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

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

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

I'm an amateur at best so please share your views.

  • 5
    $\begingroup$ What's the question? $\endgroup$ Sep 4 '16 at 17:39
  • 1
    $\begingroup$ This is indeed true, only if $0$ is not a root $\endgroup$
    – Vincent
    Sep 4 '16 at 17:44
  • $\begingroup$ 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. $\endgroup$ 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.

  • $\begingroup$ Thank you for your reply. Is this result important by any means? $\endgroup$ Sep 4 '16 at 17:56
  • $\begingroup$ @LORD_ARAVIND arxiv.org/abs/0707.0699 $\endgroup$ Sep 4 '16 at 17:59
  • $\begingroup$ 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. $\endgroup$ Sep 4 '16 at 18:06
  • $\begingroup$ 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. $\endgroup$ Sep 4 '16 at 18:14

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