Maximum of the leading coefficient when $text{deg}P=6$, $0leq P(x)leq1$ for $-1leq xleq 1$.












4












$begingroup$


Let $P(x)$ be a real polynomial of degree 6 with the following property:




For all $-1leq xleq 1$, we have $0leq P(x)leq 1$.




what is the maximum possible value of the leading coefficient of $P(x)$?



I obtained obvious answers when the degree is 1 or 2, but couldn't solve when the degree is 6.



I have a hunch that this has something to do with the Legendre polynomials, but I'm not quite sure. Any advice is welcome.






inequality polynomials real-numbers







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    Let $P(x)$ be a real polynomial of degree 6 with the following property:




    For all $-1leq xleq 1$, we have $0leq P(x)leq 1$.




    what is the maximum possible value of the leading coefficient of $P(x)$?



    I obtained obvious answers when the degree is 1 or 2, but couldn't solve when the degree is 6.



    I have a hunch that this has something to do with the Legendre polynomials, but I'm not quite sure. Any advice is welcome.






    inequality polynomials real-numbers







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      Let $P(x)$ be a real polynomial of degree 6 with the following property:




      For all $-1leq xleq 1$, we have $0leq P(x)leq 1$.




      what is the maximum possible value of the leading coefficient of $P(x)$?



      I obtained obvious answers when the degree is 1 or 2, but couldn't solve when the degree is 6.



      I have a hunch that this has something to do with the Legendre polynomials, but I'm not quite sure. Any advice is welcome.






      inequality polynomials real-numbers







      share|cite|improve this question









      $endgroup$




      Let $P(x)$ be a real polynomial of degree 6 with the following property:




      For all $-1leq xleq 1$, we have $0leq P(x)leq 1$.




      what is the maximum possible value of the leading coefficient of $P(x)$?



      I obtained obvious answers when the degree is 1 or 2, but couldn't solve when the degree is 6.



      I have a hunch that this has something to do with the Legendre polynomials, but I'm not quite sure. Any advice is welcome.






      inequality polynomials real-numbers




      inequality polynomials real-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 31 '18 at 8:23









      DilemianDilemian

      454514




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          2 Answers
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          $begingroup$

          Let $T(x)=cos 6theta$ $(xin[-1,1])$ where $x=costheta$. The leading coefficient of $T$ is $2^{6-1}$ The maximal difference to $0$ occurs at $theta_k=kpi/6,k=0,1,ldots,6$. Let $x_k=costheta_k$, when $k$ is even, $T(x_k)=1$, when $k$ is odd, $T(x_k)=-1$. Suppose $deg R=6$ and the leading coefficient of $R$ is also $2^5$, if $max|R|<1$, at $x_k$, we have $T(x_0)-R(x_0)>0$, $T(x_1)-R(x_1)<0$ ..., it changes sign at least $6$ times, contradicts with $deg T-R<6$. Hence for all polynomial $Q$ s.t. $deg Q=6$ and the leading coefficient is $2^{6-1}$,

          the minimal of $max|Q|$, which is $1$, occurs when $Q(x)=cos6arccos x$.

          Now, denote $L(x)=frac12(Q(x)+1)/max|Q|$, we can see the maximal leading coefficient of $L$ is $2^4=16$.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Hint



            The value you’re looking for is the same as the one of the problem ... For all $-1 le x le 1$, we have $-1/2 le Q(x) le 1/2$... as we’re just substracting $1/2$ to the polynomial which doesn’t change the leading coefficient.



            Now the map $f(x) = frac{1}{2^{6-1}}T_6(x)$ where $T_n$ is the n-th Chebyshev polynomials is the one having $1$ for leading coefficient and of minimal $infty$-norm (see Wikipedia article for more details). And this norm is equal to $Vert f Vert_infty =frac{1}{2^{6-1}}$. In your problem you allow the norm to be equal to $1/2$.



            Hence the value you’re looking for is equal to $frac{1/2}{1/2^{6-1}}=2^4=16$.






            share|cite|improve this answer











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              2 Answers
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              2 Answers
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              active

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              1












              $begingroup$

              Let $T(x)=cos 6theta$ $(xin[-1,1])$ where $x=costheta$. The leading coefficient of $T$ is $2^{6-1}$ The maximal difference to $0$ occurs at $theta_k=kpi/6,k=0,1,ldots,6$. Let $x_k=costheta_k$, when $k$ is even, $T(x_k)=1$, when $k$ is odd, $T(x_k)=-1$. Suppose $deg R=6$ and the leading coefficient of $R$ is also $2^5$, if $max|R|<1$, at $x_k$, we have $T(x_0)-R(x_0)>0$, $T(x_1)-R(x_1)<0$ ..., it changes sign at least $6$ times, contradicts with $deg T-R<6$. Hence for all polynomial $Q$ s.t. $deg Q=6$ and the leading coefficient is $2^{6-1}$,

              the minimal of $max|Q|$, which is $1$, occurs when $Q(x)=cos6arccos x$.

              Now, denote $L(x)=frac12(Q(x)+1)/max|Q|$, we can see the maximal leading coefficient of $L$ is $2^4=16$.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Let $T(x)=cos 6theta$ $(xin[-1,1])$ where $x=costheta$. The leading coefficient of $T$ is $2^{6-1}$ The maximal difference to $0$ occurs at $theta_k=kpi/6,k=0,1,ldots,6$. Let $x_k=costheta_k$, when $k$ is even, $T(x_k)=1$, when $k$ is odd, $T(x_k)=-1$. Suppose $deg R=6$ and the leading coefficient of $R$ is also $2^5$, if $max|R|<1$, at $x_k$, we have $T(x_0)-R(x_0)>0$, $T(x_1)-R(x_1)<0$ ..., it changes sign at least $6$ times, contradicts with $deg T-R<6$. Hence for all polynomial $Q$ s.t. $deg Q=6$ and the leading coefficient is $2^{6-1}$,

                the minimal of $max|Q|$, which is $1$, occurs when $Q(x)=cos6arccos x$.

                Now, denote $L(x)=frac12(Q(x)+1)/max|Q|$, we can see the maximal leading coefficient of $L$ is $2^4=16$.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let $T(x)=cos 6theta$ $(xin[-1,1])$ where $x=costheta$. The leading coefficient of $T$ is $2^{6-1}$ The maximal difference to $0$ occurs at $theta_k=kpi/6,k=0,1,ldots,6$. Let $x_k=costheta_k$, when $k$ is even, $T(x_k)=1$, when $k$ is odd, $T(x_k)=-1$. Suppose $deg R=6$ and the leading coefficient of $R$ is also $2^5$, if $max|R|<1$, at $x_k$, we have $T(x_0)-R(x_0)>0$, $T(x_1)-R(x_1)<0$ ..., it changes sign at least $6$ times, contradicts with $deg T-R<6$. Hence for all polynomial $Q$ s.t. $deg Q=6$ and the leading coefficient is $2^{6-1}$,

                  the minimal of $max|Q|$, which is $1$, occurs when $Q(x)=cos6arccos x$.

                  Now, denote $L(x)=frac12(Q(x)+1)/max|Q|$, we can see the maximal leading coefficient of $L$ is $2^4=16$.






                  share|cite|improve this answer











                  $endgroup$



                  Let $T(x)=cos 6theta$ $(xin[-1,1])$ where $x=costheta$. The leading coefficient of $T$ is $2^{6-1}$ The maximal difference to $0$ occurs at $theta_k=kpi/6,k=0,1,ldots,6$. Let $x_k=costheta_k$, when $k$ is even, $T(x_k)=1$, when $k$ is odd, $T(x_k)=-1$. Suppose $deg R=6$ and the leading coefficient of $R$ is also $2^5$, if $max|R|<1$, at $x_k$, we have $T(x_0)-R(x_0)>0$, $T(x_1)-R(x_1)<0$ ..., it changes sign at least $6$ times, contradicts with $deg T-R<6$. Hence for all polynomial $Q$ s.t. $deg Q=6$ and the leading coefficient is $2^{6-1}$,

                  the minimal of $max|Q|$, which is $1$, occurs when $Q(x)=cos6arccos x$.

                  Now, denote $L(x)=frac12(Q(x)+1)/max|Q|$, we can see the maximal leading coefficient of $L$ is $2^4=16$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 31 '18 at 9:03

























                  answered Dec 31 '18 at 8:45









                  Kemono ChenKemono Chen

                  3,3401844




                  3,3401844























                      1












                      $begingroup$

                      Hint



                      The value you’re looking for is the same as the one of the problem ... For all $-1 le x le 1$, we have $-1/2 le Q(x) le 1/2$... as we’re just substracting $1/2$ to the polynomial which doesn’t change the leading coefficient.



                      Now the map $f(x) = frac{1}{2^{6-1}}T_6(x)$ where $T_n$ is the n-th Chebyshev polynomials is the one having $1$ for leading coefficient and of minimal $infty$-norm (see Wikipedia article for more details). And this norm is equal to $Vert f Vert_infty =frac{1}{2^{6-1}}$. In your problem you allow the norm to be equal to $1/2$.



                      Hence the value you’re looking for is equal to $frac{1/2}{1/2^{6-1}}=2^4=16$.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        Hint



                        The value you’re looking for is the same as the one of the problem ... For all $-1 le x le 1$, we have $-1/2 le Q(x) le 1/2$... as we’re just substracting $1/2$ to the polynomial which doesn’t change the leading coefficient.



                        Now the map $f(x) = frac{1}{2^{6-1}}T_6(x)$ where $T_n$ is the n-th Chebyshev polynomials is the one having $1$ for leading coefficient and of minimal $infty$-norm (see Wikipedia article for more details). And this norm is equal to $Vert f Vert_infty =frac{1}{2^{6-1}}$. In your problem you allow the norm to be equal to $1/2$.



                        Hence the value you’re looking for is equal to $frac{1/2}{1/2^{6-1}}=2^4=16$.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Hint



                          The value you’re looking for is the same as the one of the problem ... For all $-1 le x le 1$, we have $-1/2 le Q(x) le 1/2$... as we’re just substracting $1/2$ to the polynomial which doesn’t change the leading coefficient.



                          Now the map $f(x) = frac{1}{2^{6-1}}T_6(x)$ where $T_n$ is the n-th Chebyshev polynomials is the one having $1$ for leading coefficient and of minimal $infty$-norm (see Wikipedia article for more details). And this norm is equal to $Vert f Vert_infty =frac{1}{2^{6-1}}$. In your problem you allow the norm to be equal to $1/2$.



                          Hence the value you’re looking for is equal to $frac{1/2}{1/2^{6-1}}=2^4=16$.






                          share|cite|improve this answer











                          $endgroup$



                          Hint



                          The value you’re looking for is the same as the one of the problem ... For all $-1 le x le 1$, we have $-1/2 le Q(x) le 1/2$... as we’re just substracting $1/2$ to the polynomial which doesn’t change the leading coefficient.



                          Now the map $f(x) = frac{1}{2^{6-1}}T_6(x)$ where $T_n$ is the n-th Chebyshev polynomials is the one having $1$ for leading coefficient and of minimal $infty$-norm (see Wikipedia article for more details). And this norm is equal to $Vert f Vert_infty =frac{1}{2^{6-1}}$. In your problem you allow the norm to be equal to $1/2$.



                          Hence the value you’re looking for is equal to $frac{1/2}{1/2^{6-1}}=2^4=16$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 31 '18 at 8:54

























                          answered Dec 31 '18 at 8:47









                          mathcounterexamples.netmathcounterexamples.net

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