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Question: Let $f(z) = \sum_{k\ge0} a_{k}z^k$ and $g(z) = \sum_{k\ge0} b_{k}z^k$ be power series which converge in $B_{R}(0)$ for some $R \gt 0$. Suppose there is a sequence $(w_{n})_{n\ge1} \subset B_{R}(0)$, and $w \in B_{R}(0)$, $w_{n} \to w$ as $n \to \infty$, $w_ {n} \neq w$ and

$f(w_{n}) = g(w_{n}), n\ge 1$. Show that $a_{k} = b_{k}, k\ge 0$.

Attempt: Since two function are equal with $w_{n}$, we can manipulate it as $f(w_{n}) - g(w_{n}) = 0$. Equals to $ \sum_{k\ge0}( a_{k}- b_{k})w_{n}^k = 0$.

(Informal speech warning!) I am not sure after this point, it is obvious that they are equal, at least after some point, but I am stucked. Also, we have an open ball around a singularity point $0$. It is %99 isolated zeros problem, but what approach should I prefer?

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Within their radius of convergence, complex power series are smooth functions and can be differentiated term-wise. Thus, if $f(z):=\sum\limits_{n=0}^\infty a_nz^n$, then $f^{(k)}(0)= k!\cdot a_k$, or rather $$a_k = \frac{f^{(k)}(0)}{k!}.$$ Further, within their radius of convergence, complex power series are even holomorphic, so the identity theorem together with your assumptions implies that $f^{(k)}(z)=g^{(k)}(z)$ for all $z\in B_R(0)$ and all $k\in\mathbb{N}_0$. So: $$b_k = \frac{g^{(k)}(0)}{k!} = \frac{f^{(k)}(0)}{k!} = a_k$$

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  • $\begingroup$ All derivatives are equal in the ball centered at 0 because all derivatives at 0 are 0. This is the reason, right? $\endgroup$ Commented Apr 6, 2023 at 18:04
  • $\begingroup$ And also, where did we use $w_{n} \to w$ ? $\endgroup$ Commented Apr 6, 2023 at 18:07
  • $\begingroup$ Also, $a_{0} = b_{0}$ is not guaranteed. $\endgroup$ Commented Apr 7, 2023 at 7:54
  • $\begingroup$ @amdryzen7000 First, it is a general fact about power series that within their radius convergence, so in this case in $B_R(0)$, they can be term-wise differentiated, so if $f(z)=\sum\limits_{n=0}^\infty a_n z^n$, then $f'(z) = \sum\limits_{n=0}^\infty a_{n} n z^{n-1}$ and so on. Since all $z^n$ for $n>0$ vanish if one plugs in $z=0$, this means that $f(0)=a_0$, $f'(0)=1\cdot a_1$, $f''(0)=2\cdot a_2$,...., and $f^{(k)}(0)=k! a_k$. $\endgroup$ Commented Apr 7, 2023 at 9:58
  • $\begingroup$ Secondly, within $B_R(0)$, this power series is also a holomorphic function, so the identity theorem can be applied. This means that if one has a sequence $(w_n)$ with $w_n\rightarrow w$, $w_n\neq w$, etc., so that $f(w_n)=g(w_n)$ for all $n\in\mathbb{N}$, then automatically $f(z)=g(z)$ for all $z\in B_R(0)$. Not only that, but also all of their derivatives are the same, so $f^{(k)}(0)=g^{(k)}(0)$ for all $k\in\mathbb{N}_0$: $\endgroup$ Commented Apr 7, 2023 at 9:59

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