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I am reading a collection of problems by the Russian mathematician Vladimir Arnol'd, titled A Mathematical Trivium. I am taking a stab at this one:

Calculate the $100$th derivative of the function $$\frac{x^2 + 1}{x^3 - x}.$$

The derivative is non-trivial (in the sense that I computed it for a few rounds, and it only became more assertive). My first thought was to let

$$f(x) = x^2 + 1, \text{ } g(x) = \frac{1}{x^3 - x}$$

and apply the Leibnitz rule for products,

$$fg^{(n)}(x) = \sum_{k=0}^n {n\choose k} f^{(n-k)}(x)g^{(k)}(x) .$$

Since $f$ is vanishing after the third differentiation, we get

$$fg^{(100)}(x) = {100 \choose 2}f^{(2)}g^{(98)} + {100 \choose 1}f^{(1)}g^{(99)} {100 \choose 0}f^{(0)}g^{(100)} \\= 9900g^{(98)} + 200xg^{(99)} + (x^2 + 1)g^{(100)}$$

This would be great if we could compute the last few derivatives of $g$. Indeed, we can boil this down: notice that

$$g(x) = h(x)i(x)j(x), \hspace{4mm} h(x) = \frac{1}{x-1}, \text{ } i(x) = \frac{1}{x}, \text{ } j(x) = \frac{1}{x+1};$$

further, $h, i,$ and $j$ have friendly behavior under repeated differentation, e.g. $h^{(n)}(x) = \frac{(-1)^n n!}{(x-1)^{n + 1}}$.

So overall, it is possible to use Leibnitz again to beat a lengthy derivative out of this function, (namely,

$$g^{(n)}(x) = \sum_{k=0}^n {n \choose k} h^{(n-k)}(x) \Bigl(\sum_{l=0}^k {k \choose l} i^{(k-l)}(x) j^{(l)}(x)\Bigr)$$

with the details filled in).

However, this is really pretty far from computing the derivative.


So, my question: does anyone know how to either improve the above argument, or generate a new one, which can resolve the problem?

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  • $\begingroup$ @anon I was wondering about series expansions.... But I don't see it. Could you elaborate? $\endgroup$ – Chris Aug 17 '12 at 4:17
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    $\begingroup$ Bah, series expansion would be good for evaluating a 100th derivative at $x=0$, but partial fractions is infinitely better in general. Nevermind me. $\endgroup$ – anon Aug 17 '12 at 4:19
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We have a partial fraction decomposition $$ \frac{x^2+1}{x^3-x}=\frac{-1}{x}+\frac{1}{x+1}+\frac{1}{x-1} $$

It follows that $$ \left(\frac{d}{dx}\right)^{100}\frac{x^2+1}{x^3-x}=\frac{-100!}{x^{101}}+\frac{100!}{(x+1)^{101}}+\frac{100!}{(x-1)^{101}} $$

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  • $\begingroup$ Ahhh. That's it! Thank you. :) $\endgroup$ – Chris Aug 17 '12 at 4:21

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