Big degrees

Decomposing $\frac{1}{x^3-x}$ in partial fractions, we can write

(1) $\frac{p(x)}{x^3-x} = -\frac{p(x)}{x} + \frac{1}{2}\left(\frac{p(x)}{x-1} + \frac{p(x)}{x+1}\right)$ .

In the first, second and third fractions on the RHS of (1) write $p(x)$ respectively as

(2) $p(x) = a_{2013}x^{2013} + \ldots + a_1 x + p(0)$ ,

(3) $p(x) = b_{2013}(x-1)^{2013} + \ldots + b_1(x-1) + p(1)$ , and

(4) $p(x) = c_{2013}(x+1)^{2013} + \ldots + c_1(x+1) + p(-1)$ ,

Note that, since $p(x)$ has no nonconstant factor in common with $x^3-x$ , it isn't divisible by $x$ , $x+1$ or $x-1$ , therefore $p(0)$ , $p(1)$ and $p(-1)$ are nonzero. Then

(5) $\frac{p(x)}{x^3-x} = -\left(a_{2013}x^{2012} + \ldots + a_1 + \frac{p(0)}{x}\right) + \frac{1}{2}\left(b_{2013}(x-1)^{2012} + \ldots + b_1 + \frac{p(1)}{x-1}\right) + \frac{1}{2}\left(c_{2013}(x+1)^{2012} + \ldots + c_1 + \frac{p(-1)}{x+1}\right)$ ,

so that

(6) $\frac{d^{2014}}{dx^{2014}}\left(\frac{p(x)}{x^3-x}\right) = 2014! \left[-\frac{p(0)}{x^{2015}} + \frac{1}{2}\left(\frac{p(1)}{(x-1)^{2015}} + \frac{p(-1)}{(x+1)^{2015}}\right)\right] =\frac{s(x)}{k(x)}$ ,

where

(7) $s(x) = \frac{2014!}{2}[-2p(0)(x^2-1)^{2015} + p(1)x^{2015}(x+1)^{2015} + p(-1)x^{2015}(x-1)^{2015}]$

and

(8) $k(x)=(x^3-x)^{2015}$ .

We note that $s(x)$ and $k(x)$ have no nonconstant factor in common, as the roots of $k(x)$ are $0$ , $1$ and $-1$ (each with multiplicity $2015$ ), whereas $s(0)=2014!\,p(0)$ , $s(1)=2014!\,2^{2014}p(1)$ and $s(-1)=2014!\,2^{2014}p(-1)$ are all nonzero, therefore the fraction on the RHS of (6) is in lowest terms.

Now, let's write the first few terms of $s(x)$ :

(9) $s(x) = \frac{2014!}{2}\left[(-2p(0)+p(1)+p(-1))x^{4030}+2015((p(1)-p(-1))x^{4029}+2015[2p(0)+1007((p(1)+p(-1))]x^{4028}+\ldots\right]$ .

We see that, in general, $s(x)$ is a polynomial of degree 4030, but if $p(x)$ is such that $p(0)$ , $p(1)$ and $p(-1)$ satisfy the relations $-2p(0)+p(1)+p(-1) = 0$ and $p(1) = p(-1)$ (consider, for instance, $p(x) = x^4-x^2+1$ ), its degree is reduced by 2. Besides, no further reduction is possible, as the coefficient of $x^{4028}$ is nonzero under such circumstance. We conclude, therefore, that the smallest possible degree of $s(x)$ is $\boxed{4028}\,$ .