Calculate $sum^20_k=1frac1x_k-x_k^2$ where $x_k$ are roots of $P(x)=x^20+x^10+x^5+2$Sum of cubed rootsProblem: Sum of absolute values of polynomial rootsSum of squares of roots of a polynomial $P(x)$Polynomial with real rootsHomework: Sum of the cubed roots of polynomialSum of roots rational but product irrationalPolynomial problemsCalculate sum of roots$ fracQleft(x_1right)P'left(x_1right)+fracQleft(x_2right)P'left(x_2right)+fracQleft(x_3right)P'left(x_3right)=? $If roots of $x^6=p(x)$ are given then choose the correct option

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Calculate $sum^20_k=1frac1x_k-x_k^2$ where $x_k$ are roots of $P(x)=x^20+x^10+x^5+2$


Sum of cubed rootsProblem: Sum of absolute values of polynomial rootsSum of squares of roots of a polynomial $P(x)$Polynomial with real rootsHomework: Sum of the cubed roots of polynomialSum of roots rational but product irrationalPolynomial problemsCalculate sum of roots$ fracQleft(x_1right)P'left(x_1right)+fracQleft(x_2right)P'left(x_2right)+fracQleft(x_3right)P'left(x_3right)=? $If roots of $x^6=p(x)$ are given then choose the correct option













9












$begingroup$



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




What I've 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: $sum^20_k=1frac11-x_k$.










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    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
    Mar 22 at 19:38






  • 6




    $begingroup$
    Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
    $endgroup$
    – achille hui
    Mar 22 at 19:44










  • $begingroup$
    Some of your tags have no connection with the issue : linear-algebra, symmetric polynomials (we don't have a symmetry of coefficients), abstract algebra (no group, ring, field, etc... here). I have taken the liberty to suppress them. I have added the "roots" tags.
    $endgroup$
    – Jean Marie
    Mar 23 at 16:11
















9












$begingroup$



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




What I've 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: $sum^20_k=1frac11-x_k$.










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    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
    Mar 22 at 19:38






  • 6




    $begingroup$
    Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
    $endgroup$
    – achille hui
    Mar 22 at 19:44










  • $begingroup$
    Some of your tags have no connection with the issue : linear-algebra, symmetric polynomials (we don't have a symmetry of coefficients), abstract algebra (no group, ring, field, etc... here). I have taken the liberty to suppress them. I have added the "roots" tags.
    $endgroup$
    – Jean Marie
    Mar 23 at 16:11














9












9








9


3



$begingroup$



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




What I've 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: $sum^20_k=1frac11-x_k$.










share|cite|improve this question











$endgroup$





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




What I've 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: $sum^20_k=1frac11-x_k$.







polynomials contest-math roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 23 at 16:13









Jean Marie

32k42355




32k42355










asked Mar 22 at 19:22









P. MillerP. Miller

462




462







  • 2




    $begingroup$
    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
    Mar 22 at 19:38






  • 6




    $begingroup$
    Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
    $endgroup$
    – achille hui
    Mar 22 at 19:44










  • $begingroup$
    Some of your tags have no connection with the issue : linear-algebra, symmetric polynomials (we don't have a symmetry of coefficients), abstract algebra (no group, ring, field, etc... here). I have taken the liberty to suppress them. I have added the "roots" tags.
    $endgroup$
    – Jean Marie
    Mar 23 at 16:11













  • 2




    $begingroup$
    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
    Mar 22 at 19:38






  • 6




    $begingroup$
    Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
    $endgroup$
    – achille hui
    Mar 22 at 19:44










  • $begingroup$
    Some of your tags have no connection with the issue : linear-algebra, symmetric polynomials (we don't have a symmetry of coefficients), abstract algebra (no group, ring, field, etc... here). I have taken the liberty to suppress them. I have added the "roots" tags.
    $endgroup$
    – Jean Marie
    Mar 23 at 16:11








2




2




$begingroup$
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
Mar 22 at 19:38




$begingroup$
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
Mar 22 at 19:38




6




6




$begingroup$
Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
$endgroup$
– achille hui
Mar 22 at 19:44




$begingroup$
Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
$endgroup$
– achille hui
Mar 22 at 19:44












$begingroup$
Some of your tags have no connection with the issue : linear-algebra, symmetric polynomials (we don't have a symmetry of coefficients), abstract algebra (no group, ring, field, etc... here). I have taken the liberty to suppress them. I have added the "roots" tags.
$endgroup$
– Jean Marie
Mar 23 at 16:11





$begingroup$
Some of your tags have no connection with the issue : linear-algebra, symmetric polynomials (we don't have a symmetry of coefficients), abstract algebra (no group, ring, field, etc... here). I have taken the liberty to suppress them. I have added the "roots" tags.
$endgroup$
– Jean Marie
Mar 23 at 16:11











2 Answers
2






active

oldest

votes


















11












$begingroup$

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









$endgroup$












  • $begingroup$
    Why is $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$
    $endgroup$
    – Dr. Mathva
    Mar 24 at 9:34










  • $begingroup$
    Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
    $endgroup$
    – Maria Mazur
    Mar 24 at 11:02


















4












$begingroup$

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









    11












    $begingroup$

    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









    $endgroup$












    • $begingroup$
      Why is $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$
      $endgroup$
      – Dr. Mathva
      Mar 24 at 9:34










    • $begingroup$
      Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
      $endgroup$
      – Maria Mazur
      Mar 24 at 11:02















    11












    $begingroup$

    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









    $endgroup$












    • $begingroup$
      Why is $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$
      $endgroup$
      – Dr. Mathva
      Mar 24 at 9:34










    • $begingroup$
      Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
      $endgroup$
      – Maria Mazur
      Mar 24 at 11:02













    11












    11








    11





    $begingroup$

    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 Mar 22 at 19:50









    Maria MazurMaria Mazur

    51k1362127




    51k1362127











    • $begingroup$
      Why is $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$
      $endgroup$
      – Dr. Mathva
      Mar 24 at 9:34










    • $begingroup$
      Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
      $endgroup$
      – Maria Mazur
      Mar 24 at 11:02
















    • $begingroup$
      Why is $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$
      $endgroup$
      – Dr. Mathva
      Mar 24 at 9:34










    • $begingroup$
      Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
      $endgroup$
      – Maria Mazur
      Mar 24 at 11:02















    $begingroup$
    Why is $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$
    $endgroup$
    – Dr. Mathva
    Mar 24 at 9:34




    $begingroup$
    Why is $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$
    $endgroup$
    – Dr. Mathva
    Mar 24 at 9:34












    $begingroup$
    Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
    $endgroup$
    – Maria Mazur
    Mar 24 at 11:02




    $begingroup$
    Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
    $endgroup$
    – Maria Mazur
    Mar 24 at 11:02











    4












    $begingroup$

    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









    $endgroup$

















      4












      $begingroup$

      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









      $endgroup$















        4












        4








        4





        $begingroup$

        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 Mar 22 at 19:44









        BernardBernard

        125k743119




        125k743119



























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