If $g(x) ne f(x)$ almost everywhere, then polynomial transformation of $P(g(x)) ne P(f(x))$?

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Suppose that we have two functions $f(x)$ and $g(x)$. We know that $f(x) ne g(x)$ on the interval, except for measure zero subsets of $[a,b]$. Assume that both functions are positive everywhere and both are strictly increasing almost everywhere (i.e. there is a subset $S$ of the interval $[a,b]$, where $[a,b] setminus S$ has measure $0$, and $f, g$ are strictly increasing on $S$).



Under these conditions on $f$ and $g$ (positive, strictly increasing a.e.), is it possible that there exists a non-constant polynomial transformation $P$ such that $P(f(x)) = P(g(x))$ almost everywhere on $[a,b]$?







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  • can you clarify the edited requirement? Does "postive everywhere" mean $[a,b]$ or on all of $mathbbR$?
    – 6005
    Aug 9 at 22:53










  • What about $P(f(x)) = P(g(x)) = 0$?
    – Adam Francey
    Aug 9 at 22:59






  • 1




    Are $,f,g,$ assumed to be strictly increasing, maybe?
    – dxiv
    Aug 9 at 22:59







  • 1




    @6005 everywhere on $mathbbR$.
    – user1292919
    Aug 9 at 23:02






  • 2




    I'm a bit confused. Let the interval be $[0,pi/2]$, and $f(x)=x$, $g(x)=sin(x)$. Are you suggesting that there's a non-constant polynomial $P$ such that $P(x)=P(sin(x))$? Or are you looking for some kind of additional necessary conditions on $f,g$ to guarantee such a polynomial exists?
    – Alex R.
    Aug 9 at 23:15















up vote
4
down vote

favorite
2












Suppose that we have two functions $f(x)$ and $g(x)$. We know that $f(x) ne g(x)$ on the interval, except for measure zero subsets of $[a,b]$. Assume that both functions are positive everywhere and both are strictly increasing almost everywhere (i.e. there is a subset $S$ of the interval $[a,b]$, where $[a,b] setminus S$ has measure $0$, and $f, g$ are strictly increasing on $S$).



Under these conditions on $f$ and $g$ (positive, strictly increasing a.e.), is it possible that there exists a non-constant polynomial transformation $P$ such that $P(f(x)) = P(g(x))$ almost everywhere on $[a,b]$?







share|cite|improve this question






















  • can you clarify the edited requirement? Does "postive everywhere" mean $[a,b]$ or on all of $mathbbR$?
    – 6005
    Aug 9 at 22:53










  • What about $P(f(x)) = P(g(x)) = 0$?
    – Adam Francey
    Aug 9 at 22:59






  • 1




    Are $,f,g,$ assumed to be strictly increasing, maybe?
    – dxiv
    Aug 9 at 22:59







  • 1




    @6005 everywhere on $mathbbR$.
    – user1292919
    Aug 9 at 23:02






  • 2




    I'm a bit confused. Let the interval be $[0,pi/2]$, and $f(x)=x$, $g(x)=sin(x)$. Are you suggesting that there's a non-constant polynomial $P$ such that $P(x)=P(sin(x))$? Or are you looking for some kind of additional necessary conditions on $f,g$ to guarantee such a polynomial exists?
    – Alex R.
    Aug 9 at 23:15













up vote
4
down vote

favorite
2









up vote
4
down vote

favorite
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2





Suppose that we have two functions $f(x)$ and $g(x)$. We know that $f(x) ne g(x)$ on the interval, except for measure zero subsets of $[a,b]$. Assume that both functions are positive everywhere and both are strictly increasing almost everywhere (i.e. there is a subset $S$ of the interval $[a,b]$, where $[a,b] setminus S$ has measure $0$, and $f, g$ are strictly increasing on $S$).



Under these conditions on $f$ and $g$ (positive, strictly increasing a.e.), is it possible that there exists a non-constant polynomial transformation $P$ such that $P(f(x)) = P(g(x))$ almost everywhere on $[a,b]$?







share|cite|improve this question














Suppose that we have two functions $f(x)$ and $g(x)$. We know that $f(x) ne g(x)$ on the interval, except for measure zero subsets of $[a,b]$. Assume that both functions are positive everywhere and both are strictly increasing almost everywhere (i.e. there is a subset $S$ of the interval $[a,b]$, where $[a,b] setminus S$ has measure $0$, and $f, g$ are strictly increasing on $S$).



Under these conditions on $f$ and $g$ (positive, strictly increasing a.e.), is it possible that there exists a non-constant polynomial transformation $P$ such that $P(f(x)) = P(g(x))$ almost everywhere on $[a,b]$?









share|cite|improve this question













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share|cite|improve this question








edited Aug 10 at 0:22









6005

34.9k750123




34.9k750123










asked Aug 9 at 22:32









user1292919

650512




650512











  • can you clarify the edited requirement? Does "postive everywhere" mean $[a,b]$ or on all of $mathbbR$?
    – 6005
    Aug 9 at 22:53










  • What about $P(f(x)) = P(g(x)) = 0$?
    – Adam Francey
    Aug 9 at 22:59






  • 1




    Are $,f,g,$ assumed to be strictly increasing, maybe?
    – dxiv
    Aug 9 at 22:59







  • 1




    @6005 everywhere on $mathbbR$.
    – user1292919
    Aug 9 at 23:02






  • 2




    I'm a bit confused. Let the interval be $[0,pi/2]$, and $f(x)=x$, $g(x)=sin(x)$. Are you suggesting that there's a non-constant polynomial $P$ such that $P(x)=P(sin(x))$? Or are you looking for some kind of additional necessary conditions on $f,g$ to guarantee such a polynomial exists?
    – Alex R.
    Aug 9 at 23:15

















  • can you clarify the edited requirement? Does "postive everywhere" mean $[a,b]$ or on all of $mathbbR$?
    – 6005
    Aug 9 at 22:53










  • What about $P(f(x)) = P(g(x)) = 0$?
    – Adam Francey
    Aug 9 at 22:59






  • 1




    Are $,f,g,$ assumed to be strictly increasing, maybe?
    – dxiv
    Aug 9 at 22:59







  • 1




    @6005 everywhere on $mathbbR$.
    – user1292919
    Aug 9 at 23:02






  • 2




    I'm a bit confused. Let the interval be $[0,pi/2]$, and $f(x)=x$, $g(x)=sin(x)$. Are you suggesting that there's a non-constant polynomial $P$ such that $P(x)=P(sin(x))$? Or are you looking for some kind of additional necessary conditions on $f,g$ to guarantee such a polynomial exists?
    – Alex R.
    Aug 9 at 23:15
















can you clarify the edited requirement? Does "postive everywhere" mean $[a,b]$ or on all of $mathbbR$?
– 6005
Aug 9 at 22:53




can you clarify the edited requirement? Does "postive everywhere" mean $[a,b]$ or on all of $mathbbR$?
– 6005
Aug 9 at 22:53












What about $P(f(x)) = P(g(x)) = 0$?
– Adam Francey
Aug 9 at 22:59




What about $P(f(x)) = P(g(x)) = 0$?
– Adam Francey
Aug 9 at 22:59




1




1




Are $,f,g,$ assumed to be strictly increasing, maybe?
– dxiv
Aug 9 at 22:59





Are $,f,g,$ assumed to be strictly increasing, maybe?
– dxiv
Aug 9 at 22:59





1




1




@6005 everywhere on $mathbbR$.
– user1292919
Aug 9 at 23:02




@6005 everywhere on $mathbbR$.
– user1292919
Aug 9 at 23:02




2




2




I'm a bit confused. Let the interval be $[0,pi/2]$, and $f(x)=x$, $g(x)=sin(x)$. Are you suggesting that there's a non-constant polynomial $P$ such that $P(x)=P(sin(x))$? Or are you looking for some kind of additional necessary conditions on $f,g$ to guarantee such a polynomial exists?
– Alex R.
Aug 9 at 23:15





I'm a bit confused. Let the interval be $[0,pi/2]$, and $f(x)=x$, $g(x)=sin(x)$. Are you suggesting that there's a non-constant polynomial $P$ such that $P(x)=P(sin(x))$? Or are you looking for some kind of additional necessary conditions on $f,g$ to guarantee such a polynomial exists?
– Alex R.
Aug 9 at 23:15











2 Answers
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oldest

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up vote
6
down vote



accepted










I think that the following is a counterexample:



Let $f(x)=sin(x), g(x)=-cos(x)$ on $[0, fracpi2]$ and $P(X)=X^2(1-X^2)$.



Then
$$P(f(x))=P(g(x))=sin^2(x) cos^2(x)$$






share|cite|improve this answer




















  • I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
    – 6005
    Aug 9 at 23:40


















up vote
1
down vote













It is possible to have positive, strictly increasing functions $f, g: [a,b] to mathbbR$ such that $f(x) ne g(x)$ for all $x in [a,b]$, yet there is a non-constant polynomial $P$ such that $P(f(x)) = P(g(x))$ for all $x in [a,b]$.



To find an example, let's start with the polynomial $P$: take $P(X) = (X - 2)^3 - (X - 2)$.
Then, plot the points $(X,Y)$ where $P(X) = P(Y)$. We get:
enter image description here



The key is to look for a point on the graph where there is a tangent line with positive slope in the first quadrant. Above, we see this occurs at the point $(3,1)$. So we can set $(f(t), g(t))$ to be a parametrization of the curve starting from this point and moving in the positive $x$- and $y$-directions. We note that by construction, $f$ and $g$ are increasing and positive, and they satisfy $P(f(t)) = P(g(t))$.



Explicitly, if we parametrize the above curve we get the following example, which also uses trigonometry like N. S.'s nice answer:
beginalign*
P(X) &= (x-2)^3 - (x-2) \
f(x) &= 2 + frac1sqrt3 cos x + sin x \
g(x) &= 2 + frac1sqrt3 cos x - sin x
endalign*
These are always positive and are both strictly increasing on some intervals, e.g. on the interval $[-2pi/3,-pi/3]$.
The algebra is a bit messy, but you can verify that $P(f(x)) = Q(f(x))$ with Wolfram Alpha.



In general, for any polynomial $P$ we can plot the set of $(X,Y)$ coordinates where $P(X) = P(Y)$, and anywhere where the tangent curve has positive slope we can read off a pair of functions $f, g$ satisfying $P(f(t)) = P(g(t))$.




Of course, there are also examples where no such polynomial exists. For instance, take
beginalign*
f(x) &= x^4 \
g(x) &= x^2
endalign*



on the interval $[1,2]$.




Note that for either example, you can make the function increasing on all of $mathbbR$ and not just $[a,b]$, by extending $f$ appropriately to values $x < a$ and $x > b$.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    6
    down vote



    accepted










    I think that the following is a counterexample:



    Let $f(x)=sin(x), g(x)=-cos(x)$ on $[0, fracpi2]$ and $P(X)=X^2(1-X^2)$.



    Then
    $$P(f(x))=P(g(x))=sin^2(x) cos^2(x)$$






    share|cite|improve this answer




















    • I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
      – 6005
      Aug 9 at 23:40















    up vote
    6
    down vote



    accepted










    I think that the following is a counterexample:



    Let $f(x)=sin(x), g(x)=-cos(x)$ on $[0, fracpi2]$ and $P(X)=X^2(1-X^2)$.



    Then
    $$P(f(x))=P(g(x))=sin^2(x) cos^2(x)$$






    share|cite|improve this answer




















    • I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
      – 6005
      Aug 9 at 23:40













    up vote
    6
    down vote



    accepted







    up vote
    6
    down vote



    accepted






    I think that the following is a counterexample:



    Let $f(x)=sin(x), g(x)=-cos(x)$ on $[0, fracpi2]$ and $P(X)=X^2(1-X^2)$.



    Then
    $$P(f(x))=P(g(x))=sin^2(x) cos^2(x)$$






    share|cite|improve this answer












    I think that the following is a counterexample:



    Let $f(x)=sin(x), g(x)=-cos(x)$ on $[0, fracpi2]$ and $P(X)=X^2(1-X^2)$.



    Then
    $$P(f(x))=P(g(x))=sin^2(x) cos^2(x)$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 9 at 23:37









    N. S.

    98k5105197




    98k5105197











    • I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
      – 6005
      Aug 9 at 23:40

















    • I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
      – 6005
      Aug 9 at 23:40
















    I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
    – 6005
    Aug 9 at 23:40





    I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
    – 6005
    Aug 9 at 23:40











    up vote
    1
    down vote













    It is possible to have positive, strictly increasing functions $f, g: [a,b] to mathbbR$ such that $f(x) ne g(x)$ for all $x in [a,b]$, yet there is a non-constant polynomial $P$ such that $P(f(x)) = P(g(x))$ for all $x in [a,b]$.



    To find an example, let's start with the polynomial $P$: take $P(X) = (X - 2)^3 - (X - 2)$.
    Then, plot the points $(X,Y)$ where $P(X) = P(Y)$. We get:
    enter image description here



    The key is to look for a point on the graph where there is a tangent line with positive slope in the first quadrant. Above, we see this occurs at the point $(3,1)$. So we can set $(f(t), g(t))$ to be a parametrization of the curve starting from this point and moving in the positive $x$- and $y$-directions. We note that by construction, $f$ and $g$ are increasing and positive, and they satisfy $P(f(t)) = P(g(t))$.



    Explicitly, if we parametrize the above curve we get the following example, which also uses trigonometry like N. S.'s nice answer:
    beginalign*
    P(X) &= (x-2)^3 - (x-2) \
    f(x) &= 2 + frac1sqrt3 cos x + sin x \
    g(x) &= 2 + frac1sqrt3 cos x - sin x
    endalign*
    These are always positive and are both strictly increasing on some intervals, e.g. on the interval $[-2pi/3,-pi/3]$.
    The algebra is a bit messy, but you can verify that $P(f(x)) = Q(f(x))$ with Wolfram Alpha.



    In general, for any polynomial $P$ we can plot the set of $(X,Y)$ coordinates where $P(X) = P(Y)$, and anywhere where the tangent curve has positive slope we can read off a pair of functions $f, g$ satisfying $P(f(t)) = P(g(t))$.




    Of course, there are also examples where no such polynomial exists. For instance, take
    beginalign*
    f(x) &= x^4 \
    g(x) &= x^2
    endalign*



    on the interval $[1,2]$.




    Note that for either example, you can make the function increasing on all of $mathbbR$ and not just $[a,b]$, by extending $f$ appropriately to values $x < a$ and $x > b$.






    share|cite|improve this answer


























      up vote
      1
      down vote













      It is possible to have positive, strictly increasing functions $f, g: [a,b] to mathbbR$ such that $f(x) ne g(x)$ for all $x in [a,b]$, yet there is a non-constant polynomial $P$ such that $P(f(x)) = P(g(x))$ for all $x in [a,b]$.



      To find an example, let's start with the polynomial $P$: take $P(X) = (X - 2)^3 - (X - 2)$.
      Then, plot the points $(X,Y)$ where $P(X) = P(Y)$. We get:
      enter image description here



      The key is to look for a point on the graph where there is a tangent line with positive slope in the first quadrant. Above, we see this occurs at the point $(3,1)$. So we can set $(f(t), g(t))$ to be a parametrization of the curve starting from this point and moving in the positive $x$- and $y$-directions. We note that by construction, $f$ and $g$ are increasing and positive, and they satisfy $P(f(t)) = P(g(t))$.



      Explicitly, if we parametrize the above curve we get the following example, which also uses trigonometry like N. S.'s nice answer:
      beginalign*
      P(X) &= (x-2)^3 - (x-2) \
      f(x) &= 2 + frac1sqrt3 cos x + sin x \
      g(x) &= 2 + frac1sqrt3 cos x - sin x
      endalign*
      These are always positive and are both strictly increasing on some intervals, e.g. on the interval $[-2pi/3,-pi/3]$.
      The algebra is a bit messy, but you can verify that $P(f(x)) = Q(f(x))$ with Wolfram Alpha.



      In general, for any polynomial $P$ we can plot the set of $(X,Y)$ coordinates where $P(X) = P(Y)$, and anywhere where the tangent curve has positive slope we can read off a pair of functions $f, g$ satisfying $P(f(t)) = P(g(t))$.




      Of course, there are also examples where no such polynomial exists. For instance, take
      beginalign*
      f(x) &= x^4 \
      g(x) &= x^2
      endalign*



      on the interval $[1,2]$.




      Note that for either example, you can make the function increasing on all of $mathbbR$ and not just $[a,b]$, by extending $f$ appropriately to values $x < a$ and $x > b$.






      share|cite|improve this answer
























        up vote
        1
        down vote










        up vote
        1
        down vote









        It is possible to have positive, strictly increasing functions $f, g: [a,b] to mathbbR$ such that $f(x) ne g(x)$ for all $x in [a,b]$, yet there is a non-constant polynomial $P$ such that $P(f(x)) = P(g(x))$ for all $x in [a,b]$.



        To find an example, let's start with the polynomial $P$: take $P(X) = (X - 2)^3 - (X - 2)$.
        Then, plot the points $(X,Y)$ where $P(X) = P(Y)$. We get:
        enter image description here



        The key is to look for a point on the graph where there is a tangent line with positive slope in the first quadrant. Above, we see this occurs at the point $(3,1)$. So we can set $(f(t), g(t))$ to be a parametrization of the curve starting from this point and moving in the positive $x$- and $y$-directions. We note that by construction, $f$ and $g$ are increasing and positive, and they satisfy $P(f(t)) = P(g(t))$.



        Explicitly, if we parametrize the above curve we get the following example, which also uses trigonometry like N. S.'s nice answer:
        beginalign*
        P(X) &= (x-2)^3 - (x-2) \
        f(x) &= 2 + frac1sqrt3 cos x + sin x \
        g(x) &= 2 + frac1sqrt3 cos x - sin x
        endalign*
        These are always positive and are both strictly increasing on some intervals, e.g. on the interval $[-2pi/3,-pi/3]$.
        The algebra is a bit messy, but you can verify that $P(f(x)) = Q(f(x))$ with Wolfram Alpha.



        In general, for any polynomial $P$ we can plot the set of $(X,Y)$ coordinates where $P(X) = P(Y)$, and anywhere where the tangent curve has positive slope we can read off a pair of functions $f, g$ satisfying $P(f(t)) = P(g(t))$.




        Of course, there are also examples where no such polynomial exists. For instance, take
        beginalign*
        f(x) &= x^4 \
        g(x) &= x^2
        endalign*



        on the interval $[1,2]$.




        Note that for either example, you can make the function increasing on all of $mathbbR$ and not just $[a,b]$, by extending $f$ appropriately to values $x < a$ and $x > b$.






        share|cite|improve this answer














        It is possible to have positive, strictly increasing functions $f, g: [a,b] to mathbbR$ such that $f(x) ne g(x)$ for all $x in [a,b]$, yet there is a non-constant polynomial $P$ such that $P(f(x)) = P(g(x))$ for all $x in [a,b]$.



        To find an example, let's start with the polynomial $P$: take $P(X) = (X - 2)^3 - (X - 2)$.
        Then, plot the points $(X,Y)$ where $P(X) = P(Y)$. We get:
        enter image description here



        The key is to look for a point on the graph where there is a tangent line with positive slope in the first quadrant. Above, we see this occurs at the point $(3,1)$. So we can set $(f(t), g(t))$ to be a parametrization of the curve starting from this point and moving in the positive $x$- and $y$-directions. We note that by construction, $f$ and $g$ are increasing and positive, and they satisfy $P(f(t)) = P(g(t))$.



        Explicitly, if we parametrize the above curve we get the following example, which also uses trigonometry like N. S.'s nice answer:
        beginalign*
        P(X) &= (x-2)^3 - (x-2) \
        f(x) &= 2 + frac1sqrt3 cos x + sin x \
        g(x) &= 2 + frac1sqrt3 cos x - sin x
        endalign*
        These are always positive and are both strictly increasing on some intervals, e.g. on the interval $[-2pi/3,-pi/3]$.
        The algebra is a bit messy, but you can verify that $P(f(x)) = Q(f(x))$ with Wolfram Alpha.



        In general, for any polynomial $P$ we can plot the set of $(X,Y)$ coordinates where $P(X) = P(Y)$, and anywhere where the tangent curve has positive slope we can read off a pair of functions $f, g$ satisfying $P(f(t)) = P(g(t))$.




        Of course, there are also examples where no such polynomial exists. For instance, take
        beginalign*
        f(x) &= x^4 \
        g(x) &= x^2
        endalign*



        on the interval $[1,2]$.




        Note that for either example, you can make the function increasing on all of $mathbbR$ and not just $[a,b]$, by extending $f$ appropriately to values $x < a$ and $x > b$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 10 at 0:10

























        answered Aug 9 at 23:47









        6005

        34.9k750123




        34.9k750123






















             

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