# Why are all analytic functions equivalent to their Taylor series?

"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.

• Also, I asked a related question a while ago that still has me pondering. – Leo Jun 6 '18 at 21:17
• What is your definition of analytic function? – quid Jun 6 '18 at 21:17
• 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 $x_0$. – Leo Jun 6 '18 at 21:18
• It seems to me that it is just a restatement of the definition. – arkeet Jun 6 '18 at 21:20
• 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. – Aloizio Macedo Jun 6 '18 at 21:49