[Math] The notion of “infinitely differentiable”

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Wiki takes me to the section "smoothness" which I don't entirely get, it's just too much stuff for me.

My question is, what exactly is it?

An infinitely differentiable function is one that is infinitely differentiable.

Literally, that's what my lecture note says. So I thought the notion is very obvious, so what I had in mind was

Not infinitely differentiable $\to$ those that reaches a "dead end"

$f(x)=a_0+a_1x+a_2x^2+…+a_nx^n$ since $n+1$ differetiations will make it $0$ and I thought this can be considered a "dead end" for instance.

Then,

Infinitely differentiable $\to$ doesn't have a dead end

$f(x)=\sin{x}$ or $f(x)=e^{nx}$ say. These can be differentiated as many times as one wishes, but it doesn't have a dead end. It keeps changing to another function, coming back, or staying like itself.

However, it seems like my understanding has been wrong. In a probelem's solution in class, I have been told,

$f(x)=xy$ is a polynomial so it is infinitely differentiable.

In my way of thinking, polynomials meet a dead end so they are not. But my class has claimed otherwise. So what exactly are functions that are infinitely differentiable, and what are not? Sure, $0$ differentiated then it is $0$ so we can differentiate it a million or gazzilion times in a sense.

But what functions aren't? Functions that cannot be defined at a certain point then?

Best Answer

To add to @Crostul's comment, if at some step you get $$f^{(n)}(x) = |x|$$ then you have reached a dead end and this is indeed not differentiable further (because the derivative at $x=0$ does not exist, although the function is continuous there).

Any polynomial or simple exponent is infinitely differentiable (exponents cover sines and cosines as well)...

UPDATE

For an example of a function that would reach such a dead end, consider $$ f(x) = \begin{cases} x^3, & x \ge 0 \\ -x^3, & x < 0 \end{cases} $$ and note that $$ f'(x) = \begin{cases} 3x^2, & x \ge 0 \\ -3x^2, & x < 0 \end{cases} $$ and $$ f''(x) = \begin{cases} 6x, & x \ge 0 \\ -6x, & x < 0 \end{cases} = 6|x| $$ and you have reached a dead end.

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