Positive value of function and root












0












$begingroup$


Suppose $j,kinmathbb{N}$ and $1leqslant jleqslant k$.



The polynomial
$$
f(x)=x^{2k}-sum_{i=0}^{k-j-1}x^i
$$
has exactly one positive root $x_0$ by Descartes' rule.



Do we also have that
$$
f(x)>0implies x>x_0?
$$






real-analysis roots







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Suppose $j,kinmathbb{N}$ and $1leqslant jleqslant k$.



    The polynomial
    $$
    f(x)=x^{2k}-sum_{i=0}^{k-j-1}x^i
    $$
    has exactly one positive root $x_0$ by Descartes' rule.



    Do we also have that
    $$
    f(x)>0implies x>x_0?
    $$






    real-analysis roots







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose $j,kinmathbb{N}$ and $1leqslant jleqslant k$.



      The polynomial
      $$
      f(x)=x^{2k}-sum_{i=0}^{k-j-1}x^i
      $$
      has exactly one positive root $x_0$ by Descartes' rule.



      Do we also have that
      $$
      f(x)>0implies x>x_0?
      $$






      real-analysis roots







      share|cite|improve this question









      $endgroup$




      Suppose $j,kinmathbb{N}$ and $1leqslant jleqslant k$.



      The polynomial
      $$
      f(x)=x^{2k}-sum_{i=0}^{k-j-1}x^i
      $$
      has exactly one positive root $x_0$ by Descartes' rule.



      Do we also have that
      $$
      f(x)>0implies x>x_0?
      $$






      real-analysis roots




      real-analysis roots






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 18 '18 at 10:10









      SalamoSalamo

      359412




      359412






















          1 Answer
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          1












          $begingroup$

          Since $x_0$ is the only positive root of $f$, it is the largest one. Then, there is no sign change of $f$ on $(x_0, infty)$ and since $underset{xto + infty}{lim} f = + infty$. You may indeed assert that $ x > x_0 Rightarrow f(x) > 0 $.



          However $ f(x) > 0 Rightarrow x > x_0 $ does not generally hold. Take $j=k=1$, then $f(x) = x^2$ and $f(-1)>0$ but $-1 < 0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
            $endgroup$
            – Salamo
            Dec 18 '18 at 10:42










          • $begingroup$
            That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
            $endgroup$
            – Bill O'Haran
            Dec 18 '18 at 10:54












          • $begingroup$
            And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
            $endgroup$
            – Salamo
            Dec 18 '18 at 13:13








          • 1




            $begingroup$
            Sure. I forgot that Descartes' rule took multiplicity into account.
            $endgroup$
            – Bill O'Haran
            Dec 18 '18 at 13:27











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Since $x_0$ is the only positive root of $f$, it is the largest one. Then, there is no sign change of $f$ on $(x_0, infty)$ and since $underset{xto + infty}{lim} f = + infty$. You may indeed assert that $ x > x_0 Rightarrow f(x) > 0 $.



          However $ f(x) > 0 Rightarrow x > x_0 $ does not generally hold. Take $j=k=1$, then $f(x) = x^2$ and $f(-1)>0$ but $-1 < 0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
            $endgroup$
            – Salamo
            Dec 18 '18 at 10:42










          • $begingroup$
            That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
            $endgroup$
            – Bill O'Haran
            Dec 18 '18 at 10:54












          • $begingroup$
            And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
            $endgroup$
            – Salamo
            Dec 18 '18 at 13:13








          • 1




            $begingroup$
            Sure. I forgot that Descartes' rule took multiplicity into account.
            $endgroup$
            – Bill O'Haran
            Dec 18 '18 at 13:27
















          1












          $begingroup$

          Since $x_0$ is the only positive root of $f$, it is the largest one. Then, there is no sign change of $f$ on $(x_0, infty)$ and since $underset{xto + infty}{lim} f = + infty$. You may indeed assert that $ x > x_0 Rightarrow f(x) > 0 $.



          However $ f(x) > 0 Rightarrow x > x_0 $ does not generally hold. Take $j=k=1$, then $f(x) = x^2$ and $f(-1)>0$ but $-1 < 0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
            $endgroup$
            – Salamo
            Dec 18 '18 at 10:42










          • $begingroup$
            That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
            $endgroup$
            – Bill O'Haran
            Dec 18 '18 at 10:54












          • $begingroup$
            And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
            $endgroup$
            – Salamo
            Dec 18 '18 at 13:13








          • 1




            $begingroup$
            Sure. I forgot that Descartes' rule took multiplicity into account.
            $endgroup$
            – Bill O'Haran
            Dec 18 '18 at 13:27














          1












          1








          1





          $begingroup$

          Since $x_0$ is the only positive root of $f$, it is the largest one. Then, there is no sign change of $f$ on $(x_0, infty)$ and since $underset{xto + infty}{lim} f = + infty$. You may indeed assert that $ x > x_0 Rightarrow f(x) > 0 $.



          However $ f(x) > 0 Rightarrow x > x_0 $ does not generally hold. Take $j=k=1$, then $f(x) = x^2$ and $f(-1)>0$ but $-1 < 0$.






          share|cite|improve this answer









          $endgroup$



          Since $x_0$ is the only positive root of $f$, it is the largest one. Then, there is no sign change of $f$ on $(x_0, infty)$ and since $underset{xto + infty}{lim} f = + infty$. You may indeed assert that $ x > x_0 Rightarrow f(x) > 0 $.



          However $ f(x) > 0 Rightarrow x > x_0 $ does not generally hold. Take $j=k=1$, then $f(x) = x^2$ and $f(-1)>0$ but $-1 < 0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 10:21









          Bill O'HaranBill O'Haran

          2,5431418




          2,5431418












          • $begingroup$
            Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
            $endgroup$
            – Salamo
            Dec 18 '18 at 10:42










          • $begingroup$
            That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
            $endgroup$
            – Bill O'Haran
            Dec 18 '18 at 10:54












          • $begingroup$
            And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
            $endgroup$
            – Salamo
            Dec 18 '18 at 13:13








          • 1




            $begingroup$
            Sure. I forgot that Descartes' rule took multiplicity into account.
            $endgroup$
            – Bill O'Haran
            Dec 18 '18 at 13:27


















          • $begingroup$
            Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
            $endgroup$
            – Salamo
            Dec 18 '18 at 10:42










          • $begingroup$
            That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
            $endgroup$
            – Bill O'Haran
            Dec 18 '18 at 10:54












          • $begingroup$
            And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
            $endgroup$
            – Salamo
            Dec 18 '18 at 13:13








          • 1




            $begingroup$
            Sure. I forgot that Descartes' rule took multiplicity into account.
            $endgroup$
            – Bill O'Haran
            Dec 18 '18 at 13:27
















          $begingroup$
          Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
          $endgroup$
          – Salamo
          Dec 18 '18 at 10:42




          $begingroup$
          Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
          $endgroup$
          – Salamo
          Dec 18 '18 at 10:42












          $begingroup$
          That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
          $endgroup$
          – Bill O'Haran
          Dec 18 '18 at 10:54






          $begingroup$
          That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
          $endgroup$
          – Bill O'Haran
          Dec 18 '18 at 10:54














          $begingroup$
          And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
          $endgroup$
          – Salamo
          Dec 18 '18 at 13:13






          $begingroup$
          And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
          $endgroup$
          – Salamo
          Dec 18 '18 at 13:13






          1




          1




          $begingroup$
          Sure. I forgot that Descartes' rule took multiplicity into account.
          $endgroup$
          – Bill O'Haran
          Dec 18 '18 at 13:27




          $begingroup$
          Sure. I forgot that Descartes' rule took multiplicity into account.
          $endgroup$
          – Bill O'Haran
          Dec 18 '18 at 13:27


















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