Local equality of functions implies global equality?

Question

The following question arised in my research, and I was unable to settle it after playing with it for sometime. Let $\{a^k_i\}_{i\geq 1}$ (for $k\in \{1,2,3,4\}$) be four sequences of real numbers. For each $k\in \{1,2,3,4\}$, consider the function $f_k:]-r_k,r_k[\rightarrow \mathbb{R}$ given by $x\mapsto\sum_{i=1}^{\infty}a_i^kx^i$ (Where $r_k$ is defined as $\frac{1}{\limsup_{i\rightarrow \infty}|a_i^k|^{\frac{1}{i}}}$).

It is given that for all $x$ sufficiently close to $0$, we have $f_2(f_1(x))=f_4(f_3(x))$. It is also given that for some $x_0\in\mathbb{R}$, we have that both $f_2(f_1(x_0)), f_4(f_3(x_0))$ exist. Must $f_2(f_1(x_0))= f_4(f_3(x_0))$ ? I would be very grateful if you can help, thank you.

Edit: "$f_2(f_1(x_0))$ exists" means that $x_0$ belongs to domain of $f_1$ and that $f_1(x_0)$belongs to domain of $f_2$. Similarly, "$f_4(f_3(x_0))$ exists" means that $x_0$ belongs to domain of $f_3$ and that $f_3(x_0)$belongs to domain of $f_4$

I know that the answer is yes in some special cases like $f_1$ equals identity map, or $f_1,f_2,f_3,f_4$ are rational functions (Which raises the question what if the $f_k$(s) are algebraic functions ?). I also considered extending the domain of the $f_k$(s) to an open subset of the complex plane and possibly open the route for analytic continuations and complex analysis machinery but was not successful with that.

Answer 1

Consider $$\eqalign{f_1(x) &= (x+1)^2-1\cr f_2(x) &= \sqrt{x+1}-1\cr f_3(x) &= f_4(x) = x }$$
$f_1, f_3$ and $f_4$ being polynomials, their radius of convergence is $\infty$, while $f_2(x)$ has a Maclaurin series with radius of convergence $1$. $f_2(f_1(x)) = f_4(f_3(x)) = x$ for $x \in (-1,1)$. But since $f_1(x) = f_1(-x-2)$, $f_2(f_1(x))$ is also defined for $x \in (-3,-1)$, with $f_2(f_1(x)) = -2-x$ there while $f_4(f_3(x)) = x$.