Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Asking WolframAlpha on certain functions, it happens that you get a series expansion at $\infty$. Thinking of the expansion as an approximation of the function in the vincinity of a point $a$, like in Taylor Series $$ f(x)= f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+ \cdots. \tag{$*$} $$ I wonder for which values, other than $\infty$, this expansion is a good one. Put $a=\infty$ in $(*)$ doesn't seem to make sense. I can image that something like this is used, when it comes to limit calculations, but are they of any other use?

share|improve this question
1  
Stirling's approximation for the factorial is one of the more useful asymptotic expansions that come to mind... –  J. M. Feb 14 '12 at 13:26
2  
Define a new function $\displaystyle{\large{\rm g}\left(x\right) \equiv {\rm f}\left(1 \over x\right)}$ and expands it around $\displaystyle{\large x = 0}$. That's the meaning of WA $\displaystyle{\large{\quad\rm f}\left(x\right)}$ expansion "around $\displaystyle{\large \quad x = \infty}$". –  Felix Marin Oct 1 '13 at 7:00

2 Answers 2

up vote 5 down vote accepted

One can apply a linear fractional transformation to move a point at $\infty$ to any finite point, e.g. changing variables by $z = 1/x$ transforms the series expansion from around $x= \infty$ to an ordinary Taylor (or Laurent) series around $z = 0$, e.g. see the discussion of singular points at $\infty$ in the Wiki page on regular singular points of ODEs, via Möbius transformations on the Riemann sphere.

This enables one to translate notions such as "converges in a neighborhood of $\infty$" to more standard notions such as "converges in a neighborhood of $0$". However, one will gain greater insight by instead learning about how to extend such such notions to such "completions" having ideal point(s) at infinity, since their theory is often simpler, due to having more uniform structure (so eliminating exceptional cases). For a nice introduction see the reference here on points at infinity, projective closure, compactifications, and modifications.

share|improve this answer

Your expression is a good expansion around any finite point $a$. If you look at the Alpha approximation at $\infty$, the terms are $x^{-n}$ instead of $x^{n}$ and you can often think of it informally as a standard approximation in the variable $y=\frac 1x$ around $y=0$. You could write an expansion in $y=\frac 1x$ around some non-zero value and get powers of $\frac 1x -a$, but I haven't seen it. I have used (but forget when) expansions around $\infty$ of the form $f(x)=a+\frac bx$ to approximate values.

share|improve this answer
    
Thanks for your answer. Does this relate to Watson's Lemma? –  draks ... Feb 14 '12 at 14:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.