Prove that, $nexists$ functions $f,g:Bbb{R}toBbb{R}$ such that $f(g(x))=x^{2018}$ and $g(f(x))=x^{2019}$.












11












$begingroup$


I have tried a little bit to solve the problem which goes as follows-

My intuition says $f(g(x))=x^{2018}$ and $g(f(x))=x^{2019}$, then it will violate the well-definedness of function.

Note that, $f(g(f(x)))=f(x)^{2018}implies f(x^{2019})=f(x)^{2018}$

Similarly, $g(x^{2018})=g(x)^{2019}$

Putting $x=1$ in $f(x^{2019})=f(x)^{2018}$ we get, $f(1)=f(1)^{2018}implies f(1)=0,1$

Similarly, putting $x=1$ in $g(x^{2018})=g(x)^{2019}$ we get, $g(1)=g(1)^{2019}implies g(1)=0,1,-1$

Now, I can't proceed further. Can anybody solve it? Thanks for assistance in advance.






real-analysis calculus functions







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  • 2




    $begingroup$
    From where is the question?
    $endgroup$
    – Holo
    Jan 6 at 16:40
















11












$begingroup$


I have tried a little bit to solve the problem which goes as follows-

My intuition says $f(g(x))=x^{2018}$ and $g(f(x))=x^{2019}$, then it will violate the well-definedness of function.

Note that, $f(g(f(x)))=f(x)^{2018}implies f(x^{2019})=f(x)^{2018}$

Similarly, $g(x^{2018})=g(x)^{2019}$

Putting $x=1$ in $f(x^{2019})=f(x)^{2018}$ we get, $f(1)=f(1)^{2018}implies f(1)=0,1$

Similarly, putting $x=1$ in $g(x^{2018})=g(x)^{2019}$ we get, $g(1)=g(1)^{2019}implies g(1)=0,1,-1$

Now, I can't proceed further. Can anybody solve it? Thanks for assistance in advance.






real-analysis calculus functions







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    From where is the question?
    $endgroup$
    – Holo
    Jan 6 at 16:40














11












11








11


4



$begingroup$


I have tried a little bit to solve the problem which goes as follows-

My intuition says $f(g(x))=x^{2018}$ and $g(f(x))=x^{2019}$, then it will violate the well-definedness of function.

Note that, $f(g(f(x)))=f(x)^{2018}implies f(x^{2019})=f(x)^{2018}$

Similarly, $g(x^{2018})=g(x)^{2019}$

Putting $x=1$ in $f(x^{2019})=f(x)^{2018}$ we get, $f(1)=f(1)^{2018}implies f(1)=0,1$

Similarly, putting $x=1$ in $g(x^{2018})=g(x)^{2019}$ we get, $g(1)=g(1)^{2019}implies g(1)=0,1,-1$

Now, I can't proceed further. Can anybody solve it? Thanks for assistance in advance.






real-analysis calculus functions







share|cite|improve this question









$endgroup$




I have tried a little bit to solve the problem which goes as follows-

My intuition says $f(g(x))=x^{2018}$ and $g(f(x))=x^{2019}$, then it will violate the well-definedness of function.

Note that, $f(g(f(x)))=f(x)^{2018}implies f(x^{2019})=f(x)^{2018}$

Similarly, $g(x^{2018})=g(x)^{2019}$

Putting $x=1$ in $f(x^{2019})=f(x)^{2018}$ we get, $f(1)=f(1)^{2018}implies f(1)=0,1$

Similarly, putting $x=1$ in $g(x^{2018})=g(x)^{2019}$ we get, $g(1)=g(1)^{2019}implies g(1)=0,1,-1$

Now, I can't proceed further. Can anybody solve it? Thanks for assistance in advance.






real-analysis calculus functions




real-analysis calculus functions






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asked Jan 6 at 16:33









Biswarup SahaBiswarup Saha

624110




624110








  • 2




    $begingroup$
    From where is the question?
    $endgroup$
    – Holo
    Jan 6 at 16:40














  • 2




    $begingroup$
    From where is the question?
    $endgroup$
    – Holo
    Jan 6 at 16:40








2




2




$begingroup$
From where is the question?
$endgroup$
– Holo
Jan 6 at 16:40




$begingroup$
From where is the question?
$endgroup$
– Holo
Jan 6 at 16:40










2 Answers
2






active

oldest

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10












$begingroup$

Here is a simple proof (to avoid confusion I will write $f^{y}(x)$ instead of $f(x)^{y}$):



Suppose that there are two such functions $f,g$ as in your question.




Note that $forallspace i in{-1, 0, 1}$, we have $f(i^{2019}) = f(i) = f^{2018}(i)$ and thus $label{*}tag{*} f(i) in {0, 1} quadforallspace i in{-1, 0, 1}.$




On the other hand, since $g(f(x)) = x^{2019} spaceforall xinmathbb{R}$,






  • $g(f(1)) = 1$,


  • $g(f(0)) = 0$,


  • $g(f(-1)) = -1$.




This is impossible since $f$ (and thus also $gcirc f$) only takes two (or fewer) values on ${-1, 0, 1}$ after ref{*}. $LongrightarrowLongleftarrowquadsquare$






share|cite|improve this answer











$endgroup$





















    8












    $begingroup$

    Another classic trick that can be used here is that there is a bijection between the set of fixed points of $fg$ and fixed points of $gf$. If $x$ is such that $f(g(x))=x$ then $g(f(g(x)))=g(x)$ so $g(x)$ is a fixed point of $gf$. Similarly whenever $y$ is a fixed point of $gf$, we can show that $f(y)$ is a fixed point of $fg$. Convince yourself that on the sets of fixed points, this gives a bijection.



    Can you see why the $fg$ and $gf$ you've been given fail this condition?






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      10












      $begingroup$

      Here is a simple proof (to avoid confusion I will write $f^{y}(x)$ instead of $f(x)^{y}$):



      Suppose that there are two such functions $f,g$ as in your question.




      Note that $forallspace i in{-1, 0, 1}$, we have $f(i^{2019}) = f(i) = f^{2018}(i)$ and thus $label{*}tag{*} f(i) in {0, 1} quadforallspace i in{-1, 0, 1}.$




      On the other hand, since $g(f(x)) = x^{2019} spaceforall xinmathbb{R}$,






      • $g(f(1)) = 1$,


      • $g(f(0)) = 0$,


      • $g(f(-1)) = -1$.




      This is impossible since $f$ (and thus also $gcirc f$) only takes two (or fewer) values on ${-1, 0, 1}$ after ref{*}. $LongrightarrowLongleftarrowquadsquare$






      share|cite|improve this answer











      $endgroup$


















        10












        $begingroup$

        Here is a simple proof (to avoid confusion I will write $f^{y}(x)$ instead of $f(x)^{y}$):



        Suppose that there are two such functions $f,g$ as in your question.




        Note that $forallspace i in{-1, 0, 1}$, we have $f(i^{2019}) = f(i) = f^{2018}(i)$ and thus $label{*}tag{*} f(i) in {0, 1} quadforallspace i in{-1, 0, 1}.$




        On the other hand, since $g(f(x)) = x^{2019} spaceforall xinmathbb{R}$,






        • $g(f(1)) = 1$,


        • $g(f(0)) = 0$,


        • $g(f(-1)) = -1$.




        This is impossible since $f$ (and thus also $gcirc f$) only takes two (or fewer) values on ${-1, 0, 1}$ after ref{*}. $LongrightarrowLongleftarrowquadsquare$






        share|cite|improve this answer











        $endgroup$
















          10












          10








          10





          $begingroup$

          Here is a simple proof (to avoid confusion I will write $f^{y}(x)$ instead of $f(x)^{y}$):



          Suppose that there are two such functions $f,g$ as in your question.




          Note that $forallspace i in{-1, 0, 1}$, we have $f(i^{2019}) = f(i) = f^{2018}(i)$ and thus $label{*}tag{*} f(i) in {0, 1} quadforallspace i in{-1, 0, 1}.$




          On the other hand, since $g(f(x)) = x^{2019} spaceforall xinmathbb{R}$,






          • $g(f(1)) = 1$,


          • $g(f(0)) = 0$,


          • $g(f(-1)) = -1$.




          This is impossible since $f$ (and thus also $gcirc f$) only takes two (or fewer) values on ${-1, 0, 1}$ after ref{*}. $LongrightarrowLongleftarrowquadsquare$






          share|cite|improve this answer











          $endgroup$



          Here is a simple proof (to avoid confusion I will write $f^{y}(x)$ instead of $f(x)^{y}$):



          Suppose that there are two such functions $f,g$ as in your question.




          Note that $forallspace i in{-1, 0, 1}$, we have $f(i^{2019}) = f(i) = f^{2018}(i)$ and thus $label{*}tag{*} f(i) in {0, 1} quadforallspace i in{-1, 0, 1}.$




          On the other hand, since $g(f(x)) = x^{2019} spaceforall xinmathbb{R}$,






          • $g(f(1)) = 1$,


          • $g(f(0)) = 0$,


          • $g(f(-1)) = -1$.




          This is impossible since $f$ (and thus also $gcirc f$) only takes two (or fewer) values on ${-1, 0, 1}$ after ref{*}. $LongrightarrowLongleftarrowquadsquare$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 9:41

























          answered Jan 6 at 22:35









          Maximilian JanischMaximilian Janisch

          45612




          45612























              8












              $begingroup$

              Another classic trick that can be used here is that there is a bijection between the set of fixed points of $fg$ and fixed points of $gf$. If $x$ is such that $f(g(x))=x$ then $g(f(g(x)))=g(x)$ so $g(x)$ is a fixed point of $gf$. Similarly whenever $y$ is a fixed point of $gf$, we can show that $f(y)$ is a fixed point of $fg$. Convince yourself that on the sets of fixed points, this gives a bijection.



              Can you see why the $fg$ and $gf$ you've been given fail this condition?






              share|cite|improve this answer









              $endgroup$


















                8












                $begingroup$

                Another classic trick that can be used here is that there is a bijection between the set of fixed points of $fg$ and fixed points of $gf$. If $x$ is such that $f(g(x))=x$ then $g(f(g(x)))=g(x)$ so $g(x)$ is a fixed point of $gf$. Similarly whenever $y$ is a fixed point of $gf$, we can show that $f(y)$ is a fixed point of $fg$. Convince yourself that on the sets of fixed points, this gives a bijection.



                Can you see why the $fg$ and $gf$ you've been given fail this condition?






                share|cite|improve this answer









                $endgroup$
















                  8












                  8








                  8





                  $begingroup$

                  Another classic trick that can be used here is that there is a bijection between the set of fixed points of $fg$ and fixed points of $gf$. If $x$ is such that $f(g(x))=x$ then $g(f(g(x)))=g(x)$ so $g(x)$ is a fixed point of $gf$. Similarly whenever $y$ is a fixed point of $gf$, we can show that $f(y)$ is a fixed point of $fg$. Convince yourself that on the sets of fixed points, this gives a bijection.



                  Can you see why the $fg$ and $gf$ you've been given fail this condition?






                  share|cite|improve this answer









                  $endgroup$



                  Another classic trick that can be used here is that there is a bijection between the set of fixed points of $fg$ and fixed points of $gf$. If $x$ is such that $f(g(x))=x$ then $g(f(g(x)))=g(x)$ so $g(x)$ is a fixed point of $gf$. Similarly whenever $y$ is a fixed point of $gf$, we can show that $f(y)$ is a fixed point of $fg$. Convince yourself that on the sets of fixed points, this gives a bijection.



                  Can you see why the $fg$ and $gf$ you've been given fail this condition?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 19:20









                  ChessanatorChessanator

                  2,2061412




                  2,2061412






























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