Find the smallest possible degree of a polynomial

[anemone]

Let $h(x)$ be a nonzero polynomial of degree less than 1992 having no non-constant factor in common with $x^3-x$. Let

$\dfrac{d^{1992}}{dx^{1992}}\left(\dfrac{h(x)}{x^3-x}\right)=\dfrac{m(x)}{n(x)}$

for polynomials $m(x)$ and $n(x)$. Find the smallest possible degree of $m(x)$.

 

[anemone]

Hint:

Spoiler

Rewrite $h(x)=(x^3-x)q(x)+r(x)$ and partial fraction decomposition of a certain rational function might be very useful as well.

 

[anemone]

Hint:

Spoiler

Let $h(x)=(x^3-x)q(x)+r(x)$ where $q(x)$ and $r(x)$ are polynomials, the degree of $r(x)$ is less than 3, and the degree of $q(x)$ is less than 1989, then

$\begin{align*}\dfrac{d^{1992}}{dx^{1992}}\left(\dfrac{h(x)}{x^3-x}\right)&=\dfrac{d^{1992}}{dx^{1992}}\left(\dfrac{r(x)}{x^3-x}\right)\\&=\dfrac{d^{1992}}{dx^{1992}}\left(\dfrac{A}{x-1}+\dfrac{B}{x}+\dfrac{C}{x+1}\right)\end{align*}$

That's all I can give as today's hint.:)

 

[anemone]

Solution of other:

Spoiler

Let $h(x)=(x^3-x)q(x)+r(x)$ where $q(x)$ and $r(x)$ are polynomials, the degree of $r(x)$ is less than 3, and the degree of $q(x)$ is less than 1989, then

$\begin{align*}\dfrac{d^{1992}}{dx^{1992}}\left(\dfrac{h(x)}{x^3-x}\right)&=\dfrac{d^{1992}}{dx^{1992}}\left(\dfrac{r(x)}{x^3-x}\right)\\&=\dfrac{d^{1992}}{dx^{1992}}\left(\dfrac{A}{x-1}+\dfrac{B}{x}+\dfrac{C}{x+1}\right)\end{align*}$

Because $h(x)$ and $x^3-x$ have no non-constant common factor, neither do $r(s)$ and $x^3-x$, therefore, $ABC\ne 0$.

Thus,

$\begin{align*}\dfrac{d^{1992}}{dx^{1992}}\left(\dfrac{h(x)}{x^3-x}\right)&=\dfrac{d^{1992}}{dx^{1992}}\left(\dfrac{r(x)}{x^3-x}\right)\\&=\dfrac{d^{1992}}{dx^{1992}}\left(\dfrac{A}{x-1}+\dfrac{B}{x}+\dfrac{C}{x+1}\right)\\&=1992!\left(\dfrac{A}{(x-1)^{1993}}+\dfrac{B}{x^{1993}}+\dfrac{C}{(x+1)^{1993}}\right)\\&=1992!\left(\dfrac{Ax^{1993}(x+1)^{1993}+B(x-1)^{1993}(x+1)^{1993}+Cx^{1993}(x-1)^{1993}}{(x^3-x)^{1993}}\right)\end{align*}$

Since $ABC\ne 0$, it's clear that the numerator and denominator have no common factor. Expanding the numerator gives an expression of the form

$(A+B+C)x^{3986}+1993(A-C)x^{3985}+1993(996A-B+996C)x^{3984}+\cdots$

From $A=C=1$, $B=-2$, we see the degree can be as low as 3984. A lower degree would imply $A+B+C=0$, $A-C=0$, $996A-B+996C=0$, implying that $A=B=C=0$, a contradiction.