Mathematics Asked by Joey on December 5, 2020

The question is as is, just proving that $x^2$ is analytic in $mathbb{R}$.

Is this just true because for $|x|<1$ we have by the geometric series that $sum_{n=0}^{infty}x^2 = frac{1}{1-x}$. If we can rewrite it like this, isn’t it clearly analytic, or is there more going on here.

Such coefficients exist : given $x_0$, note that $f(x_0+a) = (x_0 + a)^2 = x_0^2 + 2ax_0+a^2$.

Therefore, the power series around $x_0$ converging to $f$ near $x_0$, would just be given by $f(y) = (y-x_0)^2 + 2x_0(y-x_0) + x_0^2$ i.e. $a_0 = x_0^2, a_1 = 2x_0,a_2 = 1$ and the rest being $0$. Note that $c,d$ can be taken as large as you want : in fact, this power series converges *everywhere* because it is a polynomial.

A power series is a generalization of a polynomial. In particular, any polynomial will definitely be analytic.

Correct answer by Teresa Lisbon on December 5, 2020

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