2
$\begingroup$

I would like to expand a function as $$\frac{1}{x+1} = \frac{1}{x-1+2} = \frac{1}{x-1} \frac{1}{ 1+\frac{2}{x-1}} = \frac{1}{x-1} \left[ 1- \frac{2}{x-1} + \left(\frac{2}{x-1}\right)^2 + \cdots \right]$$

I tried

Series[1/(x + 1), {x, Infinity + 1, 3}]

apprently it does not work. Is there any robust way to realize this kind of expansion? Assume $|x|$ is sufficiently large.

$\endgroup$
2
  • 1
    $\begingroup$ I don't know if this answers your question but here is a workaround: Series[Unevaluated[1/((x - 1) + 2)] /. HoldPattern[x - 1] -> t, {t, Infinity, 3}] /. t -> (x - 1)} $\endgroup$
    – John
    Aug 2, 2014 at 15:20
  • $\begingroup$ Thanks. I managed to get a solution, 1/(x + 1) /. x -> t + 1; Series[%, {t, Infinity, 3}]; % /. t -> x - 1 $\endgroup$
    – Guest25874
    Aug 2, 2014 at 15:23

1 Answer 1

4
$\begingroup$

This is your expression:

expr1 = 1/(x + 1);

This is the the change of variables:

 sl = Solve[x - 1 == z, x][[1, 1]]

    (*  x -> 1 + z  *)

Substituted to the expression it yields this:

expr2 = expr1 /. sl

(*  1/(2 + z)   *)

Now it can be expaneded as you need:

Series[expr1 /. sl, {z, Infinity, 5}] /. z -> HoldForm[x - 1]

(*    
SeriesData[
HoldForm[x - 1], 
DirectedInfinity[1], {1, -2, 4, -8, 16}, 1, 6, 1]   *)

Just try

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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