Calculate sum of polynomial rootsSum of cubed rootsProblem: Sum of absolute values of polynomial rootsSum of squares of roots of a polynomial $P(x)$Determining polynomial from roots of another polynomialPolynomial with real rootsHow to solve this set of symmetric polynomial expressionsHomework: Sum of the cubed roots of polynomialProve an inequality of polynomialPolynomial problemsCalculate sum of roots

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Calculate sum of polynomial roots


Sum of cubed rootsProblem: Sum of absolute values of polynomial rootsSum of squares of roots of a polynomial $P(x)$Determining polynomial from roots of another polynomialPolynomial with real rootsHow to solve this set of symmetric polynomial expressionsHomework: Sum of the cubed roots of polynomialProve an inequality of polynomialPolynomial problemsCalculate sum of roots













5












$begingroup$



We have the polynomial $P=x^20+x^10+x^5+2$, which has roots $x_1,x_2,x_3,...,x_20$. Calculate the sum $$sum^20_k=1frac1x_k-x_k^2$$




What I noticed: $$sum^20_k=1frac1x_k-x_k^2=sum^20_k=1left(frac1x_k+frac11-x_kright)$$



I know how to calculate the first sum: $sum^20_k=1frac1x_k$.



Please help me calculate the second one: $sum^20_k=1frac11-x_k$.






linear-algebra abstract-algebra polynomials contest-math symmetric-polynomials







share|cite|improve this question









New contributor




P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$











  • $begingroup$
    If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
    $endgroup$
    – Sil
    2 hours ago






  • 4




    $begingroup$
    Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
    $endgroup$
    – achille hui
    2 hours ago















5












$begingroup$



We have the polynomial $P=x^20+x^10+x^5+2$, which has roots $x_1,x_2,x_3,...,x_20$. Calculate the sum $$sum^20_k=1frac1x_k-x_k^2$$




What I noticed: $$sum^20_k=1frac1x_k-x_k^2=sum^20_k=1left(frac1x_k+frac11-x_kright)$$



I know how to calculate the first sum: $sum^20_k=1frac1x_k$.



Please help me calculate the second one: $sum^20_k=1frac11-x_k$.






linear-algebra abstract-algebra polynomials contest-math symmetric-polynomials







share|cite|improve this question









New contributor




P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
    $endgroup$
    – Sil
    2 hours ago






  • 4




    $begingroup$
    Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
    $endgroup$
    – achille hui
    2 hours ago













5












5








5


1



$begingroup$



We have the polynomial $P=x^20+x^10+x^5+2$, which has roots $x_1,x_2,x_3,...,x_20$. Calculate the sum $$sum^20_k=1frac1x_k-x_k^2$$




What I noticed: $$sum^20_k=1frac1x_k-x_k^2=sum^20_k=1left(frac1x_k+frac11-x_kright)$$



I know how to calculate the first sum: $sum^20_k=1frac1x_k$.



Please help me calculate the second one: $sum^20_k=1frac11-x_k$.






linear-algebra abstract-algebra polynomials contest-math symmetric-polynomials







share|cite|improve this question









New contributor




P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





We have the polynomial $P=x^20+x^10+x^5+2$, which has roots $x_1,x_2,x_3,...,x_20$. Calculate the sum $$sum^20_k=1frac1x_k-x_k^2$$




What I noticed: $$sum^20_k=1frac1x_k-x_k^2=sum^20_k=1left(frac1x_k+frac11-x_kright)$$



I know how to calculate the first sum: $sum^20_k=1frac1x_k$.



Please help me calculate the second one: $sum^20_k=1frac11-x_k$.






linear-algebra abstract-algebra polynomials contest-math symmetric-polynomials




linear-algebra abstract-algebra polynomials contest-math symmetric-polynomials






share|cite|improve this question









New contributor




P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Robert Howard

2,2393935




2,2393935






New contributor




P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 hours ago









P. MillerP. Miller

262




262




New contributor




P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
    $endgroup$
    – Sil
    2 hours ago






  • 4




    $begingroup$
    Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
    $endgroup$
    – achille hui
    2 hours ago
















  • $begingroup$
    If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
    $endgroup$
    – Sil
    2 hours ago






  • 4




    $begingroup$
    Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
    $endgroup$
    – achille hui
    2 hours ago















$begingroup$
If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
$endgroup$
– Sil
2 hours ago




$begingroup$
If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
$endgroup$
– Sil
2 hours ago




4




4




$begingroup$
Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
$endgroup$
– achille hui
2 hours ago




$begingroup$
Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
$endgroup$
– achille hui
2 hours ago










2 Answers
2






active

oldest

votes


















6












$begingroup$

Since $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$



and $P'(x)= 20x^19+10x^9+5x^4$



we have $$sum_k=1^20frac11-x_k=fracP'(1)P(1) = 35over 5=7$$






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    Hint:



    Set $y=1-x$. If the $x_k$ satisfy the equation $;x^20+x^10+x^5+2=0$, the corresponding $:y_k$ satisfy the equation
    $$(1-y)^20+(1-y)^10+(1-y)^5+2=0.$$



    Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?






    share|cite|improve this answer









    $endgroup$












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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      Since $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$



      and $P'(x)= 20x^19+10x^9+5x^4$



      we have $$sum_k=1^20frac11-x_k=fracP'(1)P(1) = 35over 5=7$$






      share|cite|improve this answer









      $endgroup$

















        6












        $begingroup$

        Since $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$



        and $P'(x)= 20x^19+10x^9+5x^4$



        we have $$sum_k=1^20frac11-x_k=fracP'(1)P(1) = 35over 5=7$$






        share|cite|improve this answer









        $endgroup$















          6












          6








          6





          $begingroup$

          Since $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$



          and $P'(x)= 20x^19+10x^9+5x^4$



          we have $$sum_k=1^20frac11-x_k=fracP'(1)P(1) = 35over 5=7$$






          share|cite|improve this answer









          $endgroup$



          Since $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$



          and $P'(x)= 20x^19+10x^9+5x^4$



          we have $$sum_k=1^20frac11-x_k=fracP'(1)P(1) = 35over 5=7$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Maria MazurMaria Mazur

          48.1k1260120




          48.1k1260120





















              2












              $begingroup$

              Hint:



              Set $y=1-x$. If the $x_k$ satisfy the equation $;x^20+x^10+x^5+2=0$, the corresponding $:y_k$ satisfy the equation
              $$(1-y)^20+(1-y)^10+(1-y)^5+2=0.$$



              Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                Hint:



                Set $y=1-x$. If the $x_k$ satisfy the equation $;x^20+x^10+x^5+2=0$, the corresponding $:y_k$ satisfy the equation
                $$(1-y)^20+(1-y)^10+(1-y)^5+2=0.$$



                Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  Hint:



                  Set $y=1-x$. If the $x_k$ satisfy the equation $;x^20+x^10+x^5+2=0$, the corresponding $:y_k$ satisfy the equation
                  $$(1-y)^20+(1-y)^10+(1-y)^5+2=0.$$



                  Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?






                  share|cite|improve this answer









                  $endgroup$



                  Hint:



                  Set $y=1-x$. If the $x_k$ satisfy the equation $;x^20+x^10+x^5+2=0$, the corresponding $:y_k$ satisfy the equation
                  $$(1-y)^20+(1-y)^10+(1-y)^5+2=0.$$



                  Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  BernardBernard

                  123k741117




                  123k741117




















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