$f(x) = \frac{4 + x}{2 + x - x^2}$, calculate $f^{(9)}(1)$

Question

$f(x) = \frac{4 + x}{2 + x - x^2}$, calculate $f^{(9)}(1)$, where $f^{(9)}$ is the $9$-th derivative of $f$.

Domain of $f$ is $\mathbb{R} - \{-1, 2\}$. I've got that $f(x) = \frac{1}{1 - (-x)} + \frac{1}{1 - \frac{1}{2}x} = \sum_{n=0}^\infty ((-1)^n + 2^{-n})x^n$, but there is a problem that $\frac{1}{1 - (-x)} = \sum_{n=0}^\infty (-1)^nx^n$ is convergent only for $|x| < 1$, so not for $1$. How can I go about this?

Answer 1

How about:

$$f(x) = (1+x)^{-1}+2(2-x)^{-1}$$

$$f'(x) = -(1+x)^{-2}+2(2-x)^{-2}$$

$$f^{(2)}(x) = 2(1+x)^{-3}+4(2-x)^{-3}$$

. . .

$$f^{(9)}(x) = -9!(1+x)^{-10}+2\cdot 9!(2-x)^{-10}$$

Now, plug in 1:

$$f^{(9)}(1) = 2\cdot 9!-\dfrac{9!}{2^{10}}$$

You may want to check I did not miss a negative sign, but this should be correct and does not require geometric sums that have limited intervals of convergence.

Answer 2

I've solved this using Julian Mejia's suggestion by making substitution $t = x -1$. We have $g(t) = \frac{1}{2} \cdot \left(\frac{4}{1 - t} + \frac{1}{1 + \frac{1}{2}t}\right) = \sum_{n=0}^\infty (2-(-\frac{1}{2})^{n+1})t^n$ which converges for $t \in (-1, 1)$. Now $9$-th derivative is just $(2-(-\frac{1}{2})^{10}) \cdot 9!$, which is exactly what InterstellarProbe got.