Is the largest root of a parametrized family of polynomials again polynomial?












2














$newcommand{a}{alpha} newcommand{bb}{mathbb}$
Let $a = (a_0, dots, a_{n-1}) in bb R^n$ be fixed. We consider the parametrized family of monic polynomials
$$ f(r, z) = z^n + ra_{n-1} z^{n-1} + dots + r a_1 z + r a_0,$$
where $r in bb R$.



For each $r in bb R$, let $L(r)$ denote the largest root. My question is whether $L(r)$ is a polynomial in $r$. By Vieta's formula, for each $r$, we can represent all the roots as a polynomial in $r$. I am not sure if some crossing happened in the roots, this is still true.






linear-algebra abstract-algebra polynomials







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    2














    $newcommand{a}{alpha} newcommand{bb}{mathbb}$
    Let $a = (a_0, dots, a_{n-1}) in bb R^n$ be fixed. We consider the parametrized family of monic polynomials
    $$ f(r, z) = z^n + ra_{n-1} z^{n-1} + dots + r a_1 z + r a_0,$$
    where $r in bb R$.



    For each $r in bb R$, let $L(r)$ denote the largest root. My question is whether $L(r)$ is a polynomial in $r$. By Vieta's formula, for each $r$, we can represent all the roots as a polynomial in $r$. I am not sure if some crossing happened in the roots, this is still true.






    linear-algebra abstract-algebra polynomials







    share|cite|improve this question



























      2












      2








      2







      $newcommand{a}{alpha} newcommand{bb}{mathbb}$
      Let $a = (a_0, dots, a_{n-1}) in bb R^n$ be fixed. We consider the parametrized family of monic polynomials
      $$ f(r, z) = z^n + ra_{n-1} z^{n-1} + dots + r a_1 z + r a_0,$$
      where $r in bb R$.



      For each $r in bb R$, let $L(r)$ denote the largest root. My question is whether $L(r)$ is a polynomial in $r$. By Vieta's formula, for each $r$, we can represent all the roots as a polynomial in $r$. I am not sure if some crossing happened in the roots, this is still true.






      linear-algebra abstract-algebra polynomials







      share|cite|improve this question















      $newcommand{a}{alpha} newcommand{bb}{mathbb}$
      Let $a = (a_0, dots, a_{n-1}) in bb R^n$ be fixed. We consider the parametrized family of monic polynomials
      $$ f(r, z) = z^n + ra_{n-1} z^{n-1} + dots + r a_1 z + r a_0,$$
      where $r in bb R$.



      For each $r in bb R$, let $L(r)$ denote the largest root. My question is whether $L(r)$ is a polynomial in $r$. By Vieta's formula, for each $r$, we can represent all the roots as a polynomial in $r$. I am not sure if some crossing happened in the roots, this is still true.






      linear-algebra abstract-algebra polynomials




      linear-algebra abstract-algebra polynomials






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 26 at 17:47

























      asked Nov 26 at 17:10









      user1101010

      7551630




      7551630






















          1 Answer
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          5














          Consider $n = 2,$ and the case $f(r, z) = z^2 - rz = z(z-r)$, i.e., $a_1 = -1, a_0 = 0$.



          The largest root is $r$ (for $r ge 0$), but $0$ for $r < 0$. The function



          $$
          h(r) = begin{cases}
          r & r ge 0 \
          0 & r < 0
          end{cases}
          $$
          is not differentiable at zero, hence not a polynomial.






          share|cite|improve this answer





















          • Thanks for your nice answer. May I ask another question: I am looking for the zeros of $L(r) - 1$ and your example shows it is not a polynomial in $t$. But I am still hoping the number of the zeros is bounded. Could you comment on this part? Thanks.
            – user1101010
            Nov 26 at 18:36













          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5














          Consider $n = 2,$ and the case $f(r, z) = z^2 - rz = z(z-r)$, i.e., $a_1 = -1, a_0 = 0$.



          The largest root is $r$ (for $r ge 0$), but $0$ for $r < 0$. The function



          $$
          h(r) = begin{cases}
          r & r ge 0 \
          0 & r < 0
          end{cases}
          $$
          is not differentiable at zero, hence not a polynomial.






          share|cite|improve this answer





















          • Thanks for your nice answer. May I ask another question: I am looking for the zeros of $L(r) - 1$ and your example shows it is not a polynomial in $t$. But I am still hoping the number of the zeros is bounded. Could you comment on this part? Thanks.
            – user1101010
            Nov 26 at 18:36


















          5














          Consider $n = 2,$ and the case $f(r, z) = z^2 - rz = z(z-r)$, i.e., $a_1 = -1, a_0 = 0$.



          The largest root is $r$ (for $r ge 0$), but $0$ for $r < 0$. The function



          $$
          h(r) = begin{cases}
          r & r ge 0 \
          0 & r < 0
          end{cases}
          $$
          is not differentiable at zero, hence not a polynomial.






          share|cite|improve this answer





















          • Thanks for your nice answer. May I ask another question: I am looking for the zeros of $L(r) - 1$ and your example shows it is not a polynomial in $t$. But I am still hoping the number of the zeros is bounded. Could you comment on this part? Thanks.
            – user1101010
            Nov 26 at 18:36
















          5












          5








          5






          Consider $n = 2,$ and the case $f(r, z) = z^2 - rz = z(z-r)$, i.e., $a_1 = -1, a_0 = 0$.



          The largest root is $r$ (for $r ge 0$), but $0$ for $r < 0$. The function



          $$
          h(r) = begin{cases}
          r & r ge 0 \
          0 & r < 0
          end{cases}
          $$
          is not differentiable at zero, hence not a polynomial.






          share|cite|improve this answer












          Consider $n = 2,$ and the case $f(r, z) = z^2 - rz = z(z-r)$, i.e., $a_1 = -1, a_0 = 0$.



          The largest root is $r$ (for $r ge 0$), but $0$ for $r < 0$. The function



          $$
          h(r) = begin{cases}
          r & r ge 0 \
          0 & r < 0
          end{cases}
          $$
          is not differentiable at zero, hence not a polynomial.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 at 17:54









          John Hughes

          62.3k24090




          62.3k24090












          • Thanks for your nice answer. May I ask another question: I am looking for the zeros of $L(r) - 1$ and your example shows it is not a polynomial in $t$. But I am still hoping the number of the zeros is bounded. Could you comment on this part? Thanks.
            – user1101010
            Nov 26 at 18:36




















          • Thanks for your nice answer. May I ask another question: I am looking for the zeros of $L(r) - 1$ and your example shows it is not a polynomial in $t$. But I am still hoping the number of the zeros is bounded. Could you comment on this part? Thanks.
            – user1101010
            Nov 26 at 18:36


















          Thanks for your nice answer. May I ask another question: I am looking for the zeros of $L(r) - 1$ and your example shows it is not a polynomial in $t$. But I am still hoping the number of the zeros is bounded. Could you comment on this part? Thanks.
          – user1101010
          Nov 26 at 18:36






          Thanks for your nice answer. May I ask another question: I am looking for the zeros of $L(r) - 1$ and your example shows it is not a polynomial in $t$. But I am still hoping the number of the zeros is bounded. Could you comment on this part? Thanks.
          – user1101010
          Nov 26 at 18:36




















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