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It seems to me (who is quite the math novice) that a very important ‘statement’, for a lack of a better word, that is foundational to many mathematical topics is that a given curve, which is continuous and differentiable, can be built from a bunch of straight lines as long as we make those lines ‘small enough’. For a 2D case, I interpret this as being able to build a curve that traverses through a 2D plane by only using little $\Delta x$ ’s and little $\Delta y$ ’s. I am wondering how one goes about proving this statement. It seems to me a good starting point can be illustrated using the following picture:

Curve Construction from Straight Lines

I suppose I should clarify that I am simply using this circle as a starting point for this argument...this could be any arbitrary curve (not just the circumference of a circle...though I suppose there is probably a proof out there that shows tiny sections of a curve can also be approximated by an arc length of a circle with a certain radius...but that's another question for a different time).

So the question I want an answer to is the following: As $\Delta x$ becomes very small (and its corresponding $\Delta y$, based on the behavior of the curve, or, more specifically, based on the function that describes the curve, also becomes very small ), does

$(r*\Delta \theta) / (\sqrt{(\Delta y)^2+(\Delta x)^2)}$ approach 1.0?

How would one go about proving this? I feel like most arguments that I can think of are rather circular…in that I have to use a property that is based off of what I want to prove in order to prove it! Is there a proof for this limit? Or is this just an axiom we accept to be true?

Edit 1: It has been brought to my attention that including the word "differentiable" as a characteristic of a curve creates a circular argument for what I would like to prove. The logic behind that claim is "if the curve is differentiable, then of course a curve can be decomposed into line segments because that is the definition of differentiable". Assuming this is true, please disregard the word 'differentiable'. I am interested in solving the previously referred to limit as if I never knew that calculus existed!

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  • $\begingroup$ Certainly, your equation holds for the circle. However, it seems like a space-filling curve might be a counterexample to the property you're trying to convey with the equation. $\endgroup$ – Peter Shor Jun 25 '19 at 23:49
  • $\begingroup$ Hahaha, as I said in the introduction, I am not very experienced with math so I am not quite sure what that type of curve is. After looking at your hyperlink, I am not quite sure I can figure out why that is a counterexample. (I'm not quite sure my equation even 'has' a counterexample...as I am posing it as a question rather than a conclusion) $\endgroup$ – S.Cramer Jun 25 '19 at 23:52
  • $\begingroup$ Are you requiring the curve to be differentiable or smooth (infinitely differentiable)? $\endgroup$ – eyeballfrog Jun 26 '19 at 0:48
  • $\begingroup$ oh, i'm sorry. I thought that they were the same thing! I will make the edit now. I just meant differentiable (not necessarily infinitely) and continuous. $\endgroup$ – S.Cramer Jun 26 '19 at 1:02
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    $\begingroup$ @S.Cramer: Careful! Indeed, there are certain curves that inherently have this property, but they are not necessarily differentiable. That's why my above comment says "includes but is not limited to". For instance, Lipschitz-continuous curves in a Riemannian manifold have the desired property but are not necessarily differentiable. Answering your question requires a solid understanding of at least real analysis (assuming that you are interested only in what happens in $\mathbb R^n$). Otherwise add some Riemannian or even metric geometry. $\endgroup$ – Alex M. Jun 28 '19 at 18:40
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This is true for most "nice" curves but isn't true in general, at least under the usual definition of a linear continuum, particularly for space-filling curves as @PeterShor mentioned in their comment. If a space-filling curve could be approximated by straight lines, as I understand your condition, it would be differentiable everywhere, and there exists no space filling curve that is differentiable everywhere (though it is possible for it to be differentiable almost everywhere, see https://en.wikipedia.org/wiki/Sard%27s_theorem and https://mathoverflow.net/questions/201424/proof-that-no-differentiable-space-filling-curve-exists for a discussion of this.)

The notion of continuity is actually a bit tricky and relies on some general topology. For all intents and purposes in introductory calculus, this is true. I think this should be true for any curve that can be represented with a Taylor series. But it is not true for any arbitrary curve.

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  • $\begingroup$ Do you know of where I could find a proof to the limit that I am interested in? I appreciate that this is generally true (it makes my intuitive understanding of derivatives much stronger)...but seeing such a proof would really cement my complete understanding. $\endgroup$ – S.Cramer Jun 26 '19 at 1:17
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    $\begingroup$ It's a circular question when you restrict it to differentiable functions: this proposition is true if and only if the curve is differentiable, because the statement that a curve can be approximated by infinitely many line segments is (in the Euclidean plane) the definition of differentiability. en.wikipedia.org/wiki/Differentiable_function explains this well. $\endgroup$ – I. Pittenger Jun 26 '19 at 1:25
  • $\begingroup$ Ohhhh, that makes a lot of sense. What if I remove the word “differentiable” then? $\endgroup$ – S.Cramer Jun 26 '19 at 1:32
  • $\begingroup$ Then it's not true, per the mathoverflow discussion I cited. $\endgroup$ – I. Pittenger Jun 26 '19 at 1:33
  • $\begingroup$ Hmmm. This has me a little confused. If including the characteristic of ‘differentiable’ guarantees its truth (because derivatives are based off of the assumption that straight lines can be used to build curves) and if removing the characteristic of ‘differentiable’ makes it not true...how is the process of differentiation even true? I would think that proving this limit approaches 1 is the way that you conclude differentiation is actually a valid method $\endgroup$ – S.Cramer Jun 26 '19 at 1:38

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