# Given $x_1, \cdots, x_n \in (0, 1)$ such that $\sum_{i \ne j}\frac{1}{x_i - x_j} = \frac{1}{1 - x_i} - \frac{1}{x_i}$, calculate $\sum_{i = 1}^nx_i$.

Given positive integer $$n$$ and $$x_1, x_2, \cdots, x_{n - 1}, x_n \in (0, 1)$$ such that for all $$i = \overline{1, n}$$, we always have that $$\large \sum_{i \ne j}\frac{1}{x_i - x_j} = \frac{1}{1 - x_i} - \frac{1}{x_i}$$. Calculate the value of $$\displaystyle \sum_{i = 1}^nx_i$$.

We have that

$$\sum_{j \ne 1}\frac{1}{x_1 - x_j} = \frac{1}{1 - x_1} - \frac{1}{x_1}$$

$$\sum_{j \ne 2}\frac{1}{x_2 - x_j} = \frac{1}{1 - x_2} - \frac{1}{x_2}$$

$$\vdots$$

$$\sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor - 1}\frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor - 1} - x_j} = \frac{1}{1 - x_{\left\lfloor\frac{n}{2}\right\rfloor - 1}} - \frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor - 1}}$$

$$\sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor}\frac{1}{x_\left\lfloor\frac{n}{2}\right\rfloor - x_j} = \frac{1}{1 - x_\left\lfloor\frac{n}{2}\right\rfloor} - \frac{1}{x_\left\lfloor\frac{n}{2}\right\rfloor}$$

$$\implies \sum_{j \ne 1}\frac{1}{x_1 - x_j} + \sum_{j \ne 2}\frac{1}{x_2 - x_j} + \cdots + \sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor - 1}\frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor - 1} - x_j} + \sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor}\frac{1}{x_\left\lfloor\frac{n}{2}\right\rfloor - x_j}$$

$$= \sum_{i = 1}^\left\lfloor\frac{n}{2}\right\rfloor\dfrac{1}{1 - x_i} - \sum_{i = 1}^\left\lfloor\frac{n}{2}\right\rfloor\dfrac{1}{x_i}$$

Similarly, $$\implies \sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor + 1}\frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor + 1} - x_j} + \sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor + 2}\frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor + 2} - x_j} + \cdots + \sum_{j \ne n - 1}\frac{1}{x_{n - 1} - x_j} + \sum_{j \ne n}\frac{1}{x_n - x_j}$$

$$= \sum_{i = \left\lfloor\frac{n}{2}\right\rfloor + 1}^n\dfrac{1}{1 - x_i} - \sum_{i = \left\lfloor\frac{n}{2}\right\rfloor + 1}^n\dfrac{1}{x_i}$$

But that's all I got.

• I assume that your original $a_i$ then became $x_i$? Sep 11, 2019 at 18:26

Let $$P(x) = \prod_{k=1}^n (x-x_k) = x^n - Ax^{n-1} + \cdots$$ and by Vieta's formula, $$\displaystyle\;A = \sum_{i=1}^n x_i$$

In terms of $$P(x)$$, the horrible sum on LHS equals to

$$\sum_{j=1,\ne i}^n \frac{1}{x_i-x_j} = \frac12 \frac{P''(x_i)}{P'(x_i)}$$ This implies following rational function and hence its numerator vanishes at $$n$$ points $$x_1,\ldots,x_n$$.

$$\frac12\frac{P''(x)}{P'(x)} - \frac{1}{1-x} + \frac{1}{x} = \frac{(4x-2)P'(x) + (x^2-x)P''(x)}{2x(x-1)P'(x)}$$ The numerator is a polynomial of degree at most $$n$$. So it is a multiple of $$P(x)$$. Comparing coefficients of $$x^n$$, we obtain $$(4x-2)P'(x) + (x^2-x)P''(x) - n(n+3)P(x) = 0$$

Throwing the expansion $$P(x) = x^n - Ax^{n-1} + \cdots$$ into above expression and extract the coefficient of $$x^{n-1}$$, we obtain

$$(2n+2)A - (n^2+n) = 0\quad\implies\quad \sum_{i=1}^n x_i = A = \frac{n}{2}$$

Update

For a solution without calculus, multiply both sides by $$x_i(1-x_i)$$ and sum over $$i$$. One obtain:

$$\sum_{i,j;j\ne i}\frac{x_i - x_i^2}{x_i - x_j} = \sum_{i=1}^n(2 x_i - 1) = 2A - n$$ Swapping the summation index $$i,j$$ in LHS and take averages, we obtain \begin{align} {\rm LHS} = \sum_{i,j;j\ne i}\frac{x_i - x_i^2}{x_i - x_j} &= \frac12\sum_{i,j;j \ne i}\frac{(x_i - x_i^2) - (x_j - x_j^2)}{x_i - x_j}\\ &= \frac12\sum_{i,j;j \ne i}(1 - x_i - x_j)\\ &= \frac12\left[\sum_{1 \le i,j \le n}(1 - x_i - x_j) - \sum_{i=1}^n (1 - 2x_i)\right]\\ &= \frac12\left[(n^2 - 2nA) - (n - 2A)\right] \end{align} Combine with RHS $$= 2A - n$$, we get

$$n^2 - 2nA - n + 2A = 4A - 2n \iff n(n+1) = 2(n+1) A$$ This leads to $$A = \frac{n}{2}$$ again.

• I haven't learnt about derivatives though, could you provide another solution? I'm sorry. Sep 12, 2019 at 6:24
• @LêThànhĐạt I've added a solution w/o calculus, see update. Sep 12, 2019 at 7:06