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I took $3$ random polynomials with non zero roots one having even degree and two having odd degrees

  1. $f(x)=\color{red}{4}x^2-(4\sqrt3+12)x+12\sqrt3$ having roots $\color{blue}{3,\sqrt3}$ and leading coefficient $\color{red}{4}$ and calculated values of $xf'(x)$$(f'(x)$ is the derivative of $f(x))$ at both roots which are $3f'(3)$ and $\sqrt3f'(\sqrt3)$ and then sum of their reciprocals $\frac1{3f'(3)}+\frac1{\sqrt3f'(\sqrt3)}=\frac{-1}{12\sqrt3}=\frac{-1}{\color{red}{4}}\left(\frac{1}{\color{blue}{3\cdot\sqrt3}}\right)$ then repeated same thing for
  2. $g(x)=\color{red}{1}x^3-\frac{20}{3}x^2-12x+\frac{32}{3}$ having roots $\color{blue}{8,-2,\frac{2}{3}}$

    $\frac1{8g'(8)}+\frac1{-2g'(-2)}+\frac1{\frac{2}{3}g'(\frac2 3)}=\frac{1}{\color{red}{1}}\left(\frac{1}{\color{blue}{8\cdot-2\cdot\frac2 3}}\right)$

  3. $h(x)=\color{red}{1}x^5+41x^4+137x^3-1601x^2-1818x+3240 $ having roots $\color{blue}{1,-2,5,-9,-36}$

    $\frac1{1h'(1)}+\frac1{-2h'(-2)}+\frac1{5h'(5)}+\frac1{-9h'(-9)}+\frac1{-36h'(-36)}=\frac{1}{\color{red}{1}}\left(\frac{1}{\color{blue}{1\cdot-2 \cdot5\cdot-9\cdot-36}}\right)$

Is this true for all polynomials? Is there any known result?

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    $\begingroup$ If the polynomial has a double root, then it is a root of its derivative, and the expression is not well-defined. $\endgroup$ – Daniel Aug 16 '19 at 17:06
  • $\begingroup$ You want to show that $\sum\limits_{i=1}^m\frac{a_0}{\sum\limits_{j=1}^nja_jx_i^j}=(-1)^{\deg p}$ where $p(x_i)=\sum\limits_{k=0}^na_kx_i^k=0$ with $m\le n$. $\endgroup$ – TheSimpliFire Aug 16 '19 at 17:22
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This results from partial fraction decomposition.

Suppose $g(x)$ has no repeated roots, and without loss of generality is monic (leading coefficient $1$). Then it factors as $\displaystyle g(x)=\prod_{r}(x-r)$ over all roots $r$ and

$$ \frac{1}{g(x)}=\sum_{r}\frac{c(r)}{x-r} $$

for some constants $c(r)$, one for each root $r$. To find the constant $c(s)$ for a specific root $s$, first multiply the equation by the factor $(x-s)$,

$$ \frac{x-s}{g(x)}=c(s)+\sum_{r\ne s} c(r)\frac{x-s}{x-r} $$

then "evaluate" i.e. take the limit $x\to s$ to obtain

$$ \frac{1}{g'(s)}=c(s). $$

Therefore we may plug $x=0$ into

$$ \frac{1}{g(x)} =\sum_{r} \frac{1}{g'(r)(x-r)} $$

and manage negative signs to get

$$ \frac{(-1)^{\deg g}}{\displaystyle \prod r} = -\sum_{r} \frac{1}{rg'(r)}. $$

When $\deg g$ is odd, all signs can go away.

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  • $\begingroup$ Beautiful! And I managed to understand all steps! $\endgroup$ – Zamu Aug 16 '19 at 23:22
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It is true for all polynomials with non-zero simple roots.

This follows from the barycentric form of Lagrange interpolation: $$ L(x)=\ell (x)\sum _{j=1}^{n}{\frac {w_{j}}{x-x_{j}}}y_{j} $$ where $$ \ell (x)=(x-x_{1})(x-x_{1})\cdots (x-x_{n}), \quad w_{j}={\frac {1}{\ell '(x_{j})}} $$ Therefore, for the constant function $1$ evaluated at $x=0$ we have $$ 1= L(0) = -\ell (0)\sum _{j=1}^{n}{\frac {1}{x_{j}\ell '(x_{j})}} $$ For a polynomial $f$ with simple roots $x_1, \dots, x_n$, we have $$ f(x)=a(x-x_{1})(x-x_{1})\cdots (x-x_{n})=a\ell (x) $$ and so $$ \sum _{j=1}^{n}{\frac {1}{x_{j}f'(x_{j})}} =\sum _{j=1}^{n}{\frac {1}{x_{j}a\ell '(x_{j})}} =-\frac{1}{a\ell(0)} =-\frac{1}{(-1)^{n}ax_1 \cdots x_n} =\frac{(-1)^{n+1}}{ax_1 \cdots x_n} $$ This can also be written as $$ \sum _{j=1}^{n}{\frac {1}{x_{j}f'(x_{j})}} =-\frac{1}{f(0)} $$

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