Hidden Roots Part 2

Consider a polynomial function $f(x)$ with roots $1,2,3, \ldots$ . $f(x) = (x-1)(x-2)\ldots (x-100)$ Taking log both sides:- $log(f(x)) = \sum_{i=1}^{100} log(x-i)$ Differentiating wrt to $x$ :- $\dfrac{f'(x)}{f(x)} = \sum_{i=0}^{100} \dfrac{1}{x-i} = \mathfrak{F} (x)$ Question is talking about the roots of $\mathfrak{F} (x)$ which are same as the roots of $f'(x)$ .

Now, $f(x)$ is a polynomial of degree $100,$ so $f'(x)$ is a polynomial of degree $99.$ Hence $f'(x)$ has $99$ roots.

Also by Rolle's Theorem, Their lies at least $1$ root of $f'(x)$ between any two roots of $f(x).$ So we have $99$ roots of $\mathfrak{F} (x)$ each lying between two consecutive integers between $1$ to $100.$ The values of $[x_{i}]'s$ are $1,2,3,4,\ldots 99$ .

We have to arrange the values so that we get the maximum value of the given expression. The expression can be expressed as:- $\dfrac{99-49+98-48+\dots 51-1+50}{99+1}$ $=\dfrac{50\times 50}{100}$ $=\boxed{25}$

It is note that $\space F (x)=\sum _{ n=1 }^{ 100 }{ \frac { 1 }{ x-n } }\space$ has roots when not all the denominators are positive or negative. Therefore, there is no root for $\space x \le 1\space$ and $\space x \ge 100$ . And there are $\space 99\space$ roots such that: $\space \lfloor a_n \rfloor = n$ , where $\space n = 1, 2, 3... \space 99$ . See the graphs below.

Therefore, the required answer is:

$\dfrac {99-1+98-2+97-3+...+51-49+50}{99+1} = \dfrac {2500}{100} = \boxed{25}$

Adding 1st & last term, we have (2x-101)/(x-1)(x-100) Similarily 2nd & second last term = (2x-101)/(x-2)(x-99) So, function = (2x-101) [ 1/(x-1)(x-100) + 1/(x-2)(x-99)+.....+1/(x-50)(x-51)] Function has only one root , x = 50.5 GIF(x) = 50 Hence 50/2 = 25 Answer