# What is the difference between the terms smooth, analytical and continuous?

I saw the following (“roughly speaking”, like the author says) definition of a Lie group in ‘Group theory in Physics’, by Wu-Ki Tung:

“Roughly speaking, a Lie group is an infinite group whose elements can be parametrized smoothly and analytically.” After this, I was asking myself if I really already know the difference between these terms.

Because, what I know about 1 - analytic: if we say that a function is analytic at a point it means that its derivative is defined at this point and at the points of its neighborhood; 2 – continuous: a function is continuous at a point if you can write a neighborhood of this point where this function is still defined (and this is why those three conditions we learn in Calculus, including that one with the limit); 3 – smooth: I am not sure, but I think it is related to the differentiability of the function.

I think this may be a silly question, but I would thank you for answering

• In any case... You're aware you can easily look up these terms on Wikipedia, right? Asking this question here is pointless. Commented Jul 22, 2014 at 14:20
• @symplectomorphic That seems a bit unfair. Evidently Poli was misguided regarding analyticity, whether he/she got that from a poor understanding of what was going on in a calculus course or possibly from a resource on complex analysis. Commented Jul 22, 2014 at 14:27
• @Dustan: (s)he is misguided because (s)he didn't look up the precise definitions. (The first sentence of his/her explanation of continuity isn't even close.) There is nothing more to it than that. Commented Jul 22, 2014 at 16:26
• @symplectomorphic Or maybe they looked up the precise definition, and then tried to turn it into an intuitive definition, but did so incorrectly. Or they had a calculus course and obtained a flimsy understanding of these terms from that. And then when they read the precise epsilon-delta definition of continuity, and it reads greek to them (literally), they might just assume it must line up with what they thought was the definition. Only an individual who takes the time to give some thought to pedagogy instead of mere formalism can help with that problem. Commented Jul 22, 2014 at 17:04
• @symplectomorphic you seem to be one of those that take Wikipedia as the ultimate authority in information. I never trust anything from wikipedia, be it math or politics. For math, I suggest wolfram mathworld. Commented Jul 28, 2015 at 2:05

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 differentiables (all the $n$-th derivatives are thus continuous, since differentiability implies continuity).
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. 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 neighnourhood of $0$.
• What is so important about $\exp(-1/x^2)$? Why can't we have $\exp(-1/x^4)$ or $\exp(-1/x)$ or $\exp(-1/|x|)$ and keep the function continuous? Commented Jul 28, 2015 at 2:11
• @Jus12 Well, it can't be $\exp(-1/x)$ because that's not continuous. Commented Jul 28, 2015 at 3:32
• @Jus12: it's an example. There are many examples of $C^\infty$ functions that are not analytic.