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Smooth functions $f(t)$ are those such that $\frac{d^nf(t)}{dt^n}$ exists for all $n\in\Bbb{N}$.

I understand the intuition behind smoothness for functions like $f(t)=| t|$ and $f(t)=\sqrt{t}$. $f(t)$ has a "sharp" (and hence non-smooth) turn at $t=0$. Similarly, $f(t)=\sqrt{t}$ ends abruptly at $t=0$ (and hence is not "smooth").

However, functions like $f(t)=t^{\frac{1}{3}}$ seem "smooth" enough to me!

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Why does the intuitive understanding of smoothness fail here? Is this just another case of extending a definition of a term to non-intuitive cases?

Thanks in advance!

  • $t^{\frac{1}{s}}$ is a smooth function away from $0$, what is the problem here? What is your $s$? – 5xum Jan 29 '14 at 08:42
  • I don't see any "rough" edges (in the adjective sense of the word) at $t=0$ either. But it is still not mathematically smooth here. –  Jan 29 '14 at 08:44
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    I'd look at it from the other side: The concept that we are interested in is that the function is in $C^\infty$. To call these functions smooth seems justified. That does not mean that everything that might be intuitively smooth has to be in that class of functions. – Carsten S Jan 29 '14 at 08:57

2 Answers2

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The problem with the function $f(t) = t^{1/3}$ is not that it is something else. The problem is that in the point $0$, the slope of the curve is infinite as the function turns completely vertical. The curve drawn is actually smooth, as you can find a smooth parametrisation of it.

The parametrisation $$t\rightarrow (t, \sqrt[3]{t})$$ is not smooth, however, you can reparametrise the curve as $$t\rightarrow (t^3, t)$$ which is obviously smooth.

5xum
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  • What about $f(t)=t^{\frac{7}{3}}$? It does not have an infinite slope anywhere. It is still not smooth. –  Jan 29 '14 at 09:03
  • Yes, because it's second derivative is $\frac{28}{9}t^{1/3}$ which has infinite slope. it is still a smooth curve, however, with the parametrisation $t\rightarrow (t^3, t^7)$ – 5xum Jan 29 '14 at 09:23
  • That is my point exactly. The second derivative having infinite slope at $t=0$ does not take away from the fact that $f(t)=t^{\frac{7}{3}}$ is "smooth" at every point (when "smooth" is taken as an adjective). Hence, the mathematical definition of smooth seems to be non-intuitive as per its English usage. –  Jan 29 '14 at 09:25
  • @AyushKhaitan As I mention in my answer, the word smooth in relation to a function is to be interpreted as a property of a parametrization, not just of a curve. A parametrization may be thought of as like riding a roller coaster. Such a ride where you at any point have infinite speed will not be felt as smooth (indeed, there is a reason the third derivative of position is called "jerk"). When understood in this light, the English usage is in fact applied intuitively. – anon Jan 30 '14 at 01:40
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The curve itself admits a smooth parametrization, $\{(t^3,t):t\in\Bbb R\}.$ The function $f(t)=t^{1/3}$ on the other hand is not differentiable at $t=0$; one way of thinking about this is that the function $f(t)$ parametrizes the curve in such a way that we rush through the origin with infinite speed, and that isn't allowed on a smooth ride. Thus, the problem is with the parametrization, not the geometry.

anon
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