Roots Revisited

If $\displaystyle f(x)=a_0 x^n+a_1 x^{n-1}+\cdots +a_n$ has roots $x_1,x_2,\cdots,x_n$ then the substitution $y=\frac{1}{1-x}$ transforms $f(x)$ into a polynomial with roots $\displaystyle \frac{1}{1-x_1},\cdots,\frac{1}{1-x_n}$ and thus,

$\displaystyle f(x)=x^n\sum_{k=0}^{n}a_k - x^{n-1}\sum_{k=1}^{n}\binom{k}{1}a_k+\cdots+1=0$ and by Vieta's we have

$\displaystyle \sum_{i=1}^{n} \frac{1}{1-x_i}=\frac{\sum_{k=1}^{n}a_k\binom{k}{1}}{\sum_{k=0}^{n}a_k} = \frac{f'(1)}{f(1)}$

Again by Newton's identities we have $\displaystyle p_0 S_2+p_1 S_1 +2p_2=0$ where $p_i$ are the coefficients of the transformed equation and $S_m$ are the sum of m-th power of roots, Simplifying we get $\displaystyle \sum_{i=1}^{n}\frac{1}{(1-x_i)^2} = \frac{(f'(1))^2-f(1)f''(1)}{(f(1))^2} = \frac{61}{100}=\boxed{0.61}$

$\text{Note : }$ Every coefficient $p_n$ of the transformed equation can be related to $f(x)$ by $\displaystyle p_n = \frac{(-1)^n f^{(n)}(1)}{n}$ which we can observe easily.

Let $f(x)=(x-x_1)(x-x_2) \cdots (x-x_n)$

Differentiating w.r.t. x

$f'(x)=(x-x_2)(x-x_3) \cdots (x-x_n) + (x-x_1)(x-x_3) \cdots (x-x_n) + (x-x_1)(x-x_2) \cdots (x-x_{n-1})$

$f'(x)=f(x) \sum_{i=0 }^{ n } \frac{1}{x-x_i}$

Putting $x=1$ we get: $\sum_{i=0 }^{ n } \frac{1}{x-x_i}=\frac{f'(1)}{f(1)}=-1/10$

Now again differentiate w.r.t. x :

$f''(x)=\sum_{i=0}^{n} \frac{f'(x) (x-x_i)- f(x)}{(x-x_i)^2}$

$f''(x)=f'(x) \sum_{i=0}^{n} \frac{1}{x-x_i} -f(x) \sum_{i=0}^{n} \frac{1}{(x-x_i)^2)}$

$\sum_{i=0}^{n} \frac{1}{(x-x_i)^2}$ = $\dfrac{f'(x) \sum_{i=0}^{n} \frac{1}{x-x_i}-f''(x)}{f(x)}$

Putting x=1 in this equation:

$\sum_{i=0}^{n} \frac{1}{(1-x_i)^2}$ = $f'(1)(-1/10)-f''(1))/f(1)=\boxed{0.61}$

Given $x_1, x_2, \cdots x_n$ are roots.

So, $f(x) = k\prod_1^n (x-x_i)$

Differentiating once, $f'(x) = \sum_1^n \dfrac{f(x)}{x-x_i} =f(x)\sum_1^n \dfrac1{x-x_i}$

So, $\sum_1^n \dfrac1{x-x_i} = \dfrac{f'(x)}{f(x)}$

Differentiating again, $\begin{aligned}f''(x) &= \sum_1^n \left[\dfrac{f'(x)}{x-x_i} - \dfrac{f(x)}{(x-x_i)^2}\right]\\ &= f'(x)\cdot \dfrac{f'(x)}{f(x)} - f(x)\sum_1^n \dfrac1{(x-x_i)^2}\end{aligned}\\ \Rightarrow \sum_1^n \dfrac1{(x-x_i)^2} = \dfrac{(f'(x))^2 - f''(x)f(x)}{(f(x))^2}=\boxed{0.61}$