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?

  • 3
    $\begingroup$ 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

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.

  • $\begingroup$ Beautiful! And I managed to understand all steps! $\endgroup$ – Zamu Aug 16 '19 at 23:22

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 $$ \sum _{j=1}^{n}{\frac {1}{x_{j}f'(x_{j})}} =-\frac{1}{f(0)} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.