Continuous (Smooth) vs Differentiable versus Analytic

I had a lot of trouble distinguishing between continuous and differentiable and analytic functions…

Analytic

An analytic function is a function that is smooth (in the sense that it is continuous and infinitely times differentiable), and the Taylor series around a point converges to the original function in the neighbourhood of that point.

Continuous – but NOT analytic (taylor series does not converge)

A smooth function is a continuous function with a continuous derivative. Some texts use the term smooth for a continuous function that is infinitely many times differentiable (all the $n$ -th derivatives are thus continuous, since differentiability implies continuity). The existence of all derivatives doesn’t imply that the Taylor series converges. A famous example is the function

$f(x)=\exp\left(\frac{-1}{x^2}\right) \text{ if } x \neq 0$

$f(0)=0$

This function is continuous and infinitely many times differentiable in $x=0$ . The Taylor series around this point is the constant function $T(x)=0$ , so the Taylor series doesn’t converge to the function $f(x)$ in the neighborhood of $0$ .

Complex Sequences and Analytic Approximations thereof

Suppose we know the values of a complex analytic function $f$ at all $x+iy$ , for $x,y\in\mathbb{Z}$ . Can we uniquely determine $f$ ?

The Weierstrass factorization theorem asserts that for any sequence $\{a_n\}$ of nonzero complex numbers with $|a_n| \to \infty$ , there is a nontrivial analytic function $f$ whose zero set is precisely $\{a_n\}$ . Clearly we can enumerate all the nonzero points $a_n$ of the integer lattice in such a way that $|a_n| \to \infty$ , so there is a nontrivial analytic $f$ vanishing at all those points. Then the function $z f(z)$ is also not the zero function, but vanishes at all those points and 0 as well.<

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