Series expansion for expression with parameter?

Question

I would like to compute the following expansion.

Series[(A + p/x^a)^2, {x, 0, 1}] 

where $a>0$. However Mathematica simply returns the expression, unless I choose a specific value for $a$. Is it possible to somehow get a result for generic values?

Answer 1

Mma always returns the original expression if it, for whatever reason, does not know the answer. In this case, it seems to take place because depending on the values and signs of the parameters, there are too many possibilities for the value p/(A*x^a), which can be small or large. In this case, one should help Mma a bit. For example, let us first rewrite the expression as follows:

expr1=A^2 (1 + z)^2;

where z==p/(A*x^a).

To be specific, I will assume that all parameters are reals and positive, while 0<x<1. One can generalize this assumption if needed.

There are two possibilities: a) either 0<z<1, or b) z>1.

In the case a), one finds (p y)/A<1 where y==x^(-a). This splits into two cases: (i) 0<y<1 and (ii) y>1.

The case (i) 0<x^(-a)<1 can only take place at a<0. I assumed the opposite case. Thus, we conclude that within the assumption I specified above, y>1. In this case, the possibility a) fulfills at

   Reduce[(p y)/A < 1 && A > 0 && p > 0 && y > 1]

(*  A > 0 && y > 1 && 0 < p < A/y  *)

Under these conditions, one expands:

  A^2*(Series[(1 + z)^2, {z, 0, 1}] // Normal) /. 
  z -> p/(A*x^a) // Expand

yielding the following result:

The case b) one does the same way. The only difference is that in the case b) z>1 and one has to calculate the expansion around Infinity:Series[(1 + z)^2, {z, Infinity, 1}].

Have fun!