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







share|cite|improve this question









$endgroup$








  • 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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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

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











$endgroup$





















    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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3064078%2fprove-that-nexists-functions-f-g-bbbr-to-bbbr-such-that-fgx-x2%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3064078%2fprove-that-nexists-functions-f-g-bbbr-to-bbbr-such-that-fgx-x2%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Bressuire

                      Image
                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
                      Read more

                      Cabo Verde

                      Image
                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
                      Read more

                      Gyllenstierna

                      Image
                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
                      Read more