"A function is analytic if and only if its Taylor series about $x_0$ converges to the function in some neighborhood for every $x_0$ in its domain."

Clearly if its Taylor series converges to $f$ then the function is analytic, but why is the converse true?

I would really appreciate any help/thoughts.

  • $\begingroup$ Also, I asked a related question a while ago that still has me pondering. $\endgroup$ – Leo Jun 6 '18 at 21:17
  • $\begingroup$ What is your definition of analytic function? $\endgroup$ – quid Jun 6 '18 at 21:17
  • $\begingroup$ A function $f$ is analytic on an open set $D$ of the real line if for any $x_0∈D$ one can write $$f(x) = \sum_{n=1}^∞ b_n(x-x_0)^n.$$ for coefficients $b_i∈ℝ$ and the series is convergent to $f(x)$ on a neighborhood of $x_0$. $\endgroup$ – Leo Jun 6 '18 at 21:18
  • $\begingroup$ It seems to me that it is just a restatement of the definition. $\endgroup$ – arkeet Jun 6 '18 at 21:20
  • 1
    $\begingroup$ If the function is given by some power series around $x_0$, you can differentiate it $n$ times and evaluate at $x_0$ to conclude a posteriori that it must be a Taylor series by checking the values of the $b_n$ you get by doing that. $\endgroup$ – Aloizio Macedo Jun 6 '18 at 21:49

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