# Is this true for all polynomials

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?

• If the polynomial has a double root, then it is a root of its derivative, and the expression is not well-defined. – Daniel Aug 16 '19 at 17:06
• 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 $m\le n$. – 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.

• Beautiful! And I managed to understand all steps! – 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 $$\sum _{j=1}^{n}{\frac {1}{x_{j}f'(x_{j})}} =-\frac{1}{f(0)}$$