MHB Find the smallest possible degree of a polynomial

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The discussion revolves around determining the smallest possible degree of the polynomial m(x) derived from the expression d^{1992}/dx^{1992}(h(x)/(x^3-x)), where h(x) is a nonzero polynomial of degree less than 1992 and has no common non-constant factors with x^3-x. Participants analyze the implications of the degree of h(x) and the structure of the rational function formed by h(x) and x^3-x. The key focus is on the differentiation process and how it affects the degree of the resulting polynomial m(x). Ultimately, the goal is to find the minimum degree of m(x) under the given conditions. The discussion emphasizes the relationship between the degrees of the involved polynomials and the differentiation operation.
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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 $m(x)$.
 
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Hint:

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.
 
Hint:

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.:)
 
Solution of other:

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.
 
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