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.



Local equality of functions implies global equality?
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 $f\_4(f\_3(x)) = x$.


