A series in powers of $(a-z)$ instead of $(z-a)$

Sometimes it is more convenient to find a series expansion (e.g., Taylor, Laurent, Puiseux, ...) in powers of $(a-z)$ than in powers of $(z-a)$. For instance, the command

Series[(1 + Sqrt[1 - 4*z^2])/(4*z) - z, {z, 1/2, 3}]

gives $$i\sqrt{z - \frac{1}{2}} - 2\left(z - \frac{1}{2}\right) - \frac{3}{2}i\left(z - \frac{1}{2}\right)^{3/2} + \frac{23}{8}\left(z - \frac{1}{2}\right)^{5/2} - 4\left(z - \frac{1}{2}\right)^3 + O\left(z - \frac{1}{2}\right)^{7/2}.$$

The series expansion in powers of $\left(\frac{1}{2} - z\right)$ is nicer in this case, as the coefficient at the leading term becomes real. So I would like to ask what is the preferable way to obtain an expansion in terms of $\left(a - z\right)$ instead of $\left(z - a\right)$. Thank you in advance.

Comments

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Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

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