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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?

Community
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Melba1993
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    Once you differentiate a polynomial enough times you get $0$. This is not a dead end, because it is a differentiable function, whose derivative is again $0$. A dead end should be something non-differentiable. – Crostul May 12 '16 at 21:24
  • Is there an explicit and simple example of something non-differentiable? Is it something like $\frac{1}{x}$ at $x=0$? – Melba1993 May 12 '16 at 21:30
  • Consider $f(x) = |x|$. Is it differentiable (at least once) everywhere on $x \in \mathbb{R}$? – Alex May 12 '16 at 21:31
  • @Alex The absolute value function is not differentiable at $x=0$. – John Douma May 12 '16 at 22:07
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    The function $f(x)=x|x|$ is differentiable everywhere but its second derivative does not exist at $x=0$. – John Douma May 12 '16 at 22:11
  • Yes, my point was to illustrate what possibilities could arise... @JohnDouma – Alex May 12 '16 at 22:19
  • Think i got a better idea now guys, thanks so much! – Melba1993 May 12 '16 at 22:29

2 Answers2

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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.

gt6989b
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in most situations, infinitely differentiable means that you are allowed to differentiate the function as many times as you wish, since these derivatives exist (everywhere). The value of the derivatives is irrelevant, of course.

A function is not infinitely differentiable if there exist an integer $n \geq 1$ and a point $x_0$ such that the derivative of order $n$ at $x_0$ does not exist.

Siminore
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  • What about $1\over x$ at $x_0=0$? – Simply Beautiful Art May 12 '16 at 22:02
  • Thanks guys, I found this http://math.stackexchange.com/questions/991475/why-is-the-absolute-value-function-not-differentiable-at-x-0/991559 which explains it visually for the case. But yes, what about $\frac{1}{x}$ at $x=0$? Do we deem it to "not have a limit" there since it approaches infinity? – Melba1993 May 12 '16 at 22:04
  • Indeed. Although we can infinitely differentiate the function, there is a discontinuity at $x=0$ in differentials of all orders. It is not smooth. @Melba1993 – Graham Kemp May 12 '16 at 22:10
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    @SimpleArt the function $x \mapsto 1/x$ is smooth in its domain, namely $\mathbb{R} \setminus {0}$. A necessary condition to defined the derivative is that the function must be defined at the point. And, we all know, continuous. – Siminore May 13 '16 at 06:50