Confused with the use of the word smooth .

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Dembadon
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Confused with the use of the word "smooth".

[Multi-Variable Calc course]

A couple weeks ago we went over the Fundamental Theorem of Line Integrals, which requires "smooth" simple connected curves. My professor's definition of smooth was a curve having "no corners".

Now, with Green's Theorem, "smooth" curves are permitted to have corners. Are there multiple definitions for this word in mathematics? How am I supposed to justify what is smooth and what isn't?
 

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hunt_mat
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Smooth in analysis sense (and also is the differential geometric sense) means infinitely differentiable.
 
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pwsnafu
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Now, with Green's Theorem, "smooth" curves are permitted to have corners.
No, Green's theorem uses "positively oriented piecewise smooth simple closed curve".
You are allowed a finite number of corners.
 
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Dembadon
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Disclaimer: I haven't taken an analysis course yet.

Smooth in analysis sense (and also is the differential geometric sense) means infinitely differentiable.
So, [itex]f(x)=|x|[/itex] is not smooth because it is not differentiable when [itex]x=0[/itex]?

No, Green's theorem uses "positively oriented piecewise smooth simple closed curve".
You are allowed a finite number of corners.
Thank you for the correction.

I have a feeling that I'm not going to get very far in understanding the meaning of smooth until I take analysis courses.
 
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hunt_mat
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Basically yes, when I hear the word smooth, I assume that a function is infinitely differentiable on the whole interval (barring end points)

Mat
 
  • #6
HallsofIvy
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It depends upon context. Many textbooks use "smooth" simply to mean "differentiable". They then would distinguish "infinitely smooth" to mean "infinitely differentiable" or sometimes "sufficiently smooth" to mean "has enough derivatives to make this theorem work".
 
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Usually smooth means infinitely differently sometimes (implicitly) piecewise infinitely differentiable. I do not think smooth is a good term to use in describing the conditions of greens theorem since the function being integrating only has to be C1 (its first derivative is continuous) and the path only has to be piecewise continuous.
 
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Yes, the short answer is there are multiple definitions of smooth.

When a term (like smooth) is ambiguous I adopt when possible the definition that gives wikipedia, that is a function that is [tex] f \in C^{\infty} [/tex] (has derivatives of all orders).

In principle this does not have sense for a curve, because a curve is a set, and thus cannot be diferentiable. A curve is called "smooth" if it admits a smooth parametrization.

One thing that I think is wrong for the OP is the requirement for the curve to be "simply connected" (that not the same that a curve that is simple and closed)

and the path only has to be piecewise continuous.
I think that is not enough, a path can be piecewise continuous and have infinite length (not be regular), and AFAIK there is no definition of integral over such curves (example the boundary of the mandelbrot set).
 
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Hurkyl
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In principle this does not have sense for a curve, because a curve is a set, and thus cannot be diferentiable. A curve is called "smooth" if it admits a smooth parametrization.
Actually, the word "curve" is often defined so that the parametrization is actually part of what it means to be a curve. The set you mention is just the "trace" or "image" of the curve.

e.g. wikipedia's definition of a topological curve (and see the next section)
 
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Actually, the word "curve" is often defined so that the parametrization is actually part of what it means to be a curve. The set you mention is just the "trace" or "image" of the curve.

e.g. wikipedia's definition of a topological curve (and see the next section)
I read that article and agree with you it says that.
In this respect I'm used to other definition that differs from that of wikipedia. I call a curve what you would call the image of the curve, and I call path what you would call curve (the parametrization of the curve in my def.)

So, for example [tex] C = \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 = 1 \}[/tex] is a curve, because it exist a parametrization [tex] g:[0,2\pi) \to \mathbb{R}^2, g(t) = (\cos(t), \sin(t))[/tex] such that [tex] Im(g) = C [/tex].
(and I call g one of the possible paths for this curve)
 

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