Talk:Hahn series

Jumping to Conclusions

There is probably an error in the wiki page. well-ordered doesn't imply that the sum is finite (finitely many summands).

Jan Burse (talk) 23:01, 22 July 2018 (UTC)[reply]

The result is actually true: if for some

e

there are infinitely many ordered pairs

(e',e'')

with

f(e'),g(e'')\neq 0

and

e'+e''=e

, then since the supports of

f,g

are well-ordered, there must be a srictly increasing such sequence of for instance elements

e'

for

f

, which implies the existence of an infinite decreasing sequence of corresponding elements

e''=e-e'

for

g

, contradicting the hypothesis that the support of

g

is well-ordered. What is harder to justify is that the support of

fg

is well-ordered, so maybe a reference should be added there. --Vincent Bagayoko (talk) 22:10, 11 August 2018 (UTC)[reply]

Are transseries really not just iterated Hahn series?

In the section Examples, transseries are mentionned with the comment

"The construction of

\mathbb {T}

resembles (but is not literally)

T_{0}=\mathbb {R}

,

T_{n+1}=\mathbb {R} \left[\left[\varepsilon ^{T_{n}}\right]\right]

."

I understand that transseries are not constructed like that. Yet it occurs to me that if at each step $n$ , one restricts the length of the sums to be below $\omega _{n}$ where $\omega _{0}=1$ and $\forall p\omega _{p+1}=\omega ^{\omega _{p}}$ , then one obtains a field naturally isomorphic to $\mathbf {No} (\varepsilon _{0})$ : surreal numbers with birthdate below $\varepsilon _{0}$ . This is an exponential differential field (with total exp and log), so why couldn't the unbounded length construction also be, and actually be naturally isomorphic to the field of EL (not LE) transseries? I wonder if that is the case... — Preceding unsigned comment added by Vincent Bagayoko (talk • contribs) 22:30, 5 September 2018 (UTC) Vincent Bagayoko (talk) 06:28, 6 September 2018 (UTC)[reply]