How to construct a polynomial with certain restraints












3












$begingroup$


Construct a fifth-degree polynomial $p(x)$ in the interval ${cal I} = [-3,3]$ that meets these conditions:





  • $p(x)$ has 5 different roots, each in the interval ${cal I}$.

  • The coefficient of the leading term, $x^5$, must be equal to 1.

  • $max [ |p(x)| ] leq 1/5, x in {cal I}$


I am not sure if this polynomial exists.






real-analysis complex-analysis polynomials







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$endgroup$












  • $begingroup$
    Is that $E$ meant to be $in$?
    $endgroup$
    – Rhys Hughes
    Jan 22 at 23:39










  • $begingroup$
    @RhysHughes Yes, I am sorry I am not familiar with Latex.You can edit it if you want
    $endgroup$
    – Kwnstantinos Nikoloutsos
    Jan 22 at 23:41












  • $begingroup$
    @EricWofsey Yes the roots must be in [-3,3] and should be real numbers
    $endgroup$
    – Kwnstantinos Nikoloutsos
    Jan 22 at 23:42






  • 1




    $begingroup$
    Suggestion: $f(x) = (x-a)(x-b)(x-c)(x-d)(x-e)$ where ${ a, b, c, d, e } in [-3,3]$. Also, $f(0) = -a b c d e$.
    $endgroup$
    – David G. Stork
    Jan 22 at 23:56












  • $begingroup$
    @DavidG.Stork Suggestion $(x-3)(x-a)x(x+a)(x+3)$, find the maximum as a function of $a in (0,3)$ and hope there's an $a$ that works.
    $endgroup$
    – Ethan Bolker
    Jan 23 at 0:09
















3












$begingroup$


Construct a fifth-degree polynomial $p(x)$ in the interval ${cal I} = [-3,3]$ that meets these conditions:





  • $p(x)$ has 5 different roots, each in the interval ${cal I}$.

  • The coefficient of the leading term, $x^5$, must be equal to 1.

  • $max [ |p(x)| ] leq 1/5, x in {cal I}$


I am not sure if this polynomial exists.






real-analysis complex-analysis polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is that $E$ meant to be $in$?
    $endgroup$
    – Rhys Hughes
    Jan 22 at 23:39










  • $begingroup$
    @RhysHughes Yes, I am sorry I am not familiar with Latex.You can edit it if you want
    $endgroup$
    – Kwnstantinos Nikoloutsos
    Jan 22 at 23:41












  • $begingroup$
    @EricWofsey Yes the roots must be in [-3,3] and should be real numbers
    $endgroup$
    – Kwnstantinos Nikoloutsos
    Jan 22 at 23:42






  • 1




    $begingroup$
    Suggestion: $f(x) = (x-a)(x-b)(x-c)(x-d)(x-e)$ where ${ a, b, c, d, e } in [-3,3]$. Also, $f(0) = -a b c d e$.
    $endgroup$
    – David G. Stork
    Jan 22 at 23:56












  • $begingroup$
    @DavidG.Stork Suggestion $(x-3)(x-a)x(x+a)(x+3)$, find the maximum as a function of $a in (0,3)$ and hope there's an $a$ that works.
    $endgroup$
    – Ethan Bolker
    Jan 23 at 0:09














3












3








3





$begingroup$


Construct a fifth-degree polynomial $p(x)$ in the interval ${cal I} = [-3,3]$ that meets these conditions:





  • $p(x)$ has 5 different roots, each in the interval ${cal I}$.

  • The coefficient of the leading term, $x^5$, must be equal to 1.

  • $max [ |p(x)| ] leq 1/5, x in {cal I}$


I am not sure if this polynomial exists.






real-analysis complex-analysis polynomials







share|cite|improve this question











$endgroup$




Construct a fifth-degree polynomial $p(x)$ in the interval ${cal I} = [-3,3]$ that meets these conditions:





  • $p(x)$ has 5 different roots, each in the interval ${cal I}$.

  • The coefficient of the leading term, $x^5$, must be equal to 1.

  • $max [ |p(x)| ] leq 1/5, x in {cal I}$


I am not sure if this polynomial exists.






real-analysis complex-analysis polynomials




real-analysis complex-analysis polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 0:31







Kwnstantinos Nikoloutsos

















asked Jan 22 at 23:37









Kwnstantinos NikoloutsosKwnstantinos Nikoloutsos

267310




267310












  • $begingroup$
    Is that $E$ meant to be $in$?
    $endgroup$
    – Rhys Hughes
    Jan 22 at 23:39










  • $begingroup$
    @RhysHughes Yes, I am sorry I am not familiar with Latex.You can edit it if you want
    $endgroup$
    – Kwnstantinos Nikoloutsos
    Jan 22 at 23:41












  • $begingroup$
    @EricWofsey Yes the roots must be in [-3,3] and should be real numbers
    $endgroup$
    – Kwnstantinos Nikoloutsos
    Jan 22 at 23:42






  • 1




    $begingroup$
    Suggestion: $f(x) = (x-a)(x-b)(x-c)(x-d)(x-e)$ where ${ a, b, c, d, e } in [-3,3]$. Also, $f(0) = -a b c d e$.
    $endgroup$
    – David G. Stork
    Jan 22 at 23:56












  • $begingroup$
    @DavidG.Stork Suggestion $(x-3)(x-a)x(x+a)(x+3)$, find the maximum as a function of $a in (0,3)$ and hope there's an $a$ that works.
    $endgroup$
    – Ethan Bolker
    Jan 23 at 0:09


















  • $begingroup$
    Is that $E$ meant to be $in$?
    $endgroup$
    – Rhys Hughes
    Jan 22 at 23:39










  • $begingroup$
    @RhysHughes Yes, I am sorry I am not familiar with Latex.You can edit it if you want
    $endgroup$
    – Kwnstantinos Nikoloutsos
    Jan 22 at 23:41












  • $begingroup$
    @EricWofsey Yes the roots must be in [-3,3] and should be real numbers
    $endgroup$
    – Kwnstantinos Nikoloutsos
    Jan 22 at 23:42






  • 1




    $begingroup$
    Suggestion: $f(x) = (x-a)(x-b)(x-c)(x-d)(x-e)$ where ${ a, b, c, d, e } in [-3,3]$. Also, $f(0) = -a b c d e$.
    $endgroup$
    – David G. Stork
    Jan 22 at 23:56












  • $begingroup$
    @DavidG.Stork Suggestion $(x-3)(x-a)x(x+a)(x+3)$, find the maximum as a function of $a in (0,3)$ and hope there's an $a$ that works.
    $endgroup$
    – Ethan Bolker
    Jan 23 at 0:09
















$begingroup$
Is that $E$ meant to be $in$?
$endgroup$
– Rhys Hughes
Jan 22 at 23:39




$begingroup$
Is that $E$ meant to be $in$?
$endgroup$
– Rhys Hughes
Jan 22 at 23:39












$begingroup$
@RhysHughes Yes, I am sorry I am not familiar with Latex.You can edit it if you want
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:41






$begingroup$
@RhysHughes Yes, I am sorry I am not familiar with Latex.You can edit it if you want
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:41














$begingroup$
@EricWofsey Yes the roots must be in [-3,3] and should be real numbers
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:42




$begingroup$
@EricWofsey Yes the roots must be in [-3,3] and should be real numbers
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:42




1




1




$begingroup$
Suggestion: $f(x) = (x-a)(x-b)(x-c)(x-d)(x-e)$ where ${ a, b, c, d, e } in [-3,3]$. Also, $f(0) = -a b c d e$.
$endgroup$
– David G. Stork
Jan 22 at 23:56






$begingroup$
Suggestion: $f(x) = (x-a)(x-b)(x-c)(x-d)(x-e)$ where ${ a, b, c, d, e } in [-3,3]$. Also, $f(0) = -a b c d e$.
$endgroup$
– David G. Stork
Jan 22 at 23:56














$begingroup$
@DavidG.Stork Suggestion $(x-3)(x-a)x(x+a)(x+3)$, find the maximum as a function of $a in (0,3)$ and hope there's an $a$ that works.
$endgroup$
– Ethan Bolker
Jan 23 at 0:09




$begingroup$
@DavidG.Stork Suggestion $(x-3)(x-a)x(x+a)(x+3)$, find the maximum as a function of $a in (0,3)$ and hope there's an $a$ that works.
$endgroup$
– Ethan Bolker
Jan 23 at 0:09










2 Answers
2






active

oldest

votes


















1












$begingroup$

No such polynomial exists.



Let $p(x)$ be a monic polynomial of degree $5$. Let $q(t)=dfrac{1}{3^5}p(3t)$.



Then $q(t)$ is a monic polynomial of degree $5$ and
$$
max_{x in [-3,3]} |p(x)| = 3^5 max_{t in [-1,1]} |q(t)| geq dfrac{3^5}{2^{4}} > frac15
$$
because the monic polynomial of degree $5$ with minimal norm in $[-1,1]$ is $dfrac{1}{2^{4}}T_{5}(t)$, where $T_5$ is the $5$-th Chebyshev polynomial. The minimal norm is $dfrac{1}{2^{4}}$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    If the roots of $p$ are $a<b<c<d<e$, then the subintervals of $[-3,3]$ where $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ is positive are $(a,b),(c,d),$ and $(e,3]$. To make the maximum value of $p(x)$ as small as possible, the idea is to make the intervals where it is positive very short. That means we want $a$ and $b$ to be very close together, $c$ and $d$ to be very close together, and $e$ to be very close to $3$.



    So, fix $epsilon>0$ and whatever values of $a$ and $c$ you want (with $a<c-epsilon$ and $c+epsilon<3-epsilon$) and consider $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ where $b=a+epsilon$, $d=c+epsilon$, and $e=3-epsilon$. Note that then if $xin (a,b)$, $|x-a|$ and $|x-b|$ are at most $epsilon$, so $|p(x)|<6^3epsilon^2$ (the other factors in $p(x)$ have absolute value less than $6$, the diameter of $[-3,3]$). Similarly we have $|p(x)|<6^3epsilon^2$ for all $xin (c,d)$ and $|p(x)|<6^4epsilon$ for all $xin (e,3]$. So, as long as we pick $epsilon$ small enough, $|p(x)|$ will always be at most $1/5$ on any interval where $p(x)$ is positive, and so $p(x)leq 1/5$ on all of $[-3,3]$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      No such polynomial exists.



      Let $p(x)$ be a monic polynomial of degree $5$. Let $q(t)=dfrac{1}{3^5}p(3t)$.



      Then $q(t)$ is a monic polynomial of degree $5$ and
      $$
      max_{x in [-3,3]} |p(x)| = 3^5 max_{t in [-1,1]} |q(t)| geq dfrac{3^5}{2^{4}} > frac15
      $$
      because the monic polynomial of degree $5$ with minimal norm in $[-1,1]$ is $dfrac{1}{2^{4}}T_{5}(t)$, where $T_5$ is the $5$-th Chebyshev polynomial. The minimal norm is $dfrac{1}{2^{4}}$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        No such polynomial exists.



        Let $p(x)$ be a monic polynomial of degree $5$. Let $q(t)=dfrac{1}{3^5}p(3t)$.



        Then $q(t)$ is a monic polynomial of degree $5$ and
        $$
        max_{x in [-3,3]} |p(x)| = 3^5 max_{t in [-1,1]} |q(t)| geq dfrac{3^5}{2^{4}} > frac15
        $$
        because the monic polynomial of degree $5$ with minimal norm in $[-1,1]$ is $dfrac{1}{2^{4}}T_{5}(t)$, where $T_5$ is the $5$-th Chebyshev polynomial. The minimal norm is $dfrac{1}{2^{4}}$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          No such polynomial exists.



          Let $p(x)$ be a monic polynomial of degree $5$. Let $q(t)=dfrac{1}{3^5}p(3t)$.



          Then $q(t)$ is a monic polynomial of degree $5$ and
          $$
          max_{x in [-3,3]} |p(x)| = 3^5 max_{t in [-1,1]} |q(t)| geq dfrac{3^5}{2^{4}} > frac15
          $$
          because the monic polynomial of degree $5$ with minimal norm in $[-1,1]$ is $dfrac{1}{2^{4}}T_{5}(t)$, where $T_5$ is the $5$-th Chebyshev polynomial. The minimal norm is $dfrac{1}{2^{4}}$.






          share|cite|improve this answer









          $endgroup$



          No such polynomial exists.



          Let $p(x)$ be a monic polynomial of degree $5$. Let $q(t)=dfrac{1}{3^5}p(3t)$.



          Then $q(t)$ is a monic polynomial of degree $5$ and
          $$
          max_{x in [-3,3]} |p(x)| = 3^5 max_{t in [-1,1]} |q(t)| geq dfrac{3^5}{2^{4}} > frac15
          $$
          because the monic polynomial of degree $5$ with minimal norm in $[-1,1]$ is $dfrac{1}{2^{4}}T_{5}(t)$, where $T_5$ is the $5$-th Chebyshev polynomial. The minimal norm is $dfrac{1}{2^{4}}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 1:15









          lhflhf

          166k10171396




          166k10171396























              1












              $begingroup$

              If the roots of $p$ are $a<b<c<d<e$, then the subintervals of $[-3,3]$ where $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ is positive are $(a,b),(c,d),$ and $(e,3]$. To make the maximum value of $p(x)$ as small as possible, the idea is to make the intervals where it is positive very short. That means we want $a$ and $b$ to be very close together, $c$ and $d$ to be very close together, and $e$ to be very close to $3$.



              So, fix $epsilon>0$ and whatever values of $a$ and $c$ you want (with $a<c-epsilon$ and $c+epsilon<3-epsilon$) and consider $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ where $b=a+epsilon$, $d=c+epsilon$, and $e=3-epsilon$. Note that then if $xin (a,b)$, $|x-a|$ and $|x-b|$ are at most $epsilon$, so $|p(x)|<6^3epsilon^2$ (the other factors in $p(x)$ have absolute value less than $6$, the diameter of $[-3,3]$). Similarly we have $|p(x)|<6^3epsilon^2$ for all $xin (c,d)$ and $|p(x)|<6^4epsilon$ for all $xin (e,3]$. So, as long as we pick $epsilon$ small enough, $|p(x)|$ will always be at most $1/5$ on any interval where $p(x)$ is positive, and so $p(x)leq 1/5$ on all of $[-3,3]$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                If the roots of $p$ are $a<b<c<d<e$, then the subintervals of $[-3,3]$ where $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ is positive are $(a,b),(c,d),$ and $(e,3]$. To make the maximum value of $p(x)$ as small as possible, the idea is to make the intervals where it is positive very short. That means we want $a$ and $b$ to be very close together, $c$ and $d$ to be very close together, and $e$ to be very close to $3$.



                So, fix $epsilon>0$ and whatever values of $a$ and $c$ you want (with $a<c-epsilon$ and $c+epsilon<3-epsilon$) and consider $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ where $b=a+epsilon$, $d=c+epsilon$, and $e=3-epsilon$. Note that then if $xin (a,b)$, $|x-a|$ and $|x-b|$ are at most $epsilon$, so $|p(x)|<6^3epsilon^2$ (the other factors in $p(x)$ have absolute value less than $6$, the diameter of $[-3,3]$). Similarly we have $|p(x)|<6^3epsilon^2$ for all $xin (c,d)$ and $|p(x)|<6^4epsilon$ for all $xin (e,3]$. So, as long as we pick $epsilon$ small enough, $|p(x)|$ will always be at most $1/5$ on any interval where $p(x)$ is positive, and so $p(x)leq 1/5$ on all of $[-3,3]$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  If the roots of $p$ are $a<b<c<d<e$, then the subintervals of $[-3,3]$ where $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ is positive are $(a,b),(c,d),$ and $(e,3]$. To make the maximum value of $p(x)$ as small as possible, the idea is to make the intervals where it is positive very short. That means we want $a$ and $b$ to be very close together, $c$ and $d$ to be very close together, and $e$ to be very close to $3$.



                  So, fix $epsilon>0$ and whatever values of $a$ and $c$ you want (with $a<c-epsilon$ and $c+epsilon<3-epsilon$) and consider $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ where $b=a+epsilon$, $d=c+epsilon$, and $e=3-epsilon$. Note that then if $xin (a,b)$, $|x-a|$ and $|x-b|$ are at most $epsilon$, so $|p(x)|<6^3epsilon^2$ (the other factors in $p(x)$ have absolute value less than $6$, the diameter of $[-3,3]$). Similarly we have $|p(x)|<6^3epsilon^2$ for all $xin (c,d)$ and $|p(x)|<6^4epsilon$ for all $xin (e,3]$. So, as long as we pick $epsilon$ small enough, $|p(x)|$ will always be at most $1/5$ on any interval where $p(x)$ is positive, and so $p(x)leq 1/5$ on all of $[-3,3]$.






                  share|cite|improve this answer









                  $endgroup$



                  If the roots of $p$ are $a<b<c<d<e$, then the subintervals of $[-3,3]$ where $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ is positive are $(a,b),(c,d),$ and $(e,3]$. To make the maximum value of $p(x)$ as small as possible, the idea is to make the intervals where it is positive very short. That means we want $a$ and $b$ to be very close together, $c$ and $d$ to be very close together, and $e$ to be very close to $3$.



                  So, fix $epsilon>0$ and whatever values of $a$ and $c$ you want (with $a<c-epsilon$ and $c+epsilon<3-epsilon$) and consider $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ where $b=a+epsilon$, $d=c+epsilon$, and $e=3-epsilon$. Note that then if $xin (a,b)$, $|x-a|$ and $|x-b|$ are at most $epsilon$, so $|p(x)|<6^3epsilon^2$ (the other factors in $p(x)$ have absolute value less than $6$, the diameter of $[-3,3]$). Similarly we have $|p(x)|<6^3epsilon^2$ for all $xin (c,d)$ and $|p(x)|<6^4epsilon$ for all $xin (e,3]$. So, as long as we pick $epsilon$ small enough, $|p(x)|$ will always be at most $1/5$ on any interval where $p(x)$ is positive, and so $p(x)leq 1/5$ on all of $[-3,3]$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 23 at 0:21









                  Eric WofseyEric Wofsey

                  187k14216344




                  187k14216344






























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