If $a_{1}, a_{2}, ..., a_{n-1}$ are the $n$th roots of unity, find the value of $\frac{1}{2-a_{1}} + \frac{1}{2-a_{2}} + ... +\frac{1}{2-a_{n-1}}$.

Question

I took $\frac{1}{x-a_{1}} + \frac{1}{x-a_{2}} + ... +\frac{1}{x-a_{n-1}}$, tried to write it with a common denominator, I thought it would yield something useful in the numerator, the denominator would just simplify to $\frac{x^n -1}{x-1}$, but the numerator doesn't give anything useful when evaluated, it keeps going in circles.

Answer 1

To solve this question, we use $\textbf{log }$

Consider $f(x)=\log \prod_{i=1}^{n-1} (x-a_i)$ The problem is to find $f'(2)$

Now $a_i's$ are roots of $x^n-1$. so $x^n-1=(x-a_1)\cdots(x-a_{n-1})(x-1)$

so $\frac{x^n-1}{x-1}=\prod_i (x-a_i)$

Thus $$f(x)=\log \frac{x^n-1}{x-1}$$

From this we can easily find $f'(2)$.

Answer 2

Standard approach via root transformation:

Let's tweak it slightly to find $\sum_{i=1}^n \frac{1}{ 2 - a_i }$ with $a_n = 1$, then subtract $\frac{ 1}{ 2 - a_n } = 1 $ later.

The roots of $ x^n - 1$ are $\{ a_i \}_{i=1}^n$.
Let $ y = \frac{1}{2-x}$, then $ x = 2 - \frac{1}{y} = \frac{ 2y - 1 } {y}$.

So the roots of $ ( \frac{2y-1}{y} )^n - 1 = 0 \Rightarrow (2y-1)^n - y^n = 0 $ are $ \{ \frac{1}{2-a_i } \}_{i=1}^{n}$.
By Vieta, the sum of roots is $ \frac{ (-1) \text{ coefficient of } y^{n-1} } { \text{coefficient of } y^n } = \frac{ - {n\choose 1 } 2^{n-1}(-1) } { 2^n - 1 } = \frac{ n2^{n-1} } { 2^n-1 } $ (per Kolakoski54's comment).

Finally, subtract 1 and we get $ \frac{ (n-2)2^{n-1} + 1 } { 2^n - 1 } $.

Bonus: Can you use this approach to find $\sum_{i=1}^{n-1} \frac{ 1}{ 2-a_i} $ directly with my tweak of adding $ \frac { 1 } { 2 - a_n } $?
How do we modify $ \frac{ x^n - 1 } { x - 1 } $ to get at the desired expression in terms of $y$?

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We can generalize the above approach to show that for $ k ^n \neq 1$, $ \sum_{i=1}^{n-1} \frac{1}{ k - a_i } = \frac{nk^{n-1} } {k^n - 1 } - \frac{ 1 } { k - 1 } $. So, can we push through OP's idea?

Alternative solution via OP's approach:

Again, we add back the $a_n$ term and look at

$$ \sum_{i=1}^n \frac{1}{k - a_i } = \frac{ \sum_{i=1}^n \prod_{j\neq i } ( k - x_j) } { k^n - 1 }. $$

What do we get in the numerator when we expand it? Let's track the coefficients:

• $k^{n-1} $ term: $\prod_{j\neq i } ( k - x_j)$ will contribute $1$. Summing this across $i$, we get $ \sum_i 1 = n $.
• $k^{n-2} $ term: $\prod_{j\neq i } ( k - x_j)$ will contribute $ \sum_{j\neq i } - x_j $. Summing this across $i$, we get $ (n-1 ) \sum x_i = 0$.
• $k^{n-3} $ term: $\prod_{j\neq i } ( k - x_j)$ will contribute $ \sum_{j\neq i, l \neq i } (-x_j)(- x_l) $. Summing this across $i$, we get $ { n-1 \choose 2 } \sum_{j\neq l} x_j x_l = 0 $.
• $ \vdots$
• $k^0$ term: $\prod_{j\neq i } ( k - x_j)$ will contribute $ \prod_{j \neq i } (-x_j) $. Summing this across $i$, we get $ \sum_{i} (-1)^{n-1} \prod_{j\neq i } x_j = 0 $.
  Hence, the numerator is indeed just $ nk^{n-1}$. Thus, we arrive at the desired identity:

$$ \sum_{i=1}^n \frac{1}{k - a_i } = \frac{ \sum_i \prod_{j\neq i } ( k - x_j) } { k^n - 1 } = \frac{ nk^{n-1 } } { k^n - 1 }. $$

Bonus: Use this approach and carefully track the coefficients to directly show that

$$\sum_{i=1}^{n-1} \frac{ 1}{ k-a_i} = \frac{ (n-1)k^{n-2} + (n-2)k^{n-3} + \ldots + 2k + 1 } { \frac{ k^n-1 } { k-1} } \\ = \frac{(n-1)k^{n-1} - k^{n-2} - k^{n-3} - \ldots - k^1 - 1 } { k^n - 1 } = \frac{ nk^{n-1 } } { k^n -1 } - \frac{ 1}{ k - 1 } . $$