# In the Levi-Civita field, are there elements such that the standard parts of their subsequent powers produce an arbitrary sequence?

In the Levi-Civita field, are there elements such that the standard parts of their subsequent powers produce an arbitrary sequence?

Particularly, is there an element $$w$$ of the field such that the standard part (the zeroth element of the corresponding series) of $$w^n$$ is $$B_n$$ (Bernoulli numbers)?

Yes, it is true, and there even are such elements in the field of formal Laurent series. Specifically, let $$a_n,n\geq 1$$ be any sequence of real numbers. We then take a Levi-Civita series $$z=\varepsilon^{-1}+\sum_{i=0}^\infty b_i\varepsilon.$$ We want to show the $$b_i$$ can be chosen so that the constant term of $$z^n$$ is $$a_n$$ for $$n\geq 1$$. The key thing to note is that this constant term will only depend on coefficients $$b_i$$ for $$i. This lets us define $$b_i$$ recursively. Suppose we have constructed $$b_i$$ for $$i already. To determine $$b_n$$, we consider $$z^{n+1}$$, and note that the condition imposed by its constant term being $$a_{n+1}$$ involves a sum of various combinations of $$b_i,i, but $$b_n$$ only appears once, so we can always pick $$b_n$$ to make the total sum equal to $$a_{n+1}$$.
• See my other answer. Those $p_1$ and $p_2$ would satisfy $\operatorname{st}\frac1{\pi }\ln \left(\frac{p_1-\frac{z}{\pi }}{p_2+\frac{z}{\pi }}\right)=\cot z$, because of this: mathoverflow.net/questions/380142/… Commented Jan 24, 2022 at 23:20
It seems, the following elements have the prescribed property (for $$B^+(x)$$ and $$B^-(x)$$):
$$\begin{gather*} p_0=\varepsilon^{-1} -\frac{1}{2}-\frac{\varepsilon }{24}+\frac{3 \varepsilon ^3}{640}-\frac{1525 \varepsilon ^5}{580608}+\dotsb\\ p_1=\varepsilon^{-1} +\frac{1}{2}-\frac{\varepsilon }{24}+\frac{3 \varepsilon ^3}{640}-\frac{1525 \varepsilon ^5}{580608}+\dotsb . \end{gather*}$$