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 $d$:

$$ \frac{d^n}{dx}(1-x^d)^{1/2}=-\frac{d^2 x^{2 d-2}}{4 \left(1-x^d\right)^{3/2}}-\frac{(d-1) d x^{d-2}}{2 \sqrt{1-x^d}} $$ 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:

$$ 1-\frac{x^2}{2}-\frac{x^4}{6}-\frac{x^6}{12} $$

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:

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

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 $d=2$:

$$ \frac{d^n}{dx}(1-x^d)^{1/2}=-\frac{d^2 x^{2 d-2}}{4 \left(1-x^d\right)^{3/2}}-\frac{(d-1) d x^{d-2}}{2 \sqrt{1-x^d}} $$ 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:

$$ 1-\frac{x^2}{2}-\frac{x^4}{6}-\frac{x^6}{12} $$

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 $d$:

$$ \frac{d^n}{dx}(1-x^d)^{1/2}=-\frac{d^2 x^{2 d-2}}{4 \left(1-x^d\right)^{3/2}}-\frac{(d-1) d x^{d-2}}{2 \sqrt{1-x^d}} $$ 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:

$$ 1-\frac{x^2}{2}-\frac{x^4}{6}-\frac{x^6}{12} $$