# conversion of a powerseries $-3x+4x^2-5x^3+\ldots$ into $-2+\frac 1 x - 0 - \frac 1 {x^3} + \ldots$

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.

Q1:

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 $|x|>1$ .

Q2:
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]

-

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.
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