Computing a sum involving the roots of a tenth degree polynomial

Question

Let $$p(x)=\sum_{k=0}^{10}x^k=1+x+\ldots+x^{10}.$$ Let the roots of $p$ be $\alpha_i,i\in\{1,2,\ldots,10\}.$ Compute $$\sum_{i=1}^{10}\frac{1}{1-\alpha_i}.$$

My Attempt: Using the GP formula, we get that each $\alpha_i$ satisfies $\alpha_i^{11}=1.$ So, we are looking at all the eleventh roots of unity except $1.$ Since $g(x)=x^{11}-1$ has only one real root, we are working with the complex roots of $g.$ I know that I have to use Vieta's relations somewhere as this question was under the "Vieta's Relations" section. However, the expression $\frac{1}{1-\alpha_i}$ is very unwieldy to work with, and I don't know how to proceed. How to solve this question?

Answer 1

The sum is $5$. To see this, you could argue as follows. We have

$$ p(x) = 1+x+x^2 + \dots + x^{10} = \prod_{i=1}^{10} (x-\alpha_i) $$

where $\alpha_i$ are the roots of $p.$

Consider

$$ \log(p(x)) = \sum_{i=1}^{10} \log(x-\alpha_i), $$

then we have

$$ \frac{p'(x)}{p(x)} = \left(\log(p(x))\right)' = \sum_{i=1}^{10} \frac{1}{x-\alpha_i}. $$

Now note that $x=1$ is not a root of $p$. Substituting $x=1$ in the above formula yields

$$ \sum_{i=1}^{10} \frac{1}{1-\alpha_i} = \frac{p'(1)}{p(1)} = \frac{55}{11} = 5. $$

Answer 2

Outline of one possible method of solution:

1. If $\alpha_1, \ldots, \alpha_{10}$ are the roots of $p(z) = 1 + z + \cdots + z^{10}$, then what polynomial, say $q(z)$, will have roots $\beta_i = 1 - \alpha_i$?
2. If $\beta_1, \ldots, \beta_{10}$ are the roots of $q(z)$, then what polynomial, say $s(z)$, will have roots $\gamma_i = 1/\beta_i$?
3. If $\gamma_i$ are the roots of $q(z)$, then what is the sum of those roots in terms of the coefficients of $q$?

Answer 3

[This is the slight simplification of Heropup's solution that I mentioned in the comments.]

Hints towards a solution. If you're stuck, show what you've tried.

• The polynomial $ p(1-x) $has roots $ \beta_i = 1 - \alpha_i$. Let $ p(1-x) = \sum b_ i x^i$ (but do not expand as yet).
• Find $ \sum \frac{1}{ 1 - \alpha _ i } = \sum \frac{1}{ \beta_i}$ in terms of $b_i$

$$ \sum \frac{ 1}{ 1 - \alpha_i } = \sum \frac{1}{ \beta_i } = \frac { \sum \beta_1\beta_2\beta_3\beta_4\beta_5\beta_6\beta_7\beta_8\beta_9 } { \prod \beta_i } = \frac{ -b_1/ b_{10}} { b_0 / b_{10}} = -\frac{ b_1}{b_{0}}.$$

• Find just those $b_i$.

The constant term $b_0 = p(1-0) = 11$.
The linear term can be determined by binomial expansion, and we get $b_1 = -1 -2 -3 \ldots - 10 = - 55$.
Alternatively, $b_1 = -p'(1-0)$. This also shows how the answer agrees with F_M_'s calculation of $ \frac{ p'(1) } { p(1) } $.

• Hence, find $ \sum \frac{1}{1-\alpha_i }$.

$$ - \frac{-55}{11} = 5.$$