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I am given that $\dfrac{d^{100}}{dx^{100}}\left(\dfrac{p(x)}{x^3-x}\right) = \dfrac{f(x)}{g(x)}$ for some polynomials f(x) and g(x). $p(x)$ doesn't have the factor $x^3-x$ and I need to find the least possible degree of $f(x)$.

My Attempt: I am describing in short what I did. Used partial fraction to break up $\dfrac{1}{x^3-x}=\dfrac{A}{x+1}+\dfrac{B}{x}+\dfrac{C}{x-1}$

Now differentiating this $100$ times gave after simplification the denominator $[x(x+1)(x-1)]^{303}$ whle the numerator of Pairwise product of the factors of denominator.

That is, $\dfrac{d^{100}}{dx^{100}}\left(\dfrac{p(x)}{x+1}\right) =-A(100)!\left( \dfrac{a_0+a_1 x+\cdots +a_m x^m}{(x+1)^{101}}\right)$ . where degree of $p(x)$ is $m$ . The other two factors also produced a similar result and adding them the final expression had in numerator degree of $101+101+m=202+m$ .

Now the least possible degree is achieved if $m=0$ that is $p$ is a constant polynomial. So the answer is $202$.

I felt this solution was ok but I do need advices to make sure how much I would get out of 15. Thanks i Advance!

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Wrong. For example, if $p(x) = 1$, $f(x)$ has degree $200$.

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  • $\begingroup$ So where did that get wrong ? If $p(x)$ is not a constant polynomial so would it satisfy the result I proved ? $\endgroup$ – Aditya Narayan Sharma May 19 '17 at 4:06

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