This question is moved from [math stackexchange][1] which I posted several days ago without an answer.

Background(ignore this paragraph if you know finite type invariants well): Recall that a [finite type invariant][2] of degree $n$ is an invariant $V$ such that $V^{(n+1)}=0$ where $V^{(n+1)}$ is the extension of $V$ to $(n+1)$-singular knots. Fix a base field $\mathbb F$ of characteristc $0$.  Let $\mathcal V_n$ be the space of type $n$ invariants and $\mathcal K_n$ be the linear space spanned by the complete resolution of $n$-singular knots(these are commone zeros of type $n-1$ invariants), and let $\mathcal V$ and $\mathcal K$ be the corresponding filtered space of finite type invariants and knots. The relations not distinguished by the whole finite type invariants $\mathcal V=\bigcup\mathcal V_n$ is $\bigcap\mathcal K_n=\text{ker} (\mathcal K\to\hat{\mathcal K})$, and [Vassiliev conjecture][3] is that $\mathcal K\to\hat{\mathcal K}$ is injective. Note that every finite type invariant factors through the Kontsevich integral $Z:\mathcal K\to\hat{\mathcal A}$, which takes value in the completion of space of chord diagrams mod 1T/4T relations, by a weight system(a function on $\hat{\mathcal A}$ that vanishes on $\bigoplus_{k\geq n}\mathcal A_k$ for some $n$), so the whole finite type invariants are equally powerful as $Z$, and we can show that $\hat Z:\hat{\mathcal K}\cong\hat{\mathcal A}$(because you can show that $Z:\mathcal K/\mathcal K_{n+1}\cong\bigoplus_{k\leq n}\mathcal A_k$), and Vassiliev conjecture is that $Z$ injective.

My question: **What is the relation of Vassiliev's conjecture with the following conditions(at least when $\mathbb F=\mathbb R$ so that the limit is defined):**

(1) Every invariant factors through a universal finite type invariant, say $\mathcal K\to\hat{\mathcal K}$. As a result, we can write $V$ as $f(\sum_{n}V_n)$, with $f$ a linear functional on $\hat\bigoplus\mathcal V_n$ and $\sum_{n}V_n\in \hat\bigoplus\mathcal V_n$ a formal sum. 

(2)[Taylor expansion] Every invariant $V$ has an expansion $V=\sum_n V_n$ of finite type invariants for $V_n$ type $n$. This equivalent to every invariant $V$ can be wrtten as $V=\lim_{n\to \infty} V_n$ for $V_n$ type $n$.

(3)[Stone-Weierstrass]Every invariant $V$ can be approximated by finite type invariants pointwise, namely $V=\lim_{n\to \infty} V_n$ for finite type $V_n$(not necassarily type $n$)

(Rmk) I do think Vassiliev conjecture is equivalent to (1), and I don't think it implies (2),(3), even if an invariant $V$ factors through $Z:\mathcal K\to \hat{\mathcal K}=\hat {\mathcal A}=\prod_n \mathcal{A}_n$, here is an annoying problem of convergence. I have no idea of wether (2),(3) implies Vassiliev's conjecture or not.
    
 I doubt (2),(3), might be false: **are there known examples of an invariant that cannot expand as power series of finite type invariants or as a limit of finite type invariants**? 

  [2]: https://en.wikipedia.org/wiki/Finite_type_invariant
  [3]: https://mathoverflow.net/questions/452603/different-flavours-of-vassiliev-conjecture 




  [1]: https://math.stackexchange.com/questions/4917501/power-series-expansions-and-limits-of-knot-invariants