $(4,1)$, $(-1,-4)$, $(15,4)$, $(-4,-15)$. (So the upper branch passes through the reflections of these six points in the line $x=y$.) This makes the hyperbola easy to plot (in GeoGebra, for instance). Its equation is $x^2-4xy+y^2=1$. The equation of its asymptotes is $x^2-4xy+y^2=0$.
      $endgroup$
      – Calum Gilhooley
      Mar 23 at 19:52


















    16












    $begingroup$

    $$
    f(x) = x-e^{-x}
    $$
    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 $x$.



    EDIT: Since it was requested in the comments, here is a plot of this function and its inverse:
    Plot



    Note that even though the growth rate for positive $x$ is slow, the function is asymptotically linear (with slope $2-sqrt{3}approx 0.268$) and thus unbounded.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      +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
      Mar 22 at 19:44








    • 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
      Mar 22 at 22:22








    • 1




      $begingroup$
      @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
      $endgroup$
      – Apass.Jack
      Mar 22 at 22:51












    • $begingroup$
      I could’ve never come up with this.
      $endgroup$
      – Randall
      Mar 22 at 23:55










    • $begingroup$
      As in Ross Millikan's comment, $f$ is the lower branch of a hyperbola. Neatly, $f^{-1}$ is the upper branch. The lower branch passes through the points $(1,0)$, $(0,-1)$, $(4,1)$, $(-1,-4)$, $(15,4)$, $(-4,-15)$. (So the upper branch passes through the reflections of these six points in the line $x=y$.) This makes the hyperbola easy to plot (in GeoGebra, for instance). Its equation is $x^2-4xy+y^2=1$. The equation of its asymptotes is $x^2-4xy+y^2=0$.
      $endgroup$
      – Calum Gilhooley
      Mar 23 at 19:52
















    16












    16








    16





    $begingroup$

    $$
    f(x) = x-e^{-x}
    $$
    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 $x$.



    EDIT: Since it was requested in the comments, here is a plot of this function and its inverse:
    Plot



    Note that even though the growth rate for positive $x$ is slow, the function is asymptotically linear (with slope $2-sqrt{3}approx 0.268$) and thus unbounded.






    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 $x$.



    EDIT: Since it was requested in the comments, here is a plot of this function and its inverse:
    Plot



    Note that even though the growth rate for positive $x$ is slow, the function is asymptotically linear (with slope $2-sqrt{3}approx 0.268$) and thus unbounded.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 23 at 16:30

























    answered Mar 22 at 18:38









    eyeballfrogeyeballfrog

    7,044633




    7,044633








    • 1




      $begingroup$
      +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
      Mar 22 at 19:44








    • 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
      Mar 22 at 22:22








    • 1




      $begingroup$
      @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
      $endgroup$
      – Apass.Jack
      Mar 22 at 22:51












    • $begingroup$
      I could’ve never come up with this.
      $endgroup$
      – Randall
      Mar 22 at 23:55










    • $begingroup$
      As in Ross Millikan's comment, $f$ is the lower branch of a hyperbola. Neatly, $f^{-1}$ is the upper branch. The lower branch passes through the points $(1,0)$, $(0,-1)$, $(4,1)$, $(-1,-4)$, $(15,4)$, $(-4,-15)$. (So the upper branch passes through the reflections of these six points in the line $x=y$.) This makes the hyperbola easy to plot (in GeoGebra, for instance). Its equation is $x^2-4xy+y^2=1$. The equation of its asymptotes is $x^2-4xy+y^2=0$.
      $endgroup$
      – Calum Gilhooley
      Mar 23 at 19:52
















    • 1




      $begingroup$
      +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
      Mar 22 at 19:44








    • 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
      Mar 22 at 22:22








    • 1




      $begingroup$
      @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
      $endgroup$
      – Apass.Jack
      Mar 22 at 22:51












    • $begingroup$
      I could’ve never come up with this.
      $endgroup$
      – Randall
      Mar 22 at 23:55










    • $begingroup$
      As in Ross Millikan's comment, $f$ is the lower branch of a hyperbola. Neatly, $f^{-1}$ is the upper branch. The lower branch passes through the points $(1,0)$, $(0,-1)$, $(4,1)$, $(-1,-4)$, $(15,4)$, $(-4,-15)$. (So the upper branch passes through the reflections of these six points in the line $x=y$.) This makes the hyperbola easy to plot (in GeoGebra, for instance). Its equation is $x^2-4xy+y^2=1$. The equation of its asymptotes is $x^2-4xy+y^2=0$.
      $endgroup$
      – Calum Gilhooley
      Mar 23 at 19:52










    1




    1




    $begingroup$
    +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
    Mar 22 at 19:44






    $begingroup$
    +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
    Mar 22 at 19:44






    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
    Mar 22 at 22:22






    $begingroup$
    @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
    Mar 22 at 22:22






    1




    1




    $begingroup$
    @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
    $endgroup$
    – Apass.Jack
    Mar 22 at 22:51






    $begingroup$
    @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
    $endgroup$
    – Apass.Jack
    Mar 22 at 22:51














    $begingroup$
    I could’ve never come up with this.
    $endgroup$
    – Randall
    Mar 22 at 23:55




    $begingroup$
    I could’ve never come up with this.
    $endgroup$
    – Randall
    Mar 22 at 23:55












    $begingroup$
    As in Ross Millikan's comment, $f$ is the lower branch of a hyperbola. Neatly, $f^{-1}$ is the upper branch. The lower branch passes through the points $(1,0)$, $(0,-1)$, $(4,1)$, $(-1,-4)$, $(15,4)$, $(-4,-15)$. (So the upper branch passes through the reflections of these six points in the line $x=y$.) This makes the hyperbola easy to plot (in GeoGebra, for instance). Its equation is $x^2-4xy+y^2=1$. The equation of its asymptotes is $x^2-4xy+y^2=0$.
    $endgroup$
    – Calum Gilhooley
    Mar 23 at 19:52






    $begingroup$
    As in Ross Millikan's comment, $f$ is the lower branch of a hyperbola. Neatly, $f^{-1}$ is the upper branch. The lower branch passes through the points $(1,0)$, $(0,-1)$, $(4,1)$, $(-1,-4)$, $(15,4)$, $(-4,-15)$. (So the upper branch passes through the reflections of these six points in the line $x=y$.) This makes the hyperbola easy to plot (in GeoGebra, for instance). Its equation is $x^2-4xy+y^2=1$. The equation of its asymptotes is $x^2-4xy+y^2=0$.
    $endgroup$
    – Calum Gilhooley
    Mar 23 at 19:52













    5












    $begingroup$

    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









    $endgroup$


















      5












      $begingroup$

      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









      $endgroup$
















        5












        5








        5





        $begingroup$

        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 Mar 22 at 18:33









        paw88789paw88789

        29.5k12349




        29.5k12349























            3












            $begingroup$

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






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I would have preferred to incorporate the $pi x$ term into the integral, by adding a constant to the integrand. And using $pi$ is maybe not the most obvious choice of a number larger than$~fracpi2$; the essential point is having the primitive of an everywhere decreasing function of $t$ that moreover is positively bounded away from $0$ (i.e., remains ${}>c$ for some constant $c>0$).
              $endgroup$
              – Marc van Leeuwen
              Mar 25 at 17:57
















            3












            $begingroup$

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






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I would have preferred to incorporate the $pi x$ term into the integral, by adding a constant to the integrand. And using $pi$ is maybe not the most obvious choice of a number larger than$~fracpi2$; the essential point is having the primitive of an everywhere decreasing function of $t$ that moreover is positively bounded away from $0$ (i.e., remains ${}>c$ for some constant $c>0$).
              $endgroup$
              – Marc van Leeuwen
              Mar 25 at 17:57














            3












            3








            3





            $begingroup$

            $f(x) = pi x+ int_0^x arctan (-t),dt$ 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








            edited Mar 23 at 23:29

























            answered Mar 22 at 18:57









            zhw.zhw.

            74.8k43175




            74.8k43175












            • $begingroup$
              I would have preferred to incorporate the $pi x$ term into the integral, by adding a constant to the integrand. And using $pi$ is maybe not the most obvious choice of a number larger than$~fracpi2$; the essential point is having the primitive of an everywhere decreasing function of $t$ that moreover is positively bounded away from $0$ (i.e., remains ${}>c$ for some constant $c>0$).
              $endgroup$
              – Marc van Leeuwen
              Mar 25 at 17:57


















            • $begingroup$
              I would have preferred to incorporate the $pi x$ term into the integral, by adding a constant to the integrand. And using $pi$ is maybe not the most obvious choice of a number larger than$~fracpi2$; the essential point is having the primitive of an everywhere decreasing function of $t$ that moreover is positively bounded away from $0$ (i.e., remains ${}>c$ for some constant $c>0$).
              $endgroup$
              – Marc van Leeuwen
              Mar 25 at 17:57
















            $begingroup$
            I would have preferred to incorporate the $pi x$ term into the integral, by adding a constant to the integrand. And using $pi$ is maybe not the most obvious choice of a number larger than$~fracpi2$; the essential point is having the primitive of an everywhere decreasing function of $t$ that moreover is positively bounded away from $0$ (i.e., remains ${}>c$ for some constant $c>0$).
            $endgroup$
            – Marc van Leeuwen
            Mar 25 at 17:57




            $begingroup$
            I would have preferred to incorporate the $pi x$ term into the integral, by adding a constant to the integrand. And using $pi$ is maybe not the most obvious choice of a number larger than$~fracpi2$; the essential point is having the primitive of an everywhere decreasing function of $t$ that moreover is positively bounded away from $0$ (i.e., remains ${}>c$ for some constant $c>0$).
            $endgroup$
            – Marc van Leeuwen
            Mar 25 at 17:57


















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