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: $\sum^{20}_{k=1}\frac{1}{1-x_k}$.

  • 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
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    $\begingroup$ 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

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

  • $\begingroup$ 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


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?


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