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Function: $1/(1-x)$

I've already discovered the following:

  • (1) Series[1/(1-x), {x,0,10}]: expand it centered at $x_0=0$, region $|x|<1$.
  • (2) Series[1/(1-x), {x,Infinity,10}] : expand it centered at $x_0=0$, region $|x|>1$.
  • (3) Series[1/(1-x), {x,2,10}] : expand it centered at $x_0=2$, region $|x-2|<1$.

Now, how can I do the following?

  • (4) $1/(1-x)$, expand it centered at $x_0=2$, region $|x-2|>1$?
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You could try Series[1/(1 - x), {x, 2, 10}, Assumptions -> (Abs[x - 2] > 1)] but the result will be the same as without assumptions. – partial81 Feb 15 at 13:02
It would have helped to show the code you used. But how about just using Assumptions->Abs[x-2]>1 ? is this what you used for the first cases? – Nasser Feb 15 at 13:05
5  
You cannot (at least not without analytic continuation). Expansion around $2$ means you write $x = 2+y$, whence $1/(1-x)$ = $1/(1-(2+y))$ = $-1/(1+y)$. The radius of convergence of its MacLaurin series is $1$ (there's a pole at $y=-1$), so necessarily $|x-2| = |y| \lt 1$. You are, in effect, asking to expand this series in the region strictly beyond its radius of convergence. – whuber Feb 15 at 17:14
@whuber I think that there isn't going to be a better answer than that, so why not post as an answer? – Szabolcs Feb 15 at 17:15
@Szabolcs I'm curious whether the OP has mis-stated the question or if they might be interested in computing the analytic continuation of their function. – whuber Feb 15 at 17:17
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closed as too localized by Jens, m_goldberg, Sjoerd C. de Vries, Yves Klett, rm -rf♦ Feb 17 at 15:37

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