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I am working with complicated expressions expr that contains the symbolic function f[x] which I know to have a pole at x=0. I would like to carry out Series expansions of expr and therefore f[x] near x=0 in Mathematica. The default behavior of Series is to yield a SeriesData that starts at zero powers of x:

Series[f[x],{x,0,1}] // FullForm

(* SeriesData[x, 0, List[f[0], Derivative[1][f][0]], 0, 2, 1] *)

enter image description here

This output is correct as long as f[x] is regular at x=0. But the f[x] I'm using has a pole at x=0. How do I tell Series to start the expansion of f[x] at order x^-1?

I basically need the behavior:

Series[f[x],{x,0,1}] // FullForm

(* SeriesData[x, 0, List[fRef[0], Derivative[1][fReg][0]], Derivative[2][fReg][0]], -1, 2, 1] *)

enter image description here

  1. I especially need to avoid globally defining down-values of f[x] as in f[x]:=g[x]/x since I would like to keep f[x] intact throughout the rest of the evaluations in the notebook. However, if such a definition can be made to be confined within the Series function, that would be perfectly fine.

  2. I also would not like to define a whole new SeriesNew function: I'm writing a package and I prefer the user to be able to functions already familiar to him.


My attempt:

I tried to define an up-value for f[x]:

Series[f[x_], y__] ^:= Series[1/x, y] Series[fReg[x], y];
(* SeriesData[x, 0, List[f2[0], Derivative[1][f2][0]], -1, 1, 1] *)

enter image description here

Which is OK (except that the order $\mathcal{O}(x^1)$ term is hidden). But then Series reverts back to its default assumption about f[x] when I have a more complicated expresssion involving f[x]:

Series[x f[x] + Cos[x], {x, 0, 2}]

(* SeriesData[x, 0, List[1, f[0], Plus[Rational[-1, 2], Derivative[1][f][0]]], 0, 3, 1] *)

Dream: it would be perfect if I could edit the built-in Series[expr_, ...] function so that I can add at the very top, the following line:

expr /. f[x]- >1/x g[x]

and that it can see the pole at x=0, and treat the symbolic g[x] as a regular function.

enter image description here

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I feel like Series is a function that is designed and intended to give the power series (not Laurent series) expansion of a function.

It does its best, so if you did something like Series[Cos[x]/x,{x,0,10}] you will get the Laurent series though. However, if you are using unspecified f[x], it will assume f[x] has a power series expansion and give it (as you have seen).

If your function f[x] is really not specified, you may still want this expansion but you want it to include the appropriate $x^{-1}$ term in the series. You can $f(x)=g(x)/x$ and just do Series[g[x]/x,{x,0,10}]. You would then need to clean up that expression to be in terms of $f$ instead of $g$, which you can do as:

DRule[n_] :=
 If[n == 0, g[0] -> Residue[f[x], x, 0],
  Solve[
    ReplaceAll[D[x f[x] == g[x], {x, n}], x -> 0],
    ReplaceAll[D[g[x], {x, n}], x -> 0]
    ][[1, 1]]
  ]
s[x_]=Series[g[x]/x, {x, 0, 10}]
ReplaceAll[s[x], Table[DRule[n], {n, 0, 10}]]

You probably will find an easy pattern to eliminate the need for a DRule function that is quite so complicated, i.e. $g^{(n)}(0)=nf^{(n-1)}(0)$.

I'm not sure if this is what you want because it's not clear to me what you want to do with this expression once you have it. Please add a comment if there's something more that might help.

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  • $\begingroup$ I wrote answer answer to my own question here: mathematica.stackexchange.com/questions/60891/…. It looks like it is possible to define upvalues for Series. But as I mentioned there, it doesn't work for compound expressions. Is there a workaround? $\endgroup$ – QuantumDot Oct 14 '14 at 1:43
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    $\begingroup$ Series seems to be smart enough to recognize that certain special functions have a pole. For example, Series[Gamma[x]],{x,0,1} and Series[BesselY[1,x]],{x,0,1} all give a leading 1/x. So either Series or the special function have some information that has been set... $\endgroup$ – QuantumDot Oct 14 '14 at 1:56
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    $\begingroup$ I also give an example where Series recognizes a pole. However, it will not do so with expressions containing unspecified functions like f[x] -- it will assume $f(x)$ is regular at the center of your series. You've asked for a work-around. I've provided one. If this one doesn't work for you, you'll have to explain why it doesn't work so I (or someone else) can improve it. $\endgroup$ – Kellen Myers Oct 14 '14 at 14:51
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If you give UpValues for the private symbol System`Private`InternalSeries instead, you should get the behavior you want:

f /: System`Private`InternalSeries[f[x_], {y_, y0_, n_}] := System`Private`InternalSeries[
    g[x]/x,
    {y, y0, n}
]

Then:

Series[f[x], {x, 0, 1}] //TeXForm

$\frac{g(0)}{x}+g'(0)+\frac{1}{2} x g''(0)+O\left(x^2\right)$

and something more complicated:

Series[x f[x^2] + f[Cos[x]], {x, 0, 2}] //TeXForm

$\frac{g(0)}{x}+g(1)+x g'(0)+\frac{1}{2} x^2 \left(g(1)-g'(1)\right)+O\left(x^3\right)$

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