# On the sum of the reciprocals of the roots of a Polynomial

A polynomial of degree n with real coefficients may be represented as

$$P(x) = \sum_{i = 0}^n a_ix^i$$

which when factorised yields

$$P(x) = (x-r_1)(x-r_2)\cdots(x-r_n)$$

where $r_1, r_2,\cdots,r_n$ are the roots of the polynomial.

Now consider the first derivative of this polynomial. This will be

$$P^{'}(x) = \frac{(x-r_1)^{'}P(x)}{(x-r_1)} + \frac{(x-r_2)^{'}P(x)}{(x-r_2)} + \frac{(x-r_3)^{'}P(x)}{(x-r_3)}+ \cdots + \frac{(x-r_n)^{'}P(x)}{(x-r_n)}$$

On simplifying and rearranging we get

$$P^{'}(x) = \frac{P(x)}{(x-r_1)} + \frac{P(x)}{(x-r_2)} + \frac{P(x)}{(x-r_3)}+\cdots+ \frac{P(x)}{(x-r_n)}$$

$$P^{'}(x) = P(x)\left(\frac{1}{x-r_1} + \frac{1}{x-r_2} + \frac{1}{x-r_3} + \cdots+ \frac{1}{x-r_n}\right)$$

$$\frac{P^{'}(x)}{P(x)} = \frac{1}{x-r_1} + \frac{1}{x-r_2} + \frac{1}{x-r_3} + \cdots + \frac{1}{x-r_n}$$

for $x = 0$, we have

$$\frac{P^{'}(0)}{P(0)} = \frac{1}{-r_1} + \frac{1}{-r_2} + \frac{1}{-r_3} +\cdots + \frac{1}{-r_n}$$

$$-\frac{P^{'}(0)}{P(0)} = \sum_{i = 0}^n\frac{1}{r_i}$$

• What's the question? Sep 4 '16 at 17:39
• This is indeed true, only if $0$ is not a root Sep 4 '16 at 17:44
• It may be of interest for you that any symmetric rational expression in the roots of a polynomial can be expressed as a rational expression in the coefficients. Sep 4 '16 at 18:02

Seems reasonable where $P(0) \ne 0$, though a more intuitive approach exists.

If you just multiply out

$$\prod (x-r_i)$$

The ultimate term will be the product of the roots times $(-1)^{n}$ and the penultimate will be the product of the roots lacking one root times the opposite sign times $x$.

Evaluate $P(0)$ and $P'(0)$ and take their quotient, you have your result.

• Thank you for your reply. Is this result important by any means? Sep 4 '16 at 17:56
• @LORD_ARAVIND arxiv.org/abs/0707.0699 Sep 4 '16 at 17:59
• Thank you again for your reply. I'm certain that you are an experienced mathematician. I'm afraid I will have to ask you one last question. Is it possible for you to assess my mathematical ability based on this post. Sorry for wasting your time. Sep 4 '16 at 18:06
• Keep studying, researching, and stay focused, and you will do fine. "Genius is one percent inspiration, ninety nine percent perspiration" - Thomas Alva Edison. If you want to know whether something is important, think not of your solution, but rather think of other mathematicians using your techniques to solve similar problems. Sep 4 '16 at 18:14