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Can all elements of the Levi-Civita field be represented as power series of a single element

$$p=\varepsilon^{-1}-\frac{\varepsilon }{24}+\frac{3 \varepsilon ^3}{640}-\frac{1525 \varepsilon ^5}{580608}+\dotsb$$

where the numerators of the terms are given in https://oeis.org/A118050 and the denominators are in https://oeis.org/A118051?

How would look $\varepsilon$ and $\varepsilon^{-1}$ in this basis?

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    $\begingroup$ What are the coefficients in the power series? The only power series in $p$ with real coefficients that converge in the Levi-Civita field are polynomials, and, of course, all polynomials in $p$ "converge"; so we would obtain the Levi-Civita field as a quotient of $\mathbb R[p]$ (a one-variable polynomial ring in the now-formal variable $p$), but the only such field quotients are $\mathbb R$ and $\mathbb C$. (Ah, on reflection, I guess you must mean LC coefficients anyway, since otherwise obviously we can't get any non-integral powers of $\varepsilon$.) $\endgroup$
    – LSpice
    Commented Jan 26, 2022 at 23:29
  • $\begingroup$ @LSpice well, I see now that this would not converge, but let's consider generalizations of Levi-Civita field (transseries or Hardy fields). I meant the coefficients to be real. I meant Laurent series in p. $\endgroup$
    – Anixx
    Commented Jan 26, 2022 at 23:38
  • $\begingroup$ Infinite Laurent series in $p$ are not going to converge either, but Laurent series in $p^{-1}$ converge and they give all Laurent series in $\varepsilon$. They can't possibly give any elements outside it - a Laurent series in a power series can't possibly involve any fractional powers of the indeterminate. $\endgroup$
    – Wojowu
    Commented Jan 26, 2022 at 23:41
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    $\begingroup$ No, a Laurent series in $p^{-1}$ need not be a power series in $p$. Consider for instance $\sum_{n=1}^\infty p^{-n}$. If you find this to be satisfactory I can post it as an answer. $\endgroup$
    – Wojowu
    Commented Jan 26, 2022 at 23:46
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    $\begingroup$ No, it is not a Hahn series in $p$ either. The set of exponents appearing in a Hahn series has to be well-ordered, and $\{-n\mid n\in\mathbb N\}$ is not well-ordered. $\endgroup$
    – Wojowu
    Commented Jan 26, 2022 at 23:49

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Let $p$ be any Laurent series in $\varepsilon$ of the form $\varepsilon^{-1}+\sum_{n=0}^\infty a_n\varepsilon^n$, like the one in the question. Then infinite Laurent series in $p$ itself never converge (not in the ring of Laurent series, or Levi-Civita field, or Hahn series, or any related such field), because positive powers of $p$ do not converge to zero.

On the other hand, if we consider $p^{-1}=\varepsilon+\sum_{n=2}^\infty b_n\varepsilon^n$, then this element topologically generates $\mathbb R[[\varepsilon]]$ (as a ring) and $\mathbb R((\varepsilon))$ (as a field): proving this comes down to the usual method of showing that you can iteratively find coefficients of a power series in $p^{-1}$ to make it give an arbitrary element of $\mathbb R[[\varepsilon]]$.

On the other hand, there is no way such a Laurent series in $p^{-1}$ can give you anything outside the ring of formal Laurent series. There is no way to produce fractional powers of $\varepsilon$ through this procedure.

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  • $\begingroup$ Can one express $\varepsilon$ or $\varepsilon^{-1}$ via $p$? $\endgroup$
    – Anixx
    Commented Jan 27, 2022 at 0:24
  • $\begingroup$ We can express any element of $\mathbb R((\varepsilon))$ as a Laurent series in $p$, in particular we can express $\varepsilon$ and $\varepsilon^{-1}$. Given coefficients of $p^{-1}$, you can find the expression in terms of formulas described here. $\endgroup$
    – Wojowu
    Commented Jan 27, 2022 at 0:28
  • $\begingroup$ But you said it will not converge?... $\endgroup$
    – Anixx
    Commented Jan 27, 2022 at 0:42
  • $\begingroup$ Apologies, meant Laurent series in $p^{-1}$. Nontrivial Laurent series in $p$ indeed don't converge. $\endgroup$
    – Wojowu
    Commented Jan 27, 2022 at 0:42

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