let $f(x)$ be rational non constant polynomial and $fcirc f(x)=3f(x)^4-1$ then find the $f(x)$












2












$begingroup$


let $f(x)$ be rational non constant polynomial and $fcirc f(x)=3f(x)^4-1$
then find the $f(x)$ .



My Try :



$$f(f(x))=3f(x)^4-1$$



Let $f(x)=ax^n+g(x)$ so :



$$a(ax^n+g(x))^n+g(ax^n+g(x))=3(ax^n+g(x))^4-1$$



$$a^2x^{n^2}+h(x)+k(x)=3ax^{4n}+l(x)-1$$



$$3ax^{4n}-a^2x^{n^2}-1=h(x)+k(x)-l(x)$$



Now what ?






algebra-precalculus polynomials functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Formatting note: you can get the composition symbol by "circ"
    $endgroup$
    – lulu
    Jan 11 at 20:03






  • 4




    $begingroup$
    Math note: just comparing degrees, you want $n^2=4n$, so you know what $n$ is.
    $endgroup$
    – lulu
    Jan 11 at 20:07










  • $begingroup$
    @lulu why $n^2=4n $ ?
    $endgroup$
    – Almot1960
    Jan 11 at 20:09








  • 1




    $begingroup$
    Hint. If you identify $f$ with the polynomial function $f : mathbb{Q} to mathbb{Q}$, then for any $y in { f(x) : x in mathbb{Q} }$, we have $f(y) = 3y^4 - 1$. What can you say from this?
    $endgroup$
    – Sangchul Lee
    Jan 11 at 20:10






  • 2




    $begingroup$
    The degree of $fcirc f$ is $n^2$. The degree of $f^4$ is $4n$.
    $endgroup$
    – lulu
    Jan 11 at 20:10
















2












$begingroup$


let $f(x)$ be rational non constant polynomial and $fcirc f(x)=3f(x)^4-1$
then find the $f(x)$ .



My Try :



$$f(f(x))=3f(x)^4-1$$



Let $f(x)=ax^n+g(x)$ so :



$$a(ax^n+g(x))^n+g(ax^n+g(x))=3(ax^n+g(x))^4-1$$



$$a^2x^{n^2}+h(x)+k(x)=3ax^{4n}+l(x)-1$$



$$3ax^{4n}-a^2x^{n^2}-1=h(x)+k(x)-l(x)$$



Now what ?






algebra-precalculus polynomials functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Formatting note: you can get the composition symbol by "circ"
    $endgroup$
    – lulu
    Jan 11 at 20:03






  • 4




    $begingroup$
    Math note: just comparing degrees, you want $n^2=4n$, so you know what $n$ is.
    $endgroup$
    – lulu
    Jan 11 at 20:07










  • $begingroup$
    @lulu why $n^2=4n $ ?
    $endgroup$
    – Almot1960
    Jan 11 at 20:09








  • 1




    $begingroup$
    Hint. If you identify $f$ with the polynomial function $f : mathbb{Q} to mathbb{Q}$, then for any $y in { f(x) : x in mathbb{Q} }$, we have $f(y) = 3y^4 - 1$. What can you say from this?
    $endgroup$
    – Sangchul Lee
    Jan 11 at 20:10






  • 2




    $begingroup$
    The degree of $fcirc f$ is $n^2$. The degree of $f^4$ is $4n$.
    $endgroup$
    – lulu
    Jan 11 at 20:10














2












2








2





$begingroup$


let $f(x)$ be rational non constant polynomial and $fcirc f(x)=3f(x)^4-1$
then find the $f(x)$ .



My Try :



$$f(f(x))=3f(x)^4-1$$



Let $f(x)=ax^n+g(x)$ so :



$$a(ax^n+g(x))^n+g(ax^n+g(x))=3(ax^n+g(x))^4-1$$



$$a^2x^{n^2}+h(x)+k(x)=3ax^{4n}+l(x)-1$$



$$3ax^{4n}-a^2x^{n^2}-1=h(x)+k(x)-l(x)$$



Now what ?






algebra-precalculus polynomials functional-equations







share|cite|improve this question











$endgroup$




let $f(x)$ be rational non constant polynomial and $fcirc f(x)=3f(x)^4-1$
then find the $f(x)$ .



My Try :



$$f(f(x))=3f(x)^4-1$$



Let $f(x)=ax^n+g(x)$ so :



$$a(ax^n+g(x))^n+g(ax^n+g(x))=3(ax^n+g(x))^4-1$$



$$a^2x^{n^2}+h(x)+k(x)=3ax^{4n}+l(x)-1$$



$$3ax^{4n}-a^2x^{n^2}-1=h(x)+k(x)-l(x)$$



Now what ?






algebra-precalculus polynomials functional-equations




algebra-precalculus polynomials functional-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 20:45









Servaes

24.1k33893




24.1k33893










asked Jan 11 at 20:01









Almot1960Almot1960

2,288823




2,288823












  • $begingroup$
    Formatting note: you can get the composition symbol by "circ"
    $endgroup$
    – lulu
    Jan 11 at 20:03






  • 4




    $begingroup$
    Math note: just comparing degrees, you want $n^2=4n$, so you know what $n$ is.
    $endgroup$
    – lulu
    Jan 11 at 20:07










  • $begingroup$
    @lulu why $n^2=4n $ ?
    $endgroup$
    – Almot1960
    Jan 11 at 20:09








  • 1




    $begingroup$
    Hint. If you identify $f$ with the polynomial function $f : mathbb{Q} to mathbb{Q}$, then for any $y in { f(x) : x in mathbb{Q} }$, we have $f(y) = 3y^4 - 1$. What can you say from this?
    $endgroup$
    – Sangchul Lee
    Jan 11 at 20:10






  • 2




    $begingroup$
    The degree of $fcirc f$ is $n^2$. The degree of $f^4$ is $4n$.
    $endgroup$
    – lulu
    Jan 11 at 20:10


















  • $begingroup$
    Formatting note: you can get the composition symbol by "circ"
    $endgroup$
    – lulu
    Jan 11 at 20:03






  • 4




    $begingroup$
    Math note: just comparing degrees, you want $n^2=4n$, so you know what $n$ is.
    $endgroup$
    – lulu
    Jan 11 at 20:07










  • $begingroup$
    @lulu why $n^2=4n $ ?
    $endgroup$
    – Almot1960
    Jan 11 at 20:09








  • 1




    $begingroup$
    Hint. If you identify $f$ with the polynomial function $f : mathbb{Q} to mathbb{Q}$, then for any $y in { f(x) : x in mathbb{Q} }$, we have $f(y) = 3y^4 - 1$. What can you say from this?
    $endgroup$
    – Sangchul Lee
    Jan 11 at 20:10






  • 2




    $begingroup$
    The degree of $fcirc f$ is $n^2$. The degree of $f^4$ is $4n$.
    $endgroup$
    – lulu
    Jan 11 at 20:10
















$begingroup$
Formatting note: you can get the composition symbol by "circ"
$endgroup$
– lulu
Jan 11 at 20:03




$begingroup$
Formatting note: you can get the composition symbol by "circ"
$endgroup$
– lulu
Jan 11 at 20:03




4




4




$begingroup$
Math note: just comparing degrees, you want $n^2=4n$, so you know what $n$ is.
$endgroup$
– lulu
Jan 11 at 20:07




$begingroup$
Math note: just comparing degrees, you want $n^2=4n$, so you know what $n$ is.
$endgroup$
– lulu
Jan 11 at 20:07












$begingroup$
@lulu why $n^2=4n $ ?
$endgroup$
– Almot1960
Jan 11 at 20:09






$begingroup$
@lulu why $n^2=4n $ ?
$endgroup$
– Almot1960
Jan 11 at 20:09






1




1




$begingroup$
Hint. If you identify $f$ with the polynomial function $f : mathbb{Q} to mathbb{Q}$, then for any $y in { f(x) : x in mathbb{Q} }$, we have $f(y) = 3y^4 - 1$. What can you say from this?
$endgroup$
– Sangchul Lee
Jan 11 at 20:10




$begingroup$
Hint. If you identify $f$ with the polynomial function $f : mathbb{Q} to mathbb{Q}$, then for any $y in { f(x) : x in mathbb{Q} }$, we have $f(y) = 3y^4 - 1$. What can you say from this?
$endgroup$
– Sangchul Lee
Jan 11 at 20:10




2




2




$begingroup$
The degree of $fcirc f$ is $n^2$. The degree of $f^4$ is $4n$.
$endgroup$
– lulu
Jan 11 at 20:10




$begingroup$
The degree of $fcirc f$ is $n^2$. The degree of $f^4$ is $4n$.
$endgroup$
– lulu
Jan 11 at 20:10










3 Answers
3






active

oldest

votes


















4












$begingroup$

Note that $3f(x)^4 - 1 = gcirc f(x)$, where $g(x) = 3x^4 - 1$. So $f = g$, at least, is a solution.



Now, if $f$ is a solution, we need to show that $f(x) = g(x)$. But note that $f circ f(x) = gcirc f(x)$ for all $x$, so $(g - f) circ f(x) = 0$, so $g - f$ is zero on the image of $f$, which, since $f$ is non-constant, contains some interval $(a,b)$, hence $g - f$ is everywhere zero, so $f = g$ is the only solution.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    +1 This is the way to go.
    $endgroup$
    – Servaes
    Jan 11 at 20:43



















1












$begingroup$

The highest degree terms on each side of what you have are:



$a_n^{n+1}x^{n^2} = 3a_n^4x^{4n}$



$n^2 = 4n\ n = 4\ a_4^5 = 3a_4^4\
a_4 = 3$



But, there is a more direct approach.



$f(f(x)) = 3(f(x))^4 - 1$



Replace the $f(x)$ with $u$ everywhere we see it.



$f(u) = 3u^4 - 1$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Note that, because I think that you supposed deg $g(x)leq n-1$, from
    $$
    a_n(a_n x^n+g(x))^n+g(a_n x^n+g(x))=
    3(a_n x^n+g(x))^4-1$$
    the degree of the LHS is $n^2$, while on the other hand the degree of the RHS is $4n$. This is an equality between two polynomials, hence their degree must be the same. Then $n=4$. Moreover the two polynomials must have the same leading coefficient. The coefficient of $x^{16}$ in the LHS is $a_4^{5}$, while the coefficient in the RHS is $3a_4^{4}$, hence $a_4=3$.
    Now I think that you have to put $f(x)=3x^4+sum_{i=0}^{3} a_i x^i $ in the equation and compute






    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









      4












      $begingroup$

      Note that $3f(x)^4 - 1 = gcirc f(x)$, where $g(x) = 3x^4 - 1$. So $f = g$, at least, is a solution.



      Now, if $f$ is a solution, we need to show that $f(x) = g(x)$. But note that $f circ f(x) = gcirc f(x)$ for all $x$, so $(g - f) circ f(x) = 0$, so $g - f$ is zero on the image of $f$, which, since $f$ is non-constant, contains some interval $(a,b)$, hence $g - f$ is everywhere zero, so $f = g$ is the only solution.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        +1 This is the way to go.
        $endgroup$
        – Servaes
        Jan 11 at 20:43
















      4












      $begingroup$

      Note that $3f(x)^4 - 1 = gcirc f(x)$, where $g(x) = 3x^4 - 1$. So $f = g$, at least, is a solution.



      Now, if $f$ is a solution, we need to show that $f(x) = g(x)$. But note that $f circ f(x) = gcirc f(x)$ for all $x$, so $(g - f) circ f(x) = 0$, so $g - f$ is zero on the image of $f$, which, since $f$ is non-constant, contains some interval $(a,b)$, hence $g - f$ is everywhere zero, so $f = g$ is the only solution.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        +1 This is the way to go.
        $endgroup$
        – Servaes
        Jan 11 at 20:43














      4












      4








      4





      $begingroup$

      Note that $3f(x)^4 - 1 = gcirc f(x)$, where $g(x) = 3x^4 - 1$. So $f = g$, at least, is a solution.



      Now, if $f$ is a solution, we need to show that $f(x) = g(x)$. But note that $f circ f(x) = gcirc f(x)$ for all $x$, so $(g - f) circ f(x) = 0$, so $g - f$ is zero on the image of $f$, which, since $f$ is non-constant, contains some interval $(a,b)$, hence $g - f$ is everywhere zero, so $f = g$ is the only solution.






      share|cite|improve this answer









      $endgroup$



      Note that $3f(x)^4 - 1 = gcirc f(x)$, where $g(x) = 3x^4 - 1$. So $f = g$, at least, is a solution.



      Now, if $f$ is a solution, we need to show that $f(x) = g(x)$. But note that $f circ f(x) = gcirc f(x)$ for all $x$, so $(g - f) circ f(x) = 0$, so $g - f$ is zero on the image of $f$, which, since $f$ is non-constant, contains some interval $(a,b)$, hence $g - f$ is everywhere zero, so $f = g$ is the only solution.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 11 at 20:23









      user3482749user3482749

      4,266919




      4,266919








      • 1




        $begingroup$
        +1 This is the way to go.
        $endgroup$
        – Servaes
        Jan 11 at 20:43














      • 1




        $begingroup$
        +1 This is the way to go.
        $endgroup$
        – Servaes
        Jan 11 at 20:43








      1




      1




      $begingroup$
      +1 This is the way to go.
      $endgroup$
      – Servaes
      Jan 11 at 20:43




      $begingroup$
      +1 This is the way to go.
      $endgroup$
      – Servaes
      Jan 11 at 20:43











      1












      $begingroup$

      The highest degree terms on each side of what you have are:



      $a_n^{n+1}x^{n^2} = 3a_n^4x^{4n}$



      $n^2 = 4n\ n = 4\ a_4^5 = 3a_4^4\
      a_4 = 3$



      But, there is a more direct approach.



      $f(f(x)) = 3(f(x))^4 - 1$



      Replace the $f(x)$ with $u$ everywhere we see it.



      $f(u) = 3u^4 - 1$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The highest degree terms on each side of what you have are:



        $a_n^{n+1}x^{n^2} = 3a_n^4x^{4n}$



        $n^2 = 4n\ n = 4\ a_4^5 = 3a_4^4\
        a_4 = 3$



        But, there is a more direct approach.



        $f(f(x)) = 3(f(x))^4 - 1$



        Replace the $f(x)$ with $u$ everywhere we see it.



        $f(u) = 3u^4 - 1$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The highest degree terms on each side of what you have are:



          $a_n^{n+1}x^{n^2} = 3a_n^4x^{4n}$



          $n^2 = 4n\ n = 4\ a_4^5 = 3a_4^4\
          a_4 = 3$



          But, there is a more direct approach.



          $f(f(x)) = 3(f(x))^4 - 1$



          Replace the $f(x)$ with $u$ everywhere we see it.



          $f(u) = 3u^4 - 1$






          share|cite|improve this answer









          $endgroup$



          The highest degree terms on each side of what you have are:



          $a_n^{n+1}x^{n^2} = 3a_n^4x^{4n}$



          $n^2 = 4n\ n = 4\ a_4^5 = 3a_4^4\
          a_4 = 3$



          But, there is a more direct approach.



          $f(f(x)) = 3(f(x))^4 - 1$



          Replace the $f(x)$ with $u$ everywhere we see it.



          $f(u) = 3u^4 - 1$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 20:22









          Doug MDoug M

          44.9k31854




          44.9k31854























              1












              $begingroup$

              Note that, because I think that you supposed deg $g(x)leq n-1$, from
              $$
              a_n(a_n x^n+g(x))^n+g(a_n x^n+g(x))=
              3(a_n x^n+g(x))^4-1$$
              the degree of the LHS is $n^2$, while on the other hand the degree of the RHS is $4n$. This is an equality between two polynomials, hence their degree must be the same. Then $n=4$. Moreover the two polynomials must have the same leading coefficient. The coefficient of $x^{16}$ in the LHS is $a_4^{5}$, while the coefficient in the RHS is $3a_4^{4}$, hence $a_4=3$.
              Now I think that you have to put $f(x)=3x^4+sum_{i=0}^{3} a_i x^i $ in the equation and compute






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Note that, because I think that you supposed deg $g(x)leq n-1$, from
                $$
                a_n(a_n x^n+g(x))^n+g(a_n x^n+g(x))=
                3(a_n x^n+g(x))^4-1$$
                the degree of the LHS is $n^2$, while on the other hand the degree of the RHS is $4n$. This is an equality between two polynomials, hence their degree must be the same. Then $n=4$. Moreover the two polynomials must have the same leading coefficient. The coefficient of $x^{16}$ in the LHS is $a_4^{5}$, while the coefficient in the RHS is $3a_4^{4}$, hence $a_4=3$.
                Now I think that you have to put $f(x)=3x^4+sum_{i=0}^{3} a_i x^i $ in the equation and compute






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Note that, because I think that you supposed deg $g(x)leq n-1$, from
                  $$
                  a_n(a_n x^n+g(x))^n+g(a_n x^n+g(x))=
                  3(a_n x^n+g(x))^4-1$$
                  the degree of the LHS is $n^2$, while on the other hand the degree of the RHS is $4n$. This is an equality between two polynomials, hence their degree must be the same. Then $n=4$. Moreover the two polynomials must have the same leading coefficient. The coefficient of $x^{16}$ in the LHS is $a_4^{5}$, while the coefficient in the RHS is $3a_4^{4}$, hence $a_4=3$.
                  Now I think that you have to put $f(x)=3x^4+sum_{i=0}^{3} a_i x^i $ in the equation and compute






                  share|cite|improve this answer









                  $endgroup$



                  Note that, because I think that you supposed deg $g(x)leq n-1$, from
                  $$
                  a_n(a_n x^n+g(x))^n+g(a_n x^n+g(x))=
                  3(a_n x^n+g(x))^4-1$$
                  the degree of the LHS is $n^2$, while on the other hand the degree of the RHS is $4n$. This is an equality between two polynomials, hence their degree must be the same. Then $n=4$. Moreover the two polynomials must have the same leading coefficient. The coefficient of $x^{16}$ in the LHS is $a_4^{5}$, while the coefficient in the RHS is $3a_4^{4}$, hence $a_4=3$.
                  Now I think that you have to put $f(x)=3x^4+sum_{i=0}^{3} a_i x^i $ in the equation and compute







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 11 at 20:25









                  user627482user627482

                  464




                  464






























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                      How to fix TextFormField cause rebuild widget in Flutter

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                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
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                      in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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                      1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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