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This is initially a funny question, because I've found this on old notes but I do not find/recover my own derivation... But then the question is more general.


I considered the function

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

I expressed this by a powerseries $ f_1(x) = -3x + 4x^2 - 5x^3 + 6x^4 - \ldots $ and stated without the derivation that this is also $ f_2(x) = \frac {-2}{1} -\frac {-1}{x} + 0 - \frac {1} {x^3} + \frac {2}{x^4} - \ldots + \ldots $
and - well: hell, - don't see it now how I did it.

What was interesting to me was, that after looking for the fixpoints $ x_0=0, x_{1,2} =-2 $ the range of convergence in the expression by $f_1$ is obviously $ |x|<1 $ limited to the unit-interval but in that by $f_2$ it is infinity and $ |x|>1 $ .

I would like to be able to translate also other powerseries into an $f_2$-type-expression. (I remember to have read a remark of "expanding a powerseries at infinity" but have never seen an explanation of this - so this might be irrelevant for this case?) So: what is the technique to do this given a function in terms of a usual powerseries, for instance for the geometric series $ g(x)=1+x+x^2+ \ldots $ or some series $ h(x) = K + a*x + b*x^2 + c*x^3 + \ldots $ ?

[edit: minus-sign in f(x) was missing, one numerator in f2 was wrong]

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up vote 3 down vote accepted

Divide the numerator and denominator of $f(x)$ by $x^2$ and set $y=1/x$ then expand for $y$ and you have your expansion at infinity.

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Ah, thanks! so I also conclude that this is indeed the "expansion at infinity". I also just found some online-links using this term; I'll see whether there is some more general information and whereabouts for this in one of that resources. – Gottfried Helms Nov 23 '10 at 20:31
Yes, that was simply formal polynomial division $ 3x+2x^2 : 1+2x+x^2= 3x - \ldots $ and the other way round $2+3/x : 1+2/x+1/x^2 = 2 - \ldots $ where I was considering convergence w.r.t. x and tried both ways of division. Again thanks, that was helpful – Gottfried Helms Nov 23 '10 at 21:08

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