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How do you expand, say, $\frac{1}{1+x}$ at $x=\infty$? (or for those nit-pickers, as $x\rightarrow\infty$. I know it doesn't strictly make sense to say "at infinity", but I think it is standard to say it anyway).

I have a couple of interesting questions to follow... I might as well say them now.

Question 1. According to WolframAlpha, the Taylor expansion of, say, $\frac{1}{(1+x-3x^{2}+x^{3})}$ at $x=\infty$ is $\frac{1}{x^{3}}+\frac{3}{x^{4}}+\frac{8}{x^{5}}+...$ . We see that the expansion starts at $\frac{1}{x^{3}}$ and has higher order terms. I suspect this occurs for any fraction of the form 1/(polynomial in x). Why is this? (I don't see how dividing all the terms on the LHS by $\frac{1}{x^{3}}$ helps, for example).

Question 2. My motivation behind all this Taylor series stuff was originally: Can an infinite expansion $\frac{1}{a_{0}+a_{1}x+a_{1}x^{2}+...}$ be written in the form $b_{0}+\frac{b_{1}}{x}+\frac{b_{2}}{x^{2}}+...$ ? If so, when (i.e. what conditions must we have on the $a_{n}$)?

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    Also, on my screen, terms that were supposed to be 1/x^3 in fact look like 1/x^2. But if you zoom in on the screen (hold ctrl +), then you will see that some of the terms that look like 1/x^2 are actually 1/x^3.2012-04-20

2 Answers 2

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Hint:

Perform the substitution $y=x^{-1}$ and perform the Maclaurin of $y$ expansion (=Taylor expansion of $y$ around $y=0$). At the end of the day, you may substitute $x$ back...

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    O$k$ay, so using this substitution, you can use the fact that polynomial quotients (A+Bx+...Cx^n)/(K+Lx+...Mx^m) are (at most) infinitely differentiable about x=0 to ensure existence of the Taylor series. I'm pretty sure "at most" should in fact be "always", i.e. there are no a_0, ..., a_n so that 1/(a_0 + a_1x+...+a_nx^n) = b_0 + b_1(1/x) + ... + b_m(1/x^m).2012-04-20
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For question 2, I don't think so. In your first example, the series started at $\frac1{x^3}$ because the highest order term in the denominator was $x^3$. So as $x \to \infty, 1+x-3x^2+x^3 \approx x^3$ plus small corrections. If the expression in the denominator has infinitely many terms, it will grow too fast and the expansion at infinity would have to start $\frac 1{x^\infty}$ Clearly a handwave, but I think it could be made rigorous.

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    Last comment for today: Is 1/x, 1/x^2, ... dense in {continuous functions tending to 0} ? If so, we might be able to use this in a proof. Anyway I'm going to revise and might comment more tomorrow2012-04-20