If $x$ is a surreal number and $\alpha$ an ordinal, let us denote $T_\alpha(x)$ and call $\alpha$-truncation of $x$ the surreal number whose sign sequence is obtained by truncating the sign sequence of $x$ to length $\leq\alpha$. In the somewhat colorful terminology that is often used concerning surreals, this is the unique “ancestor” of $x$ whose “birthday” is $\leq\alpha$, or $x$ itself if $x$'s birthday is $\leq\alpha$.
Also recall (cf. e.g., Gonshor, An Introduction to the Theory of Surreal Numbers (1986), chap. 5, esp. theorem 5.6; or Alling, Foundations of Analysis over Surreal Number Fields (1987), §6.50–6.55) that earch surreal number $x$ has a unique expression as a “Hahn series” in $\omega$ (or more accurately, $\omega^{-1}$) with real coefficients and sureal exponents, namely one of the form $$ \sum_{\iota<\gamma} c_\iota \cdot \omega^{s_\iota} $$ the normal form of $x$, where $c_\iota$ are nonzero real numbers and $s_\iota$ is a decreasing $\gamma$-indexed sequence of surreals, and furthermore any such expression defines a surreal number (note that the sum, and in fact the expression $\omega^s$ itself, is to be interpreted in an ad hoc fashion and does not represent a convergent series in the surreals, but sums, products and order are computed on such expressions as one would expect from Hahn series).
Question: How can we compute the normal form of $T_\alpha(x)$ from that of $x$? Since the answer probably involves a lot of tedious case distinctions, feel free to assume $\alpha$ is, say, an $\varepsilon$-number (or even a cardinal, or an element of any closed unbounded subset that adequately simplifies the problem).
Note: Theorem 5.12 in Gonshor (op. cit.) states that the sign sequence of $\sum_{\iota<\gamma} c_\iota \cdot \omega^{s_\iota}$ is obtained by concatenating the sign sequences of $c_\iota \cdot \omega^{s_\iota^\circ}$ where $s_\iota^\circ$ is some modification of $s_\iota$ that only depends on the previous terms. From this it follows that the normal form of $T_\alpha(x)$ necessarily starts in the same way as that of $x$, except that it will be truncated and the final term might be modified (and possibly replaced by several terms, I'm not sure). So the question is basically how to know where to truncate and how to alter the last term.
Motivation: During a discussion on Twitter with Joel David Hamkins, he made me realize that I held the wrong belief that “for all surreal $h>0$ there exists $\alpha$ ordinal such that for all $x\in[0,1]$ surreal we have $|T_\alpha(x) - x| < h$” (to be clear, the quoted sentence is incorrect: as Joel explains, when $h$ is infinitesimal, we can find class-many disjoint open intervals of the form $\mathopen]x-h,x+h\mathclose[$ in $[0,1]$, which is more than $T_\alpha(x)$ can take values). So while $T_\alpha(x)$ eventually becomes equal to $x$ for $\alpha$ large enough, there is no kind of “uniform convergence” $T_\alpha \to \operatorname{id}$. This leads me to wonder how $T_\alpha(x)$ approximates $x$ for a given $x$ (e.g., “given $x$, how large do we have to take $\alpha$ to get $|T_\alpha(x) - x| < \omega^{-1}$ ?”), so, ultimately, how we can compute $T_\alpha(x)$.
