Series expansion not working with $\sqrt{1-x^d}$

Question

Why can't Mathematica expand simple functions like $$\sqrt{1-x^d}$$ in series? When I give particular values of $d$, then it simplifies but not otherwise. For now, I want $d \in \mathbb{N}$, and want to see the 1st few, say $3$ terms. I even tried to simplify the argument by assuming $d>0$ i.e. I ran the code:

Series[FullSimplify[Sqrt[(1 - x^d)], 
  Assumptions -> {d \[Element] PositiveIntegers, d > 0}], {x, 0, 3}]

and nothing happened: Mathematica returned the input. How to make mathematica do this?

Answer 1

Perhaps this?:

Asymptotic[Sqrt[(1 - u)], {u, 0, 2}] /. u -> x^d
(*  1 - x^d/2 - x^(2 d)/8  *)

If we examine the output of Series, we can see a limitation in the data structure used to represent series. This makes series with parameters that affect which powers appear impossible to represent. For instance,

Series[Sqrt[(1 - x^4)], {x, 0, 10}, 
  Assumptions -> {d \[Element] Reals, d > 0, x > 0}] // InputForm
(*  SeriesData[x, 0, {1, 0, 0, 0, -1/2, 0, 0, 0, -1/8}, 0, 11, 1]  *)

We see explicit coefficients (zero) for the missing powers. There are d-1 of them. For an indeterminate d, such an explicit list is impossible.

Answer 2

Since the expansion is at an analytic point of the function, the power series is a Taylor expansion: $$ S(x)=\sum_{n=0}^{\infty} \frac{\frac{d^n}{dx}(1-x^d)^{1/2}}{n!}x^n $$

However when $d$ is indeterminate, there arises indeterminate terms of the form $0^0$ which Mathematica cannot directly evaluate. Consider the first two derivatives of the expansion for indeterminate $d$:

$$ \frac{d^2}{dx}(1-x^d)^{1/2}=-\frac{d^2 x^{2 d-2}}{4 \left(1-x^d\right)^{3/2}}-\frac{(d-1) d x^{d-2}}{2 \sqrt{1-x^d}} $$ and note there is a term $x^{d-2}$ that would have to be evaluated at $x=0$ and $d=2$ producing $0^0$. We could avoid this by evaluating the limits separately with the following but inefficient code:

myExpansion[d_, n_] := 
  Module[{p, q, 
    s}, (Limit[
       Sum[(D[(1 - x^p)^(1/2), {x, v}])/(v!) s^v, {v, 0, n}, 
        Assumptions -> Element[PositiveIntegers, p]], p -> d] /. 
      x -> 0) /. s -> x
   ];
myExpansion[2, 6] 

which produces after several seconds:

$$ 1-\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^6}{16} $$