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If we have a function like:

$$\text{f[x$\_$]:=}\sum _{i=0}^{\infty } a_ix^i$$

where we can find / know the $a_i$ coefficients, but not really for which function it will converge.

How can we find $f[x]$ but using the inverse of $x$ instead? Something like this?

$$\text{f[x$\_$]:=}\sum _{i=0}^{\infty } \frac{b_i}{x^i}$$

The main problem is that the first form of $f[x]$ does not converge properly for positive values greater than one, since it comes from a Taylor series.

Edit:

I've seen an interesting strategy that we could use to find the inverse of $f[x]$, but not really the other form of $f[x]$ that I'm looking for, but maybe this could help us to find an strategy:

Series Reversion

Attempts:

Using Mathematica I tried:

$$\text{CoefficientGenerator[i$\_$]:=...}$$

$$\text{f[x$\_$]:=}\sum _{i=0}^{\infty } \text{CoefficientGenerator[i]}x^i$$

$$\text{Series[f[x], $\{$x, $\infty $, 5$\}$]}$$

but this doesn't work, I receive my input as an output. If the CoefficientGenerator function is something that it already knows (like the expansion of $e^x$), it works:

$$\text{CoefficientGenerator[i$\_$]:=}\frac{1}{\text{Gamma}[i+1]}$$

The result for this case is:

$$\exp \left(\log (e) x+O\left(\left(\frac{1}{x}\right)^4\right)\right)$$

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    $\begingroup$ The so-called connection problem is difficult in general. A lot is known if the series concerned correspond to hypergeometric functions, but there is still a lot of research necessary for e.g. solutions of Heun or Painlevé equations. Anyway, this is more a math question than a Mathematica one. $\endgroup$ May 26 '20 at 12:01
  • $\begingroup$ Hum, interesting, if not asking too much, do you remember a link from where I could read more about this? I'll also be googling as well. I've seen that Mathematica works for series that are already know, if I expand using infinity, for example, but for functions constructed like mine, not really, unfortunately. $\endgroup$
    – GarouDan
    May 26 '20 at 12:12
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This is an unfinished answer.

It looks like we can do something like this:

Define f1 as:

$$f1(x, order) = \sum _{i=0}^{order} a_ix^i$$

(where we know the $a_i$s from the CoefficientGenerator function).

Define f2 as:

$$f1(x, order) = \sum _{i=0}^{order} \frac{b_i}{x^i}$$

(where the $b_i$s are the coefficients that we want to find).

Since, $f1(x, order)=f2(x, order)$, when order tends to $\infty$, by definition, we could do:

$$f1(1-x, order)=f2(1-x, order) \implies \sum _{i=0}^{order} a_i(1-x)^i = \sum _{i=0}^{order} \frac{b_i}{(1-x)^i}$$.

Then we can expand the left and the right side and using a system of equations find the $b_i$s (since we can expand $\frac{1}{(1-x)^i}$ more easily).

But the difficulty now is, how can we implement this using Mathematica?

f1[x_, order_] := ...

f2[function_, x_, order_] := ... 

f2[f1, x, 10] (* This will give the f2 function in x until order 10. *)

One attempt of mine, a bit ugly for now, is this one, I'd like to simplify it or rewrite it in a better form.

With[
 {
  order = 10
  },
 f1[x_, order_] := \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 0\), \(order\)]\(a[i] 
\*SuperscriptBox[\(x\), \(i\)]\)\);
 f2[x_, order_] := \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 0\), \(order\)]
\*FractionBox[\(b[i]\), 
SuperscriptBox[\(x\), \(i\)]]\);
 reduce = Reduce[
   f1[1 - x, order] == Series[f2[1 - x, order], {x, 0, order}],
   Array[b, order + 1, 0]
   ];
 rules = {ToRules[reduce]}[[1]];
 f3[x_, order_] := \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 0\), \(order\)]
\*FractionBox[\(b[i] /. rules\), 
SuperscriptBox[\(x\), \(i\)]]\)
 ]
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