Find this sum involving a polynomial root

[LCSphysicist]

Homework Statement

.

Relevant Equations

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if x_{I}, I = {1,2,...,2019} is a root of P(x) = $x^{2019} +2019x - 1$

Find the value of $\sum_{1}^{2019}\frac{1}{1-\frac{1}{X_{I}}}$

I am really confused:
This polynomial jut have one root, and this root is x such that 0 < x < 1, so that each terms in the polynomial is negative. But the alternatives just give positive values.

This makes me think we need to consider the complex root of this. But i have no idea how to find them.
Maybe calling $x = re^{i\theta}$ give us:

$0 = r^{2019}e^{i2019\theta}+ 2019re^{i\theta} - 1$

?

 

[Office_Shredder]

Would you know how to compute something like
$$\sum_I x_I$$
Or
$$\sum_ {I\neq J} x_I x_J$$?

 

[anuttarasammyak]

How about trying to solve simpler case:
$$x^2+2x-1=0$$
$$x=\{x_1,x_2\}$$
Calculate
$$\sum_{I=1}^{2}\frac{1}{1-\frac{1}{x_I}}$$
to get insight.

 

[LCSphysicist]

Oh yeah, i forgot the Girard relations for polynomials :P
Now i get it, is 2017. THx

 

[fresh_42]

I got stuck at
$$
\sum_{i\in I}\dfrac{1}{1-\dfrac{1}{x_i}}=-\dfrac{1}{2019}\sum_{k=0}^{2018}\sum_{i\in I}x_i^k
$$
It is easy to see that the last sum is $2019$ for $k=0$ and $0$ for $k=1$, but I have no idea how to add the sum for $k>1$.

 

[anuttarasammyak]

A more general problem, for integer $N\geq 2$
$$x^N+Nx-1=\prod_{i=1}^N(x-x_i)$$
By expansion
$$\prod_{i=1}^N (-x_i)=-1$$
,
$$\sum_{i=1}^N \prod_{j=1,\neq i}^N (-x_i)=N$$
and
$$\sum_{1\leq i<j<...<k\leq N} x_i x_j ... x_k =0$$
for from 1 to N-2 products.
$$S:=\sum_{i=1}^N\frac{1}{1-\frac{1}{x_i}}=\sum_{i=1}^N\frac{-x_i}{1-x_i}=\{\prod_{i=1}^N (1-x_i)\}^{-1}\sum_{i=1}^N(-x_i) \prod_{j=1,j\neq i}^N (1-x_j)$$
where
$$\prod_{i=1}^N (1-x_i)=1+N-1=N$$ and
$$\sum_{i=1}^N(-x_i) \prod_{j=1,j\neq i}^N (1-x_j)=N\prod_{i=1}^N (-x_i)+(N-1)\sum_{i=1}^N \prod_{j=1,\neq i}^N (-x_i)=-N+N(N-1)=N(N-2)$$
So
$S=N-2$.

 

[Office_Shredder]

Fresh, you can actually solve that using this

https://en.m.wikipedia.org/wiki/Newton's_identities

To give an elementary example,
$$x_1^2+x_2^2 = (x_1+x_2)^2 - 2x_1x_2$$.

I wouldn't do it that way. Instead I would make some substitutions to turn the sum into something easier to compute, and get a new polynomial.

As a simple example, to compute $1/x_1+ 1/x_2$ for the polynomial $$x^2+x+3$$, you can make the substitution $y=1/x$ and you want to compute the sum of the roots of $$1/y^2 +1/y +3$$. Multiplying by $y^2$ gives you a new polynomial with the same roots since 0 is not a root of the new polynomial.

I didn't get 2017, but it's possible I made an algebra mistake.

 

[haruspex]

Fresh, you can actually solve that using this

https://en.m.wikipedia.org/wiki/Newton's_identities

To give an elementary example,
$$x_1^2+x_2^2 = (x_1+x_2)^2 - 2x_1x_2$$.

I wouldn't do it that way. Instead I would make some substitutions to turn the sum into something easier to compute, and get a new polynomial.

As a simple example, to compute $1/x_1+ 1/x_2$ for the polynomial $$x^2+x+3$$, you can make the substitution $y=1/x$ and you want to compute the sum of the roots of $$1/y^2 +1/y +3$$. Multiplying by $y^2$ gives you a new polynomial with the same roots since 0 is not a root of the new polynomial.

I didn't get 2017, but it's possible I made an algebra mistake.

Using your substitution method I get 2017.