<s>[INST] Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 - \frac{1}{8} x^4 + O \left( x^5 \right) $$
I notice powers of two in the denominator, but I'm not sure of the pattern (and to calculate the next term to confirm would entail another tedious product rule).
Any ideas? Thanks!
 [/INST] There is a much simpler way involving no contrived derivatives, binomial expansions, etc., that unfortunately does not easily generalize to any other quadratic reciprocal.
If we notice that $x^2+2x+2$ occurs in the factorization of a certain binomial:
$$ x^8 - 16 = \left( x^4 - 4 \right) \left( x^4 + 4 \right) = \left( x^4 - 4 \right) \left( x^2 - 2x + 2 \right) \left( x^2 + 2x + 2 \right) = \left( x^6 - 2x^5 + 2x^4 - 4x^2 + 8x - 8 \right) \left( x^2 + 2x + 2 \right) $$
Rearranging:
$$ \frac{1}{x^2 + 2x + 2} = \frac{x^6 - 2x^5 + 2x^4 - 4x^2 + 8x - 8}{x^8 - 16} $$
We can bring the right-hand side to the form of a simple geometric series:
$$ \frac{1}{x^2 + 2x + 2} = \frac{\frac{8}{16} - \frac{8}{16} x + \frac{4}{16} x^2 - \frac{2}{16} x^4 + \frac{2}{16} x^5 - \frac{1}{16} x^6}{\frac{16}{16} - \frac{x^8}{16}} = \frac{\frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 - \frac{1}{8} x^4 + \frac{1}{8} x^5 - \frac{1}{16} x^6}{1 - \frac{x^8}{16}} = \left( \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 - \frac{1}{8} x^4 + \frac{1}{8} x^5 - \frac{1}{16} x^6 \right) \left( 1 + \frac{x^8}{16} + \frac{x^{16}}{256} + \cdots \right) $$
Clearly, there is no overlap between the terms when multiplying out, so if we write only the coefficients, we see that they group together in sets of eight:
$$ \frac{1}{x^2 + 2x + 2} = \left[ \frac{1}{2}, -\frac{1}{2}, \frac{1}{4}, 0, -\frac{1}{8}, \frac{1}{8}, -\frac{1}{16}, 0, \frac{1}{16}, -\frac{1}{32}, \frac{1}{32}, 0, -\frac{1}{64}, \frac{1}{64}, -\frac{1}{128}, 0, \cdots \right] $$
in an elegant pattern!
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