A calculus problem by Rakshit Joshi

Relevant wiki: Remainder Factor Theorem - Advanced

Let $f(x) = A(x - r_1)(x - r_2)\cdots(x - r_n)$ , with $r_i$ the complex or real roots (with multiplicity) and $A \not = 0$ . Then

$f'(x) = f(x)\sum_i \frac{1}{x - r_i}$

If $f'(a) = 0$ but $f(a) \not= 0$ , then the sum $\sum \frac{1}{a - r_i} = 0$ . Also,

$f''(x) = f(x)\sum_{i \not = j} \frac{1}{(x-r_i)(x-r_j)} = f(x)\left(\sum_i \frac{1}{x-r_i}\right)^2 - f(x)\sum_i \frac{1}{(x-r_i)^2} = \frac{(f'(x))^2}{f(x)} - f(x)\sum_i \frac{1}{(x-r_i)^2}.$

If $f(a) \not = 0$ , $f'(a) = 0$ , and $f''(a) = 0$ then also the last term $\sum \frac{1}{(a - r_i)^2} = 0$ . But this requires some of the $(a-r_i)^2$ to have a negative real part; hence some $r_i$ are complex.

Relevant wiki: Vieta's Formula Problem Solving - Advanced

By translation, we can assume, without loss of generality, that $a = 0$ . Then we are told that $p(x) \; = \; a_nx^n + a_{n-1}x^{n-1} + \cdots + a_3x^3 + a_2x^2 + a_1x + a_0$ where $a_2 = a_1 = 0$ and $a_0 \neq 0$ . If the roots of $p(x)$ are $\alpha_1,\alpha_2,...,\alpha_n$ , then we know, using Vieta's Formulae, that the roots are all nonzero and that $\sum_{j=1}^n \alpha_j^{-1} \; = \; - \frac{a_1}{a_0} \; = \; 0 \hspace{2cm} \sum_{1 \le j < k \le n} \alpha_j^{-1}\alpha_k^{-1} \; = \; \frac{a_2}{a_0} \; = \; 0$ It follows that $\sum_{j=1}^n \alpha_j^{-2} \; = \; \left(\sum_{j=1}^n \alpha_j^{-1}\right)^2 - 2\sum_{1 \le j < k \le n}\alpha_j^{-1}\alpha_k^{-1} \; = \; 0$ which makes it impossible for all the $\alpha_j$ to be real.

A simple example would be p(x) = x^4 + x^3 + i, where x = 0