# If $\varphi$ is analytic and $t^k + \sum_{i=1}^{k-1}a_i(\varphi(t,x),x)t^i = b(\varphi(t,x),x)$ , then $a_i$ and $b$ are analytic functions.

Let $$\varphi: \mathbb{R}\times \mathbb{R}^n \to \mathbb{R}$$ be a real analytic function, such that

$$\varphi(0,0)=0,\ \frac{\partial \varphi}{\partial t}(0,0)=0,\ \ldots\ ,\frac{\partial^{k-1} \varphi}{\partial t^{k-1}}(0,0)=0\ \text{and}\ \frac{\partial^k \varphi}{\partial t^k}(0,0) \neq 0.$$

Now, suppose that there exist smooth functions $$a_1,...,a_k,b: \mathbb{R}\times U\subset \mathbb{R}\times \mathbb{R}^n\to \mathbb{R}$$ such that $$t^k + \sum_{i=1}^{k-1} t^i \cdot a_i(\varphi(t,x),x) = b(\varphi(t,x),x),\ \forall\ t\in \mathbb{R} \ \text{and}\ x\in U,$$ where $$U$$ is an open neighboorhood of $$0$$ in $$\mathbb{R}^n$$.

Question: Is it possible to guarantee that the functions $$a_1,...,a_k,b$$ are real analytic functions?

It is enough to assume that $$a_j = 0$$ for $$j >1$$ to give a counterexample. Then choosing $$a_1$$ to be non-analytic, it is "unlikely" that $$b$$ be analytic.
Take for example $$\varphi(t,x) = t^k$$, and $$a_1(t,x) = \begin{cases} e^{-\frac{1}{t^2}}, & t > 0\\ 0, & t \leq 0. \end{cases}$$
Defining $$b(t,x) = t-t^\frac{1}{k}a_1(t,x)$$, one sees that $$a_1$$ and $$b$$ are smooth, but non-analytic.