# Taylor expansions at $x=\infty$

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 $a_{n}$)?

• Better to say "expand as a Laurent series" or "expand as an asymptotic series" actually... the usual method for expanding $f(x)$ at "$x=\infty$" is to perform Maclaurin expansion of $f(1/x)$... – J. M. is a poor mathematician Apr 20 '12 at 13:41
• 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. – Adam Rubinson Apr 20 '12 at 13:43

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...

• Okay, 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). – Adam Rubinson Apr 20 '12 at 15:45

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.

• What if we consider the a_i as part of a sequence (a_n), and (a_n) is in the space l_2 for example? – Adam Rubinson Apr 20 '12 at 15:53
• You might think about trying to expand $\sin x$ around $x=\infty$. The corresponding thing is to expand $\sin \frac 1y$ around $y=0$ which doesn't work. – Ross Millikan Apr 20 '12 at 16:09
• If you are considering 1/(Taylor series of sinx), then this is of the form 1/(0+x+...). So will assuming a_0 = 0 always cause problems (or only sometimes)? Also I had in mind originally to assume a_0 =/= 0 but didn't mention it because I wasn't sure how important it was (but now I am thinking it is important) – Adam Rubinson Apr 20 '12 at 16:47
• @AdamRubinson: no, I was trying to expand $\sin 1/y$ near $y=0$ I think the failure of a derivative at 0 means there is no Taylor series. If you try to expand 1/(Taylor series for sin x up to x^n) around $\infty$, I think you will get it starting with $\frac {\pm n!}{x^n}$. Then as you take more an more terms of sine, the leading term goes further and further up. In the limit, it falls apart. – Ross Millikan Apr 20 '12 at 16:52
• Is that because eventually n! > x^n ? And you alternate signs and so get something like 1-10+100-1000+10000 etc. and it does not converge. Is that right? – Adam Rubinson Apr 20 '12 at 16:57