Calculate $sum^{20}_{k=1}frac{1}{x_k-x_k^2}$ where $x_k$ are roots of $P(x)=x^{20}+x^{10}+x^5+2$












9












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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 $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}$$




What I've noticed: $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}=sum^{20}_{k=1}left(frac{1}{x_k}+frac{1}{1-x_k}right)$$



I know how to calculate the first sum: $sum^{20}_{k=1}frac{1}{x_k}$.



Please help me calculate the second one: $sum^{20}_{k=1}frac{1}{1-x_k}$.






polynomials contest-math roots







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  • 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: $frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$
    $endgroup$
    – 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 $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}$$




What I've noticed: $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}=sum^{20}_{k=1}left(frac{1}{x_k}+frac{1}{1-x_k}right)$$



I know how to calculate the first sum: $sum^{20}_{k=1}frac{1}{x_k}$.



Please help me calculate the second one: $sum^{20}_{k=1}frac{1}{1-x_k}$.






polynomials contest-math roots







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: $frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$
    $endgroup$
    – 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 $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}$$




What I've noticed: $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}=sum^{20}_{k=1}left(frac{1}{x_k}+frac{1}{1-x_k}right)$$



I know how to calculate the first sum: $sum^{20}_{k=1}frac{1}{x_k}$.



Please help me calculate the second one: $sum^{20}_{k=1}frac{1}{1-x_k}$.






polynomials contest-math roots







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 $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}$$




What I've noticed: $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}=sum^{20}_{k=1}left(frac{1}{x_k}+frac{1}{1-x_k}right)$$



I know how to calculate the first sum: $sum^{20}_{k=1}frac{1}{x_k}$.



Please help me calculate the second one: $sum^{20}_{k=1}frac{1}{1-x_k}$.






polynomials contest-math roots




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

31.1k42255




31.1k42255










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: $frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$
    $endgroup$
    – 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: $frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$
    $endgroup$
    – 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: $frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$
$endgroup$
– achille hui
Mar 22 at 19:44




$begingroup$
Hint: $frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$
$endgroup$
– 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 $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$



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



we have $$sum_{k=1}^{20}frac{1}{1-x_k}=frac{P'(1)}{P(1)} = {35over 5}=7$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
    $endgroup$
    – 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 $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$



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



    we have $$sum_{k=1}^{20}frac{1}{1-x_k}=frac{P'(1)}{P(1)} = {35over 5}=7$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why is $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
      $endgroup$
      – 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 $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$



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



    we have $$sum_{k=1}^{20}frac{1}{1-x_k}=frac{P'(1)}{P(1)} = {35over 5}=7$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why is $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
      $endgroup$
      – 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 $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$



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



    we have $$sum_{k=1}^{20}frac{1}{1-x_k}=frac{P'(1)}{P(1)} = {35over 5}=7$$






    share|cite|improve this answer









    $endgroup$



    Since $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$



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



    we have $$sum_{k=1}^{20}frac{1}{1-x_k}=frac{P'(1)}{P(1)} = {35over 5}=7$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 22 at 19:50









    Maria MazurMaria Mazur

    49k1360122




    49k1360122












    • $begingroup$
      Why is $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
      $endgroup$
      – 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 $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
      $endgroup$
      – 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 $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
    $endgroup$
    – Dr. Mathva
    Mar 24 at 9:34




    $begingroup$
    Why is $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
    $endgroup$
    – 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

        123k741117




        123k741117






























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